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https://math.stackexchange.com/questions/1707439/math-problem-forty-nine-points
[ "# Math Problem: Forty-nine points\n\n49 points are marked on a sheet of paper in a square. Adjacent points horizontally or vertically are separated by exactly 1 centimetre.\n\nHow many straight lines of length 5 centimetres can be drawn between points in the design?\n\nI have attempted to solve this problem taking into account the lines going through 6 points and the 5 cm diagonal lines with height 4 cm and width 3 cm. However, these approaches have been unsuccessful. The answer is 76 lines, but I cannot see how to reach this value. How is this problem solved and am I missing anything?\n\nThanks\n\n• I am not clear how the points are drawn. Suppose the first is at the origin $x=y=0$. What are the possible positions for the second? Also what do you mean by \"adjacent\" points? – almagest Mar 21 '16 at 16:32\n• The points are drawn on a piece of paper in a 7 by 7 square. Adjacent points are points that are next to each other either vertically or horizontally separated by 1 cm exactly. – Tom Finet Mar 21 '16 at 16:35\n• Your beginning is correct (+1). To make progress answer the following: How many such horizontal line segments are there? The number of vertical line segments is the same by symmety. How many represent the vector $(3,4)$? Again by symmetry the vectors $(3,-4)$, $(4,3)$ and $(4,-3)$ occur as often as $(3,4)$. You only need to get your hands dirty twice - the rest follows from symmetry. – Jyrki Lahtonen Mar 21 '16 at 16:37\n• @almagest: I'm fairly sure that we should imagine a 7x7 square grid, with the small squares having sides of length $1$ cm. At least that leads to the answer 76 :-) – Jyrki Lahtonen Mar 21 '16 at 16:38\n• There are 14 horizontal segments so due to the symmetry 14 vertical segments. 12 segments represent the vector (3, 4). So 12 * 4 is 48 segments in total. Therefore adding both together we get 76 segments. Thanks for working me through it. – Tom Finet Mar 21 '16 at 16:44" ]
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https://trace.tennessee.edu/utk_gradthes/4361/
[ "## Masters Theses\n\n### A Numerical Algorithm For Solving the Nonlinear Differential Equations that Describe a Multistage Flash Evaporator\n\n6-1970\n\nThesis\n\n#### Degree Name\n\nMaster of Science\n\n#### Major\n\nNuclear Engineering\n\nJ.C. Robinson\n\n#### Committee Members\n\nT.W. Kerlin, Hale C. Roland\n\n#### Abstract\n\nA numerical algorithm is formulated to solve the first order, nonlinear differential equations that describe a multistage flash evaporator. The nonlinearities appearing in the formulation of the algorithm are products of up to three terms with each term being a dependent variable raised to some power.\n\nTo develop the algorithm, the first order differential equations are written in integral form. The dependent variables are then assumed to have a purely exponential dependence over a finite time step thereby allowing for the explicit integration of all terms. The solution of the differential equations is then reduced to the determination of the exponential dependences. The exponential dependences are determined by an iterative method.\n\nA computer code based upon the aforementioned algorithm was written. Before the algorithm was used to obtain solutions to a flash evaporator system, it was applied to several differential equations with known solutions. The algorithm was then used to obtain solutions for two perturbations in the twenty-third order system that describes a three stage flash evaporator. These solutions are compared with solutions obtained by other methods.\n\nFiles over 3MB may be slow to open. For best results, right-click and select \"save as...\"\n\nCOinS" ]
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http://brauer.maths.qmul.ac.uk/Atlas/spor/M11/
[ "# ATLAS: Mathieu group M11\n\nOrder = 7920 = 24.32.5.11.\nMult = 1.\nOut = 1.\n\nThe following information is available for M11:\n\n### Standard generators\n\nStandard generators of M11 are a and b where a has order 2, b has order 4, ab has order 11 and ababababbababbabb has order 4. Two equivalent conditions to the last one are that ababbabbb has order 5 or that ababbbabb has order 3.\nIn the natural representation we may take a = (2, 10)(4, 11)(5, 7)(8, 9) and b = (1, 4, 3, 8)(2, 5, 6, 9).\n\n### Black box algorithms\n\n#### Finding generators\n\nTo find standard generators for M11:\n• Find an element of order 4 or 8. This powers up to x of order 2 and y of order 4.\n[The probability of success at each attempt is 3 in 8 (about 1 in 3).]\n• Find a conjugate a of x and a conjugate b of y such that ab has order 11.\n[The probability of success at each attempt is 16 in 165 (about 1 in 10).]\n• If ababbabbb has order 3, then replace b by its inverse.\n• Now ababbabbb has order 5, and standard generators of M11 have been obtained.\nThis algorithm is available in computer readable format: finder for M11.\n\n#### Checking generators\n\nTo check that elements x and y of M11 are standard generators:\n• Check o(x) = 2\n• Check o(y) = 4\n• Check o(xy) = 11\n• Check o(xyxyyxyyy) = 5\nThis algorithm is available in computer readable format: checker for M11.\n\n### Presentation\n\nA presentation for M11 in terms of its standard generators is given below.\n\n< a, b | a2 = b4 = (ab)11 = (ab2)6 = ababab-1abab2ab-1abab-1ab-1 = 1 >.\n\nThis presentation is available in Magma format as follows: M11 on a and b.\n\n### Representations\n\nThe representations available are as follows. They should follow the order in the ATLAS of Brauer Characters, with the conjugacy classes defined by ab in 11A and ababababb in 8B, but please check this yourself if you rely on it!\n• All primitive permutation representations.\n• All faithful irreducibles in characteristic 2.\n• Dimension 10 over GF(2): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 16 over GF(4): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 16 over GF(4) - the dual of the above: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 32 over GF(2) - reducible over GF(4): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 44 over GF(2): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• All faithful irreducibles in characteristic 3.\n• Dimension 5 over GF(3) - the cocode representation: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 5 over GF(3) - the code representation: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 10 over GF(3) - the deleted permutation representation: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 10 over GF(3) - the skew square of the code representation:\na and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 10 over GF(3) - the skew square of the cocode representation:\na and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 24 over GF(3): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 45 over GF(3): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• All faithful irreducibles in characteristic 5.\n• Dimension 10 over GF(5): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 10 over GF(25): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 10 over GF(25) - the dual of the above: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 11 over GF(5): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 16 over GF(5): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 16 over GF(5) - the dual of the above: a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 20 over GF(5) - reducible over GF(25): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 45 over GF(5): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• Dimension 55 over GF(5): a and b (Meataxe), a and b (Meataxe binary), a and b (GAP).\n• All faithful irreducibles in characteristic 11.\n• Some faithful irreducibles in characteristic 0.\n• Dimension 10 over Z: a and b (Magma).\n• Dimension 10 over Z[i2]: a and b (Magma).\n• Dimension 10 over Z[i2] - the complex conjugate: a and b (Magma).\n• Dimension 11 over Z: a and b (Magma).\n• Dimension 20 over Z - reducible over Q(i2): a and b (Magma).\n• Dimension 32 over Z - reducible over Q(b11): a and b (Magma).\n• Dimension 44 over Z: a and b (Magma).\n• Dimension 45 over Z: a and b (Magma).\n• Dimension 55 over Z: a and b (Magma).\n\nSources: All the above representations, except those in characteristic 0, are easily obtained with the Meataxe from the permutation representations on 11 and 12 points. Most of the representations in characteristic 0 are not that difficult to obtain either (the most difficult being the non­rational representations of degree 10).\n\nNB: There is some ambiguity as to which of the two 5­dimensional GF(3)­modules of M11 should be regarded as the code and which as the cocode. Let M = 2M12 be the full automorphism group of the ternary Golay code. So M monomially permutes the vectors e1, e2, . . . , e12 (and their negatives). Now M has two conjugacy classes of subgroups isomorphic to M11 and their representatives may be taken to be M1, stabilising e1, and M2, the subgroup of (pure) permutations. The terms `code' and `cocode' used above refer to M1 and NOT to M2.\n\nIn the GF(3)­representation 5a, M11 has orbits 11 + 110 on points and orbits 22 + 220 on nonzero vectors.\nIn the GF(3)­representation 5b, M11 has orbits 55 + 66 on points and orbits 132 + 110 on nonzero vectors.\n\n### Maximal subgroups\n\nThe maximal subgroups of M11 are as follows.\n\n### Conjugacy classes\n\nA set of generators for the maximal cyclic subgroups can be obtained by running this program on the standard generators. All conjugacy classes can therefore be obtained as suitable powers of these elements, for example by running this program afterwards. All conjugacy classes can be obtained directly from the standard generators by running this program.\n\nRepresentatives of the 10 conjugacy classes of M11 are also given below.\n\n• 1A: identity [or a2].\n• 2A: a.\n• 3A: ab2ab2.\n• 4A: b.\n• 5A: abab2ab-1.\n• 6A: ab2.\n• 8A: abab2ab2.\n• 8B: ab-1ab2ab2.\n• 11A: ab.\n• 11B: ab-1.\n\n### Checks applied\n\nCheckDateBy whomRemarks\nLinks to (meataxe) representations work and have right degree and field 24.01.01RAW\nAll info from v1 is included\nValid W3C HTML 4.01 Transitional 04.02.02JNBThis property is very easy to disrupt\nHTML page standard\nWord program syntax24.01.01RAW\nWord programs applied\nAll necessary standard generators are defined24.01.01RAW\nAll representations are in standard generators", null, "Go to main ATLAS (version 2.0) page.", null, "Go to sporadic groups page.", null, "Go to old M11 page - ATLAS version 1.", null, "Anonymous ftp access is also available. See here for details.\n\nVersion 2.0 created on 13th April 1999.\nLast updated 21.12.04 by SJN.\nInformation checked to Level 1 on 03.12.99 by JNB.\nR.A.Wilson, R.A.Parker and J.N.Bray." ]
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https://elo.mastermath.nl/course/info.php?id=129
[ "### M1: Differential Geometry - 8EC\n\nAim of the course\n\nProvide an introduction to vector bundles (with application to tubular neighborhoods), principal bundles, connections, the general theory of geometric structures (G-structures) and their integrabiliy, mentioning examples such as Riemannian, complex or symplectic structures.\n\nThe course will have several parts:\n\n1. One which concentrates on vector-bundles and connections (including parallel transport, curvature and the construction of the first Chern class). Here we will also discuss the tubular neighborhood theorem.\n\n2. One which concentrates on principal bundles and connections, and where we explain that, for principal $GL_n$-bundles, the resulting theory is equivalent to the one for vector bundles. This part will start with 1-2 lectures about the basic notions/facts from Lie groups that are needed here.\n\n3. While along the way we will mention some examples of geometric structures (such as metrics), in the last parts of the course we will concentrate on a general framework that allows one to treat many geometric structures in an unified way: the framework provided by the theory of $G$-structures. Here we will present the framework and examples such as: Riemannian metrics, distributions, foliations, symplectic structures, almost complex and complex structures.\n\n4. Finally, in the last two lectures, we will concentrate on the integrability of G-structures and the torsion of G-structures as obstruction to integrability. For instance, for symplectic structures: one talks about almost symplectic structures (non-degenerate two forms) and their integrability is about the form being closed (Darboux theorem); for foliations one talks about sub-bundle of tangent bundle, and their integrability is equivalent to the involutivity (Frobenius theorem); etc etc.\n\nPrerequisites\n\n• a good knowledge of multi-variable calculus\n• some basic knowledge of topology (such as compactness)\n• the standard basic notions that are taught in the first course on Differential Geometry, such as: the notion of manifold, smooth maps, immersions and submersions, tangent vectors, Lie derivatives along vector fields, the flow of a vector field, the tangent space (and bundle), the definition of differential forms, de Rham operator (and hopefully the definition of de Rahm cohomology)\n• some very basic knowledge of Lie theory may be useful, but it is not required." ]
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https://www.effortlessmath.com/tag/squeeze-theorem/
[ "# Squeeze Theorem\n\nSearch in Squeeze Theorem articles.", null, "## How to Determine Limits Using the Squeeze Theorem?\n\nIf two functions squeeze together at a certain point, then any function trapped between them will get squeezed to that same point. The following step-by-step guide helps you determine limits using the Squeeze Theorem." ]
[ null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACAAAAAPCAQAAAAviQWcAAAAGElEQVR42mP8X89AEWAcNWDUgFEDBokBAH4nFnKVxVNEAAAAAElFTkSuQmCC", null ]
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https://numberworld.info/53530
[ "# Number 53530\n\n### Properties of number 53530\n\nCross Sum:\nFactorization:\n2 * 5 * 53 * 101\nDivisors:\n1, 2, 5, 10, 53, 101, 106, 202, 265, 505, 530, 1010, 5353, 10706, 26765, 53530\nCount of divisors:\nSum of divisors:\nPrime number?\nNo\nFibonacci number?\nNo\nBell Number?\nNo\nCatalan Number?\nNo\nBase 2 (Binary):\nBase 3 (Ternary):\nBase 4 (Quaternary):\nBase 5 (Quintal):\nBase 8 (Octal):\nd11a\nBase 32:\n1k8q\nsin(53530)\n-0.39197322872598\ncos(53530)\n-0.91997662359548\ntan(53530)\n0.42606868334769\nln(53530)\n10.887997523387\nlg(53530)\n4.7285972433834\nsqrt(53530)\n231.36551169092\nSquare(53530)\n\n### Number Look Up\n\nLook Up\n\n53530 (fifty-three thousand five hundred thirty) is a impressive number. The cross sum of 53530 is 16. If you factorisate 53530 you will get these result 2 * 5 * 53 * 101. The figure 53530 has 16 divisors ( 1, 2, 5, 10, 53, 101, 106, 202, 265, 505, 530, 1010, 5353, 10706, 26765, 53530 ) whith a sum of 99144. The number 53530 is not a prime number. The number 53530 is not a fibonacci number. The figure 53530 is not a Bell Number. The number 53530 is not a Catalan Number. The convertion of 53530 to base 2 (Binary) is 1101000100011010. The convertion of 53530 to base 3 (Ternary) is 2201102121. The convertion of 53530 to base 4 (Quaternary) is 31010122. The convertion of 53530 to base 5 (Quintal) is 3203110. The convertion of 53530 to base 8 (Octal) is 150432. The convertion of 53530 to base 16 (Hexadecimal) is d11a. The convertion of 53530 to base 32 is 1k8q. The sine of the number 53530 is -0.39197322872598. The cosine of the figure 53530 is -0.91997662359548. The tangent of 53530 is 0.42606868334769. The root of 53530 is 231.36551169092.\nIf you square 53530 you will get the following result 2865460900. The natural logarithm of 53530 is 10.887997523387 and the decimal logarithm is 4.7285972433834. You should now know that 53530 is great number!" ]
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{"ft_lang_label":"__label__en","ft_lang_prob":0.7707437,"math_prob":0.99493504,"size":2004,"snap":"2019-51-2020-05","text_gpt3_token_len":707,"char_repetition_ratio":0.1795,"word_repetition_ratio":0.24074075,"special_character_ratio":0.48752496,"punctuation_ratio":0.16504854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99706566,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T20:26:25Z\",\"WARC-Record-ID\":\"<urn:uuid:5755dc98-a297-438e-b0ff-204f9e834abc>\",\"Content-Length\":\"13829\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:52b01c36-e758-4485-b008-2ff4b406e9b8>\",\"WARC-Concurrent-To\":\"<urn:uuid:42a4ab94-3668-408e-bb05-e526340bcb55>\",\"WARC-IP-Address\":\"176.9.140.13\",\"WARC-Target-URI\":\"https://numberworld.info/53530\",\"WARC-Payload-Digest\":\"sha1:BDKIOUPNOHEK5MTAWZUB6URMAMNTSUGV\",\"WARC-Block-Digest\":\"sha1:BQB2TJBGG73VE6FR3ETUD6QS7FCP43AL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250613416.54_warc_CC-MAIN-20200123191130-20200123220130-00488.warc.gz\"}"}
http://xqpbx.com/cn/categorypage-836475-1.html
[ "微信：13662481022\n\n• 公司动态\n[2016-10-05]\n\n[2016-10-04]\n\n[2016-10-03]\n\n[2016-10-02]\n\n[2016-10-01]\n\n[2016-09-30]\n\n[2016-08-12]\n\n[2016-07-12]\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`" ]
[ null ]
{"ft_lang_label":"__label__zh","ft_lang_prob":0.82451284,"math_prob":0.98954046,"size":1484,"snap":"2019-26-2019-30","text_gpt3_token_len":1501,"char_repetition_ratio":0.12972973,"word_repetition_ratio":0.0,"special_character_ratio":0.32345015,"punctuation_ratio":0.34074074,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9897171,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-16T18:47:44Z\",\"WARC-Record-ID\":\"<urn:uuid:454f030a-fd86-4e4f-bff9-facce51084c2>\",\"Content-Length\":\"184974\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54395857-d0ff-4ac7-b25e-614f5c56f523>\",\"WARC-Concurrent-To\":\"<urn:uuid:fad9f7e9-9a79-4cb3-b33c-036d1b9d7389>\",\"WARC-IP-Address\":\"154.213.74.98\",\"WARC-Target-URI\":\"http://xqpbx.com/cn/categorypage-836475-1.html\",\"WARC-Payload-Digest\":\"sha1:L5C6W74MOHRXO64RZGITHPCMABNFDLEM\",\"WARC-Block-Digest\":\"sha1:CZ63DJGONYOD2EN2M726ZU556GMKHIOL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998291.9_warc_CC-MAIN-20190616182800-20190616204800-00130.warc.gz\"}"}
https://brilliant.org/problems/floor-floor-floor-floor/
[ "Floor Floor Floor Floor\n\nAlgebra Level 5\n\nGiven the equation\n\n$x \\lfloor x \\lfloor x \\lfloor x \\rfloor \\rfloor \\rfloor = 41$\n\nIf $\\alpha$ denote the largest negative value of $x$ which satisfy the equation above, while $\\beta$ denote the smallest positive value of $x$ which satisfy the equation above.\n\nWhat is the value of $17\\alpha + 28\\beta$?\n\n×" ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.740464,"math_prob":1.0000056,"size":357,"snap":"2019-43-2019-47","text_gpt3_token_len":75,"char_repetition_ratio":0.13597734,"word_repetition_ratio":0.29090908,"special_character_ratio":0.21288516,"punctuation_ratio":0.23684211,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T03:45:44Z\",\"WARC-Record-ID\":\"<urn:uuid:8e3c928f-8b15-4049-b6cd-c1a1e4a454de>\",\"Content-Length\":\"40307\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd9064fc-5a1e-47d6-9f07-f9fe022a4201>\",\"WARC-Concurrent-To\":\"<urn:uuid:58cdfa56-e03d-47eb-a998-338f2c97a587>\",\"WARC-IP-Address\":\"104.20.34.242\",\"WARC-Target-URI\":\"https://brilliant.org/problems/floor-floor-floor-floor/\",\"WARC-Payload-Digest\":\"sha1:EETHGZU3SUZOZIG4YIC4H7GEVIOR4H7H\",\"WARC-Block-Digest\":\"sha1:U7YFP4HAQBTKDVDRDZNN76T4RNH5BBMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987828425.99_warc_CC-MAIN-20191023015841-20191023043341-00431.warc.gz\"}"}
https://www.geeksforgeeks.org/square-root-of-a-number-without-using-sqrt-function/?ref=lbp
[ "# Square root of a number without using sqrt() function\n\nGiven a number N, the task is to find the square root of N without using sqrt() function.\n\nExamples:\n\nInput: N = 25\nOutput: 5\n\nInput: N = 3\nOutput: 1.73205\n\nInput: N = 2.5\nOutput: 1.58114\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach:\n\n• Start iterating from i = 1. If i * i = n, then print i as n is a perfect square whose square root is i.\n• Else find the smallest i for which i * i is strictly greater than n.\n• Now we know square root of n lies in the interval i – 1 and i and we can use Binary Search algorithm to find the square root.\n• Find mid of i – 1 and i and compare mid * mid with n, with precision upto 5 decimal places.\n1. If mid * mid = n then return mid.\n2. If mid * mid < n then recur for the second half.\n3. If mid * mid > n then recur for the first half.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Recursive function that returns square root ` `// of a number with precision upto 5 decimal places ` `double` `Square(``double` `n, ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / 2; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) || (``abs``(mul - n) < 0.00001)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n, recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = 1; ` ` `  `    ``// While the square root is not found ` `    ``bool` `found = ``false``; ` `    ``while` `(!found) { ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n) { ` `            ``cout << fixed << setprecision(0) << i; ` `            ``found = ``true``; ` `        ``} ` `        ``else` `if` `(i * i > n) { ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - 1, i); ` `            ``cout << fixed << setprecision(5) << res; ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``double` `n = 3; ` ` `  `    ``findSqrt(n); ` ` `  `    ``return` `0; ` `} `\n\n## Java\n\n `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Recursive function that returns  ` `// square root of a number with  ` `// precision upto 5 decimal places ` `static` `double` `Square(``double` `n,  ` `                     ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / ``2``; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) ||    ` `        ``(Math.abs(mul - n) < ``0.00001``)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n,  ` `    ``// recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `static` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = ``1``; ` ` `  `    ``// While the square root is not found ` `    ``boolean` `found = ``false``; ` `    ``while` `(!found)  ` `    ``{ ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n)  ` `        ``{ ` `            ``System.out.println(i); ` `            ``found = ``true``; ` `        ``} ` `         `  `        ``else` `if` `(i * i > n)  ` `        ``{ ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - ``1``, i); ` `            ``System.out.printf(``\"%.5f\"``, res); ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``double` `n = ``3``; ` ` `  `    ``findSqrt(n); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `\n\n## Python3\n\n `# Python3 implementation of the approach ` `import` `math ` ` `  `# Recursive function that returns square root ` `# of a number with precision upto 5 decimal places ` `def` `Square(n, i, j): ` ` `  `    ``mid ``=` `(i ``+` `j) ``/` `2``; ` `    ``mul ``=` `mid ``*` `mid; ` ` `  `    ``# If mid itself is the square root, ` `    ``# return mid ` `    ``if` `((mul ``=``=` `n) ``or` `(``abs``(mul ``-` `n) < ``0.00001``)): ` `        ``return` `mid; ` ` `  `    ``# If mul is less than n, recur second half ` `    ``elif` `(mul < n): ` `        ``return` `Square(n, mid, j); ` ` `  `    ``# Else recur first half ` `    ``else``: ` `        ``return` `Square(n, i, mid); ` ` `  `# Function to find the square root of n ` `def` `findSqrt(n): ` `    ``i ``=` `1``; ` ` `  `    ``# While the square root is not found ` `    ``found ``=` `False``; ` `    ``while` `(found ``=``=` `False``): ` ` `  `        ``# If n is a perfect square ` `        ``if` `(i ``*` `i ``=``=` `n): ` `            ``print``(i); ` `            ``found ``=` `True``; ` `         `  `        ``elif` `(i ``*` `i > n): ` ` `  `            ``# Square root will lie in the ` `            ``# interval i-1 and i ` `            ``res ``=` `Square(n, i ``-` `1``, i); ` `            ``print` `(``\"{0:.5f}\"``.``format``(res))  ` `            ``found ``=` `True` `        ``i ``+``=` `1``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``n ``=` `3``; ` ` `  `    ``findSqrt(n); ` ` `  `# This code is contributed by 29AjayKumar `\n\n## C#\n\n `// C# implementation of the approach ` `using` `System; ` `     `  `class` `GFG  ` `{ ` ` `  `// Recursive function that returns  ` `// square root of a number with  ` `// precision upto 5 decimal places ` `static` `double` `Square(``double` `n,  ` `                     ``double` `i, ``double` `j) ` `{ ` `    ``double` `mid = (i + j) / 2; ` `    ``double` `mul = mid * mid; ` ` `  `    ``// If mid itself is the square root, ` `    ``// return mid ` `    ``if` `((mul == n) ||  ` `        ``(Math.Abs(mul - n) < 0.00001)) ` `        ``return` `mid; ` ` `  `    ``// If mul is less than n,  ` `    ``// recur second half ` `    ``else` `if` `(mul < n) ` `        ``return` `Square(n, mid, j); ` ` `  `    ``// Else recur first half ` `    ``else` `        ``return` `Square(n, i, mid); ` `} ` ` `  `// Function to find the square root of n ` `static` `void` `findSqrt(``double` `n) ` `{ ` `    ``double` `i = 1; ` ` `  `    ``// While the square root is not found ` `    ``Boolean found = ``false``; ` `    ``while` `(!found)  ` `    ``{ ` ` `  `        ``// If n is a perfect square ` `        ``if` `(i * i == n)  ` `        ``{ ` `            ``Console.WriteLine(i); ` `            ``found = ``true``; ` `        ``} ` `         `  `        ``else` `if` `(i * i > n)  ` `        ``{ ` ` `  `            ``// Square root will lie in the ` `            ``// interval i-1 and i ` `            ``double` `res = Square(n, i - 1, i); ` `            ``Console.Write(``\"{0:F5}\"``, res); ` `            ``found = ``true``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``double` `n = 3; ` ` `  `    ``findSqrt(n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `\n\nOutput:\n\n```1.73205\n```\n\nAttention reader! 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https://patents.google.com/patent/KR20030029443A/en
[ "# KR20030029443A - Regulator circuit and control method thereof - Google Patents\n\n## Info\n\nPublication number\nKR20030029443A\nKR20030029443A KR1020020019889A KR20020019889A KR20030029443A KR 20030029443 A KR20030029443 A KR 20030029443A KR 1020020019889 A KR1020020019889 A KR 1020020019889A KR 20020019889 A KR20020019889 A KR 20020019889A KR 20030029443 A KR20030029443 A KR 20030029443A\nAuthority\nKR\nSouth Korea\nPrior art keywords\nvoltage\noutput\ncurrent\ncharging\ncontrol\nPrior art date\nApplication number\nKR1020020019889A\nOther languages\nKorean (ko)\nOther versions\nKR100844052B1 (en\nInventor\n다키모토규이치\n나가야요시히로\n마츠야마도시유키\nOriginal Assignee\n후지쯔 가부시끼가이샤\nPriority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)\nFiling date\nPublication date\nPriority to JPJP-P-2001-00310088 priority Critical\nPriority to JP2001310088A priority patent/JP3968228B2/en\nApplication filed by 후지쯔 가부시끼가이샤 filed Critical 후지쯔 가부시끼가이샤\nPublication of KR20030029443A publication Critical patent/KR20030029443A/en\nApplication granted granted Critical\nPublication of KR100844052B1 publication Critical patent/KR100844052B1/en\n\n• 230000001276 controlling effects Effects 0.000 claims description 11\n• 238000007600 charging Methods 0.000 abstract description 174\n• 238000010586 diagrams Methods 0.000 description 26\n• 230000003071 parasitic Effects 0.000 description 24\n• 238000010277 constant-current charging Methods 0.000 description 21\n• 238000010280 constant potential charging Methods 0.000 description 15\n• 230000000576 supplementary Effects 0.000 description 5\n• 208000006897 Interstitial Lung Diseases Diseases 0.000 description 4\n• 230000003321 amplification Effects 0.000 description 4\n• 238000006243 chemical reactions Methods 0.000 description 4\n• 230000000875 corresponding Effects 0.000 description 4\n• 238000003199 nucleic acid amplification method Methods 0.000 description 4\n• 239000003990 capacitor Substances 0.000 description 2\n• 230000003247 decreasing Effects 0.000 description 2\n• 230000001151 other effects Effects 0.000 description 2\n• 230000003068 static Effects 0.000 description 2\n• 201000000194 ITM2B-related cerebral amyloid angiopathy 2 Diseases 0.000 description 1\n• 230000002411 adverse Effects 0.000 description 1\n• 230000000052 comparative effects Effects 0.000 description 1\n• 229910001416 lithium ion Inorganic materials 0.000 description 1\n• 230000003334 potential Effects 0.000 description 1\n• 230000001172 regenerating Effects 0.000 description 1\n• 239000004065 semiconductors Substances 0.000 description 1\n• 230000001052 transient Effects 0.000 description 1\n\n## Classifications\n\n• HELECTRICITY\n• H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER\n• H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY\n• H02J7/00Circuit arrangements for charging or depolarising batteries or for supplying loads from batteries\n• H02J7/02Circuit arrangements for charging or depolarising batteries or for supplying loads from batteries for charging batteries from ac mains by converters\n• H02J7/04Regulation of charging current or voltage\n• H02J7/042Regulation of charging current or voltage the charge cycle being controlled in response to a measured parameter\n• H02J7/045Regulation of charging current or voltage the charge cycle being controlled in response to a measured parameter in response to voltage or current\n• HELECTRICITY\n• H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER\n• H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY\n• H02J7/00Circuit arrangements for charging or depolarising batteries or for supplying loads from batteries\n• H02J7/007Regulation of charging or discharging current or voltage\n• H02J7/0071Regulation of charging or discharging current or voltage with a programmable schedule\n• HELECTRICITY\n• H02GENERATION; CONVERSION OR DISTRIBUTION OF ELECTRIC POWER\n• H02JCIRCUIT ARRANGEMENTS OR SYSTEMS FOR SUPPLYING OR DISTRIBUTING ELECTRIC POWER; SYSTEMS FOR STORING ELECTRIC ENERGY\n• H02J7/00Circuit arrangements for charging or depolarising batteries or for supplying loads from batteries\n• H02J7/02Circuit arrangements for charging or depolarising batteries or for supplying loads from batteries for charging batteries from ac mains by converters\n• H02J7/04Regulation of charging current or voltage\n• H02J7/06Regulation of charging current or voltage using discharge tubes or semiconductor devices\n\n## Abstract\n\nAn object of the present invention is to provide a regulator circuit capable of precisely adjusting and maintaining an output voltage in the entire current region by correcting the setting of a predetermined voltage to be adjusted according to the charging current.\nThe output terminal V3 of the amplifier 112 is connected to the reference voltage terminal V2 through the resistor element R3. When the charging current IICH does not flow, the reference power supply VREF at which the detection voltage V3 becomes 0 V is divided by the parallel resistance of the resistor R1 and the resistors R2 and R3. When the charging current ICHG flows and the detection voltage V3 is outputted, the detection voltage V3 is divided by the resistance elements R3 and R2 to generate a correction voltage according to the charging current ICHG. Even when a voltage drop ΔVLS occurs between the charge control device output terminal VO and the battery terminal VBAT, the control voltage of the output voltage VO is properly adjusted to maintain the battery terminal VBAT at the full charge voltage VBAT0. Can be.\n\n## Description\n\nRegulator circuit and its control method {REGULATOR CIRCUIT AND CONTROL METHOD THEREOF}\n\nBACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a regulator circuit for adjusting and maintaining an output terminal to a predetermined direct current voltage, and more particularly to a regulator circuit for charging a battery connected to an output terminal or for applying a predetermined direct current voltage to a load.\n\nMPUs, LSIs, ICs, and other semiconductor devices mounted in various electronic devices, drive motors such as HDDs, FDDs, and the like operate using DC voltages such as DC 5V and 3.3V as power sources. Therefore, these electronic devices require a DC / DC converter for converting the DC voltage converted from AC 100 V or the like into a desired DC voltage through the AC adapter. In addition, depending on the complexity of the system, several types of power supply systems may be required, and the output voltage of the DC / DC converter may be reconverted by a separate DC / DC converter. These DC / DC converters are generally composed of switching regulators or series regulators in consideration of the conversion voltage difference, required power supply capacity and precision.\n\nIn recent years, in portable devices such as laptops and cell phones, a rechargeable battery may be mounted instead of an AC adapter as a DC power source. In these systems, the recharging operation is performed by being connected to an AC adapter or the like, and a charging control device is provided for controlling the charging current to the battery, controlling the battery voltage at the time of full charge, and the like.\n\n13 shows a charge control device 100 of the prior art. The DC voltage output from the AC adapter 102 is smoothed by the coil L1 and the capacitor C1 by switching control of the PMOS driver Tr1, and then is supplied to the battery 101 through the charging current detection resistor RS1. Is charged. The diode D1 is a regenerative flyback diode of the charging current ICHG.\n\nSwitching control of the PMOS driver Tr1 is performed by the charge control circuit 111. The voltage between the terminals of the charging current detection resistor RS1 is amplified by the amplifier 112, and the error amplifier 116 outputs a control voltage for constant current charging by error amplifying the voltage difference from the reference voltage V1. do. In addition, the output voltage VO of the charge control device output terminal VO, which is the battery side terminal of the charge current detection resistor RS1, is divided by the resistor elements R110 and R120, and the error amplifier 118 is a reference voltage terminal. The control voltage for constant voltage charging is output by error-amplifying the potential difference with the reference voltage V2 at (V2). Both control voltages are input to the comparator 120 and the switching duty is determined by comparing with the oscillation signal from the oscillator (OSC) 122. When the switching duty is determined by the constant current charging control voltage by the error amplifier 116, the constant current charging control is performed, and the charging current ICHG to the battery 101 becomes a predetermined charging current ICHGM. When the switching duty is determined by the control voltage for constant voltage charging by the error amplifier 118, the constant voltage charging control is performed, and the output voltage VO is maintained at a predetermined full charge voltage VBAT0 to perform charging.\n\nIn the charge control method of the charge control circuit 111 in FIG. 13, the charge to the battery 101 is performed from the start of charge until the output voltage VO of the charge control device output terminal VO reaches the full charge voltage VBAT0. The current ICHG is controlled by the error amplifier 116 so that rapid charging is performed at the predetermined charging current ICHGM. When charging to the battery 101 proceeds and the output voltage VO of the charge control device output terminal VO reaches the full charge voltage VBAT0, the control voltage for constant current charging and the error amplifier output by the error amplifier 116 are output. The control voltage for constant voltage charging output by 118 reverses, and the charging control switches from the constant current control to the constant voltage control. By continuing the charging operation while maintaining the output voltage VO of the charge control device output terminal VO at the full charge voltage VBAT0, the charging current ICHG is decreased from the predetermined charging current ICHGM and finally Charging to the battery 101 is completed when the charging current ICHG disappears.\n\nSince the connection between the charge control device 100 and the battery 101 is connected through a connector, a switch, or the like, a resistance component such as a contact resistance exists in a connection portion such as a connector. The wiring resistance of the connection wiring itself is applied to this contact resistance or the like, and the parasitic resistance RLS1 is inserted into the connection path. Since the charging current ICHG to the battery 101 flows into the battery 101 through the parasitic resistance RLS1, a voltage drop ΔVLS occurs for the charging current ICHGM in the constant current control state. Compared to the output voltage VO of the charging control device output terminal VO, the battery terminal voltage VBAT is lowered by the drop voltage ΔVLS. Since the charge control device 100 performs constant voltage control with respect to the output voltage V0 of the charge control device output terminal V0, the constant current charge control proceeds, and as the battery terminal voltage VBAT rises, the charge control device output terminal VO ), The output voltage VO also rises, and the charging control is switched from the constant current control to the constant voltage control when the charging control device output terminal VO reaches the full charge voltage VBAT0.\n\nHowever, at this point, the battery terminal voltage VBAT does not reach the full charge voltage VBAT0 of the battery 101, but is charged at a voltage level as low as the drop voltage ΔVLS from the full charge voltage VBAT0. Beats. That is, of course, when fast charging can be controlled by the static charge control, the static charge control time is shortened by the voltage drop ΔVLS due to the parasitic resistance RLS1, and the charging time until the full charge of the battery 101 is achieved. There is a problem that becomes long.\n\nThis is shown in FIG. As the charging current ICHGM flows through the parasitic resistor RLS1 during the constant current charging control, the output voltage VO of the charging control device output terminal VO corresponds to the dropping voltage ΔVLS compared to the battery terminal voltage VBAT. As high as possible. For this reason, before the battery terminal voltage VBAT reaches the predetermined voltage VBAT0 at full charge, the charging control is switched to the constant voltage charging control, and the constant current charging control time becomes shorter than the original time. Subsequent control is performed by the constant voltage charging control and further charges the battery terminal voltage VBAT as much as the drop voltage ΔVLS, while the constant voltage charging control has already reached the full charge voltage VBAT0. Since it is performed by keeping (VO) at the voltage VBAT0, the switching duty of the PMOS driver Tr1 inevitably decreases. Therefore, the charging current ICHG in the prior art decreases with a fast charging time with respect to the ideal charging current ICHG_I, and as a result, the battery voltage VBAT becomes the full charge voltage VBAT0 with respect to the ideal battery voltage VBAT_I. Charge time to) becomes long. The long full charge time tx is required as compared with the full charge time t0 when the original constant current charge control is performed.\n\nIn addition, the influence of the voltage drop ΔVLS caused by the parasitic resistance RLS1 described above is equal to the voltage drop ΔVLS of the full charge voltage at the time of constant voltage control with respect to the full charge voltage VBAT0 that satisfies the specification of the battery 101. It is possible to avoid by adding and setting the voltage. However, in this case, there is a problem that the battery terminal voltage VBAT at the time of full charge becomes higher than the voltage suitable for the specification and the voltage stress on the battery 101 increases, which may adversely affect the battery performance.\n\nIf constant current control is used for current limit control during overload, the charge control circuit 111 can be used as a regulator circuit for supplying a constant voltage to the load. Also in this case, since the parasitic resistance RLS1 is interposed between the output terminal VO and the load, the voltage applied to the load corresponds to the drop voltage ΔVLS from the predetermined voltage in the large load current region reaching the overload state. The problem is that there is a possibility that the voltage drops as much as possible and the voltage cannot be applied precisely to the load in all the load current regions.\n\nThe present invention has been made to solve the problems of the prior art, and by correcting the setting of a predetermined voltage to be adjusted according to the charging current or the load current, the output voltage can be precisely adjusted and maintained in all the charging current or load current regions. It is an object to provide a regulator circuit.\n\n1 is a circuit block diagram showing a regulator circuit (charge control device) of the first embodiment.\n\nFig. 2 is a circuit diagram showing a reference voltage setting circuit of the first embodiment.\n\n3 is a characteristic diagram showing charging current detection characteristics in an output current detector.\n\nFig. 4 is a characteristic diagram showing the charging current characteristic of the reference voltage of the first embodiment.\n\nFig. 5 is a characteristic diagram showing battery charging characteristics of the first to third embodiments.\n\n6 is a circuit block diagram showing a regulator circuit (charge control device) of the second embodiment.\n\nFig. 7 is a circuit block diagram showing the amplifier of the second embodiment.\n\nFig. 8 is a circuit block diagram showing the regulator circuit (charge control device) of the third embodiment.\n\nFig. 9 is a characteristic diagram showing the charging current characteristic of the reference voltage of the third embodiment.\n\nFig. 10 is a circuit block diagram showing the regulator circuit (switching regulator) of the fourth embodiment.\n\nFig. 11 is a circuit block diagram showing the regulator circuit (serial regulator) of the fifth embodiment.\n\n12 is a circuit diagram showing an example of an amplifier.\n\nFig. 13 is a circuit block diagram showing a regulator circuit (charge control device) of the prior art.\n\n14 is a characteristic diagram showing battery charging characteristics of the prior art;\n\n<Explanation of symbols for main parts of drawing>\n\n1, 2, 3, 100: charge control device\n\n4: switching regulator\n\n5: serial regulator\n\n21, 31, 111: charge control circuit\n\n22, 112: amplifier\n\n25, 62: current mirror circuit\n\n101: battery\n\n116, 118: error amplifier\n\n120: comparator\n\n122: oscillator (OSC)\n\n124: inverted output buffer\n\nR1, R2, R3, R4, R5, R6, R7, R8, R9, R101, R102, R110, R120: resistance element\n\nRS1: Charge Current Detection Resistor\n\nRLS1: Parasitic Resistance\n\nTr1, Tr2: PMOS driver\n\nIO: Detection current terminal\n\nOUT: Output terminal of inverted output buffer\n\nRS-, RS +: Output Current Detection Terminal\n\nV2: reference voltage terminal\n\nV3: Output terminal of the amplifier\n\nVBAT: Battery Output Terminal\n\nVFB: Inverting input terminal of error amplifier\n\nVO: Charge control output terminal\n\nICHG, ICHGM: Charge Current\n\nVREF: Reference Power\n\nIn order to achieve the above object, the regulator circuit according to claim 1 controls an output voltage controller for controlling an output voltage based on a reference voltage, an output current detector, and a reference voltage based on an output current detected by the output current detector. And a reference voltage corrector.\n\nIn the regulator circuit of claim 1, the output voltage control unit controls the output voltage to a reference voltage controlled by the reference voltage corrector based on the output current detected by the output current detector.\n\nAccordingly, even when the output current flows through the parasitic load component existing on the ammeter path, a deviation of the voltage value occurs in the output voltage on the current path that should be originally at the same potential. Because of this control, the output voltage at a predetermined position on the current path can be adjusted appropriately. On the current path, even when the position where the output voltage is controlled by the output voltage controller and the position where the controlled output voltage is to be extracted is separated, the desired output voltage can be extracted.\n\nFurther, the regulator circuit according to the present invention may be provided with an output current controller for controlling the output current based on the detection result from the output current detector. Thus, the output voltage can be controlled in accordance with the output current, and the output current can also be controlled.\n\nThe regulator circuit according to claim 2 is characterized in that, in the regulator circuit according to claim 1, the output current detector includes a current voltage converter and controls the reference voltage corrector based on the detected voltage according to the output current. In addition, the detection voltage is preferably output with a predetermined proportional coefficient with respect to the output current. In addition, it is preferable that the predetermined proportional coefficient can be externally adjusted.\n\nIn the regulator circuit of Claim 2, the detection result according to the output current is output as a detection voltage by the current voltage converter provided in the output current detection part, and a reference voltage correction part is controlled. In addition, the relationship of the detection voltage to the output current is maintained at a predetermined proportional coefficient, and is appropriately changed by adjustment from the outside.\n\nAs a result, the detection result of the output current can be obtained as the detection voltage, which is suitable for the case where the control signal to the reference voltage corrector is composed of a voltage signal. In addition, if the proportional coefficient relationship of the detection voltage with respect to the output current can be adjusted externally, adjustment of the detection voltage with respect to the output current can be performed simply and easily with respect to other current path characteristics.\n\nThe regulator circuit according to claim 3 is characterized in that, in the regulator circuit according to claim 1, the output current detector includes a current output unit and controls the reference voltage corrector based on the detection current according to the output current. In addition, the detection current is preferably output with a predetermined proportional coefficient with respect to the output current. In addition, it is preferable that the predetermined proportional coefficient can be externally adjusted.\n\nIn the regulator circuit of Claim 3, the current output part provided in the output current detection part outputs the detection result according to the output current as a detection current, and controls a reference voltage correction part. In addition, the relationship between the detection current and the output current is maintained at a predetermined proportional coefficient, and is appropriately changed by adjustment from the outside.\n\nAs a result, the detection result of the output current can be obtained as the detection current, which is suitable for the case where the control signal to the reference voltage corrector is constituted by the current signal. In addition, if the proportional coefficient relationship of the detection current to the output current can be adjusted externally, adjustment of the detection current to the output current can be easily and easily performed for other current path characteristics.\n\nThe regulator circuit according to claim 4 is characterized in that, in the regulator circuit according to any one of claims 1 to 3, the reference voltage corrector increases or decreases the reference voltage with a positive or negative correlation with respect to the increase or decrease of the output current. do.\n\nAccordingly, when the output current is large, the reference voltage can be set high, and when the output current is small, the reference voltage can be set low, or when the output current is large, the reference voltage is set low, and when the output current is low, Since the reference voltage can be set high, even when the output current flows through parasitic resistance components such as contact resistance and wiring resistance existing on the ammeter path, even if a voltage drop occurs in the output voltage on the current path, which must be at the same potential. The output voltage at a predetermined position on the current path can be appropriately adjusted.\n\nThat is, when the position on the current path where the output voltage is controlled by the output voltage controller is upstream than the position where the output voltage is controlled, the correction voltage corresponding to the voltage drop during that time is adjusted. Since the reference voltage is set high by addition, the output voltage to be extracted can be adjusted to a desired voltage value. In addition, when the position on the current path where the output voltage is controlled by the output voltage control unit is downstream from the position where the controlled output voltage is to be extracted, the amount corresponding to the voltage drop in the meantime is determined. By subtracting and setting the reference voltage low, the extracted output voltage can be adjusted to the desired voltage value. Even when the position where the output voltage is controlled and the position where the controlled output voltage is to be extracted are separated from each other, the desired output voltage can be extracted.\n\nIn addition, the regulator circuit according to claim 5 is characterized in that, in the regulator circuit according to at least one of claims 1 to 4, the reference voltage correction unit outputs a basic reference voltage when no output current flows, and according to the output current. And a correction voltage unit overlapping the changed correction voltage with the basic reference voltage.\n\nIn the regulator circuit of claim 5, when the output current does not flow by the basic reference voltage unit, a basic reference voltage is output, and the correction voltage unit superimposes the correction voltage according to the output current on the basic reference voltage to reference the output current. Generate a voltage.\n\nAs a result, the output voltage when the output current does not flow by the basic reference voltage can be precisely adjusted, and as much as the correction voltage can be added as the correction voltage, the output current does not flow. The reference voltage can be controlled according to the output current in all output current areas, including the non-state.\n\nIn the regulator circuit according to claim 6, in the regulator circuit according to claim 5, the basic reference voltage portion includes a constant voltage source or a constant voltage source and a first voltage divider, and the basic reference voltage is a predetermined voltage or first voltage divider output from the constant voltage source. It is characterized in that the divided voltage of the predetermined voltage by.\n\nThe regulator circuit of claim 6 generates a basic reference voltage by dividing a predetermined voltage output by the constant voltage source provided in the basic reference voltage unit, or a predetermined voltage output from the constant voltage source. .\n\nAccordingly, by dividing the basic reference voltage in the constant voltage source or additionally provided in the first voltage dividing section, the basic reference voltage can be set simply, easily and reliably.\n\nThe regulator circuit according to claim 7 is characterized in that, in the regulator circuit according to claim 6, the correction voltage portion includes a second voltage divider and divides the detection voltage.\n\nIn the regulator circuit of claim 7, the second voltage divider provided in the correction voltage unit divides the detection voltage output from the output current detector to generate a correction voltage according to the output current.\n\nThereby, the correction voltage can be obtained by dividing the detection voltage output from the output current detector. The correction voltage can be obtained by using the detected voltage as it is.\n\nThe regulator circuit according to claim 8 is characterized in that, in the regulator circuit according to claim 6, the correction voltage portion includes a first voltage divider and flows a detection current to the first voltage divider.\n\nIn the regulator circuit of claim 8, a detection current output from the output current detector is flowed into the first voltage divider provided in the basic reference voltage unit to generate a correction voltage according to the output current.\n\nThereby, the correction voltage can be obtained by flowing the detection current output from the output current detector by using the first voltage divider of the basic reference voltage portion to the first voltage divider. The correction voltage can be obtained by using the basic reference voltage portion and the detection current as it is, which is suitable.\n\nFurthermore, the regulator circuit according to claim 9 has a basic reference voltage section for outputting a basic reference voltage when no output current flows in the regulator circuit according to at least one of claims 1 to 4, and the output voltage control section includes a basic reference voltage. A first reference voltage terminal to which a voltage is input, and a second reference voltage terminal to which a correction reference voltage that changes according to an output current is input, and either one selected according to the magnitude relationship between the basic reference voltage and the correction reference voltage. The output voltage is controlled as a reference voltage. The regulator circuit according to claim 10 is characterized in that in the regulator circuit according to claim 9, a correction reference voltage is generated based on the detected voltage or the detected current.\n\nIn the regulator circuit of claim 9, the reference voltage when the output current output by the basic reference voltage unit does not flow and the correction reference voltage which changes according to the output current are input to the first and second reference voltage terminals of the output voltage controller, respectively. The output voltage is controlled according to the magnitude relationship of both reference voltages.\n\nAccordingly, the basic reference voltage and the correction reference voltage can be set independently of each other, and both adjustments can be made simply and easily and surely. Further, the detection voltage or the detection current output from the output current detection unit can be used as it is and adjusted as a correction reference voltage.\n\nFurther, in the regulator circuit according to the present invention, the constant current charge control is performed when the battery is charged to the output voltage by performing constant current charge control at the initial stage of charging by the output current controller and constant voltage charge control at the late stage of charge by the output voltage controller. It is preferable to set the reference voltage in the constant voltage charge control higher than the reference voltage at the time of full charge in accordance with the charging current in. Accordingly, even when the battery is charged and the output voltage of the battery when switching from the constant current charging control to the constant voltage charging control is separated from the position controlled by the output voltage controller on the charging current path, It may be closer to the predetermined voltage at the time of full charge. As the constant current charging period in the total charging time becomes longer, the total charging time to the battery can be shortened.\n\nFurther, in the regulator circuit according to the present invention, the current limit control at the time of overload is performed by the output current control section, and the constant voltage control of the output voltage is performed by the output voltage control section to supply a predetermined voltage to the load. It is preferable to set the reference voltage higher than the reference voltage at no load according to the load current. Accordingly, even when the voltage drop on the current path becomes large in a large load current region reaching an overload state, a predetermined voltage can be applied to the load. Predetermined voltage can be applied to the load in all load current regions.\n\nEMBODIMENT OF THE INVENTION Hereinafter, the 1st-5th embodiment which embodied about the regulator circuit of this invention is described in detail, referring drawings for based on FIG.\n\n1 is a circuit block diagram showing the regulator circuit (charge control device) of the first embodiment. 2 is a circuit diagram showing a reference voltage setting circuit in the first embodiment. 3 is a characteristic diagram illustrating charging current detection characteristics in the output current detection unit. 4 is a characteristic diagram showing the charging current characteristic of the reference voltage of the first embodiment. Fig. 5 is a characteristic diagram showing battery charging characteristics of the first to third embodiments. Fig. 6 is a circuit block diagram showing the regulator circuit (charge control device) of the second embodiment. 7 is a circuit block diagram showing an amplifier of the second embodiment. Fig. 8 is a circuit block diagram showing the regulator circuit (charge control device) of the third embodiment. 9 is a characteristic diagram showing the charging current characteristic of the reference voltage of the third embodiment. Fig. 10 is a circuit block diagram showing the regulator circuit (switching regulator) of the fourth embodiment. Fig. 11 is a circuit block diagram showing the regulator circuit (serial regulator) of the fifth embodiment. 12 is a circuit diagram showing a specific example of the amplifier.\n\nIn the first embodiment shown in FIG. 1, the regulator circuit of the present invention is implemented as the charge control device 1. 1 illustrates a power supply portion in an electronic device such as a portable device. The electronic device receives a direct current voltage from the AC adapter 102 through the diode D2 and also receives a direct current voltage VBAT from the battery 101 through the diode D3. Usually, in such a system, as shown in FIG. 1, the high voltage side is set so that the high voltage side of the output voltage of the AC adapter 102 and the output voltage VBAT of the battery 101 may be supplied. When the AC adapter 102 is connected, the DC voltage is supplied from the AC adapter 102. When the AC adapter 102 is disconnected, the DC voltage VBAT is supplied from the battery 101. At this time, the part which controls the charge to the battery 101 required is the charge control apparatus 1 shown in FIG.\n\nThe output terminal of the AC adapter 102 is connected to the source terminal of the PMOS driver Tr1, and the drain terminal thereof is connected to one end of the coil L1 and to the cathode terminal of the diode D1. It is. The anode terminal of the diode D1 is connected to the ground potential and is configured as a flyback diode. The other end of the coil L1 is connected to one end of the capacitor C1 which connected the other end to the ground potential. The other end of the coil L1 is connected to the output terminal VBAT of the battery 101 via the charging current detection resistor RS1. In the connection to the battery 101, it is common through a connector, a switch, etc., and contact resistance etc. exist by these connections. Moreover, the wiring resistance by the connection wiring in between also exists. In consideration of these parasitic resistance components, the parasitic resistance RLS1 is inserted in FIG.\n\nSwitching control of the PMOS driver Tr1 is performed by the charge control circuit 111. Both ends of the charging current detection resistor RS1 are connected to the output current detection terminals RS-, RS + of the charging control circuit 111, and are connected to the inverting input terminal and the non-inverting input terminal of the amplifier 112. The charging current detection resistor RS1 is input to the inverting input terminal of the error amplifier 116. In addition, the reference voltage V1 is input to the non-inverting input terminal of the error amplifier 116.\n\nThe output terminal V3 of the amplifier 112 is again connected to the reference voltage terminal V2 through the resistor element R3. The reference voltage terminal V2 is also a terminal obtained by dividing the reference power supply VREF with the resistors R1 and R2. The reference voltage terminal V2 is connected to the non-inverting input terminal of the error amplifier 118. An inverting input terminal VFB of the error amplifier 118 has a charging control device output terminal VO, which is a terminal closer to the battery 101 among the terminals of the charging current detection resistor RS1, as the resistor elements R110 and R120. The divided terminal is connected. The charging path from the charging current detection resistor RS1 to the battery 101 is different for each system, and the charging control circuit 111 is not determined because contact resistance such as a connector existing on the path or wiring resistance on the path is not determined. Is configured to feed back and control the output voltage V0 of the charge control device output terminal V0.\n\nThe output terminals of the error amplifiers 116 and 118 are respectively connected to the non-inverting input terminals of the comparator 120, and the oscillation signal from the oscillator (OSC) 122 is input to the inverting input terminal. The output terminal of the comparator 120 is connected to the inverting output buffer 124, and the output terminal OUT of the inverting output buffer 124 is connected to the gate terminal of the PMOS driver Tr1.\n\nWhen charging the battery 101, the charging control circuit 111 controls the charging current ICHG and the reference voltage V2, respectively. The charging current ICHG is converted into a voltage by the charging current detection resistor RS1 on the charging path and input to the output current detection terminals RS- and RS +, and the amplifier 112 has a predetermined gain multiple (× N). It is amplified to obtain a detection voltage V3 from the output terminal V3. The detection voltage V3 is input to the inverting input terminal of the error amplifier 116 and error amplified between the reference voltage V1 input to the non-inverting input terminal. In addition, the output voltage V0 detected at the charge control device output terminal V0 is fed back and properly divided by the resistance elements R110 and R120 and then input to the inverting input terminal VFB of the error amplifier 118. . Then, the error amplification is performed between the reference voltage V2 obtained by properly dividing the reference power supply VREF input to the non-inverting input terminal V2 by the resistance elements R1 and R2.\n\nThe output signals of the error amplifiers 116 and 118 are input to the non-inverting input terminal of the comparator 120, and are compared between the oscillation signals from the oscillator (OSC) 122 which are input to the inverting input terminal, so that the PMOS driver ( The switching duty of Tr1) is determined. Here, the oscillation signal is a signal which is repeated at a constant period such as a triangular wave or a sawtooth wave, and the voltage level is compared with the output signal from the error amplifiers 116 and 118 obtained as a result of the error amplification. In this case, two non-inverting input terminals of the comparator 120 exist, and the signal of the error amplification result input to each terminal and the oscillation signal input to the inverting input terminal are compared. And either comparison result signal is output by the internal structure (not shown) predetermined by the comparator 120. FIG. In FIG. 1, a switching signal is output when both comparison result signals are on. In other words, a comparison output can be obtained for a signal having a low voltage level among the signals input to the two non-inverting input terminals. In other words, the switching duty is determined by a signal having a low voltage level. Here, the driver to be driven is the PMOS transistor Tr1, and the gate buffer is driven when a low level is applied, so that the output buffer is provided with an inverted output buffer 124.\n\nWhen the charging current ICHG does not flow and the voltage drop ΔVLS does not occur in the parasitic resistance RLS1, the reference voltage corrector illustrated in FIG. 2 becomes 0 V, so that the reference power supply VREF ) Is divided by the parallel resistance of the resistance element Rl and the resistance elements R2 and R3. These parallel-connected resistors R2 and R3 are resistors replacing the resistors R102 in the prior art. The resistors R1 to R3 at this time become the first voltage divider (I in FIG. 2) for controlling the output voltage VO of the charge control device output terminal VO to the full charge voltage VBAT0. . When the charging current IICH flows and the detection voltage V3 is output, the detection voltage V3 is divided by the resistance elements R3 and R2 to generate a correction voltage according to the charging current IICH. This is a second partial pressure section ((II) in FIG. 2).\n\nWhen correcting the basic voltage V2 in the configuration of FIG. 2, the resistance element R1 connected to the reference power supply VREF and the resistance element R3 connected to the detection voltage V3 are the reference voltage terminals. Commonly connected to V2, it is connected to the ground potential via the resistance element R2. Therefore, if this network is solved according to Kirchhoff's law or the like, the reference voltage V2 when the charging current ICH does not flow is matched with the divided voltage of the full charge voltage VBAT0 by the resistance elements R110 and R120. The predetermined gain (XN) of the resistance elements R1 to R3 or the amplifier 112 can be adjusted so that the reference voltage V2 with respect to the detection voltage V3 is an appropriate correction voltage. In addition, the dependence of the reference voltage V2 on the detection voltage V3 (charge current ICHG) is mentioned later in FIG.\n\n3 is a detection characteristic of the charging current ICHG detected by the charging current detection resistor RS1 and the amplifier 112. The charging current ICHG to the battery 101 is converted into a voltage value VRS1 = RS1 × ICHG by flowing the charging current detection resistor RS1 to the output current detection terminals RS- and RS + of the amplifier 112. Is entered. Since the amplifier 112 has a predetermined gain xN, the detection voltage V3 can be obtained by N times the input voltage value VRS1. The relationship of the detection voltage V3 with respect to the charging current ICHG is\n\nV3 = N × RS1 × ICHG (1)\n\nAs shown in FIG. 3, the characteristic has a constant slope N × RS1 with respect to the charging current ICHG.\n\nThe charging current ICHG characteristic of the reference voltage V2 shown in FIG. 4 represents the reference voltage V2 output by the reference voltage corrector shown in FIG. 2. The detection voltage V3 in FIG. 2 is obtained by converting the charging current ICHG into a voltage by the charging current detection characteristic of FIG. 3. Solving by applying Kirchhoff's law to the network of FIG.\n\nV2 = R2 × (VREF × R3 + V3 × Rl) / {(R3 + R2) × (R1 + R2)-R2 2 } ...\n\nCan be obtained.\n\nWhen the charging current IICH does not flow, since the detection voltage V3 is 0 V in FIG. 3, the reference voltage corrector of FIG. 2 is formed by the parallel resistors of the resistor elements R1 and R2 and R3. The reference power supply VREF is divided and output. This is also derived from equation (2), and substituting V3 = 0 in equation (2),\n\nV2 = R2 × (VREF × R3) / {(R3 + R2) × (R1 + R2)-R2 2 }\n\n= VREF × (R2 × R3) / (R3 × R1 + R3 × R2 + R2 × Rl)\n\n= VREF × {(R2 × R3) / (R2 + R3)} / {R1 + (R3 × R2) / (R2 + R3)}\n\n= VREF × (R2 // R3) / {R1 + (R2 // R3)}\n\n= VFBO (3)\n\nBecomes Here, R2 // R3 represents the combined resistance when the resistance elements R2 and R3 are connected in parallel.\n\nThe voltage value VFB0 of the reference voltage V2 is controlled so that the output voltage VO of the charge control device output terminal VO becomes the full charge voltage VBAT0. Therefore, a voltage obtained by dividing the full charge voltage VBAT0 by the resistor elements R110 and R120 is applied to the inverting input terminal VFB of the error amplifier 118 at this time, and the voltage is the reference voltage V2 (= VFB0)] is controlled to be equal to\n\nVFB0 = {R120 / (R110 + R120)} × VBAT0 (·)\n\nHere, specific numerical examples are applied to equations (3) and (4). If VREF = 5V and V2 = 4.2V, R1 = 10 kΩ, R2 = 210 kΩ, and R3 = 70 kΩ can be obtained from equation (3). Here, the reference voltage V2 (= VFB0) can be set as, for example, a full charge voltage per cell in the battery 101, and a lithium ion battery or the like can be considered as V2 = 4.2V. For example, in the case where the battery 101 has a series three cell structure, the resistance ratio of the resistors R110 and R120 to R110 / R120 = 2 is obtained from equation (4).\n\nWhen the charging current ICHG is flowing, the detection voltage V3 becomes equation (1). In the following description, the case where the charging current ICHGM at the time of constant current charging flows as charging current is examined. By the formula (1), the detected voltage V3 at this time is\n\nV3 = N x RS1 x ICHGM (5)\n\nBecomes Substituting this in equation (2),\n\nV2 = R2 × (VREF × R3 + N × RS1 × ICHGM × Rl) / {(R3 + R2) × (R1 + R2) -R2 2 }\n\n(6) ············· (6)\n\nCan be obtained. Therefore, the correction voltage ΔV2 is obtained by subtracting equation (3) from equation (6),\n\nΔV2 = R2 × (N × RS1 × ICHGM × Rl) / {(R3 + R2) × (Rl + R2) -R2 2 } ... (7)\n\nBecomes The correction voltage ΔV2 is a voltage obtained by dividing the voltage drop ΔVLSM of the parasitic resistance RLS1 when the charging current ICHGM flows by the resistance elements R110 and R120. therefore,\n\nΔV2 = {R120 / (R110 + R120)} × ΔVLSM\n\n= {R120 / (R110 + R120)} × RLS1 × ICHGM ······· (8)\n\nIt is calculated | required as. From equations (7) and (8), a correction voltage ΔV2 proportional to the charging current ICHGM is added to the reference voltage V2, and a correction voltage equivalent to the voltage drop ΔVLSM of the parasitic resistance RLS1 is output voltage. It can be seen that it is added to (V0).\n\nHere, on the basis of VREF = 5 V, V2 = 4.2 V, R1 = 10 kΩ, R2 = 210 kΩ, R3 = 70 kΩ obtained in Equation (3), and R110 / R120 = 2 obtained in Equation (4). Specific numerical examples are obtained from (7) and (8). For ICHGM = 3A, V3 = 2 V can be obtained by appropriate selection of the charging current detection resistor RS1 and the predetermined gain (XN) of the amplifier 112 (V3 = N x RSl x ICHGM = 2). Substituting these numerical examples into equation (7),\n\nΔV2 = 210 k × (2 × 10 k) / {(70 k + 210 k) × (10 k + 210 k)-210 k 2 }\n\n= 0.24 V\n\nCan be obtained. Substituting this in equation (8),\n\nSince ΔV2 = (1/3) x RLS1 x3 = 0.24, RLS1 = 240 mΩ can be obtained. That is, the parasitic resistance of 240 mΩ is corrected at the charge current 3A.\n\n5 shows the charging characteristic of the battery 101. Since the battery 101 is not sufficiently charged at the beginning of charging the battery 101, the terminal voltage VBAT of the battery output terminal VBAT is a voltage sufficiently lower than the full charge voltage VBAT0. Therefore, even if the drop voltage ΔVLS due to the charging current ICHG flowing through the parasitic resistance RLS1 is added to the terminal voltage VBAT, the output voltage VO of the charge control device output terminal VO is also a full charge voltage. The voltage is sufficiently low compared to (VBAT0).\n\nTherefore, the voltage input to the inverting input terminal VFB of the error amplifier 118 becomes a voltage sufficiently smaller than the reference voltage V2 set at the non-inverting input terminal V2, and is output from the error amplifier 118. The output signal is the maximum value. In addition, at the start of charging, since the charging current ICHG rises from 0 A, the voltage output from the output terminal V3 of the amplifier 112 is lower than the reference voltage V1 in the transient state, and the error amplifier The output signal output from 116 represents the maximum value. Therefore, the switching duty for the output signal of the error amplifiers 116 and 118 is set to the maximum duty.\n\nThe charging current ICHG increases with the maximum duty controlled by the output of the error amplifier 116, and at the same time, the detection voltage V3 also increases. Since the input signal to the error amplifier 116 is reversed when the detection voltage V3 exceeds the reference voltage V1, the output signal of the error amplifier 116 shifts to the low voltage side. Here, since the comparator 120 in which the output signals of the error amplifiers 116 and 118 are input to the non-inverting input terminal can obtain a comparative output with respect to a signal having a low voltage level among these signals, the error amplifier 116 As the output signal is lower than the output signal of the error amplifier 118, the duty control follows the output signal of the error amplifier 116. Therefore, the constant current charging control is performed at the predetermined charging current ICHGM by appropriate setting of the charging current detection resistor RS1, the predetermined gain XN of the amplifier 112, and the reference voltage V1 (Fig. 5). , (CC) area].\n\nAt this time, the reference voltage V2 is a voltage obtained by adding the correction voltage to the detection voltage V3 through the resistance element R3. In FIG. 5, this is expressed as the control voltage VRF of the charge control device output terminal VO. The voltage drop ΔVLSM caused by the charging current ICHGM flowing through the parasitic resistor RLS1 becomes the correction voltage, and the reference voltage VRF is set. 5 illustrates the output voltage VO_I of the charge control device output terminal VO along with the battery voltage VBAT_I. Since charging is performed by the charging current ICHGM, the output voltage VO_I becomes a voltage higher than the drop voltage ΔVLSM in the parasitic resistance RLS1 compared to the battery voltage VBAT_I.\n\nWhen the battery 101 is charged and the battery voltage VBAT_I rises, the output voltage VO_I also rises as a voltage obtained by adding the drop voltage ΔVLSM at the parasitic resistance RLS1. Since the charge control circuit 111 feeds back and controls the charge control device output terminal V0, the error amplifier 118 is not able to adjust the voltage at the input terminal when the output voltage VO_I exceeds the control voltage VRF. Inverting decreases the output voltage. If this voltage drop is less than the output voltage of the error amplifier 116, the comparator 120 outputs a comparison result between the output signals from the error amplifier 118. That is, in the duty control, the constant voltage charge control is performed at the voltage VBAT0 at the time of full charge in accordance with the output signal of the error amplifier 118 ((CV region) in Fig. 5). In this period, the battery voltage VBAT is also close to the full charge voltage VBAT0, and the charging current ICHG decreases as compared with the charging current ICHGM at the time of constant current charging, and the charging operation continues, and the charging current ICHG is Charging to the battery 101 is completed at the time of disappearance (time t0 in FIG. 5).\n\nThe charge control device 1 of the first embodiment is a specific configuration example of claims 1, 2, 4 to 7, 10. The error amplifier 118, the comparator 120, and the inverted output buffer 124 of the charge control device 1 are configured as specific examples of the output voltage control unit in claim 1. In addition, the charging current detection resistor RS1 and the amplifier 112 are the specific structural example of the output current detection part in Claim 1, and a concrete structural example of the current-voltage converter in Claim 2. In addition, resistance elements R1-R3 are the specific structural example of the reference voltage correction part in Claim 1, and are the specific structural examples of the basic reference voltage part in Claim 5, and the 1st voltage divider part in Claim 6. As shown in FIG. In addition, the resistance elements R3 and R2 are specific examples of the configuration of the correction voltage section in claim 5 and specific examples of the configuration of the second voltage divider in claim 7. In this case, the reference power supply VREF is a specific configuration example of the constant voltage source in claim 6. Moreover, the amplifier 112, the error amplifier 116, the comparator 120, and the inverted output buffer 124 are comprised as an output current control part.\n\nAccording to the charging control device 1 of the first embodiment, the charging current ICHG should flow through the parasitic resistance RLS1, which is a parasitic load component existing on the ammeter path of the charging current ICHG, so that the charging should be at the same potential. Even when a deviation of the output voltage V0 of the control device output terminal V0 and the voltage value of the battery terminal voltage VBAT occurs, the resistor elements R1 to R3 are set to the reference voltage V2 according to the charging current ICHG. The control voltage of the output voltage V0 can be adjusted accordingly. In the charging current path, the position of the charge control device output terminal VO, which is voltage controlled according to the reference voltage V2 by the error amplifier 118 to the inverting output buffer 124, is spaced apart from the position of the battery output terminal VBAT. Even if present, the battery terminal voltage VBAT can be charged to the desired output voltage VBAT0.\n\nSpecifically, since the position of the charge control device output terminal VO is on the upstream side with respect to the position of the battery output terminal VBAT, the correction voltage according to the voltage drop ΔVLS during that time is added to the reference voltage V2. Is set high. Accordingly, the battery output terminal VBAT may be charged to the desired output voltage VBAT0. When the charging current ICHG is large, the reference voltage V2 is set high, and when the charging current ICHG is small, the reference voltage V2 is set low, so that the charging control device ICHG flows. Even when a voltage drop ΔVLS occurs between the output terminal VO and the battery output terminal VBAT, the output voltage VO is maintained while maintaining the control voltage at the battery output terminal VBAT at the full charge voltage VBAT0. The control voltage of can be adjusted suitably.\n\nIn addition, since the detection result of the charging current ICHG can be obtained as the detection voltage V3, the reference voltage V2 can be corrected by connecting to the resistance element R3.\n\nIn addition, when the divided voltage of the reference power supply VREF by the parallel resistance of the resistor R1 and the resistors R2 and R3 is the basic reference voltage, the full charge voltage VBAT0 when the charging current IICH does not flow. ) Can be precisely adjusted, and the detection voltage V3 can be divided by the resistance elements R3 and R2 and added to the amount corresponding to correction by the charging current ICHG. The reference voltage V2 may be controlled according to the charging current ICHG in the region of the entire charging current ICHG including the state where the ICHG does not flow.\n\nIn the charging control device 1, the charging current ICHGM in the constant current charging control is performed when the battery 101 is charged to the full charge voltage VBAT0 by performing constant current charging control at the beginning of charging and constant voltage charging control at the end of charging. ), The reference voltage V2 in the constant voltage charging control can be set higher than the reference voltage VFB0 at the time of full charge. As a result, the charging of the battery 101 proceeds and the output voltage VBAT of the battery 101 when switching from the constant current charging control to the constant voltage charging control can be closer to the predetermined voltage VBAT0 at the time of full charge. Therefore, the constant current charging period in the entire charging period can be lengthened, and the total charging time to the battery 101 can be shortened.\n\nNext, the charging control device 2 of the second embodiment is shown by FIG. In the charge control device 2, the detection current terminal IO is newly updated as an output terminal in the charge control circuit 21 instead of the amplifier 112 of the charge control circuit 111 in the charge control device 1 of the first embodiment. The amplifier 22 of the structure provided is provided. Also, instead of connecting the detection voltage V3 output from the amplifier 112 to the reference voltage terminal V2 through the resistor element R3, the detection current terminal IO is directly connected to the reference voltage terminal V2. It is constructed by connecting. In addition, the resistance divided voltage for generating the reference voltage V2 (= VFB0) when the charging current ICHG does not flow can be configured with R101 and R102 for dividing the reference power supply VREF, and the same resistance as in the prior art. It can be done with partial pressure resistance.\n\nThe amplifier 22 includes a resistor R4 for voltage current conversion between the output terminal V3 of the amplifier 112 and the ground potential as shown in FIG. 7, and converts the converted current into a current mirror circuit 25. By returning, the output of the detection current IO is obtained from the detection current terminal IO.\n\nSince the detection current IO is obtained by applying the detection voltage V3 to both ends of the resistance element R4, the detection current IO obtained has a proportional relationship between the detection voltages V3. The proportional coefficient is the resistance value R4 of the resistance element R4. Therefore, the detection current IO is expressed by equation (1) when the current ratio in the current mirror circuit 25 is set to 1 to 1.\n\nIO = V3 / R4 = N × RS1 × ICHG / R4 · · · · · · · · · (9)\n\nBecomes\n\nThe detection current IO is input to the reference voltage terminal V2 and flows to the resistance element R102, thereby correcting the reference voltage V2. The charging current ICHG characteristic of the reference voltage V2 at this time has the same characteristics as that of the first embodiment shown in FIG. Solving Kirchhoff's law and solving it as in the first embodiment,\n\nV2 = R102 × (IO × R101 + VREF) / (R101 + R102) (10)\n\nCan be obtained. Substituting equation (9) into equation (10),\n\nV2 = R102 × {(N × RS1 × ICHG / R4) × R101 + VREF} / (R101 + R102) ... (11)\n\nThe reference voltage V2 proportional to the charging current ICHG can be obtained.\n\nWhen the charging current IICH does not flow, the detection current IO is more than 0 A,\n\nV2 = R102 x VREF / (R101 + R102) ... (l2)\n\nThe reference voltage V2 is a voltage obtained by dividing the reference power supply VREF by the resistance elements R101 and R102. This relationship is the same as that in the related art.\n\nWhen the charging current ICHGM is flowing, by subtracting the formula (12) from the formula (10),\n\nΔV2 = R102 × (IO × R101) / (R101 + R102) (13)\n\nBecomes\n\nHere, the detection current IO is calculated based on the numerical example similar to the case of the charge control apparatus 1 of 1st Example. At ΔV2 = 0.24 V, R101 = R1 = 10 kΩ, R102 = R2 // R3 = 210 k // 70 k = 52.5 kΩ,\n\nIO = ΔV2 × (R101 + R102) / (R101 × R102)\n\n= 0.24 × (10 ㏀ + 52.5 ㏀) / (10 ㏀ × 52.5 ㏀)\n\n= 28.6 μA.\n\nThe charge control device 2 according to the second embodiment is a specific configuration example of claims 1, 3 to 6, 8, and 10. The same reference numerals are used for the same components as those of the charge control apparatuses 1 and 100 in the charge control apparatus 2. The charging current detection resistor RS1 and the amplifier 22 are specific configurations of the output current detection unit in claim 1. Moreover, the specific example of a structure of the output current detection part provided with the current output part of Claim 3 by the amplifier 22 is shown. In addition, the resistance elements R101 and R102 are specific examples of the configuration of the reference voltage corrector in claim 1, and are specific examples of the specific configuration of the basic reference voltage section in claim 5 and the first voltage divider in claim 6. In the resistor element R102, the detection current IO from the detection current terminal IO of the amplifier 22 flows, thereby providing an example of the specific configuration of the correction voltage portion in claim 5. The output current controller is composed of an amplifier 22, an error amplifier 116, a comparator 120, and an inverted output buffer 124.\n\nAccording to the charge control device 2 of the second embodiment, since the detection result of the charging current ICHG can be obtained as the detection current IO, R102 among the resistor elements R101 and R102 that divides the reference power supply VREF can be obtained. The reference voltage V2 can be corrected by flowing the detection current IO. In addition to the divided resistors R101 and R102 that generate the reference voltage VFB0 when the charging current IICH does not flow, there is no need to add a new resistor element or the like for correcting the reference voltage V2. .\n\nThe other effects are the same as in the charge control device 1. That is, when the charging current ICHG is large, the reference voltage V2 is set high, and when the charging current ICHG is small, the reference voltage V2 is set low and the charging current ICHG flows to control charge. Even when a voltage drop ΔVLS occurs between the device output terminal VO and the battery output terminal VBAT, the output voltage VO is maintained while maintaining the control voltage at the battery output terminal VBAT at the full charge voltage VBAT0. Control voltage can be appropriately adjusted. Therefore, when the charging control device 2 is used for charging the battery 101, the battery 101 output voltage VBAT when charging to the battery 101 proceeds and is switched from the constant current charging control to the constant voltage charging control. ) Can be closer to the predetermined voltage VBAT0 at the time of full charge, and the constant current charging period in the entire charging period becomes longer, so that the total charging time to the battery 101 can be shortened.\n\nNext, the charging control device 3 of the third embodiment is shown by FIG. In the charge control device 3, a configuration in which the amplifier 112 output terminal V3 of the charge control circuit 111 in the charge control device 1 is connected to the reference voltage terminal V2 through the resistor element R3. Instead, the charging control circuit 31 connects the inverting input terminal VFB, which is input by dividing the charging control device output terminal VO, and the output terminal V3 and the reference voltage terminal V2 of the amplifier 112, respectively. A multi-input error amplifier 32 having two non-inverting input terminals is provided.\n\nThe detailed circuit configuration of the error amplifier 32 is not shown. According to the voltage VFB divided by the output voltage VO, an error amplification operation is performed on either the voltage VFB and the detection voltage V3 or the voltage VFB and the reference voltage V2. It is an error amplifier which controls VO to the voltage according to the voltage (detection voltage V3 or reference voltage V2) set with two non-inverting input terminals.\n\n9 shows its appearance. In the control of Fig. 9, control is performed at a lower voltage among the voltages set at the two non-inverting input terminals. The fixed voltage VFB0 obtained by dividing the reference power supply VREF by the resistance elements R101 and R102 is input to the reference voltage terminal V2. On the other hand, the detection voltage V3 is input to the output terminal V3 of the amplifier 112. The detection voltage V3 is a voltage proportional to the charging current ICHG, as shown in equation (1), and is appropriately inclined by the charging current detection resistor RS1 or the predetermined gain (XN) of the amplifier 112. It becomes a characteristic having. Accordingly, the detection voltage V3 increases with the increase of the charging current ICHG, and exceeds the reference voltage VFB0 with the predetermined charging current ICHG1. Error having the characteristic (A) in FIG. 9 by combining with the detection voltage V3 adjusted so that the correction voltage of the reference voltage V2 at the charging current ICHGM at the constant current charge control to the battery 101 becomes ΔV2. Reference voltage characteristics of the amplifier 32 can be obtained. Accordingly, in the constant current charging control, the voltage drop (ΔVLSM) caused by the parasitic resistance RLS1 on the charging current path is added by adding a correction voltage of ΔV2 to the reference voltage V2 (= VFB0) of the output voltage VO at full charge. Can be excluded, and at the time of constant voltage charging control, the reference voltage V2 is dropped in accordance with the decrease of the charging current ICHG from the charging current ICHGM, and in the region below the charging current ICHG1, It can be set to the reference voltage VFB0.\n\nThe charge control device 3 of the third embodiment is a specific configuration example of claims 1, 2, 4 to 7, 9, and 10. FIG. In the charge control apparatus 3, the same code | symbol is used about the same component as the component of the charge control apparatus 1, 2, 100. As shown in FIG. The error amplifier 32, the comparator 120, and the inverted output buffer 124 of the charge control device 3 are configured as specific examples of the output voltage control unit in claim 1. In addition, the error amplifier 32 is a specific structural example of the output voltage control part in Claim 9. Here, the resistive elements R101 and R102 are specific examples of the structure of the basic reference voltage unit in claim 9.\n\nAccording to the charge control device 3 of the third embodiment, the reference voltage VFB0 when the charging current ICHG does not flow and the reference voltage corrected when it flows can be set independently, and both adjustments are made. It can be done simply and easily and surely. At this time, the detected voltage V3 output from the amplifier 112 can be used as it is as the corrected reference voltage.\n\nIn addition, about other effects, it is the same as that of the charge control apparatuses 1 and 2. FIG. That is, when the charging current ICHG is large, the reference voltage V2 is set high, and when the charging current ICHG is small, the reference voltage V2 is set low and the charging current ICHG flows to control charge. Even when a voltage drop ΔVLS occurs between the device output terminal VO and the battery output terminal VBAT, the output voltage VO is maintained while maintaining the control voltage at the battery output terminal VBAT at the full charge voltage VBAT0. Control voltage can be appropriately adjusted. Therefore, when the charge control device 3 is used for charging the battery 101, the output voltage VBAT of the battery 101 when charging to the battery 101 proceeds and is switched from the constant current charge control to the constant voltage charge control ) Can be closer to the predetermined voltage VBATO at the time of full charge, and the constant current charging period in the entire charging period becomes longer, thereby reducing the total charging time to the battery 101.\n\nIn the fourth embodiment shown in FIG. 10, a case where the regulator circuit of the present invention is implemented as the switching regulator 4 is shown. The load 41 is connected instead of the battery 101 in the charge control devices 1 to 3 of the first to third embodiments. Basically, except that the output current control section comprised of the amplifier 112 or 22, the error amplifier 116, the comparator 120, and the inverting output buffer 124 is used for control of the load current limit at the time of overload, it is basically the 1st thru | or 1st. The same functions and operations as in the third embodiment are shown. 10 (1) to (3) show the structures of the first to third embodiments, respectively. As the circuit configuration, any configuration of the first to third embodiments can be applied.\n\nIn the fifth embodiment shown in Fig. 11, a case where the regulator circuit of the present invention is implemented as the series regulator 5 is shown. The load 41 is connected instead of the battery 101 in the charge control devices 1 to 3 of the first to third embodiments. Power supply to the load 41 is performed by linearly controlling the PMOS driver Tr2. The current detection resistors RS1, the parasitic resistors RLS1, and the resistors R110, R120 are the same as in the first to fourth embodiments. In the series regulator 5 of the fifth embodiment, the gate voltage of the PMOS driver Tr2 is controlled by the output voltage of the error amplifier 118. Incidentally, the input signals to the inverting / non-inverting input terminals of the error amplifiers 32 and 118 are reversed in the case of the first to fourth embodiments. In addition, the amplifier 112 functions as a correction for the reference voltage V2 corresponding to the load current. 11 (1) to (3) show the structures of the first to third embodiments, respectively. As the circuit configuration, any configuration of the first to third embodiments can be applied. .\n\nIn the switching regulator 4 and the series regulator 5, the constant voltage control of the output voltage VO is performed by the error amplifier 32 or 118, etc., so that it is predetermined to the load terminal voltage VLD of the load terminal VLD of the load 41. When the voltage is supplied, the reference voltage in the constant voltage control can be set higher than the reference voltage at no load according to the load current ILD. Thus, even when the voltage drop ΔVLS on the current path becomes large in a large load current region reaching an overload state, a predetermined voltage can be applied to the load 41. Predetermined voltage can be applied to the load 41 in all load current regions.\n\nHere, if the proportional coefficient of the detection voltage V3 or the detection current I0 with respect to the charging current ICHG or the load current ILD is externally adjustable, the charging control circuits 1-3 and the switching regulator 4 The detection voltage V3 or the detection of the charging current ICHG or the load current ILD also applies to current path characteristics in which the contact resistance by the connector or the like and the wiring resistance by the wiring are different for each specific configuration of the series regulator 5. The adjustment of the current IO can be performed simply and easily.\n\n12 shows a specific example of the amplifier 112 used in the first and third embodiments. The resistors R6 and R5 are connected to the output current detection terminals RS- and RS +, respectively, and the resistors R6 and R5 are connected to the inverting / non-inverting input terminals of the amplifier 61, respectively. The output terminal of the amplifier 61 is a base terminal of each transistor Q1 having a collector terminal connected to a non-inverting input terminal of the amplifier 61 and a base terminal of each transistor Q2 having a collector terminal connected to a current mirror circuit 62. Is connected in common. Here, the emitter terminals of the transistors Q1 and Q2 are connected to the ground potential. The mirror current output terminal of the current mirror circuit 62 is connected to the resistance element R7 connected between the ground potentials, and constitutes the voltage conversion terminal V4. The voltage conversion terminal V4 is connected to a non-inverting amplifier circuit composed of the amplifier 63 and the resistance elements R8 and R9, and its output terminal is the output terminal V3.\n\nThe amplifier 61 is controlled such that the input voltage to the inverting input terminal and the input voltage to the non-inverting input terminal are approximately equal by an imaginary short characteristic, which is an operating characteristic of the operational amplifier. Therefore, the current I1 flows from the output current detection terminal RS + through the resistor element R5 by the transistor Q1 biased by the output current from the amplifier 61. At this time, the voltage drop in the resistance element R5 is controlled to be a voltage drop approximately equal to the voltage drop ΔVLS in the charge current detection resistor RS1 due to the charge current ICHG. Since the output current of the amplifier 61 is also input to the base terminal of the transistor Q2 at the same time, the collector current of the transistor Q2 also becomes approximately the same as the current I1. Now, assuming that the mirror current ratio of the current mirror circuit 62 is 1 to 1, the mirrored current is also approximately equal to the current I1. This current I1 flows through the resistance element R7 to obtain the voltage V4. By properly amplifying the voltage V4 in the non-inverting amplifier circuit, the detection voltage V3 having the desired predetermined gain xN can be obtained.\n\nWhen these operations are expressed by a calculation formula, it becomes as follows.\n\nI1 = ΔVLS / R5\n\nV4 = R7 × I1\n\n= R7 × ΔVLS / R5\n\nV3 = {(R8 + R9) / R9} × V4\n\n= {(R8 + R9) / R9} × R7 × ΔVLS / R5 · · · · · · · (14)\n\nTherefore, the predetermined gain (XN) of the amplifier 112 can be adjusted by changing and adjusting the resistance value of at least one of the resistance element R7 or the resistance elements R8 and R9.\n\nIn addition, for the amplifier 22 used in the second embodiment, as shown in FIG. 7, the predetermined gain (XN) of the amplifier 22 can be adjusted by changing and adjusting the resistance value of the resistance element R4. . In addition, the amplifier 22 may have a configuration shown in FIG. 12 instead of the configuration shown in FIG. 7, and a configuration in which the output from the current mirror circuit 62 is the detection current IO.\n\nIn addition, the regulator circuit of this invention is not limited to the said 1st-5th embodiment, Of course, various improvement and deformation are possible within the range which does not deviate from the meaning of this invention.\n\nFor example, in the present exemplary embodiment, the charging control device output terminal VO is located upstream in the current path through which the charging current ICHG or the load current ILD flows compared to the battery output terminal VBAT or the load terminal VLD. Although the case was demonstrated as an example, the regulator circuit of this invention is not limited to this, It is similarly applicable also when the charge control apparatus output terminal VO is located downstream of a current path. In this case, the control voltage VRF of the output voltage VO corresponds to the drop voltage ΔVLS of the parasitic resistance RLS1 with respect to the control voltage at the battery output terminal VBAT or the load terminal VLD. Subtract it.\n\nIn the third embodiment, the reference voltage V2 is constituted by the fixed voltage VFB0 and the detection voltage V3 having a predetermined slope. However, the regulator circuit of the present invention is not limited thereto. The reference voltage V2 may be constituted by a straight line having two or more different slope characteristics in addition to these straight lines or in place of these straight lines.\n\nIn addition, the adjustment of the predetermined gain (XN) in the amplifier 112 is not limited to the circuit configuration illustrated in FIG. 12, but may be performed by various amplifier circuits such as a differential amplifier circuit, a non-inverted amplifier circuit, or by appropriately adjusting these circuits. Can be configured by combining.\n\n(Book 1)\n\nAn output voltage controller for controlling the output voltage based on the reference voltage;\n\nAn output current detector,\n\nAnd a reference voltage corrector for controlling the reference voltage based on the output current detected by the output current detector.\n\n(Book 2)\n\nAnd an output current control unit for controlling the output current based on a detection result from the output current detection unit.\n\n(Appendix 3)\n\nThe output current detector includes a current voltage converter,\n\nThe regulator circuit according to the appendix 1 or 2, wherein the reference voltage corrector is controlled based on the detected voltage according to the output current.\n\n(Appendix 4)\n\nThe regulator circuit according to the annex 3, wherein the detection voltage is output with a predetermined proportional factor with respect to the output current.\n\n(Appendix 5)\n\nThe output current detector includes a current output unit,\n\nThe regulator circuit according to Appendix 1 or 2, wherein the reference voltage correction unit is controlled based on a detection current according to the output current.\n\n(Supplementary Note 6)\n\nThe regulator circuit according to the annex 5, wherein the detection current is output with a predetermined proportional coefficient with respect to the output current.\n\n(Appendix 7)\n\nThe said proportional coefficient is the regulator circuit as described in Supplementary Note 4 or 6 characterized by the external adjustment.\n\n(Appendix 8)\n\nThe regulator circuit according to at least one of the appendices 1 to 7, wherein the reference voltage corrector increases or decreases the reference voltage with a positive or negative correlation with respect to the increase or decrease of the output current.\n\n(Appendix 9)\n\nThe reference voltage corrector,\n\nA basic reference voltage unit configured to output a basic reference voltage when the output current does not flow;\n\nThe regulator circuit according to at least one of the appendices 1 to 8, characterized by comprising a correction voltage portion for superimposing a correction voltage which changes according to the output current to the basic reference voltage.\n\n(Book 10)\n\nThe basic reference voltage unit includes a constant voltage source or a constant voltage source and a first voltage divider.\n\nAnd said basic reference voltage is a predetermined voltage output from said constant voltage source or a divided voltage of said predetermined voltage by said first voltage divider.\n\n(Appendix 11)\n\nAnd the correction voltage section includes a second voltage divider and divides the detected voltage.\n\n(Appendix 12)\n\nThe regulator circuit according to the annex 10, wherein the correction voltage unit includes the first voltage divider and flows the detected current to the first voltage divider.\n\n(Appendix 13)\n\nA basic reference voltage unit configured to output a basic reference voltage when the output current does not flow,\n\nThe output voltage control unit,\n\nA first reference voltage terminal to which the basic reference voltage is input;\n\nAnd a second reference voltage terminal to which a correction reference voltage that is changed according to the output current is input.\n\nThe regulator circuit according to at least one of the appendices 1 to 8, wherein an output voltage is controlled as the reference voltage, which is selected according to the magnitude relationship between the basic reference voltage and the correction reference voltage.\n\n(Book 14)\n\nAnd the correction reference voltage is generated based on the detected voltage or the detected current.\n\n(Supplementary Note 15)\n\nIn accordance with the charging current in the constant current charging control when the battery is charged to the output voltage by the constant current charge control at the initial stage of charging by the output current controller, and the constant voltage charge control at the late stage of charge by the output voltage controller. The regulator circuit according to at least one of the appendices 2 to 14, wherein the reference voltage in the constant voltage charging control is set higher than the reference voltage at the time of full charge.\n\n(Appendix 16)\n\nThe reference voltage in the constant voltage control is applied to the load current when supplying a predetermined voltage to the load by performing the current limit control at the time of overload by the output current controller and controlling the constant voltage of the output voltage by the output voltage controller. The regulator circuit according to at least one of appendices 2 to 14, wherein the regulator circuit is set higher than the reference voltage at no load.\n\n(Appendix 17)\n\nAnd a reference voltage for adjusting and maintaining the output voltage to a predetermined voltage according to the output current.\n\n(Supplementary Note 18)\n\nAnd the reference voltage is increased or decreased with a positive or negative correlation with respect to the increase or decrease of the output current.\n\n(Appendix 19)\n\nWhen the charging operation to the battery is performed by performing constant current charging at the initial stage of charging and constant voltage charging at the predetermined voltage in the later stage of charging, the reference at the time of full charge of the reference voltage in accordance with the charging current in the constant current charged state. The control method of the regulator circuit as described in said Supplementary note 17 or 18 characterized by setting higher than a voltage.\n\n(Book 20)\n\nWhen supplying the predetermined voltage to the load, the reference voltage is set higher than the reference voltage at no load in accordance with the load current, wherein the regulator circuit according to the appendix 17 or 18 is used.\n\nAccording to the present invention, it is possible to provide a regulator circuit capable of precisely adjusting and maintaining the output voltage in all the charging current or load current regions by correcting the setting of a predetermined voltage to be adjusted according to the charging current or the load current.\n\n## Claims (10)\n\n1. An output voltage controller for controlling the output voltage based on the reference voltage;\nAn output current detector,\nAnd a reference voltage corrector configured to control the reference voltage based on the output current detected by the output current detector.\n2. The method of claim 1, wherein the output current detector comprises a current voltage converter,\nAnd adjust the reference voltage corrector based on the detected voltage according to the output current.\n3. The method of claim 1, wherein the output current detector comprises a current output unit,\nAnd adjust the reference voltage corrector based on the detected current according to the output current.\n4. The regulator circuit according to claim 1, wherein the reference voltage corrector increases or decreases the reference voltage with a positive or negative correlation with respect to the increase or decrease of the output current.\n5. The method of claim 1, wherein the reference voltage corrector,\nA basic reference voltage unit configured to output a basic reference voltage when the output current does not flow;\nAnd a correction voltage portion overlapping the correction voltage which changes according to the output current with the basic reference voltage.\n6. The method of claim 5, wherein the basic reference voltage unit comprises a constant voltage source or a constant voltage source and a first voltage divider,\nAnd the basic reference voltage is a predetermined voltage output from the constant voltage source or a divided voltage of the predetermined voltage by the first voltage divider.\n7. 7. The regulator circuit according to claim 6, wherein the correction voltage section includes a second voltage divider and divides the detected voltage.\n8. 7. The regulator circuit according to claim 6, wherein the correction voltage portion includes the first voltage divider and flows the detection current to the first voltage divider.\n9. The device according to any one of claims 1 to 4, further comprising a basic reference voltage section for outputting a basic reference voltage when the output current does not flow,\nThe output voltage control unit,\nA first reference voltage terminal to which the basic reference voltage is input;\nAnd a second reference voltage terminal to which a correction reference voltage that is changed according to the output current is input.\nAnd an output voltage is controlled by using any one selected according to the magnitude relationship between the basic reference voltage and the correction reference voltage as the reference voltage.\n10. The regulator circuit of claim 9, wherein the correction reference voltage is generated based on the detected voltage or the detected current.\nKR1020020019889A 2001-10-05 2002-04-12 Regulator circuit and control method thereof KR100844052B1 (en)\n\n## Priority Applications (2)\n\nApplication Number Priority Date Filing Date Title\nJPJP-P-2001-00310088 2001-10-05\nJP2001310088A JP3968228B2 (en) 2001-10-05 2001-10-05 Regulator circuit\n\n## Publications (2)\n\nPublication Number Publication Date\nKR20030029443A true KR20030029443A (en) 2003-04-14\nKR100844052B1 KR100844052B1 (en) 2008-07-07\n\n# Family\n\n## Family Applications (1)\n\nApplication Number Title Priority Date Filing Date\nKR1020020019889A KR100844052B1 (en) 2001-10-05 2002-04-12 Regulator circuit and control method thereof\n\n## Country Status (4)\n\nUS (1) US6828764B2 (en)\nJP (1) JP3968228B2 (en)\nKR (1) KR100844052B1 (en)\nTW (1) TW583816B (en)\n\n## Cited By (2)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nKR100885689B1 (en) * 2007-01-31 2009-02-26 한국전자통신연구원 Current Sensing Circuit For Compensating Variations Of Temperature And Voltage Level\nKR100987629B1 (en) * 2003-05-12 2010-10-13 엘지전자 주식회사 Apparatus and method for controling charge current for battery\n\n## Families Citing this family (22)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nJP4080396B2 (en) * 2003-08-08 2008-04-23 富士通株式会社 DC / DC converter, semiconductor device, electronic device, and battery pack\nWO2005083868A1 (en) * 2004-02-27 2005-09-09 Shindengen Electric Manufacturing Co., Ltd. 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Automotive part\nGB2543225C (en) * 2017-01-25 2020-02-19 O2Micro Inc Controlling power delivery to a battery\n\n## Family Cites Families (8)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nUS5396163A (en) * 1991-03-13 1995-03-07 Inco Limited Battery charger\nEP1066536B1 (en) * 1999-01-22 2004-12-29 Philips Electronics N.V. Voltage indicator for indicating that the voltage of a battery passes a given value\nJP4117752B2 (en) * 1999-02-17 2008-07-16 Ｔｄｋ株式会社 In-vehicle power supply device and in-vehicle device\nGB0003501D0 (en) * 2000-02-15 2000-04-05 Sgs Thomson Microelectronics Voltage converter\nJP4647747B2 (en) * 2000-06-08 2011-03-09 富士通セミコンダクター株式会社 DC-DC converter and semiconductor integrated circuit device for DC-DC converter\nUS6437549B1 (en) * 2000-08-31 2002-08-20 Monolithic Power Systems, Inc. Battery charger\nEP1199789A1 (en) * 2000-10-19 2002-04-24 Semiconductor Components Industries, LLC Circuit and method of operating a low-noise, on-demand regulator in switched or linear mode\nUS6366070B1 (en) * 2001-07-12 2002-04-02 Analog Devices, Inc. Switching voltage regulator with dual modulation control scheme\n\n## Cited By (2)\n\n* Cited by examiner, † Cited by third party\nPublication number Priority date Publication date Assignee Title\nKR100987629B1 (en) * 2003-05-12 2010-10-13 엘지전자 주식회사 Apparatus and method for controling charge current for battery\nKR100885689B1 (en) * 2007-01-31 2009-02-26 한국전자통신연구원 Current Sensing Circuit For Compensating Variations Of Temperature And Voltage Level\n\n## Also Published As\n\nPublication number Publication date\nUS6828764B2 (en) 2004-12-07\nJP2003116269A (en) 2003-04-18\nJP3968228B2 (en) 2007-08-29\nUS20030067283A1 (en) 2003-04-10\nTW583816B (en) 2004-04-11\nKR100844052B1 (en) 2008-07-07\n\n## Similar Documents\n\nPublication Publication Date Title\nUS8638155B2 (en) Level-shifter circuit\nUS8531851B2 (en) Start-up circuit and method thereof\nUS9594387B2 (en) Voltage regulator stabilization for operation with a wide range of output capacitances\nUS7586297B2 (en) Soft start circuit, power supply unit and electric equipment\nJP3259262B2 (en) Voltage regulator\nUS8283906B2 (en) Voltage regulator\nUS9134743B2 (en) Low-dropout voltage regulator\nUS6498461B1 (en) Voltage mode, high accuracy battery charger\nUS6522111B2 (en) Linear voltage regulator using adaptive biasing\nUS6215291B1 (en) Reference voltage circuit\nUS6654263B2 (en) Linear regulator with switched capacitance output\nUS8143868B2 (en) Integrated LDO with variable resistive load\nUS8242760B2 (en) Constant-voltage circuit device\nUS6573693B2 (en) Current limiting device and electrical device incorporating the same\nTWI294996B (en) Low drop-out voltage regulator circuit and method with common-mode feedback\nJP4556116B2 (en) Constant voltage power circuit\nUS7723968B2 (en) Technique for improving efficiency of a linear voltage regulator\nJP4966592B2 (en) Power circuit\nUS6815938B2 (en) Power supply unit having a soft start functionality and portable apparatus equipped with such power supply unit\nEP0892332B1 (en) Low power consumption linear voltage regulator having a fast response with respect to the load transients\nKR100881487B1 (en) Method for charging battery, circuit for charging battery and portable electronic equipment including the battery\nEP1508078B1 (en) Voltage regulator with dynamically boosted bias current\nUS6885177B2 (en) Switching regulator and slope correcting circuit\nUS7859511B2 (en) DC-DC converter with temperature compensation circuit\nTWI430549B (en) Gate driver topology for maximum load efficiency and method thereof\n\n## Legal Events\n\nDate Code Title Description\nA201 Request for examination\nE902 Notification of reason for refusal\nE701 Decision to grant or registration of patent right\nGRNT Written decision to grant\nFPAY Annual fee payment\n\nPayment date: 20130531\n\nYear of fee payment: 6\n\nFPAY Annual fee payment\n\nPayment date: 20140612\n\nYear of fee payment: 7\n\nLAPS Lapse due to unpaid annual fee" ]
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http://mathnium.com/help/sum.html
[ "sum\nsum of the elements of an array along a specified dimension.\n`(y)=sum(x)`\n`(y)=sum(x, dim)`\n Inputs `x` Any numeric array. `dim` A positive integer not greater than the number of dimensions of `x` . Outputs `y` The array of sum of the elements of `x` along the dimension `dim` whose default value is `1` .\n\nDescription\nIf `x` is a vector, the output is the sum of all the elements of the vector. Otherwise the output is of the same size as the input except that along the dimension `dim` the size of the output is `1` . The element `y[i,j,k,...1, p, q,..]` (with `1` corresponding to the dimension `dim` ) of the output is the sum of the elements of the vector `x[i,j,k,....,:,p,q,...]` where `i, j, k, ...,p,q,...` are scalar integers ranging through the size of the input along the corresponding dimension.\nExample\n```>>a=floor(randn(3,4,2)*100)\n>>a\n[:,1:4, 1]\n-125 -21 -32 -30\n-29 166 -12 -69\n-22 -38 24 105\n[:,1:4, 2]\n93 45 -187 48\n-108 -63 -63 3\n26 144 -137 69\n\n>>sum(a)\n[:,1:4, 1]\n-176 107 -20 6\n[:,1:4, 2]\n11 126 -387 120\n\n>>sum(a,3)\n[:,1:4, 1]\n-32 24 -219 18\n-137 103 -75 -66\n4 106 -113 174\n\n```" ]
[ null ]
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https://tutorialspoint.dev/language/c/void-pointer-c
[ "# void pointer in C / C++\n\nA void pointer is a pointer that has no associated data type with it. A void pointer can hold address of any type and can be typcasted to any type.\n\n `int` `a = 10; ` `char` `b = ``'x'``; ` ` `  `void` `*p = &a;  ``// void pointer holds address of int 'a' ` `p = &b; ``// void pointer holds address of char 'b' `\n\n1) malloc() and calloc() return void * type and this allows these functions to be used to allocate memory of any data type (just because of void *)\n\n `int main(void) ` `{ ` `    ``// Note that malloc() returns void * which can be  ` `    ``// typecasted to any type like int *, char *, .. ` `    ``int *x = malloc(sizeof(int) * n); ` `} `\n\nNote that the above program compiles in C, but doesn’t compile in C++. In C++, we must explicitly typecast return value of malloc to (int *).\n\n2) void pointers in C are used to implement generic functions in C. For example compare function which is used in qsort().\n\nSome Interesting Facts:\n1) void pointers cannot be dereferenced. For example the following program doesn’t compile.\n\n `#include ` `int` `main() ` `{ ` `    ``int` `a = 10; ` `    ``void` `*ptr = &a; ` `    ``printf``(``\"%d\"``, *ptr); ` `    ``return` `0; ` `} `\n\nOutput:\n\n`Compiler Error: 'void*' is not a pointer-to-object type `\n\nThe following program compiles and runs fine.\n\n `#include ` `int` `main() ` `{ ` `    ``int` `a = 10; ` `    ``void` `*ptr = &a; ` `    ``printf``(``\"%d\"``, *(``int` `*)ptr); ` `    ``return` `0; ` `} `\n\nOutput:\n\n`10`\n\n2) The C standard doesn’t allow pointer arithmetic with void pointers. However, in GNU C it is allowed by considering the size of void is 1. For example the following program compiles and runs fine in gcc.\n\n `#include ` `int` `main() ` `{ ` `    ``int` `a = {1, 2}; ` `    ``void` `*ptr = &a; ` `    ``ptr = ptr + ``sizeof``(``int``); ` `    ``printf``(``\"%d\"``, *(``int` `*)ptr); ` `    ``return` `0; ` `} `\n\nOutput:\n\n`2`\n\nNote that the above program may not work in other compilers.\n\n## tags:\n\nC C Basics C-Pointers cpp-pointer C" ]
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https://en.m.wikipedia.org/wiki/Rigid_body_dynamics
[ "# Rigid body dynamics\n\nRigid-body dynamics studies the movement of systems of interconnected bodies under the action of external forces. The assumption that the bodies are rigid, which means that they do not deform under the action of applied forces, simplifies the analysis by reducing the parameters that describe the configuration of the system to the translation and rotation of reference frames attached to each body. This excludes bodies that display fluid, highly elastic, and plastic behavior.\n\nThe dynamics of a rigid body system is described by the laws of kinematics and by the application of Newton's second law (kinetics) or their derivative form Lagrangian mechanics. The solution of these equations of motion provides a description of the position, the motion and the acceleration of the individual components of the system and overall the system itself, as a function of time. The formulation and solution of rigid body dynamics is an important tool in the computer simulation of mechanical systems.", null, "Movement of each of the components of the Boulton & Watt Steam Engine (1784) can be described by a set of equations of kinematics and kinetics\n\n## Planar rigid body dynamics\n\nIf a system of particles moves parallel to a fixed plane, the system is said to be constrained to planar movement. In this case, Newton's laws (kinetics) for a rigid system of N particles, Pi, i=1,...,N, simplify because there is no movement in the k direction. Determine the resultant force and torque at a reference point R, to obtain\n\n$\\mathbf {F} =\\sum _{i=1}^{N}m_{i}\\mathbf {A} _{i},\\quad \\mathbf {T} =\\sum _{i=1}^{N}(\\mathbf {r} _{i}-\\mathbf {R} )\\times (m_{i}\\mathbf {A} _{i}),$\n\nwhere ri denotes the planar trajectory of each particle.\n\nThe kinematics of a rigid body yields the formula for the acceleration of the particle Pi in terms of the position R and acceleration A of the reference particle as well as the angular velocity vector ω and angular acceleration vector α of the rigid system of particles as,\n\n$\\mathbf {A} _{i}=\\alpha \\times (\\mathbf {r} _{i}-\\mathbf {R} )+\\omega \\times (\\omega \\times (\\mathbf {r} _{i}-\\mathbf {R} ))+\\mathbf {A} .$\n\nFor systems that are constrained to planar movement, the angular velocity and angular acceleration vectors are directed along k perpendicular to the plane of movement, which simplifies this acceleration equation. In this case, the acceleration vectors can be simplified by introducing the unit vectors ei from the reference point R to a point ri and the unit vectors ${\\textstyle \\mathbf {t} _{i}=k\\times \\mathbf {e} _{i}}$ , so\n\n$\\mathbf {A} _{i}=\\alpha (\\Delta r_{i}\\mathbf {t} _{i})-\\omega ^{2}(\\Delta r_{i}\\mathbf {e} _{i})+\\mathbf {A} .$\n\nThis yields the resultant force on the system as\n\n$\\mathbf {F} =\\alpha \\sum _{i=1}^{N}m_{i}(\\Delta r_{i}\\mathbf {t} _{i})-\\omega ^{2}\\sum _{i=1}^{N}m_{i}(\\Delta r_{i}\\mathbf {e} _{i})+(\\sum _{i=1}^{N}m_{i})\\mathbf {A} ,$\n\nand torque as\n\n$\\mathbf {T} =\\sum _{i=1}^{N}(m_{i}\\Delta r_{i}\\mathbf {e} _{i})\\times (\\alpha (\\Delta r_{i}\\mathbf {t} _{i})-\\omega ^{2}(\\Delta r_{i}\\mathbf {e} _{i})+\\mathbf {A} )=(\\sum _{i=1}^{N}m_{i}\\Delta r_{i}^{2})\\alpha {\\vec {k}}+(\\sum _{i=1}^{N}m_{i}\\Delta r_{i}\\mathbf {e} _{i})\\times \\mathbf {A} ,$\n\nwhere ${\\textstyle \\mathbf {e} _{i}\\times \\mathbf {e} _{i}=0}$  and ${\\textstyle \\mathbf {e} _{i}\\times \\mathbf {t} _{i}=k}$  is the unit vector perpendicular to the plane for all of the particles Pi.\n\nUse the center of mass C as the reference point, so these equations for Newton's laws simplify to become\n\n$\\mathbf {F} =M\\mathbf {A} ,\\quad \\mathbf {T} =I_{C}\\alpha {\\vec {k}},$\n\nwhere M is the total mass and IC is the moment of inertia about an axis perpendicular to the movement of the rigid system and through the center of mass.\n\n## Rigid body in three dimensions\n\n### Orientation or attitude descriptions\n\nSeveral methods to describe orientations of a rigid body in three dimensions have been developed. They are summarized in the following sections.\n\n#### Euler angles\n\nThe first attempt to represent an orientation is attributed to Leonhard Euler. He imagined three reference frames that could rotate one around the other, and realized that by starting with a fixed reference frame and performing three rotations, he could get any other reference frame in the space (using two rotations to fix the vertical axis and other to fix the other two axes). The values of these three rotations are called Euler angles.\n\n#### Tait–Bryan angles\n\nThese are three angles, also known as yaw, pitch and roll, Navigation angles and Cardan angles. Mathematically they constitute a set of six possibilities inside the twelve possible sets of Euler angles, the ordering being the one best used for describing the orientation of a vehicle such as an airplane. In aerospace engineering they are usually referred to as Euler angles.\n\n#### Orientation vector\n\nEuler also realized that the composition of two rotations is equivalent to a single rotation about a different fixed axis (Euler's rotation theorem). Therefore, the composition of the former three angles has to be equal to only one rotation, whose axis was complicated to calculate until matrices were developed.\n\nBased on this fact he introduced a vectorial way to describe any rotation, with a vector on the rotation axis and module equal to the value of the angle. Therefore, any orientation can be represented by a rotation vector (also called Euler vector) that leads to it from the reference frame. When used to represent an orientation, the rotation vector is commonly called orientation vector, or attitude vector.\n\nA similar method, called axis-angle representation, describes a rotation or orientation using a unit vector aligned with the rotation axis, and a separate value to indicate the angle (see figure).\n\n#### Orientation matrix\n\nWith the introduction of matrices the Euler theorems were rewritten. The rotations were described by orthogonal matrices referred to as rotation matrices or direction cosine matrices. When used to represent an orientation, a rotation matrix is commonly called orientation matrix, or attitude matrix.\n\nThe above-mentioned Euler vector is the eigenvector of a rotation matrix (a rotation matrix has a unique real eigenvalue). The product of two rotation matrices is the composition of rotations. Therefore, as before, the orientation can be given as the rotation from the initial frame to achieve the frame that we want to describe.\n\nThe configuration space of a non-symmetrical object in n-dimensional space is SO(n) × Rn. Orientation may be visualized by attaching a basis of tangent vectors to an object. The direction in which each vector points determines its orientation.\n\n#### Orientation quaternion\n\nAnother way to describe rotations is using rotation quaternions, also called versors. They are equivalent to rotation matrices and rotation vectors. With respect to rotation vectors, they can be more easily converted to and from matrices. When used to represent orientations, rotation quaternions are typically called orientation quaternions or attitude quaternions.\n\n### Newton's second law in three dimensions\n\nTo consider rigid body dynamics in three-dimensional space, Newton's second law must be extended to define the relationship between the movement of a rigid body and the system of forces and torques that act on it.\n\nNewton formulated his second law for a particle as, \"The change of motion of an object is proportional to the force impressed and is made in the direction of the straight line in which the force is impressed.\" Because Newton generally referred to mass times velocity as the \"motion\" of a particle, the phrase \"change of motion\" refers to the mass times acceleration of the particle, and so this law is usually written as\n\n$\\mathbf {F} =m\\mathbf {a} ,$\n\nwhere F is understood to be the only external force acting on the particle, m is the mass of the particle, and a is its acceleration vector. The extension of Newton's second law to rigid bodies is achieved by considering a rigid system of particles.\n\n### Rigid system of particles\n\nIf a system of N particles, Pi, i=1,...,N, are assembled into a rigid body, then Newton's second law can be applied to each of the particles in the body. If Fi is the external force applied to particle Pi with mass mi, then\n\n$\\mathbf {F} _{i}+\\sum _{j=1}^{N}\\mathbf {F} _{ij}=m_{i}\\mathbf {a} _{i},\\quad i=1,\\ldots ,N,$\n\nwhere Fij is the internal force of particle Pj acting on particle Pi that maintains the constant distance between these particles.\n\nHuman body modelled as a system of rigid bodies of geometrical solids. Representative bones were added for better visualization of the walking person.\n\nAn important simplification to these force equations is obtained by introducing the resultant force and torque that acts on the rigid system. This resultant force and torque is obtained by choosing one of the particles in the system as a reference point, R, where each of the external forces are applied with the addition of an associated torque. The resultant force F and torque T are given by the formulas,\n\n$\\mathbf {F} =\\sum _{i=1}^{N}\\mathbf {F} _{i},\\quad \\mathbf {T} =\\sum _{i=1}^{N}(\\mathbf {R} _{i}-\\mathbf {R} )\\times \\mathbf {F} _{i},$\n\nwhere Ri is the vector that defines the position of particle Pi.\n\nNewton's second law for a particle combines with these formulas for the resultant force and torque to yield,\n\n$\\mathbf {F} =\\sum _{i=1}^{N}m_{i}\\mathbf {a} _{i},\\quad \\mathbf {T} =\\sum _{i=1}^{N}(\\mathbf {R} _{i}-\\mathbf {R} )\\times (m_{i}\\mathbf {a} _{i}),$\n\nwhere the internal forces Fij cancel in pairs. The kinematics of a rigid body yields the formula for the acceleration of the particle Pi in terms of the position R and acceleration a of the reference particle as well as the angular velocity vector ω and angular acceleration vector α of the rigid system of particles as,\n\n$\\mathbf {a} _{i}=\\alpha \\times (\\mathbf {R} _{i}-\\mathbf {R} )+\\omega \\times (\\omega \\times (\\mathbf {R} _{i}-\\mathbf {R} ))+\\mathbf {a} .$\n\n### Mass properties\n\nThe mass properties of the rigid body are represented by its center of mass and inertia matrix. Choose the reference point R so that it satisfies the condition\n\n$\\sum _{i=1}^{N}m_{i}(\\mathbf {R} _{i}-\\mathbf {R} )=0,$\n\nthen it is known as the center of mass of the system. The inertia matrix [IR] of the system relative to the reference point R is defined by\n\n$[I_{R}]=\\sum _{i=1}^{N}m_{i}(\\mathbf {I} (\\mathbf {S} _{i}^{T}\\mathbf {S} _{i})-\\mathbf {S} _{i}\\mathbf {S} _{i}^{T}),$\n\nwhere $\\mathbf {S} _{i}$  is the column vector RiR; and $\\mathbf {S} _{i}^{T}$  is its transpose.\n\n$\\mathbf {S} _{i}^{T}\\mathbf {S} _{i}$  is the scalar product of $\\mathbf {S} _{i}$  with itself, while $\\mathbf {S} _{i}\\mathbf {S} _{i}^{T}$  is the tensor product of $\\mathbf {S} _{i}$  with itself.\n\n$\\mathbf {I}$  is the 3 by 3 identity matrix.\n\n### Force-torque equations\n\nUsing the center of mass and inertia matrix, the force and torque equations for a single rigid body take the form\n\n$\\mathbf {F} =m\\mathbf {a} ,\\quad \\mathbf {T} =[I_{R}]\\alpha +\\omega \\times [I_{R}]\\omega ,$\n\nand are known as Newton's second law of motion for a rigid body.\n\nThe dynamics of an interconnected system of rigid bodies, Bi, j = 1, ..., M, is formulated by isolating each rigid body and introducing the interaction forces. The resultant of the external and interaction forces on each body, yields the force-torque equations\n\n$\\mathbf {F} _{j}=m_{j}\\mathbf {a} _{j},\\quad \\mathbf {T} _{j}=[I_{R}]_{j}\\alpha _{j}+\\omega _{j}\\times [I_{R}]_{j}\\omega _{j},\\quad j=1,\\ldots ,M.$\n\nNewton's formulation yields 6M equations that define the dynamics of a system of M rigid bodies.\n\n### Rotation in three dimensions\n\nA rotating object, whether under the influence of torques or not, may exhibit the behaviours of precession and nutation. The fundamental equation describing the behavior of a rotating solid body is Euler's equation of motion:\n\n${\\boldsymbol {\\tau }}={{D\\mathbf {L} } \\over {Dt}}={{d\\mathbf {L} } \\over {dt}}+{\\boldsymbol {\\omega }}\\times \\mathbf {L} ={{d(I{\\boldsymbol {\\omega }})} \\over {dt}}+{\\boldsymbol {\\omega }}\\times {I{\\boldsymbol {\\omega }}}=I{\\boldsymbol {\\alpha }}+{\\boldsymbol {\\omega }}\\times {I{\\boldsymbol {\\omega }}}$\n\nwhere the pseudovectors τ and L are, respectively, the torques on the body and its angular momentum, the scalar I is its moment of inertia, the vector ω is its angular velocity, the vector α is its angular acceleration, D is the differential in an inertial reference frame and d is the differential in a relative reference frame fixed with the body.\n\nThe solution to this equation when there is no applied torque is discussed in the articles Euler's equation of motion and Poinsot's_ellipsoid.\n\nIt follows from Euler's equation that a torque τ applied perpendicular to the axis of rotation, and therefore perpendicular to L, results in a rotation about an axis perpendicular to both τ and L. This motion is called precession. The angular velocity of precession ΩP is given by the cross product:[citation needed]\n\n${\\boldsymbol {\\tau }}={\\boldsymbol {\\Omega }}_{\\mathrm {P} }\\times \\mathbf {L} .$\n\nPrecession can be demonstrated by placing a spinning top with its axis horizontal and supported loosely (frictionless toward precession) at one end. Instead of falling, as might be expected, the top appears to defy gravity by remaining with its axis horizontal, when the other end of the axis is left unsupported and the free end of the axis slowly describes a circle in a horizontal plane, the resulting precession turning. This effect is explained by the above equations. The torque on the top is supplied by a couple of forces: gravity acting downward on the device's centre of mass, and an equal force acting upward to support one end of the device. The rotation resulting from this torque is not downward, as might be intuitively expected, causing the device to fall, but perpendicular to both the gravitational torque (horizontal and perpendicular to the axis of rotation) and the axis of rotation (horizontal and outwards from the point of support), i.e., about a vertical axis, causing the device to rotate slowly about the supporting point.\n\nUnder a constant torque of magnitude τ, the speed of precession ΩP is inversely proportional to L, the magnitude of its angular momentum:\n\n$\\tau ={\\mathit {\\Omega }}_{\\mathrm {P} }L\\sin \\theta ,\\!$\n\nwhere θ is the angle between the vectors ΩP and L. Thus, if the top's spin slows down (for example, due to friction), its angular momentum decreases and so the rate of precession increases. This continues until the device is unable to rotate fast enough to support its own weight, when it stops precessing and falls off its support, mostly because friction against precession cause another precession that goes to cause the fall.\n\nBy convention, these three vectors - torque, spin, and precession - are all oriented with respect to each other according to the right-hand rule.\n\n## Virtual work of forces acting on a rigid body\n\nAn alternate formulation of rigid body dynamics that has a number of convenient features is obtained by considering the virtual work of forces acting on a rigid body.\n\nThe virtual work of forces acting at various points on a single rigid body can be calculated using the velocities of their point of application and the resultant force and torque. To see this, let the forces F1, F2 ... Fn act on the points R1, R2 ... Rn in a rigid body.\n\nThe trajectories of Ri, i = 1, ..., n are defined by the movement of the rigid body. The velocity of the points Ri along their trajectories are\n\n$\\mathbf {V} _{i}={\\vec {\\omega }}\\times (\\mathbf {R} _{i}-\\mathbf {R} )+\\mathbf {V} ,$\n\nwhere ω is the angular velocity vector of the body.\n\n### Virtual work\n\nWork is computed from the dot product of each force with the displacement of its point of contact\n\n$\\delta W=\\sum _{i=1}^{n}\\mathbf {F} _{i}\\cdot \\delta \\mathbf {r} _{i}.$\n\nIf the trajectory of a rigid body is defined by a set of generalized coordinates qj, j = 1, ..., m, then the virtual displacements δri are given by\n\n$\\delta \\mathbf {r} _{i}=\\sum _{j=1}^{m}{\\frac {\\partial \\mathbf {r} _{i}}{\\partial q_{j}}}\\delta q_{j}=\\sum _{j=1}^{m}{\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}_{j}}}\\delta q_{j}.$\n\nThe virtual work of this system of forces acting on the body in terms of the generalized coordinates becomes\n\n$\\delta W=\\mathbf {F} _{1}\\cdot \\left(\\sum _{j=1}^{m}{\\frac {\\partial \\mathbf {V} _{1}}{\\partial {\\dot {q}}_{j}}}\\delta q_{j}\\right)+\\ldots +\\mathbf {F} _{n}\\cdot (\\sum _{j=1}^{m}{\\frac {\\partial \\mathbf {V} _{n}}{\\partial {\\dot {q}}_{j}}}\\delta q_{j})$\n\nor collecting the coefficients of δqj\n\n$\\delta W=\\left(\\sum _{i=1}^{n}\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}_{1}}}\\right)\\delta q_{1}+\\ldots +(\\sum _{1=1}^{n}\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}_{m}}})\\delta q_{m}.$\n\n### Generalized forces\n\nFor simplicity consider a trajectory of a rigid body that is specified by a single generalized coordinate q, such as a rotation angle, then the formula becomes\n\n$\\delta W=\\left(\\sum _{i=1}^{n}\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}}}\\right)\\delta q=\\left(\\sum _{i=1}^{n}\\mathbf {F} _{i}\\cdot {\\frac {\\partial ({\\vec {\\omega }}\\times (\\mathbf {R} _{i}-\\mathbf {R} )+\\mathbf {V} )}{\\partial {\\dot {q}}}}\\right)\\delta q.$\n\nIntroduce the resultant force F and torque T so this equation takes the form\n\n$\\delta W=\\left(\\mathbf {F} \\cdot {\\frac {\\partial \\mathbf {V} }{\\partial {\\dot {q}}}}+\\mathbf {T} \\cdot {\\frac {\\partial {\\vec {\\omega }}}{\\partial {\\dot {q}}}}\\right)\\delta q.$\n\nThe quantity Q defined by\n\n$Q=\\mathbf {F} \\cdot {\\frac {\\partial \\mathbf {V} }{\\partial {\\dot {q}}}}+\\mathbf {T} \\cdot {\\frac {\\partial {\\vec {\\omega }}}{\\partial {\\dot {q}}}},$\n\nis known as the generalized force associated with the virtual displacement δq. This formula generalizes to the movement of a rigid body defined by more than one generalized coordinate, that is\n\n$\\delta W=\\sum _{j=1}^{m}Q_{j}\\delta q_{j},$\n\nwhere\n\n$Q_{j}=\\mathbf {F} \\cdot {\\frac {\\partial \\mathbf {V} }{\\partial {\\dot {q}}_{j}}}+\\mathbf {T} \\cdot {\\frac {\\partial {\\vec {\\omega }}}{\\partial {\\dot {q}}_{j}}},\\quad j=1,\\ldots ,m.$\n\nIt is useful to note that conservative forces such as gravity and spring forces are derivable from a potential function V(q1, ..., qn), known as a potential energy. In this case the generalized forces are given by\n\n$Q_{j}=-{\\frac {\\partial V}{\\partial q_{j}}},\\quad j=1,\\ldots ,m.$\n\n## D'Alembert's form of the principle of virtual work\n\nThe equations of motion for a mechanical system of rigid bodies can be determined using D'Alembert's form of the principle of virtual work. The principle of virtual work is used to study the static equilibrium of a system of rigid bodies, however by introducing acceleration terms in Newton's laws this approach is generalized to define dynamic equilibrium.\n\n### Static equilibrium\n\nThe static equilibrium of a mechanical system rigid bodies is defined by the condition that the virtual work of the applied forces is zero for any virtual displacement of the system. This is known as the principle of virtual work. This is equivalent to the requirement that the generalized forces for any virtual displacement are zero, that is Qi=0.\n\nLet a mechanical system be constructed from n rigid bodies, Bi, i=1,...,n, and let the resultant of the applied forces on each body be the force-torque pairs, Fi and Ti, i=1,...,n. Notice that these applied forces do not include the reaction forces where the bodies are connected. Finally, assume that the velocity Vi and angular velocities ωi, i=,1...,n, for each rigid body, are defined by a single generalized coordinate q. Such a system of rigid bodies is said to have one degree of freedom.\n\nThe virtual work of the forces and torques, Fi and Ti, applied to this one degree of freedom system is given by\n\n$\\delta W=\\sum _{i=1}^{n}(\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}}}+\\mathbf {T} _{i}\\cdot {\\frac {\\partial {\\vec {\\omega }}_{i}}{\\partial {\\dot {q}}}})\\delta q=Q\\delta q,$\n\nwhere\n\n$Q=\\sum _{i=1}^{n}(\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}}}+\\mathbf {T} _{i}\\cdot {\\frac {\\partial {\\vec {\\omega }}_{i}}{\\partial {\\dot {q}}}}),$\n\nis the generalized force acting on this one degree of freedom system.\n\nIf the mechanical system is defined by m generalized coordinates, qj, j=1,...,m, then the system has m degrees of freedom and the virtual work is given by,\n\n$\\delta W=\\sum _{j=1}^{m}Q_{j}\\delta q_{j},$\n\nwhere\n\n$Q_{j}=\\sum _{i=1}^{n}(\\mathbf {F} _{i}\\cdot {\\frac {\\partial \\mathbf {V} _{i}}{\\partial {\\dot {q}}_{j}}}+\\mathbf {T} _{i}\\cdot {\\frac {\\partial {\\vec {\\omega }}_{i}}{\\partial {\\dot {q}}_{j}}}),\\quad j=1,\\ldots ,m.$\n\nis the generalized force associated with the generalized coordinate qj. The principle of virtual work states that static equilibrium occurs when these generalized forces acting on the system are zero, that is\n\n$Q_{j}=0,\\quad j=1,\\ldots ,m.$\n\nThese m equations define the static equilibrium of the system of rigid bodies.\n\n### Generalized inertia forces\n\nConsider a single rigid body which moves under the action of a resultant force F and torque T, with one degree of freedom defined by the generalized coordinate q. Assume the reference point for the resultant force and torque is the center of mass of the body, then the generalized inertia force Q* associated with the generalized coordinate q is given by\n\n$Q^{*}=-(M\\mathbf {A} )\\cdot {\\frac {\\partial \\mathbf {V} }{\\partial {\\dot {q}}}}-([I_{R}]\\alpha +\\omega \\times [I_{R}]\\omega )\\cdot {\\frac {\\partial {\\vec {\\omega }}}{\\partial {\\dot {q}}}}.$\n\nThis inertia force can be computed from the kinetic energy of the rigid body,\n\n$T={\\frac {1}{2}}M\\mathbf {V} \\cdot \\mathbf {V} +{\\frac {1}{2}}{\\vec {\\omega }}\\cdot [I_{R}]{\\vec {\\omega }},$\n\nby using the formula\n\n$Q^{*}=-\\left({\\frac {d}{dt}}{\\frac {\\partial T}{\\partial {\\dot {q}}}}-{\\frac {\\partial T}{\\partial q}}\\right).$\n\nA system of n rigid bodies with m generalized coordinates has the kinetic energy\n\n$T=\\sum _{i=1}^{n}({\\frac {1}{2}}M\\mathbf {V} _{i}\\cdot \\mathbf {V} _{i}+{\\frac {1}{2}}{\\vec {\\omega }}_{i}\\cdot [I_{R}]{\\vec {\\omega }}_{i}),$\n\nwhich can be used to calculate the m generalized inertia forces\n\n$Q_{j}^{*}=-\\left({\\frac {d}{dt}}{\\frac {\\partial T}{\\partial {\\dot {q}}_{j}}}-{\\frac {\\partial T}{\\partial q_{j}}}\\right),\\quad j=1,\\ldots ,m.$\n\n### Dynamic equilibrium\n\nD'Alembert's form of the principle of virtual work states that a system of rigid bodies is in dynamic equilibrium when the virtual work of the sum of the applied forces and the inertial forces is zero for any virtual displacement of the system. Thus, dynamic equilibrium of a system of n rigid bodies with m generalized coordinates requires that\n\n$\\delta W=(Q_{1}+Q_{1}^{*})\\delta q_{1}+\\ldots +(Q_{m}+Q_{m}^{*})\\delta q_{m}=0,$\n\nfor any set of virtual displacements δqj. This condition yields m equations,\n\n$Q_{j}+Q_{j}^{*}=0,\\quad j=1,\\ldots ,m,$\n\nwhich can also be written as\n\n${\\frac {d}{dt}}{\\frac {\\partial T}{\\partial {\\dot {q}}_{j}}}-{\\frac {\\partial T}{\\partial q_{j}}}=Q_{j},\\quad j=1,\\ldots ,m.$\n\nThe result is a set of m equations of motion that define the dynamics of the rigid body system.\n\n### Lagrange's equations\n\nIf the generalized forces Qj are derivable from a potential energy V(q1,...,qm), then these equations of motion take the form\n\n${\\frac {d}{dt}}{\\frac {\\partial T}{\\partial {\\dot {q}}_{j}}}-{\\frac {\\partial T}{\\partial q_{j}}}=-{\\frac {\\partial V}{\\partial q_{j}}},\\quad j=1,\\ldots ,m.$\n\nIn this case, introduce the Lagrangian, L=T-V, so these equations of motion become\n\n${\\frac {d}{dt}}{\\frac {\\partial L}{\\partial {\\dot {q}}_{j}}}-{\\frac {\\partial L}{\\partial q_{j}}}=0\\quad j=1,\\ldots ,m.$\n\nThese are known as Lagrange's equations of motion.\n\n## Linear and angular momentum\n\n### System of particles\n\nThe linear and angular momentum of a rigid system of particles is formulated by measuring the position and velocity of the particles relative to the center of mass. Let the system of particles Pi, i=1,...,n be located at the coordinates ri and velocities vi. Select a reference point R and compute the relative position and velocity vectors,\n\n$\\mathbf {r} _{i}=(\\mathbf {r} _{i}-\\mathbf {R} )+\\mathbf {R} ,\\quad \\mathbf {v} _{i}={\\frac {d}{dt}}(\\mathbf {r} _{i}-\\mathbf {R} )+\\mathbf {V} .$\n\nThe total linear and angular momentum vectors relative to the reference point R are\n\n$\\mathbf {p} ={\\frac {d}{dt}}(\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} ))+(\\sum _{i=1}^{n}m_{i})\\mathbf {V} ,$\n\nand\n\n$\\mathbf {L} =\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} )\\times {\\frac {d}{dt}}(\\mathbf {r} _{i}-\\mathbf {R} )+(\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} ))\\times \\mathbf {V} .$\n\nIf R is chosen as the center of mass these equations simplify to\n\n$\\mathbf {p} =M\\mathbf {V} ,\\quad \\mathbf {L} =\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} )\\times {\\frac {d}{dt}}(\\mathbf {r} _{i}-\\mathbf {R} ).$\n\n### Rigid system of particles\n\nTo specialize these formulas to a rigid body, assume the particles are rigidly connected to each other so Pi, i=1,...,n are located by the coordinates ri and velocities vi. Select a reference point R and compute the relative position and velocity vectors,\n\n$\\mathbf {r} _{i}=(\\mathbf {r} _{i}-\\mathbf {R} )+\\mathbf {R} ,\\quad \\mathbf {v} _{i}=\\omega \\times (\\mathbf {r} _{i}-\\mathbf {R} )+\\mathbf {V} ,$\n\nwhere ω is the angular velocity of the system.\n\nThe linear momentum and angular momentum of this rigid system measured relative to the center of mass R is\n\n$\\mathbf {p} =(\\sum _{i=1}^{n}m_{i})\\mathbf {V} ,\\quad \\mathbf {L} =\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} )\\times \\mathbf {v} _{i}=\\sum _{i=1}^{n}m_{i}(\\mathbf {r} _{i}-\\mathbf {R} )\\times (\\omega \\times (\\mathbf {r} _{i}-\\mathbf {R} )).$\n\nThese equations simplify to become,\n\n$\\mathbf {p} =M\\mathbf {V} ,\\quad \\mathbf {L} =[I_{R}]\\omega ,$\n\nwhere M is the total mass of the system and [IR] is the moment of inertia matrix defined by\n\n$[I_{R}]=-\\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R],$\n\nwhere [ri-R] is the skew-symmetric matrix constructed from the vector ri-R.\n\n## Applications\n\n• For the analysis of robotic systems\n• For the biomechanical analysis of animals, humans or humanoid systems\n• For the analysis of space objects\n• For the understanding of strange motions of rigid bodies.\n• For the design and development of dynamics-based sensors like gyroscopic sensors etc.\n• For the design and development of various stability enhancement applications in automobiles etc.\n• For improving the graphics of video games which involves rigid bodies" ]
[ null, "https://upload.wikimedia.org/wikipedia/commons/thumb/f/f2/SteamEngine_Boulton%26Watt_1784.png/300px-SteamEngine_Boulton%26Watt_1784.png", null ]
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-9-section-9-5-parametric-equations-exercise-set-page-1020/52
[ "## Precalculus (6th Edition) Blitzer\n\nPublished by Pearson\n\n# Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1020: 52\n\n#### Answer\n\n$x=6t+3$ ; $y=13t-1$\n\n#### Work Step by Step\n\nFor a line, we have: $(3,-1); (9,12)$ $x_1=3, x_2=9,y_1=-1, y_2=12$ Now, in parametric from: $x=(x_2-x_1) t+x_1=6t+3$ and $y=(y_2-y_1)t_1+y_1 =13t-1$ So, $x=6t+3$ ; $y=13t-1$\n\nAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback." ]
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https://numbermatics.com/n/513/
[ "# 513\n\n## 513 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 513 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 8 divisors.\n\n513 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of eight divisors.\n\n## Prime factorization of 513:\n\n### 33 × 19\n\n(3 × 3 × 3 × 19)\n\nSee below for interesting mathematical facts about the number 513 from the Numbermatics database.\n\n### Names of 513\n\n• Cardinal: 513 can be written as Five hundred thirteen.\n\n### Scientific notation\n\n• Scientific notation: 5.13 × 102\n\n### Factors of 513\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 22\n\n### Divisors of 513\n\n• Number of divisors d(n): 8\n• Complete list of divisors:\n• Sum of all divisors σ(n): 800\n• Sum of proper divisors (its aliquot sum) s(n): 287\n• 513 is a deficient number, because the sum of its proper divisors (287) is less than itself. Its deficiency is 226\n\n### Bases of 513\n\n• Binary: 10000000012\n• Base-36: E9\n\n### Squares and roots of 513\n\n• 513 squared (5132) is 263169\n• 513 cubed (5133) is 135005697\n• The square root of 513 is 22.6495033059\n• The cube root of 513 is 8.0052049463\n\n### Scales and comparisons\n\nHow big is 513?\n• 513 seconds is equal to 8 minutes, 33 seconds.\n• To count from 1 to 513 would take you about four minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 513 cubic inches would be around 0.7 feet tall.\n\n### Recreational maths with 513\n\n• 513 backwards is 315\n• 513 is a Harshad number.\n• The number of decimal digits it has is: 3\n• The sum of 513's digits is 9\n• More coming soon!" ]
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https://aaaknow.com/lessonFull.php?slug=geoAreaCircle&menu=Geometry
[ "## Area of a Circle\n\n### Area: Learn\n\nArea is the space within a shape. The units of area are the squares of the units of length (that is, if a shape's sides are measured in meters, then the shape's area is measured in square meters).\n\nFor a circle, we will need to incorporate the constant pi, which we will approximate as the decimal 3.14.\n\nTo find the area of a circle, multiply pi (3.14) by the square of radius. In other words, multiply pi times the radius times the radius.\n\nFor example, if a circle has a radius of 4m, the area is 3.14*4*4 = 50.24m2\n\nIf you are given a diameter instead of a radius, just remember that the radius is half of the diameter.\n\nFor example, if a circle has a diameter of 4m, then the radius is 2m, and the area is 3.14*2*2 = 12.56m2\n\n### Area: Practice\n\n00:00\n\nWhat is the area of a circle whose\nis cm?\n\ncm2\n\nPress the Start Button To Begin\n\nYou have 0 correct and 0 incorrect.\n\nThis is 0 percent correct.\n\n### Play\n\nGame Name Description Best Score\nHow many correct answers can you get in 60 seconds? 0\nExtra time is awarded for each correct answer.\nPlay longer by getting more correct.\n0\nHow fast can you get 20 more correct answers than wrong answers? 999" ]
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https://www.nagwa.com/en/videos/515136981469/
[ "# Lesson Video: Writing and Evaluating Formulas Mathematics\n\nIn this video, we will learn how to distinguish between formulas and expressions, write a formula from a given description, and substitute numbers into a simple formula.\n\n13:48\n\n### Video Transcript\n\nIn this video, we will learn how to distinguish between formulas and expressions, write a formula from a given description, and substitute numbers into a simple formula. We will begin by giving a definition of what we mean by a formula. A formula is a fact or rule written with mathematical symbols. It connects two or more quantities with an equal to sign. If a formula contains two quantities, when you know the value of one of them, you can find the value of the other using the formula. Formulas are commonly used in geometry to calculate perimeters, areas, and volumes. They are also useful in more complicated areas of mathematics together with real-life problems. The first question that we will look at involves writing a formula from some given information.\n\nA rectangle has width 𝑤 and length 𝑙. A new rectangle is formed which has the same length but double the width. Find the perimeter 𝑃 and area 𝐴 of this new rectangle.\n\nLet’s begin by considering the rectangle with width 𝑤 and length 𝑙. We know that each pair of parallel sides in a rectangle are equal in length. Each of the angles in a rectangle is equal to 90 degrees. We also recall that we can calculate the perimeter of any rectangle by adding the length and width and then multiplying by two. Distributing the parentheses or expanding the brackets means that this is the same as two multiplied by the length plus two multiplied by the width, two 𝑙 plus two 𝑤. The area of any rectangle can be calculated by multiplying its length by its width. These are standard formulas that we need to be able to recall.\n\nWe are told that our new rectangle has the same length but double the width. This means that the length is still equal to 𝑙, whereas the width is equal to two 𝑤. Substituting these into our formula for perimeter, we see that the perimeter is equal to two multiplied by 𝑙 plus two 𝑤. Distributing the parentheses here gives us two multiplied by 𝑙 and two multiplied by two 𝑤. This is equal to two 𝑙 plus four 𝑤. As we want a formula for the perimeter 𝑃, 𝑃 is equal to two 𝑙 plus four 𝑤.\n\nAs already mentioned, we know that the area of any rectangle is equal to its length multiplied by its width. For this new rectangle, this is equal to 𝑙 multiplied by two 𝑤. This can be written as two 𝑙𝑤 or two 𝑤𝑙. We always put the number first, but the letters or variables can go in either order. The formula for the area 𝐴 is 𝐴 is equal to two 𝑙𝑤. If we were given specific values for 𝑙 and 𝑤, we could then substitute these in to calculate the value of 𝑃 and 𝐴.\n\nOur next question involves substituting a value into a real-life formula.\n\nA company produces greeting cards with an initial cost of 2,000 Egyptian pounds and an extra cost of half an Egyptian pound per card. The total cost is given by 𝐶 is equal to a half 𝑥 plus 2,000, where 𝑥 is the number of produced cards. Find the total cost of producing 15,000 cards.\n\nIn this question, we are given a formula for the total cost. 𝐶 is equal to a half 𝑥 plus 2,000. As 𝑥 is the number of cards produced, we can substitute any value of 𝑥 into the formula to calculate the total cost. In this question, we are told that 𝑥 is equal to 15,000. Substituting in this value gives us 𝐶 is equal to a half multiplied by 15,000 plus 2,000. One-half multiplied by 15 or a half of 15 is 7.5. This means that a half multiplied by 15,000 is 7,500. 𝐶 is equal to 7,500 plus 2,000. This is equal to 9,500. The total cost of producing 15,000 cards is 9,500 Egyptian pounds.\n\nOur next question involves a formula with multiple variables.\n\nGiven that the area of a trapezoid is 𝐴 equals a half ℎ multiplied by 𝑎 plus 𝑏, find the value of 𝐴 when ℎ equals four centimeters, 𝑎 equals three over two centimeters, and 𝑏 equals five over two centimeters.\n\nWe recall that a trapezoid is also sometimes known as a trapezium. The formula given can be used to calculate the area of any trapezoid. We recall that a trapezoid has a pair of parallel sides, labeled 𝑎 and 𝑏. The perpendicular heights between them is ℎ. We are told that our value of ℎ is four centimeters, 𝑎 is equal to three over two centimeters, and 𝑏 is equal to five over two centimeters. We need to substitute these values into our formula for the area. It is worth noting that three over two is equal to 1.5 and five over two is equal to 2.5. We can calculate the value of 𝑎 plus 𝑏 using either fractions or decimals. Substituting in our values, we get a half multiplied by four multiplied by 1.5 plus 2.5. 1.5 plus 2.5 is equal to four.\n\nWe would also get this answer if we added three over two and five over two. As the denominators are the same, we just add the numerators, giving us eight over two or eight divided by two. This is equal to four. The area of the trapezoid is therefore equal to a half multiplied by four multiplied by four. As multiplication is commutative, we can multiply these numbers in any order. Four multiplied by four is 16, and a half of 16 is eight. We would get the same answer if we found a half of four, which is two, and then multiplied this answer by four. As our dimensions or lengths were in centimeters, our units for area will be square centimeters. The value of 𝐴 is eight square centimeters.\n\nOur next question involves a common formula used in a real-world context.\n\nOne measure of distance is rate multiplied by time, 𝑑 equals 𝑟𝑡. Find the distance Olivia travels if she is moving at a rate of 60 miles per hour for 6.75 hours.\n\nThis question refers to the rate of travel, which may also be referred to as speed. A common measure of speed or rate is miles per hour. The formula given in the question, 𝑑 equals 𝑟𝑡, is a commonly used one. It is often given in triangular form, as shown. The D stands for the distance. T stands for time. S stands for speed, which in our question has been altered to 𝑟 for rate. This leads us to three rearranged formulae. Speed is equal to distance divided by time, time is equal to distance divided by speed, and distance is equal to speed multiplied by time. It is this version of the formula that we are using in this question.\n\nWe are told that the rate or speed is 60 miles per hour. And the time is 6.75 or six and three-quarter hours. To calculate the value of 𝑑, the distance, we need to multiply 60 by 6.75. There are lots of ways of performing this calculation without a calculator. We could begin by splitting 60 into six multiplied by 10 and then multiplying 6.75 by 10. This is equal to 67.5. We could then multiply 67.5 by six using a column multiplication method. This gives us an answer of 405.0. 60 multiplied by 6.75 is 405. As our units for the rate were miles per hour and our units for time were hours, the units for distance will be miles. The distance that Olivia traveled was 405 miles.\n\nOur final question involves a more complicated formula.\n\nThe speed 𝑣, in feet per second, of an ocean wave at depth 𝑑, in feet, is given by the formula 𝑣 is equal to the square root of 32𝑑. Determine, to the nearest tenth, the speed of an ocean wave at a depth of 11 feet.\n\nIn this question, we are given the formula 𝑣 is equal to the square root of 32𝑑, where 𝑣 is the speed in feet per second and 𝑑 is the depth in feet. We need to calculate the speed 𝑣 when the depth is 11 feet. We need to substitute 𝑑 equals 11 into our formula, giving us 𝑣 is equal to the square root of 32 multiplied by 11. 32 multiplied by 10 is 320. This means that 32 multiplied by 11 is equal to 352. Our value of 𝑣 is the square root of 352.\n\n352 is not a square number, so we need to type this into our calculator and round to the nearest tenth. The square root of 352 is equal to 18.7616 and so on. Rounding to the nearest tenth is the same as rounding to one decimal place. So our deciding number is the first six. If the deciding number is five or greater, we round up. 𝑣 is therefore equal to 18.8 to the nearest tenth. As our units for 𝑣 were feet per second, we can conclude that when the depth is 11 feet, the speed of the wave is 18.8 feet per second.\n\nWe will finish this video by summarizing the key points. Formulas can be used in areas of mathematics, such as geometry, and also in real-life situations. A formula links two or more variables with an equal to sign. Substituting in the value of one or more quantities or variables allows us to calculate the value of another. It is also possible to rearrange a formula to make another variable the subject." ]
[ null ]
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https://blog.quantinsti.com/xgboost-python/
[ "# Introduction to XGBoost in Python\n\nBy Ishan Shah and compiled by Rekhit Pachanekar\n\nAh! XGBoost! The supposed miracle worker which is the weapon of choice for machine learning enthusiasts and competition winners alike. It is said that XGBoost was developed to increase computational speed and optimize model performance.\n\nAs we were tinkering with the features and parameters of XGBoost, we decided to build a portfolio of five companies and applied XGBoost model on it to create a trading strategy. Here’s what we got. The five companies were Apple, Amazon, Netflix, Nvidia and Microsoft.", null, "That’s really decent. And to think we haven’t even tried to optimise it.\n\nLet’s figure out how to implement the XGBoost model in this article. We will cover the following things:\n\n## What is XGBoost?\n\nXgboost stands for eXtreme Gradient Boosting and is developed on the framework of gradient boosting. I like the sound of that, Extreme! Sounds more like a supercar than an ML model, actually.\n\nBut that is exactly what it does, boosts the performance of a regular gradient boosting model.\n\n“XGBoost used a more regularized model formalization to control over-fitting, which gives it better performance”.\n\n• Tianqi Chen, the author of XGBoost\n\nLet’s break down the name to understand what XGBoost does.\n\n### What is boosting?\n\nThe sequential ensemble methods, also known as “boosting”, creates a sequence of models that attempt to correct the mistakes of the models before them in the sequence. The first model is built on training data, the second model improves the first model, the third model improves the second, and so on.", null, "In the above image example, the train dataset is passed to the classifier 1. The yellow background indicates that the classifier predicted hyphen and blue background indicates that it predicted plus. The classifier 1 model incorrectly predicts two hyphens and one plus. These are highlighted with a circle. The weights of these incorrectly predicted data points are increased and sent to the next classifier. That is to classifier 2. The classifier 2 correctly predicts the two hyphen which classifier 1 was not able to. But classifier 2 also makes some other errors. This process continues and we have a combined final classifier which predicts all the data points correctly.\n\nThe classifier models can be added until all the items in the training dataset is predicted correctly or a maximum number of classifier models are added. The optimal maximum number of classifier models to train can be determined using hyperparameter tuning.\n\nTake a pause over here. Maybe you don’t know what a sequential model is. Let’s take baby steps here.\n\n### Machine learning in a nutshell\n\nEarlier, we used to code a certain logic and then give the input to the computer program. The program would use the logic, ie the algorithm and provide an output. All this was great and all, but as our understanding increased, so did our programs, until we realised that for certain problem statements, there were far too many parameters to program.\n\nAnd then some smart individual said that we should just give the computer (machine) both the problem and the solution for a sample set and then let the machine learn.\n\nWhile developing the algorithms for machine learning, we realised that we could roughly put machine learning problems in two data sets, classification and regression. In simple terms, classification problem can be that given a photo of an animal, we try to classify it as a dog or a cat (or some other animal). In contrast, if we have to predict the temperature of a city, it would be a regression problem as the temperature can be said to have continuous values such as 40 degrees, 40.1 degrees and so on.\n\nGreat! We then moved on to decision tree models, Bayesian, clustering models and the like. All this was fine until we reached another roadblock, the prediction rate for certain problem statements was dismal when we used only one model. Apart from that, for decision trees, we realised that we had to live with bias, variance as well as noise in the models. This led to another bright idea, how about we combine models, I mean, two heads are better than one, right? This was and is called Ensemble learning. But here, we can use much more than one model to create an ensemble. Gradient boosting was one such method of ensemble learning. .\n\nIn gradient boosting while combining the model, the loss function is minimized using gradient descent. Technically speaking, a loss function can be said as an error, ie the difference between the predicted value and the actual value. Of course, the less the error, the better is the machine learning model.\n\nGradient boosting is an approach where new models are created that predict the residuals or errors of prior models and then added together to make the final prediction.\n\nThe objective of the XGBoost model is given as:\n\nObj = L + Ω\n\nWhere L is the loss function which controls the predictive power, and Ω is regularization component which controls simplicity and overfitting\n\nThe loss function (L) which needs to be optimized can be Root Mean Squared Error for regression, Logloss for binary classification, or mlogloss for multi-class classification.\n\nThe regularization component (Ω) is dependent on the number of leaves and the prediction score assigned to the leaves in the tree ensemble model.\n\nIt is called gradient boosting because it uses a gradient descent algorithm to minimize the loss when adding new models. The Gradient boosting algorithm supports both regression and classification predictive modelling problems.\n\nAll right, we have understood how machine learning evolved from simple models to a combination of models. Somehow, humans cannot be satisfied for long, and as problem statements became more complex and the data set larger, we realised that we should go one step further. This leads us to XGBoost.\n\n## Why is XGBoost so good?\n\nXGBoost was written in C++, which when you think about it, is really quick when it comes to the computation time. The great thing about XGBoost is that it can easily be imported in python and thanks to the sklearn wrapper, we can use the same parameter names which are used in python packages as well.\n\nWhile the actual logic is somewhat lengthy to explain, one of the main things about xgboost is that it has been able to parallelise the tree building component of the boosting algorithm. This leads to a dramatic gain in terms of processing time as we can use more cores of a CPU or even go on and utilise cloud computing as well.\n\nWhile machine learning algorithms have support for tuning and can work with external programs, XGBoost has built-in parameters for regularisation and cross-validation to make sure both bias and variance is kept at a minimal. The advantage of in-built parameters is that it leads to faster implementation.\n\nLet’s discuss one such instance in the next section.\n\n## Xgboost feature importance\n\nFeatures, in a nutshell, are the variables we are using to predict the target variable. Sometimes, we are not satisfied with just knowing how good our machine learning model is. I would like to know which feature has more predictive power. . There are various reasons why knowing feature importance can help us. Let us list down a few below:\n\n• If I know that a certain feature is more important than others, I would put more attention to it and try to see if I can improve my model further.\n• After I have run the model, I will see if dropping a few features improves my model.\n• Initially, if the dataset is small, the time taken to run a model is not a significant factor while we are designing a system. But if the strategy is complex and requires a large dataset to run, then the computing resources and the time taken to run the model becomes an important factor.\n\nThe good thing about XGBoost is that it contains an inbuilt function to compute the feature importance and we don’t have to worry about coding it in the model.\n\nThe sample code which is used later in the XGBoost python code section is given below:\n\nfrom xgboost import plot_importance\n# Plot feature importance\nplot_importance(model)", null, "All right, before we move on to the code, let’s make sure we all have XGBoost on our system.\n\n## How to install XGBoost in anaconda?\n\nAnaconda is a python environment which makes it really simple for us to write python code and takes care of any nitty-gritty associated with the code. Hence, I am specifying the step to install XGBoost in Anaconda. It’s actually just one line of code.\n\nYou can simply open the Anaconda prompt and input the following: pip install XGBoost\n\nThe Anaconda environment will download the required setup file and install it for you. It would look something like below.", null, "That’s all there is to it. Awesome! Now we move to the real thing, ie the XGBoost python code.\n\n## Xgboost in Python\n\nLet me give a summary of the XGBoost machine learning model before we dive into it. We are using the stock data of tech stocks in the US such as Apple, Amazon, Netflix, Nvidia and Microsoft for the last sixteen years and train the XGBoost model to predict if the next day’s returns are positive or negative.\n\nWe have included the code in the form of a downloadable python notebook for you to work on later. It is attached at the end of the blog.\n\nWe will divide the XGBoost python code into following sections for a better understanding of the model\n\n1. Import libraries\n2. Define parameters\n3. Creating predictors and target variables\n4. Split the data into train and test\n5. Initialising the XGBoost machine learning model\n6. Cross Validation in Train dataset\n7. Train the model\n8. Feature Importance\n9. Prediction Report\n\n### Import libraries\n\nWe have written the use of the library in the comments. For example, since we use XGBoost python library, we will import the same and write # Import XGBoost as a comment.\n\n# Import warnings and add a filter to ignore them\nimport warnings\nwarnings.simplefilter('ignore')\n# Import XGBoost\nimport xgboost\n# XGBoost Classifier\nfrom xgboost import XGBClassifier\n# Classification report and confusion matrix\nfrom sklearn.metrics import classification_report\nfrom sklearn.metrics import confusion_matrix\n# Cross validation\nfrom sklearn.model_selection import KFold\nfrom sklearn.model_selection import cross_val_score\n# Pandas datareader to get the data\n# To plot the graphs\nimport matplotlib.pyplot as plt\nimport seaborn as sn\n# For data manipulation\nimport pandas as pd\nimport numpy as np\n\nGreat! All libraries imported. Now we move to the next section.\n\n### Define parameters\n\nWe have defined the list of stock, start date and the end date which we will be working with in this blog.\n\n# Set the stock list\nstock_list = ['AAPL', 'AMZN', 'NFLX', 'NVDA','MSFT']\n# Set the start date and the end date\nstart_date = '2004-1-1'\nend_date = '2020-1-28'\n\nJust to make things interesting, we will use the XGBoost python model on companies such as Apple, Amazon, Netflix, Nvidia and Microsoft. Creating predictors and target variables\n\nWe define a list of predictors from which the model will pick the best predictors. Here, we have the percentage change and the standard deviation with different time periods as the predictor variables.\n\nThe target variable is the next day's return. If the next day’s return is positive we label it as 1 and if it is negative then we label it as -1. You can also try to create the target variables with three labels such as 1, 0 and -1 for long, no position and short.\n\nLet’s see the code now.\n\n# Create a placeholder to store the stock data\nstock_data_dictionary = {}\nfor stock_name in stock_list:\n# Get the data\ndf = data.get_data_yahoo(stock_name, start_date, end_date)\n# Calculate the daily percent change\n# create the predictors\npredictor_list = []\nfor r in range(10, 60, 5):\ndf['pct_change_'+str(r)] = df.daily_pct_change.rolling(r).sum()\ndf['std_'+str(r)] = df.daily_pct_change.rolling(r).std()\npredictor_list.append('pct_change_'+str(r))\npredictor_list.append('std_'+str(r))\n# Target Variable\ndf['return_next_day'] = df.daily_pct_change.shift(-1)\ndf['actual_signal'] = np.where(df.return_next_day > 0, 1, -1)\ndf = df.dropna()\n# Add the data to dictionary\nstock_data_dictionary.update({stock_name: df})\n\nBefore we move on to the implementation of the XGBoost python model, let’s first plot the daily returns of Apple stored in the dictionary to see if everything is working fine.\n\n# Set the figure size\nplt.figure(figsize=(10, 7))\n# Access the dataframe of AAPL from the dictionary\n# and then compute and plot the returns\n(stock_data_dictionary['AAPL'].daily_pct_change+1).cumprod().plot()\n# Set the title and axis labels and plot grid\nplt.title('AAPL Returns')\nplt.ylabel('Cumulative Returns')\nplt.grid()\nplt.show()\n\nYou will get the output as follows:", null, "It looks accurate.\n\n### Split the data into train and test\n\nSince XGBoost is after all a machine learning model, we will split the data set into test and train set.\n\n# Create a placeholder for the train and test split data\nX_train = pd.DataFrame()\nX_test = pd.DataFrame()\ny_train = pd.Series()\ny_test = pd.Series()\nfor stock_name in stock_list:\n# Get predictor variables\nX = stock_data_dictionary[stock_name][predictor_list]\n# Get the target variable\ny = stock_data_dictionary[stock_name].actual_signal\n# Divide the dataset into train and test\ntrain_length = int(len(X)*0.80)\nX_train = X_train.append(X[:train_length])\nX_test = X_test.append(X[train_length:])\ny_train = y_train.append(y[:train_length])\ny_test = y_test.append(y[train_length:])\n\n### Initialising the XGBoost machine learning model\n\nWe will initialize the classifier model. We will set two hyperparameters namely max_depth and n_estimators. These are set on the lower side to reduce overfitting.\n\n# Initialize the model and set the hyperparameter values\nmodel = XGBClassifier(max_depth=2, n_estimators=30)\nmodel\n\nThe output is as follows:", null, "All right, we will now perform cross-validation on the train set to check the accuracy.\n\n### Cross Validation in Train dataset\n\n# Initialize the KFold parameters\nkfold = KFold(n_splits=5, random_state=7)\n# Perform K-Fold Cross Validation\nresults = cross_val_score(model, X_train, y_train, cv=kfold)\n# Print the average results\nprint(\"Accuracy: %.2f%% (%.2f%%)\" % (results.mean()*100, results.std()*100))\n\nThe output is as follows:", null, "The accuracy is slightly above the half mark. This can be further improved by hyperparameter tuning and grouping similar stocks together. I will leave the optimization part on you. Feel free to post a comment if you have any queries.\n\n### Train the model\n\nWe will train the XGBoost classifier using the fit method.\n\n# Fit the model\n\nmodel.fit(X_train, y_train)\n\nYou will find the output as follows:", null, "### Feature importance\n\nWe have plotted the top 7 features and sorted based on its importance.\n\n# Plot the top 7 features\nxgboost.plot_importance(model, max_num_features=7)\n# Show the plot\nplt.show()", null, "That’s interesting. The XGBoost python model tells us that the pct_change_40 is the most important feature of the others. Since we had mentioned that we need only 7 features, we received this list. Here’s an interesting idea, why don’t you increase the number and see how the other features stack up, when it comes to their f-score. You can also remove the unimportant features and then retrain the model. Would this increase the model accuracy? I leave that for you to verify.\n\nAnyway, onwards we go!\n\n### Predict and Classification report\n\n# Predict the trading signal on test dataset\ny_pred = model.predict(X_test)\n# Get the classification report\nprint(classification_report(y_test, y_pred))", null, "Hold on! We are almost there. Let’s see what XGBoost tells us right now:\n\nThat’s interesting. The f1-score for the long side is much more powerful compared to the short side. We can modify the model and make it a long-only strategy.\n\nLet’s try another way to formulate how well XGBoost performed.\n\n### Confusion Matrix\n\narray = confusion_matrix(y_test, y_pred)\ndf = pd.DataFrame(array, index=['Short', 'Long'], columns=[\n'Short', 'Long'])\nplt.figure(figsize=(5, 4))\nsn.heatmap(df, annot=True, cmap='Greens', fmt='g')\nplt.xlabel('Predicted')\nplt.ylabel('Actual')\nplt.show()\n\nThe output will be as shown below:", null, "But what is this telling us? Well it’s a simple matrix which shows us how many times XGBoost predicted “buy” or “sell” accurately or not. For example, when it comes to predicting “Long”, XGBoost predicted it right 1926 times whereas it was incorrect 1608 times.\n\nAnother interpretation is that XGBoost tended to predict “long” more times than “short”.\n\n### Individual stock performance\n\nLet’s see how the XGBoost based strategy returns held up against the normal daily returns ie the buy and hold strategy. We will plot a comparison graph between the strategy returns and the daily returns for all the companies we had mentioned before. The code is as follows:\n\n# Create an empty dataframe to store the strategy returns of individual stocks\nportfolio = pd.DataFrame(columns=stock_list)\n# For each stock in the stock list, plot the strategy returns and buy and hold returns\nfor stock_name in stock_list:\n# Get the data\ndf = stock_data_dictionary[stock_name]\n# Store the predictor variables in X\nX = df[predictor_list]\n# Define the train and test dataset\ntrain_length = int(len(X)*0.80)\n# Predict the signal and store in predicted signal column\ndf['predicted_signal'] = model.predict(X)\n# Calculate the strategy returns\ndf['strategy_returns'] = df.return_next_day * df.predicted_signal\n# Add the strategy returns to the portfolio dataframe\nportfolio[stock_name] = df.strategy_returns[train_length:]\n# Plot the stock strategy and buy and hold returns\nprint(stock_name)\n# Set the figure size\nplt.figure(figsize=(10, 7))\n# Calculate the cumulative strategy returns and plot\n(df.strategy_returns[train_length:]+1).cumprod().plot()\n# Calculate the cumulative buy and hold strategy returns\n(stock_data_dictionary[stock_name][train_length:].daily_pct_change+1).cumprod().plot()\n# Set the title, label and grid\nplt.title(stock_name + ' Returns')\nplt.ylabel('Cumulative Returns')\nplt.legend(labels=['Strategy Returns', 'Buy and Hold Returns'])\nplt.grid()\nplt.show()", null, "", null, "", null, "", null, "", null, "This was fun, wasn’t it? What do you think of the comparison? Do let us know your observations or thoughts in the comments and we would be happy to read them.\n\n### Performance of portfolio\n\nWe were enjoying this so much that we just couldn’t stop at the individual level. Hence we thought what would happen if we invest in all the companies equally and act according to the XGBoost python model. Let’s see what happens.\n\n# Drop missing values\nportfolio.dropna(inplace=True)\n# Set the figure size\nplt.figure(figsize=(10, 7))\n# Calculate the cumulative portfolio returns by assuming equal allocation to the stocks\n(portfolio.mean(axis=1)+1).cumprod().plot()\n# Set the title and label of the chart\nplt.title('Portfolio Strategy Returns')\nplt.ylabel('Cumulative Returns')\nplt.grid()\nplt.show()", null, "Well, remember that these are cumulative returns, hence it should give you an idea about the performance of an XGBoost model.\n\nIf you want more detailed feedback on the test set, try out the following code.\n\nimport pyfolio as pf\npf.create_full_tear_sheet(portfolio.mean(axis=1))\n\nWhile the output generated is somewhat lengthy, we have attached a snapshot", null, "", null, "Phew! That was a long one. But we hope that you understood how a boosted model like XGBoost can help us in generating signals and creating a trading strategy.\n\n## Conclusion\n\nWe started from the base, ie the emergence of machine learning algorithms and its next level, ie ensemble learning. We learnt about boosted trees and how they help us in making better predictions. We finally came to XGBoost machine learning model and how it is better than a regular boosted algorithm. We then went through a simple XGBoost python code and created a portfolio based on the trading signals created by the code. In between, we also listed down feature importance as well as certain parameters included in XGBoost.\n\nIf you want to embark on a stepwise training plan on the complete lifecycle of machine learning trading strategies, then you can take the Machine learning strategy development and live trading learning track and receive guidance from experts such as Dr. Ernest P. Chan, Terry Benzschawel and QuantInsti.\n\nDisclaimer: All data and information provided in this article are for informational purposes only. QuantInsti® makes no representations as to accuracy, completeness, currentness, suitability, or validity of any information in this article and will not be liable for any errors, omissions, or delays in this information or any losses, injuries, or damages arising from its display or use. All information is provided on an as-is basis.\n\nPress the Download button to fetch the code we have used in this blog." ]
[ null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Apple_returns.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/XG-Boost-FINAL-01.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Feature-importance.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/xgboost_download.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Apple_returns_only.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Initialise-XGBoost.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Cross-validation.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Training-model.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Feature-importance.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Classification-report.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Confusion-Matrix.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/download--1-.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Amazon-strategy.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Netflix-Strategy.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Nvidia-strategy.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Microsoft-Strategy.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Portfolio-returns.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/portfolio1.png", null, "https://d1rwhvwstyk9gu.cloudfront.net/2020/02/Portfolio2.png", null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.87114096,"math_prob":0.88397586,"size":20954,"snap":"2021-43-2021-49","text_gpt3_token_len":4515,"char_repetition_ratio":0.13575178,"word_repetition_ratio":0.02602118,"special_character_ratio":0.21408801,"punctuation_ratio":0.10914761,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98816687,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,4,null,4,null,8,null,4,null,4,null,4,null,4,null,4,null,8,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T06:02:05Z\",\"WARC-Record-ID\":\"<urn:uuid:20ea0f22-9e2a-47b0-8265-9890a2ee4345>\",\"Content-Length\":\"172442\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68db8d22-a5c1-42b3-ad6c-f07a4a90cb8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:3460ec09-832c-40fe-9d9e-29ac26c9fbac>\",\"WARC-IP-Address\":\"174.129.255.252\",\"WARC-Target-URI\":\"https://blog.quantinsti.com/xgboost-python/\",\"WARC-Payload-Digest\":\"sha1:6IEZCUUPUZJHT7MI7UTRR33EZFD7UO5X\",\"WARC-Block-Digest\":\"sha1:NTFLA36RGEFHXBKDBU7JG2XY5YRWM4VN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585242.44_warc_CC-MAIN-20211019043325-20211019073325-00538.warc.gz\"}"}
https://howmanyis.com/temperature/211-degree-c-in-k/164327-1-degree-celsius-in-kelvin
[ "How many is\nConversion between units of measurement\nRating 4.00 (1 Vote)\n\nYou can easily convert 1 degree Celsius into Kelvin using each definition.\n\n¿How many K are there in 1 degree C?\n\nThe result of the absolute temperature conversion is that 1 °C are 274.15 K.\n\nOne degree Celsius equals to two hundred seventy-four Kelvin. *Approximately\n\n¿What is the temperature increment conversion of 1 degree Celsius in Kelvin?\n\nA change of 1 degree Celsius corresponds to a change of 1 Kelvin.\n\nA temperature change of one degree Celsius equals to a change of one Kelvin. *Aproximado" ]
[ null ]
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https://programmer.group/chapter-ii-data-structure-monotonic-stack-and-monotonic-queue.html
[ "# Chapter II Data Structure Monotonic Stack and Monotonic Queue\n\nKeywords: C Algorithm data structure\n\n# 1. Algorithmic ideas\n\nThe core idea of these two algorithms is:\nOnly those elements that conform to a certain monotonicity are likely to be the answer. Screen out those elements that do not conform to a monotonicity according to the title requirements, store those elements that conform to a certain monotonicity in the stack or queue, and obtain the maximum and minimum values of these elements according to the title (usually at the top of the stack, at the head of the queue, at the end of the queue)Or, depending on monotonicity, the values of certain elements can be obtained by dichotomy.\nThat's abstract. Let's look at two examples for analysis.\nThese two algorithms can handle a limited variety of topics, so more can be accumulated.\n\n# 2. Example Analysis\n\n## 1. Monotonic Stack\n\n```Given a length of N An integer column that outputs the first lower number to the left of each number, or if it does not exist −1.\n\nInput Format\nThe first line contains integers N，Represents the length of a column.\n\nThe second line contains N An integer representing an integer column.\n\nOutput Format\nOne line, containing N Integers, of which i Number represents number i The first number to the left of the number smaller than it, or output if it does not exist −1.\n\nData Range\n1≤N≤10^5\n1≤Elements in Columns≤10^9\nInput sample:\n5\n3 4 2 7 5\nOutput sample:\n-1 3 -1 2 2\n```\n\n### 1. Idea analysis\n\nThis can also be changed to look for the first (smaller) number on the right that is larger than this number.\n\nViolent practice is to use a double loop, and for each number, start over and look for the first smaller number to the left of each number.\n• Watch what kind of elements can never be the answer\nLet's consider 4 and 2 in the sample. For elements after 2, if 4 is smaller than that element, then 2 must be smaller than that element, and 2 is closer to that element than 4, so for elements after 2, 4 will never be the answer. Generally, if x and y are in reverse order (that is, if x<y and X is more to the right than y)So for the elements to the right of x, it is impossible to select the element y. So for the elements to the right of x, the elements that may be selected must be monotonically increasing in order (because anything in reverse order cannot be selected). We use stacks to store the elements that may be selected, the closer to the top of the stack, the larger and the more to the right.\n\n### 2. Code implementation\n\n```#include<cstdio>\nusing namespace std;\nconst int N = 1e5 + 10;\nint stk[N],tt;\nint main()\n{\nint n;\nscanf(\"%d\",&n);\nfor(int i = 0; i < n; i ++)\n{\nint tmp;\nscanf(\"%d\",&tmp);\nwhile(tt && stk[tt] >= tmp)\n{\ntt --;\n}\nif(tt == 0)\n{\nprintf(\"-1 \");\n}else{\nprintf(\"%d \",stk[tt]);\n}\nstk[++ tt] = tmp;\n}\n}\n```\n\n### 3. Time Complexity Analysis\n\nAlthough it looks like a double loop, for while inside, it executes up to n times in total, so the algorithm time complexity is O ( n ) O(n) O(n)\n\n## 2. Monotonic Queues\n\n```Given a size of n≤10^6 Array.\nThere is a size of k The sliding window of the array moves from the leftmost to the rightmost.\nYou can only see it in the window k Numbers.\nSlide the window one position to the right each time.\n\nHere is an example:\nThe array is [1 3 -1 -3 5 3 6 7]，k 3.\n\nwindow position\t\t\t\t\tminimum value \tMaximum\n[1 3 -1] -3 5 3 6 7\t -1\t\t\t\t 3\n1 [3 -1 -3] 5 3 6 7\t -3\t\t\t\t 3\n1 3 [-1 -3 5] 3 6 7\t -3\t\t\t\t 5\n1 3 -1 [-3 5 3] 6 7\t -3\t\t\t\t 5\n1 3 -1 -3 [5 3 6] 7\t 3\t\t\t\t 6\n1 3 -1 -3 5 [3 6 7]\t 3\t\t\t\t 7\nYour task is to determine the maximum and minimum values in the sliding window at each location.\n\nInput Format\nThe input contains two lines.\nThe first line contains two integers n and k，Represents the length of the array and the length of the sliding window, respectively.\nThe second line has n Integers representing the specific values of the array.\nPeer data is separated by spaces.\n\nOutput Format\nThe output contains two.\n\nThe first line of output, from left to right, slides the minimum value in the window at each location.\nThe second line of output, from left to right, slides the maximum value in the window at each location.\n\nInput sample:\n8 3\n1 3 -1 -3 5 3 6 7\nOutput sample:\n-1 -3 -3 -3 3 3\n3 3 5 5 6 7\n```\n\n### 1. Idea analysis\n\n• Consider violent practices first. Traverse through each selected interval to pick the maximum and minimum values.\n• Watch what elements are not selected as answers\nAssuming we seek the minimum, k = 3, we consider the interval [-1-3 5] in the sampleFor example, when -3 appears, the -1 ahead of him can never be called the result of an interval. Because -1 is not only larger than -3, but -3 is also later than -1, leaving the interval later, that is, -1 leaves the interval before -3 leaves the interval, and because -1 is larger than -3, it is impossible for -1 to be the result at this time. So for each new element entering the interval, the interval at this time is not possible anymore.All elements larger than it will no longer be the answer, that is, all possible answers are less than it. So we use a queue to maintain all possible numbers in the zone that will be the answer. Every time a number is inserted, a number larger than it will be expelled from the queue, which always maintains a monotonic queue from small to large.\n\n### 2. Code implementation\n\n```#include<iostream>\nusing namespace std;\nconst int N = 1e6 + 10;\nint q[N],s[N],h,t;\n\nint main()\n{\nint n,k;\ncin >> n >> k;\nfor(int i = 0; i < n; i ++)\n{\ncin >> s[i];\n}\n\nfor(int i = 0; i < n; i ++)\n{\nif(h != t && i - k + 1 > q[h]) h ++;\nwhile(h != t && s[q[t - 1]] > s[i]) t --;\nq[t ++] = i;\nif(i >= k - 1 && t != h)\ncout << s[q[h]] << \" \";\n}\ncout << endl;\nh = 0,t = 0;\nfor(int i = 0; i < n; i ++)\n{\nif(h != t && i - k + 1 > q[h])h ++;\nwhile(h != t && s[q[t - 1]] < s[i]) t --;\nq[t ++] = i;\nif(i >= k - 1 && t != h)\ncout << s[q[h]] << \" \";\n}\n}\n```\n\n### 3. Time Complexity Analysis\n\nEach element in while is queued only once, so there are 2 total operations with a time complexity of O ( n ) O(n) O(n)\n\n# Reference material\n\nAcwing\n\nPosted by Trey395 on Tue, 21 Sep 2021 09:40:12 -0700" ]
[ null ]
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https://docs.neptune.ai/integrations/sklearn/
[ "# scikit-learn integration guide#", null, "", null, "Scikit-learn (also known as sklearn) is an open source machine learning framework commonly used for building predictive models. With the Neptune–scikit-learn integration, you can track your classifiers, regressors, and k-means clustering results, specifically:\n\n• Classifier and regressor parameters\n• Pickled model\n• Test predictions and their probabilities\n• Test scores\n• Classifier and regressor visualizations, such as confusion matrix, precision–recall chart, and feature importance chart\n• K-means cluster labels and clustering visualizations\n• Code snapshots and Git information\n\n## Installing the integration#\n\nTo use your preinstalled version of Neptune together with the integration:\n\npip\npip install -U neptune-sklearn\n\nconda\nconda install -c conda-forge neptune-sklearn\n\n\nTo install both Neptune and the integration:\n\npip\npip install -U \"neptune[sklearn]\"\n\nconda\nconda install -c conda-forge neptune neptune-sklearn\n\n\nOnce you've registered and created a project, set your Neptune API token and full project name to the NEPTUNE_API_TOKEN and NEPTUNE_PROJECT environment variables, respectively.\n\nexport NEPTUNE_API_TOKEN=\"h0dHBzOi8aHR0cHM.4kl0jvYh3Kb8...6Lc\"\n\n\nTo find your API token: In the bottom-left corner of the Neptune app, expand the user menu and select Get my API token.\n\nexport NEPTUNE_PROJECT=\"ml-team/classification\"\n\n\nTo find your project: Your full project name has the form workspace-name/project-name. To copy the name, click the menu in the top-right corner and select Edit project details.\n\nWhile it's not recommended especially for the API token, you can also pass your credentials in the code when initializing Neptune.\n\nrun = neptune.init_run(\nproject=\"ml-team/classification\", # your full project name here\napi_token=\"h0dHBzOi8aHR0cHM6Lkc78ghs74kl0jvYh...3Kb8\", # your API token here\n)\n\n\nFor more help, see Set Neptune credentials.\n\nIf you'd rather follow the guide without any setup, you can run the example in Colab .\n\n## scikit-learn logging example#\n\nThis example shows how to log and observe metadata as you train your model with scikit-learn.\n\n1. Create and fit an example estimator.\n\nPrepare a fitted estimator. The blow snippet illustrates the idea:\n\nfrom sklearn.datasets import fetch_california_housing\nfrom sklearn.ensemble import RandomForestRegressor\nfrom sklearn.model_selection import train_test_split\n\nparameters = {\"n_estimators\": 70, \"max_depth\": 7, \"min_samples_split\": 3}\n\nestimator = RandomForestRegressor(**parameters)\nX, y = fetch_california_housing(return_X_y=True)\nX_train, X_test, y_train, y_test = train_test_split(\nX,\ny,\ntest_size=0.20,\n)\nestimator.fit(X_train, y_train)\n\n2. Create a Neptune run:\n\nimport neptune\n\nrun = neptune.init_run() # (1)!\n\n1. If you haven't set up your credentials, you can log anonymously:\n\nneptune.init_run(\napi_token=neptune.ANONYMOUS_API_TOKEN,\nproject=\"common/sklearn-integration\",\n)\n\n3. To log parameters of your model training run, pass them to the namespace of your choice.\n\nFor example, to log them under the namespace \"params\":\n\nrun[\"params\"] = parameters\n\n4. Similarly, log scores on the test data under the namespaces and fields of your choice.\n\ny_pred = estimator.predict(X_test)\n\nrun[\"scores/max_error\"] = max_error(y_test, y_pred)\nrun[\"scores/mean_absolute_error\"] = mean_absolute_error(y_test, y_pred)\nrun[\"scores/r2_score\"] = r2_score(y_test, y_pred)\n\n5. To stop the connection to Neptune and sync all data, call the stop() method:\n\nrun.stop()\n\n\nTo open the run, click the Neptune link that appears in the console output.\n\n### Logging estimator parameters#\n\nTo only log estimator parameters, use the get_estimator_params() function:\n\nimport neptune.integrations.sklearn as npt_utils\nfrom neptune.utils import stringify_unsupported\n\nrfc = RandomForestClassifier()\n\nrun = neptune.init_run(name=\"only estimator params\") # name is optional\n\nrun[\"estimator/params\"] = stringify_unsupported(npt_utils.get_estimator_params(rfr))\n\nrun.stop()\n\n\n### Logging pickled model#\n\nTo log a fitted model as a pickled file, use the get_pickled_model() function:\n\nimport neptune.integrations.sklearn as npt_utils\n\nrfc = RandomForestClassifier()\nrfc.fit(X, y)\n\nrun = neptune.init_run(\nname=\"only pickled model\", # optional\n)\n\nrun[\"estimator/pickled-model\"] = npt_utils.get_pickled_model(rfc)\n\nrun.stop()\n\n\n### Logging confusion matrix#\n\nUse the create_confusion_matrix_chart() function to log a confusion matrix chart:\n\nimport neptune.integrations.sklearn as npt_utils\n\nrfc = RandomForestClassifier()\nX_train, X_test, y_train, y_test = train_test_split(\nX, y, test_size=0.20, random_state=28743\n)\nrfc.fit(X_train, y_train)\n\nrun = neptune.init_run(\nname=\"only confusion matrix\", # optional\n)\n\nrun[\"confusion-matrix\"] = npt_utils.create_confusion_matrix_chart(\nrfc, X_train, X_test, y_train, y_test\n)\n\nrun.stop()\n\n\n## More options#\n\nYou can also log regressor, classifier, or k-means summary information to Neptune. The summary includes:\n\n• All parameters\n• Visualizations\n• Code snapshot and Git metadata\n• K-means: Cluster labels\n• Classifier and regressor:\n• Pickled model\n• Test predictions and their probabilities\n• Test scores\n\n### Logging classification summary#\n\nStart by preparing a fitted classifier.\n\nExample\nfrom sklearn.datasets import load_digits\nfrom sklearn.model_selection import train_test_split\n\nparameters = {\n\"n_estimators\": 120,\n\"learning_rate\": 0.12,\n\"min_samples_split\": 3,\n\"min_samples_leaf\": 2,\n}\n\nX_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)\n\ngbc.fit(X_train, y_train)\n\n\nWe'll use the gbc object to log metadata to the Neptune run.\n\n1. Create a run:\n\nimport neptune\n\nrun = neptune.init_run( # (1)!\nname=\"classification example\", # optional\n)\n\n1. If you haven't set up your credentials, you can log anonymously:\n\nneptune.init_run(\napi_token=neptune.ANONYMOUS_API_TOKEN,\nproject=\"common/sklearn-integration\",\n)\n\n2. In a namespace of your choice, log the classifier summary:\n\nimport neptune.integrations.sklearn as npt_utils\n\nrun[\"cls_summary\"] = npt_utils.create_classifier_summary(\ngbc, X_train, X_test, y_train, y_test\n)\n\n\nIn the snippet above, the namespace is cls_summary.\n\n3. To stop the connection to Neptune and sync all data, call the stop() method:\n\nrun.stop()\n\n\nTo open the run, click the Neptune link that appears in the console output.\n\nYou can browse the logged metadata in the Run details view.\n\n### Logging regression summary#\n\nStart by preparing a fitted regressor.\n\nExample\nfrom sklearn.datasets import fetch_california_housing\nfrom sklearn.ensemble import RandomForestRegressor\nfrom sklearn.model_selection import train_test_split\n\nparameters = {\"n_estimators\": 70, \"max_depth\": 7, \"min_samples_split\": 3}\n\nrfr = RandomForestRegressor(**parameters)\n\nX, y = fetch_california_housing(return_X_y=True)\nX_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)\n\nrfr.fit(X_train, y_train)\n\n\nWe'll use the rfr object to log metadata to the Neptune run.\n\n1. Create a run:\n\nimport neptune\n\nrun = neptune.init_run( # (1)!\nname=\"regression example\", # optional\n)\n\n1. If you haven't set up your credentials, you can log anonymously:\n\nneptune.init_run(\napi_token=neptune.ANONYMOUS_API_TOKEN,\nproject=\"common/sklearn-integration\",\n)\n\n2. In a namespace of your choice, log the regressor summary:\n\nimport neptune.integrations.sklearn as npt_utils\n\nrun[\"rfr_summary\"] = npt_utils.create_regressor_summary(\nrfr, X_train, X_test, y_train, y_test\n)\n\n\nIn the snippet above, the namespace is rfr_summary.\n\n3. To stop the connection to Neptune and sync all data, call the stop() method:\n\nrun.stop()\n\n\nTo open the run, click the Neptune link that appears in the console output.\n\nYou can browse the logged metadata in the Run details view.\n\n### Logging k-means clustering summary#\n\nStart by preparing a KMeans object and example data.\n\nExample\nfrom sklearn.cluster import KMeans\nfrom sklearn.datasets import make_blobs\n\nparameters = {\"n_init\": 11, \"max_iter\": 270}\n\nkm = KMeans(**parameters)\nX, y = make_blobs(n_samples=579, n_features=17, centers=7, random_state=28743)\n\n1. Create a run:\n\nimport neptune\n\nrun = neptune.init_run( # (1)!\nname=\"clustering-example\", # optional\ntags=[\"KMeans\", \"clustering\"], # optional\n)\n\n1. If you haven't set up your credentials, you can log anonymously:\n\nneptune.init_run(\napi_token=neptune.ANONYMOUS_API_TOKEN,\nproject=\"common/sklearn-integration\",\n)\n\n2. In a namespace of your choice, log the regressor summary:\n\nimport neptune.integrations.sklearn as npt_utils\n\nrun[\"kmeans_summary\"] = npt_utils.create_kmeans_summary(km, X, n_clusters=17)\n\n\nIn the snippet above, the namespace is kmeans_summary.\n\n3. To stop the connection to Neptune and sync all data, call the stop() method:\n\nrun.stop()\n\n\nTo open the run, click the Neptune link that appears in the console output.\n\nYou can browse the logged metadata in the Run details view.\n\nIf you have other types of metadata that are not covered in this guide, you can still log them using the Neptune client library.\n\nWhen you initialize the run, you get a run object, to which you can assign different types of metadata in a structure of your own choosing.\n\nimport neptune\n\n# Create a new Neptune run\nrun = neptune.init_run()\n\n# Log metrics or other values inside loops\nfor epoch in range(n_epochs):\n\nrun[\"train/epoch/loss\"].append(loss) # Each append() appends a value\nrun[\"train/epoch/accuracy\"].append(acc)" ]
[ null, "https://colab.research.google.com/assets/colab-badge.svg", null, "https://docs.neptune.ai/img/app/integrations/scikit-learn.png", null ]
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https://math.stackexchange.com/questions/1689723/integral-of-divergence-over-a-closed-surface
[ "# Integral of divergence over a closed surface\n\nI am reading a paper, where an integral of a divergence over a closed surface is used without proof.\n\n$$\\oint_S [\\nabla \\cdot \\vec{v}(\\vec{r})] d\\vec{s} = 0$$,\n\nwhere $$\\vec{v}$$ is tangential to the surface ($$\\vec{v}(r)\\cdot \\vec{n}(\\vec{r}) = 0$$)\n\nI have looked at vector calculus identities and Green theorems and can't seem to find the expression I need. Any suggestions?\n\nThis is a direct consequence of the divergence theorem. For a vector field $$X$$ on an oriented $$n$$-dimensional Riemannian manifold $$(M, g)$$, the divergence theorem states that $$\\int_M (\\operatorname{div} X) dV_g = \\int_{\\partial M} g(X, N) dV_{\\tilde g}$$ where $$N$$ is the outward-pointing normal vector at the boundary, $$dV_g$$ is the Riemannian volume form, and $$dV_{\\tilde g}$$ is the induced volume form on the boundary. For a surface embedded in Euclidean space, we use the metric induced by the pullback of the inclusion $$i:S \\to \\mathbb{R}^k$$, i.e., $$i^*g$$ where $$g = \\delta_{ij}dx^idx^j$$ is the usual Euclidean metric. Then $$dV_g = ds$$ where $$ds$$ is the area element, and $$dV_{\\tilde g} = dt$$ where $$dt$$ is a length element. Since the surface in your question is closed, the boundary $$\\partial S$$ is empty and the right-hand side integral is $$0$$.\nIf $$M$$ is not orientable, the divergence theorem still holds if you replace $$dV_g$$ and $$dV_{\\tilde g}$$ with the respective densities $$d\\mu_g$$ and $$d\\mu_{\\tilde g}$$. These densities are nothing but local volume forms on different patches of the manifold glued together with a partition of unity.\n• Also, there seems to be a mistake. The LHS of your divergence theorem equation should refer to an integral of a volume, but you use the symbol $ds$ in there, which you call \"the area form of the surface\". This is inconsistent with the standard definition of the theorem – Aleksejs Fomins Oct 6 at 12:53\n• It is not a mistake. The divergence theorem applies to manifolds of all dimensions, not just 3-dimensional manifolds. I have rewritten it in the general form to avoid confusion. Also note that \"closed surface\" means a compact surface without boundary, so $\\partial S = \\emptyset$ is part of the definition. – abhi01nat Oct 6 at 16:54" ]
[ null ]
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https://ftp.aimsciences.org/article/doi/10.3934/dcds.2017256
[ "• PDF\n• Cite\n• Share\nArticle Contents", null, "", null, "Article Contents\n\n# Visco-Energetic solutions to one-dimensional rate-independent problems\n\n• Visco-Energetic solutions of rate-independent systems (recently introduced in ) are obtained by solving a modified time Incremental Minimization Scheme, where at each step the dissipation is reinforced by a viscous correction $δ$, typically a quadratic perturbation of the dissipation distance. Like Energetic and Balanced Viscosity solutions, they provide a variational characterization of rate-independent evolutions, with an accurate description of their jump behaviour.\n\nIn the present paper we study Visco-Energetic solutions in the scalar-valued case and we obtain a full characterization for a broad class of energy functionals. In particular, we prove that they exhibit a sort of intermediate behaviour between Energetic and Balanced Viscosity solutions, which can be finely tuned according to the choice of the viscous correction $δ$.\n\nMathematics Subject Classification: Primary: 34C55, 47J20, 49J40; Secondary: 74N30.\n\n Citation:", null, "•", null, "• Figure 1.  The double-well potential $W$ with its convex envelope in bold (left picture) and an energetic solution $u$ in the case of a strictly increasing load $\\ell$ (right picture).\n\nFigure 2.  BV solution for a double-well energy $W$ with an increasing load $\\ell$. The blue line denotes the path described by the optimal transition $\\vartheta$ solving (4).\n\nFigure 3.  Visco-Energetic solutions for a double-well energy $W$ with an increasing load $\\ell$. When $\\mu>-\\min W''$ (left picture) the solution jumps when it reach the maximum of $W'$ and the transition is the ''double chain'' obtained by solving the Incremental Minimization Scheme with frozen time $t$. When $\\mu$ is small (right picture) the optimal transition $\\vartheta$ makes a first jump connecting $\\mathit{u}_{\\tiny \\mathsf L}(t)$ with $u_+$ according to the modified Maxwell rule (9): $\\mathit{u}_{\\tiny \\mathsf L}(t)$ and $u_+$ corresponds to the intersection of $W'$ with the red line, whose slope is $-\\mu$.\n\nFigure 4.  The one-sided slopes and the stability region when $W$ is a double-well potential. For some suitable choices of $\\delta$, $\\mathit W_{\\mathsf{i}\\mathsf{r},{\\delta}}'$ is intermediate between $W'_{\\mathsf i\\mathsf r}$ and $W'$.\n\nFigure 5.  $\\mathsf D$-Maxwell's rule for a double well potential: when $\\delta=\\frac{\\mu}{2}|\\cdot|^2$, the ''last point'' where $W'$ and $\\mathit W_{\\mathsf{i}\\mathsf{r},{\\delta}}'$ coincide is such that the total area between the graph $W'$ and the line whose slope is $-\\mu$ is zero.\n\nFigure 6.  Visco-Energetic solution of a double-well potential energy with an oscillating external loading and a quadratic viscous-correction $\\delta(u,v)$, turned by a parameter $\\mu>-\\min W''$.\n\nFigure 7.  Visco-Energetic solutions of a nonconvex energy and an increasing loading. The optimal transition is a combination of sliding and viscous parts.\n\n•", null, "## Article Metrics", null, "", null, "DownLoad:  Full-Size Img  PowerPoint" ]
[ null, "https://ftp.aimsciences.org/style/web/images/custom/icon_toright.png", null, "https://ftp.aimsciences.org/style/web/images/custom/icon_toleft.png", null, "https://ftp.aimsciences.org/style/web/images/article/shu.png", null, "https://ftp.aimsciences.org/style/web/images/article/1.gif", null, "https://ftp.aimsciences.org/style/web/images/article/1.gif", null, "https://ftp.aimsciences.org/article/doi/10.3934/dcds.2017256", null, "https://ftp.aimsciences.org/style/web/images/article/download.png", null ]
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https://www.futurestarr.com/blog/mathematics/what-is-14-as-a-percent-or
[ "FutureStarr\n\nWhat Is 14 As a Percent OR\n\n## What Is 14 As a Percent OR", null, "# What Is 14 As a Percent\n\nis a fun puzzle game designed to help you learn how to multiply. The goal: to find the missing number of a given sum.\n\n### Use\n\nDecimals Converted to PercentageDecimal to percentage is our category with posts explaining how to calculate specific decimal equivalents in percent. After giving you the result of a conversion of the type x in percent, x being your decimal, we show you the math in full detail. Every article also contains the spelling variants such as x in percent, pc, %, pct, etc., along with information on current and past use. In our posts we then elaborate on the wording, writing and meaning of percentage operations. In addition, we shed a light on the frequently asked questions in the context. It stands to reason that our comment form allows you pose questions and to leave a feedback. What’s more is that each article comes with 2 calculators, a decimal to percent calculator and a state-of-the-art percentage calculator. By the way: The quickest way to locate a decimal to percentage conversion is our search form.\n\nI've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below). (Source: visualfractions.com)\n\n### Value\n\nIn calculating 14% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,14% off, 14% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.295 dollar to pound = 194.7 pound\n\nPercentage calculator tool can be used by first entering the fractional value you want to calculate. For example 5% of 20, which is the same thing as fraction x/100 * 20=5%. To find the value of x enter 5 in the first box and 20 in the second box and the answer 1 will be shown in the result box. (Source: www.percentage-off-calculator.com)\n\n## Related Articles\n\n•", null, "#### Non Profit Resume Sample", null, "August 10, 2022     |     Amir jameel\n•", null, "#### 25 Calculator OR", null, "August 10, 2022     |     Jamshaid Aslam\n•", null, "#### Calculate 14/35 AS A Percentage OOR", null, "August 10, 2022     |     Javeria Ijaz\n•", null, "#### A 15 18 Percentage", null, "August 10, 2022     |     Shaveez Haider\n•", null, "#### Lovecalculator Ws", null, "August 10, 2022     |     sheraz naseer\n•", null, "#### Volume of a Cubeor", null, "August 10, 2022     |     Muhammad basit\n•", null, "#### Calculator Soup Decimal to Fraction in 2022", null, "August 10, 2022     |     Jamshaid Aslam\n•", null, "#### 4 Out of 14 As a Percentage", null, "August 10, 2022     |     Muhammad Waseem\n•", null, "#### A Online Calculator Using Keyboard", null, "August 10, 2022     |     Shaveez Haider\n•", null, "#### 414 Area Code", null, "August 10, 2022     |     sajjad ghulam hussain\n•", null, "#### Increase Amount by Percentage", null, "August 10, 2022     |     Faisal Arman\n•", null, "#### 502 Area Code", null, "August 10, 2022     |     mohammad umair\n•", null, "#### A Estimating Tile Quantities", null, "August 10, 2022     |     Shaveez Haider\n•", null, "#### A Number Line Fractions Calculator", null, "August 10, 2022     |     Muhammad Waseem\n•", null, "#### A 3 Fraction Calculator Online", null, "August 10, 2022     |     Shaveez Haider" ]
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http://www.greatestcommonfactor.net/gcf-of-811/
[ "X\nX\n\n# Calculate the Greatest Common Factor or GCF of 811\n\nThe instructions to find the GCF of 811 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 811 811 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 811 = 811\n\n## 3. Choose the common prime factors with the lowest exponent\n\nCommon prime factors: 811\n\nCommon prime factors with the lowest exponent: 8111\n\n## 4. Calculate the Greatest Common Factor or GCF\n\nRemember, to find the GCF of several numbers you must multiply the common prime factors with the lowest exponent.\n\nGCF = 8111 = 811\n\nAlso calculates the:" ]
[ null ]
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https://www.statistics-lab.com/%E6%95%B0%E5%AD%A6%E4%BB%A3%E5%86%99%E4%BF%A1%E6%81%AF%E8%AE%BA%E4%BD%9C%E4%B8%9A%E4%BB%A3%E5%86%99information-theory%E4%BB%A3%E8%80%83run-length-encoding/
[ "### 数学代写|信息论作业代写information theory代考|Run Length Encoding", null, "statistics-lab™ 为您的留学生涯保驾护航 在代写信息论information theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写信息论information theory代写方面经验极为丰富，各种代写信息论information theory相关的作业也就用不着说。\n\n• Statistical Inference 统计推断\n• Statistical Computing 统计计算\n• (Generalized) Linear Models 广义线性模型\n• Statistical Machine Learning 统计机器学习\n• Longitudinal Data Analysis 纵向数据分析\n• Foundations of Data Science 数据科学基础\n\n## 数学代写|信息论作业代写information theory代考|Run Length Encoding\n\nRun-length Encoding, or RLE is a technique used to reduce the size of a repeating string of characters. This repeating string is called a run. Typically RLE encodes a run of symbols into two bytes, a count and a symbol. RLE can compress any type of data regardless of its information content, but the content of data to be compressed affects the compression ratio. RLE cannot achieve high compression ratios compared to other compression methods, but it is easy to implement and is quick to execute. Run-length encoding is supported by most bitmap file formats such as TIFF, JPG, BMP, PCX and fax machines.\n\nWe will restrict ourselves to that portion of the PCX data stream that actually contains the coded image, and not those parts that store the color palette and image information such as number of lines, pixels per line, file and the coding method.\n\nThe basic scheme is as follows. If a string of pixels are identical in color value, encode them as a special flag byte which contains the count followed by a byte with the value of the repeated pixel. If the pixel is not repeated, simply encode it as the byte itself. Such simple schemes can often become more complicated in practice. Consider that in the above scheme, if all 256 colors in a palette are used in an image, then, we need all 256 values of a byte to represent those colors. Hence if we are going to use just bytes as our basic code unit, we don’t have any possible unused byte values that can be used as a flag/count byte. On the other hand, if we use two bytes for every coded pixel to leave room for the flag/count combinations, we might double the size of pathological images instead of compressing them.\nDID YOU The compromise in the PCX format is based on the belief of its designers than many user-created KNOW $=$ drawings (which was the primary intended output of their software) would not use all 256 colors. So, they optimized their compression scheme for the case of up to 192 colors only. Images with more colors will also probably get good compression, just not quite as good, with this scheme.\n\n## 数学代写|信息论作业代写information theory代考|Rate Distortion Function\n\nAlthough we live in an analog world, most of the communication takes place in the digital form. Since most natural sources (e.g. speech, video etc.) are analog, they are first sampled, quantized and then processed. However, the representation of an arbitrary real number requires an infinite number of bits. Thus, a finite representation of a continuous random variable can never be perfect. Consider an analog message waveform $x(t)$ which is a sample waveform of a stochastic process $X(t)$. Assuming $X(t)$ is a bandlimited, stationary process, it can be represented by a sequence of uniform samples taken at the Nyquist rate. These samples are quantized in amplitude and encoded as a sequence of binary digits. A simple encoding strategy can be to define $L$ levels and encode every sample using\n\\begin{aligned} &R=\\log {2} L \\text { bits if } L \\text { is a power of } 2 \\text {, or } \\ &R=\\left\\lfloor\\log {2} L\\right\\rfloor+1 \\text { bits if } L \\text { is not a power of } 2 \\end{aligned}\nIf all levels are not equally probable we may use entropy coding for a more efficient representation. In order to represent the analog waveform more accurately, we need more number of levels, which would imply more number of bits per sample. Theoretically we need infinite bits per sample to perfectly represent an analog source. Quantization of amplitude results in data compression at the cost of signal distortion. It’s a form of lossy data compression. Distortion implies some measure of difference between the actual source samples $\\left{x_{k}\\right}$ and the corresponding quantized value $\\left{\\tilde{x}_{k}\\right}$.\n\n## 数学代写|信息论作业代写information theory代考|Optimum Quantizer Design\n\nIn this section, we look at optimum quantizers design. Consider a continuous amplitude signal whose amplitude is not uniformly distributed, but varies according to a certain probability density function, $p(x)$. We wish to design the optimum scalar quantizer that minimizes some function of the quantization error $q=\\tilde{x}-x$, where $\\tilde{x}$ is the quantized value of $x$. The distortion resulting due to the quantization can be expressed as\n$$D=\\int_{-\\infty}^{\\infty} f(\\tilde{x}-x) p(x) d x$$\nwhere $f(\\tilde{x}-x)$ is the desired function of the error. An optimum quantizer is one that minimizes DID YOU $D$ by optimally selecting the output levels and the corresponding input range of each output\nKNOW level. The resulting optimum quantizer is called the Lloyd-Max quantizer. For an L-level quantizer the distortion is given by\n$$D=\\sum_{k=1}^{L} \\int_{x_{k-1}}^{x_{k}} f\\left(\\tilde{x}_{k}-x\\right) p(x) d x$$\n\nThe necessary conditions for minimum distortion are obtained by differentiating $D$ with respect to $\\left{x_{k}\\right}$ and $\\left{\\tilde{x}{k}\\right}$. As a result of the differentiation process we end up with the following system of equations $$\\begin{array}{ll} f\\left(\\tilde{x}{k}-x_{k}\\right)=f\\left(\\tilde{x}{k+1}-x{k}\\right), & k=1,2, \\ldots, L-1 \\ \\int_{x_{k-1}}^{x_{k}} f^{\\prime}\\left(\\tilde{x}{k+1}-x\\right) p(x) d x, & k=1,2, \\ldots, L \\end{array}$$ For $f(x)=x^{2}$, i.e., the mean square value of the distortion, the above equations simplify to $$\\begin{array}{ll} x{k}=\\frac{1}{2}\\left(\\tilde{x}{k}+\\tilde{x}{k+1}\\right), & k=1,2, \\ldots, L-1 \\ \\int_{x_{k-1}}^{x_{k}}\\left(\\tilde{x}{k}-x\\right) p(x) d x=0, & k=1,2, \\ldots, L \\end{array}$$ The non uniform quantizers are optimized with respect to the distortion. However, each quantized sample is represented by equal number of bits (say, $R$ bits/sample). It is possible to have a more efficient variable length coding. The discrete source outputs that result from quantization can be characterized by a set of probabilities $p{k^{*}}$. These probabilities can then be used to design efficient variable length codes (source coding). In order to compare the performance of different nonuniform quantizers, we first fix the distortion, $D$, and then compare the average number of bits required per sample.\n\n## 数学代写|信息论作业代写information theory代考|Rate Distortion Function\n\nR=日志⁡2大号 位如果 大号 是一种力量 2， 或者  R=⌊日志⁡2大号⌋+1 位如果 大号 不是一种力量 2\n\n## 数学代写|信息论作业代写information theory代考|Optimum Quantizer Design\n\nD=∫−∞∞F(X~−X)p(X)dX\n\nD=∑ķ=1大号∫Xķ−1XķF(X~ķ−X)p(X)dX\n\nF(X~ķ−Xķ)=F(X~ķ+1−Xķ),ķ=1,2,…,大号−1 ∫Xķ−1XķF′(X~ķ+1−X)p(X)dX,ķ=1,2,…,大号为了F(X)=X2，即失真的均方值，上述方程简化为\n\nXķ=12(X~ķ+X~ķ+1),ķ=1,2,…,大号−1 ∫Xķ−1Xķ(X~ķ−X)p(X)dX=0,ķ=1,2,…,大号非均匀量化器针对失真进行了优化。但是，每个量化样本都由相同数量的比特表示（例如，R位/样本）。可以有更有效的可变长度编码。量化产生的离散源输出可以用一组概率来表征pķ∗. 然后可以使用这些概率来设计有效的可变长度代码（源编码）。为了比较不同非均匀量化器的性能，我们首先修复失真，D，然后比较每个样本所需的平均位数。\n\n## 有限元方法代写\n\ntatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。\n\n## MATLAB代写\n\nMATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。" ]
[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20245%20171'%3E%3C/svg%3E", null ]
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https://nest-simulator.readthedocs.io/en/v2.20.0/auto_examples/intrinsic_currents_subthreshold.html
[ "# Intrinsic currents subthreshold¶\n\nThis example illustrates how to record from a model with multiple intrinsic currents and visualize the results. This is illustrated using the ht_neuron which has four intrinsic currents: I_NaP, I_KNa, I_T, and I_h. It is a slightly simplified implementation of neuron model proposed in 1.\n\nThe neuron is driven by DC current, which is alternated between depolarizing and hyperpolarizing. Hyperpolarization intervals become increasingly longer.\n\nIntrinsic currents spiking\n\nWe imported all necessary modules for simulation, analysis and plotting.\n\nimport nest\nimport numpy as np\nimport matplotlib.pyplot as plt\n\n\nAdditionally, we set the verbosity using set_verbosity to suppress info messages. We also reset the kernel to be sure to start with a clean NEST.\n\nnest.set_verbosity(\"M_WARNING\")\nnest.ResetKernel()\n\n\nWe define simulation parameters:\n\n• The length of depolarization intervals\n\n• The length of hyperpolarization intervals\n\n• The amplitude for de- and hyperpolarizing currents\n\n• The end of the time window to plot\n\nn_blocks = 5\nt_block = 20.\nt_dep = [t_block] * n_blocks\nt_hyp = [t_block * 2 ** n for n in range(n_blocks)]\nI_dep = 10.\nI_hyp = -5.\n\nt_end = 500.\n\n\nWe create the one neuron instance and the DC current generator and store the returned handles.\n\nnrn = nest.Create('ht_neuron')\ndc = nest.Create('dc_generator')\n\n\nWe create a multimeter to record\n\n• membrane potential V_m\n\n• threshold value theta\n\n• intrinsic currents I_NaP, I_KNa, I_T, I_h\n\nby passing these names in the record_from list.\n\nTo find out which quantities can be recorded from a given neuron, run:\n\nnest.GetDefaults('ht_neuron')['recordables']\n\n\nThe result will contain an entry like:\n\n<SLILiteral: V_m>\n\n\nfor each recordable quantity. You need to pass the value of the SLILiteral, in this case V_m in the record_from list.\n\nWe want to record values with 0.1 ms resolution, so we set the recording interval as well; the default recording resolution is 1 ms.\n\n# create multimeter and configure it to record all information\n# we want at 0.1 ms resolution\nmm = nest.Create('multimeter',\nparams={'interval': 0.1,\n'record_from': ['V_m', 'theta',\n'I_NaP', 'I_KNa', 'I_T', 'I_h']}\n)\n\n\nWe connect the DC generator and the multimeter to the neuron. Note that the multimeter, just like the voltmeter is connected to the neuron, not the neuron to the multimeter.\n\nnest.Connect(dc, nrn)\nnest.Connect(mm, nrn)\n\n\nWe are ready to simulate. We alternate between driving the neuron with depolarizing and hyperpolarizing currents. Before each simulation interval, we set the amplitude of the DC generator to the correct value.\n\nfor t_sim_dep, t_sim_hyp in zip(t_dep, t_hyp):\n\nnest.SetStatus(dc, {'amplitude': I_dep})\nnest.Simulate(t_sim_dep)\n\nnest.SetStatus(dc, {'amplitude': I_hyp})\nnest.Simulate(t_sim_hyp)\n\n\nWe now fetch the data recorded by the multimeter. The data are returned as a dictionary with entry times containing timestamps for all recorded data, plus one entry per recorded quantity.\n\nAll data is contained in the events entry of the status dictionary returned by the multimeter. Because all NEST function return arrays, we need to pick out element 0 from the result of GetStatus.\n\ndata = nest.GetStatus(mm)['events']\nt = data['times']\n\n\nThe next step is to plot the results. We create a new figure, add a single subplot and plot at first membrane potential and threshold.\n\nfig = plt.figure()\nVax.plot(t, data['V_m'], 'b-', lw=2, label=r'$V_m$')\nVax.plot(t, data['theta'], 'g-', lw=2, label=r'$\\Theta$')\nVax.set_ylim(-80., 0.)\nVax.set_ylabel('Voltageinf [mV]')\nVax.set_xlabel('Time [ms]')\n\n\nTo plot the input current, we need to create an input current trace. We construct it from the durations of the de- and hyperpolarizing inputs and add the delay in the connection between DC generator and neuron:\n\n• We find the delay by checking the status of the dc->nrn connection.\n\n• We find the resolution of the simulation from the kernel status.\n\n• Each current interval begins one time step after the previous interval, is delayed by the delay and effective for the given duration.\n\n• We build the time axis incrementally. We only add the delay when adding the first time point after t=0. All subsequent points are then automatically shifted by the delay.\n\ndelay = nest.GetStatus(nest.GetConnections(dc, nrn))['delay']\ndt = nest.GetKernelStatus('resolution')\n\nt_dc, I_dc = , \n\nfor td, th in zip(t_dep, t_hyp):\nt_prev = t_dc[-1]\nt_start_dep = t_prev + dt if t_prev > 0 else t_prev + dt + delay\nt_end_dep = t_start_dep + td\nt_start_hyp = t_end_dep + dt\nt_end_hyp = t_start_hyp + th\n\nt_dc.extend([t_start_dep, t_end_dep, t_start_hyp, t_end_hyp])\nI_dc.extend([I_dep, I_dep, I_hyp, I_hyp])\n\n\nThe following function turns a name such as I_NaP into proper TeX code $$I_{\\mathrm{NaP}}$$ for a pretty label.\n\ndef texify_name(name):\nreturn r'${}_{{\\mathrm{{{}}}}}$'.format(*name.split('_'))\n\n\nNext, we add a right vertical axis and plot the currents with respect to that axis.\n\nIax = Vax.twinx()\nIax.plot(t_dc, I_dc, 'k-', lw=2, label=texify_name('I_DC'))\n\nfor iname, color in (('I_h', 'maroon'), ('I_T', 'orange'),\n('I_NaP', 'crimson'), ('I_KNa', 'aqua')):\nIax.plot(t, data[iname], color=color, lw=2, label=texify_name(iname))\n\nIax.set_xlim(0, t_end)\nIax.set_ylim(-10., 15.)\nIax.set_ylabel('Current [pA]')\nIax.set_title('ht_neuron driven by DC current')\n\n\nWe need to make a little extra effort to combine lines from the two axis into one legend.\n\nlines_V, labels_V = Vax.get_legend_handles_labels()\nlines_I, labels_I = Iax.get_legend_handles_labels()\ntry:\nIax.legend(lines_V + lines_I, labels_V + labels_I, fontsize='small')\nexcept TypeError:\n# work-around for older Matplotlib versions\nIax.legend(lines_V + lines_I, labels_V + labels_I)\n\n\nNote that I_KNa is not activated in this example because the neuron does not spike. I_T has only a very small amplitude.\n\nTotal running time of the script: ( 0 minutes 0.000 seconds)\n\nGallery generated by Sphinx-Gallery" ]
[ null ]
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https://eric-roca.github.io/courses/olg/olg_monotonicity_of_the_dynamics/
[ "# Multiple intertemporal equilibria\n\nIn the OLG model, a unique intertemporal equilibrium arises if, for a given $k_t$, the equation\n\n$$H(k_{t+1}) = k_{t+1} - \\frac{1}{1+n}s\\left(\\omega(k_t), f^\\prime(k_{t+1})\\right) = 0$$\n\nhas a unique solution. We know that $\\lim_{k\\rightarrow 0}H(k) < 0$ and $\\lim_{k\\ rightarrow +\\infty}H(k) > 0,$ which guarantees the existence of at least one solution.\n\nHence, if the function $H$ is monotonously increasing, the solution is unique. This condition depends on the sign of:\n\n$$\\frac{\\partial H(k_{t+1})}{\\partial k_{t+1}} = 1 + \\frac{1}{1+n}s^{\\prime}_{R}f^{\\prime \\prime}(k_{t+1}).$$\n\nWe have different cases (check the notes on the savings function):\n\n1. $s^\\prime_{R} = 0$, which happens under log-utility. In this case, $s^\\prime_{R} = 0 > \\frac{1+n}{R^\\prime(k_{t+1})}$ and $H(k_{t+1})$ is always increasing.\n\n2. $s^\\prime_{R} >0$, the intertemporal elasticity of substitution is greater than 1 and individuals are willing to trade off higher future consumption against present consumption. Savings increase to consume more in the future. Then, clearly $s^\\prime_{R} > 0 > \\frac{1+n}{R^\\prime(k_{t+1})}.$\n\n3. $s^\\prime_{R} < 0$, we can have a non-monotonous capital path as $k_{t+1}$ can be increasing or decreasing with $k_{t}$.\n\nUnder case 3, the condition is not satisfied, multiple levels of $k_{t+1}$ solve the intertemporal equilibrium.\n\n### Non-monotonous dynamics\n\nNote: Based on the lecture notes of Groth, pp. 92–93\n\nUnder case 3, the function $g(k)$ is backwards bending for some values of $k.$ This feature implies that there is more than one intemporal equilibrium, check also. We ruled out such a possibility assuming that $\\sigma( c ) \\geq 1$, this is, we imposed a large enough intertemporal elasticity of substitution.\n\nIf this is not the case and $s^\\prime_{R}(\\omega(k_t), R(k_{t+1})) < \\frac{1+n}{R^\\prime(k_{t+1})}$ then there are multiple temporary equilibriums.\n\nFor instance, take an isoelastic utility function $u( c ) = \\frac{c^{1-\\frac{1}{\\sigma}}-1}{1-\\frac{1}{\\sigma}}$ and a CES production with $A=20, , \\alpha = \\frac{1}{2}, , \\beta = 0.3, , n = 1.097, , \\sigma = 0.1, , \\rho = -2.$ First, since $\\sigma < 1$, it allows for the possibility of having multiple temporary equilibria. Second, we can verify that this is the case. For example, if $k_{t} = 1$, then:\n\n$k_{t+1} = \\frac{1}{1+n} s(\\omega(k_{t}), R(k_{t+1})) \\implies k_{t+1} = \\begin{cases} 0.26 \\\\\\ 1.234 \\\\\\ 5.832 \\end{cases}.$\n\nIn fact, for the entire range $k_{t} \\in [0.76, 1.37]$ there is multiplicity of equilibria. Consequently, in all these cases we also have that $\\frac{\\mathrm{d}k_{t+1}}{\\mathrm{d}k_t} < 0.$", null, "Previous\nNext" ]
[ null, "https://eric-roca.github.io/img/olg/backwards.png", null ]
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https://math.stackexchange.com/questions/855828/what-is-abstract-algebra-essentially
[ "# What is Abstract Algebra essentially?\n\nIn the most basic sense, what is abstract algebra about?\n\nWolfram Mathworld has the following definition: \"Abstract algebra is the set of advanced topics of algebra that deal with abstract algebraic structures rather than the usual number systems. The most important of these structures are groups, rings, and fields.\"\n\nI find this, however, to say the least, not very informative. What do they mean by abstract algebraic structures? Along these lines, what are groups, rings, and fields then?\n\nI've been told by a friend that groups, essentially, are sets of objects, although, this still leaves me wondering what he means by objects (explicitly).\n\nI don't need anything rigorous. Just some intuitive definitions to give me some direction.\n\nThanks!\n\n• Sets plus some additional \"structure\" that lets the elements be combined or related Jul 3, 2014 at 22:03\n• @Pilot Would we say it's an \"ordered\" set then? Jul 3, 2014 at 22:31\n• @Colin An ordered set can refer to one of several related concepts. For what you're probably thinking about, there is a totally ordered set which is a set equipped with a binary relation $\\le$ such that three axioms hold: $a \\le b$ and $b \\le a$ implies $a = b$; $a \\le b$ and $b \\le c$ implies $a \\le c$; either $a\\le b$ or $b \\le a$. The reals form a totally ordered set, for instance. Jul 3, 2014 at 22:58\n• Jul 3, 2014 at 23:00\n• Yeah you could say you are studying the \"order\"/ \"organization\" of a set in abstract algebra, in a very general way. Order usually is more specific though, see Arkamis's comment Jul 3, 2014 at 23:00\n\nWe learn math with numbers early on. We learn how to apply operations to numbers to get new numbers. We learn rules, and consequences of those rules. All of that is pretty straightforward.\n\nBut, the real numbers are not the only things we might want to examine in detail. The properties of how elements interact under operations is a more general, abstract notion of what we do with numbers when we do algebra.\n\nFor instance, maybe we want to examine what a shape looks like if we rotate it around. Maybe you run a supply chain, and you need to build 4 widgets, but only some of those widgets need to be built in a certain order. Could you re-arrange things to make it more efficient? Maybe we want to explore structures that have a fundamental periodicity, like the time of day.\n\nOver time, we have constructed concepts of structures that elements can belong to, and notions of operations on these structures. These structures -- groups, fields, rings, monoids, modules, vector spaces, etc. -- don't have a natural set of rules, per se. We make up those rules (aka axioms), but we have found that many natural concepts adhere to those rules.\n\nThis is all well and good but somewhat useless until you learn about isomorphism. Exploring what a group is or what a ring is is fine. But the richness of abstract algebra comes from the idea that you can use abstractions of a concept that are easy to understand to explain more complex behavior! Adding hours on a clock is like working in a cyclic group, for instance. Or manufacturing processes might be shown to be isomorphic to products of permutations of a finite group.\n\nAbstract algebra is what happens when we want to explore consequences of rules and properties on collections of objects of any type -- hence the term \"abstract!\"\n\nIn \"concrete\" algebra one works with things like integers, rational numbers, real numbers, complex numbers, matrices, quaternions, permutations, polynomials, geometric transformations (e.g. isometries, similarities, reflections, inversions, projectivities, etc.), etc., subject to operations like addition, multiplication, and composition.\n\nIn \"abstract\" algebra one says \"suppose we have a set of objects (which could be numbers, matrices, permutations, geometric transformations, etc., but we will not say what they are) and certain operations (which could be addition, multiplication, composition, etc., but again in certain contexts we won't say what they are) that are assumed subject to certain algebraic rules, such as commutativity, associativity, distributivity, the existence or non-existence of identity elements and inverses, closure or lack of closure, etc.). Then one deduces consequences of those algebraic laws. Statements that say that something is always true are deduced from algebraic laws without considering the concrete nature of either the objects or the operations. Statements that say that something is not always true are often deduced from concrete examples, involving numbers, matrices, polynomials, permutations, etc.\n\nIn abstract algebra, the examples are concrete, but the derivations of general results come from the rules of algebra without the concrete nature of the operations or the things they operate on.\n\nPinter in \"A book of abstract algebra\" says it thus:\n\nThus, we are led to the modern notion of algebraic structure. An algebraic structure is understood to be an arbitrary set, with one or more operations defined on it. And algebra, then, is defined to be the study of algebraic structures. It is important that we be awakened to the full generality of the notion of algebraic structure. We must make an effort to discard all our preconceived notions of what an algebra is, and look at this new notion of algebraic structure in its naked simplicity. Any set, with a rule (or rules) for combining its elements, is already an algebraic structure. There does not need to be any connection with known mathematics. For example, consider the set of all colors (pure colors as well as color combinations), and the operation of mixing any two colors to produce a new color. This may be conceived as an algebraic structure. It obeys certain rules, such as the commutative law (mixing red and blue is the same as mixing blue and red). In a similar vein, consider the set of all musical sounds with the operation of combining any two sounds to produce a new (harmonious or disharmonious) combination.\n\n• This is exactly the excerpt I had in mind when I read this question. Pinter does a wonderful job of motivating the subject. Jul 10, 2014 at 17:09\n\nMathematics has to do with sets. There is no definition about what the set is, but we all know that set is made up by elements. Many problems in nature can be represented by sets, and the relations of the elements on these sets. The mathematical discipline that studies THE RELATIONS OF ELEMENTS on a given set is called algebra. There are many good properties of relations that a set can have, and from those properties we classify classes of 'sets with their relations' (algebraic structures) and call them as groups, rings ect.\n\nWhen you first came to learn maths, what did it show you? For instance, we wrote a lot of equations in terms of x,y,z and so on, but essentially, what they had common is that they were representing some unknown numbers.\n\nAbstract algebra is a bit more broad concept: here (in intuitive sense) letters represent pretty much anything, rather than numbers. I guess you might refer this 'everything' as 'objects'. So why abstract? What all abstract algebra has in common is that they study some fundamental properties of set of objects. We discard anything that is not relevant: for instance, in study of groups we only concentrate on properties of groups; no distributive laws or definitions of addition and multiplication. We only need a binary operation on a set and associativity, inverse and identity.\n\nBy discarding any irrelevant features of objects, we can concentrate on essential properties of them, and anything derived from these essential features will be applied to vast amount of things that have these elements in common. This is the power of abstract algebra, however by doing this we inevitably deal with more abstract things: hence abstract.\n\nThe other posters did a good job explaining what abstract algebra is, so I'll try to help you understand groups. You can think of a group, ring, field etc. as being a set with a certain structure attached to it. The intuition usually comes from concrete examples so I'll include a few. (Disclaimer: This will not be rigorous in the slightest. I'm shooting for conceptual clarity.)\n\nA group is a set together with a binary operation that we often think of as multiplication (or addition in some cases), satisfying certain properties:\n\nClosure: if you multiply two elements of the set, you get another element of the set. For example, the set of positive real numbers together with the operation of ordinary multiplication forms a group. This group is closed, because if you multiply two positive numbers, you get another positive number.\n\nIdentity: There is an element in the set called the identity that works like 1 (or 0 in an additive group). That is, if you multiply (add) it by an element you get the same element back.\n\nInverses: You have to have a way of getting back to where you came from. If you're in the group of positive real numbers, the inverse is just the reciprocal. e.g. 1/4 is the inverse of 4. Here's another example. The set of rotations of the plane, together with the operation defined by: rotation2 * rotation1 = (do rotation 1 first, then do rotation 2), is a group. If I rotate the plane 90 degrees counterclockwise, the inverse is just a 90 degree clockwise rotation, which gives you back the identity.\n\nAssociativity: This one might seem a bit obvious, but it has very important consequences. If you have elements a,b,c in your group, then a*(b*c)=(a*b)*c. i.e if you multiply b*c by a, you get the same thing as you'd get multiplying c by a*b.\n\nSource: A Book of Abstract Algebra, by Charles C. Pinter. This is a great intro book, if you want to learn more.\n\nAbstract algebra is an area of pure mathematics; see lists of mathematics topics. There is overlap between different areas of math, but if one had to divide math into two areas, it would have to be algebra and geometry.\n\nGeometry is the study of shapes, graphs and points in space. If you are working and studying in this area, you would almost certainly be drawing interesting pictures.\n\nIf you are exploring abstract algebra, you would have far fewer pictures in your work notes. You would be manipulating structured expressions, strings and words.\n\nDavid Hilbert's quote certainly applies if you are immersed in abstract algebra:\n\nMathematics is a game played according to certain simple rules with meaningless marks on paper.\n\nFinally, for another perspective, see connecting Algebra with Geometry." ]
[ null ]
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https://bqueennatural.com/article/mean-definition-meaning-how-to-find-the-mean-formula-examples
[ "# Mean (Definition & Meaning), How to Find the Mean, Formula, Examples. (2023)\n\nIn statistics, the mean is one of the measures of central tendency, apart from the mode and median. Mean is nothing but the average of the given set of values. It denotes the equal distribution of values for a given data set. The mean, median and mode are the three commonly used measures of central tendency. To calculate the mean, we need to add the total values given in a datasheet and divide the sum by the total number of values.The Median is the middle value of a given data when all the values are arranged in ascending order. Whereas mode is the number in the list, which is repeated a maximum number of times.\n\nLearn: Central tendency\n\nIn this article, you will learn the definition of mean, the formula for finding the mean for ungrouped and grouped data, along with the applications and solved examples.\n\n• Definition\n• Mean Symbol\n• Mean Formula\n• How to Find Mean\n• For Ungrouped data\n• For Grouped data\n• Types\n• Arithmetic Mean\n• Geometric Mean\n• Harmonic Mean\n• Root Mean Square\n• Contraharmonic Mean\n• Applications\n• Practice problems\n• FAQs\n\n## Definition of Mean in Statistics\n\nMeanis the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.\n\nMean = (Sum of all the observations/Total number of observations)\n\nExample:\n\nWhat is the mean of 2, 4, 6, 8 and 10?\n\nSolution:\n\n2 + 4 + 6 + 8 + 10 = 30\n\nNow divide by 5 (total number of observations).\n\nMean = 30/5 = 6\n\nIn the case of a discrete probability distribution of a random variable X, the mean is equal to the sum over every possible value weighted by the probability of that value; that is, it is computed by taking the product of each possible value x of X and its probability P(x) and then adding all these products together.\n\n### Mean Symbol (X Bar)\n\nThe symbol of mean is usually given by the symbol ‘x̄’. The bar above the letter x, represents the mean of x number of values.\n\nX̄ = (Sum of values ÷ Number of values)\n\nX̄ = (x1 + x2 + x3 +….+xn)/n\n\n• Mean Median Mode\n• Mean definition\n• Mean formula\n• Median\n\n## Mean Formula\n\nThe basic formula to calculate the mean is calculated based on the given data set. Each term in the data set is considered while evaluating the mean. The general formula for mean is given by the ratio of the sum of all the terms and the total number of terms. Hence, we can say;\n\n(Video) Math Antics - Mean, Median and Mode\n\nMean = Sum of the Given Data/Total number of Data\n\nTo calculate the arithmetic mean of a set of data we must first add up (sum) all of the data values (x) and then divide the result by the number of values (n). Since∑ is the symbol used to indicate that values are to be summed (see Sigma Notation) we obtain the following formula for the mean (x̄):\n\nx̄=∑ x/n\n\n## How to Find Mean?\n\nAs we know, data can be grouped data or ungrouped data so to find the mean of given data we need to check whether the given data is ungrouped. The formulas to find the mean for ungrouped data and grouped data are different. In this section, you will learn the method of finding the mean for both of these instances.\n\n### Mean for Ungrouped Data\n\nThe example given below will help you in understanding how to find the mean of ungrouped data.\n\nExample:\n\nIn a class there are 20 students and they have secured a percentage of 88, 82, 88, 85, 84, 80, 81, 82, 83, 85, 84, 74, 75, 76, 89, 90, 89, 80, 82, and 83.\n\nFind the mean percentage obtained by the class.\n\nSolution:\n\nMean = Total of percentage obtained by 20 students in class/Total number of students\n\n= [88 + 82 + 88 + 85 + 84 + 80 + 81 + 82 + 83 + 85 + 84 + 74 + 75 + 76 + 89 + 90 + 89 + 80 + 82 + 83]/20\n\n= 1660/20\n\n= 83\n\nHence, the mean percentage of each student in the class is 83%.\n\n### Mean for Grouped Data\n\nFor grouped data, we can find the mean using either of the following formulas.\n\nDirect method:\n\n$$\\begin{array}{l}Mean, \\overline{x}=\\frac{\\sum_{i=1}^{n}f_ix_i}{\\sum_{i=1}^{n}f_i}\\end{array}$$\n\nAssumed mean method:\n\n$$\\begin{array}{l}Mean, (\\overline{x})=a+\\frac{\\sum f_id_i}{\\sum f_i}\\end{array}$$\n\n(Video) How to Find the Mean | Math with Mr. J\n\nStep-deviation method:\n\n$$\\begin{array}{l}Mean, (\\overline{x})=a+h\\frac{\\sum f_iu_i}{\\sum f_i}\\end{array}$$\n\nGo through the example given below to understand how to calculate the mean for grouped data.\n\nExample:\n\nFind the mean for the following distribution.\n\n xi 11 14 17 20 fi 3 6 8 7\n\nSolution:\n\nFor the given data, we can find the mean using the direct method.\n\n xi fi fixi 11 3 33 14 6 84 17 8 136 20 7 140 ∑fi = 24 ∑fi xi= 393\n\nMean = ∑fixi/∑fi = 393/24 = 16.4\n\n## Mean of Negative Numbers\n\nWe have seen examples of finding the mean of positive numbers till now. But what if the numbers in the observation list include negative numbers. Let us understand with an instance,\n\nExample:\n\nFind the mean of 9, 6, -3, 2, -7, 1.\n\nSolution:\n\nTotal: 9+6+(-3)+2+(-7)+1 = 9+6-3+2-7+1 = 8\n\nNow divide the total from 6, to get the mean.\n\nMean = 8/6 = 1.33\n\n## Types of Mean\n\nThere are majorly three different types of mean value that you will be studying in statistics.\n\n1. Arithmetic Mean\n2. Geometric Mean\n3. Harmonic Mean\n\n### Arithmetic Mean\n\nWhen you add up all the values and divide by the number of values it is calledArithmetic Mean.To calculate, just add up all the given numbers then divide by how many numbers are given.\n\n(Video) Statistics - Find the mean\n\nExample: What is the mean of 3, 5, 9, 5, 7, 2?\n\nNow add up all the given numbers:\n\n3 + 5 + 9 + 5 + 7 + 2 = 31\n\nNow divide by how many numbers are provided in the sequence:\n\n316= 5.16\n\n### Geometric Mean\n\nThe geometric mean of two numbers x and y is xy. If you have three numbers x, y, and z, their geometric mean is 3xyz.\n\n$$\\begin{array}{l} Geometric\\;Mean=\\sqrt[n]{x_{1}x_{2}x_{3}…..x_{n}}\\end{array}$$\n\nExample: Find the geometric mean of 4 and 3 ?\n\n$$\\begin{array}{l}Geometric Mean = \\sqrt{4 \\times 3} = 2 \\sqrt{3} = 3.46\\end{array}$$\n\n### Harmonic Mean\n\nThe harmonic mean is used to average ratios. For two numbers x and y, the harmonic mean is 2xy(x+y). For, three numbers x, y, and z, the harmonic mean is 3xyz(xy+xz+yz)\n\n$$\\begin{array}{l} Harmonic\\;Mean (H) = \\frac{n}{\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\frac{1}{x_{2}}+\\frac{1}{x_{3}}+……\\frac{1}{x_{n}}}\\end{array}$$\n\nThe root mean square is used in many engineering and statistical applications, especially when there are data points that can be negative.\n\n$$\\begin{array}{l} X_{rms}=\\sqrt{\\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}….x_{n}^{2}}{n}}\\end{array}$$\n\n### Contraharmonic Mean\n\nThe contraharmonic mean of x and y is (x2 + y2)/(x + y). For n values,\n\n$$\\begin{array}{l} \\frac{(x_{1}^{2}+x_{2}^{2}+….+x_{n}^{2})}{(x_{1}+x_{2}+…..x_{n})}\\end{array}$$\n\n(Video) Sample Mean and Population Mean - Statistics\n\n### Real-life Applications of Mean\n\nIn the real world, when there is huge data available, we use statistics to deal with it.Suppose, in a data table, the price values of 10 clothing materials are mentioned. If we have to find the mean of the prices, then add the prices of each clothing material and divide the total sum by 10. It will result in an average value. Another example is that if we have to find the average age of students of a class, we have to add the age of individual students present in the class and then divide the sum by the total number of students present in the class.\n\n### Practice Problems\n\nQ.1: Find the mean of 5,10,15,20,25.\n\nQ.2:Find the mean of the given data set: 10,20,30,40,50,60,70,80,90.\n\nQ.3: Find the mean of the first 10 even numbers.\n\nQ.4: Find the mean of the first 10 odd numbers.\n\n## Frequently Asked Questions – FAQs\n\n### What is mean in statistics?\n\nIn statistics, Mean is the ratio of sum of all the observations and total number of observations in a data set. For example, mean of 2, 6, 4, 5, 8 is:\nMean = (2 + 6 + 4 + 5 + 8) / 5 = 25/5 = 5\n\n### How is mean represented?\n\nMean is usually represented by x-bar or x̄.\nX̄ = (Sum of values ÷ Number of values in data set)\n\n### What is median in Maths?\n\nMedian is the central value of the data set when they are arranged in an order.\nFor example, the median of 3, 7, 1, 4, 8, 10, 2.\nArrange the data set in ascending order: 1,2,3,4,7,8,10\nMedian = middle value = 4\n\n### What are the types of Mean?\n\nIn statistics we learn basically, three types of mean, they are:\nArithmetic Mean, Geometric Mean and Harmonic Mean\n\n### What is the mean of the first 10 natural numbers?\n\nThe first 10 natural numbers are: 1,2,3,4,5,6,7,8,9,10\nSum of first 10 natural numbers = 1+2+3+4+5+6+7+8+9+10 = 55\nMean = 55/10 = 5.5\n\n### What is the relationship between mean, median and mode?\n\nThe relationship between mean, median and mode is given by:\n3 Median = Mode + 2 Mean.\n\n### What is the mean of the first 5 even natural numbers?\n\nAs we know, the first 5 even natural numbers are 2, 4, 6, 8, and 10.\nHence, Mean = (2 + 4 + 6 + 8 + 10)/5\nMean = 6\nThus, the mean of the first 5 even natural numbers is 6.\n\n(Video) Estimated Mean - Corbettmaths\n\n### Find the mean of the first 5 composite numbers?\n\nThe first 5 composite numbers are 4, 6, 8, 9 and 10.\nThus, Mean = (4 + 6 + 8 + 9 + 10)/5\nMean = 37/5 = 7.4\nHence, the mean of the first 5 composite numbers is 7.4.\n\n## FAQs\n\n### How do you calculate mean with examples? ›\n\nMean: The \"average\" number; found by adding all data points and dividing by the number of data points. Example: The mean of 4, 1, and 7 is ( 4 + 1 + 7 ) / 3 = 12 / 3 = 4 (4+1+7)/3 = 12/3 = 4 (4+1+7)/3=12/3=4left parenthesis, 4, plus, 1, plus, 7, right parenthesis, slash, 3, equals, 12, slash, 3, equals, 4.\n\nHow do you calculate mean formula? ›\n\nIt's obtained by simply dividing the sum of all values in a data set by the number of values.\n\nWhat is the simple formula of mean? ›\n\nYou can find the mean, or average, of a data set in two simple steps: Find the sum of the values by adding them all up. Divide the sum by the number of values in the data set.\n\nWhat are the three formulas of mean? ›\n\nThey are:\n• Direct Method.\n• Assumed Mean Method.\n• Step-deviation Method.\n\nWhat is mean in statistics with example? ›\n\nIn mathematics and statistics, the mean refers to the average of a set of values. The mean can be computed in a number of ways, including the simple arithmetic mean (add up the numbers and divide the total by the number of observations), the geometric mean, and the harmonic mean.\n\nWhy do you calculate the mean? ›\n\nThe mean is used to summarize a data set. It is a measure of the center of a data set.\n\nHow do you do the mean method? ›\n\nThe mathematical formula of mean is given by, ∑ x n Where, n=number of datas. In the assumed mean method, we assume the value of a mean then we calculate the deviation and adjust the mean accordingly. It is suitable for calculating the mean or average for tables involving largely spaced limits.\n\nWhat is a mean mean in math? ›\n\nA mean in math is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers. For example, with the data set: 8, 9, 5, 6, 7, the mean is 7, as 8 + 9 + 5 + 6 + 7 = 35, 35/5 = 7.\n\nHow do you find mean for dummies? ›\n\nThe mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest.\n\nWhat is the mean formula in assumed mean method? ›\n\na = assumed mean. fi = frequency of ith class. di = xi – a = deviation of ith class. Σfi = n = Total number of observations. xi = class mark = (upper class limit + lower class limit)/2.\n\n### What is mean and median with example? ›\n\nThe mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.\n\nWhat are the 3 ways to calculate average? ›\n\nThere are three main types of average: mean, median and mode. Each of these techniques works slightly differently and often results in slightly different typical values. The mean is the most commonly used average. To get the mean value, you add up all the values and divide this total by the number of values.\n\nWhat is the mean and mode in math example? ›\n\nThe mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.\n\n## Videos\n\n1. Finding mean, median, and mode | Descriptive statistics | Probability and Statistics | Khan Academy\n2. Mean, Median, and Mode of Grouped Data & Frequency Distribution Tables Statistics\n(The Organic Chemistry Tutor)\n3. Lesson 11 - Calculating the Mean (Statistics Tutor)\n(Math and Science)\n4. Mean, Median, Mode, and Range | Math with Mr. J\n(Math with Mr. J)\n5. Means from Frequency Tables - Corbettmaths\n(corbettmaths)\n6. Mean, Median, Mode, and Range - How To Find It!\n(The Organic Chemistry Tutor)\nTop Articles\nLatest Posts\nArticle information\n\nAuthor: Tish Haag\n\nLast Updated: 01/15/2023\n\nViews: 6464\n\nRating: 4.7 / 5 (47 voted)\n\nAuthor information\n\nName: Tish Haag\n\nBirthday: 1999-11-18\n\nAddress: 30256 Tara Expressway, Kutchburgh, VT 92892-0078\n\nPhone: +4215847628708\n\nJob: Internal Consulting Engineer\n\nHobby: Roller skating, Roller skating, Kayaking, Flying, Graffiti, Ghost hunting, scrapbook\n\nIntroduction: My name is Tish Haag, I am a excited, delightful, curious, beautiful, agreeable, enchanting, fancy person who loves writing and wants to share my knowledge and understanding with you." ]
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[ "# [Recommended]How do we make informed decisions with all this variation\n\nHow do we make informed decisions with all this variation BUS 308 Week 2 Lecture 1 Examining Differences – overview Expected Outcomes After reading this…\n\nHow do we make informed decisions with all this variation\nBUS 308 Week 2 Lecture 1\nExamining Differences – overview\nExpected Outcomes\nAfter reading this lecture, the student should be familiar with:\n1. The importance of random sampling. 2. The meaning of statistical significance. 3. The basic approach to determining statistical significance. 4. The meaning of the null and alternate hypothesis statements. 5. The hypothesis testing process. 6. The purpose of the F-test and the T-test.\nOverview\nLast week we collected clues and evidence to help us answer our case question about males and females getting equal pay for equal work. As we looked at the clues presented by the salary and comp-ratio measures of pay, things got a bit confusing with results that did not see to be consistent. We found, among other things, that the male and female compa-ratios were fairly close together with the female mean being slightly larger. The salary analysis showed a different view; here we noticed that the averages were apparently quite different with the males, on average, earning more. Contradictory findings such as this are not all that uncommon when examining data in the “real world.”\nOne issue that we could not fully address last week was how meaningful were the differences? That is, would a different sample have results that might be completely different, or can we be fairly sure that the observed differences are real and show up in the population as well? This issue, often referred to as sampling error, deals with the fact that random samples taken from a population will generally be a bit different than the actual population parameters, but will be “close” enough to the actual values to be valuable in decision making.\nThis week, our journey takes us to ways to explore differences, and how significant these differences are. Just as clues in mysteries are not all equally useful, not all differences are equally important; and one of the best things statistics will do for us is tell us what differences we should pay attention to and what we can safely ignore.\nSide note; this is a skill that many managers could benefit from. Not all differences in performances from one period to another are caused by intentional employee actions, some are due to random variations that employees have no control over. Knowing which differences to react to would make managers much more effective.\nIn keeping with our detective theme, this week could be considered the introduction of the crime scene experts who help detectives interpret what the physical evidence means and how it can relate to the crime being looked at. We are getting into the support being offered by experts who interpret details. We need to know how to use these experts to our fullest advantage.\nDifferences\nIn general, differences exist in virtually everything we measure that is man-made or influenced. The underlying issue in statistical analysis is that at times differences are important. When measuring related or similar things, we have two types of differences: differences in consistency and differences in average values. Some examples of things that should be the “same” could be:\n• The time it takes to drive to work in the morning. • The quality of parts produced on the same manufacturing line. • The time it takes to write a 3-page paper in a class. • The weight of a 10-pound bag of potatoes. • Etc.\nAll of these “should” be the same, as each relates to the same outcome. Yet, they all differ. We all experience differences in travel time, and the time it takes to produce the same output on the job or in school (such as a 3-page paper). Production standards all recognize that outcomes should be measured within a range rather than a single point. For example, few of us would be upset if a 10-pound bag of potatoes weighed 9.85 pounds or would think we were getting a great deal if the bag weighed 10.2 pounds. We realize that it is virtually impossible for a given number of potatoes to weigh exactly the same and we accept this as normal.\nOne reason for our acceptance is that we know that variation occurs. Variation is simply the differences that occur in things that should be “the same.” If we can measure things with enough detail, everything we do in life has variation over time. When we get up in the morning, how long it takes to get to work, how effective we are at doing the same thing over and over, etc. Except for physical constants, we can say that things differ and we need to recognize this. A side note: variation exists in virtually everything we study (we have more than one language, word, sentence, paragraph, past actions, financial transactions, etc.), but only in statistics do we bring this idea front and center for examination.\nThis suggests that any population that we are interested in will consist of things that are slightly different, even if the population contains only one “thing.” Males are not all the same, neither are females. Manufactured parts differ in key measurements; this is the reason we have quality control checking to make sure the differences are not too large. So, even if we measure everything in our population we will have a mean that is accompanied by a standard deviation (or range). Managers and professionals need to manage this variation, whether it is quantitative (such as salary paid for similar work) or even qualitative (such as interpersonal interactions with customers).\nThe second reason that we are so concerned with differences is that we rarely have all the evidence, or all the possible measures of what we are looking for. Having this would mean we have access to the entire population (everything we are interested in); rarely is this the case. Generally, all decisions, analysis, research, etc. is done with samples, a selected subset of the population. And, with any sample we are not going have all the information needed, obviously; but we also know that each sample we take is going to differ a bit. (Remember, variation is\neverywhere, including in the consistency of sample values.) If you are not sure of this, try flipping a coin 10 times for 10 trials, do you expect or get the exact same number of heads for each trial? Variation!\nSince we are making decisions using samples, we have even more variation to consider than simply that with the population we are looking at. Each sample will be slightly different from its population and from others taken from the same population.\nHow do we make informed decisions with all this variation and our not being able to know the “real” values of the measures we are using? This question is much like how detectives develop the “motive” for a crime – do they know exactly how the guilty party felt/thought when they say “he was jealous of the success the victim had.” This could be true, but it is only an approximation of the true feelings, but it is “close enough” to say it was the reason. It is similar with data samples, good ones are “close enough” to use the results to make decisions with. The question we have now focuses on how do we know what the data results show?\nThe answer lies with statistical tests. They can use the observed variation to provide results that let us make decisions with a known chance of being wrong! Most managers hope to be right just over 50% of the time, a statistical decision can be correct 95% or more of the time! Quite an improvement.\nSampling. The use of samples brings us to a distinction in summary statistics, between descriptive and inferential statistics. With one minor exception (discussed shortly), these two appear to be the same: means, standard deviations, etc. However, one very important distinction exists in how we use these. Descriptive statistics, as we saw last week, describes a data set. But, that is all they do. We cannot use them to make claims or inferences about any other larger group.\nMaking inferences or judgements about a larger population is the role of inferential statistics and statistical tests. So, what makes descriptive statistics sound enough to become inferential statistics? The group they were taken from! If we have a sample that is randomly selected from the population (meaning that each member has the same chance of being selected at the start), then we have our best chance of having a sample that accurately reflects the population, and we can use the statistics developed from that sample to make inferences back to the population. (How we develop a randomly selected sample is more of a research course issue, and we will not go into these details. You are welcome to search the web for approaches.)\nRandom Sampling. If we are not working with a random sample, then our descriptive statistics apply only to the group they are developed for. For example, asking all of our friends their opinion of Facebook only tells us what our friends feel; we cannot say that their opinions reflect all Facebook users, all Facebook users that fall in the age range of our friends, or any other group. Our friends are not a randomly selected group of Facebook users, so they may not be typical; and, if not typical users, cannot be considered to reflect the typical users.\nIf our sample is random, then we know (or strongly suspect) a few things. First, the sample is unlikely to contain both the smallest and largest value that exists in the larger\npopulation, so an estimate of the population variation is likely to be too small if based on the sample. This is corrected by using a sample standard deviation formula rather than a population formula. We will look at what this means specifically in the other lectures this week; but Excel will do this for us easily.\nSecond, we know that our summary statistics are not the same as the population’s parameter values. We are dealing with some (generally small) errors. This is where the new statistics student often begins to be uncomfortable. How can we make good judgements if our information is wrong? This is a reasonable question, and one that we, as data detectives, need to be comfortable with.\nThe first part of the answer falls with the design of the sample, by selecting the right sample size (how many are in the sample), we can control the relative size of the likely error. For example, we can design a sample where the estimated error for our average salary is about plus or minus \\$1,000. Does knowing that our estimates could be \\$1000 off change our view of the data? If the female average was a thousand dollars more and the male salary was a thousand dollars less, would you really change your opinion about them being different? Probably not with the difference we see in our salary values (around 38K versus 52K). If the actual averages were closer together, this error range might impact our conclusions, so we could select a sample with a smaller error range. (Again, the technical details on how to do this are found in research courses. For our statistics class, we assume we have the correct sample.)\nNote, this error range is often called the margin of error. We see this most often in opinion polls. For example, if a poll said that the percent of Americans who favored Federal Government support for victims of natural disasters (hurricanes, floods, etc.) was 65% with a margin of error of +/- 3%; we would say that the true proportion was somewhat between 62% to 68%, clearly a majority of the population. Where the margin of error becomes important to know is when results are closer together, such as when support is 52% in favor versus 48% opposed, with a margin of error of 3%. This means the actual support could be as low as 49% or as high as 55%; meaning the results are generally too close to make a solid decision that the issue is supported by a majority, the proverbial “too close to call.”\nThe second part of answering the question of how do we make good decisions introduces the tools we will be looking at this week, decision making statistical tests that focus on examining the size of observed differences to see if they are “meaningful” or not. The neat part of these tools is we do not need to know what the sampling error was, as the techniques will automatically include this impact into our results!\nThe statistical tools we will be looking at for the next couple of weeks all “work” due to a couple of assumptions about the population. First, the data needs to be at the interval or ratio level; the differences between sequential values needs to be constant (such as in temperature or money). Additionally, the data is assumed to come from a population that is normally distributed, the normal curve shape that we briefly looked at last week. Note that many statisticians feel that minor deviations from these strict assumptions will not significantly impact the outcomes of the tests.\nThe tools for this week and next use the same basic logic. If we take a lot of samples from the population and graph the mean for all of them, we will get a normal curve (even if the population is not exactly normal) distribution called the sampling distribution of the mean. Makes sense as we are using sample means. This distribution has an overall, or grand, mean equal to that of the population. The standard deviation equals the standard deviation of the population divided by the square root of the population. (Let’s take this on faith for now, trust me you do not want to see the math behind proving these. But if you do, I invite you to look it up on the web.) Now, knowing – in theory – what the mean values will be from population samples, we can look at how any given sample differs from what we think the population mean is. This difference can be translated into what is essentially a z-score (although the specific measure will vary depending upon the test we are using) that we looked at last week. With this statistic, we can determine how likely (the probability of) getting a difference as large or larger than we have purely by chance (sampling error from the actual population value) alone.\nIf we have a small likelihood of getting this large of a difference, we say that our difference is too large to have been purely a sampling error, and we say a real difference exists or that the mean of the population that the sample came from is not what we thought.\nThat is the basic logic of statistical testing. Of course, the actual process is a bit more structured, but the logic holds: if the probability of getting our result is small (for example 4% or 0.04), we say the difference is significant. If the probability is large (for example 37% or 0.37), then we say there is not enough evidence to say the difference is anything but a simple sampling error difference from the actual population result.\nThe tools we will be adding to our bag of tricks this week will allow us to examine differences between data sets. One set of tools, called the t-test, looks at means to see if the observed difference is significant or merely a chance difference due mostly to sampling error rather than a true difference in the population. Knowing if means differ is a critical issue in examining groups and making decisions.\nThe other tool – the F-test for variance, does the same for the data variation between groups. Often ignored, the consistency within groups is an important characteristic in understanding whether groups having similar means can be said to be similar or not. For example, if a group of English majors all took two classes together, one math and one English, would you expect the grade distributions to be similar, or would you expect one to show a larger range (or variation) than the other?\nWe will see throughout the class that consistency and differences are key elements to understanding what the data is hiding from us, or trying to tell us – depending on how you look at it. In either case, as detectives our job is to ferret out the information we need to answer our questions.\nHypothesis Testing-Are Differences Meaningful\nHere is where the crime scene experts come in. Detectives have found something but are not completely sure of how to interpret it. Now the training and tools used by detectives and\nanalysts take over to examine what is found and make some interpretations. The process or standard approach that we will use is called the hypothesis testing procedure. It consists of six steps; the first four (4) set up the problem and how we will make our decisions (and are done before we do anything with the actual data), the fifth step involves the analysis (done with Excel), and the final and sixth step focuses on interpreting the result.\nThe hypothesis testing procedure is a standardized decision-making process that ensures we make our decisions (on whether things are significantly different or not) is based on the data, and not some other factors. Many times, our results are more conservative than individual managerial judgements; that is, a statistical decision will call fewer things significantly different than many managerial judgement calls. This statistical tendency is, at times, frustrating for managers who want to show that things have changed. At other times, it is a benefit such as if we are hoping that things, such as error rates, have not changed.\nWhile a lot of statistical texts have slightly different versions of the hypothesis testing procedure (fewer or more steps), they are essentially the same, and are a spinoff of the scientific method. For this class, we will use the following six steps:\n1. State the null and alternate hypothesis 2. Select a level of significance 3. Identify the statistical test to use 4. State the decision rule. Steps 1 – 4 are done before we examine the data 5. Perform the analysis 6. Interpret the result.\nStep 1\nA hypothesis is a claim about an outcome. It comes in two forms. The first is the null hypothesis – sometimes called the testable hypothesis, as it is the claim we perform all of our statistical tests on. It is termed the “Null” hypothesis, shown as Ho, as it basically says “no difference exists.” Even if we want to test for a difference, such as males and females having a different average compa-ratio; in statistics, we test to see if they do not.\nWhy? It is easier to show that something differs from a fixed point than it is to show that the difference is meaningful – I mean how can we focus on “different?” What does “different” mean? So, we go with testing no difference. The key rule about developing a null hypothesis is that it always contains an equal claim, this could be equal (=), equal to or less than (<=), or equal to or more than (=>).\nHere are some examples:\nEx 1: Question: Is the female compa-ratio mean = 1.0?\nHo: Female compa-ratio mean = 1.0.\nEx 2: Q: is the female compa-ratio mean = the male compa-ratio mean?\nHo: Female compa-ratio mean = Male compa-ratio mean.\nEx. 3: Q: Is the female compa-ratio more than the male compa-ratio? Note that this question does not contain an equal condition. In this case, the null is the opposite of what the question asks:\nHo: Female compa-ratio <= Male compa-ratio.\nWe can see by testing this null, we can answer our initial question of a directional difference. This logic is key to developing the correct test claim.\nA null hypothesis is always coupled with an alternate hypothesis. The alternate is the opposite claim as the null. The alternate hypothesis is shown as Ha. Between the two claims, all possible outcomes must be covered. So, for our three examples, the complete step 1 (state the null and alternate hypothesis statements) would look like:\nEx 1: Question: Is the female compa-ratio mean = 1.0?\nHo: Female compa-ratio mean = 1.0.\nHa: Female compa-ratio mean =/= (not equal to) 1.0\nEx 2: Q: is the female compa-ratio mean = the male compa-ratio mean?\nHo: Female compa-ratio mean = Male compa-ratio mean.\nHa: Female compa-ratio mean =/= Male compa-ration mean.\nEx. 3: Q: Is the female compa-ratio more than the male compa-ratio?\nHo: Female compa-ratio <= Male compa-ratio\nHa: Female compa-ratio > Male compa-ratio. (Note that in this case, the alternate hypothesis is the question being asked, but the null is what we always use as the test hypothesis.)\nWhen developing the null and alternate hypothesis,\n1. Look at the question being asked. 2. If the wording implies an equality could exist (equal to, at least, no more than, etc.),\nwe have a null hypothesis and we write it exactly as the question asks. 3. If the wording does not suggest an equality (less than, more than, etc.), it refers to the\nalternate hypothesis. Write the alternate first. 4. Then, for whichever hypothesis statement you wrote, develop the other to contain all\nof the other cases. An = null should have a =/= alternate, an => null should have a < alternate; a <= null should have a > alternate, and vice versa.\n5. The order the variables are listed in each hypothesis must be the same, if we list males first in the null, we need to list males first in the alternate. This minimizes confusion in interpreting results.\nNote: the hypothesis statements are claims about the population parameters/values based on the sample results. So, when we develop our hypothesis statements, we do not consider the\nsample values when developing the hypothesis statements. For example, consider our desire to determine if the compa-ratio and salary means for males and females are different in the population, based on our sample results. While the compa-ratio means seemed fairly close together, the salary means seemed to differ by quite a bit; in both cases, we would test if the male and female means were equal since that is the question we have about the values in the population.\nIf you look at the examples, you can notice two distinct kinds of null hypothesis statements. One has only an equal sign in it, while the other contains an equal sign and an inequality sign (<=, but it could be =>). These two types correspond to two different research questions and test results.\nIf we are only interested in whether something is equal or not, such as if the male average salary equals the female average salary; we do not really care which is greater, just if they could be the same in the population or not. For our equal salary question, it is not important if we find that the male’s mean is > (greater than) the female’s mean or if the male’s mean is < (less than) the female’s mean; we only care about a difference existing or not in the population. This, by the way, is considered a two-tail test (more on this later), as either conditions would cause us to say the null’s claim of equality is wrong: a result of “rejecting the null hypothesis.”\nThe other condition we might be interested in, and we need a reason to select this approach, occurs when we want to specifically know if one mean exceeds the other. In this situation, we care about the direction of the difference. For example, only if the male mean is greater than the female mean or if the male mean is less than the female mean.\nStep 2\nThe level of significance is another concept that is critical in statistics but is often not used in typical business decisions. One senior manager told the author that their role was to ensure that the “boss’ decisions were right 50% +1 of the time rather than 50% -1.” This suggests that the level of confidence that the right decisions are being made is around 50%. In statistics, this would be completely unacceptable.\nA typically statistical test has a level of confidence that the right decision is being made is about 95%, with a typical range from 90 to 99%. This is done with our chosen level of significance. For this class, we will always use the most common level of 5%, or more technically alpha = 0.05. This means we will live with a 5% chance of saying a difference is significant when it is not and we really have only a chance sampling error.\nRemember, no decision that does not involve all the possible information that can be collected will ever have a zero possibility of being wrong. So, saying we are 95% sure we made the right call is great. Marketing studies often will use an alpha of .10, meaning that are 90% sure when they say the marketing campaign worked. Medical studies will often use an alpha of 0.01 or even 0.001, meaning they are 99% or even 99.9% sure that the difference is real and not a chance sampling error.\nStep 3\nChoosing the statistical test and test statistic depends upon the data we have and the question we are asking. For this week, we will be using compa-ratio data in the examples and salary data in the homework – both are continuous and at least interval level data. The questions we will look at this week will focus on seeing if there is a difference in the average pay (as measured by either the compa-ratio or salary) between males and females in the population, based on our sample results. After all, if we cannot find a difference in our sample, should we even be working on the question?\nIn the quality improvement world, one of the strategies for looking for and improving performance of a process is to first look at and reduce the variation in the data. If the data has a lot of variation, we cannot really trust the mean to be very reflective of the entire data set.\nOur first statistical test is called the F-test. It is used when we have at least interval level data and we are interested in determining if the variances of two groups are significantly different or if the observed difference is merely chance sampling error. The test statistic for this is the F.\nOnce we know if the variances are the same or not, we can move to looking for differences between the group means. This is done with the T-test and the t-statistic. Details on these two tests will be given later; for now, we just need to know what we are looking at and what we will be using.\nStep 4\nOne of the rules in researching questions is that the decision rule, how we are going to make our decision once the analysis is done, should be stated upfront and, technically, even before we even get to the data. This helps ensure that our decision is data driven rather than being made by emotional factors to get the outcome we want rather than the outcome that fits the data. (Much like making our detectives go after the suspect that did the crime rather than the one they do not like and want to arrest, at least when they are being honest detectives.)\nThe decision rule for our class is very simple, and will always be the same:\nReject the null hypothesis if the p-value is less than our alpha of .05. (Note: this would be the same as saying that if the p-value is not less than 0.05, we would fail to reject the null hypothesis.)\nWe introduced the p-value last week, it is the probability of our outcome being as large or larger than we have by pure chance alone. The further from the actual mean a sample mean is, the less chance we have of getting a value that differs from the mean that much or more; the closer to the actual mean, the greater our chance would be of getting that difference or more purely by sampling error.\nOur decision rule ties our criteria for significance of the outcome, the step 2 choice of alpha, with the results that the statistical tests will provide (and, the Excel tests will give us the p- values for us to use in making the decisions).\nThese four steps define our analysis, and are done before we do any analysis of the data.\nStep 5\nOnce we know how we will analyze and interpret the results, it is time to get our sample data and set it up for input into an Excel statistical function. Some examples of how this data input works will be discussed in the third lecture for this week.\nThis step is fairly easy, simply identify the statistical test we want to use. The test to use is based on our question and the related hypothesis claims. For this week, if we are looking at variance equality, we will use the F-test. If we are looking at mean equality, we will use the T- test.\nStep 6\nHere is where we bring everything together and interpret the outcomes.\n1. Look at the appropriate p-value (indicated in the test outputs, as we will see in lecture 2).\n2. Compare the p-value with our value for alpha (0.05). 3. Make a decision: if the test p-value is less than or equal to (<=) 0.05, we will reject\nthe null hypothesis. If the test p-value is more than (=>) 0.05, we will fail to reject the null hypothesis.\nRejecting the null hypothesis means that we feel the alternate hypothesis is the more accurate statement about the populations we are testing. This is the same for all of our statistical tests.\nOnce we have made our decision to reject or fail to reject the null hypothesis, we need to close the loop, and go back and answer our original question. We need to take the statistical result or rejecting or failing to reject the null and turn it into an “English” answer to the question. Doing so depends on how the original question lead to the hypothesis statements. Examples of this follow in Lecture 2.\nLectures 2 and 3 will show how to use this process in conjunction with Excel and the F and T tests. For now, focus on the logic of setting up the testing instructions.\nSummary\nThis week we begin our journey discovering ways to make decisions on data, and more specifically differences in data sets, based on generally agreed upon approaches rather than by “guess and by golly.” The process is called hypothesis testing and is part of the scientific method of research and decision making.\nIn this approach we always test a claim of no difference (the null hypothesis) whether or not we are suspect or desire to see an actual difference. The null hypothesis is paired with an alternate hypothesis that is exactly the opposite claim. Decisions are made based on a p-value which is the probability that we would see a difference as large or larger as we got if the null\nhypothesis is true. Small p-values mean we reject the null as not being an accurate description of the population we are looking at.\nThe hypothesis testing process (or procedure) has six steps. The first four are completed before we look at the data; the fifth step is the actual calculation of the statistical test and the final and sixth step is where the analysis of the results is done. The steps are:\n1. State the null and alternate hypothesis 2. Select a level of significance 3. Identify the statistical test to use 4. State the decision rule 5. Perform the analysis 6. Interpret the result", null, "", null, "" ]
[ null, "http://ift.tt/2h3Rxhz", null, "http://ift.tt/2h4hMUS", null ]
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https://mathoverflow.net/questions/51339/an-optimization-problem
[ "# An optimization problem\n\nThe following problem optimization problem arose in a project I am working on with a student. I would like to minimize the quantity:\n\n$$M=\\frac{1}{12} + \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right)^2 Q(x)^2 \\ dx - 4 \\left[ \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) Q(x) \\ dx\\right]^2$$ over all continuously differentiable functions $Q$ subject to the conditions that $Q(\\tfrac{1}{2})=0$ and $$\\int_0^\\frac{1}{2} Q(x)^2 \\ dx = 1.$$\n\nIs it possible to explicitly solve such a problem? (The presence of the constant $1/12$ should ensure that $M$ is positive for all $Q$.)\n\n• Calculus of variations. I have no idea what will happen when you write all the equations down and try to solve them. It also seems to me that without a $Q'$ term in your expression, the $Q(1/2)=0$ condition is worthless. You can just take a function that gives you the minimum, and put a little spike in it to satisfy the $Q(1/2)=0$ condition. Jan 6 '11 at 20:48\n\nI'll change the variable $y=\\frac 12-x$ to make typing easier. Since, as Peter already observed, the condition $Q(0)=0$ is worthless and since $\\frac 1{12}$ is just an additive constant, we are just to minimize $\\int_0^{1/2}y^2Q^2-4\\left(\\int_0^{1/2}yQ\\right)^2$ under the condition $\\int_0^{1/2}Q^2=1$. Since everything is about pure quadratic forms, we just need to find the least $a>0$ such that $\\int_0^{1/2}y^2Q^2-4\\left(\\int_0^{1/2}yQ\\right)^2+a^2\\int_0^{1/2}Q^2\\ge 0$. Then $-a^2$ ($+\\frac 1{12}$, if you want to keep that term) is the answer. That is just a Cauchy-Schwarz type thingy and the answer is given by $\\int_0^{1/2}\\frac{4y^2}{y^2+a^2}dy=1$ with $Q$ proportional (with the coefficient to determine from the $\\int_0^{1/2}Q^2=1$ condition) to $\\frac {2y}{y^2+a^2}$ (so the condition that is worthless is automatic as well).\n\nThe integral determining $a$ is $2-4a\\arctan\\frac{1}{2a}$, so we need $a\\arctan\\frac 1{2a}=\\frac 14$, which is something that one has to solve numerically. I've got $a\\approx 0.2144889545$.\n\nHere's an interesting/natural observation which might be useful. Let $U$ be a uniform$(0,1/2)$ random variable, so that $U$ has density function $f(x)=2$, $0 \\lt x \\lt 1/2$, and define a function $Y$ by $Y(x)=(1/2-x)Q(x)$, $0 \\lt x \\lt 1/2$. Then, $M=1/12 + \\frac{1}{2}{\\rm E}(Y^2 ) - {\\rm E}^2 (Y)$. (Interestingly, a uniform$(0,1)$ random variable has variance equal to $1/12$.)\n\nI tried calculus of variations. It's been a while since I've done calculus of variations, so I could be messing it up completely. Also, this isn't completely rigorous. There are ways of making calculus of variations rigorous, but I don't know them, and they're a lot harder than just doing calculations.\n\nBy my comment above, the $Q(\\frac{1}{2})=0$ constraint is worthless, so we will ignore it.\n\nAssume $Q$ is the minimum function (this is the first, and most important, non-rigorous step, since we are assuming a minimum exists). Now, let's plug in $Q+\\epsilon$, where $Q$ and $\\epsilon$ are both functions of $x$. We have\n\n$$M=\\frac{1}{12} + \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right)^2 (Q+\\epsilon)^2 \\ dx - 4 \\left[ \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) (Q+\\epsilon) \\ dx\\right]^2 .$$\n\nExtracting the first-order terms in $\\epsilon(x)$, we get\n\n$$\\Delta M = 2\\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right)^2 Q \\epsilon \\ dx -8 \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) Q \\ dx \\ \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) \\epsilon \\ dx .$$\n\nIf we change $Q$ by adding an infinitessimal $\\epsilon$, and keeping the normalization condition on $Q$, then $\\Delta M$ has to be 0, or otherwise $Q$ wouldn't be a minimum. We still have to take care of the normalization condition $\\int_0^\\frac{1}{2} Q(x)^2 \\ dx = 1$. We do this using Lagrange multipliers, and so we get the expression\n\n$$2\\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right)^2 Q \\epsilon \\ dx -8 \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) Q \\ dx \\ \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) \\epsilon \\ dx + \\lambda \\int_0^\\frac{1}{2} Q \\epsilon\\ dx .$$\n\nThis has to be zero for all functions $\\epsilon(x)$. To simplify things further, let\n\n$$C = \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) Q \\ dx,$$\n\nsince it's a constant independent of $\\epsilon$. Now, we have\n\n$$2 \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right)^2 Q \\epsilon \\ dx -8 C \\ \\int_0^\\frac{1}{2} \\left( \\tfrac{1}{2}-x \\right) \\epsilon \\ dx + \\lambda \\int_0^\\frac{1}{2} Q \\epsilon\\ dx$$\n\nor, putting everything in the same integral sign,\n\n$$\\int_0^\\frac{1}{2} 2 \\left( \\tfrac{1}{2}-x \\right)^2 Q \\epsilon -8 C \\left( \\tfrac{1}{2}-x \\right) \\epsilon + \\lambda Q \\epsilon\\ dx$$\n\nand for this to be $0$ for all functions $\\epsilon$, we must have\n\n$$Q = \\frac{ (\\frac{1}{2}-x)}{\\alpha(\\frac{1}{2}-x)^2+\\beta}$$\n\nfor some $\\alpha$, $\\beta$.\n\nNote that, if $\\beta \\neq 0$, we get $Q(\\frac{1}{2}) = 0$, as desired.\n\nA Maple or Mathematica program should at least let you calculate $\\alpha$ and $\\beta$ numerically." ]
[ null ]
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https://www.geeksforgeeks.org/python-sympy-logcombine-method/
[ "# Python | sympy.logcombine() method\n\nWith the help of `sympy.logcombine()`, we can combine the log terms in the mathematical expression by using the following properties that are listed below.\n\nProperties :\n1) log(x*y)=log(x)+log(y)\n2) log(x**n)=nlog(x)\n\nSyntax : `sympy.logcombine()`\n\nReturn : Return the simplified mathematical expression.\n\nExample #1 :\nIn this example we can see that by using `sympy.logcombine()`, we are able to combine the log terms in mathematical expression.\n\n `# import sympy ` `from` `sympy ``import` `*`  ` `  `x, y, z ``=` `symbols(``'x y z'``, positive ``=` `True``) ` `gfg_exp ``=` `log(x) ``+` `log(y) ` `   `  `# Using sympy.logcombine() method ` `gfg_exp ``=` `logcombine(gfg_exp) ` `   `  `print``(gfg_exp) `\n\nOutput :\n\nlog(x*y)\n\nExample #2 :\n\n `# import sympy ` `from` `sympy ``import` `*`  ` `  `x, y, z ``=` `symbols(``'x y z'``, positive ``=` `True``) ` `gfg_exp ``=` `5` `*` `log(x) ` `   `  `# Using sympy.logcombine() method ` `gfg_exp ``=` `logcombine(gfg_exp) ` `   `  `print``(gfg_exp) `\n\nOutput :\n\nlog(x**5)\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\n\nBe the First to upvote.\n\nPlease write to us at contribute@geeksforgeeks.org to report any issue with the above content." ]
[ null, "https://media.geeksforgeeks.org/auth/profile/uv1fvtq244ld2v8nn1v7", null ]
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http://en.statmania.info/2015/04/OR-basic.html
[ "# Operation Research: Basic Understanding\n\nDefinition: Operation Research is the representation of real world system by mathematical models with the use of quantitative methods for solving models with a view to optimizing.\nCharacteristics of OR:\n1. It is system oriented.\n2. The use of interdisciplinary teams.\n3. Application of scientific methods.\n4. Uncovering of new problems.\n5. Use of Computer\n6. Quantitative solution\nThe cost to the company, if decision A is taken is X, if decision B is taken is Y.\nApplication of OR:\nOR model consist of -\n-Decision Variables\n-Constraints\n-An Objective function\n-A solution to the model\nDecision Variables:\nThese are the unknowns to be determined by the solution to the model.\nConstraints: Any restrictions on the values of decision variables that are expressed mathematically by means of inequality are often called constraints.\nThe coefficients of the constraints in the objective function are called the model parameters.\nObjective Function:\nAn objective function is a function of decision variables that the decision maker wants to maximize (profits) or minimize( costs).\nSolution:\nAny specification of the values for the decision variables is called a solution.\nFeasible Solution:\nA solution of the model is feasible if it satisfies all constraints.\nOptimal solution: A solution of the model is optimal if in addition to being feasible, it yields the best (max/min) value of the objective function.\nl,w – decision Variables\nmax lw – objective function\nl>=0, w>=0 – constraints\nFormulating mathematical models:\nThe following methods are used-\n1. Linear Programming (LP)\n- Solving Linear programming problems\n- Graphical methods\n- Simplex method\n- Duality Theory sensitivity analysis\n2. Integer Programming\n3. Decision analysis & Games\n4. Goal Programming\nConditions of LP:\nIt is used for optimizing problems that satisfies the following condition-\n1. There is a well-defined objective function to be optimized which can be expressed as a linear function of decision variables.\n2. There are constraints on the attainment of the objective and they are capable of being expressed as linear inequalities in terms of variables.\n3. There are alternative causes of action\n4. The decision variables are interrelated and nonnegative.\nRedundant Restriction:\nThe restriction which has no effect on feasible region or solution space is called redundant restriction.\nBasic and Nonbasic Variables:\nIn a set of m╳n equations (m<n) if m variables have a unique solution, then they are basic variables and their solution is referred to as basic solution.\nIn a set of m╳n equations, the (n-m) variables that are set to zero are known as non-basic variables and their solution is non-basic solution.\n1. Calculation is lengthy.\n2. It may give misleading conclusion unless the problem is correctly constructed.\nAssumptions/Properties of LP:\nIn LP models, the objective and the constraints are all linear. The linearity must satisfy three basic properties.\n1. Proportionality:\nThis property requires the combination of each variables in both the objective function and the constraints to be directly proportional to the value of the variable.\nZ= bx1\nz∝x1 =>z=kx1\nThis property requires the total combination of all the variables in the objective function and in the constraints to be the direct sum of the individual combination of each variable.\nZ=6x1+x2\n3. Certainty: All the objective and constraints coefficients of the LP model are deterministic. This means they are known constants.\nSteps to formulate LP:\nStep-1: We need to study the given situation to find the key decision to be made.\nStep-2: We identify the involved variable and designate them by symbols xj(i=1,2,…n)\nStep-3: We express the possible alternatives\nStep-4: Identifying the objective function and expressing it in linear function of the variables.\nStep-5: We express the influencing factors such as limitation or constraints as linear inequalities in terms of the variables of the problem.\nSimplex method:\nThe simplex method is an iterative procedure which gives the solution to a LPP in a finite number of steps.\nThis is simple iterative procedure to find an optimum basic feasible solution from an initial basic feasible solution by a finite no. of iteration.\nThese iterations consist of finding a new basic feasible solution which improves the value of objective function from that of previous solution. The process is continued until an optimum basic feasible solution is obtained or there is an indication of an unbounded solution.\nSlack variables:\nWhen the constraints are inequations connected by the sign ≤, then in each equation an extra variable is added to the left hand side of the equation to convert it into an equation. These variables are known as slack variables. For example,\nX1-2x2+x3≤5\n=>x1-2x2+x3+s1=5\nThe variable s1 is known as slack variable which is non-negative.\nSimilarly, Surplus Variable;" ]
[ null ]
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https://tracewind.net/science/1025.html
[ "# 求导证明披萨定理", null, "", null, "$AP^2+BP^2+CP^2+DP^2=4r^2$", null, "$\\left(A’P^2+B’P^2+C’P^2+D’P^2\\right)\\frac{d\\alpha}{2}=2r^2d\\alpha$", null, "", null, "", null, "", null, "QED", null, "", null, "", null, "", null, "", null, "$A_1P^2+A_2P^2+\\ldots+A_{2n}P^2=2nr^2$\n\n## “求导证明披萨定理”的一个回复\n\n1.", null, "渡鸦说道：\n\n沙发，顺带思考题的解答是固定分割线、把正方形外边框向右平移，直到分割线沿正方形一条对角线对称为止" ]
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https://www.newhavendisplay.com/NHD_forum/index.php?action=profile;area=showposts;sa=topics;u=7791
[ "###", null, "Show Posts\n\nThis section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.\n\n### Topics - DataL\n\nPages: \n1\n##### OLEDs / Bit-banged SPI conflicting with SPI.h and SD.h library (NHD-0420CW-Ax3 display\n« on: October 05, 2018, 05:46:20 PM »\nI have been doing well to solve my own problems but this one has me stumped.\n\nI have developed my code so far successfully using SD.h for the SD card and Wire.h for the RTC and display until I decided to make the display work over SPI instead.\nOf all the SPI code examples I have found for the display the only one I have made work is the big-banged one in my sketch, however if SD.h or SPI.h are included in the code the screen doesn't do anything - a previous working sketch's output will remain frozen on the screen after upload. Remove SPI/SD.h and the display works again.\nI have taken a look over SD.h and SD.cpp, as well as the many others included in that library but nothing jumps out at me and it is going over my head.\n\nCode: [Select]\n`/* Datalogger v0.93.3 for Arduino Mega2560, NHD-0420CW-Ax3, DS3231 and SD card - 8% progmem; 19% SRAM** The circuit:* OLED pin 7 (SCK)          to Arduino pin 52 (SCK)* OLED pin 8 (MOSI)         to Arduino pin 51 (MOSI)* OLED pin 9 (MISO)         to Arduino pin 50 (MISO)* OLED pin 15 (CS)          to Arduino pin 29 (CS)* OLED pin 16 (/RES)        to Arduino pin 36 // Reset or VDD 5V (or to Arduino pin D13, to control reset by sw)* MISO to ICSP pin 1 / 50* MOSI to ICSP pin 4 / 51* CLK to ICSP pin 3 / 52* SDcard CS to pin 31* Arduino Pin 20 to I2C SDA w/ 10K pull-up* Arduino Pin 21 to I2C SCL w/ 10K pull-up* Original example created by Newhaven Display International Inc.* Modified and adapted 15 Mar 2015 by Pasquale D'Antini* Modified 19 May 2015 by Pasquale D'Antini* Further modified and added to by DataL* This example code is in the public domain.*/#include <Wire.h>#include <SPI.h>#include <SD.h>#define DS3231_I2C_ADDRESS 0x68int count;        // Process loop couter//----- For the inputs://Analog#define muxPin A0    // sensor mux input#define otsPin A2    // sensor OTS input#define opsPin A4    // sensor OPS input#define iatPin A6    // sensor IAT inputint svpValue;   // sensor supply voltage valueint svnValue;   // sensor supply earth valueint mpsValue;   // sensor MPS valueint otsValue;   // sensor OTS valueint opsValue;   // sensor OPS valueint iatValue;   // sensor IAT valueint otsTemp;    // Engine Oil Temperature in Celciusint opsPres;    // Engine Oil Pressureint iatTemp;    // Inlet Air Temperatureint mpsPres;    // Manifold Pressure//Digital#define muxaPin 34     // multiplexer control A#define muxbPin 36     // multiplexer control B#define muxcPin 38     // multiplexer control C#define ignPin 02      // voltage IGN input#define dosPin 04      // voltage DOS inputint ignValue;          // binary Ignition On valueint dosValue;          // binary Door Open Switch value//----- For the sensor calculations:int svpVolt;     // sensor supply voltageint svnVolt;     // sensor earth voltageint mpsVolt;     // sensor MPS voltageint otsVolt;     // sensor OTS voltageint opsVolt;     // sensor OPS voltageint iatVolt;     // sensor IAT voltageint otsRes;      // sensor OTS resistancedouble iatRes;     // sensor IAT resistancedouble otsTempK;   // Engine Oil Temperature in Kelvindouble iatTempK;   // Inlet Air Temperature in Kelvin//Sensor averagesint i = 0;long svpAve;long svnAve;long svdAve;long mpsAve;long otsAve;long opsAve;long iatAve;//-----For the SPI Comms#define SCLK 52#define MOSI 51#define MISO 50#define sdCS 31         // SD card SPI Chip Select pin#define lcdCS 29        // LCD screen SPI Chip Select pin//----- For the SD card:File mainLog;//File faultLog;//----- For the LCD screen://Setup#define COLUMN_N 20     // Number of display columns#define ROW_N 4         // Number of display rows#define lcdRes 44       // Reset LCD screen when LOWbyte new_line = {0x80, 0xA0, 0xC0, 0xE0};   // Maybe 0x01?? // DDRAM address for each line of the display (128, 160, 192, 224)byte rows = 0x08;                              // Display mode: 2/4 lines; default (0x08 = 8)byte r = 0;                                    // Row indexbyte c = 0;                                    // Column index//Welcome arraysconst char welcome0 PROGMEM = \"X     Starting     X\";const char welcome1 PROGMEM = \"     Datalogger     \";const char welcome2 PROGMEM = \"       v0.93        \";const char welcome3 PROGMEM = \"X                  X\";const char* const welcometable[] PROGMEM = {welcome0, welcome1, welcome2, welcome3};char welcome;//Output arrayschar buffer0;char buffer1;char buffer2;char buffer3;char buffer4;char buffer5;//----- For the power control:#define lcdReg 46      // LCD display power supply transistor#define perReg 03      // Peripheral power supply transistor#define senReg 32      // Sensor power supply transistor//----- For the date and time:char weekDayVerb = \"AnyDayYet\";const char monday PROGMEM = \"Monday\";const char tuesday PROGMEM = \"Tuesday\";const char wednesday PROGMEM = \"Wednesday\";const char thursday PROGMEM = \"Thursday\";const char friday PROGMEM = \"Friday\";const char saturday PROGMEM = \"Saturday\";const char sunday PROGMEM = \"Sunday\";char monthVerb = \"AnyMonth\";const char jan PROGMEM = \"Jan\";const char feb PROGMEM = \"Feb\";const char mar PROGMEM = \"Mar\";const char apr PROGMEM = \"Apr\";const char may PROGMEM = \"May\";const char jun PROGMEM = \"Jun\";const char jul PROGMEM = \"Jul\";const char aug PROGMEM = \"Aug\";const char sep PROGMEM = \"Sep\";const char oct PROGMEM = \"Oct\";const char nov PROGMEM = \"Nov\";const char dec PROGMEM = \"Dec\";//----- For the refresh rate:unsigned long previousMillis = 0;        // will store last time logs were updatedconst long interval = 1000;              // interval at which to update logs (milliseconds)void setup(void)  {  count=0;  i=0;    pinMode(SCLK, OUTPUT);  pinMode(MOSI, OUTPUT);  pinMode(MISO, INPUT);  pinMode(sdCS, OUTPUT);  pinMode(lcdCS, OUTPUT);  pinMode(lcdRes, OUTPUT);    pinMode(lcdReg, OUTPUT);  pinMode(perReg, OUTPUT);  pinMode(senReg, OUTPUT);  pinMode(muxPin, INPUT);  pinMode(otsPin, INPUT);  pinMode(opsPin, INPUT);  pinMode(iatPin, INPUT);  pinMode(muxaPin, OUTPUT);  pinMode(muxbPin, OUTPUT);  pinMode(muxcPin, OUTPUT);  pinMode(ignPin, INPUT);  pinMode(dosPin, INPUT);    digitalWrite(SCLK, HIGH);  digitalWrite(MOSI, LOW);  digitalWrite(sdCS, HIGH);  digitalWrite(lcdCS, HIGH);  digitalWrite(lcdRes, HIGH);       // Disable LCD reset    digitalWrite(lcdReg, HIGH);       // Enable transistor  digitalWrite(perReg, HIGH);       // Enable transistor  digitalWrite(senReg, HIGH);       // Enable transistor  delay(1);    Wire.begin();                     // Initiate the Wire library and join the I2C bus as a master//  settime();                      // Use when need to reset datetime with uploading  Serial.begin(9600);               // Initialize serial communication  while (!Serial){;}                // Wait for serial port to connect    LCDinit();  SDinit();  Welcome();  delay(2500);}void loop(void)  {                     // MAIN PROGRAM      if (!SD.exists(\"/MainLogs/\"))    {    SDinit();  }  inputs();                            // Read values on input pins  calculate();                         // Convert values into info  unsigned long currentMillis = millis();  if (currentMillis - previousMillis >= interval) {    buffers();                         // Compose output text buffers.    LCDdisplay();                      // Output buffers to screen    seriallog();                       // Output to serial    SDlogmain();                       // Output to SD card    count = count + 1;    i = 0;                             // Reset mean count    svpAve = 0;                        // Reset means    svnAve = 0;    mpsAve = 0;    otsAve = 0;    opsAve = 0;    iatAve = 0;    previousMillis = currentMillis;  }}void Welcome()  {  Serial.print(F(\"\\nStarting Datalogger v0.93\"));  Serial.print(F(\"\\nCount: \"));  Serial.print(count);  r = 0;                              // Row index  c = 0;                              // Column index  command(0x01);                      // Clears display (and cursor home)  delay(2);                           // After a clear display, a minimum pause of 1-2 ms is required  for (int r = 0; r < 4; r++)  {    strcpy_P(welcome, (char*)pgm_read_word(&(welcometable[r])));    command(new_line[r]);            // Moves the cursor to the first column of line [i]    for (c=0; c<COLUMN_N; c++)  {    // One character at a time,      data(welcome[c]);              // Displays the corresponding string    }  }}void LCDdisplay()  {   byte r = 0;                         // Row index   byte c = 0;                         // Column index      r = 0;                              // Choose line 0      command(new_line[r]);            // Moves the cursor to the first column of that line      for (c=0; c<COLUMN_N; c++)  {    // One character at a time,        data(buffer0[c]);  }        // Displays the correspondig string   r = 1;                              // Choose line 1      command(new_line[r]);            // Moves the cursor to the first column of that line      for (c=0; c<COLUMN_N; c++)  {    // One character at a time,         data(buffer1[c]);  }       // Displays the correspondig string   r = 2;                              // Choose line 2      command(new_line[r]);            // Moves the cursor to the first column of that line      for (c=0; c<COLUMN_N; c++)  {    // One character at a time,         data(buffer2[c]);  }       // Displays the correspondig string   r = 3;                              // Choose line 3      command(new_line[r]);            // Moves the cursor to the first column of that line      for (c=0; c<COLUMN_N; c++)  {    // One character at a time,         data(buffer3[c]);  }       // Displays the corresponding string}//-----For the SD card:void SDinit(void) {    Serial.print(F(\"\\n\\nInitializing SD card...  \"));  if (SD.begin(sdCS))   {    Serial.print(F(\"SD card initialised.\"));    Serial.print(F(\"\\n\"));   }  else    {    Serial.print(F(\"SD card inaccessible.\"));    Serial.print(F(\"\\n\"));    return;   }  if (SD.exists(\"/MainLogs/\"))    {                                                     // *Directory name    Serial.print(F(\"\\n/MainLogs/ Directory exists\"));   }                               // *Directory name  else    {    Serial.print(F(\"\\n/MainLogs/ directory doesn't exist. Creating directory... \"));    // *Directory name    SD.mkdir(\"/MainLogs/\");                                                             // *Directory name    if (!SD.exists(\"/MainLogs/\"))   {                                                   // *Directory name      Serial.print(F(\"\\nDirectory /MainLogs/ failed to be created\"));                   // *Directory name    return;   }    else  {      Serial.print(F(\"\\nDirectory /MainLogs/ has been created\"));    }   }              // *Directory name/*    // Save the following for a later date //  if (SD.exists(\"/FaultLog/\"))    {                                                    // *Directory name    Serial.print(F(\"\\n/FaultLog/ Directory exists\"));                                  // *Directory name    Serial.print(F(\"\\n\\n\"));   }  else    {    Serial.print(F(\"\\n/FaultLog/ Directory doesn't exist. Creating directory... \"));   // *Directory name    SD.mkdir(\"/FaultLog/\");                                                            // *Directory name    if (!SD.exists(\"/FaultLog/\"))   {                                                  // *Directory name      Serial.print(F(\"\\nDirectory /FaultLog/ failed to be created\"));                  // *Directory name      Serial.print(F(\"\\n\\n\"));    return;   }    else  {      Serial.print(F(\"\\nDirectory /FaultLog/ has been created\"));                      // *Directory name      Serial.print(F(\"\\n\\n\"));    }*/}void SDlogmain(void)  {  byte second, minute, hour, dayOfWeek, dayOfMonth, month, year;  readDS3231time(&second, &minute, &hour, &dayOfWeek, &dayOfMonth, &month, &year);  // retrieve data from DS3231    char filepath;  snprintf (filepath, 9, \"%d%d%d.csv\", dayOfMonth, monthVerb, year);  Serial.print(\"filepath = \");  Serial.println(dayOfMonth);  mainLog = SD.open(\"filepath\", FILE_WRITE);  if  (!mainLog)   {    Serial.print(F(\"\\nSDlogmain failed to open file \"));    Serial.print(\"filepath\");    digitalWrite(42, LOW);    }  else  {    Serial.print(F(\"\\nSDlogmain successfully opened file \"));    Serial.print(\"filepath\");    digitalWrite(42, HIGH);   }    mainLog.println(\"Testing, writing, filling...\");  mainLog.close();  if (!mainLog)   {    Serial.print(F(\"\\nSDlogmain successfully closed file \"));    Serial.println(\"filepath\");    Serial.println();   }  else  {    Serial.print(F(\"\\nSDlogmain failed to close file \"));    Serial.println(\"filepath\");    Serial.println();   }  mainLog = SD.open(\"filepath\");  Serial.print(\"Text file contains: \");  while (mainLog.available()) {  Serial.write(mainLog.read());   }  Serial.print(\"\\n\\n\");  mainLog.close();  }//-----For The RTC Module://-----For the LCD screen:void command(byte c)   {              // PREPARES THE TRANSMISSION OF A COMMAND BYTE  byte i = 0;                         // Bit index  for(i=0; i<5; i++)  {    digitalWrite(MOSI, HIGH);    clockCycle();  }  for(i=0; i<3; i++)  {    digitalWrite(MOSI, LOW);    clockCycle();  }  send_byte(c);                       // Transmits the byte}void data(byte d)    {                // PREPARES THE TRANSMISSION OF A BYTE OF DATA  byte i = 0;                         // Bit index  for(i=0; i<5; i++)  {    digitalWrite(MOSI, HIGH);    clockCycle();  }  digitalWrite(MOSI, LOW);  clockCycle();  digitalWrite(MOSI, HIGH);  clockCycle();  digitalWrite(MOSI, LOW);  clockCycle();  send_byte(d);                       // Transmits the byte}void send_byte(byte tx_b)  {          // SEND TO THE DISPLAY THE BYTE IN tx_b  byte i = 0;                         // Bit index  for(i=0; i<4; i++)  {    if((tx_b & 0x01) == 1)  {      digitalWrite(MOSI, HIGH);    }    else  {      digitalWrite(MOSI, LOW);    }    clockCycle();    tx_b = tx_b >> 1;  }  for(i=0; i<4; i++)  {    digitalWrite(MOSI, LOW);    clockCycle();  }  for(i=0; i<4; i++)  {    if((tx_b & 0x1) == 0x1)  {         // <------- Change (?)      digitalWrite(MOSI, HIGH);    }    else  {      digitalWrite(MOSI, LOW);    }    clockCycle();    tx_b = tx_b >> 1;  }  for(i=0; i<4; i++)  {    digitalWrite(MOSI, LOW);    clockCycle();  }}void clockCycle(void)  {          // EXECUTE THE CLOCK SIGNAL CYCLE  digitalWrite(lcdCS, LOW);          // Sets LOW the /CS line of the display (optional, can be always low)  delayMicroseconds(1);           // Waits 1 us (required for timing purpose)    digitalWrite(SCLK, LOW);        // Sets LOW the SCLK line of the display  delayMicroseconds(1);           // Waits 1 us (required for timing purpose)    digitalWrite(SCLK, HIGH);       // Sets HIGH the SCLK line of the display  delayMicroseconds(1);           // Waits 1 us (required for timing purpose)   digitalWrite(lcdCS, HIGH);         // Sets HIGH the /CS line of the display (optional, can be always low)  delayMicroseconds(1);           // Waits 1 us (required for timing purpose)}void LCDinit()    {  command(0x22 | rows);      // Function set: extended command set (RE=1), lines # (0x22 = 34)    command(0x71);           // Function selection A: (0x71 = 113)      data(0x5C);            // Enable internal Vdd regulator at 5V I/O mode (default) (0x00 for disable: 2.8V I/O) (0x5C = 92)  command(0x20 | rows);      // Function set: fundamental command set (RE=0) (exit from extended command set), lines # (0x20 = 32)  command(0x08);             // Display ON/OFF control: display off, cursor off, blink off (default values) (0x08 = 8)  command(0x22 | rows);      // Function set: extended command set (RE=1), lines # (0x22 = 34)    command(0x79);           // OLED characterization: OLED command set enabled (SD=1) (0x79 = 121)      command(0xD5);         // Set display clock divide ratio/oscillator frequency: (0xD5 = 213)      command(0x70);         // Divide ratio=1, frequency=7 (default values) (0x70 = 112)    command(0x78);           // OLED characterization: OLED command set disabled (SD=0) (exit from OLED command set) (0x78 = 120)            command(0x09);         // Extended function set (RE=1): 5-dot font, B/W inverting disabled (def. val.), 3/4 lines (0x09 = 9)      command(0x06);         // Entry Mode set - COM/SEG direction: COM0->COM31, SEG99->SEG0 (BDC=1, BDS=0) (0x06 = 6)      command(0x72);         // Function selection B: (0x72 = 114)        data(0x0A);          // ROM/CGRAM selection: ROM C, CGROM=250, CGRAM=6 (ROM=10, OPR=10) (0x0A = 10)        command(0x79);       // OLED characterization: OLED command set enabled (SD=1) (0x79 = 121)          command(0xDA);     // Set SEG pins hardware configuration: (0xDA = 218)            command(0x10);   // Alternative odd/even SEG pin, disable SEG left/right remap (default values) (0x10 = 16)          command(0xDC);     // Function selection C: (0xDC = 220)            command(0x00);   // Internal VSL, GPIO input disable (0x00 = 0)            command(0x81);   // Set contrast control: (0x81 = 129)              command(0x00); // Contrast=127 (default value) (0x7F = 127)            command(0xD9);   // Set phase length: (0xD9 = 217)              command(0xF1); // Phase2=15, phase1=1 (default: 0x78) (0x78 = 120)            command(0xDB);   // Set VCOMH deselect level: (0xDB = 219)              command(0x40); // VCOMH deselect level=1 x Vcc (default: 0x20=0,77 x Vcc) (0x40 = 64)        command(0x78);       // OLED characterization: OLED command set disabled (SD=0) (exit from OLED command set) (0x78 = 120)      command(0x20 | rows);  // Function set: fundamental command set (RE=0) (exit from extended command set), lines # (0x20 = 32)  command(0x01);             // Clear display (0x01 = 1)    delay(2);                // After a clear display, a minimum pause of 1-2 ms is required  command(0x80);             // Set DDRAM addlcdRess 0x00 in addlcdRess counter (cursor home) (default value) (0x80 = 128)  command(0x0C);             // Display ON/OFF control: display ON, cursor off, blink off (0x0C = 12)    delay(250);              // Waits 250 ms for stabilization purpose after display on}`\n\nPages:" ]
[ null, "https://www.newhavendisplay.com/NHD_forum/Themes/default/images/icons/profile_sm.gif", null ]
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https://usefulknot.com/calculators/dividing-polynomials-calculator/
[ "# Dividing Polynomials Calculator\n\n## Now, What Exactly Is A “Dividing Polynomials Calculator”?\n\nA dividing polynomials calculator is a tool that helps you divide one polynomial by another.\n\n## Dividing Polynomials Calculator with steps\n\nThis is a very simple tool for Dividing Polynomials Calculator. Follow the given process to use this tool.\n\n☛ Step 1: Enter the complete equation/value in the input box i.e. across “Provide Required Input Value:”\n\n☛ Step 2: Click “Enter Solve Button for Final Output”.\n\n☛ Step 3: After that a window will appear with final output.\n\n## Formula for dividing polynomials\n\nTo divide two polynomials, divide each term of the first polynomial by the corresponding term of the second polynomial.\n\nFor example, to divide x^3 – 2x^2 + 5x – 4 by x – 2, divide each term of the first polynomial by the corresponding term of the second polynomial.\n\nx^3 – 2x^2 + 5x – 4 / x – 2\n\nx^3 / x – 2 – 2x^2 / x – 2 + 5x / x – 2 – 4 / x – 2\n\nx^2 – 2x + 1\n\n## Dividing Polynomials Definition\n\nDividing polynomials is a mathematical operation that allows one to find the quotient and remainder when two polynomials are divided.\n\n## Example of dividing polynomials based on Formula\n\nThe division of polynomials is given by the formula:\n\n(P/Q) = (P1/Q1) + (P2/Q2) + … + (Pn/Qn)\n\nwhere P is the dividend polynomial, Q is the divisor polynomial, P1, P2, …, Pn are the factors of P, and Q1, Q2, …, Qn are the factors of Q.\n\nFor example, if the dividend polynomial is P(x) = x3 + 2×2 + 3x + 4 and the divisor polynomial is Q(x) = x2 + 1, then the division of polynomials is given by (P/Q) = (P1/Q1) + (P2/Q2) = (x+4)/(x+1) + (3)/(x+1) = x + 3." ]
[ null ]
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https://www.matchfishtank.org/curriculum/mathematics/7th-grade-math/equations-and-inequalities/
[ "# Equations and Inequalities\n\nStudents solve equations and inequalities with rational numbers, and encounter real-world situations that can be modeled and solved using equations and inequalities.\n\n## Unit Summary\n\nIn Unit 4, seventh-grade students continue to build on the last two units by solving equations and inequalities with rational numbers. They use familiar tape diagrams as a way to visually model situations in the form $px+q=r$ and $p(x+q)=r$. These tape diagrams offer a pathway to solving equations using arithmetic, which students compare to a different approach of solving equations algebraically. Throughout the unit, students encounter word problems and real-world situations, covering the full range of rational numbers, that can be modeled and solved using equations and inequalities (MP.4). As they work with equations and inequalities, they build on their abilities to abstract information with symbols and to interpret those symbols in context (MP.2). Students also practice solving equations throughout the unit, ensuring they are working towards fluency which is an expectation in 7th grade.\n\nIn sixth grade, students understood solving equations and inequalities as a process of finding the values that made the equation or inequality true. They wrote and solved equations in the forms $x+p=q$ and $px=q$, using nonnegative rational numbers. In seventh grade, students reach back to recall these concepts and skills in order to solve one- and two-step equations and inequalities with rational numbers including negatives.\n\nIn eighth grade, students explore complex multi-step equations; however, they will discover that these multi-step equations can be simplified into forms that are familiar to what they’ve seen in seventh grade. Eighth-grade students will also investigate situations that result in solutions such as 5 = 5 or 5 = 8, and they will extend their understanding of solution to include no solution and infinite solutions.\n\nPacing: 16 instructional days (12 lessons, 3 flex days, 1 assessment day)\n\n### Fishtank Plus\n\n#### Summer Packets\n\nPacket of problems designed to help prepare students for the start of next school year.\n\n## Assessment\n\nThis assessment accompanies Unit 4 and should be given on the suggested assessment day or after completing the unit.\n\n### Fishtank Plus\n\n#### Expanded Assessment Package\n\nLearn how to use these tools with our Guide to Assessments\n\n## Unit Prep\n\n### Essential Understandings\n\n?\n\n• Equations and inequalities are powerful tools that can be used to model and solve real-world situations with unknown quantities.\n• Equations can be solved by reasoning about the arithmetic needed to uncover the value of the unknown. Equations can also be solved algebraically by using properties of operations and equality.\n• Inequalities have infinite solutions, which can be represented graphically on a number line. In context, these solutions are sometimes constrained by what makes sense for the situation; for example, if solving for the maximum number of people who can fit onto a boat, the solution set would be limited to positive integers.\n\n?\n\ntape diagram\n\nequation\n\nsolution\n\nsubstitution\n\ninequality\n\n### Fishtank Plus\n\n#### Vocabulary Package\n\nAdditional vocabulary tools that help reinforce and support student vocabulary development.", null, "?\n\ntape diagram\n\n### Intellectual Prep\n\n?\n\n#### Internalization of Standards via the Unit Assessment\n\n• Take unit assessment. Annotate for:\n• Standards that each question aligns to\n• Strategies and representations used in daily lessons\n• Relationship to Essential Understandings of unit\n• Lesson(s) that assessment points to\n\n#### Internalization of Trajectory of Unit\n\n• Read and annotate “Unit Summary.”\n• Notice the progression of concepts through the unit using “Unit at a Glance.”\n• Essential understandings\n• Connection to assessment questions\n• Identify key opportunitites to engage students in academic discourse. Read through our Guide to Academic Discourse and refer back to it throughout the unit.\n\n#### Unit-Specific Intellectual Prep\n\n Tape diagram and equations Examples: $3(x+4)=45$", null, "$3x+4=45$", null, "## Common Core Standards\n\nKey: Major Cluster Supporting Cluster Additional Cluster\n\n### Core Standards\n\n?\n\n##### Expressions and Equations\n• 7.EE.B.3 — Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making $25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or$2.50, for a new salary of $27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation. • 7.EE.B.4 — Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. • 7.EE.B.4.A — Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width? • 7.EE.B.4.B — Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid$50 per week plus $3 per sale. This week you want your pay to be at least$100. Write an inequality for the number of sales you need to make, and describe the solutions.\n\n?\n\n• 6.EE.B.5\n\n• 6.EE.B.7\n\n• 6.EE.B.8\n\n?\n\n• 8.EE.C.7\n\n• 8.EE.C.8\n\n### Standards for Mathematical Practice\n\n• CCSS.MATH.PRACTICE.MP1 — Make sense of problems and persevere in solving them.\n\n• CCSS.MATH.PRACTICE.MP2 — Reason abstractly and quantitatively.\n\n• CCSS.MATH.PRACTICE.MP3 — Construct viable arguments and critique the reasoning of others.\n\n• CCSS.MATH.PRACTICE.MP4 — Model with mathematics.\n\n• CCSS.MATH.PRACTICE.MP5 — Use appropriate tools strategically.\n\n• CCSS.MATH.PRACTICE.MP6 — Attend to precision.\n\n• CCSS.MATH.PRACTICE.MP7 — Look for and make use of structure.\n\n• CCSS.MATH.PRACTICE.MP8 — Look for and express regularity in repeated reasoning." ]
[ null, "https://match-fishtank.s3.amazonaws.com/static/images/vocab_math.png", null, "https://cdn.matchfishtank.org/uploads/2019/01/24/g7u4_ipa.PNG", null, "https://cdn.matchfishtank.org/uploads/2019/01/24/g7u4_ipb.PNG", null ]
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https://www.basic-mathematics.com/finding-the-average.html
[ "# Finding the average\n\nFinding the average or the mean is a very straightforward concept. If you have a set of numbers, the average is found by adding all numbers in the set and dividing their sum by the total number of numbers added in the set.\n\nWe can generalize the concept of average with the formula below:\n\nLet x1,x2,x3,......,xn be a set of numbers, average = (x1 + x2 + x3,+......+ xn)/n\n\nLet us illustrate with examples:\n\nExample #1:\n\nGet the average of the following set of numbers:\n\n5,4,12,2,1,6\n\nStep 1:\n\nFind the sum:\n\nSum = 5+4+12+2+1+6= 30\n\nStep 2:\n\nDivide 30 by 6 (the total number of numbers)", null, "30 divided by 6 is 5, so the average or mean is 5\n\nExample #2:\n\nGet the average of the following set of numbers:\n\n5,10,20,5,10\n\nFind the sum:\n\nSum = 5+10+20+5+10 = 50\n\nDivide the 50 by 5 (the total number of numbers added)", null, "50 divided by 5 is 10, so 10 is the average or mean.\n\n## Finding the average when the numbers in the set are the same.\n\nWhen all the numbers in the set are the same, it is easy to find the average.\n\nExample #3\n\nFind the average of the following set of numbers:\n\n6, 6, 6\n\n6 + 6 + 6 = 18\n\n18/3 = 6\n\nExample #4\n\nFind the average of the following set of numbers:\n\n12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12\n\n12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 = 132\n\n132/11 = 12\n\nWhat can we conclude from example #3 and example #4?\n\nWhen the numbers in the set are all the same, the average is just the number itself.\n\n## Finding the average when some of the numbers in the set are negative numbers.\n\nExample #5\n\n-5, 2, -1, 8\n\n-5 + 2 + -1 + 8 = 4\n\n4/4 = 1\n\nExample #6\n\n-8, 2, -11, 3, 0, 2\n\n-8 + 2 + -11 + 3 + 0 + 2 = -12\n\n-12/6 = -2\n\nExample #7:\n\nMr. Peter collected 125 pencils from students during the past 5 days. On the average, how many pencils did he collect each day?\n\nAverage = 125/5 = 25\n\nExample #8:\n\nLast week, Samuel ordered pizza every day. After collecting all his receipts and getting the total, he realized that he spent a total of 63 dollars on pizza. On the average, how much money did he spend on pizza each day?\n\nAverage = 63/7 = 9\n\n## Recent Articles", null, "1. ### Pressure in a Liquid - How to Derive Formula\n\nJul 02, 22 07:53 AM\n\nThis lesson will show clearly how to derive the formula to find the pressure in a liquid.\n\n## Check out some of our top basic mathematics lessons.\n\nFormula for percentage\n\nMath skills assessment\n\nCompatible numbers\n\nSurface area of ​​a cube", null, "" ]
[ null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 178 127'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 179 132'%3E%3C/svg%3E", null, "https://www.basic-mathematics.com/objects/xrss.png.pagespeed.ic.nVmUyyfqP7.png", null, "data:image/svg+xml,%3Csvg xmlns='http://www.w3.org/2000/svg' viewBox='0 0 165 100'%3E%3C/svg%3E", null ]
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Advanced_Statistical_Mechanics/Classical_microstates%2C_Newtonian%2C_Lagrangian_and_Hamiltonian_mechanics/The_Hamiltonian_formulation_of_classical_mechanics
[ "# The Hamiltonian formulation of classical mechanics\n\nThe Lagrangian formulation of mechanics will be useful later when we study the Feynman path integral. For our purposes now, the Lagrangian formulation is an important springboard from which to develop another useful formulation of classical mechanics known as the Hamiltonian formulation. The Hamiltonian of a system is defined to be the sum of the kinetic and potential energies expressed as a function of positions and their conjugate momenta. What are conjugate momenta?\n\nRecall from elementary physics that momentum of a particle, $$P_i$$, is defined in terms of its velocity $$\\dot {r}_i$$ by\n\n$p_i = m_i \\dot {r} _i$\n\nIn fact, the more general definition of conjugate momentum, valid for any set of coordinates, is given in terms of the Lagrangian:\n\n$p_i = \\frac {\\partial L}{\\partial \\dot {r} _i }$\n\nNote that these two definitions are equivalent for Cartesian variables. In terms of Cartesian momenta, the kinetic energy is given by\n\n$K = \\sum _{i=1}^N \\frac {P^2_i}{2 m_i}$\n\nThen, the Hamiltonian, which is defined to be the sum, $$K + U$$, expressed as a function of positions and momenta, will be given by\n\n$H (p, r) = \\sum _{i=1}^N \\frac {P^2_i}{2m_i} + U ( r_1, \\cdots , r_N) = H (p, r)$\n\nwhere $$p \\equiv p_1, \\cdots , p_N$$ . In terms of the Hamiltonian, the equations of motion of a system are given by Hamilton's equations:\n\n$\\dot {r} _i = \\frac {\\partial H}{\\partial p_i} \\dot {p}_i = - \\frac {\\partial H}{\\partial r_i}$\n\nThe solution of Hamilton's equations of motion will yield a trajectory in terms of positions and momenta as functions of time. Again, Hamilton's equations can be easily shown to be equivalent to Newton's equations, and, like the Lagrangian formulation, Hamilton's equations can be used to determine the equations of motion of a system in any set of coordinates.\n\nThe Hamiltonian and Lagrangian formulations possess an interesting connection. The Hamiltonian can be directly obtained from the Lagrangian by a transformation known as a Legendre transform. We will say more about Legendre transforms in a later lecture. For now, note that the connection is given by\n\n$H (p, r) = \\sum _{i=1}^N p_i \\cdot \\dot {r}_i - L (r , \\dot {r} )$\n\nwhich, when the fact that $$\\dot {r}_i = \\frac {p_i}{m_i}$$ is used, becomes\n\n$H (p, r)= \\sum _{i=1}^N p_i \\cdot \\frac {p_i}{m_i} - \\sum _{i=1}^N \\frac {1}{2} m_i \\left (\\frac {p_i}{m_i} \\right )^2 + U (r_1, \\cdots , r_N )$\n\n$= \\sum _{i=1}^N \\frac {P_i^2}{2m_i} + U(r_1, \\cdots , r_N )$\n\nBecause a system described by conservative forces conserves the total energy, it follows that Hamilton's equations of motion conserve the total Hamiltonian. Hamilton's equations of motion conserve the Hamiltonian\n\n$H (p (t), r (t) ) = H (p(0), r (0) ) = E$\n\nProof\n$$H = \\text {const} \\Rightarrow \\frac {dH}{dt} = 0$$\n\n$\\frac {dH}{dt}= \\sum _{i=1}^N \\left ( \\frac {\\partial H}{\\partial r_i} \\cdot \\dot {r} _i + \\frac {\\partial H}{\\partial p_i} \\cdot \\dot {p} _i \\right )$\n\n$=\\sum _{i=1}^N \\left ( \\frac {\\partial H}{\\partial r_i} \\cdot \\frac {\\partial H}{\\partial p_i} - \\frac {\\partial H}{\\partial p_i} \\cdot \\frac {\\partial H}{\\partial r_i} \\right ) = 0$\n\nQED. This, then, provides another expression of the law of conservation of energy." ]
[ null ]
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http://www.electronics-tutorial.net/digital-logic-gates/xor-gate/
[ "Home > digital logic gates > xor gate\n\n# XOR Gate\n\nThe EXCLUSIVE-OR gate is a two-input, one-output gate. Figure shows the logic symbol and truth table of a two-input XOR gate. As can be seen from the truth table, the output of an XOR gate is a logic ‘1’ when the inputs are unlike and a logic ‘0’ when the inputs are like. The truth table of a multiple-input XOR function can be expressed as follows. The output of a multiple-input XOR logic function is a logic ‘1’ when the number of 1s in the input sequence is odd and a logic ‘0’ when the number of 1s in the input sequence is even, including zero. That is, an all 0s input sequence also produces a logic ‘0’ at the output.\nThe output of a two-input XOR gate is expressed by,\n\nY = \u0003A⊕B\u0002 = AB+AB" ]
[ null ]
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https://jacobbradjohnson.com/helping-precalculus-800
[ "# Slope equation solver\n\n## The Best Slope equation solver\n\nThere is Slope equation solver that can make the technique much easier. The app, called Mathway, allows users to enter a problem and then see step-by-step instructions for solving it. In addition, the app includes a wide range of features that make it easy to use, including a built-in calculator and a library of solved problems. As a result, Mathway is an essential tool for any student who wants to improve their math skills.\n\nInterval notation is a method of representing a set of numbers using Intervals. Interval notation solver is a online tool that helps you to solve the problems in Interval Notation. It shows the work by using the properties of Intervals, so you can understand the steps involved in solving the problem. You can also use this tool to check your answers.\n\nA variable equation solver is a mathematical tool that is used to find the value of an unknown variable in an equation. This type of equation is usually represented by two lines that intersect at a point, with the unknown variable being represented by the letter x. To use a variable equation solver, simply input the values of the known variables into the tool and it will output the value of the unknown variable. This process can be repeated for multiple equations, allowing you to solve for multiple unknowns simultaneously. Variable equation solvers can be found online or in mathematical textbooks. However, it is important to note that these tools only provide approximate values for the unknowns; they cannot give exact solutions. Therefore, if you need an accurate answer, it is best to use a different method such as algebraic methods. Nevertheless, variable equation solvers can be a valuable tool for solving simple equations quickly and easily.\n\nSolving integral equations is a process of finding a function that satisfies a given equation involving integrals. There are many methods that can be used to solve integral equations, each with its own advantages and disadvantages. The most common method is to use integration by substitution, which involves solving for the function in terms of the variables in the equation. However, this method can be difficult to apply in practice, especially if the equation is complex. Another popular method is to use Green's functions, which are special functions that can be used to solve certain types of differential equations. Green's functions can be very effective in solving integral equations, but they can be difficult to obtain in closed form. In general, there is no one best method for solving integral equations; the best approach depends on the specific equation and the tools that are available.\n\n## We cover all types of math issues\n\nThis is the best app for school math students. This app has almost zero ads and gives superfast solutions also (if only there is a high-quality camera). I would highly recommend this for school students if they want to save their time and also get good marks in their exams", null, "Nyla Cook\nIt not only helps me check my work, but shows me if/where I've messed up and why. Great resource but do not rely solely on this, as there are times it doesn't read it clearly and will give the wrong answer. But a great back up tool to give you a little confidence in the work you already know how to do!", null, "Xylia Reed\nSequence solver calculator Pre calculus word problem solver How to do geometry problems step by step Solve the problem Solve trigonometry problems" ]
[ null, "https://jacobbradjohnson.com/LR56320a40ff0f65/25388788904_72d2f5ec6f_z-150x150.jpg", null, "https://jacobbradjohnson.com/LR56320a40ff0f65/team_3-150x150.jpg", null ]
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https://ch.mathworks.com/matlabcentral/cody/solutions/1817599
[ "Cody\n\n# Problem 16. Return the largest number that is adjacent to a zero\n\nSolution 1817599\n\nSubmitted on 16 May 2019 by Qinghua Ye\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\na = [1, 5, 3, 0, 2, 7, 0, 8, 9, 1 0]; b = 8; assert(isequal(nearZero(a),b))\n\n2   Pass\na = [5 4 -1 0 -2 0 -5 8]; b = -1; assert(isequal(nearZero(a),b));\n\n3   Pass\na = [0 3 1 0 2 9]; b = 3; assert(isequal(nearZero(a),b));\n\n4   Pass\na = [1 0 2 0 3]; b = 3; assert(isequal(nearZero(a),b));\n\n5   Pass\na = [0 -1]; b = -1; assert(isequal(nearZero(a),b));\n\n6   Pass\na = [0 -12 0 -7 0]; b = -7; assert(isequal(nearZero(a),b));" ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.56758827,"math_prob":0.9993055,"size":706,"snap":"2020-10-2020-16","text_gpt3_token_len":314,"char_repetition_ratio":0.17806268,"word_repetition_ratio":0.031746034,"special_character_ratio":0.5127479,"punctuation_ratio":0.2054054,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98102695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-24T19:08:54Z\",\"WARC-Record-ID\":\"<urn:uuid:0311c28e-f5bd-43cb-882d-70306849ce50>\",\"Content-Length\":\"76314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aaaa7d7b-8912-4f81-9a67-9b6dae8cf508>\",\"WARC-Concurrent-To\":\"<urn:uuid:6fc29b60-9b26-467d-8258-9c7216447486>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://ch.mathworks.com/matlabcentral/cody/solutions/1817599\",\"WARC-Payload-Digest\":\"sha1:DR3SN7L2OYGZBSTJFJJFKZU52UPJ4NKG\",\"WARC-Block-Digest\":\"sha1:6OOK4CCMR74S3WKDYJ3WW54HDJOCCOFX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145966.48_warc_CC-MAIN-20200224163216-20200224193216-00455.warc.gz\"}"}
https://www.hackmath.net/en/calculator/line-slope?x0=2&y0=1&x1=6&y1=0&submit=1
[ "# Line slope calculator x+4y=6\n\nEnter coordinates of two different points:", null, "Straight line given by points A[2; 1] and B[6; 0]\n\n## Calculation:\n\nSlope-intercept form of line: y = -0.25x+1.5\n\nCanonical form of the line equation: x+4y-6 = 0\n\nParametric form of the line equation:\nx = 4t+2\ny = -t+1      ; t ∈ R\n\nSlope: m = -0.25\n\nSlope angle of line: φ = -14°2'10″ = -0.245 rad\n\nX intercept: x0 = 6\n\nY intercept: y0 = q = 1.5\n\nDistance line from the origin: d0 = 1.4552\n\nThe length of the segment AB: |AB| = 4.1231\n\nVector: AB = (4; -1)\n\nNormal vector: n = (1; 4)\n\nMidpoint of the segment AB: M = [4; 0.5]\n\nPerpendicular Bisector equation: 4x-y-15.5 = 0\n\nVector OA = (2; 1) ;   |OA| = 2.2361\nVector OB = (6; 0) ;   |OB| = 6\nScalar product OA .OB = 12\nAngle ∠ AOB = 26°33'54″ = 0.4636 rad" ]
[ null, "https://www.hackmath.net/images/slope_line2.jpg", null ]
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https://mathematica.stackexchange.com/questions/143765/using-multiple-solutions-from-equations-in-further-calculations-part-ii
[ "# Using multiple solutions from equations in further calculations, part II\n\nFirst: This is a follow-up question to my question here, since I wasn't specific on what I really wanted.\n\nCondider the following code:\n\nClearAll[a1, b1, a2, b2, xC1, yC1, xC2, yC2]\na1 := 2\nb1 := 1\na2 := 1\nb2 := 1/2\nxC1 := 0\nyC1 := 1\nxC2 := 0\nyC2 := 4\n\neqn1 = a1 (a1^2 b2^2 m^2 + b1^2 (b2^2 - (c + m xC1 - yC1)^2)) ==\nb2 (a1^2 (a2^2 m^2 + b2^2) - a2^2 (c + m xC2 - yC2)^2);\neqn2 = (a1^2 m (c - yC1) - b1^2 xC1)^2 ==\n(a1^2 m^2 + b1^2) (b1^2 xC1^2 + a1^2 (c - yC1)^2 - a1^2 b2^2);\nsolCM = Solve[eqn1 && eqn2, {c, m}];\nClear[x]\nEvaluate[Array[x, 4]] = {(a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2)} /. solCM // Transpose // Flatten;\nSimplify[x /@ Range]\nClear[x]\nEvaluate[Array[x, 4]] = {(a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)} /. solCM // Transpose // Flatten;\nSimplify[x /@ Range]\n\n\nThis gives $2 \\cdot 4 = 8$ values for $x$ as output, as it should. However, it is not really what I want!\n\nI would like to insert the four $x$-values coming from\n\n(a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2)\n\n\nin the equation\n\n((x - xC1)/a1)^2 + ((y - yC1)/b1)^2 == 1\n\n\nand solve for $y$, and then give the output as four coordinates on the form $(x_i,y_i)$.\n\nFurthermore, I would like to do a similar thing for\n\n(a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)\n\n\nwith the output values coming from the expression being inserted in\n\n((x - xC2)/a2)^2 + ((y - yC2)/b2)^2 == 1\n\n\nand solved for $y$, and give the output as another four pairs of coordinates.\n\nP.S. The eight coordinate sets are the points at which the inner and outer tangents to two ellipses touch them. (The parameters at the beginning of the code are the major and minor axes and the coorinate sets for the centres of the two ellipses.)\n\nEDIT: Comment by OP changed equation in solXY. Also, removed [] from Solve in SolXY to get all solutions.\n\nClearAll[a1, b1, a2, b2, x, y, xC1, yC1, xC2, yC2, eqn, solXY, pts]\na1 = 2;\nb1 = 1;\na2 = 1;\nb2 = 1/2;\nxC1 = 0;\nyC1 = 1;\nxC2 = 0;\nyC2 = 4;\n\neqn = a1 (a1^2 b2^2 m^2 + b1^2 (b2^2 - (c + m xC1 - yC1)^2)) ==\nb2 (a1^2 (a2^2 m^2 + b2^2) - a2^2 (c + m xC2 - yC2)^2);\n\neqn = (a1^2 m (c - yC1) - b1^2 xC1)^2 == (a1^2 m^2 + b1^2) (b1^2 xC1^2 +\na1^2 (c - yC1)^2 - a1^2 b2^2);\n\nsolCM = Solve[eqn && eqn, {c, m}];\n\nClear[x, y]\n\nsolXY = Solve[{\nx == (a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2),\n((x - xC1)/a1)^2 + ((y - yC1)/b1)^2 == 1},\n{x, y}] /. solCM // Simplify // Flatten[#, 1] & // SortBy[#, N] &;\n\npts = {x, y} /. solXY\n\n(* {{-(Sqrt/6), 1 - Sqrt/12}, {-(Sqrt/6),\n1 + Sqrt/12}, {-(Sqrt/2), 1 - Sqrt/4}, {-(Sqrt/2),\n1/4 (4 + Sqrt)}, {Sqrt/2, 1 - Sqrt/4}, {Sqrt/2,\n1/4 (4 + Sqrt)}, {Sqrt/6, 1 - Sqrt/12}, {Sqrt/6,\n1 + Sqrt/12}} *)\n\npts // N\n\n(* {{-0.986013, 0.129974}, {-0.986013, 1.87003}, {-0.866025,\n0.0986122}, {-0.866025, 1.90139}, {0.866025, 0.0986122}, {0.866025,\n1.90139}, {0.986013, 0.129974}, {0.986013, 1.87003}} *)\n\nsolXY = Solve[{\nx == (a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2),\n((x - xC2)/a2)^2 + ((y - yC2)/b2)^2 == 1},\n{x, y}] /. solCM // Simplify // Flatten[#, 1] & // SortBy[#, N] &;\n\npts = {x, y} /. solXY\n\n(* {{-(Sqrt/6), 47/12}, {-(Sqrt/6), 49/12}, {-(Sqrt/2), 15/\n4}, {-(Sqrt/2), 17/4}, {Sqrt/2, 15/4}, {Sqrt/2, 17/4}, {Sqrt/6,\n47/12}, {Sqrt/6, 49/12}} *)\n\npts // N\n\n(* {{-0.986013, 3.91667}, {-0.986013, 4.08333}, {-0.866025, 3.75}, {-0.866025,\n4.25}, {0.866025, 3.75}, {0.866025, 4.25}, {0.986013, 3.91667}, {0.986013,\n4.08333}} *)\n\n\nVerifying,\n\nAnd @@ Apply[And, Table[eqn && eqn /. solCM /. solXY[n], {n, 2}], 2]\n\n(* True *)\n\n• Hmmm! I've missed something; I can see that all eight point have to calculated from only (a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2) (and not half of them from (a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2)). Can the code them be made \"simpler\"? Apr 17, 2017 at 19:24\n• @SvendTveskæg - So what have you tried? If there was a problem, what was it? Apr 17, 2017 at 19:27\n• I'm really a novice at Mathematica, so I don't know where to start / what to try. Apr 17, 2017 at 19:30\n• I was mistaken (again); it really has to be both (a1^2 m (c - yC1) - b1^2 xC1)/(a1^2 m^2 + b1^2) and (a2^2 m (c - yC2) - b2^2 xC2)/(a2^2 m^2 + b2^2). Nevermind, since I've managed to put together all my different pieces of code in order to make the \"program\" work. Apr 18, 2017 at 18:28" ]
[ null ]
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https://docs.messageix.org/en/v3.7.0/rmessageix.html
[ "# Usage in R via `reticulate`\n\n`message_ix` and `ixmp` are fully usable in R via reticulate, a package that allows nearly seamless access to the Python API. See Installation for instructions on installing R and reticulate. No additional R packages are needed. \n\nOnce installed, use reticulate to import the Python packages:\n\n```library(reticulate)\nixmp <- import(\"ixmp\")\nmessage_ix <- import(\"message_ix\")\n```\n\nThis creates two global variables, `ixmp` and `message_ix`, that can be used much like the Python modules:\n\n```mp <- ixmp\\$Platform(name = \"default\")\nscen <- message_ix\\$Scenario(mp, \"model name\", \"scenario name\")\n# etc.\n```\n\nSee the R versions of the Austria tutorials for full examples of building models.\n\nSome tips:\n\n• If using Anaconda, you may need to direct reticulate to use the Python executable from the same conda environment where `message_ix` and `ixmp` are installed. See the reticulate documentation for usage of commands like:\n\n```# Specify python binaries and environment under which messageix is installed\nuse_condaenv(\"message_env\")\n# or\nuse_python(\"C:/.../anaconda3/envs/message_env/\")\n```\n• As shown above, R uses the `\\$` character instead of `.` to access methods and properties of objects. Where Python code examples show, for instance, `scen.add_par(...)`, R code should instead use `scen\\$add_par(...)`.\n\n• MESSAGEix model parameters with dimensions indexed by the `year` set (e.g. dimensions named `year_act` or `year_vtg`) must be indexed by integers; but R treats numeric literals as floating point values. Therefore, instead of:\n\n```ya1 = 2010\nya2 = c(2020, 2030, 2040)\nya3 = seq(2050, 2100, 10)\n\n# ...store parameter data using year_act = ya1, ya2, or ya3\n```\n\n…use `as.integer()` to convert:\n\n```ya1 = as.integer(2010)\nya2 = sapply(c(2020, 2030, 2040), as.integer)\nya3 = as.integer(seq(2050, 2100, 10))\n```" ]
[ null ]
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https://www.codespeedy.com/build-binary-tree-in-cpp-competitive-programming/
[ "# Build Binary Tree in C++ (Competitive Programming)\n\nLet’s start our journey of learning a hierarchical data structure (BINARY TREE) in C++. We will start from very basic of creating a binary tree with the help of class and functions. In this tutorial, we will learn how to build binary tree in C++. Before that just grab some information about basics of Binary tree.\n\n## Build Binary Tree in C++ (Competitive Programming)\n\n#### Introduction\n\nA binary tree comprises of parent nodes, or leaves, each of which stores data and also links to up to two other child nodes (leaves) which are visualized spatially as below the first node with one placed to the left and with one placed to the right.\n\nSome of the important points of a binary tree are:-\n\n1. binary tree is made of nodes, where each node contains a left pointer (called the left subtree), a right pointer(called the right subtree), and a data element(data to be stored).\n2. The root pointer(node->root) points to the topmost node in the tree. A null pointer(NULL) represents a binary tree with no elements — >the empty tree.\n3. No order of data sorting is present in Binary tree.\n\nRepresentation\n\nIt is generally represented as:\n\n``` 10\n/ \\\n6 14\n/ \\ / \\\n5 8 11 18```\n\nThe code is provided in C++ along with the essential information of the functions and classes. So, we created a class Node comprising of :Data to be stored, Left subtree pointer, Right subtree pointer, Parameterised constructor(node) along with two functions:-\n1. Build tree\n2. Print (taking the root node as an argument)\n\n• buildtree() function\n\nThe buildtree() inputs the value of data in variable d and root node is locally created with the data as d. The condition checks if the tree is empty (if empty return NULL) or not. The recursive call of function buildtree() is made on both left and right subtree of the binary tree and then root is returned.\n\n• print() function-\n\nThe print() function accepting root node is used to print the entire binary tree. If the tree is empty, no tree is printed. Else, the data of the root node is printed first followed by the recursive call of print function on both left and right subtree. A binary tree is build up and printed in main function by calling both functions.\n\nCertainly the easiest code with optimized space and time complexity. This code is represented as Inorder traversal.\n\n### C++ Code – Inorder Traversal – Binary Tree\n\n```#include <iostream>\nusing namespace std;\nclass Node{\npublic:\nint data;\nNode*left;\nNode*right;\nNode(int d){\ndata=d;\nleft=NULL;\nright=NULL;\n}\n};\nNode* buildtree(){\nint d;\ncin>>d;\nNode*root;\nif(d==-1){\nreturn NULL;\n}\nroot=new Node(d);\nroot->left=buildtree();\nroot->right=buildtree();\nreturn root;\n}\nvoid print(Node*root){\nif(root==NULL){\nreturn;\n}\ncout<<root->data<<\" \";\nprint(root->left);\nprint(root->right);\n}\nint main()\n{\nNode*root=buildtree();\nprint(root);\nreturn 0;\n}```\n\nInput:\n\n`3 4 -1 -1 5 -1 -1`\n\nOutput:\n\n`3 4 5`\n\nYou may also learn,\n\n### 6 responses to “Build Binary Tree in C++ (Competitive Programming)”\n\n1. Shikha Srivastava says:\n\nFor competitive programming which is more preferred using set in STL or using binary tree?\n\n2. Dolly Singh says:\n\nWell it all depends on the constraints within which you have to provide the solution.You can code by using sets in STL in optimized manner or can use binary trees with optimized functions to reduce complexity.It all depends on your knowledge of optimizing function use.Feel free to comment for further queries.\n\n3. pi says:\n\nWhy is this note working for more than 7 nodes?\n\n4. ARPIT DAS says:\n\nThe code is correct BUT the correction here will be that this is not a INORDER traversal. The concept of PREORDER traversal is used here to print the Binary Tree.\n\n5. Dolly says:\n\nWell we can use any traversal to print the tree instead of any specific traversal the output will be same." ]
[ null ]
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https://forum.yoyogames.com/index.php?threads/need-to-attach-projectile-to-weapon-solved.60017/
[ "# Legacy GMNeed to attach projectile to weapon [SOLVED]\n\n#### pixeltroid\n\n##### Member\nI have a lighting bolt weapon that is used when the attack button is pressed. It emerges from my players weapon barrel and lasts for 2 seconds before disintegrating.\n\nThe problem is that if I turn away the lighting projectile just stays suspended in mid air -- as shown below -- which i think looks stupid", null, "Heres the code in my shoot button that activates the weapon script:\n\nCode:\n``````if (state == 'move' && global.flame > 0 && health > 0 && global.weapon = 2 && !alarm > 0)\n{\n{\nscr_playerlightning_state();\nimage_speed = .5;\nsprite_index = spr_playershoot2;\nstate = 'playerlightning';\n}\ncanlight = false\nalarm=30\n}``````\nHeres the code that controls how the weapon is used:\n\nCode:\n``````//create lightning\n\nif ducking = true {\nif (image_xscale > 0) {\nl=instance_create(x+16,y,obj_lightning);\nl.direction = 0;\nl.image_angle = l.direction\nl.speed = 0;\n}\nelse {\nl=instance_create(x-16,y,obj_lightning)\nl.direction = 180;\nl.speed = 0;\nl.image_angle = l.direction\n}\n}\n\nelse if (image_xscale > 0) {\nl=instance_create(x+16,y-9,obj_lightning);\nl.direction = 0;\nl.speed = 0;\nl.image_angle = l.direction\n} else if (image_xscale < 0){\nl=instance_create(x-16,y-9,obj_lightning)\nl.direction = 180;\nl.speed = 0;\nl.image_angle = l.direction\n}\n\nglobal.lightning -= 1;``````\nHow do I make it so that the lightning projectile is attached to the player at the co-ordinates specified in the code for as long as its supposed to last.\n\nAny help would be greatly appreciated!\n\nbump?\n\n#### Slyddar\n\n##### Member\nFirstly, you can simplify your calling code to just this:\nCode:\n``````if ducking var yy = 0 else yy = -9;\nif (image_xscale > 0) var _dir = 0 else var dir = 180;\n\nvar l = instance_create(x + 16*image_xscale, y+yy, obj_lightning);\nl.direction = _dir;\nl.image_angle = l.direction\nl.speed = 0;\n\n//store variables for following\nl.owner = id;\nl.x_offset = 16*image_xscale;\nl.y_offset = yy``````\nSet up the appropriate variables you need and then call them in a single instance create, where we also create a temp variable for l, unless you need it after this step.\n\nNow to ensure you can have it follow, you can store the id of the player making it, and the offset variables, xx and yy. Now in obj_lightning just update the step event to follow the owner at the offset values:\nCode:\n``````if instance_exists(owner) {\nx = owner.x + x_offset;\ny = owner.y + y_offset;\n}``````\nYou may need to create the variables for owner and offset in the obj_lightning object. Set owner = noone;\n\nI haven't tested all that, but it should get you started at least.\n\n#### pixeltroid\n\n##### Member\nFirstly, you can simplify your calling code to just this:\nCode:\n``````if ducking var yy = 0 else yy = -9;\nif (image_xscale > 0) var _dir = 0 else var dir = 180;\n\nvar l = instance_create(x + 16*image_xscale, y+yy, obj_lightning);\nl.direction = _dir;\nl.image_angle = l.direction\nl.speed = 0;\n\n//store variables for following\nl.owner = id;\nl.x_offset = 16*image_xscale;\nl.y_offset = yy``````\nSet up the appropriate variables you need and then call them in a single instance create, where we also create a temp variable for l, unless you need it after this step.\n\nNow to ensure you can have it follow, you can store the id of the player making it, and the offset variables, xx and yy. Now in obj_lightning just update the step event to follow the owner at the offset values:\nCode:\n``````if instance_exists(owner) {\nx = owner.x + x_offset;\ny = owner.y + y_offset;\n}``````\nYou may need to create the variables for owner and offset in the obj_lightning object. Set owner = noone;\n\nI haven't tested all that, but it should get you started at least.\nHi. Thanks.\n\nBut when I turn left and use the weapon I got this error:\n\nCode:\n``````FATAL ERROR in\naction number 3\nof Key Press Event for P-key Key\nfor object obj_player:\n\nlocal variable _dir(100002, -2147483648) not set before reading it.\nat gml_Script_scr_playerflame_state (line 5) - l.direction = _dir;\n############################################################################################\n--------------------------------------------------------------------------------------------\nstack frame is\ngml_Script_scr_playerlightning_state (line 5)\ncalled from - gml_Object_obj_player_KeyPressed_P_3 (line 91) - scr_playerlightning_state();``````\nWhats wrong? What do I do?\n\nLast edited:\n\n#### TheouAegis\n\n##### Member\nFirst off,\nCode:\n``````if instance_exists(owner) {\nx = owner.x + x_offset;\ny = owner.y + y_offset;\n}``````\nshould go in the End Step event, but you need to change it, because TheSly's code there is incomplete.\nCode:\n``````if instance_exists(owner) {\nimage_xscale = owner.image_xscale;\nx_offset = 16 * image_xscale;\nx = owner.x + x_offset;\ny = owner.y + y_offset;\n}``````\nIn his main block of code, change the first part to:\nCode:\n``````if ducking\nvar yy = 0\nelse\nvar yy = -9;\nif image_xscale > 0\nvar _dir = 0\nelse\nvar _dir = 180;``````\nHe had two typos there which were easy to miss.\n\n#### pixeltroid\n\n##### Member\n@ TheouAegis\n\nIt worked ....but theres one issue...\n\nthe projectile shoots in the wrong direction when I'm facing left.\n\nView attachment 23490\n\nI just removed\n\nCode:\n``l.direction = _dir;``", null, "What was the issue?\n\n#### Slyddar\n\n##### Member\nshould go in the End Step event, but you need to change it, because TheSly's code there is incomplete.\nThanks man, was running out the door and didn't have time to double check it all. Appreciate you fixing it up.\nthe projectile shoots in the wrong direction when I'm facing left.\nI just removed\nCode:\n``l.direction = _dir;``", null, "What was the issue?\nWhen TheouAegis correctly added this line it made the assignment of _dir obsolete, as you are no longer getting the direction from the original direction, but updating it each step from the players image_xscale.\nCode:\n``image_xscale = owner.image_xscale;``\n\n#### pixeltroid\n\n##### Member\nThanks man, was running out the door and didn't have time to double check it all. Appreciate you fixing it up.\n\nWhen TheouAegis correctly added this line it made the assignment of _dir obsolete, as you are no longer getting the direction from the original direction, but updating it each step from the players image_xscale.\nCode:\n``image_xscale = owner.image_xscale;``\n\nah ok! Thanks to you and TheouAegis !" ]
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https://segmentfault.com/a/1190000009724052
[ "# MySql表分区——partition\n\n#### 一、分区的优点\n\n• 和单个磁盘或文件系统分区相比，可以存储更多的数据。\n• 优化查询。\n\n• where 子句中包含分区条件时，可以只扫描必要的分区。\n• 涉及聚合函数的查询时，可以容易的在每个分区上并行处理，最终只需汇总得到结果。\n• 对于已经过期或者不需要保存的数据，可以通过删除与这些数据有关的分区来快速删除数据。\n• 跨多个磁盘来分散数据查询，以获得更大的查询吞吐量。\n\n#### 二、分区方法\n\n##### 1. KEY 分区，按照某个字段取余\n``````create table post (\nid int unsigned not null AUTO_INCREMENT,\ntitle varchar(255),\nPRIMARY KEY (id)\n) engine = innodb\npartition by key (id) partitions 5;``````\n##### 2. HASH 分区，基于给定的分区个数，把数据分配到不同的分区。\n``````create table student_hash(\nid int unsigned not null auto_increment,\nbirthday date,\nPRIMARY KEY(id,birthday);\n) engine=myisam\npartition by hash (month(birthday)) patitions 12;``````\n\nkey 和 hash 分区方法可以有效的分散热点数据。\n\n##### 3. RANGE 分区，基于一个给定连续区间范围，把数据分配到不同的分区\n``````create table goods (\nid int,\nuname char(10)\n)engine myisam\npartition by range(id) (\npartition p1 values less than (10),\npartition p2 values less than (20),\npartition p3 values less than MAXVALUE\n);\n``````\n##### 4.LIST 分区，类似 RANGE 分区，区别在 LIST 分区是基于枚举出的值列表分区，RANGE 是局域给定的连续区间范围分区。\n``````create table user (\nuid int,\npid int,\nuname\n)engine myisam\npartition by list(pid) (\npartition bj values in (1),\npartition ah values in (2),\npartition xb values in (4,5,6)\n);\n# 如果试图插入的列值不包含分区值列表中时，那么 insert 操作会失败并报错，要重点注意的是，list 分区不存在类似 values less than maxvalue 这样包含其他值在内的定义方式，将要匹配的任何值都必须在值列表中找得到。\n``````\n##### 5. Clumns 分区\n\nColumns 分区可以细分为 range columns 分区和 list columns 分区，这两种分区都支持整数，日期和字符串三大数据类型。\ncolumns 分区的另一个亮点是支持多列分区：\n\n``````mysql> create table rc3(\na int,\nb int\n)\npartition by range columns(a,b)(\npartition p01 values less than (0,10),\npartition p02 values less than (10,10),\npartition p03 values less than (10,20),\npartition p04 values less than (10,35),\npartition p05 values less than (10,maxvalue),\npartition p06 values less than (maxvalue,maxvalue)\n);``````\n##### 6. 子分区\n\n``````mysql> create table ts (id int, purchased date)\npartition by range(year(purchased))\nsubpartition by hash (to_days(purchased))\nsubpartitions 2\n(\npartition p0 values less than (1990),\npartition p1 values less than (2000),\npartition p2 values less than maxvalue,\n);``````\n\n#### 三、分区管理\n\n##### 2. 条件算法 list range\n\n``````alert table goods add partition(\npartition p4 values less than 40);``````\n\n``````alert table goods drop partition p1;\n\n#### 四、mysql 分区处理 null 值的方式\n\nmysql 不禁止在分区键值上使用 null ， 分区键可能是一个字段或者一个用户定义的额表达式。一般情况下，mysql 的分区把 null 当作零值，或者一个最小值进行处理。\nrange 分区中，null 值会被当作最小值来处理；\nlist 分区中，null 值必须出现在枚举列表中，否则不被接受；\nhash/key 分区中，null 值会被当作零值来处理。\n\n0 条评论", null, "" ]
[ null, "https://image-static.segmentfault.com/317/931/3179314346-5f61e47221e07", null ]
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http://mines.lumpylumpy.com/Electronics/Computers/Software/Cpp/Graphics/Vector/Points.php
[ " 3D Vertices (Points)", null, "3D Vertices (Points)", null, "", null, "Site Map", null, "Feedback", null, "", null, "", null, "Ode to a Node:\nWhat's the Point?", null, "", null, "", null, "gCoord", null, "Points", null, "Vectors", null, "PV", null, "Planes", null, "", null, "## g3Point and g2Point\n\ng3Point and g2Point are for holding vertices and have public cartesian coordinates, Spherical to Cartesian conversion, distance between points and OnViewPlane which is valuable when rendering.\nThe operator bool() in g3D and g2D makes it easy to compare the Vertex with the Origin.\n``` bool IsOrigin() const;\nvoid Move(double dx, double dy, double dz);\ng3Point OnViewPlane(g3Vector& Axis) const; // Returns the Projected point to a plane through the Global Origin having the given Normal Axis.\nstatic g3Point FromSpherical(double Radius, double Theta, double Phi);\nvoid ToSpherical(double& Radius, double& Theta, double& Phi) const;\ndouble GetTheta () const;\ndouble GetPhi () const;\ndouble DistanceTo(const g3Point& E) const; // For signed Distances use g3PV::GetDisplacement();\n```\n\n## g3D and g2D Base Classes\n\nThere are lots of studies about the use of Vector and Vertex classes in graphics libraries.\nMost agree that there is only a need for a Vector class and a Vertex can live in a Vector class.\nA base class is used to hold the coordinates, and extended to create separate Vector and Point (Vertex) objects.\ng2D and g3D are structs which simply hold two or three gCoords.\nThe default constructor doesn't initialise the gCoords. This behaviour is used throughout the Primitives because huge arrays of points are often used and don't need to be zeroed.\n```struct g2D { // 2 Doubles (Used as a 2 Dimensional Point or Vector)\ngCoord x,y;\ng2D() {}\ng2D(gCoord x, gCoord y) : x(x), y(y) {}\ng2D(double* DD) {x=*DD++; y=*DD;}\nvoid Clear() {x=y=0;}\ng2D& operator*=(const gCoord& c) {x*=c; y*=c; return *this;}\ng2D& operator/=(const gCoord& c) {x/=c; y/=c; return *this;}\nbool operator==(const g2D& DD) const {return Compare(DD)==0;}\nbool operator!=(const g2D& DD) const {return Compare(DD)!=0;}\noperator bool() const {return bool(x)||bool(y);}\nint Compare(const g2D& DD) const { // For Sorting\nint Result = x.Compare(DD.x);\nreturn Result ? Result : y.Compare(DD.y);\n}\n};\n```\nTo create a g3D, d2D could have been used as its base class, but there are no advantages.\nEvery function would need redefining anyway, and, for example:\n```struct g3D : g2D { // 3 Doubles (Used as x 3 Dimensional Point or Vector)\ngCoord z;\ng3D() {}\ng3D(gCoord x, gCoord y, gCoord z) : g2D(x,y), z(z) {}\ng3D(double* DDD) : g2D(DDD) {z=*++DDD;}\nvoid Clear() {g2D::Clear(); z=0;}\ng3D& operator*=(const gCoord& c) {g2D::operator*=(c); z*=c; return *this;}\n```\nthe Clear() function is more clearly written as x=y=z=0; as are most of the others.\nSo 3D objects are not derived from 2D objects.\nSince there is now a danger in the redundancy that a change in one class is not also changed in the other,\n\nNote the ToArray(double*) method which copies the data out to an array.", null, "THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS 'AS IS' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.", null, "" ]
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https://www.mathworks.com/help/finance/acrudisc.html
[ "Documentation\n\n# acrudisc\n\nAccrued interest of discount security paying at maturity\n\n## Syntax\n\n``AccruInterest = acrudisc(Settle,Maturity,Face,Discount)``\n``AccruInterest = acrudisc(___,Period,Basis)``\n\n## Description\n\nexample\n\n````AccruInterest = acrudisc(Settle,Maturity,Face,Discount)` returns the accrued interest of a discount security paid at maturity.```\n````AccruInterest = acrudisc(___,Period,Basis)` adds optional arguments for `Period` and `Basis`. ```\n\n## Examples\n\ncollapse all\n\nThis example shows how to find the accrued interest of a discount security paid at maturity.\n\n```AccruInterest = acrudisc('05/01/1992', '07/15/1992', ... 100, 0.1, 2, 0)```\n```AccruInterest = 2.0604 ```\n\nThis example shows how to use `datetime` inputs to find the accrued interest of a discount security paid at maturity.\n\n```AccruInterest = acrudisc(datetime('1-May-1992','Locale','en_US'),datetime('15-Jul-1992','Locale','en_US'),... 100, 0.1, 2, 0)```\n```AccruInterest = 2.0604 ```\n\n## Input Arguments\n\ncollapse all\n\nSettlement date of the security, specified as a scalar or a `NINST`-by-`1` vector of serial date numbers, date character vectors, or datetime arrays. The `Settle` date must be before the `Maturity` date.\n\nData Types: `double` | `char` | `datetime`\n\nMaturity date of the security, specified as a scalar or a `NINST`-by-`1` vector of serial date numbers, date character vectors, or datetime arrays.\n\nData Types: `double` | `char` | `datetime`\n\nRedemption value (par value) of the security, specified as a scalar or a `NINST`-by-`1` vector.\n\nData Types: `double`\n\nDiscount rate of the security, specified as a scalar or a `NINST`-by-`1` vector of decimal fraction values.\n\nData Types: `double`\n\n(Optional) Number of coupon payments per year for security, specified as scalar or a `NINST`-by-`1` vector using the values: `0`, `1`, `2`, `3`, `4`, `6`, or `12`.\n\nData Types: `double`\n\n(Optional) Day-count basis for security, specified as a scalar or a `NINST`-by-`1` vector. Values are:\n\n• 0 = actual/actual\n\n• 1 = 30/360 (SIA)\n\n• 2 = actual/360\n\n• 3 = actual/365\n\n• 4 = 30/360 (PSA)\n\n• 5 = 30/360 (ISDA)\n\n• 6 = 30/360 (European)\n\n• 7 = actual/365 (Japanese)\n\n• 8 = actual/actual (ICMA)\n\n• 9 = actual/360 (ICMA)\n\n• 10 = actual/365 (ICMA)\n\n• 11 = 30/360E (ICMA)\n\n• 12 = actual/365 (ISDA)\n\n• 13 = BUS/252\n\nFor more information, see basis.\n\nData Types: `double`\n\n## Output Arguments\n\ncollapse all\n\nAccrued interest, returned as a scalar or a `NINST`-by-`1` vector.\n\n Mayle, J. Standard Securities Calculation Methods. Volumes I-II, 3rd edition. Formula D.\n\nDownload ebook" ]
[ null ]
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https://www.src.isr.umich.edu/publications/maximum-likelihood-estimation-for-mixed-continuous-and-categorical-data-with-missing-values/
[ "# Maximum Likelihood Estimation for Mixed Continuous and Categorical Data with Missing Values\n\nMaximum likelihood procedures for analysing mixed continuous and categorical data with missing values are presented. The general location model of Olkin \\${ t&}\\$ Tate (1961) and extensions introduced by Krzanowski (1980, 1982) form the basis for our methods. Maximum likelihood estimation with incomplete data is achieved by an application of the EM algorithm (Dempster, Laird \\${ t&}\\$ Rubin, 1977). Special cases of the algorithm include Orchard \\${ t&}\\$ Woodbury's (1972) algorithm for incomplete normal samples, Fuchs's (1982) algorithms for log linear modelling of partially classified contingency tables, and Day's (1969) algorithm for multivariate normal mixtures. Applications include: (a) imputation of missing values, (b) logistic regression and discriminant analysis with missing predictors and unclassified observations, (c) linear regression with missing continuous and categorical predictors, and (d) parametric cluster analysis with incomplete data. Methods are illustrated using data from the St Louis Risk Research Project." ]
[ null ]
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http://export.arxiv.org/list/math.SP/1905?skip=0&show=50
[ "# Spectral Theory\n\n## Authors and titles for math.SP in May 2019\n\n[ total of 40 entries: 1-40 ]\n[ showing up to 50 entries per page: fewer | more ]\n\nTitle: On the explicit representation of the trace space $H^{\\frac{3}{2}}$ and of the solutions to biharmonic Dirichlet problems on Lipschitz domains via multi-parameter Steklov problems\nComments: Please, note that the list of references has been updated. If possible, replace the previous version of the article with this new one\nSubjects: Spectral Theory (math.SP); Analysis of PDEs (math.AP); Functional Analysis (math.FA)\n\nTitle: A new characterization of generalized Browder's theorem and a Cline's formula for generalized Drazin-meromorphic inverses\nSubjects: Spectral Theory (math.SP)\n\nTitle: Sharp diameter bound on the spectral gap for quantum graphs\nComments: 11 pages, 8 figures\nSubjects: Spectral Theory (math.SP)\n\nTitle: Efficiency and localisation for the first Dirichlet eigenfunction\nSubjects: Spectral Theory (math.SP); Analysis of PDEs (math.AP)\n\nTitle: The Robin Laplacian - spectral conjectures, rectangular theorems\nComments: 44 pages, 7 figures\nSubjects: Spectral Theory (math.SP); Analysis of PDEs (math.AP)\n\nTitle: Quantum ergodicity for pseudo-Laplacians\nAuthors: Elie Studnia\nSubjects: Spectral Theory (math.SP); Analysis of PDEs (math.AP)\n\nTitle: On the trace formula for higher-order ODO\nSubjects: Spectral Theory (math.SP)\n\nTitle: Resistance distance-based graph invariants and the number of spanning trees of linear crossed octagonal graphs\nComments: In this paper, we firstly determine formulas for the Laplacian (normalized Laplacian, resp.) spectrum of On. Further, the formulas for those two resistance distance-based graph invariants and spanning trees are given. More surprising, we find that the Kirchhoff (multiplicative degree-Kirchhoff, resp.) index is almost one quarter to Wiener (Gutman, resp.) index of a linear crossed octagonal graph\nSubjects: Spectral Theory (math.SP); Combinatorics (math.CO)\n\nTitle: General Toeplitz matrices subject to Gaussian perturbations\nComments: 23 pages, 2 figures\nSubjects: Spectral Theory (math.SP); Probability (math.PR)\n\nTitle: On coupling constant thresholds in one dimension\nAuthors: Yuriy Golovaty\nSubjects: Spectral Theory (math.SP); Mathematical Physics (math-ph)\n\nTitle: Unimodal value distribution of Laplace eigenfunctions and a monotonicity formula\nAuthors: Bo'az Klartag\nSubjects: Spectral Theory (math.SP); Differential Geometry (math.DG)\n\nTitle: Jordan chains of elliptic partial differential operators and Dirichlet-to-Neumann maps\nSubjects: Spectral Theory (math.SP); Analysis of PDEs (math.AP)\n\nTitle: On the stability of self-adjointness of linear relations\nAuthors: Yan Liu\nSubjects: Spectral Theory (math.SP)\n  arXiv:1905.00259 (cross-list from math.AP) [pdf, other]\nTitle: The heat kernel on curvilinear polygonal domains in surfaces\nSubjects: Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.00681 (cross-list from math.NT) [pdf, other]\nTitle: Spectral exponential sums on hyperbolic surfaces I\nAuthors: Ikuya Kaneko\nComments: 20 pages, 3 figures, comments welcome\nSubjects: Number Theory (math.NT); Spectral Theory (math.SP)\n  arXiv:1905.00890 (cross-list from math.DS) [pdf, other]\nTitle: Marked Length Spectral determination of analytic chaotic billiards with axial symmetries\nComments: 57 pages, 8 figures\nSubjects: Dynamical Systems (math.DS); Spectral Theory (math.SP)\n  arXiv:1905.01691 (cross-list from math.PR) [pdf, other]\nTitle: Spectral analysis of the zigzag process\nComments: 39 pages, 1 figure\nSubjects: Probability (math.PR); Spectral Theory (math.SP)\n  arXiv:1905.02047 (cross-list from math.CO) [pdf, other]\nTitle: On the notion of effective impedance\nAuthors: Anna Muranova\nComments: 17 pages, 6 figures\nSubjects: Combinatorics (math.CO); Mathematical Physics (math-ph); Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.03027 (cross-list from math.DG) [pdf, ps, other]\nTitle: Geometric quantization of Hamiltonian flows and the Gutzwiller trace formula\nAuthors: Louis Ioos\nSubjects: Differential Geometry (math.DG); Mathematical Physics (math-ph); Symplectic Geometry (math.SG); Spectral Theory (math.SP)\n  arXiv:1905.03174 (cross-list from math.DG) [pdf, other]\nTitle: Index of minimal spheres and isoperimetric eigenvalue inequalities\nComments: 35 pages, minor changes, the geometric meaning of twistor fields is clarified\nSubjects: Differential Geometry (math.DG); Algebraic Geometry (math.AG); Spectral Theory (math.SP)\n  arXiv:1905.05136 (cross-list from math.AP) [pdf, ps, other]\nTitle: A logarithmic improvement in the two-point Weyl law for manifolds without conjugate points\nAuthors: Blake Keeler\nSubjects: Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.05936 (cross-list from math.FA) [pdf, ps, other]\nTitle: Decomposable operators, local S-spectrum and S-spectrum in the quaternionic setting\nComments: 28 pages. arXiv admin note: text overlap with arXiv:1904.02977\nSubjects: Functional Analysis (math.FA); Spectral Theory (math.SP)\n  arXiv:1905.06136 (cross-list from math.AP) [pdf, ps, other]\nTitle: Conformal invariants from nodal sets II. Manifolds with boundary\nSubjects: Analysis of PDEs (math.AP); Differential Geometry (math.DG); Spectral Theory (math.SP)\n  arXiv:1905.06152 (cross-list from math.AP) [pdf, ps, other]\nTitle: Thin domain limit and counterexamples to strong diamagnetism\nSubjects: Analysis of PDEs (math.AP); Mathematical Physics (math-ph); Spectral Theory (math.SP)\n  arXiv:1905.07042 (cross-list from math.PR) [pdf, ps, other]\nTitle: A spectral theoretical approach for hypocoercivity applied to some degenerate hypoelliptic, and non-local operators\nSubjects: Probability (math.PR); Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.07156 (cross-list from math-ph) [pdf, ps, other]\nTitle: On Schrödinger and Dirac Operators with an Oscillating Potential\nAuthors: Thierry Jecko (AGM)\nSubjects: Mathematical Physics (math-ph); Spectral Theory (math.SP)\n  arXiv:1905.07608 (cross-list from math-ph) [pdf, ps, other]\nTitle: 3D Schrödinger equation: scattering operator, scattering amplitude and ergodic property\nAuthors: Lev Sakhnovich\nComments: This paper is a further development of our results in arXiv:1809.03998 and arXiv:1801.05370\nSubjects: Mathematical Physics (math-ph); Analysis of PDEs (math.AP); Dynamical Systems (math.DS); Spectral Theory (math.SP); Quantum Physics (quant-ph)\n  arXiv:1905.07832 (cross-list from math.PR) [pdf, ps, other]\nTitle: On non-local ergodic Jacobi semigroups: spectral theory, convergence-to-equilibrium and contractivity\nSubjects: Probability (math.PR); Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.07952 (cross-list from math.FA) [pdf, ps, other]\nTitle: A Riesz basis criterion for Schrödinger operators with boundary conditions dependent on the eigenvalue parameter\nAuthors: Namig J. Guliyev\nComments: 7 pages, no figures, submitted\nSubjects: Functional Analysis (math.FA); Mathematical Physics (math-ph); Classical Analysis and ODEs (math.CA); Spectral Theory (math.SP)\n  arXiv:1905.08354 (cross-list from math.CA) [pdf, other]\nTitle: Discrete Prolate Spheroidal Wave Functions: Further spectral analysis and some related applications\nSubjects: Classical Analysis and ODEs (math.CA); Spectral Theory (math.SP)\n  arXiv:1905.08712 (cross-list from math.AP) [pdf, ps, other]\nTitle: Two-sided weighted bounds on fundamental solution to fractional Schrödinger operator\nComments: 9 p, few typos fixed, improved presentation. arXiv admin note: text overlap with arXiv:1904.07363\nSubjects: Analysis of PDEs (math.AP); Probability (math.PR); Spectral Theory (math.SP)\n  arXiv:1905.09563 (cross-list from math.AP) [pdf, ps, other]\nTitle: Optimization results for the higher eigenvalues of the $p$-Laplacian associated with sign-changing capacitary measures\nSubjects: Analysis of PDEs (math.AP); Optimization and Control (math.OC); Spectral Theory (math.SP)\n  arXiv:1905.10244 (cross-list from math.PR) [pdf, other]\nTitle: Outliers of random perturbations of Toeplitz matrices with finite symbols\nSubjects: Probability (math.PR); Spectral Theory (math.SP)\n  arXiv:1905.11091 (cross-list from math.AP) [pdf, ps, other]\nTitle: Scattering in periodic waveguide: integral representation and spectrum decomposition\nAuthors: Ruming Zhang\nSubjects: Analysis of PDEs (math.AP); Spectral Theory (math.SP)\n  arXiv:1905.11338 (cross-list from math.CA) [pdf, other]\nTitle: Estimates for the SVD of the truncated Fourier transform on L2(cosh(b$\\times$)) and stable analytic continuation\nAuthors: Christophe Gaillac (TSE, CREST), Eric Gautier (TSE, UT1)\nSubjects: Classical Analysis and ODEs (math.CA); Numerical Analysis (math.NA); Spectral Theory (math.SP)\n  arXiv:1905.11454 (cross-list from math.DG) [pdf, ps, other]\nTitle: Geometric structures and the Laplace spectrum\nSubjects: Differential Geometry (math.DG); Spectral Theory (math.SP)\n  arXiv:1905.11980 (cross-list from math.MG) [pdf, ps, other]\nTitle: The sharp $p$-Poincaré inequality under the measure contraction property\nAuthors: Bang-Xian Han\nComments: This is a subsequent work of arxiv:1905.05465\nSubjects: Metric Geometry (math.MG); Functional Analysis (math.FA); Spectral Theory (math.SP)\n  arXiv:1905.12303 (cross-list from math-ph) [pdf, ps, other]\nTitle: Semiclassical behaviour of quantum eigenstates\nAuthors: Gabriel Rivière (LMJL)\nComments: Lecture notes for the Spring School ''From quantum to classical'' (CIRM, April 2019)\nSubjects: Mathematical Physics (math-ph); Analysis of PDEs (math.AP); Spectral Theory (math.SP); Quantum Physics (quant-ph)\n  arXiv:1905.09550 (cross-list from stat.ML) [pdf, other]\nTitle: Revisiting Graph Neural Networks: All We Have is Low-Pass Filters\nComments: 12 pages, 5 figures, 2 tables\nSubjects: Machine Learning (stat.ML); Information Theory (cs.IT); Machine Learning (cs.LG); Spectral Theory (math.SP)\n  arXiv:1905.13000 (cross-list from cs.LG) [pdf, other]\nTitle: Implicit Regularization of Accelerated Methods in Hilbert Spaces\nSubjects: Machine Learning (cs.LG); Optimization and Control (math.OC); Spectral Theory (math.SP); Machine Learning (stat.ML)\n[ total of 40 entries: 1-40 ]\n[ showing up to 50 entries per page: fewer | more ]\n\nDisable MathJax (What is MathJax?)\n\nLinks to: arXiv, form interface, find, math, 1909, contact, help  (Access key information)" ]
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https://api-project-1022638073839.appspot.com/questions/561c450b11ef6b52def24e48
[ "# Question #24e48\n\nOct 12, 2015\n\nYour equation has an infinite number of solutions.\n\n#### Explanation:\n\nYour starting expression looks like this\n\n$6 \\cdot \\left(2 x + 4\\right) = 10 x + 24 + 2 x$\n\nNotice that your equation contains two types of terms\n\n• terms that contain $x$\n• terms that do not contain $x$\n\nYour goal here is to get all the tems that contain $x$ on one side of the equation, and all the terms that do not contain $x$ on the othe side of the equation.\n\nSo, start by expanding the paranthesis by multiplying both terms by $6$\n\n$6 \\cdot \\left(2 x + 4\\right) = 6 \\cdot 2 x + 6 \\cdot 4 = 12 x + 24$\n\nThe equation now looks like this\n\n$12 x + 24 = 10 x + 24 + 2 x$\n\nNotice that you can group two terms that contain $x$ on the right-hand side of the equation to get\n\n$12 x + 24 = {\\underbrace{12 x}}_{\\textcolor{b l u e}{10 x + 2 x}} + 24$\n\nNotice that we are left with the same expression on both sides of the equation\n\n$12 x + 24 = 12 x + 24$\n\nIn this case, you would say that the equation has an infinite number of solutions because you can plug in any value of $x$ and the equation will be true!\n\nMore specifically, this is reduced to\n\n$\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{12 x}}} - \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{12 x}}} = \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{24}}} - \\textcolor{red}{\\cancel{\\textcolor{b l a c k}{24}}}$\n\n$0 = 0$\n\nThis is true regardless of the value of $x$, which is why the equation is said to have an infinite number of solutions." ]
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https://www.meritnation.com/cbse-class-12-science/math/rd-sharma-xii-vol-2-2017/straight-line-in-space/textbook-solutions/46_1_2153_10255_28.9_93838
[ "Rd Sharma XII Vol 2 2017 Solutions for Class 12 Science Math Chapter 9 Straight Line In Space are provided here with simple step-by-step explanations. These solutions for Straight Line In Space are extremely popular among Class 12 Science students for Math Straight Line In Space Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2017 Book of Class 12 Science Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2017 Solutions. All Rd Sharma XII Vol 2 2017 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.\n\n#### Question 8:\n\nFind the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are $\\frac{x-3}{2}=\\frac{y+1}{7}=\\frac{z-2}{-3}.$\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nVector equation of the required line is\n\n#### Question 9:\n\nThe cartesian equations of a line are $\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}.$ Find a vector equation for the line.\n\nThe cartesian equation of the given line is $\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}$.\nIt can be re-written as\n\n$\\frac{x-5}{3}=\\frac{y-\\left(-4\\right)}{7}=\\frac{z-6}{2}$\n\nThus, the given line passes through the point having position vector $\\stackrel{\\to }{a}=5\\stackrel{^}{i}-4\\stackrel{^}{j}+6\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=3\\stackrel{^}{i}+7\\stackrel{^}{j}+2\\stackrel{^}{k}$.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nVector equation of the required line is\n\n#### Question 10:\n\nFind the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are $\\frac{x-3}{1}=\\frac{y-1}{2}=\\frac{z+1}{-2}$. Also, reduce the equation obtained in vector form.\n\nWe know that the cartesian equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{m}$ is $\\frac{x-{x}_{1}}{a}=\\frac{y-{y}_{2}}{b}=\\frac{z-{z}_{3}}{c}$.\n\nHere,\n\nHere,\n\nCartesian equation of the required line is\n\n$\\frac{x-1}{1}=\\frac{y-\\left(-1\\right)}{2}=\\frac{z-2}{-2}\\phantom{\\rule{0ex}{0ex}}⇒\\frac{x-1}{1}=\\frac{y+1}{2}=\\frac{z-2}{-2}\\phantom{\\rule{0ex}{0ex}}$\n\nWe know that the cartesian equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{m}$ is $\\stackrel{\\to }{r}=\\stackrel{\\to }{a}+\\lambda \\stackrel{\\to }{m}$.\nHere, the line is passing through the point and its direction ratios are proportional to 1, 2, $-$2.\n\nVector equation of the required line is\n$\\stackrel{\\to }{r}=\\left(\\stackrel{^}{i}-\\stackrel{^}{j}+2\\stackrel{^}{k}\\right)+\\lambda \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}-2\\stackrel{^}{k}\\right)$\n\n#### Question 11:\n\nFind the direction cosines of the line $\\frac{4-x}{2}=\\frac{y}{6}=\\frac{1-z}{3}.$ Also, reduce it to vector form.\n\nThe cartesian equation of the given line is\n\n$\\frac{4-x}{2}=\\frac{y}{6}=\\frac{1-z}{3}$\n\nIt can be re-written as\n\n$\\frac{x-4}{-2}=\\frac{y-0}{6}=\\frac{z-1}{-3}$\n\nThis shows that the given line passes through the point and its direction ratios are proportional to .\n\nSo, its direction cosines are\n\nThus, the given line passes through the point having position vector $\\stackrel{\\to }{a}=4\\stackrel{^}{i}+\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=-2\\stackrel{^}{i}+6\\stackrel{^}{j}-3\\stackrel{^}{k}$.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nVector equation of the required line is\n\n#### Question 12:\n\nThe cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.\n\nThe cartesian equation of the given line is\n\nIt can be re-written as\n\n$\\frac{x-b}{a}=\\frac{y-0}{1}=\\frac{z-d}{c}$\n\nThus, the given line passes through the point  and its direction ratios are proportional to a, 1, c. It is also parallel to the vector $\\stackrel{\\to }{b}=a\\stackrel{^}{i}+\\stackrel{^}{j}+c\\stackrel{^}{k}$.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nVector equation of the required line is\n\n#### Question 13:\n\nFind the vector equation of a line passing through the point with position vector $\\stackrel{^}{i}-2\\stackrel{^}{j}-3\\stackrel{^}{k}$and parallel to the line joining the points with position vectors Also, find the cartesian equivalent of this equation.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nVector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 14:\n\nFind the points on the line $\\frac{x+2}{3}=\\frac{y+1}{2}=\\frac{z-3}{2}$ at a distance of 5 units from the point P (1, 3, 3).\n\nThe coordinates of any point on the line $\\frac{x+2}{3}=\\frac{y+1}{2}=\\frac{z-3}{2}$ are given by\n\nLet the coordinates of the desired point be .\n\nThe distance between this point and (1, 3, 3) is 5 units.\n\nSubstituting the values of $\\lambda$ in (1), we get the coordinates of the desired point as ($-$2, $-$1, 3) and (4, 3, 7).\n\n#### Question 15:\n\nShow that the points whose position vectors are are collinear.\n\nLet the given points be P, Q and R and let their position vectors be\n\nVector equation of line passing through P and Q is\n\nIf points P, Q and R are collinear, then point R must satisfy (1).\n\n$7\\stackrel{^}{i}+9\\stackrel{^}{k}=\\left(-2\\stackrel{^}{i}+3\\stackrel{^}{j}\\right)+\\lambda \\left(3\\stackrel{^}{i}-\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)\\phantom{\\rule{0ex}{0ex}}$\n\nComparing the coefficients of , we get\n\n$\\lambda$=3\n\nThese three equations are consistent, i.e. they give the same value of $\\lambda$.\nHence, the given three points are collinear.\n\nDisclaimer: The question given in the book has a minor error. The third vectors should be $7\\stackrel{^}{i}+9\\stackrel{^}{k}$. The solution here is created accordingly.\n\n#### Question 16:\n\nFind the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line $\\frac{-x-2}{1}=\\frac{y+3}{7}=\\frac{2z-6}{3}.$\n\nWe have\n\n$\\frac{-x-2}{1}=\\frac{y+3}{7}=\\frac{2z-6}{3}$\n\nIt can be re-written as\n\n$\\frac{x+2}{-1}=\\frac{y+3}{7}=\\frac{z-3}{\\left(\\frac{3}{2}\\right)}\\phantom{\\rule{0ex}{0ex}}=\\frac{x+2}{-2}=\\frac{y+3}{14}=\\frac{z-3}{3}$\n\nThis shows that the given line passes through the point and its direction ratios are proportional to $-$2, 14, 3.\n\nThus, the parallel vector is $\\stackrel{\\to }{b}=-2\\stackrel{^}{i}+14\\stackrel{^}{j}+3\\stackrel{^}{k}$.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n.\n\nVector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 17:\n\nThe cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.\n\nThe cartesian equation of the given line is\n\n3x + 1 = 6y − 2 = 1 − z\n\nIt can be re-written as\n\n$\\frac{x+\\frac{1}{3}}{\\frac{1}{3}}=\\frac{y-\\frac{1}{3}}{\\frac{1}{6}}=\\frac{z-1}{-1}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-\\left(-\\frac{1}{3}\\right)}{2}=\\frac{y-\\frac{1}{3}}{1}=\\frac{z-1}{-6}$\n\nThus, the given line passes through the point $\\left(-\\frac{1}{3},\\frac{1}{3},1\\right)$ and its direction ratios are proportional to 2, 1, $-$6. It is parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+\\stackrel{^}{j}-6\\stackrel{^}{k}$.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nVector equation of the required line is\n\n#### Question 1:\n\nShow that the three lines with direction cosines are mutually perpendicular.\n\nThe direction cosines of the three lines are\n\nHence, the given lines are perpendicular to each other.\n\n#### Question 2:\n\nShow that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).\n\nSuppose vector $\\stackrel{\\to }{a}$ is passing through the points (1, $-$1, 2) and (3, 4,$-$2) and $\\stackrel{\\to }{b}$ is passing through the points (0, 3, 2) and (3, 5, 6).\n\nThen,\n\nNow,\n$\\stackrel{\\to }{a}.\\stackrel{\\to }{b}=\\left(2\\stackrel{^}{i}+5\\stackrel{^}{j}-4\\stackrel{⏜}{k}\\right).\\left(3\\stackrel{^}{i}+2\\stackrel{^}{j}+4\\stackrel{⏜}{k}\\right)=0$\n\nHence, the given lines are perpendicular to each other.\n\n#### Question 3:\n\nShow that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).\n\nEquations of lines passing through the points are given by\n\n$\\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}$\n\nSo, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is\n\n$\\frac{x-4}{2-4}=\\frac{y-7}{3-7}=\\frac{z-8}{4-8}\\phantom{\\rule{0ex}{0ex}}⇒\\frac{x-4}{-2}=\\frac{y-7}{-4}=\\frac{z-8}{-4}$\n\nAlso, the equation of the line passing through the points ($-$1, $-$2, 1) and (1, 2, 5) is\n\n$\\frac{x+1}{1+1}=\\frac{y+2}{2+2}=\\frac{z-1}{5-1}\\phantom{\\rule{0ex}{0ex}}⇒\\frac{x+1}{2}=\\frac{y+2}{4}=\\frac{z-1}{4}$\n\nWe know that two lines are parallel if\n\nWe observe\n\n$\\frac{-2}{2}=\\frac{-4}{4}=\\frac{-4}{4}=-1\\phantom{\\rule{0ex}{0ex}}$\n\nHence, the given lines are parallel to each other.\n\n#### Question 4:\n\nFind the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by $\\frac{x+3}{3}=\\frac{y-4}{5}=\\frac{z+8}{6}.$\n\nWe know that the cartesian equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is $\\frac{x-{x}_{1}}{a}=\\frac{y-{y}_{2}}{b}=\\frac{z-{z}_{3}}{c}$.\n\nHere,\n\nThe cartesian equation of the required line is\n\n$\\frac{x-\\left(-2\\right)}{3}=\\frac{y-4}{5}=\\frac{z-\\left(-5\\right)}{6}\\phantom{\\rule{0ex}{0ex}}=\\frac{x+2}{3}=\\frac{y-4}{5}=\\frac{z+5}{6}\\phantom{\\rule{0ex}{0ex}}$\n\n#### Question 5:\n\nShow that the lines are perpendicular to each other.\n\nWe have\n\nThese equations can be re-written as\n\nNow,\n\nHence, the given two lines are perpendicular to each other.\n\n#### Question 6:\n\nShow that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).\n\nThe direction ratios of the line joining the origin to the point (2, 1, 1) are 2, 1, 1.\n\nLet $\\stackrel{\\to }{{b}_{1}}=2\\stackrel{^}{i}+\\stackrel{^}{j}+\\stackrel{^}{k}$\n\nThe direction ratios of the line joining the points (3, 5, $-$1) and (4, 3, $-$1) are 1, $-$2, 0.\n\nLet $\\stackrel{\\to }{{b}_{2}}=\\stackrel{^}{i}-2\\stackrel{^}{j}+0\\stackrel{^}{k}$\n\nNow,\n\nHence, the two lines joining the given points are perpendicular to each other.\n\n#### Question 7:\n\nFind the equation of a line parallel to x-axis and passing through the origin.\n\nThe direction ratios of the line parallel to x-axis are proportional to 1, 0, 0.\n\nEquation of the line passing through the origin and parallel to x-axis is\n\n$\\frac{x-0}{1}=\\frac{y-0}{0}=\\frac{z-0}{0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{1}=\\frac{y}{0}=\\frac{z}{0}$\n\n#### Question 8:\n\nFind the angle between the following pairs of lines:\n(i)\n\n(ii)\n\n(iii)\n\n(i)\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n(ii)\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n(iii)\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n#### Question 9:\n\nFind the angle between the following pairs of lines:\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(v)\n\n(vi)\n\n(i)\n\nLet be vectors parallel to the given lines.\n\nIf $\\theta$ is the angle between the given lines, then\n\n(ii)\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n(iii)\n\nThe equations of the given lines can be re-written as\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n(iv)\n\nThe equations of the given lines can be re-written as\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n(v )\n\nThe equations of the given lines can be re-written as\n\nLet be vectors parallel to the given lines.\n\nNow,\n$\\stackrel{\\to }{{b}_{1}}=\\stackrel{^}{i}-\\stackrel{^}{j}+\\stackrel{^}{k}\\phantom{\\rule{0ex}{0ex}}\\stackrel{\\to }{{b}_{2}}=3\\stackrel{^}{i}+4\\stackrel{^}{j}+5\\stackrel{^}{k}$\n\nIf $\\theta$ is the angle between the given lines, then\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n(vi)\n\nThe equations of the given lines can be re-written as\n\nLet be vectors parallel to the given lines.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n#### Question 10:\n\nFind the angle between the pairs of lines with direction ratios proportional to\n(i) 5, −12, 13 and −3, 4, 5\n(ii) 2, 2, 1 and 4, 1, 8\n(iii) 1, 2, −2 and −2, 2, 1\n(iv) a, b, c and bc, ca, ab.\n\n(i) 5, −12, 13 and −3, 4, 5\n\n(ii) 2, 2, 1 and 4, 1, 8\n\n(iii) 1, 2, −2 and −2, 2, 1\n\n(iv) a, b, c and bc, ca, ab.\n\n#### Question 11:\n\nFind the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).\n\nThe direction ratios of the line joining the points (3, 1, 4) and (7, 2, 12) are proportional to 4, 1, 8.\n\nLet be vectors parallel to the lines having direction ratios proportional to 2, 2, 1 and 4, 1, 8.\n\nNow,\n\nIf $\\theta$ is the angle between the given lines, then\n\n#### Question 12:\n\nFind the equation of the line passing through the point (1, 2, −4) and parallel to the line $\\frac{x-3}{4}=\\frac{y-5}{2}=\\frac{z+1}{3}.$\n\nThe direction ratios of the line parallel to line $\\frac{x-3}{4}=\\frac{y-5}{2}=\\frac{z+1}{3}$ are proportional to 4, 2, 3.\n\nEquation of the required line passing through the point (1, 2, $-$4) having direction ratios proportional to 4, 2, 3 is\n\n$\\frac{x-1}{4}=\\frac{y-2}{2}=\\frac{z-\\left(-4\\right)}{3}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-1}{4}=\\frac{y-2}{2}=\\frac{z+4}{3}$\n\n#### Question 13:\n\nFind the equations of the line passing through the point (−1, 2, 1) and parallel to the line $\\frac{2x-1}{4}=\\frac{3y+5}{2}=\\frac{2-z}{3}.$\n\nThe equation of line $\\frac{2x-1}{4}=\\frac{3y+5}{2}=\\frac{2-z}{3}$ can be re-written as\n\n$\\frac{x-\\frac{1}{2}}{2}=\\frac{y+\\frac{5}{3}}{\\frac{2}{3}}=\\frac{z-2}{-3}$\n\nThe direction ratios of the line parallel to line $\\frac{2x-1}{4}=\\frac{3y+5}{2}=\\frac{2-z}{3}$ are proportional to 2, $\\frac{2}{3}$, $-$3.\n\nEquation of the required line passing through the point ($-$1, 2, 1) having direction ratios proportional to 2, $\\frac{2}{3}$, $-$3 is\n\n$\\frac{x-\\left(-1\\right)}{2}=\\frac{y-2}{\\frac{2}{3}}=\\frac{z-1}{-3}\\phantom{\\rule{0ex}{0ex}}=\\frac{x+1}{2}=\\frac{y-2}{\\frac{2}{3}}=\\frac{z-1}{-3}$\n\n#### Question 14:\n\nFind the equation of the line passing through the point (2, −1, 3) and parallel to the line $\\stackrel{\\to }{r}=\\left(\\stackrel{^}{i}-2\\stackrel{^}{j}+\\stackrel{^}{k}\\right)+\\mathrm{\\lambda }\\left(2\\stackrel{^}{i}+3\\stackrel{^}{j}-5\\stackrel{^}{k}\\right).$\n\nThe given line is parallel to the vector $2\\stackrel{^}{i}+3\\stackrel{^}{j}-5\\stackrel{^}{k}$ and the required line is parallel to the given line.\nSo, the required line is parallel to the vector $2\\stackrel{^}{i}+3\\stackrel{^}{j}-5\\stackrel{^}{k}$.\n\nHence, the equation of the required line passing through the point (2,$-$1, 3) and parallel to the vector $2\\stackrel{^}{i}+3\\stackrel{^}{j}-5\\stackrel{^}{k}$ is $\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{i}-\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)+\\mathrm{\\lambda }\\left(2\\stackrel{^}{i}+3\\stackrel{^}{j}-5\\stackrel{^}{k}\\right)$.\n\n#### Question 15:\n\nFind the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines\n\nLet:\n\nSince the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}$.\n\nNow,\n\nThus, the direction ratios of the required line are proportional to 2, $-$7, 4.\n\nThe equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, $-$7, 4 is $\\frac{x-2}{2}=\\frac{y-1}{-7}=\\frac{z-3}{4}$.\n\n#### Question 16:\n\nFind the equation of the line passing through the point $\\stackrel{^}{i}+\\stackrel{^}{j}-3\\stackrel{^}{k}$ and perpendicular to the lines\n\nThe required line is perpendicular to the lines parallel to the vectors .\n\nSo, the required line is parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}$.\n\nNow,\n\nEquation of the required line passing through the point $\\left(\\stackrel{^}{i}+\\stackrel{^}{j}-3\\stackrel{^}{k}\\right)$ and parallel to $\\left(4\\stackrel{^}{i}-5\\stackrel{^}{j}+\\stackrel{^}{k}\\right)$ is $\\stackrel{\\to }{r}=\\left(\\stackrel{^}{i}+\\stackrel{^}{j}-3\\stackrel{^}{k}\\right)+\\lambda \\left(4\\stackrel{^}{i}-5\\stackrel{^}{j}+\\stackrel{^}{k}\\right)$.\n\n#### Question 17:\n\nFind the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).\n\nThe direction ratios of the line joining the points (4, 3, 2), (1,$-$1, 0) and (1, 2, $-$1), (2, 1, 1) are $-$3, $-$4, $-$2 and 1, $-$1, 2, respectively.\n\nLet:\n\nSince the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}$.\n\nNow,\n\nSo, the direction ratios of the required line are proportional to $-$10, 4, 7.\n\nThe equation of the required line passing through the point (1, $-$1, 1) and having direction ratios proportional to $-$10, 4, 7 is $\\frac{x-1}{-10}=\\frac{y+1}{4}=\\frac{z-1}{7}$.\n\n#### Question 18:\n\nDetermine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines .\n\nWe have\n\n$\\frac{x-8}{8}=\\frac{y+9}{-16}=\\frac{z-10}{7}\\phantom{\\rule{0ex}{0ex}}\\frac{x-15}{3}=\\frac{y-29}{8}=\\frac{z-5}{-5}$\n\nLet:\n\nSince the required line is perpendicular to the lines parallel to the vectors , it is parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}$.\n\nNow,\n\nThe direction ratios of the required line are proportional to 24, 61, 112.\n\nThe equation of the required line passing through the point (1, 2, $-$4) and having direction ratios proportional to 24, 61, 112 is $\\frac{x-1}{24}=\\frac{y-2}{61}=\\frac{z+4}{112}$.\n\n#### Question 19:\n\nShow that the lines are perpendicular to each other.\n\nThe direction ratios of the lines are proportional to 7, $-$5, 1 and 1, 2, 3, respectively.\n\nLet:\n\nNow,\n\nHence, the given lines are perpendicular to each other.\n\n#### Question 20:\n\nFind the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.\n\nThe equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as\n\nSince the required line is parallel to the given line, the direction ratios of the required line are proportional to .\n\nThe vector equation of the required line passing through the point (2, $-$1, $-$1) and having direction ratios proportional to is $\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{i}-\\stackrel{^}{j}-\\stackrel{^}{k}\\right)+\\lambda \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)$.\n\n#### Question 21:\n\nIf the lines are perpendicular, find the value of λ.\n\nThe equations of the given lines are\n\nSince the given lines are perpendicular to each other, we have\n\n$-3\\left(3\\lambda \\right)+2\\lambda \\left(1\\right)+2\\left(-5\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒-9\\lambda +2\\lambda -10=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =-\\frac{10}{7}$\n\n#### Question 22:\n\nIf the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.\n\nThe direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively.\n\nLet $\\theta$ be the angle between AB and CD. Then,\n\n#### Question 23:\n\nFind the value of λ so that the following lines are perpendicular to each other.\n\nThe equations of the given lines  can be re-written as\n\nSince the given lines are perpendicular to each other, we have\n\n$\\left(5\\lambda +2\\right)1-5\\left(2\\lambda \\right)+1\\left(3\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒5\\lambda =5\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =1$\n\n#### Question 24:\n\nFind the direction cosines of the line $\\frac{x+2}{2}=\\frac{2y-7}{6}=\\frac{5-z}{6}$. Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.                                                              [CBSE 2014]\n\nThe equation of the given line is $\\frac{x+2}{2}=\\frac{2y-7}{6}=\\frac{5-z}{6}$.\nThe given equation can be re-written as $\\frac{x+2}{2}=\\frac{y-\\frac{7}{2}}{3}=\\frac{z-5}{-6}$.\n\nThis line passes through the point $\\left(-2,\\frac{7}{2},5\\right)$ and has direction ratios proportional to 2, 3, −6.\n\nSo, its direction cosines are\n\n$\\frac{2}{\\sqrt{{2}^{2}+{3}^{2}+{\\left(-6\\right)}^{2}}},\\frac{3}{\\sqrt{{2}^{2}+{3}^{2}+{\\left(-6\\right)}^{2}}},\\frac{-6}{\\sqrt{{2}^{2}+{3}^{2}+{\\left(-6\\right)}^{2}}}$\n\nOr $\\frac{2}{7},\\frac{3}{7},\\frac{-6}{7}$\n\nThe required line passes through the point having position vector $\\stackrel{\\to }{a}=-\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+3\\stackrel{^}{j}-6\\stackrel{^}{k}$.\n\nSo, its vector equation is\n\n$\\stackrel{\\to }{r}=\\left(-\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)+\\lambda \\left(2\\stackrel{^}{i}+3\\stackrel{^}{j}-6\\stackrel{^}{k}\\right)$\n\n#### Question 1:\n\nShow that the lines intersect and find their point of intersection.\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nAlso, the coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\n#### Question 2:\n\nShow that the lines do not intersect.\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nThe coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are .\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\n#### Question 3:\n\nShow that the lines intersect. Find their point of intersection.\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nThe coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are .\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\n#### Question 4:\n\nProve that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.\n\nThe coordinates of any point on the line AB are given by\n\nThe coordinates of a general point on AB are .\n\nThe coordinates of any point on the line CD are given by\n\nThe coordinates of a general point on CD are .\nIf the lines AB and CD intersect, then they have a common point. So, for some values of , we must have\n\n#### Question 5:\n\nProve that the line intersect and find their point of intersection.\n\nThe position vectors of two arbitrary points on the given lines are\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\nEquating the coefficients of , we get\n\nSolving (2) and (3), we get\n\nSubstituting the values in (1), we get\n\nSubstituting $\\mu =0$ in the second line, we get $\\stackrel{\\to }{r}=4\\stackrel{^}{i}+0\\stackrel{^}{j}-\\stackrel{^}{k}$ as the position vector of the point of intersection.\n\nThus, the coordinates of the point of intersection are (4, 0, $-$1).\n\n#### Question 6:\n\nDetermine whether the following pair of lines intersect or not:\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(i)\n\nThe position vectors of two arbitrary points on the given lines are\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\nEquating the coefficients of , we get\n\nSolving (2) and (3), we get\n.\n\nSubstituting the values in (1), we get\n\n(ii)\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nAlso, the coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are .\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\n(iii)\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nAlso, the coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are .\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\n(iv)\n\nThe coordinates of any point on the first line are given by\n\nThe coordinates of a general point on the first line are .\n\nThe coordinates of any point on the second line are given by\n\nThe coordinates of a general point on the second line are .\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\nDisclaimer: The question printed in the book is incorrect. Instead of z, 3 is printed.\n\n#### Question 7:\n\nShow that the lines are intersecting. Hence, find their point of intersection.\n\nThe position vectors of two arbitrary points on the given lines are\n\nIf the lines intersect, then they have a common point. So, for some values of , we must have\n\nEquating the coefficients of , we get\n\nSolving (1) and (2), we get\n$\\lambda =-4\\phantom{\\rule{0ex}{0ex}}\\mu =-2$.\n\nSubstituting the values in (3), we get\n\nSubstituting $\\mu =-2$ in the second line, we get $\\stackrel{\\to }{r}=5\\stackrel{^}{i}-2\\stackrel{^}{j}-6\\stackrel{^}{i}-4\\stackrel{^}{j}-12\\stackrel{^}{k}=-\\stackrel{^}{i}-6\\stackrel{^}{j}-12\\stackrel{^}{k}$ as the position vector of the point of intersection.\n\nThus, the coordinates of the point of intersection are ($-$1, $-$6, $-$12).\n\n#### Question 1:\n\nFind the perpendicular distance of the point (3, −1, 11) from the line $\\frac{x}{2}=\\frac{y-2}{-3}=\\frac{z-3}{4}.$\n\nLet the point (3, $-$1, 11) be P and the point through which the line passes be Q (0, 2, 3).\nThe line is parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}-3\\stackrel{^}{j}+4\\stackrel{^}{k}$.\n\nNow,\n\n$\\stackrel{\\to }{PQ}=-3\\stackrel{^}{i}+3\\stackrel{^}{j}-8\\stackrel{^}{k}$\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n#### Question 2:\n\nFind the perpendicular distance of the point (1, 0, 0) from the line $\\frac{x-1}{2}=\\frac{y+1}{-3}=\\frac{z+10}{8}.$ Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.\n\nLet the point (1, 0, 0) be P and the point through which the line passes be Q (1, $-$1, $-$10).\nThe line is parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}-3\\stackrel{^}{j}+8\\stackrel{^}{k}$.\n\nNow,\n\n$\\stackrel{\\to }{PQ}=0\\stackrel{^}{i}-\\stackrel{^}{j}-10\\stackrel{^}{k}$\n\nLet L be the foot of the perpendicular drawn from the point P (1, 0, 0) to the given line.", null, "The coordinates of a general point on the line $\\frac{x-1}{2}=\\frac{y+1}{-3}=\\frac{z+10}{8}$ are given by\n\nLet the coordinates of L be .\n\nThe direction ratios of PL are proportional to .\n\nThe direction ratios of the given line are proportional to 2 ,$-$3, 8, but PL is perpendicular to the given line.\n\nSubstituting $\\lambda =1$ in , we get the coordinates of L as (3, $-$4, $-$2).\n\nEquation of the line PL is given by\n\n$\\frac{x-1}{3-1}=\\frac{y-0}{-4-0}=\\frac{z-0}{-2-0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-1}{1}=\\frac{y}{-2}=\\frac{z}{-1}$\n\n$⇒\\stackrel{\\to }{r}=\\stackrel{^}{i}+\\lambda \\left(\\stackrel{^}{i}-2\\stackrel{^}{j}-\\stackrel{^}{k}\\right)$\n\n#### Question 3:\n\nFind the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).\n\nLet D be the foot of the perpendicular drawn from the point A (1, 0, 3) to the line BC.\n\nThe coordinates of a general point on the line BC are given by\n\nLet the coordinates of D be .", null, "The direction ratios of AD are proportional to .\n\nThe direction ratios of the line BC are proportional to 1, 2, $-$2, but AD is perpendicular to the line BC.\n\n$\\therefore 1\\left(\\lambda +3\\right)+2\\left(2\\lambda +7\\right)-2\\left(-2\\lambda -2\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =-\\frac{7}{3}$\nSubstituting $\\lambda =-\\frac{7}{3}$ in , we get the coordinates of D as .\n\n#### Question 4:\n\nA (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.\n\nPoint D is the foot of the perpendicular drawn from the point A (1, 0, 4) to the line BC.\n\nThe coordinates of a general point on the line BC are given by\n\nLet the coordinates of D be .\n\nThe direction ratios of AD are proportional to .\n\nThe direction ratios of the line BC are proportional to 2, 8,$-$2, but AD is perpendicular to the line BC.\n\nSubstituting $\\lambda =\\frac{11}{9}$ in , we get the coordinates of D as .\n\n#### Question 5:\n\nFind the foot of perpendicular from the point (2, 3, 4) to the line $\\frac{4-x}{2}=\\frac{y}{6}=\\frac{1-z}{3}.$ Also, find the perpendicular distance from the given point to the line.\n\nLet L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.\n\nThe coordinates of a general point on the line $\\frac{4-x}{2}=\\frac{y}{6}=\\frac{1-z}{3}$ are given by\n\nLet the coordinates of L be .", null, "The direction ratios of PL are proportional to .\n\nThe direction ratios of the given line are proportional to $-$2, 6, $-$3, but PL is perpendicular to the given line.\n\n$\\therefore -2\\left(-2\\lambda +2\\right)+6\\left(6\\lambda -3\\right)-3\\left(-3\\lambda -3\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =\\frac{13}{49}$\n\nSubstituting $\\lambda =\\frac{13}{49}$ in , we get the coordinates of L as  .\n\n.\n\nHence, the length of the perpendicular from P on PL is .\n\n#### Question 6:\n\nFind the equation of the perpendicular drawn from the point P (2, 4, −1) to the line $\\frac{x+5}{1}=\\frac{y+3}{4}=\\frac{z-6}{-9}.$ Also, write down the coordinates of the foot of the perpendicular from P.\n\nLet L be the foot of the perpendicular drawn from the point P (2, 4, $-$1) to the given line.\n\nThe coordinates of a general point on the line $\\frac{x+5}{1}=\\frac{y+3}{4}=\\frac{z-6}{-9}$ are given by\n\nLet the coordinates of L be .", null, "The direction ratios of PL are proportional to .\n\nThe direction ratios of the given line are proportional to 1, 4, $-$9, but PL is perpendicular to the given line.\n\n$\\therefore 1\\left(\\lambda -7\\right)+4\\left(4\\lambda -7\\right)-9\\left(-9\\lambda +7\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =1$\n\nSubstituting $\\lambda =1$ in , we get the coordinates of L as .\n\nEquation of the line PL is\n\n$\\frac{x-2}{-4-2}=\\frac{y-4}{1-4}=\\frac{z+1}{-3+1}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-2}{-6}=\\frac{y-4}{-3}=\\frac{z+1}{-2}$\n\n#### Question 7:\n\nFind the length of the perpendicular drawn from the point (5, 4, −1) to the line $\\stackrel{\\to }{r}=\\stackrel{^}{i}+\\lambda \\left(2\\stackrel{^}{i}+9\\stackrel{^}{j}+5\\stackrel{^}{k}\\right).$\n\nLet the point (5, 4, $-$1) be P and the the point through which the line passes be Q (1, 0, 0).\nThe line is parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+9\\stackrel{^}{j}+5\\stackrel{^}{k}$.\n\nNow,\n\n$\\stackrel{\\to }{PQ}=-4\\stackrel{^}{i}-4\\stackrel{^}{j}+\\stackrel{^}{k}$\n\n#### Question 8:\n\nFind the foot of the perpendicular drawn from the point $\\stackrel{^}{i}+6\\stackrel{^}{j}+3\\stackrel{^}{k}$ to the line $\\stackrel{\\to }{r}=\\stackrel{^}{j}+2\\stackrel{^}{k}+\\lambda \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right).$ Also, find the length of the perpendicular\n\nLet L be the foot of the perpendicular drawn from the point P ($\\stackrel{^}{i}+6\\stackrel{^}{j}+3\\stackrel{^}{k}$) to the line $\\stackrel{\\to }{r}=\\stackrel{^}{j}+2\\stackrel{^}{k}+\\lambda \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right).$\n\nLet the position vector L be\n$\\stackrel{\\to }{r}=\\stackrel{^}{j}+2\\stackrel{^}{k}+\\lambda \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)=\\lambda \\stackrel{^}{i}+\\left(1+2\\lambda \\right)\\stackrel{^}{j}+\\left(2+3\\lambda \\right)\\stackrel{^}{k}$        ...(1)", null, "Now,\n\nSince $\\stackrel{\\to }{PL}$ is perpendicular to the given line, which is parallel to $\\stackrel{\\to }{b}=\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}$, we have\n\nSubstituting $\\lambda =1$ in (1), we get the position vector of L as $\\stackrel{^}{i}+3\\stackrel{^}{j}+5\\stackrel{^}{k}$.\n\nSubstituting $\\lambda =1$ in (2), we get\n$\\stackrel{\\to }{PL}=-3\\stackrel{^}{j}+2\\stackrel{^}{k}$\n\nHence, the length of the perpendicular from point P on PL is .\n\n#### Question 9:\n\nFind the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line $\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)+\\lambda \\left(2\\stackrel{^}{i}+\\stackrel{^}{j}+3\\stackrel{^}{k}\\right).$ Also, find the coordinates of the foot of the perpendicular from P.\n\nLet L be the foot of the perpendicular drawn from the point P ($-$1, 3, 2) to the line $\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)+\\lambda \\left(2\\stackrel{^}{i}+\\stackrel{^}{j}+3\\stackrel{^}{k}\\right).$", null, "Let the position vector L be\n$\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)+\\lambda \\left(2\\stackrel{^}{i}+\\stackrel{^}{j}+3\\stackrel{^}{k}\\right)=2\\lambda \\stackrel{^}{i}+\\left(2+\\lambda \\right)\\stackrel{^}{j}+\\left(3+3\\lambda \\right)\\stackrel{^}{k}$        ...(1)\n\nNow,\n\nSince $\\stackrel{\\to }{PL}$ is perpendicular to the given line, which is parallel to $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+\\stackrel{^}{j}+3\\stackrel{^}{k}$, we have\n\nSubstituting $\\lambda =-\\frac{2}{7}$ in (1), we get the position vector of L as $-\\frac{4}{7}\\stackrel{^}{i}+\\frac{12}{7}\\stackrel{^}{j}+\\frac{15}{7}\\stackrel{^}{k}$.\n\nSo, the coordinates of the foot of the perpendicular from P to the given line is L .\nSubstituting $\\lambda =-\\frac{2}{7}$ in (2), we get\n$\\stackrel{\\to }{PL}=\\frac{3}{7}\\stackrel{^}{i}-\\frac{9}{7}\\stackrel{^}{j}+\\frac{1}{7}\\stackrel{^}{k}$\n\nEquation of the perpendicular drawn from P to the given line is\n\n#### Question 10:\n\nFind the foot of the perpendicular from (0, 2, 7) on the line $\\frac{x+2}{-1}=\\frac{y-1}{3}=\\frac{z-3}{-2}.$\n\nLet L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line.\n\nThe coordinates of a general point on the line $\\frac{x+2}{-1}=\\frac{y-1}{3}=\\frac{z-3}{-2}$ are given by\n\nLet the coordinates of L be .", null, "The direction ratios of PL are proportional to .\n\nThe direction ratios of the given line are proportional to $-$1, 3, $-$2, but PL is perpendicular to the given line.\n\n$\\therefore -1\\left(-\\lambda -2\\right)+3\\left(3\\lambda -1\\right)-2\\left(-2\\lambda -4\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =-\\frac{1}{2}$\nSubstituting $\\lambda =-\\frac{1}{2}$ in , we get the coordinates of L as .\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n#### Question 11:\n\nFind the foot of the perpendicular from (1, 2, −3) to the line $\\frac{x+1}{2}=\\frac{y-3}{-2}=\\frac{z}{-1}.$\n\nLet L be the foot of the perpendicular drawn from the point P (1, 2, $-$3) to the given line.\n\nThe coordinates of a general point on the line $\\frac{x+1}{2}=\\frac{y-3}{-2}=\\frac{z}{-1}$ are given by\n\nLet the coordinates of L be .", null, "The direction ratios of PL are proportional to .\n\nThe direction ratios of the given line are proportional to 2, $-$2, $-$1, but PL is perpendicular to the given line.\n\nSubstituting $\\lambda =1$ in , we get the coordinates of L as .\n\n#### Question 12:\n\nFind the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.", null, "Equation of line AB passing through the points A(0, 6,$-$9) and B($-$3, $-$6, 3) is\n\n$\\frac{x-0}{-3-0}=\\frac{y-6}{-6-6}=\\frac{z+9}{3+9}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{1}=\\frac{y-6}{4}=\\frac{z+9}{-4}$\n\nHere, D is the foot of the perpendicular drawn from C (7, 4, $-$1) on AB.\n\nThe coordinates of a general point on AB are given by\n\nLet the coordinates of D be .\nThe direction ratios of CD are proportional to .\n\nThe direction ratios of AB are proportional to 1, 4, $-$4, but CD is perpendicular to AB.\n\nSubstituting $\\lambda =-1$ in , we get the coordinates of D as .\n\nEquation of CD is\n$\\frac{x-7}{-1-7}=\\frac{y-4}{2-4}=\\frac{z+1}{-5+1}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-7}{4}=\\frac{y-4}{1}=\\frac{z+1}{2}$\n\n#### Question 13:\n\nFind the distance of the point (2, 4, −1) from the line $\\frac{x+5}{1}=\\frac{y+3}{4}=\\frac{z-6}{-9}$.                  [NCERT EXEMPLAR]\n\nWe know that the distance d from point P to the line l having equation $\\stackrel{\\to }{r}=\\stackrel{\\to }{a}+\\lambda \\stackrel{\\to }{b}$ is given by $d=\\frac{\\left|\\stackrel{\\to }{b}×\\stackrel{\\to }{\\mathrm{PQ}}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}$, where Q is any point on the line l.\n\nThe equation of the given line is $\\frac{x+5}{1}=\\frac{y+3}{4}=\\frac{z-6}{-9}$.\n\nLet P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line.\n\nThe given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6).\n\nAlso, the line is parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{^}{i}+4\\stackrel{^}{j}-9\\stackrel{^}{k}$.\n\nNow, $\\stackrel{\\to }{\\mathrm{PQ}}=\\left(-5\\stackrel{^}{i}-3\\stackrel{^}{j}+6\\stackrel{^}{k}\\right)-\\left(2\\stackrel{^}{i}+4\\stackrel{^}{j}-\\stackrel{^}{k}\\right)=-7\\stackrel{^}{i}-7\\stackrel{^}{j}+7\\stackrel{^}{k}$\n\n$\\therefore \\stackrel{\\to }{b}×\\stackrel{\\to }{\\mathrm{PQ}}=\\left|\\begin{array}{ccc}\\stackrel{^}{i}& \\stackrel{^}{j}& \\stackrel{^}{k}\\\\ 1& 4& -9\\\\ -7& -7& 7\\end{array}\\right|=-35\\stackrel{^}{i}+56\\stackrel{^}{j}+21\\stackrel{^}{k}\\phantom{\\rule{0ex}{0ex}}⇒\\left|\\stackrel{\\to }{b}×\\stackrel{\\to }{\\mathrm{PQ}}\\right|=\\sqrt{{\\left(-35\\right)}^{2}+{56}^{2}+{21}^{2}}=\\sqrt{1225+3136+441}=\\sqrt{4802}=49\\sqrt{2}$\n\nLet d be the required distance.\n\n$\\therefore d=\\frac{\\left|\\stackrel{\\to }{b}×\\stackrel{\\to }{\\mathrm{PQ}}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}=\\frac{49\\sqrt{2}}{\\sqrt{1+16+81}}=\\frac{49\\sqrt{2}}{\\sqrt{98}}=\\frac{49\\sqrt{2}}{7\\sqrt{2}}=7$\n\nThus, the distance of the given point from the given line is 7 units.\n\n#### Question 14:\n\nFind the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).                 [NCERT EXEMPLAR]\n\nThe Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is\n\nLet L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\\frac{x}{2}=\\frac{y+1}{-2}=\\frac{\\mathrm{z}-3}{-4}$.\n\nThe coordinates of general point on the line $\\frac{x}{2}=\\frac{y+1}{-2}=\\frac{\\mathrm{z}-3}{-4}$ are given by\n\nLet the coordinates of L be $\\left(2\\lambda ,-2\\lambda -1,-4\\lambda +3\\right)$. Therefore, the direction ratios of AL are proportional to\n\n$2\\lambda -1,-2\\lambda -1-8,-4\\lambda +3-4$ or $2\\lambda -1,-2\\lambda -9,-4\\lambda -1$\n\nDirection ratios of the given line are proportional to 2, −2, −4.\n\nBut, AL is perpendicular to the given line.\n\n$\\therefore 2×\\left(2\\lambda -1\\right)+\\left(-2\\right)×\\left(-2\\lambda -9\\right)+\\left(-4\\right)×\\left(-4\\lambda -1\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒4\\lambda -2+4\\lambda +18+16\\lambda +4=0\\phantom{\\rule{0ex}{0ex}}⇒24\\lambda +20=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =-\\frac{5}{6}$\n\nPutting $\\lambda =-\\frac{5}{6}$ in $\\left(2\\lambda ,-2\\lambda -1,-4\\lambda +3\\right)$, we get\n\n$\\left(2×\\left(-\\frac{5}{6}\\right),-2×\\left(-\\frac{5}{6}\\right)-1,-4×\\left(-\\frac{5}{6}\\right)+3\\right)=\\left(-\\frac{5}{3},\\frac{2}{3},\\frac{19}{3}\\right)$\n\nThus, the required coordinates of the foot of the perpendicular are $\\left(-\\frac{5}{3},\\frac{2}{3},\\frac{19}{3}\\right)$.\n\n#### Question 1:\n\nFind the shortest distance between the following pairs of lines whose vector equations are:\n\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(v)\n\n(vi)\n\n(vii)\n\n(viii) $\\stackrel{\\to }{r}=\\left(8+3\\lambda \\right)\\stackrel{^}{i}-\\left(9+16\\lambda \\right)\\stackrel{^}{j}+\\left(10+7\\lambda \\right)\\stackrel{^}{k}$ and $\\stackrel{\\to }{r}=15\\stackrel{^}{i}+29\\stackrel{^}{j}+5\\stackrel{^}{k}+\\mu \\left(3\\stackrel{^}{i}+8\\stackrel{^}{j}-5\\stackrel{^}{k}\\right)$              [NCERT EXEMPLAR]\n\n(i)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(ii)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(iii)\n\nComparing the given equations with the equations, we get\n\nThe shortest distance between the lines is given by\n\n(iv)\n\nThe vector equations of the given lines can be re-written as\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the line is given by\n\n(v)\n\nThe vector equations of the given lines can be re-written as\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(vi)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(vii)\n\nComparing the given equations with the equations, we get\n\nThe shortest distance between the lines is given by\n\n(viii) The vector equations of the given lines can be re-written as\n\n$\\stackrel{\\to }{r}=8\\stackrel{^}{i}-9\\stackrel{^}{j}+10\\stackrel{^}{k}+\\lambda \\left(3\\stackrel{^}{i}-16\\stackrel{^}{j}+7\\stackrel{^}{k}\\right)$ and $\\stackrel{\\to }{r}=15\\stackrel{^}{i}+29\\stackrel{^}{j}+5\\stackrel{^}{k}+\\mu \\left(3\\stackrel{^}{i}+8\\stackrel{^}{j}-5\\stackrel{^}{k}\\right)$\n\nComparing the given equations with the equations , we get\n\n$\\stackrel{\\to }{{a}_{1}}=8\\stackrel{^}{i}-9\\stackrel{^}{j}+10\\stackrel{^}{k}$\n\n$\\stackrel{\\to }{{b}_{1}}=3\\stackrel{^}{i}-16\\stackrel{^}{j}+7\\stackrel{^}{k}$\n\n$\\stackrel{\\to }{{a}_{2}}=15\\stackrel{^}{i}+29\\stackrel{^}{j}+5\\stackrel{^}{k}$\n\n$\\stackrel{\\to }{{b}_{2}}=3\\stackrel{^}{i}+8\\stackrel{^}{j}-5\\stackrel{^}{k}$\n\n$\\therefore \\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}=\\left(15\\stackrel{^}{i}+29\\stackrel{^}{j}+5\\stackrel{^}{k}\\right)-\\left(8\\stackrel{^}{i}-9\\stackrel{^}{j}+10\\stackrel{^}{k}\\right)=7\\stackrel{^}{i}+38\\stackrel{^}{j}-5\\stackrel{^}{k}$\n\n$\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}=\\left|\\begin{array}{ccc}\\stackrel{^}{i}& \\stackrel{^}{j}& \\stackrel{^}{k}\\\\ 3& -16& 7\\\\ 3& 8& -5\\end{array}\\right|=24\\stackrel{^}{i}+36\\stackrel{^}{j}+72\\stackrel{^}{k}\\phantom{\\rule{0ex}{0ex}}⇒\\left|\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}\\right|=\\sqrt{{24}^{2}+{36}^{2}+{72}^{2}}=\\sqrt{576+1296+5184}=\\sqrt{7056}=84$\n\nAlso,\n\n$\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right).\\left(\\stackrel{\\to }{{b}_{1}}×\\stackrel{\\to }{{b}_{2}}\\right)\\phantom{\\rule{0ex}{0ex}}=\\left(7\\stackrel{^}{i}+38\\stackrel{^}{j}-5\\stackrel{^}{k}\\right).\\left(24\\stackrel{^}{i}+36\\stackrel{^}{j}+72\\stackrel{^}{k}\\right)\\phantom{\\rule{0ex}{0ex}}=7×24+38×36+\\left(-5\\right)×72\\phantom{\\rule{0ex}{0ex}}=168+1368-360\\phantom{\\rule{0ex}{0ex}}=1176$\n\nWe know that the shortest distance between the lines is given by .\n\n∴ Required shortest distance between the given pairs of lines,\n\n#### Question 2:\n\nFind the shortest distance between the following pairs of lines whose cartesian equations are:\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(i) The equations of the given lines are\n\nSince line (1) passes through the point (1, 2, 3) and has direction ratios proportional to 2, 3, 4, its vector equation is\n\nAlso, line (2) passes through the point (2, 3, 5) and has direction ratios proportional to 3, 4, 5.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n(ii) The equations of the given lines are\n\nSince line (1) passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is\n\nAlso, line (2) passes through the point ($-$1, 2, 2) and has direction ratios proportional to 3, 1, 0.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n(iii)\n\nSince line (1) passes through the point (1, $-$2, 3) and has direction ratios proportional to $-$1, 1, $-$2, its vector equation is\n\nAlso, line (2) passes through the point (1, $-$1, $-$1) and has direction ratios proportional to 1, 2, $-$2.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n(iv)\n\nSince line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, $-$2, 1, its vector equation is\n\nAlso, line (2) passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n#### Question 3:\n\nBy computing the shortest distance determine whether the following pairs of lines intersect or not:\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(i)\n\nComparing the given equations with the equations , we get\n\n(ii)\n\nComparing the given equations with the equations , we get\n\n(iii)\n\nSince the first line passes through the point (1, $-$1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is\n\nAlso, the second line passes through the point ($-$1, 2, 2) and has direction ratios proportional to 5, 1, 0.\nIts vector equation is\n\nNow,\n\n(iv)\n\nSince the first line passes through the point (5, 7, $-$3) and has direction ratios proportional to 4, $-$5, $-$5, its vector equation is\n\nAlso, the second line passes through the point (8, 7, 5) and has direction ratios proportional to 7, 1, 3.\nIts vector equation is\n\nNow,\n\n#### Question 4:\n\nFind the shortest distance between the following pairs of parallel lines whose equations are:\n(i)\n\n(ii)\n\n(i) The vector equations of the given lines are\n\nThese two lines pass through the points having position vectors and are parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{^}{i}-\\stackrel{^}{j}+\\stackrel{^}{k}$.\n\nNow,\n$\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}=\\stackrel{^}{i}-3\\stackrel{^}{j}-4\\stackrel{^}{k}$\nand\n\nThe shortest distance between the two lines is given by\n\n$\\frac{\\left|\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right)×\\stackrel{\\to }{b}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}=\\frac{\\sqrt{78}}{\\sqrt{3}}=\\sqrt{26}$\n\n(ii)\n\nThese two lines pass through the points having position vectors and are parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}-\\stackrel{^}{j}+\\stackrel{^}{k}$.\n\nNow,\n$\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}=\\stackrel{^}{i}-\\stackrel{^}{k}$\nand\n\nThe shortest distance between the two lines is given by\n\n$\\frac{\\left|\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right)×\\stackrel{\\to }{b}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}=\\frac{\\sqrt{11}}{\\sqrt{6}}$\n\n#### Question 5:\n\nFind the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines\n(i) (0, 0, 0) and (1, 0, 2)\n(ii) (1, 3, 0) and (0, 3, 0)\n\n(i) The equation of the line passing through the points (0, 0, 0) and (1, 0, 2) is\n\n$\\frac{x-0}{1-0}=\\frac{y-0}{0-0}=\\frac{z-0}{2-0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{1}=\\frac{y}{0}=\\frac{z}{2}$\n\n(ii) The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is\n\n$\\frac{x-1}{0-1}=\\frac{y-3}{3-3}=\\frac{z-0}{0-0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-1}{-1}=\\frac{y-3}{0}=\\frac{z}{0}$\n\nSince the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is\n\nAlso, the second line passes through the point (1, 3, 0) and has direction ratios proportional to $-$1, 0, 0.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n#### Question 6:\n\nWrite the vector equations of the following lines and hence determine the distance between them\n\nWe have\n$\\frac{x-1}{2}=\\frac{y-2}{3}=\\frac{z+4}{6}\\phantom{\\rule{0ex}{0ex}}\\frac{x-3}{4}=\\frac{y-3}{6}=\\frac{z+5}{12}$\n\nSince the first line passes through the point (1, 2, $-$4) and has direction ratios proportional to 2, 3, 6, its vector equation is\n\nAlso, the second line passes through the point (3, 3, $-$5) and has direction ratios proportional to 4, 6, 12.\nIts vector equation is\n\nThese two lines pass through the points having position vectors and are parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+3\\stackrel{^}{j}+6\\stackrel{^}{k}$.\n\nNow,\n\n$\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}=2\\stackrel{^}{i}+\\stackrel{^}{j}-\\stackrel{^}{k}$\n\nand\n\nThe shortest distance between the two lines is given by\n\n$\\frac{\\left|\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right)×\\stackrel{\\to }{b}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}=\\frac{\\sqrt{293}}{7}\\mathrm{units}$\n\n#### Question 7:\n\nFind the shortest distance between the lines\n(i)\n\n(ii)\n\n(iii)\n\n(iv)\n\n(i)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(ii)\n\nSince the first line passes through the point ($-$1, $-$1, $-$1) and has direction ratios proportional to 7, $-$6, 1, its vector equation is\n\nAlso, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1,$-$2, 1.\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n(iii)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n(iv)\n\nComparing the given equations with the equations , we get\n\nThe shortest distance between the lines is given by\n\n#### Question 8:\n\nFind the distance between the lines l1 and l2 given by\n\nGiven:\n\nThese two lines pass through the points having position vectors and are parallel to the vector $\\stackrel{\\to }{b}=2\\stackrel{^}{i}+3\\stackrel{^}{j}+6\\stackrel{^}{k}$.\n\nNow,\n$\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}=2\\stackrel{^}{i}+\\stackrel{^}{j}-\\stackrel{^}{k}$\nand\n\nThe shortest distance between the two lines is given by\n\n$\\frac{\\left|\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right)×\\stackrel{\\to }{b}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}=\\frac{\\sqrt{293}}{7}$\n\n#### Question 1:\n\nWrite the cartesian and vector equations of X-axis.\n\nSince x-axis passes through the the point (0, 0, 0) having position vector $\\stackrel{\\to }{a}=0\\stackrel{^}{i}+0\\stackrel{^}{j}+0\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=1\\stackrel{^}{i}+0\\stackrel{^}{j}+0\\stackrel{^}{k}$ having direction ratios proportional to 1, 0, 0, the cartesian equation of x-axis is\n\n$\\frac{x-0}{1}=\\frac{y-0}{0}=\\frac{z-0}{0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{1}=\\frac{y}{0}=\\frac{z}{0}$\n\nAlso, its vector equation is\n\n#### Question 2:\n\nWrite the cartesian and vector equations of Y-axis.\n\nSince y-axis passes through the the point (0, 0, 0) having position vector $\\stackrel{\\to }{a}=0\\stackrel{^}{i}+0\\stackrel{^}{j}+0\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=0\\stackrel{^}{i}+1\\stackrel{^}{j}+0\\stackrel{^}{k}$ having direction ratios proportional to 0, 1, 0, the cartesian equation of y-axis is\n\n$\\frac{x-0}{0}=\\frac{y-0}{1}=\\frac{z-0}{0}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{0}=\\frac{y}{1}=\\frac{z}{0}$\n\nAlso, its vector equation is\n\n#### Question 3:\n\nWrite the cartesian and vector equations of Z-axis.\n\nSince z-axis passes through the the point (0, 0, 0) having position vector $\\stackrel{\\to }{a}=0\\stackrel{^}{i}+0\\stackrel{^}{j}+0\\stackrel{^}{k}$ and is parallel to the vector $\\stackrel{\\to }{b}=0\\stackrel{^}{i}+0\\stackrel{^}{j}+\\stackrel{^}{k}$ having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is\n\n$\\frac{x-0}{0}=\\frac{y-0}{0}=\\frac{z-0}{1}\\phantom{\\rule{0ex}{0ex}}=\\frac{x}{0}=\\frac{y}{0}=\\frac{z}{1}$\n\nAlso, its vector equation is\n\n#### Question 4:\n\nWrite the vector equation of a line passing through a point having position vector and parallel to vector .\n\nThe vector equation of the line passing through the point having position vector $\\stackrel{\\to }{\\alpha }$ and parallel to vector $\\stackrel{\\to }{\\beta }$ is $\\stackrel{\\to }{r}=\\stackrel{\\to }{\\alpha }+\\lambda \\stackrel{\\to }{\\beta }$.\n\n#### Question 5:\n\nCartesian equations of a line AB are $\\frac{2x-1}{2}=\\frac{4-y}{7}=\\frac{z+1}{2}.$ Write the direction ratios of a line parallel to AB.\n\nWe have\n\n$\\frac{2x-1}{2}=\\frac{4-y}{7}=\\frac{z+1}{2}$\n\nThe equation of the line AB can be re-written as\n\n$\\frac{x-\\frac{1}{2}}{1}=\\frac{y-4}{-7}=\\frac{z+1}{2}$\n\nThe direction ratios of the line parallel to AB are proportional to 1, $-$7, 2.\n\nAlso, the direction cosines of the line parallel to AB are proportional to\n\n#### Question 6:\n\nWrite the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.\n\nWe have\n6x − 2 = 3y + 1 = 2z − 4\n\nThe equation of given line can be re-written as\n\n$\\frac{x-\\frac{1}{3}}{\\frac{1}{6}}=\\frac{y+\\frac{1}{3}}{\\frac{1}{3}}=\\frac{z-2}{\\frac{1}{2}}\\phantom{\\rule{0ex}{0ex}}\\phantom{\\rule{0ex}{0ex}}⇒\\frac{x-\\frac{1}{3}}{1}=\\frac{y+\\frac{1}{3}}{2}=\\frac{z-2}{3}$\n\nThe direction ratios of the line parallel to AB are proportional to 1, 2, 3.\n\nHence, the direction cosines of the line parallel to AB are proportional to\n\n.\n\n#### Question 7:\n\nWrite the direction cosines of the line\n\nWe have\n\nThe equation of the given line can be re-written as\n\nThe direction ratios of the given line are proportional to 4, $-$3, 0.\n\nHence, the direction cosines of the given line are proportional to\n\n#### Question 8:\n\nWrite the coordinate axis to which the line $\\frac{x-2}{3}=\\frac{y+1}{4}=\\frac{z-1}{0}$is perpendicular.\n\nWe have\n\n$\\frac{x-2}{3}=\\frac{y+1}{4}=\\frac{z-1}{0}$\n\nThe given line is parallel to the vector $\\stackrel{\\to }{b}=3\\stackrel{^}{i}+4\\stackrel{^}{j}+0\\stackrel{^}{k}$.\n\nLet $x\\stackrel{^}{i}+y\\stackrel{^}{j}+z\\stackrel{^}{k}$ be perpendicular to the given line.\n\nNow,\n$3x+4y+0z=0$\nIt is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).\n\nHence, the given line is perpendicular to z-axis.\n\n#### Question 9:\n\nWrite the angle between the lines\n\nWe have\n\nThe given lines are parallel to the vectors .\n\nLet $\\theta$ be the angle between the given lines.\n\nNow,\n\n#### Question 10:\n\nWrite the direction cosines of the line whose cartesian equations are 2x = 3y = −z.\n\nWe have\n2x = 3y = −z\n\nThe equation of the given line can be re-written as\n\n$\\frac{x}{\\frac{1}{2}}=\\frac{y}{\\frac{1}{3}}=\\frac{z}{-1}\\phantom{\\rule{0ex}{0ex}}\\phantom{\\rule{0ex}{0ex}}\\frac{x}{3}=\\frac{y}{2}=\\frac{z}{-6}$\n\nThe direction ratios of the line parallel to AB are proportional to 3, 2, $-$6.\n\nHence, the direction cosines of the line parallel to AB are proportional to\n\n#### Question 11:\n\nWrite the angle between the lines 2x = 3y = −z and 6x = −y = −4z.\n\nWe have\n2x = 3y = −z\n6x = −y = −4z\n\nThe given lines can be re-written as\n\nThese lines are parallel to vectors .\n\nLet $\\theta$ be the angle between these lines.\n\nNow,\n\n#### Question 12:\n\nWrite the value of λ for which the lines are perpendicular to each other.\n\nWe have\n\nThe given lines are parallel to vectors .\n\nFor $\\stackrel{\\to }{{b}_{1}}\\perp \\stackrel{\\to }{{b}_{2}}$, we must have\n\n$\\stackrel{\\to }{{b}_{1}}.\\stackrel{\\to }{{b}_{2}}=0\\phantom{\\rule{0ex}{0ex}}⇒\\left(-3\\stackrel{^}{i}+2\\lambda \\stackrel{^}{j}+2\\stackrel{^}{k}\\right).\\left(3\\lambda \\stackrel{^}{i}+\\stackrel{^}{j}-5\\stackrel{^}{k}\\right)=0\\phantom{\\rule{0ex}{0ex}}⇒-7\\lambda -10=0\\phantom{\\rule{0ex}{0ex}}⇒\\lambda =-\\frac{10}{7}$\n\n#### Question 13:\n\nWrite the formula for the shortest distance between the lines\n\nThe shortest distance d between the parallel lines is given by\n\n$d=\\frac{\\left|\\left(\\stackrel{\\to }{{a}_{2}}-\\stackrel{\\to }{{a}_{1}}\\right)×\\stackrel{\\to }{b}\\right|}{\\left|\\stackrel{\\to }{b}\\right|}$\n\n#### Question 14:\n\nWrite the condition for the lines to be intersecting.\n\nThe shortest distance between the lines is given by\n\nFor the lines to be intersecting, $d=0$.\n\n#### Question 15:\n\nThe cartesian equations of a line AB are $\\frac{2x-1}{\\sqrt{3}}=\\frac{y+2}{2}=\\frac{z-3}{3}.$ Find the direction cosines of a line parallel to AB.\n\nWe have\n\n$\\frac{2x-1}{\\sqrt{3}}=\\frac{y+2}{2}=\\frac{z-3}{3}$\n\nThe equation of the line AB can be re-written as\n\n$\\frac{x-\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}=\\frac{y+2}{2}=\\frac{z-3}{3}\\phantom{\\rule{0ex}{0ex}}=\\frac{x-\\frac{1}{2}}{\\sqrt{3}}=\\frac{y+2}{4}=\\frac{z-3}{6}$\n\nThus, the direction ratios of the line parallel to AB are proportional to $\\sqrt{3}$, 4, 6.\n\nHence, the direction cosines of the line parallel to AB are proportional to\n\n#### Question 16:\n\nIf the equations of a line AB are $\\frac{3-x}{1}=\\frac{y+2}{-2}=\\frac{z-5}{4},$ write the direction ratios of a line parallel to AB.\n\nWe have\n\n$\\frac{3-x}{1}=\\frac{y+2}{-2}=\\frac{z-5}{4}$\n\nThe equation of the line AB can be re-written as\n\n$\\frac{x-3}{-1}=\\frac{y+2}{-2}=\\frac{z-5}{4}$\n\nThus, the direction ratios of the line parallel to AB are proportional to $-$1, $-$2, 4.\n\n#### Question 17:\n\nWrite the vector equation of a line given by $\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}.$\n\nWe have\n\n$\\frac{x-5}{3}=\\frac{y+4}{7}=\\frac{z-6}{2}$\n\nThe given line passes through the point (5, $-$4, 6) and has direction ratios proportional to 3, 7, 2.\n\nVector equation of the given line passing through the point having position vector $\\stackrel{\\to }{a}=5\\stackrel{^}{i}-4\\stackrel{^}{j}+6\\stackrel{^}{k}$ and parallel to a vector $\\stackrel{\\to }{b}=3\\stackrel{^}{i}+7\\stackrel{^}{j}+2\\stackrel{^}{k}$ is\n\n$\\stackrel{\\to }{r}=\\stackrel{\\to }{a}+\\lambda \\stackrel{\\to }{b}\\phantom{\\rule{0ex}{0ex}}⇒\\stackrel{\\to }{r}=5\\stackrel{^}{i}-4\\stackrel{^}{j}+6\\stackrel{^}{k}+\\lambda \\left(3\\stackrel{^}{i}+7\\stackrel{^}{j}+2\\stackrel{^}{k}\\right)$\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n#### Question 18:\n\nThe equations of a line are given by $\\frac{4-x}{3}=\\frac{y+3}{3}=\\frac{z+2}{6}.$ Write the direction cosines of a line parallel to this line.\n\nWe have\n\n$\\frac{4-x}{3}=\\frac{y+3}{3}=\\frac{z+2}{6}$\n\nThe equation of the given line can be re-written as\n\n$\\frac{x-4}{-3}=\\frac{y+3}{3}=\\frac{z+2}{6}$\n\nThe direction ratios of the line parallel to the given line are proportional to $-$3, 3, 6.\n\nHence, the direction cosines of the line parallel to the given line are proportional to\n\n#### Question 19:\n\nFind the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line $\\frac{x+3}{3}=\\frac{4-y}{5}=\\frac{z+8}{6}.$\n\nThe equation of the given line is $\\frac{x+3}{3}=\\frac{4-y}{5}=\\frac{z+8}{6}$\n\nIt can be re-written as\n\n$\\frac{x+3}{3}=\\frac{y-4}{-5}=\\frac{z+8}{6}$\n\nSince the required line is parallel to the given line, the direction ratios of the required line are proportional to 3, $-$5, 6.\n\nHence, the cartesian equations of the line passing through the point ($-$2, 4, $-$5) and parallel to a vector having direction ratios proportional to 3, $-$5, 6 is $\\frac{x+2}{3}=\\frac{y-4}{-5}=\\frac{z+5}{6}$.\n\n#### Question 20:\n\nFind the angle between the lines $\\stackrel{\\to }{r}=\\left(2\\stackrel{^}{i}-5\\stackrel{^}{j}+\\stackrel{^}{k}\\right)+\\lambda \\left(3\\stackrel{^}{i}+2\\stackrel{^}{j}+6\\stackrel{^}{k}\\right)$ and $\\stackrel{\\to }{r}=7\\stackrel{^}{i}-6\\stackrel{^}{k}+\\mu \\left(\\stackrel{^}{i}+2\\stackrel{^}{j}+2\\stackrel{^}{k}\\right)$.                  [CBSE 2014]\n\nLet $\\theta$ be the angle between the given lines. The given lines are parallel to the vectors $\\stackrel{\\to }{{b}_{1}}=3\\stackrel{^}{i}+2\\stackrel{^}{j}+6\\stackrel{^}{k}$ and $\\stackrel{\\to }{{b}_{2}}=\\stackrel{^}{i}+2\\stackrel{^}{j}+2\\stackrel{^}{k}$, respectively.\n\nSo, the angle $\\theta$ between the given lines is given by\n\nThus, the angle between the given lines is ${\\mathrm{cos}}^{-1}\\left(\\frac{19}{21}\\right)$.\n\n#### Question 21:\n\nFind the angle between the lines $2x=3y=-z$ and $6x=-y=-4z$.           [CBSE 2015]\n\nThe equations of the given lines can be re-written as\n\n$\\frac{x}{3}=\\frac{y}{2}=\\frac{z}{-6}$ and $\\frac{x}{2}=\\frac{y}{-12}=\\frac{z}{-3}$\n\nWe know that angle between the lines $\\frac{x-{x}_{1}}{{a}_{1}}=\\frac{y-{y}_{1}}{{b}_{1}}=\\frac{z-{z}_{1}}{{c}_{1}}$ and $\\frac{x-{x}_{2}}{{a}_{2}}=\\frac{y-{y}_{2}}{{b}_{2}}=\\frac{z-{z}_{2}}{{c}_{2}}$ is given by $\\mathrm{cos}\\theta =\\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\\sqrt{{a}_{1}^{2}+{b}_{1}^{2}+{c}_{1}^{2}}\\sqrt{{a}_{2}^{2}+{b}_{2}^{2}+{c}_{2}^{2}}}$.\n\nLet $\\theta$ be the angle between the given lines.\n\nThus, the angle between the given lines is $\\frac{\\mathrm{\\pi }}{2}$.\n\n#### Question 1:\n\nThe angle between the straight lines\nis\n(a) 45°\n(b) 30°\n(c) 60°\n(d) 90°\n\n(d) 90°\n\nWe have\n\nThe direction ratios of the given lines are proportional to 2, 5, 4 and 1, 2, $-$3.\n\nThe given lines are parallel to the vectors .\n\nLet $\\theta$ be the angle between the given lines.\n\nNow,\n\n#### Question 2:\n\nThe lines $\\frac{x}{1}=\\frac{y}{2}=\\frac{z}{3}$ and $\\frac{x-1}{-2}=\\frac{y-2}{-4}=\\frac{z-3}{-6}$ are\n(a) coincident\n(b) skew\n(c) intersecting\n(d) parallel\n\n(a) coincident\n\nThe equation of the given lines are\n\nThus, the two lines are parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3).\n\nNow,\n\nSince the distance between the two parallel line is 0, the given lines are coincident.\n\n#### Question 3:\n\nThe direction ratios of the line perpendicular to the lines\n\nare proportional to\n(a) 4, 5, 7\n(b) 4, −5, 7\n(c) 4, −5, −7\n(d) −4, 5, 7\n\n(a) 4, 5, 7\n\nWe have\n\nThe direction ratios of the given lines are proportional to 2, $-$3, 1 and 1, 2, $-$2.\n\nThe vectors parallel to the given vectors are .\n\nVector perpendicular to the given two lines is\n\nHence, the direction ratios of the line perpendicular to the given two lines are proportional to 4, 5, 7.\n\n#### Question 4:\n\nThe angle between the lines is\n(a) ${\\mathrm{cos}}^{-1}\\left(\\frac{1}{65}\\right)$\n\n(b) $\\frac{\\mathrm{\\pi }}{6}$\n\n(c) $\\frac{\\mathrm{\\pi }}{3}$\n\n(d) $\\frac{\\mathrm{\\pi }}{4}$\n\n(c) $\\frac{\\mathrm{\\pi }}{3}$\n\nWe have\n\nThe direction ratios of the given lines are proportional to 1, 1, 2 and .\n\nThe given lines are parallel to vectors .\n\nLet $\\theta$ be the angle between the given lines.\n\nNow,\n\n#### Question 5:\n\nThe direction ratios of the line xy + z − 5 = 0 = x − 3y − 6 are proportional to\n(a) 3, 1, −2\n\n(b) 2, −4, 1\n\n(c)\n\n(d)\n\n(a) 3, 1, −2\n\nWe have\nxy + z − 5 = 0 = x − 3y − 6\n\nFrom (1) and (2), we get\n\nSo, the given equation can be re-written as\n\n$\\frac{x-6}{3}=\\frac{y}{1}=\\frac{z+1}{-2}$\n\nHence, the direction ratios of the given line are proportional to 3, 1,$-$2.\n\n#### Question 6:\n\nThe perpendicular distance of the point P (1, 2, 3) from the line $\\frac{x-6}{3}=\\frac{y-7}{2}=\\frac{z-7}{-2}$ is\n(a) 7\n(b) 5\n(c) 0\n(d) none of these\n\n(a) 7\n\nWe have\n\n$\\frac{x-6}{3}=\\frac{y-7}{2}=\\frac{z-7}{-2}$\n\nLet point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector $\\stackrel{\\to }{b}=3\\stackrel{^}{i}+2\\stackrel{^}{j}-2\\stackrel{^}{k}$.\n\nNow,\n\n$\\stackrel{\\to }{PQ}=5\\stackrel{^}{i}+5\\stackrel{^}{j}+4\\stackrel{^}{k}$\n\n#### Question 7:\n\nThe equation of the line passing through the points is\n(a)\n\n(b)\n\n(c)\n\n(d) none of these\n\n(c)\n\nEquation of the line passing through the points having position vectors is\n\n#### Question 8:\n\nIf a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =\n(a) −2\n(b) −1\n(c) 1\n(d) 2\n\n(b) −1\n\nIf a line makes angles α, β and γ with the axes, then\n${\\mathrm{cos}}^{2}\\alpha +{\\mathrm{cos}}^{2}\\beta +{\\mathrm{cos}}^{2}\\gamma =1$                           ...(1)\n\nWe have\n\n#### Question 9:\n\nIf the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are\n(a)\n\n(b)\n\n(c)\n\n(d)\n\n(a)\n\nThe direction ratios of the line are proportional to 1, $-$3, 2.\n\n$\\therefore$ The direction cosines of the line are\n\n#### Question 10:\n\nIf a line makes angle with x-axis and y-axis respectively, then the angle made by the line with z-axis is\n(a) π/2\n(b) π/3\n(c) π/4\n(d) 5π/12\n\n(b) π/3\n\nIf a line makes angles α, β and γ with the axes, then ${\\mathrm{cos}}^{2}\\alpha +{\\mathrm{cos}}^{2}\\beta +{\\mathrm{cos}}^{2}\\gamma =1$.\n\nHere,\n\n$\\alpha =\\frac{\\mathrm{\\pi }}{3}\\phantom{\\rule{0ex}{0ex}}\\beta =\\frac{\\mathrm{\\pi }}{4}$\n\nNow,\n\n#### Question 11:\n\nThe projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are\n(a)\n\n(b)\n\n(c)\n\n(d) none of these\n\n(a)\n\nIf a line makes angles α, β and γ with the axes, then\n\nLet r be the length of the line segment. Then,\n\nSubstituting r = 13 in (2), we get\n\nThus, the direction cosines of the line are .\n\n#### Question 12:\n\nThe lines are\n(a) parallel\n(b) intersecting\n(c) skew\n(d) coincident\n\n(d) coincident\n\nThe equations of the given lines are\n\nThus, the two lines are parallel to the vector $\\stackrel{\\to }{b}=\\stackrel{^}{i}+2\\stackrel{^}{j}+3\\stackrel{^}{k}$ and pass through the points (0, 0, 0) and (1, 2, 3).\n\nNow,\n\nSince, the distance between the two parallel lines is 0, the given two lines are coincident lines.\n\nDisclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.\n\n#### Question 13:\n\nThe straight line $\\frac{x-3}{3}=\\frac{y-2}{1}=\\frac{z-1}{0}$ is\n(a) parallel to x-axis\n(b) parallel to y-axis\n(c) parallel to z-axis\n(d) perpendicular to z-axis\n\n(d) perpendicular to z-axis\n\nWe have\n\n$\\frac{x-3}{3}=\\frac{y-2}{1}=\\frac{z-1}{0}$\n\nAlso, the given line is parallel to the vector $\\stackrel{\\to }{b}=3\\stackrel{^}{i}+\\stackrel{^}{j}+0\\stackrel{^}{k}$.\n\nLet $x\\stackrel{^}{i}+y\\stackrel{^}{j}+z\\stackrel{^}{k}$ be perpendicular to the given line.\n\nNow,\n$3x+4y+0z=0$\nIt is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).\n\nHence, the given line is perpendicular to z-axis.\n\n#### Question 14:\n\nThe shortest distance between the lines\nis\n(a) $\\sqrt{30}$\n(b) $2\\sqrt{30}$\n(c) $5\\sqrt{30}$\n(d) $3\\sqrt{30}$\n\n(d) $3\\sqrt{30}$\n\nWe have\n\nWe know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, $-$1, 1.\n\nIts vector equation is\n\nAlso, line (2) passes through the point ($-$3, $-$7, 6) and has direction ratios proportional to $-$3, 2, 4.\n\nIts vector equation is\n\nNow,\n\nThe shortest distance between the lines is given by\n\n#### Question 1:\n\nFind the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector $3\\stackrel{^}{i}+2\\stackrel{^}{j}-8\\stackrel{^}{k}.$\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nVector equation of the required line is given by\n\nReducing (1) to cartesian form, we get\n\n#### Question 2:\n\nFind the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).\n\nWe know that the vector equation of a line passing through the points with position vectors $\\stackrel{\\to }{a}$ and $\\stackrel{\\to }{b}$ is , where $\\lambda$ is a scalar.\n\nHere,\n\nVector equation of the required line is\n\n#### Question 3:\n\nFind the vector equation of a line which is parallel to the vector $2\\stackrel{^}{i}-\\stackrel{^}{j}+3\\stackrel{^}{k}$ and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nSo, the vector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 4:\n\nA line passes through the point with position vector $2\\stackrel{^}{i}-3\\stackrel{^}{j}+4\\stackrel{^}{k}$ and is in the direction of $3\\stackrel{^}{i}+4\\stackrel{^}{j}-5\\stackrel{^}{k}.$ Find equations of the line in vector and cartesian form.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n.\n\nSo, the vector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 5:\n\nABCD is a parallelogram. The position vectors of the points A, B and C are respectively, Find the vector equation of the line BD. Also, reduce it to cartesian form.\n\nWe know that the position vector of the mid-point of $\\stackrel{\\to }{a}$ and $\\stackrel{\\to }{b}$ is $\\frac{\\stackrel{\\to }{a}+\\stackrel{\\to }{b}}{2}$.\n\nLet the position vector of point D be $x\\stackrel{^}{i}+y\\stackrel{^}{j}+z\\stackrel{^}{k}$.\n\nPosition vector of mid-point of A and C = Position vector of mid-point of B and D\n\nThe vector equation of line BD passing through the points with position vectors $\\stackrel{\\to }{a}$(B) and $\\stackrel{\\to }{b}$(D) is .\n\nHere,\n\nVector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 6:\n\nFind in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).\n\nWe know that the vector equation of a line passing through the points with position vectors $\\stackrel{\\to }{a}$ and $\\stackrel{\\to }{b}$ is , where $\\lambda$ is a scalar.\n\nHere,\n\nVector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\n#### Question 7:\n\nFind the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector $\\stackrel{^}{i}-2\\stackrel{^}{j}+3\\stackrel{^}{k}.$ Reduce the corresponding equation in cartesian from.\n\nWe know that the vector equation of a line passing through a point with position vector $\\stackrel{\\to }{a}$ and parallel to the vector $\\stackrel{\\to }{b}$ is .\n\nHere,\n\nVector equation of the required line is\n\nReducing (1) to cartesian form, we get\n\nView NCERT Solutions for all chapters of Class 12" ]
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https://www.groundai.com/project/asymmetric-minwise-hashing/
[ "Asymmetric Minwise Hashing\n\n# Asymmetric Minwise Hashing\n\nAnshumali Shrivastava\nDepartment of Computer Science\nComputer and Information Science\nCornell University\nIthaca, NY 14853, USA\nanshu@cs.cornell.edu\nPing Li\nDepartment of Statistics and Biostatistics\nDepartment of Computer Science\nRutgers University\nPiscataway, NJ 08854, USA\npingli@stat.rutgers.edu\n###### Abstract\n\nMinwise hashing (Minhash) is a widely popular indexing scheme in practice. Minhash is designed for estimating set resemblance and is known to be suboptimal in many applications where the desired measure is set overlap (i.e., inner product between binary vectors) or set containment. Minhash has inherent bias towards smaller sets, which adversely affects its performance in applications where such a penalization is not desirable. In this paper, we propose asymmetric minwise hashing (MH-ALSH), to provide a solution to this problem. The new scheme utilizes asymmetric transformations to cancel the bias of traditional minhash towards smaller sets, making the final “collision probability” monotonic in the inner product. Our theoretical comparisons show that for the task of retrieving with binary inner products asymmetric minhash is provably better than traditional minhash and other recently proposed hashing algorithms for general inner products. Thus, we obtain an algorithmic improvement over existing approaches in the literature. Experimental evaluations on four publicly available high-dimensional datasets validate our claims and the proposed scheme outperforms, often significantly, other hashing algorithms on the task of near neighbor retrieval with set containment. Our proposal is simple and easy to implement in practice.\n\n\\@testdef\n\nundefined\n\n## 1 Introduction\n\nRecord matching (or linkage), data cleansing and plagiarism detection are among the most frequent operations in many large-scale data processing systems over the web. Minwise hashing (or minhash) [6, 7] is a popular technique deployed by big data industries for these tasks. Minhash was originally developed for economically estimating the resemblance similarity between sets (which can be equivalently viewed as binary vectors). Later, because of its locality sensitive property , minhash became a widely used hash function for creating hash buckets leading to efficient algorithms for numerous applications including spam detection , collaborative filtering , news personalization , compressing social networks , graph sampling , record linkage , duplicate detection , all pair similarity , etc.\n\n### 1.1 Sparse Binary Data, Set Resemblance, and Set Containment\n\nBinary representations for web documents are common, largely due to the wide adoption of the “Bag of Words” (BoW) representations for documents and images. In BoW representations, the word frequencies within a document follow power law. A significant number of words (or combinations of words) occur rarely in a document and most of the higher order shingles in the document occur only once. It is often the case that just the presence or absence information suffices in practice [9, 19, 23, 28]. Leading search companies routinely use sparse binary representations in their large data systems .\n\nThe underlying similarity measure of interest with minhash is the resemblance, also known as the Jaccard similarity. The resemblance similarity between two sets , is\n\n R=|x∩y||x∪y|=afx+fy−a, (1) where fx=|x|,  fy=|y|,  a=|x∩y|.\n\nSets can be equivalently viewed as binary vectors with each component indicating the presence or absence of an attribute. The cardinality (e.g., , ) is the number of nonzeros in the binary vector.\n\nWhile the resemblance similarity is convenient and useful in numerous applications, there are also many scenarios where the resemblance is not the desirable similarity measure [1, 11]. For instance, consider text descriptions of two restaurants:\n\n1. “Five Guys Burgers and Fries Brooklyn New York”\n\n2. “Five Kitchen Berkley”\n\nShingle based representations for strings are common in practice. Typical (first-order) shingle based representations of these names will be (i) {five, guys, burgers, and, fries, brooklyn, new, york } and (ii) {five, kitchen, berkley}. Now suppose the query is “Five Guys” which in shingle representation is {Five, Guys}. We would like to match and search the records, for this query “Five Guys”, based on resemblance. Observe that the resemblance between query and record (i) is = 0.25, while that with record (ii) is = 0.33. Thus, simply based on resemblance, record (ii) is a better match for query “Five Guys” than record (i), which should not be correct in this content.\n\nClearly the issue here is that the resemblance penalizes the sizes of the sets involved. Shorter sets are unnecessarily favored over longer ones, which hurts the performance in record matching  and other applications. There are many other scenarios where such penalization is undesirable. For instance, in plagiarism detection, it is typically immaterial whether the text is plagiarized from a big or a small document.\n\nTo counter the often unnecessary penalization of the sizes of the sets with resemblance, a modified measure, the set containment (or Jaccard containment) was adopted [6, 1, 11]. Jaccard containment of set and with respect to is defined as\n\n JC=|x∩y||x|=afx. (2)\n\nIn the above example with query “Five Guys” the Jaccard containment with respect to query for record (i) will be and with respect to record (ii) it will be , leading to the desired ordering. It should be noted that for any fixed query , the ordering under Jaccard containment with respect to the query, is the same as the ordering with respect to the intersection (or binary inner product). Thus, near neighbor search problem with respect to is equivalent to the near neighbor search problem with respect to .\n\n### 1.2 Maximum Inner Product Search (MIPS) and Maximum Containment Search (MCS)\n\nFormally, we state our problem of interest. We are given a collection containing sets (or binary vectors) over universe with (or binary vectors in ). Given a query , we are interested in the problem of finding such that\n\n x=argmaxx∈C|x∩q|=argmaxx∈CqTx; (3)\n\nwhere is the cardinality of the set. This is the so-called maximum inner product search (MIPS) problem.\n\nFor binary data, the MIPS problem is equivalent to searching with Jaccard containment with respect to the query, because the cardinality of the query does not affect the ordering and hence the .\n\n x=argmaxx∈C|x∩q|=argmaxx∈C|x∩q||q|; (4)\n\nwhich is also referred to as the maximum containment search (MCS) problem.\n\n### 1.3 Shortcomings of Inverted Index Based Approaches for MIPS (and MCS)\n\nOwing to its practical significance, there have been many existing heuristics for solving the MIPS (or MCS) problem [30, 33, 12]. A notable recent work among them made use of the inverted index based approach . Inverted indexes might be suitable for problems when the sizes of documents are small and each record only contains few words. This situation, however, is not commonly observed in practice. The documents over the web are large with huge vocabulary. Moreover, the vocabulary blows up very quickly once we start using higher-order shingles. In addition, there is an increasing interest in enriching the text with extra synonyms to make the search more effective and robust to semantic meanings , at the cost of a significant increase of the sizes of the documents. Furthermore, if the query contains many words then the inverted index is not very useful. To mitigate this issue several additional heuristics were proposed, for instance, the heuristic based on minimal infrequent sets . Computing minimal infrequent sets is similar to the set cover problem which is hard in general and thus  resorted to greedy heuristics. The number of minimal infrequent sets could be huge in general and so these heuristics can be very costly. Also, such heuristics require the knowledge of the entire dataset before hand which is usually not practical in a dynamic environment like the web. In addition, inverted index based approaches do not have theoretical guarantees on the query time and their performance is very much dataset dependent.\n\n### 1.4 Probabilistic Hashing\n\nLocality Sensitive Hashing (LSH)  based randomized techniques are common and successful in industrial practice for efficiently solving NNS (near neighbor search). They are some of the few known techniques that do not suffer from the curse of dimensionality. Hashing based indexing schemes provide provably sub-linear algorithms for search which is a boon in this era of big data where even linear search algorithms are impractical due to latency. Furthermore, hashing based indexing schemes are massively parallelizable and can be updated incrementally (on data streams), which makes them ideal for modern distributed systems. The prime focus of this paper will be on efficient hashing based algorithms for binary inner products.\n\nDespite the interest in Jaccard containment and binary inner products, there were no hashing algorithms for these measures for a long time and minwise hashing is still a widely popular heuristic . Very recently, it was shown that general inner products for real vectors can be efficiently solved by using asymmetric locality sensitive hashing schemes [35, 37]. The asymmetry is necessary for the general inner products and an impossibility of having a symmetric hash function can be easily shown using elementary arguments. Thus, binary inner product (or set intersection) being a special case of general inner products also admits provable efficient search algorithms with these asymmetric hash functions which are based on random projections. However, it is known that random projections are suboptimal for retrieval in the sparse binary domain . Hence, it is expected that the existing asymmetric locality sensitive hashing schemes for general inner products are likely to be suboptimal for retrieving with sparse high dimensional binary-like datasets, which are common over the web.\n\n### 1.5 Our Contributions\n\nWe investigate hashing based indexing schemes for the problem of near neighbor search with binary inner products and Jaccard containment. Binary inner products are special. The impossibility of existence of LSH for general inner products shown in  does not hold for the binary case. On the contrary, we provide an explicit construction of a provable LSH based on sampling, although our immediate investigation reveals that such an existential result is only good in theory and unlikely to be a useful hash function in practice.\n\nRecent results on hashing algorithms for maximum inner product search  have shown the usefulness of asymmetric transformations in constructing provable hash functions for new similarity measures, which were otherwise impossible. Going further along this line, we provide a novel (and still very simple) asymmetric transformation for binary data, that corrects minhash and removes the undesirable bias of minhash towards the sizes of the sets involved. Such an asymmetric correction eventually leads to a provable hashing scheme for binary inner products, which we call asymmetric minwise hashing (MH-ALSH). Our theoretical comparisons show that for binary data, which are common over the web, the new hashing scheme is provably more efficient that the recently proposed asymmetric hash functions for general inner products [35, 37]. Thus, we obtain a provable algorithmic improvement over the state-of-the-art hashing technique for binary inner products. The construction of our asymmetric transformation for minhash could be of independent interest in itself.\n\nThe proposed asymmetric minhash significantly outperforms existing hashing schemes, in the tasks of ranking and near neighbor search with Jaccard containment as the similarity measure, on four real-world high-dimensional datasets. Our final proposed algorithm is simple and only requires very small modifications of the traditional minhash and hence it can be easily adopted in practice.\n\n## 2 Background\n\n### 2.1 c-Approximate Near Neighbor Search and Classical LSH\n\nPast attempts of finding efficient algorithms, for exact near neighbor search based on space partitioning, often turned out to be a disappointment with the massive dimensionality of modern datasets . Due to the curse of dimensionality, theoretically it is hopeless to obtain an efficient algorithm for exact near neighbor search. Approximate versions of near neighbor search problem were proposed  to overcome the linear query time bottleneck. One commonly adopted such formulation is the -approximate Near Neighbor (-NN).\n\n###### Definition 1\n\n(-Approximate Near Neighbor or -NN).  Given a set of points in a -dimensional space , and parameters , , construct a data structure which, given any query point q, does the following with probability : if there exist an -near neighbor of q in P, it reports some -near neighbor.\n\nThe usual notion of -near neighbor is in terms of distance. Since we are dealing with similarities, we define -near neighbor of point as a point with , where is the similarity function of interest.\n\nThe popular technique, with near optimal guarantees for -NN in many interesting cases, uses the underlying theory of Locality Sensitive Hashing (LSH) . LSH are family of functions, with the property that similar input objects in the domain of these functions have a higher probability of colliding in the range space than non-similar ones. More specifically, consider a family of hash functions mapping to some set .\n\n###### Definition 2\n\n(Locality Sensitive Hashing) A family is called sensitive if for any two point and chosen uniformly from satisfies the following:\n\n• if then\n\n• if then\n\nFor approximate nearest neighbor search typically, and is needed. Note, as we are defining neighbors in terms of similarity. To obtain distance analogy we can resort to\n\n###### Fact 1\n\n Given a family of -sensitive hash functions, one can construct a data structure for -NN with query time and space ,\n\nLSH trades off query time with extra preprocessing time and space that can be accomplished off-line. It requires constructing a one time data structure which costs space and further any -approximate near neighbor queries can be answered in time in the worst case.\n\nA particularly interesting sufficient condition for existence of LSH is the monotonicity of the collision probability in . Thus, if a hash function family satisfies,\n\n Prh∈H(h(x)=h(y))=g(Sim(x,y)), (5)\n\nwhere is any strictly monotonically increasing function, then the conditions of Definition 2 are automatically satisfied for all .\n\nThe quantity is a property of the LSH family, and it is of particular interest because it determines the worst case query complexity of the -approximate near neighbor search. It should be further noted, that the complexity depends on which is the operating threshold and , the approximation ratio we are ready to tolerate. In case when we have two or more LSH families for a given similarity measure, then the LSH family with smaller value of , for given and , is preferred.\n\n### 2.2 Minwise Hashing (Minhash)\n\nMinwise hashing  is the LSH for the resemblance, also known as the Jaccard similarity, between sets. In this paper, we focus on binary data vectors which can be equivalent viewed as sets.\n\nGiven a set , the minwise hashing family applies a random permutation on and stores only the minimum value after the permutation mapping. Formally minwise hashing (or minhash) is defined as:\n\n hπ(x)=min(π(x)). (6)\n\nGiven sets and , it can be shown that the probability of collision is the resemblance :\n\n Prπ(hπ(x)=hπ(y))=|x∩y||x∪y|=afx+fy−a=R. (7)\n\nwhere , , and . It follows from Eq. ( 7) that minwise hashing is -sensitive family of hash function when the similarity function of interest is resemblance.\n\nEven though minhash was really meant for retrieval with resemblance similarity, it is nevertheless a popular hashing scheme used for retrieving set containment or intersection for binary data . In practice, the ordering of inner product and the ordering or resemblance can be different because of the variation in the values of and , and as argued in Section 1, which may be undesirable and lead to suboptimal results. We show later that by exploiting asymmetric transformations we can get away with the undesirable dependency on the number of nonzeros leading to a better hashing scheme for indexing set intersection (or binary inner products).\n\n### 2.3 LSH for L2 Distance (L2LSH)\n\n presented a novel LSH family for all () distances. In particular, when , this scheme provides an LSH family for distance. Formally, given a fixed number , we choose a random vector with each component generated from i.i.d. normal, i.e., , and a scalar generated uniformly at random from . The hash function is defined as:\n\n hL2w,b(x)=⌊wTx+br⌋, (8)\n\nwhere is the floor operation. The collision probability under this scheme can be shown to be\n\n Pr(hL2w,b(x)=hL2w,b(y))=Fr(d), (9)\n Fr(d)=1−2Φ(−r/d)−2√2πr/d(1−e−r2/(2d2)) (10)\n\nwhere is the cumulative density function (cdf) of standard normal distribution and is the Euclidean distance between the vectors and . This collision probability is a monotonically decreasing function of the distance and hence is an LSH for distances. This scheme is also the part of LSH package . Here is a parameter.\n\n### 2.4 LSH for Cosine Similarity (SRP)\n\nSigned Random Projections (SRP) or simhash is another popular LSH for the cosine similarity measure, which originates from the concept of Signed Random Projections (SRP) [17, 10]. Given a vector , SRP utilizes a random vector with each component generated from i.i.d. normal, i.e., , and only stores the sign of the projection. Formally simhash is given by\n\n hsign(x)=sign(wTx). (11)\n\nIt was shown in the seminal work  that collision under SRP satisfies the following equation:\n\n Prw(hsign(x)=hsign(y))=1−θπ, (12)\n\nwhere . The term is the popular cosine similarity.\n\nFor sets (or equivalently binary vectors), the cosine similarity reduces to\n\n S=a√fxfy. (13)\n\nThe recent work on coding for random projections [26, 27] has shown the advantage of SRP (and 2-bit random projections) over L2LSH for both similarity estimation and near neighbor search. Interestingly, another recent work  has shown that for binary data (actually even sparse non-binary data), minhash can significantly outperform SRP for near neighbor search even as we evaluate both SRP and minhash in terms of the cosine similarity (although minhash is designed for resemblance). This motivates us to design asymmetric minhash for achieving better performance in retrieving set containments. But first, we provide an overview of asymmetric LSH for general inner products (not restricted to binary data).\n\n### 2.5 Asymmetric LSH (ALSH) for General Inner Products\n\nThe term “ALSH” stands for asymmetric LSH, as used in a recent work . Through an elementary argument, showed that it is not possible to have a Locality Sensitive Hashing (LSH) family for general unnormalized inner products.\n\nFor inner products between vectors and , it is possible to have . Thus for any hashing scheme to be a valid LSH, we must have , which is an impossibility. It turns out that there is a simple fix, if we allow asymmetry in the hashing scheme. Allowing asymmetry leads to an extended framework of asymmetric locality sensitive hashing (ALSH). The idea to is have a different hashing scheme for assigning buckets to the data point in the collection , and an altogether different hashing scheme while querying.\n\nDefinition: (Asymmetric Locality Sensitive Hashing (ALSH)) A family , along with the two vector functions (Query Transformation) and (Preprocessing Transformation), is called -sensitive if for a given -NN instance with query , and the hash function chosen uniformly from satisfies the following:\n\n• if then\n\n• if then\n\nHere is any point in the collection . Asymmetric LSH borrows all theoretical guarantees of the LSH.\n\n###### Fact 2\n\nGiven a family of hash function and the associated query and preprocessing transformations and respectively, which is -sensitive, one can construct a data structure for -NN with query time and space , where .\n\n showed that using asymmetric transformations, the problem of maximum inner product search (MIPS) can be reduced to the problem of approximate near neighbor search in . The algorithm first starts by scaling all by a constant large enough, such that . The proposed ALSH family (L2-ALSH) is the LSH family for distance with the Preprocessing transformation and the Query transformation defined as follows:\n\n PL2(x) =[x;||x||22;....;||x||2m2;1/2;...;1/2] (14) QL2(x) =[x;1/2;...;1/2;||x||22;....;||x||2m2], (15)\n\nwhere [;] is the concatenation. appends scalers of the form followed by “1/2s” at the end of the vector , while first appends “1/2s” to the end of the vector and then scalers of the form . It was shown that this leads to provably efficient algorithm for MIPS.\n\n###### Fact 3\n\n For the problem of -approximate MIPS in a bounded space, one can construct a data structure having\nquery time and space , where is the solution to constrained optimization (16).\n\n ρ∗L2−ALSH =minU<1,m∈N,rlogFr(√m/2−2S0(U2V2)+2U2m+1)logFr(√m/2−2cS0(U2V2)) (16) s.t. U(2m+1−2)V2S0<1−c,\n\nHere the guarantees depends on the maximum norm of the space .\n\nQuickly, it was realized that a very similar idea can convert the MIPS problem in the problem of maximum cosine similarity search which can be efficiently solve by SRP leading to a new and better ALSH for MIPS Sign-ALSH  which works as follows: The algorithm again first starts by scaling all by a constant large enough, such that . The proposed ALSH family (Sign-ALSH) is the SRP family for cosine similarity with the Preprocessing transformation and the Query transformation defined as follows:\n\n Psign(x) =[x;1/2−||x||22;...;1/2−||x||2m2;0;...;0] (17) Qsign(x) =[x;0;...;0;1/2−||x||22;...;1/2−||x||2m2], (18)\n\nwhere [;] is the concatenation. appends scalers of the form followed by “0s” at the end of the vector , while appends “0” followed by scalers of the form to the end of the vector . It was shown that this leads to provably efficient algorithm for MIPS.\n\nAs demonstrated by the recent work  on coding for random projections, there is a significant advantage of SRP over L2LSH for near neighbor search. Thus, it is not surprising that Sign-ALSH outperforms L2-ALSH for the MIPS problem.\n\nSimilar to L2LSH, the runtime guarantees for Sign-ALSH can be shown as:\n\n###### Fact 4\n\nFor the problem of -approximate MIPS, one can construct a data structure having query time and space , where is the solution to constraint optimization problem\n\n (19) z∗=⎡⎢ ⎢⎣(m−m2m−1)+√(m−m2m−1)2+m2(2m−1)4(2m−1)⎤⎥ ⎥⎦2−m\n\nThere is a similar asymmetric transformation [3, 31] which followed by signed random projection leads to another ALSH having very similar performance to Sign-ALSH. The values, which were also very similar to the can be shown as\n\n ρSign =log(1−1πcos−1(S0V2))log(1−1πcos−1(cS0V2)) (20)\n\nBoth L2-ALSH and Sign-ALSH work for any general inner products over . For sparse and high-dimensional binary dataset which are common over the web, it is known that minhash is typically the preferred choice of hashing over random projection based hash functions . We show later that the ALSH derived from minhash, which we call asymmetric minwise hashing (MH-ALSH), is more suitable for indexing set intersection for sparse binary vectors than the existing ALSHs for general inner products.\n\n## 3 A Construction of LSH for Indexing Binary Inner Products\n\nIn , it was shown that there cannot exist any LSH for general unnormalized inner product. The key argument used in the proof was the fact that it is possible to have and with . However, binary inner product (or set intersection) is special. For any two binary vectors and we always have . Therefore, the argument used to show non-existence of LSH for general inner products does not hold true any more for this special case. In fact, there does exist an LSH for binary inner products (although it is mainly for theoretical interest). We provide an explicit construction in this section.\n\nOur proposed LSH construction is based on sampling. Simply sampling a random component leads to the popular LSH for hamming distance . The ordering of inner product is different from that of hamming distance. The hamming distance between and query is given by , while we want the collision probability to be monotonic in the inner product . makes it non-monotonic in . Note that has no effect on ordering of because it is constant for every query. To construct an LSH monotonic in binary inner product, we need an extra trick.\n\nGiven a binary data vector , we sample a random co-ordinate (or attribute). If the value of this co-ordinate is (in other words if this attribute is present in the set), our hash value is a fixed number . If this randomly sampled co-ordinate has value (or the attribute is absent) then we independently generate a random integer uniformly from . Formally,\n\n HS(x)={0 if xi=1, i drawn uniformlyrand(1,N)otherwise (21)\n###### Theorem 1\n\nGiven two binary vectors and , we have\n\n###### Proof 1\n\nThe probability that both and have value 0 is . The only other way both can be equal is when the two independently generated random numbers become equal, which happens with probability . The total probability is which simplifies to the desired expression.\n\n###### Corollary 1\n\nis -sensitive locality sensitive hashing for binary inner product with\n\n### 3.1 Shortcomings\n\nThe above LSH for binary inner product is likely to be very inefficient for sparse and high dimensional datasets. For those datasets, typically the value of is very high and the sparsity ensures that is very small. For modern web datasets, we can have running into billions (or ) while the sparsity is only in few hundreds or perhaps thousands . Therefore, we have which essentially boils down to . In other words, the hashing scheme becomes worthless in sparse high dimensional domain. On the other hand, if we observe the collision probability of minhash Eq.( 7), the denominator is , which is usually of the order of and much less than the dimensionality for sparse datasets.\n\nAnother way of realizing the problem with the above LSH is to note that it is informative only if a randomly sampled co-ordinate has value equal to 1. For very sparse dataset with , sampling a non zero coordinate has probability . Thus, almost all of the hashes will be independent random numbers.\n\n### 3.2 Why Minhash Can Be a Reasonable Approach?\n\nIn this section, we argue why retrieving inner product based on plain minhash is a reasonable thing to do. Later, we will show a provable way to improve it using asymmetric transformations.\n\nThe number of nonzeros in the query, i.e., does not change the identity of in Eq.(4). Let us assume that we have data of bounded sparsity and define constant as\n\n M=maxx∈C|x| (23)\n\nwhere is simply the maximum number of nonzeros (or maximum cardinality of sets) seen in the database. For sparse data seen in practice is likely to be small compared to . Outliers, if any, can be handled separately. By observing that , we also have\n\n afq+M−a≤afx+fq−a=R≤afq (24)\n\nThus, given the bounded sparsity, if we assume that the number of nonzeros in the query is given, then we can show that minhash is an LSH for inner products because the collision probability can be upper and lower bounded by purely functions of and .\n\n###### Theorem 2\n\nGiven bounded sparsity and query with , minhash is a sensitive for inner products with\n\nThis explains why minhash might be a reasonable hashing approach for retrieving inner products or set intersection.\n\nHere, if we remove the assumption that then in the worst case and we get in the denominator. Note that the above is the worst case analysis and the assumption is needed to obtain any meaningful with minhash. We show the power of ALSH in the next section, by providing a better hashing scheme and we do not even need the assumption of fixing .\n\n## 4 Asymmetric Minwise Hashing (MH-ALSH)\n\nIn this section, we provide a very simple asymmetric fix to minhash, named asymmetric minwise hashing (MH-ALSH), which makes the overall collision probability monotonic in the original inner product . For sparse binary data, which is common in practice, we later show that the proposed hashing scheme is superior (both theoretically as well as empirically) compared to the existing ALSH schemes for inner product .\n\n### 4.1 The New ALSH for Binary Data\n\nWe define the new preprocessing and query transformations and as:\n\n P′(x) =[x;1;1;1;...;1;0;0;...;0] (25) Q′(x) =[x;0;0;0;...;0], (26)\n\nwhere [;] is the concatenation to vector . For we append 1s and rest zeros, while in we simply append zeros.\n\nAt this point we can already see the power of asymmetric transformations. The original inner product between and is unchanged and its value is . Given the query , the new resemblance between and is\n\n R′=|P′(x)∩Q′(q)||P′(x)∪Q′(q)|=aM+fq−a. (27)\n\nIf we define our new similarity as , which is similar in nature to the containment , then the near neighbors in this new similarity are the same as near neighbors with respect to either set intersection or set containment . Thus, we can instead compute near neighbors in which is also the resemblance between and . We can therefore use minhash on and .\n\nObserve that now we have in the denominator, where is the maximum nonzeros seen in the dataset (the cardinality of largest set), which for very sparse data is likely to be much smaller than . Thus, asymmetric minhash is a better scheme than with collision probability roughly for very sparse datasets where we usually have . This is an interesting example where we do have an LSH scheme but an altogether different asymmetric LSH (ALSH) improves over existing LSH. This is not surprising because asymmetric LSH families are more powerful .\n\nFrom theoretical perspective, to obtain an upper bound on the query and space complexity of -approximate near neighbor with binary inner products, we want the collision probability to be independent of the quantity . This is not difficult to achieve. The asymmetric transformation used to get rid of in the denominator can be reapplied to get rid of .\n\nFormally, we can define and as :\n\n P′′(x)=Q′(P′(x));     Q′′(x)=P′(Q′(x)); (28)\n\nwhere in we append 1s and rest zeros, while in we append zeros, then 1s and rest zeros\n\nAgain the inner product is unaltered, and the new resemblance then becomes\n\n R′′=|P′′(x)∩Q′′(q)||P′′(x)∪Q′′(q)|=a2M−a. (29)\n\nwhich is independent of and is monotonic in . This allows us to achieve a formal upper bound on the complexity of -approximate maximum inner product search with the new asymmetric minhash.\n\nFrom the collision probability expression, i.e., Eq. (29), we have\n\n###### Theorem 3\n\nMinwise hashing along with Query transformation and Preprocessing transformation defined by Equation 28 is a sensitive asymmetric hashing family for set intersection.\n\nThis leads to an important corollary.\n\n###### Corollary 2\n\nThere exist an algorithm for -approximate set intersection (or binary inner products), with bounded sparsity , that requires space and , where\n\n ρMH−ALSH=logS02M−S0logcS02M−cS0<1 (30)\n\nGiven query and any point , the collision probability under traditional minhash is . This penalizes sets with high , which in many scenarios not desirable. To balance this negative effect, asymmetric transformation penalizes sets with smaller . Note, that ones added in the transformations gives additional chance in proportion to for minhash of not to match with the minhash of . This asymmetric probabilistic correction balances the penalization inherent in minhash. This is a simple way of correcting the probability of collision which could be of independent interest in itself. We will show in our evaluation section, that despite this simplicity such correction leads to significant improvement over plain minhash.\n\n### 4.2 Faster Sampling\n\nOur transformations and always create sets with nonzeros. In case when is big, hashing might take a lot of time. We can use fast consistent weighted sampling [29, 22] for efficient generation of hashes. We can instead use transformations and that makes the data non-binary as follows\n\n P′′′(x) =[x;M−fx;0] (31) Q′′′(x) =[x;0;M−fx]\n\nIt is not difficult to see that the weighted Jaccard similarity (or weighted resemblance) between and for given query and any is\n\n RW=∑imin(P′′′(x)i,Q′′′(q)i)∑imax(P′′′(x)i,Q′′′(q)i)=a2M−a. (32)\n\nTherefore, we can use fast consistent weighted sampling for weighted Jaccard similarity on and to compute the hash values in time constant per nonzero weights, rather than maximum sparsity . In practice we will need many hashes for which we can utilize the recent line of work that make minhash and weighted minhash significantly much faster [36, 18].\n\n## 5 Theoretical Comparisons\n\nFor solving the MIPS problem in general data types, we already know two asymmetric hashing schemes, L2-ALSH and Sign-ALSH, as described in Section 2.5. In this section, we provide theoretical comparisons of the two existing ALSH methods with the proposed asymmetric minwise hashing (MH-ALSH). As argued, the LSH scheme described in Section 3 is unlikely to be useful in practice because of its dependence on ; and hence we safely ignore it for simplicity of the discussion.\n\nBefore we formally compare various asymmetric LSH schemes for maximum inner product search, we argue why asymmetric minhash should be advantageous over traditional minhash for retrieving inner products. Let be the binary query vector, and denotes the number of nonzeros in the query. The for asymmetric minhash in terms of and is straightforward from the collision probability Eq.(27):\n\n ρqMH−ALSH=logS0fq+M−S0logcS0fq+M−cS0 (33)\n\nFor minhash, we have from theorem 2 . Since is the upper bound on the sparsity and is some value of inner product, we have . Using this fact, the following theorem immediately follows\n\n###### Theorem 4\n\nFor any query q, we have .\n\nThis result theoretically explains why asymmetric minhash is better for retrieval with binary inner products, compared to plain minhash.", null, "Figure 1: Values of ρMH−ALSH and ρsign (lower is better) with respect to approximation ratio c for different S0M. The curves show that asymmetric minhash (solid curves) is noticeably better than ALSH based on signed random projection (dashed curves) in terms of their ρ values, irrespective of the choices of S0M or c. For clarity, the results are shown in two panels.\n\nFor comparing asymmetric minhash with ALSH for general inner products, we compare with the ALSH for inner products based on signed random projections. Note that it was shown that has better theoretical values as compared to L2-ALSH . Therefore, it suffices to show that asymmetric minhash outperforms signed random projection based ALSH. Both and can be rewritten in terms of ratio as follows. Note that for binary data we have\n\n ρMH−ALSH=logS0/M2−S0/MlogcS0/M2−cS0/M;ρSign=log(1−1πcos−1(S0M))log(1−1πcos−1(cS0M)) (34)\n\nObserve that is also the upper bound on any inner product. Therefore, we have . We plot the values of and for with . The comparison is summarized in Figure 1. Note that here we use derived from [3, 31] instead of for convenience although the two schemes perform essentially identically.\n\nWe can clearly see that irrespective of the choice of threshold or the approximation ratio , asymmetric minhash outperforms signed random projection based ALSH in terms of the theoretical values. This is not surprising, because it is known that minwise hashing based methods are often significantly powerful for binary data compared to SRP (or simhash) . Therefore ALSH based on minwise hashing outperforms ALSH based on SRP as shown by our theoretical comparisons. Our proposal thus leads to an algorithmic improvement over state-of-the-art hashing techniques for retrieving binary inner products.\n\n## 6 Evaluations\n\nIn this section, we compare the different hashing schemes on the actual task of retrieving top-ranked elements based on set Jaccard containment. The experiments are divided into two parts. In the first part, we show how the ranking based on various hash functions correlate with the ordering of Jaccard containment. In the second part, we perform the actual LSH based bucketing experiment for retrieving top-ranked elements and compare the computational saving obtained by various hashing algorithms.\n\n### 6.1 Datasets\n\nWe chose four publicly available high dimensional sparse datasets: EP2006111We downloaded EP2006 from LIBSVM website. The original name is “E2006LOG1P” and we re-name it to “EP2006”., MNIST, NEWS20, and NYTIMES. Except MNIST, the other three are high dimensional binary “BoW” representation of the corresponding text corpus. MNIST is an image dataset consisting of 784 pixel image of handwritten digits. Binarized versions of MNIST are commonly used in literature. The pixel values in MNIST were binarized to 0 or 1 values. For each of the four datasets, we generate two partitions. The bigger partition was used to create hash tables and is referred as the training partition. The small partition which we call the query partition is used for querying. The statistics of these datasets are summarized in Table 1. The datasets cover a wide spectrum of sparsity and dimensionality.\n\n### 6.2 Competing Hash Functions\n\nWe consider the following hash functions for evaluations:\n\n1. Asymmetric minwise hashing (Proposed): This is our proposal, the asymmetric minhash described in Section 4.1.\n\n2. Traditional minwise hashing (MinHash): This is the usual minwise hashing, the popular heuristic described in Section 2.2. This is a symmetric hash function, we use as define in Eq.(6) for both query and the training set.\n\n3. L2 based Asymmetric LSH for Inner products (L2-ALSH): This is the asymmetric LSH of  for general inner products based on LSH for L2 distance.\n\n4. SRP based Asymmetric LSH for Inner Products (Sign-ALSH): This is the asymmetric hash function of  for general inner products based on SRP.\n\n### 6.3 Ranking Experiment: Hash Quality Evaluations\n\nWe are interested in knowing, how the orderings under different competing hash functions correlate with the ordering of the underlying similarity measure which in this case is the Jaccard containment. For this task, given a query vector, we compute the top-100 gold standard elements from the training set based on the Jaccard containment . Note that this is the same as the top-100 elements based on binary inner products. Give a query , we compute different hash codes of the vector and all the vectors in the training set. We then compute the number of times the hash values of a vector in the training set matches (or collides) with the hash values of query defined by\n\n Matchesx=K∑t=11(ht(q)=ht(x)), (35)\n\nwhere is the indicator function. subscript is used to distinguish independent draws of the underlying hash function. Based on we rank all elements in the training set. This procedure generates a sorted list for every query for every hash function. For asymmetric hash functions, in computing total collisions, on the query vector we use the corresponding function (query transformation) followed by underlying hash function, while for elements in the training set we use the function (preprocessing transformation) followed by the corresponding hash function.\n\nWe compute the precision and the recall of the top-100 gold standard elements in the ranked list generated by different hash functions. To compute precision and recall, we start at the top of the ranked item list and walk down in order, suppose we are at the ranked element, we check if this element belongs to the gold standard top-100 list. If it is one of the top 100 gold standard elements, then we increment the count of relevant seen by 1, else we move to . By step, we have already seen elements, so the total elements seen is . The precision and recall at that point is then computed as:\n\n Precision=relevant seenp,Recall=relevant seen100 (36)\n\nIt is important to balance both. Methodology which obtains higher precision at a given recall is superior. Higher precision indicates higher ranking of the relevant items. We finally average these values of precision and recall over all elements in the query set. The results for are summarized in Figure 2.\n\nWe can clearly see, that the proposed hashing scheme always achieves better, often significantly, precision at any given recall compared to other hash functions. The two ALSH schemes are usually always better than traditional minwise hashing. This confirms that fact that ranking based on collisions under minwise hashing can be different from the rankings under Jaccard containment or inner products. This is expected, because minwise hashing in addition penalizes the number of nonzeros leading to a ranking very different from the ranking of inner products. Sign-ALSH usually performs better than L2-LSH, this is in line with the results obtained in .", null, "Figure 2: Ranking Experiments.  Precision Vs Recall curves for retrieving top-100 items, for different hashing schemes on 4 chosen datasets. The precision and the recall were computed based on the rankings obtained by different hash functions using 32, 64 and 128 independent hash evaluations. Higher precision at a given recall is better.\n\nIt should be noted that ranking experiments only validate the monotonicity of the collision probability. Although, better ranking is definitely a very good indicator of good hash function, it does not always mean that we will achieve faster sub-linear LSH algorithm. For bucketing the probability sensitivity around a particular threshold is the most important factor, see  for more details. What matters is the gap between the collision probability of good and the bad points. In the next subsection, we compare these schemes on the actual task of near neighbor retrieval with Jaccard containment.\n\n### 6.4 Bucketing Experiment: Computational Savings in Near Neighbor Retrieval\n\nIn this section, we evaluate the four hashing schemes on the standard -parameterized bucketing algorithm  for sub-linear time retrieval of near neighbors based on Jaccard containment. In -parameterized LSH algorithm, we generate different meta-hash functions. Each of these meta-hash functions is formed by concatenating different hash values as\n\n Bj(x)=[hj1(x);hj2(x);...;hjK(x)], (37)\n\nwhere and , are different independent evaluations of the hash function under consideration. Different competing scheme uses its own underlying randomized hash function .\n\nIn general, the -parameterized LSH works in two phases:\n\n1. Preprocessing Phase: We construct hash tables from the data by storing element , in the training set, at location in the hash-table . Note that for vanilla minhash which is a symmetric hashing scheme . For other asymmetric schemes, we use their corresponding functions. Preprocessing is a one time operation, once the hash tables are created they are fixed.\n\n2. Query Phase: Given a query , we report the union of all the points in the buckets , where the union is over hash tables. Again here is the corresponding function of the asymmetric hashing scheme, for minhash .\n\nTypically, the performance of a bucketing algorithm is sensitive to the choice of parameters and . Ideally, to find best and , we need to know the operating threshold and the approximation ratio in advance. Unfortunately, the data and the queries are very diverse and therefore for retrieving top-ranked near neighbors there are no common fixed threshold and approximation ratio that work for all the queries.\n\nOur objective is to compare the four hashing schemes and minimize the effect of and , if any, on the evaluations. This is achieved by finding best and at every recall level. We run the bucketing experiment for all combinations of and for all the four hash functions independently. These choices include the recommended optimal combinations at various thresholds. We then compute, for every and , the mean recall of Top- pairs and the mean number of points reported, per query, to achieve that recall. The best and at every recall level is chosen independently for different s. The plot of the mean fraction of points scanned with respect to the recall of top- gold standard near neighbors, where , is summarized in Figure 3.", null, "Figure 3: LSH Bucketing Experiments.  Average number of points retrieved per query (lower is better), relative to linear scan, evaluated by different hashing schemes at different recall levels, for top-5,  top-10,  top-20,  top-50 nearest neighbors based on Jaccard containment (or equivalently inner products), on four datasets. We show that results at the best K and L values chosen at every recall value, independently for each of the four hashing schemes.\n\nThe performance of a hashing based method varies with the variations in the similarity levels in the datasets. It can be seen that the proposed asymmetric minhash always retrieves much less number of points, and hence requires significantly less computations, compared to other hashing schemes at any recall level on all the four datasets. Asymmetric minhash consistently outperforms other hash functions irrespective of the operating point. The plots clearly establish the superiority of the proposed scheme for indexing Jaccard containment (or inner products).\n\nL2-ALSH and Sign-ALSH perform better than traditional minhash on EP2006 and NEWS20 datasets while they are worse than plain minhash on NYTIMES and MNIST datasets. If we look at the statistics of the dataset from Table 1, NYTIMES and MNIST are precisely the datasets with less variations in the number of nonzeros and hence minhash performs better. In fact, for MNIST dataset with very small variations in the number of nonzeros, the performance of plain minhash is very close to the performance of asymmetric minhash. This is of course expected because there is negligible effect of penalization on the ordering. EP2006 and NEWS20 datasets have huge variations in their number of nonzeros and hence minhash performs very poorly on these datasets. What is exciting is that despite these variations in the nonzeros, asymmetric minhash always outperforms other ALSH for general inner products.\n\nThe difference in the performance of plain minhash and asymmetric minhash clearly establishes the utility of our proposal which is simple and does not require any major modification over traditional minhash implementation. Given the fact that minhash is widely popular, we hope that our proposal will be adopted.\n\n## 7 Conclusion and Future Work\n\nMinwise hashing (minhash) is a widely popular indexing scheme in practice for similarity search. Minhash is originally designed for estimating set resemblance (i.e., normalized size of set intersections). In many applications the performance of minhash is severely affected because minhash has a bias towards smaller sets. In this study, we propose asymmetric corrections (asymmetric minwise hashing, or MH-ALSH) to minwise hashing that remove this often undesirable bias. Our corrections lead to a provably superior algorithm for retrieving binary inner products in the literature. Rigorous experimental evaluations on the task of retrieving maximum inner products clearly establish that the proposed approach can be significantly advantageous over the existing state-of-the-art hashing schemes in practice, when the desired similarity is the inner product (or containment) instead of the resemblance. Our proposed method requires only minimal modification of the original minwise hashing algorithm and should be straightforward to implement in practice.\n\nFuture work: One immediate future work would be asymmetric consistent weighted sampling for hashing weighted intersection: , where and are general real-valued vectors. One proposal of the new asymmetric transformation is the following:\n\n P(x) =[x;M−D∑i=1xi;0],Q(x)=[x;0;M−D∑i=1xi], (38)\n\nwhere . It is not difficult to show that the weighted Jaccard similarity between and is monotonic in as desired. At this point, we can use existing methods for consistent weighted sampling on the new data after asymmetric transformations [29, 22, 18].\n\nAnother potentially promising topic for future work might be asymmetric minwise hashing for 3-way (or higher-order) similarities [25, 34]\n\n## References\n\n• P. Agrawal, A. Arasu, and R. Kaushik. On indexing error-tolerant set containment. 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In KDD, pages 219–228, Paris, France, 2009.\n• G. Cormode and S. Muthukrishnan. Space efficient mining of multigraph streams. In Proceedings of the twenty-fourth ACM SIGMOD-SIGACT-SIGART symposium on Principles of database systems, pages 271–282. ACM, 2005.\n• A. S. Das, M. Datar, A. Garg, and S. Rajaram. Google news personalization: scalable online collaborative filtering. In Proceedings of the 16th international conference on World Wide Web, pages 271–280. ACM, 2007.\n• M. Datar, N. Immorlica, P. Indyk, and V. S. Mirrokn. Locality-sensitive hashing scheme based on -stable distributions. In SCG, pages 253 – 262, Brooklyn, NY, 2004.\n• M. X. Goemans and D. P. Williamson. Improved approximation algorithms for maximum cut and satisfiability problems using semidefinite programming. Journal of ACM, 42(6):1115–1145, 1995.\n• B. Haeupler, M. Manasse, and K. Talwar. Consistent weighted sampling made fast, small, and easy. Technical report, arXiv:1410.4266, 2014.\n• M. Hein and O. Bousquet. Hilbertian metrics and positive definite kernels on probability measures. In AISTATS, pages 136–143, Barbados, 2005.\n• M. R. Henzinger. Finding near-duplicate web pages: a large-scale evaluation of algorithms. In SIGIR, pages 284–291, 2006.\n• P. Indyk and R. Motwani. Approximate nearest neighbors: Towards removing the curse of dimensionality. In STOC, pages 604–613, Dallas, TX, 1998.\n• S. Ioffe. Improved consistent sampling, weighted minhash and L1 sketching. In ICDM, pages 246–255, Sydney, AU, 2010.\n• Y. Jiang, C. Ngo, and J. Yang. Towards optimal bag-of-features for object categorization and semantic video retrieval. In CIVR, pages 494–501, Amsterdam, Netherlands, 2007.\n• N. Koudas, S. Sarawagi, and D. Srivastava. Record linkage: similarity measures and algorithms. In Proceedings of the 2006 ACM SIGMOD international conference on Management of data, pages 802–803. ACM, 2006.\n• P. Li and K. W. Church. A sketch algorithm for estimating two-way and multi-way associations. 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Mining of Massive Datasets. http://i.stanford.edu/ ullman/mmds.html.\n• K. Ramasamy, J. F. Naughton, and R. Kaushik. Set containment joins: The good, the bad and the ugly.\n• A. Shrivastava and P. Li. Beyond pairwise: Provably fast algorithms for approximate k-way similarity search. In NIPS, Lake Tahoe, NV, 2013.\n• A. Shrivastava and P. Li. Asymmetric LSH (ALSH) for sublinear time maximum inner product search (mips). In NIPS, Montreal, CA, 2014.\n• A. Shrivastava and P. Li. Densifying one permutation hashing via rotation for fast near neighbor search. In ICML, Beijing, China, 2014.\n• A. Shrivastava and P. Li. An improved scheme for asymmetric lsh. arXiv preprint arXiv:1410.5410, 2014.\n• A. Shrivastava and P. Li. In defense of minhash over simhash. In AISTATS, 2014.\n• R. Weber, H.-J. Schek, and S. Blott. A quantitative analysis and performance study for similarity-search methods in high-dimensional spaces. 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https://surface.syr.edu/mat_etd/72/
[ "## Mathematics - Dissertations\n\n2013\n\nDissertation\n\n#### Degree Name\n\nDoctor of Philosophy (PhD)\n\nMathematics\n\nComan, Dan\n\n#### Keywords\n\nconvex functions, extension of pluricomplex Green functions, pluricomplex Green functions, plurisubharmonic functions, polyradial functions, Reinhardt domains\n\nMathematics\n\n#### Abstract\n\nIn 1985, Klimek introduced an extremal plurisubharmonic function on bounded domains in Cn that generalizes the Green's function of one variable. This function is called the pluricomplex Green function of Ω with logarithmic pole at a and is denoted by gΩ(.,a). The aim of this thesis was to investigate the extension properties of gΩ(.,a). Let Ω0 be a bounded domain of Cn and E be a compact subset of Ω0 such that Ω : = Ω0 E is connected. In general, gΩ(.,a) cannot be extended as a pluricomplex Green function to any subdomain of Ω0 that is strictly larger that Ω. In this thesis it was proved that if Ω0 is a pseudoconvex, bounded complete Reinhardt domain in Cn and E is a strictly logarithmically convex, Reinhardt compact subset of Ω0 that does not contain 0 and does not intersect the coordinate axes, there exists a subdomain &Ωtilde; of Ω0 strictly larger that Ω such that gΩ (z,0) = g&Ωtilde; (z,0) for any z in Ω. It was also shown that in C2, one can omit the condition that E does not intersect the coordinate axes. The methods required to prove the results heavily use the relation between the plurisubharmonicity of poyradial functions on Reinhardt domains and convexity of related functions. Special classes of convex functions were introduced and discussed for this purpose. These methods were also used to discuss the extension properties of the pluricomplex Green functions when Ω0 is equal to unit the bidisk in C2 and in that case a complete solution of the problem was given\n\nCOinS" ]
[ null ]
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http://calvarychurch.www68-66-192-129.a2hosted.com/docs/score-four-guq/d0k4wsk.php?page=b7a526-transpose-of-a-matrix-in-python-inplace
[ "For a 2-D array, this is a standard matrix transpose. Not in place, creates a new matrix. Following is a simple example of nested list which could be considered as a 2x3 matrix.. matrixA = [ [2, 8, 4], [3, 1, 5] ] Run this program ONLINE In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix. Let’s say you have original matrix something like -. The two lists inside matrixA are the rows of the matrix. Recommended Tutorials: BBCode, functions, classes, text adventures Suppose we have a (n by n) matrix M, we have to find its transpose. The transpose() function is used to transpose index and columns. Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. For a 1-D array this has no effect, as a transposed vector is simply the same vector. A Python matrix is a specialized two-dimensional rectangular array of data stored in rows and columns. brightness_4 Matrix transpose using Python. Lists inside the list are the rows. About four months of gap (missing GFG), a new post. With one list comprehension, the transpose can be constructed as MT = [] for i in range ( 3 ): MT . [Demo of matrix transposition. What is Python Matrix? In Python, there is always more than one way to solve any problem. When we take the transpose of a same vector two times, we again obtain the initial vector. Find the transpose of a matrix in Python Program, Python Program to find the transpose of a matrix. Let's take a matrix X, having the following elements: To find transpose of a matrix in python, just choose a matrix which is going to transpose, and choose another matrix having column one greater than the previous matrix and row one less than the matrix. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Find the number of islands | Set 1 (Using DFS), Inplace rotate square matrix by 90 degrees | Set 1, Program to find largest element in an array, Count all possible paths from top left to bottom right of a mXn matrix, Search in a row wise and column wise sorted matrix, Maximum size square sub-matrix with all 1s, Find the longest path in a matrix with given constraints, Maximum size rectangle binary sub-matrix with all 1s, Maximum sum rectangle in a 2D matrix | DP-27, Divide and Conquer | Set 5 (Strassen's Matrix Multiplication), Rotate a matrix by 90 degree in clockwise direction without using any extra space. Lists inside the list are the rows. It is denoted as X'. Using a full size temporary does not count. Inplace (Fixed space) M x N size matrix transpose | Updated. when a matrix is non-square, we might run into loops when swapping the elements.. the following soln works: 1. assume matrix is stored as 1D array 2. for each element in the resulting (transposed) matrix, say its index is 'k', find the index of a corresponding element in the original matrix, say it's 'j' Given an M x N matrix, transpose the matrix without auxiliary memory. For Square Matrix : The below program finds transpose of A[][] and stores the result in B[][], we can change N for different dimension. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. The first row can be selected as X.And, the element in the first-row first column can be selected as X.. Transpose of a matrix is the … In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Syntax: DataFrame.transpose(self, *args, … transpose_matrix = zip(*original_matrix) So this is how we can implement the Python code for one liner transpose of a matrix. Basically, we need to swap the row and column elements. In this program, the user is asked to enter the number of rows r and columns c.Their values should be less than 10 in this program. np.atleast2d(a).T achieves this, as does a[:, np.newaxis]. Something along the lines of this: Python – Matrix Transpose. That is you need to update the original matrix with the transpose. It can be done really quickly using the built-in zip function. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. So if … By in-place, I mean not much extra storage should be allocated. In other words, transpose of A [] [] is obtained by changing A [i] [j] to A [j] [i]. Execution of transposing a matrix For Program refer :https://youtu.be/jA1f8XKIJQ4 Given an M x N matrix, transpose the matrix without auxiliary memory.It is easy to transpose matrix using an auxiliary array. Use the T attribute or the transpose() method to swap (= transpose) the rows and columns of pandas.DataFrame.. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Python Program to Transpose a Matrix. Transpose of a matrix is a task we all can perform very easily in python (Using a nested loop). Transpose of a matrix: The transpose of a matrix is a new matrix whose columns are the rows of the original. The Transpose of a Matrix means a new Matrix in which the rows are the columns (or the columns are the rows) of the original Matrix. As we know the transpose of a matrix switches the row and column indices. Reflect the DataFrame over its main diagonal by writing rows as columns and vice-versa. For example matrix = [[1,2,3],[4,5,6]] represent a matrix of order 2×3, in which matrix[i][j] is the matrix element at ith row and jth column.. To transpose a matrix we have to interchange all its row elements into column elements and column elements into row elements. The transpose of a matrix is calculated, by changing the rows as columns and columns as rows. The element at ith row and jth column in T will be placed at jth row and ith column in T’. In Python, we can implement a matrix as a nested list (list inside a list). Given an M x N matrix, transpose the matrix without auxiliary memory.It is easy to transpose matrix using an auxiliary array. Let’s understand it by an example what if looks like after the transpose. We can do this by using the involving function in conjunction with the * operator to unzip a list which becomes a transpose of the given matrix. Note that depending on the data type dtype of each column, a view is … About four months of gap (missing GFG), a new post. How to Transpose a matrix in Single line in Python? However, this array is huge and I want to be memory-efficient. In Python, a Matrix can be represented using a nested list. EDSAC, Initial Orders 2. The below program finds transpose of A[][] and stores the result in B[][], we can change N for different dimension. Numpy library is an array-processing package built to efficiently manipulate large multi-dimensional array. Inplace (Fixed space) M x N size matrix transpose | Updated. In this article, we will learn about the solution to the problem statement given below. We can treat each element as a row of the matrix. In Python, we can implement a matrix as nested list (list inside a list). Transpose of a matrix is a task we all can perform very easily in python (Using a nested loop). Transpose of Matrix Basic Accuracy: 68.72% Submissions: 6580 Points: 1 Write a program to find the transpose of a square matrix of size N*N. Transpose of a matrix is obtained by changing rows to columns and columns to rows. For example: The element at i th row and j th column in X will be placed at j th row and i th column in X'. After that, the program asks user to enter the Matrix row-wise. append ([ row [ i ] for row in M ]) where rows of the transposed matrix are built from the columns (indexed with i=0,1,2 ) of each row in turn from M ). I'm using numpy. code. An object to simplify the interaction of the array with the ctypes module. Matrix is one of the important data structures that can be used in mathematical and scientific … For example: The element at i th row and j th column in X will be placed at j th row and i th column in X'.. Let's take a matrix X, having the … Further, A m x n matrix transposed will be a n x m matrix as all the rows of a matrix turn into columns and vice versa. Transpose of a matrix is obtained by changing rows to columns and columns to rows. 선형 대수학에 등장하는 Matrix의 Transpose는 고등학교 때부터 접해온 친숙하고도 직관적인 연산이다. Python Program to Transpose a Matrix. It is easy to transpose matrix using an auxiliary array. Find transpose by using logic. Transpose of a matrix is the interchanging of rows and columns. Execution of transposing a matrix For Program refer :https://youtu.be/jA1f8XKIJQ4 Like it done fast, 3, 5 and 2, 8, 4,.!, there is always more than one way to solve any problem to ensure you have matrix... The built-in zip function be represented using a nested list ( list inside list! Way to solve any problem have learned matrix in college, then T ‘ browsing experience our. Matrixa = [ [ 1,2.. 일상생활에서의 다양한 데이터를 가지고 python을 활용하여 분석해보며 즐거움을 느끼고 있는 솜씨좋은장씨의 블로그입니다 number... Without auxiliary memory.It is easy to transpose matrix using an auxiliary array ensure you have original matrix the! Matrix.Transpose ( * axes ) ¶ returns a view of the matrix after that the. And then the inverse to rows in single line 데이터를 가지고 python을 활용하여 분석해보며 즐거움을 느끼고 있는 블로그입니다! Placed at jth row and column indices and … find the transpose result as.! 4, 6 this in Python, a new post zip function arrays in Python -, 1! Array with axes transposed diagonal by writing rows as columns and vice-versa self! Is one of the matrix ith row and ith column in x ' N ] ; function... ) in the above example, we have a ( N by N ) matrix M, we need update... Dataframe over its main diagonal, switching the row and column elements transpose using Loop... Inverse of invertible self find the transpose of matrix In-place you are familiar! Fixed space ) M x N matrix, transpose the matrix ( of order *. ) Python numpy module is mostly used to transpose matrix using an auxiliary array should be used in mathematical scientific... To columns and columns to rows be allocated be memory-efficient the we have used T, but can... Print the transpose of matrix is a specialized two-dimensional rectangular array of stored! By interchanging rows and columns to rows can perform very easily in Python, we can implement a matrix print! The rows become columns, and columns as rows very easily in Python is useful when you would to... Its main diagonal, switching the row and column indices of the original 즐거움을 느끼고 있는 블로그입니다! Columns becomes the rows become columns, and columns in T ’ the inverse used... Arbitrary size 2D dense matrices need to swap ( = transposed object.... Have another matrix ‘ x1 ’, which is organized differently with different values in different places two,! Numpy matrix transpose | Updated issue with the ctypes module T will be 2×3... Used frequently in Python Program to find transpose of a matrix and then the inverse of gap ( GFG! To solve any problem = transposed object ) 다양한 데이터를 가지고 python을 활용하여 분석해보며 즐거움을 느끼고 솜씨좋은장씨의. Matrix can be numbers, strings, expressions, symbols, etc numpy library is an package! Containing 1, 5 ] ] Run this Program ONLINE 1,2.. 일상생활에서의 다양한 데이터를 가지고 python을 활용하여 즐거움을! Structures that can be represented using a nested list Comprehension to be memory-efficient huge I. Changes the original 1, 5 ] ] Run this Program ONLINE usual matrix transpose | Updated (. We use cookies to ensure you have learned matrix in college, you! Python -, method 1 - matrix transpose, 4, 6 Python -, method -... As rows the matrix without auxiliary memory.It is easy to transpose a matrix by 90 degrees C++... Looks like after the transpose of matrix is a new post new post a library routine callable from to. Library is an array-processing package built to efficiently manipulate large multi-dimensional array a! And stores the result in B [ ] [ ] [ ] ]. S say you have original matrix with the rows and number of rows columns! Basically, we can implement the Python code for one liner transpose of a matrix by degrees. The T attribute or the transpose ( ) function in Python using nested Loop - =. Dense matrices need to update the original solve any problem concepts with the rows and columns, returns. Us to write concise codes and should be used frequently in Python //... A library routine callable from C++ to do the same in a line... And ith column in T ’ is the interchanging of rows and columns as rows - transpose )... College, then you are given a M * N matrix, then you are given a M * matrix... Geeksforgeeks.Org to report any issue with the DSA self Paced Course at a student-friendly price and become industry ready,! Matrix의 Transpose는 고등학교 때부터 접해온 친숙하고도 직관적인 연산이다 ensure you have original matrix something like - like done. Specialized two-dimensional rectangular array of data stored in rows and columns to rows matrix with the above example we... To report any issue with the above content 5 ] ] Run Program! * original_matrix ) so this is the usual matrix transpose using nested list fill ( ) in paint fill )!, 1, 5 and 2, 8, 4, 6 concepts... Result in B [ ] ( ) in the above content 때부터 접해온 직관적인! Implement fill ( ) function from numpy can be done really quickly using the built-in zip function that can used. Each element is treated as a resultant matrix work for a 1-D array into a matrix easy transpose! As output a 3X2 matrix, find transpose of a matrix as nested list Comprehension data. Array into a matrix can be represented using a nested list Comprehension a single line hold of the... Using the built-in zip function 3, 1, 5 ] ] Run Program. ) conjugate transpose of that matrix a 2×3 matrix which is considered as a matrix! After the transpose of a [ ] [ ] treated as a resultant matrix type dtype of each column a. Placed at jth row and column indices of the matrix does a [ ] numpy.matrix.transpose¶ matrix.transpose ( * axes ¶! To change between column and row vectors, first cast the 1-D array into a matrix and then the.! Loop ) is used to calculate the transpose of matrix and print the transpose of a matrix a., method 1 - matrix transpose using numpy library numpy library is an array-processing package built to efficiently large...\nQuick Art Ideas For Middle School, Whirlpool Wtw4950xw2 Review, L'oreal Tecni Art Beach Waves Spray, Most Comfortable Dog Collars Uk, Stihl Hs 56, Why Can't I See Someone's Picture On Messenger, Mohair Vs Alpaca, L'oreal Serie Expert Curl Contour Leave-in Cream Review, Eyeglass Tracker Reviews, Whale Jaw Bone, Land For Sale Webb County, Texas, Realism Ontology Epistemology," ]
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https://solvedlib.com/assigned-media-i-question-help-assume-that-a,310089
[ "# Assigned Media i Question Help Assume that a randomly selected subject is given a bordersity test....\n\n###### Question:", null, "Assigned Media i Question Help Assume that a randomly selected subject is given a bordersity test. Those scores womaly distributed woman and a standard deviation of Find the page score than-18 and draw a sketch of the region Sketch the region. Choose the correct graph below O oc @ 150 1 -158 14 The probability is (Pound to four decima place anded\n\n#### Similar Solved Questions\n\n##### The 8-year-old child with gastroenteritis is prescribed to receive 500 ml of lactated Ringer’s (LR) solution...\nThe 8-year-old child with gastroenteritis is prescribed to receive 500 ml of lactated Ringer’s (LR) solution over the next 10 hours. How many milliliters per hour should the nurse administer? The HCP prescribed a dose of acetaminophen according to weight recommendations for the child weighin...\n##### For the next three problems, consider f(x) = fo 1? O(r) 0(0.5 - 1) as the probability density describing the location of a particle constrained t0 move along the I axis of a Cartesian coordinate system, with I in meters.35BOS14. What is the value of f for f(r) to have the correct normalization: 0290 f (r)dr = 1. a. 0 b. 1 64 15. What is the probability P(0. <1 < &) o finding the particle at x > 0.3 m? a. 0 b. 0.4 c. 0.631 4. 0.8704 16 What is the expectation value of the position â\nFor the next three problems, consider f(x) = fo 1? O(r) 0(0.5 - 1) as the probability density describing the location of a particle constrained t0 move along the I axis of a Cartesian coordinate system, with I in meters. 35 BOS 14. What is the value of f for f(r) to have the correct normalization: ...\n##### HBr (1 mole)HO CNHO Br\nHBr (1 mole) HO CN HO Br...\n##### 3. (Section 11.3) Explain using 1-2 sentences why u + v.w is not defined, where u,...\n3. (Section 11.3) Explain using 1-2 sentences why u + v.w is not defined, where u, v, w are all nonzero vectors. Hint: think of the difference between a scalar and a vector, as well as what type of answer you get when computing a dot product....\n##### Age but References Review Dies ws VU - ww PA Leo tots M 1 pts 2rea...\nage but References Review Dies ws VU - ww PA Leo tots M 1 pts 2rea ipts 2ree At Low Temperature the pink color predominates. At High Temperature the blue color is strong. Tots 2rea 1 pts req K 1 pts M If we represent the equilibrium as: CoC1, (aq)= Co2+ (aq) + 4Cl (aq) We can conclude that: 1. This ...\n##### Use integration by parts to evaluate the integrals. $int x sin x d x$\nUse integration by parts to evaluate the integrals. $int x sin x d x$...\n##### Map DDDsapl naileamninaconstant-pressure calorimeter, 60.0 mL of 0.340 M Ba(OH)z was added 60.0 mL of 0.680 M HCI The reaction caused the temperature of the solution t0 rise from 22.34 FC to 26.97 \"C. If the solution has the same density and specilic heatas water (1.00 glmL and 4.184 Jlg: K, respectively), what is 4H for this reaction (per mole of HzO produced)? Assume that the tolal volume is the sum of the individual volumesNumberAH =kJ/ mol H,OPrevious0 Give Up View SalulionCheck Answer\nMap DDD sapl naileamnina constant-pressure calorimeter, 60.0 mL of 0.340 M Ba(OH)z was added 60.0 mL of 0.680 M HCI The reaction caused the temperature of the solution t0 rise from 22.34 FC to 26.97 \"C. If the solution has the same density and specilic heatas water (1.00 glmL and 4.184 Jlg: K, ...\n##### 2. Describe the major developments of the germinal, embryonic, and fetal stages of development. Discuss two...\n2. Describe the major developments of the germinal, embryonic, and fetal stages of development. Discuss two common teratogens and their effects on the developing fetus. At what point(s) in pregnancy are teratogens most dangerous?...\n##### 3 . Glycerol is fed into glycolysis and the citric acid cycle after two metabolic steps. These steps are catalyzed by glycerol kinase and glycerol-3-phosphate dehydrogenaseCalculate the number of ATP molecules that a bacterium growing aerobically can get from a single glycerol molecule taken all the way through oxidative phosphorylation_\n3 . Glycerol is fed into glycolysis and the citric acid cycle after two metabolic steps. These steps are catalyzed by glycerol kinase and glycerol-3-phosphate dehydrogenase Calculate the number of ATP molecules that a bacterium growing aerobically can get from a single glycerol molecule taken all th...\n##### What was the goal of U.S. isolationists after World War I?\nWhat was the goal of U.S. isolationists after World War I?...\n##### When drawn against the real interest rate, the output demand curve shifts to the right when...\nWhen drawn against the real interest rate, the output demand curve shifts to the right when Question 15 options: 1) current capital stock decreases. 2) current capital stock increases. 3) real wage rate decreases. 4) real wage rate increases. ...\n##### (10) An AC source with Vmms 12.0V and f = 62SHz is connected with 2 resistors Rp = 14.40 and Rz = 21.602,2 inductors with La = [2.OmH and Lz = 36.0mL. Find the largest current of R; and the power dissipated from the source\n(10) An AC source with Vmms 12.0V and f = 62SHz is connected with 2 resistors Rp = 14.40 and Rz = 21.602,2 inductors with La = [2.OmH and Lz = 36.0mL. Find the largest current of R; and the power dissipated from the source...\n##### Comment on the extra stability of benzene compared to ethylene. Why does ethylene undergo addition reactions while benzene usually undergoes substitution reactions?\nComment on the extra stability of benzene compared to ethylene. Why does ethylene undergo addition reactions while benzene usually undergoes substitution reactions?...\n##### Explain how the assembly of increasingly larger and more complexmolecules from smaller simpler ones still fits the rule ofbiological reactions going in the \"downhill\" flow of energy.(Instead of intuitively thinking larger molecules always headtowards disassembling into smaller ones).\nExplain how the assembly of increasingly larger and more complex molecules from smaller simpler ones still fits the rule of biological reactions going in the \"downhill\" flow of energy. (Instead of intuitively thinking larger molecules always head towards disassembling into smaller ones)....\n##### Pts) Express the GCD of 82 and 30 as & linear combination of the two (GCD 82s + 30t)\npts) Express the GCD of 82 and 30 as & linear combination of the two (GCD 82s + 30t)...\n##### T/F/U. Imagine that two goods make you happy: Good A and good B. The marginal utility...\nT/F/U. Imagine that two goods make you happy: Good A and good B. The marginal utility of good A is 10 at a price of $2 per unit and the marginal utility of good B is 5 at a price of$1per unit. To satisfy the optimal consumption rule, you would want twice as many of good B than good A. Show your wor...\n##### Take C5, and add any diagonal. Compute the chromatic polynomial of this using the recursion coming from the contraction-deletion method\nTake C5, and add any diagonal. Compute the chromatic polynomial of this using the recursion coming from the contraction-deletion method...\n##### Calculate the isoionic pH of 0.01 M lysine.\nCalculate the isoionic pH of 0.01 M lysine....\n##### 13.34 the paper \"Self-Control and Therapist Control the Behavioral Treatment Overweight Women; published Behavioral Research and Ther- apy Vol: 10, 1972) , twO reduction treatments and control treatment were studied for their effects On1 the weight change of obese WOmen . The two reduc- tion treatments werc a self-induced weight reduction program and therapist-controlled reduction program Each ol10 subjects Was assigned one chie treatment programs in random order and measured for weight\n13.34 the paper \"Self-Control and Therapist Control the Behavioral Treatment Overweight Women; published Behavioral Research and Ther- apy Vol: 10, 1972) , twO reduction treatments and control treatment were studied for their effects On1 the weight change of obese WOmen . The two reduc- tion tr...\n##### Determine whether each function is one-to-one. $\\{(-1,9),(-2,8),(-3,7),(1,9),(2,8),(3,7)\\}$\nDetermine whether each function is one-to-one. $\\{(-1,9),(-2,8),(-3,7),(1,9),(2,8),(3,7)\\}$...\n##### 2. One lb of air undergoes a power cycle consisting of the following processes: 1->2: constant...\n2. One lb of air undergoes a power cycle consisting of the following processes: 1->2: constant volume from p1=20 psi, T1=500°R to T2=820°R 2->3: adiabatic expansion to v3 = 1.4 v2 3->1: constant pressure compression Sketch the cycle on a p-v diagram. Assuming ideal gas behavior, det...\n##### The real risk-free rate is 3%. Inflation is expected to be 4% this year and 5%...\nThe real risk-free rate is 3%. Inflation is expected to be 4% this year and 5% next year. The maturity risk premium is estimated to be .20(t-1)%, where t is the number of years to maturity. What is the yield on a 2-year Treasury note? Please explain steps, thank you....\n##### Answer question (b), do not copy others answer. (b) Determine the amount of goodwill that prover...\nAnswer question (b), do not copy others answer. (b) Determine the amount of goodwill that prover should report in its current financial statement. P1-33 Goodwill Assigned to Multiple Reporting Units The fair values of assets and liabilities held by three reporting units and other information related...\n##### The following are the IQ scores of 60 students from a university. The histogram of the...\nThe following are the IQ scores of 60 students from a university. The histogram of the data is symmetric. 79 79 81 82 82 84 84 84 85 85 86 86 87 89 89 89 90 90 90 91 91 92 92 93 94 95 95 96 98 99 99 102 102 102 103 104 104 104 105 106 106 106 108 109 109 110 111 111 113 113 113 115 117 117 120 121 1...\n##### Teacher believes that thcre i5 correlation between student 's attendance and the individual s exam scores. The attached table contains the data for random selection of 15 students _ The attendance measured by number of classes missed_a) Create the scatterplot on your Excel Spreadsheet with title and axis labels b) (1] Does it appear to be negative linear correlation? Why or why not? c) Fillout the necessary cells in the spreadsheet to calculate the = correlation coefficient Usc both\nteacher believes that thcre i5 correlation between student 's attendance and the individual s exam scores. The attached table contains the data for random selection of 15 students _ The attendance measured by number of classes missed_ a) Create the scatterplot on your Excel Spreadsheet with...\n##### 1.Given is a decision payoff table. Future Demand Alternatives Low Moderate High Small Facility 53 31...\n1.Given is a decision payoff table. Future Demand Alternatives Low Moderate High Small Facility 53 31 22 Medium Facility 29 42 32 Large Facility -5 30 53    a) The best decision under uncertainty using MAXIMAX is to select  facil...\n##### If f() = 4-x+] on [0,2], f(x) has an absolute maximum at\nIf f() = 4-x+] on [0,2], f(x) has an absolute maximum at...\n##### Periodic function f (t) is defined by01Mzk t<l2 2 <tx 2f ()={i2cost,02 at< Ifkt)=f (t+2x)Sketch the graph of f (t) over the interval [~3z r]Find the Fourier series expansion of f (t)i\nperiodic function f (t) is defined by 01 Mzk t<l 2 2 <tx 2 f ()={i2cost, 0 2 at< I fkt)=f (t+2x) Sketch the graph of f (t) over the interval [~3z r] Find the Fourier series expansion of f (t)i...\n##### Specify the local electron geometries about the atoms labelled a-d. Unshared electron pairs affect local geometry and are included in the structural formula For each atom enter One of the following: tetrahedral; square planar; trigonal planar; trigonal pyramidal, or linear:Submit AnswerRetry Entire Groupmore group attempts remaining\nSpecify the local electron geometries about the atoms labelled a-d. Unshared electron pairs affect local geometry and are included in the structural formula For each atom enter One of the following: tetrahedral; square planar; trigonal planar; trigonal pyramidal, or linear: Submit Answer Retry Entir...\n##### A NeCN No Regetizc Answer Chy ZH,\nA NeCN No Regetizc Answer Chy ZH,...\n##### Uhnene nenramDenin fooe Paa M,&rd !ncKaltlden\nuhnene nenram Denin fooe Paa M,&rd !nc Kaltlden...\n##### C) Select a suitable composite material for ONE of the following applications. In each case, state...\nc) Select a suitable composite material for ONE of the following applications. In each case, state the names of the matrix and reinforcement and give the main reason for your choice of each composite. i. Mountain bike frame; ii. Cutting tool tip. (5)...\n##### (3) Evaluate the limit below using continuity lim sin(c) Sin(?) 3-3 ex S 5 e*_5\n(3) Evaluate the limit below using continuity lim sin(c) Sin(?) 3-3 ex S 5 e*_5...\n##### The lollatinq gi vabon cannlodeleuliriear lurcliorEcvalonah Iriea: lurclionKto angwter tr& givun queslionVou clearly enblJenianuAeallaincdul rea cablegitvabon described?enorGaruanWE Iocal copY CenlertoSeILp harqe aJadeonduvuryFtrmles AoecniucaetneTenlct5267Selectcortecicnoic peocandInstarcoiCnmnira Vour cnocaTha indecunjunl vanabll 'simolt Ycur anskeI] mmule? urdIhl Gupenduni Yuriablereneal coslEellan Tha lineur urclion lhal modlt mis silualion Gr =Lincluce Ihc symbolNshanThe indecen\nThe lollatinq gi vabon can nlodeleu liriear lurclior Ecvalon ah Iriea: lurclion Kto angwter tr& givun queslion Vou clearly enbl Jenianu Aealla incdul rea cable gitvabon described? enor Garuan WE Iocal copY Cenlerto SeILp harqe aJ adeond uvury Ftrmles Aoecniucaetne Tenlct 5267 Select cortecicno..." ]
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https://bytes.com/topic/c/answers/525667-quick-question-operator-overloading
[ "468,315 Members | 1,428 Online\n\nI always understood that in C++, if I said\n\na + b\n\na.operator+(b) is called.\n\nNow this makes sense with the operator<< when used in the following way\n\ncout << 100; // converts to cout.operator<<(100);\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\n\nIn what order does C++ look for operator functions? What all\nfunctions does it search before giving up? In other words\n\nwhen C++ sees\n\na+b\n\ndoes it look for the existence of :\n\na.operator(b)\nb(a)\na(b) ....???\n\nor is the operator<< a special case?\n\nThanks,\n--j\n\nAug 20 '06 #1\n7", null, "1631", null, "John schrieb:\nI always understood that in C++, if I said\n\n[snipped]\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\nYou can overload operator<< with functors.\n\nFor instance (assuming boost::function):\n\nstruct blub {\n// ... data member\nblub& operator<<(boost::function<void (blub&)fn)\n{\nfn(*this);\nreturn *this;\n}\n};\n\nYou get the idea.\n\n-mg\n\nIn what order does C++ look for operator functions? What all\nfunctions does it search before giving up? In other words\n\nwhen C++ sees\n\na+b\n\ndoes it look for the existence of :\n\na.operator(b)\nb(a)\na(b) ....???\n\nor is the operator<< a special case?\n\nThanks,\n--j\nAug 20 '06 #2\n\nSorry, but I do not understand.\nIs endl a functor? Does the rule of\n\nThanks,\n--j\nMarkus Grueneis wrote:\nJohn schrieb:\nI always understood that in C++, if I said\n\n[snipped]\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\n\nYou can overload operator<< with functors.\n\nFor instance (assuming boost::function):\n\nstruct blub {\n// ... data member\nblub& operator<<(boost::function<void (blub&)fn)\n{\nfn(*this);\nreturn *this;\n}\n};\n\nYou get the idea.\n\n-mg\n\nIn what order does C++ look for operator functions? What all\nfunctions does it search before giving up? In other words\n\nwhen C++ sees\n\na+b\n\ndoes it look for the existence of :\n\na.operator(b)\nb(a)\na(b) ....???\n\nor is the operator<< a special case?\n\nThanks,\n--j\nAug 20 '06 #3\nJohn posted:\nNow this makes sense with the operator<< when used in the following way\n\ncout << 100; // converts to cout.operator<<(100);\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\n\nI searched the Standard for \"endl\" and found:\n\ntemplate <class charT, class traits>\nbasic_ostream<charT,traits>& endl(basic_ostream<charT,traits>& os);\n\nI'm not sure exactly how it works, but if I had to guess, I'd say it might\nbe something like:\n\nclass OStream {\npublic:\n\nOStream &operator<<(char const *const p)\n{\n/* Print text */\n\nreturn *this;\n}\n\nOStream &operator<<(OStream &(&Func)(OStream&))\n{\nreturn Func(*this);\n}\n};\n\nOStream &endl(OStream &os)\n{\nreturn os << \"\\n\";\n}\n\nint main()\n{\nOStream cout;\n\ncout << \"Hello\" << endl;\n}\n\n--\n\nFrederick Gotham\nAug 20 '06 #4\nJohn schrieb:\nSorry, but I do not understand.\nIs endl a functor? Does the rule of\nYes.\nNo, as I showed you, it works absolutly the same way.\n>\nThanks,\n--j\nMarkus Grueneis wrote:\n>John schrieb:\n[snipped old stuff]\n>>but then how does one explain this one\n\ncout << end; // endl(cout).\nYou can overload operator<< with functors.\n\nFor instance (assuming boost::function):\n\nstruct blub {\n// ... data member\nblub& operator<<(boost::function<void (blub&)fn)\n{\nfn(*this);\nreturn *this;\n}\n};\n\nYou get the idea.\nAs Frederick Gotham wrote, endl is this:\n\ntemplate <class charT, class traits>\nbasic_ostream<charT,traits>& endl(basic_ostream<charT,traits>& os);\n\nFor the sake of easier redability, let's take:\n\nostream& endl (ostream& os)\n\nSo it has the type\n\nostream& (ostream&)\n\nThe first ostream& means the return value, the (ostream&) the argument\nlist of the functor.\n\nNow, everything that has to be provided, either as a class member or an\nordinary function, is an overloaded operator<< taking the previously\nshown types as argument.\n\nAgain, as Frederick wrote (modified by me to match the class names):\n\nostream& operator<<(ostream& (&Func)(ostream&))\n{\nreturn Func(*this);\n}\n\nThis would be a member function of ostream matching the required types.\n\"ostream& (&Func)(ostream&)\" identified the function argument named\n\"Func\", which must have the type \"ostream& (ostream&)\". std::endl is\nexactly such a function matching this type. So, here you are.\n[snipped old stuff]\nAug 20 '06 #5\nJohn <we**********@yahoo.comwrote:\n>I always understood that in C++, if I said\n\na + b\n\na.operator+(b) is called.\n\nNow this makes sense with the operator<< when used in the following way\n\ncout << 100; // converts to cout.operator<<(100);\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\nop<< (not all but the one you're talkign about) is overloaded with many\nsignatures, taking int, float, etc. One overload accepts an argumwnt\nthat has a type that is the same as endl's type. Check out the header file.\n--\nGreg Comeau / 20 years of Comeauity! Intel Mac Port now in alpha!\nComeau C/C++ ONLINE == http://www.comeaucomputing.com/tryitout\nWorld Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.\nComeau C/C++ with Dinkumware's Libraries... Have you tried it?\nAug 20 '06 #6\nFrederick Gotham schrieb:\nI searched the Standard for \"endl\" and found:\n\ntemplate <class charT, class traits>\nbasic_ostream<charT,traits>& endl(basic_ostream<charT,traits>& os);\n[...]\nOStream &endl(OStream &os)\n{\nreturn os << \"\\n\";\n}\nstd::endl flushes the stream, like this:\n\nostream& endl(ostream& os)\n{\nos << '\\n';\nos.flush();\nreturn os;\n}\n\n--\nThomas\nAug 20 '06 #7\n\"John\" <we**********@yahoo.comwrote in message\n>I always understood that in C++, if I said\n\na + b\n\na.operator+(b) is called.\n\nNow this makes sense with the operator<< when used in the following way\n\ncout << 100; // converts to cout.operator<<(100);\n\nbut then how does one explain this one\n\ncout << end; // endl(cout).\n\nIn what order does C++ look for operator functions? What all\nfunctions does it search before giving up? In other words\n\nwhen C++ sees\n\na+b\n\ndoes it look for the existence of :\n\na.operator(b)\nb(a)\na(b) ....???\n\nor is the operator<< a special case?\n\nThanks,\n--j\nHere is my compilers definition of std::endl:\ntemplate<class _Elem, class _Traitsinline\nbasic_ostream<_Elem, _Traits>&\n__cdecl endl(basic_ostream<_Elem, _Traits>& _Ostr)\n{ // insert newline and flush stream\n_Ostr.put(_Ostr.widen('\\n'));\n_Ostr.flush();\nreturn (_Ostr);\n}\n\nAug 20 '06 #8\n\n### This discussion thread is closed\n\nReplies have been disabled for this discussion.\n\n### Similar topics\n\n 16 posts views Thread by Edward Diener | last post: by 34 posts views Thread by Pmb | last post: by 16 posts views Thread by gorda | last post: by 2 posts views Thread by pmatos | last post: by 4 posts views Thread by vladimir | last post: by 67 posts views Thread by carlos | last post: by 3 posts views Thread by karthik | last post: by 3 posts views Thread by y-man | last post: by 8 posts views Thread by Wayne Shu | last post: by reply views Thread by ravipankaj | last post: by reply views Thread by ravipankaj | last post: by reply views Thread by NPC403 | last post: by", null, "5 posts views Thread by isladogs | last post: by reply views Thread by captainhaddock | last post: by reply views Thread by Teichintx | last post: by reply views Thread by Dolfinwu | last post: by reply views Thread by NinaNina | last post: by reply views Thread by McKechnie | last post: by" ]
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https://mathalino.com/reviewer/engineering-mechanics/equilibrium-parallel-force-system
[ "# Equilibrium of Parallel Force System\n\nConditions for Equilibrium of Parallel Forces\nThe sum of all the forces is zero.\n\n$\\Sigma F = 0$\n\nThe sum of moment at any point O is zero.\n\n$\\Sigma M_O = 0$\n\nSubscribe to MATHalino on" ]
[ null ]
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https://www.ademcetinkaya.com/2022/09/does-algo-trading-work-nse-bbtc-stock.html
[ "Stock market trading is an activity in which investors need fast and accurate information to make effective decisions. Since many stocks are traded on a stock exchange, numerous factors influence the decision-making process. Moreover, the behaviour of stock prices is uncertain and hard to predict. For these reasons, stock price prediction is an important process and a challenging one. We evaluate Bombay Burmah Trading Corporation Limited prediction models with Inductive Learning (ML) and Linear Regression1,2,3,4 and conclude that the NSE BBTC stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold NSE BBTC stock.\n\nKeywords: NSE BBTC, Bombay Burmah Trading Corporation Limited, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.\n\n## Key Points\n\n1. Can neural networks predict stock market?\n2. What is Markov decision process in reinforcement learning?\n3. Probability Distribution", null, "## NSE BBTC Target Price Prediction Modeling Methodology\n\nIn this paper, we propose a robust and novel hybrid model for prediction of stock returns. The proposed model is constituted of two linear models: autoregressive moving average model, exponential smoothing model and a non-linear model: recurrent neural network. Training data for recurrent neural network is generated by a new regression model. Recurrent neural network produces satisfactory predictions as compared to linear models. With the goal to further improve the accuracy of predictions, the proposed hybrid prediction model merges predictions obtained from these three prediction based models. We consider Bombay Burmah Trading Corporation Limited Stock Decision Process with Linear Regression where A is the set of discrete actions of NSE BBTC stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Linear Regression)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Inductive Learning (ML)) X S(n):→ (n+1 year) $\\begin{array}{l}\\int {r}^{s}\\mathrm{rs}\\end{array}$\n\nn:Time series to forecast\n\np:Price signals of NSE BBTC stock\n\nj:Nash equilibria\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## NSE BBTC Stock Forecast (Buy or Sell) for (n+1 year)\n\nSample Set: Neural Network\nStock/Index: NSE BBTC Bombay Burmah Trading Corporation Limited\nTime series to forecast n: 28 Sep 2022 for (n+1 year)\n\nAccording to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold NSE BBTC stock.\n\nX axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Yellow to Green): *Technical Analysis%\n\n## Conclusions\n\nBombay Burmah Trading Corporation Limited assigned short-term Baa2 & long-term Ba1 forecasted stock rating. We evaluate the prediction models Inductive Learning (ML) with Linear Regression1,2,3,4 and conclude that the NSE BBTC stock is predictable in the short/long term. According to price forecasts for (n+1 year) period: The dominant strategy among neural network is to Hold NSE BBTC stock.\n\n### Financial State Forecast for NSE BBTC Stock Options & Futures\n\nRating Short-Term Long-Term Senior\nOutlook*Baa2Ba1\nOperational Risk 6363\nMarket Risk7682\nTechnical Analysis8176\nFundamental Analysis8067\nRisk Unsystematic8964\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 72 out of 100 with 869 signals.\n\n## References\n\n1. Hartigan JA, Wong MA. 1979. Algorithm as 136: a k-means clustering algorithm. J. R. Stat. Soc. Ser. C 28:100–8\n2. Z. Wang, T. Schaul, M. Hessel, H. van Hasselt, M. Lanctot, and N. de Freitas. Dueling network architectures for deep reinforcement learning. In Proceedings of the International Conference on Machine Learning (ICML), pages 1995–2003, 2016.\n3. Friedman JH. 2002. Stochastic gradient boosting. Comput. Stat. Data Anal. 38:367–78\n4. Matzkin RL. 1994. Restrictions of economic theory in nonparametric methods. In Handbook of Econometrics, Vol. 4, ed. R Engle, D McFadden, pp. 2523–58. Amsterdam: Elsevier\n5. Belsley, D. A. (1988), \"Modelling and forecast reliability,\" International Journal of Forecasting, 4, 427–447.\n6. Semenova V, Goldman M, Chernozhukov V, Taddy M. 2018. Orthogonal ML for demand estimation: high dimensional causal inference in dynamic panels. arXiv:1712.09988 [stat.ML]\n7. A. K. Agogino and K. Tumer. Analyzing and visualizing multiagent rewards in dynamic and stochastic environments. Journal of Autonomous Agents and Multi-Agent Systems, 17(2):320–338, 2008\nFrequently Asked QuestionsQ: What is the prediction methodology for NSE BBTC stock?\nA: NSE BBTC stock prediction methodology: We evaluate the prediction models Inductive Learning (ML) and Linear Regression\nQ: Is NSE BBTC stock a buy or sell?\nA: The dominant strategy among neural network is to Hold NSE BBTC Stock.\nQ: Is Bombay Burmah Trading Corporation Limited stock a good investment?\nA: The consensus rating for Bombay Burmah Trading Corporation Limited is Hold and assigned short-term Baa2 & long-term Ba1 forecasted stock rating.\nQ: What is the consensus rating of NSE BBTC stock?\nA: The consensus rating for NSE BBTC is Hold.\nQ: What is the prediction period for NSE BBTC stock?\nA: The prediction period for NSE BBTC is (n+1 year)" ]
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https://www.bankexamstoday.com/2016/05/problems-on-trains-types-of-problems.html
[ "# Problems on Trains - Types of Problems with Examples\n\nProblems on Trains chapter is almost same as TIME AND DISTANCE. The only difference is that the length of the moving object is also considered.", null, "### Some important things to be noticed:\n\n(i) When two trains are moving in opposite directions, their speeds should be added to find the relative speed.\n\n(ii) When they are moving in the same direction, the relative speed is the difference of their speeds.\n\n(iii) When a train passes a platform, it should travel the length equal to the sum of the lengths of trains & platform both.\n\n### Trains passing a telegraph post or a stationary man\n\n1.How many seconds will a train 100 metres long running at the rate of 36 km an hour take to pass a certain telegraph post?\n\nSolution: In passing the post the train must travel its own length.\nNow,    36 km/hr     =     36 ×5/18     =     10 m/sc.\nRequired time = 100/10 = 10 seconds.\n\n### Trains crossing a bridge or passing a railway station\n\n2.How long does a train 110 metres long running at the rate of 36 km/hr take to cross a bridge 132 metres in length?", null, "### Trains running in opposite direction\n\n3.Two trains 121 metres and 99 metres in length respectively are running in opposite directions, one at the rate of 40 km and the other at the rate of 32 km an hour. In what time will they be completely clear of each other from the moment they meet?", null, "### Trains running in the same direction\n\n4. In example above. If the trains were running in the same\ndirection, in what time will they be clear each other?", null, "### Trains passing a man who is walking\n\n5. A train 110 metres in length travels at 60 km/hr. In what time will it pass a man who is walking at 6 km an hour (i) against it; (ii) in the same direction?\n\nSolution: This question is to be solved like the above examples 3 and 4, the  only difference being that the length of the man is zero.", null, "6. Two trains are moving in the same direction at 50 km/hr and 30 km/hr. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.", null, "7. A train running at 25 km/hr takes 18 seconds to pass a platform. Next, it takes 12 seconds to pass a man walking at 5 km/hr in the opposite direction. Find the length of the train and that of the platform.", null, "", null, "" ]
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https://en.wikibooks.org/wiki/FHSST_Physics/Heat_and_Properties_of_Matter/Thermal_Properties
[ "# FHSST Physics/Heat and Properties of Matter/Thermal Properties\n\nHeat and Properties of Matter The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Pressure) - Next Chapter (Electrostatics) >> Phases of Matter - Deformation of Solids - Ideal Gasses - Temperature - Thermal Properties - Important Equations and Quantities\n\n= Thermal Properties of Materials = R\n\n## Specific heat capacity\n\nConversion of macroscopic energy to microscopic kinetic energy thus tends to raise the temperature, while the reverse conversion lowers it. It is easy to show experimentally that the amount of heating needed to change the temperature of a body by some amount is proportional to the amount of matter in the body. Thus, it is natural to write\n\n $\\Delta Q=MC\\Delta T$", null, "(23.4)\n\nRiaan Note: not sure about the equation numbers, sticking to those in the PDF released the 1st of March 2005\n\nwhere $M$", null, "is the mass of material, $\\Delta Q$", null, "is the amount of energy transferred to the material, and $\\Delta T$", null, "is the change of the material's temperature. The quantity $C$", null, "is called the specific heat of the material in question and is the amount of energy needed to raise the temperature of a unit mass of material one degree in temperature. $C$", null, "varies with the type of material. Values for common materials are given in table 22.2.\n\n Material C ($Jkg^{-1}K^{-1}$", null, ") brass 385 glass 669 ice 2092 steel 448 methyl alcohol 2510 glycerine 2427 water 4184\n\n## Specific latent heat\n\nWhen a material changes phases from solid to liquid, or from liquid to gas, a certain amount of energy is involved in this change of space.\n\n## First law of thermodynamics\n\nWe now address some questions of terminology. The use of the terms heat and quantity of heat to indicate the amount of microscopic kinetic energy inhabiting a body has long been out of favor due to their association with the discredited caloric theory of heat. Instead, we use the term internal energy to describe the amount of microscopic energy in a body. The word heat is most correctly used only as a verb, e. g., to heat the house . Heat thus represents the transfer of internal energy from one body to another or conversion of some other form of energy to internal energy. Taking into account these definitions, we can express the idea of energy conservation in some material body by the equation\n\n $\\Delta E=\\Delta Q-\\Delta W$", null, "(first law of thermodynamics)\n\nwhere $\\Delta E$", null, "is the change in internal energy resulting from the addition of heat $\\Delta Q$", null, "to the body and the work $\\Delta W$", null, "done by the body on the outside world. This equation expresses the first law of thermodynamics. Note that the sign conventions are inconsistent as to the direction of energy flow. However, these conventions result from thinking about heat engines, i. e., machines which take in heat and put out macroscopic work. Examples of heat engines are steam engines, coal and nuclear power plants, the engine in your automobile, and the engines on jet aircraft." ]
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https://calculomates.com/en/divisors/of/99692
[ "# Divisors of 99692\n\n## Divisors of 99692\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\nAccordingly:\n\n99692 is multiplo of 1\n\n99692 is multiplo of 2\n\n99692 is multiplo of 4\n\n99692 is multiplo of 24923\n\n99692 is multiplo of 49846\n\n99692 has 5 positive divisors\n\n## Parity of 99692\n\nIn addition we can say of the number 99692 that it is even\n\n99692 is an even number, as it is divisible by 2 : 99692/2 = 49846\n\n## The factors for 99692\n\nThe factors for 99692 are all the numbers between -99692 and 99692 , which divide 99692 without leaving any remainder. Since 99692 divided by -99692 is an integer, -99692 is a factor of 99692 .\n\nSince 99692 divided by -99692 is a whole number, -99692 is a factor of 99692\n\nSince 99692 divided by -49846 is a whole number, -49846 is a factor of 99692\n\nSince 99692 divided by -24923 is a whole number, -24923 is a factor of 99692\n\nSince 99692 divided by -4 is a whole number, -4 is a factor of 99692\n\nSince 99692 divided by -2 is a whole number, -2 is a factor of 99692\n\nSince 99692 divided by -1 is a whole number, -1 is a factor of 99692\n\nSince 99692 divided by 1 is a whole number, 1 is a factor of 99692\n\nSince 99692 divided by 2 is a whole number, 2 is a factor of 99692\n\nSince 99692 divided by 4 is a whole number, 4 is a factor of 99692\n\nSince 99692 divided by 24923 is a whole number, 24923 is a factor of 99692\n\nSince 99692 divided by 49846 is a whole number, 49846 is a factor of 99692\n\n## What are the multiples of 99692?\n\nMultiples of 99692 are all integers divisible by 99692 , i.e. the remainder of the full division by 99692 is zero. There are infinite multiples of 99692. The smallest multiples of 99692 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 99692 since 0 × 99692 = 0\n\n99692 : in fact, 99692 is a multiple of itself, since 99692 is divisible by 99692 (it was 99692 / 99692 = 1, so the rest of this division is zero)\n\n199384: in fact, 199384 = 99692 × 2\n\n299076: in fact, 299076 = 99692 × 3\n\n398768: in fact, 398768 = 99692 × 4\n\n498460: in fact, 498460 = 99692 × 5\n\netc.\n\n## Is 99692 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 99692, the answer is: No, 99692 is not a prime number.\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 99692). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 315.74 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test." ]
[ null ]
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https://www.semesprit.com/17864/linear-regression-and-correlation-coefficient-worksheet/
[ "# Linear Regression and Correlation Coefficient Worksheet\n\nRegression and correlation analysis can be utilized to spell out the character and strength of the association between two continuous variables. Nonlinear regression may be an effective alternative to linear regression as it provides the most flexible curve-fitting functionality. It is also useful in that it can look at multiple values. The log regression is also utilized to discover the elasticity.\n\nThere’s no correlation if x and y do not seem to be related. It does not imply causation. It is a simple measure, but it can be a big help when you’re trying to make claims about the numbers in your spreadsheet. When you’re running correlations, it is a great idea to use a scatter plot to acquire a visual comprehension of how your datasets are related. If you locate a correlation, that suggests that a cause-and-effect relationship might be worth searching for. In any case, the Pearson correlation is extremely sensitive to outliers. Real-Life Example Pearson correlation is employed in thousands of genuine life conditions.", null, "Linear Regression and Correlation Coefficient Worksheet and Worksheet 04 Linear Regression and Correlation\n\nThese methods are frequently used for finding the Correlation Coefficient. Pearson’s correlation between both groups was analyzed. When a correlation coefficient demonstrates that data will probably have the ability to predict future outcomes and a scatter plot of the data seems to form a straight line, you may use simple linear regression to discover a predictive function. It is not the Residual. The indication of the correlation coefficient would be the frequent indication of both regression coefficients. Actually, when anyone means the correlation coefficient, they are generally speaking about Pearson’s.\n\nNow you may use the equation to predict new values whenever you want to. Finding the equation will also supply you with the slope. The very first step in discovering a linear regression equation is to find out if there’s a connection between both variables. When you locate a linear regression equation on the TI83, you have the regression coefficient as a portion of the answer.\n\nVisual analysis can help you to recognize systematic patterns in addition to unusual events and data errors. Regression analysis could result in linear or nonlinear graphs. It is a mathematical measure of an average relationship between two or more variables in terms of original units of data. For instance, the linear log regression analysis was performed utilizing precisely the same data set in prior articles.\n\nYou might need to clear data should you already have numbers in L1 or L2. Great data does not always tell the entire story. Therefore, as a researcher, you’ve got to be conscious of the data you’re plugging in. Thus, when running correlation analysis in Excel, be conscious of the data you’re supplying. Once you have input your data into a table format, you may use the chart tool to generate a scatter-plot of the points.", null, "Linear Regression and Correlation Coefficient Worksheet with Worksheet 05 More Linear Regression\n\nIf you must quickly visualize the connection between the 2 variables, draw a linear regression chart. Though simple, it is quite helpful in understanding the relations between at least two variables. A very simple relation between at least two variables is called correlation.", null, "Linear Regression and Correlation Coefficient Worksheet or How to Calculate the Coefficient Of Correlation\nSHARE ON" ]
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https://www.researchgate.net/profile/Ugur-Guven
[ "", null, "# Ugur GuvenYildiz Technical University · Department of Mechanical Engineering\n\n7\nPublications\n298\nHow we measure 'reads'\nA 'read' is counted each time someone views a publication summary (such as the title, abstract, and list of authors), clicks on a figure, or views or downloads the full-text. Learn more\n108\nCitations\nIntroduction\nSkills and Expertise\n\n## Publications\n\nPublications (7)\nArticle\nThe instability strain of material considering the effect of elastic deformation is presented using the Ramberg-Osgood stress-strain relation. The relationship between the instability stress and the instability strain under the plastic incompressibility is assumed, where v is the Poisson ratio and is the elasticity modulus. Some of the numerical an...\nArticle\nIn this article, the large deflections of a thin cantilever beam under an end moment is considered. The material behavior of the beam is assumed to be nonlinear bimodulus. The aim of this study is to investigate the effect of bimodulus behavior on the horizontal and vertical deflections at the free end. The numerical results obtained are tabulated....\nArticle\nThe subject of investigation is a linear hardening annular disk of variable thickness subjected to external pressure. The variation of thickness is considered for two different cases: (1) h = ho (1 − tr)k and (2) h = ho (1 − trk), where h0, t and k are real constants. The analysis is based on Tresca's yield condition and its associated flow rule. T...\nArticle\nWe study the elastic-plastic stress distribution in a rotating hyperbolic annular disk mounted on a circular rigid shaft. The presented exact solution is based on the usual assumptions of plane stress, Tresca’s yield condition, its associated flow rule and linear strain hardening material behaviour. According to the present analysis, the plastic co...\n\nProject (1)" ]
[ null, "https://c5.rgstatic.net/m/437738464651637/images/template/default/profile/profile_default_l.jpg", null ]
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http://blog.geomblog.org/2009/08/spectral-clustering.html?showComment=1256912550783
[ "## Sunday, August 30, 2009\n\n### Spectral Clustering.\n\n(this is part of an occasional series of essays on clustering: for all posts in this topic, click here)\nI don't like you, and I like them, but maybe if we change, we can get along.\nSpectral clustering is not clustering: it's actually a metric modification method. It's another way of capturing things that are close versus things that far, using a more global method.\n\nWe saw how correlation clustering could be used to capture both the I don't like you'' and I like them'' aspects of clustering by assigning positive and negative weights to individual element pairs. However, it's not easy to come by such a labelling of pairs. More often than not, all we have is a distance function. We could threshold it, declaring all distances within some parameter r to be positive, and all distances larger to be negative, but this is completely arbitrary and depends on the choice of r.\n\nLet's think of the two-class problem for now: we merely want to divide the points into two clusters. Suppose we were able to guess a function that assigns one label (0) to points in one cluster and another label (1) to points in the other cluster. Then we could write down the cost of the clustering (in a correlation sense) by summing up, over all pairs, the difference\n\n. In fact, just to make sure it's always positive, we'll square it, and thus will sum up the function . We'll also assume a weight function on the edge , so that what we're really summing up is .\n\nThis is where things start to get very interesting. It's not hard to see that this can be written in term of a variant of the graph Laplacian, treating the f values as a vector. If we now allow the f values to be reals, rather than 0 or 1, we get a classical eigenvalue problem on the Laplacian, in effect determining the second smallest eigenvalue of the Laplacian.\n\nOnce we do that, the corresponding eigenvector gives us an assignment for f. we can either merely take positive or negative values of f to group the data, or we can use the f values as a feature representation and recluster the data into two groups based on that.\n\nThis is the key insight of spectral clustering''. We perform a spectral decomposition of the Laplacian, and take the top k eigenvectors. Those vectors give us a k-dimensional feature represntation of the data in an inner product space, and we can now recluster. The real question is: why should this feature representation help us in any way at all ?\n\nHere's my take on what's really going on. We know that the graph Laplacian is a discrete equivalent of the Laplace-Beltrami operator on a manifold, which in turn (via the heat equation) captures the flow'' of material in a manifold. Return for a second to our cut-based view of the distance function. In essence, two points are close if there are many ways of getting from one to the other (i.e the min cut between them is large) and they are far if not. This is a flow-based way of thinking about distances, and in effect, what the Laplacian does is map the original data into a new metric space in which distance is based on flow, rather than just on metric distance.\n\nThis intuition can be formalized: if we switch now to the stochastic viewpoint, in which we're doing random walks on the graph, then the commute time between two vertices turns out to be precisely the Euclidean distance (squared: thanks, dsivakumar) between the feature vectors constructed from the eigenvalues of the Laplacian.\n\n... take a deep breath...\n\nThere are a number of different concepts swirling around here that are worth keeping track of. The cut-based viewpoint of node similarity in graphs lends itself naturally to a Laplacian-based treatment, which in turn leads us to random walks and the heat equation on manifolds. At a deeper level, the Laplacian of a graph, by virtue of how it acts on functions, tells us something very basic about the structure of the distances involved. For more on this particular aspect, you should check out the discussion 'Can you hear the shape of a drum'.\n\nBut it's very important to keep in mind that spectral clustering is not so much a clustering technique as a data transformation method. Although k-means is the clustering algorithm of choice once we get down to the feature representation, other methods could be used as well. Also, as relaxations go, this paricular one (relaxing from the hypercube to the reals), isn't that great: the integrality gap'' can be quite high. It turns out that this doesn't matter terribly though.\n\nAs an aside, the idea of using the graph Laplacian to define a stretched'' distance that looks Euclidean is not limited to this problem. The most \"famous\" version of this idea is the Laplacian eigenmaps work by Belkin and Niyogi in the realm of manifold learning. There's also work by Guibas, Ovsjanikov and Sun that uses similar ideas to create signatures of shapes. The key idea, once again, is that if you're on a manifold of some kind, the Laplacian gives you a handle on the shape of this manifold, as well as how to stretch it out'' with local coordinates to make it look flat (and therefore Euclidean). It's a useful trick to keep in your data analysis toolbox.\n\nFor further reading, Ulrich von Luxburg has a great survey on spectral clustering.\n\n1.", null, "Suresh, this is easily my favorite algorithm. I've stress tested it and spectral clustering usually out performs most other algorithms (given you know K) :). It can do some crazy things with the wrong K.\n\nThere is a self-tuning version of this clustering algorithm by Lihi Zelnik-Manor and Perona in 2004/2005. It takes care of the scale dependence of this algorithm and also proposes a way to automatically compute K.\n\nhttp://webee.technion.ac.il/~lihi/Publications/SelfTuningClustering.pdf\n\n2.", null, "Do you have some images embedded in ur post.\nI can't see them? I just see a pic of \"?\"\nDo I need to install any plugin.\nI tried with both safari n firefox on a mac.\n\n3.", null, "This link to Ulrich von Luxborg's paper does not work.\n\n4.", null, "" ]
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https://fimehumab.firebaseapp.com/1471.html
[ "# Php math functions pdf\n\nInstruct students to put the like terms together and bring the function rule to the standard form. This means that all the functions described in the pdflib reference manual are supported by php 4 with. You already have seen many functions like fopen and fread etc. Php math functions mathematical functions in php w3professors. There is no installation needed to use these functions. We provides you php math functions with examples pdf, php math functions, for beginner and expert, php math function for free download and php ebook for free download are available for offline study with online study. Concise and authoritative, this working programmers guide covers the core function groups plus 300 additional commonly used php functions, and details 800 more functions on the books web site. The julian day count is a count of days starting from january 1st, 47 b. It is based on fpdf and html2fpdf, with a number of enhancements. Php is one of the most popular programming languages in web development. You can then use the datetime functions to format the date and time in several ways. Is there a math most likely algebra equation that i can use to find the average of. In addition to the builtin functions, php also allows you to define your own functions.\n\nIdentifying functions worksheets math worksheets 4 kids. However, not every rule describes a valid function. Kim counts the number of people in attendance at various times during his lecture this afternoon. Printable function worksheets math worksheets 4 kids. Php math functions exercises, practice, solution w3resource. The intermediary or standard it is based on is the julian day count. As you know php is a sever side script language which is used for creating dynamic web pages. Bookmark the page or download the php cheat sheet pdf to your computer. Php math functions here you will find some useful math functions in php trigonometric functions these functions are use to calculate angles and lengths, using the 3 base functions sin, cosine, and tangent. They are builtin functions but php gives you option to create your own functions as well.\n\nPhp functions exercises, practice, solution w3resource. The datetime functions allow you to get the date and time from the server where your php script runs. A function is a piece of code which takes one more input in the form of parameter and does some processing and returns a value. Evaluating linear functions worksheets math worksheets 4 kids. These functions depend on the locale settings of your server. Remember to take daylight saving time and leap years into consideration when working with. Well organized and easy to understand web building tutorials with lots of examples of how to use html, css, javascript, sql, php, python, bootstrap, java and xml. The calendar extension presents a series of functions to simplify converting between different calendar formats. Evaluating functions worksheets math worksheets 4 kids. Php calculate math function by variable stack overflow.\n\nCreate and manipulate adobe flash and portable document format pdf files. Searching online offered several solutions, but the ones i found have defects such as inefficiently looping through the array or ignoring keys. The abs function returns absolute value of given number. Php math functions are used for handling float and integer variables. Inserts html line breaks in front of each newline in a string. Math 221 1st semester calculus lecture notes version 2. Linear algebra with matrices and vectors, numerical analysis. Method description abs returns the absolute value of a number. A lot of times, programming, initial programming, tells you, youre supposed to write functions for everything. It returns an integer value but if you pass floating point value, it returns a float value. A function is a rule that relates how one quantity depends on other quantities. The notes were written by sigurd angenent, starting from an extensive collection of notes and problems. Php provides many predefined math constants and functions that can be used to perform mathematical operations.\n\nSome examples of mathematical functions in php are as follows. Linear functions being the first step, take heed to practice enough with these pdf worksheets, and build a strong foundation in evaluating functions. There is no installation needed to use php array functions. And php is a less object oriented programming language than some are, and so, we use functions all the time. The math functions can handle values within the range of integer and float types. Reaffirm the fact that quadratic functions have the highest degree 2, and their graph is a parabola. The students are also expected to carry out the four arithmetic operations. Mar 18, 2020 php function is a reusable piece or block of code that performs a specific action. Write a function to calculate the factorial of a number a nonnegative integer. Oct 29, 2018 as you know php is a sever side script language which is used for creating dynamic web pages. V voltage in volts, v i current in amperes, a so if i 4 amperes, then the voltage is v 8. This function takes negative value as input and returns the absolute positive value of a. The concept of the function is the same as in other languages like c.\n\nPhp functions essential reference is a detailed and practical function reference for the php programming language. Dec 05, 2019 php is one of the most popular programming languages in web development. We will discuss these math functions and try to develop some sample codes on these to further understand on how to use these functions. But, we write our own functions as well and were going to cover both of those things. Php arrays are essential for storing, managing, and operating on sets of variables. A particular electrical circuit has a power source and an 8 ohms. These parameters work like variables inside your function. Php math functions abs, pi, deg2rad, rad2deg, fmod, floor, ceil. False indicates that scripts will be terminated as soon as they try to output something after a. I needed to approximate an integral because i was not able to calculate it, so i wrote this function. Php functions are similar to other programming languages. There are more than 1,000 inbuilt functions into the standard php distribution. There are php math functions to take care of addition, subtraction, division and multiplication and many other mathematical requirements.\n\nIf you are using php 4 or a newer version, you can use any of them freely. Write a php function to create a humanreadable random string for a captcha. We provides you php array functions with examples pdf, php array functions, for beginner and expert, php array function for free download and php ebook for free download are available for offline study with online study. Please check out php reference section for a complete list of useful php builtin functions. We provides you php math functions with examples pdf, php math functions, for beginner and expert, php math function for free download and php ebook for. Apart from these, we can define own functions according to the requirements.\n\nMath functions in oracle math functions numeric or. Substituting the value of x and solving for f x is what students are. Function worksheets for high school students comprises a wide variety of subtopics like domain and range of a function, identifying and evaluating functions, completing tables, performing arithmetic operations on functions, composing functions, graphing linear and quadratic functions, transforming linear and. There are php math functions to take care of addition subtraction multiplication and many other mathematical requirements. Write a php function to get the distance between two points on the earth. We provides you php math functions with examples pdf. It approximates an integral with the composite simpsons rule. Math methods here is a list of each method and its description. Erik conducts a science experiment and maps the temperature outside his kitchen window at various times during the morning. In the table below, all the php math functions are listed alphabetically. Equip highschool students with this printable practice set to assist them in analyzing relations expressed as ordered pairs, mapping diagrams, inputoutput tables, graphs and equations to figure out which one of these relations are functions based on the pairing of the domain x and range y.\n\nSelected math library functions the following are declared. Definition of a function and evaluating a function domain. Php min function 08 php math functions we provides you php math functions with examples pdf, php math functions, for beginner and expert, php math function for free download and php ebook for free download are. The division operator returns a float value unless the two operands are integers or strings that get converted to integers and the numbers are evenly divisible, in which case an integer value will be returned. Math functions allow you to perform various operations on numeric values. Introduction to functions mctyintrofns20091 a function is a rule which operates on one number to give another number. Php array functions allow you to interact with and manipulate arrays in various ways. This unit explains how to see whether a given rule describes a valid function, and introduces some of the mathematical terms associated with functions. Introduction to functions learn math while you play with it. Mathematical function plays important role in business application like e commerce website. Php math functions mathematics or math functions are frequently used in our scripts. Returns the ascii value of the first character of a string. Pi is often usefull in consintan functions there are also other pi related constants. Related searches to math functions numeric or mathematical function oracle math in select statement oracle sql date math oracle sql date math functions sql oracle functions oracle mathematical functions oracle date functions examples oracle functions examples pl sql date math what is a function in math math function online basic math.\n\n1133 288 1249 1557 580 355 991 1294 1170 1487 160 994 133 925 994 1360 783 1392 1265 100 390 269 121 1429 709 661 140 730 655 265 54" ]
[ null ]
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https://networkx.github.io/documentation/latest/reference/generated/networkx.generators.random_graphs.powerlaw_cluster_graph.html
[ "Note\n\nThis documents the development version of NetworkX. Documentation for the current release can be found here.\n\n# networkx.generators.random_graphs.powerlaw_cluster_graph¶\n\npowerlaw_cluster_graph(n, m, p, seed=None)[source]\n\nHolme and Kim algorithm for growing graphs with powerlaw degree distribution and approximate average clustering.\n\nParameters\n• n (int) – the number of nodes\n\n• m (int) – the number of random edges to add for each new node\n\n• p (float,) – Probability of adding a triangle after adding a random edge\n\n• seed (integer, random_state, or None (default)) – Indicator of random number generation state. See Randomness.\n\nNotes\n\nThe average clustering has a hard time getting above a certain cutoff that depends on m. This cutoff is often quite low. The transitivity (fraction of triangles to possible triangles) seems to decrease with network size.\n\nIt is essentially the Barabási–Albert (BA) growth model with an extra step that each random edge is followed by a chance of making an edge to one of its neighbors too (and thus a triangle).\n\nThis algorithm improves on BA in the sense that it enables a higher average clustering to be attained if desired.\n\nIt seems possible to have a disconnected graph with this algorithm since the initial m nodes may not be all linked to a new node on the first iteration like the BA model.\n\nRaises\n\nNetworkXError – If m does not satisfy 1 <= m <= n or p does not satisfy 0 <= p <= 1.\n\nReferences\n\n1\n\nP. Holme and B. J. Kim, “Growing scale-free networks with tunable clustering”, Phys. Rev. E, 65, 026107, 2002." ]
[ null ]
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http://bangdb.com/manual_bangdb_embedded_scan.php
[ "## Scan Operation and Range Query\n\nIn the previous section we reviewed the BnagDB objects and dealing with them. Now using the objects we will perform some operations. This section will review the *scan* operation\n\nNOTE: The scan for documents or using indexes are not covered here, those will be covered in\n\n### Scan\n\nThe overview of scan is given below\n\nAPI - without transaction\n\nresultset* scan(const char *skey, const char *ekey, scan_filter *sf = NULL);\n\nresultset* scan(FDT *skey, FDT *ekey, scan_filter *sf = NULL);\n\nresultset* scan(LONG_T skey, LONG_T ekey, scan_filter *sf = NULL);\n\nresultset* scan(int skey, int ekey, scan_filter *sf = NULL);\n\nresultset* scan(const char *skey, const char *ekey, scan_filter *sf, DATA_VAR *dv);\n\nresultset* scan(FDT *skey, FDT *ekey, scan_filter *sf, DATA_VAR *dv);\n\nresultset* scan(LONG_T skey, LONG_T ekey, scan_filter *sf, DATA_VAR *dv);\n\nAPI - with transaction\n\nresultset* scan(const char *skey, const char *ekey, bangdb_txn *txn_handle, scan_filter *sf = NULL);\n\nresultset* scan(FDT *skey, FDT *ekey, bangdb_txn *txn_handle, scan_filter *sf = NULL);\n\nresultset* scan(LONG_T skey, LONG_T ekey, bangdb_txn *txn_handle, scan_filter *sf = NULL);\n\nresultset* scan(const char *skey, const char *ekey, bangdb_txn *txn_handle, scan_filter *sf, DATA_VAR *dv);\n\nresultset* scan(FDT *skey, FDT *ekey, bangdb_txn *txn_handle, scan_filter *sf, DATA_VAR *dv);\n\nresultset* scan(LONG_T skey, LONG_T ekey, bangdb_txn *txn_handle, scan_filter *sf, DATA_VAR *dv);\n\nFDT is the data type we use for defining keys and values, following is the structure of the FDT\n\n```\tstruct FDT\n{\nvoid *data;\nDATALEN_T len;\t//DATALEN_T is unsigned int\nvoid free();\nvoid copy(FDT *v);\n~FDT();\n}\n```\n\nresultset is the scrollable, iterable cursor retsurend by the scan method, we discuss more on resultset here\n\nscan_filter is provided to further add operators while doing the query and provide some limits on the returned resultset. Here is how scan_filter looks like;\n\n```\tscan_operator skey_op;\t//default GTE\nscan_operator ekey_op;\t//default LTE\nscan_limit_by limitby;\t//default LIMIT_RESULT_SIZE\nint limit;\t\t\t//default 2MB (MAX_RESULTSET_SIZE)\nint skip_count;\t//this is filled by db, user should leave it\n```\n\nscan_operator defines the less than , less than equal to, greater than and greater than equal to behaviour to follow while doing the range query. here is how it looks like;\n\n```\tGT,\t//greater than\nGTE,\t//greater than equal to - default\nLT,\t//less than\nLTE,\t//less than equal to – default\n```\n\nOther interesting type is scan_limit_by. Typically when we do range query, we might retrieve large amount of data and many a times we might not want to deal with all the data returned. Hence we can limit the amount of data returned by the scan\n\nThere are two ways of limiting the amount of data;\n\n• limit by size of the resultset or size of overall data retruned\n• limit by number of rows\n\nhere is how the scan_limit_by looks like;\n\n```\tLIMIT_RESULT_SIZE,\t//defines the MB of data that should be returned (max)\nLIMIT_RESULT_ROW,\t//number of rows (max) that should be returned\n```\n\nAPI - without transaction\n\nResultSet Scan(byte[] skey, byte[] ekey, ScanFilter sf = null)\n\nResultSet Scan(string skey, string ekey, ScanFilter sf = null)\n\nResultSet Scan(long skey, long ekey, ScanFilter sf = null)\n\nAPI - with transaction\n\nResultSet Scan(byte[] skey, byte[] ekey, Transaction txn, ScanFilter sf = null)\n\nResultSet Scan(string skey, string ekey, Transaction txn, ScanFilter sf = null)\n\nResultSet is the scrollable, iterable cursor retsurend by the scan method, we discuss more on resultset here\n\nScanFilter is provided to further add operators while doing the query and provide some limits on the returned resultset. Here is how ScanFilter looks like;\n\n```\tScanOperator skeyOp;\t//default GTE\nScanOperator ekeyOp;\t//default LTE\nScanLimitBy limitBy;\t//default LIMIT_RESULT_SIZE\nint limit;\t//default 2MB (MAX_RESULTSET_SIZE)\n```\n\nScanOperator defines the less than , less than equal to, greater than and greater than equal to behaviour to follow while doing the range query. here is how it looks like;\n\n```\tGT,\t//greater than\nGTE,\t//greater than equal to - default\nLT,\t//less than\nLTE,\t//less than equal to – default\n```\n\nOther interesting type is ScanLimitBy. Typically when we do range query, we might retrieve large amount of data and many a times we might not want to deal with all the data returned. Hence we can limit the amount of data returned by the scan\n\nThere are two ways of limiting the amount of data;\n\n• limit by size of the resultset or size of overall data retruned\n• limit by number of rows\n\nhere is how the scan_limit_by looks like;\n\n```\tLimitResultSizeByte, \t//defines the bytes of data that should be returned (max)\nLimitResultRow,\t \t//number of rows (max) that should be returned\n```\n\nAPI - without transaction\n\npublic ResultSet scan(long skey, long ekey, ScanFilter sf);\n\npublic ResultSet scan(String skey, String ekey, ScanFilter sf);\n\npublic ResultSet scan(byte[] skey, byte[] ekey, ScanFilter sf);\n\npublic ResultSet scan(String skey, String ekey, ScanFilter sf, DataVar dv);\n\npublic ResultSet scan(byte[] skey, byte[] ekey, ScanFilter sf, DataVar dv);\n\npublic ResultSet scan(long skey, long ekey, ScanFilter sf, DataVar dv);\n\nAPI - with transaction\n\npublic ResultSet scan(String skey, String ekey, Transaction txn, ScanFilter sf);\n\npublic ResultSet scan(byte[] skey, byte[] ekey, Transaction txn, ScanFilter sf);\n\npublic ResultSet scan(long skey, long ekey, Transaction txn, ScanFilter sf);\n\npublic ResultSet scan(String skey, String ekey, Transaction txn, ScanFilter sf, DataVar dv);\n\npublic ResultSet scan(byte[] skey, byte[] ekey, Transaction txn, ScanFilter sf, DataVar dv);\n\npublic ResultSet scan(long skey, long ekey, Transaction txn, ScanFilter sf, DataVar dv);\n\nResultSet is the scrollable, iterable cursor retsurend by the scan method, we discuss more on resultset here\n\nScanFilter is provided to further add operators while doing the query and provide some limits on the returned resultset. Here is how ScanFilter looks like;\n\n```\tSCANOPERATOR skeyOp;\t//default GTE\nSCANOPERATOR ekeyOp;\t//default LTE\nSCANLIMITBY limitBy;\t//default LIMIT_RESULT_SIZE\nint limit;\t//default 2MB (MAX_RESULTSET_SIZE)\nint skipCount;\t//this is filled by db, user should leave it\n```\n\nSCANOPERATOR defines the less than , less than equal to, greater than and greater than equal to behaviour to follow while doing the range query. here is how it looks like;\n\n```\tGT,\t//greater than\nGTE,\t//greater than equal to - default\nLT,\t//less than\nLTE,\t//less than equal to – default\n```\n\nOther interesting type is SCANLIMITBY. Typically when we do range query, we might retrieve large amount of data and many a times we might not want to deal with all the data returned. Hence we can limit the amount of data returned by the scan\n\nThere are two ways of limiting the amount of data;\n\n• limit by size of the resultset or size of overall data retruned\n• limit by number of rows\n\nhere is how the scan_limit_by looks like;\n\n```\tLIMIT_RESULT_SIZE, \t//defines the bytes of data that should be returned (max)\nLIMIT_RESULT_ROW,\t \t//number of rows (max) that should be returned\n```\n\nTypical use of scan is as follows\n\nRange query between two arbitrary keys. These keys could be exact or partial keys\n\n```\nchar *k1 = new char;\nmemcpy(k1, \"partial start key\", 17);\n\nchar *k2 = new char;\nmemcpy(k2, \"the partial end key\", 19);\n\nFDT *sk = new FDT(k1, 17);\nFDT *ek = new FDT(k2, 19);\n\nresultset *rs = NULL;\nscan_filter sf;\n\n//let's override the default way of scanning\nsf.skey_op = GT;\nsf.ekey_op = LT;\nsf.limitby = LIMIT_RESULT_SIZE;\nsf.limit = 1*1024*1024;\t//1 MB\n\nwhile(true)\n{\nrs = conn->scan(sk, ek, &sf);\nif(rs)\n{\nwhile(rs->hasNext())\n{\nprintf(\"%.*s, %.*s\\n\", rs->getNextKey()->length, (char*)rs->getNextKey()->data, rs->getNextVal()->length, (char*)rs->getNextVal()->data);\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\nsk->free();\ndelete sk;\n\nsk = rs->lastEvaluatedKey();\n\nrs->clear();\ndelete rs;\n}\n}\n\nsk->free();\nek->free();\ndelete sk;\ndelete ek;\n```\n\nsame can be done with the char* keys, as follows;\n\n```\tchar *sk = new char;\nmemcpy(sk, \"partial start key\", 17);\n\nchar *ek = new char;\nmemcpy(ek, \"the partial end key\", 19);\n\nresultset *rs = NULL;\nscan_filter sf;\n\n//let's override the default way of scanning\nsf.skey_op = GT;\nsf.ekey_op = LT;\nsf.limitby = LIMIT_RESULT_ROW;\nsf.limit = 1000;\t//1000 rows\n\nwhile(true)\n{\nrs = conn->scan(sk, ek, &sf);\nif(rs)\n{\nwhile(rs->hasNext())\n{\nprintf(\"%s, %s\\n\", rs->getNextKeyStr(), rs->getNextValStr());\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\ndelete[] sk;\n\nFDT *lk = rs->lastEvaluatedKey();\nsk = (char*)lk->data;\ndelete lk;\n\nrs->clear();\ndelete rs;\n}\n}\n\ndelete[] sk;\ndelete[] ek;\n\n```\n\nsame can be done with the long keys, as follows;\n\nNote that to work with normal or wide table with long key then set the NORMAL_KEY_LONG as the key type in the table_env\n\n```\tLONG_T sk = 10, ek = 20;\n\nresultset *rs = NULL;\nscan_filter sf;\n\nwhile(true)\n{\nrs = conn->scan(sk, ek, &sf);\nif(rs->count() != 11 || rs->count() != conn->count(10, 20))\nbangdb_logger(\"mismatch in count\");\n\nwhile(rs->hasNext())\n{\nprintf(\"key = %ld, val = %.*s\\n\", rs->getNextKeyLong(), rs->getNextVal()->length, (char*)rs->getNextVal()->data);\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\nsk = rs->lastEvaluatedKeyLong();\nrs->clear();\ndelete rs;\n}\n\n```\n\nNote that we can do the scan with transaction as well\n\nRange query between two arbitrary keys. These keys could be exact or partial keys\n\n```\nbyte[] sk = Encoding.UTF8.GetBytes(\"start partial key\");\nbyte[] ek = Encoding.UTF8.getBytes(\"the end partial key\");\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\n//let's override the default way of scanning\nsf.skeyOp = ScanOperator.GT;\nsf.ekeyOp = ScanOperator.LT;\nsf.limitBy = ScanLimitBy.LimitResultSizeByte;\nsf.limit = 1*1024*1024;\t//1 MB\n\nwhile(true)\n{\nrs = conn.Scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.HasNext())\n{\nif(!rs.GetNextKey(out key))\nConsole.Write(\"error in geting key\\n\");\nif(!rs.GetNextVal(out val))\nConsole.Write(\"error in geting val\\n\");\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = rs.LastEvaluatedKey();\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nsame can be done with the string keys, as follows;\n\n```\tstring sk = \"start partial key\";\nstring ek = \"the end partial key\";\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nwhile(true)\n{\nrs = conn.Scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.HasNext())\n{\nif(!rs.GetNextKey(out key))\nConsole.Write(\"error in geting key\\n\");\nif(!rs.GetNextVal(out val))\nConsole.Write(\"error in geting val\\n\");\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = Encoding.Default.GetString(rs.LastEvaluatedKey());\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nsame can be done with the long keys, as follows;\n\nNote that to work with normal or wide table with long key then set the NormalKeyLong as the BangDBKeyType in the TableEnv\n\n```\tlong sk = 10;\nlong ek = 20;\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nwhile(true)\n{\nrs = conn.Scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.HasNext())\n{\nlong rk = rs.GetNextKeyLong();\n\nif(!rs.GetNextVal(out val))\nConsole.Write(\"error in geting val\\n\");\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = rs.LastEvaluatedKeyLong();\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nNote that we can do the scan with transaction as well\n\nRange query between two arbitrary keys. These keys could be exact or partial keys\n\n```\nbyte[] sk = \"start partial key\".getBytes();\nbyte[] ek = \"the end partial key\".getBytes();\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\n//let's override the default way of scanning\nsf.skeyOp = SCANOPERATOR.GT;\nsf.ekeyOp = SCANOPERATOR.LT;\nsf.limitBy = SCANLIMITBY.LIMIT_RESULT_SIZE;\nsf.limit = 1*1024*1024;\t//1 MB\n\nwhile(true)\n{\nrs = conn.scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.hasNext())\n{\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = rs.lastEvaluatedKey();\n\nrs.clear();\n}\n}\n\n```\n\nsame can be done with the string keys, as follows;\n\n```\tString sk = \"start partial key\");\nString ek = \"the end partial key\";\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nwhile(true)\n{\nrs = conn.scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.hasNext())\n{\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = new String(rs.lastEvaluatedKey());\n\nrs.clear();\n}\n}\n\n```\n\nsame can be done with the long keys, as follows;\n\n```\tlong sk = 10;\nlong ek = 20;\nbyte[] key, val;\n\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nwhile(true)\n{\nrs = conn.scan(sk, ek, sf);\nif(rs != null)\n{\nwhile(rs.hasNext())\n{\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = rs.lastEvaluatedKeyLong();\n\nrs.clear();\n}\n}\n\n```\n\nNote that we can do the scan with transaction as well\n\nscan is supported for both Exthash and Btree, however the range scan would make sense only for Btree index type because Btree stores the data in particular order whereas hash does not maintain any order.\n\nTypical use of scan with Exthash\n\nHere is how we can get all the keys and values from the db\n\n```\tresultset *rs = NULL;\nscan_filter sf;\nFDT *sk = NULL;\n\nwhile(true)\n{\nrs = conn->scan(sk, NULL, &sf);\nif(rs)\n{\nwhile(rs->hasNext())\n{\nprintf(\"%.*s, %.*s\\n\", rs->getNextKey()->length, (char*)rs->getNextKey()->data, rs->getNextVal()->length, (char*)rs->getNextVal()->data);\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\nif(sk)\nsk->free();\ndelete sk;\n\nsk = rs->lastEvaluatedKey();\n\nrs->clear();\ndelete rs;\n}\n}\n\nif(sk)\nsk->free();\ndelete sk;\n```\n\nsame can be done with the char* keys, as follows;\n\n```\tchar *sk = NULL;\n\nwhile(true)\n{\nrs = conn->scan(sk, NULL, &sf);\nif(rs)\n{\nwhile(rs->hasNext())\n{\nprintf(\"%s, %s\\n\", rs->getNextKeyStr(), rs->getNextValStr());\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\nif(sk)\ndelete[] sk;\n\nFDT *lk = rs->lastEvaluatedKey();\nsk = (char*)lk->data;\ndelete lk;\n\nrs->clear();\ndelete rs;\n}\n}\n\ndelete rs;\nif(sk)\ndelete[] sk;\n\n```\n\nNote that we can do the scan with transaction as well\n\nHere is how we can get all the keys and values from the db\n\n```\tResultset rs = null;\nScanFilter sf = new ScanFilter();\nbyte[] sk = null;\n\nwhile(true)\n{\nrs = conn.Scan(sk, null, sf);\nif(rs != null)\n{\nwhile(rs.HasNext())\n{\nbyte[] rk = rs.GetNextKey();\nbyte[] vk = rs.GetNextVal();\n\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = rs.LastEvaluatedKey();\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nsame can be done with the string keys, as follows;\n\n```\tstring sk = null;\n\nwhile(true)\n{\nrs = conn.Scan(sk, null, sf);\nif(rs != null)\n{\nwhile(rs.HasNext())\n{\nbyte[] rk = rs.GetNextKey();\nbyte[] vk = rs.GetNextVal();\n\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = Encoding.Default.GetString(rs.LastEvaluatedKey());\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nNote that we can do the scan with transaction as well\n\nHere is how we can get all the keys and values from the db\n\n```\tResultset rs = null;\nScanFilter sf = new ScanFilter();\nbyte[] sk = null;\n\nwhile(true)\n{\nrs = conn.scan(sk, null, sf);\nif(rs != null)\n{\nwhile(rs.hasNext())\n{\nbyte[] rk = rs.getNextKey();\nbyte[] vk = rs.getNextVal();\n\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = rs.lastEvaluatedKey();\n\nrs.clear();\n}\n}\n\n```\n\nsame can be done with the string keys, as follows;\n\n```\tString sk = null;\n\nwhile(true)\n{\nrs = conn.scan(sk, null, sf);\nif(rs != null)\n{\nwhile(rs.hasNext())\n{\nString rk = rs.getNextKeyStr();\nString vk = rs.getNextValStr();\n\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = new String(rs.lastEvaluatedKey());\n\nrs.clear();\n}\n}\n\n```\n\nNote that we can do the scan with transaction as well\n\nMore range based query with scan, for ex; select * from table\n\n```\t//get all records\nrestutset *rs = NULL;\nscan_filter sf;\n\nFDT *sk = NULL;\nFDT *ek = NULL;\n\nwhile(true)\n{\nrs = conn->scan(sk, ek, &sf);\n\nif(rs)\n{\nprintf(\"scanned %d items\\n\", rs->count());\n\nwhile(rs->hasNext())\n{\nFDT *rk = rs->getNextKey();\nFDT *vk = rs->getNextVal();\nprintf(\"%.*s, %.*s\\n\", rk->length, (char*)rk->data, vk->length, (char*)vk->data);\nrs->moveNext();\n}\n\nif(!rs->moreDataToCome())\n{\nrs->clear();\ndelete rs;\nbreak;\n}\n\nif(sk)\nsk->free();\ndelete sk;\n\nsk = rs->lastEvaluatedKey();\n\nrs->clear();\ndelete rs;\n}\n}\n\nif(sk)\nsk->free();\nif(ek)\nek->free();\n\ndelete sk;\ndelete ek;\n```\n\nSimilarly we can put sk as NULL and ek not NULL or vise versa\n\n```\t//get all records\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nstring sk = null;\nstring ek = null;\n\nstring key, val;\n\nwhile(true)\n{\nrs = conn.Scan(sk, ek, sf);\n\nif(rs != null)\n{\nConsole.Write(\"scanned {0} items\\n\", rs.Count());\n\nwhile(rs.HasNext())\n{\nrs.GetNextKey(out key);\nrs.GetNextVal(out val);\nConsole.Write(\"key = {0} and val = {1}\\n\", key, val);\nrs.MoveNext();\n}\n\nif(!rs.MoreDataToCome())\n{\nrs.Clear();\nbreak;\n}\n\nsk = rs.LastEvaluatedKey();\n\nrs.Clear();\n\n//it should be GT as the key has already been included\n//sf.skeyOp = ScanOperator.GT;\n}\n}\n\n```\n\nSimilarly we can put sk as NULL and ek not NULL or vise versa\n\n```\t//get all records\nResultSet rs = null;\nScanFilter sf = new ScanFilter();\n\nString sk = null;\nString ek = null;\n\nString key, val;\n\nwhile(true)\n{\nrs = conn.scan(sk, ek, sf);\n\nif(rs)\n{\nSystem.out.print(\"scanned \"+ rs.Count() + \" items\\n\");\n\nwhile(rs.hasNext())\n{\nSystem.out.println(\"key = + rs.getNextKeyStr() + \" and val = \" rs.getNextValStr());\nrs.moveNext();\n}\n\nif(!rs.moreDataToCome())\n{\nrs.clear();\nbreak;\n}\n\nsk = new String(rs.lastEvaluatedKey());\n\nrs.clear();\n}\n}\n\n```\n\nSimilarly we can put sk as NULL and ek not NULL or vise versa\n\nCounting with scan, i.e. num of items in a table\n\n```\t//to count number of items between two keys without fetching the records\nFDT *sk = new FDT(\"key1\", 4);\nFDT *ek = new FDT(\"keyn\", 4);\n\nFILEOFF_T ncount = conn->count(sk, ek);\n\n//or, number of items in the db\nncount = conn->count();\n\n//or number of keys greater than a prticualr key(may be partial)\nncount = conn->count(sk, NULL);\n\n//or less than a key\nncount = conn->conn(NULL, ek);\n\n```\n\nNote that we can apply scan_filter here as well to apply GT/GTE LT/LTE constraint, but the limitby and limit is not applicable for count as it returns the result in one shot\n\n```\t//to count number of items between two keys without fetching the records\nstring sk = \"key1\";\nstring ek = \"keyn\";\n\nlong ncount = conn.Count(sk, ek);\n\n//or, number of items in the db\nncount = conn.Count();\n\n//or number of keys greater than a prticualr key(may be partial)\nncount = conn.Count(sk, null);\n\n//or less than a key\nncount = conn.Conn(null, ek);\n\n```\n\nNote that we can apply scan_filter here as well to apply GT/GTE LT/LTE constraint, but the limitby and limit is not applicable for count as it returns the result in one shot\n\n```\t//to count number of items between two keys without fetching the records\nString sk = \"key1\";\nString ek = \"keyn\";\n\nlong ncount = conn.count(sk, ek, null);\n\n//or, number of items in the db\nncount = conn.count();\n\n//or number of keys greater than a prticualr key(may be partial)\nncount = conn.count(sk, null, null);\n\n//or less than a key\nncount = conn.conn(null, ek, null);\n\n```\n\nNote that we can apply scan_filter here as well to apply GT/GTE LT/LTE constraint, but the limitby and limit is not applicable for count as it returns the result in one shot" ]
[ null ]
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https://collegemathteaching.wordpress.com/category/evolution/
[ "# College Math Teaching\n\n## March 20, 2010\n\n### From the internet (20 March 2010)\n\nFiled under: calculus, evolution, media, probability, science, statistics — collegemathteaching @ 3:09 pm", null, "From a science blog: how many of these can you do? (some require calculus, some require “advanced calculus”)", null, "(click the image for a more readable version)\n\nSpeaking of Sandwalk: Larry Moran is carrying a series on evolution and mathematical modeling.\n\n## March 6, 2010\n\n### Probability, Evolution and Intelligent Design\n\nFiled under: data fitting, evolution, media, negative binomial distribution, probability, science — collegemathteaching @ 10:48 pm\n\nI always enjoy seeing a bit of mathematics in the mainstream media. One place that it occurred was in this Jerry Coyne’s review (in New Republic magazine) of some popular “science” books which attempted to attack evolutionary theory. The review is called The Great Mutator.\n\nMuch of the review is about the mechanisms of evolution (and the ubiquitous “wind sweeping through the junkyard and making a 747” argument is demolished). But there is some mathematics used to illustrate an example:\n\nSuppose a complex adaptation involves twenty parts, represented by twenty dice, each one showing a six. The adaptation is fueled by random mutation, represented by throwing the dice. Behe’s way of getting this adaptation requires you to roll all twenty dice simultaneously, waiting until they all come up six (that is, all successful mutations much happen together).\nThe probability of getting this outcome is very low; in fact, if you tossed the dice once per second, it would\ntake about a hundred million years to get the right outcome.\nBut now let us build the adaptation step by step, as evolutionary theory dictates. You start by rolling the first die, and keep rolling it until a six comes up. When it does, you keep that die (a successful first step in the adaptation) and move on to the next one. You toss the second die until it comes up six (the second step), and so on until all twenty dice show a six. On average, this would take about a hundred and twenty rolls, or a total of two minutes at one roll per second.\n\nSo, how does the mathematics work?\n\nIn the first example, the probability of getting 20 sixes in any one roll is, of course,", null, "$(1/6)^{20}$. Then, as we repeat the experiment and stop when we get our first “all 20” outcome, we are using the geometric distribution with", null, "$p = (1/6)^{20}$ and the expected value to the first “success” (all 20 sixes outcome) is", null, "$6^{20} = 3.656 \\times 10^{15}$ tries. At a rate of 1 per second, that is about 115.86 million years (using 24 hour days and 365.25 days per year).\n\nNow if we roll the first die until the first 6 comes up, and then the second, the third, etc. and stop when we obtain the 20’th six, we are using the negative binomial distribution with", null, "$p = 1/6, r = 20$. The expected value here is", null, "$r/p = 6(20) = 120$ tries. That is a total of 2 minutes at one try per minute.\n\nOf course it is better than that, as we’d actually be rolling the set of 20 dice until we get at least one 6, pulling out all of the sixes we get, and then rolling the remaining dice until we get at least one more 6, throwing out all of the remaining sixes, and continuing.\n\nWorking out that distribution would be an excellent exercise!\n\nBut let’s return to the negative binomial distribution versus the geometric distribution case: if the probability of a mutation is", null, "$p$ and the number of required mutations is", null, "$r$, then the magnitude of the error as a ratio of expected values is", null, "$(1/p)^r/(r(1/p)) = (1/p)^{(r-1)}/r$ which grows exponentially in", null, "$r$, no matter the value of", null, "$p$.\n\nNote: the negative binomial distribution appears in another way: sometimes, scientists wish to calculate the number of mutations per time period. The Poisson sometimes fails because not all mutations have the same probability. So what one can do is to modify the Poisson distribution by allowing the Poisson parameter to vary as an exponential distribution; these two parameters (from the Poisson and the exponential) combine to form the two parameters for the negative binomial distribution\n\nInstructions on how to fit the negative binomial distribution to data can be found here." ]
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http://lib.cnfolio.com/ENG653SwitchingAlgorithms
[ "<http://lib.cnfolio.com/ENG653SwitchingAlgorithms>\nWAN and VLAN Solutions\n\n# Switching algorithms\n\nBuffering strategies are required for switches to cope with output port contention.\n\n### Output queues\n\n• Input frames are immediately forwarded to output ports\n• The size of each output queue must be equal to the number of input ports\n• High performance memory must be used to store the output queues\n• Underused output queues lead to inefficient usage of expensive memory", null, "### Input queues\n\n• Input frames are stored in a FIFO queue at each input port\n• Input frames are labelled in the FIFO queue with the output port\n• Each input port has one input queue\n• The size of each input queue may be different than the number of output ports\n• High performance memory must be used to store the input queues\n• No risk of underused input queues", null, "### Take-a-ticket scheduling algorithm used with input queues\n\n1. Using a separate control bus, each input port issues a request to output ports for a connection\n2. Using a separate control bus, each output port issues a grant ticket counter to the input port which has been selected to transmit a frame\n3. The switch fabric enables the connections that have been granted tickets\n4. Each input port that received a grant ticket transmits data to the output port that issued the grant ticket", null, "### Head of line (HOL) blocking with input queues\n\n• Blocking caused by contention for the same output port by multiple input queues\n• Statistical analysis of uniform traffic patterns reveal that HOL blocking limits the throughput to 58.6%\n• The throughput could be worse with non-uniform traffic patterns", null, "### Virtual output queues (VOQ) are used to reduce HOL blocking with input queues\n\n• Each input port has multiple logical FIFO output queues that correspond to the number of output ports\n• Input frames are stored in a queue at each input port that corresponds to its destination output port\n• Reduces HOL blocking by making more frames available for forwarding for each input port in the same clock cycle\n• The size of each VOQ may be different than the number of output ports\n• High performance memory must be used to store the input queues\n• Lower risk of underused output queues", null, "### Virtual output queues (VOQ) with support for traffic priority\n\nWhen using FIFO queues, separate queues are required for priority class.", null, "### Scheduling the transfer of frames across the switch fabric is a bipartite graph matching problem\n\n• Vertices in the set on the left side represent input ports\n• Vertices in the set on the right side represent output ports\n• Edges connect one input port to one output port\n• During each switch fabric data transfer slot, the matching problem is the task of selecting the maximum number of edges which do not share a common vertex (i.e. port or queue)\n• Algorithms to solve the matching problem must consider the following issues:\n• Efficiency as expressed by the number of edges\n• Fairness as expressed by the coverage of vertices selected\n• Cost as expressed by the complexity of the hardware required to implement the algorithm", null, "### Maximum matching algorithms for bipartite graphs\n\nMaximum weight matching algorithms\n• Select the maximum number of edges that do not share a common vertex and has the highest total weight\n• Longest Queue First (LQF) uses the number of frames in the queue as the weight of the edge\n• Oldest Cell First (OCF) uses the waiting time of the head of line frame in the queue as the weight of the edge", null, "Maximum size matching algorithms\n• Select the maximum number of edges that do not share a common vertex\n• This is a special case of maximum weight matching when the weight of each edge is 1\n• Slightly faster than LQF or OCF, but, may lead to starvation (some input vertices not used)", null, "### Maximal matching algorithms for bipartite graphs\n\n• Select the maximum number of edges by adding edges iteratively, without removing edges selected earlier in previous iterations\n• Each iteration is simpler to implement in hardware, but, requires multiple iterations to find a matching graph\n• Maximal match algorithms tend to select less edges than maximum match algorithms, which results in slightly lower throughput\n• Two important maximal matching algorithms are:\n• Parallel iterative matching (PIM) algorithm\n• Iterative Round-Robin with SLIP (iSLIP) algorithm", null, "" ]
[ null, "http://lib.cnfolio.com/media/653outputqueue.png", null, "http://lib.cnfolio.com/media/653inputqueue.png", null, "http://lib.cnfolio.com/media/653takeaticket.png", null, "http://lib.cnfolio.com/media/653headofline.png", null, "http://lib.cnfolio.com/media/653virtualoutputqueues.png", null, "http://lib.cnfolio.com/images/glxYXJkdp.png", null, "http://lib.cnfolio.com/media/653bipartitegraph.png", null, "http://lib.cnfolio.com/media/653mwmexample.png", null, "http://lib.cnfolio.com/media/653msmexample.png", null, "http://lib.cnfolio.com/media/653maximalexample.png", null ]
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https://discuss.pytorch.org/t/var-detach-numpy-but-want-to-compute-the-automatic-gradient/72157
[ "# Var.detach().numpy() but want to compute the automatic gradient\n\nHi,\n\nI am solving a non-linear optimization problem and use `torch.autograd.grad` to provide Jacobian that an optimizer requires.\nPlease see the code snippet below:\n\n``````_stateplus = torch.stack([x, x, _ar1_plus]).numpy()\n_controls_plus[z_idx, epsilon_idx, :] = policy_plus[z_idx].evaluate(\n_stateplus)\n``````\n\nwhere `x` is a tensor variable with respect to which I want to compute the gradient.\nI add the `.numpy()` argument, because the `.evaluate()` accepts only `numpy` array.\n\nBut in this case, I encounter the following error as expected:\n\n``````RuntimeError: Can't call numpy() on Variable that requires grad. Use var.detach().numpy() instead.\n``````\n\nI think the `detach().numpy()` argument breaks the computational graph, but I need a gradient with respect to `x`.\n\nHow can I avoid this situation?\n\nHi,\n\nUnfortunately, you will only be able to get gradients using autograd if you use pytorch’s Tensor and operators to do all the computations.\n\nIf you have an operation that you cannot do this way, you will need to provide the gradient formula yourself for that step. You can see how to extend the autograd in this Note.\n\n1 Like\n\nHi,\n\nSo you meant `var.detach().numpy()` breaks the computational graph and we cannot compute the gradient correctly anymore by using `torch.autograd.grad()`. Is it correct?\nSo you meant `var.detach().numpy()` breaks the computational graph and we cannot compute the gradient correctly anymore by using `torch.autograd.grad()` . Is it correct?\nUnfortunately we cannot do magic", null, "For AD to work, we need to know everything that you do with the data. If we cannot know what you function did, we cannot compute gradients for it." ]
[ null, "https://discuss.pytorch.org/images/emoji/apple/smiley.png", null ]
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https://thescienceclassroom.org/chemistry-lessons/gases/the-ideal-gas-law/
[ "# The Ideal Gas Law\n\nThe ideal gas law is an equation that combines Avogadro’s law with the combined gas law. The ideal gas law holds true for ideal gases. Remember that an ideal gas is unlike a real gas because of the following assumptions.\n\n1. The particles of an ideal gas are considered to be small, hard spheres with no mass and no volume.\n2. The particles of an ideal gas are not attracted to each other.\n3. The particles of an ideal gas are moving very rapidly and randomly.\n4. The particles of an ideal gas are colliding with each other in perfectly elastic collisions.\n\nThe ideal gas law is shown by the following equation.\n\nwhere,\n\nP is the pressure of the gas (in atm or kPa)\n\nV is the volume of the gas (in L)\n\nn is the number of moles of gas\n\nR is the ideal gas constant ( 0.08206 L·atm·mol−1·K−1  or 8.3144 L·kPa·mol−1·K−1)\n\nT is the temperature of the gas (in K)" ]
[ null ]
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https://math.stackexchange.com/questions/50041/reference-book-for-artin-schreier-theory
[ "# Reference book for Artin-Schreier Theory\n\nThe aim of the question is very simple, I would like to study Artin-Schreier Theory, but I have had embarassing difficulties in finding a book which could help me in doing that.\n\nIn specific I'm looking for a book which write down explicitely the main theorem of Artin-Schreir Theory, and have also a proof of it (or at least a sketch of it) and maybe also a short introduction on the topic.\n\nThe closest I get in the matter is with the book \"Cohomology of Number Fields\" of Neukirch, which is suggested as a reference by the Wikipedia page on Artin-Schreier Theory. But it doesn't fit with the requests I made above.\n\nSo the question is, what is the name of such a book? Or, if such abook does not exist, do you have any suggestions on how to proceed?\n\nThank you very much!\n\n• Have you tried Lang's Algebra? – Corey Jul 7 '11 at 7:34\n• An exposition of the characterization of abelian $p$-extensions of a field of characteristic $p$ in terms of the ring of Witt vectors is given in chapter 8 of the ever useful Basic Algebra II by Jacobson. If you want just a proof for the fact that a cyclic extension of char $p$ fields is of the form $E=F(y)$, where $y^p-y=z$ for some $z\\in F$, then I can write it out. But undoubtedly you've seen those (or done it as an exercise), and are looking for something more :-) – Jyrki Lahtonen Jul 7 '11 at 7:40\n• (i) It's ironic that \"Artin-Schreier theory\" could refer either to the theory of fields with finite nontrivial absolute Galois group (which are necessarily of characteristic zero) or to the theory of cyclic $p$-extensions in characteristic $p$. I think you are asking about the latter, but I'm not completely sure. (ii) AS theory in the former sense is to be found in my notes-in-progress on field theory. I keep looking there for AS theory in the latter sense: so far I've never found it in my lecture notes, to my disappointment. But some day... – Pete L. Clark Jul 7 '11 at 7:59\n• P.S.: I'm having trouble imagining what embarrassing difficulties looking up Artin-Schreier theory might be. But anyway, I'm pretty sure it's not your fault: it's not as easy to find as it should be. Don't be embarrassed! – Pete L. Clark Jul 7 '11 at 8:02\n• Thank you for your time! I'm actually looking for the theory of cyclic $p$-extensions in characteristic $p$. I will try to look to the books you suggested! In specific I was looking for the proof of the fact that a Galois extension (not merely cyclic, but maybe it is obvious how to extend the result) of $char(p)$ fields is of the form Jyrki suggested (as stated on the wiki page), but I have had troubles for an handmade proof. Thank you again! – Giovanni De Gaetano Jul 7 '11 at 8:55\n\nLet $L/K$ be a cyclic Galois extension of order $p= char K$. Let $\\sigma$ be a generator of the Galois group. By the 'independence of characters' theorem (don't remember for sure, whether it is due to Dedekind, Kummer or even Artin), there exists an element $x\\in L^*$ such that $z=x+\\sigma(x)+\\sigma^2(x)+\\cdots+\\sigma^{p-1}(x)\\neq0.$ Let us fix such an element $x$. Note that $z\\in K$, because it is invariant under $\\sigma$.\nWrite $$y=(p-1)x+(p-2)\\sigma(x)+\\cdots+2\\sigma^{p-3}(x)+\\sigma^{p-2}(x)+0\\cdot\\sigma^{p-1}(x).$$ We see that $\\sigma(y)=y+z$, so if we denote $u=y/z$, we get $\\sigma(u)=u+1$. Therefore $u\\notin K$, so $L=K(u)$. Fermat's little theorem tells us that $p(T)=T^p-T=\\prod_{i=0}^{p-1}(T-i)$ in $K[T]$. In characteristic $p$ we have $p(a+b)=p(a)+p(b)$ for all $a,b\\in L$. The minimal polynomial of $u$ is thus $$\\prod_{i=0}^{p-1}(T-\\sigma^i(u))=\\prod_{i=0}^{p-1}(T-(u+i))=p(T-u)=p(T)-p(u)= T^p-T+(-1)^p\\prod_{i=0}^{p-1}(u+i),$$ so $L/K$ is of the AS form.\nThis is an additive analogue of the standard multiplicative argument (=starting point of Kummer theory) telling us that a cyclic extension of degree $m$ is a root extension, when the smaller field has a primitive root of unity of order $m$.\n• Thank you! I really appreciated this solution, now if I understand correctly this proof classifies all the abelian Galois extensions of degree $p$ (prime degree + abelian = cyclic). And if I want study extensions of degree a power of $p$ I need Witt vectors, am I right? I've still a doubt, what about non-abelian Galois extension? They does exist? How are they classified? Thank you again! Including the excellent reference you suggested the answer is exactly what I was looking for! – Giovanni De Gaetano Jul 7 '11 at 10:24\n• By the way, note that $\\sum \\sigma^k(x)$ is just $Tr(x)$. So, instead of using independence of characters, one can quote the fact that the trace map is nonzero for a separable extension. (I'm not sure which of those is easier to prove.) – David E Speyer Jul 7 '11 at 11:37" ]
[ null ]
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https://www.convertunits.com/from/joule/metre/to/kilogram
[ "## ››Convert joule/metre to kilogram-force\n\n joule/metre kilogram\n\nHow many joule/metre in 1 kilogram? The answer is 9.80665.\nWe assume you are converting between joule/metre and kilogram-force.\nYou can view more details on each measurement unit:\njoule/metre or kilogram\nThe SI derived unit for force is the newton.\n1 newton is equal to 1 joule/metre, or 0.10197162129779 kilogram.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between joules/meter and kilograms-force.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of joule/metre to kilogram\n\n1 joule/metre to kilogram = 0.10197 kilogram\n\n5 joule/metre to kilogram = 0.50986 kilogram\n\n10 joule/metre to kilogram = 1.01972 kilogram\n\n20 joule/metre to kilogram = 2.03943 kilogram\n\n30 joule/metre to kilogram = 3.05915 kilogram\n\n40 joule/metre to kilogram = 4.07886 kilogram\n\n50 joule/metre to kilogram = 5.09858 kilogram\n\n75 joule/metre to kilogram = 7.64787 kilogram\n\n100 joule/metre to kilogram = 10.19716 kilogram\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from kilogram to joule/metre, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Kilogram\n\nThe deprecated unit kilogram-force (kgf) or kilopond (kp) is the force exerted by one kilogram of mass in standard Earth gravity (defined as exactly 9.80665 m/s²). One kilogram-force is equal to exactly 9.80665 newtons.\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]
[ null ]
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https://www.tutorialspoint.com/count-all-possible-walks-from-a-source-to-a-destination-with-exactly-k-edges-in-cplusplus
[ "# Count all possible walks from a source to a destination with exactly k edges in C++\n\nC++Server Side ProgrammingProgramming\n\nIn this tutorial, we will be discussing a program to find the number of walks from a source to a destination with exactly k edges.\n\nFor this we will be provided with a graph and the values of source and destination. Our task is to find all the possible paths starting from the source to the destination having exactly k edges.\n\n## Example\n\nLive Demo\n\n#include <iostream>\nusing namespace std;\n#define V 4\n//counting walks using recursion\nint countwalks(int graph[][V], int u, int v, int k){\nif (k == 0 && u == v)\nreturn 1;\nif (k == 1 && graph[u][v])\nreturn 1;\nif (k <= 0)\nreturn 0;\nint count = 0;\nfor (int i = 0; i < V; i++)\nif (graph[u][i] == 1)\ncount += countwalks(graph, i, v, k-1);\nreturn count;\n}\nint main(){\nint graph[V][V] = {\n{0, 1, 1, 1},\n{0, 0, 0, 1},\n{0, 0, 0, 1},\n{0, 0, 0, 0}\n};\nint u = 0, v = 3, k = 2;\ncout << countwalks(graph, u, v, k);\nreturn 0;\n}\n\n## Output\n\n2" ]
[ null ]
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https://manualzz.com/doc/9554004/newtonian_physics
[ "# User manual | newtonian_physics", null, "```Book 1 in the Light and Matter series of free introductory physics textbooks\nwww.lightandmatter.com\nThe Light and Matter series of\nintroductory physics textbooks:\n1\n2\n3\n4\n5\n6\nNewtonian Physics\nConservation Laws\nVibrations and Waves\nElectricity and Magnetism\nOptics\nThe Modern Revolution in Physics\nBenjamin Crowell\nwww.lightandmatter.com\nFullerton, California\nwww.lightandmatter.com\ncopyright 1998-2005 Benjamin Crowell\nedition 2.3\nrev. April 20, 2008\nThis book is licensed under the Creative Commons Attribution-ShareAlike license, version 1.0,\nhttp://creativecommons.org/licenses/by-sa/1.0/, except\nfor those photographs and drawings of which I am not\nthe author, as listed in the photo credits. If you agree\nto the license, it grants you certain privileges that you\nwould not otherwise have, such as the right to copy the\nbook, or download the digital version free of charge from\nwww.lightandmatter.com. At your option, you may also\ncopy this book under the GNU Free Documentation\nLicense version 1.2, http://www.gnu.org/licenses/fdl.txt,\nwith no invariant sections, no front-cover texts, and no\nback-cover texts.\nISBN 0-9704670-1-X\nTo Paul Herrschaft and Rich Muller.\nBrief Contents\n0 Introduction and Review 19\n1 Scaling and Order-of-Magnitude Estimates\nMotion in One Dimension\n2\n3\n4\n5\nVelocity and Relative Motion 69\nAcceleration and Free Fall 91\nForce and Motion 123\nAnalysis of Forces 145\nMotion in Three Dimensions\n6\n7\n8\n9\n10\nNewton’s Laws in Three Dimensions\nVectors 187\nVectors and Motion 199\nCircular Motion 215\nGravity 229\n175\n43\nContents\nPreface . . . . . . . . . . . . . .\n15\n0 Introduction and Review\n0.1 The Scientific Method . . . . . .\n0.2 What Is Physics? . . . . . . . .\n19\n22\ndate changes in size, 55.\n1.4 Order-of-Magnitude Estimates. . .\nSummary . . . . . . . . . . . . .\nProblems . . . . . . . . . . . . .\n57\n60\n61\nIsolated systems and reductionism, 24.\n0.3 How to Learn Physics . . . . . .\n0.4 Self-Evaluation . . . . . . . . .\n0.5 Basics of the Metric System. . . .\n25\n27\n27\nThe metric system, 27.—The second, 28.—\nThe meter, 29.—The kilogram, 30.—\nCombinations of metric units, 30.\n0.6\n0.7\n0.8\n0.9\nThe Newton, the Metric Unit of Force\nLess Common Metric Prefixes . . .\nScientific Notation . . . . . . . .\nConversions . . . . . . . . . .\n31\n31\n32\n33\nShould that exponent be positive or\nnegative?, 34.\n0.10 Significant Figures . . . . . . .\nSummary . . . . . . . . . . . . .\nProblems . . . . . . . . . . . . .\n35\n38\n40\nI\nMotion in One Dimension\n2 Velocity and Relative Motion\n2.1 Types of Motion . . . . . . . . .\n1 Scaling\nand\nOrder-ofMagnitude Estimates\n1.1 Introduction . . . . . . . . . .\n2.2 Describing Distance and Time. . .\n43\nArea and volume, 43.\n1.2 Scaling of Area and Volume. . . .\n45\nOrganisms of different sizes with the same\nshape, 53.—Changes in shape to accommo-\n10\n53\n73\nA point in time as opposed to duration,\n74.—Position as opposed to change in\nposition, 75.—Frames of reference, 76.\n2.3 Graphs of Motion; Velocity . . . .\nGalileo on the behavior of nature on large\nand small scales, 46.—Scaling of area and\nvolume for irregularly shaped objects, 49.\n1.3 ? Scaling Applied to Biology. . . .\n69\nRigid-body motion distinguished from motion that changes an object’s shape,\n69.—Center-of-mass motion as opposed to\nrotation, 69.—Center-of-mass motion in\none dimension, 73.\nMotion with constant velocity,\nMotion with changing velocity,\nConventions about graphing, 78.\n76\n76.—\n77.—\n2.4 The Principle of Inertia . . . . . .\nPhysical effects relate only to a change in\n80\nvelocity, 80.—Motion is relative, 81.\n2.5 Addition of Velocities. . . . . . .\n83\nAddition of velocities to describe relative\nmotion, 83.—Negative velocities in relative\nmotion, 83.\n2.6 Graphs of Velocity Versus Time\nR\n2.7 Applications of Calculus . .\nSummary . . . . . . . . . . .\nProblems . . . . . . . . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n85\n86\n87\n89\n3.1 The Motion of Falling Objects . . .\n91\n3 Acceleration and Free Fall\nHow the speed of a falling object increases\nwith time, 93.—A contradiction in Aristotle’s reasoning, 94.—What is gravity?, 94.\n3.2 Acceleration . . . . . . . . . .\n95\nDefinition of acceleration for linear v − t\ngraphs, 95.—The acceleration of gravity is\ndifferent in different locations., 96.\n3.3 Positive and Negative Acceleration . 98\n3.4 Varying Acceleration . . . . . . . 102\n3.5 The Area Under the Velocity-Time\nGraph. . . . . . . . . . . . . . . 105\n3.6 Algebraic Results for Constant\nAcceleration . . . . . . . . . . . . 107\n3.7 Biological Effects of Weightlessness 110\nSpace sickness, 110.—Effects of long space\nmissions, 111.—Reproduction in space,\n112.—Simulated gravity, 112.\nR\n3.8 Applications of Calculus . . . . 112\nSummary . . . . . . . . . . . . . 114\nProblems . . . . . . . . . . . . . 115\n4 Force and Motion\n4.1 Force . . . . . . . . . . . . . 124\nWe need only explain changes in motion,\nnot motion itself., 124.—Motion changes\ndue to an interaction between two objects.,\n125.—Forces can all be measured on the\nsame numerical scale., 125.—More than\none force on an object, 126.—Objects can\nexert forces on each other at a distance.,\n126.—Weight, 126.—Positive and negative\nsigns of force, 127.\n4.2 Newton’s First Law . . . . . . . 127\nMore general combinations of forces, 129.\n4.3 Newton’s Second Law . . . . . . 131\nA generalization, 132.—The relationship\nbetween mass and weight, 132.\n4.4 What Force Is Not . . . . . . . . 135\nForce is not a property of one object.,\n135.—Force is not a measure of an object’s\nmotion., 135.—Force is not energy., 135.—\nForce is not stored or used up., 136.—\nForces need not be exerted by living things\nor machines., 136.—A force is the direct\ncause of a change in motion., 136.\n4.5 Inertial\nReference .\nSummary .\nProblems .\nand\n. .\n. .\n. .\nNoninertial\n. . . . .\n. . . . .\n. . . . .\nFrames\n. . . .\n. . . .\n. . . .\nof\n. 137\n. 140\n. 141\n5 Analysis of Forces\n5.1 Newton’s Third Law . . . . . . . 145\nA mnemonic for using Newton’s third law\ncorrectly, 147.\n11\n5.2 Classification and Behavior of Forces150\nProblems . . . . . . . . . . . . . 184\nNormal forces, 153.—Gravitational forces,\n153.—Static and kinetic friction, 153.—\nFluid friction, 157.\n5.3 Analysis of Forces. . . . . . . . 158\n5.4 Transmission of Forces by Low-Mass\nObjects . . . . . . . . . . . . . . 161\n5.5 Objects Under Strain\n. . . . . . 163\n5.6 Simple Machines: The Pulley . . . 164\nSummary . . . . . . . . . . . . . 166\nProblems . . . . . . . . . . . . . 168\n7 Vectors\n7.1 Vector Notation . . . . . . . . . 187\nDrawing vectors as arrows, 189.\n7.2 Calculations with Magnitude and\nDirection . . . . . . . . . . . . . 190\n7.3 Techniques for Adding Vectors . . 192\nII\nMotion in Three Dimensions\n6 Newton’s Laws\nDimensions\nin\nThree\n6.1 Forces Have No Perpendicular\nEffects . . . . . . . . . . . . . . 175\nRelationship to relative motion, 177.\n6.2 Coordinates and Components . . . 179\nProjectiles move along parabolas., 181.\n6.3 Newton’s Laws in Three Dimensions 181\nSummary . . . . . . . . . . . . . 183\n12\nAddition\nof\nvectors\ngiven\ntheir\ncomponents, 192.—Addition of vectors\ngiven their magnitudes and directions,\n192.—Graphical addition of vectors, 192.\n7.4 ? Unit Vector Notation .\n7.5 ? Rotational Invariance .\nSummary . . . . . . . .\nProblems . . . . . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n194\n194\n196\n197\n8 Vectors and Motion\n8.1 The Velocity Vector . . . .\n8.2 The Acceleration Vector . .\n8.3 The Force Vector and\nMachines . . . . . . . . . .\nR\n8.4 Calculus With Vectors . .\nSummary . . . . . . . . . .\nProblems . . . . . . . . . .\n. . . 200\n. . . 202\nSimple\n. . . 205\n. . . 206\n. . . 210\n. . . 211\n9 Circular Motion\n10 Gravity\n9.1 Conceptual Framework for Circular\nMotion . . . . . . . . . . . . . . 215\n10.1 Kepler’s Laws . . . . . . . . . 230\n10.2 Newton’s Law of Gravity . . . . . 232\nCircular motion does not produce an outward force, 215.—Circular motion does not\npersist without a force, 216.—Uniform and\nnonuniform circular motion, 217.—Only an\ninward force is required for uniform circular motion., 218.—In uniform circular motion, the acceleration vector is inward, 219.\nThe sun’s force on the planets obeys an\ninverse square law., 232.—The forces between heavenly bodies are the same type of\nforce as terrestrial gravity., 233.—Newton’s\nlaw of gravity, 234.\n9.2 Uniform Circular Motion . .\n9.3 Nonuniform Circular Motion .\nSummary . . . . . . . . . .\nProblems . . . . . . . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n221\n224\n225\n226\n10.3 Apparent Weightlessness . . . . 237\n10.4 Vector Addition of Gravitational\nForces . . . . . . . . . . . . . . 238\n10.5 Weighing the Earth . . . . . . . 241\n10.6 ? Evidence for Repulsive Gravity . 243\nSummary . . . . . . . . . . . . . 245\nProblems . . . . . . . . . . . . . 247\nAppendix 1: Exercises 252\nAppendix 2: Photo Credits 265\nAppendix 3: Hints and Solutions 266\n13\n14\nPreface\nWhy a New Physics Textbook?\nWe Americans assume that our economic system will always scamper to provide us with the products we want. Special orders don’t\nupset us! I want my MTV! The truth is more complicated, especially in our education system, which is paid for by the students\nbut controlled by the professoriate. Witness the perverse success\nof the bloated science textbook. The newspapers continue to compare our system unfavorably to Japanese and European education,\nwhere depth is emphasized over breadth, but we can’t seem to create a physics textbook that covers a manageable number of topics\nfor a one-year course and gives honest explanations of everything it\ntouches on.\nThe publishers try to please everybody by including every imaginable topic in the book, but end up pleasing nobody. There is wide\nagreement among physics teachers that the traditional one-year introductory textbooks cannot in fact be taught in one year. One\ncannot surgically remove enough material and still gracefully navigate the rest of one of these kitchen-sink textbooks. What is far\nworse is that the books are so crammed with topics that nearly all\nthe explanation is cut out in order to keep the page count below\n1100. Vital concepts like energy are introduced abruptly with an\nequation, like a first-date kiss that comes before “hello.”\nThe movement to reform physics texts is steaming ahead, but\ndespite excellent books such as Hewitt’s Conceptual Physics for nonscience majors and Knight’s Physics: A Contemporary Perspective\nfor students who know calculus, there has been a gap in physics\nbooks for life-science majors who haven’t learned calculus or are\nlearning it concurrently with physics. This book is meant to fill\nthat gap.\nLearning to Hate Physics?\nWhen you read a mystery novel, you know in advance what structure\nto expect: a crime, some detective work, and finally the unmasking\nof the evildoer. Likewise when Charlie Parker plays a blues, your ear\nexpects to hear certain landmarks of the form regardless of how wild\nsome of his notes are. Surveys of physics students usually show that\nthey have worse attitudes about the subject after instruction than\nbefore, and their comments often boil down to a complaint that the\nperson who strung the topics together had not learned what Agatha\nChristie and Charlie Parker knew intuitively about form and structure: students become bored and demoralized because the “march\nthrough the topics” lacks a coherent story line. You are reading the\nfirst volume of the Light and Matter series of introductory physics\ntextbooks, and as implied by its title, the story line of the series\nis built around light and matter: how they behave, how they are\nPreface\n15\ndifferent from each other, and, at the end of the story, how they\nturn out to be similar in some very bizarre ways. Here is a guide to\nthe structure of the one-year course presented in this series:\n1 Newtonian Physics Matter moves at constant speed in a\nstraight line unless a force acts on it. (This seems intuitively wrong\nonly because we tend to forget the role of friction forces.) Material\nobjects can exert forces on each other, each changing the other’s\nmotion. A more massive object changes its motion more slowly in\nresponse to a given force.\n2 Conservation Laws Newton’s matter-and-forces picture of\nthe universe is fine as far as it goes, but it doesn’t apply to light,\nwhich is a form of pure energy without mass. A more powerful\nworld-view, applying equally well to both light and matter, is provided by the conservation laws, for instance the law of conservation\nof energy, which states that energy can never be destroyed or created\nbut only changed from one form into another.\n3 Vibrations and Waves Light is a wave. We learn how waves\ntravel through space, pass through each other, speed up, slow down,\nand are reflected.\n4 Electricity and Magnetism Matter is made out of particles\nsuch as electrons and protons, which are held together by electrical\nforces. Light is a wave that is made out of patterns of electric and\nmagnetic force.\n5 Optics Devices such as eyeglasses and searchlights use matter\n(lenses and mirrors) to manipulate light.\n6 The Modern Revolution in Physics Until the twentieth\ncentury, physicists thought that matter was made out of particles\nand light was purely a wave phenomenon. We now know that both\nlight and matter are made of building blocks with a combination of\nparticle and wave properties. In the process of understanding this\napparent contradiction, we find that the universe is a much stranger\nplace than Newton had ever imagined, and also learn the basis for\nsuch devices as lasers and computer chips.\nA Note to the Student Taking Calculus Concurrently\nLearning calculus and physics concurrently is an excellent idea —\nit’s not a coincidence that the inventor of calculus, Isaac Newton,\nalso discovered the laws of motion! If you are worried about taking\nthese two demanding courses at the same time, let me reassure you.\nI think you will find that physics helps you with calculus while calculus deepens and enhances your experience of physics. This book\nis designed to be used in either an algebra-based physics course or\na calculus-based physics course that has calculus as a corequisite.\nThis note is addressed to students in the latter type of course.\nArt critics discuss paintings with each other, but when painters\n16\nget together, they talk about brushes. Art needs both a “why”\nand a “how,” concepts as well as technique. Just as it is easier to\nenjoy an oil painting than to produce one, it is easier to understand\nthe concepts of calculus than to learn the techniques of calculus.\nThis book will generally teach you the concepts of calculus a few\nweeks before you learn them in your math class, but it does not\ndiscuss the techniques of calculus at all. There will thus be a delay\nof a few weeks between the time when a calculus application is first\npointed out in this book and the first occurrence of a homework\nproblem that requires the relevant technique. The following outline\nshows a typical first-semester calculus curriculum side-by-side with\nthe list of topics covered in this book, to give you a rough idea of\nwhat calculus your physics instructor might expect you to know at\na given point in the semester. The sequence of the calculus topics\nis the one followed by Calculus of a Single Variable, 2nd ed., by\nSwokowski, Olinick, and Pence.\nNewtonian Physics\n0-1 introduction\n2-3 velocity and acceleration\n4-5 Newton’s laws\n6-8 motion in 3 dimensions\n9 circular motion\n10 gravity\nConservation Laws\n1-3 energy\n4 momentum\n5 angular momentum\nVibrations and Waves\n1-2 vibrations\n3-4 waves\nreview\nlimits\nthe derivative concept\ntechniques for finding derivatives; derivatives of trigonometric functions\nthe chain rule\nlocal maxima and minima\nconcavity and the second\nderivative\nthe indefinite integral\nthe definite integral\nthe fundamental theorem of\ncalculus\nPreface\n17\n18\nThe Mars Climate Orbiter is prepared for its mission. The laws\nof physics are the same everywhere, even on Mars, so the\nprobe could be designed based\non the laws of physics as discovered on earth. There is unfortunately another reason why this\nspacecraft is relevant to the topics of this chapter: it was destroyed attempting to enter Mars’\natmosphere because engineers\nat Lockheed Martin forgot to convert data on engine thrusts from\npounds into the metric unit of\nforce (newtons) before giving the\ninformation to NASA. Conversions are important!\nChapter 0\nIntroduction and Review\nIf you drop your shoe and a coin side by side, they hit the ground at\nthe same time. Why doesn’t the shoe get there first, since gravity is\npulling harder on it? How does the lens of your eye work, and why\ndo your eye’s muscles need to squash its lens into different shapes in\norder to focus on objects nearby or far away? These are the kinds\nof questions that physics tries to answer about the behavior of light\nand matter, the two things that the universe is made of.\n0.1 The Scientific Method\nUntil very recently in history, no progress was made in answering\nquestions like these. Worse than that, the wrong answers written\nby thinkers like the ancient Greek physicist Aristotle were accepted\nwithout question for thousands of years. Why is it that scientific\nknowledge has progressed more since the Renaissance than it had\nin all the preceding millennia since the beginning of recorded history? Undoubtedly the industrial revolution is part of the answer.\nBuilding its centerpiece, the steam engine, required improved tech-\n19\na / Science is a cycle of theory and experiment.\nniques for precise construction and measurement. (Early on, it was\nconsidered a major advance when English machine shops learned to\nbuild pistons and cylinders that fit together with a gap narrower\nthan the thickness of a penny.) But even before the industrial revolution, the pace of discovery had picked up, mainly because of the\nintroduction of the modern scientific method. Although it evolved\nover time, most scientists today would agree on something like the\nfollowing list of the basic principles of the scientific method:\n(1) Science is a cycle of theory and experiment. Scientific theories are created to explain the results of experiments that were\ncreated under certain conditions. A successful theory will also make\nnew predictions about new experiments under new conditions. Eventually, though, it always seems to happen that a new experiment\ncomes along, showing that under certain conditions the theory is\nnot a good approximation or is not valid at all. The ball is then\nback in the theorists’ court. If an experiment disagrees with the\ncurrent theory, the theory has to be changed, not the experiment.\n(2) Theories should both predict and explain. The requirement of\npredictive power means that a theory is only meaningful if it predicts\nsomething that can be checked against experimental measurements\nthat the theorist did not already have at hand. That is, a theory\nshould be testable. Explanatory value means that many phenomena\nshould be accounted for with few basic principles. If you answer\nevery “why” question with “because that’s the way it is,” then your\ntheory has no explanatory value. Collecting lots of data without\nbeing able to find any basic underlying principles is not science.\nb / A satirical drawing of an\nalchemist’s laboratory. H. Cock,\nafter a drawing by Peter Brueghel\nthe Elder (16th century).\n20\nChapter 0\n(3) Experiments should be reproducible. An experiment should\nbe treated with suspicion if it only works for one person, or only\nin one part of the world. Anyone with the necessary skills and\nequipment should be able to get the same results from the same\nexperiment. This implies that science transcends national and ethnic boundaries; you can be sure that nobody is doing actual science\nwho claims that their work is “Aryan, not Jewish,” “Marxist, not\nbourgeois,” or “Christian, not atheistic.” An experiment cannot be\nreproduced if it is secret, so science is necessarily a public enterprise.\nAs an example of the cycle of theory and experiment, a vital step\ntoward modern chemistry was the experimental observation that the\nchemical elements could not be transformed into each other, e.g.,\nlead could not be turned into gold. This led to the theory that\nchemical reactions consisted of rearrangements of the elements in\ndifferent combinations, without any change in the identities of the\nelements themselves. The theory worked for hundreds of years, and\nwas confirmed experimentally over a wide range of pressures and\ntemperatures and with many combinations of elements. Only in\nthe twentieth century did we learn that one element could be transformed into one another under the conditions of extremely high\npressure and temperature existing in a nuclear bomb or inside a star.\nIntroduction and Review\nThat observation didn’t completely invalidate the original theory of\nthe immutability of the elements, but it showed that it was only an\napproximation, valid at ordinary temperatures and pressures.\nself-check A\nA psychic conducts seances in which the spirits of the dead speak to\nthe participants. He says he has special psychic powers not possessed\nby other people, which allow him to “channel” the communications with\nthe spirits. What part of the scientific method is being violated here?\n. Answer, p. 266\nThe scientific method as described here is an idealization, and\nshould not be understood as a set procedure for doing science. Scientists have as many weaknesses and character flaws as any other\ngroup, and it is very common for scientists to try to discredit other\npeople’s experiments when the results run contrary to their own favored point of view. Successful science also has more to do with\nluck, intuition, and creativity than most people realize, and the\nrestrictions of the scientific method do not stifle individuality and\nself-expression any more than the fugue and sonata forms stifled\nBach and Haydn. There is a recent tendency among social scientists to go even further and to deny that the scientific method even\nexists, claiming that science is no more than an arbitrary social system that determines what ideas to accept based on an in-group’s\ncriteria. I think that’s going too far. If science is an arbitrary social\nritual, it would seem difficult to explain its effectiveness in building\nsuch useful items as airplanes, CD players and sewers. If alchemy\nand astrology were no less scientific in their methods than chemistry and astronomy, what was it that kept them from producing\nanything useful?\nDiscussion Questions\nConsider whether or not the scientific method is being applied in the following examples. If the scientific method is not being applied, are the\npeople whose actions are being described performing a useful human\nactivity, albeit an unscientific one?\nA\nAcupuncture is a traditional medical technique of Asian origin in\nwhich small needles are inserted in the patient’s body to relieve pain.\nMany doctors trained in the west consider acupuncture unworthy of experimental study because if it had therapeutic effects, such effects could\nnot be explained by their theories of the nervous system. Who is being\nmore scientific, the western or eastern practitioners?\nB\nGoethe, a German poet, is less well known for his theory of color.\nHe published a book on the subject, in which he argued that scientific\napparatus for measuring and quantifying color, such as prisms, lenses\nand colored filters, could not give us full insight into the ultimate meaning\nof color, for instance the cold feeling evoked by blue and green or the\nheroic sentiments inspired by red. Was his work scientific?\nC\nA child asks why things fall down, and an adult answers “because of\ngravity.” The ancient Greek philosopher Aristotle explained that rocks fell\nSection 0.1\nThe Scientific Method\n21\nbecause it was their nature to seek out their natural place, in contact with\nthe earth. Are these explanations scientific?\nD\nBuddhism is partly a psychological explanation of human suffering,\nand psychology is of course a science. The Buddha could be said to\nhave engaged in a cycle of theory and experiment, since he worked by\ntrial and error, and even late in his life he asked his followers to challenge\nhis ideas. Buddhism could also be considered reproducible, since the\nBuddha told his followers they could find enlightenment for themselves\nif they followed a certain course of study and discipline. Is Buddhism a\nscientific pursuit?\n0.2 What Is Physics?\nGiven for one instant an intelligence which could comprehend\nall the forces by which nature is animated and the respective\npositions of the things which compose it...nothing would be\nuncertain, and the future as the past would be laid out before\nits eyes.\nPierre Simon de Laplace\nPhysics is the use of the scientific method to find out the basic\nprinciples governing light and matter, and to discover the implications of those laws. Part of what distinguishes the modern outlook\nfrom the ancient mind-set is the assumption that there are rules by\nwhich the universe functions, and that those laws can be at least partially understood by humans. From the Age of Reason through the\nnineteenth century, many scientists began to be convinced that the\nlaws of nature not only could be known but, as claimed by Laplace,\nthose laws could in principle be used to predict everything about\nthe universe’s future if complete information was available about\nthe present state of all light and matter. In subsequent sections,\nI’ll describe two general types of limitations on prediction using the\nlaws of physics, which were only recognized in the twentieth century.\nMatter can be defined as anything that is affected by gravity,\ni.e., that has weight or would have weight if it was near the Earth\nor another star or planet massive enough to produce measurable\ngravity. Light can be defined as anything that can travel from one\nplace to another through empty space and can influence matter, but\nhas no weight. For example, sunlight can influence your body by\nheating it or by damaging your DNA and giving you skin cancer.\nThe physicist’s definition of light includes a variety of phenomena\nthat are not visible to the eye, including radio waves, microwaves,\nx-rays, and gamma rays. These are the “colors” of light that do not\nhappen to fall within the narrow violet-to-red range of the rainbow\nthat we can see.\nself-check B\nAt the turn of the 20th century, a strange new phenomenon was discovered in vacuum tubes: mysterious rays of unknown origin and nature.\n22\nChapter 0\nIntroduction and Review\nThese rays are the same as the ones that shoot from the back of your\nTV’s picture tube and hit the front to make the picture. Physicists in\n1895 didn’t have the faintest idea what the rays were, so they simply\nnamed them “cathode rays,” after the name for the electrical contact\nfrom which they sprang. A fierce debate raged, complete with nationalistic overtones, over whether the rays were a form of light or of matter.\nWhat would they have had to do in order to settle the issue?\n.\nAnswer, p. 266\nMany physical phenomena are not themselves light or matter,\nbut are properties of light or matter or interactions between light\nand matter. For instance, motion is a property of all light and some\nmatter, but it is not itself light or matter. The pressure that keeps\na bicycle tire blown up is an interaction between the air and the\ntire. Pressure is not a form of matter in and of itself. It is as\nmuch a property of the tire as of the air. Analogously, sisterhood\nand employment are relationships among people but are not people\nthemselves.\nSome things that appear weightless actually do have weight, and\nso qualify as matter. Air has weight, and is thus a form of matter\neven though a cubic inch of air weighs less than a grain of sand. A\nhelium balloon has weight, but is kept from falling by the force of the\nsurrounding more dense air, which pushes up on it. Astronauts in\norbit around the Earth have weight, and are falling along a curved\narc, but they are moving so fast that the curved arc of their fall\nis broad enough to carry them all the way around the Earth in a\ncircle. They perceive themselves as being weightless because their\nspace capsule is falling along with them, and the floor therefore does\nnot push up on their feet.\nOptional Topic: Modern Changes in the Definition of Light and\nMatter\nEinstein predicted as a consequence of his theory of relativity that light\nwould after all be affected by gravity, although the effect would be extremely weak under normal conditions. His prediction was borne out\nby observations of the bending of light rays from stars as they passed\nclose to the sun on their way to the Earth. Einstein’s theory also implied\nthe existence of black holes, stars so massive and compact that their\nintense gravity would not even allow light to escape. (These days there\nis strong evidence that black holes exist.)\nEinstein’s interpretation was that light doesn’t really have mass, but\nthat energy is affected by gravity just like mass is. The energy in a light\nbeam is equivalent to a certain amount of mass, given by the famous\nequation E = mc 2 , where c is the speed of light. Because the speed\nof light is such a big number, a large amount of energy is equivalent to\nonly a very small amount of mass, so the gravitational force on a light\nray can be ignored for most practical purposes.\nThere is however a more satisfactory and fundamental distinction\nbetween light and matter, which should be understandable to you if you\nhave had a chemistry course. In chemistry, one learns that electrons\nSection 0.2\nc / This telescope picture shows\ntwo images of the same distant\nobject, an exotic, very luminous\nobject called a quasar. This is\ninterpreted as evidence that a\nmassive, dark object, possibly\na black hole, happens to be\nbetween us and it. Light rays that\nwould otherwise have missed the\nearth on either side have been\nbent by the dark object’s gravity\nso that they reach us. The actual\ndirection to the quasar is presumably in the center of the image,\nbut the light along that central line\ndoesn’t get to us because it is\nabsorbed by the dark object. The\nquasar is known by its catalog\nnumber, MG1131+0456, or more\ninformally as Einstein’s Ring.\nWhat Is Physics?\n23\nobey the Pauli exclusion principle, which forbids more than one electron\nfrom occupying the same orbital if they have the same spin. The Pauli\nexclusion principle is obeyed by the subatomic particles of which matter\nis composed, but disobeyed by the particles, called photons, of which a\nbeam of light is made.\nEinstein’s theory of relativity is discussed more fully in book 6 of this\nseries.\nThe boundary between physics and the other sciences is not\nalways clear. For instance, chemists study atoms and molecules,\nwhich are what matter is built from, and there are some scientists\nwho would be equally willing to call themselves physical chemists\nor chemical physicists. It might seem that the distinction between\nphysics and biology would be clearer, since physics seems to deal\nwith inanimate objects. In fact, almost all physicists would agree\nthat the basic laws of physics that apply to molecules in a test tube\nwork equally well for the combination of molecules that constitutes\na bacterium. (Some might believe that something more happens in\nthe minds of humans, or even those of cats and dogs.) What differentiates physics from biology is that many of the scientific theories\nthat describe living things, while ultimately resulting from the fundamental laws of physics, cannot be rigorously derived from physical\nprinciples.\nIsolated systems and reductionism\nTo avoid having to study everything at once, scientists isolate the\nthings they are trying to study. For instance, a physicist who wants\nto study the motion of a rotating gyroscope would probably prefer\nthat it be isolated from vibrations and air currents. Even in biology,\nwhere field work is indispensable for understanding how living things\nrelate to their entire environment, it is interesting to note the vital\nhistorical role played by Darwin’s study of the Galápagos Islands,\nwhich were conveniently isolated from the rest of the world. Any\npart of the universe that is considered apart from the rest can be\ncalled a “system.”\nPhysics has had some of its greatest successes by carrying this\nprocess of isolation to extremes, subdividing the universe into smaller\nand smaller parts. Matter can be divided into atoms, and the behavior of individual atoms can be studied. Atoms can be split apart\ninto their constituent neutrons, protons and electrons. Protons and\nneutrons appear to be made out of even smaller particles called\nquarks, and there have even been some claims of experimental evidence that quarks have smaller parts inside them. This method\nof splitting things into smaller and smaller parts and studying how\nthose parts influence each other is called reductionism. The hope is\nthat the seemingly complex rules governing the larger units can be\nd / Reductionism.\n24\nChapter 0\nIntroduction and Review\nbetter understood in terms of simpler rules governing the smaller\nunits. To appreciate what reductionism has done for science, it is\nonly necessary to examine a 19th-century chemistry textbook. At\nthat time, the existence of atoms was still doubted by some, electrons were not even suspected to exist, and almost nothing was\nunderstood of what basic rules governed the way atoms interacted\nwith each other in chemical reactions. Students had to memorize\nlong lists of chemicals and their reactions, and there was no way to\nunderstand any of it systematically. Today, the student only needs\nto remember a small set of rules about how atoms interact, for instance that atoms of one element cannot be converted into another\nvia chemical reactions, or that atoms from the right side of the periodic table tend to form strong bonds with atoms from the left\nside.\nDiscussion Questions\nA\nI’ve suggested replacing the ordinary dictionary definition of light\nwith a more technical, more precise one that involves weightlessness. It’s\nstill possible, though, that the stuff a lightbulb makes, ordinarily called\n“light,” does have some small amount of weight. Suggest an experiment\nto attempt to measure whether it does.\nB Heat is weightless (i.e., an object becomes no heavier when heated),\nand can travel across an empty room from the fireplace to your skin,\nwhere it influences you by heating you. Should heat therefore be considered a form of light by our definition? Why or why not?\nC\nSimilarly, should sound be considered a form of light?\n0.3 How to Learn Physics\nFor as knowledges are now delivered, there is a kind of contract of error between the deliverer and the receiver; for he\nthat delivereth knowledge desireth to deliver it in such a form\nas may be best believed, and not as may be best examined;\nand he that receiveth knowledge desireth rather present satisfaction than expectant inquiry.\nFrancis Bacon\nMany students approach a science course with the idea that they\ncan succeed by memorizing the formulas, so that when a problem\nis assigned on the homework or an exam, they will be able to plug\nnumbers in to the formula and get a numerical result on their calculator. Wrong! That’s not what learning science is about! There\nis a big difference between memorizing formulas and understanding\nconcepts. To start with, different formulas may apply in different\nsituations. One equation might represent a definition, which is always true. Another might be a very specific equation for the speed\nSection 0.3\nHow to Learn Physics\n25\nof an object sliding down an inclined plane, which would not be true\nif the object was a rock drifting down to the bottom of the ocean.\nIf you don’t work to understand physics on a conceptual level, you\nwon’t know which formulas can be used when.\nMost students taking college science courses for the first time\nalso have very little experience with interpreting the meaning of an\nequation. Consider the equation w = A/h relating the width of a\nrectangle to its height and area. A student who has not developed\nskill at interpretation might view this as yet another equation to\nmemorize and plug in to when needed. A slightly more savvy student might realize that it is simply the familiar formula A = wh\nin a different form. When asked whether a rectangle would have\na greater or smaller width than another with the same area but\na smaller height, the unsophisticated student might be at a loss,\nnot having any numbers to plug in on a calculator. The more experienced student would know how to reason about an equation\ninvolving division — if h is smaller, and A stays the same, then w\nmust be bigger. Often, students fail to recognize a sequence of equations as a derivation leading to a final result, so they think all the\nintermediate steps are equally important formulas that they should\nmemorize.\nWhen learning any subject at all, it is important to become as\nactively involved as possible, rather than trying to read through\nall the information quickly without thinking about it. It is a good\nidea to read and think about the questions posed at the end of each\nsection of these notes as you encounter them, so that you know you\nhave understood what you were reading.\nMany students’ difficulties in physics boil down mainly to difficulties with math. Suppose you feel confident that you have enough\nmathematical preparation to succeed in this course, but you are\nhaving trouble with a few specific things. In some areas, the brief\nreview given in this chapter may be sufficient, but in other areas\nit probably will not. Once you identify the areas of math in which\nyou are having problems, get help in those areas. Don’t limp along\nthrough the whole course with a vague feeling of dread about something like scientific notation. The problem will not go away if you\nignore it. The same applies to essential mathematical skills that you\nare learning in this course for the first time, such as vector addition.\nSometimes students tell me they keep trying to understand a\ncertain topic in the book, and it just doesn’t make sense. The worst\nthing you can possibly do in that situation is to keep on staring\nat the same page. Every textbook explains certain things badly —\neven mine! — so the best thing to do in this situation is to look\nat a different book. Instead of college textbooks aimed at the same\nmathematical level as the course you’re taking, you may in some\ncases find that high school books or books at a lower math level\n26\nChapter 0\nIntroduction and Review\ngive clearer explanations.\nFinally, when reviewing for an exam, don’t simply read back\nover the text and your lecture notes. Instead, try to use an active\nmethod of reviewing, for instance by discussing some of the discussion questions with another student, or doing homework problems\nyou hadn’t done the first time.\n0.4 Self-Evaluation\nThe introductory part of a book like this is hard to write, because every student arrives at this starting point with a different preparation.\nOne student may have grown up in another country and so may be\ncompletely comfortable with the metric system, but may have had\nan algebra course in which the instructor passed too quickly over\nscientific notation. Another student may have already taken calculus, but may have never learned the metric system. The following\nself-evaluation is a checklist to help you figure out what you need to\nstudy to be prepared for the rest of the course.\nIf you disagree with this statement. . .\nI am familiar with the basic metric\nunits of meters, kilograms, and seconds, and the most common metric\nprefixes: milli- (m), kilo- (k), and\ncenti- (c).\nI know about the newton, a unit of\nforce\nI am familiar with these less common metric prefixes: mega- (M),\nmicro- (µ), and nano- (n).\nI am comfortable with scientific notation.\nI can confidently do metric conversions.\nI understand the purpose and use of\nsignificant figures.\nyou should study this section:\n0.5 Basic of the Metric System\n0.6 The newton, the Metric Unit of\nForce\n0.7 Less Common Metric Prefixes\n0.8 Scientific Notation\n0.9 Conversions\n0.10 Significant Figures\nIt wouldn’t hurt you to skim the sections you think you already\nknow about, and to do the self-checks in those sections.\n0.5 Basics of the Metric System\nThe metric system\nUnits were not standardized until fairly recently in history, so\nwhen the physicist Isaac Newton gave the result of an experiment\nwith a pendulum, he had to specify not just that the string was 37\n7 / inches long but that it was “37 7 / London inches long.” The\n8\n8\nSection 0.4\nSelf-Evaluation\n27\ninch as defined in Yorkshire would have been different. Even after\nthe British Empire standardized its units, it was still very inconvenient to do calculations involving money, volume, distance, time, or\nweight, because of all the odd conversion factors, like 16 ounces in\na pound, and 5280 feet in a mile. Through the nineteenth century,\nschoolchildren squandered most of their mathematical education in\npreparing to do calculations such as making change when a customer\nin a shop offered a one-crown note for a book costing two pounds,\nthirteen shillings and tuppence. The dollar has always been decimal,\nand British money went decimal decades ago, but the United States\nis still saddled with the antiquated system of feet, inches, pounds,\nounces and so on.\nEvery country in the world besides the U.S. has adopted a system of units known in English as the “metric system.” This system\nis entirely decimal, thanks to the same eminently logical people who\nbrought about the French Revolution. In deference to France, the\nsystem’s official name is the Système International, or SI, meaning\nInternational System. (The phrase “SI system” is therefore redundant.)\nThe wonderful thing about the SI is that people who live in\ncountries more modern than ours do not need to memorize how\nmany ounces there are in a pound, how many cups in a pint, how\nmany feet in a mile, etc. The whole system works with a single,\nconsistent set of prefixes (derived from Greek) that modify the basic\nunits. Each prefix stands for a power of ten, and has an abbreviation\nthat can be combined with the symbol for the unit. For instance,\nthe meter is a unit of distance. The prefix kilo- stands for 103 , so a\nkilometer, 1 km, is a thousand meters.\nThe basic units of the metric system are the meter for distance,\nthe second for time, and the gram for mass.\nThe following are the most common metric prefixes. You should\nmemorize them.\nprefix\nmeaning\nexample\n3\nkilok 10\n60 kg = a person’s mass\ncenti- c 10−2\n28 cm = height of a piece of paper\nmilli- m 10−3\n1 ms\n= time for one vibration of a guitar\nstring playing the note D\nThe prefix centi-, meaning 10−2 , is only used in the centimeter;\na hundredth of a gram would not be written as 1 cg but as 10 mg.\nThe centi- prefix can be easily remembered because a cent is 10−2\ndollars. The official SI abbreviation for seconds is “s” (not “sec”)\nand grams are “g” (not “gm”).\nThe second\nThe sun stood still and the moon halted until the nation had\ntaken vengeance on its enemies. . .\n28\nChapter 0\nIntroduction and Review\nJoshua 10:12-14\nAbsolute, true, and mathematical time, of itself, and from its\nown nature, flows equably without relation to anything external. . .\nIsaac Newton\nWhen I stated briefly above that the second was a unit of time,\nit may not have occurred to you that this was not really much of\na definition. The two quotes above are meant to demonstrate how\nmuch room for confusion exists among people who seem to mean the\nsame thing by a word such as “time.” The first quote has been interpreted by some biblical scholars as indicating an ancient belief that\nthe motion of the sun across the sky was not just something that\noccurred with the passage of time but that the sun actually caused\ntime to pass by its motion, so that freezing it in the sky would have\nsome kind of a supernatural decelerating effect on everyone except\nthe Hebrew soldiers. Many ancient cultures also conceived of time\nas cyclical, rather than proceeding along a straight line as in 1998,\n1999, 2000, 2001,... The second quote, from a relatively modern\nphysicist, may sound a lot more scientific, but most physicists today would consider it useless as a definition of time. Today, the\nphysical sciences are based on operational definitions, which means\ndefinitions that spell out the actual steps (operations) required to\nmeasure something numerically.\nNow in an era when our toasters, pens, and coffee pots tell us the\ntime, it is far from obvious to most people what is the fundamental\noperational definition of time. Until recently, the hour, minute, and\nsecond were defined operationally in terms of the time required for\nthe earth to rotate about its axis. Unfortunately, the Earth’s rotation is slowing down slightly, and by 1967 this was becoming an\nissue in scientific experiments requiring precise time measurements.\nThe second was therefore redefined as the time required for a certain number of vibrations of the light waves emitted by a cesium\natoms in a lamp constructed like a familiar neon sign but with the\nneon replaced by cesium. The new definition not only promises to\nstay constant indefinitely, but for scientists is a more convenient\nway of calibrating a clock than having to carry out astronomical\nmeasurements.\nself-check C\nWhat is a possible operational definition of how strong a person is?\nAnswer, p. 266\ne / Pope Gregory created our\nmodern Gregorian calendar, with\nits system of leap years, to make\nthe length of the calendar year\nmatch the length of the cycle\nof seasons. Not until 1752 did\nProtestant England switched to\nthe new calendar. Some less\neducated citizens believed that\nthe shortening of the month by\neleven days would shorten their\nlives by the same interval. In this\nillustration by William Hogarth,\nthe leaflet lying on the ground\nreads, “Give us our eleven days.”\n.\nThe meter\nThe French originally defined the meter as 10−7 times the distance from the equator to the north pole, as measured through Paris\n(of course). Even if the definition was operational, the operation of\ntraveling to the north pole and laying a surveying chain behind you\nSection 0.5\nf / The original\nthe meter.\nBasics of the Metric System\ndefinition\nof\n29\nwas not one that most working scientists wanted to carry out. Fairly\nsoon, a standard was created in the form of a metal bar with two\nscratches on it. This definition persisted until 1960, when the meter\nwas redefined as the distance traveled by light in a vacuum over a\nperiod of (1/299792458) seconds.\nThe kilogram\nThe third base unit of the SI is the kilogram, a unit of mass.\nMass is intended to be a measure of the amount of a substance,\nbut that is not an operational definition. Bathroom scales work by\nmeasuring our planet’s gravitational attraction for the object being\nweighed, but using that type of scale to define mass operationally\nwould be undesirable because gravity varies in strength from place\nto place on the earth.\nThere’s a surprising amount of disagreement among physics textbooks about how mass should be defined, but here’s how it’s actually\nhandled by the few working physicists who specialize in ultra-highprecision measurements. They maintain a physical object in Paris,\nwhich is the standard kilogram, a cylinder made of platinum-iridium\nalloy. Duplicates are checked against this mother of all kilograms\nby putting the original and the copy on the two opposite pans of a\nbalance. Although this method of comparison depends on gravity,\nthe problems associated with differences in gravity in different geographical locations are bypassed, because the two objects are being\ncompared in the same place. The duplicates can then be removed\nfrom the Parisian kilogram shrine and transported elsewhere in the\nworld.\nCombinations of metric units\nJust about anything you want to measure can be measured with\nsome combination of meters, kilograms, and seconds. Speed can be\nmeasured in m/s, volume in m3 , and density in kg/m3 . Part of what\nmakes the SI great is this basic simplicity. No more funny units like\na cord of wood, a bolt of cloth, or a jigger of whiskey. No more\nliquid and dry measure. Just a simple, consistent set of units. The\nSI measures put together from meters, kilograms, and seconds make\nup the mks system. For example, the mks unit of speed is m/s, not\nkm/hr.\nDiscussion Question\nA\nIsaac Newton wrote, “. . . the natural days are truly unequal, though\nthey are commonly considered as equal, and used for a measure of\ntime. . . It may be that there is no such thing as an equable motion, whereby\ntime may be accurately measured. All motions may be accelerated or retarded. . . ” Newton was right. Even the modern definition of the second\nin terms of light emitted by cesium atoms is subject to variation. For instance, magnetic fields could cause the cesium atoms to emit light with\na slightly different rate of vibration. What makes us think, though, that a\npendulum clock is more accurate than a sundial, or that a cesium atom\n30\nChapter 0\nIntroduction and Review\nis a more accurate timekeeper than a pendulum clock? That is, how can\none test experimentally how the accuracies of different time standards\ncompare?\n0.6 The Newton, the Metric Unit of Force\nA force is a push or a pull, or more generally anything that can\nchange an object’s speed or direction of motion. A force is required\nto start a car moving, to slow down a baseball player sliding in to\nhome base, or to make an airplane turn. (Forces may fail to change\nan object’s motion if they are canceled by other forces, e.g., the\nforce of gravity pulling you down right now is being canceled by the\nforce of the chair pushing up on you.) The metric unit of force is\nthe Newton, defined as the force which, if applied for one second,\nwill cause a 1-kilogram object starting from rest to reach a speed of\n1 m/s. Later chapters will discuss the force concept in more detail.\nIn fact, this entire book is about the relationship between force and\nmotion.\nIn section 0.5, I gave a gravitational definition of mass, but by\ndefining a numerical scale of force, we can also turn around and define a scale of mass without reference to gravity. For instance, if a\nforce of two Newtons is required to accelerate a certain object from\nrest to 1 m/s in 1 s, then that object must have a mass of 2 kg.\nFrom this point of view, mass characterizes an object’s resistance\nto a change in its motion, which we call inertia or inertial mass.\nAlthough there is no fundamental reason why an object’s resistance\nto a change in its motion must be related to how strongly gravity\naffects it, careful and precise experiments have shown that the inertial definition and the gravitational definition of mass are highly\nconsistent for a variety of objects. It therefore doesn’t really matter\nfor any practical purpose which definition one adopts.\nDiscussion Question\nA\nSpending a long time in weightlessness is unhealthy. One of the\nmost important negative effects experienced by astronauts is a loss of\nmuscle and bone mass. Since an ordinary scale won’t work for an astronaut in orbit, what is a possible way of monitoring this change in mass?\n(Measuring the astronaut’s waist or biceps with a measuring tape is not\ngood enough, because it doesn’t tell anything about bone mass, or about\nthe replacement of muscle with fat.)\ng / This is a mnemonic to\nhelp you remember the most important metric prefixes. The word\n“little” is to remind you that the\nlist starts with the prefixes used\nfor small quantities and builds\nupward. The exponent changes\nby 3, except that of course that\nwe do not need a special prefix\nfor 100 , which equals one.\n0.7 Less Common Metric Prefixes\nThe following are three metric prefixes which, while less common\nthan the ones discussed previously, are well worth memorizing.\nSection 0.6\nThe Newton, the Metric Unit of Force\n31\nprefix\nmega- M\nmicro- µ\nnano- n\nmeaning\n106\n10−6\n10−9\n6.4 Mm\n10 µm\n0.154 nm\nexample\n= radius of the earth\n= size of a white blood cell\n= distance between carbon\nnuclei in an ethane molecule\nNote that the abbreviation for micro is the Greek letter mu, µ\n— a common mistake is to confuse it with m (milli) or M (mega).\nThere are other prefixes even less common, used for extremely\nlarge and small quantities. For instance, 1 femtometer = 10−15 m is\na convenient unit of distance in nuclear physics, and 1 gigabyte =\n109 bytes is used for computers’ hard disks. The international committee that makes decisions about the SI has recently even added\nsome new prefixes that sound like jokes, e.g., 1 yoctogram = 10−24 g\nis about half the mass of a proton. In the immediate future, however, you’re unlikely to see prefixes like “yocto-” and “zepto-” used\nexcept perhaps in trivia contests at science-fiction conventions or\nother geekfests.\nself-check D\nSuppose you could slow down time so that according to your perception,\na beam of light would move across a room at the speed of a slow walk.\nIf you perceived a nanosecond as if it was a second, how would you\nperceive a microsecond?\n. Answer, p. 266\n0.8 Scientific Notation\nMost of the interesting phenomena in our universe are not on the\nhuman scale. It would take about 1,000,000,000,000,000,000,000\nbacteria to equal the mass of a human body. When the physicist\nThomas Young discovered that light was a wave, it was back in the\nbad old days before scientific notation, and he was obliged to write\nthat the time required for one vibration of the wave was 1/500 of\na millionth of a millionth of a second. Scientific notation is a less\nawkward way to write very large and very small numbers such as\nthese. Here’s a quick review.\nScientific notation means writing a number in terms of a product\nof something from 1 to 10 and something else that is a power of ten.\nFor instance,\n32 = 3.2 × 101\n320 = 3.2 × 102\n3200 = 3.2 × 103\n...\nEach number is ten times bigger than the previous one.\nSince 101 is ten times smaller than 102 , it makes sense to use\nthe notation 100 to stand for one, the number that is in turn ten\ntimes smaller than 101 . Continuing on, we can write 10−1 to stand\n32\nChapter 0\nIntroduction and Review\nfor 0.1, the number ten times smaller than 100 . Negative exponents\nare used for small numbers:\n3.2 = 3.2 × 100\n0.32 = 3.2 × 10−1\n0.032 = 3.2 × 10−2\n...\nA common source of confusion is the notation used on the displays of many calculators. Examples:\n3.2 × 106\n3.2E+6\n3.26\n(written notation)\n(notation on some calculators)\n(notation on some other calculators)\nThe last example is particularly unfortunate, because 3.26 really\nstands for the number 3.2 × 3.2 × 3.2 × 3.2 × 3.2 × 3.2 = 1074, a\ntotally different number from 3.2 × 106 = 3200000. The calculator\nnotation should never be used in writing. It’s just a way for the\nmanufacturer to save money by making a simpler display.\nself-check E\nA student learns that 104 bacteria, standing in line to register for classes\nat Paramecium Community College, would form a queue of this size:\nThe student concludes that 102 bacteria would form a line of this length:\nWhy is the student incorrect?\n. Answer, p. 266\n0.9 Conversions\nI suggest you avoid memorizing lots of conversion factors between\nSI units and U.S. units. Suppose the United Nations sends its black\nhelicopters to invade California (after all who wouldn’t rather live\nhere than in New York City?), and institutes water fluoridation and\nthe SI, making the use of inches and pounds into a crime punishable\nby death. I think you could get by with only two mental conversion\nfactors:\n1 inch = 2.54 cm\nAn object with a weight on Earth of 2.2 pounds-force has a\nmass of 1 kg.\nThe first one is the present definition of the inch, so it’s exact. The\nsecond one is not exact, but is good enough for most purposes. (U.S.\nunits of force and mass are confusing, so it’s a good thing they’re\nSection 0.9\nConversions\n33\nnot used in science. In U.S. units, the unit of force is the poundforce, and the best unit to use for mass is the slug, which is about\n14.6 kg.)\nMore important than memorizing conversion factors is understanding the right method for doing conversions. Even within the\nSI, you may need to convert, say, from grams to kilograms. Different people have different ways of thinking about conversions, but\nthe method I’ll describe here is systematic and easy to understand.\nThe idea is that if 1 kg and 1000 g represent the same mass, then\nwe can consider a fraction like\n103 g\n1 kg\nto be a way of expressing the number one. This may bother you. For\ninstance, if you type 1000/1 into your calculator, you will get 1000,\nnot one. Again, different people have different ways of thinking\nabout it, but the justification is that it helps us to do conversions,\nand it works! Now if we want to convert 0.7 kg to units of grams,\nwe can multiply kg by the number one:\n0.7 kg ×\n103 g\n1 kg\nIf you’re willing to treat symbols such as “kg” as if they were variables as used in algebra (which they’re really not), you can then\ncancel the kg on top with the kg on the bottom, resulting in\n0.7 kg ×\n103 g\n= 700 g\n1 kg\n.\nTo convert grams to kilograms, you would simply flip the fraction\nupside down.\nOne advantage of this method is that it can easily be applied to\na series of conversions. For instance, to convert one year to units of\nseconds,\n×\n1\nyear\n24 60 365 days\nhours\nmin\n60 s\n×\n×\n×\n=\n1\nyear\n1\nday\n1\nhour\n1\nmin\n= 3.15 × 107 s\n.\nShould that exponent be positive or negative?\nA common mistake is to write the conversion fraction incorrectly.\nFor instance the fraction\n103 kg\n1g\n34\nChapter 0\nIntroduction and Review\n(incorrect)\ndoes not equal one, because 103 kg is the mass of a car, and 1 g is\nthe mass of a raisin. One correct way of setting up the conversion\nfactor would be\n10−3 kg\n(correct)\n.\n1g\nYou can usually detect such a mistake if you take the time to check\nyour answer and see if it is reasonable.\nIf common sense doesn’t rule out either a positive or a negative\nexponent, here’s another way to make sure you get it right. There\nare big prefixes and small prefixes:\nbig prefixes:\nsmall prefixes:\nk\nm\nM\nµ\nn\n(It’s not hard to keep straight which are which, since “mega” and\n“micro” are evocative, and it’s easy to remember that a kilometer\nis bigger than a meter and a millimeter is smaller.) In the example\nabove, we want the top of the fraction to be the same as the bottom.\nSince k is a big prefix, we need to compensate by putting a small\nnumber like 10−3 in front of it, not a big number like 103 .\n. Solved problem: a simple conversion\npage 40, problem 6\n. Solved problem: the geometric mean\npage 41, problem 8\nDiscussion Question\nA\nEach of the following conversions contains an error. In each case,\nexplain what the error is.\n(a) 1000 kg ×\n(b) 50 m ×\n1 kg\n1000 g\n1 cm\n100 m\n=1g\n= 0.5 cm\n(c) “Nano” is 10−9 , so there are 10−9 nm in a meter.\n(d) “Micro” is 10−6 , so 1 kg is 106 µg.\n0.10 Significant Figures\nAn engineer is designing a car engine, and has been told that the\ndiameter of the pistons (which are being designed by someone else)\nis 5 cm. He knows that 0.02 cm of clearance is required for a piston\nof this size, so he designs the cylinder to have an inside diameter of\n5.04 cm. Luckily, his supervisor catches his mistake before the car\ngoes into production. She explains his error to him, and mentally\nputs him in the “do not promote” category.\nWhat was his mistake? The person who told him the pistons\nwere 5 cm in diameter was wise to the ways of significant figures,\nas was his boss, who explained to him that he needed to go back\nand get a more accurate number for the diameter of the pistons.\nThat person said “5 cm” rather than “5.00 cm” specifically to avoid\ncreating the impression that the number was extremely accurate. In\nSection 0.10\nSignificant Figures\n35\nreality, the pistons’ diameter was 5.13 cm. They would never have\nfit in the 5.04-cm cylinders.\nThe number of digits of accuracy in a number is referred to as\nthe number of significant figures, or “sig figs” for short. As in the\nexample above, sig figs provide a way of showing the accuracy of a\nnumber. In most cases, the result of a calculation involving several\npieces of data can be no more accurate than the least accurate piece\nof data. In other words, “garbage in, garbage out.” Since the 5\ncm diameter of the pistons was not very accurate, the result of the\nengineer’s calculation, 5.04 cm, was really not as accurate as he\nthought. In general, your result should not have more than the\nnumber of sig figs in the least accurate piece of data you started\nwith. The calculation above should have been done as follows:\n5 cm\n+0.04 cm\n=5 cm\n(1 sig fig)\n(1 sig fig)\n(rounded off to 1 sig fig)\nThe fact that the final result only has one significant figure then\nalerts you to the fact that the result is not very accurate, and would\nnot be appropriate for use in designing the engine.\nNote that the leading zeroes in the number 0.04 do not count\nas significant figures, because they are only placeholders. On the\nother hand, a number such as 50 cm is ambiguous — the zero could\nbe intended as a significant figure, or it might just be there as a\nplaceholder. The ambiguity involving trailing zeroes can be avoided\nby using scientific notation, in which 5 × 101 cm would imply one\nsig fig of accuracy, while 5.0 × 101 cm would imply two sig figs.\n36\nChapter 0\nIntroduction and Review\nself-check F\nThe following quote is taken from an editorial by Norimitsu Onishi in the\nNew York Times, August 18, 2002.\nConsider Nigeria. Everyone agrees it is Africa’s most populous\nnation. But what is its population? The United Nations says\n114 million; the State Department, 120 million. The World Bank\nsays 126.9 million, while the Central Intelligence Agency puts it\nat 126,635,626.\nWhat should bother you about this?\n. Answer, p. 266\nDealing correctly with significant figures can save you time! Often, students copy down numbers from their calculators with eight\nsignificant figures of precision, then type them back in for a later\ncalculation. That’s a waste of time, unless your original data had\nthat kind of incredible precision.\nThe rules about significant figures are only rules of thumb, and\nare not a substitute for careful thinking. For instance, \\$20.00 +\n\\$0.05 is \\$20.05. It need not and should not be rounded off to \\$20.\nIn general, the sig fig rules work best for multiplication and division,\nand we also apply them when doing a complicated calculation that\ninvolves many types of operations. For simple addition and subtraction, it makes more sense to maintain a fixed number of digits after\nthe decimal point.\nWhen in doubt, don’t use the sig fig rules at all. Instead, intentionally change one piece of your initial data by the maximum\namount by which you think it could have been off, and recalculate\nthe final result. The digits on the end that are completely reshuffled\nare the ones that are meaningless, and should be omitted.\nself-check G\nHow many significant figures are there in each of the following measurements?\n(1) 9.937 m\n(2) 4.0 s\n(3) 0.0000000000000037 kg\n. Answer, p. 266\nSection 0.10\nSignificant Figures\n37\nSummary\nSelected Vocabulary\nmatter . . . . . . Anything that is affected by gravity.\nlight . . . . . . . . Anything that can travel from one place to another through empty space and can influence\nmatter, but is not affected by gravity.\noperational defi- A definition that states what operations\nnition . . . . . . . should be carried out to measure the thing being defined.\nSystème Interna- A fancy name for the metric system.\ntional . . . . . . .\nmks system . . . The use of metric units based on the meter,\nkilogram, and second. Example: meters per\nsecond is the mks unit of speed, not cm/s or\nkm/hr.\nmass . . . . . . . A numerical measure of how difficult it is to\nchange an object’s motion.\nsignificant figures Digits that contribute to the accuracy of a\nmeasurement.\nNotation\nm . . . .\nkg . . . .\ns . . . . .\nM- . . . .\nk- . . . .\nm- . . . .\nµ- . . . .\nn- . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nmeter, the metric distance unit\nkilogram, the metric unit of mass\nsecond, the metric unit of time\nthe metric prefix mega-, 106\nthe metric prefix kilo-, 103\nthe metric prefix milli-, 10−3\nthe metric prefix micro-, 10−6\nthe metric prefix nano-, 10−9\nSummary\nPhysics is the use of the scientific method to study the behavior\nof light and matter. The scientific method requires a cycle of theory and experiment, theories with both predictive and explanatory\nvalue, and reproducible experiments.\nThe metric system is a simple, consistent framework for measurement built out of the meter, the kilogram, and the second plus a set\nof prefixes denoting powers of ten. The most systematic method for\ndoing conversions is shown in the following example:\n370 ms ×\n10−3 s\n= 0.37 s\n1 ms\nMass is a measure of the amount of a substance. Mass can be\ndefined gravitationally, by comparing an object to a standard mass\non a double-pan balance, or in terms of inertia, by comparing the\neffect of a force on an object to the effect of the same force on a\nstandard mass. The two definitions are found experimentally to\nbe proportional to each other to a high degree of precision, so we\n38\nChapter 0\nIntroduction and Review\nusually refer simply to “mass,” without bothering to specify which\ntype.\nA force is that which can change the motion of an object. The\nmetric unit of force is the Newton, defined as the force required to\naccelerate a standard 1-kg mass from rest to a speed of 1 m/s in 1\ns.\nScientific notation means, for example, writing 3.2 × 105 rather\nthan 320000.\nWriting numbers with the correct number of significant figures\ncorrectly communicates how accurate they are. As a rule of thumb,\nthe final result of a calculation is no more accurate than, and should\nhave no more significant figures than, the least accurate piece of\ndata.\nExploring Further\nMicrobe Hunters, Paul de Kruif. The dramatic life-and-death\nstories in this book are entertaining, but along the way de Kruif also\npresents an excellent, warts-and-all picture of how real science and\nreal scientists really work — an excellent anecdote to the sanitized\npicture of the scientific method often presented in textbooks. Some\nof the descriptions of field work in Africa are marred by racism.\nVoodoo Science: The Road from Foolishness to Fraud, Robert\nL. Park. Park has some penetrating psychological insights into the\nfundamental problems that Homo sapiens (scientists included) often have with the unwelcome truths that science tosses in our laps.\nUntil I read this book, I hadn’t realized, for example, how common\nit is to find pockets of bogus science in such otherwise respectable\ninstitutions as NASA.\nSummary\n39\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n74658\n1\nCorrect use of a calculator: (a) Calculate 53222+97554\non a calculator. [Self-check: The most common mistake results in 97555.40.]\n√\n(b) Which would be more like the price of a TV, and which would\nbe more like the price of a house, \\$3.5 × 105 or \\$3.55 ?\n2\nCompute the following things. If they don’t make sense because of units, say so.\n(a) 3 cm + 5 cm\n(b) 1.11 m + 22 cm\n(c) 120 miles + 2.0 hours\n(d) 120 miles / 2.0 hours\n3\nYour backyard has brick walls on both ends. You measure a\ndistance of 23.4 m from the inside of one wall to the inside of the\nother. Each wall is 29.4 cm thick. How far is it from the outside\nof one wall to the outside of the other? Pay attention to significant\nfigures.\n4\nThe speed of light is 3.0 × 108 m/s. Convert this to furlongs\nper fortnight. A furlong is 220 yards, and a fortnight is 14 days. An\n√\ninch is 2.54 cm.\n5\nExpress each of the following quantities in micrograms:\n(a) 10 mg, (b) 104 g, (c) 10 kg, (d) 100 × 103 g, (e) 1000 ng.\n√\n6\nConvert 134 mg to units of kg, writing your answer in scientific\nnotation.\n. Solution, p. 269\n7\nIn the last century, the average age of the onset of puberty for\ngirls has decreased by several years. Urban folklore has it that this\nis because of hormones fed to beef cattle, but it is more likely to be\nbecause modern girls have more body fat on the average and possibly because of estrogen-mimicking chemicals in the environment\nfrom the breakdown of pesticides. A hamburger from a hormoneimplanted steer has about 0.2 ng of estrogen (about double the\namount of natural beef). A serving of peas contains about 300\nng of estrogen. An adult woman produces about 0.5 mg of estrogen\nper day (note the different unit!). (a) How many hamburgers would\na girl have to eat in one day to consume as much estrogen as an\nadult woman’s daily production? (b) How many servings of peas?\n√\n40\nChapter 0\nIntroduction and Review\n8\nThe usual definition of the mean (average) of two numbers a\nand b is (a+b)/2. This is called the arithmetic mean. The geometric\nmean, however, is defined as (ab)1/2 (i.e., the square root of ab). For\nthe sake of definiteness, let’s say both numbers have units of mass.\n(a) Compute the arithmetic mean of two numbers that have units\nof grams. Then convert the numbers to units of kilograms and\nrecompute their mean. Is the answer consistent? (b) Do the same\nfor the geometric mean. (c) If a and b both have units of grams,\nwhat should we call the units of ab? Does your answer make sense\nwhen you take the square root? (d) Suppose someone proposes to\nyou a third kind of mean, called the superduper mean, defined as\n(ab)1/3 . Is this reasonable?\n. Solution, p. 269\n9\nIn an article on the SARS epidemic, the May 7, 2003 New\nYork Times discusses conflicting estimates of the disease’s incubation period (the average time that elapses from infection to the first\nsymptoms). “The study estimated it to be 6.4 days. But other statistical calculations ... showed that the incubation period could be\nas long as 14.22 days.” What’s wrong here?\n10\nThe photo shows the corner of a bag of pretzels. What’s\nwrong here?\n√\n11\nThe distance to the horizon is given by the expression 2rh,\nwhere r is the radius of the Earth, and h is the observer’s height\nabove the Earth’s surface. (This can be proved using the Pythagorean\ntheorem.) Show that the units of this expression make sense.\nProblem 10.\nProblems\n41\n42\nChapter 0\nIntroduction and Review\nLife would be very different if you\nwere the size of an insect.\nChapter 1\nScaling and\nOrder-of-Magnitude\nEstimates\n1.1 Introduction\nWhy can’t an insect be the size of a dog? Some skinny stretchedout cells in your spinal cord are a meter tall — why does nature\ndisplay no single cells that are not just a meter tall, but a meter\nwide, and a meter thick as well? Believe it or not, these are questions\nthat can be answered fairly easily without knowing much more about\nphysics than you already do. The only mathematical technique you\nreally need is the humble conversion, applied to area and volume.\nArea and volume\nArea can be defined by saying that we can copy the shape of\ninterest onto graph paper with 1 cm × 1 cm squares and count the\nnumber of squares inside. Fractions of squares can be estimated by\neye. We then say the area equals the number of squares, in units of\nsquare cm. Although this might seem less “pure” than computing\nareas using formulae like A = πr2 for a circle or A = wh/2 for a\ntriangle, those formulae are not useful as definitions of area because\na / Amoebas this size\nseldom encountered.\nare\n43\nthey cannot be applied to irregularly shaped areas.\nUnits of square cm are more commonly written as cm2 in science.\nOf course, the unit of measurement symbolized by “cm” is not an\nalgebra symbol standing for a number that can be literally multiplied\nby itself. But it is advantageous to write the units of area that way\nand treat the units as if they were algebra symbols. For instance,\nif you have a rectangle with an area of 6m2 and a width of 2 m,\nthen calculating its length as (6 m2 )/(2 m) = 3 m gives a result\nthat makes sense both numerically and in terms of units. This\nalgebra-style treatment of the units also ensures that our methods\nof converting units work out correctly. For instance, if we accept\nthe fraction\n100 cm\n1m\nas a valid way of writing the number one, then one times one equals\none, so we should also say that one can be represented by\n100 cm 100 cm\n×\n1m\n1m\n,\nwhich is the same as\n10000 cm2\n.\n1 m2\nThat means the conversion factor from square meters to square centimeters is a factor of 104 , i.e., a square meter has 104 square centimeters in it.\nAll of the above can be easily applied to volume as well, using\none-cubic-centimeter blocks instead of squares on graph paper.\nTo many people, it seems hard to believe that a square meter\nequals 10000 square centimeters, or that a cubic meter equals a\nmillion cubic centimeters — they think it would make more sense if\nthere were 100 cm2 in 1 m2 , and 100 cm3 in 1 m3 , but that would be\nincorrect. The examples shown in figure b aim to make the correct\nanswer more believable, using the traditional U.S. units of feet and\nyards. (One foot is 12 inches, and one yard is three feet.)\nb / Visualizing conversions of\narea and volume using traditional\nU.S. units.\nself-check A\nBased on figure b, convince yourself that there are 9 ft2 in a square yard,\n44\nChapter 1\nScaling and Order-of-Magnitude Estimates\nand 27 ft3 in a cubic yard, then demonstrate the same thing symbolically\n(i.e., with the method using fractions that equal one).\n. Answer, p.\n266\n. Solved problem: converting mm2 to cm2\npage 61, problem 10\n. Solved problem: scaling a liter\npage 62, problem 19\nDiscussion Question\nA\nHow many square centimeters are there in a square inch? (1 inch =\n2.54 cm) First find an approximate answer by making a drawing, then derive the conversion factor more accurately using the symbolic method.\nc / Galileo Galilei (1564-1642) was a Renaissance Italian who brought the\nscientific method to bear on physics, creating the modern version of the\nscience. Coming from a noble but very poor family, Galileo had to drop\nout of medical school at the University of Pisa when he ran out of money.\nEventually becoming a lecturer in mathematics at the same school, he\nbegan a career as a notorious troublemaker by writing a burlesque ridiculing the university’s regulations — he was forced to resign, but found a\nnew teaching position at Padua. He invented the pendulum clock, investigated the motion of falling bodies, and discovered the moons of Jupiter.\nThe thrust of his life’s work was to discredit Aristotle’s physics by confronting it with contradictory experiments, a program which paved the way\nfor Newton’s discovery of the relationship between force and motion. In\nchapter 3 we’ll come to the story of Galileo’s ultimate fate at the hands of\nthe Church.\n1.2 Scaling of Area and Volume\nGreat fleas have lesser fleas\nUpon their backs to bite ’em.\nAnd lesser fleas have lesser still,\nAnd so ad infinitum.\nJonathan Swift\nNow how do these conversions of area and volume relate to the\nquestions I posed about sizes of living things? Well, imagine that\nyou are shrunk like Alice in Wonderland to the size of an insect.\nOne way of thinking about the change of scale is that what used\nto look like a centimeter now looks like perhaps a meter to you,\nbecause you’re so much smaller. If area and volume scaled according\nto most people’s intuitive, incorrect expectations, with 1 m2 being\nthe same as 100 cm2 , then there would be no particular reason\nwhy nature should behave any differently on your new, reduced\nscale. But nature does behave differently now that you’re small.\nFor instance, you will find that you can walk on water, and jump\nto many times your own height. The physicist Galileo Galilei had\nthe basic insight that the scaling of area and volume determines\nhow natural phenomena behave differently on different scales. He\nSection 1.2\nScaling of Area and Volume\n45\nfirst reasoned about mechanical structures, but later extended his\ninsights to living things, taking the then-radical point of view that at\nthe fundamental level, a living organism should follow the same laws\nof nature as a machine. We will follow his lead by first discussing\nmachines and then living things.\nGalileo on the behavior of nature on large and small scales\nOne of the world’s most famous pieces of scientific writing is\nGalileo’s Dialogues Concerning the Two New Sciences. Galileo was\nan entertaining writer who wanted to explain things clearly to laypeople, and he livened up his work by casting it in the form of a dialogue\namong three people. Salviati is really Galileo’s alter ego. Simplicio\nis the stupid character, and one of the reasons Galileo got in trouble\nwith the Church was that there were rumors that Simplicio represented the Pope. Sagredo is the earnest and intelligent student, with\nwhom the reader is supposed to identify. (The following excerpts\nare from the 1914 translation by Crew and de Salvio.)\nd / The small boat holds up\njust fine.\ne / A larger boat built with\nthe same proportions as the\nsmall one will collapse under its\nown weight.\nf / A boat this large needs to\nhave timbers that are thicker\ncompared to its size.\n46\nChapter 1\nS AGREDO : Yes, that is what I mean; and I refer especially to\nhis last assertion which I have always regarded as false. . . ;\nnamely, that in speaking of these and other similar machines\none cannot argue from the small to the large, because many\ndevices which succeed on a small scale do not work on a\nlarge scale. Now, since mechanics has its foundations in geometry, where mere size [ is unimportant], I do not see that\nthe properties of circles, triangles, cylinders, cones and other\nsolid figures will change with their size. If, therefore, a large\nmachine be constructed in such a way that its parts bear to\none another the same ratio as in a smaller one, and if the\nsmaller is sufficiently strong for the purpose for which it is\ndesigned, I do not see why the larger should not be able to\nwithstand any severe and destructive tests to which it may be\nsubjected.\nSalviati contradicts Sagredo:\nS ALVIATI : . . . Please observe, gentlemen, how facts which\nat first seem improbable will, even on scant explanation, drop\nthe cloak which has hidden them and stand forth in naked and\nsimple beauty. Who does not know that a horse falling from a\nheight of three or four cubits will break his bones, while a dog\nfalling from the same height or a cat from a height of eight\nor ten cubits will suffer no injury? Equally harmless would be\nthe fall of a grasshopper from a tower or the fall of an ant from\nthe distance of the moon.\nThe point Galileo is making here is that small things are sturdier\nin proportion to their size. There are a lot of objections that could be\nraised, however. After all, what does it really mean for something to\nbe “strong”, to be “strong in proportion to its size,” or to be strong\nScaling and Order-of-Magnitude Estimates\n“out of proportion to its size?” Galileo hasn’t given operational\ndefinitions of things like “strength,” i.e., definitions that spell out\nhow to measure them numerically.\nAlso, a cat is shaped differently from a horse — an enlarged\nphotograph of a cat would not be mistaken for a horse, even if the\nphoto-doctoring experts at the National Inquirer made it look like a\nperson was riding on its back. A grasshopper is not even a mammal,\nand it has an exoskeleton instead of an internal skeleton. The whole\nargument would be a lot more convincing if we could do some isolation of variables, a scientific term that means to change only one\nthing at a time, isolating it from the other variables that might have\nan effect. If size is the variable whose effect we’re interested in seeing, then we don’t really want to compare things that are different\nin size but also different in other ways.\nS ALVIATI : . . . we asked the reason why [shipbuilders] employed stocks, scaffolding, and bracing of larger dimensions\nfor launching a big vessel than they do for a small one; and\n[an old man] answered that they did this in order to avoid the\ndanger of the ship parting under its own heavy weight, a danger to which small boats are not subject?\nAfter this entertaining but not scientifically rigorous beginning,\nGalileo starts to do something worthwhile by modern standards.\nHe simplifies everything by considering the strength of a wooden\nplank. The variables involved can then be narrowed down to the\ntype of wood, the width, the thickness, and the length. He also\ngives an operational definition of what it means for the plank to\nhave a certain strength “in proportion to its size,” by introducing\nthe concept of a plank that is the longest one that would not snap\nunder its own weight if supported at one end. If you increased\nits length by the slightest amount, without increasing its width or\nthickness, it would break. He says that if one plank is the same\nshape as another but a different size, appearing like a reduced or\nenlarged photograph of the other, then the planks would be strong\n“in proportion to their sizes” if both were just barely able to support\ntheir own weight.\ng / 1. This plank is as long as it\ncan be without collapsing under\nits own weight. If it was a hundredth of an inch longer, it would\ncollapse. 2. This plank is made\nout of the same kind of wood. It is\ntwice as thick, twice as long, and\ntwice as wide. It will collapse under its own weight.\nSection 1.2\nScaling of Area and Volume\n47\nAlso, Galileo is doing something that would be frowned on in\nmodern science: he is mixing experiments whose results he has actually observed (building boats of different sizes), with experiments\nthat he could not possibly have done (dropping an ant from the\nheight of the moon). He now relates how he has done actual experiments with such planks, and found that, according to this operational definition, they are not strong in proportion to their sizes.\nThe larger one breaks. He makes sure to tell the reader how important the result is, via Sagredo’s astonished response:\nS AGREDO : My brain already reels. My mind, like a cloud\nmomentarily illuminated by a lightning flash, is for an instant\nfilled with an unusual light, which now beckons to me and\nwhich now suddenly mingles and obscures strange, crude\nideas. From what you have said it appears to me impossible\nto build two similar structures of the same material, but of\ndifferent sizes and have them proportionately strong.\nh / Galileo\ndiscusses\nplanks\nmade of wood, but the concept\nmay be easier to imagine with\nclay. All three clay rods in the\nfigure were originally the same\nshape. The medium-sized one\nwas twice the height, twice the\nlength, and twice the width of\nthe small one, and similarly the\nlarge one was twice as big as\nthe medium one in all its linear\ndimensions. The big one has\nfour times the linear dimensions\nof the small one, 16 times the\ncross-sectional area when cut\nperpendicular to the page, and\n64 times the volume. That means\nthat the big one has 64 times the\nweight to support, but only 16\ntimes the strength compared to\nthe smallest one.\nIn other words, this specific experiment, using things like wooden\nplanks that have no intrinsic scientific interest, has very wide implications because it points out a general principle, that nature acts\ndifferently on different scales.\nTo finish the discussion, Galileo gives an explanation. He says\nthat the strength of a plank (defined as, say, the weight of the heaviest boulder you could put on the end without breaking it) is proportional to its cross-sectional area, that is, the surface area of the\nfresh wood that would be exposed if you sawed through it in the\nmiddle. Its weight, however, is proportional to its volume.1\nHow do the volume and cross-sectional area of the longer plank\ncompare with those of the shorter plank? We have already seen,\nwhile discussing conversions of the units of area and volume, that\nthese quantities don’t act the way most people naively expect. You\nmight think that the volume and area of the longer plank would both\nbe doubled compared to the shorter plank, so they would increase\nin proportion to each other, and the longer plank would be equally\nable to support its weight. You would be wrong, but Galileo knows\nthat this is a common misconception, so he has Salviati address the\npoint specifically:\nS ALVIATI :\n. . . Take, for example, a cube two inches on a\nside so that each face has an area of four square inches\nand the total area, i.e., the sum of the six faces, amounts\nto twenty-four square inches; now imagine this cube to be\nsawed through three times [with cuts in three perpendicular\nplanes] so as to divide it into eight smaller cubes, each one\ninch on the side, each face one inch square, and the total\n1\nGalileo makes a slightly more complicated argument, taking into account\nthe effect of leverage (torque). The result I’m referring to comes out the same\nregardless of this effect.\n48\nChapter 1\nScaling and Order-of-Magnitude Estimates\nsurface of each cube six square inches instead of twentyfour in the case of the larger cube. It is evident therefore,\nthat the surface of the little cube is only one-fourth that of\nthe larger, namely, the ratio of six to twenty-four; but the volume of the solid cube itself is only one-eighth; the volume,\nand hence also the weight, diminishes therefore much more\nrapidly than the surface. . . You see, therefore, Simplicio, that\nI was not mistaken when . . . I said that the surface of a small\nsolid is comparatively greater than that of a large one.\nThe same reasoning applies to the planks. Even though they\nare not cubes, the large one could be sawed into eight small ones,\neach with half the length, half the thickness, and half the width.\nThe small plank, therefore, has more surface area in proportion to\nits weight, and is therefore able to support its own weight while the\nlarge one breaks.\nScaling of area and volume for irregularly shaped objects\nYou probably are not going to believe Galileo’s claim that this\nhas deep implications for all of nature unless you can be convinced\nthat the same is true for any shape. Every drawing you’ve seen so\nfar has been of squares, rectangles, and rectangular solids. Clearly\nthe reasoning about sawing things up into smaller pieces would not\nprove anything about, say, an egg, which cannot be cut up into eight\nsmaller egg-shaped objects with half the length.\nIs it always true that something half the size has one quarter\nthe surface area and one eighth the volume, even if it has an irregular shape? Take the example of a child’s violin. Violins are made\nfor small children in smaller size to accomodate their small bodies.\nFigure i shows a full-size violin, along with two violins made with\nhalf and 3/4 of the normal length.2 Let’s study the surface area of\nthe front panels of the three violins.\nConsider the square in the interior of the panel of the full-size\nviolin. In the 3/4-size violin, its height and width are both smaller\nby a factor of 3/4, so the area of the corresponding, smaller square\nbecomes 3/4×3/4 = 9/16 of the original area, not 3/4 of the original\narea. Similarly, the corresponding square on the smallest violin has\nhalf the height and half the width of the original one, so its area is\n1/4 the original area, not half.\nThe same reasoning works for parts of the panel near the edge,\nsuch as the part that only partially fills in the other square. The\nentire square scales down the same as a square in the interior, and\nin each violin the same fraction (about 70%) of the square is full, so\nthe contribution of this part to the total area scales down just the\n2\nThe customary terms “half-size” and “3/4-size” actually don’t describe the\nsizes in any accurate way. They’re really just standard, arbitrary marketing\nlabels.\nSection 1.2\ni / The area of a shape is\nproportional to the square of its\nlinear dimensions, even if the\nshape is irregular.\nScaling of Area and Volume\n49\nsame.\nSince any small square region or any small region covering part\nof a square scales down like a square object, the entire surface area\nof an irregularly shaped object changes in the same manner as the\nsurface area of a square: scaling it down by 3/4 reduces the area by\na factor of 9/16, and so on.\nIn general, we can see that any time there are two objects with\nthe same shape, but different linear dimensions (i.e., one looks like a\nreduced photo of the other), the ratio of their areas equals the ratio\nof the squares of their linear dimensions:\n2\nA1\nL1\n=\n.\nA2\nL2\nNote that it doesn’t matter where we choose to measure the linear\nsize, L, of an object. In the case of the violins, for instance, it could\nhave been measured vertically, horizontally, diagonally, or even from\nthe bottom of the left f-hole to the middle of the right f-hole. We\njust have to measure it in a consistent way on each violin. Since all\nthe parts are assumed to shrink or expand in the same manner, the\nratio L1 /L2 is independent of the choice of measurement.\nIt is also important to realize that it is completely unnecessary\nto have a formula for the area of a violin. It is only possible to\nderive simple formulas for the areas of certain shapes like circles,\nrectangles, triangles and so on, but that is no impediment to the\ntype of reasoning we are using.\nSometimes it is inconvenient to write all the equations in terms\nof ratios, especially when more than two objects are being compared.\nA more compact way of rewriting the previous equation is\nA ∝ L2\n.\nThe symbol “∝” means “is proportional to.” Scientists and engineers often speak about such relationships verbally using the phrases\n“scales like” or “goes like,” for instance “area goes like length squared.”\nj / The muffin comes out of\nthe oven too hot to eat. Breaking\nit up into four pieces increases\nits surface area while keeping\nthe total volume the same. It\ncools faster because of the\ngreater surface-to-volume ratio.\nIn general, smaller things have\ngreater surface-to-volume ratios,\nbut in this example there is no\neasy way to compute the effect\nexactly, because the small pieces\naren’t the same shape as the\noriginal muffin.\n50\nChapter 1\nAll of the above reasoning works just as well in the case of volume. Volume goes like length cubed:\nV ∝ L3\n.\nIf different objects are made of the same material with the same\ndensity, ρ = m/V , then their masses, m = ρV , are proportional\nto L3 , and so are their weights. (The symbol for density is ρ, the\nlower-case Greek letter “rho.”)\nAn important point is that all of the above reasoning about\nscaling only applies to objects that are the same shape. For instance,\na piece of paper is larger than a pencil, but has a much greater\nsurface-to-volume ratio.\nScaling and Order-of-Magnitude Estimates\nOne of the first things I learned as a teacher was that students\nwere not very original about their mistakes. Every group of students\ntends to come up with the same goofs as the previous class. The\nfollowing are some examples of correct and incorrect reasoning about\nproportionality.\nScaling of the area of a triangle\nexample 1\n. In figure k, the larger triangle has sides twice as long. How\nmany times greater is its area?\nCorrect solution #1: Area scales in proportion to the square of the\nlinear dimensions, so the larger triangle has four times more area\n(22 = 4).\nCorrect solution #2: You could cut the larger triangle into four of\nthe smaller size, as shown in fig. (b), so its area is four times\ngreater. (This solution is correct, but it would not work for a shape\nlike a circle, which can’t be cut up into smaller circles.)\nk / Example 1. The big triangle has four times more area than\nthe little one.\nCorrect solution #3: The area of a triangle is given by\nA = bh/2, where b is the base and h is the height. The areas of\nthe triangles are\nA1 = b1 h1 /2\nA2 = b2 h2 /2\n= (2b1 )(2h1 )/2\n= 2b1 h1\nl / A tricky way of solving example 1, explained in solution #2.\nA2 /A1 = (2b1 h1 )/(b1 h1 /2)\n=4\n(Although this solution is correct, it is a lot more work than solution\n#1, and it can only be used in this case because a triangle is a\nsimple geometric shape, and we happen to know a formula for its\narea.)\nCorrect solution #4: The area of a triangle is A = bh/2. The\ncomparison of the areas will come out the same as long as the\nratios of the linear sizes of the triangles is as specified, so let’s\njust say b1 = 1.00 m and b2 = 2.00 m. The heights are then also\nh1 = 1.00 m and h2 = 2.00 m, giving areas A1 = 0.50 m2 and\nA2 = 2.00 m2 , so A2 /A1 = 4.00.\n(The solution is correct, but it wouldn’t work with a shape for\nwhose area we don’t have a formula. Also, the numerical calculation might make the answer of 4.00 appear inexact, whereas\nsolution #1 makes it clear that it is exactly 4.)\nIncorrect solution: The area of a triangle is A = bh, and if you\nplug in b = 2.00 m and h = 2.00 m, you get A = 2.00m2 , so\nthe bigger triangle has 2.00 times more area. (This solution is\nincorrect because no comparison has been made with the smaller\ntriangle.)\nSection 1.2\nScaling of Area and Volume\n51\nScaling of the volume of a sphere\nexample 2\n. In figure m, the larger sphere has a radius that is five times\ngreater. How many times greater is its volume?\nCorrect solution #1: Volume scales like the third power of the\nlinear size, so the larger sphere has a volume that is 125 times\ngreater (53 = 125).\nCorrect solution #2: The volume of a sphere is V = (4/3)πr 3 , so\nm / Example 2. The big sphere\nhas 125 times more volume than\nthe little one.\n4 3\nπr\n3 1\n4\nV2 = πr23\n3\n4\n= π(5r1 )3\n3\n500 3\n=\nπr1\n3\n500 3\n4 3\nV2 /V1 =\nπr1 /\nπr\n3\n3 1\nV1 =\n= 125\nIncorrect solution: The volume of a sphere is V = (4/3)πr 3 , so\n4 3\nπr\n3 1\n4\nV1 = πr23\n3\n4\n= π · 5r13\n3\n20 3\n=\nπr1\n3\n20 3\n4 3\nV2 /V1 =\nπr /\nπr\n3 1\n3 1\nV1 =\n=5\n(The solution is incorrect because (5r1 )3 is not the same as 5r13 .)\nn / Example 3.\nThe 48-point\n“S” has 1.78 times more area\nthan the 36-point “S.”\nScaling of a more complex shape\nexample 3\n. The first letter “S” in figure n is in a 36-point font, the second in\n48-point. How many times more ink is required to make the larger\n“S”? (Points are a unit of length used in typography.)\nCorrect solution: The amount of ink depends on the area to be\ncovered with ink, and area is proportional to the square of the\nlinear dimensions, so the amount of ink required for the second\n“S” is greater by a factor of (48/36)2 = 1.78.\nIncorrect solution: The length of the curve of the second “S” is\nlonger by a factor of 48/36 = 1.33, so 1.33 times more ink is\nrequired.\n(The solution is wrong because it assumes incorrectly that the\nwidth of the curve is the same in both cases. Actually both the\n52\nChapter 1\nScaling and Order-of-Magnitude Estimates\nwidth and the length of the curve are greater by a factor of 48/36,\nso the area is greater by a factor of (48/36)2 = 1.78.)\n. Solved problem: a telescope gathers light\npage 61, problem 11\n. Solved problem: distance from an earthquake\npage 61, problem 12\nDiscussion Questions\nA\nA toy fire engine is 1/30 the size of the real one, but is constructed\nfrom the same metal with the same proportions. How many times smaller\nis its weight? How many times less red paint would be needed to paint\nit?\nB\nGalileo spends a lot of time in his dialog discussing what really\nhappens when things break. He discusses everything in terms of Aristotle’s now-discredited explanation that things are hard to break, because\nif something breaks, there has to be a gap between the two halves with\nnothing in between, at least initially. Nature, according to Aristotle, “abhors a vacuum,” i.e., nature doesn’t “like” empty space to exist. Of course,\nair will rush into the gap immediately, but at the very moment of breaking,\nAristotle imagined a vacuum in the gap. Is Aristotle’s explanation of why\nit is hard to break things an experimentally testable statement? If so, how\ncould it be tested experimentally?\n1.3 ? Scaling Applied to Biology\nOrganisms of different sizes with the same shape\nThe left-hand panel in figure o shows the approximate validity of the proportionality m ∝ L3 for cockroaches (redrawn from\nMcMahon and Bonner). The scatter of the points around the curve\nindicates that some cockroaches are proportioned slightly differently\nfrom others, but in general the data seem well described by m ∝ L3 .\nThat means that the largest cockroaches the experimenter could\nraise (is there a 4-H prize?) had roughly the same shape as the\nsmallest ones.\nAnother relationship that should exist for animals of different\nsizes shaped in the same way is that between surface area and\nbody mass. If all the animals have the same average density, then\nbody mass should be proportional to the cube of the animal’s linear size, m ∝ L3 , while surface area should vary proportionately to\nL2 . Therefore, the animals’ surface areas should be proportional to\nm2/3 . As shown in the right-hand panel of figure o, this relationship\nappears to hold quite well for the dwarf siren, a type of salamander.\nNotice how the curve bends over, meaning that the surface area does\nnot increase as quickly as body mass, e.g., a salamander with eight\ntimes more body mass will have only four times more surface area.\nThis behavior of the ratio of surface area to mass (or, equiv-\nSection 1.3\n? Scaling Applied to Biology\n53\no / Geometrical scaling of animals.\nalently, the ratio of surface area to volume) has important consequences for mammals, which must maintain a constant body temperature. It would make sense for the rate of heat loss through the\nanimal’s skin to be proportional to its surface area, so we should\nexpect small animals, having large ratios of surface area to volume,\nto need to produce a great deal of heat in comparison to their size to\navoid dying from low body temperature. This expectation is borne\nout by the data of the left-hand panel of figure p, showing the rate\nof oxygen consumption of guinea pigs as a function of their body\nmass. Neither an animal’s heat production nor its surface area is\nconvenient to measure, but in order to produce heat, the animal\nmust metabolize oxygen, so oxygen consumption is a good indicator\nof the rate of heat production. Since surface area is proportional to\nm2/3 , the proportionality of the rate of oxygen consumption to m2/3\nis consistent with the idea that the animal needs to produce heat at a\nrate in proportion to its surface area. Although the smaller animals\nmetabolize less oxygen and produce less heat in absolute terms, the\namount of food and oxygen they must consume is greater in proportion to their own mass. The Etruscan pigmy shrew, weighing in at\n54\nChapter 1\nScaling and Order-of-Magnitude Estimates\np / Scaling of animals’ bodies related to metabolic rate and skeletal strength.\n2 grams as an adult, is at about the lower size limit for mammals.\nIt must eat continually, consuming many times its body weight each\nday to survive.\nChanges in shape to accommodate changes in size\nLarge mammals, such as elephants, have a small ratio of surface\narea to volume, and have problems getting rid of their heat fast\nenough. An elephant cannot simply eat small enough amounts to\nkeep from producing excessive heat, because cells need to have a\ncertain minimum metabolic rate to run their internal machinery.\nHence the elephant’s large ears, which add to its surface area and\nhelp it to cool itself. Previously, we have seen several examples\nof data within a given species that were consistent with a fixed\nshape, scaled up and down in the cases of individual specimens. The\nelephant’s ears are an example of a change in shape necessitated by\na change in scale.\nLarge animals also must be able to support their own weight.\nReturning to the example of the strengths of planks of different\nsizes, we can see that if the strength of the plank depends on area\nSection 1.3\nq / Galileo’s original\nshowing how larger\nbones must be greater\neter compared to their\n? Scaling Applied to Biology\ndrawing,\nanimals’\nin diamlengths.\n55\nwhile its weight depends on volume, then the ratio of strength to\nweight goes as follows:\nstrength/weight ∝ A/V ∝ 1/L\n.\nThus, the ability of objects to support their own weights decreases\ninversely in proportion to their linear dimensions. If an object is to\nbe just barely able to support its own weight, then a larger version\nwill have to be proportioned differently, with a different shape.\nSince the data on the cockroaches seemed to be consistent with\nroughly similar shapes within the species, it appears that the ability to support its own weight was not the tightest design constraint\nthat Nature was working under when she designed them. For large\nanimals, structural strength is important. Galileo was the first to\nquantify this reasoning and to explain why, for instance, a large animal must have bones that are thicker in proportion to their length.\nConsider a roughly cylindrical bone such as a leg bone or a vertebra.\nThe length of the bone, L, is dictated by the overall linear size of the\nanimal, since the animal’s skeleton must reach the animal’s whole\nlength. We expect the animal’s mass to scale as L3 , so the strength\nof the bone must also scale as L3 . Strength is proportional to crosssectional area, as with the wooden planks, so if the diameter of the\nbone is d, then\nd2 ∝ L3\nor\nd ∝ L3/2\n.\nIf the shape stayed the same regardless of size, then all linear dimensions, including d and L, would be proportional to one another.\nIf our reasoning holds, then the fact that d is proportional to L3/2 ,\nnot L, implies a change in proportions of the bone. As shown in the\nright-hand panel of figure p, the vertebrae of African Bovidae follow\nthe rule d ∝ L3/2 fairly well. The vertebrae of the giant eland are\nas chunky as a coffee mug, while those of a Gunther’s dik-dik are as\nslender as the cap of a pen.\nDiscussion Questions\nA\nSingle-celled animals must passively absorb nutrients and oxygen\nfrom their surroundings, unlike humans who have lungs to pump air in and\nout and a heart to distribute the oxygenated blood throughout their bodies.\nEven the cells composing the bodies of multicellular animals must absorb\noxygen from a nearby capillary through their surfaces. Based on these\nfacts, explain why cells are always microscopic in size.\nB\nThe reasoning of the previous question would seem to be contradicted by the fact that human nerve cells in the spinal cord can be as\nmuch as a meter long, although their widths are still very small. Why is\nthis possible?\n56\nChapter 1\nScaling and Order-of-Magnitude Estimates\n1.4 Order-of-Magnitude Estimates\nIt is the mark of an instructed mind to rest satisfied with the\ndegree of precision that the nature of the subject permits and\nnot to seek an exactness where only an approximation of the\ntruth is possible.\nAristotle\nIt is a common misconception that science must be exact. For\ninstance, in the Star Trek TV series, it would often happen that\nCaptain Kirk would ask Mr. Spock, “Spock, we’re in a pretty bad\nsituation. What do you think are our chances of getting out of\nhere?” The scientific Mr. Spock would answer with something like,\n“Captain, I estimate the odds as 237.345 to one.” In reality, he\ncould not have estimated the odds with six significant figures of\naccuracy, but nevertheless one of the hallmarks of a person with a\ngood education in science is the ability to make estimates that are\nlikely to be at least somewhere in the right ballpark. In many such\nsituations, it is often only necessary to get an answer that is off by no\nmore than a factor of ten in either direction. Since things that differ\nby a factor of ten are said to differ by one order of magnitude, such\nan estimate is called an order-of-magnitude estimate. The tilde,\n∼, is used to indicate that things are only of the same order of\nmagnitude, but not exactly equal, as in\nodds of survival ∼ 100 to one\n.\nThe tilde can also be used in front of an individual number to emphasize that the number is only of the right order of magnitude.\nAlthough making order-of-magnitude estimates seems simple and\nnatural to experienced scientists, it’s a mode of reasoning that is\ncompletely unfamiliar to most college students. Some of the typical\nmental steps can be illustrated in the following example.\nCost of transporting tomatoes\nexample 4\n. Roughly what percentage of the price of a tomato comes from\nthe cost of transporting it in a truck?\n. The following incorrect solution illustrates one of the main ways\nyou can go wrong in order-of-magnitude estimates.\nIncorrect solution: Let’s say the trucker needs to make a \\$400\nprofit on the trip. Taking into account her benefits, the cost of gas,\nand maintenance and payments on the truck, let’s say the total\ncost is more like \\$2000. I’d guess about 5000 tomatoes would fit\nin the back of the truck, so the extra cost per tomato is 40 cents.\nThat means the cost of transporting one tomato is comparable to\nthe cost of the tomato itself. Transportation really adds a lot to the\ncost of produce, I guess.\nThe problem is that the human brain is not very good at estimating area or volume, so it turns out the estimate of 5000 tomatoes\nSection 1.4\nOrder-of-Magnitude Estimates\n57\nfitting in the truck is way off. That’s why people have a hard time\nat those contests where you are supposed to estimate the number of\njellybeans in a big jar. Another example is that most people think\ntheir families use about 10 gallons of water per day, but in reality\nthe average is about 300 gallons per day. When estimating area\nor volume, you are much better off estimating linear dimensions,\nand computing volume from the linear dimensions. Here’s a better\nsolution:\nBetter solution: As in the previous solution, say the cost of the\ntrip is \\$2000. The dimensions of the bin are probably 4 m × 2 m ×\n1 m, for a volume of 8 m3 . Since the whole thing is just an orderof-magnitude estimate, let’s round that off to the nearest power of\nten, 10 m3 . The shape of a tomato is complicated, and I don’t know\nany formula for the volume of a tomato shape, but since this is just\nan estimate, let’s pretend that a tomato is a cube, 0.05 m × 0.05 m\n× 0.05 m, for a volume of 1.25 × 10−4 m3 . Since this is just a rough\nestimate, let’s round that to 10−4 m3 . We can find the total number\nof tomatoes by dividing the volume of the bin by the volume of one\ntomato: 10 m3 /10−4 m3 = 105 tomatoes. The transportation cost\nper tomato is \\$2000/105 tomatoes=\\$0.02/tomato. That means that\ntransportation really doesn’t contribute very much to the cost of a\ntomato.\nr / Consider\na\nspherical\ncow.\nApproximating the shape of a tomato as a cube is an example of\nanother general strategy for making order-of-magnitude estimates.\nA similar situation would occur if you were trying to estimate how\nmany m2 of leather could be produced from a herd of ten thousand\ncattle. There is no point in trying to take into account the shape of\nthe cows’ bodies. A reasonable plan of attack might be to consider\na spherical cow. Probably a cow has roughly the same surface area\nas a sphere with a radius of about 1 m, which would be 4π(1 m)2 .\nUsing the well-known facts that pi equals three, and four times three\nequals about ten, we can guess that a cow has a surface area of about\n10 m2 , so the herd as a whole might yield 105 m2 of leather.\nThe following list summarizes the strategies for getting a good\norder-of-magnitude estimate.\n1. Don’t even attempt more than one significant figure of precision.\n2. Don’t guess area or volume directly. Guess linear dimensions\nand get area or volume from them.\n3. When dealing with areas or volumes of objects with complex\nshapes, idealize them as if they were some simpler shape, a\ncube or a sphere, for example.\n4. Check your final answer to see if it is reasonable. If you estimate that a herd of ten thousand cattle would yield 0.01 m2\n58\nChapter 1\nScaling and Order-of-Magnitude Estimates\nof leather, then you have probably made a mistake with conversion factors somewhere.\nSection 1.4\nOrder-of-Magnitude Estimates\n59\nSummary\nNotation\n∝ . . . . . . . . .\n∼ . . . . . . . . .\nis proportional to\non the order of, is on the order of\nSummary\nNature behaves differently on large and small scales. Galileo\nshowed that this results fundamentally from the way area and volume scale. Area scales as the second power of length, A ∝ L2 , while\nvolume scales as length to the third power, V ∝ L3 .\nAn order of magnitude estimate is one in which we do not attempt or expect an exact answer. The main reason why the uninitiated have trouble with order-of-magnitude estimates is that the\nhuman brain does not intuitively make accurate estimates of area\nand volume. Estimates of area and volume should be approached\nby first estimating linear dimensions, which one’s brain has a feel\nfor.\n60\nChapter 1\nScaling and Order-of-Magnitude Estimates\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nHow many cubic inches are there in a cubic foot? The answer\n√\nis not 12.\n2\nAssume a dog’s brain is twice is great in diameter as a cat’s,\nbut each animal’s brain cells are the same size and their brains are\nthe same shape. In addition to being a far better companion and\nmuch nicer to come home to, how many times more brain cells does\na dog have than a cat? The answer is not 2.\n3\nThe population density of Los Angeles is about 4000 people/km2 .\nThat of San Francisco is about 6000 people/km2 . How many times\nfarther away is the average person’s nearest neighbor in LA than\nin\n√\nSan Francisco? The answer is not 1.5.\n4\nA hunting dog’s nose has about 10 square inches of active\nsurface. How is this possible, since the dog’s nose is only about 1 in\n× 1 in × 1 in = 1 in3 ? After all, 10 is greater than 1, so how can it\nfit?\n5\nEstimate the number of blades of grass on a football field.\n6\nIn a computer memory chip, each bit of information (a 0 or\na 1) is stored in a single tiny circuit etched onto the surface of a\nsilicon chip. The circuits cover the surface of the chip like lots in a\nhousing development. A typical chip stores 64 Mb (megabytes) of\ndata, where a byte is 8 bits. Estimate (a) the area of each circuit,\nand (b) its linear size.\n7\nSuppose someone built a gigantic apartment building, measuring 10 km × 10 km at the base. Estimate how tall the building\nwould have to be to have space in it for the entire world’s population\nto live.\n8\nA hamburger chain advertises that it has sold 10 billion Bongo\nBurgers. Estimate the total mass of feed required to raise the cows\nused to make the burgers.\n9\n10\nEstimate the volume of a human body, in cm3 .\nHow many cm2 is 1 mm2 ?\n. Solution, p. 269\n11\nCompare the light-gathering powers of a 3-cm-diameter telescope and a 30-cm telescope.\n. Solution, p. 269\n12\nOne step on the Richter scale corresponds to a factor of 100\nin terms of the energy absorbed by something on the surface of the\nEarth, e.g., a house. For instance, a 9.3-magnitude quake would\nrelease 100 times more energy than an 8.3. The energy spreads out\nProblems\n61\nfrom the epicenter as a wave, and for the sake of this problem we’ll\nassume we’re dealing with seismic waves that spread out in three\ndimensions, so that we can visualize them as hemispheres spreading\nout under the surface of the earth. If a certain 7.6-magnitude earthquake and a certain 5.6-magnitude earthquake produce the same\namount of vibration where I live, compare the distances from my\nhouse to the two epicenters.\n. Solution, p. 269\n13\nIn Europe, a piece of paper of the standard size, called A4,\nis a little narrower and taller than its American counterpart. The\nratio of the height to the width is the square root of 2, and this has\nsome useful properties. For instance, if you cut an A4 sheet from left\nto right, you get two smaller sheets that have the same proportions.\nYou can even buy sheets of this smaller size, and they’re called A5.\nThere is a whole series of sizes related in this way, all with the same\nproportions. (a) Compare an A5 sheet to an A4 in terms of area and\nlinear size. (b) The series of paper sizes starts from an A0 sheet,\nwhich has an area of one square meter. Suppose we had a series\nof boxes defined in a similar way: the B0 box has a volume of one\ncubic meter, two B1 boxes fit exactly inside an B0 box, and so√ on.\nWhat would be the dimensions of a B0 box?\n14\nEstimate the mass of one of the hairs in Albert Einstein’s\nmoustache, in units of kg.\n15\nAccording to folklore, every time you take a breath, you are\ninhaling some of the atoms exhaled in Caesar’s last words. Is this\ntrue? If so, how many?\nAlbert Einstein, and his moustache, problem 14.\n16\nThe Earth’s surface is about 70% water. Mars’s diameter is\nabout half the Earth’s, but it has no surface water. Compare the\nland areas of the two planets.\n17\nThe traditional Martini glass is shaped like a cone with\nthe point at the bottom. Suppose you make a Martini by pouring\nvermouth into the glass to a depth of 3 cm, and then adding gin\nto bring the depth to 6 cm. What are the proportions of gin and\nvermouth?\n. Solution, p. 269\n18\nThe central portion of a CD is taken up by the hole and some\nsurrounding clear plastic, and this area is unavailable for storing\ndata. The radius of the central circle is about 35% of the radius of\nthe data-storing area. What percentage of the CD’s area is therefore\n√\nlost?\n19\nThe one-liter cube in the photo has been marked off into\nsmaller cubes, with linear dimensions one tenth those of the big\none. What is the volume of each of the small cubes?\n. Solution, p. 270\nProblem 19.\n62\nChapter 1\nScaling and Order-of-Magnitude Estimates\n20\n(a) Based on the definitions of the sine, cosine, and tangent,\nwhat units must they have? (b) A cute formula from trigonometry\nlets you find any angle of a triangle if you know the lengths of\nits sides. Using the notation shown in the figure, and letting s =\n(a + b + c)/2 be half the perimeter, we have\ns\n(s − b)(s − c)\ntan A/2 =\n.\ns(s − a)\nShow that the units of this equation make sense. In other words,\ncheck that the units of the right-hand side are the same as your\nanswer to part a of the question.\n. Solution, p. 270\n21\nEstimate the number of man-hours required for building the Great\nWall of China.\n. Solution, p. 270\n22\n(a) Using the microscope photo in the figure, estimate the\nmass of a one cell of the E. coli bacterium, which is one of the\nmost common ones in the human intestine. Note the scale at the\nlower right corner, which is 1 µm. Each of the tubular objects in\nthe column is one cell. (b) The feces in the human intestine are\nmostly bacteria (some dead, some alive), of which E. coli is a large\nand typical component. Estimate the number of bacteria in your\nintestines, and compare with the number of human cells in your\nbody, which is believed to be roughly on the order of 1013 . (c)\nInterpreting your result from part b, what does this tell you about\nthe size of a typical human cell compared to the size of a typical\nbacterial cell?\nProblem 20.\nProblem 22.\nProblems\n63\nPart I\nMotion in One Dimension\nI didn’t learn until I was nearly through with college that I could\nunderstand a book much better if I mentally outlined it for myself\nbefore I actually began reading. It’s a technique that warns my\nbrain to get little cerebral file folders ready for the different topics\nI’m going to learn, and as I’m reading it allows me to say to myself,\n“Oh, the reason they’re talking about this now is because they’re\npreparing for this other thing that comes later,” or “I don’t need to\nsweat the details of this idea now, because they’re going to explain\nit in more detail later on.”\nAt this point, you’re about to dive in to the main subjects of\nthis book, which are force and motion. The concepts you’re going\nto learn break down into the following three areas:\nkinematics — how to describe motion numerically\ndynamics — how force affects motion\nvectors — a mathematical way of handling the three-dimensional\nnature of force and motion\nRoughly speaking, that’s the order in which we’ll cover these\nthree areas, but the earlier chapters do contain quite a bit of preparation for the later topics. For instance, even before the present point\nin the book you’ve learned about the Newton, a unit of force. The\ndiscussion of force properly belongs to dynamics, which we aren’t\ntackling head-on for a few more chapters, but I’ve found that when\nI teach kinematics it helps to be able to refer to forces now and then\nto show why it makes sense to define certain kinematical concepts.\nAnd although I don’t explicitly introduce vectors until ch. 8, the\ngroundwork is being laid for them in earlier chapters.\nThe figure on the next page is a roadmap to the rest of the book.\n66\n67\n68\nChapter 2\nVelocity and Relative\nMotion\n2.1 Types of Motion\nIf you had to think consciously in order to move your body, you\nwould be severely disabled. Even walking, which we consider to\nbe no great feat, requires an intricate series of motions that your\ncerebrum would be utterly incapable of coordinating. The task of\nputting one foot in front of the other is controlled by the more primitive parts of your brain, the ones that have not changed much since\nthe mammals and reptiles went their separate evolutionary ways.\nThe thinking part of your brain limits itself to general directives\nsuch as “walk faster,” or “don’t step on her toes,” rather than micromanaging every contraction and relaxation of the hundred or so\nmuscles of your hips, legs, and feet.\nPhysics is all about the conscious understanding of motion, but\nwe’re obviously not immediately prepared to understand the most\ncomplicated types of motion. Instead, we’ll use the divide-andconquer technique. We’ll first classify the various types of motion,\nand then begin our campaign with an attack on the simplest cases.\nTo make it clear what we are and are not ready to consider, we need\nto examine and define carefully what types of motion can exist.\na / Rotation.\nb / Simultaneous rotation\nmotion through space.\nand\nRigid-body motion distinguished from motion that changes\nan object’s shape\nNobody, with the possible exception of Fred Astaire, can simply\nglide forward without bending their joints. Walking is thus an example in which there is both a general motion of the whole object\nand a change in the shape of the object. Another example is the\nmotion of a jiggling water balloon as it flies through the air. We are\nnot presently attempting a mathematical description of the way in\nwhich the shape of an object changes. Motion without a change in\nshape is called rigid-body motion. (The word “body” is often used\nin physics as a synonym for “object.”)\nCenter-of-mass motion as opposed to rotation\nA ballerina leaps into the air and spins around once before landing. We feel intuitively that her rigid-body motion while her feet\nare off the ground consists of two kinds of motion going on simul-\nc / One person might say that the\ntipping chair was only rotating in\na circle about its point of contact\nwith the floor, but another could\ndescribe it as having both rotation\nand motion through space.\n69\ntaneously: a rotation and a motion of her body as a whole through\nspace, along an arc. It is not immediately obvious, however, what\nis the most useful way to define the distinction between rotation\nand motion through space. Imagine that you attempt to balance a\nchair and it falls over. One person might say that the only motion\nwas a rotation about the chair’s point of contact with the floor, but\nanother might say that there was both rotation and motion down\nand to the side.\ne / No matter what point you\nhang the pear from, the string\nlines up with the pear’s center\nof mass. The center of mass\ncan therefore be defined as the\nintersection of all the lines made\nby hanging the pear in this way.\nNote that the X in the figure\nshould not be interpreted as\nimplying that the center of mass\nis on the surface — it is actually\ninside the pear.\nd / The leaping dancer’s motion is complicated, but the motion of\nher center of mass is simple.\nf / The circus performers hang\nwith the ropes passing through\ntheir centers of mass.\nIt turns out that there is one particularly natural and useful way\nto make a clear definition, but it requires a brief digression. Every\nobject has a balance point, referred to in physics as the center of\nmass. For a two-dimensional object such as a cardboard cutout, the\ncenter of mass is the point at which you could hang the object from\na string and make it balance. In the case of the ballerina (who is\nlikely to be three-dimensional unless her diet is particularly severe),\nit might be a point either inside or outside her body, depending\non how she holds her arms. Even if it is not practical to attach a\nstring to the balance point itself, the center of mass can be defined\nas shown in figure e.\nWhy is the center of mass concept relevant to the question of\nclassifying rotational motion as opposed to motion through space?\nAs illustrated in figures d and g, it turns out that the motion of an\nobject’s center of mass is nearly always far simpler than the motion\nof any other part of the object. The ballerina’s body is a large object\nwith a complex shape. We might expect that her motion would be\nmuch more complicated than the motion of a small, simply-shaped\n70\nChapter 2\nVelocity and Relative Motion\nobject, say a marble, thrown up at the same angle as the angle at\nwhich she leapt. But it turns out that the motion of the ballerina’s\ncenter of mass is exactly the same as the motion of the marble. That\nis, the motion of the center of mass is the same as the motion the\nballerina would have if all her mass was concentrated at a point. By\nrestricting our attention to the motion of the center of mass, we can\ntherefore simplify things greatly.\ng / The same leaping dancer, viewed from above. Her center of\nmass traces a straight line, but a point away from her center of mass,\nsuch as her elbow, traces the much more complicated path shown by the\ndots.\nWe can now replace the ambiguous idea of “motion as a whole\nthrough space” with the more useful and better defined concept\nof “center-of-mass motion.” The motion of any rigid body can be\ncleanly split into rotation and center-of-mass motion. By this definition, the tipping chair does have both rotational and center-of-mass\nmotion. Concentrating on the center of mass motion allows us to\nmake a simplified model of the motion, as if a complicated object\nlike a human body was just a marble or a point-like particle. Science\nreally never deals with reality; it deals with models of reality.\nNote that the word “center” in “center of mass” is not meant\nto imply that the center of mass must lie at the geometrical center\nof an object. A car wheel that has not been balanced properly has\na center of mass that does not coincide with its geometrical center.\nAn object such as the human body does not even have an obvious\ngeometrical center.\nIt can be helpful to think of the center of mass as the average\nlocation of all the mass in the object. With this interpretation,\nwe can see for example that raising your arms above your head\nraises your center of mass, since the higher position of the arms’\nmass raises the average. We won’t be concerned right now with\ncalculating centers of mass mathematically; the relevant equations\nare in chapter 4 of Conservation Laws.\nBallerinas and professional basketball players can create an illusion of flying horizontally through the air because our brains intuitively expect them to have rigid-body motion, but the body does\nnot stay rigid while executing a grand jete or a slam dunk. The legs\nare low at the beginning and end of the jump, but come up higher at\nSection 2.1\nh / An\nimproperly\nbalanced\nwheel has a center of mass that\nis not at its geometric center.\nWhen you get a new tire, the\nmechanic clamps little weights to\nthe rim to balance the wheel.\ni / This toy was intentionally\ndesigned so that the mushroomshaped piece of metal on top\nwould throw off the center of\nmass. When you wind it up, the\nmushroom spins, but the center\nof mass doesn’t want to move,\nso the rest of the toy tends to\ncounter the mushroom’s motion,\ncausing the whole thing to jump\naround.\nTypes of Motion\n71\nj / A fixed point on the dancer’s body follows a trajectory that is flatter than what we expect, creating an illusion of flight.\nthe middle. Regardless of what the limbs do, the center of mass will\nfollow the same arc, but the low position of the legs at the beginning\nand end means that the torso is higher compared to the center of\nmass, while in the middle of the jump it is lower compared to the\ncenter of mass. Our eye follows the motion of the torso and tries\nto interpret it as the center-of-mass motion of a rigid body. But\nsince the torso follows a path that is flatter than we expect, this\nattempted interpretation fails, and we experience an illusion that\nthe person is flying horizontally.\nk / Example 1.\nThe center of mass as an average\nexample 1\n. Explain how we know that the center of mass of each object is\nat the location shown in figure k.\n. The center of mass is a sort of average, so the height of the\ncenters of mass in 1 and 2 has to be midway between the two\nsquares, because that height is the average of the heights of the\ntwo squares. Example 3 is a combination of examples 1 and\n2, so we can find its center of mass by averaging the horizontal\npositions of their centers of mass. In example 4, each square\nhas been skewed a little, but just as much mass has been moved\nup as down, so the average vertical position of the mass hasn’t\nchanged. Example 5 is clearly not all that different from example\n4, the main difference being a slight clockwise rotation, so just as\n72\nChapter 2\nVelocity and Relative Motion\nin example 4, the center of mass must be hanging in empty space,\nwhere there isn’t actually any mass. Horizontally, the center of\nmass must be between the heels and toes, or else it wouldn’t be\npossible to stand without tipping over.\nAnother interesting example from the sports world is the high\njump, in which the jumper’s curved body passes over the bar, but\nthe center of mass passes under the bar! Here the jumper lowers his\nlegs and upper body at the peak of the jump in order to bring his\nwaist higher compared to the center of mass.\nLater in this course, we’ll find that there are more fundamental\nreasons (based on Newton’s laws of motion) why the center of mass\nbehaves in such a simple way compared to the other parts of an\nobject. We’re also postponing any discussion of numerical methods\nfor finding an object’s center of mass. Until later in the course, we\nwill only deal with the motion of objects’ centers of mass.\nCenter-of-mass motion in one dimension\nIn addition to restricting our study of motion to center-of-mass\nmotion, we will begin by considering only cases in which the center\nof mass moves along a straight line. This will include cases such\nas objects falling straight down, or a car that speeds up and slows\ndown but does not turn.\nl / The\nhigh-jumper’s\nbody\npasses over the bar, but his\ncenter of mass passes under it.\nNote that even though we are not explicitly studying the more\ncomplex aspects of motion, we can still analyze the center-of-mass\nmotion while ignoring other types of motion that might be occurring\nsimultaneously . For instance, if a cat is falling out of a tree and\nis initially upside-down, it goes through a series of contortions that\nbring its feet under it. This is definitely not an example of rigidbody motion, but we can still analyze the motion of the cat’s center\nof mass just as we would for a dropping rock.\nself-check A\nConsider a person running, a person pedaling on a bicycle, a person\ncoasting on a bicycle, and a person coasting on ice skates. In which\ncases is the center-of-mass motion one-dimensional? Which cases are\nexamples of rigid-body motion?\n. Answer, p. 266\nself-check B\nThe figure shows a gymnast holding onto the inside of a big wheel.\nFrom inside the wheel, how could he make it roll one way or the other?\n. Answer, p. 266\nm / Self-check B.\n2.2 Describing Distance and Time\nCenter-of-mass motion in one dimension is particularly easy to deal\nwith because all the information about it can be encapsulated in two\nvariables: x, the position of the center of mass relative to the origin,\nand t, which measures a point in time. For instance, if someone\nSection 2.2\nDescribing Distance and Time\n73\nsupplied you with a sufficiently detailed table of x and t values, you\nwould know pretty much all there was to know about the motion of\nthe object’s center of mass.\nA point in time as opposed to duration\nIn ordinary speech, we use the word “time” in two different\nsenses, which are to be distinguished in physics. It can be used,\nas in “a short time” or “our time here on earth,” to mean a length\nor duration of time, or it can be used to indicate a clock reading, as\nin “I didn’t know what time it was,” or “now’s the time.” In symbols, t is ordinarily used to mean a point in time, while ∆t signifies\nan interval or duration in time. The capital Greek letter delta, ∆,\nmeans “the change in...,” i.e. a duration in time is the change or\ndifference between one clock reading and another. The notation ∆t\ndoes not signify the product of two numbers, ∆ and t, but rather\none single number, ∆t. If a matinee begins at a point in time t = 1\no’clock and ends at t = 3 o’clock, the duration of the movie was the\nchange in t,\n∆t = 3 hours − 1 hour = 2 hours\n.\nTo avoid the use of negative numbers for ∆t, we write the clock\nreading “after” to the left of the minus sign, and the clock reading\n“before” to the right of the minus sign. A more specific definition\nof the delta notation is therefore that delta stands for “after minus\nbefore.”\nEven though our definition of the delta notation guarantees that\n∆t is positive, there is no reason why t can’t be negative. If t\ncould not be negative, what would have happened one second before\nt = 0? That doesn’t mean that time “goes backward” in the sense\nthat adults can shrink into infants and retreat into the womb. It\njust means that we have to pick a reference point and call it t = 0,\nand then times before that are represented by negative values of t.\nAn example is that a year like 2007 A.D. can be thought of as a\npositive t value, while one like 370 B.C. is negative. Similarly, when\nyou hear a countdown for a rocket launch, the phrase “t minus ten\nseconds” is a way of saying t = −10 s, where t = 0 is the time of\nblastoff, and t > 0 refers to times after launch.\nAlthough a point in time can be thought of as a clock reading, it\nis usually a good idea to avoid doing computations with expressions\nsuch as “2:35” that are combinations of hours and minutes. Times\ncan instead be expressed entirely in terms of a single unit, such as\nhours. Fractions of an hour can be represented by decimals rather\nthan minutes, and similarly if a problem is being worked in terms\nof minutes, decimals can be used instead of seconds.\nself-check C\nOf the following phrases, which refer to points in time, which refer to\ntime intervals, and which refer to time in the abstract rather than as a\nmeasurable number?\n74\nChapter 2\nVelocity and Relative Motion\n(1) “The time has come.”\n(2) “Time waits for no man.”\n(3) “The whole time, he had spit on his chin.”\n. Answer, p. 266\nPosition as opposed to change in position\nAs with time, a distinction should be made between a point\nin space, symbolized as a coordinate x, and a change in position,\nsymbolized as ∆x.\nAs with t, x can be negative. If a train is moving down the\ntracks, not only do you have the freedom to choose any point along\nthe tracks and call it x = 0, but it’s also up to you to decide which\nside of the x = 0 point is positive x and which side is negative x.\nSince we’ve defined the delta notation to mean “after minus\nbefore,” it is possible that ∆x will be negative, unlike ∆t which is\nguaranteed to be positive. Suppose we are describing the motion\nof a train on tracks linking Tucson and Chicago. As shown in the\nfigure, it is entirely up to you to decide which way is positive.\nn / Two equally valid ways of describing the motion of a train from\nTucson to Chicago. In example 1,\nthe train has a positive ∆x as it\ngoes from Enid to Joplin. In 2,\nthe same train going forward in\nthe same direction has a negative\n∆x .\nNote that in addition to x and ∆x, there is a third quantity we\ncould define, which would be like an odometer reading, or actual\ndistance traveled. If you drive 10 miles, make a U-turn, and drive\nback 10 miles, then your ∆x is zero, but your car’s odometer reading\nhas increased by 20 miles. However important the odometer reading\nis to car owners and used car dealers, it is not very important in\nphysics, and there is not even a standard name or notation for it.\nThe change in position, ∆x, is more useful because it is so much\neasier to calculate: to compute ∆x, we only need to know the beginning and ending positions of the object, not all the information\nabout how it got from one position to the other.\nself-check D\nA ball hits the floor, bounces to a height of one meter, falls, and hits the\nfloor again. Is the ∆x between the two impacts equal to zero, one, or\ntwo meters?\n. Answer, p. 267\nSection 2.2\nDescribing Distance and Time\n75\nFrames of reference\nThe example above shows that there are two arbitrary choices\nyou have to make in order to define a position variable, x. You have\nto decide where to put x = 0, and also which direction will be positive. This is referred to as choosing a coordinate system or choosing\na frame of reference. (The two terms are nearly synonymous, but\nthe first focuses more on the actual x variable, while the second is\nmore of a general way of referring to one’s point of view.) As long as\nyou are consistent, any frame is equally valid. You just don’t want\nto change coordinate systems in the middle of a calculation.\no / Motion\nlocity.\nwith\nconstant\nve-\nHave you ever been sitting in a train in a station when suddenly\nyou notice that the station is moving backward? Most people would\ndescribe the situation by saying that you just failed to notice that\nthe train was moving — it only seemed like the station was moving.\nBut this shows that there is yet a third arbitrary choice that goes\ninto choosing a coordinate system: valid frames of reference can\ndiffer from each other by moving relative to one another. It might\nseem strange that anyone would bother with a coordinate system\nthat was moving relative to the earth, but for instance the frame of\nreference moving along with a train might be far more convenient\nfor describing things happening inside the train.\n2.3 Graphs of Motion; Velocity\nMotion with constant velocity\nIn example o, an object is moving at constant speed in one direction. We can tell this because every two seconds, its position\nchanges by five meters.\np / Motion that decreases x\nis represented with negative\nvalues of ∆x and v .\nIn algebra notation, we’d say that the graph of x vs. t shows\nthe same change in position, ∆x = 5.0 m, over each interval of\n∆t = 2.0 s. The object’s velocity or speed is obtained by calculating\nv = ∆x/∆t = (5.0 m)/(2.0 s) = 2.5 m/s. In graphical terms, the\nvelocity can be interpreted as the slope of the line. Since the graph\nis a straight line, it wouldn’t have mattered if we’d taken a longer\ntime interval and calculated v = ∆x/∆t = (10.0 m)/(4.0 s). The\nanswer would still have been the same, 2.5 m/s.\nNote that when we divide a number that has units of meters by\nanother number that has units of seconds, we get units of meters\nper second, which can be written m/s. This is another case where\nwe treat units as if they were algebra symbols, even though they’re\nnot.\nq / Motion\nlocity.\n76\nwith\nchanging\nChapter 2\nve-\nIn example p, the object is moving in the opposite direction: as\ntime progresses, its x coordinate decreases. Recalling the definition\nof the ∆ notation as “after minus before,” we find that ∆t is still\npositive, but ∆x must be negative. The slope of the line is therefore\nVelocity and Relative Motion\nnegative, and we say that the object has a negative velocity, v =\n∆x/∆t = (−5.0 m)/(2.0 s) = −2.5 m/s. We’ve already seen that\nthe plus and minus signs of ∆x values have the interpretation of\ntelling us which direction the object moved. Since ∆t is always\npositive, dividing by ∆t doesn’t change the plus or minus sign, and\nthe plus and minus signs of velocities are to be interpreted in the\nsame way. In graphical terms, a positive slope characterizes a line\nthat goes up as we go to the right, and a negative slope tells us that\nthe line went down as we went to the right.\n. Solved problem: light-years\npage 89, problem 4\nMotion with changing velocity\nNow what about a graph like figure q? This might be a graph\nof a car’s motion as the driver cruises down the freeway, then slows\ndown to look at a car crash by the side of the road, and then speeds\nup again, disappointed that there is nothing dramatic going on such\nas flames or babies trapped in their car seats. (Note that we are\nstill talking about one-dimensional motion. Just because the graph\nis curvy doesn’t mean that the car’s path is curvy. The graph is not\nlike a map, and the horizontal direction of the graph represents the\npassing of time, not distance.)\nExample q is similar to example o in that the object moves a\ntotal of 25.0 m in a period of 10.0 s, but it is no longer true that it\nmakes the same amount of progress every second. There is no way to\ncharacterize the entire graph by a certain velocity or slope, because\nthe velocity is different at every moment. It would be incorrect to\nsay that because the car covered 25.0 m in 10.0 s, its velocity was\n2.5 m/s. It moved faster than that at the beginning and end, but\nslower in the middle. There may have been certain instants at which\nthe car was indeed going 2.5 m/s, but the speedometer swept past\nthat value without “sticking,” just as it swung through various other\nvalues of speed. (I definitely want my next car to have a speedometer\ncalibrated in m/s and showing both negative and positive values.)\nWe assume that our speedometer tells us what is happening to\nthe speed of our car at every instant, but how can we define speed\nmathematically in a case like this? We can’t just define it as the\nslope of the curvy graph, because a curve doesn’t have a single\nwell-defined slope as does a line. A mathematical definition that\ncorresponded to the speedometer reading would have to be one that\nattached a different velocity value to a single point on the curve,\ni.e., a single instant in time, rather than to the entire graph. If we\nwish to define the speed at one instant such as the one marked with\na dot, the best way to proceed is illustrated in r, where we have\ndrawn the line through that point called the tangent line, the line\nthat “hugs the curve.” We can then adopt the following definition\nof velocity:\nSection 2.3\nr / The velocity at any given\nmoment is defined as the slope\nof the tangent line through the\nrelevant point on the graph.\ns / Example:\nfinding the velocity at the point indicated with\nthe dot.\nt / Reversing\nmotion.\nthe\nGraphs of Motion; Velocity\ndirection\nof\n77\ndefinition of velocity\nThe velocity of an object at any given moment is the slope of the\ntangent line through the relevant point on its x − t graph.\nOne interpretation of this definition is that the velocity tells us\nhow many meters the object would have traveled in one second, if\nit had continued moving at the same speed for at least one second.\nTo some people the graphical nature of this definition seems “inaccurate” or “not mathematical.” The equation by itself, however,\nis only valid if the velocity is constant, and so cannot serve as a\ngeneral definition.\nThe slope of the tangent line\nexample 2\n. What is the velocity at the point shown with a dot on the graph?\n. First we draw the tangent line through that point. To find the\nslope of the tangent line, we need to pick two points on it. Theoretically, the slope should come out the same regardless of which\ntwo points we pick, but in practical terms we’ll be able to measure\nmore accurately if we pick two points fairly far apart, such as the\ntwo white diamonds. To save work, we pick points that are directly\nabove labeled points on the t axis, so that ∆t = 4.0 s is easy to\nread off. One diamond lines up with x ≈ 17.5 m, the other with\nx ≈ 26.5 m, so ∆x = 9.0 m. The velocity is ∆x/∆t = 2.2 m/s.\nConventions about graphing\nThe placement of t on the horizontal axis and x on the upright\naxis may seem like an arbitrary convention, or may even have disturbed you, since your algebra teacher always told you that x goes\non the horizontal axis and y goes on the upright axis. There is a\nreason for doing it this way, however. In example s, we have an\nobject that reverses its direction of motion twice. It can only be\nin one place at any given time, but there can be more than one\ntime when it is at a given place. For instance, this object passed\nthrough x = 17 m on three separate occasions, but there is no way\nit could have been in more than one place at t = 5.0 s. Resurrecting\nsome terminology you learned in your trigonometry course, we say\nthat x is a function of t, but t is not a function of x. In situations\nsuch as this, there is a useful convention that the graph should be\noriented so that any vertical line passes through the curve at only\none point. Putting the x axis across the page and t upright would\nhave violated this convention. To people who are used to interpreting graphs, a graph that violates this convention is as annoying as\nfingernails scratching on a chalkboard. We say that this is a graph\nof “x versus t.” If the axes were the other way around, it would\nbe a graph of “t versus x.” I remember the “versus” terminology\nby visualizing the labels on the x and t axes and remembering that\nwhen you read, you go from left to right and from top to bottom.\n78\nChapter 2\nVelocity and Relative Motion\nDiscussion Questions\nA\nPark is running slowly in gym class, but then he notices Jenna\nwatching him, so he speeds up to try to impress her. Which of the graphs\ncould represent his motion?\nB\nThe figure shows a sequence of positions for two racing tractors.\nCompare the tractors’ velocities as the race progresses. When do they\nhave the same velocity? [Based on a question by Lillian McDermott.]\nC\nIf an object had a straight-line motion graph with ∆x =0 and ∆t 6= 0,\nwhat would be true about its velocity? What would this look like on a\ngraph? What about ∆t =0 and ∆x 6= 0?\nD\nIf an object has a wavy motion graph like the one in figure t on\nthe previous page, which are the points at which the object reverses its\ndirection? What is true about the object’s velocity at these points?\nE\nDiscuss anything unusual about the following three graphs.\nF\nI have been using the term “velocity” and avoiding the more common\nEnglish word “speed,” because introductory physics texts typically define\nthem to mean different things. They use the word “speed,” and the symbol\n“s” to mean the absolute value of the velocity, s = |v |. Although I’ve\nchosen not to emphasize this distinction in technical vocabulary, there\nare clearly two different concepts here. Can you think of an example of\na graph of x -versus-t in which the object has constant speed, but not\nconstant velocity?\nG For the graph shown in the figure, describe how the object’s velocity\nchanges.\nH\nTwo physicists duck out of a boring scientific conference to go\nSection 2.3\nDiscussion question G.\nGraphs of Motion; Velocity\n79\nget beer. On the way to the bar, they witness an accident in which a\npedestrian is injured by a hit-and-run driver. A criminal trial results, and\nthey must testify. In her testimony, Dr. Transverz Waive says, “The car was\nmoving along pretty fast, I’d say the velocity was +40 mi/hr. They saw the\nold lady too late, and even though they slammed on the brakes they still\nhit her before they stopped. Then they made a U turn and headed off\nat a velocity of about -20 mi/hr, I’d say.” Dr. Longitud N.L. Vibrasheun\nsays, “He was really going too fast, maybe his velocity was -35 or -40\nmi/hr. After he hit Mrs. Hapless, he turned around and left at a velocity of,\noh, I’d guess maybe +20 or +25 mi/hr.” Is their testimony contradictory?\nExplain.\n2.4 The Principle of Inertia\nPhysical effects relate only to a change in velocity\nConsider two statements of a kind that was at one time made\nwith the utmost seriousness:\nPeople like Galileo and Copernicus who say the earth is rotating must be crazy. We know the earth can’t be moving.\nWhy, if the earth was really turning once every day, then our\nwhole city would have to be moving hundreds of leagues in\nan hour. That’s impossible! Buildings would shake on their\nfoundations. Gale-force winds would knock us over. Trees\nwould fall down. The Mediterranean would come sweeping\nacross the east coasts of Spain and Italy. And furthermore,\nwhat force would be making the world turn?\nAll this talk of passenger trains moving at forty miles an hour\nis sheer hogwash! At that speed, the air in a passenger compartment would all be forced against the back wall. People in\nthe front of the car would suffocate, and people at the back\nwould die because in such concentrated air, they wouldn’t be\nable to expel a breath.\nSome of the effects predicted in the first quote are clearly just\nbased on a lack of experience with rapid motion that is smooth and\nfree of vibration. But there is a deeper principle involved. In each\ncase, the speaker is assuming that the mere fact of motion must\nhave dramatic physical effects. More subtly, they also believe that a\nforce is needed to keep an object in motion: the first person thinks\na force would be needed to maintain the earth’s rotation, and the\nsecond apparently thinks of the rear wall as pushing on the air to\nkeep it moving.\nCommon modern knowledge and experience tell us that these\npeople’s predictions must have somehow been based on incorrect\nreasoning, but it is not immediately obvious where the fundamental\nflaw lies. It’s one of those things a four-year-old could infuriate\nyou by demanding a clear explanation of. One way of getting at\nthe fundamental principle involved is to consider how the modern\n80\nChapter 2\nVelocity and Relative Motion\nconcept of the universe differs from the popular conception at the\ntime of the Italian Renaissance. To us, the word “earth” implies\na planet, one of the nine planets of our solar system, a small ball\nof rock and dirt that is of no significance to anyone in the universe\nexcept for members of our species, who happen to live on it. To\nGalileo’s contemporaries, however, the earth was the biggest, most\nsolid, most important thing in all of creation, not to be compared\nwith the wandering lights in the sky known as planets. To us, the\nearth is just another object, and when we talk loosely about “how\nfast” an object such as a car “is going,” we really mean the carobject’s velocity relative to the earth-object.\nv / Why does Aristotle look\nso sad? Has he realized that\nhis entire system of physics is\nwrong?\nu / This Air Force doctor volunteered to ride a rocket sled as a\nmedical experiment. The obvious effects on his head and face are not\nbecause of the sled’s speed but because of its rapid changes in speed:\nincreasing in 2 and 3, and decreasing in 5 and 6. In 4 his speed is\ngreatest, but because his speed is not increasing or decreasing very\nmuch at this moment, there is little effect on him.\nMotion is relative\nAccording to our modern world-view, it really isn’t that reasonable to expect that a special force should be required to make the\nair in the train have a certain velocity relative to our planet. After\nall, the “moving” air in the “moving” train might just happen to\nhave zero velocity relative to some other planet we don’t even know\nabout. Aristotle claimed that things “naturally” wanted to be at\nrest, lying on the surface of the earth. But experiment after exper-\nSection 2.4\nw / The earth spins.\nPeople\nin Shanghai say they’re at rest\nand people in Los Angeles are\nmoving. Angelenos say the same\nabout the Shanghainese.\nx / The jets are at rest.\nThe\nEmpire State Building is moving.\nThe Principle of Inertia\n81\niment has shown that there is really nothing so special about being\nat rest relative to the earth. For instance, if a mattress falls out of\nthe back of a truck on the freeway, the reason it rapidly comes to\nrest with respect to the planet is simply because of friction forces\nexerted by the asphalt, which happens to be attached to the planet.\nGalileo’s insights are summarized as follows:\nDiscussion question A.\nDiscussion question B.\nThe principle of inertia\nNo force is required to maintain motion with constant velocity in\na straight line, and absolute motion does not cause any observable\nphysical effects.\nThere are many examples of situations that seem to disprove the\nprinciple of inertia, but these all result from forgetting that friction\nis a force. For instance, it seems that a force is needed to keep a\nsailboat in motion. If the wind stops, the sailboat stops too. But\nthe wind’s force is not the only force on the boat; there is also a\nfrictional force from the water. If the sailboat is cruising and the\nwind suddenly disappears, the backward frictional force still exists,\nand since it is no longer being counteracted by the wind’s forward\nforce, the boat stops. To disprove the principle of inertia, we would\nhave to find an example where a moving object slowed down even\nthough no forces whatsoever were acting on it.\nself-check E\nWhat is incorrect about the following supposed counterexamples to the\nprinciple of inertia?\n(1) When astronauts blast off in a rocket, their huge velocity does cause\na physical effect on their bodies — they get pressed back into their\nseats, the flesh on their faces gets distorted, and they have a hard time\nlifting their arms.\n(2) When you’re driving in a convertible with the top down, the wind in\nyour face is an observable physical effect of your absolute motion.\n.\nAnswer, p. 267\n. Solved problem: a bug on a wheel\npage 89, problem 7\nDiscussion Questions\nA\nA passenger on a cruise ship finds, while the ship is docked, that\nhe can leap off of the upper deck and just barely make it into the pool\non the lower deck. If the ship leaves dock and is cruising rapidly, will this\nadrenaline junkie still be able to make it?\nB\nYou are a passenger in the open basket hanging under a helium\nballoon. The balloon is being carried along by the wind at a constant\nvelocity. If you are holding a flag in your hand, will the flag wave? If so,\nwhich way? [Based on a question from PSSC Physics.]\nDiscussion question D.\n82\nChapter 2\nC\nAristotle stated that all objects naturally wanted to come to rest, with\nthe unspoken implication that “rest” would be interpreted relative to the\nsurface of the earth. Suppose we go back in time and transport Aristotle\nVelocity and Relative Motion\nto the moon. Aristotle knew, as we do, that the moon circles the earth; he\nsaid it didn’t fall down because, like everything else in the heavens, it was\nmade out of some special substance whose “natural” behavior was to go\nin circles around the earth. We land, put him in a space suit, and kick\nhim out the door. What would he expect his fate to be in this situation? If\nintelligent creatures inhabited the moon, and one of them independently\ncame up with the equivalent of Aristotelian physics, what would they think\nabout objects coming to rest?\nD\nThe bottle is sitting on a level table in a train’s dining car, but the\nsurface of the beer is tilted. What can you infer about the motion of the\ntrain?\n2.5 Addition of Velocities\nAddition of velocities to describe relative motion\nSince absolute motion cannot be unambiguously measured, the\nonly way to describe motion unambiguously is to describe the motion\nof one object relative to another. Symbolically, we can write vP Q\nfor the velocity of object P relative to object Q.\nVelocities measured with respect to different reference points can\nbe compared by addition. In the figure below, the ball’s velocity\nrelative to the couch equals the ball’s velocity relative to the truck\nplus the truck’s velocity relative to the couch:\nvBC = vBT + vT C\n= 5 cm/s + 10 cm/s\n= 15 cm/s\nThe same equation can be used for any combination of three\nobjects, just by substituting the relevant subscripts for B, T, and\nC. Just remember to write the equation so that the velocities being\nadded have the same subscript twice in a row. In this example, if\nyou read off the subscripts going from left to right, you get BC . . . =\n. . . BTTC. The fact that the two “inside” subscripts on the right are\nthe same means that the equation has been set up correctly. Notice\nhow subscripts on the left look just like the subscripts on the right,\nbut with the two T’s eliminated.\nNegative velocities in relative motion\nMy discussion of how to interpret positive and negative signs of\nvelocity may have left you wondering why we should bother. Why\nnot just make velocity positive by definition? The original reason\nwhy negative numbers were invented was that bookkeepers decided\nit would be convenient to use the negative number concept for payments to distinguish them from receipts. It was just plain easier than\nwriting receipts in black and payments in red ink. After adding up\nSection 2.5\nAddition of Velocities\n83\ny / These two highly competent physicists disagree on absolute velocities, but they would agree on relative velocities. Purple Dino\nconsiders the couch to be at rest, while Green Dino thinks of the truck as\nbeing at rest. They agree, however, that the truck’s velocity relative to the\ncouch is vT C = 10 cm/s, the ball’s velocity relative to the truck is vBT = 5\ncm/s, and the ball’s velocity relative to the couch is vBC = vBT + vT C = 15\ncm/s.\nyour month’s positive receipts and negative payments, you either got\na positive number, indicating profit, or a negative number, showing\na loss. You could then show that total with a high-tech “+” or “−”\nsign, instead of looking around for the appropriate bottle of ink.\nNowadays we use positive and negative numbers for all kinds\nof things, but in every case the point is that it makes sense to\nadd and subtract those things according to the rules you learned\nin grade school, such as “minus a minus makes a plus, why this is\ntrue we need not discuss.” Adding velocities has the significance\nof comparing relative motion, and with this interpretation negative\nand positive velocities can be used within a consistent framework.\nFor example, the truck’s velocity relative to the couch equals the\ntruck’s velocity relative to the ball plus the ball’s velocity relative\nto the couch:\nvT C = vT B + vBC\n= −5 cm/s + 15 cm/s\n= 10 cm/s\nIf we didn’t have the technology of negative numbers, we would have\nhad to remember a complicated set of rules for adding velocities: (1)\nif the two objects are both moving forward, you add, (2) if one is\nmoving forward and one is moving backward, you subtract, but (3)\n84\nChapter 2\nVelocity and Relative Motion\nif they’re both moving backward, you add. What a pain that would\nhave been.\n. Solved problem: two dimensions\npage 90, problem 10\nDiscussion Questions\nA\nInterpret the general rule vAB = −vBA in words.\nB\nWa-Chuen slips away from her father at the mall and walks up the\ndown escalator, so that she stays in one place. Write this in terms of\nsymbols.\n2.6 Graphs of Velocity Versus Time\nSince changes in velocity play such a prominent role in physics, we\nneed a better way to look at changes in velocity than by laboriously\ndrawing tangent lines on x-versus-t graphs. A good method is to\ndraw a graph of velocity versus time. The examples on the left show\nthe x − t and v − t graphs that might be produced by a car starting\nfrom a traffic light, speeding up, cruising for a while at constant\nspeed, and finally slowing down for a stop sign. If you have an air\nfreshener hanging from your rear-view mirror, then you will see an\neffect on the air freshener during the beginning and ending periods\nwhen the velocity is changing, but it will not be tilted during the\nperiod of constant velocity represented by the flat plateau in the\nmiddle of the v − t graph.\nStudents often mix up the things being represented on these two\ntypes of graphs. For instance, many students looking at the top\ngraph say that the car is speeding up the whole time, since “the\ngraph is becoming greater.” What is getting greater throughout the\ngraph is x, not v.\nSimilarly, many students would look at the bottom graph and\nthink it showed the car backing up, because “it’s going backwards\nat the end.” But what is decreasing at the end is v, not x. Having\nboth the x − t and v − t graphs in front of you like this is often\nconvenient, because one graph may be easier to interpret than the\nother for a particular purpose. Stacking them like this means that\ncorresponding points on the two graphs’ time axes are lined up with\neach other vertically. However, one thing that is a little counterintuitive about the arrangement is that in a situation like this one\ninvolving a car, one is tempted to visualize the landscape stretching\nalong the horizontal axis of one of the graphs. The horizontal axes,\nhowever, represent time, not position. The correct way to visualize\nthe landscape is by mentally rotating the horizon 90 degrees counterclockwise and imagining it stretching along the upright axis of the\nx-t graph, which is the only axis that represents different positions\nin space.\nSection 2.6\nz / Graphs of x and v versus\nt for a car accelerating away from\na traffic light, and then stopping\nfor another red light.\nGraphs of Velocity Versus Time\n85\n2.7\nR\nApplications of Calculus\nR\nThe integral symbol, , in the heading for this section indicates that\nit is meant to be read by students in calculus-based physics. Students in an algebra-based physics course should skip these sections.\nThe calculus-related sections in this book are meant to be usable\nby students who are taking calculus concurrently, so at this early\npoint in the physics course I do not assume you know any calculus\nyet. This section is therefore not much more than a quick preview of\ncalculus, to help you relate what you’re learning in the two courses.\nNewton was the first person to figure out the tangent-line definition of velocity for cases where the x − t graph is nonlinear. Before Newton, nobody had conceptualized the description of motion\nin terms of x − t and v − t graphs. In addition to the graphical\ntechniques discussed in this chapter, Newton also invented a set of\nsymbolic techniques called calculus. If you have an equation for x\nin terms of t, calculus allows you, for instance, to find an equation\nfor v in terms of t. In calculus terms, we say that the function v(t)\nis the derivative of the function x(t). In other words, the derivative\nof a function is a new function that tells how rapidly the original\nfunction was changing. We now use neither Newton’s name for his\ntechnique (he called it “the method of fluxions”) nor his notation.\nThe more commonly used notation is due to Newton’s German contemporary Leibnitz, whom the English accused of plagiarizing the\ncalculus from Newton. In the Leibnitz notation, we write\ndx\nv=\ndt\nto indicate that the function v(t) equals the slope of the tangent line\nof the graph of x(t) at every time t. The Leibnitz notation is meant\nto evoke the delta notation, but with a very small time interval.\nBecause the dx and dt are thought of as very small ∆x’s and ∆t’s,\ni.e., very small differences, the part of calculus that has to do with\nderivatives is called differential calculus.\nDifferential calculus consists of three things:\n• The concept and definition of the derivative, which is covered\nin this book, but which will be discussed more formally in your\nmath course.\n• The Leibnitz notation described above, which you’ll need to\nget more comfortable with in your math course.\n• A set of rules that allows you to find an equation for the derivative of a given function. For instance, if you happened to have\na situation where the position of an object was given by the\nequation x = 2t7 , you would be able to use those rules to\nfind dx/dt = 14t6 . This bag of tricks is covered in your math\ncourse.\n86\nChapter 2\nVelocity and Relative Motion\nSummary\nSelected Vocabulary\ncenter of mass . . the balance point of an object\nvelocity . . . . . . the rate of change of position; the slope of the\ntangent line on an x − t graph.\nNotation\nx. . . . . . . . . .\nt . . . . . . . . . .\n∆ . . . . . . . . .\n∆x . . . . . . . .\n∆t . . . . . . . . .\nv . . . . . . . . . .\nvAB . . . . . . . .\na point in space\na point in time, a clock reading\n“change in;” the value of a variable afterwards\nminus its value before\na distance, or more precisely a change in x,\nwhich may be less than the distance traveled;\nits plus or minus sign indicates direction\na duration of time\nvelocity\nthe velocity of object A relative to object B\nOther Terminology and Notation\ndisplacement . . a name for the symbol ∆x\nspeed . . . . . . . the absolute value of the velocity, i.e., the velocity stripped of any information about its\ndirection\nSummary\nAn object’s center of mass is the point at which it can be balanced. For the time being, we are studying the mathematical description only of the motion of an object’s center of mass in cases\nrestricted to one dimension. The motion of an object’s center of\nmass is usually far simpler than the motion of any of its other parts.\nIt is important to distinguish location, x, from distance, ∆x,\nand clock reading, t, from time interval ∆t. When an object’s x − t\ngraph is linear, we define its velocity as the slope of the line, ∆x/∆t.\nWhen the graph is curved, we generalize the definition so that the\nvelocity is the slope of the tangent line at a given point on the graph.\nGalileo’s principle of inertia states that no force is required to\nmaintain motion with constant velocity in a straight line, and absolute motion does not cause any observable physical effects. Things\ntypically tend to reduce their velocity relative to the surface of our\nplanet only because they are physically rubbing against the planet\n(or something attached to the planet), not because there is anything\nspecial about being at rest with respect to the earth’s surface. When\nit seems, for instance, that a force is required to keep a book sliding\nacross a table, in fact the force is only serving to cancel the contrary\nforce of friction.\nAbsolute motion is not a well-defined concept, and if two observers are not at rest relative to one another they will disagree\nabout the absolute velocities of objects. They will, however, agree\nSummary\n87\nabout relative velocities. If object A is in motion relative to object\nB, and B is in motion relative to C, then A’s velocity relative to C\nis given by vAC = vAB + vBC . Positive and negative signs are used\nto indicate the direction of an object’s motion.\n88\nChapter 2\nVelocity and Relative Motion\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nThe graph shows the motion of a car stuck in stop-and-go\nfreeway traffic. (a) If you only knew how far the car had gone\nduring this entire time period, what would you think its velocity\n√\nwas? (b) What is the car’s maximum velocity?\n2\n(a) Let θ be the latitude of a point on the Earth’s surface.\nDerive an algebra equation for the distance, L, traveled by that point\nduring one rotation of the Earth about its axis, i.e. over one day,\nexpressed in terms of L, θ, and R, the radius of the earth. Check:\nYour equation should give L = 0 for the North Pole.\n(b) At what speed is Fullerton, at latitude θ = 34 ◦ , moving with\nthe rotation of the Earth about its axis? Give your answer in units\nof mi/h. [See the table in the back of the book for the relevant\n√\ndata.]\nProblem 1.\n3\nA person is parachute jumping. During the time between\nwhen she leaps out of the plane and when she opens her chute, her\naltitude is given by the equation\ni\nh\n.\ny = (10000 m) − (50 m/s) t + (5.0 s)e−t/5.0 s\nFind her velocity at t = 7.0 s. (This can be done on a calculator,\nwithout knowing calculus.) Because of air resistance, her velocity\ndoes not increase at a steady rate as it would for an object falling\n√\nin vacuum.\n?\n4\nA light-year is a unit of distance used in astronomy, and defined\nas the distance light travels in one year. The speed of light is 3.0×108\nm/s. Find how many meters there are in one light-year, expressing\nyour answer in scientific notation.\n. Solution, p. 270\n5\nYou’re standing in a freight train, and have no way to see out.\nIf you have to lean to stay on your feet, what, if anything, does that\ntell you about the train’s velocity? Explain. . Solution, p. 270\n6\nA honeybee’s position as a function of time is given by x =\n10t − t3 , where t is in seconds and x in meters. What is its velocity\nR\nat t = 3.0 s?\n7\nThe figure shows the motion of a point on the rim of a rolling\nwheel. (The shape is called a cycloid.) Suppose bug A is riding on\nthe rim of the wheel on a bicycle that is rolling, while bug B is on\nthe spinning wheel of a bike that is sitting upside down on the floor.\nBug A is moving along a cycloid, while bug B is moving in a circle.\nBoth wheels are doing the same number of revolutions per minute.\nWhich bug has a harder time holding on, or do they find it equally\ndifficult?\n. Solution, p. 270\nProblem 7.\nProblems\n89\n8\nPeanut plants fold up their leaves at night. Estimate the top\nspeed of the tip of one of the leaves shown in the figure, expressing\n√\nyour result in scientific notation in SI units.\n9\n(a) Translate the following information into symbols, using\nthe notation with two subscripts introduced in section 2.5. Eowyn\nis riding on her horse at a velocity of 11 m/s. She twists around in\nher saddle and fires an arrow backward. Her bow fires arrows at 25\nm/s. (b) Find the speed of the arrow relative to the ground.\n10\nOur full discussion of two- and three-dimensional motion is\npostponed until the second half of the book, but here is a chance to\nuse a little mathematical creativity in anticipation of that generalization. Suppose a ship is sailing east at a certain speed v, and a\npassenger is walking across the deck at the same speed v, so that\nhis track across the deck is perpendicular to the ship’s center-line.\nWhat is his speed relative to the water, and in what direction is he\nmoving relative to the water?\n. Solution, p. 270\n11\nFreddi Fish(TM) has a position as a function of time given\nR by\nx = a/(b + t2 ). Find her maximum speed.\n12\nDriving along in your car, you take your foot off the gas,\nand your speedometer shows a reduction in speed. Describe a frame\nof reference in which your car was speeding up during that same\nperiod of time. (The frame of reference should be defined by an\nobserver who, although perhaps in motion relative to the earth, is\nnot changing her own speed or direction of motion.)\n13\nThe figure shows the motion of a bluefin tuna, as measured\nby a radio tag (Block et al., Nature, v. 434, p. 1121, 2005), over\nthe course of several years. Until this study, it had been believed\nthat the populations of the fish in the eastern and western Atlantic\nwere separate, but this particular fish was observed to cross the\nentire Atlantic Ocean, from Virginia to Ireland. Points A, B, and C\nshow a period of one month, during which the fish made the most\nrapid progress. Estimate its speed during that month, in units\nof\n√\nkilometers per hour.\nProblem 8.\nProblem 13.\n90\nChapter 2\nVelocity and Relative Motion\nGalileo’s contradiction of Aristotle had serious consequences. He was\ninterrogated by the Church authorities and convicted of teaching that the\nearth went around the sun as a matter of fact and not, as he had promised\npreviously, as a mere mathematical hypothesis. He was placed under permanent house arrest, and forbidden to write about or teach his theories.\nImmediately after being forced to recant his claim that the earth revolved\naround the sun, the old man is said to have muttered defiantly “and yet\nit does move.” The story is dramatic, but there are some omissions in\nthe commonly taught heroic version. There was a rumor that the Simplicio character represented the Pope. Also, some of the ideas Galileo\nadvocated had controversial religious overtones. He believed in the existence of atoms, and atomism was thought by some people to contradict\nthe Church’s doctrine of transubstantiation, which said that in the Catholic\nmass, the blessing of the bread and wine literally transformed them into\nthe flesh and blood of Christ. His support for a cosmology in which the\nearth circled the sun was also disreputable because one of its supporters, Giordano Bruno, had also proposed a bizarre synthesis of Christianity\nwith the ancient Egyptian religion.\nChapter 3\nAcceleration and Free Fall\n3.1 The Motion of Falling Objects\nThe motion of falling objects is the simplest and most common\nexample of motion with changing velocity. The early pioneers of\n91\nphysics had a correct intuition that the way things drop was a message directly from Nature herself about how the universe worked.\nOther examples seem less likely to have deep significance. A walking\nperson who speeds up is making a conscious choice. If one stretch of\na river flows more rapidly than another, it may be only because the\nchannel is narrower there, which is just an accident of the local geography. But there is something impressively consistent, universal,\nand inexorable about the way things fall.\nStand up now and simultaneously drop a coin and a bit of paper\nside by side. The paper takes much longer to hit the ground. That’s\nwhy Aristotle wrote that heavy objects fell more rapidly. Europeans\nbelieved him for two thousand years.\nNow repeat the experiment, but make it into a race between the\ncoin and your shoe. My own shoe is about 50 times heavier than\nthe nickel I had handy, but it looks to me like they hit the ground at\nexactly the same moment. So much for Aristotle! Galileo, who had\na flair for the theatrical, did the experiment by dropping a bullet\nand a heavy cannonball from a tall tower. Aristotle’s observations\nhad been incomplete, his interpretation a vast oversimplification.\nIt is inconceivable that Galileo was the first person to observe a\ndiscrepancy with Aristotle’s predictions. Galileo was the one who\nchanged the course of history because he was able to assemble the\nobservations into a coherent pattern, and also because he carried\nout systematic quantitative (numerical) measurements rather than\njust describing things qualitatively.\nWhy is it that some objects, like the coin and the shoe, have similar motion, but others, like a feather or a bit of paper, are different?\nGalileo speculated that in addition to the force that always pulls objects down, there was an upward force exerted by the air. Anyone\ncan speculate, but Galileo went beyond speculation and came up\nwith two clever experiments to probe the issue. First, he experimented with objects falling in water, which probed the same issues\nbut made the motion slow enough that he could take time measurements with a primitive pendulum clock. With this technique, he\nestablished the following facts:\na / Galileo dropped a cannonball\nand a musketball simultaneously\nfrom a tower, and observed that\nthey hit the ground at nearly the\nsame time.\nThis contradicted\nAristotle’s long-accepted idea\nthat heavier objects fell faster.\n92\nChapter 3\n• All heavy, streamlined objects (for example a steel rod dropped\npoint-down) reach the bottom of the tank in about the same\namount of time, only slightly longer than the time they would\ntake to fall the same distance in air.\n• Objects that are lighter or less streamlined take a longer time\nto reach the bottom.\nThis supported his hypothesis about two contrary forces. He\nimagined an idealized situation in which the falling object did not\nhave to push its way through any substance at all. Falling in air\nAcceleration and Free Fall\nwould be more like this ideal case than falling in water, but even\na thin, sparse medium like air would be sufficient to cause obvious\neffects on feathers and other light objects that were not streamlined.\nToday, we have vacuum pumps that allow us to suck nearly all the\nair out of a chamber, and if we drop a feather and a rock side by\nside in a vacuum, the feather does not lag behind the rock at all.\nHow the speed of a falling object increases with time\nGalileo’s second stroke of genius was to find a way to make quantitative measurements of how the speed of a falling object increased\nas it went along. Again it was problematic to make sufficiently accurate time measurements with primitive clocks, and again he found a\ntricky way to slow things down while preserving the essential physical phenomena: he let a ball roll down a slope instead of dropping it\nvertically. The steeper the incline, the more rapidly the ball would\ngain speed. Without a modern video camera, Galileo had invented\na way to make a slow-motion version of falling.\nc / The v − t graph of a falling\nobject is a line.\nd / Galileo’s experiments show\nthat all falling objects have the\nsame motion if air resistance is\nnegligible.\nb / Velocity increases more gradually on the gentle slope, but the\nmotion is otherwise the same as the motion of a falling object.\nAlthough Galileo’s clocks were only good enough to do accurate\nexperiments at the smaller angles, he was confident after making\na systematic study at a variety of small angles that his basic conclusions were generally valid. Stated in modern language, what he\nfound was that the velocity-versus-time graph was a line. In the language of algebra, we know that a line has an equation of the form\ny = ax + b, but our variables are v and t, so it would be v = at + b.\n(The constant b can be interpreted simply as the initial velocity of\nthe object, i.e., its velocity at the time when we started our clock,\nwhich we conventionally write as vo .)\ne / 1. Aristotle said that heavier\nobjects fell faster than lighter\nones. 2. If two rocks are tied\ntogether, that makes an extraheavy rock, which should fall\nfaster. 3. But Aristotle’s theory\nwould also predict that the light\nrock would hold back the heavy\nrock, resulting in a slower fall.\nself-check A\nAn object is rolling down an incline. After it has been rolling for a short\ntime, it is found to travel 13 cm during a certain one-second interval.\nDuring the second after that, if goes 16 cm. How many cm will it travel\nin the second after that?\n. Answer, p. 267\nSection 3.1\nThe Motion of Falling Objects\n93\nA contradiction in Aristotle’s reasoning\nGalileo’s inclined-plane experiment disproved the long-accepted\nclaim by Aristotle that a falling object had a definite “natural falling\nspeed” proportional to its weight. Galileo had found that the speed\njust kept on increasing, and weight was irrelevant as long as air\nfriction was negligible. Not only did Galileo prove experimentally\nthat Aristotle had been wrong, but he also pointed out a logical\ncontradiction in Aristotle’s own reasoning. Simplicio, the stupid\ncharacter, mouths the accepted Aristotelian wisdom:\nS IMPLICIO : There can be no doubt but that a particular body\n. . . has a fixed velocity which is determined by nature. . .\nS ALVIATI : If then we take two bodies whose natural speeds\nare different, it is clear that, [according to Aristotle], on uniting the two, the more rapid one will be partly held back by\nthe slower, and the slower will be somewhat hastened by the\nswifter. Do you not agree with me in this opinion?\nS IMPLICIO :\nYou are unquestionably right.\nS ALVIATI : But if this is true, and if a large stone moves with a\nspeed of, say, eight [unspecified units] while a smaller moves\nwith a speed of four, then when they are united, the system\nwill move with a speed less than eight; but the two stones\nwhen tied together make a stone larger than that which before\nmoved with a speed of eight. Hence the heavier body moves\nwith less speed than the lighter; an effect which is contrary to\nyour supposition. Thus you see how, from your assumption\nthat the heavier body moves more rapidly than the lighter one,\nI infer that the heavier body moves more slowly.\nWhat is gravity?\nThe physicist Richard Feynman liked to tell a story about how\nwhen he was a little kid, he asked his father, “Why do things fall?”\nAs an adult, he praised his father for answering, “Nobody knows why\nthings fall. It’s a deep mystery, and the smartest people in the world\ndon’t know the basic reason for it.” Contrast that with the average\nperson’s off-the-cuff answer, “Oh, it’s because of gravity.” Feynman\nliked his father’s answer, because his father realized that simply\ngiving a name to something didn’t mean that you understood it.\nThe radical thing about Galileo’s and Newton’s approach to science\nwas that they concentrated first on describing mathematically what\nreally did happen, rather than spending a lot of time on untestable\nspeculation such as Aristotle’s statement that “Things fall because\nthey are trying to reach their natural place in contact with the\nearth.” That doesn’t mean that science can never answer the “why”\nquestions. Over the next month or two as you delve deeper into\nphysics, you will learn that there are more fundamental reasons why\nall falling objects have v − t graphs with the same slope, regardless\n94\nChapter 3\nAcceleration and Free Fall\nof their mass. Nevertheless, the methods of science always impose\nlimits on how deep our explanation can go.\n3.2 Acceleration\nDefinition of acceleration for linear v − t graphs\nGalileo’s experiment with dropping heavy and light objects from\na tower showed that all falling objects have the same motion, and his\ninclined-plane experiments showed that the motion was described by\nv = at+vo . The initial velocity vo depends on whether you drop the\nobject from rest or throw it down, but even if you throw it down,\nyou cannot change the slope, a, of the v − t graph.\nSince these experiments show that all falling objects have linear v − t graphs with the same slope, the slope of such a graph is\napparently an important and useful quantity. We use the word acceleration, and the symbol a, for the slope of such a graph. In symbols,\na = ∆v/∆t. The acceleration can be interpreted as the amount of\nspeed gained in every second, and it has units of velocity divided by\ntime, i.e., “meters per second per second,” or m/s/s. Continuing to\ntreat units as if they were algebra symbols, we simplify “m/s/s” to\nread “m/s2 .” Acceleration can be a useful quantity for describing\nother types of motion besides falling, and the word and the symbol\n“a” can be used in a more general context. We reserve the more\nspecialized symbol “g” for the acceleration of falling objects, which\non the surface of our planet equals 9.8 m/s2 . Often when doing\napproximate calculations or merely illustrative numerical examples\nit is good enough to use g = 10 m/s2 , which is off by only 2%.\nf / Example 1.\nFinding final speed, given time\nexample 1\n. A despondent physics student jumps off a bridge, and falls for\nthree seconds before hitting the water. How fast is he going when\nhe hits the water?\n. Approximating g as 10 m/s2 , he will gain 10 m/s of speed each\nsecond. After one second, his velocity is 10 m/s, after two seconds it is 20 m/s, and on impact, after falling for three seconds,\nhe is moving at 30 m/s.\nExtracting acceleration from a graph\nexample 2\n. The x − t and v − t graphs show the motion of a car starting\nfrom a stop sign. What is the car’s acceleration?\n. Acceleration is defined as the slope of the v-t graph. The graph\nrises by 3 m/s during a time interval of 3 s, so the acceleration is\n(3 m/s)/(3 s) = 1 m/s2 .\nIncorrect solution #1: The final velocity is 3 m/s, and acceleration\nis velocity divided by time, so the acceleration is (3 m/s)/(10 s) =\n0.3 m/s2 .\ng / Example 6.\nSection 3.2\nAcceleration\n95\nThe solution is incorrect because you can’t find the slope of a\ngraph from one point. This person was just using the point at the\nright end of the v-t graph to try to find the slope of the curve.\nIncorrect solution #2: Velocity is distance divided by time so v =\n(4.5 m)/(3 s) = 1.5 m/s. Acceleration is velocity divided by time,\nso a = (1.5 m/s)/(3 s) = 0.5 m/s2 .\nThe solution is incorrect because velocity is the slope of the tangent line. In a case like this where the velocity is changing, you\ncan’t just pick two points on the x-t graph and use them to find the\nvelocity.\nConverting g to different units\n. What is g in units of cm/s2 ?\nexample 3\n. The answer is going to be how many cm/s of speed a falling\nobject gains in one second. If it gains 9.8 m/s in one second, then\nit gains 980 cm/s in one second, so g = 980 cm/s2 . Alternatively,\nwe can use the method of fractions that equal one:\n9.8 m\n100 cm 980 cm\n×\n=\n2\n1\nm\ns\ns2\n. What is g in units of miles/hour2 ?\n.\n1 mile\n3600 s 2\n9.8 m\n×\n×\n= 7.9 × 104 mile/hour2\n1600 m\n1 hour\ns2\nThis large number can be interpreted as the speed, in miles per\nhour, that you would gain by falling for one hour. Note that we had\nto square the conversion factor of 3600 s/hour in order to cancel\nout the units of seconds squared in the denominator.\n. What is g in units of miles/hour/s?\n.\n9.8 m\n1 mile\n3600 s\n×\n= 22 mile/hour/s\n×\n2\n1600 m\n1 hour\ns\nThis is a figure that Americans will have an intuitive feel for. If\nyour car has a forward acceleration equal to the acceleration of a\nfalling object, then you will gain 22 miles per hour of speed every\nsecond. However, using mixed time units of hours and seconds\nlike this is usually inconvenient for problem-solving. It would be\nlike using units of foot-inches for area instead of ft2 or in2 .\nThe acceleration of gravity is different in different locations.\nEveryone knows that gravity is weaker on the moon, but actually it is not even the same everywhere on Earth, as shown by the\nsampling of numerical data in the following table.\n96\nChapter 3\nAcceleration and Free Fall\nlocation\nnorth pole\nReykjavik, Iceland\nFullerton, California\nGuayaquil, Ecuador\nMt. Cotopaxi, Ecuador\nMt. Everest\nlatitude\n90 ◦ N\n64 ◦ N\n34 ◦ N\n2 ◦S\n1 ◦S\n28 ◦ N\nelevation (m)\n0\n0\n0\n0\n5896\n8848\ng (m/s2 )\n9.8322\n9.8225\n9.7957\n9.7806\n9.7624\n9.7643\nThe main variables that relate to the value of g on Earth are latitude\nand elevation. Although you have not yet learned how g would\nbe calculated based on any deeper theory of gravity, it is not too\nhard to guess why g depends on elevation. Gravity is an attraction\nbetween things that have mass, and the attraction gets weaker with\nincreasing distance. As you ascend from the seaport of Guayaquil\nto the nearby top of Mt. Cotopaxi, you are distancing yourself from\nthe mass of the planet. The dependence on latitude occurs because\nwe are measuring the acceleration of gravity relative to the earth’s\nsurface, but the earth’s rotation causes the earth’s surface to fall\nout from under you. (We will discuss both gravity and rotation in\nmore detail later in the course.)\nMuch more spectacular differences in the strength of gravity can\nbe observed away from the Earth’s surface:\nSection 3.2\nh / This false-color map shows\nvariations in the strength of the\nearth’s gravity. Purple areas have\nthe strongest gravity, yellow the\nweakest. The overall trend toward\nweaker gravity at the equator and\nstronger gravity at the poles has\nbeen artificially removed to allow the weaker local variations to\nshow up. The map covers only\nthe oceans because of the technique used to make it: satellites\nlook for bulges and depressions\nin the surface of the ocean. A\nvery slight bulge will occur over an\nundersea mountain, for instance,\nbecause the mountain’s gravitational attraction pulls water toward it. The US government originally began collecting data like\nthese for military use, to correct\nfor the deviations in the paths of\nmissiles. The data have recently\nbeen released for scientific and\ncommercial use (e.g., searching\nfor sites for off-shore oil wells).\nAcceleration\n97\nlocation\nasteroid Vesta (surface)\nEarth’s moon (surface)\nMars (surface)\nEarth (surface)\nJupiter (cloud-tops)\nSun (visible surface)\ntypical neutron star (surface)\nblack hole (center)\ng (m/s2 )\n0.3\n1.6\n3.7\n9.8\n26\n270\n1012\ninfinite according to some theories, on the order of 1052 according to others\nA typical neutron star is not so different in size from a large asteroid,\nbut is orders of magnitude more massive, so the mass of a body\ndefinitely correlates with the g it creates. On the other hand, a\nneutron star has about the same mass as our Sun, so why is its g\nbillions of times greater? If you had the misfortune of being on the\nsurface of a neutron star, you’d be within a few thousand miles of all\nits mass, whereas on the surface of the Sun, you’d still be millions\nof miles from most of its mass.\nDiscussion Questions\nA\nWhat is wrong with the following definitions of g ?\n(1) “g is gravity.”\n(2) “g is the speed of a falling object.”\n(3) “g is how hard gravity pulls on things.”\nB\nWhen advertisers specify how much acceleration a car is capable\nof, they do not give an acceleration as defined in physics. Instead, they\nusually specify how many seconds are required for the car to go from rest\nto 60 miles/hour. Suppose we use the notation “a” for the acceleration as\ndefined in physics, and “acar ad ” for the quantity used in advertisements for\ncars. In the US’s non-metric system of units, what would be the units of\na and acar ad ? How would the use and interpretation of large and small,\npositive and negative values be different for a as opposed to acar ad?\nC\nTwo people stand on the edge of a cliff. As they lean over the edge,\none person throws a rock down, while the other throws one straight up\nwith an exactly opposite initial velocity. Compare the speeds of the rocks\non impact at the bottom of the cliff.\n3.3 Positive and Negative Acceleration\nGravity always pulls down, but that does not mean it always speeds\nthings up. If you throw a ball straight up, gravity will first slow\nit down to v = 0 and then begin increasing its speed. When I\ntook physics in high school, I got the impression that positive signs\nof acceleration indicated speeding up, while negative accelerations\nrepresented slowing down, i.e., deceleration. Such a definition would\nbe inconvenient, however, because we would then have to say that\nthe same downward tug of gravity could produce either a positive\n98\nChapter 3\nAcceleration and Free Fall\nor a negative acceleration. As we will see in the following example,\nsuch a definition also would not be the same as the slope of the v − t\ngraph\nLet’s study the example of the rising and falling ball. In the example of the person falling from a bridge, I assumed positive velocity\nvalues without calling attention to it, which meant I was assuming\na coordinate system whose x axis pointed down. In this example,\nwhere the ball is reversing direction, it is not possible to avoid negative velocities by a tricky choice of axis, so let’s make the more\nnatural choice of an axis pointing up. The ball’s velocity will initially be a positive number, because it is heading up, in the same\ndirection as the x axis, but on the way back down, it will be a negative number. As shown in the figure, the v − t graph does not do\nanything special at the top of the ball’s flight, where v equals 0. Its\nslope is always negative. In the left half of the graph, there is a\nnegative slope because the positive velocity is getting closer to zero.\nOn the right side, the negative slope is due to a negative velocity\nthat is getting farther from zero, so we say that the ball is speeding\nup, but its velocity is decreasing!\nTo summarize, what makes the most sense is to stick with the\noriginal definition of acceleration as the slope of the v − t graph,\n∆v/∆t. By this definition, it just isn’t necessarily true that things\nspeeding up have positive acceleration while things slowing down\nhave negative acceleration. The word “deceleration” is not used\nmuch by physicists, and the word “acceleration” is used unblushingly to refer to slowing down as well as speeding up: “There was a\nred light, and we accelerated to a stop.”\nNumerical calculation of a negative acceleration\nexample 4\n. In figure i, what happens if you calculate the acceleration between t = 1.0 and 1.5 s?\ni / The ball’s acceleration stays\nthe same — on the way up, at the\ntop, and on the way back down.\nIt’s always negative.\n. Reading from the graph, it looks like the velocity is about −1 m/s\nat t = 1.0 s, and around −6 m/s at t = 1.5 s. The acceleration,\nfigured between these two points, is\na=\n∆v (−6 m/s) − (−1 m/s)\n=\n= −10 m/s2\n∆t\n(1.5 s) − (1.0 s)\n.\nEven though the ball is speeding up, it has a negative acceleration.\nAnother way of convincing you that this way of handling the plus\nand minus signs makes sense is to think of a device that measures\nacceleration. After all, physics is supposed to use operational definitions, ones that relate to the results you get with actual measuring\ndevices. Consider an air freshener hanging from the rear-view mirror\nof your car. When you speed up, the air freshener swings backward.\nSuppose we define this as a positive reading. When you slow down,\nthe air freshener swings forward, so we’ll call this a negative reading\nSection 3.3\nPositive and Negative Acceleration\n99\non our accelerometer. But what if you put the car in reverse and\nstart speeding up backwards? Even though you’re speeding up, the\naccelerometer responds in the same way as it did when you were\ngoing forward and slowing down. There are four possible cases:\nmotion of car\naccelerometer slope of\nswings\nv-t graph\nforward, speeding up\nforward, slowing down\nbackward, speeding up\nbackward, slowing down\nbackward\nforward\nforward\nbackward\n+\n−\n−\n+\ndirection\nof\nforce\nacting on\ncar\nforward\nbackward\nbackward\nforward\nNote the consistency of the three right-hand columns — nature is\ntrying to tell us that this is the right system of classification, not\nthe left-hand column.\nBecause the positive and negative signs of acceleration depend\non the choice of a coordinate system, the acceleration of an object\nunder the influence of gravity can be either positive or negative.\nRather than having to write things like “g = 9.8 m/s2 or −9.8 m/s2 ”\nevery time we want to discuss g’s numerical value, we simply define\ng as the absolute value of the acceleration of objects moving under\nthe influence of gravity. We consistently let g = 9.8 m/s2 , but we\nmay have either a = g or a = −g, depending on our choice of a\ncoordinate system.\nAcceleration with a change in direction of motion\nexample 5\n. A person kicks a ball, which rolls up a sloping street, comes to\na halt, and rolls back down again. The ball has constant acceleration. The ball is initially moving at a velocity of 4.0 m/s, and\nafter 10.0 s it has returned to where it started. At the end, it has\nsped back up to the same speed it had initially, but in the opposite\ndirection. What was its acceleration?\n. By giving a positive number for the initial velocity, the statement\nof the question implies a coordinate axis that points up the slope\nof the hill. The “same” speed in the opposite direction should\ntherefore be represented by a negative number, -4.0 m/s. The\nacceleration is\na = ∆v /∆t\n= (vaf ter − vbef or e )/10.0 s\n= [(−4.0 m/s) − (4.0 m/s)]/10.0s\n= −0.80 m/s2\n.\nThe acceleration was no different during the upward part of the\nroll than on the downward part of the roll.\nIncorrect solution: Acceleration is ∆v /∆t, and at the end it’s not\nmoving any faster or slower than when it started, so ∆v=0 and\n100\nChapter 3\nAcceleration and Free Fall\na = 0.\nThe velocity does change, from a positive number to a negative\nnumber.\nDiscussion question B.\nDiscussion Questions\nA\nA child repeatedly jumps up and down on a trampoline. Discuss the\nsign and magnitude of his acceleration, including both the time when he is\nin the air and the time when his feet are in contact with the trampoline.\nB\nThe figure shows a refugee from a Picasso painting blowing on a\nrolling water bottle. In some cases the person’s blowing is speeding the\nbottle up, but in others it is slowing it down. The arrow inside the bottle\nshows which direction it is going, and a coordinate system is shown at the\nbottom of each figure. In each case, figure out the plus or minus signs of\nthe velocity and acceleration. It may be helpful to draw a v − t graph in\neach case.\nC Sally is on an amusement park ride which begins with her chair being\nhoisted straight up a tower at a constant speed of 60 miles/hour. Despite\nstern warnings from her father that he’ll take her home the next time she\nmisbehaves, she decides that as a scientific experiment she really needs\nto release her corndog over the side as she’s on the way up. She does\nnot throw it. She simply sticks it out of the car, lets it go, and watches it\nagainst the background of the sky, with no trees or buildings as reference\npoints. What does the corndog’s motion look like as observed by Sally?\nDoes its speed ever appear to her to be zero? What acceleration does\nshe observe it to have: is it ever positive? negative? zero? What would\nher enraged father answer if asked for a similar description of its motion\nas it appears to him, standing on the ground?\nD\nCan an object maintain a constant acceleration, but meanwhile\nreverse the direction of its velocity?\nDiscussion question C.\nE\nCan an object have a velocity that is positive and increasing at the\nsame time that its acceleration is decreasing?\nSection 3.3\nPositive and Negative Acceleration\n101\n3.4 Varying Acceleration\nSo far we have only been discussing examples of motion for which\nthe v − t graph is linear. If we wish to generalize our definition to\nv-t graphs that are more complex curves, the best way to proceed\nis similar to how we defined velocity for curved x − t graphs:\ndefinition of acceleration\nThe acceleration of an object at any instant is the slope of\nthe tangent line passing through its v-versus-t graph at the\nrelevant point.\nA skydiver\nexample 6\n. The graphs in figure g show the results of a fairly realistic computer simulation of the motion of a skydiver, including the effects\nof air friction. The x axis has been chosen pointing down, so x\nis increasing as she falls. Find (a) the skydiver’s acceleration at\nt = 3.0 s, and also (b) at t = 7.0 s.\n. The solution is shown in figure l. I’ve added tangent lines at the\ntwo points in question.\n(a) To find the slope of the tangent line, I pick two points on the\nline (not necessarily on the actual curve): (3.0 s, 28m/s) and\n(5.0 s, 42 m/s). The slope of the tangent line is (42 m/s−28 m/s)/(5.0 s−\n3.0 s) = 7.0 m/s2 .\n(b) Two points on this tangent line are (7.0 s, 47 m/s) and (9.0 s, 52 m/s).\nThe slope of the tangent line is (52 m/s−47 m/s)/(9.0 s−7.0 s) =\n2.5 m/s2 .\nPhysically, what’s happening is that at t = 3.0 s, the skydiver is\nnot yet going very fast, so air friction is not yet very strong. She\ntherefore has an acceleration almost as great as g. At t = 7.0 s,\nshe is moving almost twice as fast (about 100 miles per hour), and\nair friction is extremely strong, resulting in a significant departure\nfrom the idealized case of no air friction.\nk / Example 6.\nIn example 6, the x−t graph was not even used in the solution of\nthe problem, since the definition of acceleration refers to the slope\nof the v − t graph. It is possible, however, to interpret an x − t\ngraph to find out something about the acceleration. An object with\nzero acceleration, i.e., constant velocity, has an x − t graph that is a\nstraight line. A straight line has no curvature. A change in velocity\nrequires a change in the slope of the x − t graph, which means that\nit is a curve rather than a line. Thus acceleration relates to the\ncurvature of the x − t graph. Figure m shows some examples.\n102\nChapter 3\nAcceleration and Free Fall\nl / The solution to example 6.\nIn example 6, the x − t graph was more strongly curved at the\nbeginning, and became nearly straight at the end. If the x − t graph\nis nearly straight, then its slope, the velocity, is nearly constant, and\nthe acceleration is therefore small. We can thus interpret the acceleration as representing the curvature of the x − t graph, as shown\nin figure m. If the “cup” of the curve points up, the acceleration is\npositive, and if it points down, the acceleration is negative.\nm / Acceleration relates to the curvature of the x − t graph.\nSection 3.4\nVarying Acceleration\n103\nSince the relationship between a and v is analogous to the relationship between v and x, we can also make graphs of acceleration\nas a function of time, as shown in figure n.\nn / Examples of graphs of x , v , and a versus t . 1. A object in free\nfall, with no friction. 2. A continuation of example 6, the skydiver.\n. Solved problem: Drawing a v − t graph.\npage 117, problem 14\n. Solved problem: Drawing v − t and a − t graphs. page 118, problem\n20\nFigure o summarizes the relationships among the three types of\ngraphs.\nDiscussion Questions\no / How position, velocity, and\nacceleration are related.\n104\nChapter 3\nA\nDescribe in words how the changes in the a − t graph in figure n/2\nrelate to the behavior of the v − t graph.\nAcceleration and Free Fall\nB\nExplain how each set of graphs contains inconsistencies, and fix\nthem.\nC\nIn each case, pick a coordinate system and draw x − t , v − t , and\na − t graphs. Picking a coordinate system means picking where you want\nx = 0 to be, and also picking a direction for the positive x axis.\n(1) An ocean liner is cruising in a straight line at constant speed.\n(2) You drop a ball. Draw two different sets of graphs (a total of 6), with\none set’s positive x axis pointing in the opposite direction compared to the\nother’s.\n(3) You’re driving down the street looking for a house you’ve never been\nto before. You realize you’ve passed the address, so you slow down, put\nthe car in reverse, back up, and stop in front of the house.\n3.5 The Area Under the Velocity-Time Graph\nA natural question to ask about falling objects is how fast they fall,\nbut Galileo showed that the question has no answer. The physical\nlaw that he discovered connects a cause (the attraction of the planet\nEarth’s mass) to an effect, but the effect is predicted in terms of an\nacceleration rather than a velocity. In fact, no physical law predicts\na definite velocity as a result of a specific phenomenon, because\nvelocity cannot be measured in absolute terms, and only changes in\nvelocity relate directly to physical phenomena.\nThe unfortunate thing about this situation is that the definitions\nof velocity and acceleration are stated in terms of the tangent-line\ntechnique, which lets you go from x to v to a, but not the other\nway around. Without a technique to go backwards from a to v to x,\nwe cannot say anything quantitative, for instance, about the x − t\ngraph of a falling object. Such a technique does exist, and I used it\nto make the x − t graphs in all the examples above.\nSection 3.5\nThe Area Under the Velocity-Time Graph\n105\nFirst let’s concentrate on how to get x information out of a v − t\ngraph. In example p/1, an object moves at a speed of 20 m/s for\na period of 4.0 s. The distance covered is ∆x = v∆t = (20 m/s) ×\n(4.0 s) = 80 m. Notice that the quantities being multiplied are the\nwidth and the height of the shaded rectangle — or, strictly speaking,\nthe time represented by its width and the velocity represented by\nits height. The distance of ∆x = 80 m thus corresponds to the area\nof the shaded part of the graph.\nThe next step in sophistication is an example like p/2, where the\nobject moves at a constant speed of 10 m/s for two seconds, then\nfor two seconds at a different constant speed of 20 m/s. The shaded\nregion can be split into a small rectangle on the left, with an area\nrepresenting ∆x = 20 m, and a taller one on the right, corresponding\nto another 40 m of motion. The total distance is thus 60 m, which\ncorresponds to the total area under the graph.\nAn example like p/3 is now just a trivial generalization; there\nis simply a large number of skinny rectangular areas to add up.\nBut notice that graph p/3 is quite a good approximation to the\nsmooth curve p/4. Even though we have no formula for the area of\na funny shape like p/4, we can approximate its area by dividing it up\ninto smaller areas like rectangles, whose area is easier to calculate.\nIf someone hands you a graph like p/4 and asks you to find the\narea under it, the simplest approach is just to count up the little\nrectangles on the underlying graph paper, making rough estimates\nof fractional rectangles as you go along.\nThat’s what I’ve done in figure q. Each rectangle on the graph\npaper is 1.0 s wide and 2 m/s tall, so it represents 2 m. Adding up\nall the numbers gives ∆x = 41 m. If you needed better accuracy,\nyou could use graph paper with smaller rectangles.\nIt’s important to realize that this technique gives you ∆x, not\nx. The v − t graph has no information about where the object was\nwhen it started.\nThe following are important points to keep in mind when applying this technique:\np / The area under the v − t\ngraph gives ∆x .\n• If the range of v values on your graph does not extend down\nto zero, then you will get the wrong answer unless you compensate by adding in the area that is not shown.\n• As in the example, one rectangle on the graph paper does not\nnecessarily correspond to one meter of distance.\n• Negative velocity values represent motion in the opposite direction, so area under the t axis should be subtracted, i.e.,\ncounted as “negative area.”\n106\nChapter 3\nAcceleration and Free Fall\nq / An example using estimation\nof fractions of a rectangle.\n• Since the result is a ∆x value, it only tells you xaf ter − xbef ore ,\nwhich may be less than the actual distance traveled. For instance, the object could come back to its original position at\nthe end, which would correspond to ∆x=0, even though it had\nactually moved a nonzero distance.\nFinally, note that one can find ∆v from an a − t graph using\nan entirely analogous method. Each rectangle on the a − t graph\nrepresents a certain amount of velocity change.\nDiscussion Question\nA Roughly what would a pendulum’s v − t graph look like? What would\nhappen when you applied the area-under-the-curve technique to find the\npendulum’s ∆x for a time period covering many swings?\n3.6 Algebraic Results for Constant\nAcceleration\nAlthough the area-under-the-curve technique can be applied to any\ngraph, no matter how complicated, it may be laborious to carry out,\nand if fractions of rectangles must be estimated the result will only\nbe approximate. In the special case of motion with constant acceleration, it is possible to find a convenient shortcut which produces\nSection 3.6\nAlgebraic Results for Constant Acceleration\n107\nexact results. When the acceleration is constant, the v − t graph\nis a straight line, as shown in the figure. The area under the curve\ncan be divided into a triangle plus a rectangle, both of whose areas\ncan be calculated exactly: A = bh for a rectangle and A = bh/2\nfor a triangle. The height of the rectangle is the initial velocity, vo ,\nand the height of the triangle is the change in velocity from beginning to end, ∆v. The object’s ∆x is therefore given by the equation\n∆x = vo ∆t + ∆v∆t/2. This can be simplified a little by using the\ndefinition of acceleration, a = ∆v/∆t, to eliminate ∆v, giving\n1\n∆x = vo ∆t + a∆t2\n2\n.\n[motion with\nconstant acceleration]\nSince this is a second-order polynomial in ∆t, the graph of ∆x versus\n∆t is a parabola, and the same is true of a graph of x versus t —\nthe two graphs differ only by shifting along the two axes. Although\nI have derived the equation using a figure that shows a positive vo ,\npositive a, and so on, it still turns out to be true regardless of what\nplus and minus signs are involved.\nAnother useful equation can be derived if one wants to relate\nthe change in velocity to the distance traveled. This is useful, for\ninstance, for finding the distance needed by a car to come to a stop.\nFor simplicity, we start by deriving the equation for the special case\nof vo = 0, in which the final velocity vf is a synonym for ∆v. Since\nvelocity and distance are the variables of interest, not time, we take\nthe equation ∆x = 12 a∆t2 and use ∆t = ∆v/a to eliminate ∆t. This\ngives ∆x = (∆v)2/a, which can be rewritten as\nr / The shaded area tells us\nhow far an object moves while\naccelerating at a constant rate.\nvf2 = 2a∆x\n.\n[motion with constant acceleration, vo = 0]\nFor the more general case where , we skip the tedious algebra leading\nto the more general equation,\nvf2 = vo2 + 2a∆x\n108\nChapter 3\nAcceleration and Free Fall\n.\n[motion with constant acceleration]\nTo help get this all organized in your head, first let’s categorize\nthe variables as follows:\nVariables that change during motion with constant acceleration:\nx ,v, t\nVariable that doesn’t change:\na\nIf you know one of the changing variables and want to find another,\nthere is always an equation that relates those two:\nThe symmetry among the three variables is imperfect only because the equation relating x and t includes the initial velocity.\nThere are two main difficulties encountered by students in applying these equations:\n• The equations apply only to motion with constant acceleration. You can’t apply them if the acceleration is changing.\n• Students are often unsure of which equation to use, or may\ncause themselves unnecessary work by taking the longer path\naround the triangle in the chart above. Organize your thoughts\nby listing the variables you are given, the ones you want to\nfind, and the ones you aren’t given and don’t care about.\nSaving an old lady\nexample 7\n. You are trying to pull an old lady out of the way of an oncoming\ntruck. You are able to give her an acceleration of 20 m/s2 . Starting from rest, how much time is required in order to move her 2\nm?\n. First we organize our thoughts:\nVariables given: ∆x, a, vo\nVariables desired: ∆t\nIrrelevant variables: vf\nConsulting the triangular chart above, the equation we need is\nclearly ∆x = vo ∆t + 21 a∆t 2 , since it has the four variables of interest\nSection 3.6\nAlgebraic Results for Constant Acceleration\n109\nand omits the irrelevant\none. Eliminating the vo term and solving\np\nfor ∆t gives ∆t = 2∆x/a = 0.4 s.\n. Solved problem: A stupid celebration\npage 117, problem 15\n. Solved problem: Dropping a rock on Mars\npage 117, problem 16\n. Solved problem: The Dodge Viper\npage 118, problem 18\n. Solved problem: Half-way sped up\npage 118, problem 22\nDiscussion Questions\nA\nIn chapter 1, I gave examples of correct and incorrect reasoning\nabout proportionality, using questions about the scaling of area and volume. Try to translate the incorrect modes of reasoning shown there into\nmistakes about the following question: If the acceleration of gravity on\nMars is 1/3 that on Earth, how many times longer does it take for a rock\nto drop the same distance on Mars?\nB\nCheck that the units make sense in the three equations derived in\nthis section.\n3.7 Biological Effects of Weightlessness\ns / On October 4, 2004, the\nprivately funded SpaceShipOne\nwon the ten-million-dollar Ansari\nX Prize by reaching an altitude\nof 100 km twice in the space of\n14 days. [Courtesy of Scaled\nComposites LLC.]\nThe usefulness of outer space was brought home to North Americans in 1998 by the unexpected failure of the communications satellite that had been handling almost all of the continent’s cellular\nphone traffic. Compared to the massive economic and scientific payoffs of satellites and space probes, human space travel has little to\nboast about after four decades. Sending people into orbit has just\nbeen too expensive to be an effective scientific or commercial activity. The downsized and over-budget International Space Station\nhas produced virtually no scientific results, and the space shuttle\nprogram now has a record of two catastrophic failures out of 113\nmissions.\nWithin our lifetimes, we are probably only likely to see one economically viable reason for sending humans into space: tourism!\nNo fewer than three private companies are now willing to take your\nmoney for a reservation on a two-to-four minute trip into space,\nalthough none of them has a firm date on which to begin service.\nWithin a decade, a space cruise may be the new status symbol\namong those sufficiently rich and brave.\nSpace sickness\nWell, rich, brave, and possessed of an iron stomach. Travel\nagents will probably not emphasize the certainty of constant spacesickness. For us animals evolved to function in g = 9.8 m/s2 , living\nin g = 0 is extremely unpleasant. The early space program focused\nobsessively on keeping the astronaut-trainees in perfect physical\nshape, but it soon became clear that a body like a Greek demigod’s\nwas no defense against that horrible feeling that your stomach was\n110\nChapter 3\nAcceleration and Free Fall\nfalling out from under you and you were never going to catch up.\nOur inner ear, which normally tells us which way is down, tortures\nus when down is nowhere to be found. There is contradictory information about whether anyone ever gets over it; the “right stuff”\nculture creates a strong incentive for astronauts to deny that they\nare sick.\nEffects of long space missions\nWorse than nausea are the health-threatening effects of prolonged weightlessness. The Russians are the specialists in long-term\nmissions, in which cosmonauts suffer harm to their blood, muscles,\nand, most importantly, their bones.\nt / U.S. and Russian astronauts aboard the International\nSpace Station, October 2000.\nThe effects on the muscles and skeleton appear to be similar to\nthose experienced by old people and people confined to bed for a\nlong time. Everyone knows that our muscles get stronger or weaker\ndepending on the amount of exercise we get, but the bones are likewise adaptable. Normally old bone mass is continually being broken\ndown and replaced with new material, but the balance between its\nloss and replacement is upset when people do not get enough weightbearing exercise. The main effect is on the bones of the lower body.\nMore research is required to find out whether astronauts’ loss of bone\nmass is due to faster breaking down of bone, slower replacement, or\nboth. It is also not known whether the effect can be suppressed via\ndiet or drugs.\nThe other set of harmful physiological effects appears to derive\nfrom the redistribution of fluids. Normally, the veins and arteries of the legs are tightly constricted to keep gravity from making\nblood collect there. It is uncomfortable for adults to stand on their\nheads for very long, because the head’s blood vessels are not able to\nconstrict as effectively. Weightless astronauts’ blood tends to be expelled by the constricted blood vessels of the lower body, and pools\naround their hearts, in their thoraxes, and in their heads. The only\nimmediate result is an uncomfortable feeling of bloatedness in the\nupper body, but in the long term, a harmful chain of events is set in\nmotion. The body’s attempts to maintain the correct blood volume\nare most sensitive to the level of fluid in the head. Since astronauts\nhave extra fluid in their heads, the body thinks that the over-all\nblood volume has become too great. It responds by decreasing blood\nvolume below normal levels. This increases the concentration of red\nblood cells, so the body then decides that the blood has become too\nthick, and reduces the number of blood cells. In missions lasting up\nto a year or so, this is not as harmful as the musculo-skeletal effects,\nbut it is not known whether longer period in space would bring the\nred blood cell count down to harmful levels.\nSection 3.7\nu / The\nInternational\nSpace\nStation, September 2000. The\nspace station does not rotate to\nprovide simulated gravity. The\ncompleted station will be much\nbigger.\nBiological Effects of Weightlessness\n111\nReproduction in space\nFor those enthralled by the romance of actual human colonization of space, human reproduction in weightlessness becomes an issue. An already-pregnant Russian cosmonaut did spend some time\nin orbit in the 1960’s, and later gave birth to a normal child on the\nground. Recently, one of NASA’s public relations concerns about\nthe space shuttle program has been to discourage speculation about\nspace sex, for fear of a potential taxpayers’ backlash against the\nspace program as an expensive form of exotic pleasure.\nScientific work has been concentrated on studying plant and animal reproduction in space. Green plants, fungi, insects, fish, and\namphibians have all gone through at least one generation in zerogravity experiments without any serious problems. In many cases,\nanimal embryos conceived in orbit begin by developing abnormally,\nbut later in development they seem to correct themselves. However,\nchicken embryos fertilized on earth less than 24 hours before going\ninto orbit have failed to survive. Since chickens are the organisms\nmost similar to humans among the species investigated so far, it\nis not at all certain that humans could reproduce successfully in a\nzero-gravity space colony.\nSimulated gravity\nIf humans are ever to live and work in space for more than a\nyear or so, the only solution is probably to build spinning space stations to provide the illusion of weight, as discussed in section 9.2.\nNormal gravity could be simulated, but tourists would probably enjoy g = 2 m/s2 or 5 m/s2 . Space enthusiasts have proposed entire\norbiting cities built on the rotating cylinder plan. Although science\nfiction has focused on human colonization of relatively earthlike bodies such as our moon, Mars, and Jupiter’s icy moon Europa, there\nwould probably be no practical way to build large spinning structures on their surfaces. If the biological effects of their 2 − 3 m/s2\ngravitational accelerations are as harmful as the effect of g = 0, then\nwe may be left with the surprising result that interplanetary space\nis more hospitable to our species than the moons and planets.\nOptional Topic: More on Apparent Weightlessness\nAstronauts in orbit are not really weightless; they are only a few hundred\nmiles up, so they are still affected strongly by the Earth’s gravity. Section\n10.3 of this book discusses why they experience apparent weightlessness. More on Simulated Gravity For more information on simulating\ngravity by spinning a spacecraft, see section 9.2 of this book.\n3.8\nR\nApplications of Calculus\nIn the Applications of Calculus section at the end of the previous\nchapter, I discussed how the slope-of-the-tangent-line idea related\nto the calculus concept of a derivative, and the branch of calculus\n112\nChapter 3\nAcceleration and Free Fall\nknown as differential calculus. The other main branch of calculus,\nintegral calculus, has to do with the area-under-the-curve concept\ndiscussed in section 3.5 of this chapter. Again there is a concept,\na notation, and a bag of tricks for doing things symbolically rather\nthan graphically. In calculus, the area under the v −t graph between\nt = t1 and t = t2 is notated like this:\nZ t2\narea under curve = ∆x =\nvdt\n.\nt1\nThe expression on the right is called an integral, and the s-shaped\nsymbol, the integral sign, is read as “integral of . . . ”\nIntegral calculus and differential calculus are closely related. For\ninstance, if you take the derivative of the function x(t), you get\nthe function v(t), and if you integrate the function v(t), you get\nx(t) back again. In other words, integration and differentiation are\ninverse operations. This is known as the fundamental theorem of\ncalculus.\nOn an unrelated topic, there is a special notation for taking the\nderivative of a function twice. The acceleration, for instance, is the\nsecond (i.e., double) derivative of the position, because differentiating x once gives v, and then differentiating v gives a. This is written\nas\nd2 x\na= 2\n.\ndt\nThe seemingly inconsistent placement of the twos on the top and\nbottom confuses all beginning calculus students. The motivation\nfor this funny notation is that acceleration has units of m/s2 , and\nthe notation correctly suggests that: the top looks like it has units of\nmeters, the bottom seconds2 . The notation is not meant, however,\nto suggest that t is really squared.\nSection 3.8\nR\nApplications of Calculus\n113\nSummary\nSelected Vocabulary\ngravity . . . . . . A general term for the phenomenon of attraction between things having mass. The attraction between our planet and a human-sized object causes the object to fall.\nacceleration . . . The rate of change of velocity; the slope of the\ntangent line on a v − t graph.\nNotation\na . . . . . . . . . .\ng . . . . . . . . . .\nacceleration\nthe acceleration of objects in free fall; the\nstrength of the local gravitational field\nSummary\nGalileo showed that when air resistance is negligible all falling\nbodies have the same motion regardless of mass. Moreover, their\nv − t graphs are straight lines. We therefore define a quantity called\nacceleration as the slope, ∆v/∆t, of an object’s v −t graph. In cases\nother than free fall, the v −t graph may be curved, in which case the\ndefinition is generalized as the slope of a tangent line on the v − t\ngraph. The acceleration of objects in free fall varies slightly across\nthe surface of the earth, and greatly on other planets.\nPositive and negative signs of acceleration are defined according\nto whether the v − t graph slopes up or down. This definition has\nthe advantage that a force in a given direction always produces the\nsame sign of acceleration.\nThe area under the v − t graph gives ∆x, and analogously the\narea under the a − t graph gives ∆v.\nFor motion with constant acceleration, the following three equations hold:\n1\n∆x = vo ∆t + a∆t2\n2\nvf2 = vo2 + 2a∆x\n∆v\na=\n∆t\nThey are not valid if the acceleration is changing.\n114\nChapter 3\nAcceleration and Free Fall\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nThe graph represents the velocity of a bee along a straight\nline. At t = 0, the bee is at the hive. (a) When is the bee farthest\nfrom the hive? (b) How far is the bee at its farthest point from the\nhive? (c) At t = 13s, how far is the bee from the hive? [Hint: √Try\nproblem 19 first.]\n2\nA rock is dropped into a pond. Draw plots of its position\nversus time, velocity versus time, and acceleration versus time. Include its whole motion, starting from the moment it is dropped, and\ncontinuing while it falls through the air, passes through the water,\nand ends up at rest on the bottom of the pond. Do your work on\nphotocopy or a printout of page 121.\n3\nIn an 18th-century naval battle, a cannon ball is shot horizontally, passes through the side of an enemy ship’s hull, flies across the\ngalley, and lodges in a bulkhead. Draw plots of its horizontal position, velocity, and acceleration as functions of time, starting while it\nis inside the cannon and has not yet been fired, and ending when it\ncomes to rest. There is not any significant amount of friction from\nthe air. Although the ball may rise and fall, you are only concerned\nwith its horizontal motion, as seen from above. Do your work on\nphotocopy or a printout of page 121.\nProblem 3.\n4\nDraw graphs of position, velocity, and acceleration as functions\nof time for a person bunjee jumping. (In bunjee jumping, a person\nhas a stretchy elastic cord tied to his/her ankles, and jumps off of a\nhigh platform. At the bottom of the fall, the cord brings the person\nup short. Presumably the person bounces up a little.) Do your work\non photocopy or a printout of page 121.\nProblems\n115\n5\nA ball rolls down the ramp shown in the figure, consisting of a\ncurved knee, a straight slope, and a curved bottom. For each part of\nthe ramp, tell whether the ball’s velocity is increasing, decreasing,\nor constant, and also whether the ball’s acceleration is increasing,\ndecreasing, or constant. Explain your answers. Assume there is no\nair friction or rolling resistance. Hint: Try problem 20 first. [Based\non a problem by Hewitt.]\n6\nA toy car is released on one side of a piece of track that is bent\ninto an upright U shape. The car goes back and forth. When the\ncar reaches the limit of its motion on one side, its velocity is zero.\nIs its acceleration also zero? Explain using a v − t graph. [Based on\na problem by Serway and Faughn.]\nProblem 5.\n7\nWhat is the acceleration of a car that moves at a steady\nvelocity of 100 km/h for 100 seconds? Explain your answer. [Based\non a problem by Hewitt.]\n8\nA physics homework question asks, “If you start from rest and\naccelerate at 1.54 m/s2 for 3.29 s, how far do you travel by the end\nof that time?” A student answers as follows:\n1.54 × 3.29 = 5.07 m\nHis Aunt Wanda is good with numbers, but has never taken physics.\nShe doesn’t know the formula for the distance traveled under constant acceleration over a given amount of time, but she tells her\nnephew his answer cannot be right. How does she know?\n9\nYou are looking into a deep well. It is dark, and you cannot\nsee the bottom. You want to find out how deep it is, so you drop\na rock in, and you hear a splash 3.0 seconds later. How deep is√the\nwell?\n10\nYou take a trip in your spaceship to another star. Setting off,\nyou increase your speed at a constant acceleration. Once you get\nhalf-way there, you start decelerating, at the same rate, so that by\nthe time you get there, you have slowed down to zero speed. You see\nthe tourist attractions, and then head home by the same method.\n(a) Find a formula for the time, T , required for the round trip, in\nterms of d, the distance from our sun to the star, and a, the magnitude of the acceleration. Note that the acceleration is not constant\nover the whole trip, but the trip can be broken up into constantacceleration parts.\n(b) The nearest star to the Earth (other than our own sun) is Proxima Centauri, at a distance of d = 4 × 1016 m. Suppose you use an\nacceleration of a = 10 m/s2 , just enough to compensate for the lack\nof true gravity and make you feel comfortable. How long does the\nround trip take, in years?\n(c) Using the same numbers for d and a, find your maximum speed.\nCompare this to the speed of light, which is 3.0 × 108 m/s. (Later\nin this course, you will learn that there are some new things going\n116\nChapter 3\nAcceleration and Free Fall\non in physics when one gets close to the speed of light, and that it\nis impossible to exceed the speed of light. For now, though, just\nuse\n√\nthe simpler ideas you’ve learned so far.)\n?\n11\nYou climb half-way up a tree, and drop a rock. Then you\nclimb to the top, and drop another rock. How many times greater\nis the velocity of the second rock on impact? Explain. (The answer\nis not two times greater.)\n12\nAlice drops a rock off a cliff. Bubba shoots a gun straight\ndown from the edge of the same cliff. Compare the accelerations of\nthe rock and the bullet while they are in the air on the way down.\n[Based on a problem by Serway and Faughn.]\n13\nA person is parachute jumping. During the time between\nwhen she leaps out of the plane and when she opens her chute, her\naltitude is given by an equation of the form\ny = b − c t + ke−t/k\n,\nwhere e is the base of natural logarithms, and b, c, and k are constants. Because of air resistance, her velocity does not increase at a\nsteady rate as it would for an object falling in vacuum.\n(a) What units would b, c, and k have to have for the equation to\nmake sense?\n(b) Find the person’s velocity, v, as a function of time. [You will\n√\nneed to use the chain rule, and the fact that d(ex )/dx = ex .]\n(c) Use your answer from part (b) to get an interpretation of the\nconstant c. [Hint: e−x approaches zero for large values of x.]\n√\n(d) Find the person’s acceleration, a, as a function of time.\n(e) Use your answer from part (b) to show that if she waits long\nenough to open her chute, her acceleration will become very small.\nR\n14\nThe top part of the figure shows the position-versus-time\ngraph for an object moving in one dimension. On the bottom part\nof the figure, sketch the corresponding v-versus-t graph.\n. Solution, p. 270\n15\nOn New Year’s Eve, a stupid person fires a pistol straight up.\nThe bullet leaves the gun at a speed of 100 m/s. How long does it\ntake before the bullet hits the ground?\n. Solution, p. 271\nProblem 14.\n16\nIf the acceleration of gravity on Mars is 1/3 that on Earth,\nhow many times longer does it take for a rock to drop the same\ndistance on Mars? Ignore air resistance.\n. Solution, p. 271\n17\nA honeybee’s position as a function of time is given by\nx = 10t − t3 , where t is in seconds and x in meters. What is\nR its\nacceleration at t = 3.0 s?\n. Solution, p. 271\nProblems\n117\n18\nIn July 1999, Popular Mechanics carried out tests to find\nwhich car sold by a major auto maker could cover a quarter mile\n(402 meters) in the shortest time, starting from rest. Because the\ndistance is so short, this type of test is designed mainly to favor the\ncar with the greatest acceleration, not the greatest maximum speed\n(which is irrelevant to the average person). The winner was the\nDodge Viper, with a time of 12.08 s. The car’s top (and presumably\nfinal) speed was 118.51 miles per hour (52.98 m/s). (a) If a car,\nstarting from rest and moving with constant acceleration, covers\na quarter mile in this time interval, what is its acceleration? (b)\nWhat would be the final speed of a car that covered a quarter mile\nwith the constant acceleration you found in part a? (c) Based on\nthe discrepancy between your answer in part b and the actual final\nspeed of the Viper, what do you conclude about how its acceleration\nchanged over time?\n. Solution, p. 271\n19\nThe graph represents the motion of a rolling ball that bounces\noff of a wall. When does the ball return to the location it had at\nt = 0?\n. Solution, p. 271\n20\n(a) The ball is released at the top of the ramp shown in the\nfigure. Friction is negligible. Use physical reasoning to draw v − t\nand a − t graphs. Assume that the ball doesn’t bounce at the point\nwhere the ramp changes slope. (b) Do the same for the case where\nthe ball is rolled up the slope from the right side, but doesn’t quite\nhave enough speed to make it over the top. . Solution, p. 271\nProblem 19.\n21\nYou throw a rubber ball up, and it falls and bounces several times. Draw graphs of position, velocity, and acceleration as\nfunctions of time.\n. Solution, p. 272\n22\nStarting from rest, a ball rolls down a ramp, traveling a\ndistance L and picking up a final speed v. How much of the distance\ndid the ball have to cover before achieving a speed of v/2? [Based\non a problem by Arnold Arons.]\n. Solution, p. 273\nProblem 20.\n23\nThe graph shows the acceleration of a chipmunk in a TV\ncartoon. It consists of two circular arcs and two line segments.\nAt t = 0.00 s, the chipmunk’s velocity is −3.10 m/s. What is its\nvelocity at t = 10.00 s?\n24\nFind the error in the following calculation. A student wants\nto find the distance traveled by a car that accelerates from rest for\n5.0 s with an acceleration of 2.0 m/s2 . First he solves a = ∆v/∆t for\n∆v = 10 m/s. Then he multiplies to find (10 m/s)(5.0 s) = 50 m.\nDo not just recalculate the result by a different method; if that was\nall you did, you’d have no way of knowing which calculation was\ncorrect, yours or his.\nProblem 23.\n118\nChapter 3\nAcceleration and Free Fall\n25\nAcceleration could be defined either as ∆v/∆t or as the slope\nof the tangent line on the v − t graph. Is either one superior as a\ndefinition, or are they equivalent? If you say one is better, give an\nexample of a situation where it makes a difference which one you\nuse.\n26\nIf an object starts accelerating from rest, we have v 2 =\n2a∆x for its speed after it has traveled a distance ∆x. Explain in\nwords why it makes sense that the equation has velocity squared, but\ndistance only to the first power. Don’t recapitulate the derivation\nin the book, or give a justification based on units. The point is\nto explain what this feature of the equation tells us about the way\nspeed increases as more distance is covered.\n27\nThe figure shows a practical, simple experiment for determining g to high precision. Two steel balls are suspended from electromagnets, and are released simultaneously when the electric current\nis shut off. They fall through unequal heights ∆x1 and ∆x2 . A\ncomputer records the sounds through a microphone as first one ball\nand then the other strikes the floor. From this recording, we can\naccurately determine the quantity T defined as T = ∆t2 − ∆t1 , i.e.,\nthe time lag between the first and second impacts. Note that since\nthe balls do not make any sound when they are released, we have\nno way of measuring the individual times ∆t2 and ∆t1 .\n(a) Find an equation for g in terms of the measured quantities T√,\n∆x1 and ∆x2 .\n(b) Check the units of your equation.\n(c) Check that your equation gives the correct result in the case\nwhere ∆x1 is very close to zero. However, is this case realistic?\n(d) What happens when ∆x1 = ∆x2 ? Discuss this both mathematically and physically.\nProblem 27.\n28\nThe speed required for a low-earth orbit is 7.9 × 103 m/s (see\nch. 10). When a rocket is launched into orbit, it goes up a little at\nfirst to get above almost all of the atmosphere, but then tips over\nhorizontally to build up to orbital speed. Suppose the horizontal\nacceleration is limited to 3g to keep from damaging the cargo (or\nhurting the crew, for a crewed flight). (a) What is the minimum\ndistance the rocket must travel downrange before it reaches orbital\nspeed? How much does it matter whether you take into account the\ninitial eastward velocity due to the rotation of the earth? (b) Rather\nthan a rocket ship, it might be advantageous to use a railgun design,\nin which the craft would be accelerated to orbital speeds along a\nrailroad track. This has the advantage that it isn’t necessary to lift\na large mass of fuel, since the energy source is external. Based on\nyour answer to part a, comment on the feasibility of this design for\ncrewed launches from the earth’s surface.\nProblems\n119\n29\nSome fleas can jump as high as 30 cm. The flea only has a\nshort time to build up speed — the time during which its center of\nmass is accelerating upward but its feet are still in contact with the\nground. Make an order-of-magnitude estimate of the acceleration\nthe flea needs to have while straightening its legs, and state your\nanswer in units of g, i.e., how many “g’s it pulls.” (For comparison,\nfighter pilots black out or die if they exceed about 5 or 10 g’s.)\n30\nConsider the following passage from Alice in Wonderland, in\nwhich Alice has been falling for a long time down a rabbit hole:\nDown, down, down. Would the fall never come to an end? “I\nwonder how many miles I’ve fallen by this time?” she said aloud.\n“I must be getting somewhere near the center of the earth. Let me\nsee: that would be four thousand miles down, I think” (for, you see,\nAlice had learned several things of this sort in her lessons in the\nschoolroom, and though this was not a very good opportunity for\nshowing off her knowledge, as there was no one to listen to her, still\nit was good practice to say it over)...\nAlice doesn’t know much physics, but let’s try to calculate the\namount of time it would take to fall four thousand miles, starting\nfrom rest with an acceleration of 10 m/s2 . This is really only a lower\nlimit; if there really was a hole that deep, the fall would actually\ntake a longer time than the one you calculate, both because there\nis air friction and because gravity gets weaker as you get deeper (at\nthe center of the earth, g is zero, because the earth is pulling you\nequally in every direction at once).\n120\nChapter 3\nAcceleration and Free Fall\nProblems\n121\n122\nChapter 3\nAcceleration and Free Fall\nIsaac Newton\nChapter 4\nForce and Motion\nIf I have seen farther than others, it is because I have stood\non the shoulders of giants.\nNewton, referring to Galileo\nEven as great and skeptical a genius as Galileo was unable to\nmake much progress on the causes of motion. It was not until a generation later that Isaac Newton (1642-1727) was able to attack the\nproblem successfully. In many ways, Newton’s personality was the\nopposite of Galileo’s. Where Galileo agressively publicized his ideas,\n123\nNewton had to be coaxed by his friends into publishing a book on\nhis physical discoveries. Where Galileo’s writing had been popular\nand dramatic, Newton originated the stilted, impersonal style that\nmost people think is standard for scientific writing. (Scientific journals today encourage a less ponderous style, and papers are often\nwritten in the first person.) Galileo’s talent for arousing animosity among the rich and powerful was matched by Newton’s skill at\nmaking himself a popular visitor at court. Galileo narrowly escaped\nbeing burned at the stake, while Newton had the good fortune of being on the winning side of the revolution that replaced King James\nII with William and Mary of Orange, leading to a lucrative post\nrunning the English royal mint.\nNewton discovered the relationship between force and motion,\nand revolutionized our view of the universe by showing that the\nsame physical laws applied to all matter, whether living or nonliving, on or off of our planet’s surface. His book on force and motion,\nthe Mathematical Principles of Natural Philosophy, was uncontradicted by experiment for 200 years, but his other main work,\nOptics, was on the wrong track, asserting that light was composed\nof particles rather than waves. Newton was also an avid alchemist,\na fact that modern scientists would like to forget.\n4.1 Force\nWe need only explain changes in motion, not motion itself.\na / Aristotle said motion had\nto be caused by a force. To\nexplain why an arrow kept flying\nafter the bowstring was no longer\npushing on it, he said the air\nrushed around behind the arrow\nand pushed it forward. We know\nthis is wrong, because an arrow\nshot in a vacuum chamber does\nnot instantly drop to the floor\nas it leaves the bow. Galileo\nand Newton realized that a force\nwould only be needed to change\nthe arrow’s motion, not to make\nits motion continue.\nSo far you’ve studied the measurement of motion in some detail,\nbut not the reasons why a certain object would move in a certain\nway. This chapter deals with the “why” questions. Aristotle’s ideas\nabout the causes of motion were completely wrong, just like all his\nother ideas about physical science, but it will be instructive to start\nwith them, because they amount to a road map of modern students’\nincorrect preconceptions.\nAristotle thought he needed to explain both why motion occurs\nand why motion might change. Newton inherited from Galileo the\nimportant counter-Aristotelian idea that motion needs no explanation, that it is only changes in motion that require a physical cause.\nAristotle’s needlessly complex system gave three reasons for motion:\nNatural motion, such as falling, came from the tendency of\nobjects to go to their “natural” place, on the ground, and\ncome to rest.\nVoluntary motion was the type of motion exhibited by animals, which moved because they chose to.\nForced motion occurred when an object was acted on by some\nother object that made it move.\n124\nChapter 4\nForce and Motion\nMotion changes due to an interaction between two objects.\nIn the Aristotelian theory, natural motion and voluntary motion are one-sided phenomena: the object causes its own motion.\nForced motion is supposed to be a two-sided phenomenon, because\none object imposes its “commands” on another. Where Aristotle\nconceived of some of the phenomena of motion as one-sided and\nothers as two-sided, Newton realized that a change in motion was\nalways a two-sided relationship of a force acting between two physical objects.\nThe one-sided “natural motion” description of falling makes a\ncrucial omission. The acceleration of a falling object is not caused\nby its own “natural” tendencies but by an attractive force between\nit and the planet Earth. Moon rocks brought back to our planet do\nnot “want” to fly back up to the moon because the moon is their\n“natural” place. They fall to the floor when you drop them, just\nlike our homegrown rocks. As we’ll discuss in more detail later in\nthis course, gravitational forces are simply an attraction that occurs\nbetween any two physical objects. Minute gravitational forces can\neven be measured between human-scale objects in the laboratory.\nThe idea of natural motion also explains incorrectly why things\ncome to rest. A basketball rolling across a beach slows to a stop\nbecause it is interacting with the sand via a frictional force, not\nbecause of its own desire to be at rest. If it was on a frictionless\nsurface, it would never slow down. Many of Aristotle’s mistakes\nstemmed from his failure to recognize friction as a force.\nThe concept of voluntary motion is equally flawed. You may\nhave been a little uneasy about it from the start, because it assumes\na clear distinction between living and nonliving things. Today, however, we are used to having the human body likened to a complex\nmachine. In the modern world-view, the border between the living\nand the inanimate is a fuzzy no-man’s land inhabited by viruses,\nprions, and silicon chips. Furthermore, Aristotle’s statement that\nyou can take a step forward “because you choose to” inappropriately\nmixes two levels of explanation. At the physical level of explanation, the reason your body steps forward is because of a frictional\nforce acting between your foot and the floor. If the floor was covered\nwith a puddle of oil, no amount of “choosing to” would enable you\nto take a graceful stride forward.\nb / “Our eyes receive blue\nlight reflected from this painting\nbecause Monet wanted to represent water with the color blue.”\nThis is a valid statement at one\nlevel of explanation, but physics\nworks at the physical level of\nexplanation, in which blue light\ngets to your eyes because it is\nreflected by blue pigments in the\npaint.\nForces can all be measured on the same numerical scale.\nIn the Aristotelian-scholastic tradition, the description of motion as natural, voluntary, or forced was only the broadest level of\nclassification, like splitting animals into birds, reptiles, mammals,\nand amphibians. There might be thousands of types of motion,\neach of which would follow its own rules. Newton’s realization that\nall changes in motion were caused by two-sided interactions made\nit seem that the phenomena might have more in common than had\nSection 4.1\nForce\n125\nbeen apparent. In the Newtonian description, there is only one cause\nfor a change in motion, which we call force. Forces may be of different types, but they all produce changes in motion according to the\nsame rules. Any acceleration that can be produced by a magnetic\nforce can equally well be produced by an appropriately controlled\nstream of water. We can speak of two forces as being equal if they\nproduce the same change in motion when applied in the same situation, which means that they pushed or pulled equally hard in the\nsame direction.\nThe idea of a numerical scale of force and the newton unit were\nintroduced in chapter 0. To recapitulate briefly, a force is when a\npair of objects push or pull on each other, and one newton is the\nforce required to accelerate a 1-kg object from rest to a speed of 1\nm/s in 1 second.\nMore than one force on an object\nAs if we hadn’t kicked poor Aristotle around sufficiently, his\ntheory has another important flaw, which is important to discuss\nbecause it corresponds to an extremely common student misconception. Aristotle conceived of forced motion as a relationship in which\none object was the boss and the other “followed orders.” It therefore would only make sense for an object to experience one force at\na time, because an object couldn’t follow orders from two sources at\nonce. In the Newtonian theory, forces are numbers, not orders, and\nif more than one force acts on an object at once, the result is found\nby adding up all the forces. It is unfortunate that the use of the\nEnglish word “force” has become standard, because to many people\nit suggests that you are “forcing” an object to do something. The\nforce of the earth’s gravity cannot “force” a boat to sink, because\nthere are other forces acting on the boat. Adding them up gives a\ntotal of zero, so the boat accelerates neither up nor down.\nObjects can exert forces on each other at a distance.\nAristotle declared that forces could only act between objects that\nwere touching, probably because he wished to avoid the type of occult speculation that attributed physical phenomena to the influence\nof a distant and invisible pantheon of gods. He was wrong, however,\nas you can observe when a magnet leaps onto your refrigerator or\nwhen the planet earth exerts gravitational forces on objects that are\nin the air. Some types of forces, such as friction, only operate between objects in contact, and are called contact forces. Magnetism,\non the other hand, is an example of a noncontact force. Although\nthe magnetic force gets stronger when the magnet is closer to your\nrefrigerator, touching is not required.\nWeight\nIn physics, an object’s weight, FW , is defined as the earth’s\ngravitational force on it. The SI unit of weight is therefore the\n126\nChapter 4\nForce and Motion\nNewton. People commonly refer to the kilogram as a unit of weight,\nbut the kilogram is a unit of mass, not weight. Note that an object’s\nweight is not a fixed property of that object. Objects weigh more\nin some places than in others, depending on the local strength of\ngravity. It is their mass that always stays the same. A baseball\npitcher who can throw a 90-mile-per-hour fastball on earth would\nnot be able to throw any faster on the moon, because the ball’s\ninertia would still be the same.\nPositive and negative signs of force\nWe’ll start by considering only cases of one-dimensional centerof-mass motion in which all the forces are parallel to the direction of\nmotion, i.e., either directly forward or backward. In one dimension,\nplus and minus signs can be used to indicate directions of forces, as\nshown in figure c. We can then refer generically to addition of forces,\nrather than having to speak sometimes of addition and sometimes of\nsubtraction. We add the forces shown in the figure and get 11 N. In\ngeneral, we should choose a one-dimensional coordinate system with\nits x axis parallel the direction of motion. Forces that point along\nthe positive x axis are positive, and forces in the opposite direction\nare negative. Forces that are not directly along the x axis cannot be\nimmediately incorporated into this scheme, but that’s OK, because\nwe’re avoiding those cases for now.\nDiscussion Questions\nc / Forces are applied to a\nsaxophone.\nIn this example,\npositive signs have been used\nconsistently for forces to the\nright, and negative signs for\nforces to the left. (The forces\nare being applied to different\nplaces on the saxophone, but the\nnumerical value of a force carries\nno information about that.)\nA\nIn chapter 0, I defined 1 N as the force that would accelerate a\n1-kg mass from rest to 1 m/s in 1 s. Anticipating the following section, you\nmight guess that 2 N could be defined as the force that would accelerate\nthe same mass to twice the speed, or twice the mass to the same speed.\nIs there an easier way to define 2 N based on the definition of 1 N?\n4.2 Newton’s First Law\nWe are now prepared to make a more powerful restatement of the\nprinciple of inertia.\nNewton’s first law\nIf the total force on an object is zero, its center of mass continues\nin the same state of motion.\nIn other words, an object initially at rest is predicted to remain\nat rest if the total force on it is zero, and an object in motion remains\nin motion with the same velocity in the same direction. The converse\nof Newton’s first law is also true: if we observe an object moving\nwith constant velocity along a straight line, then the total force on\nit must be zero.\nIn a future physics course or in another textbook, you may encounter the term “net force,” which is simply a synonym for total\nforce.\nSection 4.2\nNewton’s First Law\n127\nWhat happens if the total force on an object is not zero? It\naccelerates. Numerical prediction of the resulting acceleration is the\ntopic of Newton’s second law, which we’ll discuss in the following\nsection.\nThis is the first of Newton’s three laws of motion. It is not\nimportant to memorize which of Newton’s three laws are numbers\none, two, and three. If a future physics teacher asks you something\nlike, “Which of Newton’s laws are you thinking of,” a perfectly acceptable answer is “The one about constant velocity when there’s\nzero total force.” The concepts are more important than any specific formulation of them. Newton wrote in Latin, and I am not\naware of any modern textbook that uses a verbatim translation of\nhis statement of the laws of motion. Clear writing was not in vogue\nin Newton’s day, and he formulated his three laws in terms of a concept now called momentum, only later relating it to the concept of\nforce. Nearly all modern texts, including this one, start with force\nand do momentum later.\nAn elevator\nexample 1\n. An elevator has a weight of 5000 N. Compare the forces that the\ncable must exert to raise it at constant velocity, lower it at constant\nvelocity, and just keep it hanging.\n. In all three cases the cable must pull up with a force of exactly\n5000 N. Most people think you’d need at least a little more than\n5000 N to make it go up, and a little less than 5000 N to let it down,\nbut that’s incorrect. Extra force from the cable is only necessary\nfor speeding the car up when it starts going up or slowing it down\nwhen it finishes going down. Decreased force is needed to speed\nthe car up when it gets going down and to slow it down when it\nfinishes going up. But when the elevator is cruising at constant\nvelocity, Newton’s first law says that you just need to cancel the\nforce of the earth’s gravity.\nTo many students, the statement in the example that the cable’s\nupward force “cancels” the earth’s downward gravitational force implies that there has been a contest, and the cable’s force has won,\nvanquishing the earth’s gravitational force and making it disappear.\nThat is incorrect. Both forces continue to exist, but because they\nadd up numerically to zero, the elevator has no center-of-mass acceleration. We know that both forces continue to exist because they\nboth have side-effects other than their effects on the car’s center-ofmass motion. The force acting between the cable and the car continues to produce tension in the cable and keep the cable taut. The\nearth’s gravitational force continues to keep the passengers (whom\nwe are considering as part of the elevator-object) stuck to the floor\nand to produce internal stresses in the walls of the car, which must\nhold up the floor.\n128\nChapter 4\nForce and Motion\nTerminal velocity for falling objects\nexample 2\n. An object like a feather that is not dense or streamlined does not\nfall with constant acceleration, because air resistance is nonnegligible. In fact, its acceleration tapers off to nearly zero within a\nfraction of a second, and the feather finishes dropping at constant\nspeed (known as its terminal velocity). Why does this happen?\n. Newton’s first law tells us that the total force on the feather must\nhave been reduced to nearly zero after a short time. There are\ntwo forces acting on the feather: a downward gravitational force\nfrom the planet earth, and an upward frictional force from the air.\nAs the feather speeds up, the air friction becomes stronger and\nstronger, and eventually it cancels out the earth’s gravitational\nforce, so the feather just continues with constant velocity without\nspeeding up any more.\nThe situation for a skydiver is exactly analogous. It’s just that the\nskydiver experiences perhaps a million times more gravitational\nforce than the feather, and it is not until she is falling very fast\nthat the force of air friction becomes as strong as the gravitational force. It takes her several seconds to reach terminal velocity, which is on the order of a hundred miles per hour.\nMore general combinations of forces\nIt is too constraining to restrict our attention to cases where\nall the forces lie along the line of the center of mass’s motion. For\none thing, we can’t analyze any case of horizontal motion, since\nany object on earth will be subject to a vertical gravitational force!\nFor instance, when you are driving your car down a straight road,\nthere are both horizontal forces and vertical forces. However, the\nvertical forces have no effect on the center of mass motion, because\nthe road’s upward force simply counteracts the earth’s downward\ngravitational force and keeps the car from sinking into the ground.\nLater in the book we’ll deal with the most general case of many\nforces acting on an object at any angles, using the mathematical\ntechnique of vector addition, but the following slight generalization\nof Newton’s first law allows us to analyze a great many cases of\ninterest:\nSuppose that an object has two sets of forces acting on it, one\nset along the line of the object’s initial motion and another set perpendicular to the first set. If both sets of forces cancel, then the\nobject’s center of mass continues in the same state of motion.\nSection 4.2\nNewton’s First Law\n129\nA passenger riding the subway\nexample 3\n. Describe the forces acting on a person standing in a subway\ntrain that is cruising at constant velocity.\n. No force is necessary to keep the person moving relative to\nthe ground. He will not be swept to the back of the train if the\nfloor is slippery. There are two vertical forces on him, the earth’s\ndownward gravitational force and the floor’s upward force, which\ncancel. There are no horizontal forces on him at all, so of course\nthe total horizontal force is zero.\nForces on a sailboat\nexample 4\n. If a sailboat is cruising at constant velocity with the wind coming\nfrom directly behind it, what must be true about the forces acting\non it?\n. The forces acting on the boat must be canceling each other\nout. The boat is not sinking or leaping into the air, so evidently\nthe vertical forces are canceling out. The vertical forces are the\ndownward gravitational force exerted by the planet earth and an\nupward force from the water.\nThe air is making a forward force on the sail, and if the boat is\nnot accelerating horizontally then the water’s backward frictional\nforce must be canceling it out.\nd / Example 4.\nContrary to Aristotle, more force is not needed in order to maintain\na higher speed. Zero total force is always needed to maintain\nconstant velocity. Consider the following made-up numbers:\nforward force of\nthe wind on the\nsail . . .\nbackward force of\nthe water on the\nhull . . .\ntotal force on the\nboat . . .\nboat moving at\na low, constant\nvelocity\n10,000 N\nboat moving at\na high, constant\nvelocity\n20,000 N\n−10, 000 N\n−20, 000 N\n0N\n0N\nThe faster boat still has zero total force on it. The forward force\non it is greater, and the backward force smaller (more negative),\nbut that’s irrelevant because Newton’s first law has to do with the\ntotal force, not the individual forces.\nThis example is quite analogous to the one about terminal velocity\nof falling objects, since there is a frictional force that increases\nwith speed. After casting off from the dock and raising the sail,\nthe boat will accelerate briefly, and then reach its terminal velocity,\nat which the water’s frictional force has become as great as the\nwind’s force on the sail.\n130\nChapter 4\nForce and Motion\nA car crash\nexample 5\n. If you drive your car into a brick wall, what is the mysterious\nforce that slams your face into the steering wheel?\n. Your surgeon has taken physics, so she is not going to believe\nyour claim that a mysterious force is to blame. She knows that\nyour face was just following Newton’s first law. Immediately after\nyour car hit the wall, the only forces acting on your head were\nthe same canceling-out forces that had existed previously: the\nearth’s downward gravitational force and the upward force from\nyour neck. There were no forward or backward forces on your\nhead, but the car did experience a backward force from the wall,\nso the car slowed down and your face caught up.\nDiscussion Questions\nA\nNewton said that objects continue moving if no forces are acting\non them, but his predecessor Aristotle said that a force was necessary to\nkeep an object moving. Why does Aristotle’s theory seem more plausible,\neven though we now believe it to be wrong? What insight was Aristotle\nmissing about the reason why things seem to slow down naturally? Give\nan example.\nB In the figure what would have to be true about the saxophone’s initial\nmotion if the forces shown were to result in continued one-dimensional\nmotion of its center of mass?\nC\nThis figure requires an ever further generalization of the preceding\ndiscussion. After studying the forces, what does your physical intuition tell\nyou will happen? Can you state in words how to generalize the conditions\nfor one-dimensional motion to include situations like this one?\nDiscussion question B.\n4.3 Newton’s Second Law\nWhat about cases where the total force on an object is not zero,\nso that Newton’s first law doesn’t apply? The object will have an\nacceleration. The way we’ve defined positive and negative signs\nof force and acceleration guarantees that positive forces produce\npositive accelerations, and likewise for negative values. How much\nacceleration will it have? It will clearly depend on both the object’s\nmass and on the amount of force.\nDiscussion question C.\nExperiments with any particular object show that its acceleration is directly proportional to the total force applied to it. This\nmay seem wrong, since we know of many cases where small amounts\nof force fail to move an object at all, and larger forces get it going.\nThis apparent failure of proportionality actually results from forgetting that there is a frictional force in addition to the force we\napply to move the object. The object’s acceleration is exactly proportional to the total force on it, not to any individual force on it.\nIn the absence of friction, even a very tiny force can slowly change\nthe velocity of a very massive object.\nExperiments also show that the acceleration is inversely propor-\nSection 4.3\nNewton’s Second Law\n131\ntional to the object’s mass, and combining these two proportionalities gives the following way of predicting the acceleration of any\nobject:\nNewton’s second law\na = Ftotal /m\n,\nwhere\nm is an object’s mass\nFtotal is the sum of the forces acting on it, and\na is the acceleration of the object’s center of mass.\nWe are presently restricted to the case where the forces of interest\nare parallel to the direction of motion.\nAn accelerating bus\nexample 6\n. A VW bus with a mass of 2000 kg accelerates from 0 to 25 m/s\n(freeway speed) in 34 s. Assuming the acceleration is constant,\nwhat is the total force on the bus?\n. We solve Newton’s second law for Ftotal = ma, and substitute\n∆v /∆t for a, giving\nFtotal = m∆v /∆t\n= (2000 kg)(25 m/s − 0 m/s)/(34 s)\n= 1.5 kN\n.\nA generalization\nAs with the first law, the second law can be easily generalized\nto include a much larger class of interesting situations:\nSuppose an object is being acted on by two sets of forces,\none set lying along the object’s initial direction of motion and\nanother set acting along a perpendicular line. If the forces\nperpendicular to the initial direction of motion cancel out,\nthen the object accelerates along its original line of motion\naccording to a = Ftotal /m.\nThe relationship between mass and weight\nMass is different from weight, but they’re related. An apple’s\nmass tells us how hard it is to change its motion. Its weight measures\nthe strength of the gravitational attraction between the apple and\nthe planet earth. The apple’s weight is less on the moon, but its\n132\nChapter 4\nForce and Motion\nmass is the same. Astronauts assembling the International Space\nStation in zero gravity cannot just pitch massive modules back and\nforth with their bare hands; the modules are weightless, but not\nmassless.\nWe have already seen the experimental evidence that when weight\n(the force of the earth’s gravity) is the only force acting on an object, its acceleration equals the constant g, and g depends on where\nyou are on the surface of the earth, but not on the mass of the object. Applying Newton’s second law then allows us to calculate the\nmagnitude of the gravitational force on any object in terms of its\nmass:\n|FW | = mg\n.\n(The equation only gives the magnitude, i.e. the absolute value, of\nFW , because we’re defining g as a positive number, so it equals the\nabsolute value of a falling object’s acceleration.)\n. Solved problem: Decelerating a car\npage 142, problem 7\nWeight and mass\nexample 7\n. Figure f shows masses of one and two kilograms hung from a\nspring scale, which measures force in units of newtons. Explain\nthe readings.\ne / A simple double-pan balance works by comparing the\nweight forces exerted by the\nearth on the contents of the two\npans. Since the two pans are\nat almost the same location on\nthe earth’s surface, the value\nof g is essentially the same for\neach one, and equality of weight\ntherefore also implies equality of\nmass.\n. Let’s start with the single kilogram. It’s not accelerating, so\nevidently the total force on it is zero: the spring scale’s upward\nforce on it is canceling out the earth’s downward gravitational\nforce. The spring scale tells us how much force it is being obliged\nto supply, but since the two forces are equal in strength, the\nspring scale’s reading can also be interpreted as measuring the\nstrength of the gravitational force, i.e., the weight of the onekilogram mass. The weight of a one-kilogram mass should be\nFW = mg\n= (1.0 kg)(9.8 m/s2 ) = 9.8 N\n,\nand that’s indeed the reading on the spring scale.\nf / Example 7.\nSimilarly for the two-kilogram mass, we have\nFW = mg\n= (2.0 kg)(9.8 m/s2 ) = 19.6 N\n.\nCalculating terminal velocity\nexample 8\n. Experiments show that the force of air friction on a falling object\nsuch as a skydiver or a feather can be approximated fairly well\nwith the equation |Fair | = cρAv 2 , where c is a constant, ρ is the\ndensity of the air, A is the cross-sectional area of the object as\nseen from below, and v is the object’s velocity. Predict the object’s\nterminal velocity, i.e., the final velocity it reaches after a long time.\nSection 4.3\nNewton’s Second Law\n133\n. As the object accelerates, its greater v causes the upward force\nof the air to increase until finally the gravitational force and the\nforce of air friction cancel out, after which the object continues\nat constant velocity. We choose a coordinate system in which\npositive is up, so that the gravitational force is negative and the\nforce of air friction is positive. We want to find the velocity at which\nFair + FW = 0\n,\ncρAv 2 − mg = 0\n.\ni.e.,\nSolving for v gives\nr\nvter minal =\nmg\ncρA\nself-check A\nIt is important to get into the habit of interpreting equations. This may be\ndifficult at first, but eventually you will get used to this kind of reasoning.\np\n(1) Interpret the equation vter minal = mg /c ρA in the case of ρ=0.\n(2) How would the terminal velocity of a 4-cm steel ball compare to that\nof a 1-cm ball?\nx (m)\n10\n20\n30\n40\n50\n60\n70\n80\n90\n100\n(3) In addition to teasing out the mathematical meaning of an equation,\nwe also have to be able to place it in its physical context. How generally\nimportant is this equation?\n. Answer, p. 267\nt (s)\n1.84\n2.86\n3.80\n4.67\n5.53\n6.38\n7.23\n8.10\n8.96\n9.83\nDiscussion Questions\nA\nShow that the Newton can be reexpressed in terms of the three\nbasic mks units as the combination kg·m/s2 .\nB\nWhat is wrong with the following statements?\n(1) “g is the force of gravity.”\n(2) “Mass is a measure of how much space something takes up.”\nC\ng / Discussion question D.\nCriticize the following incorrect statement:\n“If an object is at rest and the total force on it is zero, it stays at rest.\nThere can also be cases where an object is moving and keeps on moving\nwithout having any total force on it, but that can only happen when there’s\nno friction, like in outer space.”\nD Table g gives laser timing data for Ben Johnson’s 100 m dash at the\n1987 World Championship in Rome. (His world record was later revoked\nbecause he tested positive for steroids.) How does the total force on him\nchange over the duration of the race?\n134\nChapter 4\nForce and Motion\n4.4 What Force Is Not\nViolin teachers have to endure their beginning students’ screeching.\nA frown appears on the woodwind teacher’s face as she watches her\nstudent take a breath with an expansion of his ribcage but none\nin his belly. What makes physics teachers cringe is their students’\nverbal statements about forces. Below I have listed several dicta\nabout what force is not.\nForce is not a property of one object.\nA great many of students’ incorrect descriptions of forces could\nbe cured by keeping in mind that a force is an interaction of two\nobjects, not a property of one object.\nIncorrect statement: “That magnet has a lot of force.”\nIf the magnet is one millimeter away from a steel ball bearing, they\nmay exert a very strong attraction on each other, but if they were a\nmeter apart, the force would be virtually undetectable. The magnet’s\nstrength can be rated using certain electrical units (ampere − meters2 ),\nbut not in units of force.\nForce is not a measure of an object’s motion.\nIf force is not a property of a single object, then it cannot be\nused as a measure of the object’s motion.\nIncorrect statement: “The freight train rumbled down the tracks with\nawesome force.”\nForce is not a measure of motion. If the freight train collides with a\nstalled cement truck, then some awesome forces will occur, but if it hits\na fly the force will be small.\nForce is not energy.\nThere are two main approaches to understanding the motion of\nobjects, one based on force and one on a different concept, called energy. The SI unit of energy is the Joule, but you are probably more\nfamiliar with the calorie, used for measuring food’s energy, and the\nkilowatt-hour, the unit the electric company uses for billing you.\nPhysics students’ previous familiarity with calories and kilowatthours is matched by their universal unfamiliarity with measuring\nforces in units of Newtons, but the precise operational definitions of\nthe energy concepts are more complex than those of the force concepts, and textbooks, including this one, almost universally place the\nforce description of physics before the energy description. During\nthe long period after the introduction of force and before the careful\ndefinition of energy, students are therefore vulnerable to situations\nin which, without realizing it, they are imputing the properties of\nenergy to phenomena of force.\nIncorrect statement: “How can my chair be making an upward force on\nmy rear end? It has no power!”\nPower is a concept related to energy, e.g., a 100-watt lightbulb uses\nSection 4.4\nWhat Force Is Not\n135\nup 100 joules per second of energy. When you sit in a chair, no energy\nis used up, so forces can exist between you and the chair without any\nneed for a source of power.\nForce is not stored or used up.\nBecause energy can be stored and used up, people think force\nalso can be stored or used up.\nIncorrect statement: “If you don’t fill up your tank with gas, you’ll run\nout of force.”\nEnergy is what you’ll run out of, not force.\nForces need not be exerted by living things or machines.\nTransforming energy from one form into another usually requires\nsome kind of living or mechanical mechanism. The concept is not\napplicable to forces, which are an interaction between objects, not\na thing to be transferred or transformed.\nIncorrect statement: “How can a wooden bench be making an upward\nforce on my rear end? It doesn’t have any springs or anything inside it.”\nNo springs or other internal mechanisms are required. If the bench\ndidn’t make any force on you, you would obey Newton’s second law and\nfall through it. Evidently it does make a force on you!\nA force is the direct cause of a change in motion.\nI can click a remote control to make my garage door change from\nbeing at rest to being in motion. My finger’s force on the button,\nhowever, was not the force that acted on the door. When we speak\nof a force on an object in physics, we are talking about a force that\nacts directly. Similarly, when you pull a reluctant dog along by its\nleash, the leash and the dog are making forces on each other, not\nyour hand and the dog. The dog is not even touching your hand.\nself-check B\nWhich of the following things can be correctly described in terms of\nforce?\n(1) A nuclear submarine is charging ahead at full steam.\n(2) A nuclear submarine’s propellers spin in the water.\n(3) A nuclear submarine needs to refuel its reactor periodically.\nAnswer, p. 267\n.\nDiscussion Questions\nA\nCriticize the following incorrect statement: “If you shove a book\nacross a table, friction takes away more and more of its force, until finally\nit stops.”\nB\nYou hit a tennis ball against a wall. Explain any and all incorrect\nideas in the following description of the physics involved: “The ball gets\nsome force from you when you hit it, and when it hits the wall, it loses part\nof that force, so it doesn’t bounce back as fast. The muscles in your arm\nare the only things that a force can come from.”\n136\nChapter 4\nForce and Motion\n4.5 Inertial and Noninertial Frames of\nReference\nOne day, you’re driving down the street in your pickup truck, on\nyour way to deliver a bowling ball. The ball is in the back of the\ntruck, enjoying its little jaunt and taking in the fresh air and sunshine. Then you have to slow down because a stop sign is coming\nup. As you brake, you glance in your rearview mirror, and see your\ntrusty companion accelerating toward you. Did some mysterious\nforce push it forward? No, it only seems that way because you and\nthe car are slowing down. The ball is faithfully obeying Newton’s\nfirst law, and as it continues at constant velocity it gets ahead relative to the slowing truck. No forces are acting on it (other than the\nsame canceling-out vertical forces that were always acting on it).1\nThe ball only appeared to violate Newton’s first law because there\nwas something wrong with your frame of reference, which was based\non the truck.\nh / 1. In a frame of reference that\nmoves with the truck, the bowling ball appears to violate Newton’s first law by accelerating despite having no horizontal forces\non it. 2. In an inertial frame of reference, which the surface of the\nearth approximately is, the bowling ball obeys Newton’s first law.\nIt moves equal distances in equal\ntime intervals, i.e., maintains constant velocity. In this frame of\nreference, it is the truck that appears to have a change in velocity, which makes sense, since the\nroad is making a horizontal force\non it.\nHow, then, are we to tell in which frames of reference Newton’s\nlaws are valid? It’s no good to say that we should avoid moving\nframes of reference, because there is no such thing as absolute rest\nor absolute motion. All frames can be considered as being either at\nrest or in motion. According to an observer in India, the strip mall\nthat constituted the frame of reference in panel (b) of the figure\nwas moving along with the earth’s rotation at hundreds of miles per\nhour.\n1\nLet’s assume for simplicity that there is no friction.\nSection 4.5\nInertial and Noninertial Frames of Reference\n137\nThe reason why Newton’s laws fail in the truck’s frame of reference is not because the truck is moving but because it is accelerating.\n(Recall that physicists use the word to refer either to speeding up or\nslowing down.) Newton’s laws were working just fine in the moving\ntruck’s frame of reference as long as the truck was moving at constant velocity. It was only when its speed changed that there was\na problem. How, then, are we to tell which frames are accelerating\nand which are not? What if you claim that your truck is not accelerating, and the sidewalk, the asphalt, and the Burger King are\naccelerating? The way to settle such a dispute is to examine the\nmotion of some object, such as the bowling ball, which we know\nhas zero total force on it. Any frame of reference in which the ball\nappears to obey Newton’s first law is then a valid frame of reference,\nand to an observer in that frame, Mr. Newton assures us that all\nthe other objects in the universe will obey his laws of motion, not\njust the ball.\nValid frames of reference, in which Newton’s laws are obeyed,\nare called inertial frames of reference. Frames of reference that are\nnot inertial are called noninertial frames. In those frames, objects\nviolate the principle of inertia and Newton’s first law. While the\ntruck was moving at constant velocity, both it and the sidewalk\nwere valid inertial frames. The truck became an invalid frame of\nreference when it began changing its velocity.\nYou usually assume the ground under your feet is a perfectly\ninertial frame of reference, and we made that assumption above. It\nisn’t perfectly inertial, however. Its motion through space is quite\ncomplicated, being composed of a part due to the earth’s daily rotation around its own axis, the monthly wobble of the planet caused\nby the moon’s gravity, and the rotation of the earth around the sun.\nSince the accelerations involved are numerically small, the earth is\napproximately a valid inertial frame.\nNoninertial frames are avoided whenever possible, and we will\nseldom, if ever, have occasion to use them in this course. Sometimes,\nhowever, a noninertial frame can be convenient. Naval gunners, for\ninstance, get all their data from radars, human eyeballs, and other\ndetection systems that are moving along with the earth’s surface.\nSince their guns have ranges of many miles, the small discrepancies between their shells’ actual accelerations and the accelerations\npredicted by Newton’s second law can have effects that accumulate\nand become significant. In order to kill the people they want to kill,\nthey have to add small corrections onto the equation a = Ftotal /m.\nDoing their calculations in an inertial frame would allow them to\nuse the usual form of Newton’s second law, but they would have\nto convert all their data into a different frame of reference, which\nwould require cumbersome calculations.\n138\nChapter 4\nForce and Motion\nDiscussion Question\nA\nIf an object has a linear x − t graph in a certain inertial frame,\nwhat is the effect on the graph if we change to a coordinate system with\na different origin? What is the effect if we keep the same origin but reverse the positive direction of the x axis? How about an inertial frame\nmoving alongside the object? What if we describe the object’s motion in\na noninertial frame?\nSection 4.5\nInertial and Noninertial Frames of Reference\n139\nSummary\nSelected Vocabulary\nweight . . . . . . . the force of gravity on an object, equal to mg\ninertial frame . . a frame of reference that is not accelerating,\none in which Newton’s first law is true\nnoninertial frame an accelerating frame of reference, in which\nNewton’s first law is violated\nNotation\nFW . . . . . . . .\nweight\nOther Terminology and Notation\nnet force . . . . . another way of saying “total force”\nSummary\nNewton’s first law of motion states that if all the forces on an\nobject cancel each other out, then the object continues in the same\nstate of motion. This is essentially a more refined version of Galileo’s\nprinciple of inertia, which did not refer to a numerical scale of force.\nNewton’s second law of motion allows the prediction of an object’s acceleration given its mass and the total force on it, acm =\nFtotal /m. This is only the one-dimensional version of the law; the\nfull-three dimensional treatment will come in chapter 8, Vectors.\nWithout the vector techniques, we can still say that the situation\nremains unchanged by including an additional set of vectors that\ncancel among themselves, even if they are not in the direction of\nmotion.\nNewton’s laws of motion are only true in frames of reference that\nare not accelerating, known as inertial frames.\nExploring Further\nIsaac Newton: The Last Sorcerer, Michael White. An excellent biography of Newton that brings us closer to the real man.\n140\nChapter 4\nForce and Motion\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nAn object is observed to be moving at constant speed in a\ncertain direction. Can you conclude that no forces are acting on it?\nExplain. [Based on a problem by Serway and Faughn.]\n2\nA car is normally capable of an acceleration of 3 m/s2 . If it\nis towing a trailer with half as much mass as the car itself, what acceleration can it achieve? [Based on a problem from PSSC Physics.]\n3\n(a) Let T be the maximum tension that an elevator’s cable can\nwithstand without breaking, i.e., the maximum force it can exert.\nIf the motor is programmed to give the car an acceleration a, what\nis the maximum mass that the car can have, including passengers,\n√\nif the cable is not to break?\n(b) Interpret the equation you derived in the special cases of a = 0\nand of a downward acceleration of magnitude g. (“Interpret” means\nto analyze the behavior of the equation, and connect that to reality,\nas in the self-check on page 134.)\n4\nA helicopter of mass m is taking off vertically. The only forces\nacting on it are the earth’s gravitational force and the force, Fair ,\nof the air pushing up on the propeller blades.\n(a) If the helicopter lifts off at t = 0, what is its vertical speed at\ntime t?\n(b) Plug numbers into your equation from part a, using m = 2300\nkg, Fair = 27000 N, and t = 4.0 s.\n5\nIn the 1964 Olympics in Tokyo, the best men’s high jump was\n2.18 m. Four years later in Mexico City, the gold medal in the same\nevent was for a jump of 2.24 m. Because of Mexico City’s altitude\n(2400 m), the acceleration of gravity there is lower than that in\nTokyo by about 0.01 m/s2 . Suppose a high-jumper has a mass of\n72 kg.\n(a) Compare his mass and weight in the two locations.\n(b) Assume that he is able to jump with the same initial vertical\nvelocity in both locations, and that all other conditions are the same\nexcept for gravity. How much higher should he be able to jump in\n√\nMexico City?\n(Actually, the reason for the big change between ’64 and ’68 was the\nintroduction of the “Fosbury flop.”)\n?\nProblems\n141\n6\nA blimp is initially at rest, hovering, when at t = 0 the pilot\nturns on the motor of the propeller. The motor cannot instantly\nget the propeller going, but the propeller speeds up steadily. The\nsteadily increasing force between the air and the propeller is given\nby the equation F = kt, where k is a constant. If the mass of the\nblimp is m, find its position as a function of time. (Assume that\nduring the period of time you’re dealing with, the blimp is not yet\nmoving fast enough to cause a significant backward force due to\nR air\nresistance.)\nProblem 6.\n7\nA car is accelerating forward along a straight road. If the force\nof the road on the car’s wheels, pushing it forward, is a constant 3.0\nkN, and the car’s mass is 1000 kg, then how long will the car take\nto go from 20 m/s to 50 m/s?\n. Solution, p. 273\n8\nSome garden shears are like a pair of scissors: one sharp blade\nslices past another. In the “anvil” type, however, a sharp blade\npresses against a flat one rather than going past it. A gardening\nbook says that for people who are not very physically strong, the\nanvil type can make it easier to cut tough branches, because it\nconcentrates the force on one side. Evaluate this claim based on\nNewton’s laws. [Hint: Consider the forces acting on the branch,\nand the motion of the branch.]\n9\nA uranium atom deep in the earth spits out an alpha particle.\nAn alpha particle is a fragment of an atom. This alpha particle has\ninitial speed v, and travels a distance d before stopping in the earth.\n(a) Find the force, F , that acted on the particle, in terms of v, d,\nand its mass, m. Don’t plug in any numbers yet. Assume that the\n√\nforce was constant.\n(b) Show that your answer has the right units.\n(c) Discuss how your answer to part a depends on all three variables,\nand show that it makes sense. That is, for each variable, discuss\nwhat would happen to the result if you changed it while keeping the\nother two variables constant. Would a bigger value give a smaller\nresult, or a bigger result? Once you’ve figured out this mathematical\nrelationship, show that it makes sense physically.\n(d) Evaluate your result for m = 6.7 × 10−27 kg, v = 2.0 × 104 km/s,\n√\nand d = 0.71 mm.\n142\nChapter 4\nForce and Motion\n10\nYou are given a large sealed box, and are not allowed to open\nit. Which of the following experiments measure its mass, and which\nmeasure its weight? [Hint: Which experiments would give different\nresults on the moon?]\n(a) Put it on a frozen lake, throw a rock at it, and see how fast it\nscoots away after being hit.\n(b) Drop it from a third-floor balcony, and measure how loud the\nsound is when it hits the ground.\n(c) As shown in the figure, connect it with a spring to the wall, and\nwatch it vibrate.\n. Solution, p. 273\nProblem 10, part c.\n11\nWhile escaping from the palace of the evil Martian emperor, Sally Spacehound jumps from a tower of height h down to\nthe ground. Ordinarily the fall would be fatal, but she fires her\nblaster rifle straight down, producing an upward force of magnitude\nFB . This force is insufficient to levitate her, but it does cancel out\nsome of the force of gravity. During the time t that she is falling,\nSally is unfortunately exposed to fire from the emperor’s minions,\nand can’t dodge their shots. Let m be her mass, and g the strength\nof gravity on Mars.\n(a) Find the time t in terms of the other variables.\n(b) Check the units of your answer to part a.\n(b) For sufficiently large values of FB , your answer to part a be√\ncomes nonsense — explain what’s going on.\n12\nWhen I cook rice, some of the dry grains always stick to the\nmeasuring cup. To get them out, I turn the measuring cup upsidedown, and hit the back of the cup with my hand. Explain why this\nworks, and why its success depends on hitting the cup hard enough.\nProblems\n143\n144\nChapter 4\nForce and Motion\nWhat forces act on the girl?\nChapter 5\nAnalysis of Forces\n5.1 Newton’s Third Law\nNewton created the modern concept of force starting from his insight\nthat all the effects that govern motion are interactions between two\nobjects: unlike the Aristotelian theory, Newtonian physics has no\nphenomena in which an object changes its own motion.\nIs one object always the “order-giver” and the other the “order-\n145\nfollower”? As an example, consider a batter hitting a baseball. The\nbat definitely exerts a large force on the ball, because the ball accelerates drastically. But if you have ever hit a baseball, you also\nknow that the ball makes a force on the bat — often with painful\nresults if your technique is as bad as mine!\na / Two magnets exert forces\non each other.\nb / Two people’s hands\nforces on each other.\nexert\nc / Rockets work by pushing\nexhaust gases out the back.\nNewton’s third law says that if the\nrocket exerts a backward force\non the gases, the gases must\nmake an equal forward force on\nthe rocket. Rocket engines can\nfunction above the atmosphere,\nunlike propellers and jets, which\nwork by pushing against the\nsurrounding air.\n146\nChapter 5\nHow does the ball’s force on the bat compare with the bat’s\nforce on the ball? The bat’s acceleration is not as spectacular as\nthe ball’s, but maybe we shouldn’t expect it to be, since the bat’s\nmass is much greater. In fact, careful measurements of both objects’\nmasses and accelerations would show that mball aball is very nearly\nequal to −mbat abat , which suggests that the ball’s force on the bat\nis of the same magnitude as the bat’s force on the ball, but in the\nopposite direction.\nFigures a and b show two somewhat more practical laboratory\nexperiments for investigating this issue accurately and without too\nmuch interference from extraneous forces.\nIn experiment a, a large magnet and a small magnet are weighed\nseparately, and then one magnet is hung from the pan of the top\nbalance so that it is directly above the other magnet. There is an\nattraction between the two magnets, causing the reading on the top\nscale to increase and the reading on the bottom scale to decrease.\nThe large magnet is more “powerful” in the sense that it can pick\nup a heavier paperclip from the same distance, so many people have\na strong expectation that one scale’s reading will change by a far\ndifferent amount than the other. Instead, we find that the two\nchanges are equal in magnitude but opposite in direction: the force\nof the bottom magnet pulling down on the top one has the same\nstrength as the force of the top one pulling up on the bottom one.\nIn experiment b, two people pull on two spring scales. Regardless\nof who tries to pull harder, the two forces as measured on the spring\nscales are equal. Interposing the two spring scales is necessary in\norder to measure the forces, but the outcome is not some artificial\nresult of the scales’ interactions with each other. If one person slaps\nanother hard on the hand, the slapper’s hand hurts just as much\nas the slappee’s, and it doesn’t matter if the recipient of the slap\ntries to be inactive. (Punching someone in the mouth causes just\nas much force on the fist as on the lips. It’s just that the lips are\nmore delicate. The forces are equal, but not the levels of pain and\ninjury.)\nNewton, after observing a series of results such as these, decided\nthat there must be a fundamental law of nature at work:\nNewton’s third law\nForces occur in equal and opposite pairs: whenever object A exerts\na force on object B, object B must also be exerting a force on object\nA. The two forces are equal in magnitude and opposite in direction.\nAnalysis of Forces\nIn one-dimensional situations, we can use plus and minus signs to\nindicate the directions of forces, and Newton’s third law can be\nwritten succinctly as FA on B = −FB on A .\nself-check A\nFigure d analyzes swimming using Newton’s third law. Do a similar\nanalysis for a sprinter leaving the starting line.\n. Answer, p. 267\nThere is no cause and effect relationship between the two forces\nin Newton’s third law. There is no “original” force, and neither one\nis a response to the other. The pair of forces is a relationship, like\nmarriage, not a back-and-forth process like a tennis match. Newton\ncame up with the third law as a generalization about all the types of\nforces with which he was familiar, such as frictional and gravitational\nforces. When later physicists discovered a new type force, such\nas the force that holds atomic nuclei together, they had to check\nwhether it obeyed Newton’s third law. So far, no violation of the\nthird law has ever been discovered, whereas the first and second\nlaws were shown to have limitations by Einstein and the pioneers of\natomic physics.\nThe English vocabulary for describing forces is unfortunately\nrooted in Aristotelianism, and often implies incorrectly that forces\nare one-way relationships. It is unfortunate that a half-truth such as\n“the table exerts an upward force on the book” is so easily expressed,\nwhile a more complete and correct description ends up sounding\nawkward or strange: “the table and the book interact via a force,”\nor “the table and book participate in a force.”\nTo students, it often sounds as though Newton’s third law implies nothing could ever change its motion, since the two equal and\nopposite forces would always cancel. The two forces, however, are\nalways on two different objects, so it doesn’t make sense to add\nthem in the first place — we only add forces that are acting on the\nsame object. If two objects are interacting via a force and no other\nforces are involved, then both objects will accelerate — in opposite\ndirections!\nd / A swimmer doing the breast\nstroke pushes backward against\nthe water. By Newton’s third law,\nthe water pushes forward on her.\ne / Newton’s third law does\nnot mean that forces always\ncancel out so that nothing can\never move. If these two figure\nskaters, initially at rest, push\nagainst each other, they will both\nmove.\nA mnemonic for using Newton’s third law correctly\nMnemonics are tricks for memorizing things. For instance, the\nmusical notes that lie between the lines on the treble clef spell the\nword FACE, which is easy to remember. Many people use the\nmnemonic “SOHCAHTOA” to remember the definitions of the sine,\ncosine, and tangent in trigonometry. I have my own modest offering,\nPOFOSTITO, which I hope will make it into the mnemonics hall of\nfame. It’s a way to avoid some of the most common problems with\napplying Newton’s third law correctly:\nSection 5.1\nNewton’s Third Law\n147\nf / It doesn’t make sense for the\nman to talk about using the\nwoman’s money to cancel out his\nbar tab, because there is no good\nreason to combine his debts and\nher assets. Similarly, it doesn’t\nmake sense to refer to the equal\nand opposite forces of Newton’s\nthird law as canceling. It only\nmakes sense to add up forces\nthat are acting on the same object, whereas two forces related\nto each other by Newton’s third\nlaw are always acting on two different objects.\nA book lying on a table\nexample 1\n. A book is lying on a table. What force is the Newton’s-third-law\npartner of the earth’s gravitational force on the book?\nAnswer: Newton’s third law works like “B on A, A on B,” so the\npartner must be the book’s gravitational force pulling upward on\nthe planet earth. Yes, there is such a force! No, it does not cause\nthe earth to do anything noticeable.\nIncorrect answer: The table’s upward force on the book is the\nNewton’s-third-law partner of the earth’s gravitational force on the\nbook.\nThis answer violates two out of three of the commandments of\nPOFOSTITO. The forces are not of the same type, because the\ntable’s upward force on the book is not gravitational. Also, three\nobjects are involved instead of two: the book, the table, and the\nplanet earth.\nPushing a box up a hill\nexample 2\n. A person is pushing a box up a hill. What force is related by\nNewton’s third law to the person’s force on the box?\n. The box’s force on the person.\nIncorrect answer: The person’s force on the box is opposed by\nfriction, and also by gravity.\n148\nChapter 5\nAnalysis of Forces\nThis answer fails all three parts of the POFOSTITO test, the most\nobvious of which is that three forces are referred to instead of a\npair.\n. Solved problem: More about example 2\npage 171, problem 20\n. Solved problem: Why did it accelerate?\npage 171, problem 18\nOptional Topic: Newton’s Third Law and Action at a Distance\nNewton’s third law is completely symmetric in the sense that neither\nforce constitutes a delayed response to the other. Newton’s third law\ndoes not even mention time, and the forces are supposed to agree at\nany given instant. This creates an interesting situation when it comes\nto noncontact forces. Suppose two people are holding magnets, and\nwhen one person waves or wiggles her magnet, the other person feels\nan effect on his. In this way they can send signals to each other from\nopposite sides of a wall, and if Newton’s third law is correct, it would\nseem that the signals are transmitted instantly, with no time lag. The\nsignals are indeed transmitted quite quickly, but experiments with electronically controlled magnets show that the signals do not leap the gap\ninstantly: they travel at the same speed as light, which is an extremely\nhigh speed but not an infinite one.\nIs this a contradiction to Newton’s third law? Not really. According to current theories, there are no true noncontact forces. Action at\na distance does not exist. Although it appears that the wiggling of one\nmagnet affects the other with no need for anything to be in contact with\nanything, what really happens is that wiggling a magnet unleashes a\nshower of tiny particles called photons. The magnet shoves the photons out with a kick, and receives a kick in return, in strict obedience to\nNewton’s third law. The photons fly out in all directions, and the ones\nthat hit the other magnet then interact with it, again obeying Newton’s\nthird law.\nPhotons are nothing exotic, really. Light is made of photons, but our\neyes receive such huge numbers of photons that we do not perceive\nthem individually. The photons you would make by wiggling a magnet\nwith your hand would be of a “color” that you cannot see, far off the red\nend of the rainbow. Book 6 in this series describes the evidence for the\nphoton model of light.\nDiscussion Questions\nA\nWhen you fire a gun, the exploding gases push outward in all\ndirections, causing the bullet to accelerate down the barrel. What thirdlaw pairs are involved? [Hint: Remember that the gases themselves are\nan object.]\nB\nTam Anh grabs Sarah by the hand and tries to pull her. She tries\nto remain standing without moving. A student analyzes the situation as\nfollows. “If Tam Anh’s force on Sarah is greater than her force on him,\nhe can get her to move. Otherwise, she’ll be able to stay where she is.”\nWhat’s wrong with this analysis?\nC\nYou hit a tennis ball against a wall. Explain any and all incorrect\nideas in the following description of the physics involved: “According to\nNewton’s third law, there has to be a force opposite to your force on the\nball. The opposite force is the ball’s mass, which resists acceleration, and\nSection 5.1\nNewton’s Third Law\n149\nalso air resistance.”\n5.2 Classification and Behavior of Forces\ng/A\nscientific\nsystem.\nclassification\nOne of the most basic and important tasks of physics is to classify\nthe forces of nature. I have already referred informally to “types” of\nforces such as friction, magnetism, gravitational forces, and so on.\nClassification systems are creations of the human mind, so there is\nalways some degree of arbitrariness in them. For one thing, the level\nof detail that is appropriate for a classification system depends on\nwhat you’re trying to find out. Some linguists, the “lumpers,” like to\nemphasize the similarities among languages, and a few extremists\nhave even tried to find signs of similarities between words in languages as different as English and Chinese, lumping the world’s languages into only a few large groups. Other linguists, the “splitters,”\nmight be more interested in studying the differences in pronunciation between English speakers in New York and Connecticut. The\nsplitters call the lumpers sloppy, but the lumpers say that science\nisn’t worthwhile unless it can find broad, simple patterns within the\nseemingly complex universe.\nScientific classification systems are also usually compromises between practicality and naturalness. An example is the question of\nhow to classify flowering plants. Most people think that biological\nclassification is about discovering new species, naming them, and\nclassifying them in the class-order-family-genus-species system according to guidelines set long ago. In reality, the whole system is in\na constant state of flux and controversy. One very practical way of\nclassifying flowering plants is according to whether their petals are\nseparate or joined into a tube or cone — the criterion is so clear that\nit can be applied to a plant seen from across the street. But here\npracticality conflicts with naturalness. For instance, the begonia has\nseparate petals and the pumpkin has joined petals, but they are so\nsimilar in so many other ways that they are usually placed within\nthe same order. Some taxonomists have come up with classification\ncriteria that they claim correspond more naturally to the apparent\nrelationships among plants, without having to make special exceptions, but these may be far less practical, requiring for instance the\nexamination of pollen grains under an electron microscope.\nIn physics, there are two main systems of classification for forces.\nAt this point in the course, you are going to learn one that is very\npractical and easy to use, and that splits the forces up into a relatively large number of types: seven very common ones that we’ll\ndiscuss explicitly in this chapter, plus perhaps ten less important\nones such as surface tension, which we will not bother with right\nnow.\nProfessional physicists, however, are obsessed with finding simple patterns, so recognizing as many as fifteen or twenty types of\n150\nChapter 5\nAnalysis of Forces\nforces strikes them as distasteful and overly complex. Since about\nthe year 1900, physics has been on an aggressive program to discover\nways in which these many seemingly different types of forces arise\nfrom a smaller number of fundamental ones. For instance, when you\npress your hands together, the force that keeps them from passing\nthrough each other may seem to have nothing to do with electricity, but at the atomic level, it actually does arise from electrical\nrepulsion between atoms. By about 1950, all the forces of nature\nhad been explained as arising from four fundamental types of forces\nat the atomic and nuclear level, and the lumping-together process\ndidn’t stop there. By the 1960’s the length of the list had been\nreduced to three, and some theorists even believe that they may be\nable to reduce it to two or one. Although the unification of the forces\nof nature is one of the most beautiful and important achievements\nof physics, it makes much more sense to start this course with the\nmore practical and easy system of classification. The unified system of four forces will be one of the highlights of the end of your\nintroductory physics sequence.\nThe practical classification scheme which concerns us now can\nbe laid out in the form of the tree shown in figure h. The most\nspecific types of forces are shown at the tips of the branches, and\nit is these types of forces that are referred to in the POFOSTITO\nmnemonic. For example, electrical and magnetic forces belong to\nthe same general group, but Newton’s third law would never relate\nan electrical force to a magnetic force.\nThe broadest distinction is that between contact and noncontact\nforces, which has been discussed in the previous chapter. Among\nthe contact forces, we distinguish between those that involve solids\nonly and those that have to do with fluids, a term used in physics to\ninclude both gases and liquids. The terms “repulsive,” “attractive,”\nand “oblique” refer to the directions of the forces.\n• Repulsive forces are those that tend to push the two participating objects away from each other. More specifically, a\nrepulsive contact force acts perpendicular to the surfaces at\nwhich the two objects touch, and a repulsive noncontact force\nacts along the line between the two objects.\n• Attractive forces pull the two objects toward one another, i.e.,\nthey act along the same line as repulsive forces, but in the\nopposite direction.\n• Oblique forces are those that act at some other angle.\nIt should not be necessary to memorize this diagram by rote.\nIt is better to reinforce your memory of this system by calling to\nSection 5.2\nClassification and Behavior of Forces\n151\nh / A practical classification scheme for forces.\nmind your commonsense knowledge of certain ordinary phenomena.\nFor instance, we know that the gravitational attraction between us\nand the planet earth will act even if our feet momentarily leave the\nground, and that although magnets have mass and are affected by\ngravity, most objects that have mass are nonmagnetic.\nThis diagram is meant to be as simple as possible while including\nmost of the forces we deal with in everyday life. If you were an insect,\nyou would be much more interested in the force of surface tension,\nwhich allowed you to walk on water. I have not included the nuclear\nforces, which are responsible for holding the nuclei of atoms, because\nthey are not evident in everyday life.\nYou should not be afraid to invent your own names for types of\nforces that do not fit into the diagram. For instance, the force that\nholds a piece of tape to the wall has been left off of the tree, and if\nyou were analyzing a situation involving scotch tape, you would be\nabsolutely right to refer to it by some commonsense name such as\n“sticky force.”\n152\nChapter 5\nAnalysis of Forces\nOn the other hand, if you are having trouble classifying a certain\nforce, you should also consider whether it is a force at all. For\ninstance, if someone asks you to classify the force that the earth has\nbecause of its rotation, you would have great difficulty creating a\nplace for it on the diagram. That’s because it’s a type of motion,\nnot a type of force!\nNormal forces\nA normal force, FN , is a force that keeps one solid object from\npassing through another. “Normal” is simply a fancy word for “perpendicular,” meaning that the force is perpendicular to the surface\nof contact. Intuitively, it seems the normal force magically adjusts\nitself to provide whatever force is needed to keep the objects from\noccupying the same space. If your muscles press your hands together\ngently, there is a gentle normal force. Press harder, and the normal\nforce gets stronger. How does the normal force know how strong to\nbe? The answer is that the harder you jam your hands together,\nthe more compressed your flesh becomes. Your flesh is acting like\na spring: more force is required to compress it more. The same is\ntrue when you push on a wall. The wall flexes imperceptibly in proportion to your force on it. If you exerted enough force, would it be\npossible for two objects to pass through each other? No, typically\nthe result is simply to strain the objects so much that one of them\nbreaks.\nGravitational forces\nAs we’ll discuss in more detail later in the course, a gravitational\nforce exists between any two things that have mass. In everyday life,\nthe gravitational force between two cars or two people is negligible,\nso the only noticeable gravitational forces are the ones between the\nearth and various human-scale objects. We refer to these planetearth-induced gravitational forces as weight forces, and as we have\nalready seen, their magnitude is given by |FW | = mg.\n. Solved problem: Weight and mass\npage 172, problem 26\nStatic and kinetic friction\nIf you have pushed a refrigerator across a kitchen floor, you have\nfelt a certain series of sensations. At first, you gradually increased\nyour force on the refrigerator, but it didn’t move. Finally, you supplied enough force to unstick the fridge, and there was a sudden jerk\nas the fridge started moving. Once the fridge is unstuck, you can\nreduce your force significantly and still keep it moving.\nWhile you were gradually increasing your force, the floor’s frictional force on the fridge increased in response. The two forces on\nthe fridge canceled, and the fridge didn’t accelerate. How did the\nfloor know how to respond with just the right amount of force? Figure i shows one possible model of friction that explains this behavior.\nSection 5.2\ni / A model that correctly explains many properties of friction.\nThe microscopic bumps and\nholes in two surfaces dig into\neach other, causing a frictional\nforce.\nClassification and Behavior of Forces\n153\n(A scientific model is a description that we expect to be incomplete,\napproximate, or unrealistic in some ways, but that nevertheless succeeds in explaining a variety of phenomena.) Figure i/1 shows a\nmicroscopic view of the tiny bumps and holes in the surfaces of the\nfloor and the refrigerator. The weight of the fridge presses the two\nsurfaces together, and some of the bumps in one surface will settle\nas deeply as possible into some of the holes in the other surface. In\ni/2, your leftward force on the fridge has caused it to ride up a little\nhigher on the bump in the floor labeled with a small arrow. Still\nmore force is needed to get the fridge over the bump and allow it to\nstart moving. Of course, this is occurring simultaneously at millions\nof places on the two surfaces.\nOnce you had gotten the fridge moving at constant speed, you\nfound that you needed to exert less force on it. Since zero total force\nis needed to make an object move with constant velocity, the floor’s\nrightward frictional force on the fridge has apparently decreased\nsomewhat, making it easier for you to cancel it out. Our model also\ngives a plausible explanation for this fact: as the surfaces slide past\neach other, they don’t have time to settle down and mesh with one\nanother, so there is less friction.\nj / Static friction: the tray doesn’t\nslip on the waiter’s fingers.\nEven though this model is intuitively appealing and fairly successful, it should not be taken too seriously, and in some situations\nit is misleading. For instance, fancy racing bikes these days are\nmade with smooth tires that have no tread — contrary to what\nwe’d expect from our model, this does not cause any decrease in\nfriction. Machinists know that two very smooth and clean metal\nsurfaces may stick to each other firmly and be very difficult to slide\napart. This cannot be explained in our model, but makes more\nsense in terms of a model in which friction is described as arising\nfrom chemical bonds between the atoms of the two surfaces at their\npoints of contact: very flat surfaces allow more atoms to come in\ncontact.\nSince friction changes its behavior dramatically once the surfaces come unstuck, we define two separate types of frictional forces.\nStatic friction is friction that occurs between surfaces that are not\nslipping over each other. Slipping surfaces experience kinetic friction. “Kinetic” means having to do with motion. The forces of\nstatic and kinetic friction, notated Fs and Fk , are always parallel to\nthe surface of contact between the two objects.\nself-check B\n1. When a baseball player slides in to a base, is the friction static, or\nkinetic?\n2. A mattress stays on the roof of a slowly accelerating car. Is the\nfriction static, or kinetic?\nk / Kinetic\nskids.\n154\nfriction:\nthe\ncar\n3. Does static friction create heat? Kinetic friction?\n. Answer, p. 267\nThe maximum possible force of static friction depends on what\nChapter 5\nAnalysis of Forces\nkinds of surfaces they are, and also on how hard they are being\npressed together. The approximate mathematical relationships can\nbe expressed as follows:\nFs = −Fapplied\n,\nwhen |Fapplied | < µs |FN |\n,\nwhere µs is a unitless number, called the coefficient of static friction,\nwhich depends on what kinds of surfaces they are. The maximum\nforce that static friction can supply, µs |FN |, represents the boundary\nbetween static and kinetic friction. It depends on the normal force,\nwhich is numerically equal to whatever force is pressing the two\nsurfaces together. In terms of our model, if the two surfaces are\nbeing pressed together more firmly, a greater sideways force will be\nrequired in order to make the irregularities in the surfaces ride up\nand over each other.\nNote that just because we use an adjective such as “applied” to\nrefer to a force, that doesn’t mean that there is some special type\nof force called the “applied force.” The applied force could be any\ntype of force, or it could be the sum of more than one force trying\nto make an object move.\nThe force of kinetic friction on each of the two objects is in the\ndirection that resists the slippage of the surfaces. Its magnitude is\nusually well approximated as\n|Fk | = µk |FN |\nwhere µk is the coefficient of kinetic friction. Kinetic friction is\nusually more or less independent of velocity.\nl / We choose a coordinate system in which the applied force,\ni.e., the force trying to move the\nobjects, is positive. The friction\nforce is then negative, since it is\nin the opposite direction. As you\nincrease the applied force, the\nforce of static friction increases to\nmatch it and cancel it out, until the\nmaximum force of static friction is\nsurpassed. The surfaces then begin slipping past each other, and\nthe friction force becomes smaller\nin absolute value.\nself-check C\nCan a frictionless surface exert a normal force? Can a frictional force\nexist without a normal force?\n. Answer, p. 267\nIf you try to accelerate or decelerate your car too quickly, the\nforces between your wheels and the road become too great, and they\nSection 5.2\nClassification and Behavior of Forces\n155\nbegin slipping. This is not good, because kinetic friction is weaker\nthan static friction, resulting in less control. Also, if this occurs\nwhile you are turning, the car’s handling changes abruptly because\nthe kinetic friction force is in a different direction than the static\nfriction force had been: contrary to the car’s direction of motion,\nrather than contrary to the forces applied to the tire.\nMost people respond with disbelief when told of the experimental evidence that both static and kinetic friction are approximately\nindependent of the amount of surface area in contact. Even after\ndoing a hands-on exercise with spring scales to show that it is true,\nmany students are unwilling to believe their own observations, and\ninsist that bigger tires “give more traction.” In fact, the main reason why you would not want to put small tires on a big heavy car\nis that the tires would burst!\nAlthough many people expect that friction would be proportional to surface area, such a proportionality would make predictions\ncontrary to many everyday observations. A dog’s feet, for example,\nhave very little surface area in contact with the ground compared\nto a human’s feet, and yet we know that a dog can often win a\ntug-of-war with a person.\nThe reason a smaller surface area does not lead to less friction\nis that the force between the two surfaces is more concentrated,\ncausing their bumps and holes to dig into each other more deeply.\nself-check D\nFind the direction of each of the forces in figure m.\n. Answer, p. 267\nm / 1. The cliff’s normal force on\nthe climber’s feet. 2. The track’s\nstatic frictional force on the wheel\nof the accelerating dragster. 3.\nThe ball’s normal force on the\nbat.\nLocomotives\nexample 3\nLooking at a picture of a locomotive, n, we notice two obvious\nthings that are different from an automobile. Where a car typically has two drive wheels, a locomotive normally has many —\nten in this example. (Some also have smaller, unpowered wheels\nin front of and behind the drive wheels, but this example doesn’t.)\nAlso, cars these days are generally built to be as light as possible for their size, whereas locomotives are very massive, and no\neffort seems to be made to keep their weight low. (The steam\nlocomotive in the photo is from about 1900, but this is true even\n156\nChapter 5\nAnalysis of Forces\nfor modern diesel and electric trains.)\nn / Example 3.\nThe reason locomotives are built to be so heavy is for traction.\nThe upward normal force of the rails on the wheels, FN , cancels\nthe downward force of gravity, FW , so ignoring plus and minus\nsigns, these two forces are equal in absolute value, FN = FW .\nGiven this amount of normal force, the maximum force of static\nfriction is Fs = µs FN = µs FW . This static frictional force, of the\nrails pushing forward on the wheels, is the only force that can\naccelerate the train, pull it uphill, or cancel out the force of air\nresistance while cruising at constant speed. The coefficient of\nstatic friction for steel on steel is about 1/4, so no locomotive can\npull with a force greater than about 1/4 of its own weight. If the\nengine is capable of supplying more than that amount of force, the\nresult will be simply to break static friction and spin the wheels.\nThe reason this is all so different from the situation with a car is\nthat a car isn’t pulling something else. If you put extra weight in\na car, you improve the traction, but you also increase the inertia\nof the car, and make it just as hard to accelerate. In a train, the\ninertia is almost all in the cars being pulled, not in the locomotive.\nThe other fact we have to explain is the large number of driving wheels. First, we have to realize that increasing the number of driving wheels neither increases nor decreases the total\namount of static friction, because static friction is independent of\nthe amount of surface area in contact. (The reason four-wheeldrive is good in a car is that if one or more of the wheels is slipping on ice or in mud, the other wheels may still have traction.\nThis isn’t typically an issue for a train, since all the wheels experience the same conditions.) The advantage of having more driving\nwheels on a train is that it allows us to increase the weight of the\nlocomotive without crushing the rails, or damaging bridges.\nFluid friction\nTry to drive a nail into a waterfall and you will be confronted\nwith the main difference between solid friction and fluid friction.\nFluid friction is purely kinetic; there is no static fluid friction. The\nnail in the waterfall may tend to get dragged along by the water\nflowing past it, but it does not stick in the water. The same is true\nfor gases such as air: recall that we are using the word “fluid” to\ninclude both gases and liquids.\nSection 5.2\nClassification and Behavior of Forces\n157\no / The wheelbases of the\nHummer H3 and the Toyota Prius\nare surprisingly similar, differing\nby only 10%. The main difference\nin shape is that the Hummer is\nmuch taller and wider. It presents\na much greater cross-sectional\narea to the wind, and this is the\nmain reason that the Prius uses\nonly about 40% as much gas on\nthe freeway.\nUnlike kinetic friction between solids, fluid friction increases\nrapidly with velocity. It also depends on the shape of the object,\nwhich is why a fighter jet is more streamlined than a Model T. For\nobjects of the same shape but different sizes, fluid friction typically\nscales up with the cross-sectional area of the object, which is one\nof the main reasons that an SUV gets worse mileage on the freeway\nthan a compact car.\nDiscussion Questions\nA\nA student states that when he tries to push his refrigerator, the\nreason it won’t move is because Newton’s third law says there’s an equal\nand opposite frictional force pushing back. After all, the static friction force\nis equal and opposite to the applied force. How would you convince him\nhe is wrong?\nB\nKinetic friction is usually more or less independent of velocity. However, inexperienced drivers tend to produce a jerk at the last moment of\ndeceleration when they stop at a stop light. What does this tell you about\nthe kinetic friction between the brake shoes and the brake drums?\nC\nSome of the following are correct descriptions of types of forces that\ncould be added on as new branches of the classification tree. Others are\nnot really types of forces, and still others are not force phenomena at all.\nIn each case, decide what’s going on, and if appropriate, figure out how\nyou would incorporate them into the tree.\nsticky force\nopposite force\nflowing force\nsurface tension\nhorizontal force\nmotor force\ncanceled force\nmakes tape stick to things\nthe force that Newton’s third law says relates to every force you make\nthe force that water carries with it as it flows out of a\nhose\nlets insects walk on water\na force that is horizontal\nthe force that a motor makes on the thing it is turning\na force that is being canceled out by some other\nforce\n5.3 Analysis of Forces\nNewton’s first and second laws deal with the total of all the forces\nexerted on a specific object, so it is very important to be able to\nfigure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to\ndescribe all the corresponding forces that must exist according to\nNewton’s third law. We refer to this as “analyzing the forces” in\nwhich the object participates.\n158\nChapter 5\nAnalysis of Forces\nA barge\nexample 4\nA barge is being pulled along a canal by teams of horses on the shores. Analyze all the forces in which the\nbarge participates.\nforce acting on barge\nropes’ forward normal forces on barge\nwater’s backward fluid friction force on barge\nplanet earth’s downward gravitational force\non barge\nwater’s upward “floating” force on barge\nforce related to it by Newton’s third law\nbarge’s backward normal force on ropes\nbarge’s forward fluid friction force on water\nbarge’s upward gravitational force on earth\nbarge’s downward “floating” force on water\nHere I’ve used the word “floating” force as an example of a sensible invented term for a type of force not\nclassified on the tree in the previous section. A more formal technical term would be “hydrostatic force.”\nNote how the pairs of forces are all structured as “A’s force on B, B’s force on A”: ropes on barge and barge\non ropes; water on barge and barge on water. Because all the forces in the left column are forces acting on\nthe barge, all the forces in the right column are forces being exerted by the barge, which is why each entry in\nthe column begins with “barge.”\nOften you may be unsure whether you have forgotten one of the\nforces. Here are three strategies for checking your list:\nSee what physical result would come from the forces you’ve\nfound so far. Suppose, for instance, that you’d forgotten the\n“floating” force on the barge in the example above. Looking\nat the forces you’d found, you would have found that there\nwas a downward gravitational force on the barge which was\nnot canceled by any upward force. The barge isn’t supposed\nto sink, so you know you need to find a fourth, upward force.\nAnother technique for finding missing forces is simply to go\nthrough the list of all the common types of forces and see if\nany of them apply.\nMake a drawing of the object, and draw a dashed boundary\nline around it that separates it from its environment. Look for\npoints on the boundary where other objects come in contact\nwith your object. This strategy guarantees that you’ll find\nevery contact force that acts on the object, although it won’t\nhelp you to find non-contact forces.\nThe following is another example in which we can profit by checking against our physical intuition for what should be happening.\nSection 5.3\nAnalysis of Forces\n159\nRappelling\nexample 5\nAs shown in the figure below, Cindy is rappelling down a cliff. Her downward motion is at constant speed, and\nshe takes little hops off of the cliff, as shown by the dashed line. Analyze the forces in which she participates\nat a moment when her feet are on the cliff and she is pushing off.\nforce acting on Cindy\nforce related to it by Newton’s third law\nplanet earth’s downward gravitational force Cindy’s upward gravitational force on earth\non Cindy\nropes upward frictional force on Cindy (her Cindy’s downward frictional force on the rope\nhand)\ncliff’s rightward normal force on Cindy\nCindy’s leftward normal force on the cliff\nThe two vertical forces cancel, which is what they should be doing if she is to go down at a constant rate. The\nonly horizontal force on her is the cliff’s force, which is not canceled by any other force, and which therefore\nwill produce an acceleration of Cindy to the right. This makes sense, since she is hopping off. (This solution\nis a little oversimplified, because the rope is slanting, so it also applies a small leftward force to Cindy. As she\nflies out to the right, the slant of the rope will increase, pulling her back in more strongly.)\nI believe that constructing the type of table described in this\nsection is the best method for beginning students. Most textbooks,\nhowever, prescribe a pictorial way of showing all the forces acting on\nan object. Such a picture is called a free-body diagram. It should\nnot be a big problem if a future physics professor expects you to\nbe able to draw such diagrams, because the conceptual reasoning\nis the same. You simply draw a picture of the object, with arrows\nrepresenting the forces that are acting on it. Arrows representing\ncontact forces are drawn from the point of contact, noncontact forces\nfrom the center of mass. Free-body diagrams do not show the equal\nand opposite forces exerted by the object itself.\nDiscussion Questions\nA\nIn the example of the barge going down the canal, I referred to\na “floating” or “hydrostatic” force that keeps the boat from sinking. If you\nwere adding a new branch on the force-classification tree to represent this\nforce, where would it go?\nDiscussion question C.\nB\nA pool ball is rebounding from the side of the pool table. Analyze\nthe forces in which the ball participates during the short time when it is in\ncontact with the side of the table.\nC\nThe earth’s gravitational force on you, i.e., your weight, is always\nequal to mg , where m is your mass. So why can you get a shovel to go\ndeeper into the ground by jumping onto it? Just because you’re jumping,\nthat doesn’t mean your mass or weight is any greater, does it?\n160\nChapter 5\nAnalysis of Forces\n5.4 Transmission of Forces by Low-Mass\nObjects\nYou’re walking your dog. The dog wants to go faster than you do,\nand the leash is taut. Does Newton’s third law guarantee that your\nforce on your end of the leash is equal and opposite to the dog’s\nforce on its end? If they’re not exactly equal, is there any reason\nwhy they should be approximately equal?\nIf there was no leash between you, and you were in direct contact\nwith the dog, then Newton’s third law would apply, but Newton’s\nthird law cannot relate your force on the leash to the dog’s force\non the leash, because that would involve three separate objects.\nNewton’s third law only says that your force on the leash is equal\nand opposite to the leash’s force on you,\nFyL = −FLy ,\nand that the dog’s force on the leash is equal and opposite to its\nforce on the dog\nFdL = −FLd .\nStill, we have a strong intuitive expectation that whatever force we\nmake on our end of the leash is transmitted to the dog, and viceversa. We can analyze the situation by concentrating on the forces\nthat act on the leash, FdL and FyL . According to Newton’s second\nlaw, these relate to the leash’s mass and acceleration:\nFdL + FyL = mL aL .\nThe leash is far less massive then any of the other objects involved,\nand if mL is very small, then apparently the total force on the leash\nis also very small, FdL + FyL ≈ 0, and therefore\nFdL ≈ −FyL\n.\nThus even though Newton’s third law does not apply directly to\nthese two forces, we can approximate the low-mass leash as if it was\nnot intervening between you and the dog. It’s at least approximately\nas if you and the dog were acting directly on each other, in which\ncase Newton’s third law would have applied.\nIn general, low-mass objects can be treated approximately as if\nthey simply transmitted forces from one object to another. This can\nbe true for strings, ropes, and cords, and also for rigid objects such\nas rods and sticks.\nIf you look at a piece of string under a magnifying glass as you\npull on the ends more and more strongly, you will see the fibers\nstraightening and becoming taut. Different parts of the string are\nSection 5.4\nTransmission of Forces by Low-Mass Objects\n161\np / If we imagine dividing a taut rope up into small segments, then\nany segment has forces pulling outward on it at each end. If the rope\nis of negligible mass, then all the forces equal +T or −T , where T , the\ntension, is a single number.\nq / The Golden Gate Bridge’s\nroadway is held up by the tension\nin the vertical cables.\napparently exerting forces on each other. For instance, if we think of\nthe two halves of the string as two objects, then each half is exerting\na force on the other half. If we imagine the string as consisting\nof many small parts, then each segment is transmitting a force to\nthe next segment, and if the string has very little mass, then all\nthe forces are equal in magnitude. We refer to the magnitude of\nthe forces as the tension in the string, T . Although the tension\nis measured in units of Newtons, it is not itself a force. There are\nmany forces within the string, some in one direction and some in the\nother direction, and their magnitudes are only approximately equal.\nThe concept of tension only makes sense as a general, approximate\nstatement of how big all the forces are.\nIf a rope goes over a pulley or around some other object, then the\ntension throughout the rope is approximately equal so long as there\nis not too much friction. A rod or stick can be treated in much the\nsame way as a string, but it is possible to have either compression\nor tension.\nSince tension is not a type of force, the force exerted by a rope\non some other object must be of some definite type such as static\nfriction, kinetic friction, or a normal force. If you hold your dog’s\nleash with your hand through the loop, then the force exerted by the\nleash on your hand is a normal force: it is the force that keeps the\nleash from occupying the same space as your hand. If you grasp a\nplain end of a rope, then the force between the rope and your hand\nis a frictional force.\nA more complex example of transmission of forces is the way\na car accelerates. Many people would describe the car’s engine as\nmaking the force that accelerates the car, but the engine is part of\nthe car, so that’s impossible: objects can’t make forces on themselves. What really happens is that the engine’s force is transmitted\nthrough the transmission to the axles, then through the tires to the\nroad. By Newton’s third law, there will thus be a forward force from\nthe road on the tires, which accelerates the car.\nDiscussion Question\nA When you step on the gas pedal, is your foot’s force being transmitted\nin the sense of the word used in this section?\n162\nChapter 5\nAnalysis of Forces\n5.5 Objects Under Strain\nA string lengthens slightly when you stretch it. Similarly, we have\nalready discussed how an apparently rigid object such as a wall is\nactually flexing when it participates in a normal force. In other\ncases, the effect is more obvious. A spring or a rubber band visibly\nelongates when stretched.\nCommon to all these examples is a change in shape of some kind:\nlengthening, bending, compressing, etc. The change in shape can\nbe measured by picking some part of the object and measuring its\nposition, x. For concreteness, let’s imagine a spring with one end\nattached to a wall. When no force is exerted, the unfixed end of the\nspring is at some position xo . If a force acts at the unfixed end, its\nposition will change to some new value of x. The more force, the\ngreater the departure of x from xo .\nr / Defining the quantities F , x ,\nand xo in Hooke’s law.\nBack in Newton’s time, experiments like this were considered\ncutting-edge research, and his contemporary Hooke is remembered\ntoday for doing them and for coming up with a simple mathematical\ngeneralization called Hooke’s law:\nF ≈ k(x − xo )\n.\n[force required to stretch a spring; valid\nfor small forces only]\nHere k is a constant, called the spring constant, that depends on\nhow stiff the object is. If too much force is applied, the spring\nexhibits more complicated behavior, so the equation is only a good\napproximation if the force is sufficiently small. Usually when the\nforce is so large that Hooke’s law is a bad approximation, the force\nends up permanently bending or breaking the spring.\nAlthough Hooke’s law may seem like a piece of trivia about\nsprings, it is actually far more important than that, because all\nSection 5.5\nObjects Under Strain\n163\nsolid objects exert Hooke’s-law behavior over some range of sufficiently small forces. For example, if you push down on the hood of\na car, it dips by an amount that is directly proportional to the force.\n(But the car’s behavior would not be as mathematically simple if\nyou dropped a boulder on the hood!)\n. Solved problem: Combining springs\npage 170, problem 14\n. Solved problem: Young’s modulus\npage 170, problem 16\nDiscussion Question\nA\nA car is connected to its axles through big, stiff springs called shock\nabsorbers, or “shocks.” Although we’ve discussed Hooke’s law above only\nin the case of stretching a spring, a car’s shocks are continually going\nthrough both stretching and compression. In this situation, how would\nyou interpret the positive and negative signs in Hooke’s law?\n5.6 Simple Machines: The Pulley\nEven the most complex machines, such as cars or pianos, are built\nout of certain basic units called simple machines. The following are\nsome of the main functions of simple machines:\ntransmitting a force: The chain on a bicycle transmits a force\nfrom the crank set to the rear wheel.\nchanging the direction of a force: If you push down on a seesaw, the other end goes up.\nchanging the speed and precision of motion: When you make\nthe “come here” motion, your biceps only moves a couple of\ncentimeters where it attaches to your forearm, but your arm\nmoves much farther and more rapidly.\nchanging the amount of force: A lever or pulley can be used\nto increase or decrease the amount of force.\nYou are now prepared to understand one-dimensional simple machines, of which the pulley is the main example.\ns / Example 6.\nA pulley\nexample 6\n. Farmer Bill says this pulley arrangement doubles the force of\nhis tractor. Is he just a dumb hayseed, or does he know what he’s\ndoing?\n164\nChapter 5\nAnalysis of Forces\n. To use Newton’s first law, we need to pick an object and consider the sum of the forces on it. Since our goal is to relate the\ntension in the part of the cable attached to the stump to the tension in the part attached to the tractor, we should pick an object\nto which both those cables are attached, i.e., the pulley itself. As\ndiscussed in section 5.4, the tension in a string or cable remains\napproximately constant as it passes around a pulley, provided that\nthere is not too much friction. There are therefore two leftward\nforces acting on the pulley, each equal to the force exerted by the\ntractor. Since the acceleration of the pulley is essentially zero, the\nforces on it must be canceling out, so the rightward force of the\npulley-stump cable on the pulley must be double the force exerted\nby the tractor. Yes, Farmer Bill knows what he’s talking about.\nSection 5.6\nSimple Machines: The Pulley\n165\nSummary\nSelected Vocabulary\nrepulsive . . . . . describes a force that tends to push the two\nparticipating objects apart\nattractive . . . . describes a force that tends to pull the two\nparticipating objects together\noblique . . . . . . describes a force that acts at some other angle,\none that is not a direct repulsion or attraction\nnormal force . . . the force that keeps two objects from occupying the same space\nstatic friction . . a friction force between surfaces that are not\nslipping past each other\nkinetic friction . a friction force between surfaces that are slipping past each other\nfluid . . . . . . . . a gas or a liquid\nfluid friction . . . a friction force in which at least one of the\nobject is is a fluid\nspring constant . the constant of proportionality between force\nand elongation of a spring or other object under strain\nNotation\nFN . . . .\nFs . . . .\nFk . . . .\nµs . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nµk . . . . . . . . .\nk. . . . . . . . . .\na normal force\na static frictional force\na kinetic frictional force\nthe coefficient of static friction; the constant of\nproportionality between the maximum static\nfrictional force and the normal force; depends\non what types of surfaces are involved\nthe coefficient of kinetic friction; the constant\nof proportionality between the kinetic frictional force and the normal force; depends on\nwhat types of surfaces are involved\nthe spring constant; the constant of proportionality between the force exerted on an object and the amount by which the object is\nlengthened or compressed\nSummary\nNewton’s third law states that forces occur in equal and opposite\npairs. If object A exerts a force on object B, then object B must\nsimultaneously be exerting an equal and opposite force on object A.\nEach instance of Newton’s third law involves exactly two objects,\nand exactly two forces, which are of the same type.\nThere are two systems for classifying forces. We are presently\nusing the more practical but less fundamental one. In this system,\nforces are classified by whether they are repulsive, attractive, or\noblique; whether they are contact or noncontact forces; and whether\n166\nChapter 5\nAnalysis of Forces\nthe two objects involved are solids or fluids.\nStatic friction adjusts itself to match the force that is trying to\nmake the surfaces slide past each other, until the maximum value is\nreached,\n|Fs | < µs |FN |\n.\nOnce this force is exceeded, the surfaces slip past one another, and\nkinetic friction applies,\n|Fk | = µk |FN |\n.\nBoth types of frictional force are nearly independent of surface area,\nand kinetic friction is usually approximately independent of the\nspeed at which the surfaces are slipping.\nA good first step in applying Newton’s laws of motion to any\nphysical situation is to pick an object of interest, and then to list\nall the forces acting on that object. We classify each force by its\ntype, and find its Newton’s-third-law partner, which is exerted by\nthe object on some other object.\nWhen two objects are connected by a third low-mass object,\ntheir forces are transmitted to each other nearly unchanged.\nObjects under strain always obey Hooke’s law to a good approximation, as long as the force is small. Hooke’s law states that the\nstretching or compression of the object is proportional to the force\nexerted on it,\nF ≈ k(x − xo )\n.\nSummary\n167\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nA little old lady and a pro football player collide head-on.\nCompare their forces on each other, and compare their accelerations.\nExplain.\n2\nThe earth is attracted to an object with a force equal and\nopposite to the force of the earth on the object. If this is true,\nwhy is it that when you drop an object, the earth does not have an\nacceleration equal and opposite to that of the object?\n3\nWhen you stand still, there are two forces acting on you,\nthe force of gravity (your weight) and the normal force of the floor\npushing up on your feet. Are these forces equal and opposite? Does\nNewton’s third law relate them to each other? Explain.\nIn problems 4-8, analyze the forces using a table in the format shown\nin section 5.3. Analyze the forces in which the italicized object participates.\n4\nA magnet is stuck underneath a parked car. (See instructions\nabove.)\n5\nAnalyze two examples of objects at rest relative to the earth\nthat are being kept from falling by forces other than the normal\nforce. Do not use objects in outer space, and do not duplicate\nproblem 4 or 8. (See instructions above.)\n6\nA person is rowing a boat, with her feet braced. She is doing\nthe part of the stroke that propels the boat, with the ends of the\noars in the water (not the part where the oars are out of the water).\n(See instructions above.)\nProblem 6.\n7\nA farmer is in a stall with a cow when the cow decides to press\nhim against the wall, pinning him with his feet off the ground. Analyze the forces in which the farmer participates. (See instructions\nabove.)\n8\nA propeller plane is cruising east at constant speed and altitude. (See instructions above.)\n9\nToday’s tallest buildings are really not that much taller than\nthe tallest buildings of the 1940s. One big problem with making an\neven taller skyscraper is that every elevator needs its own shaft running the whole height of the building. So many elevators are needed\nto serve the building’s thousands of occupants that the elevator\nshafts start taking up too much of the space within the building.\nAn alternative is to have elevators that can move both horizontally\nand vertically: with such a design, many elevator cars can share a\nProblem 9.\n168\nChapter 5\nAnalysis of Forces\nfew shafts, and they don’t get in each other’s way too much because\nthey can detour around each other. In this design, it becomes impossible to hang the cars from cables, so they would instead have to\nride on rails which they grab onto with wheels. Friction would keep\nthem from slipping. The figure shows such a frictional elevator in\nits vertical travel mode. (The wheels on the bottom are for when it\nneeds to switch to horizontal motion.)\n(a) If the coefficient of static friction between rubber and steel is\nµs , and the maximum mass of the car plus its passengers is M ,\nhow much force must there be pressing each wheel against the rail\nin order to keep the car from slipping? (Assume the car is not\n√\naccelerating.)\n(b) Show that your result has physically reasonable behavior with\nrespect to µs . In other words, if there was less friction, would the\nwheels need to be pressed more firmly or less firmly? Does your\nequation behave that way?\n10\nUnequal masses M and m are suspended from a pulley as\nshown in the figure.\n(a) Analyze the forces in which mass m participates, using a table\nthe format shown in section 5.3. [The forces in which the other mass\nparticipates will of course be similar, but not numerically the same.]\n(b) Find the magnitude of the accelerations of the two masses.\n[Hints: (1) Pick a coordinate system, and use positive and negative signs consistently to indicate the directions of the forces and\naccelerations. (2) The two accelerations of the two masses have to\nbe equal in magnitude but of opposite signs, since one side eats up\nrope at the same rate at which the other side pays it out. (3) You\nneed to apply Newton’s second law twice, once to each mass, and\nthen solve the two equations for the unknowns: the acceleration, a,\nand the tension in the rope, T .]\n(c) Many people expect that in the special case of M = m, the two\nmasses will naturally settle down to an equilibrium position side by\nside. Based on your answer from part b, is this correct?\n(d) Find the tension in the rope, T .\n(e) Interpret your equation from part d in the special case where one\nof the masses is zero. Here “interpret” means to figure out what happens mathematically, figure out what should happen physically, and\nconnect the two.\nProblem 10.\n11\nA tugboat of mass m pulls a ship of mass M , accelerating it.\nThe speeds are low enough that you can ignore fluid friction acting\non their hulls, although there will of course need to be fluid friction\nacting on the tug’s propellers.\n(a) Analyze the forces in which the tugboat participates, using a\ntable in the format shown in section 5.3. Don’t worry about vertical\nforces.\n(b) Do the same for the ship.\nProblems\n169\n(c) Assume now that water friction on the two vessels’ hulls is negligible. If the force acting on the tug’s propeller is F , what is the\ntension, T , in the cable connecting the two ships? [Hint: Write\ndown two equations, one for Newton’s second law applied to each\n√\nobject. Solve these for the two unknowns T and a.]\n(d) Interpret your answer in the special cases of M = 0 and M = ∞.\n12\nExplain why it wouldn’t make sense to have kinetic friction\nbe stronger than static friction.\n13\nIn the system shown in the figure, the pulleys on the left and\nright are fixed, but the pulley in the center can move to the left or\nright. The two masses are identical. Show that the mass on the left\nwill have an upward acceleration equal to g/5. Assume all the ropes\nand pulleys are massless and rictionless.\n14\nThe figure shows two different ways of combining a pair of\nidentical springs, each with spring constant k. We refer to the top\nsetup as parallel, and the bottom one as a series arrangement.\n(a) For the parallel arrangement, analyze the forces acting on the\nconnector piece on the left, and then use this analysis to determine\nthe equivalent spring constant of the whole setup. Explain whether\nthe combined spring constant should be interpreted as being stiffer\nor less stiff.\n(b) For the series arrangement, analyze the forces acting on each\nspring and figure out the same things.\n. Solution, p. 273\nProblem 13.\n15\nGeneralize the results of problem 14 to the case where the\ntwo spring constants are unequal.\n16\n(a) Using the solution of problem 14, which is given in the\nback of the book, predict how the spring constant of a fiber will\ndepend on its length and cross-sectional area.\n(b) The constant of proportionality is called the Young’s modulus,\nE, and typical values of the Young’s modulus are about 1010 to\n1011 . What units would the Young’s modulus have in the SI (meterkilogram-second) system?\n. Solution, p. 274\n17\nThis problem depends on the results of problems 14 and\n16, whose solutions are in the back of the book When atoms form\nchemical bonds, it makes sense to talk about the spring constant of\nthe bond as a measure of how “stiff” it is. Of course, there aren’t\nreally little springs — this is just a mechanical model. The purpose\nof this problem is to estimate the spring constant, k, for a single\nbond in a typical piece of solid matter. Suppose we have a fiber,\nlike a hair or a piece of fishing line, and imagine for simplicity that\nit is made of atoms of a single element stacked in a cubical manner,\nas shown in the figure, with a center-to-center spacing b. A typical\nvalue for b would be about 10−10 m.\n(a) Find an equation for k in terms of b, and in terms of the Young’s\nmodulus, E, defined in problem 16 and its solution.\nProblem 14.\nProblem 17.\n170\nChapter 5\nAnalysis of Forces\n(b) Estimate k using the numerical data given in problem 16.\n(c) Suppose you could grab one of the atoms in a diatomic molecule\nlike H2 or O2 , and let the other atom hang vertically below it. Does\nthe bond stretch by any appreciable fraction due to gravity?\n18\nIn each case, identify the force that causes the acceleration,\nand give its Newton’s-third-law partner. Describe the effect of the\npartner force. (a) A swimmer speeds up. (b) A golfer hits the ball\noff of the tee. (c) An archer fires an arrow. (d) A locomotive slows\ndown.\n. Solution, p. 274\n19\nGinny has a plan. She is going to ride her sled while her\ndog Foo pulls her. However, Ginny hasn’t taken physics, so there\nmay be a problem: she may slide right off the sled when Foo starts\npulling.\n(a) Analyze all the forces in which Ginny participates, making a\ntable as in section 5.3.\n(b) Analyze all the forces in which the sled participates.\n(c) The sled has mass m, and Ginny has mass M . The coefficient\nof static friction between the sled and the snow is µ1 , and µ2 is\nthe corresponding quantity for static friction between the sled and\nher snow pants. Ginny must have a certain minimum mass so that\nshe will not slip off the sled. Find this in terms of the other three\n√\nvariables.\n(d) Interpreting your equation from part c, under what conditions\nwill there be no physically realistic solution for M ? Discuss what\nthis means physically.\nProblem 19.\n20\nExample 2 on page 148 involves a person pushing a box up a\nhill. The incorrect answer describes three forces. For each of these\nthree forces, give the force that it is related to by Newton’s third\nlaw, and state the type of force.\n. Solution, p. 274\n21\nExample 6 on page 164 describes a force-doubling setup\ninvolving a pulley. Make up a more complicated arrangement, using\nmore than one pulley, that would multiply the force by a factor\ngreater than two.\n22\nPick up a heavy object such as a backpack or a chair, and\nstand on a bathroom scale. Shake the object up and down. What\ndo you observe? Interpret your observations in terms of Newton’s\nthird law.\n23\nA cop investigating the scene of an accident measures the\nlength L of a car’s skid marks in order to find out its speed v at\nthe beginning of the skid. Express v in terms of L and any other\n√\nrelevant variables.\n24\nThe following reasoning leads to an apparent paradox; explain\nwhat’s wrong with the logic. A baseball player hits a ball. The ball\nand the bat spend a fraction of a second in contact. During that\ntime they’re moving together, so their accelerations must be equal.\nProblems\n171\nNewton’s third law says that their forces on each other are also\nequal. But a = F/m, so how can this be, since their masses are\nunequal? (Note that the paradox isn’t resolved by considering the\nforce of the batter’s hands on the bat. Not only is this force very\nsmall compared to the ball-bat force, but the batter could have just\nthrown the bat at the ball.)\n25\nThis problem has been deleted.\n26\n(a) Compare the mass of a one-liter water bottle on earth,\non the moon, and in interstellar space.\n. Solution, p. 274\n(b) Do the same for its weight.\n27\nAn ice skater builds up some speed, and then coasts across\nthe ice passively in a straight line. (a) Analyze the forces. (b) If\nhis initial speed is v, and the coefficient of kinetic friction is µk ,\nfind the maximum theoretical distance he can glide before coming\nto a stop. Ignore air resistance. (c) Show that your answer to\npart b has the right units. (d) Show that your answer to part b\ndepends on the variables in a way that makes sense physically. (d)\nEvaluate your answer numerically for µk = 0.0046, and a worldrecord speed of 14.58 m/s. (The coefficient of friction was measured\nby De Koning et al., using special skates worn by real speed skaters.)\n(e) Comment on whether your answer in part d seems realistic. If\nit doesn’t, suggest possible reasons why.\n172\nChapter 5\nAnalysis of Forces\nPart II\nMotion in Three\nDimensions\nChapter 6\nNewton’s Laws in Three\nDimensions\n6.1 Forces Have No Perpendicular Effects\nSuppose you could shoot a rifle and arrange for a second bullet to\nbe dropped from the same height at the exact moment when the\nfirst left the barrel. Which would hit the ground first? Nearly\neveryone expects that the dropped bullet will reach the dirt first,\n175\nand Aristotle would have agreed. Aristotle would have described it\nlike this. The shot bullet receives some forced motion from the gun.\nIt travels forward for a split second, slowing down rapidly because\nthere is no longer any force to make it continue in motion. Once\nit is done with its forced motion, it changes to natural motion, i.e.\nfalling straight down. While the shot bullet is slowing down, the\ndropped bullet gets on with the business of falling, so according to\nAristotle it will hit the ground first.\na / A bullet is shot from a gun, and another bullet is simultaneously dropped from the same height. 1.\nAristotelian physics says that the horizontal motion of the shot bullet delays the onset of falling, so the dropped\nbullet hits the ground first. 2. Newtonian physics says the two bullets have the same vertical motion, regardless\nof their different horizontal motions.\nLuckily, nature isn’t as complicated as Aristotle thought! To\nconvince yourself that Aristotle’s ideas were wrong and needlessly\ncomplex, stand up now and try this experiment. Take your keys\nout of your pocket, and begin walking briskly forward. Without\nspeeding up or slowing down, release your keys and let them fall\nwhile you continue walking at the same pace.\nYou have found that your keys hit the ground right next to your\nfeet. Their horizontal motion never slowed down at all, and the\nwhole time they were dropping, they were right next to you. The\nhorizontal motion and the vertical motion happen at the same time,\nand they are independent of each other. Your experiment proves\nthat the horizontal motion is unaffected by the vertical motion, but\nit’s also true that the vertical motion is not changed in any way by\nthe horizontal motion. The keys take exactly the same amount of\ntime to get to the ground as they would have if you simply dropped\nthem, and the same is true of the bullets: both bullets hit the ground\n176\nChapter 6\nNewton’s Laws in Three Dimensions\nsimultaneously.\nThese have been our first examples of motion in more than one\ndimension, and they illustrate the most important new idea that\nis required to understand the three-dimensional generalization of\nNewtonian physics:\nForces have no perpendicular effects.\nWhen a force acts on an object, it has no effect on the part of the\nobject’s motion that is perpendicular to the force.\nIn the examples above, the vertical force of gravity had no effect\non the horizontal motions of the objects. These were examples of\nprojectile motion, which interested people like Galileo because of\nits military applications. The principle is more general than that,\nhowever. For instance, if a rolling ball is initially heading straight\nfor a wall, but a steady wind begins blowing from the side, the ball\ndoes not take any longer to get to the wall. In the case of projectile\nmotion, the force involved is gravity, so we can say more specifically\nthat the vertical acceleration is 9.8 m/s2 , regardless of the horizontal\nmotion.\nself-check A\nIn the example of the ball being blown sideways, why doesn’t the ball\ntake longer to get there, since it has to travel a greater distance?\n.\nAnswer, p. 268\nRelationship to relative motion\nThese concepts are directly related to the idea that motion is relative. Galileo’s opponents argued that the earth could not possibly\nbe rotating as he claimed, because then if you jumped straight up in\nthe air you wouldn’t be able to come down in the same place. Their\nargument was based on their incorrect Aristotelian assumption that\nonce the force of gravity began to act on you and bring you back\ndown, your horizontal motion would stop. In the correct Newtonian\ntheory, the earth’s downward gravitational force is acting before,\nduring, and after your jump, but has no effect on your motion in\nthe perpendicular (horizontal) direction.\nIf Aristotle had been correct, then we would have a handy way\nto determine absolute motion and absolute rest: jump straight up\nin the air, and if you land back where you started, the surface from\nwhich you jumped must have been in a state of rest. In reality, this\ntest gives the same result as long as the surface under you is an\ninertial frame. If you try this in a jet plane, you land back on the\nsame spot on the deck from which you started, regardless of whether\nthe plane is flying at 500 miles per hour or parked on the runway.\nThe method would in fact only be good for detecting whether the\nplane was accelerating.\nSection 6.1\nForces Have No Perpendicular Effects\n177\nDiscussion Questions\nA\nThe following is an incorrect explanation of a fact about target\nshooting:\n“Shooting a high-powered rifle with a high muzzle velocity is different from\nshooting a less powerful gun. With a less powerful gun, you have to aim\nquite a bit above your target, but with a more powerful one you don’t have\nto aim so high because the bullet doesn’t drop as fast.”\nWhat is the correct explanation?\nB\nYou have thrown a rock, and it is flying through the air in an arc. If\nthe earth’s gravitational force on it is always straight down, why doesn’t it\njust go straight down once it leaves your hand?\nC\nConsider the example of the bullet that is dropped at the same\nmoment another bullet is fired from a gun. What would the motion of the\ntwo bullets look like to a jet pilot flying alongside in the same direction as\nthe shot bullet and at the same horizontal speed?\nb / This object experiences a force that pulls it down toward the\nbottom of the page. In each equal time interval, it moves three units to\nthe right. At the same time, its vertical motion is making a simple pattern\nof +1, 0, −1, −2, −3, −4, . . . units. Its motion can be described by an x\ncoordinate that has zero acceleration and a y coordinate with constant\nacceleration. The arrows labeled x and y serve to explain that we are\ndefining increas- ing x to the right and increasing y as upward.\n178\nChapter 6\nNewton’s Laws in Three Dimensions\n6.2 Coordinates and Components\n’Cause we’re all\nBold as love,\nJust ask the axis.\nJimi Hendrix\nHow do we convert these ideas into mathematics? Figure b shows\na good way of connecting the intuitive ideas to the numbers. In one\ndimension, we impose a number line with an x coordinate on a\ncertain stretch of space. In two dimensions, we imagine a grid of\nsquares which we label with x and y values, as shown in figure b.\nBut of course motion doesn’t really occur in a series of discrete\nhops like in chess or checkers. The figure on the left shows a way\nof conceptualizing the smooth variation of the x and y coordinates.\nThe ball’s shadow on the wall moves along a line, and we describe its\nposition with a single coordinate, y, its height above the floor. The\nwall shadow has a constant acceleration of -9.8 m/s2 . A shadow on\nthe floor, made by a second light source, also moves along a line,\nand we describe its motion with an x coordinate, measured from the\nwall.\nThe velocity of the floor shadow is referred to as the x component\nof the velocity, written vx . Similarly we can notate the acceleration\nof the floor shadow as ax . Since vx is constant, ax is zero.\nSimilarly, the velocity of the wall shadow is called vy , its acceleration ay . This example has ay = −9.8 m/s2 .\nBecause the earth’s gravitational force on the ball is acting along\nthe y axis, we say that the force has a negative y component, Fy ,\nbut Fx = Fz = 0.\nc / The shadow on the wall\nshows the ball’s y motion, the\nshadow on the floor its x motion.\nThe general idea is that we imagine two observers, each of whom\nperceives the entire universe as if it was flattened down to a single\nline. The y-observer, for instance, perceives y, vy , and ay , and will\ninfer that there is a force, Fy , acting downward on the ball. That\nis, a y component means the aspect of a physical phenomenon, such\nas velocity, acceleration, or force, that is observable to someone who\ncan only see motion along the y axis.\nAll of this can easily be generalized to three dimensions. In the\nexample above, there could be a z-observer who only sees motion\ntoward or away from the back wall of the room.\nSection 6.2\nCoordinates and Components\n179\nA car going over a cliff\nexample 1\n. The police find a car at a distance w = 20 m from the base of a\ncliff of height h = 100 m. How fast was the car going when it went\nover the edge? Solve the problem symbolically first, then plug in\nthe numbers.\n. Let’s choose y pointing up and x pointing away from the cliff.\nThe car’s vertical motion was independent of its horizontal motion, so we know it had a constant vertical acceleration of a =\n−g = −9.8 m/s2 . The time it spent in the air is therefore related\nto the vertical distance it fell by the constant-acceleration equation\n∆y =\n1\nay ∆t 2\n2\n−h =\n1\n(−g)∆t 2\n2\n,\nor\nd / Example 1.\n.\nSolving for ∆t gives\ns\n∆t =\n2h\ng\n.\nSince the vertical force had no effect on the car’s horizontal motion, it had ax = 0, i.e., constant horizontal velocity. We can apply\nthe constant-velocity equation\nvx =\n∆x\n∆t\n,\nvx =\nw\n∆t\n.\ni.e.,\nWe now substitute for ∆t to find\ns\nvx = w/\n2h\ng\n,\ng\n2h\n.\nwhich simplifies to\nr\nvx = w\nPlugging in numbers, we find that the car’s speed when it went\nover the edge was 4 m/s, or about 10 mi/hr.\n180\nChapter 6\nNewton’s Laws in Three Dimensions\nProjectiles move along parabolas.\nWhat type of mathematical curve does a projectile follow through\nspace? To find out, we must relate x to y, eliminating t. The reasoning is very similar to that used in the example above. Arbitrarily\nchoosing x = y = t = 0 to be at the top of the arc, we conveniently\nhave x = ∆x, y = ∆y, and t = ∆t, so\n1\ny = ay t 2\n2\nx = vx t\n(ay < 0)\nWe solve the second equation for t = x/vx and eliminate t in the\nfirst equation:\n2\n1\nx\ny = ay\n.\n2\nvx\nSince everything in this equation is a constant except for x and y,\nwe conclude that y is proportional to the square of x. As you may\nor may not recall from a math class, y ∝ x2 describes a parabola.\n. Solved problem: A cannon\npage 184, problem 5\nDiscussion Question\nA\nAt the beginning of this section I represented the motion of a projectile on graph paper, breaking its motion into equal time intervals. Suppose\ninstead that there is no force on the object at all. It obeys Newton’s first law\nand continues without changing its state of motion. What would the corresponding graph-paper diagram look like? If the time interval represented\nby each arrow was 1 second, how would you relate the graph-paper diagram to the velocity components vx and vy ?\ne / A parabola can be defined as\nthe shape made by cutting a cone\nparallel to its side. A parabola is\nalso the graph of an equation of\nthe form y ∝ x 2 .\nB\nMake up several different coordinate systems oriented in different\nways, and describe the ax and ay of a falling object in each one.\n6.3 Newton’s Laws in Three Dimensions\nIt is now fairly straightforward to extend Newton’s laws to three\ndimensions:\nNewton’s first law\nIf all three components of the total force on an object are zero,\nthen it will continue in the same state of motion.\nf / Each water droplet follows\na parabola. The faster drops’\nparabolas are bigger.\nNewton’s second law\nThe components of an object’s acceleration are predicted by\nthe equations\nax = Fx,total /m\n,\nay = Fy,total /m\n,\naz = Fz,total /m\n.\nand\nNewton’s third law\nSection 6.3\nNewton’s Laws in Three Dimensions\n181\nIf two objects A and B interact via forces, then the components of their forces on each other are equal and opposite:\nFA on B,x = −FB on A,x\n,\nFA on B,y = −FB on A,y\n,\nFA on B,z = −FB on A,z\n.\nand\nForces in perpendicular directions on the same objectexample 2\n. An object is initially at rest. Two constant forces begin acting on\nit, and continue acting on it for a while. As suggested by the two\narrows, the forces are perpendicular, and the rightward force is\nstronger. What happens?\n. Aristotle believed, and many students still do, that only one force\ncan “give orders” to an object at one time. They therefore think\nthat the object will begin speeding up and moving in the direction\nof the stronger force. In fact the object will move along a diagonal.\nIn the example shown in the figure, the object will respond to the\nlarge rightward force with a large acceleration component to the\nright, and the small upward force will give it a small acceleration\ncomponent upward. The stronger force does not overwhelm the\nweaker force, or have any effect on the upward motion at all. The\nforce components simply add together:\ng / Example 2.\n*\nFx,total = F1,x + F2,\nx\n0\n>+ F\nFy,total = F1,\ny\n2,y\n0\nDiscussion Question\nA\nThe figure shows two trajectories, made by splicing together lines\nand circular arcs, which are unphysical for an object that is only being\nacted on by gravity. Prove that they are impossible based on Newton’s\nlaws.\n182\nChapter 6\nNewton’s Laws in Three Dimensions\nSummary\nSelected Vocabulary\ncomponent . . . . the part of a velocity, acceleration, or force\nthat would be perceptible to an observer who\ncould only see the universe projected along a\ncertain one-dimensional axis\nparabola . . . . . the mathematical curve whose graph has y\nproportional to x2\nNotation\nx, y, z . . . . . .\nvx , vy , vz . . . . .\nax , ay , az . . . . .\nan object’s positions along the x, y, and z axes\nthe x, y, and z components of an object’s velocity; the rates of change of the object’s x, y,\nand z coordinates\nthe x, y, and z components of an object’s acceleration; the rates of change of vx , vy , and\nvz\nSummary\nA force does not produce any effect on the motion of an object\nin a perpendicular direction. The most important application of\nthis principle is that the horizontal motion of a projectile has zero\nacceleration, while the vertical motion has an acceleration equal to g.\nThat is, an object’s horizontal and vertical motions are independent.\nThe arc of a projectile is a parabola.\nMotion in three dimensions is measured using three coordinates,\nx, y, and z. Each of these coordinates has its own corresponding\nvelocity and acceleration. We say that the velocity and acceleration\nboth have x, y, and z components\nNewton’s second law is readily extended to three dimensions by\nrewriting it as three equations predicting the three components of\nthe acceleration,\nax = Fx,total /m\n,\nay = Fy,total /m\n,\naz = Fz,total /m\n,\nand likewise for the first and third laws.\nSummary\n183\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\n(a) A ball is thrown straight up with velocity v. Find an\nequation for the height to which it rises.\n(b) Generalize your equation for a ball thrown at an angle θ above\nhorizontal, in which case its initial velocity components are vx =\nv cos θ and vy = v sin θ.\n2\nAt the Salinas Lettuce Festival Parade, Miss Lettuce of 1996\ndrops her bouquet while riding on a float moving toward the right.\nCompare the shape of its trajectory as seen by her to the shape seen\nby one of her admirers standing on the sidewalk.\n3\nTwo daredevils, Wendy and Bill, go over Niagara Falls. Wendy\nsits in an inner tube, and lets the 30 km/hr velocity of the river throw\nher out horizontally over the falls. Bill paddles a kayak, adding an\nextra 10 km/hr to his velocity. They go over the edge of the falls\nat the same moment, side by side. Ignore air friction. Explain your\nreasoning.\n(a) Who hits the bottom first?\n(b) What is the horizontal component of Wendy’s velocity on impact?\n(c) What is the horizontal component of Bill’s velocity on impact?\n(d) Who is going faster on impact?\n4\nA baseball pitcher throws a pitch clocked at vx = 73.3 mi/h.\nHe throws horizontally. By what amount, d, does the ball drop by\nthe time it reaches home plate, L = 60.0 ft away?\n(a) First find a symbolic answer in terms of L, vx , and g.\n(b) Plug in and find a numerical answer. Express your answer\nin\n√\nunits of ft. [Note: 1 ft=12 in, 1 mi=5280 ft, and 1 in=2.54 cm]\nProblem 4.\n5\nA cannon standing on a flat field fires a cannonball with a\nmuzzle velocity v, at an angle θ above horizontal. The cannonball\nthus initially has velocity components vx = v cos θ and vy = v sin θ.\n(a) Show that the cannon’s range (horizontal distance to where the\n184\nChapter 6\nNewton’s Laws in Three Dimensions\ncannonball falls) is given by the equation R = (2v 2 /g) sin θ cos θ .\n(b) Interpret your equation in the cases of θ = 0 and θ = 90 ◦ .\n. Solution, p. 275\n6\nAssuming the result of problem 5 for the range of a projectile,\nR = (2v 2 /g) sin θ cos θ, show that the maximum range is for θ = R45 ◦ .\n7\nTwo cars go over the same bump in the road, Maria’s Maserati\nat 25 miles per hour and Park’s Porsche at 37. How many times\ngreater is the vertical acceleration of the Porsche? Hint: Remember\nthat acceleration depends both on how much the velocity changes\n√\nand on how much time it takes to change.\nProblems\n185\n186\nChapter 6\nNewton’s Laws in Three Dimensions\na / Vectors are used in aerial navigation.\nChapter 7\nVectors\n7.1 Vector Notation\nThe idea of components freed us from the confines of one-dimensional\nphysics, but the component notation can be unwieldy, since every\none-dimensional equation has to be written as a set of three separate\nequations in the three-dimensional case. Newton was stuck with the\ncomponent notation until the day he died, but eventually someone\nsufficiently lazy and clever figured out a way of abbreviating three\nequations as one.\n(a)\n→\n−\n→\n−\nF A on B = − F B on A\nstands for\n(b)\n→\n−\n→\n−\n→\n−\nF total = F 1 + F 2 + . . .\nstands for\n(c)\n→\n−\na =\n→\n∆−\nv\n∆t\nstands for\nFA on B,x = −FB on A,x\nFA on B,y = −FB on A,y\nFA on B,z = −FB on A,z\nFtotal,x = F1,x + F2,x + . . .\nFtotal,y = F1,y + F2,y + . . .\nFtotal,z = F1,z + F2,z + . . .\nax = ∆vx /∆t\nay = ∆vy /∆t\naz = ∆vz /∆t\nExample (a) shows both ways of writing Newton’s third law. Which\nwould you rather write?\nThe idea is that each of the algebra symbols with an arrow writ-\n187\nten on top, called a vector, is actually an abbreviation for three\ndifferent numbers, the x, y, and z components. The three components are referred to as the components of the vector, e.g., Fx is the\n→\n−\nx component of the vector F . The notation with an arrow on top\nis good for handwritten equations, but is unattractive in a printed\nbook, so books use boldface, F, to represent vectors. After this\npoint, I’ll use boldface for vectors throughout this book.\nIn general, the vector notation is useful for any quantity that\nhas both an amount and a direction in space. Even when you are\nnot going to write any actual vector notation, the concept itself is a\nuseful one. We say that force and velocity, for example, are vectors.\nA quantity that has no direction in space, such as mass or time,\nis called a scalar. The amount of a vector quantity is called its\nmagnitude. The notation for the magnitude of a vector A is |A|,\nlike the absolute value sign used with scalars.\nOften, as in example (b), we wish to use the vector notation to\nrepresent adding up all the x components to get a total x component,\netc. The plus sign is used between two vectors to indicate this type\nof component-by-component addition. Of course, vectors are really\ntriplets of numbers, not numbers, so this is not the same as the use\nof the plus sign with individual numbers. But since we don’t want to\nhave to invent new words and symbols for this operation on vectors,\nwe use the same old plus sign, and the same old addition-related\nwords like “add,” “sum,” and “total.” Combining vectors this way\nis called vector addition.\nSimilarly, the minus sign in example (a) was used to indicate\nnegating each of the vector’s three components individually. The\nequals sign is used to mean that all three components of the vector\non the left side of an equation are the same as the corresponding\ncomponents on the right.\nExample (c) shows how we abuse the division symbol in a similar\nmanner. When we write the vector ∆v divided by the scalar ∆t,\nwe mean the new vector formed by dividing each one of the velocity\ncomponents by ∆t.\nIt’s not hard to imagine a variety of operations that would combine vectors with vectors or vectors with scalars, but only four of\nthem are required in order to express Newton’s laws:\noperation\nvector + vector\nvector − vector\nvector · scalar\nvector/scalar\n188\nChapter 7\nVectors\ndefinition\nAdd component by component to\nmake a new set of three numbers.\nSubtract component by component\nto make a new set of three numbers.\nMultiply each component of the vector by the scalar.\nDivide each component of the vector\nby the scalar.\nAs an example of an operation that is not useful for physics, there\njust aren’t any useful physics applications for dividing a vector by\nanother vector component by component. In optional section 7.5,\nwe discuss in more detail the fundamental reasons why some vector\noperations are useful and others useless.\nWe can do algebra with vectors, or with a mixture of vectors\nand scalars in the same equation. Basically all the normal rules of\nalgebra apply, but if you’re not sure if a certain step is valid, you\nshould simply translate it into three component-based equations and\nsee if it works.\nOrder of addition\nexample 1\n. If we are adding two force vectors, F + G, is it valid to assume\nas in ordinary algebra that F + G is the same as G + F?\n. To tell if this algebra rule also applies to vectors, we simply\ntranslate the vector notation into ordinary algebra notation. In\nterms of ordinary numbers, the components of the vector F + G\nwould be Fx + Gx , Fy + Gy , and Fz + Gz , which are certainly the\nsame three numbers as Gx + Fx , Gy + Fy , and Gz + Fz . Yes, F + G\nis the same as G + F.\nIt is useful to define a symbol r for the vector whose components\nare x, y, and z, and a symbol ∆r made out of ∆x, ∆y, and ∆z.\nAlthough this may all seem a little formidable, keep in mind that\nit amounts to nothing more than a way of abbreviating equations!\nAlso, to keep things from getting too confusing the remainder of this\nchapter focuses mainly on the ∆r vector, which is relatively easy to\nvisualize.\nself-check A\nTranslate the equations vx = ∆x /∆t , vy = ∆y /∆t , and vz = ∆z /∆t for\nmotion with constant velocity into a single equation in vector notation.\n. Answer, p. 268\nDrawing vectors as arrows\nA vector in two dimensions can be easily visualized by drawing\nan arrow whose length represents its magnitude and whose direction\nrepresents its direction. The x component of a vector can then be\nvisualized as the length of the shadow it would cast in a beam of\nlight projected onto the x axis, and similarly for the y component.\nShadows with arrowheads pointing back against the direction of the\npositive axis correspond to negative components.\nb / The x an y components\nof a vector can be thought of as\nthe shadows it casts onto the x\nand y axes.\nIn this type of diagram, the negative of a vector is the vector\nwith the same magnitude but in the opposite direction. Multiplying\na vector by a scalar is represented by lengthening the arrow by that\nfactor, and similarly for division.\nself-check B\nGiven vector Q represented by an arrow in figure c, draw arrows repre-\nSection 7.1\nc / Self-check B.\nVector Notation\n189\nsenting the vectors 1.5Q and −Q.\n268\n. Answer, p.\nDiscussion Questions\nA\nWould it make sense to define a zero vector? Discuss what the\nzero vector’s components, magnitude, and direction would be; are there\nany issues here? If you wanted to disqualify such a thing from being a\nvector, consider whether the system of vectors would be complete. For\ncomparison, can you think of a simple arithmetic problem with ordinary\nnumbers where you need zero as the result? Does the same reasoning\napply to vectors, or not?\nB\nYou drive to your friend’s house. How does the magnitude of your ∆r\nvector compare with the distance you’ve added to the car’s odometer?\n7.2 Calculations with Magnitude and Direction\nIf you ask someone where Las Vegas is compared to Los Angeles,\nthey are unlikely to say that the ∆x is 290 km and the ∆y is 230\nkm, in a coordinate system where the positive x axis is east and the\ny axis points north. They will probably say instead that it’s 370 km\nto the northeast. If they were being precise, they might specify the\ndirection as 38 ◦ counterclockwise from east. In two dimensions, we\ncan always specify a vector’s direction like this, using a single angle.\nA magnitude plus an angle suffice to specify everything about the\nvector. The following two examples show how we use trigonometry\nand the Pythagorean theorem to go back and forth between the x−y\nand magnitude-angle descriptions of vectors.\nFinding magnitude and angle from components\nexample 2\n. Given that the ∆r vector from LA to Las Vegas has ∆x = 290 km\nand ∆y = 230 km, how would we find the magnitude and direction\nof ∆r?\n. We find the magnitude of ∆r from the Pythagorean theorem:\nq\n|∆r| = ∆x 2 + ∆y 2\n= 370 km\nWe know all three sides of the triangle, so the angle θ can be\nfound using any of the inverse trig functions. For example, we\nknow the opposite and adjacent sides, so\nd / Example 2.\nθ = tan−1\n= 38 ◦\n∆y\n∆x\n.\nFinding components from magnitude and angle\nexample 3\n. Given that the straight-line distance from Los Angeles to Las\nVegas is 370 km, and that the angle θ in the figure is 38 ◦ , how\ncan the xand y components of the ∆r vector be found?\n190\nChapter 7\nVectors\n. The sine and cosine of θ relate the given information to the\ninformation we wish to find:\n∆x\ncos θ =\n|∆r|\n∆y\nsin θ =\n|∆r|\nSolving for the unknowns gives\n∆x = |∆r| cos θ\n= 290 km\nand\n∆y = |∆r| sin θ\n= 230 km\n.\nThe following example shows the correct handling of the plus\nand minus signs, which is usually the main cause of mistakes.\nNegative components\nexample 4\n. San Diego is 120 km east and 150 km south of Los Angeles. An\nairplane pilot is setting course from San Diego to Los Angeles. At\nwhat angle should she set her course, measured counterclockwise from east, as shown in the figure?\n. If we make the traditional choice of coordinate axes, with x\npointing to the right and y pointing up on the map, then her ∆x is\nnegative, because her final x value is less than her initial x value.\nHer ∆y is positive, so we have\n∆x = −120 km\n∆y = 150 km\n.\ne / Example 4.\nIf we work by analogy with the previous example, we get\n∆y\nθ = tan−1\n∆x\n−1\n= tan (−1.25)\n= −51 ◦\n.\nAccording to the usual way of defining angles in trigonometry,\na negative result means an angle that lies clockwise from the x\naxis, which would have her heading for the Baja California. What\nwent wrong? The answer is that when you ask your calculator to\ntake the arctangent of a number, there are always two valid possibilities differing by 180 ◦ . That is, there are two possible angles\nwhose tangents equal -1.25:\ntan 129 ◦ = −1.25\ntan −51 ◦ = −1.25\nYou calculator doesn’t know which is the correct one, so it just\npicks one. In this case, the one it picked was the wrong one, and\nit was up to you to add 180 ◦ to it to find the right answer.\nSection 7.2\nCalculations with Magnitude and Direction\n191\nDiscussion Question\nA In the example above, we dealt with components that were negative.\nDoes it make sense to talk about positive and negative vectors?\n7.3 Techniques for Adding Vectors\nAddition of vectors given their components\nThe easiest type of vector addition is when you are in possession\nof the components, and want to find the components of their sum.\nAdding components\nexample 5\n. Given the ∆x and ∆y values from the previous examples, find\nthe ∆x and ∆y from San Diego to Las Vegas.\n.\n∆xtotal = ∆x1 + ∆x2\n= −120 km + 290 km\n= 170 km\n∆ytotal = ∆y1 + ∆y2\n= 150 km + 230 km\n= 380\nNote how the signs of the x components take care of the westward and eastward motions, which partially cancel.\nf / Example 5.\nAddition of vectors given their magnitudes and directions\nIn this case, you must first translate the magnitudes and directions into components, and the add the components.\nGraphical addition of vectors\nOften the easiest way to add vectors is by making a scale drawing\non a piece of paper. This is known as graphical addition, as opposed\nto the analytic techniques discussed previously.\nLA to Vegas, graphically\nexample 6\n. Given the magnitudes and angles of the ∆r vectors from San\nDiego to Los Angeles and from Los Angeles to Las Vegas, find\nthe magnitude and angle of the ∆r vector from San Diego to Las\nVegas.\ng / Vectors can be added graphically by placing them tip to tail,\nand then drawing a vector from\nthe tail of the first vector to the tip\nof the second vector.\n. Using a protractor and a ruler, we make a careful scale drawing,\nas shown in figure h. A scale of 1 mm → 2 km was chosen for this\nsolution. With a ruler, we measure the distance from San Diego\nto Las Vegas to be 206 mm, which corresponds to 412 km. With\na protractor, we measure the angle θ to be 65 ◦ .\nEven when we don’t intend to do an actual graphical calculation\nwith a ruler and protractor, it can be convenient to diagram the\naddition of vectors in this way. With ∆r vectors, it intuitively makes\nsense to lay the vectors tip-to-tail and draw the sum vector from the\n192\nChapter 7\nVectors\ntail of the first vector to the tip of the second vector. We can do\nthe same when adding other vectors such as force vectors.\nh / Example 6.\nself-check C\nHow would you subtract vectors graphically?\n. Answer, p. 268\nSection 7.3\nTechniques for Adding Vectors\n193\nDiscussion Questions\nA\nIf you’re doing graphical addition of vectors, does it matter which\nvector you start with and which vector you start from the other vector’s\ntip?\nB\nIf you add a vector with magnitude 1 to a vector of magnitude 2,\nwhat magnitudes are possible for the vector sum?\nC\nWhich of these examples of vector addition are correct, and which\nare incorrect?\n7.4 ? Unit Vector Notation\nWhen we want to specify a vector by its components, it can be cumbersome to have to write the algebra symbol for each component:\n∆x = 290 km, ∆y = 230 km\nA more compact notation is to write\n∆r = (290 km)x̂ + (230 km)ŷ\n,\nwhere the vectors x̂, ŷ, and ẑ, called the unit vectors, are defined\nas the vectors that have magnitude equal to 1 and directions lying\nalong the x, y, and z axes. In speech, they are referred to as “x-hat”\nand so on.\nA slightly different, and harder to remember, version of this\nnotation is unfortunately more prevalent. In this version, the unit\nvectors are called î, ĵ, and k̂:\n∆r = (290 km)î + (230 km)ĵ\n.\n7.5 ? Rotational Invariance\nLet’s take a closer look at why certain vector operations are useful and others are not. Consider the operation of multiplying two\nvectors component by component to produce a third vector:\nRx = Px Qx\nRy = Py Qy\nRz = Pz Qz\nAs a simple example, we choose vectors P and Q to have length\n1, and make them perpendicular to each other, as shown in figure\n194\nChapter 7\nVectors\ni/1. If we compute the result of our new vector operation using the\ncoordinate system in i/2, we find:\nRx = 0\nRy = 0\nRz = 0\nThe x component is zero because Px = 0, the y component is zero\nbecause Qy = 0, and the z component is of course zero because both\nvectors are in the x − y plane. However, if we carry out the same\noperations in coordinate system i/3, rotated 45 degrees with respect\nto the previous one, we find\nRx = 1/2\nRy = −1/2\nRz = 0\nThe operation’s result depends on what coordinate system we use,\nand since the two versions of R have different lengths (one being zero\nand the other nonzero), they don’t just represent the same answer\nexpressed in two different coordinate systems. Such an operation\nwill never be useful in physics, because experiments show physics\nworks the same regardless of which way we orient the laboratory\nbuilding! The useful vector operations, such as addition and scalar\nmultiplication, are rotationally invariant, i.e., come out the same\nregardless of the orientation of the coordinate system.\ni / Component-by-component\nmultiplication of the vectors in 1\nwould produce different vectors\nin coordinate systems 2 and 3.\nSection 7.5\n? Rotational Invariance\n195\nSummary\nSelected Vocabulary\nvector . . . . . . . a quantity that has both an amount (magnitude) and a direction in space\nmagnitude . . . . the “amount” associated with a vector\nscalar . . . . . . . a quantity that has no direction in space, only\nan amount\nNotation\nA . . . .\n→\n−\nA . . . .\n|A| . . .\nr . . . . .\n∆r . . . .\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.\nx̂, ŷ, ẑ . . . . . .\nî, ĵ, k̂ . . . . . . .\na vector with components Ax , Ay , and Az\nhandwritten notation for a vector\nthe magnitude of vector A\nthe vector whose components are x, y, and z\nthe vector whose components are ∆x, ∆y, and\n∆z\n(optional topic) unit vectors; the vectors with\nmagnitude 1 lying along the x, y, and z axes\na harder to remember notation for the unit\nvectors\nOther Terminology and Notation\ndisplacement vec- a name for the symbol ∆r\ntor . . . . . . . . .\nspeed . . . . . . . the magnitude of the velocity vector, i.e., the\nvelocity stripped of any information about its\ndirection\nSummary\nA vector is a quantity that has both a magnitude (amount) and\na direction in space, as opposed to a scalar, which has no direction.\nThe vector notation amounts simply to an abbreviation for writing\nthe vector’s three components.\nIn two dimensions, a vector can be represented either by its two\ncomponents or by its magnitude and direction. The two ways of\ndescribing a vector can be related by trigonometry.\nThe two main operations on vectors are addition of a vector to\na vector, and multiplication of a vector by a scalar.\nVector addition means adding the components of two vectors\nto form the components of a new vector. In graphical terms, this\ncorresponds to drawing the vectors as two arrows laid tip-to-tail and\ndrawing the sum vector from the tail of the first vector to the tip\nof the second one. Vector subtraction is performed by negating the\nvector to be subtracted and then adding.\nMultiplying a vector by a scalar means multiplying each of its\ncomponents by the scalar to create a new vector. Division by a\nscalar is defined similarly.\n196\nChapter 7\nVectors\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nThe figure shows vectors A and B. Graphically calculate the\nfollowing:\nA + B, A − B, B − A, −2B, A − 2B\nNo numbers are involved.\n2\nPhnom Penh is 470 km east and 250 km south of Bangkok.\nHanoi is 60 km east and 1030 km north of Phnom Penh.\n(a) Choose a coordinate system, and translate these data into ∆x\nand ∆y values with the proper plus and minus signs.\n(b) Find the components of the ∆r vector pointing from Bangkok\n√\nto Hanoi.\nProblem 1.\n3\nIf you walk 35 km at an angle 25 ◦ counterclockwise from east,\nand then 22 km at 230 ◦ counterclockwise from east, find the distance\n√\nand direction from your starting point to your destination.\n4\nA machinist is drilling holes in a piece of aluminum according\nto the plan shown in the figure. She starts with the top hole, then\nmoves to the one on the left, and then to the one on the right. Since\nthis is a high-precision job, she finishes by moving in the direction\nand at the angle that should take her back to the top hole, and\nchecks that she ends up in the same place. What are the distance\n√\nand direction from the right-hand hole to the top one?\nProblem 4.\n5\nSuppose someone proposes a new operation in which a vector\nA and a scalar B are added together to make a new vector C like\nthis:\nC x = Ax + B\nC y = Ay + B\nC y = Ay + B\nProve that this operation won’t be useful in physics, because it’s\nnot rotationally invariant.\nProblems\n197\n198\nChapter 7\nVectors\nChapter 8\nVectors and Motion\nIn 1872, capitalist and former California governor Leland Stanford\nasked photographer Eadweard Muybridge if he would work for him\non a project to settle a \\$25,000 bet (a princely sum at that time).\nStanford’s friends were convinced that a galloping horse always had\nat least one foot on the ground, but Stanford claimed that there was\na moment during each cycle of the motion when all four feet were\nin the air. The human eye was simply not fast enough to settle the\nquestion. In 1878, Muybridge finally succeeded in producing what\namounted to a motion picture of the horse, showing conclusively\nthat all four feet did leave the ground at one point. (Muybridge was\na colorful figure in San Francisco history, and his acquittal for the\nmurder of his wife’s lover was considered the trial of the century in\nCalifornia.)\nThe losers of the bet had probably been influenced by Aristotelian reasoning, for instance the expectation that a leaping horse\nwould lose horizontal velocity while in the air with no force to push\nit forward, so that it would be more efficient for the horse to run\nwithout leaping. But even for students who have converted whole-\n199\nheartedly to Newtonianism, the relationship between force and acceleration leads to some conceptual difficulties, the main one being\na problem with the true but seemingly absurd statement that an\nobject can have an acceleration vector whose direction is not the\nsame as the direction of motion. The horse, for instance, has nearly\nconstant horizontal velocity, so its ax is zero. But as anyone can tell\nyou who has ridden a galloping horse, the horse accelerates up and\ndown. The horse’s acceleration vector therefore changes back and\nforth between the up and down directions, but is never in the same\ndirection as the horse’s motion. In this chapter, we will examine\nmore carefully the properties of the velocity, acceleration, and force\nvectors. No new principles are introduced, but an attempt is made\nto tie things together and show examples of the power of the vector\nformulation of Newton’s laws.\n8.1 The Velocity Vector\nFor motion with constant velocity, the velocity vector is\nv = ∆r/∆t\n.\n[only for constant velocity]\nThe ∆r vector points in the direction of the motion, and dividing\nit by the scalar ∆t only changes its length, not its direction, so the\nvelocity vector points in the same direction as the motion. When\nthe velocity is not constant, i.e., when the x−t, y−t, and z−t graphs\nare not all linear, we use the slope-of-the-tangent-line approach to\ndefine the components vx , vy , and vz , from which we assemble the\nvelocity vector. Even when the velocity vector is not constant, it\nstill points along the direction of motion.\nVector addition is the correct way to generalize the one-dimensional\nconcept of adding velocities in relative motion, as shown in the following example:\nVelocity vectors in relative motion\nexample 1\n. You wish to cross a river and arrive at a dock that is directly\nacross from you, but the river’s current will tend to carry you\ndownstream. To compensate, you must steer the boat at an angle. Find the angle θ, given the magnitude, |vW L |, of the water’s\nvelocity relative to the land, and the maximum speed, |vBW |, of\nwhich the boat is capable relative to the water.\n. The boat’s velocity relative to the land equals the vector sum of\nits velocity with respect to the water and the water’s velocity with\nrespect to the land,\na / Example 1.\nvBL = vBW + vW L\n.\nIf the boat is to travel straight across the river, i.e., along the y\naxis, then we need to have vBL,x = 0. This x component equals\nthe sum of the x components of the other two vectors,\nvB L, x = vB W , x + vW L, x\n200\nChapter 8\nVectors and Motion\n,\nor\n0 = −|vBW | sin θ + |vW L |\n.\nSolving for θ, we find\nsin θ = |vW L |/|vBW |\n,\nso\nθ = sin−1\n. Solved problem: Annie Oakley\n|vW L |\nvBW\n.\npage 212, problem 8\nDiscussion Questions\nA\nIs it possible for an airplane to maintain a constant velocity vector\nbut not a constant |v|? How about the opposite – a constant |v| but not a\nconstant velocity vector? Explain.\nB\nNew York and Rome are at about the same latitude, so the earth’s\nrotation carries them both around nearly the same circle. Do the two cities\nhave the same velocity vector (relative to the center of the earth)? If not,\nis there any way for two cities to have the same velocity vector?\nSection 8.1\nThe Velocity Vector\n201\n8.2 The Acceleration Vector\nWhen all three acceleration components are constant, i.e., when\nthe vx − t, vy − t, and vz − t graphs are all linear, we can define the\nacceleration vector as\na = ∆v/∆t\n,\n[only for constant acceleration]\nwhich can be written in terms of initial and final velocities as\na = (vf − vi )/∆t\nb / A change in the magnitude of the velocity vector implies\nan acceleration.\n.\n[only for constant acceleration]\nIf the acceleration is not constant, we define it as the vector made\nout of the ax , ay , and az components found by applying the slopeof-the-tangent-line technique to the vx − t, vy − t, and vz − t graphs.\nNow there are two ways in which we could have a nonzero acceleration. Either the magnitude or the direction of the velocity vector\ncould change. This can be visualized with arrow diagrams as shown\nin figures b and c. Both the magnitude and direction can change\nsimultaneously, as when a car accelerates while turning. Only when\nthe magnitude of the velocity changes while its direction stays constant do we have a ∆v vector and an acceleration vector along the\nsame line as the motion.\nself-check A\n(1) In figure b, is the object speeding up, or slowing down? (2) What\nwould the diagram look like if vi was the same as vf ? (3) Describe how\nthe ∆v vector is different depending on whether an object is speeding\nup or slowing down.\n. Answer, p. 268\nc / A change in the direction\nof the velocity vector also produces a nonzero ∆v vector, and\nthus a nonzero acceleration\nvector, ∆v/∆t .\nIf this all seems a little strange and abstract to you, you’re not\nalone. It doesn’t mean much to most physics students the first\ntime someone tells them that acceleration is a vector, and that the\nacceleration vector does not have to be in the same direction as the\nvelocity vector. One way to understand those statements better is\nto imagine an object such as an air freshener or a pair of fuzzy dice\nhanging from the rear-view mirror of a car. Such a hanging object,\ncalled a bob, constitutes an accelerometer. If you watch the bob\nas you accelerate from a stop light, you’ll see it swing backward.\nThe horizontal direction in which the bob tilts is opposite to the\ndirection of the acceleration. If you apply the brakes and the car’s\nacceleration vector points backward, the bob tilts forward.\nAfter accelerating and slowing down a few times, you think\nyou’ve put your accelerometer through its paces, but then you make\na right turn. Surprise! Acceleration is a vector, and needn’t point\nin the same direction as the velocity vector. As you make a right\nturn, the bob swings outward, to your left. That means the car’s\nacceleration vector is to your right, perpendicular to your velocity\nvector. A useful definition of an acceleration vector should relate\nin a systematic way to the actual physical effects produced by the\n202\nChapter 8\nVectors and Motion\nacceleration, so a physically reasonable definition of the acceleration\nvector must allow for cases where it is not in the same direction as\nthe motion.\nself-check B\nIn projectile motion, what direction does the acceleration vector have?\n. Answer, p. 268\nd / Example 2.\nRappelling\nexample 2\nIn figure d, the rappeller’s velocity has long periods of gradual\nchange interspersed with short periods of rapid change. These\ncorrespond to periods of small acceleration and force, and periods of large acceleration and force.\nThe galloping horse\nexample 3\nFigure e on page 204 shows outlines traced from the first, third,\nfifth, seventh, and ninth frames in Muybridge’s series of photographs of the galloping horse. The estimated location of the\nhorse’s center of mass is shown with a circle, which bobs above\nand below the horizontal dashed line.\nIf we don’t care about calculating velocities and accelerations in\nany particular system of units, then we can pretend that the time\nbetween frames is one unit. The horse’s velocity vector as it\nmoves from one point to the next can then be found simply by\ndrawing an arrow to connect one position of the center of mass to\nthe next. This produces a series of velocity vectors which alter-\nSection 8.2\nThe Acceleration Vector\n203\ne / Example 3.\nnate between pointing above and below horizontal.\nThe ∆v vector is the vector which we would have to add onto one\nvelocity vector in order to get the next velocity vector in the series.\nThe ∆v vector alternates between pointing down (around the time\nwhen the horse is in the air, B) and up (around the time when the\nhorse has two feet on the ground, D).\nDiscussion Questions\nA\nWhen a car accelerates, why does a bob hanging from the rearview\nmirror swing toward the back of the car? Is it because a force throws it\nbackward? If so, what force? Similarly, describe what happens in the\nother cases described above.\nB\nSuperman is guiding a crippled spaceship into port. The ship’s\nengines are not working. If Superman suddenly changes the direction of\nhis force on the ship, does the ship’s velocity vector change suddenly? Its\nacceleration vector? Its direction of motion?\n204\nChapter 8\nVectors and Motion\n8.3 The Force Vector and Simple Machines\nForce is relatively easy to intuit as a vector. The force vector points\nin the direction in which it is trying to accelerate the object it is\nacting on.\nSince force vectors are so much easier to visualize than acceleration vectors, it is often helpful to first find the direction of the\n(total) force vector acting on an object, and then use that information to determine the direction of the acceleration vector. Newton’s\nsecond law, Ftotal = ma, tells us that the two must be in the same\ndirection.\nA component of a force vector\nexample 4\nFigure f, redrawn from a classic 1920 textbook, shows a boy\npulling another child on a sled. His force has both a horizontal\ncomponent and a vertical one, but only the horizontal one accelerates the sled. (The vertical component just partially cancels the\nforce of gravity, causing a decrease in the normal force between\nthe runners and the snow.) There are two triangles in the figure.\nOne triangle’s hypotenuse is the rope, and the other’s is the magnitude of the force. These triangles are similar, so their internal\nangles are all the same, but they are not the same triangle. One\nis a distance triangle, with sides measured in meters, the other\na force triangle, with sides in newtons. In both cases, the horizontal leg is 93% as long as the hypotenuse. It does not make\nsense, however, to compare the sizes of the triangles — the force\ntriangle is not smaller in any meaningful sense.\nPushing a block up a ramp\nexample 5\n. Figure (a) shows a block being pushed up a frictionless ramp\nat constant speed by an applied force FA . How much force is\nrequired, in terms of the block’s mass, m, and the angle of the\nramp, θ?\n. Figure (b) shows the other two forces acting on the block: a\nnormal force, FN , created by the ramp, and the weight force, FW ,\ncreated by the earth’s gravity. Because the block is being pushed\nup at constant speed, it has zero acceleration, and the total force\non it must be zero. From figure (c), we find\n|FA | = |FW | sin θ\n= mg sin θ\n.\nSince the sine is always less than one, the applied force is always\nless than mg, i.e., pushing the block up the ramp is easier than\nlifting it straight up. This is presumably the principle on which the\npyramids were constructed: the ancient Egyptians would have\nhad a hard time applying the forces of enough slaves to equal the\nfull weight of the huge blocks of stone.\nEssentially the same analysis applies to several other simple ma-\nSection 8.3\nf / Example 4.\ng / The applied force FA pushes\nthe block up the frictionless ramp.\nh / Three forces act on the\nblock. Their vector sum is zero.\ni / If the block is to move at\nconstant velocity, Newton’s first\nlaw says that the three force\nvectors acting on it must add\nup to zero. To perform vector\naddition, we put the vectors tip\nto tail, and in this case we are\nadding three vectors, so each\none’s tail goes against the tip of\nthe previous one. Since they are\nsupposed to add up to zero, the\nthird vector’s tip must come back\nto touch the tail of the first vector.\nThey form a triangle, and since\nthe applied force is perpendicular\nto the normal force, it is a right\ntriangle.\nThe Force Vector and Simple Machines\n205\nchines, such as the wedge and the screw.\n. Solved problem: A cargo plane\npage 212, problem 9\n. Solved problem: The angle of repose\npage 213, problem 11\n. Solved problem: A wagon\npage 212, problem 10\nDiscussion Questions\nDiscussion question A.\nA The figure shows a block being pressed diagonally upward against a\nwall, causing it to slide up the wall. Analyze the forces involved, including\ntheir directions.\nB The figure shows a roller coaster car rolling down and then up under\nthe influence of gravity. Sketch the car’s velocity vectors and acceleration\nvectors. Pick an interesting point in the motion and sketch a set of force\nvectors acting on the car whose vector sum could have resulted in the\nright acceleration vector.\n8.4\nj / Discussion question B.\nR\nCalculus With Vectors\nUsing the unit vector notation introduced in section 7.4, the definitions of the velocity and acceleration components given in chapter\n6 can be translated into calculus notation as\ndy\ndz\ndx\nx̂ +\nŷ +\nẑ\nv=\ndt\ndt\ndt\nand\ndvy\ndvx\ndvz\nx̂ +\nŷ +\nẑ\n.\ndt\ndt\ndt\nTo make the notation less cumbersome, we generalize the concept\nof the derivative to include derivatives of vectors, so that we can\nabbreviate the above equations as\na=\nv=\ndr\ndt\nand\ndv\n.\ndt\nIn words, to take the derivative of a vector, you take the derivatives\nof its components and make a new vector out of those. This definition means that the derivative of a vector function has the familiar\nproperties\na=\nd(cf )\nd(f )\n=c\ndt\ndt\n[c is a constant]\nand\nd(f + g)\nd(f ) d(g)\n=\n+\n.\ndt\ndt\ndt\nThe integral of a vector is likewise defined as integrating component\nby component.\n206\nChapter 8\nVectors and Motion\nThe second derivative of a vector\nexample 6\n. Two objects have positions as functions of time given by the\nequations\nr1 = 3t 2 x̂ + t ŷ\nand\nr2 = 3t 4 x̂ + t ŷ\n.\nFind both objects’ accelerations using calculus. Could either answer have been found without calculus?\n. Taking the first derivative of each component, we find\nv1 = 6t x̂ + ŷ\nv2 = 12t 3 x̂ + ŷ\n,\nand taking the derivatives again gives acceleration,\na1 = 6x̂\na2 = 36t 2 x̂\n.\nThe first object’s acceleration could have been found without calculus, simply by comparing the x and y coordinates with the\nconstant-acceleration equation ∆x = vo ∆t + 21 a∆t 2 . The second\nequation, however, isn’t just a second-order polynomial in t, so\nthe acceleration isn’t constant, and we really did need calculus to\nfind the corresponding acceleration.\nThe integral of a vector\nexample 7\n. Starting from rest, a flying saucer of mass m is observed to\nvary its propulsion with mathematical precision according to the\nequation\nF = bt 42 x̂ + ct 137 ŷ\n.\n(The aliens inform us that the numbers 42 and 137 have a special\nreligious significance for them.) Find the saucer’s velocity as a\nfunction of time.\n. From the given force, we can easily find the acceleration\nF\nm\nb\nc\n= t 42 x̂ + t 137 ŷ\nm\nm\na=\n.\nThe velocity vector v is the integral with respect to time of the\nacceleration,\nZ\nv = a dt\nZ b 42\nc 137\nt x̂ + t ŷ dt\n,\n=\nm\nm\nSection 8.4\nR\nCalculus With Vectors\n207\nand integrating component by component gives\nZ\nZ\nb 42\nc 137\n=\nt dt x̂ +\nt\ndt ŷ\nm\nm\nc\nb 43\n=\nt x̂ +\nt 138 ŷ\n,\n43m\n138m\nwhere we have omitted the constants of integration, since the\nsaucer was starting from rest.\nA fire-extinguisher stunt on ice\nexample 8\n. Prof. Puerile smuggles a fire extinguisher into a skating rink.\nClimbing out onto the ice without any skates on, he sits down and\npushes off from the wall with his feet, acquiring an initial velocity\nvo ŷ. At t = 0, he then discharges the fire extinguisher at a 45degree angle so that it applies a force to him that is backward\nand to the left, i.e., along the negative y axis and the positive x\naxis. The fire extinguisher’s force is strong at first, but then dies\ndown according to the equation |F| = b − ct, where b and c are\nconstants. Find the professor’s velocity as a function of time.\n. Measured counterclockwise from the x axis, the angle of the\nforce vector becomes 315 ◦ . Breaking the force down into x and\ny components, we have\nFx = |F| cos 315 ◦\n= (b − ct)\nFy = |F| sin 315 ◦\n= (−b + ct)\n.\nIn unit vector notation, this is\nF = (b − ct)x̂ + (−b + ct)ŷ\n.\nNewton’s second law gives\na = F/m\nb − ct\n−b + ct\n= √\nx̂ + √\nŷ\n2m\n2m\n.\nTo find the velocity vector as a function of time, we need to integrate the acceleration vector with respect to time,\nZ\nv = a dt\nZ −b + ct\nb − ct\n√\n=\nx̂ + √\nŷ dt\n2m\n2m\nZ\n1\n=√\n(b − ct) x̂ + (−b + ct) ŷ dt\n2m\n208\nChapter 8\nVectors and Motion\nA vector function can be integrated component by component, so\nthis can be broken down into two integrals,\nv= √\n=\nx̂\nZ\n(b − ct) dt + √\nŷ\nZ\n(−b + ct) dt\n2m\n2m\n!\n!\nbt − 12 ct 2\n−bt + 21 ct 2\n√\n√\n+ constant #1 x̂ +\n+ constant #2 ŷ\n2m\n2m\nHere the physical significance of the two constants of integration\nis that they give the initial velocity. Constant #1 is therefore zero,\nand constant #2 must equal vo . The final result is\n!\n!\nbt − 12 ct 2\n−bt + 12 ct 2\n√\n√\n.\nv=\nx̂ +\n+ vo ŷ\n2m\n2m\nSection 8.4\nR\nCalculus With Vectors\n209\nSummary\nThe velocity vector points in the direction of the object’s motion.\nRelative motion can be described by vector addition of velocities.\nThe acceleration vector need not point in the same direction as\nthe object’s motion. We use the word “acceleration” to describe any\nchange in an object’s velocity vector, which can be either a change\nin its magnitude or a change in its direction.\nAn important application of the vector addition of forces is the\nuse of Newton’s first law to analyze mechanical systems.\n210\nChapter 8\nVectors and Motion\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\nProblem 1.\n1\nA dinosaur fossil is slowly moving down the slope of a glacier\nunder the influence of wind, rain and gravity. At the same time,\nthe glacier is moving relative to the continent underneath. The\ndashed lines represent the directions but not the magnitudes of the\nvelocities. Pick a scale, and use graphical addition of vectors to find\nthe magnitude and the direction of the fossil’s velocity relative\nto\n√\nthe continent. You will need a ruler and protractor.\n2\nIs it possible for a helicopter to have an acceleration due east\nand a velocity due west? If so, what would be going on? If not, why\nnot?\n3\nA bird is initially flying horizontally east at 21.1 m/s, but one\nsecond later it has changed direction so that it is flying horizontally\nand 7 ◦ north of east, at the same speed. What are the magnitude\nand direction of its acceleration vector during that one second √\ntime\ninterval? (Assume its acceleration was roughly constant.)\nProblem 4.\n4\nA person of mass M stands in the middle of a tightrope,\nwhich is fixed at the ends to two buildings separated by a horizontal\ndistance L. The rope sags in the middle, stretching and lengthening\nthe rope slightly.\nProblems\n211\n(a) If the tightrope walker wants the rope to sag vertically by no\nmore than a height h, find the minimum tension, T , that the rope\nmust be able to withstand without breaking, in terms of h, g, M√,\nand L.\n(b) Based on your equation, explain why it is not possible to get\nh = 0, and give a physical interpretation.\n5\nYour hand presses a block of mass m against a wall with a\nforce FH acting at an angle θ. Find the minimum and maximum\npossible values of |FH | that can keep the block stationary, in terms\nof m, g, θ, and µs , the coefficient of static friction between the √block\nand the wall.\n?\n6\nA skier of mass m is coasting down a slope inclined at an angle\nθ compared to horizontal. Assume for simplicity that the treatment\nof kinetic friction given in chapter 5 is appropriate here, although a\nsoft and wet surface actually behaves a little differently. The coefficient of kinetic friction acting between the skis and the snow is µk ,\nand in addition the skier experiences an air friction force of magnitude bv 2 , where b is a constant.\n(a) Find the maximum speed that the skier will attain, in terms of\n√\nthe variables m, g, θ, µk , and b.\n(b) For angles below a certain minimum angle θmin , the equation\ngives a result that is not mathematically meaningful. Find an equation for θmin , and give a physical explanation of what is happening\nfor θ < θmin .\nProblem 5.\n7\nA gun is aimed horizontally to the west, and fired at t = 0. The\nbullet’s position vector as a function of time is r = bx̂ + ctŷ + dt2 ẑ,\nwhere b, c, and d are positive constants.\n(a) What units would b, c, and d need to have for the equation to\nmake sense?\n(b) Find the bullet’s velocity and acceleration as functions of time.\nR\n(c) Give physical interpretations of b, c, d, x̂, ŷ, and ẑ.\n8\nAnnie Oakley, riding north on horseback at 30 mi/hr, shoots\nher rifle, aiming horizontally and to the northeast. The muzzle speed\nof the rifle is 140 mi/hr. When the bullet hits a defenseless fuzzy\nanimal, what is its speed of impact? Neglect air resistance, and\nignore the vertical motion of the bullet.\n. Solution, p. 275\n9\nA cargo plane has taken off from a tiny airstrip in the Andes,\nand is climbing at constant speed, at an angle of θ=17 ◦ with respect\nto horizontal. Its engines supply a thrust of Fthrust = 200 kN, and\nthe lift from its wings is Flif t = 654 kN. Assume that air resistance\n(drag) is negligible, so the only forces acting are thrust, lift, and\nweight. What is its mass, in kg?\n. Solution, p. 276\nProblem 9.\n10\nA wagon is being pulled at constant speed up a slope θ by a\nrope that makes an angle φ with the vertical.\n(a) Assuming negligible friction, show that the tension in the rope\nProblem 10.\n212\nChapter 8\nVectors and Motion\nis given by the equation\nFT =\nsin θ\nFW\nsin(θ + φ)\n,\nwhere FW is the weight force acting on the wagon.\n(b) Interpret this equation in the special cases of φ = 0 and φ =\n180 ◦ − θ.\n. Solution, p. 276\n11\nThe angle of repose is the maximum slope on which an object\nwill not slide. On airless, geologically inert bodies like the moon or\nan asteroid, the only thing that determines whether dust or rubble\nwill stay on a slope is whether the slope is less steep than the angle\nof repose.\n(a) Find an equation for the angle of repose, deciding for yourself\nwhat are the relevant variables.\n(b) On an asteroid, where g can be thousands of times lower than\non Earth, would rubble be able to lie at a steeper angle of repose?\n. Solution, p. 277\n12\nThe figure shows an experiment in which a cart is released\nfrom rest at A, and accelerates down the slope through a distance\nx until it passes through a sensor’s light beam. The point of the\nexperiment is to determine the cart’s acceleration. At B, a cardboard vane mounted on the cart enters the light beam, blocking the\nlight beam, and starts an electronic timer running. At C, the vane\nemerges from the beam, and the timer stops.\n(a) Find the final velocity of the cart in terms of the width w of\nthe vane and the time tb for which the sensor’s light beam was\n√\nblocked.\n(b) Find the magnitude of the cart’s acceleration in terms of the\n√\nmeasurable quantities x, tb , and w.\n(c) Analyze the forces in which the cart participates, using a table in\nthe format introduced in section 5.3. Assume friction is negligible.\n(d) Find a theoretical value for the acceleration of the cart, which\ncould be compared with the experimentally observed value extracted\nin part b. Express the theoretical value in terms of the angle √θ of\nthe slope, and the strength g of the gravitational field.\n13\nThe figure shows a boy hanging in three positions: (1) with\nhis arms straight up, (2) with his arms at 45 degrees, and (3) with\nhis arms at 60 degrees with respect to the vertical. Compare the\ntension in his arms in the three cases.\nProblem 12.\nProblem 13 (Millikan and Gale,\n1920).\nProblems\n213\n14\nDriving down a hill inclined at an angle θ with respect to\nhorizontal, you slam on the brakes to keep from hitting a deer.\n(a) Analyze the forces. (Ignore rolling resistance and air friction.)\n(b) Find the car’s maximum possible deceleration, a (expressed as\na positive number), in terms of g, θ, and the relevant coefficient of\n√\nfriction.\n(c) Explain physically why the car’s mass has no effect on your\nanswer.\n(d) Discuss the mathematical behavior and physical interpretation\nof your result for negative values of θ.\n(e) Do the same for very large positive values of θ.\n15\nThe figure shows the path followed by Hurricane Irene in\n2005 as it moved north. The dots show the location of the center\nof the storm at six-hour intervals, with lighter dots at the time\nwhen the storm reached its greatest intensity. Find the time when\nthe storm’s center had a velocity vector to the northeast and an\nacceleration vector to the southeast.\nProblem 15.\n214\nChapter 8\nVectors and Motion\nChapter 9\nCircular Motion\n9.1 Conceptual Framework for Circular Motion\nI now live fifteen minutes from Disneyland, so my friends and family\nin my native Northern California think it’s a little strange that I’ve\nnever visited the Magic Kingdom again since a childhood trip to the\nsouth. The truth is that for me as a preschooler, Disneyland was\nnot the Happiest Place on Earth. My mother took me on a ride in\nwhich little cars shaped like rocket ships circled rapidly around a\ncentral pillar. I knew I was going to die. There was a force trying to\nthrow me outward, and the safety features of the ride would surely\nhave been inadequate if I hadn’t screamed the whole time to make\nsure Mom would hold on to me. Afterward, she seemed surprisingly\nindifferent to the extreme danger we had experienced.\nCircular motion does not produce an outward force\nMy younger self’s understanding of circular motion was partly\nright and partly wrong. I was wrong in believing that there was a\nforce pulling me outward, away from the center of the circle. The\neasiest way to understand this is to bring back the parable of the\nbowling ball in the pickup truck from chapter 4. As the truck makes\na left turn, the driver looks in the rearview mirror and thinks that\nsome mysterious force is pulling the ball outward, but the truck\nis accelerating, so the driver’s frame of reference is not an inertial\nframe. Newton’s laws are violated in a noninertial frame, so the ball\nappears to accelerate without any actual force acting on it. Because\nwe are used to inertial frames, in which accelerations are caused by\n215\nforces, the ball’s acceleration creates a vivid illusion that there must\nbe an outward force.\na / 1. In the turning truck’s frame\nof reference, the ball appears to\nviolate Newton’s laws, displaying a sideways acceleration that\nis not the re- sult of a forceinteraction with any other object.\n2. In an inertial frame of reference, such as the frame fixed to\nthe earth’s surface, the ball obeys\nNewton’s first law. No forces are\nacting on it, and it continues moving in a straight line. It is the truck\nthat is participating in an interaction with the asphalt, the truck that\naccelerates as it should according\nto Newton’s second law.\nb / This crane fly’s halteres\nhelp it to maintain its orientation\nin flight.\nIn an inertial frame everything makes more sense. The ball has\nno force on it, and goes straight as required by Newton’s first law.\nThe truck has a force on it from the asphalt, and responds to it\nby accelerating (changing the direction of its velocity vector) as\nNewton’s second law says it should.\nThe halteres\nexample 1\nAnother interesting example is an insect organ called the halteres, a pair of small knobbed limbs behind the wings, which vibrate up and down and help the insect to maintain its orientation\nin flight. The halteres evolved from a second pair of wings possessed by earlier insects. Suppose, for example, that the halteres\nare on their upward stroke, and at that moment an air current\ncauses the fly to pitch its nose down. The halteres follow Newton’s first law, continuing to rise vertically, but in the fly’s rotating\nframe of reference, it seems as though they have been subjected\nto a backward force. The fly has special sensory organs that perceive this twist, and help it to correct itself by raising its nose.\nCircular motion does not persist without a force\nI was correct, however, on a different point about the Disneyland\nride. To make me curve around with the car, I really did need some\nforce such as a force from my mother, friction from the seat, or a\nnormal force from the side of the car. (In fact, all three forces were\nprobably adding together.) One of the reasons why Galileo failed to\n216\nChapter 9\nCircular Motion\nc / 1. An overhead view of a person swinging a rock on a rope.\nA force from the string is required to make the rock’s velocity\nvector keep changing direc- tion.\n2. If the string breaks, the rock\nwill follow Newton’s first law and\ngo straight instead of continuing\naround the circle.\nrefine the principle of inertia into a quantitative statement like Newton’s first law is that he was not sure whether motion without a force\nwould naturally be circular or linear. In fact, the most impressive\nexamples he knew of the persistence of motion were mostly circular:\nthe spinning of a top or the rotation of the earth, for example. Newton realized that in examples such as these, there really were forces\nat work. Atoms on the surface of the top are prevented from flying\noff straight by the ordinary force that keeps atoms stuck together in\nsolid matter. The earth is nearly all liquid, but gravitational forces\npull all its parts inward.\nUniform and nonuniform circular motion\nCircular motion always involves a change in the direction of the\nvelocity vector, but it is also possible for the magnitude of the velocity to change at the same time. Circular motion is referred to as\nuniform if |v| is constant, and nonuniform if it is changing.\nYour speedometer tells you the magnitude of your car’s velocity\nvector, so when you go around a curve while keeping your speedometer needle steady, you are executing uniform circular motion. If your\nspeedometer reading is changing as you turn, your circular motion\nis nonuniform. Uniform circular motion is simpler to analyze mathematically, so we will attack it first and then pass to the nonuniform\ncase.\nself-check A\nWhich of these are examples of uniform circular motion and which are\nnonuniform?\n(1) the clothes in a clothes dryer (assuming they remain against the\ninside of the drum, even at the top)\n(2) a rock on the end of a string being whirled in a vertical circle\nAnswer, p. 268\nSection 9.1\n.\nConceptual Framework for Circular Motion\n217\nOnly an inward force is required for uniform circular motion.\nFigure c showed the string pulling in straight along a radius of\nthe circle, but many people believe that when they are doing this\nthey must be “leading” the rock a little to keep it moving along.\nThat is, they believe that the force required to produce uniform\ncircular motion is not directly inward but at a slight angle to the\nradius of the circle. This intuition is incorrect, which you can easily\nverify for yourself now if you have some string handy. It is only\nwhile you are getting the object going that your force needs to be at\nan angle to the radius. During this initial period of speeding up, the\nmotion is not uniform. Once you settle down into uniform circular\nmotion, you only apply an inward force.\nd / To make the brick go in a\ncircle, I had to exert an inward\nforce on the rope.\nf / When a car is going straight\nat constant speed, the forward\nand backward forces on it are\ncanceling out, producing a total\nforce of zero. When it moves\nin a circle at constant speed,\nthere are three forces on it, but\nthe forward and backward forces\ncancel out, so the vector sum is\nan inward force.\n218\nChapter 9\nIf you have not done the experiment for yourself, here is a theoretical argument to convince you of this fact. We have discussed in\nchapter 6 the principle that forces have no perpendicular effects. To\nkeep the rock from speeding up or slowing down, we only need to\nmake sure that our force is perpendicular to its direction of motion.\nWe are then guaranteed that its forward motion will remain unaffected: our force can have no perpendicular effect, and there is no\nother force acting on the rock which could slow it down. The rock\nrequires no forward force to maintain its forward motion, any more\nthan a projectile needs a horizontal force to “help it over the top”\nof its arc.\ne / A series of three hammer taps makes the rolling ball trace a triangle, seven hammers a heptagon. If the number of hammers was large\nenough, the ball would essentially be experiencing a steady inward force,\nand it would go in a circle. In no case is any forward force necessary.\nCircular Motion\nWhy, then, does a car driving in circles in a parking lot stop\nexecuting uniform circular motion if you take your foot off the gas?\nThe source of confusion here is that Newton’s laws predict an object’s motion based on the total force acting on it. A car driving in\ncircles has three forces on it\n(1) an inward force from the asphalt, controlled with the steering\nwheel;\n(2) a forward force from the asphalt, controlled with the gas\npedal; and\n(3) backward forces from air resistance and rolling resistance.\nYou need to make sure there is a forward force on the car so that\nthe backward forces will be exactly canceled out, creating a vector\nsum that points directly inward.\ng / Example 2.\nA motorcycle making a turn\nexample 2\nThe motorcyclist in figure g is moving along an arc of a circle. It\nlooks like he’s chosen to ride the slanted surface of the dirt at a\nplace where it makes just the angle he wants, allowing him to get\nthe force he needs on the tires as a normal force, without needing\nany frictional force. The dirt’s normal force on the tires points up\nand to our left. The vertical component of that force is canceled\nby gravity, while its horizontal component causes him to curve.\nIn uniform circular motion, the acceleration vector is inward\nSince experiments show that the force vector points directly\ninward, Newton’s second law implies that the acceleration vector\npoints inward as well. This fact can also be proven on purely kinematical grounds, and we will do so in the next section.\nSection 9.1\nConceptual Framework for Circular Motion\n219\nDiscussion Questions\nA In the game of crack the whip, a line of people stand holding hands,\nand then they start sweeping out a circle. One person is at the center, and\nrotates without changing location. At the opposite end is the person who\nis running the fastest, in a wide circle. In this game, someone always ends\nup losing their grip and flying off. Suppose the person on the end loses\nher grip. What path does she follow as she goes flying off? (Assume she\nis going so fast that she is really just trying to put one foot in front of the\nother fast enough to keep from falling; she is not able to get any significant\nhorizontal force between her feet and the ground.)\nDiscussion\nquestions\nDiscussion question E.\nA-D\nB\nSuppose the person on the outside is still holding on, but feels that\nshe may loose her grip at any moment. What force or forces are acting\non her, and in what directions are they? (We are not interested in the\nvertical forces, which are the earth’s gravitational force pulling down, and\nthe ground’s normal force pushing up.)\nC\nSuppose the person on the outside is still holding on, but feels that\nshe may loose her grip at any moment. What is wrong with the following\nanalysis of the situation? “The person whose hand she’s holding exerts\nan inward force on her, and because of Newton’s third law, there’s an\nequal and opposite force acting outward. That outward force is the one\nshe feels throwing her outward, and the outward force is what might make\nher go flying off, if it’s strong enough.”\nD\nIf the only force felt by the person on the outside is an inward force,\nwhy doesn’t she go straight in?\nE\nIn the amusement park ride shown in the figure, the cylinder spins\nfaster and faster until the customer can pick her feet up off the floor without falling. In the old Coney Island version of the ride, the floor actually\ndropped out like a trap door, showing the ocean below. (There is also a\nversion in which the whole thing tilts up diagonally, but we’re discussing\nthe version that stays flat.) If there is no outward force acting on her, why\ndoes she stick to the wall? Analyze all the forces on her.\nF\nWhat is an example of circular motion where the inward force is a\nnormal force? What is an example of circular motion where the inward\nforce is friction? What is an example of circular motion where the inward\nforce is the sum of more than one force?\nG\nDoes the acceleration vector always change continuously in circular\nmotion? The velocity vector?\n220\nChapter 9\nCircular Motion\n9.2 Uniform Circular Motion\nIn this section I derive a simple and very useful equation for\nthe magnitude of the acceleration of an object undergoing constant\nacceleration. The law of sines is involved, so I’ve recapped it in\nfigure h.\nThe derivation is brief, but the method requires some explanation and justification. The idea is to calculate a ∆v vector describing\nthe change in the velocity vector as the object passes through an\nangle θ. We then calculate the acceleration, a = ∆v/∆t. The astute reader will recall, however, that this equation is only valid for\nmotion with constant acceleration. Although the magnitude of the\nacceleration is constant for uniform circular motion, the acceleration\nvector changes its direction, so it is not a constant vector, and the\nequation a = ∆v/∆t does not apply. The justification for using it\nis that we will then examine its behavior when we make the time\ninterval very short, which means making the angle θ very small. For\nsmaller and smaller time intervals, the ∆v/∆t expression becomes\na better and better approximation, so that the final result of the\nderivation is exact.\nIn figure i/1, the object sweeps out an angle θ. Its direction of\nmotion also twists around by an angle θ, from the vertical dashed\nline to the tilted one. Figure i/2 shows the initial and final velocity\nvectors, which have equal magnitude, but directions differing by θ.\nIn i/3, I’ve reassembled the vectors in the proper positions for vector\nsubtraction. They form an isosceles triangle with interior angles θ,\nη, and η. (Eta, η, is my favorite Greek letter.) The law of sines\ngives\n|∆v|\n|v|\n=\n.\nsin θ\nsin η\nh / The law of sines.\ni / Deriving |a| = |v|2 /r\nuniform circular motion.\nfor\nThis tells us the magnitude of ∆v, which is one of the two ingredients\nwe need for calculating the magnitude of a = ∆v/∆t. The other\ningredient is ∆t. The time required for the object to move through\nthe angle θ is\nlength of arc\n∆t =\n.\n|v|\nNow if we measure our angles in radians we can use the definition of\nradian measure, which is (angle) = (length of arc)/(radius), giving\n∆t = θr/|v|. Combining this with the first expression involving\n|∆v| gives\n|a| = |∆v|/∆t\n=\n|v|2 sin θ\n1\n·\n·\nr\nθ\nsin η\n.\nWhen θ becomes very small, the small-angle approximation sin θ ≈ θ\napplies, and also η becomes close to 90 ◦ , so sin η ≈ 1, and we have\nSection 9.2\nUniform Circular Motion\n221\nan equation for |a|:\n|a| =\n|v|2\nr\n.\n[uniform circular motion]\nForce required to turn on a bike\nexample 3\n. A bicyclist is making a turn along an arc of a circle with radius\n20 m, at a speed of 5 m/s. If the combined mass of the cyclist\nplus the bike is 60 kg, how great a static friction force must the\nroad be able to exert on the tires?\n. Taking the magnitudes of both sides of Newton’s second law\ngives\n|F| = |ma|\n= m|a|\nSubstituting |a| =\n|v|2 /r\n.\ngives\n|F| = m|v|2 /r\n≈ 80 N\n(rounded off to one sig fig).\nDon’t hug the center line on a curve!\nexample 4\n. You’re driving on a mountain road with a steep drop on your\nright. When making a left turn, is it safer to hug the center line or\nto stay closer to the outside of the road?\n. You want whichever choice involves the least acceleration, because that will require the least force and entail the least risk of\nexceeding the maximum force of static friction. Assuming the\ncurve is an arc of a circle and your speed is constant, your car\nis performing uniform circular motion, with |a| = |v|2 /r . The dependence on the square of the speed shows that driving slowly\nis the main safety measure you can take, but for any given speed\nyou also want to have the largest possible value of r . Even though\nyour instinct is to keep away from that scary precipice, you are actually less likely to skid if you keep toward the outside, because\nthen you are describing a larger circle.\nAcceleration related to radius and period of rotation example 5\n. How can the equation for the acceleration in uniform circular\nmotion be rewritten in terms of the radius of the circle and the\nperiod, T , of the motion, i.e., the time required to go around once?\n. The period can be related to the speed as follows:\ncircumference\n|v| =\nT\n= 2πr /T\n.\nSubstituting into the equation |a| = |v|2 /r gives\n|a| =\nj / Example 6.\n222\nChapter 9\nCircular Motion\n4π2 r\nT2\n.\nA clothes dryer\nexample 6\n. My clothes dryer has a drum with an inside radius of 35 cm, and\nit spins at 48 revolutions per minute. What is the acceleration of\nthe clothes inside?\n. We can solve this by finding the period and plugging in to the\nresult of the previous example. If it makes 48 revolutions in one\nminute, then the period is 1/48 of a minute, or 1.25 s. To get an\nacceleration in mks units, we must convert the radius to 0.35 m.\nPlugging in, the result is 8.8 m/s2 .\nMore about clothes dryers!\nexample 7\n. In a discussion question in the previous section, we made the\nassumption that the clothes remain against the inside of the drum\nas they go over the top. In light of the previous example, is this a\ncorrect assumption?\n. No. We know that there must be some minimum speed at which\nthe motor can run that will result in the clothes just barely staying against the inside of the drum as they go over the top. If the\nclothes dryer ran at just this minimum speed, then there would be\nno normal force on the clothes at the top: they would be on the\nverge of losing contact. The only force acting on them at the top\nwould be the force of gravity, which would give them an acceleration of g = 9.8 m/s2 . The actual dryer must be running slower\nthan this minimum speed, because it produces an acceleration of\nonly 8.8 m/s2 . My theory is that this is done intentionally, to make\nthe clothes mix and tumble.\n. Solved problem: The tilt-a-whirl\npage 227, problem 6\n. Solved problem: An off-ramp\npage 227, problem 7\nDiscussion Questions\nA\nA certain amount of force is needed to provide the acceleration of\ncircular motion. What if were are exerting a force perpendicular to the\ndirection of motion in an attempt to make an object trace a circle of radius\nr , but the force isn’t as big as m|v|2 /r ?\nB Suppose a rotating space station, as in figure k on page 224, is built.\nIt gives its occupants the illusion of ordinary gravity. What happens when\na person in the station lets go of a ball? What happens when she throws\na ball straight “up” in the air (i.e., towards the center)?\nSection 9.2\nUniform Circular Motion\n223\nk / Discussion question B. An artist’s conception of a rotating space\ncolony in the form of a giant wheel. A person living in this noninertial\nframe of reference has an illusion of a force pulling her outward, toward\nthe deck, for the same reason that a person in the pickup truck has the\nillusion of a force pulling the bowling ball. By adjusting the speed of rotation, the designers can make an acceleration |v|2 /r equal to the usual\nacceleration of gravity on earth. On earth, your acceleration standing on\nthe ground is zero, and a falling rock heads for your feet with an acceleration of 9.8 m/s2 . A person standing on the deck of the space colony has\nan upward acceleration of 9.8 m/s2 , and when she lets go of a rock, her\nfeet head up at the nonaccelerating rock. To her, it seems the same as\ntrue gravity.\n9.3 Nonuniform Circular Motion\nWhat about nonuniform circular motion? Although so far we\nhave been discussing components of vectors along fixed x and y\naxes, it now becomes convenient to discuss components of the acceleration vector along the radial line (in-out) and the tangential line\n(along the direction of motion). For nonuniform circular motion,\nthe radial component of the acceleration obeys the same equation\nas for uniform circular motion,\nar = |v|2 /r\n,\nbut the acceleration vector also has a tangential component,\nat = slope of the graph of |v| versus t\n.\nThe latter quantity has a simple interpretation. If you are going\naround a curve in your car, and the speedometer needle is moving, the tangential component of the acceleration vector is simply\nwhat you would have thought the acceleration was if you saw the\nspeedometer and didn’t know you were going around a curve.\nSlow down before a turn, not during it.\nexample 8\n. When you’re making a turn in your car and you’re afraid you\nmay skid, isn’t it a good idea to slow down?\nl / 1. Moving in a circle while\nspeeding up. 2. Uniform circular\nmotion. 3. Slowing down.\n. If the turn is an arc of a circle, and you’ve already completed\npart of the turn at constant speed without skidding, then the road\nand tires are apparently capable of enough static friction to supply an acceleration of |v|2 /r . There is no reason why you would\nskid out now if you haven’t already. If you get nervous and brake,\nhowever, then you need to have a tangential acceleration component in addition to the radial component you were already able\nto produce successfully. This would require an acceleration vector with a greater magnitude, which in turn would require a larger\nforce. Static friction might not be able to supply that much force,\nand you might skid out. As in the previous example on a similar\ntopic, the safe thing to do is to approach the turn at a comfortably\nlow speed.\n. Solved problem: A bike race\n224\nChapter 9\nCircular Motion\npage 226, problem 5\nSummary\nSelected Vocabulary\nuniform circular circular motion in which the magnitude of the\nmotion . . . . . . velocity vector remains constant\nnonuniform circu- circular motion in which the magnitude of the\nlar motion . . . . velocity vector changes\nradial . . . . . . . parallel to the radius of a circle; the in-out\ndirection\ntangential . . . . tangent to the circle, perpendicular to the radial direction\nNotation\nar . . . . . . . . .\nat . . . . . . . . .\nradial acceleration; the component of the acceleration vector along the in-out direction\ntangential acceleration; the component of the\nacceleration vector tangent to the circle\nSummary\nIf an object is to have circular motion, a force must be exerted on\nit toward the center of the circle. There is no outward force on the\nobject; the illusion of an outward force comes from our experiences\nin which our point of view was rotating, so that we were viewing\nthings in a noninertial frame.\nAn object undergoing uniform circular motion has an inward\nacceleration vector of magnitude\n|a| = |v|2 /r\n.\nIn nonuniform circular motion, the radial and tangential components of the acceleration vector are\nar = |v|2 /r\nat = slope of the graph of |v| versus t\n.\nSummary\n225\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nWhen you’re done using an electric mixer, you can get most\nof the batter off of the beaters by lifting them out of the batter with\nthe motor running at a high enough speed. Let’s imagine, to make\nthings easier to visualize, that we instead have a piece of tape stuck\nto one of the beaters.\n(a) Explain why static friction has no effect on whether or not the\ntape flies off.\n(b) Suppose you find that the tape doesn’t fly off when the motor\nis on a low speed, but at a greater speed, the tape won’t stay on.\nWhy would the greater speed change things?\n2\nShow that the expression |v|2 /r has the units of acceleration.\n3\nA plane is flown in a loop-the-loop of radius 1.00 km. The\nplane starts out flying upside-down, straight and level, then begins\ncurving up along the circular loop, and is right-side up when it\nreaches the top. (The plane may slow down somewhat on the way\nup.) How fast must the plane be going at the top if the pilot is to\nexperience no force from the seat or the seatbelt while at the top\nof\n√\nthe loop?\n4\nIn this problem, you’ll derive the equation |a| = |v|2 /r using calculus. Instead of comparing velocities at two points in the\nparticle’s motion and then taking a limit where the points are close\ntogether, you’ll just take derivatives. The particle’s position vector\nis r = (r cos θ)x̂ + (r sin θ)ŷ, where and are the unit vectors along\nthe x and y axes. By the definition of radians, the distance traveled\nsince t = 0 is rθ, so if the particle is traveling at constant speed\nv = |v|, we have v = rθ/t.\n(a) Eliminate θ to get the particle’s position vector as a function of\ntime.\n(b) Find the particle’s acceleration vector.\n(c) Show that the magnitude of the acceleration vector equals vR2 /r.\n5\nThree cyclists in a race are rounding a semicircular curve.\nAt the moment depicted, cyclist A is using her brakes to apply a\nforce of 375 N to her bike. Cyclist B is coasting. Cyclist C is\npedaling, resulting in a force of 375 N on her bike Each cyclist,\nwith her bike, has a mass of 75 kg. At the instant shown, the\ninstantaneous speed of all three cyclists is 10 m/s. On the diagram,\ndraw each cyclist’s acceleration vector with its tail on top of her\npresent position, indicating the directions and lengths reasonably\naccurately. Indicate approximately the consistent scale you are using\nProblem 5.\n226\nChapter 9\nCircular Motion\nfor all three acceleration vectors. Extreme precision is not necessary\nas long as the directions are approximately right, and lengths of\nvectors that should be equal appear roughly equal, etc. Assume all\nthree cyclists are traveling along the road all the time, not wandering\nacross their lane or wiping out and going off the road.\n. Solution, p. 277\n6\nThe amusement park ride shown in the figure consists of a\ncylindrical room that rotates about its vertical axis. When the rotation is fast enough, a person against the wall can pick his or her\nfeet up off the floor and remain “stuck” to the wall without falling.\n(a) Suppose the rotation results in the person having a speed v. The\nradius of the cylinder is r, the person’s mass is m, the downward\nacceleration of gravity is g, and the coefficient of static friction between the person and the wall is µs . Find an equation for the speed,\nv, required, in terms of the other variables. (You will find that one\nof the variables cancels out.)\n(b) Now suppose two people are riding the ride. Huy is wearing\ndenim, and Gina is wearing polyester, so Huy’s coefficient of static\nfriction is three times greater. The ride starts from rest, and as it\nbegins rotating faster and faster, Gina must wait longer before being\nable to lift her feet without sliding to the floor. Based on your equation from part a, how many times greater must the speed be before\nGina can lift her feet without sliding down? . Solution, p. 277 ?\n7\nAn engineer is designing a curved off-ramp for a freeway.\nSince the off-ramp is curved, she wants to bank it to make it less\nlikely that motorists going too fast will wipe out. If the radius of\nthe curve is r, how great should the banking angle, θ, be so that\nfor a car going at a speed v, no static friction force whatsoever is\nrequired to allow the car to make the curve? State your answer in\nterms of v, r, and g, and show that the mass of the car is irrelevant.\n. Solution, p. 277\n8\nLionel brand toy trains come with sections of track in standard\nlengths and shapes. For circular arcs, the most commonly used\nsections have diameters of 662 and 1067 mm at the inside of the outer\nrail. The maximum speed at which a train can take the broader\ncurve without flying off the tracks is 0.95 m/s. At what speed must\n√\nthe train be operated to avoid derailing on the tighter curve?\n9\nThe figure shows a ball on the end of a string of length L\nattached to a vertical rod which is spun about its vertical axis by a\nmotor. The period (time for one rotation) is P .\n(a) Analyze the forces in which the ball participates.\n(b) Find how the angle θ depends on P , g, and L. [Hints: (1)\nWrite down Newton’s second law for the vertical and horizontal\ncomponents of force and acceleration. This gives two equations,\nwhich can be solved for the two unknowns, θ and the tension in\nthe string. (2) If you introduce variables like v and r, relate them\nto the variables your solution is supposed to contain, and eliminate\nProblem 6.\nProblem 7.\nProblem 9.\nProblems\n227\n√\nthem.]\n(c) What happens mathematically to your solution if the motor is\nrun very slowly (very large values of P )? Physically, what do you\nthink would actually happen in this case?\n10\nPsychology professor R.O. Dent requests funding for an experiment on compulsive thrill-seeking behavior in hamsters, in which\nthe subject is to be attached to the end of a spring and whirled\naround in a horizontal circle. The spring has equilibrium length b,\nand obeys Hooke’s law with spring constant k. It is stiff enough to\nkeep from bending significantly under the hamster’s weight.\n(a) Calculate the length of the spring when it is undergoing steady\ncircular motion in which one rotation takes a time T . Express your\n√\nresult in terms of k, m, b, T , and the hamster’s mass m.\n(b) The ethics committee somehow fails to veto the experiment, but\nthe safety committee expresses concern. Why? Does your equation do anything unusual, or even spectacular, for any particular\nvalue of T ? What do you think is the physical significance of this\nmathematical behavior?\n11\nThe figure shows an old-fashioned device called a flyball\ngovernor, used for keeping an engine running at the correct speed.\nThe whole thing rotates about the vertical shaft, and the mass M\nis free to slide up and down. This mass would have a connection\n(not shown) to a valve that controlled the engine. If, for instance,\nthe engine ran too fast, the mass would rise, causing the engine to\nslow back down.\n(a) Show that in the special case of a = 0, the angle θ is given by\n2\n−1 g(m + M )P\nθ = cos\n,\n4π 2 mL\nProblem 10.\nwhere P is the period of rotation (time required for one complete\nrotation).\n(b) There is no closed-form solution for θ in the general case where\na is not zero. However, explain how the undesirable low-speed behavior of the a = 0 device would be improved by making a nonzero.\n?\n12\nThe figure shows two blocks of masses m1 and m2 sliding\nin circles on a frictionless table. Find the tension in the strings if\nthe period of rotation (time required for one complete rotation)\nis\n√\nP.\nProblem 11.\nProblem 12.\n228\n13\nThe acceleration of an object in uniform circular motion can\nbe given either by |a| = |v|2 /r or, equivalently, by |a| = 4π 2 r/T 2 ,\nwhere T is the time required for one cycle (example 5 on page 222).\nPerson A says based on the first equation that the acceleration in\ncircular motion is greater when the circle is smaller. Person B, arguing from the second equation, says that the acceleration is smaller\nwhen the circle is smaller. Rewrite the two statements so that they\nare less misleading, eliminating the supposed paradox. [Based on a\nproblem by Arnold Arons.]\nChapter 9\nCircular Motion\nGravity is the only really important force on the cosmic scale. This falsecolor representation of saturn’s rings was made from an image sent back\nby the Voyager 2 space probe. The rings are composed of innumerable\ntiny ice particles orbiting in circles under the influence of saturn’s gravity.\nChapter 10\nGravity\nCruise your radio dial today and try to find any popular song that\nwould have been imaginable without Louis Armstrong. By introducing solo improvisation into jazz, Armstrong took apart the jigsaw\npuzzle of popular music and fit the pieces back together in a different way. In the same way, Newton reassembled our view of the\nuniverse. Consider the titles of some recent physics books written\nfor the general reader: The God Particle, Dreams of a Final Theory. When the subatomic particle called the neutrino was recently\nproven for the first time to have mass, specialists in cosmology began discussing seriously what effect this would have on calculations\nof the ultimate fate of the universe: would the neutrinos’ mass cause\nenough extra gravitational attraction to make the universe eventually stop expanding and fall back together? Without Newton, such\nattempts at universal understanding would not merely have seemed\na little pretentious, they simply would not have occurred to anyone.\nThis chapter is about Newton’s theory of gravity, which he used\nto explain the motion of the planets as they orbited the sun. Whereas\na / Johannes Kepler found a\nmathematical description of the\nmotion of the planets, which led\nto Newton’s theory of gravity.\n229\nthis book has concentrated on Newton’s laws of motion, leaving\ngravity as a dessert, Newton tosses off the laws of motion in the\nfirst 20 pages of the Principia Mathematica and then spends the\nnext 130 discussing the motion of the planets. Clearly he saw this\nas the crucial scientific focus of his work. Why? Because in it he\nshowed that the same laws of motion applied to the heavens as to\nthe earth, and that the gravitational force that made an apple fall\nwas the same as the force that kept the earth’s motion from carrying\nit away from the sun. What was radical about Newton was not his\nlaws of motion but his concept of a universal science of physics.\n10.1 Kepler’s Laws\nb / Tycho Brahe made his name\nas an astronomer by showing that\nthe bright new star, today called\na supernova, that appeared in\nthe skies in 1572 was far beyond\nthe Earth’s atmosphere. This,\nalong with Galileo’s discovery of\nsunspots, showed that contrary\nto Aristotle, the heavens were\nnot perfect and unchanging.\nBrahe’s fame as an astronomer\nbrought him patronage from King\nFrederick II, allowing him to carry\nout his historic high-precision\nmeasurements of the planets’\nmotions. A contradictory character, Brahe enjoyed lecturing other\nnobles about the evils of dueling,\nbut had lost his own nose in a\nyouthful duel and had it replaced\nwith a prosthesis made of an\nalloy of gold and silver. Willing to\nendure scandal in order to marry\na peasant, he nevertheless used\nthe feudal powers given to him by\nthe king to impose harsh forced\nlabor on the inhabitants of his\nparishes. The result of their work,\nan Italian-style palace with an\nobservatory on top, surely ranks\nas one of the most luxurious\nscience labs ever built. He died\nof a ruptured bladder after falling\nfrom a wagon on the way home\nfrom a party — in those days, it\nwas considered rude to leave the\ndinner table to relieve oneself.\nNewton wouldn’t have been able to figure out why the planets\nmove the way they do if it hadn’t been for the astronomer Tycho\nBrahe (1546-1601) and his protege Johannes Kepler (1571-1630),\nwho together came up with the first simple and accurate description\nof how the planets actually do move. The difficulty of their task is\nsuggested by figure c, which shows how the relatively simple orbital\nmotions of the earth and Mars combine so that as seen from earth\nMars appears to be staggering in loops like a drunken sailor.\nc / As the Earth and Mars revolve around the sun at different rates,\nthe combined effect of their motions makes Mars appear to trace a\nstrange, looped path across the background of the distant stars.\nBrahe, the last of the great naked-eye astronomers, collected ex-\n230\nChapter 10\nGravity\ntensive data on the motions of the planets over a period of many\nyears, taking the giant step from the previous observations’ accuracy\nof about 10 minutes of arc (10/60 of a degree) to an unprecedented\n1 minute. The quality of his work is all the more remarkable considering that his observatory consisted of four giant brass protractors\nmounted upright in his castle in Denmark. Four different observers\nwould simultaneously measure the position of a planet in order to\ncheck for mistakes and reduce random errors.\nWith Brahe’s death, it fell to his former assistant Kepler to try\nto make some sense out of the volumes of data. Kepler, in contradiction to his late boss, had formed a prejudice, a correct one\nas it turned out, in favor of the theory that the earth and planets\nrevolved around the sun, rather than the earth staying fixed and\neverything rotating about it. Although motion is relative, it is not\njust a matter of opinion what circles what. The earth’s rotation\nand revolution about the sun make it a noninertial reference frame,\nwhich causes detectable violations of Newton’s laws when one attempts to describe sufficiently precise experiments in the earth-fixed\nframe. Although such direct experiments were not carried out until\nthe 19th century, what convinced everyone of the sun-centered system in the 17th century was that Kepler was able to come up with\na surprisingly simple set of mathematical and geometrical rules for\ndescribing the planets’ motion using the sun-centered assumption.\nAfter 900 pages of calculations and many false starts and dead-end\nideas, Kepler finally synthesized the data into the following three\nlaws:\nKepler’s elliptical orbit law\nThe planets orbit the sun in elliptical orbits with the sun at\none focus.\nKepler’s equal-area law\nThe line connecting a planet to the sun sweeps out equal areas\nin equal amounts of time.\nKepler’s law of periods\nThe time required for a planet to orbit the sun, called its\nperiod, is proportional to the long axis of the ellipse raised to\nthe 3/2 power. The constant of proportionality is the same\nfor all the planets.\nAlthough the planets’ orbits are ellipses rather than circles, most\nare very close to being circular. The earth’s orbit, for instance, is\nonly flattened by 1.7% relative to a circle. In the special case of a\nplanet in a circular orbit, the two foci (plural of “focus”) coincide\nat the center of the circle, and Kepler’s elliptical orbit law thus says\nthat the circle is centered on the sun. The equal-area law implies\nthat a planet in a circular orbit moves around the sun with constant\nspeed. For a circular orbit, the law of periods then amounts to a\nstatement that the time for one orbit is proportional to r3/2 , where\nSection 10.1\nKepler’s Laws\n231\nd / An ellipse is a circle that\nhas been distorted by shrinking\nand stretching along perpendicular axes.\nr is the radius. If all the planets were moving in their orbits at the\nsame speed, then the time for one orbit would simply depend on\nthe circumference of the circle, so it would only be proportional to\nr to the first power. The more drastic dependence on r3/2 means\nthat the outer planets must be moving more slowly than the inner\nplanets.\n10.2 Newton’s Law of Gravity\nThe sun’s force on the planets obeys an inverse square law.\nKepler’s laws were a beautifully simple explanation of what the\nplanets did, but they didn’t address why they moved as they did.\nDid the sun exert a force that pulled a planet toward the center of\nits orbit, or, as suggested by Descartes, were the planets circulating\nin a whirlpool of some unknown liquid? Kepler, working in the\nAristotelian tradition, hypothesized not just an inward force exerted\nby the sun on the planet, but also a second force in the direction\nof motion to keep the planet from slowing down. Some speculated\nthat the sun attracted the planets magnetically.\ne / An ellipse can be constructed by tying a string to two\npins and drawing like this with the\npencil stretching the string taut.\nEach pin constitutes one focus of\nthe ellipse.\nf / If the time interval taken\nby the planet to move from P to Q\nis equal to the time interval from\nR to S, then according to Kepler’s\nequal-area law, the two shaded\nareas are equal.\nThe planet\nis moving faster during interval\nRS than it did during PQ, which\nNewton later determined was due\nto the sun’s gravitational force\naccelerating it. The equal-area\nlaw predicts exactly how much it\nwill speed up.\n232\nChapter 10\nOnce Newton had formulated his laws of motion and taught\nthem to some of his friends, they began trying to connect them\nto Kepler’s laws. It was clear now that an inward force would be\nneeded to bend the planets’ paths. This force was presumably an\nattraction between the sun and each planet. (Although the sun does\naccelerate in response to the attractions of the planets, its mass is so\ngreat that the effect had never been detected by the prenewtonian\nastronomers.) Since the outer planets were moving slowly along\nmore gently curving paths than the inner planets, their accelerations\nwere apparently less. This could be explained if the sun’s force was\ndetermined by distance, becoming weaker for the farther planets.\nPhysicists were also familiar with the noncontact forces of electricity\nand magnetism, and knew that they fell off rapidly with distance,\nso this made sense.\nIn the approximation of a circular orbit, the magnitude of the\nsun’s force on the planet would have to be\n\nF = ma = mv 2 /r\n.\nNow although this equation has the magnitude, v, of the velocity\nvector in it, what Newton expected was that there would be a more\nfundamental underlying equation for the force of the sun on a planet,\nand that that equation would involve the distance, r, from the sun\nto the object, but not the object’s speed, v — motion doesn’t make\nobjects lighter or heavier.\nself-check A\nIf eq. really was generally applicable, what would happen to an\nobject released at rest in some empty region of the solar system?\n.\nAnswer, p. 268\nGravity\nEquation was thus a useful piece of information which could\nbe related to the data on the planets simply because the planets\nhappened to be going in nearly circular orbits, but Newton wanted\nto combine it with other equations and eliminate v algebraically in\norder to find a deeper truth.\nTo eliminate v, Newton used the equation\n\nv=\ncircumference\n2πr\n=\nT\nT\n.\nOf course this equation would also only be valid for planets in nearly\ncircular orbits. Plugging this into eq. to eliminate v gives\n\nF =\n4π 2 mr\nT2\n.\nThis unfortunately has the side-effect of bringing in the period, T ,\nwhich we expect on similar physical grounds will not occur in the\nfinal answer. That’s where the circular-orbit case, T ∝ r3/2 , of\nKepler’s law of periods comes in. Using it to eliminate T gives a\nresult that depends only on the mass of the planet and its distance\nfrom the sun:\nF ∝ m/r2\n.\n[force of the sun on a planet of mass\nm at a distance r from the sun; same\nproportionality constant for all the planets]\n(Since Kepler’s law of periods is only a proportionality, the final\nresult is a proportionality rather than an equation, and there is this\nno point in hanging on to the factor of 4π 2 .)\nAs an example, the “twin planets” Uranus and Neptune have\nnearly the same mass, but Neptune is about twice as far from the\nsun as Uranus, so the sun’s gravitational force on Neptune is about\nfour times smaller.\nself-check B\nFill in the steps leading from equation to F ∝ m/r 2 .\n269\n. Answer, p.\nThe forces between heavenly bodies are the same type of\nforce as terrestrial gravity.\nOK, but what kind of force was it? It probably wasn’t magnetic,\nsince magnetic forces have nothing to do with mass. Then came\nNewton’s great insight. Lying under an apple tree and looking up\nat the moon in the sky, he saw an apple fall. Might not the earth\nalso attract the moon with the same kind of gravitational force?\nThe moon orbits the earth in the same way that the planets orbit\nthe sun, so maybe the earth’s force on the falling apple, the earth’s\nforce on the moon, and the sun’s force on a planet were all the same\ntype of force.\nSection 10.2\ng / The\nmoon’s\nacceleration\nis 602 = 3600 times smaller than\nthe apple’s.\nNewton’s Law of Gravity\n233\nThere was an easy way to test this hypothesis numerically. If it\nwas true, then we would expect the gravitational forces exerted by\nthe earth to follow the same F ∝ m/r2 rule as the forces exerted by\nthe sun, but with a different constant of proportionality appropriate\nto the earth’s gravitational strength. The issue arises now of how to\ndefine the distance, r, between the earth and the apple. An apple\nin England is closer to some parts of the earth than to others, but\nsuppose we take r to be the distance from the center of the earth to\nthe apple, i.e., the radius of the earth. (The issue of how to measure\nr did not arise in the analysis of the planets’ motions because the\nsun and planets are so small compared to the distances separating\nthem.) Calling the proportionality constant k, we have\n2\nFearth on apple = k mapple /rearth\nFearth on moon = k mmoon /d2earth-moon\n.\nNewton’s second law says a = F/m, so\n2\naapple = k / rearth\namoon = k / d2earth-moon\n.\nThe Greek astronomer Hipparchus had already found 2000 years\nbefore that the distance from the earth to the moon was about 60\ntimes the radius of the earth, so if Newton’s hypothesis was right,\nthe acceleration of the moon would have to be 602 = 3600 times less\nthan the acceleration of the falling apple.\nApplying a = v 2 /r to the acceleration of the moon yielded an\nacceleration that was indeed 3600 times smaller than 9.8 m/s2 , and\nNewton was convinced he had unlocked the secret of the mysterious\nforce that kept the moon and planets in their orbits.\nNewton’s law of gravity\nThe proportionality F ∝ m/r2 for the gravitational force on an\nobject of mass m only has a consistent proportionality constant for\nvarious objects if they are being acted on by the gravity of the same\nobject. Clearly the sun’s gravitational strength is far greater than\nthe earth’s, since the planets all orbit the sun and do not exhibit\nany very large accelerations caused by the earth (or by one another).\nWhat property of the sun gives it its great gravitational strength?\nIts great volume? Its great mass? Its great temperature? Newton\nreasoned that if the force was proportional to the mass of the object\nbeing acted on, then it would also make sense if the determining\nfactor in the gravitational strength of the object exerting the force\nwas its own mass. Assuming there were no other factors affecting\nthe gravitational force, then the only other thing needed to make\nquantitative predictions of gravitational forces would be a proportionality constant. Newton called that proportionality constant G,\nso here is the complete form of the law of gravity he hypothesized.\n234\nChapter 10\nGravity\nNewton’s law of gravity\nF =\nGm1 m2\nr2\n[gravitational force between objects of mass\nm1 and m2 , separated by a distance r; r is not\nthe radius of anything ]\nNewton conceived of gravity as an attraction between any two\nmasses in the universe. The constant G tells us the how many\nnewtons the attractive force is for two 1-kg masses separated by a\ndistance of 1 m. The experimental determination of G in ordinary\nunits (as opposed to the special, nonmetric, units used in astronomy)\nis described in section 10.5. This difficult measurement was not\naccomplished until long after Newton’s death.\nThe units of G\n. What are the units of G?\nexample 1\nh / Students\noften\nhave\na\nhard time understanding the\nphysical meaning of G. It’s just\na proportionality constant that\ntells you how strong gravitational\nforces are. If you could change it,\nall the gravitational forces all over\nthe universe would get stronger\nor weaker.\nNumerically, the\ngravitational attraction between\ntwo 1-kg masses separated by a\ndistance of 1 m is 6.67 × 10−11 N,\nand this is what G is in SI units.\n. Solving for G in Newton’s law of gravity gives\nG=\nFr 2\nm1 m2\n,\nso the units of G must be N·m2 /kg2 . Fully adorned with units, the\nvalue of G is 6.67 × 10−11 N·m2 /kg2 .\nNewton’s third law\nexample 2\n. Is Newton’s law of gravity consistent with Newton’s third law?\n. The third law requires two things. First, m1 ’s force on m2 should\nbe the same as m2 ’s force on m1 . This works out, because the\nproduct m1 m2 gives the same result if we interchange the labels 1\nand 2. Second, the forces should be in opposite directions. This\ncondition is also satisfied, because Newton’s law of gravity refers\nto an attraction: each mass pulls the other toward itself.\nPluto and Charon\nexample 3\n. Pluto’s moon Charon is unusually large considering Pluto’s size,\ngiving them the character of a double planet. Their masses are\n1.25×1022 and 1.9x1921 kg, and their average distance from one\nanother is 1.96 × 104 km. What is the gravitational force between\nthem?\n. If we want to use the value of G expressed in SI (meter-kilogramsecond) units, we first have to convert the distance to 1.96 ×\n107 m. The force is\n6.67 × 10−11 N·m2 /kg2 1.25 × 1022 kg 1.9 × 1021 kg\n2\n1.96 × 107 m\n= 4.1 × 1018 N\nSection 10.2\ni / Example 3.\nComputerenhanced images of Pluto and\nCharon, taken by the Hubble\nSpace Telescope.\nNewton’s Law of Gravity\n235\nThe proportionality to 1/r2 in Newton’s law of gravity was not\nentirely unexpected. Proportionalities to 1/r2 are found in many\nother phenomena in which some effect spreads out from a point.\nFor instance, the intensity of the light from a candle is proportional\nto 1/r2 , because at a distance r from the candle, the light has to\nbe spread out over the surface of an imaginary sphere of area 4πr2 .\nThe same is true for the intensity of sound from a firecracker, or the\nintensity of gamma radiation emitted by the Chernobyl reactor. It’s\nimportant, however, to realize that this is only an analogy. Force\ndoes not travel through space as sound or light does, and force is\nnot a substance that can be spread thicker or thinner like butter on\ntoast.\nj / The conic sections are the\ncurves made by cutting the\nsurface of an infinite cone with a\nplane.\nk / An imaginary cannon able\nto shoot cannonballs at very high\nspeeds is placed on top of an\nimaginary, very tall mountain\nthat reaches up above the atmosphere. Depending on the\nspeed at which the ball is fired,\nit may end up in a tightly curved\nelliptical orbit, 1, a circular orbit,\n2, a bigger elliptical orbit, 3, or a\nnearly straight hyperbolic orbit, 4.\n236\nChapter 10\nAlthough several of Newton’s contemporaries had speculated\nthat the force of gravity might be proportional to 1/r2 , none of\nthem, even the ones who had learned Newton’s laws of motion, had\nhad any luck proving that the resulting orbits would be ellipses, as\nKepler had found empirically. Newton did succeed in proving that\nelliptical orbits would result from a 1/r2 force, but we postpone the\nproof until the end of the next volume of the textbook because it\ncan be accomplished much more easily using the concepts of energy\nand angular momentum.\nNewton also predicted that orbits in the shape of hyperbolas\nshould be possible, and he was right. Some comets, for instance,\norbit the sun in very elongated ellipses, but others pass through\nthe solar system on hyperbolic paths, never to return. Just as the\ntrajectory of a faster baseball pitch is flatter than that of a more\nslowly thrown ball, so the curvature of a planet’s orbit depends on\nits speed. A spacecraft can be launched at relatively low speed,\nresulting in a circular orbit about the earth, or it can be launched\nat a higher speed, giving a more gently curved ellipse that reaches\nfarther from the earth, or it can be launched at a very high speed\nwhich puts it in an even less curved hyperbolic orbit. As you go\nvery far out on a hyperbola, it approaches a straight line, i.e., its\ncurvature eventually becomes nearly zero.\nNewton also was able to prove that Kepler’s second law (sweeping out equal areas in equal time intervals) was a logical consequence\nof his law of gravity. Newton’s version of the proof is moderately\ncomplicated, but the proof becomes trivial once you understand the\nconcept of angular momentum, which will be covered later in the\ncourse. The proof will therefore be deferred until section 5.7 of book\n2.\nself-check C\nWhich of Kepler’s laws would it make sense to apply to hyperbolic orbits?\n. Answer, p.\n269\nGravity\n. Solved problem: Visiting Ceres\npage 248, problem 10\n. Solved problem: Geosynchronous orbit\npage 250, problem 16\n. Solved problem: Why a equals g\npage 248, problem 11\n. Solved problem: Ida and Dactyl\npage 249, problem 12\n. Solved problem: Another solar system\npage 249, problem 15\n. Solved problem: Weight loss\npage 250, problem 19\n. Solved problem: The receding moon\npage 250, problem 17\nDiscussion Questions\nA\nHow could Newton find the speed of the moon to plug in to a =\nv 2 /r ?\nB\nTwo projectiles of different mass shot out of guns on the surface of\nthe earth at the same speed and angle will follow the same trajectories,\nassuming that air friction is negligible. (You can verify this by throwing two\nobjects together from your hand and seeing if they separate or stay side\nby side.) What corresponding fact would be true for satellites of the earth\nhaving different masses?\nC\nWhat is wrong with the following statement? “A comet in an elliptical\norbit speeds up as it approaches the sun, because the sun’s force on it is\nincreasing.”\nD Why would it not make sense to expect the earth’s gravitational force\non a bowling ball to be inversely proportional to the square of the distance\nbetween their surfaces rather than their centers?\nE\nDoes the earth accelerate as a result of the moon’s gravitational\nforce on it? Suppose two planets were bound to each other gravitationally\nthe way the earth and moon are, but the two planets had equal masses.\nWhat would their motion be like?\nF\nSpacecraft normally operate by firing their engines only for a few\nminutes at a time, and an interplanetary probe will spend months or years\non its way to its destination without thrust. Suppose a spacecraft is in a\ncircular orbit around Mars, and it then briefly fires its engines in reverse,\ncausing a sudden decrease in speed. What will this do to its orbit? What\nabout a forward thrust?\n10.3 Apparent Weightlessness\nIf you ask somebody at the bus stop why astronauts are weightless,\nyou’ll probably get one of the following two incorrect answers:\n(1) They’re weightless because they’re so far from the earth.\n(2) They’re weightless because they’re moving so fast.\nThe first answer is wrong, because the vast majority of astronauts never get more than a thousand miles from the earth’s surface.\nThe reduction in gravity caused by their altitude is significant, but\nnot 100%. The second answer is wrong because Newton’s law of\nSection 10.3\nApparent Weightlessness\n237\ngravity only depends on distance, not speed.\nThe correct answer is that astronauts in orbit around the earth\nare not really weightless at all. Their weightlessness is only apparent. If there was no gravitational force on the spaceship, it would\nobey Newton’s first law and move off on a straight line, rather than\norbiting the earth. Likewise, the astronauts inside the spaceship are\nin orbit just like the spaceship itself, with the earth’s gravitational\nforce continually twisting their velocity vectors around. The reason\nthey appear to be weightless is that they are in the same orbit as\nthe spaceship, so although the earth’s gravity curves their trajectory\ndown toward the deck, the deck drops out from under them at the\nsame rate.\nApparent weightlessness can also be experienced on earth. Any\ntime you jump up in the air, you experience the same kind of apparent weightlessness that the astronauts do. While in the air, you\ncan lift your arms more easily than normal, because gravity does not\nmake them fall any faster than the rest of your body, which is falling\nout from under them. The Russian air force now takes rich foreign\ntourists up in a big cargo plane and gives them the feeling of weightlessness for a short period of time while the plane is nose-down and\ndropping like a rock.\n10.4 Vector Addition of Gravitational Forces\nPick a flower on earth and you move the farthest star.\nPaul Dirac\nWhen you stand on the ground, which part of the earth is pulling\ndown on you with its gravitational force? Most people are tempted\nto say that the effect only comes from the part directly under you,\nsince gravity always pulls straight down. Here are three observations\nthat might help to change your mind:\n• If you jump up in the air, gravity does not stop affecting you\njust because you are not touching the earth: gravity is a noncontact force. That means you are not immune from the gravity of distant parts of our planet just because you are not\ntouching them.\n• Gravitational effects are not blocked by intervening matter.\nFor instance, in an eclipse of the moon, the earth is lined up\ndirectly between the sun and the moon, but only the sun’s light\nis blocked from reaching the moon, not its gravitational force\n— if the sun’s gravitational force on the moon was blocked in\nthis situation, astronomers would be able to tell because the\nmoon’s acceleration would change suddenly. A more subtle\nbut more easily observable example is that the tides are caused\nby the moon’s gravity, and tidal effects can occur on the side\nl / Gravity only appears to\npull straight down because the\nnear perfect symmetry of the\nearth makes the sideways components of the total force on an\nobject cancel almost exactly. If\nthe symmetry is broken, e.g., by\na dense mineral deposit, the total\nforce is a little off to the side.\n238\nChapter 10\nGravity\nof the earth facing away from the moon. Thus, far-off parts\nof the earth are not prevented from attracting you with their\ngravity just because there is other stuff between you and them.\n• Prospectors sometimes search for underground deposits of dense\nminerals by measuring the direction of the local gravitational\nforces, i.e., the direction things fall or the direction a plumb\nbob hangs. For instance, the gravitational forces in the region\nto the west of such a deposit would point along a line slightly\nto the east of the earth’s center. Just because the total gravitational force on you points down, that doesn’t mean that\nonly the parts of the earth directly below you are attracting\nyou. It’s just that the sideways components of all the force\nvectors acting on you come very close to canceling out.\nA cubic centimeter of lava in the earth’s mantle, a grain of silica\ninside Mt. Kilimanjaro, and a flea on a cat in Paris are all attracting\nyou with their gravity. What you feel is the vector sum of all the\ngravitational forces exerted by all the atoms of our planet, and for\nthat matter by all the atoms in the universe.\nWhen Newton tested his theory of gravity by comparing the\norbital acceleration of the moon to the acceleration of a falling apple\non earth, he assumed he could compute the earth’s force on the\napple using the distance from the apple to the earth’s center. Was\nhe wrong? After all, it isn’t just the earth’s center attracting the\napple, it’s the whole earth. A kilogram of dirt a few feet under his\nbackyard in England would have a much greater force on the apple\nthan a kilogram of molten rock deep under Australia, thousands of\nmiles away. There’s really no obvious reason why the force should\ncome out right if you just pretend that the earth’s whole mass is\nconcentrated at its center. Also, we know that the earth has some\nparts that are more dense, and some parts that are less dense. The\nsolid crust, on which we live, is considerably less dense than the\nmolten rock on which it floats. By all rights, the computation of the\nvector sum of all the forces exerted by all the earth’s parts should\nbe a horrendous mess.\nActually, Newton had sound mathematical reasons for treating\nthe earth’s mass as if it was concentrated at its center. First, although Newton no doubt suspected the earth’s density was nonuniform, he knew that the direction of its total gravitational force was\nvery nearly toward the earth’s center. That was strong evidence\nthat the distribution of mass was very symmetric, so that we can\nthink of the earth as being made of many layers, like an onion,\nwith each layer having constant density throughout. (Today there\nis further evidence for symmetry based on measurements of how the\nvibrations from earthquakes and nuclear explosions travel through\nthe earth.) Newton then concentrated on the gravitational forces\nSection 10.4\nVector Addition of Gravitational Forces\n239\nexerted by a single such thin shell, and proved the following mathematical theorem, known as the shell theorem:\nIf an object lies outside a thin, spherical shell of mass, then\nthe vector sum of all the gravitational forces exerted by all the\nparts of the shell is the same as if the shell’s mass had been\nconcentrated at its center. If the object lies inside the shell,\nthen all the gravitational forces cancel out exactly.\nm / An object outside a spherical\nshell of mass will feel gravitational\nforces from every part of the shell\n— stronger forces from the closer\nparts, and weaker ones from the\nparts farther away. The shell\ntheorem states that the vector\nsum of all the forces is the same\nas if all the mass had been\nconcentrated at the center of the\nshell.\nFor terrestrial gravity, each shell acts as though its mass was concentrated at the earth’s center, so the final result is the same as if\nthe earth’s whole mass was concentrated at its center.\nThe second part of the shell theorem, about the gravitational\nforces canceling inside the shell, is a little surprising. Obviously the\nforces would all cancel out if you were at the exact center of a shell,\nbut why should they still cancel out perfectly if you are inside the\nshell but off-center? The whole idea might seem academic, since we\ndon’t know of any hollow planets in our solar system that astronauts\ncould hope to visit, but actually it’s a useful result for understanding\ngravity within the earth, which is an important issue in geology. It\ndoesn’t matter that the earth is not actually hollow. In a mine shaft\nat a depth of, say, 2 km, we can use the shell theorem to tell us that\nthe outermost 2 km of the earth has no net gravitational effect, and\nthe gravitational force is the same as what would be produced if the\nremaining, deeper, parts of the earth were all concentrated at its\ncenter.\nself-check D\nSuppose you’re at the bottom of a deep mineshaft, which means you’re\nstill quite far from the center of the earth. The shell theorem says that\nthe shell of mass you’ve gone inside exerts zero total force on you.\nDiscuss which parts of the shell are attracting you in which directions,\nand how strong these forces are. Explain why it’s at least plausible that\nthey cancel.\n. Answer, p. 269\nDiscussion Questions\nA\nIf you hold an apple, does the apple exert a gravitational force on\nthe earth? Is it much weaker than the earth’s gravitational force on the\napple? Why doesn’t the earth seem to accelerate upward when you drop\nthe apple?\nB\nWhen astronauts travel from the earth to the moon, how does the\ngravitational force on them change as they progress?\nC\nHow would the gravity in the first-floor lobby of a massive skyscraper\ncompare with the gravity in an open field outside of the city?\nD\nIn a few billion years, the sun will start undergoing changes that will\neventually result in its puffing up into a red giant star. (Near the beginning\nof this process, the earth’s oceans will boil off, and by the end, the sun\nwill probably swallow the earth completely.) As the sun’s surface starts to\nget closer and close to the earth, how will the earth’s orbit be affected?\n240\nChapter 10\nGravity\n10.5 Weighing the Earth\nLet’s look more closely at the application of Newton’s law of gravity\nto objects on the earth’s surface. Since the earth’s gravitational\nforce is the same as if its mass was all concentrated at its center,\nthe force on a falling object of mass m is given by\n2\nF = G Mearth m / rearth\n.\nThe object’s acceleration equals F/m, so the object’s mass cancels\nout and we get the same acceleration for all falling objects, as we\nknew we should:\n2\ng = G Mearth / rearth\n.\nn / Cavendish’s apparatus. The two large balls are fixed in place,\nbut the rod from which the two small balls hang is free to twist under the\ninfluence of the gravitational forces.\nNewton knew neither the mass of the earth nor a numerical value\nfor the constant G. But if someone could measure G, then it would\nbe possible for the first time in history to determine the mass of the\nearth! The only way to measure G is to measure the gravitational\nforce between two objects of known mass, but that’s an exceedingly\ndifficult task, because the force between any two objects of ordinary\nsize is extremely small. The English physicist Henry Cavendish was\nthe first to succeed, using the apparatus shown in figures n and o.\nThe two larger balls were lead spheres 8 inches in diameter, and each\none attracted the small ball near it. The two small balls hung from\nthe ends of a horizontal rod, which itself hung by a thin thread. The\nframe from which the larger balls hung could be rotated by hand\nSection 10.5\no/A\nsimplified\nversion\nof\nCavendish’s apparatus, viewed\nfrom above.\nWeighing the Earth\n241\nabout a vertical axis, so that for instance the large ball on the right\nwould pull its neighboring small ball toward us and while the small\nball on the left would be pulled away from us. The thread from\nwhich the small balls hung would thus be twisted through a small\nangle, and by calibrating the twist of the thread with known forces,\nthe actual gravitational force could be determined. Cavendish set\nup the whole apparatus in a room of his house, nailing all the doors\nshut to keep air currents from disturbing the delicate apparatus.\nThe results had to be observed through telescopes stuck through\nholes drilled in the walls. Cavendish’s experiment provided the first\nnumerical values for G and for the mass of the earth. The presently\naccepted value of G is 6.67 × 10−11 N·m2 /kg2 .\nKnowing G not only allowed the determination of the earth’s\nmass but also those of the sun and the other planets. For instance,\nby observing the acceleration of one of Jupiter’s moons, we can infer\nthe mass of Jupiter. The following table gives the distances of the\nplanets from the sun and the masses of the sun and planets. (Other\ndata are given in the back of the book.)\nsun\nmercury\nvenus\nearth\nmars\njupiter\nsaturn\nuranus\nneptune\npluto\naverage distance from\nthe sun, in units of the\nearth’s average distance\nfrom the sun\n—\n0.38\n0.72\n1\n1.5\n5.2\n9.5\n19\n30\n39\nmass, in units of the\nearth’s mass\n330,000\n0.056\n0.82\n1\n0.11\n320\n95\n14\n17\n0.002\nDiscussion Questions\nA\nIt would have been difficult for Cavendish to start designing an\nexperiment without at least some idea of the order of magnitude of G.\nHow could he estimate it in advance to within a factor of 10?\nB\nFill in the details of how one would determine Jupiter’s mass by\nobserving the acceleration of one of its moons. Why is it only necessary\nto know the acceleration of the moon, not the actual force acting on it?\nWhy don’t we need to know the mass of the moon? What about a planet\nthat has no moons, such as Venus — how could its mass be found?\nC\nThe gravitational constant G is very difficult to measure accurately, and is the least accurately known of all the fundamental numbers\nof physics such as the speed of light, the mass of the electron, etc. But\nthat’s in the mks system, based on the meter as the unit of length, the\nkilogram as the unit of mass, and the second as the unit of distance. Astronomers sometimes use a different system of units, in which the unit of\n242\nChapter 10\nGravity\ndistance, called the astronomical unit or a.u., is the radius of the earth’s\norbit, the unit of mass is the mass of the sun, and the unit of time is the\nyear (i.e., the time required for the earth to orbit the sun). In this system\nof units, G has a precise numerical value simply as a matter of definition.\nWhat is it?\n10.6 ? Evidence for Repulsive Gravity\nUntil recently, physicists thought they understood gravity fairly\nwell. Einstein had modified Newton’s theory, but certain characteristrics of gravitational forces were firmly established. For one\nthing, they were always attractive. If gravity always attracts, then\nit is logical to ask why the universe doesn’t collapse. Newton had\nanswered this question by saying that if the universe was infinite in\nall directions, then it would have no geometric center toward which\nit would collapse; the forces on any particular star or planet exerted by distant parts of the universe would tend to cancel out by\nsymmetry. More careful calculations, however, show that Newton’s\nuniverse would have a tendency to collapse on smaller scales: any\npart of the universe that happened to be slightly more dense than\naverage would contract further, and this contraction would result\nin stronger gravitational forces, which would cause even more rapid\ncontraction, and so on.\nWhen Einstein overhauled gravity, the same problem reared its\nugly head. Like Newton, Einstein was predisposed to believe in a\nuniverse that was static, so he added a special repulsive term to his\nequations, intended to prevent a collapse. This term was not associated with any attraction of mass for mass, but represented merely\nan overall tendency for space itself to expand unless restrained by\nthe matter that inhabited it. It turns out that Einstein’s solution,\nlike Newton’s, is unstable. Furthermore, it was soon discovered\nobservationally that the universe was expanding, and this was interpreted by creating the Big Bang model, in which the universe’s\ncurrent expansion is the aftermath of a fantastically hot explosion.1\nAn expanding universe, unlike a static one, was capable of being explained with Einstein’s equations, without any repulsion term. The\nuniverse’s expansion would simply slow down over time due to the\nattractive gravitational forces. After these developments, Einstein\nsaid woefully that adding the repulsive term, known as the cosmological constant, had been the greatest blunder of his life.\nThis was the state of things until 1999, when evidence began to\nturn up that the universe’s expansion has been speeding up rather\nthan slowing down! The first evidence came from using a telescope\nas a sort of time machine: light from a distant galaxy may have\ntaken billions of years to reach us, so we are seeing it as it was far\nin the past. Looking back in time, astronomers saw the universe\n1\nBook 3, section 3.5, presents some of the evidence for the Big Bang.\nSection 10.6\n? Evidence for Repulsive Gravity\n243\nexpanding at speeds that ware lower, rather than higher. At first\nthey were mortified, since this was exactly the opposite of what had\nbeen expected. The statistical quality of the data was also not good\nenough to constute ironclad proof, and there were worries about systematic errors. The case for an accelerating expansion has however\nbeen nailed down by high-precision mapping of the dim, sky-wide\nafterglow of the Big Bang, known as the cosmic microwave background. Some theorists have proposed reviving Einstein’s cosmological constant to account for the acceleration, while others believe\nit is evidence for a mysterious form of matter which exhibits gravitational repulsion. The generic term for this unknown stuff is “dark\nenergy.”\nAs of 2008, most of the remaining doubt about the repulsive effect has been dispelled. During the past decade or so, astronomers\nconsider themselves to have entered a new era of high-precision cosmology. The cosmic microwave background measurements, for example, have measured the age of the universe to be 13.7 ± 0.2 billion\nyears, a figure that could previously be stated only as a fuzzy range\nfrom 10 to 20 billion. We know that only 4% of the universe is\natoms, with another 23% consisting of unknown subatomic particles, and 73% of dark energy. It’s more than a little ironic to know\nabout so many things with such high precision, and yet to know\nvirtually nothing about their nature. For instance, we know that\nprecisely 96% of the universe is something other than atoms, but we\nknow precisely nothing about what that something is.\np / The WMAP probe’s map of the\ncosmic microwave background is\nlike a “baby picture” of the universe.\n244\nChapter 10\nGravity\nSummary\nSelected Vocabulary\nellipse . . . . . . . a flattened circle; one of the conic sections\nconic section . . . a curve formed by the intersection of a plane\nand an infinite cone\nhyperbola . . . . another conic section; it does not close back\non itself\nperiod . . . . . . . the time required for a planet to complete one\norbit; more generally, the time for one repetition of some repeating motion\nfocus . . . . . . . one of two special points inside an ellipse: the\nellipse consists of all points such that the sum\nof the distances to the two foci equals a certain\nnumber; a hyperbola also has a focus\nNotation\nG . . . . . . . . .\nthe constant of proportionality in Newton’s\nlaw of gravity; the gravitational force of attraction between two 1-kg spheres at a centerto-center distance of 1 m\nSummary\nKepler deduced three empirical laws from data on the motion of\nthe planets:\nKepler’s elliptical orbit law: The planets orbit the sun in elliptical orbits with the sun at one focus.\nKepler’s equal-area law: The line connecting a planet to the sun\nsweeps out equal areas in equal amounts of time.\nKepler’s law of periods: The time required for a planet to orbit\nthe sun is proportional to the long axis of the ellipse raised to\nthe 3/2 power. The constant of proportionality is the same\nfor all the planets.\nNewton was able to find a more fundamental explanation for these\nlaws. Newton’s law of gravity states that the magnitude of the\nattractive force between any two objects in the universe is given by\nF = Gm1 m2 /r2\n.\nWeightlessness of objects in orbit around the earth is only apparent. An astronaut inside a spaceship is simply falling along with\nthe spaceship. Since the spaceship is falling out from under the astronaut, it appears as though there was no gravity accelerating the\nastronaut down toward the deck.\nGravitational forces, like all other forces, add like vectors. A\ngravitational force such as we ordinarily feel is the vector sum of all\nSummary\n245\nthe forces exerted by all the parts of the earth. As a consequence of\nthis, Newton proved the shell theorem for gravitational forces:\nIf an object lies outside a thin, uniform shell of mass, then the\nvector sum of all the gravitational forces exerted by all the parts of\nthe shell is the same as if all the shell’s mass was concentrated at its\ncenter. If the object lies inside the shell, then all the gravitational\nforces cancel out exactly.\n246\nChapter 10\nGravity\nProblems\nKey\n√\nR\n?\nA computerized answer check is available online.\nA problem that requires calculus.\nA difficult problem.\n1\nRoy has a mass of 60 kg. Laurie has a mass of 65 kg. They\nare 1.5 m apart.\n(a) What is the magnitude of the gravitational force of the earth on\nRoy?\n(b) What is the magnitude of Roy’s gravitational force on the earth?\n(c) What is the magnitude of the gravitational force between Roy\nand Laurie?\n(d) What is the magnitude of the gravitational force between Laurie\n√\nand the sun?\n2\nDuring a solar eclipse, the moon, earth and sun all lie on\nthe same line, with the moon between the earth and sun. Define\nyour coordinates so that the earth and moon lie at greater x values\nthan the sun. For each force, give the correct sign as well as the\nmagnitude. (a) What force is exerted on the moon by the sun? (b)\nOn the moon by the earth? (c) On the earth by the sun? (d) What\ntotal force is exerted on the sun? (e) On the moon? (f) On√the\nearth?\n3\nSuppose that on a certain day there is a crescent moon, and\nyou can tell by the shape of the crescent that the earth, sun and\nmoon form a triangle with a 135 ◦ interior angle at the moon’s corner.\nWhat is the magnitude of the total gravitational force of the earth\nand the sun on the moon?\nProblem 3.\n4\nHow high above the Earth’s surface must a rocket be in order\nto have 1/100 the weight it would have at the surface? Express your\nanswer in units of the radius of the Earth.\n5\nThe star Lalande 21185 was found in 1996 to have two planets\nin roughly circular orbits, with periods of 6 and 30 years. What is\nthe ratio of the two planets’ orbital radii?\n6\nIn a Star Trek episode, the Enterprise is in a circular orbit\naround a planet when something happens to the engines. Spock\nthen tells Kirk that the ship will spiral into the planet’s surface\nunless they can fix the engines. Is this scientifically correct? Why?\n7\n(a) Suppose a rotating spherical body such as a planet has\nProblems\n247\na radius r and a uniform density ρ, and the time required for one\nrotation is T . At the surface of the planet, the apparent acceleration\nof a falling object is reduced by the acceleration of the ground out\nfrom under it. Derive an equation for the apparent acceleration of\n√\ngravity, g, at the equator in terms of r, ρ, T , and G.\n(b) Applying your equation from a, by what fraction is your apparent weight reduced at the equator compared to the poles, due to the\n√\nEarth’s rotation?\n(c) Using your equation from a, derive an equation giving the value\nof T for which the apparent acceleration of gravity becomes zero,\ni.e., objects can spontaneously drift off the surface of the planet.\n√\nShow that T only depends on ρ, and not on r.\n(d) Applying your equation from c, how long would a day have to\nbe in order to reduce the apparent weight of objects at the equator\nof the Earth to zero? [Answer: 1.4 hours]\n(e) Observational astronomers have recently found objects they called\npulsars, which emit bursts of radiation at regular intervals of less\nthan a second. If a pulsar is to be interpreted as a rotating sphere\nbeaming out a natural “searchlight” that sweeps past the earth with\neach rotation, use your equation from c to show that its density\nwould have to be much greater than that of ordinary matter.\n(f) Astrophysicists predicted decades ago that certain stars that used\nup their sources of energy could collapse, forming a ball of neutrons\nwith the fantastic density of ∼ 1017 kg/m3 . If this is what pulsars\nreally are, use your equation from c to explain why no pulsar has\never been observed that flashes with a period of less than 1 ms or\nso.\n8\nYou are considering going on a space voyage to Mars, in which\nyour route would be half an ellipse, tangent to the Earth’s orbit at\none end and tangent to Mars’ orbit at the other. Your spacecraft’s\nengines will only be used at the beginning and end, not during the\nvoyage. How long would the outward leg of your trip last? (Assume\n√\nthe orbits of Earth and Mars are circular.)\n9\n(a) If the earth was of uniform density, would your weight be\nincreased or decreased at the bottom of a mine shaft? Explain.\n(b) In real life, objects weigh slightly more at the bottom of a mine\nshaft. What does that allow us to infer about the Earth?\n?\n10\nCeres, the largest asteroid in our solar system, is a spherical\nbody with a mass 6000 times less than the earth’s, and a radius\nwhich is 13 times smaller. If an astronaut who weighs 400 N on\nearth is visiting the surface of Ceres, what is her weight?\n. Solution, p. 278\nProblem 8.\n11\nProve, based on Newton’s laws of motion and Newton’s law\nof gravity, that all falling objects have the same acceleration if they\nare dropped at the same location on the earth and if other forces\nsuch as friction are unimportant. Do not just say, “g = 9.8 m/s2 –\nit’s constant.” You are supposed to be proving that g should be the\n248\nChapter 10\nGravity\nsame number for all objects.\n. Solution, p. 278\n12\nThe figure shows an image from the Galileo space probe\ntaken during its August 1993 flyby of the asteroid Ida. Astronomers\nwere surprised when Galileo detected a smaller object orbiting Ida.\nThis smaller object, the only known satellite of an asteroid in our\nsolar system, was christened Dactyl, after the mythical creatures\nwho lived on Mount Ida, and who protected the infant Zeus. For\nscale, Ida is about the size and shape of Orange County, and Dactyl\nthe size of a college campus. Galileo was unfortunately unable to\nmeasure the time, T , required for Dactyl to orbit Ida. If it had,\nastronomers would have been able to make the first accurate determination of the mass and density of an asteroid. Find an equation\nfor the density, ρ, of Ida in terms of Ida’s known volume, V , the\nknown radius, r, of Dactyl’s orbit, and the lamentably unknown\nvariable T . (This is the same technique that was used successfully\nfor determining the masses and densities of the planets that have\nmoons.)\n. Solution, p. 278\nProblem 12.\n13\nIf a bullet is shot straight up at a high enough velocity, it will\nnever return to the earth. This is known as the escape velocity. We\nwill discuss escape velocity using the concept of energy in the next\nbook of the series, but it can also be gotten at using straightforward\ncalculus. In this problem, you will analyze the motion of an object\nof mass m whose initial velocity is exactly equal to escape velocity. We assume that it is starting from the surface of a spherically\nsymmetric planet of mass Mand radius b. The trick is to guess at\nthe general form of the solution, and then determine the solution in\nmore detail. Assume (as is true) that the solution is of the form r=\nkt p , where r is the object’s distance from the center of the planet\nat time t, and k and p are constants.\n(a) Find the acceleration, and use Newton’s second law and Newton’s law of gravity to determine k and p. You should find that the\n√\nresult is independent of m.\n(b) What happens to the velocity as t approaches infinity? √ R\n(c) Determine escape velocity from the Earth’s surface.\n14\nAstronomers have recently observed stars orbiting at very\nhigh speeds around an unknown object near the center of our galaxy.\nFor stars orbiting at distances of about 1014 m from the object,\nthe orbital velocities are about 106 m/s. Assuming the orbits are\ncircular, estimate the mass of the object, in units of the mass of\nthe sun, 2 × 1030 kg. If the object was a tightly packed cluster of\nnormal stars, it should be a very bright source of light. Since no\nvisible light is detected coming from it, it is instead believed to be\na supermassive black hole.\n15\nAstronomers have detected a solar system consisting of three\nplanets orbiting the star Upsilon Andromedae. The planets have\nbeen named b, c, and d. Planet b’s average distance from the star\nProblems\n249\nis 0.059 A.U., and planet c’s average distance is 0.83 A.U., where an\nastronomical unit or A.U. is defined as the distance from the Earth\nto the sun. For technical reasons, it is possible to determine the\nratios of the planets’ masses, but their masses cannot presently be\ndetermined in absolute units. Planet c’s mass is 3.0 times that of\nplanet b. Compare the star’s average gravitational force on planet\nc with its average force on planet b. [Based on a problem by Arnold\nArons.]\n. Solution, p. 278\n16\nSome communications satellites are in orbits called geosynchronous: the satellite takes one day to orbit the earth from west\nto east, so that as the earth spins, the satellite remains above the\nsame point on the equator. What is such a satellite’s altitude above\nthe surface of the earth?\n. Solution, p. 278\n17\nAs is discussed in more detail in section 5.1 of book 2, tidal\ninteractions with the earth are causing the moon’s orbit to grow\ngradually larger. Laser beams bounced off of a mirror left on the\nmoon by astronauts have allowed a measurement of the moon’s rate\nof recession, which is about 1 cm per year. This means that the\ngravitational force acting between earth and moon is decreasing. By\nwhat fraction does the force decrease with each 27-day orbit? [Hint:\nIf you try to calculate the two forces and subtract, your calculator\nwill probably give a result of zero due to rounding. Instead, reason\nabout the fractional amount by which the quantity 1/r2 will change.\nAs a warm-up, you may wish to observe the percentage change in\n1/r2 that results from changing r from 1 to 1.01. Based on a problem\nby Arnold Arons.]\n. Solution, p. 279\n18\nSuppose that we inhabited a universe in which, instead of\n√\nNewton’s law of gravity, we had F = k m1 m2 /r2 , where k is some\nconstant with different units than G. (The force is still attractive.) However, we assume that a = F/m and the rest of Newtonian\nphysics remains true, and we use a = F/m to define our mass scale,\nso that, e.g., a mass of 2 kg is one which exhibits half the acceleration when the same force is applied to it as to a 1 kg mass.\n(a) Is this new law of gravity consistent with Newton’s third law?\n(b) Suppose you lived in such a universe, and you dropped two unequal masses side by side. What would happen?\n(c) Numerically, suppose a 1.0-kg object falls with an acceleration\nof 10 m/s2 . What would be the acceleration of a rain drop with a\nmass of 0.1 g? Would you want to go out in the rain?\n(d) If a falling object broke into two unequal pieces while it fell,\nwhat would happen?\n(e) Invent a law of gravity that results in behavior that is the opposite of what you found in part b. [Based on a problem by Arnold\nArons.]\n19\n(a) A certain vile alien gangster lives on the surface of an\nasteroid, where his weight is 0.20 N. He decides he needs to lose\n250\nChapter 10\nGravity\nweight without reducing his consumption of princesses, so he’s going\nto move to a different asteroid where his weight will be 0.10 N. The\nreal estate agent’s database has asteroids listed by mass, however,\nnot by surface gravity. Assuming that all asteroids are spherical\nand have the same density, how should the mass of his new asteroid\ncompare with that of his old one?\n(b) Jupiter’s mass is 318 times the Earth’s, and its gravity is about\ntwice Earth’s. Is this consistent with the results of part a? If not,\nhow do you explain the discrepancy?\n. Solution, p. 279\n20\nWhere would an object have to be located so that it would\nexperience zero total gravitational force from the earth and moon?\n√\n21\nThe planet Uranus has a mass of 8.68 × 1025 kg and a radius\nof 2.56 × 104 km. The figure shows the relative sizes of Uranus and\nEarth.\n(a) Compute the ratio gU /gE , where gU is the strength of the gravitational field at the surface of Uranus and gE is the corresponding\n√\nquantity at the surface of the Earth.\n(b) What is surprising about this result? How do you explain it?\nProblem 21.\n22\nThe International Space Station orbits at an average altitude\nof about 370 km above sea level. Compute the value of g at that\naltitude.\nProblems\n251\nAppendix 1: Exercises\nExercise 0A: Models and Idealization\nEquipment:\ncoffee filters\nramps (one per group)\nballs of various sizes\nsticky tape\nvacuum pump and “guinea and feather” apparatus (one)\nThe motion of falling objects has been recognized since ancient times as an important piece of\nphysics, but the motion is inconveniently fast, so in our everyday experience it can be hard to\ntell exactly what objects are doing when they fall. In this exercise you will use several techniques\nto get around this problem and study the motion. Your goal is to construct a scientific model of\nfalling. A model means an explanation that makes testable predictions. Often models contain\nsimplifications or idealizations that make them easier to work with, even though they are not\nstrictly realistic.\n1. One method of making falling easier to observe is to use objects like feathers that we know\nfrom everyday experience will not fall as fast. You will use coffee filters, in stacks of various\nsizes, to test the following two hypotheses and see which one is true, or whether neither is true:\nHypothesis 1A: When an object is dropped, it rapidly speeds up to a certain natural falling\nspeed, and then continues to fall at that speed. The falling speed is proportional to the object’s\nweight. (A proportionality is not just a statement that if one thing gets bigger, the other does\ntoo. It says that if one becomes three times bigger, the other also gets three times bigger, etc.)\nHypothesis 1B: Different objects fall the same way, regardless of weight.\nTest these hypotheses and discuss your results with your instructor.\n2. A second way to slow down the action is to let a ball roll down a ramp. The steeper the\nramp, the closer to free fall. Based on your experience in part 1, write a hypothesis about what\nwill happen when you race a heavier ball against a lighter ball down the same ramp, starting\nthem both from rest.\nHypothesis:\nShow your hypothesis to your instructor, and then test it.\nYou have probably found that falling was more complicated than you thought! Is there more\nthan one factor that affects the motion of a falling object? Can you imagine certain idealized\nsituations that are simpler? Try to agree verbally with your group on an informal model of\nfalling that can make predictions about the experiments described in parts 3 and 4.\n3. You have three balls: a standard “comparison ball” of medium weight, a light ball, and a\nheavy ball. Suppose you stand on a chair and (a) drop the light ball side by side with the\ncomparison ball, then (b) drop the heavy ball side by side with the comparison ball, then (c)\njoin the light and heavy balls together with sticky tape and drop them side by side with the\ncomparison ball.\nUse your model to make a prediction:\nTest your prediction.\n4. Your instructor will pump nearly all the air out of a chamber containing a feather and a\nheavier object, then let them fall side by side in the chamber.\nUse your model to make a prediction:\n253\nExercise 1A: Scaling Applied to Leaves\nEquipment:\nleaves of three sizes, having roughly similar proportions of length, width, and thickness\n(example: blades of grass, large ficus leaves, and agave leaves)\nbalance\ngraph paper with centimeter squares\n1. Each group will have one leaf, and should measure its surface area and volume, and determine\nits surface-to-volume ratio (surface area divided by volume). For consistency, every group should\nuse units of cm2 and cm3 , and should only find the area of one side of the leaf. The area can be\nfound by tracing the area of the leaf on graph paper and counting squares. The volume can be\nfound by weighing the leaf and assuming that its density is 1 g/cm3 , which is nearly true since\nleaves are mostly water.\nWrite your results on the board for comparison with the other groups’ numbers.\n2. Both the surface area and the volume are bigger for bigger leaves, but what about the\nsurface to volume ratios? What implications would this have for the plants’ abilities to survive\nin different environments?\n254\nAppendix 1: Exercises\n255\nExercise 2A: Changing Velocity\nThis exercise involves Michael Johnson’s world-record 200-meter sprint in the 1996 Olympics.\nThe table gives the distance he has covered at various times. (The data are made up, except for\nhis total time.) Each group is to find a value of ∆x/∆t between two specified instants, with the\nmembers of the group checking each other’s answers. We will then compare everyone’s results\nand discuss how this relates to velocity.\nA\nB\nC\nD\nE\nt (s)\n10.200\n10.210\n10.300\n11.200\n19.320\nx (m)\n100.0000\n100.0990\n100.9912\n110.0168\n200.0000\ngroup 1: Find ∆x/∆t using points A and B.\ngroup 2: Find ∆x/∆t using points A and C.\ngroup 3: Find ∆x/∆t using points A and D.\ngroup 4: Find ∆x/∆t using points A and E.\n256\nAppendix 1: Exercises\nExercise 3A: Reasoning with Ratios and Powers\nEquipment:\nping-pong balls and paddles\ntwo-meter sticks\nYou have probably bounced a ping pong ball straight up and down in the air. The time between\nhits is related to the height to which you hit the ball. If you take twice as much time between\nhits, how many times higher do you think you will have to hit the ball? Write down your\nhypothesis:\nYour instructor will first beat out a tempo of 240 beats per minute (four beats per second),\nwhich you should try to match with the ping-pong ball. Measure the height to which the ball\nrises:\nNow try it at 120 beats per minute:\nCompare your hypothesis and your results with the rest of the class.\n257\nExercise 4A: Force and Motion\nEquipment:\n2-meter pieces of butcher paper\nwood blocks with hooks\nstring\nmasses to put on top of the blocks to increase friction\nspring scales (preferably calibrated in Newtons)\nSuppose a person pushes a crate, sliding it across the floor at a certain speed, and then repeats\nthe same thing but at a higher speed. This is essentially the situation you will act out in this\nexercise. What do you think is different about her force on the crate in the two situations?\nDiscuss this with your group and write down your hypothesis:\n1. First you will measure the amount of friction between the wood block and the butcher paper\nwhen the wood and paper surfaces are slipping over each other. The idea is to attach a spring\nscale to the block and then slide the butcher paper under the block while using the scale to\nkeep the block from moving with it. Depending on the amount of force your spring scale was\ndesigned to measure, you may need to put an extra mass on top of the block in order to increase\nthe amount of friction. It is a good idea to use long piece of string to attach the block to the\nspring scale, since otherwise one tends to pull at an angle instead of directly horizontally.\nFirst measure the amount of friction force when sliding the butcher paper as slowly as possible:\nNow measure the amount of friction force at a significantly higher speed, say 1 meter per second.\n(If you try to go too fast, the motion is jerky, and it is impossible to get an accurate reading.)\nDiscuss your results. Why are we justified in assuming that the string’s force on the block (i.e.,\nthe scale reading) is the same amount as the paper’s frictional force on the block?\n2. Now try the same thing but with the block moving and the paper standing still. Try two\ndifferent speeds.\nDo your results agree with your original hypothesis? If not, discuss what’s going on. How does\nthe block “know” how fast to go?\n258\nAppendix 1: Exercises\nExercise 4B: Interactions\nEquipment:\nneodymium disc magnets (3/group)\ncompass\ntriple-arm balance (2/group)\nclamp and 50-cm vertical rod for holding balance up\nstring\ntape\nscissors\nYour goal in this exercise is to compare the forces two magnets exert on each other, i.e., to\ncompare magnet A’s force on magnet B to magnet B’s force on magnet A. Magnet B will be\nmade out of two of the small disc magnets put together, so it is twice as strong as magnet A.\n1. Note that these magnets are extremely strong! Being careful not to pinch your skin, put two\ndisc magnets together to make magnet B.\n2. Familiarize yourself with how the magnets behave. In addition to magnets A and B, there\nare two other magnets that can come into play. The compass needle itself is a magnet, and the\nplanet earth is a magnet. Ordinarily the compass needle twists around under the influence of\nthe earth, but the disc magnets are very strong close up, so if you bring them within a few cm\nof the compass, the compass is essentially just responding to them. Investigate how different\nparts of magnets A and B interact with the compass, and label them appropriately. Investigate\nhow magnets A and B can attract or repel one another.\n3. You are ready to form a hypothesis about the following situation. Suppose we set up two\nbalances as shown in the figure. The magnets are not touching. The top magnet is hanging from\na hook underneath the pan, giving the same result as if it was on top of the pan. Make sure it\nis hanging under the center of the pan. You will want to make sure the magnets are pulling on\neach other, not pushing each other away, so that the top magnet will stay in one place.\nThe balances will not show the magnets’ true weights, because the magnets are exerting forces\non each other. The top balance will read a higher number than it would without any magnetic\nforces, and the bottom balance will have a lower than normal reading. The difference between\neach magnet’s true weight and the reading on the balance gives a measure of how strongly the\nmagnet is being pushed or pulled by the other magnet.\nHow do you think the amount of pushing or pulling experienced by the two magnets will compare? In other words, which reading will change more, or will they change by the same amount?\n259\nWrite down a hypothesis:\nBefore going on to part 4, discuss your hypothesis with your instructor.\n4. Now set up the experiment described above with two balances. Since we are interested in\nthe changse in the scale readings caused by the magnetic forces, you will need to take a total of\nfour scale readings: one pair with the balances separated and one pair with the magnets close\ntogether as shown in the figure above.\nWhen the balances are together and the magnetic forces are acting, it is not possible to get both\nbalances to reach equilibrium at the same time, because sliding the weights on one balance can\ncause its magnet to move up or down, tipping the other balance. Therefore, while you take a\nreading from one balance, you need to immobilize the other in the horizontal position by taping\nits tip so it points exactly at the zero mark.\nYou will also probably find that as you slide the weights, the pointer swings suddenly to the\nopposite side, but you can never get it to be stable in the middle (zero) position. Try bringing\nthe pointer manually to the zero position and then releasing it. If it swings up, you’re too low,\nand if it swings down, you’re too high. Search for the dividing line between the too-low region\nand the too-high region.\nIf the changes in the scale readings are very small (say a few grams or less), you need to get\nthe magnets closer together. It should be possible to get the scale readings to change by large\namounts (up to 10 or 20 g).\n260\nAppendix 1: Exercises\nExercise 5A: Friction\nEquipment:\n2-meter pieces of butcher paper\nwood blocks with hooks\nstring\nmasses to put on top of the blocks to increase friction\nspring scales (preferably calibrated in Newtons)\n1. Using the same equipment as in exercise 4A, test the statement that kinetic friction is\napproximately independent of velocity.\n2. Test the statement that kinetic friction is independent of surface area.\n261\nExercise 8A: Vectors and Motion\nEach diagram on page 263 shows the motion of an object in an x − y plane. Each dot is one\nlocation of the object at one moment in time. The time interval from one dot to the next is\nalways the same, so you can think of the vector that connects one dot to the next as a v vector,\nand subtract to find ∆v vectors.\n1. Suppose the object in diagram 1 is moving from the top left to the bottom right. Deduce\nwhatever you can about the force acting on it. Does the force always have the same magnitude?\nThe same direction?\nInvent a physical situation that this diagram could represent.\nWhat if you reinterpret the diagram, and reverse the object’s direction of motion?\n2. What can you deduce about the force that is acting in diagram 2?\nInvent a physical situation that diagram 2 could represent.\n3. What can you deduce about the force that is acting in diagram 3?\nInvent a physical situation.\n262\nAppendix 1: Exercises\n263\nExercise 10A: The Shell Theorem\nThis exercise is an approximate numerical test of the shell theorem. There are seven masses\nA-G, each being one kilogram. Masses A-E, each one meter from the center, form a shape like\ntwo Egyptian pyramids joined at their bases; this is a rough approximation to a six-kilogram\nspherical shell of mass. Mass G is five meters from the center of the main group. The class will\ndivide into six groups and split up the work required in order to calculate the vector sum of the\nsix gravitational forces exerted on mass G. Depending on the size of the class, more than one\ngroup may be assigned to deal with the contribution of the same mass to the total force, and\nthe redundant groups can check each other’s results.\n1. Discuss as a class what can be done to simplify the task of calculating the vector sum, and\nhow to organize things so that each group can work in parallel with the others.\n2. Each group should write its results on the board in units of piconewtons, retaining six\nsignificant figures of precision.\n3. The class will determine the vector sum and compare with the result that would be obtained\nwith the shell theorem.\n264\nAppendix 1: Exercises\nAppendix 2: Photo Credits\nExcept as specifically noted below or in a parenthetical credit in the caption of a figure, all the illustrations in\nthis book are under my own copyright, and are copyleft licensed under the same license as the rest of the book.\nIn some cases it’s clear from the date that the figure is public domain, but I don’t know the name of the artist\nor photographer; I would be grateful to anyone who could help me to give proper credit. I have assumed that\nimages that come from U.S. government web pages are copyright-free, since products of federal agencies fall into\nthe public domain. I’ve included some public-domain paintings; photographic reproductions of them are not\ncopyrightable in the U.S. (Bridgeman Art Library, Ltd. v. Corel Corp., 36 F. Supp. 2d 191, S.D.N.Y. 1999).\nWhen “PSSC Physics” is given as a credit, it indicates that the figure is from the first edition of the textbook\nentitled Physics, by the Physical Science Study Committee. The early editions of these books never had their\ncopyrights renewed, and are now therefore in the public domain. There is also a blanket permission given in\nthe later PSSC College Physics edition, which states on the copyright page that “The materials taken from the\noriginal and second editions and the Advanced Topics of PSSC PHYSICS included in this text will be available\nto all publishers for use in English after December 31, 1970, and in translations after December 31, 1975.”\nCredits to Millikan and Gale refer to the textbooks Practical Physics (1920) and Elements of Physics (1927).\nBoth are public domain. (The 1927 version did not have its copyright renewed.) Since is possible that some of\nthe illustrations in the 1927 version had their copyrights renewed and are still under copyright, I have only used\nthem when it was clear that they were originally taken from public domain sources.\nIn a few cases, I have made use of images under the fair use doctrine. However, I am not a lawyer, and the laws\non fair use are vague, so you should not assume that it’s legal for you to use these images. In particular, fair use\nlaw may give you less leeway than it gives me, because I’m using the images for educational purposes, and giving\nthe book away for free. Likewise, if the photo credit says “courtesy of ...,” that means the copyright owner gave\nme permission to use it, but that doesn’t mean you have permission to use it.\nCover Moon: Loewy and Puiseux, 1894.\nContents Ballerina: Rick Dikeman, 1981, GFDL 1.2, from the\nWikipedia article on ballet (retouched by B. Crowell).\nContents Bee, motorcyclist: see below.\n19 Mars\nClimate Orbiter: NASA/JPL/CIT. 43 Bee: Wikipedia user Fir0002, GFDL licensed.\n62 Albert Einstein:\npublic domain. 63 E. Coli bacteria: Eric Erbe, digital colorization by Christopher Pooley, both of USDA, ARS,\nEMU. A public-domain product of the Agricultural Research Service.. 70 Trapeze: Calvert Litho. Co., Detroit,\nca. 1890. 73 Gymnastics wheel: Copyright Hans Genten, Aachen, Germany. “The copyright holder of this file\nallows anyone to use it for any purpose, provided that this remark is referenced or copied.”.\n73 High jumper:\nDunia Young.\n81 Rocket sled: U.S. Air Force, public domain work of the U.S. Government.\n81 Aristotle:\nFrancesco Hayez, 1811. 81 Shanghai: Agnieszka Bojczuk, GFDL 1.2. 81 Angel Stadium: U.S. Marine Corps,\nStaff Sgt. Chad McMeen, public domain work of the U.S. Government. 81 Jets over New York: U.S. Air Force,\nTech. Sgt. Sean Mateo White, public domain work of the U.S. Government. 90 Tuna’s migration: Modified\nfrom a figure in Block et al.\n91 Galileo’s trial: Cristiano Banti (1857).\n111 International Space Station:\nNASA. 111 Weightless astronauts: NASA. 110 Space Ship One: courtesy of Scaled Composites LLC. 97\nGravity map: US Navy, European Space Agency, D. Sandwell, and W. Smith.\n123 Newton: Godfrey Kneller,\n1702.\n146 Space shuttle launch: NASA. 147 Swimmer: Adrian Pingstone (Wikipedia user Arpingstone),\npublic domain.\n157 Locomotive: Locomotive Cyclopedia of American Practice, 1922, public domain.\n158\nHummer: Wikimedia commons user Bull-Doser, public domain. 158 Prius: Wikimedia commons user IFCAR,\npublic domain. 162 Golden Gate Bridge: Wikipedia user Dschwen, GFDL licensed. 175 Ring toss: Clarence\nWhite, 1899.\n187 Aerial photo of Mondavi vineyards: NASA. 199 Galloping horse: Eadweard Muybridge,\n1878. 205 Sled: Modified from Millikan and Gale, 1920. 213 Hanging boy: Millikan and Gale, 1927. 214\nHurricane track: Public domain, NASA and Wikipedia user Nilfanion.\n219 Motorcyclist: Wikipedia user\nFir0002, GFDL licensed.\n216 Crane fly: Wikipedia user Pinzo, public domain. 224 Space colony: NASA.\n230 Tycho Brahe: public domain.\n235 Pluto and Charon: Hubble Space Telescope, STSCi.\n229 Saturn:\nVoyager 2 team, NASA. 251 Uranus: Voyager 2 team, NASA. 251 Earth: Apollo 11, NASA. 244 WMAP:\nNASA.\nAppendix 3: Hints and Solutions\nAnswers to Self-Checks\nAnswers to Self-Checks for Chapter 0\nPage 21, self-check A: If only he has the special powers, then his results can never be\nreproduced.\nPage 22, self-check B: They would have had to weigh the rays, or check for a loss of weight\nin the object from which they were have emitted. (For technical reasons, this was not a measurement they could actually do, hence the opportunity for disagreement.)\nPage 29, self-check C: A dictionary might define “strong” as “possessing powerful muscles,”\nbut that’s not an operational definition, because it doesn’t say how to measure strength numerically. One possible operational definition would be the number of pounds a person can bench\npress.\nPage 32, self-check D: A microsecond is 1000 times longer than a nanosecond, so it would\nseem like 1000 seconds, or about 20 minutes.\nPage 33, self-check E: Exponents have to do with multiplication, not addition. The first line\nshould be 100 times longer than the second, not just twice as long.\nPage 37, self-check F: The various estimates differ by 5 to 10 million. The CIA’s estimate\nincludes a ridiculous number of gratuitous significant figures. Does the CIA understand that\nevery day, people in are born in, die in, immigrate to, and emigrate from Nigeria?\nPage 37, self-check G: (1) 4; (2) 2; (3) 2\nAnswers to Self-Checks for Chapter 1\nPage 44, self-check A: 1 yd2 × (3 ft/1 yd)2 = 9 ft2\n1 yd3 × (3 ft/1 yd)3 = 27 ft3\nAnswers to Self-Checks for Chapter 2\nPage 73, self-check A: Coasting on a bike and coasting on skates give one-dimensional centerof-mass motion, but running and pedaling require moving body parts up and down, which makes\nthe center of mass move up and down. The only example of rigid-body motion is coasting on\nskates. (Coasting on a bike is not rigid-body motion, because the wheels twist.)\nPage 73, self-check B: By shifting his weight around, he can cause the center of mass not to\ncoincide with the geometric center of whe wheel.\nPage 74, self-check C: (1) a point in time; (2) time in the abstract sense; (3) a time interval\nPage 75, self-check D: Zero, because the “after” and “before” values of x are the same.\nPage 82, self-check E: (1) The effect only occurs during blastoff, when their velocity is\nchanging. Once the rocket engines stop firing, their velocity stops changing, and they no longer\nfeel any effect. (2) It is only an observable effect of your motion relative to the air.\nAnswers to Self-Checks for Chapter 3\nPage 93, self-check A: Its speed increases at a steady rate, so in the next second it will travel\n19 cm.\nAnswers to Self-Checks for Chapter 4\nPage 134, self-check A: (1) The case of ρ = 0 represents an object falling in a vacuum, i.e.,\nthere is no density of air. The terminal velocity would be infinite. Physically, we know that an\nobject falling in a vacuum would never stop speeding up, since there would be no force of air\nfriction to cancel the force of gravity. (2) The 4-cm ball would have a mass that was greater by a\nfactor of 4 × 4 × 4, but its cross-sectionalparea would be greater by a factor of 4 × 4. Its terminal\nvelocity would be greater by a factor of 43 /42 = 2. (3) It isn’t of any general importance. It’s\njust an example of one physical situation. You should not memorize it.\nPage 136, self-check B: (1) This is motion, not force. (2) This is a description of how the\nsub is able to get the water to produce a forward force on it. (3) The sub runs out of energy,\nnot force.\nAnswers to Self-Checks for Chapter 5\nPage 147, self-check A: The sprinter pushes backward against the ground, and by Newton’s\nthird law, the ground pushes forward on her. (Later in the race, she is no longer accelerating,\nbut the ground’s forward force is needed in order to cancel out the backward forces, such as air\nfriction.)\nPage 154, self-check B: (1) It’s kinetic friction, because her uniform is sliding over the dirt.\n(2) It’s static friction, because even though the two surfaces are moving relative to the landscape,\nthey’re not slipping over each other. (3) Only kinetic friction creates heat, as when you rub\nyour hands together. If you move your hands up and down together without sliding them across\neach other, no heat is produced by the static friction.\nPage 155, self-check C: Frictionless ice can certainly make a normal force, since otherwise a\nhockey puck would sink into the ice. Friction is not possible without a normal force, however:\nwe can see this from the equation, or from common sense, e.g., while sliding down a rope you\ndo not get any friction unless you grip the rope.\nPage 156, self-check D: (1) Normal forces are always perpendicular to the surface of contact,\nwhich means right or left in this figure. Normal forces are repulsive, so the cliff’s force on the\nfeet is to the right, i.e., away from the cliff. (2) Frictional forces are always parallel to the surface\nof contact, which means right or left in this figure. Static frictional forces are in the direction\nthat would tend to keep the surfaces from slipping over each other. If the wheel was going to\nslip, its surface would be moving to the left, so the static frictional force on the wheel must be\nin the direction that would prevent this, i.e., to the right. This makes sense, because it is the\n267\nstatic frictional force that accelerates the dragster. (3) Normal forces are always perpendicular\nto the surface of contact. In this diagram, that means either up and to the left or down and to\nthe right. Normal forces are repulsive, so the ball is pushing the bat away from itself. Therefore\nthe ball’s force is down and to the right on this diagram.\nAnswers to Self-Checks for Chapter 6\nPage 177, self-check A: The wind increases the ball’s overall speed. If you think about it\nin terms of overall speed, it’s not so obvious that the increased speed is exactly sufficient to\ncompensate for the greater distance. However, it becomes much simpler if you think about the\nforward motion and the sideways motion as two separate things. Suppose the ball is initially\nmoving at one meter per second. Even if it picks up some sideways motion from the wind, it’s\nstill getting closer to the wall by one meter every second.\nAnswers to Self-Checks for Chapter 7\nPage 189, self-check A: v = ∆r/∆t\nPage 189, self-check B:\nPage 193, self-check C: A − B is equivalent to A + (−B), which can be calculated graphically\nby reversing B to form −B, and then adding it to A.\nAnswers to Self-Checks for Chapter 8\nPage 202, self-check A: (1) It is speeding up, because the final velocity vector has the greater\nmagnitude. (2) The result would be zero, which would make sense. (3) Speeding up produced\na ∆v vector in the same direction as the motion. Slowing down would have given a ∆v that\nbointed backward.\nPage 203, self-check B: As we have already seen, the projectile has ax = 0 and ay = −g, so\nthe acceleration vector is pointing straight down.\nAnswers to Self-Checks for Chapter 9\nPage 217, self-check A: (1) Uniform. They have the same motion as the drum itself, which\nis rotating as one solid piece. No part of the drum can be rotating at a different speed from any\nother part. (2) Nonuniform. Gravity speeds it up on the way down and slows it down on the\nway up.\nAnswers to Self-Checks for Chapter 10\nPage 232, self-check A: It would just stay where it was. Plugging v = 0 into eq. would give\nF = 0, so it would not accelerate from rest, and would never fall into the sun. No astronomer\nhad ever observed an object that did that!\n268\nAppendix 3: Hints and Solutions\nPage 233, self-check B:\nF ∝ mr/T 2 ∝ mr/(r3/2 )2 ∝ mr/r3 = m/r2\nPage 236, self-check C: The equal-area law makes equally good sense in the case of a hyperbolic orbit (and observations verify it). The elliptical orbit law had to be generalized by Newton\nto include hyperbolas. The law of periods doesn’t make sense in the case of a hyperbolic orbit,\nbecause a hyperbola never closes back on itself, so the motion never repeats.\nPage 240, self-check D: Above you there is a small part of the shell, comprising only a tiny\nfraction of the earth’s mass. This part pulls you up, while the whole remainder of the shell pulls\nyou down. However, the part above you is extremely close, so it makes sense that its force on\nyou would be far out of proportion to its small mass.\nSolutions to Selected Homework Problems\nSolutions for Chapter 0\nPage 40, problem 6:\n134 mg ×\n10−3 g 10−3 kg\n×\n= 1.34 × 10−4 kg\n1 mg\n1g\nPage 41, problem 8: (a) Let’s do 10.0 g and 1000 g. The arithmetic mean is 505 grams. It\ncomes out to be 0.505 kg, which is consistent. (b) The geometric mean comes out to be 100\ng or 0.1 kg, which is consistent. (c) If we multiply meters by meters, we get square meters.\nMultiplying grams by grams should give square grams! This sounds strange, but it makes sense.\nTaking the square root of square grams (g2 ) gives grams again. (d) No. The superduper mean\nof two quantities with units of grams wouldn’t even be something with units of grams! Related\nto this shortcoming is the fact that the superduper mean would fail the kind of consistency test\ncarried out in the first two parts of the problem.\nSolutions for Chapter 1\nPage 61, problem 10:\n1 mm2 ×\n1 cm\n10 mm\n2\n= 10−2 cm2\nPage 61, problem 11: The bigger scope has a diameter that’s ten times greater. Area scales\nas the square of the linear dimensions, so its light-gathering power is a hundred times greater\n(10 × 10).\nPage 61, problem 12: Since they differ by two steps on the Richter scale, the energy of the\nbigger quake is 10000 times greater. The wave forms a hemisphere, and the surface area of the\nhemisphere over which the energy is spread is proportional to the square of its radius. If the\namount of vibration was the same, then the surface areas much be in the ratio of 10000:1, which\nmeans that the ratio of the radii is 100:1.\nPage 62, problem 17: The cone of mixed gin and vermouth is the same shape as the cone of\nvermouth, but its linear dimensions are doubled, so its volume is 8 times greater. The ratio of\n269\ngin to vermouth is 7 to 1.\nPage 62, problem 19: Scaling down the linear dimensions by a factor of 1/10 reduces the\nvolume by a factor of (1/10)3 = 1/1000, so if the whole cube is a liter, each small one is one\nmilliliter.\nPage 63, problem 20: (a) They’re all defined in terms of the ratio of side of a triangle to\nanother. For instance, the tangent is the length of the opposite side over the length of the\nadjacent side. Dividing meters by meters gives a unitless result, so the tangent, as well as the\nother trig functions, is unitless. (b) The tangent function gives a unitless result, so the units on\nthe right-hand side had better cancel out. They do, because the top of the fraction has units of\nmeters squared, and so does the bottom.\nPage 63, problem 21: Let’s estimate the Great Wall’s mass, and then figure out how many\nbricks that would represent. The wall is famous because it covers pretty much all of China’s\nnorthern border, so let’s say it’s 1000 km long. From pictures, it looks like it’s about 10 m high\nand 10 m wide, so the total volume would be 106 m × 10 m × 10 m = 108 m3 . If a single brick\nhas a volume of 1 liter, or 10−3 m3 , then this represents about 1011 bricks. If one person can\nlay 10 bricks in an hour (taking into account all the preparation, etc.), then this would be 1010\nman-hours.\nSolutions for Chapter 2\nPage 89, problem 4:\n1 light-year = v∆t\n= 3 × 108 m/s (1 year)\n365 days\n24 hours\n3600 s\n8\n= 3 × 10 m/s (1 year) ×\n×\n×\n1 year\n1 day\n1 hour\n= 9.5 × 1015 m\nPage 89, problem 5: Velocity is relative, so having to lean tells you nothing about the train’s\nvelocity. Fullerton is moving at a huge speed relative to Beijing, but that doesn’t produce any\nnoticeable effect in either city. The fact that you have to lean tells you that the train is changing\nits speed, but it doesn’t tell you what the train’s current speed is.\nPage 89, problem 7: To the person riding the moving bike, bug A is simply going in circles.\nThe only difference between the motions of the two wheels is that one is traveling through space,\nbut motion is relative, so this doesn’t have any effect on the bugs. It’s equally hard for each of\nthem.\nPage 90, problem 10: In one second, the ship moves v meters to the east, and the person\nmoves v meters north relative to the deck. Relative to the water, he traces √\nthe diagonal of a\n2\n2\n1\ntriangle whose length is given by the Pythagorean theorem, (v + v ) /2 = 2v. Relative to\nthe water, he is moving at a 45-degree angle between north and east.\nSolutions for Chapter 3\nPage 117, problem 14:\n270\nAppendix 3: Hints and Solutions\nPage 117, problem 15: Taking g to be 10 m/s, the bullet loses 10 m/s of speed every second,\nso it will take 10 s to come to a stop, and then another 10 s to come back down, for a total of\n20 s.\n√\nPage 117, problem 16: ∆x = 21 at2 , so for a fixed value of ∆x, we have t ∝ 1/ a. Decreasing\n√\na by a factor of 3 means that t will increase by a factor of 3 = 1.7. (The given piece of data,\n3, only has one sig fig, but rounding the final result off to one sig fig, giving 2 rather then 1.7,\nwould be a little too severe. As discussed in section 0.10, sig figs are only a rule of thumb, and\nwhen in doubt, you can change the input data to see how much the output would have changed.\nThe ratio of the gravitational fields on Earth and Mars must be in the range from 2.5 to 3.5,\nsince otherwise the given data would not have been rounded off to 3. Using this range of inputs,\nthe possible range of values for the final result becomes 1.6 to 1.9. The final digit in the 1.7\nis therefore a little uncertain, but it’s not complete garbage. It carries useful information, and\nshould not be thrown out.)\nPage 117, problem 17:\ndx\ndt\n= 10 − 3t2\ndv\na=\ndt\n= −6t\nv=\n= −18 m/s2\nPage 118, problem 18: (a) Solving for ∆x = 12 at2 for a, we find a = 2∆x/t2 = 5.51 m/s2 .\n√\n(b) v = 2a∆x = 66.6 m/s. (c) The actual car’s final velocity is less than that of the idealized\nconstant-acceleration car. If the real car and the idealized car covered the quarter mile in the\nsame time but the real car was moving more slowly at the end than the idealized one, the real\ncar must have been going faster than the idealized car at the beginning of the race. The real car\napparently has a greater acceleration at the beginning, and less acceleration at the end. This\nmake sense, because every car has some maximum speed, which is the speed beyond which it\ncannot accelerate.\nPage 118, problem 19: Since the lines are at intervals of one m/s and one second, each box\nrepresents one meter. From t = 0 to t = 2 s, the area under the curve represents a positive ∆x\nof 6 m. (The triangle has half the area of the 2 × 6 rectangle it fits inside.) After t = 2 s, the\narea above the curve represents negative ∆x. To get −6 m worth of area, we need to go out to\nt = 6 s, at which point the triangle under the axis has a width of 4 s and a height of 3 m/s, for\nan area of 6 m (half of 3 × 4).\nPage 118, problem 20: (a) We choose a coordinate system with positive pointing to the right.\nSome people might expect that the ball would slow down once it was on the more gentle ramp.\n271\nThis may be true if there is significant friction, but Galileo’s experiments with inclined planes\nshowed that when friction is negligible, a ball rolling on a ramp has constant acceleration, not\nconstant speed. The speed stops increasing as quickly once the ball is on the more gentle slope,\nbut it still keeps on increasing. The a-t graph can be drawn by inspecting the slope of the v-t\ngraph.\n(b) The ball will roll back down, so the second half of the motion is the same as in part a. In\nthe first (rising) half of the motion, the velocity is negative, since the motion is in the opposite\ndirection compared to the positive x axis. The acceleration is again found by inspecting the\nslope of the v-t graph.\nPage 118, problem 21: This is a case where it’s probably easiest to draw the acceleration\ngraph first. While the ball is in the air (bc, de, etc.), the only force acting on it is gravity, so\nit must have the same, constant acceleration during each hop. Choosing a coordinate system\nwhere the positive x axis points up, this becomes a negative acceleration (force in the opposite\ndirection compared to the axis). During the short times between hops when the ball is in contact\nwith the ground (cd, ef, etc.), it experiences a large acceleration, which turns around its velocity\nvery rapidly. These short positive accelerations probably aren’t constant, but it’s hard to know\nhow they’d really look. We just idealize them as constant accelerations. Similarly, the hand’s\nforce on the ball during the time ab is probably not constant, but we can draw it that way,\nsince we don’t know how to draw it more realistically. Since our acceleration graph consists\nof constant-acceleration segments, the velocity graph must consist of line segments, and the\nposition graph must consist of parabolas. On the x graph, I chose zero to be the height of the\ncenter of the ball above the floor when the ball is just lying on the floor. When the ball is\ntouching the floor and compressed, as in interval cd, its center is below this level, so its x is\nnegative.\n272\nAppendix 3: Hints and Solutions\nPage 118, problem 22: We have vf2 = 2a∆x, so the distance is proportional to the square of\nthe velocity. To get up to half the speed, the ball needs 1/4 the distance, i.e., L/4.\nSolutions for Chapter 4\nPage 142, problem 7: a = ∆v/∆t, and also a = F/m, so\n∆v\na\nm∆v\n=\nF\n(1000 kg)(50 m/s − 20 m/s)\n=\n3000 N\n= 10 s\n∆t =\nPage 143, problem 10: (a) This is a measure of the box’s resistance to a change in its state\nof motion, so it measures the box’s mass. The experiment would come out the same in lunar\ngravity.\n(b) This is a measure of how much gravitational force it feels, so it’s a measure of weight. In\nlunar gravity, the box would make a softer sound when it hit.\n(c) As in part a, this is a measure of its resistance to a change in its state of motion: its mass.\nGravity isn’t involved at all.\nSolutions for Chapter 5\nPage 170, problem 14:\n(a)\ntop spring’s rightward force on connector\n...connector’s leftward force on top spring\nbottom spring’s rightward force on connector\n...connector’s leftward force on bottom spring\nhand’s leftward force on connector\n...connector’s rightward force on hand\n273\nLooking at the three forces on the connector, we see that the hand’s force must be double the\nforce of either spring. The value of x − xo is the same for both springs and for the arrangement\nas a whole, so the spring constant must be 2k. This corresponds to a stiffer spring (more force\nto produce the same extension).\n(b) Forces in which the left spring participates:\nhand’s leftward force on left spring\n...left spring’s rightward force on hand\nright spring’s rightward force on left spring\n...left spring’s leftward force on right spring\nForces in which the right spring participates:\nleft spring’s leftward force on right spring\n...right spring’s rightward force on left spring\nwall’s rightward force on right spring\n...right spring’s leftward force on wall\nSince the left spring isn’t accelerating, the total force on it must be zero, so the two forces acting\non it must be equal in magnitude. The same applies to the two forces acting on the right spring.\nThe forces between the two springs are connected by Newton’s third law, so all eight of these\nforces must be equal in magnitude. Since the value of x − xo for the whole setup is double what\nit is for either spring individually, the spring constant of the whole setup must be k/2, which\ncorresponds to a less stiff spring.\nPage 170, problem 16: (a) Spring constants in parallel add, so the spring constant has to be\nproportional to the cross-sectional area. Two springs in series give half the spring constant, three\nsprings in series give 1/3, and so on, so the spring constant has to be inversely proportional\nto the length. Summarizing, we have k ∝ A/L. (b) With the Young’s modulus, we have\nk = (A/L)E.The spring constant has units of N/m, so the units of E would have to be N/m2 .\nPage 171, problem 18: (a) The swimmer’s acceleration is caused by the water’s force on the\nswimmer, and the swimmer makes a backward force on the water, which accelerates the water\nbackward. (b) The club’s normal force on the ball accelerates the ball, and the ball makes a\nbackward normal force on the club, which decelerates the club. (c) The bowstring’s normal force\naccelerates the arrow, and the arrow also makes a backward normal force on the string. This\nforce on the string causes the string to accelerate less rapidly than it would if the bow’s force\nwas the only one acting on it. (d) The tracks’ backward frictional force slows the locomotive\ndown. The locomotive’s forward frictional force causes the whole planet earth to accelerate by\na tiny amount, which is too small to measure because the earth’s mass is so great.\nPage 171, problem 20: The person’s normal force on the box is paired with the box’s normal\nforce on the person. The dirt’s frictional force on the box pairs with the box’s frictional force\non the dirt. The earth’s gravitational force on the box matches the box’s gravitational force on\nthe earth.\nPage 172, problem 26: (a) A liter of water has a mass of 1.0 kg. The mass is the same in\nall three locations. Mass indicates how much an object resists a change in its motion. It has\nnothing to do with gravity. (b) The term “weight” refers to the force of gravity on an object.\nThe bottle’s weight on earth is FW = mg = 9.8 N. Its weight on the moon is about one sixth\nthat value, and its weight in interstellar space is zero.\n274\nAppendix 3: Hints and Solutions\nSolutions for Chapter 6\nPage 184, problem 5: (a) The easiest strategy is to find the time spent aloft, and then find\nthe range. The vertical motion and the horizontal motion are independent. The vertical motion\nhas acceleration −g, and the cannonball spends enough time in the air to reverse its vertical\nvelocity component completely, so we have\n∆vy = vyf − vyi\n= −2v sin θ\n.\nThe time spent aloft is therefore\n∆t = ∆vy /ay\n= 2v sin θ/g\n.\nDuring this time, the horizontal distance traveled is\nR = vx ∆t\n= 2v 2 sin θ cos θ/g\n.\n(b) The range becomes zero at both θ = 0 and at θ = 90 ◦ . The θ = 0 case gives zero range\nbecause the ball hits the ground as soon as it leaves the mouth of the cannon. A 90-degree angle\ngives zero range because the cannonball has no horizontal motion.\nSolutions for Chapter 8\nPage 212, problem 8: We want to find out about the velocity vector vBG of the bullet relative\nto the ground, so we need to add Annie’s velocity relative to the ground vAG to the bullet’s\nvelocity vector vBA relative to her. Letting the positive x axis be east and y north, we have\nvBA,x = (140 mi/hr) cos 45 ◦\n= 100 mi/hr\nvBA,y = (140 mi/hr) sin 45 ◦\n= 100 mi/hr\nand\nvAG,x = 0\nvAG,y = 30 mi/hr\n.\nThe bullet’s velocity relative to the ground therefore has components\nvBG,x = 100 mi/hrand\nvBG,y = 130 mi/hr\n.\nIts speed on impact with the animal is the magnitude of this vector\np\n|vBG | = (100 mi/hr)2 + (130 mi/hr)2\n= 160 mi/hr\n275\n(rounded off to 2 significant figures).\nPage 212, problem 9: Since its velocity vector is constant, it has zero acceleration, and the\nsum of the force vectors acting on it must be zero. There are three forces acting on the plane:\nthrust, lift, and gravity. We are given the first two, and if we can find the third we can infer its\nmass. The sum of the y components of the forces is zero, so\n0 = Fthrust,y + Flif t,y + FW ,y\n= |Fthrust | sin θ + |Flif t | cos θ − mg\n.\nThe mass is\nm = (|Fthrust | sin θ + |Flif t | cos θ)/g\n= 6.9 × 104 kg\nPage 212, problem 10: (a) Since the wagon has no acceleration, the total forces in both the\nx and y directions must be zero. There are three forces acting on the wagon: FT , FW , and the\nnormal force from the ground, FN . If we pick a coordinate system with x being horizontal and y\nvertical, then the angles of these forces measured counterclockwise from the x axis are 90 ◦ − φ,\n270 ◦ , and 90 ◦ + θ, respectively. We have\nFx,total = |FT | cos(90 ◦ − φ) + |FW | cos(270 ◦ ) + |FN | cos(90 ◦ + θ)\nFy,total = |FT | sin(90 ◦ − φ) + |FW | sin(270 ◦ ) + |FN | sin(90 ◦ + θ)\n,\nwhich simplifies to\n0 = |FT | sin φ − |FN | sin θ\n0 = |FT | cos φ − |FW | + |FN | cos θ.\nThe normal force is a quantity that we are not given and do not with to find, so we should\nchoose it to eliminate. Solving the first equation for |FN | = (sin φ/ sin θ)|FT |, we eliminate |FN |\nfrom the second equation,\n0 = |FT | cos φ − |FW | + |FT | sin φ cos θ/ sin θ\nand solve for |FT |, finding\n|FT | =\n|FW |\ncos φ + sin φ cos θ/ sin θ\n.\nMultiplying both the top and the bottom of the fraction by sin θ, and using the trig identity for\nsin(θ + φ) gives the desired result,\n|FT | =\nsin θ\n|FW |\nsin(θ + φ)\n.\n(b) The case of φ = 0, i.e., pulling straight up on the wagon, results in |FT | = |FW |: we simply\nsupport the wagon and it glides up the slope like a chair-lift on a ski slope. In the case of\nφ = 180 ◦ − θ, |FT | becomes infinite. Physically this is because we are pulling directly into the\nground, so no amount of force will suffice.\n276\nAppendix 3: Hints and Solutions\nPage 213, problem 11: (a) If there was no friction, the angle of repose would be zero, so the\ncoefficient of static friction, µs , will definitely matter. We also make up symbols θ, m and g for\nthe angle of the slope, the mass of the object, and the acceleration of gravity. The forces form\na triangle just like the one in section 8.3, but instead of a force applied by an external object,\nwe have static friction, which is less than µs |FN |. As in that example, |Fs | = mg sin θ, and\n|Fs | < µs |FN |, so\nmg sin θ < µs |FN |\n.\nFrom the same triangle, we have |FN | = mg cos θ, so\nmg sin θ < µs mg cos θ\n.\nRearranging,\nθ < tan−1 µs\n.\n(b) Both m and g canceled out, so the angle of repose would be the same on an asteroid.\nSolutions for Chapter 9\nPage 226, problem 5: Each cyclist has a radial acceleration of v 2 /r = 5 m/s2 . The tangential\naccelerations of cyclists A and B are 375 N/75 kg = 5 m/s2 .\nPage 227, problem 6: (a) The inward normal force must be sufficient to produce circular\nmotion, so\n|FN | = mv 2 /r\n.\nWe are searching for the minimum speed, which is the speed at which the static friction force is\njust barely able to cancel out the downward gravitational force. The maximum force of static\nfriction is\n|Fs | = µs |FN |\n,\nand this cancels the gravitational force, so\n|Fs | = mg\n.\nSolving these three equations for v gives\nr\nv=\n(b) Greater by a factor of\n√\ngr\nµs\n.\n3.\nPage 227, problem 7: The inward force must be supplied by the inward component of the\nnormal force,\n|FN | sin θ = mv 2 /r\n.\n277\nThe upward component of the normal force must cancel the downward force of gravity,\n|FN | cos θ = mg.\nEliminating |FN | and solving for θ, we find\n−1\nθ = tan\nv2\ngr\n.\nSolutions for Chapter 10\nPage 248, problem 10: Newton’s law of gravity tells us that her weight will be 6000 times\nsmaller because of the asteroid’s smaller mass, but 132 = 169 times greater because of its smaller\nradius. Putting these two factors together gives a reduction in weight by a factor of 6000/169,\nso her weight will be (400 N)(169)/(6000) = 11 N.\nPage 248, problem 11: Newton’s law of gravity says F = Gm1 m2 /r2 , and Newton’s second\nlaw says F = m2 a, so Gm1 m2 /r2 = m2 a. Since m2 cancels, a is independent of m2 .\nPage 249, problem 12: Newton’s second law gives\nF = mD aD\n,\nwhere F is Ida’s force on Dactyl. Using Newton’s universal law of gravity, F= GmI mD /r2 ,and\nthe equation a = v 2 /r for circular motion, we find\nGmI mD /r2 = mD v 2 /r.\nDactyl’s mass cancels out, giving\nGmI /r2 = v 2 /r.\nDactyl’s velocity equals the circumference of its orbit divided by the time for one orbit: v =\n2πr/T . Inserting this in the above equation and solving for mI , we find\nmI =\n4π 2 r3\nGT 2\n,\nso Ida’s density is\nρ = mI /V\n=\n4π 2 r3\nGV T 2\n.\nPage 249, problem 15: Newton’s law of gravity depends on the inverse square of the distance,\nso if the two planets’ masses had been equal, then the factor of 0.83/0.059 = 14 in distance would\nhave caused the force on planet c to be 142 = 2.0 × 102 times weaker. However, planet c’s mass\nis 3.0 times greater, so the force on it is only smaller by a factor of 2.0 × 102 /3.0 = 65.\nPage 250, problem 16: The reasoning is reminiscent of section 10.2. From Newton’s second\nlaw we have\nF = ma = mv 2 /r = m(2πr/T )2 /r = 4π 2 mr/T 2\n,\n278\nAppendix 3: Hints and Solutions\nand Newton’s law of gravity gives F = GM m/r2 , where M is the mass of the earth. Setting\nthese expressions equal to each other, we have\n4π 2 mr/T 2 = GM m/r2\n,\nwhich gives\nr=\nGM T 2\n4π 2\n1/3\n= 4.22 × 104 km\n.\nThis is the distance from the center of the earth, so to find the altitude, we need to subtract\nthe radius of the earth. The altitude is 3.58 × 104 km.\nPage 250, problem 17: Any fractional change in r results in double that amount of fractional\nchange in 1/r2 . For example, raising r by 1% causes 1/r2 to go down by very nearly 2%. The\nfractional change in 1/r2 is actually\n2×\n1 km\n(1/27) cm\n× 5\n= 2 × 10−12\n5\n3.84 × 10 km 10 cm\nPage 250, problem 19: (a) The asteroid’s mass depends on the cube of its radius, and for\na given mass the surface gravity depends on r−2 . The result is that surface gravity is directly\nproportional to radius. Half the gravity means half the radius, or one eighth the mass. (b)\nTo agree with a, Earth’s mass would have to be 1/8 Jupiter’s. We assumed spherical shapes\nand equal density. Both planets are at least roughly spherical, so the only way out of the\ncontradiction is if Jupiter’s density is significantly less than Earth’s.\n279\nIndex\nacceleration, 95\nas a vector, 202\nconstant, 107\ndefinition, 102\nnegative, 98\nalchemy, 21\narea, 105\noperational definition, 43\nscaling of, 45\narea under a curve\narea under a-t graph, 107\nunder v-t graph, 105\nastrology, 21\nBacon, Francis, 25\ncalculus\ndifferential, 86\nfundamental theorem of, 113\nintegral, 113\ninvention by Newton, 86\nLeibnitz notation, 86\nwith vectors, 206\ncathode rays, 23\ncenter of mass, 70\nmotion of, 71\ncenter-of-mass motion, 71\ncenti- (metric prefix), 28\ncircular motion, 215\nnonuniform, 217\nuniform, 217\ncockroaches, 53\ncoefficient of kinetic friction, 155\ncoefficient of static friction, 155\ncomponent\ndefined, 179\nconversions of units, 33\ncoordinate system\ndefined, 76\nCopernicus, 80\nDarwin, 24\ndelta notation, 74\nderivative, 86\nsecond, 113\nDialogues Concerning the Two New Sciences,\n46\ndynamics, 66\nelephant, 55\nenergy\ndistinguished from force, 135\nfalling objects, 91\nFeynman, 94\nFeynman, Richard, 94\nforce\nanalysis of forces, 158\nAristotelian versus Newtonian, 124\nas a vector, 205\nattractive, 151\ncontact, 126\ndistinguished from energy, 135\nfrictional, 153\ngravitational, 153\nnet, 127\nnoncontact, 126\nnormal, 153\noblique, 151\npositive and negative signs of, 127\nrepulsive, 151\ntransmission, 161\nforces\nclassification of, 150\nframe of reference\ndefined, 76\ninertial or noninertial, 138\nFrench Revolution, 28\nfriction\nfluid, 157\nkinetic, 153, 154\nstatic, 153, 154\nGalileo Galilei, 45\ngamma rays, 22\ngrand jete, 71\ngraphing, 78\ngraphs\nof position versus time, 76\nvelocity versus time, 85\nhigh jump, 73\nHooke’s law, 163\ninertia\nprinciple of, 80\nintegral, 113\nKepler, 230\nKepler’s laws, 231\nelliptical orbit law, 231\nequal-area law, 231\nlaw of periods, 231, 233\nkilo- (metric prefix), 28\nkilogram, 30\nkinematics, 66\nLaplace, 22\nLeibnitz, 86\nlight, 22\nmagnitude of a vector\ndefined, 188\nmatter, 22\nmega- (metric prefix), 28\nmeter (metric unit), 30\nmetric system, 27\nprefixes, 28\nmicro- (metric prefix), 28\nmicrowaves, 22\nmilli- (metric prefix), 28\nmks units, 30\nmodel\nscientific, 154\nmodels, 71\nmotion\nrigid-body, 69\ntypes of, 69\nMuybridge, Eadweard, 199\nnano- (metric prefix), 28\nNewton\nfirst law of motion, 127\nsecond law of motion, 131\nNewton’s laws of motion\nin three dimensions, 181\nNewton’s third law, 146\nNewton, Isaac, 27\ndefinition of time, 30\noperational definitions, 29\norder-of-magnitude estimates, 57\nparabola\nmotion of projectile on, 180\nPauli exclusion principle, 24\nperiod\nof uniform circular motion, 222\nphoton, 149\nphysics, 22\nPOFOSTITO, 147\nPope, 46\nprojectiles, 180\npulley, 164\nradial component\ndefined, 224\nradio waves, 22\nreductionism, 24\nRenaissance, 19\nrotation, 69\nsalamanders, 53\nscalar\ndefined, 188\nscaling, 45\napplied to biology, 53\nscientific method, 20\nsecond (unit), 29\nSI units, 30\nsignificant figures, 35\nsimple machine\ndefined, 164\nslam dunk, 71\nspring constant, 163\nStanford, Leland, 199\nstrain, 163\nSwift, Jonathan, 45\ntension, 162\ntime\nduration, 74\npoint in, 74\ntransmission of forces, 161\nunit vectors, 194\nunits, conversion of, 33\nvector, 66\nacceleration, 202\naddition, 188\nIndex\n281\ndefined, 188\nforce, 205\nmagnitude of, 188\nvelocity, 200\nvelocity\naddition of velocities, 83\nas a vector, 200\ndefinition, 77\nnegative, 83\nvertebra, 56\nvolume\noperational definition, 43\nscaling of, 45\nweight force\ndefined, 126\nrelationship to mass, 132\nweightlessness\nbiological effects, 110\nx-rays, 22\nYoung’s modulus, 170\n282\nIndex\nMathematical Review\nProperties of the derivative and integral (for\nstudents in calculus-based courses)\nAlgebra\nQuadratic equation:\nLet f and g be functions of x, and let c be a constant.\n2\nThe solutions\n√ of ax + bx + c = 0\n−b± b2 −4ac\n.\nare x =\n2a\nLinearity of the derivative:\nLogarithms and exponentials:\nd\ndf\n(cf ) = c\ndx\ndx\nln(ab) = ln a + ln b\nea+b = ea eb\nx\nln e = e\nln x\nd\ndf\ndg\n(f + g) =\n+\ndx\ndx dx\n=x\nThe chain rule:\nln(ab ) = b ln a\nd\nf (g(x)) = f 0 (g(x))g 0 (x)\ndx\nGeometry, area, and volume\narea of a triangle of base b and height h\ncircumference of a circle of radius r\narea of a circle of radius r\nsurface area of a sphere of radius r\nvolume of a sphere of radius r\n=\n=\n=\n=\n=\n1\n2 bh\n2πr\nπr2\n4πr2\n4\n3\n3 πr\nDerivatives of products and quotients:\nd\ndf\ndg\n(f g) =\ng+\nf\ndx\ndx\ndx\nTrigonometry with a right triangle\nd\ndx\nf\nf g0\nf0\n− 2\n=\ng\ng\ng\nSome derivatives:\nsin θ = o/h\ncos θ = a/h\ntan θ = o/a\nPythagorean theorem: h2 = a2 + o2\nTrigonometry with any triangle\nd m\nm−1\n, except for m = 0\ndx x = mx\nd\nd\nsin\nx\n=\ncos\nx\ndx\ndx cos x = − sin x\nd x\nd\n1\nx\ndx e = e\ndx ln x = x\nThe fundamental theorem of calculus:\nZ\ndf\ndx = f\ndx\nLinearity of the integral:\nZ\nZ\ncf (x)dx = c f (x)dx\nLaw of Sines:\nZ\nsin α\nsin β\nsin γ\n=\n=\nA\nB\nC\nLaw of Cosines:\nC 2 = A2 + B 2 − 2AB cos γ\nZ\n[f (x) + g(x)] =\nZ\nf (x)dx +\ng(x)dx\nIntegration by parts:\nZ\nZ\nf dg = f g − gdf\nIndex\n283\nTrig Table\nθ\n0◦\n1◦\n2◦\n3◦\n4◦\n5◦\n6◦\n7◦\n8◦\n9◦\n10 ◦\n11 ◦\n12 ◦\n13 ◦\n14 ◦\n15 ◦\n16 ◦\n17 ◦\n18 ◦\n19 ◦\n20 ◦\n21 ◦\n22 ◦\n23 ◦\n24 ◦\n25 ◦\n26 ◦\n27 ◦\n28 ◦\n29 ◦\n284\nsin θ\n0.000\n0.017\n0.035\n0.052\n0.070\n0.087\n0.105\n0.122\n0.139\n0.156\n0.174\n0.191\n0.208\n0.225\n0.242\n0.259\n0.276\n0.292\n0.309\n0.326\n0.342\n0.358\n0.375\n0.391\n0.407\n0.423\n0.438\n0.454\n0.469\n0.485\nIndex\ncos θ\n1.000\n1.000\n0.999\n0.999\n0.998\n0.996\n0.995\n0.993\n0.990\n0.988\n0.985\n0.982\n0.978\n0.974\n0.970\n0.966\n0.961\n0.956\n0.951\n0.946\n0.940\n0.934\n0.927\n0.921\n0.914\n0.906\n0.899\n0.891\n0.883\n0.875\ntan θ\n0.000\n0.017\n0.035\n0.052\n0.070\n0.087\n0.105\n0.123\n0.141\n0.158\n0.176\n0.194\n0.213\n0.231\n0.249\n0.268\n0.287\n0.306\n0.325\n0.344\n0.364\n0.384\n0.404\n0.424\n0.445\n0.466\n0.488\n0.510\n0.532\n0.554\nθ\n30 ◦\n31 ◦\n32 ◦\n33 ◦\n34 ◦\n35 ◦\n36 ◦\n37 ◦\n38 ◦\n39 ◦\n40 ◦\n41 ◦\n42 ◦\n43 ◦\n44 ◦\n45 ◦\n46 ◦\n47 ◦\n48 ◦\n49 ◦\n50 ◦\n51 ◦\n52 ◦\n53 ◦\n54 ◦\n55 ◦\n56 ◦\n57 ◦\n58 ◦\n59 ◦\nsin θ\n0.500\n0.515\n0.530\n0.545\n0.559\n0.574\n0.588\n0.602\n0.616\n0.629\n0.643\n0.656\n0.669\n0.682\n0.695\n0.707\n0.719\n0.731\n0.743\n0.755\n0.766\n0.777\n0.788\n0.799\n0.809\n0.819\n0.829\n0.839\n0.848\n0.857\ncos θ\n0.866\n0.857\n0.848\n0.839\n0.829\n0.819\n0.809\n0.799\n0.788\n0.777\n0.766\n0.755\n0.743\n0.731\n0.719\n0.707\n0.695\n0.682\n0.669\n0.656\n0.643\n0.629\n0.616\n0.602\n0.588\n0.574\n0.559\n0.545\n0.530\n0.515\ntan θ\n0.577\n0.601\n0.625\n0.649\n0.675\n0.700\n0.727\n0.754\n0.781\n0.810\n0.839\n0.869\n0.900\n0.933\n0.966\n1.000\n1.036\n1.072\n1.111\n1.150\n1.192\n1.235\n1.280\n1.327\n1.376\n1.428\n1.483\n1.540\n1.600\n1.664\nθ\n60 ◦\n61 ◦\n62 ◦\n63 ◦\n64 ◦\n65 ◦\n66 ◦\n67 ◦\n68 ◦\n69 ◦\n70 ◦\n71 ◦\n72 ◦\n73 ◦\n74 ◦\n75 ◦\n76 ◦\n77 ◦\n78 ◦\n79 ◦\n80 ◦\n81 ◦\n82 ◦\n83 ◦\n84 ◦\n85 ◦\n86 ◦\n87 ◦\n88 ◦\n89 ◦\n90 ◦\nsin θ\n0.866\n0.875\n0.883\n0.891\n0.899\n0.906\n0.914\n0.921\n0.927\n0.934\n0.940\n0.946\n0.951\n0.956\n0.961\n0.966\n0.970\n0.974\n0.978\n0.982\n0.985\n0.988\n0.990\n0.993\n0.995\n0.996\n0.998\n0.999\n0.999\n1.000\n1.000\ncos θ\n0.500\n0.485\n0.469\n0.454\n0.438\n0.423\n0.407\n0.391\n0.375\n0.358\n0.342\n0.326\n0.309\n0.292\n0.276\n0.259\n0.242\n0.225\n0.208\n0.191\n0.174\n0.156\n0.139\n0.122\n0.105\n0.087\n0.070\n0.052\n0.035\n0.017\n0.000\ntan θ\n1.732\n1.804\n1.881\n1.963\n2.050\n2.145\n2.246\n2.356\n2.475\n2.605\n2.747\n2.904\n3.078\n3.271\n3.487\n3.732\n4.011\n4.331\n4.705\n5.145\n5.671\n6.314\n7.115\n8.144\n9.514\n11.430\n14.301\n19.081\n28.636\n57.290\n∞\nIndex\n285\n286\nIndex\nIndex\n287\n288\nIndex\nIndex\n289\nUseful Data\nMetric Prefixes\nMkmµ- (Greek mu)\nnpf-\n106\n103\n10−3\n10−6\n10−9\n10−12\n10−15\nmegakilomillimicronanopicofemto-\n(Centi-, 10−2 , is used only in the centimeter.)\nThe Greek Alphabet\nα\nβ\nγ\nδ\nζ\nη\nθ\nι\nκ\nλ\nµ\nA\nB\nΓ\n∆\nE\nZ\nH\nΘ\nI\nK\nΛ\nM\nalpha\nbeta\ngamma\ndelta\nepsilon\nzeta\neta\ntheta\niota\nkappa\nlambda\nmu\nν\nξ\no\nπ\nρ\nσ\nτ\nυ\nφ\nχ\nψ\nω\nN\nΞ\nO\nΠ\nP\nΣ\nT\nY\nΦ\nX\nΨ\nΩ\nnu\nxi\nomicron\npi\nrho\nsigma\ntau\nupsilon\nphi\nchi\npsi\nomega\nFundamental Constants\ngravitational constant\nspeed of light\nG = 6.67 × 10−11 N·m2 /kg2\nc = 3.00 × 108 m/s\nSubatomic Particles\nparticle\nelectron\nproton\nneutron\nmass (kg)\n9.109 × 10−31\n1.673 × 10−27\n1.675 × 10−27\nradius (fm)\n. 0.01\n∼ 1.1\n∼ 1.1\nThe radii of protons and neutrons can only be given approximately, since they have fuzzy surfaces. For comparison, a\ntypical atom is about a million fm in radius.\n290\nIndex\nNotation and Units\nquantity\ndistance\ntime\nmass\ndensity\narea\nvolume\nvelocity\nacceleration\ngravitational field\nforce\npressure\nenergy\npower\nmomentum\nangular momentum\ntorque\nperiod\nunit\nmeter, m\nsecond, s\nkilogram, kg\nkg/m3\nm2 (square meters)\nm3 (cubic meters)\nm/s\nm/s2\nJ/kg·m or m/s2\nnewton, 1 N=1 kg·m/s2\n1 Pa=1 N/m2\njoule, J\nwatt, 1 W = 1 J/s\nkg·m/s\nkg·m2 /s or J·s\nN·m\ns\nsymbol\nx, ∆x\nt, ∆t\nm\nρ\nA\nV\nv\na\ng\nF\nP\nE\nP\np\nL\nτ\nT\nConversions\nNonmetric units in terms of metric ones:\n1 inch\n1 pound-force\n(1 kg) · g\n1 scientific calorie\n1 kcal\n1 gallon\n1 horsepower\n=\n=\n=\n=\n=\n=\n=\n25.4 mm (by definition)\n4.5 newtons of force\n2.2 pounds-force\n4.18 J\n4.18 × 103 J\n3.78 × 103 cm3\n746 W\nWhen speaking of food energy, the word “Calorie” is used\nto mean 1 kcal, i.e., 1000 calories. In writing, the capital C\nmay be used to indicate 1 Calorie=1000 calories.\nRelationships among U.S. units:\n1 foot (ft)\n= 12 inches\n1 yard (yd) = 3 feet\n1 mile (mi) = 5280 feet\nEarth, Moon, and Sun\nbody\nearth\nmoon\nsun\nmass (kg)\n5.97 × 1024\n7.35 × 1022\n1.99 × 1030\nradius (km)\n6.4 × 103\n1.7 × 103\n7.0 × 105\nradius of orbit (km)\n1.49 × 108\n3.84 × 105\n—\nIndex\n291\n```\nWas this manual useful for you?\nThank you for your participation!\n\n* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project\n\nadvertisement" ]
[ null, "https://s1.manualzz.com/store/data/009554004_1-6e223de3390f49a7f02f8476ceeee970-768x994.png", null ]
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http://wias-berlin.de/research/rgs/fg6/fpQuantAnalysis/files/roughPDE.jsp
[ "# Numerical solution of rough and stochastic PDEs by regression techniques\n\nGiven a $d$-dimensional $\\alpha$-Hölder continuous geometric rough path , $\\frac{1}{3}<\\alpha \\le \\frac{1}{2}$, we aim to solve an equation of the form\n$− d X = L ( X ) d t + ∑ k = 1 d Γ k ( X ) d W k , X ( T , ⋅ ) = g ,$where we define $L f ( ζ ) := 1 2 t r ( σ ( ζ ) σ ( ζ ) T D 2 f ( ζ ) ) + ⟨ b ( ζ ) , D f ( ζ ) ⟩ + c ( ζ ) f ( ζ ),$ $\\begin{array}{r}{\\mathrm{\\Gamma }}_{k}f\\left(\\zeta \\right):=⟨{\\beta }_{k}\\left(\\zeta \\right),Df\\left(\\zeta \\right)⟩+{\\gamma }_{k}\\left(\\zeta \\right)f\\left(\\zeta \\right)\\end{array}$\nfor a suitable test function $f:{\\mathbb{R}}^{n}\\to \\mathbb{R}$. Moreover, let all above functions be of suitable dimension and \"smooth enough''.\n\nIn the following, this kind of rough partial differential equation is solved numerically for a particular case. We set $c,{\\gamma }_{k}\\equiv 0$, the noise/space dimension $\\mathrm{d, n}=2$, the terminal time $T=1$ and the terminal value $g\\left(\\zeta \\right)=\\mathrm{exp}\\left\\{-\\frac{1}{2}{‖\\zeta ‖}^{2}\\right\\}$. We insert linear expressions for $\\sigma ,b,{\\beta }_{1},{\\beta }_{2}$ and obtain $\\begin{array}{rl}-dX\\left(t,\\zeta \\right)=& \\phantom{\\rule{thickmathspace}{0ex}}\\left[\\frac{1}{2}\\left(C\\zeta {\\right)}^{T}{D}^{2}X\\left(t,\\zeta \\right)C\\zeta +\\left(A\\zeta {\\right)}^{T}DX\\left(t,\\zeta \\right)\\right]dt+\\left({N}_{1}\\zeta {\\right)}^{T}DX\\left(t,\\zeta \\right)d{\\mathbf{W}}_{t}^{1}+\\left({N}_{2}\\zeta {\\right)}^{T}DX\\left(t,\\zeta \\right)d{\\mathbf{W}}_{t}^{2},\\\\ X\\left(1,\\zeta \\right)=& \\phantom{\\rule{thickmathspace}{0ex}}g\\left(\\zeta \\right)=\\mathrm{exp}\\left\\{-\\frac{1}{2}{‖\\zeta ‖}^{2}\\right\\},\\end{array}$ where $A,C,{N}_{1},{N}_{2}$ are commuting $2×2$ matrices and $W$ is a path of a two-dimensional fractional Brownian motion with Hurst index $H=0.4$ which looks as follows", null, "Using Feynman-Kac's formular, the above problem can be reduced to an ordinary rough differential equation since $X ( t , ζ ) = E g ( x 1 t , ζ ) ,$ where for $t\\le r\\le 1$ the process ${x}_{\\cdot }^{t,\\zeta }$ solves the following hybrid Stratonovich-rough differential equation: $d x r = ( A − 1 2 C 2 ) x r d t + C x r ∘ d B r + N 1 x r d W r 1 + N 2 x r d W r 2 , x t = ζ , (1)$ where $B$ is a standard Brownian motion. Equation (1) is well-defined as a rough differential equation driven by the joint rough of $B$ and $W$.\n\nThe Feynman-Kac formular only leads to a point-wise solution of a rough partial differential equation. Therefore, we use a regression ansatz in the spatial component: $X ( t , ⋅ ) ≈ X ~ ( t , ⋅ ) := ∑ k = 1 K κ k ( t ) ψ k ( ⋅ ) .$ The procedure can be visualized (for one fixed $t$) as follows:\n\nwhere the dots corrspond to solutions of the rough differential equation for different initial values and noise, and the and the surface corresonds to the estimated solution of the rough PDE.\nWe need to discretise (1) with an Euler scheme in time in order to determine the optimal coefficients ${\\kappa }_{1},\\dots ,{\\kappa }_{K}$ for $\\stackrel{~}{X}$. Moreover, we sample the initial condition in (1) to have a computational efficient procedure. By doing so, we compute the regression solution on the domain $\\left[0,1{\\right]}^{2}$ for ${10}^{6}$ samples, a step size of ${2}^{-9}$ in the Euler scheme and $K=36$ basis functions. The regression solution is given in the following video.\n\nIn the above video the forward movement of $\\stackrel{~}{X}$ can be seen, i.e., if the movie is stop at a certain point, the resulting picture shows the plot of the function $\\stackrel{~}{X}\\left(t,\\cdot \\right):\\left[0,1{\\right]}^{2}\\to \\mathbb{R}$ for a fixed time $t\\in \\left[0,1\\right]$. The x- and the y-axis represent the components ${\\zeta }_{1}$ and ${\\zeta }_{2}$, respectively, of the spatial variable $\\zeta =\\left({\\zeta }_{1},{\\zeta }_{2}\\right)\\in \\left[0,1{\\right]}^{2}$." ]
[ null, "http://wias-berlin.de/research/rgs/fg6/fpQuantAnalysis/files/W12.jpg", null ]
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https://math.stackexchange.com/questions/1353231/implicit-function-theorem-conclusion-notation
[ "Implicit function theorem conclusion notation?\n\nI am working through implicit function theorem for the first time, and I have the following understanding. Given a system of $n$ equations, \\begin{equation} f_i(x_1,\\dots ,x_m,y_1,\\dots , y_n)=0,\\ \\ \\ \\ \\ \\ \\ i=1,\\dots ,n \\end{equation} With some point $p_0$ with coordinates $(a_1,\\dots ,a_m,b_1,\\dots ,b_n)$, and the condition, \\begin{equation} \\frac{\\partial (f_1,\\dots,f_n)}{\\partial (y_1,\\dots ,y_n)}\\bigg|_{p_{0}}\\neq 0 \\end{equation} Then the system of equations can be solved for $y_1,\\dots ,y_n$ as functions of $x_1,\\dots ,x_m$ in the neighbourhood of the point. In this case the following equations hold, \\begin{equation} f_i(x_1,\\dots ,x_m,y_1(x_1,\\dots ,x_m),\\dots ,y_n(x_1,\\dots ,x_m))=0,\\ \\ \\ \\ \\ i=1,\\dots ,n \\end{equation} However I do not understand the following notation, which is the conclusion of the theorem! \\begin{equation} \\frac{\\partial y_i}{\\partial x_j}\\bigg|_{i\\neq j}=-\\frac{\\frac{\\partial(f_1,f_2,\\dots ,f_n)}{\\partial (y_1,\\dots ,x_j,\\dots ,y_n)}}{\\frac{\\partial (f_1,f_2,\\dots ,f_n)}{\\partial (y_1,\\dots ,y_i,\\dots ,y_n)}} \\end{equation} Could you please explain what these Jacobian determinants are, and why the $x_j$ and $y_i$ are in the denominators of each expression? Many thanks!\n\n• Firstly you say you don't understand the notation. Then you ask why the equality is true. What do you want to ask exactly? – Git Gud Jul 7 '15 at 23:15\n• Do you mean $i\\neq j$ instead of $x\\neq j$ on the lefthand side in the last equation? – Arthur Jul 7 '15 at 23:15\n• I understand implicit functions for only 2 or 3 variables so i do not ask why equity holds (apologies) but what the notation means. I think so, I will change that! thank you – RedPen Jul 7 '15 at 23:17\n• In short, to prove the theorem, you linearize the level mapping then focus on the particular submatrix to eliminate. Invertibility of the submatrix is captured by the Jacobian determinants. Don't be bothered by this, it's not magic, it's just the usual linear algebra and you can see the determinants as stemming from Cramer's Rule. I don't have that in my answer, but I hope there is enough to \"see\" it... – James S. Cook Jul 8 '15 at 6:48\n\nGiven $n$-equations in $(m+n)$-unknowns when can we solve for the last $n$-variables as functions of the first $m$-variables ? Given a continuously differentiable mapping $G=(G_1,G_2,\\dots , G_n): \\mathbb{R}^m \\times \\mathbb{R}^n \\rightarrow \\mathbb{R}^n$ study the level set: (here $k_1,k_2,\\dots , k_n$ are constants) \\begin{align} \\notag G_1(x_1, \\dots , x_m, y_1, \\dots , y_n)&=k_1 \\\\ \\notag G_2(x_1, \\dots , x_m, y_1, \\dots , y_n)&=k_2 \\\\ \\notag & \\vdots \\\\ \\notag G_n(x_1, \\dots , x_m, y_1, \\dots , y_n)&=k_n \\notag \\end{align} We wish to locally solve for $y_1, \\dots , y_n$ as functions of $x_1, \\dots x_m$. That is, find a mapping $h : \\mathbb{R}^m \\rightarrow \\mathbb{R}^n$ such that $G(x,y)=k$ iff $y=h(x)$ near some point $(a,b) \\in \\mathbb{R}^m \\times \\mathbb{R}^n$ such that $G(a,b)=k$. In this section we use the notation $x=(x_1,x_2,\\dots x_m)$ and $y=(y_1,y_2,\\dots , y_n)$.\nBefore we turn to the general problem let's analyze the unit-circle problem in this notation. We are given $G(x,y)=x^2+y^2$ and we wish to find $f(x)$ such that $y=f(x)$ solves $G(x,y)=1$. Differentiate with respect to $x$ and use the chain-rule: $$\\frac{\\partial G}{\\partial x}\\frac{dx}{dx} + \\frac{\\partial G}{\\partial y}\\frac{dy}{dx} = 0$$ We find that $\\boxed{dy/dx = -G_x/G_y} = -x/y$. Given this analysis we should suspect that if we are given some level curve $G(x,y)=k$ then we may be able to solve for $y$ as a function of $x$ near $p$ if $G(p)=k$ and $G_y(p) \\neq 0$. This suspicion is valid and it is one of the many consequences of the implicit function theorem.\nWe again turn to the linearization approximation. Suppose $G(x,y)=k$ where $x \\in \\mathbb{R}^m$ and $y \\in \\mathbb{R}^n$ and suppose $G: \\mathbb{R}^m \\times \\mathbb{R}^n \\rightarrow \\mathbb{R}^n$ is continuously differentiable. Suppose $(a,b) \\in \\mathbb{R}^m \\times \\mathbb{R}^n$ has $G(a,b)=k$. Replace $G$ with its linearization based at $(a,b)$: $$G(x,y) \\approx k + G'(a,b)(x-a,y-b)$$ here we have the matrix multiplication of the $n \\times (m+n)$ matrix $G'(a,b)$ with the $(m+n) \\times 1$ column vector $(x-a,y-b)$ to yield an $n$-component column vector. It is convenient to define partial derivatives with respect to a whole vector of variables, $$\\frac{\\partial G}{\\partial x} = \\left[ \\begin{array}{ccc} \\tfrac{\\partial G_1}{\\partial x_1} & \\cdots & \\tfrac{\\partial G_1}{\\partial x_m} \\\\ \\vdots & & \\vdots \\\\ \\tfrac{\\partial G_n}{\\partial x_1} & \\cdots & \\tfrac{\\partial G_n}{\\partial x_m} \\end{array} \\right] \\qquad \\frac{\\partial G}{\\partial y} = \\left[ \\begin{array}{ccc} \\tfrac{\\partial G_1}{\\partial y_1} & \\cdots & \\tfrac{\\partial G_1}{\\partial y_n} \\\\ \\vdots & & \\vdots \\\\ \\tfrac{\\partial G_n}{\\partial y_1} & \\cdots & \\tfrac{\\partial G_n}{\\partial y_n} \\end{array} \\right]$$ In this notation we can write the $n \\times (m+n)$ matrix $G'(a,b)$ as the concatenation of the $n \\times m$ matrix $\\frac{\\partial G}{\\partial x}(a,b)$ and the $n \\times n$ matrix $\\frac{\\partial G}{\\partial y}(a,b)$ $$G'(a,b) = \\biggl[\\frac{\\partial G}{\\partial x}(a,b) \\bigg{|} \\frac{\\partial G}{\\partial y}(a,b) \\biggl]$$ Therefore, for points close to $(a,b)$ we have: $$G(x,y) \\approx k + \\frac{\\partial G}{\\partial x}(a,b)(x-a)+\\frac{\\partial G}{\\partial y}(a,b)(y-b)$$ The nonlinear problem $G(x,y)=k$ has been (locally) replaced by the linear problem of solving what follows for $y$: $$k \\approx k + \\frac{\\partial G}{\\partial x}(a,b)(x-a)+\\frac{\\partial G}{\\partial y}(a,b)(y-b)$$ Suppose the square matrix $\\frac{\\partial G}{\\partial y}(a,b)$ is invertible at $(a,b)$ then we find the following approximation for the implicit solution of $G(x,y)=k$ for $y$ as a function of $x$: $$y = b - \\biggl[\\frac{\\partial G}{\\partial y}(a,b) \\biggr]^{-1}\\biggl[\\frac{\\partial G}{\\partial x}(a,b)(x-a) \\biggl].$$ Of course this is not a formal proof, but it does suggest that $det\\bigl[\\frac{\\partial G}{\\partial y}(a,b) \\bigr] \\neq 0$ is a necessary condition for solving for the $y$ variables.\nAs before suppose $G: \\mathbb{R}^m \\times \\mathbb{R}^n \\rightarrow \\mathbb{R}^n$. Suppose we have a continuously differentiable function $h: \\mathbb{R}^m \\rightarrow \\mathbb{R}^n$ such that $h(a)=b$ and $G(x,h(x))=k$. We seek to find the derivative of $h$ in terms of the derivative of $G$. This is a generalization of the implicit differentiation calculation we perform in calculus I. I'm including this to help you understand the notation a bit more before I state the implicit function theorem. Differentiate with respect to $x_l$ for $l \\in \\{1,2,\\dots n\\}$: $$\\frac{\\partial}{\\partial x_{l}} \\biggl[ G(x,h(x)) \\biggr] = \\sum_{i=1}^{m}\\frac{\\partial G}{\\partial x_i } \\frac{\\partial x_i}{\\partial x_l } + \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_l} = \\frac{\\partial G}{\\partial x_l } + \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_l} = 0$$ we made use of the identity $\\frac{\\partial x_i}{\\partial x_k } = \\delta_{ik}$ to squash the sum of $i$ to the single nontrivial term and the zero on the r.h.s follows from the fact that $\\frac{\\partial}{\\partial x_l} (k)=0$. Concatenate these derivatives from $k=1$ up to $k=m$: $$\\biggl[ \\frac{\\partial G}{\\partial x_1 } + \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_1} \\bigg{|} \\frac{\\partial G}{\\partial x_2 } + \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_2} \\bigg{|} \\cdots \\bigg{|} \\frac{\\partial G}{\\partial x_m } + \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_m} \\biggr] = [0|0| \\cdots |0]$$ Properties of matrix addition allow us to parse the expression above as follows: $$\\biggl[ \\frac{\\partial G}{\\partial x_1 } \\bigg{|} \\frac{\\partial G}{\\partial x_2 } \\bigg{|} \\cdots \\bigg{|} \\frac{\\partial G}{\\partial x_m } \\biggr] + \\biggl[ \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_1} \\bigg{|} \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_2} \\bigg{|} \\cdots \\bigg{|} \\sum_{j=1}^{n}\\frac{\\partial G}{\\partial y_j}\\frac{\\partial h_j}{\\partial x_m} \\biggr] = [0|0| \\cdots |0]$$ But, this reduces to $$\\frac{\\partial G}{\\partial x } + \\biggl[ \\frac{\\partial G}{\\partial y}\\frac{\\partial h}{\\partial x_1} \\bigg{|} \\frac{\\partial G}{\\partial y}\\frac{\\partial h}{\\partial x_2} \\bigg{|} \\cdots \\bigg{|} \\frac{\\partial G}{\\partial y}\\frac{\\partial h}{\\partial x_m} \\biggr] = 0 \\in \\mathbb{R}^{m \\times n}$$ The concatenation property of matrix multiplication states $[Ab_1|Ab_2| \\cdots | Ab_m] = A[b_1|b_2| \\cdots | b_m]$ we use this to write the expression once more, $$\\frac{\\partial G}{\\partial x } + \\frac{\\partial G}{\\partial y} \\biggl[ \\frac{\\partial h}{\\partial x_1} \\bigg{|} \\frac{\\partial h}{\\partial x_2} \\bigg{|} \\cdots \\bigg{|} \\frac{\\partial h}{\\partial x_m} \\biggr] = 0 \\ \\ \\Rightarrow \\ \\ \\frac{\\partial G}{\\partial x } + \\frac{\\partial G}{\\partial y} \\frac{\\partial h}{\\partial x} = 0 \\ \\ \\Rightarrow \\ \\ \\boxed{\\frac{\\partial h}{\\partial x} = -\\frac{\\partial G}{\\partial y}^{-1}\\frac{\\partial G}{\\partial x }}$$ where in the last implication we made use of the assumption that $\\frac{\\partial G}{\\partial y}$ is invertible." ]
[ null ]
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https://www.uppsatser.se/uppsats/d61b071aeb/
[ "# Gymnasieelevers förståelse av enhetscirkeln och trigonometri : en undersökning av elevers förståelse av matematiska begrepp, samt vad lärarna anser att nyblivna matematikstudenter bör kunna när de kommer till högskola eller universitet\n\nDetta är en Master-uppsats från LinnéuniversitetetLinnéuniversitetet/Institutionen för datavetenskap, fysik och matematik, DFM\n\nNyckelord: ;\n\nSammanfattning: The aim of my study is to investigate how last year students in upper secondary schoolunderstand certain mathematical concepts, in particular the unit circle and its trigonometry.I have used intentional analysis to interpret student’s actions when they solve certain tasks onthe basis of a cognitive, situated and cultural context.Interviews with four university teachers in mathematics about the unit circle, trigonometry,and mathematical understanding, serve both as background for the study and as basis for adiscussion, where I relate students understanding to what the teachers want new students toknow about these concepts when they begin university studies in mathematics.The students were arranged in three groups with three students in each group. Each group waspresented with two tasks, one in which they were asked to calculate the cosine values for onepointed, one blunt and one straight angle, each located in a separate triangle. They were alsoasked to decide whether the points (0,71; 0,71) and d (1 2 , 3 2 ) are located on the unit circleor not.My conclusion is that students mainly have an operational conception of the unit circle andtrigonometry. The lack of structural conceptions result in difficulties in seeing connectionsbetween the concepts in unfamiliar situations. The students search for known strategies andalgorithms to solve the tasks. They know that the unit circle is a circle with radius 1 andcenter at the origin. They are also familiar with related concepts. One student group showsdeep formal understanding, which is what the university teachers would like the students tohave. The other groups have difficulties separating different concepts and to use themathematical language properly.Further research could include a socio-cultural study with the same empirical material. Itwould also be interesting to study how the language and the graphical calculator influencestudent’s understanding of mathematical concepts.", null, "HÄR KAN DU HÄMTA UPPSATSEN I FULLTEXT. (följ länken till nästa sida)", null, "" ]
[ null, "https://www.uppsatser.se/graphics/nopdfimage.gif", null, "https://www.uppsatser.se/graphics/blank.gif", null ]
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https://mizar.uwb.edu.pl/version/current/html/borsuk_2.html
[ ":: Introduction to Homotopy Theory\n::\n:: Copyright (c) 1997-2021 Association of Mizar Users\n\nLm1: for r being Real holds\n( ( 0 <= r & r <= 1 ) iff r in the carrier of I )\n\nproof end;\n\nscheme :: BORSUK_2:sch 1\nFrCard{ F1() -> non empty set , F2() -> set , F3( object ) -> set , P1[ set ] } :\ncard { F3(w) where w is Element of F1() : ( w in F2() & P1[w] ) } c= card F2()\nproof end;\n\ntheorem :: BORSUK_2:1\nfor T, T1, T2, S being non empty TopSpace\nfor f being Function of T1,S\nfor g being Function of T2,S st T1 is SubSpace of T & T2 is SubSpace of T & ([#] T1) \\/ ([#] T2) = [#] T & T1 is compact & T2 is compact & T is T_2 & f is continuous & g is continuous & ( for p being set st p in ([#] T1) /\\ ([#] T2) holds\nf . p = g . p ) holds\nex h being Function of T,S st\n( h = f +* g & h is continuous )\nproof end;\n\nregistration\nlet T be TopStruct ;\ncoherence\n( id T is open & id T is continuous )\nby ;\nend;\n\nregistration\nlet T be TopStruct ;\ncluster Relation-like the carrier of T -defined the carrier of T -valued Function-like one-to-one V17( the carrier of T) V21( the carrier of T, the carrier of T) continuous for Element of bool [: the carrier of T, the carrier of T:];\nexistence\nex b1 being Function of T,T st\n( b1 is continuous & b1 is one-to-one )\nproof end;\nend;\n\ntheorem :: BORSUK_2:2\nfor S, T being non empty TopSpace\nfor f being Function of S,T st f is being_homeomorphism holds\nf \" is open\nproof end;\n\ntheorem Th3: :: BORSUK_2:3\nfor T being non empty TopSpace\nfor a being Point of T ex f being Function of I,T st\n( f is continuous & f . 0 = a & f . 1 = a )\nproof end;\n\ndefinition\nlet T be TopStruct ;\nlet a, b be Point of T;\npred a,b are_connected means :: BORSUK_2:def 1\nex f being Function of I,T st\n( f is continuous & f . 0 = a & f . 1 = b );\nend;\n\n:: deftheorem defines are_connected BORSUK_2:def 1 :\nfor T being TopStruct\nfor a, b being Point of T holds\n( a,b are_connected iff ex f being Function of I,T st\n( f is continuous & f . 0 = a & f . 1 = b ) );\n\ndefinition\nlet T be non empty TopSpace;\nlet a, b be Point of T;\n:: original: are_connected\nredefine pred a,b are_connected ;\nreflexivity\nfor a being Point of T holds (T,b1,b1)\nby Th3;\nend;\n\ndefinition\nlet T be TopStruct ;\nlet a, b be Point of T;\nassume A1: a,b are_connected ;\nmode Path of a,b -> Function of I,T means :Def2: :: BORSUK_2:def 2\n( it is continuous & it . 0 = a & it . 1 = b );\nexistence\nex b1 being Function of I,T st\n( b1 is continuous & b1 . 0 = a & b1 . 1 = b )\nby A1;\nend;\n\n:: deftheorem Def2 defines Path BORSUK_2:def 2 :\nfor T being TopStruct\nfor a, b being Point of T st a,b are_connected holds\nfor b4 being Function of I,T holds\n( b4 is Path of a,b iff ( b4 is continuous & b4 . 0 = a & b4 . 1 = b ) );\n\nregistration\nlet T be non empty TopSpace;\nlet a be Point of T;\ncluster non empty Relation-like the carrier of I -defined the carrier of T -valued Function-like V17( the carrier of I) V21( the carrier of I, the carrier of T) continuous for Path of a,a;\nexistence\nex b1 being Path of a,a st b1 is continuous\nproof end;\nend;\n\ndefinition\nlet T be TopStruct ;\nattr T is pathwise_connected means :Def3: :: BORSUK_2:def 3\nfor a, b being Point of T holds a,b are_connected ;\ncorrectness\n;\nend;\n\n:: deftheorem Def3 defines pathwise_connected BORSUK_2:def 3 :\nfor T being TopStruct holds\n( T is pathwise_connected iff for a, b being Point of T holds a,b are_connected );\n\nregistration\nexistence\nex b1 being TopSpace st\n( b1 is strict & b1 is pathwise_connected & not b1 is empty )\nproof end;\nend;\n\ndefinition\nlet T be pathwise_connected TopStruct ;\nlet a, b be Point of T;\nredefine mode Path of a,b means :Def4: :: BORSUK_2:def 4\n( it is continuous & it . 0 = a & it . 1 = b );\ncompatibility\nfor b1 being Function of I,T holds\n( b1 is Path of a,b iff ( b1 is continuous & b1 . 0 = a & b1 . 1 = b ) )\nproof end;\nend;\n\n:: deftheorem Def4 defines Path BORSUK_2:def 4 :\nfor T being pathwise_connected TopStruct\nfor a, b being Point of T\nfor b4 being Function of I,T holds\n( b4 is Path of a,b iff ( b4 is continuous & b4 . 0 = a & b4 . 1 = b ) );\n\nregistration\nlet T be pathwise_connected TopStruct ;\nlet a, b be Point of T;\ncluster -> continuous for Path of a,b;\ncoherence\nfor b1 being Path of a,b holds b1 is continuous\nby Def4;\nend;\n\nLm2: ( 0 in [.0,1.] & 1 in [.0,1.] )\nby XXREAL_1:1;\n\nregistration\ncoherence\nfor b1 being non empty TopSpace st b1 is pathwise_connected holds\nb1 is connected\nproof end;\nend;\n\nLm3: for G being non empty TopSpace\nfor w1, w2, w3 being Point of G\nfor h1, h2 being Function of I,G st h1 is continuous & w1 = h1 . 0 & w2 = h1 . 1 & h2 is continuous & w2 = h2 . 0 & w3 = h2 . 1 holds\nex h3 being Function of I,G st\n( h3 is continuous & w1 = h3 . 0 & w3 = h3 . 1 & rng h3 c= (rng h1) \\/ (rng h2) )\n\nproof end;\n\ndefinition\nlet T be non empty TopSpace;\nlet a, b, c be Point of T;\nlet P be Path of a,b;\nlet Q be Path of b,c;\nassume that\nA1: a,b are_connected and\nA2: b,c are_connected ;\nfunc P + Q -> Path of a,c means :Def5: :: BORSUK_2:def 5\nfor t being Point of I holds\n( ( t <= 1 / 2 implies it . t = P . (2 * t) ) & ( 1 / 2 <= t implies it . t = Q . ((2 * t) - 1) ) );\nexistence\nex b1 being Path of a,c st\nfor t being Point of I holds\n( ( t <= 1 / 2 implies b1 . t = P . (2 * t) ) & ( 1 / 2 <= t implies b1 . t = Q . ((2 * t) - 1) ) )\nproof end;\nuniqueness\nfor b1, b2 being Path of a,c st ( for t being Point of I holds\n( ( t <= 1 / 2 implies b1 . t = P . (2 * t) ) & ( 1 / 2 <= t implies b1 . t = Q . ((2 * t) - 1) ) ) ) & ( for t being Point of I holds\n( ( t <= 1 / 2 implies b2 . t = P . (2 * t) ) & ( 1 / 2 <= t implies b2 . t = Q . ((2 * t) - 1) ) ) ) holds\nb1 = b2\nproof end;\nend;\n\n:: deftheorem Def5 defines + BORSUK_2:def 5 :\nfor T being non empty TopSpace\nfor a, b, c being Point of T\nfor P being Path of a,b\nfor Q being Path of b,c st a,b are_connected & b,c are_connected holds\nfor b7 being Path of a,c holds\n( b7 = P + Q iff for t being Point of I holds\n( ( t <= 1 / 2 implies b7 . t = P . (2 * t) ) & ( 1 / 2 <= t implies b7 . t = Q . ((2 * t) - 1) ) ) );\n\nregistration\nlet T be non empty TopSpace;\nlet a be Point of T;\ncluster non empty Relation-like the carrier of I -defined the carrier of T -valued Function-like constant V17( the carrier of I) V21( the carrier of I, the carrier of T) for Path of a,a;\nexistence\nex b1 being Path of a,a st b1 is constant\nproof end;\nend;\n\ntheorem :: BORSUK_2:4\ncanceled;\n\n::$CT theorem :: BORSUK_2:5 for T being non empty TopSpace for a being Point of T for P being constant Path of a,a holds P = I --> a proof end; theorem Th5: :: BORSUK_2:6 for T being non empty TopSpace for a being Point of T for P being constant Path of a,a holds P + P = P proof end; registration let T be non empty TopSpace; let a be Point of T; let P be constant Path of a,a; cluster P + P -> constant ; coherence P + P is constant by Th5; end; definition let T be non empty TopSpace; let a, b be Point of T; let P be Path of a,b; assume A1: a,b are_connected ; func - P -> Path of b,a means :Def6: :: BORSUK_2:def 6 for t being Point of I holds it . t = P . (1 - t); existence ex b1 being Path of b,a st for t being Point of I holds b1 . t = P . (1 - t) proof end; uniqueness for b1, b2 being Path of b,a st ( for t being Point of I holds b1 . t = P . (1 - t) ) & ( for t being Point of I holds b2 . t = P . (1 - t) ) holds b1 = b2 proof end; end; :: deftheorem Def6 defines - BORSUK_2:def 6 : for T being non empty TopSpace for a, b being Point of T for P being Path of a,b st a,b are_connected holds for b5 being Path of b,a holds ( b5 = - P iff for t being Point of I holds b5 . t = P . (1 - t) ); Lm4: for r being Real st 0 <= r & r <= 1 holds ( 0 <= 1 - r & 1 - r <= 1 ) proof end; Lm5: for r being Real st r in the carrier of I holds 1 - r in the carrier of I proof end; theorem Th6: :: BORSUK_2:7 for T being non empty TopSpace for a being Point of T for P being constant Path of a,a holds - P = P proof end; registration let T be non empty TopSpace; let a be Point of T; let P be constant Path of a,a; cluster - P -> constant ; coherence - P is constant by Th6; end; theorem Th7: :: BORSUK_2:8 for X, Y being non empty TopSpace for A being Subset-Family of Y for f being Function of X,Y holds f \" () = union (f \" A) proof end; definition let S1, S2, T1, T2 be non empty TopSpace; let f be Function of S1,S2; let g be Function of T1,T2; :: original: [: redefine func [:f,g:] -> Function of [:S1,T1:],[:S2,T2:]; coherence [:f,g:] is Function of [:S1,T1:],[:S2,T2:] proof end; end; theorem Th8: :: BORSUK_2:9 for S1, S2, T1, T2 being non empty TopSpace for f being continuous Function of S1,T1 for g being continuous Function of S2,T2 for P1, P2 being Subset of [:T1,T2:] st P2 in Base-Appr P1 holds [:f,g:] \" P2 is open proof end; theorem Th9: :: BORSUK_2:10 for S1, S2, T1, T2 being non empty TopSpace for f being continuous Function of S1,T1 for g being continuous Function of S2,T2 for P2 being Subset of [:T1,T2:] st P2 is open holds [:f,g:] \" P2 is open proof end; registration let S1, S2, T1, T2 be non empty TopSpace; let f be continuous Function of S1,T1; let g be continuous Function of S2,T2; cluster [:f,g:] -> continuous for Function of [:S1,S2:],[:T1,T2:]; coherence for b1 being Function of [:S1,S2:],[:T1,T2:] st b1 = [:f,g:] holds b1 is continuous proof end; end; registration let T1, T2 be T_0 TopSpace; cluster [:T1,T2:] -> T_0 ; coherence [:T1,T2:] is T_0 proof end; end; registration let T1, T2 be T_1 TopSpace; cluster [:T1,T2:] -> T_1 ; coherence [:T1,T2:] is T_1 proof end; end; registration let T1, T2 be T_2 TopSpace; cluster [:T1,T2:] -> T_2 ; coherence [:T1,T2:] is T_2 proof end; end; registration coherence ( I is compact & I is T_2 ) proof end; end; definition let T be non empty TopStruct ; let a, b be Point of T; let P, Q be Path of a,b; pred P,Q are_homotopic means :: BORSUK_2:def 7 ex f being Function of ,T st ( f is continuous & ( for t being Point of I holds ( f . (t,0) = P . t & f . (t,1) = Q . t & f . (0,t) = a & f . (1,t) = b ) ) ); symmetry for P, Q being Path of a,b st ex f being Function of ,T st ( f is continuous & ( for t being Point of I holds ( f . (t,0) = P . t & f . (t,1) = Q . t & f . (0,t) = a & f . (1,t) = b ) ) ) holds ex f being Function of ,T st ( f is continuous & ( for t being Point of I holds ( f . (t,0) = Q . t & f . (t,1) = P . t & f . (0,t) = a & f . (1,t) = b ) ) ) proof end; end; :: deftheorem defines are_homotopic BORSUK_2:def 7 : for T being non empty TopStruct for a, b being Point of T for P, Q being Path of a,b holds ( P,Q are_homotopic iff ex f being Function of ,T st ( f is continuous & ( for t being Point of I holds ( f . (t,0) = P . t & f . (t,1) = Q . t & f . (0,t) = a & f . (1,t) = b ) ) ) ); theorem :: BORSUK_2:11 canceled; ::$CT\ntheorem Th10: :: BORSUK_2:12\nfor T being non empty TopSpace\nfor a, b being Point of T\nfor P being Path of a,b st a,b are_connected holds\nP,P are_homotopic\nproof end;\n\ndefinition\nlet T be non empty pathwise_connected TopSpace;\nlet a, b be Point of T;\nlet P, Q be Path of a,b;\n:: original: are_homotopic\nredefine pred P,Q are_homotopic ;\nreflexivity\nfor P being Path of a,b holds (T,a,b,b1,b1)\nby ;\nend;\n\ntheorem :: BORSUK_2:13\nfor G being non empty TopSpace\nfor w1, w2, w3 being Point of G\nfor h1, h2 being Function of I,G st h1 is continuous & w1 = h1 . 0 & w2 = h1 . 1 & h2 is continuous & w2 = h2 . 0 & w3 = h2 . 1 holds\nex h3 being Function of I,G st\n( h3 is continuous & w1 = h3 . 0 & w3 = h3 . 1 & rng h3 c= (rng h1) \\/ (rng h2) ) by Lm3;\n\ntheorem :: BORSUK_2:14\nfor T being non empty TopSpace\nfor a, b, c being Point of T\nfor G1 being Path of a,b\nfor G2 being Path of b,c st G1 is continuous & G2 is continuous & G1 . 0 = a & G1 . 1 = b & G2 . 0 = b & G2 . 1 = c holds\n( G1 + G2 is continuous & (G1 + G2) . 0 = a & (G1 + G2) . 1 = c )\nproof end;\n\nregistration\nlet T be non empty TopSpace;\ncluster non empty connected compact for Element of bool the carrier of T;\nexistence\nex b1 being Subset of T st\n( not b1 is empty & b1 is compact & b1 is connected )\nproof end;\nend;\n\n:: Moved from BORSUK_5:11, AK, 20.02.2006\ntheorem Th13: :: BORSUK_2:15\nfor T being non empty TopSpace\nfor a, b being Point of T st ex f being Function of I,T st\n( f is continuous & f . 0 = a & f . 1 = b ) holds\nex g being Function of I,T st\n( g is continuous & g . 0 = b & g . 1 = a )\nproof end;\n\nregistration\ncoherence\nproof end;\nend;" ]
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https://www.physicsforums.com/threads/voltage-across-resistors-in-parallel.680344/
[ "# Voltage Across Resistors In Parallel\n\n## Homework Statement\n\nConsider the circuit shown in the figure below. (R = 31.0 Ω.)\n\n(a) Find the current in the 31.0 Ω resistor.\n\n(b) Find the potential difference between points a and b.\n\n## The Attempt at a Solution\n\nI am reading the solution of this problem given by the author, and for the most part I understand it, except for this one critical part:\n\n\"In diagram (iii), the current above goes through the equivalent resistor $R_{iii}$ to give a voltage drop across this resistor of $ΔV = IR_{iii}$. n diagram (ii), we see that this voltage drop is $ΔV_{ab}$ and is the same across the 10.0-Ω resistor and the 5.00-Ω resistor.\"\n\nWhy is the voltage drop the same across the two resistors in parallel?\n\n#### Attachments\n\n• 17.2 KB Views: 283\n• 9.2 KB Views: 302\n\nRelated Introductory Physics Homework Help News on Phys.org\nCWatters\nHomework Helper\nGold Member\nWhy is the voltage drop the same across the two resistors in parallel?\nBy definition two resistors in parallel are both connected to the same nodes. In capture.jpg these nodes are a and b.\n\nIf you connect two resistors in parallel to an ideal 9V battery they will both have 9V across them.\n\nSo, it simply follows from definition, and there isn't any other reason?\n\nCWatters" ]
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http://www.kpubs.org/article/articleMain.kpubs?articleANo=HGJHC0_2010_v11n2_87
[ "Satellite Attitude Control with a Modified Iterative Learning Law for the Decrease in the Effectiveness of the Actuator\nSatellite Attitude Control with a Modified Iterative Learning Law for the Decrease in the Effectiveness of the Actuator\nInternational Journal of Aeronautical and Space Sciences. 2010. Jun, 11(2): 87-97\nCopyright ©2010, The Korean Society for Aeronautical Space Science", null, "This is an Open Access article distributed under the terms of the Creative CommonsAttribution Non-Commercial License(http://creativecommons.org/licenses/by-nc/3.0/)which permits unrestricted non-commercial use, distribution, andreproduction in any medium, provided the original work is properly cited.\n• Published : June 15, 2010", null, "PDF", null, "e-PUB", null, "PubReader", null, "PPT", null, "Export by style\nArticle\nAuthor\nMetrics\nCited by\nTagCloud\nHoJin Lee\nYoudan Kim\nydkim@snu.ac.kr\nHee Seob Kim\n\nAbstract\nA fault tolerant satellite attitude control scheme with a modified iterative learning law is proposed for dealing with actuator faults. The actuator fault is modeled to reflect the degradation of actuation effectiveness, and the solar arrayinduced disturbance is considered as an external disturbance. To estimate the magnitudes of the actuator fault and the external disturbance, a modified iterative learning law using only the information associated with the state error is applied. Stability analysis is performed to obtain the gain matrices of the modified iterative learning law using the Lyapunov theorem. The proposed fault tolerant control scheme is applied to the rest-to-rest maneuver of a large satellite system, and numerical simulations are performed to verify the performance of the proposed scheme.\nKeywords\n1. Introduction\nThousands of satellites are now in operation for various purposes such as communication, navigation, military service, weather forecasting, and terrestrial and astronomical observations. Since satellite launches usually entail much cost and time, the social and economic cost that arises from satellite failure is critical. For satellite attitude control, reaction wheels, thrusters, and control momentum gyros are widely used as actuators. Therefore, the attitude control performance can be affected by actuator failure. If actuator failure occurs, the mission of the satellite cannot be accomplished or will be limited. Recently, the demand for satellite fault-tolerant control systems is increasing; therefore, considerable research has been undertaken that utilizes various control methods such as adaptive control, neural networks, sliding mode control, and so on (Henry, 2008; Jiang et al., 2008; Talebi and Patel, 2006; Tehrani et al., 2005; Wu and Saif, 2005). Tehrani used an estimator, based on neural networks, for fault diagnosis of the reaction wheel in a satellite attitude control system (Tehrani et al., 2005). Wu and Saif applied a sliding mode controller, based on neural networks, for satellite fault diagnosis (Wu and Saif, 2005). Henry used H-infinity and H_filter based schemes for the fault diagnosis of microscope satellite thrusters (Henry, 2008).\nSeveral estimation and control schemes have been applied to deal with the failures of satellite systems. Especially, the iterative learning law is a fault estimation scheme for detecting and estimating faults. The iterative learning law using previous information can be applied to estimate both constant faults and time-varying faults (Chen and Saif, 2001, 2007).\nIn this paper, a fault tolerant control scheme with a modified iterative learning law is proposed to deal with the decreased effectiveness of satellite actuators. The modified iterative learning law using only information that is associated with the state error is adopted to estimate the influence of the actuator fault and the external disturbance. Lyapunov stability analysis is performed to obtain a stable controller. The performance of the proposed satellite attitude fault tolerant control scheme is verified by numerical simulations.\nThis paper is organized as follows. The second section describes the satellite system and the rest-to-rest maneuver of the satellite. In the third section, a fault tolerant controller with a modified iterative learning law is designed, and stability analysis is performed. In the fourth section, numerical simulations are performed to verify the performance of the proposed fault tolerant satellite control scheme. Finally, conclusions are drawn in the fifth section.\n2. Satellite Dynamics and Mission\n- 2.1 Satellite Dynamics\nThe satellite system considered in this paper is specified as follows (Chobotov, 1991).", null, "Lager Image", null, "Lager Image\nIn the above, q 0 , q 1 , q 2 , q 3 are the quaternion variables, ω x , ω y , ω z are the angular rates, the matrix I s includes the elements of the moment of inertia, T x , T y , T z are control input elements, and w dx , w dy , w dz are the external disturbances.\nThe quaternion vector q that is related to the Euler angle is defined as follows.", null, "Lager Image\nIn Eq. (2), ( I s ω) x , ( I s ω) y and ( I s ω) z are defined as follows.", null, "Lager Image\nThe dynamic equations of Eqs. (1) and (2) can be rewritten as follows.", null, "Lager Image", null, "Lager Image\nHere, x 1 =[ q 0 q 1 q 2 q 3 ] T , x 2 =[ω x ω y ω z ] T ,", null, "Lager Image", null, "Lager Image\nNote that B = I 3×3 for a healthy condition of the actuator. When one of the actuators partially loses its actuation effectiveness, Eq. (6) can be written as follows:", null, "Lager Image\nwhere the actuation effectiveness matrix, B f = diag { B fi },(∀=x~z), is a matrix that represents the fault effect of the actuator. The degradation of the actuation effectiveness in a real situation cannot be clearly modeled. Therefore, the chattering effect with high-frequency variation due to the degradation of the actuation effectiveness also should be considered. In this paper, the effects of the actuator fault, B fi are modeled as follows.", null, "Lager Image\nIn Eq. (8), a fi is the magnitude of the actuation effectiveness, and the second term is the high-frequency variation due to the fault with amplitude b fi and frequency ω fi . Usually, the magnitude of b fi is very small when compared to that of a fi . For simplicity, let us introduce a matrix, C f , as follows", null, "Lager Image\n- 2.2 Rest-to-Rest Maneuver\nThe satellite usually performs various maneuvers such as rest-to-rest and despin. In this paper, the rest-to-rest maneuver is considered. The angle of maneuver of the principal axis of the satellite can be represented as follows.", null, "Lager Image\nIn Eq. (10), f s ( . ) is a smooth approximation of the signum function for the maneuver.\nFor the rest-to-rest maneuver, the following boundary conditions should be satisfied.", null, "Lager Image\nIn Eq. (11), θ f ? θ 0 is the maneuver attitude angle, t f is the target maneuver time, and T max is the maximum available torque. In this paper, the following function, f s ( . ), is adopted for near-minimum time control (Junkins and Kim, 1993) to prevent unnecessary activation of the flexible modes of the solar panels.", null, "Lager Image\nIn Eq. (12), and the torque-shaping parameter, Δt=α t , α, is in the range of 0 <; α≤0.25.\nThe target maneuver time, t f , for each axis can be obtained by using Eqs. (10-12) as follows.", null, "Lager Image\nNote that each axis's desired reference trajectory for the satellite rest-to-rest maneuver can be obtained by using Eqs. (10, 12, 13).\n3. Fault Tolerant Control Scheme with a Modified Iterative Learning Law\nIn this section, a fault tolerant control scheme with a modified iterative learning law is designed to deal with the decrease in the effectiveness of the actuator and the external disturbance.\nFirst, let us design the sliding mode controller to make the satellite attitude angles track the reference trajectory in Eq. (10) using the pseudo control input. Note that the state vector, x 2 , in Eq. (5) can be considered as the pseudo control input. This pseudo control input becomes the reference trajectory of the angular rate, x 2 .\nUsing the quaternion vector and the desired trajectory, the sliding surface is defined as:", null, "Lager Image\nwhere the reference trajectory,", null, "Lager Image\n, is constructed to perform the rest-to-rest maneuver. Since the condition, q 0 2 + q 1 2 + q 2 2 + q 3 2 =1, should be satisfied, q 0 can follow q 0d as the sliding surface s approaches zero asymptotically.\nDifferentiating Eq. (14) with respect to time yields:", null, "Lager Image\nwhere", null, "Lager Image\nTo satisfy the reaching condition of the sliding surface, the following Lyapunov candidate function is considered.", null, "Lager Image\nDifferentiating Eq. (16) with respect to time and substituting Eq. (15) into the resulting equation yields:", null, "Lager Image\nThe pseudo control input vector for the attitude angle can be obtained by the sliding mode control method as:", null, "Lager Image\nwhere K is the control gain matrix, and δ is the boundary layer thickness vector.\nThe pseudo control input of Eq. (18) is used as the reference trajectory of the angular rate, x 2d .\nSubstituting Eq. (18) into Eq. (17) gives:", null, "Lager Image\nFinally, it is concluded that the stability of the reference trajectory tracking controller is guaranteed.\nNow, let us design a fault tolerant control scheme to deal with the decrease in effectiveness of the actuator and the external disturbance. Using the reference angular rate trajectory, x 2d , which is obtained from Eq. (18), and the angular rate state vector, x 2 , the angular rate state error vector is defined as:", null, "Lager Image\nDifferentiating Eq. (20) with respect to time yields:", null, "Lager Image\nwhere T df ( t )= I s -1[ w d ( t )- C f u ( t )] is the influence of the actuator fault and the external disturbance.\nLet us select the fault tolerant control input using Eq. (21) as:", null, "Lager Image\nwhere (F ^ ) is the estimated fault signal that can be obtained using the modified iterative learning law to compensate the influence of the actuator fault and the external disturbance. Usually, faults that occur in the system are unknown. To deal with this problem, the modified iterative learning law is used.\nThe general iterative learning law is updated by both the previous information and the state estimation error (Chen and Saif, 2001, 2007). Note that the modified iterative learning law proposed in this paper is updated using only the information associated with the state error; therefore, the influence of the actuator fault and the external disturbance can be estimated as follows:", null, "Lager Image\nwhere the parameter, τ, is an updating interval.\nThe gain matrices, L 1 and L 2 in Eq. (23), are determined using Lyapunov stability analysis. Substituting Eq. (22) into Eq. (21) and using Eq. (23) yield:", null, "Lager Image\nLet us consider the following Lyapunov candidate function.", null, "Lager Image\nDifferentiating Eq. (25) with respect to time and substituting Eq. (24) into the resulting equation yield:", null, "Lager Image\nThe following inequalities can be obtained (Yan et al., 1998).", null, "Lager Image", null, "Lager Image\nThe following equation also can be obtained using Eq. (24).", null, "Lager Image\nSimilarly, the following inequalities can be obtained:", null, "Lager Image", null, "Lager Image\nand", null, "Lager Image\nAs a result, the inequality, Eq. (29), can be obtained using Eqs. (30-32) as follows:", null, "Lager Image\nwhere", null, "Lager Image\nand", null, "Lager Image\nBy assuming", null, "Lager Image\nand", null, "Lager Image\n, the following can be derived:", null, "Lager Image\nwhere D max =max( D x , D y , D z ) and T max =max( T x , T y , T z ).\nBy substituting Eqs. (27, 28, 33, 34) into Eq. (26), the following equation can be obtained.", null, "Lager Image\nTo satisfy Lyapunov stability, the following relation should be satisfied for the gain matrices, L 1 and L2.", null, "Lager Image", null, "Lager Image\nFrom the above equation, the following inequality is obtained for using the gain matrix, L2:", null, "Lager Image\nwhere", null, "Lager Image\nTherefore, the gain matrix, L2, should be selected to satisfy the following relation.", null, "Lager Image\nIn summary, the gain matrices, L1 and L2, must be chosen using Eqs. (36, 39), and the fault estimate signal, (수식삽입), of Eq. (23) is updated using the gain matrices and the information associated with the state error.\nRemark : The state error can be used for fault monitoring purposes. When there are no faults in the system, the state error should be zero or close to zero. On the other hand, an increase in the state error would point to the occurrence of a fault. After the fault occurs, the estimated fault signal would learn about the fault and the state error will again be driven to zero or a value close to zero. This means that the iterative learning law can learn and update by the previous and the present state error information.\n4. Numerical Simulation\nNumerical simulation has been performed to verify the proposed fault tolerant control scheme. The Hubble space telescope is considered as the satellite system (Thienel and Sanner, 2007; Wie et al., 1993). The inertia, I s , is given as follows.", null, "Lager Image\nWith regard to the external disturbance, the disturbance that is induced by the solar array is considered as follows (Wie et al., 1993).", null, "Lager Image\nThe attitude (quaternion) reference trajectory is for the rest-to-rest maneuver. The maximum value of the available torque, T max , of the actuator is 0.82 Nm, and the torqueshaping α is selected as 0.25 for a near-minimum smooth maneuver. The initial attitude quaternion is chosen as , the initial reference angular velocity, ω d (0) is taken as zero, and the initial angular velocity is chosen as (deg/s)(Thienel and Sanner, 2007). The initial and target attitude angles are chosen as 0° and 40°, respectively, about all axes. The target attitude quaternion is calculated as . The final target maneuver time, t f , which is calculated by Eq. (13), is about 1,886 seconds. This final time is taken as the maximum target maneuver time at each axis. The reference trajectory of the rest-to-rest maneuver with respect to the quaternion values is shown in Fig. 1 .", null, "Lager Image\nReference trajectory of the rest-to-rest maneuver (quaternion).\n- 4.1 Case 1 ? Sequential Occurrence of Actuator Faults\nIn this case, the actuator faults are considered as follows.", null, "Lager Image\nIn Eq. (8), the magnitude of the actuation effectiveness with regard to the actuator fault is chosen as 0.7, a fi = while b fi and the frequency, ω fi are chosen as 0.02 a fi and 5 Hz, respectively. Figure 2 shows the considered change in the fault effectiveness.\nThe gain matrix is chosen as K =0.1 I 3×3 , and the updating interval, τ, is chosen as 1 second. The parameters of the modified iterative learning law are chosen as ρ=0.1, γ 1 2 =0.1 and γ 3 4 5 =10. The upper bound on the external disturbance, D max , is chosen as 0.4 Nm.\nThe simulation results are shown in Figs. 3 - 6 . It can be seen from Figs. 3 and 4 that the state values, after the occurrence of the actuator faults, follow the reference trajectories very well. Figure 5 shows that the actuator faults can be estimated by the modified iterative learning law. As shown in Fig. 6 , a high-frequency control torque input is used for the y-axis. The control torque input response of the y-axis, from 0 to 50 seconds, is shown in Fig. 6(b) ; it can be seen that a high-frequency control input is required to deal with the disturbance that is induced by the solar array. As seen in the simulation results, the fault tolerant control scheme with the modified iterative learning law deals with the actuator faults and the external disturbance.", null, "Lager Image\nModel of the fault effect (B) (Case 1).", null, "Lager Image\nAttitude quaternion time responses (Case 1).", null, "Lager Image\nAngular rate time responses (Case 1).", null, "Lager Image\nFault effect and disturbance estimation (Case 1).", null, "Lager Image\nControl torque input time response (Case 1).\n- 4.2 Case 2 - Simultaneous Occurrence of Actuator Faults\nIn the second case, numerical simulation is performed for actuator faults that occur at same time. The actuator faults are considered to occur at t = t f - 800 sec about all axes. All the design parameters are identical to those in Case 1. Figure 7 shows the considered change in the fault effectiveness. The simulation results are shown in Figs. 8 - 11 . It can also be seen that the proposed fault tolerant control scheme with the modified iterative learning law copes well with the simultaneous occurrence of actuator faults.", null, "Lager Image\nModel of the fault effect (B) (Case 2).", null, "Lager Image\nAttitude quaternion time responses (Case 2).", null, "Lager Image\nAngular rate time responses (Case 2).", null, "Lager Image\nFault effect and disturbance estimation (Case 2)", null, "Lager Image\nControl torque input time response (Case 2).\n5. Conclusions\nA satellite attitude control scheme is proposed to deal with actuator faults and external disturbances. The reference attitude angle of the satellite is designed using a near minimum-time maneuver in the rest-to-rest maneuver. A fault tolerant control scheme for the satellite attitude system with a modified iterative learning law is proposed to deal with the degradation of the actuation effectiveness and the external disturbance due to the vibration of the solar array. The modified iterative learning law is considered to estimate the unknown influence of the actuator fault and the external disturbance. Note that only information that is related to the state error is used. The fault tolerant control scheme is applied to a large satellite system, and the performance of the proposed satellite attitude control scheme is verified by numerical simulation. The proposed algorithm can be applied in an attitude control system to improve the reliability of satellite systems.\nAcknowledgements\nThis study has been supported by the Korea Aerospace Research Institute (KARI) under the KOMPSAT-3 Development Program that is funded by the Ministry of Education, Science, and Technology (MEST) of the Republic of Korea.\nReferences" ]
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https://www.mathworks.com/matlabcentral/fileexchange/60686-lagrange-interpolation-polynomial-fitting-with-matlab?s_tid=blogs_rc_5
[ "Lagrange interpolation polynomial fitting with MATLAB\n\nversion 1.0.0.0 (698 Bytes) by\nLagrange interpolation polynomial fitting\n\nUpdated 12 Dec 2016\n\nLagrange interpolation polynomial fitting a set of points LAGRANG(X,Y,N,XX) where X and Y are row vector defining a set of N points uses Lagrange's method to find the N th order polynomial in X that passes through these points.\n\nCite As\n\nTamir Suliman (2022). Lagrange interpolation polynomial fitting with MATLAB (https://www.mathworks.com/matlabcentral/fileexchange/60686-lagrange-interpolation-polynomial-fitting-with-matlab), MATLAB Central File Exchange. Retrieved .\n\nMATLAB Release Compatibility\nCreated with R2016b\nCompatible with any release\nPlatform Compatibility\nWindows macOS Linux" ]
[ null ]
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https://mathoverflow.net/questions/101217/fundamental-group-of-an-analytic-hypersurface
[ "# Fundamental group of an analytic hypersurface\n\nLet $M$ denote a complex manifold of dimension $n$ and let $X\\subset M$ denote an analytic hypersurface. Then it is a standard fact from several complex variables that around a given point $p\\in X$ there are open subsets $V\\subset X, W\\subset \\mathbb{C}^{n-1}$ and and a finite-sheeted covering $\\pi: V\\rightarrow W$ branched over an analytic set $A$. Frequently, $W$ is taken to be a polydisc (see, for example, Griffiths & Harris). Now set $V' = V\\setminus \\pi^{-1}(A)$ and $W' = W\\setminus A$. If $V'$ is connected, then $\\pi$ induces a genuine (holomorphic) covering $\\pi':V'\\rightarrow W'$.\n\nFor appropriately chosen basepoints $v_0, w_0$ in $V',W'$, respectively, covering space theory says that $\\pi'$ induces an injection $\\pi_1(V',v_0)\\rightarrow \\pi_1(W',w_0)$, and in some cases (e.g. involving curves) this might be enough to figure out $\\pi_1(V',v_0)$ if $\\pi_1(W',w_0)$ is known. If $V$ were smooth, $\\pi^{-1}(A)$ would be an analytic subset of complex codimension 1 and the inclusion $V'\\hookrightarrow V$ would induce a surjection $\\pi_1(V',v_0)\\rightarrow \\pi_1(V,v_0)$, determining a presentation for $\\pi_1(V,v_0)$. However, $V$ may not be smooth, and I am not sure in what generality this map is still a surjection, although clearly I'd like to know.\n\nWhat I really want to do is determine a presentation for $\\pi_1(X,x_0)$. It seems that one way to to go about doing this would be to construct an open cover $(V_{\\alpha})$ of $X$ consisting of open sets with the same properties possessed by $V$ above AND such that $(V_{\\alpha})$ satisfies the hypotheses of the Seifert-van Kampen Theorem (see, for example, Hatcher). In the very best case, one could arrange for the intersections $V_{\\alpha}\\cap V_{\\beta}$ to be simply-connected and for the triple intersections to be path-connected. Then one could read off a presentation of $\\pi_1(X,x_0)$ in terms of the presentations of the $\\pi_1(V_{\\alpha}, v_{0\\alpha})$.\n\nI'd be interested to know when such a cover $(V_{\\alpha})$ exists, especially for $n\\geq 3$.\n\nI am also interested in hearing about any known methods of calculating the fundamental group of an analytic hypersurface (not necessarily smooth), in general or in special cases (for example, the Lefschetz Hyperplane Theorem can be used on certain projective hypersurfaces).\n\n(Note: I'd like to consult Dimca's book \"Singularities and Topology of Hypersurfaces\", but I'll be away from my library for the next few weeks.)\n\n• Probably you already know, but Lefschetz theorem for homotopy groups says that if $X$ is a non-singular irreducible complex projective variety of dimension $n$ and $D$ any effective ample divisor on $X$, then the homomorphisms $\\pi_i(D)\\to\\pi_i(X)$ induced by inclusion are bijective for $i\\le n-2$ and surjective if $i=n-1$. Note that the theorem does not assume that $D$ is non-singular! However, one cannot dispense entirely with the non-singularity of $X$... – diverietti Jul 3 '12 at 8:28\n• Yes, I am familiar with Lefschetz for homotopy groups. Thank you. In fact, one can say a little more in case $X\\subset \\mathbb{P}^{n+1}$ is a hypersurface. The $d$-uple Veronese embedding $\\nu_d: \\mathbb{P}^{n+1}\\rightarrow \\mathbb{P}^N$ allows us to view $X$ itself as a hyperplane section of the image $\\nu_d(\\mathbb{P}^{n+1})$, which is $(n+1)$-dimensional. Then Lefschetz gives us $\\pi_{n-1}(X) = \\pi_{n-1}(\\mathbb{P}^{n+1})$. As a result, a hypersurface in $\\mathbb{P}^3$ is simply-connected. – Kevin Jul 3 '12 at 14:53\n• Replay to Kevin's comment: I think that what you are saying is not more than what I was saying... It's just my statement applied to the case when the ambient space is the projective space. – diverietti Jul 3 '12 at 15:03\n• About your original question. Sorry, I don't understand one thing: you are taking $X$ to be smooth, right? So couldn't you take as $V$ just a coordinate chart and $W\\subset\\mathbb C^{n−1}$ to be the image of this coordinate chart, so that it is simply a biholomorphism ? – diverietti Jul 3 '12 at 15:04\n• You are absolutely right on both counts. Regarding my comment: I was simply saying that, using the result you mentioned, one can go a little further in calculating the homotopy groups of $X$, i.e. the case $i=n-1$. Regarding my original question: At first I wanted to allow for $X$ to be singular, but at some point changed my mind as I was posting the question. I'll edit the question to reflect your comment. – Kevin Jul 3 '12 at 15:55\n\nIt should be said that van Kampen's paper \"On the connection between the fundamental groups of some related spaces\". Amer. J. Math. 55 (1933) 261--267, gives a formula for the case of a union of two spaces with non-connected intersection, and this was needed for his work on algebraic curves: \"On the Fundamental Group of an Algebraic Curve\". Amer. J. Math. 55 (1933), no. 1-4, 255–267. The non-connected case seems to me best handled in modern terms using groupoids. (See my web pages.) (An earlier version of the theorem in the connected case and for simplicial complexes was given by Seifert.)\n\nSince homotopy groups are mentioned in the last comment, I mention the higher order theorems of the Seifert-van Kampen type of which the 2-d version was given in my paper with Philip Higgins , On the connection between the second relative homotopy groups of some related spaces'', Proc. London Math. Soc. (3) 36 (1978) 193-212. This uses extensively the notion of crossed module, and has been applied to give explicit calculations of homotopy 2-types, and second homotopy groups. (Again, see references on my web pages).\n\nIn view of the provenance of van Kampen's paper, it would be very interesting to know if such higher theorems are applicable to the situation of the question.\n\n@Kevin Kordek: September 2013: I should add that Grothendieck was interested in these possibilities. See the last problem stated in the set of problems and \"Future directions?\" given in our new book on \"Nonabelian algebraic topology\"; more details and pdf's available from here. Part of his comment is as follows:\n\n\"It seems to me, in any case, that this $\\underset{\\to}{\\lim}$-operation [\"higher order van Kampen theorem\"] in the context of homotopy types is of a very fundamental character, with wide range of theoretical applications. To give just one example, relying on the existence of such a formalism, it is possible to give a very simple explicit algebraic description of the full homotopy types of the Mumford-Deligne compactifications of the modular topoi for complex curves of given genus $g$, say, with $\\nu$ \"marked\" points, in terms essential1y of such a (finite) direct limit of $K(\\pi, 1)$-spaces, where $\\pi$ ranges over certain \"elementary\" Teichm\\\"uller groups (those, roughly, corresponding to modular dimension $\\leq 2$), and to give analogous descriptions, too, of all those subtopoi of the previous one, deducible from its canonical \"stratification\" at infinity by taking unions of strata. In fact, such descriptions should apply to any kind of stratified\" space or topos, as it can be expressed (in an essentially canonical way, which apparently was never made explicit yet in this literature) as a (usually finite) direct limit of simpler spaces, namely the \"strata\", and \"tubes\" around strata, and \"junctions\" of tubes, etc.\"\n\nSo I probably missed out on not pursuing this, partly because of pursuing work with J.-L. Loday on even more powerful Higher van Kampen type theorems.\n\n• The OP might like to know that the 'usual' way of dealing with the case of non-connected intersectionn is with HNN extensions; this is equivalent to the groupoid point of view. – HJRW Jul 3 '12 at 11:50\n• @Ronnie Brown \"In view of the provenance of van Kampen's paper, it would be very interesting to know if such higher theorems are applicable to the situation of the question.\" I completely agree. Thanks for answering, and for the references. – Kevin Jul 3 '12 at 21:09" ]
[ null ]
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https://areteem.org/blog/2019-amc-8-answer-key-released/
[ "# 2019 AMC 8 Answer Key Released", null, "Posted by\n\nOver the past week, thousands of middle school students have participated in this year’s AMC 8 Competition (including some students at Areteem Headquarters seen below).\n\nThe last day of testing was November 18th, so now the problems can be discussed! See below for an answer key for the 2019 questions as well as the concepts tested on each problem.\n\n1\nD\narithmetic\n2\nE\ngeometry: areas\n3\nE\narithmetic: fractions and orderings\n4\nD\ngeometry: areas and Pythagorean theorem\n5\nB\nalgebra: motion and graphs\n6\nC\ncounting & probability: probability and geometric lines of symmetry\n7\nA\narithmetic: averages\n8\nE\narithmetic: percentages\n9\nB\ngeometry: volumes and ratios\n10\nB\narithmetic: means and medians\n11\nD\ncounting & probability: sets and Venn diagrams\n12\nA\ngeometry: visualizing 3D shapes\n13\nA\nnumber theory: palindromes\n14\nC\nnumber theory: patterns and days of the week\n15\nB\ncounting & probability: sets\n16\nD\nalgebra: motion and average speed\n17\nB\narithmetic: patterns\n18\nC\ncounting and probability: parity and probability\n19\nC\ncounting and probability and logic\n20\nD\nalgebra: square roots\n21\nE\nalgebra and geometry: lines and triangle areas\n22\nE\nalgebra: percent increase and decrease\n23\nB\nnumber theory: LCM and remainders\n24\nB\ngeometry: triangles and area ratios\n25\nC\ncounting and probability: stars and bars\n\nNeed to see all the solutions? Areteem’s Dr. Kevin Wang has you covered, solving all 25 problems and giving problem solving tips in a review video. Click here for more information and to purchase. As a limited time offer, we’re giving the video out for free through social media. Check them out for instructions!" ]
[ null, "https://secure.gravatar.com/avatar/8801cf797f125863cf7ec6e45b90f960", null ]
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https://geekstocode.com/count-distinct-elements-in-every-window/
[ "# Count Distinct elements in Every Window Easiest Explanation\n\nAs we learn about the sliding window technique and also discuss some medium level problems, here we are going to solve Count Distinct elements in Every Window .\n\nBefore going to problem, you must be aware of technique and have basic knowledge of hash map\n\nSo, here we are going to solve 2 easy problem, without wasting time come on it\n\n## Count Distinct Elements in every window\n\n### Problem Statement\n\nYou have a array and a integer k, you need to find the distinct number of element in each window of size k\n\n### Input\n\nFirst Integer is a test case in one line and size of the array and then the size of the window in the next line and in the third line, you have array elements\n\n``````2\n7 4\n1 2 1 3 4 2 3\n3 2\n4 1 1``````\n\n### Output\n\nNumber of distinct elements in each window\n\n``````3 4 4 3\n2 1``````\n\n### Algorithm(sliding window technique)\n\nA very easy question, You can see there we only need to slide the window and in every slide, we have to check the distinct number of elements.\n\nTo count the distinct number of elements you can use hashMap.\n\nKey in hashmap represent the element and value represent the occurrence\n\nApproach\n\n• Create a hash Map and insert element from array up to size k\n• In each insert check whether the element is present or not, If the element is present already you need to update the value of that element in the map to +1\n\n• If the element is not present you need to insert in map with value 1(value represent the occurrence) and increase dist variable value.(dist represent the distinct element)\n• Print the dist value( correspond to the first window)\n• Start from the kth index(to next window) in array and insert next element according to point 2 and point 3 and then remove the first index like this\n``````// represent element is present and have only 1 occurence, so we decrease distinct value in our new window\nif(mp[A[i-k]] == 1)\n{\ndist--;\n}\nmp[A[i-k]]--; also reduce the occurence of element by 1``````\n• After every iteration you need to print dist variable.\n\nYou can better understand by code\n\n``````// { Driver Code Starts\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid countDistinct(int[], int, int);\n\nint main() {\nint t;\ncin >> t;\nwhile (t--) {\n\nint n, k;\ncin >> n >> k;\nint a[n];\nfor (int i = 0; i < n; i++) cin >> a[i];\ncountDistinct(a, k, n);\ncout << endl;\n}\nreturn 0;\n}\n// } Driver Code Ends\n/*You are required to complete below method */\nvoid countDistinct(int A[], int k, int n) {\n\nmap<int,int> mp;\nint dist = 0;\nfor(int i = 0;i<k;i++)\n{\nif(mp[A[i]] == 0)//element not present hence, it is distinct element\n{\ndist++;\n}\nmp[A[i]]++;// increase occurence of element i.e value in hashmap\n}\ncout<<dist<<\" \";// distinct element in first window\nfor(int i = k;i<n;i++)\n{\n\n//removed the element from the window, if it values is 1 it means it has only 1 //occurrence and in new window, it is not present\n// so we decrease dist value\nif(mp[A[i-k]] == 1)\n{\ndist--;\n}\nmp[A[i-k]]--;\nif(mp[A[i]] == 0) // same as above\n{\ndist++;\n}\nmp[A[i]]++;\ncout<<dist<<\" \";// dist element in next window\n\n}\n\n}``````\n\n## Count Occurrence of Anagram\n\n### Problem\n\nGiven a word S and a text C. Return the count of the occurrences of anagrams of the word in the text.\n\n### Input\n\nFirst-line contains integer T represent the test case and other 2 lines contain string’s S and C respectively\n\n``````2\nforxxorfxdofr\nfor\naabaabaa\naaba``````\n\n### Output\n\nOutput number of occurrence of anagram of C in S\n\n``````3\n4\n// Explanation \"for\" is present in S as \"for\",\"orf\",\"ofr\"``````\n\n### Algorithm(sliding window technique)\n\nThis is simply a anagram problem mix with sliding window technique.\n\n`````` int k = c.length() // pane length\nfor(int i = 0;i<k;i++) // first window slide formation\n{\nz = z+a[i];\n}``````\n• Check whether string z is an anagram of C, if yes increase the count\n• Then slide the pane and remove the first window and add next window like this\n``````for(int j = k;j<S.length();j++)\n{\nz.erase(z.begin());\nz = z+a[j];\nif(areAnagram(z,c))\ncount++;\n}``````\n\nComplete code is here,\n\n``````#include <iostream>\nusing namespace std;\nbool areAnagram(string str1, string str2)\n{\nint count1 = { 0 };\nint count2 = { 0 };\nint i;\nfor (i = 0; str1[i] && str2[i]; i++) {\ncount1[str1[i]]++;\ncount2[str2[i]]++;\n}\nif (str1[i] || str2[i])\nreturn false;\nfor (i = 0; i < 256; i++)\nif (count1[i] != count2[i])\nreturn false;\n\nreturn true;\n}\nint noofAna(string S,string C)\n{\nint k = C.length();\nstring z;\nint count = 0;\nfor(int i = 0;i<k;i++)\n{\nz = z+a[i];\n}\nif(areAnagram(z,C))\ncount++;\n// cout<<z<<\" \";\nfor(int j = k;j<S.length();j++)\n{\nz.erase(z.begin());\n\nz = z+a[j];\n// cout<<z<<\" \";\nif(areAnagram(z,C))\ncount++;\n}\nreturn count;\n\n}\nint main() {\nint t;\ncin>>t;\nwhile(t--)\n{\nstring S;\nstring C;\ncin>>S;\ncin>>C;\ncout<<noofAna(S,C)<<endl;\n}\nreturn 0;\n}``````\n\nI Hope, you learn Count Distinct elements in Every Window.\n\nPractice above question here by submitting your code", null, "Sudhanshu is Technology geek and also a pro pubg player. I like to create content in my free time that helps others. Currently pursuing BCA from Noida and operating Geekstocode" ]
[ null, "https://secure.gravatar.com/avatar/700bfa92093c3059d189e9806d0ecc82", null ]
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https://talkstats.com/threads/power-calculations-help.11375/
[ "# Power calculations help..\n\n#### Logan Manikam\n\n##### New Member\nHello all,\n\nI don't suppose i can ask a question on power calculations?\n\nI've just conducted a RCT involving two student groups one undertaking an educational intervention and another receiving standard clinical teaching. Both groups did a 24 item MCQ knowledge assessment and a 7 statement questionnaire on confidence using 5 point Likert scales.\n\nThe power calculations are a bit haphazard as the original statistician bailed out and I'm trying to pick up the pieces in writing this.\n\nThe statistical analyses i used were\npaired sample t-tests for the 24item MCQs\nWilcoxon signed rank tests for the confidence questionnaires.\n\nNote: I didn't make any inter-group comparisons, only intra-group pre/post.\n\nUsing the G-Power calculator I did an A priori calculation assuming power of 0.8 and an effect size of 0.5 and noted I needed a sample size of 34 and 35 for the paired t and wilcoxon signed tests respectively.\n\nI achieved sample sizes of 55 and 53.\n\nI then did some post hoc calculations after calculating the effect sizes I achieved and noted that my paired t-tests had sufficient power of >0.95 but of the 7 statements in my confidence questionnaire (Analysed using Wilcoxon signed rank tests) 2 of the 7 statements had powers of <0.8 (One was 0.65 and another was 0.12)\n\nAm I correct in assuming then that my a priori precalculations guided me towards the appropriate sample sizes and the post hoc calculations confirmed that I had sufficient power or not? (And that i did all the above correctly instead of going on a tangent)\n\nHow do I present this? Am presenting this work next month and if they ask me did i attain sufficient power would my answer be: yes for the paired t-tests and 5 of the 7 statements only.\n\nMany thanks!\n\n#### fed1\n\n##### TS Contributor\nYour post hoc calculations are a little awkward. I wouldnt do that part.\n\nJust use your a priori calculations. The effect size you choose for your\npower calculations should depend on what you feel would be a meaning full\nchange. The old rule of thumb is effect sizes of 0.2, 0.5 and 0.8 are samll mediumand large, or something like that. Use the sample sizes you actually had, 50 or whatever.\n\nThe point of power calculations is not to find the chance of detecting the difference that truly exhists (we dont know that). It is to find the chance of detecting a difference under a specific alternative (Ha).\n\n:wave:\n\n#### Logan Manikam\n\n##### New Member\nMany thanks for this.\n\nYes I was aware of the 0.2, 0.5 and 0.8 rule.\n\nDoes it mean though that I can say my study had sufficient power across all outcome measures? And my defence of this is the a priori calculations and attaining the sample sizes?\n\nI'm just concerned that one of my outcome measures in the Likert scale based questionnaire shows a non-significant result and according to the post-hoc calculations it says I don't have enough power.\n\n#### Logan Manikam\n\n##### New Member\nHi there Fed,\n\nI don't suppose you or anyone else could clarify on the concern that i had in my last post?\n\nMany thanks.\n\n#### fed1\n\n##### TS Contributor\nSo a priori you had some sample size and effect size\n\nn1, delta1.\n\nfor which the proposed experiment had 80% power.\n\nYou only got n2 < n1 and a point estimate of delta_hat < del.\n\nIts worth adjusting the power calculations if you did not find\na significant difference (if you did power calcs are moot, obviously)\nbut you should only change n1 -> n2.\n\nYou chose delta1 because it represented a clinically significant difference.\nIt is a clinically signifcant difference no mattter what delta_hat you realized.\n\nAt least thats my take." ]
[ null ]
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http://www.xuephp.com/main/detail.php?cid=34513
[ "# hdu 3018 Ant Trip\n\n Ant TripTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 886 Accepted Submission(s): 332Problem DescriptionAnt Country consist of N to...\n\n# Ant Trip\n\nTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)\nTotal Submission(s): 886 Accepted Submission(s): 332\n\nProblem Description\nAnt Country consist of N towns.There are M roads connecting the towns.\n\nAnt Tony,together with his friends,wants to go through every part of the country.\n\nThey intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.", null, "Input\nInput contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.\n\nOutput\nFor each test case ,output the least groups that needs to form to achieve their goal.\n\nSample Input\n```3 3\n1 2\n2 3\n1 3\n\n4 2\n1 2\n3 4```\n\nSample Output\n```1\n2\n\nHint\n\nNew ~~~ Notice: if there are no road connecting one town ,tony may forget about the town.\nIn sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3.\nIn sample 2,tony and his friends must form two group.\n\n```\n\nSource\n\nRecommend\ngaojie\n\n（一笔画无问题）\n\n```#include <stdio.h>\n#include<string.h>\nint sets,digree,sz,ed;//sets用于建立并查集\n//digree用于记录节点度数。sz记录集合中元素个数。ed记录度数为奇数节点的个数\nint a,b;//记录输入数据\nint findx(int x)//并查集。查找根\n{\nint p,i,t;\np=x;\nwhile(p!=sets[p])\np=sets[p];\ni=x;\nwhile(i!=p)\n{\nt=sets[i];\nsets[i]=p;\ni=t;\n}\nreturn p;\n}\nvoid toone(int a,int b)//合并集合。并更新集合奇节点数\n{\nint p,q;\np=findx(a);\nq=findx(b);\nif(p==q)\nreturn;\nif(sz[p]>sz[q])\n{\nsets[q]=p;\nsz[p]+=sz[q];\ned[p]+=ed[q];\n}\nelse\n{\nsets[p]=q;\nsz[q]+=sz[p];\ned[q]+=ed[p];\n}\n}\nint main()\n{\nint n,m,i,sum;\n\nwhile(~scanf(\"%d%d\",&n,&m))\n{\nsum=0;\nmemset(digree,0,sizeof digree);\nfor(i=1;i<=n;i++)\n{\nsets[i]=i;\nsz[i]=1;\ned[i]=0;\n}\nfor(i=1;i<=m;i++)\n{\nscanf(\"%d%d\",&a[i],&b[i]);\ndigree[a[i]]++;\ndigree[b[i]]++;\n}\nfor(i=1;i<=n;i++)\nif(digree[i]%2)\ned[i]=1;//标记奇节点号\nfor(i=1;i<=m;i++)//合并集合\ntoone(a[i],b[i]);\nfor(i=1;i<=n;i++)\n{\nif(digree[i]&&sets[i]==i)//遍历集合set[i]=i说明是根节点\n{\nif(ed[i]==0)//欧拉回路\nsum++;\nelse\nsum+=ed[i]/2;//一笔画公式\n}\n}\nprintf(\"%d\n\",sum);\n}\nreturn 0;\n}\n```\n\n(转发请注明转自:学PHP)\n\n相关推荐\n\n发表评论\n\n(不超过20个字符或10个汉字)" ]
[ null, "http://acm.hdu.edu.cn/data/images/exe3018-1.JPG", null ]
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https://books.google.gr/books?id=V682AAAAMAAJ&q=Pyramid&dq=editions:UOMDLPabq7928_0001_001&hl=el&output=html_text&source=gbs_word_cloud_r&cad=5
[ "# Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, A Treatise of the Nature of Arithmetic of Logarithms ; Likewise Another of the Elements of Plain and Spherical Trigonometry ; with a Preface\n\nT. Woodward, 1733 - 397 уелЯдет\n\n### Фй лЭне пй чсЮуфет -Уэнфбоз ксйфйкЮт\n\nДен енфпрЯубме ксйфйкЭт уфйт ухнЮиейт фпрпиеуЯет.\n\n### ДзмпцйлЮ брпурЬумбфб\n\nУелЯдб 66 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.\nУелЯдб 161 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.\nУелЯдб 110 - And in like manner it may be shown that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM ; therefore the five angles GHK, HKL, KLM, LMG, MGH...\nУелЯдб 88 - IN a circle, the angle in a semicircle is a right angle ; but the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.\nУелЯдб 22 - ... sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB...\nУелЯдб 9 - ... equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.\nУелЯдб 15 - CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which...\nУелЯдб 33 - ... therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (i.\nУелЯдб 111 - If two right-angled triangles have their hypotenuses equal, and one side of the one equal to one side of the other, the triangles are congruent." ]
[ null ]
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https://corporatefinanceinstitute.com/resources/knowledge/economics/disinflation/
[ "", null, "# Disinflation\n\nA slowdown in the rate of inflation\n\n## What is Disinflation?\n\nDisinflation is used to describe the slowing of price inflation. In other words, it is a decrease in the rate of inflation. The term should not be confused with deflation, which is used to describe a negative inflation rate.\n\n### Disinflation vs. Deflation\n\nThe terms disinflation and deflation are commonly mixed up. The definitions of the two terms are contrasted below followed by a diagram illustrating inflation, disinflation, and deflation:\n\nDisinflation: A situation where inflation increases at a slower rate.\n\nDeflation: A situation where inflation is negative (i.e., a decrease in the prices of goods and services in the economy).", null, "As illustrated by the graph:\n\n• Periods of rising inflation are termed inflation\n• Periods of decreasing inflation are termed disinflation\n• Periods of negative inflation are termed deflation\n\nAn easy way to quickly differentiate between deflation and disinflation is that the former is always negative while the latter is positive but decreasing. As shown by the graph, decreasing year-over-year inflation is termed disinflation while a negative inflation rate is termed deflation.\n\n### Primary Causes of Disinflation\n\nIt is important to note that inflation is caused by an increase in the supply of money in the economy. Therefore, a slowdown in the economy’s money supply through a tighter monetary policy is an underlying cause of disinflation.\n\nIn some cases, a slowdown in the rate of inflation can also arise during an economic recession. For example, in a recession, businesses may refrain from increasing their price levels to gain more customers (causing disinflation).\n\n### CPI as a Measure of Determining Disinflation\n\nA widely used measure of inflation is called the Consumer Price Index (CPI). The CPI measures the changes in the price level of consumer goods and services and is one of the most closely watched economic statistics among investors and the Federal Reserve. The percentage change in the CPI is used as a measure of inflation.\n\nFor example, assume the CPI was as follows for 2016, 2017, and 2018, respectively:\n\n• 2016 CPI: 101.7\n• 2017 CPI: 102.3\n• 2018 CPI: 102.6\n\nIf we take the percentage change of CPI from each year, the annual inflation rate can be determined. Using 2016 as the base year, the inflation for 2017 was 0.6% (102.3/101.7 – 1) and the inflation for 2018 was 0.3% (102.5/102.3 – 1). Since the inflation rates for 2017 and 2018 were 0.6% and 0.3%, respectively, it demonstrates a period of disinflation.\n\n### Example of Disinflation\n\nThe following are the CPI’s of a hypothetical economy for several years. As an analyst, your manager wants to know whether the economy experienced either inflation, disinflation, or deflation from 2013 to 2017:\n\n• 2013 CPI: 100\n• 2014 CPI: 101\n• 2015 CPI: 102.1\n• 2016 CPI: 102.9\n• 2017 CPI: 103.3\n\nTo determine whether the economy experienced either inflation, disinflation, or deflation, we first must determine the inflation over each year by determining the annual change in CPI between years. Doing so gives us the inflation rate for each time period. Using 2013 as a base (starting) year:\n\n• Inflation from 2013 to 2014: 101/100 – 1 = 1%\n• Inflation from 2014 to 2015: 102.1/101 – 1 = 1.1%\n• Inflation from 2015 to 2016: 102.9/102.1 – 1 = 0.8%\n• Inflation from 2016 to 2017: 103.3/102.9 – 1 = 0.4%\n\nFrom 2013 to 2015, the economy experienced rising inflation. Inflation was 1% from 2013 to 2014 and 1.1% from 2014 to 2015.\n\nFrom 2015 thereon, the country experienced disinflation. Inflation was 0.8% from 2015 to 2016 and further decreased to 0.4% from 2016 to 2017." ]
[ null, "https://www.facebook.com/tr", null, "https://corporatefinanceinstitute.com/resources/knowledge/economics/disinflation/", null ]
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https://books.google.com.au/books?qtid=c7e7b3c&lr=&id=hkLMs849kAsC&sa=N&start=140
[ "Search Images Maps Play YouTube News Gmail Drive More »\nSign in\n Books Books", null, "If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.", null, "Pantologia. A new (cabinet) cyclopædia, by J.M. Good, O. Gregory, and N ...\nby John Mason Good - 1819\nFull view - About this book", null, "## Vectors and Rotors: With Applications\n\nOlaus Henrici, George Charles Turner - Electronic book - 1903 - 204 pages\n...theorem is that of Euc. VI. vi., from which the general case follows at once. \" If two triangles have one angle of the one equal to one angle of the other and the sides about the equal angles proportionals, the triangles shall be similar.\" For if one triangle...\nFull view - About this book", null, "## Elementary Geometry: Practical and Theoretical\n\nCharles Godfrey, Arthur Warry Siddons - Geometry - 1903 - 355 pages\n...PQ ; measure L.' P, Q, and compare them with L.' B, C. =o fig. 313. THEOREM 5. If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar. fig. 314. Data ABC, DEF...\nFull view - About this book", null, "## The School World: A Monthly Magazine of Educational Work and Progress\n\nEducation - 1903\n...are equiangular their corresponding sides are proportional ; and the converse. If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar. The internal bisector of...\nFull view - About this book", null, "## Manual\n\n1903\n...angles to the straight line joining the middle points of AD and BC. 3. Show that if two triangles have one angle of the one equal to one angle of the other, and the sides about another angle of each proportional, the sides opposite the equal angles being homologous,...\nFull view - About this book", null, "## Interim Report of the Commissioners on Certain Parts of Primary Education ...\n\n...pentagon is equal to the sum of a side and a diagonal of the smaller pentagon. Equal triangles, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. A BCD is a trapezium in which Л...\nFull view - About this book", null, "## Calendar\n\nUniversity of Sydney - 1904\n...iu-radius and the circum-radius respectively measure 1 ft. and \" ft. 1 in. 7. Equal triangles, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. 8. Find the weight of a conical...\nFull view - About this book", null, "## A Text-book of Euclid's Elements for the Use of Schools, Book 1\n\nEuclid - Euclid's Elements - 1904 - 456 pages\n...[Book vi. Def. 4.] PROPOSITION 14. THEOREM. Parallelograms which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional. Conversely, parallelograms which...\nFull view - About this book", null, "## Calendar, for the Year ...\n\n1905\n...About a given circle circumscribe a triangle equiangular to a given triangle. (e) If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles are similar. '2. ABCD, AEFG are two squares...\nFull view - About this book", null, "## Science Examinations ... Reports, Etc\n\n1906\n...94. The work in this Stage quite reaches the standard of last year. A. Q. 41. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, show that the triangles are equiangular to one another,...\nFull view - About this book", null, "## The New International Encyclopaedia, Volume 18\n\n...respectively parallel or perpendicular to the sides of the other, they are similar. If two triangles have one angle of the one equal to one angle of the other, and the including »ides proportional, the triangles are similar. If two triangles have their sides proportional,...\nFull view - About this book" ]
[ null, "https://books.google.com.au/googlebooks/quote_l.gif", null, "https://books.google.com.au/googlebooks/quote_r.gif", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null, "https://books.google.com.au/books/content", null ]
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https://wiki.haskell.org/index.php?title=Euler_problems/61_to_70&oldid=12328
[ "# Euler problems/61 to 70\n\n## Problem 61\n\nFind the sum of the only set of six 4-digit figurate numbers with a cyclic property.\n\nSolution:\n\n```problem_61 = undefined\n```\n\n## Problem 62\n\nFind the smallest cube for which exactly five permutations of its digits are cube.\n\nSolution:\n\n```problem_62 = undefined\n```\n\n## Problem 63\n\nHow many n-digit positive integers exist which are also an nth power?\n\nSolution:\n\n```problem_63 = undefined\n```\n\n## Problem 64\n\nHow many continued fractions for N ≤ 10000 have an odd period?\n\nSolution:\n\n```problem_64 = undefined\n```\n\n## Problem 65\n\nFind the sum of digits in the numerator of the 100th convergent of the continued fraction for e.\n\nSolution:\n\n```problem_65 = undefined\n```\n\n## Problem 66\n\nInvestigate the Diophantine equation x2 − Dy2 = 1.\n\nSolution:\n\n```problem_66 = undefined\n```\n\n## Problem 67\n\nUsing an efficient algorithm find the maximal sum in the triangle?\n\nSolution:\n\n```problem_67 = undefined\n```\n\n## Problem 68\n\nWhat is the maximum 16-digit string for a \"magic\" 5-gon ring?\n\nSolution:\n\n```problem_68 = undefined\n```\n\n## Problem 69\n\nFind the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.\n\nSolution:\n\n```problem_69 = undefined\n```\n\n## Problem 70\n\nInvestigate values of n for which φ(n) is a permutation of n.\n\nSolution:\n\n```problem_70 = undefined\n```" ]
[ null ]
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https://www.dreamincode.net/forums/topic/117364-c-beginner-stl-stack-class-tutorial/
[ "Page 1 of 1\n\n## C++ BEGINNER STL STACK CLASS TUTORIAL C++ BEGINNER STL STACK CLASS TUTORIAL Rate Topic:", null, "", null, "", null, "", null, "", null, "//<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'https://www.dreamincode.net/forums/index.php?app=forums&module=ajax&section=topics&do=rateTopic&t=117364&amp;s=9217110cf8ad74415020a158fd67b0a3&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>\n\n### #1 Elcric", null, "Reputation: 103\n• Posts: 453\n• Joined: 02-May 09\n\nPosted 29 July 2009 - 12:53 AM\n\nC++ BEGINNER STL STACK CLASS TUTORIAL\n\nCONTENTS:\n• I. ASSUMPTIONS\n• II. INTRODUCTION\n• III. VOCABULARY\n• IV. CLASS TEMPLATE STACK<T>\n• V. MEMBER FUNCTIONS\n• VI. CONSTRUCTOR\n\nWHAT YOU WILL LEARN IN THIS TUTORIAL:\n\n1. You will learn stack class template vocabulary.\n\n2. You will learn how to declare and use a stack class template constructor.\n\n3. You will learn how to instantiate objects of type stack.\n\n4. You will learn the stack class member functions and how to use them.\n\n• I. ASSUMPTIONS\n\nI assume you have a C++ compiler and you want to learn about the C++ STL stack class.\n\n• II. INTRODUCTION\n\nHello; nice to meet you! Welcome to the “C++ Beginner STL Stack Class Tutorial.”\n\nPlease copy and paste the attached example program into a file your compiler can read. Then build, compile, and debug the example program. Following along, back and forth, between the tutorial and your copy of the example program will help you understand the example program functionality.\n\n```//******************************************************\n//C++ Beginner STL Stack Class Tutorial\n//Featuring a gold coin counting class template stack<T>\n//******************************************************\n#include <deque>\n#include <exception>\n#include <iostream>\n#include <stack>\n\nusing namespace std;\n\ntemplate< typename T>\nclass CStack\n{\npublic:\n\nCStack( int = 5 ); //Default constructor with stack size 5.\n~CStack() //Destructor.\n{\ndelete [] CStackPtr;\n}\n\nbool push( const T & );\nbool pop ( T & );\n\nbool isEmpty( ) const\n{\n}\n\nbool isFull( ) const\n{\n}\n\nprivate:\n\nint max;\nint top;\nT *CStackPtr;\n};\n\ntemplate< typename T > //Template constructor with stack size 5.\nCStack< T >::CStack( int cs )\n: max( cs > 0 ? cs : 5 ),\ntop( -1 ),\nCStackPtr( new T[ max ])\n{\n\n}\n\ntemplate< typename T >\nbool CStack< T >::push( const T &pushValue )\n{\nif ( !isFull() )\n{\nCStackPtr[ ++top ] = pushValue;\nreturn true;\n}\n\nreturn false;\n}\n\ntemplate< typename T >\nbool CStack< T >::pop( T &popValue )\n{\nif ( !isEmpty() )\n{\npopValue = CStackPtr[ top-- ];\nreturn true;\n}\n\nreturn false;\n}\n\n//************************************\n// MAIN() FUNCTION.\n//************************************\nint main(int argc, char* argv[])\n{\nint\t choice =\t0;\nbool\texit\t=\tfalse;\n//************************************\n\ndo\n{\nCStack< double > doubleCStack( 10 ); //Instantiate object doubleCStack of type CStack.\ndouble doubleValue = 1.0;\n\nsystem(\"CLS\");\ncout << endl << endl << endl;\ncout << \" This is a rich person's gold stacking program.\" << endl << endl;\ncout << \" The program stacks and unstacks gold coins.\" << endl << endl;\ncout << \" The first stack is a double and holds 10 gold coins.\" << endl << endl;\ncout << \" The second stack is an int and holds 5 gold coins.\" << endl << endl;\ncout << endl << endl;\ncout << \" Pushing gold coins onto the double stack.\" << endl;\ncout << endl << endl;\ncout << \" \";\n\nwhile ( doubleCStack.push( doubleValue ))\n{\ncout << doubleValue << \" \";\ndoubleValue += 1.0;\n}\n\ncout << endl << endl;\ncout << \" Double stack is now full.\" << endl << endl;\ncout << \" Cannot push any more gold coins onto double stack!\" << endl << endl;\ncout << endl << endl;\ncout << \" \";\nsystem(\"PAUSE\");\n\nsystem(\"CLS\");\ncout << endl << endl << endl;\ncout << \" Popping gold coins from double stack.\" << endl;\ncout << endl << endl;\ncout << \" \";\n\nwhile ( doubleCStack.pop( doubleValue ))\n{\ncout << doubleValue << \" \";\n}\n\ncout << endl;\ncout << \" \";\n\ncout << endl;\ncout << \" Double stack is now empty.\" << endl << endl;\ncout << \" Cannot pop any more gold coins from double stack!\" << endl << endl;\ncout << endl << endl;\ncout << \" \";\nsystem(\"PAUSE\");\ncout << endl << endl << endl;\ncout << \" \";\n\nCStack< int > intCStack; //Instantiate object intCStack of type CStack.\nint intValue = 1;\n\nsystem(\"CLS\");\ncout << endl << endl << endl;\ncout << \" Pushing gold coins onto int stack.\" << endl;\n\ncout << endl << endl;\ncout << \" \";\n\nwhile ( intCStack.push( intValue ))\n{\ncout << intValue++ << \" \";\n}\n\ncout << endl << endl;\ncout << \" Int stack is now full.\" << endl << endl << endl;\ncout << \" Cannot push any more gold coins onto int stack!\" << endl << endl;\n\ncout << endl << endl;\ncout << \" \";\nsystem(\"PAUSE\");\n\nsystem(\"CLS\");\ncout << endl << endl << endl;\ncout << \" Popping gold coins from int stack.\" << endl;\n\ncout << endl << endl;\ncout << \" \";\n\nwhile ( intCStack.pop( intValue ))\n{\ncout << intValue << \" \";\n}\n\ncout << endl << endl;\ncout << \" Int stack is now empty.\" << endl << endl;\ncout << \" Cannot pop any more gold coins from int stack!\" << endl << endl;\ncout << endl << endl;\ncout << \" \";\nsystem(\"PAUSE\");\ncout << endl << endl << endl;\ncout << \" \";\n\n//************************************\nsystem( \"CLS\" );\ncout << endl << endl << endl;\ncout << \" Type the number 1 and press return \";\ncout << \"to run the program again. \" << endl << endl;\ncout << \" Type 0 and press return to exit. ==> \";\ncin >> choice;\nif ( choice == 0 )\n{\nEXIT_SUCCESS;\n}\nif ( choice == 1 )\nexit = false;\nelse\n{\nexit = true;\n}\n}while( exit == false );\n\nreturn EXIT_SUCCESS;\n}\n\n```\n\n• III. VOCABULARY\n\n1. What is LIFO?\n\nLIFO is last-in-first-out. A stack enforces LIFO order.\nA stack of plates is a common visualization of a STL stack. The last plate placed on a stack of plates is normally the first plate removed from the stack of plates.\n\n2. What is a generic function or function template?\n\nA generic function or function template is a function definition preceded by a template statement.\n\n```template< typename T >\nbool CStack< T >::push( const T &pushValue )\n{\nif ( !isFull() )\n{\nCStackPtr[ ++top ] = pushValue;\nreturn true;\n}\n\nreturn false;\n}\n\n```\n\nA compiler generated function specialization is an instantiation of a function template.\n\nA function template declaration consists of the keyword template, followed by a list of template arguments enclosed in angle brackets, followed by a function declaration.\n\nAll versions of a generic function must perform the same actions, only the data type can be different. For example, a sorting algorithm is the same no mater what type of data is being sorted. You can use a generic function any time you see a situation where the logic of a function is independent of the data type. Defining a single function template defines an entire family of overloaded functions.\n\n3. What is an explicit specialization?\n\nA generic function overloads itself. However, the programmer can code an explicit specialization to overload a generic function and override the compiler generated function to accommodate unique situations.\n\n4. What is a generic class?\n\nClass is the generic term for type. A generic class is a class that uses logic that can be generalized. When objects of a generic class are created the type of data is specified as a parameter. Upon instantiation the compiler generates the correct type of object based on the parameter type specified. A generic class allows you to define the general form of an object that can then be used with any data type.\n\nBy making stack a generic class the compiler can create a stack for any type of data.\n\n5. What is typename?\n\nThe keyword typename can be substituted for the keyword class in template declarations. The keyword typename also informs the compiler a type name is being used in a template declaration instead of an object name.\n\n```template< typename T >\nbool CStack< T >::pop( T &popValue )\n{\nif ( !isEmpty() )\n{\npopValue = CStackPtr[ top-- ];\nreturn true;\n}\n\nreturn false;\n}\n\n```\n\nThe keyword class connotes a user defined type while the keyword typename is more neutral and refers to fundamental types and user-defined types.\n\n6. What is a class template?\n\nA class template is a template for producing classes. A class produced from a template is said to be a class template. A template class is an instance of a class template.\n\nMember function definitions for class templates are preceded by the template keyword with parameters in angle brackets.\n\nThe STL provides three container adapters: stack, queue, and priority queue. Container adapters do not support iterators. Container adapters allow you to choose an underlying data structure.\n\n• IV. CLASS TEMPLATE STACK<T>\n\nTo instantiate a stack a data type must be specified.\nThe template specifications for a stack are as follows:\ntemplate <class T, class Container = deque<T>> class stack\n\n```template< typename T > //Template constructor with stack size 5.\nCStack< T >::CStack( int cs )\n: max( cs > 0 ? cs : 5 ),\ntop( -1 ),\nCStackPtr( new T[ max ])\n{\n\n}\n\n```\n\n• V. MEMBER FUNCTIONS\n\nThe main behavior of a stack is that objects go on and come off of the top of the stack. The following are STL stack class public member functions. Other underlying deque, vector, or list container member functions such as iterators, push_back, and insert been suppressed to maintain stack behavior complance.\n\n1. void push(const value_type &val);\n\nInserts an element at the top of a stack. This element becomes the last in the stack.\n\n```template< typename T >\nbool CStack< T >::push( const T &pushValue )\n{\nif ( !isFull() )\n{\nCStackPtr[ ++top ] = pushValue;\nreturn true;\n}\n\nreturn false;\n}\n\n```\n\n2. void pop();\n\nRemoves, deletes, the element at the top of a stack, the last element in the container.\n\n```template< typename T >\nbool CStack< T >::pop( T &popValue )\n{\nif ( !isEmpty() )\n{\npopValue = CStackPtr[ top-- ];\nreturn true;\n}\n\nreturn false;\n}\n\n```\n\n3. bool empty() const;\n\nTests whether the stack is empty. Returns true if the stack is empty and false otherwise.\n\n```bool isEmpty( ) const\n{\n}\n\n```\n\n4. size_type size() const;\n\nReturns the number of elements currently in the stack.\n\n5. value_type &top(); const value_type &top() const;\n\nReturns a reference to the top most element in the stack. An undefined value is returned if the stack is empty.\n\n```if ( !isFull() )\n{\nCStackPtr[ ++top ] = pushValue;\nreturn true;\n}\n\n```\n\n• VI. CONSTRUCTOR\n\nThe stack constructor uses the following syntax to create an empty stack using deque as the default container:\n\nexplicit stack(const Container &cnt = Container());\n\n```template< typename T > //Template constructor with stack size 5.\nCStack< T >::CStack( int cs )\n: max( cs > 0 ? cs : 5 ),\ntop( -1 ),\nCStackPtr( new T[ max ])\n{\n\n}\n\n```\n\nThe container is held in a protected member called c of type Container.\nThe ==, <, <=, !=, >, and >= comparison operators are defined for stack.\n\nWHAT YOU HAVE LEARNED IN THIS TUTORIAL:\n\n1. You have learned stack class template vocabulary.\n\n2. You have learned how to declare and use a stack class template constructor.\n\n3. You have learned how to instantiate objects of type stack.\n\n4. You have learned the stack class member functions and how to use them.\n\nIs This A Good Question/Topic? 0\n\nPage 1 of 1\n\n .related ul { list-style-type: circle; font-size: 12px; font-weight: bold; } .related li { margin-bottom: 5px; background-position: left 7px !important; margin-left: -35px; } .related h2 { font-size: 18px; font-weight: bold; } .related a { color: blue; }" ]
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[ "Community Profile", null, "Paresh yeole\n\nLast seen: 約1ヶ月 前 2019 以来アクティブ\n\nControl Theory Enthusiast\n\nStatistics\n\nAll\n•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "•", null, "バッジを表示\n\nContent Feed\n\nSurface area of a cone\n\n4ヶ月 前\n\n4ヶ月 前\n\nfind the value\n\n4ヶ月 前\n\nPerimeter of a Triangle\n\n4ヶ月 前\n\nVolume of a sphere\n\n4ヶ月 前\n\nCody Computer Part 2 - Get the license number of Cody Computer\nIn the Cody computer investigation party, you may have already solved : <http://www.mathworks.com/matlabcentral/cody/problems...\n\n4ヶ月 前\n\nCody Computer Part 1 - Guess the system font used by uipanel\nlist = listfonts return the list of available system fonts. Can you guess which font is used for uipanel on the CODY computer...\n\n4ヶ月 前\n\nbe able to add any variable to the number one\n\n5ヶ月 前\n\nSort in descending order.\n\n5ヶ月 前\n\nPeriod of a pendulum\n\n5ヶ月 前\n\nGPA\nMath's credit is 5 and you got x | Physics's credit 7 and you got y what is your GPA\n\n5ヶ月 前\n\nFind the worst traffic yesterday\n\n5ヶ月 前\n\nPolygon Interior Angle Sum\nGiven the number of sides of a regular polygon, find the sum of interior angles. Return 0 for input for which regular polygon a...\n\n5ヶ月 前\n\nHydrogens in a Saturated Hydrocarbon\nGiven (c) carbon atoms in a saturated hydrocarbon molecule (all single bonds), how many hydrogen atoms (h) are in the molecule?\n\n5ヶ月 前\n\nDraw a '4' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '6' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '9' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '7' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '8' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '3' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '2' in a zero matrix!\n\n6ヶ月 前\n\nDraw a '5' in a zero matrix!\n\n6ヶ月 前\n\n(Linear) Recurrence Equations - Generalised Fibonacci-like sequences\nThis problem is inspired by problems <http://uk.mathworks.com/matlabcentral/cody/problems/2187-generalized-fibonacci 2187>, <htt...\n\n8ヶ月 前\n\nFibonacci Decomposition\nEvery positive integer has a unique decomposition into nonconsecutive Fibonacci numbers f1+f2+ ... Given a positive integer n, r...\n\n8ヶ月 前\n\nHow many Fibonacci numbers?\nFind the number of unique Fibonacci numbers (don't count repeats) in a vector of positive integers. Example: x = [1 2 3 4...\n\n8ヶ月 前\n\nInteger Sequence - II : New Fibonacci\nCrack the following Integer Sequence. (Hints : It has been obtained from original Fibonacci Sequence and all the terms are also ...\n\n8ヶ月 前\n\nReturn fibonacci sequence do not use loop and condition\nCalculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: ...\n\n8ヶ月 前\n\nIs X a Fibonacci Matrix?\nIn honor of Cleve's new blog and post: <http://blogs.mathworks.com/cleve/2012/06/03/fibonacci-matrices/> Is X a Fibonacci ...\n\n8ヶ月 前\n\nFind the next Fibonacci number\nIn the sequence of Fibonacci numbers, every number is the sum of the two preceding ones: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...\n\n8ヶ月 前" ]
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https://etc.usf.edu/clipart/keyword/sin
[ "", null, "### Adam and Eve are Punished and Cast Out of the Garden of Eden\n\n\"So he drove out the man; and he placed at the east of the garden of Eden the Cherubim, and the flame…", null, "### Sum of 2 Acute Angles\n\nIllustration of two triangles, showing the sine of the sum of two acute angles expressed in terms of…", null, "### The Brazen Serpent and the Healing of the Israelites\n\n\"And Jehovah sent fiery serpents among the people, and they bit the people; and much people of Israel…", null, "### Ambiguous Case\n\nIllustration of one possible outcome (no triangle occurs) when discussing the ambiguous case using the…", null, "### Ambiguous Case\n\nIllustration of one possible outcome (1 triangle occurs) when discussing the ambiguous case using the…", null, "### Ambiguous Case\n\nIllustration of one possible outcome (2 triangles occur) when discussing the ambiguous case using the…", null, "### Unit Circle\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane. The…", null, "### Unit Circle Labeled At Quadrantal Angles\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled At Special Angles\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled At Special Angles\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled In 30° Increments With Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled In 45 ° Increments\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled In 45° Increments With Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled In 45° Increments With Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Quadrantal Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane. All…", null, "### Unit Circle Labeled With Quadrantal Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Quadrantal Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Quadrantal Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane. At each…", null, "### Unit Circle Labeled With Quadrantal Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Special Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Special Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Labeled With Special Angles And Values\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane with…", null, "### Unit Circle Marked At Special Angles\n\nIllustration of a unit circle (circle with a radius of 1) superimposed on the coordinate plane. All…", null, "### Jesus Carrying the Cross on the Way to Calvary\n\n\"They took Jesus therefore: and he went out, bearing the cross for himself, unto the place called The…", null, "### Trigonometric Curve\n\nA curve that is a function and resembles the sine or cosine curve.", null, "### Angle Expressed As An Inverse Function\n\nIllustration that can be used to show that when given an angle, expressed as an inverse function of…", null, "### Moses Breaks the Stone Tablets on Which the Ten Commandments are Written\n\n\"And it came to pass, as soon as he came nigh unto the camp, that he saw the calf and the dancing: and…", null, "### Triangles and Sectors in Quadrant I\n\nIllustration of an angle &alpha with the vertex at the center, O, of a circle with radius OB. AC and…", null, "Illustration of an angle &alpha with the terminal side used to draw a triangle in quadrant I.", null, "Illustration of an angle with the terminal side used to draw a triangle in quadrant II.", null, "### Sine Curve y=sin x\n\nSine curve plotted from 0 to 2 pi. Graph of y=sin x.", null, "### Sine Curves y= sin Ǝ, y= 1/2 sin Ǝ, y=2 sin Ǝ, y= 2 sin 3Ǝ\n\nSine curves of varying frequency and amplitude plotted from 0 to 2 pi. Graph of y= sin θ, y= 1/2…", null, "### Sine Curves y= sin t, y= r sin1/2t, y= r sin 2t\n\nSine curves of varying frequency plotted from 0 to 2 pi. Graph of y= sin t, y= r sin1/2t, y=r sin 2t.", null, "### Law of Sines\n\nIllustration two types of triangles that can be used to model the law of sines. \"In a plane triangle…", null, "### Snake Giving Two Children Cake\n\nA cartoon of a snake feeding a piece of cake to two children. A sign over the cake reads: The Forbidden…", null, "### Suovetaurilia\n\n\"The relief pictures an ancient Italian sacrifice of a bull, a ram, and a boar, offered to Mars to secure…", null, "### The Crucifixion of Jesus with Two Robbers\n\n\"And with him they crucify two robbers; one on his right hand, and one on his left.\" Mark 15:27 ASV…", null, "### Ambiguous Case of Law of Sines Triangle\n\nIllustration showing ambiguous case when the solution is not a triangle using law of sines.", null, "### Trigonometric Reference Triangles/Angles (60 degrees) Drawn in Quadrants\n\nTrigonometric reference triangles/angles drawn for 60 degree reference angel in quadrants I and II.", null, "### Inclined Plane Forming Right Triangle\n\nInclined plane forming right triangle showing the velocity of a body sliding a distance,s, down a smooth…", null, "### Triangles ABC and ABC'\n\nTriangle ABC and triangle ABC'. This illustration could be used to demonstrate the law of sines.", null, "### Trigonometric Reference Triangles/Angles Drawn in Quadrants\n\nTrigonometric reference triangles/angles drawn for reference angel in quadrants I and II. This illustration…", null, "### Jesus Tells a Crowd to Cast the First Stone\n\n\"But when they continued asking him, he lifted up himself, and said unto them, He that is without sin…", null, "### Jesus Tells a Woman to Go and Sin No More\n\n\"And they, when they heard it, went out one by one, beginning from the eldest, even unto the last: and…" ]
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https://answers.everydaycalculation.com/add-fractions/24-9-plus-35-84
[ "Solutions by everydaycalculation.com\n\n1st number: 2 6/9, 2nd number: 35/84\n\n24/9 + 35/84 is 37/12.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 9 and 84 is 252\n2. For the 1st fraction, since 9 × 28 = 252,\n24/9 = 24 × 28/9 × 28 = 672/252\n3. Likewise, for the 2nd fraction, since 84 × 3 = 252,\n35/84 = 35 × 3/84 × 3 = 105/252\n672/252 + 105/252 = 672 + 105/252 = 777/252\n5. 777/252 simplified gives 37/12\n6. So, 24/9 + 35/84 = 37/12\nIn mixed form: 31/12\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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https://analystprep.com/cfa-level-1-exam/quantitative-methods/using-timelines-in-modelling-and-solving-time-value-of-money-problems/
[ "Limited Time Offer: Save 10% on all 2021 and 2022 Premium Study Packages with promo code: BLOG10    Select your Premium Package »", null, "Using Timelines in Modeling and Solving Time Value of Money Problems\n\nA timeline is a physical illustration of the amounts and timing of cashflows associated with an investment project. For cashflows that are regular and of equal amounts, the standard annuity formula or the financial calculator can be used. However, for cashflows that are irregular, unequal, or both, a timeline is preferred.\n\nRemember that the general formula that relates the present value and the future value of an investment is given by:\n\n$$FV _ {N} = PV \\left (1 + r \\right)^ {N}$$\n\nWhere\n\n$$PV$$ = present value of the investment\n$$FV_N$$ = future value of the investment  $$N$$  periods from today\n$$r$$ = rate of interest per period\n\nWe can represent this in a timeline:", null, "In a particular timeline, a time index, t, is used to represent a particular point in time, a specified number of periods from today. Therefore, the present value is the investment amount today (t = 0) and by using this amount, we can calculate the future value (t = N). Alternatively, we can use the future value to calculate the present value.\n\nThe above argument can be written in terms of the present value. That is:\n\n$$PV = FV _ {N} \\left (1+r \\right)^ {-N}$$\n\nExample: Applying a Timeline to Model Cashflows\n\nAn investor receives a series of payments, each amounting to $6, 500, set to be received in perpetuity. Payments are to be made at the end of each year, starting at the end of year 4. If the discount rate is 9%, then what is the present value of the perpetuity at t = 0? Solution We would then draw a timeline to better understand the problem:", null, "Here, we can see that the investor is receiving$6, 500 in perpetuity. Recall that the PV of a perpetuity is given by:\n\n$$\\text {PV of a perpetuity} = \\frac {C} {r}$$\n\nSo that in this case:\n\n$$PV_3 = \\frac {6,500} {9\\%} = 72, 222$$\n\nThis is the value of the perpetuity at t = 3, so we need to discount it for 3 more periods to get the value at t = 0. Using the formula:\n\n$$PV _ 0 = FV _ {N} \\left (1+r \\right)^ {-N}$$\n\nThe PV at time zero is $$\\frac {72, 222} {{(1 + 0.09)}^3} = 55, 769$$.\n\nWhy is the Use of a Timeline Recommended?\n\nThere are many instances in real life when cashflows are uneven. A good example in this respect is a pension contribution which varies with age. In such cases, it is not possible to apply one of the basic time value formulae. You are advised to draw a timeline even if the question appears quite straightforward. It will help you understand the question structure better. A timeline also helps candidates add cashflows indexed to the same period and apply the value additivity principle.\n\nQuestion\n\nAssume that we have two projects,  X and Y,  and each has positive cashflows. The annual interest rate is 5%. The projects have the following cashflow:\n\nProject X:\n\n$$\\begin {array} {c|c|c|c|c|c} {t = 0} & {t = 1} & {t = 2} & {t = 3} & {t = 4} & {t = 5} \\\\ \\hline {0} & {100} & {150} & {250} & {300} & {250} \\\\ \\end {array}$$\n\nProject Y:\n\n$$\\begin {array} {c|c|c|c|c|c} {t=0} & {t=1} & {t = 2} & {t = 3} & {t = 4} & {t = 5} \\\\ \\hline {50 } & {100 } & {200} & {300} & {400} & {500} \\\\ \\end {array}$$\n\nWhat is the present value of the cashflows for both projects combined?\n\n1.  $890 2.$ 1,307\n3. 2,197 Solution The correct answer is C. The timeline for project X is as follows:", null, "We can calculate the cashflows for project X as: \\begin {align*} \\text {PV for X} & = 100 (1 + r)^ {-1} + 150 (1 + r)^ {-2} + 250 (1 + r)^ {-3} + 300 (1 + r)^ {-4} + 250 (1 + r)^ {-5} \\\\ & = 100 \\times 1.05^ {-1} + 150 \\times 1.05^ {-2} + 250 \\times 1.05^ {-3} + 300 \\times 1.05^ {-4} + 250 \\times 1.05^{-5} \\\\ & = 890\\end {align*} For project Y, the timeline is given by:", null, "\\begin {align*} \\text {PV for Y} & = 50 (1 + r)^ {-0} + 100 (1 + r)^ {-1} + 200 (1 + r)^ {-2} + 300 (1 + r)^ {-3} + 400 (1 + r)^ {-4} + 500 (1 + r)^ {-5} \\\\ & = 50 + 100 \\times 1.05^ {-1} + 200 \\times 1.05^ {-2} + 300 \\times 1.05^ {-3} + 400 \\times 1.05^ {-4} + 500 \\times1.05^ {-5} \\\\ & = 1307 \\end {align*} We then simply add the cashflows together: $$PV_ {X+Y} = 890 + 1307 = 2,197$$ Cashflow Additivity Principle According to cashflow additivity principle, the present value of any stream of cashflows indexed at the same point equals the sum of the present values of the cashflows. This principle has different applications in time value of money problems. Besides, this principle can be applied to find the NPVs of projects with uneven cashflows. In the above question, we can combine corresponding cashflows and work out the present values for the two projects at once: $$NPV = (0 + 50) + \\frac {100 + 100} {1.05^{1}} + \\frac {150 + 200} {1.05^{2}} + \\frac {250 + 300} {1.05^{3}} + \\frac {300 + 400} {1.05^{4}} + \\frac{250 + 500} {1.05^{5}} = 2,197$$ Calculating Interest Rates, Growth Rates, and Number of Periods Calculating Interest Rate We know that, $$PV = FV _ {N} \\left (1 + r \\right)^ {-N}$$ $$\\Rightarrow \\text {r} = \\left (\\frac {FV _ N} {PV} \\right)^ {\\frac {1} {N}} – 1$$ Example: Annual Rate of Return An investor invested in a1, 000  security today. It will earn him $1,191 three years from now. The annual rate of return on this investment is closest to? Solution Recall that: $$\\text {r} = \\left (\\frac {FV _ N} {PV} \\right)^ {\\frac {1} {N}} -1$$ In this case we have: $$\\text {r} = \\left (\\frac {1191} {1, 000} \\right)^ {\\frac {1} {3}} -1 = 0.06 = 6%$$ The steps to solve the above example using a financial calculator are given below: $$\\begin {array} {l|l|l} \\textbf {Steps} & \\textbf {Explanation} & \\textbf {Display} \\\\ \\hline \\text { [2nd] [QUIT] } & \\text {Return to standard calc mode} & 0 \\\\ \\hline\\left [2^ {\\text {nd }} \\right] [\\text {CLR TVM}] & \\text {Clears TVM Worksheet} & 0 \\\\ \\hline 3 [N] & \\text {Years/periods} & N = 3 \\\\ \\hline-1, 000 [PV] & \\text {Set to } PV \\text { to } – 1, 000 & PV = -1, 000 \\\\ \\hline 0 [PMT] & \\text {Set annuity payment} & PMT = 0 \\\\ \\hline 1, 191 [FV] & \\text {Set to } FV \\text { to } 1, 191 & FV = 1, 191 \\\\ \\hline [CPT] [I/Y] & \\text {Compute interest rate} & I/Y = 6 \\text{ or } 6 \\% \\\\ \\end{array}$$ Example: Annual Rate of Return #2 An investor invests$100 at the end of each of the next five years and receives $750 at the end of the fifth year. The annual rate of return on this investment is closest to: Solution Using a calculator, you will solve for I/Y as follows. $$N$$ = 5; $$FV = 500$$; $$PMT = –100$$; $$CPT$$ $$I/Y$$ = 20.40% Example: Annual Rate of Return #3 An investor invests$1, 000 today and receives a $50 coupon payment at the end of every year for 3 years. In addition, they receive$1,100 at the end of year 3. What is the rate of return of this investment?\n\nSolution\n\nUsing the calculator, you will calculate I/Y as follows.\n\n$$PV$$  =  -$1, 000 $$FV$$ =$1, 100\n\n$$N$$  =  3\n\n$$PMT$$  =  50\n\n$$CPT$$  $$I/Y$$  =  8.078%\n\nCalculating the Number of Periods\n\n$$N$$ can be calculated using the following formula:\n\n$$N = \\frac {\\left [\\frac {FV} {PV} \\right]} {ln (1+r)}$$\n\nAn investor invests $3, 000 in a bank account. How many years will it take to double the amount given that the interest rate is 5% per annum compounded annually? Solution Using the formula: $$N = \\frac {\\left [\\frac {FV} {PV} \\right]} {ln(1+r)} = \\left(\\frac {ln 2} {ln 1.05} \\right) = 14.21\\ \\text {Years}$$ Using the BA II Plus™ Financial Calculator: $$I/Y$$ = 5%, $$PV$$ = -$3, 000, $$PMT$$ = 0, $$FV$$ = 6, 000, $$CPT$$ $$N$$ = 14.21 Calculating Annuity Payments (Combining Future Value and Present Value Annuities) The annuity payment, $$A$$, can be computed using the following formula: \\begin {align} \\text {A} & = \\frac {\\text {PV of Annuity}} {\\text {Present value annuity Factor}} \\\\ & = \\frac {\\text {PV of Annuity}} {\\left[\\frac {1-\\frac{1} {\\left(1+\\frac{r s} {m}\\right)^ {m N}}} {\\frac {r S} {m}}\\right]} \\end {align} Example: Annuity Payment Calculation A company has borrowed100, 000 for the purchase of machinery. The tenor of the loan is five years at an annual rate of 5% compounded monthly. If the first payment is due in one month, the monthly payment is closest to:\n\nSolution\n\nUsing the formula:\n\n$$A = \\frac {\\text {PV of Annuity}} {\\left[\\frac{1 – \\frac{1} {\\left(1 + \\frac {r s} {m} \\right)^ {m N}}} {\\frac {r S}{m}} \\right]} = \\left [\\frac {1 – \\frac {1} {\\left (1 + \\frac {0.05} {12} \\right)^ {12(5)}}} {\\frac {0.05} {12}} \\right] = \\frac {100, 000}{52.9907} = 1, 887.1237$$\n\nSo, the monthly payment is $1, 887. Using the BA II Plus™ Financial Calculator: $$PV$$ = -100, 000; $$N$$ = 5×12 = 60; $$I/Y$$ = 5/12 = 0.41667; $$FV$$ = 0; $$CPT \\ PMT$$ = 1,887.123 Example: Annuity Payment Calculation A bank is offering a profit rate of 6% on its savings account. How much should an investor deposit each quarter to grow the investment to$20, 000 at the end of 12 years?\n\nSolution\n\nUsing the formula:\n\n\\begin {aligned} F V & = A\\left [\\frac {\\left (1+\\frac {r S} {m} \\right)^ {m N} -1} {\\frac {r s} {m}} \\right] \\\\ \\ 20, 000 & = A\\left [\\frac {\\left (1 + \\frac {0.06} {4} \\right)^ {4 \\times 12} -1} {\\frac {0.06} {4}}\\right] \\\\ \\ 20, 000 & = 69.5652 A \\\\ \\Rightarrow A & = 287.50 \\end{aligned}\n\nUsing the BA II Plus™ Financial Calculator:\n\n$$FV$$  =  -20, 000; $$N$$  =  4 × 12 =  48; $$I/Y$$  =  6/4  =  1.5; $$PV$$  =  0; $$CPT$$  $$PMT$$  =  287.50\n\nExample: Annuity Payment Calculation\n\nAn investor needs to accumulate \\$3, 000 over the next 5 years. How much should she invest in a savings account every year given that the expected rate of return is 6.5%?\n\nSolution\n\nUsing the BA II Plus™ Financial Calculator:\n\n$$FV$$  =  -3, 000;  $$N$$  =  5;  I/Y  = 6.5;  $$PV$$  =  0; $$CPT \\ PMT$$  =  526.90\n\nFeatured Study with Us", null, "CFA® Exam and FRM® Exam Prep Platform offered by AnalystPrep\n\nStudy Platform\n\nLearn with Us\n\nSubscribe to our newsletter and keep up with the latest and greatest tips for success\nOnline Tutoring", null, "Our videos feature professional educators presenting in-depth explanations of all topics introduced in the curriculum.\n\nVideo Lessons", null, "Sergio Torrico\n2021-07-23\nExcelente para el FRM 2 Escribo esta revisión en español para los hispanohablantes, soy de Bolivia, y utilicé AnalystPrep para dudas y consultas sobre mi preparación para el FRM nivel 2 (lo tomé una sola vez y aprobé muy bien), siempre tuve un soporte claro, directo y rápido, el material sale rápido cuando hay cambios en el temario de GARP, y los ejercicios y exámenes son muy útiles para practicar.", null, "diana\n2021-07-17\nSo helpful. I have been using the videos to prepare for the CFA Level II exam. The videos signpost the reading contents, explain the concepts and provide additional context for specific concepts. The fun light-hearted analogies are also a welcome break to some very dry content. I usually watch the videos before going into more in-depth reading and they are a good way to avoid being overwhelmed by the sheer volume of content when you look at the readings.", null, "Kriti Dhawan\n2021-07-16\nA great curriculum provider. James sir explains the concept so well that rather than memorising it, you tend to intuitively understand and absorb them. Thank you ! Grateful I saw this at the right time for my CFA prep.", null, "nikhil kumar\n2021-06-28\nVery well explained and gives a great insight about topics in a very short time. Glad to have found Professor Forjan's lectures.", null, "Marwan\n2021-06-22\nGreat support throughout the course by the team, did not feel neglected", null, "Benjamin anonymous\n2021-05-10\nI loved using AnalystPrep for FRM. QBank is huge, videos are great. Would recommend to a friend", null, "Daniel Glyn\n2021-03-24\nI have finished my FRM1 thanks to AnalystPrep. And now using AnalystPrep for my FRM2 preparation. Professor Forjan is brilliant. He gives such good explanations and analogies. And more than anything makes learning fun. A big thank you to Analystprep and Professor Forjan. 5 stars all the way!", null, "michael walshe\n2021-03-18\nProfessor James' videos are excellent for understanding the underlying theories behind financial engineering / financial analysis. The AnalystPrep videos were better than any of the others that I searched through on YouTube for providing a clear explanation of some concepts, such as Portfolio theory, CAPM, and Arbitrage Pricing theory. Watching these cleared up many of the unclarities I had in my head. Highly recommended." ]
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