CALCULUS 
 
F. M. AND C. 
 
AN ELEMENTARY TREATISE 
 
 ON 
 
 CALCULUS 
 
 A TEXT BOOK FOR 
 COLLEGES AND TECHNICAL SCHOOLS 
 
 WILLIAM S. FRANKLIN, BARRY MacNUTT 
 AND ROLLIN L. CHARLES 
 
 OT LEHIGH XJNIVBHSITT 
 
 SOUTH BETHLEHEM, PA. 
 PUBLISHED BY THE AUTHORS 
 
 1913 
 
 All rights reserved 
 
o(\ 
 
 o 
 
 03 
 
 ^7 
 
 Copyright, 1913 
 By Franklin, MacNutt and Charles 
 
 Set up and electrotyped. Published April, 1913 
 
 PRESS or 
 
 THE NEW ERA PRINTING COMPANY 
 LANCAbTER. PA. 
 
PREFACE. 
 
 1. We believe that the most important thing in the teaching of 
 calculus is to lead the student to a clear understanding of principles. 
 Therefore our chief endeavor has been to develop the subject as 
 simply and as directly as possible. 
 
 2. We fully appreciate the importance of extensive practice in 
 the handling of algebraic transformations. Therefore we have 
 given an adequate collection of formal problems in differentiation 
 and integration. 
 
 3. We are convinced, however, that it is a pedagogical mistake, 
 in a text book on calculus for young men, to break the thread of 
 the textual discussion by unnecessary algebraic developments 
 and by large and frequent groups of purely formal problems. 
 We believe indeed that the proverbial unintelligibility of calculus 
 is to a very great extent a psychological consequence of this almost 
 universal and really hideous feature. Therefore we have arranged 
 the greater portion of our formal problems in an appendix. 
 
 4. Nearly every elementary science text in existence carries a 
 false suggestion of completeness and finality, and there are two 
 things which young men should understand in connection with 
 their study of the mathematical sciences. The first is that such 
 study is exacting beyond all compromise, involving as it does a 
 degree of constraint which it is beyond the power of any teacher 
 greatly to mitigate. And the second is that the completest science 
 stands abashed before the infinitely complicated and fluid array 
 of phenomena of the material world, except only in the assurance 
 which method gives. We hope that this elementary treatise on 
 calculus may prove to be sufficiently definite and exacting to be 
 useful; next to this there is nothing we coul3Vish to have more in 
 evidence from beginning to end than its incompleteness. 
 
 273389 
 
VI PREFACE. 
 
 In order to emphasize the incomplete character of this text we 
 have introduced references to more complete treatises throughout 
 the text and we have given in Appendix C a carefully selected list of 
 treatises on mathematics and on mathematical physics. Teachers 
 who use this text should, we believe, direct the students* attention 
 to this appendix. 
 
 In our brief reference to the infinitesimal method in article 20 
 we do not wish to be thought of as taking sides in the old contro- 
 versy as between the " method of limits " and the " method of 
 infinitesimals" in calculus. Article 20 is unquestionably falla- 
 cious as it stands ; and the same may be said of the discussion of 
 divergence and curl in articles 126 and 129. Indeed articles 20, 
 126 and 129 may be characterized as mere plausibilities, and the 
 harder one tries to understand them the more vague and unintel- 
 ligible they become. The fact remains, however, that the infini- 
 tesimal method contributes very greatly to directness and sim- 
 plicity of speech in the discussion of physical problems, and the 
 idea of infinitesimals is therefore used throughout this text. Any 
 one who is disturbed by the element of easy plausibility that is 
 involved in the s raight-forward use of the infinitesimal method 
 in the discussion of physical problems should heed the advice 
 given by D'Alembert to a young student, "Go ahead, young man, 
 go ahead! Conviction will come to you later." 
 
 "The absolute requisites for the study of this work are: a 
 knowledge of elementary algebra to the binomial theorem (accord- 
 ing to the usual arrangement), plane and solid geometry, trigono- 
 metry, and the most simple parts of the usual apphcations of 
 algebra to geometry." 
 
 This was said by De Morgan in the preface to his great treatise 
 on Differential and Integral Calculus (London, 1842), in com- 
 parison with which this book is a primer. 
 
 Franklin, MacNutt and Charles. 
 South Bethlehem, Pa., March 22, 1913. 
 
CALCULUS. 
 
 In the study of phenomena which depend upon conditions 
 which vary in time, that is, upon conditions which vary from 
 instant to instant, it is necessary to direct the attention to what 
 is taking place at an instant; or, in other words, to direct the 
 attention to what takes place during a very short interval of 
 time; or, borrowing a phrase from the photographer, to make 
 a snap-shot, as it were, of the varying conditions. In the study 
 of phenomena which depend upon conditions which vary from 
 point to point in space, the attention must be directed to what 
 takes place in a very small region. 
 
 This paying attention to what takes place during a very short 
 interval of time or in a very small region of space does not refer 
 to observation but to thinking, it is a mathematical method and it 
 is called calculus. Thus the density of a body is its mass divided 
 by its volume. This definition, and indeed the idea of density, 
 appHes only to a homogeneous substance. To apply the idea 
 of density to a non-homogeneous substance one must think of a 
 very small portion of the substance. The same thing is true 
 of every measurable property of a substance. Thus the idea of 
 elastic strain as a measurable effect is easily established when 
 every part of a body is similarly strained as in the case of a 
 stretched rod, but to apply this precise idea of strain to a bent 
 beam or to a twisted rod one must think of a small portion of 
 the beam or rod. 
 
 Two distinct methods are involved in the directing of the 
 attention to what takes place during very short intervals of time 
 or in very small regions of space, as follows: 
 
 (a) The meth gdj:iljii^[srmtial cakiihisv - A phenomenon may be 
 prescribed as a pure assumption, and the successive instantaneous 
 
viii CALCULUS. 
 
 aspects may then be derived from this prescription. Thus we 
 may prescribe uniform motion of a particle in a circular path, 
 and then determine the acceleration of the particle at each instant. 
 
 Or, a condition in space may be prescribed as a pure assumption, 
 and the minute aspects may then be derived from this prescrip- 
 tion. Thus we may prescribe the distribution of temperature 
 along a rod, and then determine the temperature gradient at 
 each point from this prescription. 
 
 (6) The method of integral calculus. It frequently happens that 
 we know and can easily formulate the action which takes place 
 at a given instant or in a small region of space. The problem 
 then is to build up an idea of the result of this action throughout 
 a finite interval of time or throughout a finite region of space. 
 Thus the acceleration of a body may be known at each instant, 
 and from this knowledge we may find the velocity gained and 
 the distance traveled in a finite interval of time. Or, consider a 
 disk rotating at a given speed. It is easy to establish a formula 
 for the energy of a very small particle of this rotating disk, and 
 it is then possible to derive an expression for the energy of the 
 entire disk. 
 
 **% 
 
TABLE OF CONTENTS. 
 
 PAGE 
 
 Preface and Introduction v-x 
 
 Chapter I. A General Survey of Differential and 
 
 Integral Calculus 1-37 
 
 Chapter II. Formulas for Differentiation and Inte- 
 gration 38r 73 
 
 Chapter III. Integration 74- 86 
 
 Chapter IV. Partial Differentiation and Integration 87-108 
 
 Chapter V. Miscellaneous Applications of Calculus 109-152 
 
 Chapter VI. Expansions in Series 153-167 
 
 Chapter VII. Some Ordinary Differential Equations 168-189 
 
 Chapter VIII. The Partial Differential Equation of 
 
 Wave Motion 190-209 
 
 Chapter IX. Vector Analysis 210-253 
 
 Appendix A. Problems 1-20 
 
 Appendix B. Table of Integrals 21- 28 
 
 Appendix C. Some Important Books on Mathematical 
 
 Theory 29- 38 
 
 ix 
 
" In view of the acknowledged difficulty of calculus 
 
 \the student must be willing to stop in his course 
 
 luntil he can form exact notions and acquire precise 
 
 ndeas.^' 
 
 Augustus De Morgan. 
 
CHAPTER I. 
 
 GENERAL SURVEY OF DIFFERENTIAL AND INTEGRAL 
 CALCULUS. 
 
 1. Constant quantities and variable quantities. — Elementa ry 
 
 algebra deals only wil^h quantities which do not change in value 
 (constant quantities). For example let x be the quantity which 
 subtracted from 12 leaves a remainder equal to Jx. That is, 
 12 — a; = ^x, whence x = S. The quantity x has the same 
 value throughout this discussion and it is therefore called a 
 constant quantity. Calculus is a branch of algebra and it deals 
 not only with constant quantities but also and especially with 
 quantities which change in value {variable quantities). Thus one 
 of the simplest problems of calculus is to consider how rapidly x^ 
 grows in value when x is imagined to grow at a definite rate. 
 
 Variable quantities fall intQjthree^ fajxl^jd i s t i nct .classES,. naiaely : 
 (a) Quantities which vary in time, (h) quantities which vary in 
 space, and (c) quantities which are arbitrarily assumed to vary. 
 An example of the first class is the varying temperature of the air 
 at a given place as the seasons come and go. An example of the 
 second class is the varying temperature of the air from place to 
 place at a given time. 
 
 Variables of the first and second classes are sometimes called 
 natural variables to distinguish them from variables of the third 
 class, examples of which are considered in Arts. 10 and 11. 
 
 2. Value at a given instant. Value at a given point. — Let y 
 
 represent the amount of water in a pail into which a small stream 
 of water is flowing. Evidently y is sl changing quantity. If the 
 flow of water were to be shut off at any given instant there would 
 be a definite quantity of water left in the pail, and, therefore, there 
 is a definite amount of water in the pail at each instant even while 
 2 1 
 
2 /'V r -' :.•:».• ^ . ■ CALCULUS. 
 
 the inflow continues. Any changing quantity y has a definite 
 value at each instant* 
 
 A quantity which varies in space has a definite value at each 
 point.* Thus everyone understands that there is a definite tem- 
 perature at Philadelphia although it may be different from the 
 temperature at New York or Washington. An iron rod with one 
 end in the fire has a definite temperature at each point. 
 
 3. Increments and decrements. — It is frequently desirable to 
 consider an increase (or a decrease) in the value of a variable 
 quantity. Such an increase is called an increment, A decrease is 
 called a decrement. For example, let y he & variable quantity. 
 An increment of y is usually represented by the symbol Aj/. 
 When Ay is negative it is a decrement. This symbol Ay does 
 NO T mean A multiplied by y, it is a single symbol, and it may 
 be read increment-of-y or delta-y. The latter is preferable because 
 of its brevity. The Greek letter A used as a prefix always 
 means increment of. 
 
 4. Rate of change of a quantity which varies in time. — Consider 
 a pail into which water is flowing in a small stream. Let y be 
 the amount of water in the pail. Then y is evidently a changing 
 quantity. Consider the amount of water that flows into the pail 
 during a short interval of time At; this amount of water is evi- 
 dently the increment of y during the short intervaj.-of timCj and 
 
 Ay 
 it is to be represented by the symbol Ay. The quotient -j 
 
 At 
 
 is called the average rate of change of y duHng_ihe -interval — ^."f^ 
 
 If the inflowing stream of water is constant, say, always 2 cubic 
 
 inches of water per second, then if At is chosen smaller and smaller, 
 
 the value of Ay will be proportionately smaller and smaller, and 
 
 Ay 
 the value of the quotient -^ will be exactly 2 cubic inches 
 
 per second whatever the duration or length of the time-interval A^. 
 
 * This may not be true of a discontinuous variable. 
 
 t Thus if y is the amount of money a man has saved, then, if he saves 
 $20 more during 30 days, we have Ay = $20 and At = 30 days, and the aver- 
 age rate of saving during the 30 days will be 66 1 cents per day. 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 3 
 
 If the inflowing stream is variable, then the value of the quotient 
 
 ~ will not be the same for different values of Ai, but the amount 
 
 of water which flows into the pail during a very short interval of time 
 will be very nearly proportional to the interval. For example, 
 a certain amount of water W flows into the pail during a particular 
 one-thousandth of a second. Imagine this particular one-thou- 
 sandth of a second to be divided in halves; then only a very Httle 
 more than ^W will flow into the pail during one of the half- 
 thousandths of a second and a very httle less than |TF will flow 
 into the pail during the other half-thousandth of a second. // 
 a shorter and shorter interval of time he taken the amount of inflow 
 of water will be more and more nearly in EXACT proportion 
 to the duration of the interval. This means that the quotient 
 
 -Tj- approaches a definite limiting value as At and Ay both ap- 
 
 All 
 proach zero; and this limiting value of -rf is called the rate of 
 
 change of y at a certain instant or the instantaneous rate of change 
 
 of y. 
 
 All dii 
 
 The limiting value of -^ is always represented by -^ which 
 
 means the rate of change of y at a given instant. 
 
 Nearly everyone falls into the idea that such an expression as 
 10 feet per second means 10 feet of actual movement in an actual 
 second of time, but a body moving at a velocity of 10 feet per 
 second might not continue to move for a whole second or its 
 velocity might change before a whole second has clasped. Three 
 cubic inches per second is the same rate of inflow of water into a 
 pail as 2,070 cubic yards per year, but to specify the rate of inflow 
 in cubic yards per year does not mean that a whole cubic yard of 
 water flows into the pail nor does it mean that the inflow continues 
 for a year. A man does not need to work for a whole month to 
 earn money at the rate of $60.00 per month, nor for a whole day 
 to earn money at the rate of $2.00 per day. A falling body has 
 
4 CALCULUS. 
 
 a velocity of 19,130,000 miles per century after it has been falling 
 for one second, but to specify its velocity in miles per century 
 does not mean that it falls as far as a mile nor that it continues to 
 fall for a century ! The units of length and time which appear in the 
 specification of a velocity are completely swallowed up, as it were, 
 in the idea of velocity; and the same thing is true of the specifica- 
 tion of any rate. 
 
 5. Continuous variables and discontinuous variables. — A quan- 
 tity which changes by sudden jumps is called a discontinuous 
 variable. For example the amount of money one has is a dis- 
 continuous variable, because a debt is made on the instant that 
 one decides to accept a purchase; that is to say, money is spent in 
 lumps, and a lump of money is spent during an indefinitely short 
 time. The amount of money spent during an interval of time does 
 NOT become more and more nearly proportional to the interval as the 
 interval grows shorter and shorter, and consequently it is meaningless 
 to speak of the rate of spending money at a given instant. If y is 
 a discontinuous variable and if the time-interval A^ happens to 
 include a jump in the value of y, then the value of Ay remains 
 
 finite as At approaches zero and the quotient -r^ approaches in- 
 finity. The rate of change of a discontinuous variable at a given 
 instant is unthinkable. 
 
 The amount of water in a pail, as considered in Art. 4, is an 
 example of what is called a continuous variable. If y is a con- 
 tinuous variable and if Ay is the increment of y during the time- 
 interval At, then Ay becomes more and more nearly proportional 
 
 to At as A^ approaches zero; that is to say, the quotient -J 
 
 a,pproaches a definite limiting value as At approaches zero, and this 
 
 limiting value of -~ is called the instantaneous rate of change 
 zir 
 
 of y. 
 
 In the study of calculus we deal almost exclusively with con- 
 tinuous variables. 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 5 
 
 6. Example showing determination of instantaneous rate of 
 change. — Let us assume that the amount of water in the pail in 
 the discussion of Art. 4 is proportional to the square of the time 
 t which has elapsed since a chosen instant. Then we may write 
 
 y = kt' (1) 
 
 where /: is a constant. A moment later t has increased and of 
 course y has increased also. Let us represent the new value of t 
 by t + At and the new value of y by y -\- Ay. Then, since 
 equation (1) is assumed to be true for all values of t and y, we 
 
 have 
 
 y + Ay = k{t + Aty 
 or 
 
 y + Ay = kt^ + 2kt'At + h{AtY (2) 
 
 Subtracting equation (1) from equation (2) member by member, 
 
 we have: 
 
 Ay = 2kt'At -f- k{AtY (3) 
 
 Whence, dividing both members by Aty we have: 
 
 ^ ^2kt + k'At (4) 
 
 Now it is evident that k-At approaches zero as At approaches 
 
 Av 
 zero, and therefore the limiting value of -ry (as At approaches 
 
 dv Av 
 
 zero) is 2kt. Hence, writing -~ for tY.e limiting value of -r^, 
 
 we have : 
 
 | = 2« (5) 
 
 That is, the rate of change of y is at each instant equal to 2kt. 
 
 Observation or thinking; which? — The discussion, in Art. 4, 
 of the rate of increase of the amount of water in a pail presents a 
 serious difl&culty. How is one to know the increment of water 
 which occurs during an indefinitely short interval of time? One 
 
6 CALCULUS. 
 
 certainly cannot measure it. Observation and measurement are 
 entirely usele!<s in the consideration of an indefinitely small change 
 which takes place during an indefinitely short interval of time. 
 One cannot observe such things, one can only think about them. 
 Indeed the whole matter at present under discussion is a matter 
 of mathematical reasoning. 
 
 7. Gradient of a quantity which varies in space. — Consider an 
 iron rod which stands with one end in a fire. Then the tempera- 
 ture of the rod varies from point to point along the rod. Consider 
 two points very near together, let AT be their difference in 
 temperature and let Ax be their distance apart. The quotient 
 
 AT 
 
 — - approaches a definite limiting value when Ax is chosen smaller 
 
 i\X 
 
 AT 
 and smaller, and this limiting value of — is called the temperature 
 
 gradient at the point (degrees per inch). In this example T 
 varies continuously along the rod, it does not vary by jumps, other- 
 wise the gradient of T at a point would be unthinkable. 
 
 AT 
 The limiting value of the quotient — is always represented by 
 
 m^X 
 
 dT 
 
 j-j and it means the gradient of T at a given point. 
 
 Another quantity which varies from point to point in space is 
 the pressure of the atmosphere; the higher one goes the less the 
 pressure. Consider two points A and B, one of which is at a 
 distance Ax above the other, and let Ap be the amount by which 
 
 the pressure at A exceeds the pressure at B. Then -—^ the 
 
 Av 
 limiting vahie of the quotient -^, is called the pressure gradient at 
 
 A.* The pressure gradient of the atmosphere at sea level is 
 usually about 2.3 pounds-per-square-inch per mile. 
 
 The word gradient as used in the above discussion comes from 
 the word grade meaning the steepness of a slope. Thus if the 
 
 * Or at B) the two points A and B approach coincidence as Ax ap- 
 proaches zero. 
 
 iJ&M^^ 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 7 
 
 slope of a hill at a given place rises A^ feet in a horizontal distance 
 
 of Aa; feet, then the limiting value of — when Ax is made smaller 
 
 and smaller is called the grade of the hill at the given place. 
 The grade of a hill at a point might be one foot of rise per ten 
 feet of horizontal distance, or ten feet of rise per hundred feet of 
 horizontal distance, or 10 per cent, as it is usually expressed. 
 It is evident that 10 feet of rise per 100 feet of horizontal distance 
 does not mean necessarily 10 feet of actual rise in 100 feet of hori- 
 zontal distance, because the slope may not be 100 feet long and 
 the grade may vary from point to point. 
 
 Similarly a pressure gradient of 2.3 pounds-per-square-inch per 
 mile, which is the upward gradient of atmospheric pressure at sea 
 level, does not mean that the atmospheric pressure at a point one 
 mile above sea level is 2.3 pounds per square inch less than at sea 
 level, because the pressure gradient becomes less and less at points 
 higher and higher above the sea. The units which appear in the 
 specification of a grade or a gradient are completely swallowed up, 
 as it were, in the idea of grade or gradient just as the units which 
 appear in the specification of a rate are swallowed up in the idea of 
 rate. 
 
 8. Example showing the determination of a gradient at a point. 
 
 — Consider a metal bar AB, Fig. 1. Suppose the temperature 
 
 e6td T h&i 
 
 k X — -^ 
 
 Fig. 1. 
 
 of the bar to be zero at the end A, and let us assume the tempera- 
 ture T at the point p to be: 
 
 T = hx" (1) 
 
 where x is the distance from A to p, and A; is a constant. 
 Under these conditions let it be required to find the temperature 
 gradient at the point p. Now equation (1) is true for every value 
 of X and T. Therefore writing x -\- Ax for x and writing 
 
8 CALCULUS. 
 
 T + AT for T we have 
 
 T + AT = ik(x + Ax)* (2) 
 
 or 
 
 T + AT = A:^* + 2fcc-Aa; + A:(Ax)* (3) 
 
 Subtracting equation (1) from equation (3) member by member, 
 
 we have: 
 
 AT = 2ib:-A:t + A:(Ax)* (4) 
 
 whence 
 
 ^ = 2A:x + k-^x (5) 
 
 Now A; -Ax approaches zero as Ax approaches zero, and it is 
 
 AT 
 therefore evident that the Umiting value of — is 2kx, Therefore, 
 
 /iX 
 
 writing -r- for the limiting value of -— , we have: 
 
 g = 2fcx (6) 
 
 It is evident from this equation that the temperature gradient is 
 
 zero at the end A of the bar where x = 0; and it is evident that 
 
 the temperature gradient grows greater and greater (steeper and 
 
 steeper) at points farther and farther from A where x is larger 
 
 . and larger. 
 J PROBLEMS. 
 
 1. The amount of water in a pail is assumed to be proportional 
 to the square of the time which has clasped since a chosen instant, 
 according to equation (1) of Art. 6. It is evident from this equa- 
 tion that the amount of water in the pail is zero when i = 0. 
 After the lapse of 20 seconds there is 600 cubic inches of water in 
 the pail. Find the value of k and state the unit in terms of which 
 k is expressed. Ans. 1.5 cubic inches per second per second. 
 
 2. Using the value of k from problem 1, find the rate at which 
 water flows into the pail, (a) when t = 5 seconds, and (6) when 
 t = 20 seconds. Ans. (a) 15 cubic inches per second, (6) 60 
 cubic inches per second. 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 9 
 
 3. Find the average rate of inflow of water into the pail during 
 the time from t =^ to t = 20 seconds. Ans. 30 cubic inches per 
 second. 
 
 4. The temperature of an iron rod increases along the rod in 
 accordance with equation (1) of Art. 8. At a point 20 inches from 
 the end of the rod (x = 20 inches), the value of jT is 240° C. ; 
 Find the value of k and state the unit in terms of which it is 
 expressed. Ans. 0.6 degree per inch per inch. 
 
 5. Using the value of k from problem 4, find the temperature 
 gradient (a) at the point where x = 5 inches and (b) at the point 
 where x = 20 inches. Ans. (a) ^ degrees per inch, (6) 24 d^rees 
 p>er inch. 
 
 6. (o) Find the average temperature gradient along the iron 
 rod of problem 4 between x = and x = 20 inches. (6) Find 
 the average temperature gradient between x = 10 inches and 
 X = 11 inches. Ans. (a) 12 d^;rees per inch, (6) 12.6 degrees per 
 inch. 
 
 9. Arbitrary variations. — One of the most important things in 
 calculus is to consider how fast such an expression as x* or x* or 
 sin X changes when x is arbitrarily assumed to grow steadily in 
 value. Such -variations may. be called arbitrary variations. The 
 ver>' great importance of arbitrary- variations will be understood 
 when the student reaches Arts. 22 and 23. 
 
 10. A purely algebraic example of an arbitrary variation. — 
 Having given the equation 
 
 y = aa^ (1) 
 
 let it be required to find the limiting value of the quotient — 
 
 when the arbitrary increment of x, namely Ax, approaches zero. 
 Writing y -{- Ay for y and writing x + Ax for x in equation 
 (1) we have 
 
 y + Ay = a(x + Ax)* (2) 
 
 or 
 
 y + Ay = ax» 4- 2axAx + a(Ax)* (3) 
 
 whence, subtracting equation (1) from equation (3) member by 
 
10 CALCULUS. 
 
 memoer, we have 
 
 Ai/ = 2aa:-Aa; + a(Ax)« (4) 
 
 and dividing both members by Ax we have 
 
 ^ = 2ax + a-Ax (6) 
 
 Ax 
 
 But a 'Ax approaches zero when Ax approaches zero, and there- 
 fore it is evident that the limiting value of -r^ is 2ax. The 
 
 Ax 
 
 . Ay dv 
 
 limitmg value of -r^ is always represented by ~. Therefore we 
 i^c dx 
 
 have: 
 
 g = ^ (6) 
 
 11. Two more purely algebraic examples of arbitrary variations. 
 
 — (a) Having given the equation 
 
 y = as? (1) 
 
 let it be required to find the limiting value of -^ when the arbi- 
 trary increment of x approaches zero. Writing y -\- Ay for y 
 and writing x + Ax for a; in equation (1) we have 
 
 y-^Ay = a{x + AxY (2) 
 
 or 
 
 y-\- Ay ^ a3? -\- Sax^-Ax + 3ax(Ax)2 + a{^xY (3) 
 
 whence, subtracting equation (1) from equation (3) member by 
 member, we have 
 
 Ay = 3ax2.Ax + 3ax(Ax)2 + a(Ax)« (4) 
 
 or 
 
 ^ = 3ax2 + 3ax-Ax + a{Axy (5) 
 
 Now 3ax-Ax and a{Axy both approach zero when Ax approaches 
 
 Ay 
 eero, and therefore it is evident that the limiting value of ~ is 
 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 11 
 
 Sax^; that is, 
 
 # _ 
 
 dx 
 (6) Having the equation 
 
 3ax2 (6) 
 
 y = l (7) 
 
 let it be required to find the limiting value of ~ when the arbi- 
 trary increment of x approaches zero. Writing y -\- Ay for y 
 and writing x + Ax for x in equation (7) we have 
 
 y + ^y-^^ («) 
 
 Subtracting equation (7) from equation (8) member by member, 
 we have 
 
 A2/ = -^ - ? (9) 
 
 ^ x + Ax X 
 
 Reducing the two fractions to a common denominator we have: 
 
 
 whence 
 
 Ay 
 
 (11) 
 
 Ax x^ -\- x-Ax 
 but x^ -\- X'Ax approaches x'^ as its limit when Ax approaches 
 zero, and therefore the limiting value of ■— is ^ ; that is : 
 
 dx X' ^^^^ 
 
 dy 
 The derivative — is sometimes called the rate of change of y 
 
 with respect to x. 
 
12 CALCULUS. 
 
 PROBLEMS. 
 
 The following problems can be solved by the methods employed 
 in Arts. 10 and 11. In each case £uid the rate of change of y 
 with respect to x. 
 
 l.y^ax, g = a. 
 
 2. 2/ = ax», ^ = 5ax^, 
 
 3. y = ms? + w, -p = 27nx. 
 
 4. 2/ = oar^ + 6a; + c, ^ = 2ax + 6. 
 
 5. ^ = ax^ + 6x2, ^ = 3ac' + 2&a;. 
 - a di/ 2a 
 
 _ g dy _ — a(2hx + 1) 
 
 **^~6a;'» + x' 5i 
 
 n 10.2/ = =^, * 
 
 V '^ 6 — x' da; 
 
 11. y ^ 2X^-5 ^ 
 
 12. 1/ 
 
 X dy ^ 
 
 (x-l)2' d!x {x-iy 
 
 (6X2 + 3.)2 
 
 9 
 
 (x + sr 
 
 ab 
 
 (6 - x)2- 
 
 9 
 
 (x + 2)2- 
 
 x + 1 
 
 12. Functions. — When a spring is subjected to a stretching force 
 Ff a certain elongation e is produced as shown in Fig. 2. For 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 
 
 13 
 
 every value of e there is a definite corresponding value of F. When 
 two quantities are associated in this way either one is said to be a 
 
 un stretched spring 
 
 Fig. 2. 
 The elongation of the spring is a function of the stretching force. 
 
 function of the other. Thus the elongation e produced by a 
 given force F in Fig. 2 is a function of F; or the force F re- 
 quired to produce a given elongation e is a function of e. The 
 
 1 
 
 1 
 
 K -^ i j| 
 
 
 w 
 
 1 
 1 
 1 
 1 
 
 area^lw 
 
 Fig. 3. 
 
 The area of a square is 
 function of length of side. 
 
 Fig. 4. 
 
 The area of a rectangle is a 
 function of length and width. 
 
 meaning of the word function is further illustrated by Figs. 3, 4, 5, 
 6, 7 and 8. 
 
 13. Dependent and independent variables. — If I in Fig. 3 is 
 assumed to be arbitrarily variable it is called an independent 
 variable, and the area of the square is called a dependent variable 
 because its value is determined or fixed when the value of I is 
 given. The quantity of air pumped into the cylinder in Fig. 7 
 may be as much as one pleases, the temperature of the cylinder 
 
14 
 
 . CALCULUS. 
 
 and air may be increased or decreased at will, and the volume of 
 the cylinder may be changed by moving the piston. Therefore 
 
 tr- thermometer 
 
 ^ 
 
 "l-cloaed • 
 '. : vessel •'•' 
 containing 
 ': air . • - 
 
 Fig. 5. 
 
 The pressure of the air is a function of 
 the amount of air pumped into the tire. 
 
 Fig. 6. 
 
 rhe pressure of the air in the 
 vessel is a function of the amount 
 of au- pumped into the vessel and of 
 the temperature. 
 
 the three quantities, (a) quantity of air, (6) temperature of the air, 
 and (c) volume of space occupied by the air are called independent 
 variables, and the pressure of the air is called a dependent variable, 
 
 thermometer 
 
 pressure 
 gauge 
 
 piston 
 
 cylinder 
 containing \ air 
 
 SiLnp 
 
 Fig. 7. 
 
 The pressure of the air in the cylinder is a function of (a) the amount of 
 air pumped into the cylinder, (6) the temperature, and (c) the volume of 
 the space occupied by the air. 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 
 
 15 
 
 given curve 
 
 because its value is fixed or determined when the quantities (a), 
 (6) and (c) are given. 
 
 From this discussion it is evident that a dependent variable 
 may be a function of one, two, three or more independent variables. 
 
 14. Natural functions and artificial functions. — We have hereto- 
 fore taken the point of view that two quantities must be connected 
 physically if either is to be a function of the other. Such a function 
 may be called a natural func- 
 tion. It is, however, of very 
 great importance to consider func- 
 tional relations which grow out 
 of Or are expressed by algebraic 
 equations. Such functions may 
 be called artificial functions. Some 
 idea of the importance of artificial 
 functions may be obtained by 
 looking back at Articles 4 and 6, 
 and 7 and 8. Natural functions 
 can be in many cases expressed 
 algebraically^; indeed no function can be handled in calculus unless 
 the function is expressed algebraically. 
 
 15. Tabulation of a function. — ^When one quantity is a function 
 of another it is often convenient to express the functional relation 
 by means of a table giving pairs of corresponding values of the 
 two quantities. An ordinary table of logarithms is such a table, 
 a table of sines or cosines is such a table, the mortality table* 
 used in life insurance is such a table. 
 
 When a functional relation is determined by experiment the 
 
 * If one were to consider all men in the United States who are now forty 
 years of age and keep a future record of their deaths one would find that their 
 average age at death would be, say, sixty-five years. ^This is expressed by 
 saying that the expectancy of fife at forty years is twenty-five years, because 
 on the average a man forty years old may expect to live twenty-five years. 
 The expectancy of life is, of course, less and less with increasing age, and there 
 is a definite expectancy of life corresponding to each age. I That is, expectancy 
 of life is a function of age. 7 
 
 Fig. 8. 
 
 The ordinate of a point on a curve is 
 a function of the abscissa of the point. 
 
16 
 
 CALCULUS. 
 
 Fig. 9. 
 
 observed quantities are always 
 arranged in tabular form. For 
 example Fig. 9 represents an am- 
 meter arranged to measure the 
 electric current flowing through a 
 coil of wire, and a spring scale ar- 
 ranged to measure the force with 
 which an iron plunger is pulled into 
 the coil; and the accompanying 
 table shows a series of observed 
 values of current in amperes and the 
 corresponding pulls in pounds. 
 
 OBSERVED RELATION BETWEEN CURRENT AND PULL IN 
 
 FIG. 9. 
 
 Current in 
 
 Pull on Plunger 
 
 Amperes. 
 
 in Pounds. 
 
 1.0 
 
 3.9 
 
 2.0 
 
 23.6 
 
 S.0 
 
 40.0 
 
 10 
 
 48.1 
 
 a. 
 
 16. Graphical representation of a fimction. — The relation be- 
 tween the pull of the spring 
 and the reading of the am- 
 meter in Fig. 9, as shown in 
 the table of Art. 15, may be 
 represented graphically by 
 a curve of which the ab- 
 scissas represent ammeter 
 readings and of which the 
 ordinates represent corre- 
 sponding readings of the 
 spring scale. Such a curve 
 is shown in Fig. 10. In the 
 same way an algebraic func- 
 tion may be represented graphically, 
 represents the function y = x^. 
 
 50 
 40 
 30 
 
 20 
 
 10 
 
 
 
 
 
 ^ 
 
 ^ 
 
 
 
 
 
 
 /^ 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 / 
 
 / 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 ^ 
 
 [/ 
 
 
 
 
 
 
 ' 
 
 345 
 
 amperes 
 
 Fig. 10. 
 Thus the curve in Fig. 11 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 
 
 17 
 
 17. Derivative of a function. 
 
 let A?/ be the increment of 
 y due to an arbitrary incre- 
 ment of X. Then the lim- 
 iting value of the quotient 
 
 -^ as Ax approaches zero 
 
 is called the derivative of y 
 mth respect to x or the rate 
 of change of y with respect 
 to X* 
 
 When one wishes to speak 
 in general terms of any func- 
 tion of X and its derivative, 
 the function may be repre- 
 sented by f{x) or by <^(x) 
 and the derivative of the 
 
 -Let 2/ be a function of x and 
 
 I X'OxiiB 
 
 Fig. 11. 
 
 function with respect to x may be represented hy f'{x) or by <l>'{x). 
 
 Meaning of a derivative. — Let i/ be a function of x and let x 
 
 dy 
 be assumed to grow steadily at a definite rate, then y grows -r- 
 
 times as fast as x at each instant. For example, let y = x^; then 
 
 dif 
 
 ~ = 2x, according to Art. 10, and the following relations exist: 
 
 y increases 2x{= 20) times as fast as x 
 
 value X — 10, 
 y increases 2a; (= 22) times as fast as x 
 
 value X = 11, 
 y increases 2a: (= 24) times as fast as x 
 
 value X — 12, 
 ^ etc., etc., 
 
 while ; 
 while ; 
 while ; 
 
 As another example, let y = -; then -^ 
 
 X dx 
 
 is passing through the 
 is passing through the 
 is passing through the 
 
 etc. 
 
 1_ 
 
 x^' 
 
 according to 
 
 Art. 11, and the following relations exist: 
 
 * Sometimes also called the differential coefficient of y with respect to x. 
 
18 
 
 CALCULUS. 
 
 y decreases ^ ( ~ 4 ) as fast as x increases when x is passing through the 
 
 value 2, 
 
 y decreases -j ( = qJ as fast as x increases when x is passing through the 
 
 value 3, 
 
 etc., etc., etc. --> 
 
 A negative value of ^ shows that y decreases as x increases. 
 ax 
 
 The derivative of a function of x is itself a function of x. 
 
 — If y = dx^. then ^ = 2ax, according to Art. 10. In this case 
 ax 
 
 dti 
 it is evident that -p is a function of x because it is equal to 2ax 
 
 and it has a definite value for every value of x. If /(x) represents 
 any function of x, then its derivative }'{x) is in general a function 
 of X also. 
 
 18. Graphical representation of derivative. Slope of a curve 
 at a point. — Consider the curve CCy Fig. 12. This curve represents 
 
 Fig. 12. 
 
 a definite function, that is, the ordinate ?/ is a definite function of 
 the abscissa x, the point p being anywhere on the curve. If x 
 is increased by the amount Ax, then y will be increased by the 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 19 
 
 amount A^ as shown, and the ratio -f- is equal to the tangent of 
 
 the angle a. 
 
 Draw the hne tt touching the curve at p as shown. As Aa; 
 approaches zero the angle a becomes more and more nearly 
 equal to the angle 6, in fact 6 is the limiting value of a, and, 
 
 of course, the limiting value of -r^ is ~. Therefore, since 
 
 tan a = — ^ , we must have tan ^ = -r^- That is, the function y 
 
 dti 
 being represented by the ordinates of a curve, the derivative -r- is 
 
 represented by the steepness or grade of the curve at each point. 
 
 19. Derivative notation and differential notation. — Consider the 
 function 
 
 y = ax^ (1) 
 
 of which the derivative is: 
 
 I = ^ (2) 
 
 dv 
 Heretofore we have looked upon ^- as a single symbol the mean- 
 ing of which is explained in Art. 17. Convenience of notation 
 sometimes makes it desirable to write equation (2) thus: 
 
 dy = 2ax'dx (3) 
 
 This equation expresses the limiting relation between the incre- 
 ments A^ and Ax, that is to say, the relation which is approached 
 as Ay and Ax both approach zero; dy is called the differential of 
 y (that is to say, the differential of ax^, because y here stands for 
 ax^), and dx is called the differential of x. These two differentials 
 may be thought of as indefinitely small increments of y and x 
 respectively.* 
 
 * When y = ax^ we have 
 
 Ay = 2ax • Ax + ^(Ax)^ 
 
 according to Art. 10. That is to say, the increment of y is not equal to 
 2ax times the increment of x unless both increments are indefinitely smaU. 
 
20 
 
 CALCULUS. 
 
 Or, if one wishes to think of dy and dx as having a physical 
 
 dt 
 
 meaning they may be thought of as abbreviated expressions for 
 
 dx 
 and -rr respectively, that is, dy may be thought of as the rate of 
 
 change of y and dx may be thought of as the rate of change of x, 
 
 PROBLEMS. 
 
 1. A pail is 10 inches in diameter, so that the volume of water in 
 
 the pail is y = —^ Xj where x is the depth of the water in the 
 
 pail. Find how fast water must be poured into the pail to cause 
 the water level to rise at a velocity of 4 inches per second. Ans. 
 IOOtt cubic inches per second. 
 
 */t2. Water flows at a constant rate of 30 cubic inches per second 
 into a metal cone of which the dimensions are as shown in Fig. p2. 
 Find the velocity at which the water level rises 
 (a) when x = b inches, and (6) when a; = 15 
 inches. Ans. (a) 2.715 inches per second, (6) 
 0.302 inch per second. €m t'c-- ^ 
 
 3. The sides of a square are growing at the 
 rate of 5 inches per second. Find the rate of 
 growth of the area of ^ the square (a) when 
 the sides of the square are 10 inches and (6) 
 when the sides of the square are 20 inches. 
 Ans. (a) 100 square inches per second, (6) 200 
 square inches per second. 
 
 4. The radius of a circle is growing at the 
 rate of 5 inches per second. Find the rate of 
 growth of the area of the circle (a) when the 
 
 radius is 10 inches and (h) when the radius is 20 inches. Ans. (a) 
 lOOir square inches per second, (h) 2007r square inches per second. 
 
 5. The edges of a cube are growing at the rate of 5 inches per 
 second. Find the rate of growth of the volume of the cube when 
 the edges are 10 inches long. Ans. 1,500 cubic inches per second. 
 
 Fig. p2. 
 
DIFFERENTIAL i^fe INTEGRAL CALCULUS. 21 
 
 m 
 
 9 of iR] 
 
 - 6. Air is blown into a soapi«3ble at the rate of 10 cubic inches 
 per second. Find the rate of Slrease of the radius of the bubble 
 when the radius has the values (^ 1 inch and (6) 4 inches. Ans. 
 (a) 0.795 inch per second, (6) 0.050 inch per second. 
 
 -7. A man 6 feet tall walks at a speed of 4 miles per hour under 
 a lamp which is 10 feetfrgmthe ground. Find how fast ^€-tip • 
 ef- the man's shadow wa^eS-mien the horizontal distance from 
 the lamp to the man is 20 feet. Ans. 6 miles per hour. 
 
 8. The two sides of a right angled triangle are 40 inches and x^ 
 and the hypothenuse is y. The side z is growing at the rate of 
 5 inches per second. How fast is y growing when x = 30 inches? 
 Ans. 3 inches per second. 
 
 9. The two sides of a right angled triangle are x and y, and 
 the hypothenuse is 50 inches. The side x is growing at the rate 
 of 4 inches per second. How fast is y growing (a) when x = 30 
 inches, and (6) when x = 40 inches. Ans. (a) 3 inches per 
 second. (6) 5.33 inches per second. 
 
 10. The observed temperatures of a vessel of cooling water after 
 1 minute, after 2 minutes, and so forth are as follows: 
 
 Elapsed Time in Minutes, ObserTed Temperaturea, 
 
 t T. 
 
 92.0 
 
 1 85.3 
 
 2 79.5 
 
 3 74.5 
 5 67.0 
 7 60.5 
 
 10 53.5 
 
 15 45.0 
 
 20 39.5 
 
 Plot these values of t and T, using the best grade of squared 
 paper, draw a smooth curve through the plotted points, draw 
 tangents to the curve at the points corresponding to (a) T = 80°,, 
 (h) T = 65°, and (c) T = 50°, and determine the rate of decrease 
 of the temperature at each point by measuring the intercepts; 
 of these tangents on the axes of reference. 
 
22 CALCULUS. 
 
 11. Find for what values of x the function ?/ = Gx^ — 12x is 
 an increasing function and for what values of x it is a decreasing 
 function. Ans. 2/ is an increasing function when x is greater 
 than 1, and a decreasing function when x is less than 1. 
 
 12. Plot the values of x and y^ where y = x^, using the best 
 grade of squared paper, draw a smooth curve through the plotted 
 points, draw tangents to the curve at the points corresponding to 
 X = 1, x = 2 and x = 3, and determine the corresponding 
 
 values of -p by measuring the intercepts of the tangents on the 
 
 dii 
 axes of reference. Compare the values of -— thus found with 
 
 dxi 
 the values as calculated from the formula -— = Sx^. 
 
 ox 
 
 13. Plot the values of x and y where y = logio x. Take the 
 
 values of logio x from an ordinary table of logarithms and use the 
 
 following values of x: 2, 3, 4, 5, 6, 7, 8, 9 and 10. Draw a smooth 
 
 curve through the plotted points, draw tangents to the curve at 
 
 the points corresponding to x = 4, x = 6 and x = 8, and deter- 
 
 dv 
 mine the corresponding values of -^ by measuring the intercepts 
 
 of these tangents on the axes of reference. Compare the values of 
 
 dti 
 
 -^ thus found with 
 
 dx 
 
 dy ^ logioe ^ 0.4343 ^ 
 
 dx X X ' 
 
 20. Determination of limiting relation between Ay and Ax by 
 
 Aiy 
 consideration of infinitesimals. — The value of -~ is found in 
 
 Ax 
 
 Arts. 10 and 11 as the sum of a finite qvxintity and a quantity which 
 
 approaches zero as Ax approaches zero. Thus when y = as? we 
 
 found that -r^ = Zax^ + 3ax-Ax + a(Ax)2, in which Sax^ is a 
 
 ZaX 
 
 finite quantity and 3ax-Ax + a(Ax)^ is a quantity which ap- 
 proaches zero as Ax approaches zero. In such a case it is very 
 
 dti 
 
 -^ thus found with the values as calculated by the formula 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 23 
 
 A?/ 
 easy to see what the limiting value of -~- must be. 
 
 In many cases, >oweve% it is desirable to find the limiting rela- 
 tion between ^y and Ax when Ax approaches zero without 
 
 deriving an expression for ~ . For example let y = a^. Then 
 
 from Art. 11 we have: - - - -^ ^ 
 
 _/■' ^ 
 , Ay = Sax^'Ax + Zax{Axy + a{Axy (1) 
 
 Both members of this equation approach zero when Ax approaches 
 zero, and it would therefore seem rather difficult to determine 
 the hmiting relation between Ay and Ax (when Ax approaches 
 zero) from this equation as it stands. But when Ax is made as 
 small as you please, thejM|(Aa;)2 is infinitely smaller than Ax, 
 and {AxY is infi]||||fily sn^^r than (Ax)^. For example let Ax 
 be a millionth of^Mpait^i^K (Ax)^ is a million-millionth of a 
 unit, and (Ax)^ is^3fcpioB;^Blion-millionth of aunit. Therefore 
 the terms 3ax • (Ax)^and ' ^J\x)^ become more and more nearly 
 negligible in comparison with Sax^ • Ax as Ax grows smaller 
 and smaller in equation (1), The limiting relation between Ay 
 and Ax may be found by writing dx and dy for Ax and Ay 
 to indicate that we have proceeded to the limit, and by dropping 
 every term which contains the square or any higher power of 
 Ax (or the square or any higher power of Ay). This gives 
 
 dy = Sax^-c^x (2) 
 
 In this equation dy and dx are as small as you please and they 
 are called infinitesimals; but {dyY and {dxY being infinitely 
 smaller than dy and dx are called infinitesimals of the second 
 order J {dyY and {dxY being infinitely smaller than {dyY and 
 {dxY are called infinitesimals of the third order, and so on. In any 
 differential expression the lowest order infinitesimals, only, are 
 significant; all terms containing higher order infinitesimals as factors 
 may he dropped. 
 
24 
 
 CALCULUS. 
 
 Example 1. — Let it be required to differentiate the expression 
 
 2/2 = ax» (3) 
 
 that is to say, let it be required to determine the relation between 
 Ay and Ax when Ax is as small as you please Writing {y -{• Ay) 
 for y and writing (x + Ax) for x, we have 
 
 y^-\-2yAy-i- {Ay)^ = ax^ + Sax^-Ax + 3ax(Ax)2 + a(Ax)» (4) 
 
 whence, subtracting equation (3) from equation (4) member by 
 member, we have 
 
 2y'Ay -\- (Ay)^ = 3ax2-Ax + 3ax(Ax)2 -f- a{Ax)^ 
 
 (5) 
 
 from which the desired limiting relation is found by dropping 
 second and third order infinitesimals, so that we have: 
 
 2y-dy = dax^-dx 
 
 (6) 
 
 The value of y from equation (3) may be substituted in equation 
 (6), giving: 
 
 2^[a^-dy = Zax^'dx (7) 
 
 which can be simplified by cancellation, giving: 
 dy = j-^ax-dx 
 
 (8) 
 
 This dropping of higher order infinitesimals from differential expres- 
 sions does not lead to merely 
 approximate results, because in 
 every case it is the limiting rela- 
 tion between Ay and Ax which 
 is to be determined, that is, the 
 relation when Ay and Ax are both 
 as small as you please. Equation 
 (8), for example, is rigorously true. 
 
 Example 2. Differential of the 
 arc of a parabola. — The length s of 
 the heavy portion of the curve cc, 
 
'X^' 
 
 DIFFERENTIAL AND INTEGRAL CALCULUS. 
 
 25) 
 
 Fig. 13, from x = b tox = x is a function of x, and from the in- 
 finitesimal triangle whose sides are dx, dy and ds we have 
 
 ds = ^l{dxy + {dyy 
 
 or 
 
 *= >r+ ity-^ 
 
 (9) 
 (10) 
 
 For example suppose the curve cc, Fig. 13, to be a parabola whose 
 equation is: 
 
 y = kx^ (11) 
 
 then 
 
 dy _ 
 
 dx 
 
 = 2kx 
 
 which, substituted in equation 10, gives: 
 
 ds = ^ll-\-4:k^x^'dx 
 
 (12) 
 
 (13) 
 
 21. Functions which have the same derivative. — The curve A, 
 Fig. 14, defines i/' as a function of a;, and the curve B defines y 
 
 y-axiB 
 
 Fig. 14. 
 
 as a function of x. The steepness or grade of curve A is every- 
 where the same as the steepness or grade of curve B, that is, the 
 
 derivative -j- is equal to the derivative — for each value of a;, 
 
26 CALCULUS, 
 
 or expressed as an equation we have 
 
 dx dx 
 
 (1) 
 
 From the figure it is evident that the difference 1/ — y is 
 everywhere equal to the constant quantity C. Consequently 
 when equation (1) is true we have 
 
 y' — y = a. constant (2) 
 
 That is to say, two functions whose derivatives are equal have a 
 constant difference. 
 
 Examples. — ^When two men save money at the same rate there 
 is a constant difference between the amounts they have saved; 
 or, if they start even, their savings are equal. When two trains 
 travel at the same speed they remain at a constant distance apart. 
 The two functions ax^ and ax^ + h have the same derivative 
 with respect to x, h being a constant. 
 
 22. Example showing the use of calculus. Work required to 
 
 stretch a spring. — Let W be the work required to stretch a spring 
 
 from condition A to condition B, Fig. 2. It is evident that W 
 
 is a function of e, and it is desired to find an algebraic expression 
 
 for this function; that is, it is desired to find an equation expressing 
 
 W in terms of e. To do this the first step is to find an expression for 
 
 dW 
 the derivative -t~ by considering the amount of work ATF that 
 
 "^ "'- /^ 
 
 must be done to produce a slight additional elongation Ae. 
 
 The force F in Fig. 2 is, according to Hooke's law, proportional 
 to e; that is, 
 
 F = ke (1) 
 
 where A; is a constant. 
 
 Imagine the force F to be increased by the amount AF so 
 that the elongation would be increased by the amount Ae. 
 Then the work done would be greater than F-Ae and less than 
 {F -\- AF)-Ae because the average value of the force which acts 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 27 
 
 on the spring while the added elongation Ae is being produced is 
 greater than F and less than F + AF. Therefore 
 
 AW is greater than F • Ae (2) 
 
 and 
 
 AW is less than (F -\- AF) - Ae (3) 
 
 whence, dividing by Ae, we have : 
 
 aw 
 
 -^ is greater than F and less than F + AF (4) 
 
 Ae 
 
 AW 
 Therefore —— approaches F as its limit as AF (and also Ae) 
 
 approaches zero. That is, 
 
 dW 
 
 or, using the value of F from equation (1), we have 
 
 f=^ ,,, 
 
 Now in Art. 10 it is shown that ax^ is a function whose derivative 
 is 2ax. Therefore: 
 
 ae2 is a function whose derivative is 2ae (7) 
 
 Therefore, substituting k = 2a (or a = \k) in (7) we have: 
 
 \ke^ is a function whose derivative is ke (8) 
 
 The work W is also a function whose derivative is ke, according 
 to equation (6). Therefore, according to Art. 21, we must have 
 
 W = W + C (9) 
 
 where C is a constant. To determine the value of the constant 
 C we must know the value of W for some particular value of e, 
 and of course we know that TT = when e = 0, that is, no work 
 is required when the spring is not stretched at all. Therefore, 
 
28 
 
 CALCULUS. 
 
 substituting this pair of values of W and e in equation (9), we 
 have 
 
 = C (10) 
 
 Substituting this value of C in equation (9) we have 
 
 W = ike'' (11) 
 
 which is the desired expression for the work W. 
 
 23. Another example showing the use of calculus. The area 
 under a parabola. — The curve cc. Fig. 15, represents a parabola 
 of which the equation is: 
 
 y = px^ (1) 
 
 and it is required to find an expression for the area A. 
 Imagine the value of x in Fig. 15 to increase by the amount Ax^ 
 
 X 
 
 X — 
 
 Fig 15. Fig. 16. 
 
 then the increment of A is the area of the narrow strip in Fig. 16. 
 The width of this strip is Ax and its height is y (= px^). There- 
 
 fore 
 whence 
 
 A A = px^'Ax 
 AA 
 
 (2) 
 (3) 
 
 Now the width Ax in Fig. 16 must be infinitely small in order 
 that the height of the strip may be considered to be equal to 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 29 
 
 y {= px^) . That is, we have already proceeded to the limit in writing 
 
 down equation (2), and it is evident that equation (3) gives the 
 
 AA 
 limiting value of — . Therefore we have 
 
 In Art. 11 it was shown that: 
 
 ac^ is a function whose derivative is Soi* (5) 
 
 Therefore substituting ^ for a we have : 
 
 Ipa^ is a function whose derivative is px'^ (6) 
 
 The area A is also a function whose derivative with respect to 
 X is px^ according to equation (4), and therefore, according to 
 Art. 21, we must have: 
 
 A = ip:x^+C (7) 
 
 where C is a constant. To determine the value of the constant 
 C we must know the value of A for some particular value of x. 
 An inspection of Fig. 15 shows that A = when x = L There- 
 fore, substituting this pair of values of A and x in equation (7) 
 we have: 
 
 = ipP + C (8) 
 
 so that 
 
 C = - ipP (9) 
 
 Substituting this value of C in equation (7) we have: 
 
 A = ipa^ - \pP (10) 
 
 which is the desired expression for the area A in Fig. 15. 
 
 It is to be noted that a^ is a variable in the above discussion 
 because we have arbitrarily made it vary in the derivation of 
 equation (4), whereas p and I are both constant. 
 
30 CALCULUS. 
 
 PROBLEMS. 
 
 In each of the first six following problems, find the rate of 
 change of y with respect to x. 
 
 1. y^ = 3x, 
 
 2. y^ = ax^ — 3x, 
 
 3. y^ = 2ax^ + 3x, 
 
 4. y^ = a^ - x\ 
 
 dx 2 >x* 
 d^ _ 3(ax^ - 1) 
 d^ ~X^ax^~^Zx ' 
 dy _ 4:ax + 3 
 dx ~i (2aa;2 + 3x)« 
 
 5 .2 = 1 ^=-,J. 
 
 ^* ^ x' <te 2a;f 
 
 °' ^ x' dx 3x1* 
 
 7. The volume V of a cone of which the half-angle at the apex 
 is 30° is a function of the altitude x of the cone. Set up the 
 
 derivative -j-, Ans. -j- = \irx^. 
 
 8. Set up the derivative of the area A of the curved surface of 
 
 dA 
 the cone of problem 7 with respect to x. Ans. y- = ^irx. 
 
 9. The volume F of a cone of which the half-angle at the apex 
 is 30° is a function of the slant height x of the cone. Set up the 
 
 , . ,. dV . dV V3 2 
 derivative -,-. Ans. -5- = -^-ttx^. 
 ox ax o 
 
 10. The kinetic energy TT of a disk of steel rotating at n 
 
 revolutions per second is a function of the radius r of the disk. 
 
 dW 
 Set up the derivative -7- , the thickness of the disk being 10 
 dr 
 
 dW 
 centimeters. Ans. -3— = 787rnV 
 dr 
 
 ■m^ 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 31 
 
 Note. — The density of steel is 7.8 grams per cubic centiraeter and the 
 kinetic energy of a particle in ergs is \mv^ where m is the mass of the particle 
 in grams and v is the velocity of the particle in centimeters per second. 
 
 11. The shaded area in Fig. pll is a function of the angle 6. 
 Find the derivative of the area with respect to B, the angle being 
 expressed in radios and the equation of the curve cc being 
 r = kB -\-hf where k and h are constants. 
 
 Ans. ^ = \{kB + by. 
 
 12. The length of the arc s in Fig. pl2 is a function of the 
 
 Fig. pll. Fig. pl2. 
 
 angle B. Find the derivative of s with respect to B, the equation 
 
 d s I 
 
 of the curve cc being as given in problem 11. Ans. -^ = V fc^ -f ^^. 
 
 au 
 
 24. The differential equation. Law of growth of a function. — 
 
 Let y be an undetermined function of x concerning which it is 
 known that y increases 2ax times as fast as x. This property 
 of the unknown function may be expressed by the equation: 
 
 I = 2« « 
 
 This equation expresses the law of growth of the function y, the 
 independent variable x being assumed to grow steadily in value. 
 Any equation which expresses the law of growth of a function is 
 called a differential equation. 
 
 Examples, (a) A man A saves money at a constant rate.* 
 This statement is a verbal expression of a differential equation. 
 
 * Money saved is here assumed to be a continuous variable. See Art. 5. 
 
32 CALCULUS. 
 
 To reduce the statement to algebra, let y be the amount of money 
 saved by A, then the above statement is equivalent to 
 
 f = . (2) 
 
 where A; is a constant. 
 
 (6) One man A saves money twice as fast as another man B, 
 This statement also is a verbal expression of a differential equation. 
 To reduce it to algebra let y be the amount of money saved by A 
 and let x be the amount of money saved by B. Then while B 
 saves Ax dollars A must save twice as much according to the 
 
 above statement. Therefore ^y — 2Ax or -r^ = 2, or 
 
 i=2 (3) 
 
 (c) Equation (6), Art. 22, is a differential equation, and equation 
 (4), Art. 23, is a differential equation. 
 
 25. Differentiation and integration. — The finding of the deriva- 
 tive of a function as exemplified in Arts. 10 and 11 is called dif- 
 ferentiation. 
 
 Let y be an unknown function of x which is known to in- 
 crease 2ax times as fast as x. That is to say, it is known concern- 
 ing y that 
 
 1 = 20. (1) 
 
 Now when one has previously found that ax^ is a function of x 
 whose derivative is eqvxil to 2ax one is able to recognize that the 
 unknown function y whose law of growth is given by equation (1) 
 must be equal to ax^ -f C, where C is a constant, as explained in 
 Art. 21. This recognition of a function from its derivative is 
 called integration. 
 Symbol of integration. — Let y stand for ax^ + C, then 
 
 dy 
 
 dx 
 
 2ax (2a) 
 
 .^vV 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 33 
 
 that is to say, 
 
 the derivative of (ax^ + C) is equal to 2ax (26) 
 
 Or, using the differential notation which is explained in Art. 19, 
 we have 
 
 dy = 2aX'dx (2c) 
 
 that is to say, 
 
 the differential of (ax^ + C) is equal to 2ax • dx (2d) 
 
 All of these equations (2a), (26), (2c) and {2d) express the same 
 thing or the same relation, and this relation may be expressed in 
 an inverse manner by saying: 
 
 (ax^ + C) is the integral of 2aX'dx (3a) 
 
 or, in symboUc notation: 
 
 ax'^- C = J2ax'dx (36) 
 
 The symbol J means integral of. The undetermined constant 
 C is called the constant of integration. 
 
 A differential equation does not completely determine the func- 
 tion whose law of growth it expresses. — This is evident when we 
 consider that the function y -{- C has the same derivative (the 
 same law of growth) whatever the value of the constant C may be. 
 
 Thus the amount of a man's savings cannot be calculated when 
 his rate of saving alone is known; one must know also how much 
 money the man had on a given date. Thus a man who had $1,000 
 on January 1, 1912, and who saves $10 per month would have 
 1,000 + 10m dollars at any time m months after January 1. 
 The amount of a man's savings is completely determined as a 
 function of elapsed time by knowing (a) his rate of saving and 
 (6) how much he had on a given date. 
 
 In general a function y is completely determined* by knowing 
 
 * This statement will have to be modified when we come to consider dif- 
 ferential equations of higher orders, and when we come to consider partial 
 differential equations. 
 4 
 
34 CALCULUS. 
 
 (o) its law of growth and (b) its value corresponding to any given 
 value of the independent variable x or < as the case may be. 
 This matter is fully illustrated in Arts. 22 and 23. 
 
 26. Indefinite integrals and definite integrals. — Let y be a 
 
 dv 
 function of x concerning which nothing is known except that -^ 
 
 is equal, say, to 2ax; or, using differential notation, we know that 
 
 dy = 2ax'dx (1) 
 
 Then according to Arts. 10 and 21 we know that y must be given 
 by the equation: 
 
 2/ = 0x2 + C (2) 
 
 where C is a constant which may have any value whatever. This 
 equation (2) expresses what is called the indefinite integral of equa- 
 tion (1). 
 
 Let it he required to find the growth of y while x grows from any 
 given value I to any other given value L. This growth of y is 
 called a definite integral of equation (1), and the two given values 
 of X are called the boundaries or limits* of the definite integral. 
 
 Now, according to equation (2), we have y = aP -]- C when 
 X = I, and y = aL^ -{• C when x = L; and the difference 
 between these two values of y, namely, aU — al^, is the desired 
 growth of y from x = I to x = L. That is, the definite integral 
 of equation (1) between the boundaries or Hmits x = I and x = L 
 is equal to aL^ — aP. 
 
 Examples of definite integrals. — Consider the area A under 
 the parabola in Fig. 15. The differential of this area is 
 
 dA = px^'dx (3) 
 
 and the value of the area is 
 
 A = ips^- JpP (4) 
 
 * The word limit as here used has a very di£ferent meaning from the word 
 limit as used in Arts. 4 and 7, 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 35 
 
 as explained in Art. 23. Now equation (4) is the definite integral 
 of equation (3) from x = I to x = x (meaning any value of x 
 whatever). 
 
 Sjrmbolic representation of a definite integral. — The usual 
 method of expressing a definite integral is 
 
 = j^'^px^'dx (5) 
 
 where A is the area under the parabola in Fig. 15, the differential 
 of A being given by equation (3) above. 
 
 Similarly, the work W required to stretch the spring in Fig. 2 
 is found as a definite integral in Art. 22, and to express W sym- 
 bolically as a definite integral we write 
 
 W^f' 
 
 ke-de j[6) 
 
 The value of this definite integral is Jfce^ as explained in Art. 22. 
 
 27. A definite integral interpreted as the limit of a sum. — Imagine the 
 area A of Fig. 15 to be divided into narrow strips as shown in Fig. 17, the 
 width of successive strips being represented by successive increments Ax 
 of X, starting at x = I and extending to any given value of x. This given 
 value of X is represented by L in Fig. 17, because it is now desirable to 
 think of X as having various intermediate values (between x = I and 
 x = L). 
 
 The width of any strip is Ax and its altitude is px^, where x is the 
 distance of the particular strip from the y-axis. Therefore the area of the 
 strip is px^ • Ax and the sum of the areas of all the strips is more and more 
 nearly equal to the area A of Fig. 15 as the number of the strips is made greater 
 and greater. 
 
 The area A is equal to J^px^ • dx as explained in Art. 26, and the sum 
 
 L 
 
 of all the strips in Fig. 17 may be represented symbolically as ^px^ • Ax. 
 When the number of strips is made greater and greater (successive increments 
 
 L PL 
 
 Ax made smaller and smaller) then "^px^ • Ax approaches f , px^ • dx as 
 
 I * 
 
 . . ^ . C^ 
 
 a limit. That is, the limit of ^px^ 'Ax is J. px^ • dx. 
 
36 
 
 CALCULUS. 
 
 28. Plan of this treatise. — The application of calculus involves 
 
 three important things. Thus in Art. 22 the following three things 
 
 dW 
 are involved: (a) The finding of the derivative -t- by making use 
 
 of fundamental principles in physics, (6) the becoming familiar 
 with the derivative of ax^ in Art. 10, and (c) the substitution of 
 Jfc for a and e for z so as to get a known function whose deriva- 
 tive is ke. 
 
 The application of calculus 
 
 in engineering involves these 
 three steps in nearly every 
 case, and the following devel- 
 opment of calculus consists of 
 three parts (more or less inter- 
 mingled) as follows: 
 
 (o) The setting up of de- 
 rivative expressions (differen- 
 tial equations) with the help 
 of fundamental principles in 
 physics. 
 
 (b) The study of derivatives 
 of algebraic functions; and 
 (c) Practice in transforming derivative expressions to standard 
 forms for purposes of recognition. 
 
 The most important branches of calculus which do not fall into 
 this outline are the discussion of maximum and minimum values of 
 functions and the expansion of functions in series 
 
 A table of functions Q,nd their derivatives* is given in Appendix B. 
 The student should get into the habit of referring to this table, 
 
 PROBLEMS. 
 
 1. A force of 100 pounds produces 4 inches of elongation of a 
 spring. Find the value of k in equation (1) of Article 22, and 
 express the unit in terms of which k is found. Ans. 25 pounds 
 per inch. 
 
 * Or of differential expressions and their integrals. 
 
DIFFERENTIAL AND INTEGRAL CALCULUS. 37 
 
 2. Using the spring referred to in problem 1, find the work 
 required to produce (a) an elongation of 2 inches, and (6) an elonga- 
 tion of 4 inches, Ans. (a) 50 pound-inches, (6) 200 pound-inches. 
 
 3. Using the spring referred to in problem 1, find how much work 
 is required to increase the elongation from 3 inches to 4 inches. 
 Ans. 87.5 pound-inches. 
 
 4. Consider the parabola y = 2x^. Find the area under this 
 curve between ordinates at a; = 2 and x = 10. Ans. 66^J. 
 
 5. Find an expression for the area under the curve y = qx 
 between an ordinate at x = a and the ordinate corresponding to 
 any abscissa x. Ans. A = ^qx^ — fga^. 
 
CHAPTER II. 
 FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 
 
 29. Differentiation by development and differentiation by rule. 
 — Throughout the previous chapter the derivative of a function 
 (as in Arts. 10 and 11) or the differential of a function (as in Art. 
 
 20) is determined by developing an expression for — ^ or for Ay, 
 
 Ax 
 
 and considering the limiting form of this expression as Ay and 
 Ax approach zero. This is the fundamental method of differentia- 
 tion. It is possible, however, by means of this fundamental method 
 to establish a few rules which enable one to write down the derivative 
 or differential of almost any algebraic function. It is the object of 
 this chapter to establish these rules and give examples of their use. 
 Differentiation may be defined in general as the finding of the 
 rate of change of a function. Thus problem 10 on page 21 illus- 
 trates what may be called graphical differentiation. And to de- 
 termine a rate by actually measuring the change which, takes 
 place in a given time may be called physical differentiation. Thus 
 the speed of a runner is determined by measuring the distance 
 traveled by the runner in a given time, or rather by measuring 
 the time required for the runner to travel a given distance. 
 To determine the rate of supply of water by a small spring is 
 to "differentiate the spring" and the simplest method is to 
 make the differentiation by means of a dipper and a dollar watch. 
 
 30. Differentiation of ax". — Let 
 
 y = ax" (1) 
 
 Then 
 
 dy = nax"-i-dx * (2) 
 
 for any value of n. We shall here prove equation (2) when n 
 is a positive integer.* 
 
 *A general proof is given in Art. 39. 
 
 38 
 
 £J^^ 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 39 
 
 Proof. — Writing y + Ay for y and writing x + Ax for x in 
 equation (1) we have: 
 
 y + Ay^a{x + Ax^ (3) 
 
 When n is a positive integer {x + AxY can be expanded by the 
 binomial theorem,* and we have 
 
 y -\- Ay = ax"" + nax""-^ • Ax -\ — ^ ^^ — ^ — - + etc. (4) 
 
 Subtracting equation (1) from equation (4), member by member, 
 we have: 
 
 Ay = nax^'-^'Ax + -^ ^ — - — - + etc. (5) 
 
 J. * z 
 
 Whence, writing dy for Ay and dx for Ax, and dropping in- 
 finitesimals of second and higher orders, we have : 
 
 dy = nax''~^'dx (2) 
 
 31. Differentiation of the sum of two or more functions. — Let 
 
 u and V be any two functions of x, and let 
 
 y = u + V (1) 
 
 Then 
 
 dy = du + dv (2) 
 
 or, using derivative notation, 
 
 dx dx ^ dx ^^^ 
 
 Proof. — It is evident that the increment of y is equal to the 
 sum of the increments of u and v, that is. Ay = Au -\- Av; and 
 this is true however small the increments may be. Therefore 
 when the increments are infinitesimal we have dy — du ■\- dv. 
 
 * The binomial theorem is: 
 
 (a + 6)« = a« + -J- + — + —— + etc. 
 
 When n is not an integer this becomes an infinite series, and an infinite series 
 cannot be used in a mathematical argument unless the question of its con- 
 vergence is carefully considered. See Art. 35. 
 
40 CALCULUS. 
 
 Also it is evident that 
 
 Ay»_AuAv 
 Ax~ Ax Ax 
 
 and this is true however small the increments may be. Therefore 
 when the increments are infinitesimal we have 
 
 d]i _ du dv 
 dx dx dx 
 
 Explanation. — The fact which is expressed by equation (3) is 
 extremely simple and it is self-evident when it is expressed in 
 familiar terms as follows: The amount of money earned by one 
 
 man is u and his rate of earning is 2 dollars per day ( = 75: ) J 
 
 the amount of money earned by another man is v and his rate 
 
 of earning is 3 dollars per day ( = ;77 ) • The money earned by 
 
 both men together is u -\- v and their combined rate of earning is 
 5 dollars per day, which is the rate of change of {u -^ v). 
 
 32. The differentiation of the product of two functions. — Let 
 u and V be any two functions of x, and let 
 
 y = uv (1) 
 
 Then 
 
 dy = udv + v-du (2) 
 
 Proof. — Writing y ■}- Ay for y, 
 
 u -\- Au for u and 
 
 V -\- Av for V in equation (1) we have: 
 y + Ay = {u + Au){v + Av) (3) 
 
 or 
 
 y -^ Ay = uv -\- u-Av + v-Au + Au-Av (4) 
 
 Whence, subtracting equation (1) from equation (3) member by 
 member, we have 
 
 Ay = u- Av -\- V ' Au ■}- Au' Av (5) 
 
 But Au'Av is a second order infinitesimal, and therefore when 
 
FORMULAS FOR DIFFERENTIATION AND -INTEGRATION. 41 
 
 Au and Av approach zero we may drop Au-Av and we have the 
 limiting relation: 
 
 dy = U'dv + V'du 
 
 Example. — Let u = ax^ + hx, let v = cx^ + 1 and let 
 y = {ax^ + hx){cx^ + !)• To differentiate this expression substi- 
 tute the values of u and v, and the values of 
 
 du (= 2ax'dx + h-dx) and dv (= Zcx'^-dx) 
 in equation (2), and we have 
 
 dy = {ax^ + 5a;) X ?>cx^'dx -\- {cx^ + l)(2ax -\-h)'dx 
 
 33. Differentiation of the quotient of two functions. — Let u 
 
 and V be any two functions of Xy and let 
 
 y = - (1) 
 
 Then 
 
 ^^^vdu-u-dv (2) 
 
 Proof. — From equation (1) we have 
 
 . . u-\- Au .^. 
 
 whence, subtracting equation (1) from equation (3) member by 
 member, we have 
 
 . u -\- Au u ,.. 
 
 Ay = — — (4) 
 
 ^ V + Av V 
 
 Reducing ; — -— and - to a common denominator, equation 
 
 V -\- Av V * ^ 
 
 (4) becomes: 
 
 V'Au — u-Av ,.v 
 
 v^ + v-Av 
 
 But as Au and Av approach zero the denominator approaches 
 v^ as its limit. Therefore equation (5) reduces to equation (2). 
 
42 CALCULUS. 
 
 Example. — Let it be required to differentiate the function: 
 
 Let u = as? -^rhx and let v = cx^ -\- d. Substitute these values 
 of u and v, and the values of du{= 2ax-dx -\-h'dx) and 
 d/o{ — Zcx^ ' dx) in equation (2) and we get the desired result. 
 
 34. Differentiation of a function of a function. — Let y he a 
 
 Junction of x and let z he a function of y. Then y changes -^ 
 
 dz 
 times as fast as x, and z changes j- times as fast as y. There- 
 
 dz dv 
 fore it is evident that z changes t" X -t^ times as fast as x. 
 
 That is, 
 
 ^ = - X ^ (I) 
 
 dx dy ^ dx ^^l 
 
 dxi dz 
 
 The symbols -^ and -7- are so cumbersome that the proposition 
 
 upon which this equation is based is not easy to understand when 
 it is stated as above. Reduced to its simplest terms the proposi- 
 tion is as follows: If z changes 5 times as fast as y and if y 
 changes 7 times as fast as x, then z changes 5X7 times as fast 
 as x\ 
 
 Example. — Let 
 
 2 = a(x2 -h a; -f- l)^ 
 
 To differentiate this expression one can write y for a;^ + a; -}- 1. 
 
 Then z = ay^ and j- = Zay"". But ^ = 2x-\-l. Therefore, 
 
 dz 
 using equation (1), we have t~ = Sa{x^ + x -\- l)2(2x + 1). 
 
 For problems see group 1 in the Appendix. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 43 
 DIFFERENTIATION OF LOGARITHM. 
 
 35. Convergent series. — Before the derivative of logarithm of x 
 can be found it is helpful to consider the two series of fractions: 
 
 
 a 
 
 h 
 
 c 
 
 d 
 
 e 
 
 . . . 
 
 n 
 
 1st series 
 
 1 
 2 
 
 1 
 4 
 
 1 
 
 8 
 
 1 
 16 
 
 1 
 32 
 
 . . . 
 
 1 
 
 2» 
 
 2d series 
 
 1 
 
 II 
 
 1 
 1 
 
 1 
 
 II 
 
 1 
 
 1 
 
 
 1 
 
 where |n stands for the product of all positive integers from 1 
 to n inclusive, factorial n, as it is sometimes called. 
 
 The first series may be brought before the reader most distinctly 
 by means of the following schedule; 
 
 Fraction of original sum 
 which is spent each day. 
 
 Half of a sum of money is spent the first day ^ 
 
 Half of the remainder is spent the second day i 
 
 Half of the remainder is spent the third day i 
 
 Half of the remainder is spent the fourth day . . . tV 
 etc., etc., etc. 
 
 If one were to follow this schedule indefinitely the original sum 
 of money would never be entirely spent, because there is always 
 an unspent remainder; but the remainder would grow smaller 
 and smaller, and the amount spent would approach, as nearly as 
 you please, to the entire original sum. This is equivalent to 
 saying that i + J + i + T^ + etc. can never be equal to unity, 
 but it can be made to approach unity, as nearly as you please, by 
 adding together a larger and larger number of the successive 
 fractions of the series. Such a series of fractions is called a 
 convergent series, and the limit of J + i + i + etc., as a greater 
 and greater number of the successive fractions are included, is 
 called the sum of the series. The sum of the series ^ + i + i + 
 xV + etc. is unity. 
 
 A series need not be convergent merely because the successive fractions of 
 the series grow smaller and smaller. Thus | + i + i+i + i+etc. grows 
 
44 
 
 CALCULUS. 
 
 
 a 
 
 h 
 
 c 
 
 d 
 
 1st series 
 
 i 
 
 hoi a 
 
 iof6 
 
 ioic 
 
 2d series 
 
 J 
 
 J of a 
 
 iof6 
 
 ioic 
 
 larger and larger without limit as a greater and greater number of the successive 
 fractions are included. Such a series of fractions is called a divergent aeries. 
 
 To show that the series ,7:, , o > rj » ,-^) etc. is a convergent series 
 
 |2 |3 14 [5 
 
 it is sufficient to arrange the two series, as in the following table, 
 showing how each fraction is related to the preceding fraction 
 in each series: 
 
 etc. 
 etc. 
 
 Each fraction of the second series is equal to or smaller than 
 the corresponding fraction of the first series. Therefore, if the 
 first series is convergent, the second series must also be convergent. 
 
 The particular series which is used in finding the derivative of a 
 logarithm is 
 
 ^=^+i+L2+ii+ii+ii+^*''- ^'^ 
 
 and this series is convergent because it is identical to the second 
 series in the above table with the exception of the first two terms. 
 The sum of this series, correct to seven decimal places, is: 
 
 e = 2.7182818 (2) 
 
 36. The limit of ( 1 + - j as z approaches infinity. — This limit 
 
 is important in finding the derivative of a logarithm. Let us assume 
 
 that z is always a positive integer. In this case ( 1 -f - j 
 
 be expanded by the binomial theorem,* giving: 
 
 (■+»'-+,l'©+,i*-'>©' 
 
 can 
 
 + ^z{z-l)(z-2)(^iy+ etc. 
 
 (3) 
 
 * See foot-note to Art. 30. Substitute o = 1, 6 = - and n = z, and we 
 
 z 
 
 have equation (3). 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 45 
 
 But as 2 approaches infinity (1 j, (1 J, etc., approach 
 
 unity, and therefore the limiting value of ( 1 H — ) , as 2 approaches 
 infinity, is found by writing unity for each of the expressions 
 (l ), (l ), etc., in equation (4), and when this is done the 
 
 second member of equation (4) reduces to 1 + il "^ 1 9 "^ I q "^ fl "^ 
 
 etc. Therefore the limiting value of ( 1 + - ) as z approaches 
 infinity is 2.7182818, or e, as explained in the previous article. 
 
 To prove that the limit o/ f 1 + - ] is e when z approaches infinity, hiU 
 
 when z is not always a positive integer. Let s be any positive value whatever 
 of z, and let w be a positive integer such that z lies between m and m + 1. 
 
 Then ( 1 + - ) lies between [ 1 -\ — ) and { 1 + — r— r ) . Now as 
 \ zj \ mj V ' m + 1/ 
 
 (I \m / 1 \m+l 
 
 1 -\ j and {l-\ ^— r 1 both approach 
 
 e as a Umit; and therefore f 1 + - J , which is always between ( 1 -^ — j 
 
 (1 \m+i 
 1 4 ^— r j , also approaches e as a limit. 
 
 To show that ( 1 H — ) approaches e as a limit when z approaches 
 
 negative infinity. Let z = — r. Then fl-1 — j =fl j where z is 
 
 negative and r is positive. Now 
 
 but (l+^^J approaches unity as r approaches infinity and ( H — ^ ] 
 
46 CALCULUS. 
 
 approaches e as r approaches infinity. Therefore every member of equation 
 (5) approaches e as r approaches infinity. Consequently f 1 H — ] ap- 
 proaches e when z approaches negative infinity. 
 
 PROBLEM. 
 
 One dollar placed at compound interest for 10 years at 6 per 
 
 cent, amounts to (1 + 0.06)^^ dollars, when accrued interest is 
 
 added to the principal once a year. If accrued interest is added 
 
 to the principle n times per year, one dollar after 10 years would 
 
 / 0.06 \ ^^ 
 amount to ( 1 + — — I . Find what one dollar would amount 
 
 ('+"-?) 
 
 to after 10 years at 6 per cent, compound interest (a) when accrued 
 interest is added to principle once per year, that is, when n = 1, 
 (6) when n = 2, and (c) when accrued interest bears interest 
 without any delay whatever, that is, when n = oo. Ans. (a) 
 $1,791, (6) $1,803, (c) $1,822. 
 
 Note.- ( 1 + M§ )^"= [(l +\yj where . = ^ and a = 0.6. 
 
 37. Differentiation of the logarithm of x. — Let y be the number 
 of times that a given number a must be multiplied by itself to 
 give X. Then x = a^, and y is called the logarithm of x to the 
 base a. The base of the common system of logarithms is 10, and 
 the base of the Napierian system of logarithms is 2.7182818 or e, 
 Napierian logarithms are used throughout this treatise except where 
 it is otherwise expressly stated.* 
 
 Let 
 
 y = log X (1) 
 
 Then 
 
 dy = | (2) 
 
 Proof. — ^Writing y + Ay for y and writing x + Ax for x in 
 
 * Napierian log x = 
 
 logio X 
 logxo e 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 47 
 
 equation (1) we have: 
 
 y + Ay = log (x + Ax) (3) 
 
 whence, subtracting equation (1) from equation (3) member by- 
 member, we have: 
 
 Ay = log (x + Ax) — log X (4) 
 
 or 
 
 Ay _ log(a; + Ax) - log x , . 
 
 (X + Ax\ 
 J or to 
 
 log ll-j-~y, and ^log (l+"^) is equal to logfl+-^j^. 
 
 Therefore equation (5) reduces to 
 
 £ = '-( 
 
 l+f)^ (6) 
 
 and this may be written 
 
 A^ 
 Ax 
 
 = l,og(i+^)£ (7) 
 
 X 
 
 or, writing z for -— , we have: 
 Ax 
 
 :=ilog(l + y' (8) 
 
 But, as Ax approaches zero ^ {='ir-) approaches infinity, and 
 
 the limiting value of ( 1 H — 1 as z approaches infinity is e, as 
 
 explained in Art. 36. Therefore as z approaches infinity equation 
 (8) becomes: 
 
 But the Napierian logarithm of e is unity because e is equal to eK 
 
48 CALCULUS. 
 
 Therefore equation (9) reduces to 
 
 ^= - d =— 
 
 dx X X 
 
 Differentiation of loga x. — Let 
 
 y = loga X (10) 
 
 This means that 
 
 x = a« (11) 
 
 Therefore, taking Napierian logarithm of each member of this 
 equation, we have 
 
 log X = 2/ log a 
 or 
 
 where * is, of course, a constant. Therefore, according to 
 
 loga 
 
 equation (1) and (2) we have 
 
 dy = ,-i-.^ = l5i^l:^ (13) 
 
 •' log^ X ^ X ^ 
 
 38. Differentiation of the exponential function. — Let 
 
 y = Ae^- (1) 
 
 where A and k are constants and e = 2.7182818. Then: 
 
 g = kAe^- = ky (2) 
 
 Proof. — Taking logarithms of both members of equation (1) 
 
 we have 
 
 log y = \ogA + kx (3) 
 
 But the differential* of log y is — according to Art. 37. There- 
 
 * It is very important that the student keep in mind what a differential is. 
 The differential of log y is the infinitesimal increment of log y due to an 
 infinitesimal increment, dy, of y. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 49 
 fore, by differentiating, both members of equation (3) we have: 
 
 ^ = k-dx (5) 
 
 y 
 
 dv 
 whence equation (3) is obtained by solving for ^; and the value of 
 
 y{= Ae^") may be substituted for y in the resulting equation. 
 
 Differentiation of a**. — Let 
 
 y = a*' (6) 
 
 where a and k are constants. Taking logarithm of each member 
 of (6), we have 
 
 log y = kx log a (7) 
 
 Differentiating we have 
 
 '^^kloga-dx (8) 
 
 dv 
 Therefore, solving for -~ and substituting the value of y from 
 
 equation (6), we have: 
 
 dv 
 
 ^ = k log a-a** = k log a-y (9) 
 
 PROBLEM. 
 
 Given the logarithm of 2 to base 10 find the approximate 
 value of logarithm of 2.01. Ans. 0.3032. 
 
 Note. — Consider logio x. The derivative of logio x multiplied by a small 
 increment of x gives the approximate value of the increment of logio x. 
 
 39. General proof of equation (2) of Art. 30. — Let 
 
 y = ax"* (1) 
 
 where n has any value, integral or fractional, positive or negative. Taking 
 logarithms of both members of (1) we have: 
 
 log y = log o + n log a: (2) 
 
 Differentiating, we have 
 
 ^^ = n^ (3) 
 
 y X 
 
 5 
 
50 
 
 CALCULUS. 
 
 log o being a constant. Substitute ox" for y in this equation and solve for 
 dy, and we have 
 
 dy = ruix'*~^'dx (4) 
 
 For problems see group 2 in the Appendix. 
 
 DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS. 
 
 40. Limit of 
 
 sin 
 
 2r sin p 
 
 Fig. 18. 
 chord ob is equal to 2r sin 4>. 
 
 as approaches zero. — It is a fundamental 
 
 principle in geometry that if a 
 polygon of n sides be inscribed 
 in a circle,* the sum of the 
 sides of the polygon approaches 
 the length of the circular arc as 
 a limit when n approaches in- 
 finity. The idea that a circu- 
 lar arc has a definite length de- 
 pends upon this principle. 
 
 Consider Fig. 18. The arc 
 ah is equal to 2r0, <f) being 
 expressed in radians, and the 
 Therefore : 
 
 chord ah 2r sin <^ sin <f> 
 
 arc ah 
 
 2r0 
 
 But approaches imity as its limit when approaches 
 
 arc 
 
 zero; and therefore 
 
 sm (f> 
 
 must also approach unity as its limit 
 
 when (f) approaches zero. 
 41. Differentiation of sin x. — Let 
 
 y = sin X (1) 
 
 dy = cos x-dx (2) 
 
 Proof. — Writing y -\- Ay for y and writing x + Ax for x, we 
 
 Then 
 
 * Or any continuous curve. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 51 
 
 have : 
 
 y -{• Ay = sin(x + Ax) (3) 
 
 Whence, subtracting equation (1) from equation (3) member by- 
 member and solving for -^ , we have 
 Ax 
 
 Ay _ sm(x + Ax) — sin x 
 Ax ~ Ax 
 
 (4) 
 
 Using the trigonometric formula: 
 
 sin a — sin 6 = 2 sin ^{a — h) cos ^{a + 6) (5) 
 
 equation (3) may be reduced to: 
 
 Ay _ sin jAx 
 
 •cos(x + J Ax) (6) 
 
 Ax |Ax 
 
 but as Ax approaches zero cos(x + J Arc) approaches cos x as 
 its limit, and -^~ — approaches unity as its limit. Therefore, 
 from equation (6) we have : 
 
 -^ = cosx or dy = cosx-dx 
 
 42. Differentiation of cos x. — Let 
 
 y = cos X (1) 
 
 Then 
 
 dy = — sinx-dx (2) 
 
 Proof. — Proceeding as in the previous article we get 
 
 Ay _ cos(a; + Ax) — cos x 
 Ax Ax 
 
 (3) 
 
 and using the trigonometric formula: 
 
 cos a — cos h — — 2 sin i(a — 6)sin i(a + h) (4) 
 
 we obtain 
 
 Ay sinjArc . / , i. n /kx 
 
52 CALCULUS, 
 
 and from this we obtain 
 
 -^ = — emx or dy — — Qmx'dx 
 
 Far problems see group 3 in the Appendix. 
 
 EXAMPLES SHOWING USE OF FORMULAS FOR 
 DIFFERENTIATION. 
 
 43. Differentiation of tan x. — In order to differentiate tan x 
 
 sin X 
 
 the familiar formula tan x = may be used as follows: 
 
 cos a; 
 
 , sinx ,^. 
 
 y = tan x = (1) 
 
 cos a; 
 
 Proceeding according to Art. 33, let w = sin x so that du = 
 cos x • dx, and let v = cos x so that dv = — sin x • dx. Then 
 using equation (2) of Art. 33, we have: 
 
 J cos^ X'dx + sin^ x-dx dx /„^ 
 
 dy = ^ = — ^ (2) 
 
 COS^ X COS^ X 
 
 Any trigonometric function can he differentiated hy expressing the 
 function in terms of sine and cosine and using the formulas of Arts, 
 41 and 4^. 
 As a further example, consider 
 
 i = I sin (at (3) 
 
 in which I and to are constants and t is clasped time reckoned 
 
 from any chosen instant, and let it be required to find the rate of 
 
 di 
 change of i, namely, -jz. Now cot is a function of t and sin <at 
 
 is a function of (at. Therefore we may use the formula for the 
 differentiation of a function of a function, as explained in Art. 34. 
 Let 
 
 z = (at (4) 
 
 then 
 
 dz = (a-dt (5) 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 53 
 
 Writing z for wt in equation (3) we have : 
 
 i = 7 sin 2 (6) 
 
 which, by differentiation gives: 
 
 di = I cos Z'dz (7) 
 
 Whence, substituting ot for z and 03 -dt for dz we have 
 
 di = 0)1 cos cot'dt (8) 
 
 or 
 
 -r: = co7 cos co< (9) 
 
 at 
 
 For problems see group 4 ^^ t^^ Appendix. 
 
 44. Differentiation of sin*^ x. — The inverse trigonometric func- 
 tions can all be differentiated with the help of the formulas al- 
 ready established. For example let 
 
 y = sin-^ X (1) 
 
 this means that 
 
 siny = X (2) 
 
 which, by differentiation, gives: 
 
 cos y-dy = dx (3) 
 
 but cos 2/ = Vl — sm^y = V 1 — x^, according to equation (2). 
 Therefore, substituting this value for cos y in equation (3) and 
 solving for dy, we have 
 
 For problems see group 6 in the Appendix. 
 
 45. Differentiation of u^. — Let u and v be any two functions 
 of Xf and let 
 
 y = u- (1) 
 
 To differentiate this expression take logarithms of both members 
 and we have 
 
 \ogy -= V log u (2) 
 
54 CALCULUS. 
 
 This is a product of two functions v and logw and their 
 respective differentials are dv and — . Therefore, using the 
 method of Art. 32, we may differentiate equation (2), and we have: 
 
 dy V'du , I , /ON 
 
 — = + logU'dv (3) 
 
 y u '=' 
 
 whence, substituting u" for y and solving for dy we have 
 
 dy = vu*'~^du + log uu^-dv (4) 
 
 When u and v are known as functions of x, then du and dv 
 are known in terms of x and dx, and the known values of w, r, 
 du and dv may be substituted in equation (4) thus giving dy in 
 terms of x and dx. 
 For problems see group 6 in the Appendix. 
 
 SUCCESSIVE DIFFERENTIATION AND INTEGRATION. 
 
 46. Successive derivatives of a function. — Consider a function 
 of x. For example: 
 
 y = ax^ (a) 
 
 then 
 
 I = '^ (« 
 
 But this derivative is itself a function of x and its derivative is 
 
 12ax^; this is also a function of x and its derivative is 24aa;; 
 
 and so on. These successive derivatives of the original function 
 
 du CbU d u 
 y are usually represented by the symbols —, -~, -^ and so on; 
 
 so that we may write 
 
 dx^ 
 
 Uax" (2)^ 
 
 g = 24ax (3) 
 
 etc. etc. 
 
 *kM}' 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 55 
 
 Equations (1), (2), (3), etc., express what are called the first 
 
 derivative^ the second derivative, the third derivative, etc., of y. 
 
 The exact meaning of the successive derivatives of y may be 
 
 most easily expressed in words when x represents elapsed time. 
 
 du d^ij 
 
 Then ~ is the rate of change of y, -t\ is the rate of change of 
 
 dv d^y d^ij 
 
 ^, -7-^ is the rate of change of ~, and so on. In the particular 
 
 example, namely y = ax^, the fourth derivative is a constant 
 (= 24a), and the fifth and all higher derivatives are equal to zero. 
 
 47. Velocity and acceleration. Use of first and second deriva- 
 tives. — In the formulation of problems in mechanics one must 
 frequently express the velocity and acceleration of a moving body 
 in terms of its coordinates. Thus x and y are the varying 
 coordinates of a moving particle B in Fig. 19; x and y have 
 definite values at each instant as the body moves, and they are 
 therefore functions of elapsed time; hence we have the following 
 important relations: 
 
 dx 
 
 -^ is the x-component of the velocity of B. 
 
 dv 
 
 -jj is the ^/-component of the velocity of B. 
 
 -77^ is the a;-component of the acceleration of B, 
 
 -^ is the ^/-component of the acceleration of B, 
 
 These relations are evident from the following considerations: 
 
 Let Ax and Ay be the increments of x and y, respectively, during 
 
 Ax 
 a short interval of time. Then — is the average velocity of the 
 
 body in the direction of the a:-axis during the interval At, and 
 
 dx Ax 
 
 -r. is the limiting value of — when A^ and Ax approach zero. 
 
 Av 
 Similarly -^ is the average velocity of the body in the direction 
 At 
 
56 
 
 CALCULUS. 
 
 of the j/-axis, and -^ is the limiting value of -^ when At and Ay 
 approach zero. Furthermore the x-component of the acceleration 
 
 iHixtti; 
 
 baseball 
 
 B 
 
 x-axU 
 
 X 
 
 Fig. 19. 
 
 Fig. 20. 
 
 of B is equal to the rate of change of the a;-component of the 
 
 velocity of 5, that is, the a;-component of the acceleration of B 
 
 dx d/^x 
 
 is the rate of change of -^r which is -^. 
 
 Example. — ^A ball travels at uniform angular velocity (w radians 
 per second) around a circle of radius A as shown in Fig. 20. 
 In this case the coordinates x and y are simple functions of the 
 elapsed time t; indeed from Fig. 20 we have: 
 
 a; = A cos wt 
 y =^ A sin <t)t 
 
 Differentiating these expressions with respect to < we have: 
 
 dx 
 and 
 
 dt 
 
 = — w A sin ui 
 
 -^ = (aA cos (i)t 
 
 (1) 
 
 (2) 
 
 (3) 
 
 (4) 
 
 These equations express the x and y components of the velocity 
 of the ball B at each instant. Differentiating (3) and (4) with 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 57 
 respect to < we have: 
 
 (5) 
 
 (6) 
 
 -7i5 = — 03^ A COS (d 
 
 -^ = — o)^A sin (at 
 ar 
 
 These equations express the x and y components of the acceler- 
 ation of the ball B at each instant. 
 
 PROBLEMS 
 
 1 . If X = tan 6 -f sec 0, prove that 
 
 cos 
 
 cPx 
 
 dff" (1 - sin ey 
 
 2. If 2/ = x^ log X, prove that j:J = ~« 
 
 3. If 1/ = e-loga;, prove that g = e*^^og «+~|+|-|)- 
 
 4. If 2/ = c~* cos a;, prove that -7^ + ^2/ = 0. 
 
 d^x 
 
 5. If a; = A sin n< + B cos n<, prove that -77^ + n^a; = 0. 
 
 6. For what values of a will y = e*^ satisfy the equation 
 
 da^ da;^ da; 
 Ans. 0, - 1, + 2. 
 
 7. A ball moves so that its x and y coordinates are At cos d 
 and At sin 0, respectively, as shown in Fig. p7, where A and 
 are constants and t is elapsed time. Find (a) the a;-component 
 
58 CALCULUS. 
 
 and (6) the y-component of the velocity of the ball, and (c) the 
 a;-component and (d) the y-component of its acceleration. 
 Ans. (a) A cos d, (b) A sin d, (c) 0, (d) 
 8. A ball moves so that its x and y coordinates are A^ cos ^ 
 and Bt^ sin 6, respectively, where A, B and are constants 
 and t is elapsed time. Find (a) the x-component and (6) the 
 ^/-component of the velocity of the ball, and (c) the x-component 
 and (d) the ^/-component of its acceleration. 
 Ans. (a) A cos 6, (h) 2Bt sin d, (c) 0, (d) 2B sin 6. 
 
 9. The resultant acceleration of a body is 
 
 mHM' 
 
 dt^ 
 as shown in Fig. p9, where </> is the angle whose tangent is ^- * 
 
 dt^ 
 Show from equations (1) to (6) of Art. 47 that the resultant 
 acceleration (or total acceleration) of the ball B in Fig. 20 is 
 
 equal to -j , and show that it is parallel to the radius A at each 
 
 instant, v being the resultant velocity \/ ( ;^ ) + ( 7^ ) • 
 
 10. If the distance traveled by a body is s = ae*' + 6e~**, 
 show that the acceleration is a^s. If t is expressed in seconds 
 and s in feet, (a) in what units is a expressed, and (6) in what 
 units is ah expressed? 
 
 Ans. (a) reciprocal seconds, (6) feet per second per second. 
 
 48. Harmonic motion. Example showing the use of a second 
 derivative. — A body m is suspended by a spring and the body 
 stands in equilibrium in the position shown in Fig. 21. If the 
 body is displaced x feet downwards (or upwards) an unbalanced 
 force F will act upon the body, and this unbalanced force will 
 be proportional to x. Therefore we may write 
 
 F = -kx (1) 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION- 59 
 
 The negative sign is chosen because F is upwards when x is 
 
 downwards, and F is downwards when x is upwards. 
 
 If the body is pulled upwards or downwards and 
 
 released, it will vibrate up and down, and x will 
 
 dx 
 vary with the time t. Then -rr is the velocity 
 
 dt 
 
 d^x 
 of the body at each instant, and -^ (the rate of 
 
 dx\ 
 change of -77 ) is the acceleration of the body at 
 
 each instant. But the unbalanced force which 
 acts upon a body is equal* to mass X acceleration. 
 
 Therefore F = m > -p, or, substituting the value 
 
 of F from equation (1), we have: 
 
 spring 
 
 df 
 
 m 
 
 (2) 
 
 To determine the motion of the body in Fig. 21 
 it is necessary to find a; as a function of t which 
 will satisfy the law of growth as expressed by equa- 
 tion (2), and the problem is solved if we find a 
 
 k 
 
 function of t whose second derivative is equal to multiplied 
 
 by the function itself. This function would be at once recognized 
 if one were familiar with the derivatives of all kinds of functions. 
 Thus we have another example showing how important it is to 
 be familiar with the derivatives of functions. Compare Arts. 
 22 and 23. 
 
 Consider the function 
 
 a; = a sin (coi + 6) (3) 
 
 where a, w and are undetermined constants. Differentiatingf 
 * When mass is expressed in pounds, acceleration in feet per second per 
 
 second, and force in poundals; or when C.G.S. units are used throughout, 
 t Let z = Oil ■\- B and use the formula for the differentiation of a function 
 
 of a function as explained in Art. 34. 
 
60 CALCULUS, 
 
 equation (3) with respect to t we have: 
 
 ^ = coa cos (o)t -f 0) 
 
 and differentiating again, we have: 
 
 '«! 
 
 cPx 
 
 ^ = - co^a sin (ut + 6) (6) 
 
 or, substituting x for a sin {cot -\- 0), we have: 
 
 <Px 
 
 ^, = -<^ (6) 
 
 Therefore, if 
 
 „ k 
 
 m 
 
 (7) 
 
 equation (3) gives the required function, and a and B can have 
 
 any values whatever. 
 
 If the vibrations are started by drawing the body down to 
 
 a? = A, and if time is reckoned from the instant of release, then 
 
 dx 
 X — A when < = 0, and also -r: = when f = because the 
 
 at 
 
 body has no velocity at the instant of release. Using the values 
 :^ = and « = : 
 equation (3) becomes; 
 
 dx 
 
 -ji — ^ and i = in equation (4), we find = 0. Therefore 
 
 aj = a sin w< (8) 
 
 Using the values x = A and f = in equation (8), we find 
 a — A» Therefore, jor the case in which the body is pulled down to 
 X = A and released when t = 0, equation (8) becomes: 
 
 a; = A sin cot (9) 
 
 The motion which is defined by equation (9), or in general by 
 equation (3), is called simple harmonic motion. The distance x 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 61 
 
 Fig. 22. 
 
 of the vibrating body from its equilibrium position is at each 
 instant proportional to the sine of a 
 uniformly growing angle o)t; or, the 
 position P of the vibrating body is 
 at each instant the projection on the 
 straight line ah of a point P' which 
 moves round the circle CC at con- 
 stant angular velocity w, the radius 
 of the circle being A, as shown in 
 Fig. 22. 
 
 One complete vibration of the body in 
 Fig. 21 takes place while the point P' in 
 Fig. 22 makes one complete revolution, 
 that is, while the angle (at increases from zero to 2t radians,or while 
 
 the time t increases from zero to — seconds, to being expressed 
 
 CO 
 
 in radians per second. Therefore — , or 27r -^ a/— [see equation 
 
 (7)], is the period of one complete vibration. That is 27r -^ -W— 
 
 fk 
 is the number of seconds per vibration, or -yl— 4- 2ir is the number 
 
 of vibrations per second, or frequency, of the vibrating body. 
 
 In this discussion the mass of the spring and the effects of friction 
 are ignored. 
 
 49. Curvature. — Consider the curve cc in Fig. 23. The 
 tangent line a touches the curve at p and the tangent line h 
 touches the curve at q. Therefore the tangent line turns through 
 the angle ^ when the point of tangency moves from p to q. 
 The sharper the bend or curvature of the line cc the greater the 
 value of j8 for a given length of arc pq. Dividing the angle 
 iS (expressed in radians) by the length of the arc pq we have a 
 quotient which is used as a measure of the average curvature of cc 
 between p and q, and the limiting value of this quotient when 
 
62 
 
 CALCULUS, 
 
 the arc pq approaches zero is the measure of the curvature of cc 
 at the point p.* 
 
 It is required to find an expression for the limiting value of the 
 
 quotient — - — in Figs. 23 and 24 when the arc pq approaches 
 arc pq 
 
 zero. Let the coordinates of p be x and y, and let the coordi- 
 
 Fig. 23. Fig. 24. 
 
 nates of q be x -\- Ax and y + Ay as shown in Fig. 24. Then: 
 
 chord pq = V(Ax)2 + {AyY (1) 
 
 and when the arc pq approaches zero the arc becomes more and 
 more nearly equal to the chord so that the limiting form of equation 
 (1) is: 
 
 arc pq = V(dx)2 + {dy^ (2) 
 
 Furthermore we have: 
 
 X dy 
 
 tan a = -5^ 
 
 ax 
 
 (3) 
 
 according to Art. 18, where a is the angle shown in Fig. 23; and 
 
 dv 
 of course tan a' (see Fig. 23) is the value of ~ at the point q. 
 
 But ^ changes ^ times as fast as x, and therefore the change 
 
 * Or at q, because p and q approach coincidence as the arc approaches 
 zero. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 63 
 
 dv . d^v 
 
 in the value of — from p to q is ~' ^^» when Ax is infinitely 
 
 small. Therefore the value of -r at q is -~ + -^K ' dx, and 
 
 ax dx dx^ 
 
 consequently we have: 
 
 Now the angle jS in Fig. 23 is equal to a! — a, and, since a! — a 
 is an infinitesimal angle, we may write : 
 
 tan {a' —a) = a' — a 
 
 Therefore we have: 
 
 
 ^ = a' — a ^ tan {a! — a) 
 But 
 
 (5) 
 
 . , , s tan a' — tan a 
 
 tan (a' — a) = ., , . ^-r 
 
 1 + tan a' tan a 
 
 (6) 
 
 Therefore, substituting the values of tan a and tan a! from 
 equations (3) and (4), and writing j8 for tan {a! — a) according 
 to equation (5), we have: 
 
 ^•dx 
 dx^ 
 
 ^~^^x\dx'^d^^'^y 
 But the infinitesimal term in the denominator ( -~ • dx J may be 
 
 discarded because it is added to a finite quantity ( -r j- Therefore 
 
 equation (7) becomes: 
 
 ^-^ . dx 
 dx^ 
 
 P = /...x. (8) 
 
 1 + 
 
 \dx) 
 
64 CALCULUS. 
 
 Therefore, using the value of arc pq from equation (2) we have 
 
 fi 
 
 dx* 
 
 dx 
 
 arc pq 
 
 '^(si 
 
 ^(dxy + {dyy 
 
 (9) 
 
 This is the required limiting value of 
 
 i8 
 
 because equations (2) 
 
 arc pq 
 
 and (5) are limiting forms and because an infinitesimal has been 
 
 discarded in going from equation (7) to equation (8). Dividing 
 
 dx 
 
 by dx this factor 
 
 numerator and denominator of 
 1 
 
 becomes 
 
 Mty 
 
 ^{dxY +(d2/)2 
 and equation (9) becomes: 
 
 Curvature at p = 
 
 dx" 
 
 [■ - (2)T 
 
 (10) 
 
 Curvature of a circle. — The idea of curvature is necessarily 
 vague until we imderstand clearly the unit in terms of which 
 
 curvature is expressed. Therefore 
 let us apply the above definition 
 of curvature to a circle. The 
 line a, Fig. 25, is tangent to the 
 circle at p and h is tangent to the 
 ^ circle at 5, and the measure of the 
 
 curvature of the circle is — - — -, 
 arc pq 
 
 But the angle j8, between the 
 
 tangent lines is equal to the 
 
 angle between the radii Cp and 
 
 Cq, Therefore the angle i3, in radians, is equal to arc pq divided 
 
 by the radius r of the circle, and consequently the quotient 
 
 Fig. 25. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 65 
 
 — is equal to -. That is, the curvature of a circle is meas- 
 
 arc pq r 
 
 ured by the reciprocal of its radius, or the reciprocal of the curva- 
 ture of a circle is equal to the radius of the circle. Therefore the 
 reciprocal of the curvature of any curve at a point is called the radius 
 of curvature of the curve at the point. In accordance with this 
 statement equation (10) may be written: 
 
 [i + (^YV 
 
 {radius of curvature of any "I _ L \dx/ J .^ . ^ 
 
 curve at a point J d^y 
 
 1^ 
 
 the coordinates of the point p being substituted in the general 
 
 £ dy , d-y 
 expressions for -^ and -r%. 
 
 Remark. — At a point where a curve is horizontal (parallel to 
 
 du 
 the X-axis) the first derivative -p is equal to zero, and at such a 
 
 point the expression for radius of curvature reduces to -j^. Where 
 
 a curve is very nearly horizontal the first derivative -^ is small 
 
 as compared with unity, the expression 1 + ( ;p ) is sensibly 
 
 equal to unity, and the radius of curvature is sensibly equal to 
 1 
 
 dx' 
 
 Example. — Consider the parabola whose equation is y = x^, 
 
 dv d^xi 
 
 Then -^ = 2x and ■— = 2. To find the radius of curvature at 
 
 the origin of coordinates, substitute the values -p = 2x = and 
 6 
 
66 CALCULUS. 
 
 ~ = 2 in equation (11) which gives R= 2. To find the radius 
 of curvature at the point x = 1 and 2/ = 1, substitute the values 
 _| = 2x = 2 and^J = 2 in equation (10) which gives R = 5.6.* 
 
 Further discussion of curvatiwe. The osculating circle. — A 
 
 straight Une may coincide with a curve at a point and have the same 
 
 direction as the curve at the point. Such a straight line is called 
 
 a tangent line. 
 
 A circle may coincide with any curve at a point, have the same 
 
 direction as the curve at the point, and have the same curvature as 
 
 the curve at the point. To coincide with the curve at a point 
 
 means that the ordinate y of the circle is equal to the ordinate y 
 
 of the curve for a certain abscissa x. To have the same direction 
 
 dy dy 
 
 as the curve at the point means that -^ for the circle is equal to -~ 
 
 for the curve for the same value of x. To have the same curva- 
 
 ture as the curve at the point means that -v^ for the circle is 
 
 equal to —^ for the curve (if the condition of tangency is satisfied) 
 
 for the same value of x. The circle which satisfies these three 
 
 * Whenever an equation between x and y is to be plotted as a curve^ 
 and whenever definite values are to be determined for the derivatives of y 
 with respect to x, the question arises as to what units are to be used in the 
 expression of y and its various derivatives. Quantities such as 2 inches, 
 5 pounds and 6 hours, are called denominate quantities. The ratio of two 
 denominate quantities of the same kind is called a pure number. Both mem- 
 bers of an equation and every separate term in each member must have the 
 same denomination. Consider the equation y = ax^. In order to plot this 
 equation as a curve, x and y must both be expressed in inches, let us say. 
 Therefore the coefficient o is a denominate quantity which gives inches when 
 it is multiplied by inches squared. That is, the reciprocal of a is expressed in 
 inches. When an equation such as ?/ = 4x' is to be plotted as a curve, x 
 and y must both be expressed in inches, and the coefficient 4 must be thought 
 of as a denominate number, the denomination, or unit, being such as will give 
 y in inches when x is expressed in inches. 
 
 Hi 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 67 
 
 conditions at a point on a curve is called the osculating circle at 
 the point, and the radius of the osculating circle is called the 
 radius of curvature of the curve at the point. 
 
 Determination of the osculating circle. — One can easily find 
 
 the values of -^ and J at a given point p on a given curve cc 
 
 (see Fig. 26) by differentiating the equation of the curve. 
 Consider any circle whatever as shown in Fig. 27. The equation 
 
 y-axi8 
 
 ( X >i 
 
 Fig. 26. 
 of this circle is 
 
 ■a — 
 ~x- 
 
 Fig. 27. 
 
 {x - ay + (2/ - hf 
 
 (12) 
 
 where a and h are the coordinates of the center of the circle 
 and r is the radius of the circle. Differentiating equation (12), 
 we have 
 
 2{x - a) ' dx + 2{y -h) ' dy = 
 or 
 
 {x-a) + {y- h)u = (13) 
 
 du 
 where u is written for -~. Differentiating equation (13), re- 
 membering that y and u are both functions of x, we have: 
 dx -\- {y — h) • du -{- u * dy = Q 
 
68 CALCULUS. 
 
 Dividing through by dx and remembering that w • ^ = w^ we 
 have: 
 
 1 + (2/ - 6) • g + w^ = 
 
 du (Pv 
 
 But ^ is the second derivative ~. Therefore, representing 
 
 ^ by V, we have: 
 
 l-\-{y-h)v-\-u' = 
 or 
 
 2/ - 6 = -— (14) 
 
 and, substituting this value of {y — h) in equation (13), we have: 
 
 V 
 
 Substituting the values of (x — a) and {y — h) from equations 
 (15) and (14) in equation (12), we get: 
 
 , = a±j^' (16) 
 
 Now u and v and x and y are all known at a given point p 
 of the curve cc of Fig. 26, and these four quantities u, v, x and y 
 have the same values for the osculating circle where it touches the 
 curve cc at the point p. Therefore a and h can be found 
 from equations (14) and (15), and r can be determined from 
 equation (16) so that the osculating circle at the point p is com- 
 pletely determined. 
 
 PROBLEMS. 
 
 1. Find the radius of curvature at the point p of the parabola 
 shown in Fig. pi. 
 Ans. 40.5 inches. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 69 
 
 2. Find the radius of curvature of the ellipse shown in Fig. p2, 
 (a) at the point p, and (6) at the point p'. 
 
 Fig. pi. 
 
 Ans. (a) 2.67 inches, (b) 18 inches. 
 
 dy . 
 
 Fig. p2. 
 
 Note. — The value of -^ is infinite where a curve is parallel to the i/-axis 
 
 and generally the value of ^ is also infinite at such a point. Therefore the 
 
 expression for radius of curvature at such a point of a curve usually becomes 
 
 indeterminate. It is evident, however, that x and y may be interchanged 
 
 throughout the entire discussion of Art 49. This interchange gives an ex- 
 
 dx d^x 
 
 pression for radius of curvature in terms of t- and ^ where x is thought 
 
 of as a function of y] and this expression can be used to determine the radius 
 of curvature at a point on the curve where the curve is parallel to the y-axis. 
 
 Fig. p3. 
 
 18 inches 
 
 Fig. p5. 
 
 3. A stone arch is to be made as shown in Fig. p3, the two 
 dotted circles having the same curvature as the ellipse at the 
 junction points pp. Find the radius of the circles. 
 
 Ans. 14.07 feet. 
 
70 
 
 CALCULUS. 
 
 center of curve 
 
 4. Find the radius of curvature of the curve y = sinx at the 
 points (a) a; = and (h) a: = ^. 
 
 Ans. (a) infinity, (6) 2.6. 
 
 5. Find the radius of curvature at the point p of the sine 
 
 curve shown in Fig. p5. 
 
 Ans. 5.47 inches. 
 
 6. A straight portion of 
 railway track merges gradually 
 into a circular curve as indi- 
 cated by the dotted line in 
 Fig. p6. The dotted portion 
 is called a transition curve. 
 The curvature of the track 
 
 (which is sensibly equal to ^~ 
 
 RM. 
 
 100 feet 
 
 Fig. pQ. 
 
 because 
 
 dy . , 
 
 ~ IS very nearly 
 
 zero) increases uniformly from zero at x = to t^ftf: at a: = 100. 
 
 Find the equation of the dotted transition curve. 
 Ans. 900,000 y = a^. 
 
 Note. — If the curvature v 2 increases uniformly its rate of change, ^-j, 
 is a constant. 
 
 50. Geometric differentiation. The acceleration of a particle 
 which travels in a circular orbit. — There are cases in which the 
 instantaneous rate of change of a given quantity may be found 
 when the quantity is specified in geometric terms. An important 
 example is the following: A particle travels round and round a 
 circular path or orbit at velocity v, and it is required to find the 
 
 deceleration ^ of the particle. 
 
 At a certain instant the particle is at P, Fig. 28, and its velocity 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 71 
 
 at this instant is represented in direction and magnitude by the 
 Hne vi. During a short interval of time the particle will travel 
 
 Fig. 28. 
 
 Fig. 29. 
 
 the distance v.dt, the particle will then be at Q, and its velocity 
 will be as represented by the line v^ which has the same length as Vi. 
 
 From any fixed point 0' draw two lines parallel and equal to 
 vi and V2 respectively, as shown in Fig. 29. Then the line 
 dv represents the velocity which must be added* to Vi to give i;2, 
 that is the line dv represents the increment of the velocity of the 
 particle during the interval dt. 
 
 Now when dt is chosen smaller and smaller, the angle <(> ap- 
 proaches zero, and the angles at P' and Q' approach 90° so that 
 dv becomes more and more nearly parallel to the radius OP. 
 Also when dt is very small the arc PQ ( = v-dt) is sensibly a 
 straight line and the two triangles OPQ and O'P'Q' are similar. 
 Therefore from these similar triangles we have : 
 
 r : v.dt : :v:dv 
 
 (1) 
 
 In this proportion v is written instead of Vi or V2, both of which 
 
 * We are here concerned with what is called vector addition or geometric 
 addition. The most familiar example of this kind of addition is the addition 
 of two forces by the principle of the "parallelogram of forces." 
 
72 
 
 CALCULUS. 
 
 have the same numerical value. Solving this proportion we have; 
 
 dv 
 dt 
 
 (2) 
 
 We have already proceeded to the limit in considering the arc 
 PQ a straight line. That is, the acceleration of a particle moving 
 in a circular orbit is equal to the square of the velocity of the 
 particle divided by the radius of the orbit, and the acceleration is 
 at each instant towards the center of the circle because the infin- 
 itesimal increment of velocity dv in Fig. 29 is in the direction of 
 PO in Fig. 28. 
 
 51. Geometric differentiation. The inward force per unit 
 length of a barrel hoop. — Another important case of geometric 
 differentiation is the following: Figure 30 represents a hoop 
 
 around a tank. It is required to find the value of -r where dF 
 
 is the force with which a short length ds of the hoop pushes inwards 
 on the tank. 
 
 The radius r of the tank and the tension T of the hoop are 
 given. The tension of the hoop is the force with which a portion 
 of the hoop pulls on an adjoining portion. Thus the two forces 
 
 tank 
 
 Fig. 30. 
 
 Fig. 31. 
 
FORMULAS FOR DIFFERENTIATION AND INTEGRATION. 73 
 
 Ti and T2, Fig. 30, which pull on the portion ds of the hoop 
 in Fig. 30, are each equal to the tension T, and the resultant 
 of the two forces Ti and T2 is the total inward force dF exerted 
 on the portion; and this total inward force acting on ds is the 
 force with which ds pushes against the tank. 
 
 The resultant of Ti and T2 is shown in Fig. 31, and the triangle 
 aVh' in Fig. 31 is similar to the triangle aOb in Fig. 30. There- 
 fore we have 
 
 r :ds::T :dF 
 so that 
 
 ? = ^ (1) 
 
 ds r 
 
 We have already proceeded to the limit in considering aOh to 
 be a triangle, that is, in considering ds to be a straight line, 
 which is only true when ds is infinitely small. 
 
 The practical meaning of equation (1) can be best understood 
 
 dP 
 by thinking of -p as the really important quantity under con- 
 sideration, not by thinking of the meaning of F* 
 
 * Indeed one gets into unfamiliar vector theory when one tries to consider 
 the meaning of F. 
 
CHAPTER III. 
 INTEGRATION. 
 
 52. Integration as an arithmetical argument. — ^Any argument 
 which leads to a proposition concerning total change when a rate 
 of change is given may properly be called integration. Thus the 
 following arguments are the simplest existing examples of inte- 
 gration: 
 
 (a) A man earns money at the constant rate of two dollars per 
 day. Therefore the man earns two dollars each day, and in 
 100 days he would earn 200 dollars. Or, to get the amount of 
 money earned in 100 days multiply two dollars per day by 100 
 days.* 
 
 (h) One man A earns money twice as fast as another man B. 
 Therefore in any interval of time whatever A earns twice as 
 much as B. 
 
 53. Integration as a mechanical process. Integrating machines. 
 
 — The simplest and most familiar integrating machine is the 
 ordinary cyclometer. The wheel of a bicycle turns at a speed which 
 is proportional to the velocity of travel of the bicycle. Therefore 
 the number of revolutions of the wheel is proportional to the 
 distance traveled, and the cyclometer is a revolution counter 
 arranged to iesA one additional unit for each mile traveled. 
 
 The ordinary "electricity meter "f which is installed in con- 
 nection with electric lamps is an integrating machine. The 
 
 * If one can get 200 dollars by multiplying two dollars per day by 100 days, 
 one might well ask how to perform such a profitable operation? The answer 
 is: Work hard for 100 days! That is what this particular multiplication 
 means. Nearly every mathematical process rightly understood relates to 
 physical reality in a manner as definite and as exacting as this particular case 
 of multiplication. 
 
 t Properly called a watt-hour meter. 
 
 74 
 
 1-lfer- 
 
INTEGRATION. 75 
 
 spindle of the meter turns at a speed which is proportional to the 
 rate of delivery of energy (in watts) to a customer. Therefore the 
 number of revolutions of the spindle is proportional to the total 
 amount of energy delivered. The clock-work of the watt-hour 
 meter is a revolution counter which is arranged to read one addi- 
 tional unit for each watt-hour of energy delivered. 
 
 54. The planimeter.* — The planimeter is an integrating machine 
 for measuring area, and it depends upon the following kinematical 
 principle : 
 
 Consider a line AB, Fig. 32, moving in any manner whatever 
 in the plane of the paper. The motion of the line at each instant 
 may be thought of as a motion of translation combined with a motion 
 of rotation about an arbitrary point p.f 
 
 Let V be the component at right angles to AB of the velocity 
 of translation, as shown in Fig. 32, and let to be the angular 
 velocity of AB about the point p. It is desired to find the rate 
 at which the line AB sweeps over area, area swept over by the 
 line being considered as positive when the line sweeps over it from 
 right to left to an observer looking from A towards B. 
 
 During a short interval of time, A^, the motion of translation 
 carries the line from AB to A'B' as shown in Fig. 34, and the 
 
 * The planimeter which is here described is the Amsler planimeter. 
 
 The student wishing to become familiar with various kinds of integrating 
 machines should consult Les Integraphes Abdank-Abakanowicz, Paris, Gauth- 
 ier-Villars. 
 
 A discussion of integrating machines is also given in Encyclopddie der Mathe- 
 matischen Wissenschaften, Vol. II. This great work is now appearing in a re- 
 vised form in a French translation. 
 
 See also Report on Planimeter s, British Association Annual Report, 1894, 
 pages 496-523. 
 
 See also Mechanical integrators, Shaw, Proceedings of the Institution of Civil 
 Engineers, London, 1885, pages 75-143. 
 
 The earhest type of integrating machine for use in harmonic analysis is 
 that of Lord Kelvin. This machine is described in Franklin's Electric Waves, 
 pages 240-242, The Macmillan Co., New York, 1909. 
 
 t See Franklin and MacNutt's Mechanics and Heat, Arts. 74-77, pages 
 174 and 175, The Macmillan Co., N. Y., 1910. 
 
76 
 
 CALCULUS. 
 
 area swept over by the line is I X v-At. Dividing this area by 
 At gives the rate at which the Une sweeps over area because of its 
 motion of translation. 
 
 / Sf center of 
 / < I 
 
 Fig. 32. 
 
 During a short interval of time, A<, the motion of rotation 
 carries the Une from AB to A"B" ^ as shown in Fig. 35, and the 
 
 Fig. 35. 
 
 \t ■ 
 area swept over consists of two parts as shown. The positive 
 
 area 
 
 \(l \2 
 
 in Fig. 35 is equal to 2(2 + '^) X w-A<, the negative 
 
 area 
 
 in Fig. 35 is equal to 
 
 \%-) 
 
 X oo'At, and the total area swept 
 
 over is equal to IDo) • At. Dividing this area by At gives the rate 
 at which the line sweeps over area because of its motion of rotation. 
 
INTEGRATION. 77 
 
 Therefore the total rate at which the line sweeps over area 
 
 /dA\ . 
 
 is equal to Iv + IDoo, or, using -77 for w, we have 
 
 Let AB be a metal bar carrying a wheel mounted as shown in 
 Fig. 36 (which is a side view), and let the rim of this wheel roll on 
 the paper as the bar AB moves, as shown in Figs. 36 and 37. 
 
 y^middle of bar W^u„^ ao 
 
 "\|— 4— U paper 
 
 j.,,roUinff 
 I * \ '^y/ wheel 
 
 side view paper 
 
 Fig. 36. Fig. 37. 
 
 Let ^ be the total angle turned by the wheel in a given time; then 
 ^- is the angular velocity of the rolling wheel, and it is equal to -, 
 where r is the radius of the wheel, as shown in Fig. 37. There- 
 fore we may substitute r -^ for v in equation (1) and we have: 
 
 in which Ir -^ is the rate of sweeping over area because of the 
 
 do 
 translatory motion of AB, and ID -r. is the rate of sweeping over 
 
 area because of the rotatory motion of AB. 
 
 Now if the rate of sweeping area due to the translatory motion 
 of AB is Ir times the rate of turning of the wheel, then the 
 total area swept over because of the translatory motion of AB is 
 Ir times the total angle rp turned by the wheel. 
 
 Similarly, if the rate of sweeping area due to the rotatory motion 
 
78 
 
 CALCULUS. 
 
 of AB is ID times the rate of turning of AB, then the total 
 area swept over because of the turning of AB is ID times the 
 total angle 6 turned by AB. 
 
 Therefore the total area swept over by AB is Irrj/ + IDd.* 
 That is, 
 
 A = H-\- IDS (3) 
 
 If the bar AB comes back to its initial position without turning 
 completely round, then = and equation (3) becomes 
 
 A =^lri 
 
 (4) 
 
 That is, the total area swept over by AB is proportional to the 
 angle ^ turned by the rolling wheel, and the value of Ir may 
 
 be so chosen that one square inch 
 corresponds to each revolution of 
 the wheel. In this case square 
 inches of area are indicated by 
 whole revolutions of the wheel, 
 and the circumference of the wheel 
 may be divided so as to indicate 
 fractions of a square inch. 
 
 In the actual planimeter one end 
 
 of the bar AB is constrained to 
 
 move back and forth along a circlef 
 
 CC, Fig. 38, while the other end 
 
 A travels once round the closed 
 
 curve CC which bounds the area to be 
 
 measured. Then any area outside of 
 
 CC which is swept over by AB at all is swept over as many times 
 
 to the left as to the right, but every portion of the shaded area is 
 
 swept over once more to the left than to the right. Therefore, 
 
 * How much more easily miderstood is this argument than to say integrating 
 equation {2) we have equation {3) ! and yet this brief statement means exactly 
 what is given in the above argument. 
 
 t Along a straight line in some forms of planimeter. 
 
 Fig. 38. 
 
 .m^ 
 
INTEGRATION. 
 
 79 
 
 in view of the agreement as to algebraic signs, the total area 
 A (= lr\J/) swept over by the line AB is equal to the shaded area. 
 
 55. Integration by steps. — In the laying out of a new electric 
 railway the engineer usually makes a study of the probable 
 schedule speed of the car in order to be able to estimate the traffic 
 income and to be able to choose suitable driving motors. In the 
 
 i€ 
 
 « 8 
 
 
 
 
 
 
 
 
 
 1^' 
 
 — 
 
 
 / 
 
 
 
 "*^ 
 
 
 r^ 
 
 /" 
 
 
 
 ^ 
 
 
 
 
 / 
 
 / 
 
 ■^ 
 
 
 
 
 
 
 
 / 
 
 / 
 / 
 
 
 
 
 
 
 
 
 9' 
 
 
 
 
 
 V 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 1/ ^^ 
 
 
 
 
 
 
 
 
 
 1 
 
 24 
 
 4 •«: 
 
 10 20 30 40 50 60 70 80 <y3 
 
 time in seconds 
 
 Fig. 39. 
 
 study of a particular run of a car, the curve A, Fig. 39, is deter- 
 mined from the known characteristics of a motor* (tentatively 
 chosen), the weight of the car, and the curvature and grade of 
 the track. The curve A gives the speed of the car as a function 
 of elapsed time and it is called a speed-time curve. In the particular 
 case represented in Fig. 39 the car is being accelerated from to 
 45 seconds; at 45 seconds the power is turned off and the car is 
 allowed to coast; and at 78 seconds the brakes are applied and the 
 car is rapidly brought to rest. The ordinate of curve A is the 
 velocity v of the car, the product v-dt is the distance traveled 
 
 during the infinitesimal interval dt, and I v-dt is the total 
 
 distance traveled by the car from the start up to any given instant 
 * These characteristics are furnished by the manufacturing company. 
 
80 CALCULUS. 
 
 L This distance is shown as a function of t in the accompanying 
 
 SPEED AND DISTANCE TABLE OF AN ELECTRIC STREET CAR 
 
 
 (See Fig. 39). 
 
 
 Time in 
 
 Speed: Miles 
 Fer Hour. 
 
 DiBiance 
 
 Seconds. 
 
 . In Feet 
 
 
 
 
 
 
 
 10 
 
 15.2 
 
 111 
 
 20 
 
 22.0 
 
 384 
 
 30 
 
 23.6 
 
 719 
 
 40 
 
 24.2 
 
 1,071 
 
 45 
 
 24.3 
 
 1,246 
 
 60 
 
 21.6 
 
 1,751 
 
 78 
 
 19.4 
 
 2,292 
 
 91 
 
 
 
 2,476 
 
 table and also it is shown by the ordinate of the curve BB in 
 Fig. 39. The tabulated values of the distance traveled by the 
 car are calculated approximately as follows : From to 10 seconds 
 
 the average speed of the car is approximately ^ — - miles per 
 
 hour, and the distance traveled during this 10-second interval is 
 found by multiplying the average speed (reduced to feet per 
 second) by 10 seconds which gives 111 feet. From 10 seconds 
 to 20 seconds the average speed of the car is approximately 
 
 — ■ — ^ — miles per hour, and the distance traveled during 
 
 this 10-second interval is found by multiplying the average speed 
 (reduced to feet per second) by 10 seconds, which gives 273 feet, 
 so that the total distance traveled during 20 seconds is 111 feet + 
 273 feet, which is 384 feet; and so on. 
 
 PROBLEMS. 
 
 1. The accompanying figure, pi, shows the stretching force 
 per square inch of section and the fractional elongation of a ] 
 sample of steel under test. The sample is one inch in diameter ] 
 and the portion which is stretched is 4 inches long initially. Find 
 
INTEGRATION. 
 
 81 
 
 the work in foot-pounds expended on the specimen up to the 
 breaking point h. 
 Ans. 53.1 foot-pounds. 
 
 Note. — Multiplying abscissas in Fig. pi by 4 inches gives actual elongations 
 in inches, and multiplying ordinates by square inches of section of sample 
 gives actual stretching forces in pounds. Let elongation in inches be e and 
 let stretching force be F. Then work done is the integral of F-de. 
 
 2. Find the work (in ergs and in foot-pounds) required to 
 magnetize a wrought iron bar 3 inches by 3 inches by 20 inches 
 
 <s 
 
 m <tm 
 
 
 
 
 
 
 
 
 
 ■■ 
 
 
 
 
 ■■ 
 
 ■■ 
 
 
 
 ~ 
 
 ■" 
 
 " 
 
 ■" 
 
 "" 
 
 
 ■" 
 
 "■ 
 
 ■■ 
 
 "" 
 
 
 ■" 
 
 "" 
 
 
 "■ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 - 
 
 
 - 
 
 
 
 
 
 
 
 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 1^ 
 
 
 
 — 
 
 ■* 
 
 ^' 
 
 
 
 
 
 1 
 
 ■^ 
 
 .^ 
 
 
 
 /- 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 J 
 
 X 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 \t 
 
 
 
 
 
 
 , ' 
 
 s 
 
 s 
 
 
 
 
 ^ 
 
 
 
 
 
 
 y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 s 
 
 
 
 
 
 
 
 / 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 1 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 . 
 
 « 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \\ 
 
 S; 
 
 f> J 
 
 f 
 
 
 
 
 
 
 
 
 
 
 
 
 
 P 
 
 y 
 
 
 v 
 
 /\ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 - 
 
 Kr 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ? 
 
 
 
 
 
 
 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 ^45 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 g 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 § 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Vj 
 
 
 
 
 
 
 
 
 
 
 
 <. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 B 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^•'" 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 g 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 «7r 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 % 
 
 
 /^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 _ 
 
 
 
 
 
 
 
 
 
 
 
 
 5, JO 15 .20 25 30 
 
 ebieUsi, bandredtba tot curve A, tea-tbouaandUi9 for curv» B- 
 
 Fig. pi. 
 
 from a neutral condition to B = 16,000 lines per square centi- 
 meter. 
 
 Ans. 326 X 10^ ergs or 2.406 foot-pounds. 
 
 Note, — ^The work done in magnetizing iron is: 
 
 '^ = If^ 
 
 dB 
 
82 CALCULUS. 
 
 where W is expressed in ergs and V is the volume of the iron in cubic cenf i- 
 meters. There are 30.5 X 453.6 X 980 ergs in one foot-pound. 
 
 TABLE OF B AND H FOR WROUGHT IRON. 
 
 H B 
 
 10 12,400 
 
 20 14,330 
 
 30 15,100 
 
 40 15,550 
 
 50 15,950 
 
 60 16,280 
 
 70 16,500 
 
 3. The force F which acts on a body is given as a function of 
 the time in the accompanying table. Force being expressed in 
 poundals and time in seconds. The mass of the body is 1,000 
 pounds. Find the velocity produced from ^ = to i = 40 
 seconds. 
 
 Ans. 27.1 feet per second. 
 
 dv F F 
 
 Note. — The acceleration -j-. = — or dv = —'dl or the velocity v is equal 
 dt m m J -i 
 
 to — times the integral of F-dt; velocity being expressed in feet per second. 
 
 TABLE OF F AND L 
 
 t in Second!. F in Poundali. 
 
 500 
 
 10 625 
 
 20 700 
 
 30 745 
 
 40 785 
 
 56. Algebraic integration. — Integration, in the sense in which 
 
 this word is used throughout this text, is the recognition of a 
 
 given differential expression as the differential of a known function, 
 
 and the rules for differentiation which are given in Chapter II 
 
 are rules for integration also. Thus the differential of log x + C 
 
 dx dx 
 
 is — , and therefore, by definition, the integral of — is loga;+C, 
 
 X X 
 
INTEGRATION. 83 
 
 where C is any constant. The rules given in Chapter II cover 
 differentiation completely, or nearly so, but they do not cover 
 integration completely. Indeed complete rules cannot be given 
 for integration.* 
 
 In the recognition of a given differential expression as the 
 differential of a known function, two things are helpful, namely, 
 (a) A table of standard forms or functions with their differentials 
 (such a table is usually called a table of integrals), and (6) some 
 degree of familiarity with the transformations which may be 
 used to reduce a given differential expression to a standard form 
 or to a combination of standard forms. 
 
 A table of integrals is given in Appendix B. 
 
 57. Transformation of differential expressions. — When a differ- 
 ential equation has been found by differentiating a given function, 
 the integral of the differential equation is of course known. Many 
 of the forms in the table of integrals have been established in 
 this way as problems in differentiation in Chapter II. When a 
 given differential equation has not been found by differentiating 
 a known function it is in many casesf possible to invent a scheme 
 for transforming the given differential expression so as to reduce 
 it to one or more of the standard forms in the table of integrals. 
 The most important^ of these transformations are embodied in 
 the following rules, and a few of the lesser important transforma- 
 tions are illustrated by problems. 
 
 As a very simple example consider the differential equation 
 
 * Integration by series is a universal and systematic method of integration. 
 
 dv 
 Any given algebraic expression for -^ can be expanded by Maclaurin s 
 
 theorem, each term of the series so obtained can be integrated according to 
 form 1 in the table of integrals, and the resulting series, when it is convergent 
 
 is the integral of the given expression for -p . 
 
 t Not in every case, by any means; many integrals cannot be expressed in 
 terms of the elementary functions which are used in the table of integrals. 
 
 X Transformations depending upon the use of imaginaries are also very 
 important. See Art* 93 in Chap. VI. 
 
84 CALCULUS. 
 
 which was found in the discussion of the arc of a parabola in 
 Art. 20, namely: 
 
 ds = VT+lfc2^2 . dx (1) 
 
 This expression is easily reduced to: 
 
 ds=2ky]^,-hx''dx (2) 
 
 and according to rule I, below, we have: 
 
 h^li' + *= • dx = 2kf^± +x^.dx (3) 
 
 Therefore, writing a^ for — , we get an expression which is 
 
 identical to form 57 in the table of integrals. 
 
 Rule I. — Any constant factor may be removed from under the 
 integral sign. Thus: 
 
 J a ' du = ajdu 
 
 where a is a constant and u is any differential expression what- 
 ever. 
 
 Rule n. — The integral of the sum of two differential expressions 
 is equal to the sum of the integrals of the respective expressions. 
 Thus: 
 
 J(du + dv) = f du -\- jdv 
 
 where du and dv are any differential expressions whatever. 
 
 Rule in. — When a differential expression is of the form w" • du 
 its integral is, of course, given by form 1 or by form 2 of the table 
 of integrals, whatever the function u may be. 
 
 For example consider the differential equation: 
 
 dy — sin^ x cos x * dx (4) 
 
 Let mix = u then sin^ x = u^ and co^x • dx = duj and 
 equation (4) becomes: 
 
 dy = u^ • du 
 
 ."ii^ii- 
 
INTEGRATION. 85 
 
 whence 
 
 y^iu' + C 
 
 or, substituting sin a; for u we have. 
 
 y = i sin^ x -j- C (5) 
 
 For problems see group 7 in the appendix. 
 
 Rule IV. Integration by parts. — This rule is expressed thus: 
 
 fu'dv = uv —fv • du (6) 
 
 This formula is sometimes useful when a given differential ex- 
 pression can be resolved into two factors, namely, (a) any function 
 u, and (6) a familiar differential expression dv. 
 
 This rule is based on Art. 32 where it is shown that: 
 
 d(uv) = u • dv -\- V ' du 
 which is to say: 
 
 uv =fu • dv +fv ' du 
 
 so that, by transposing we get equation (6). 
 For example consider the differential equation: 
 
 dy = X log X ' dx (7) 
 
 In this case x - dx(= dv) is a familiar differential expression, 
 that is, it is the differential of ix^{= v). Therefore, using equa- 
 tion (6), we have: 
 
 J u • dv = u ' v —f V ' du (6) 
 
 f{\ogx)(x'dx) = {\ogx)(ix')-fiix')(j) (8) 
 
 But (ia;2) f — j is Jx • dx, and its integral is ix^ + C. Therefore 
 equation (8) becomes: 
 
 fx log X ' dx = ix^ log X - ix^ - C (9) 
 
 For problems see group 8 in the appendix. 
 
86 CALCULUS. 
 
 Special methods. Some differential expressions require more 
 or less elaborate transformations or substitutions to get them into 
 forms whose integrals can be recognized, or easily obtained by the 
 simple transformations of rules III and IV above. Among these 
 special methods are the following: 
 
 (a) Integration of rational fractions by resolution into partial 
 fractions. For 'problems illustrating this method see Group 9 in the 
 Appendix. 
 
 (6) Integra tion of expressions involving Va^ —x ^, V a;^ + a* 
 or V x^ — a^. Such expressions can usually be reduced to recog- 
 nizable ^ormsJ)y making the following substitutions. 
 
 In Va^ — x^ substitute a sin d for x. 
 
 In V^^^J^ substitute a tan d for x. 
 
 In Vx^ -f a^ substitute a sec $ for x. 
 
 Problems illustrating these trigonometric substitutions are given in 
 Group 10 of the Appendix. 
 
 (c) Miscellaneous Substitutions. A few problems involving mis- 
 cellaneous substitutions are given in Group 11 in the Appendix. 
 
 A good discussion of integration transformations is given on 
 pages 32-70 of W. E. Byerly's Integral Calculus ^ Ginn & Co., 
 Boston, 1881 J second e dition^ 1^ 
 
CHAPTER IV. 
 PARTIAL DIFFERENTIATION AND INTEGRATION. 
 
 58. Differentiation of a function of two variables. — Consider a 
 rectangle of length x and breadth y as shown in Fig. 40. The 
 
 area of the rectangle is: 
 
 A = xy (1) 
 
 that is, the area is a function of the two variables x and y. 
 Writing A -\- AA for A, 
 
 X + Ax for X, and 
 
 y -\- Ay for y, we have: 
 
 A + AA = (x + Ax){y + Ay) 
 or 
 
 A-\-AA='xy-\-X'Ay-\-y'Ax-{-Ax'Ay (2) 
 
 whence, subtracting equation (1) from equation (2) member by- 
 member, we have 
 
 AA ==^ X ' Ay ■\- y ' Ax ■\- Ax ^ Ay (3) 
 
 But when Ax and Ay are infinitely small we may drop the 
 second order infinitesimal Ax • Ay, and 
 we get: 
 
 dA = X • dy ■{• y ' dx (4) 
 
 Now X ' dy is the infinitesimal incre- 
 ment of A when y alone increases, 
 and y ' dx \& the infinitesimal incre- 
 ment of A when x alone increases. 
 Therefore, according to equation (4), the infinitesimal increment 
 of A when x and y both increase is equal to the sum of the two 
 infinitesimal increments: — (a) that which is due to the increase of x 
 alone and (6) that which is due to the increase of y alone. 
 
 87 
 
88 CALCULUS. 
 
 This proposition is a special case of a general theorem which is 
 expressed by the equation: 
 
 dz 
 where — is the derivative of z with respect to x on the assump- 
 
 dz 
 tion that y is constanty and — is the derivative of z with respect 
 
 dz 
 Uyyonthe assumption that x is constant; that is -^ * dx is the 
 
 ox 
 
 infinitesimal increment of z due to an increment of x alone, 
 
 IT ' dy is the infinitesimal increment of z due to an increment 
 dy 
 
 of y alone, and dz I =-r~ ' dx + — - dy ) is the infinitesimal 
 
 increment of z when x and y both increase. 
 
 It is customary to use the symbol d instead of d in partial 
 differentiation, but to do so is apt to be misleading because it 
 gives the impression that there is an unexplained and mysterious 
 difference between ordinary and partial differentiation. It is 
 indeed allowable to use d instead of d because one always knows 
 when one is dealing with a function of more than one independent 
 variable. 
 
 59. Successive partial differentiation. — Let y he a. function 
 of a single independent variable x. Then the rate of change of y 
 with respect to x is called the derivative of y with respect to x. 
 But this derivative is itself a function of x and its rate of change 
 with respect to x is called the second derivative of y with respect 
 to x; and so on as explained in Art. 46. 
 
 Let 2 be a function of two independent variables x and y. 
 
 Then the two first derivatives r- and t- are themselves functions 
 
 ax dy 
 
 of x and y, and each of these first derivatives has its derivative 
 
 with respect to x and its derivative with respect to y. For 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 89 
 
 example, if 2 = T^y^y then — = 3xy(=a), and — =3a;y(=/3); 
 
 and each of these first derivatives is a function of x and y. The 
 •\ »\ 
 
 derivative t~ (= ") ^^s two derivatives, namely, t- ( = 6x2/^) 
 ox ox 
 
 and ^(= OxV); and the derivative — (= /S) has two deriva- 
 tives, namely, ^ (= 9xV) and 7- (= 6x^1/), 
 
 dz d^Z 
 
 The derivative of — with respect to x is indicated as r^. 
 
 62 d^Z 
 
 The derivative of — with respect to y is indicated as —. 
 
 uX oy ' ox 
 
 dz d^z 
 
 The derivative of — with respect to x is indicated as -r r-. 
 
 dy ^ dx ' dy 
 
 The derivative of — with respect to y is indicated as r-r. 
 
 In the above example, namely, when z = oi^y^j the two second 
 
 d^z d^z 
 
 derivatives, -r-^- and r-r-, are each equal to 9xV. That is 
 dydx dxdy ^ ^ 
 
 these two second derivatives are identical. Indeed the two deriva- 
 tives T-^ and ^-T- are always identical when z is a function of 
 
 X and y. This proposition is very important in mathematical 
 physics. 
 
 60. Partial differential equation.* — Let 2 be a function of x 
 and y concerning which it is known that 
 
 1= ax^ (1) 
 
 and let it be required to find an expression for z. Now y is 
 
 dz 
 
 assumed to be a constant when the derivative r- is found from a 
 
 OX 
 
 * A very interesting example of a partial differential equation is discussed 
 in Art. 90. 
 
90 
 
 CALCULUS. 
 
 given function, and therefore y is to be looked upon as a constant 
 in equation (1) when one tries to think of the function from which 
 equation (1) is derived, that is, ay in equation (1) is to be con- 
 sidered as a constant, so that we may write 
 
 ay = h 
 
 (2) 
 
 and consider what function of x has a derivative equal to hx^. 
 Evidently the desired function of x is [^bx^ + C]. But in this 
 argument y is assumed to be constant, so that the constant of 
 integration C can be any function whatever of y. Therefore, 
 writing f(y) for C and using ay for h, we have Hayx^ -{- f{y)] 
 as the most general algebraic expression which gives ax^y when 
 differentiated with respect to x, and this is therefore the desired 
 expression for z. That is: 
 
 z = iai^y+f{y) (3) 
 
 where f(y) represents any function of y whatever. Of course 
 this function f{y) may include a term which does not contain y, 
 as in ay^ + hy + c, or indeed it may be a simple constant which 
 does not contain y at all. 
 
 Equation (1) expresses the law of growth of z with respect 
 to X, and it is called a partial differential equation because z is 
 a function of more than one independent variable. 
 
 Example of an ordinary differ- 
 ential equation. 
 
 If 
 
 then 
 
 dz 
 dx 
 
 ax^ 
 
 (4) 
 
 (5) 
 
 z==iax^-{-C 
 
 where C is a constant which 
 
 may have any value whatever. 
 
 If we place x = inequation 
 
 Example of a partial differ- 
 ential eqvxition. 
 
 If 
 
 then 
 
 52 
 
 dx 
 
 = ax"^ 
 
 (6) 
 
 z^\a:>?-\-f{y) (7) 
 
 where f{y) is any function 
 whatever of y which does not 
 depend on x. 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 
 
 91 
 
 (5) we have z = C. Therefore 
 the constant C is determined 
 if we know the value of z when 
 a; = 0. 
 
 The constant C is called the 
 \ constant of integration. 
 
 If we place x = in equa- 
 tion (7) we have z = f{y). 
 Therefore the function f{y) is 
 determined if we know 2 as a 
 function of y when a; = 0. 
 
 The function f{y) is called 
 the function of integration. 
 
 The above discussion refers to very simple partial differential 
 equations. As another example let us set up the partial differ- 
 ential equation which characterizes a given function. Let z be 
 any function whatever of s where s stands for x + ay. Then 
 ds 
 dx 
 
 = V 
 
 , ds 
 and -y- = a. 
 dy 
 
 Also, according to Art. 34, we have: 
 
 dz 
 dx 
 
 dz 
 ds 
 
 -r = -r • -7- and -r = -i- 
 
 dx 
 
 Therefore, using the values, 
 dz 
 
 ds 
 dx 
 
 = 1 and 
 
 d2 
 dx 
 
 and 
 
 dz _ dz 
 dy ds 
 ds 
 dy 
 dz dz 
 dy 
 
 dy 
 
 = a. 
 
 we have 
 
 ds 
 
 and consequently: 
 
 dz 
 
 dy 
 
 = a 
 
 dz 
 
 dx 
 
 (8) 
 
 This is, of course, a partial differential equation, and it is satisfied 
 by any function whatever of {x -\- ay). For example it is satis- 
 fied by z = X -{- ay, by z = {x -\- ayy, by sin {x + ^y), by 
 log {x + ay), by e^^^', etc. 
 
 6L Differentiation of an implicit function. — Any equation 
 between x and y defines 2/ as a function of x^. If the equa- 
 tion is solved for y, we have y as an explicit function of x; 
 otherwise the equation defines y as an implicit function of x. 
 
 * In this discussion the symbol d is used instead of d for partial differentia- 
 tion. 
 
 t Or X as a function of y. 
 
92 CALCULUS. 
 
 Thus 
 
 6X2 ^ IQ^yS + 2/4 + 5 == (1) 
 
 defines y as an implicit function of x. It is convenient to repre- 
 sent the entire left-hand member of equation (1) by the letter /. 
 Imagine for a moment that / need not be equal to zero as required 
 
 by equation (1), then ^ • dx would be the increment of / due 
 
 to an increment of x, and ^ • dy would be the increment of / 
 
 due to an increment of y. But according to equation (1) / is 
 always equal to zero. Therefore the increments dx and dy 
 must be so related to each other as to make the sum of the two 
 increments of / equal to zero. That is we must have: — 
 
 ) (2) 
 
 (3) 
 
 when y is an implicit 
 
 function of a; as defined by the equation / = 0, where / is any 
 algebraic expression involving x and y. 
 
 In the above expression the two differentials dx and d-^ are 
 called increments for the sake of brevity and clearness. A differ- 
 ential is of course not an increment because it is infinitely small. 
 
 PROBLEMS. 
 
 dv 
 Using the method of Art. 61, find -^ in the following 
 
 expressions. 
 
 1 ?!_L.^'_i =n dy^_^ 
 
 *• a^'^b^ ^' dx a'y 
 
 
 ' 
 
 df 
 dx 
 
 dx + 
 
 dy 
 
 'dy 
 
 or, solving for 
 
 dy 
 dx' 
 
 we have 
 
 
 
 
 
 
 dy 
 
 
 dx 
 
 
 
 
 d^ " 
 
 
 ^1 
 dy 
 
 This formula gives the value of 
 
 dy 
 dx 
 
dy 
 dx 
 dy 
 
 x^ - ay 
 ' ax - t' 
 
 6a: + 52^ 
 
 dx 
 dy 
 
 y\\hx + 22/) 
 x^ + y^x^ - y^ 
 
 dx 
 
 dy _ 
 dx 
 
 x{y + Vx2 - 2/2) 
 y - X 
 ' y + x' 
 
 PARTIAL DIFFERENTIATION AND INTEGRATION. 93 
 
 2. a:^ + 2/' - ^axy -3 = 0, 
 
 3. Qx^ + lOxy^ + 2/' + 5 = 0, 
 
 4. log (x^ - 2/2) - 2 sin-i - = 0, 
 
 X 
 
 5. 2 tan-i - - log {x^ + y^) = 0, 
 
 y 
 
 62. Slope of a hill. — Imagine a hill built upon the plane which 
 contains the x and y axes of reference, and let z be the height 
 of the hill above the point on the plane whose coordinates are 
 X and y. Then an equation expressing z as a function of x 
 and 2/ is the equation of the surface of the hill. Thus 
 
 z = Vr2 - a:2 - 2/2 (1) 
 
 is the equation of a hemispherical hill of radius r. 
 
 Consider the derivatives -r- and r- ; these derivatives are both 
 dx dy 
 
 we have 
 
 functions of x and y. Thus if 2 = Vr^ — x^ - 
 dz X 
 
 -f, 
 
 dx Vr2 - x^ - y^ 
 and 
 
 dz y 
 
 
 dy Vr2-a:2-2/2* 
 
 
 Let a:', 2/' and 2' be the coordinates of a particular point on the 
 surface of the hill as shown in Fig. 41. If the values x' and 2/' 
 
 are substituted for x and y in the general expressions for — 
 
 ox 
 
 dz 
 and — we get the values of these derivatives at the point p, and we 
 
 will represent these values by ( t" ) and ( t- ) • Then ( — 1 
 
94 
 
 CALCULUS. 
 
 is the slope of the tangent Hne qq and it is equal to tan a where 
 a is the angle shown in Figs. 41a and 416; and ( 7- I is the slope 
 
 \2-axi8 
 
 >^ x^axht 
 
 y-axis 
 
 Fig. 41a. 
 
 A 
 
 ~S^^ 
 
 
 ^ 
 
 ^S 
 
 
 y 
 
 
 ^r 
 
 
 X 
 
 
 / X-axis 
 
 y 
 
 '«\ / 
 
 X' 
 
 /f*' 
 
 
 // 
 
 >' 
 
 
 /v-axi» 
 
 
 Fig. 
 
 416. 
 
 of the tangent line rr and it is equal to tan /3 where /3 is the 
 angle shown in Figs. 41a and 416. 
 
 In Fig. 41a — and — are both negative; that is, z decreases 
 
 as X increases, and z decreases as y increases. In Fig. 416 
 
 ■r- and -r- are both positive. 
 dx dy 
 
 63. Equation of the tangent plane. — The derivation of the 
 equation of the tangent plane which touches a surface at a point 
 is somewhat similar to the derivation of the equation of a tangent 
 line which touches a curve at a point. Therefore it is worth while 
 to give the derivation of the equation of a tangent line before 
 considering the equation of a tangent plane. 
 
 Let a;' and z' be the coordinates of the point p where the 
 tangent line touches the plane curve cc, as shown in Fig. 42a. 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 
 
 95 
 
 Starting at the point p in Fig. 42a, which is at a height z' above 
 the base Hne, travel along the tangent line tt covering the hori- 
 
 Fig. 42&. 
 
 zontal distance {x — x') and the rise will be {x — x') tan a or 
 
 dz 
 (x — x') -J-. We thus reach any given point B in the tangent 
 
 line of which the coordinates are x and z, and the value of z is 
 s = 2' + (^ - a;') S (1) 
 
 which is the desired equation of the tangent line tt. 
 
 Let x', y' and z' be the coordinates of the point p where the tan- 
 gent plane touches the given surface 
 (see Fig. 426) . Starting at the point p, 
 which is at a distance z' above the 
 base plane, travel along the tangent 
 line rr (which lies in the tangent 
 plane) covering the horizontal dis- t^ 
 tance (y — y') towards the reader 
 in Fig. 426, and the rise will be 
 
 (2/-2/')tan/3=(2/-2/')(|)^ 
 
96 
 
 CALCULUS. 
 
 Then continue along the Hne q'q' parallel to the tangent line qq 
 (the line q'q' lies in the tangent plane) covering the horizontal 
 distance (x — x') to the right in Fig. 426, and the rise will be 
 
 {x - x') tan a = (x - ^')\r£) 
 
 We thus reach any given point B in the tangent plane of which 
 the coordinates are x, y and z, and the value of z is: 
 
 2 = 0' + 
 
 which is the desired equation of the tangent plane. Of course the 
 derivatives in this equation refer to the equation of the hill. 
 
 64. Component slopes and resultant slope. — Let X be the 
 
 slope of a hill at the point p, Fig. 41, in a direction parallel to the 
 
 the rise* per unit-horizontal-distance- 
 parallel-to-the-x-axis, and it is equal to 
 
 Also let Y be the slope of 
 
 the hill at p in a direction parallel to 
 
 theiz-axis. Then Y = (^) . Let X 
 
 x-axis. 
 
 That 
 
 IS, 
 
 X is 
 
 x-axia 
 
 
 Wp 
 
 and Y be represented to a chosen 
 scale by the lines X and Y in Fig. 
 43. Then the diagonal R in Fig, 43 
 represents what may be called the result- 
 ant or actual slope of the hill at p. 
 That is, the line R points directly up 
 hill, and the length of R represents (to 
 the chosen scale) the rise of the hill per 
 unit-of-horizontal-distance-in-the-direction-of-/2. This is a funda- 
 mental and important theorem in the mathematical theory of 
 electricity and magnetism, and the proof of the theorem is as 
 follows: 
 
 * If the slope is negative, X is the drop per unit horizontal distance, etc. 
 
 y-axis 
 
 Fig 43. 
 
 The X and y axes are shown 
 as they would appear if seen 
 from above in rigs.41a and 416. 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 
 
 97 
 
 Let the line AB in Fig. 44 be the line of intersection of a 
 horizontal plane through p with the plane which is tangent to 
 the hill at p. Then the line AB is evidently at right angles to 
 
 y-axia 
 
 Fig. 44. 
 
 R. Let the line ah he the line in the tangent plane at every point of 
 which the tangent plane is unit distance above the horizontal plane 
 through p. Then, by definition: 
 
 and 
 
 R = 
 
 X = 
 
 Y = 
 
 pr 
 1 
 
 ps 
 J_ 
 pq 
 
 (1) 
 (2) 
 (3) 
 
 where pr, ps and pq are the distances shown in Fig. 44. But 
 1 , — vr 1 
 
 ps 
 
 pr _ 
 cos 6 R-cosO 
 
 , and pq — 
 
 sin R ' sin 0' 
 
 Therefore 
 
 and 
 
 X = RcosB 
 Y = Rsind 
 
 Z2 + F2 = R^ 
 
 (4) 
 (5) 
 
 (6) 
 
98 
 
 CALCULUS. 
 
 These equations establish the proposition that R is the diagonal 
 of a rectangle of which X and Y are the sides as shown in Fig. 44. 
 
 65. Example of gradient in two dimensions. — The observed 
 pull in Fig. 10 is a function of one independent variable, the current 
 as indicated by the ammeter, and in such a case it is helpful to 
 plot a curve of which the abscissas represent observed currents 
 and of which the corresponding ordinates represent observed 
 values of pull. This curve is shown in Fig. 11. 
 
 Frequently one is concerned with a quantity which depends 
 upon two independent variables. For example one might be 
 
 concerned with the distri- 
 bution of temperature over 
 a flat metal plate. In such 
 a case it is helpful to think 
 of the temperature at each 
 point p as represented by 
 the height at p of a hill. 
 Then the component gradi- 
 ents of temperature are rep- 
 resented everywhere by the 
 component slopes of the 
 hill, and the resultant gra- 
 dient of temperature is rep- 
 resented everywhere by the 
 resultant slope of the hill, as indicated by the sides and diagonal 
 of the rectangle in Fig. 45. 
 
 66. Example of gradient in three dimensions. — One might be 
 concerned with the distribution of temperature throughout a solid 
 body. Of course, the temperature would have a definite valiie at each 
 point in a body, or, as expressed in mathematical language, the 
 temperature T at each point would be a definite function of the 
 
 coordinates x, y and z of the point, and the derivatives ^— , -r- 
 
 
 x-axi8 
 
 metal 
 plate 
 
 '< 
 
 ^ i 
 
 x" 
 
 Y 
 
 U^axis 
 
 H 
 
 Fig 45. 
 
 and -^r- would also be functions of x, y and z. 
 oz 
 
 dx^ dy 
 That is, the 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 99 
 
 three derivatives would have definite values at each point in the 
 
 body, and the three derivatives, --, -^- and -r-, are the com- 
 
 ponent gradients of the temperature at the point in directions 
 parallel to the respective axes of reference. Representing the 
 resultant temperature gradient at a point by R and the three 
 components of R by X, Y and Z, we have 
 
 R^ = X'+Y^ + Z2 (1) 
 
 2 = S (4) 
 
 These equations constitute an extension to three dimensions of 
 the theorem of Art. 64. 
 
 In Fig. 45 the direction at right angles to the surface of the 
 plate is available for purposes of geometrical representation, and 
 one can think of an actual hill being built upon the plate so that 
 the temperature of the plate at each point is represented by the 
 height of the hill at that point, and of course this " hill" becomes 
 a "valley" dipping below the plate where the temperature of the 
 plate is below zero. 
 
 The temperature at each point of a solid body, however, cannot 
 be represented geometrically as a height because space is filled 
 in every direction by the body itself. That is, every direction 
 in space is used for representing the independent variables x, y 
 and z, and no direction is left for the representation of T. It is 
 convenient, however, even in this case, to speak of the "temperature 
 hill,^' the "height" of the hill at each point of the body being the 
 temperature itself, high or low as the case may be. 
 
 67. Partial integrations.* — In many cases the derivative of a 
 
 * What is here called partial integration is usually called multiple integra- 
 tion, double, triple, quadruple, etc., as the case may be. 
 
100 
 
 CALCULUS. 
 
 function of one independent variable can be set up or established 
 by the use of elementary principles of physics and arithmetic, 
 and the function itself can be found by one integration, as exempli- 
 fied in Arts. 22 and 23. l 
 
 But when the function to be established is a function, say, of 
 two independent variables {x and y for example), then it is 
 usually necessary to perform an integration with respect to x 
 in setting up the derivative of the desired function with respect 
 to 2/, or to perform an integration with respect to y in setting 
 up the derivative of the desired function with respect to x. Such 
 integrations may be called partial integrations because they are 
 related to partial differentiation in the same way that ordinary 
 integration (with respect to one independent variable) is related 
 to ordinary dijfferentiation (with respect to one independent 
 variable). 
 
 68. Volume of a hill. Example of partial integrations. — For 
 
 the sake of simplicity let us consider a particular case, namely, a 
 hemispherical hill with its center at the origin of coordinates. 
 Then the equation of the surface of the hill is 
 
 = Vr2 - x2 - 
 
 r 
 
 (1) 
 
 Thus Fig. 41a represents one quarter of a hemispherical hill, 
 and it is required to find its volume. 
 
 Iz-axU 
 
 I 
 
 ^ff-axis 
 
 Fig. 46a. 
 
 Fig. 466. 
 
 i 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 101 
 
 Let V be the volume of the portion ahcABC of the hill, as 
 shown in Fig. 46a. Then y is a function of y (see figure), and 
 the increment of v due to an infinitesimal increment of y is the 
 volume of the thin slab in Fig. 466. Therefore we have: 
 
 dv = A ' dy (2) 
 
 where A is the area of the face of the slab as shown in Fig. 466. 
 Therefore we must find an expression for A before we have a 
 
 known expression for -7-. 
 
 Now ape in Fig. 46a is a plane curve, and equation (1) is the 
 equation of this curve (ordinate z, abscissa x) if y is constant. 
 Let a be the shaded area in Fig. 46a; then the strip with double 
 shading is da and it is equal to z • dx. Therefore using the 
 value of z from equation (1) we have: 
 
 da = Vr2 - ^2 - 2/2 . dx ■ (3) 
 
 and the total area A of the face abcp is the integral of this 
 expression from a; = to x = ha = Vr^ — y^* Therefore: 
 
 Vr2 - x^- ?/2 . dx (4) 
 
 
 
 In performing this integration y is a. constant. This integral 
 gives an expression for A, and the value of A so found can be 
 substituted in equation (2). Then equation (2) may be integrated 
 between the limits y= to y = r to give the desired expression 
 for the volume V of the hill. That is 
 
 v=r 
 
 *Jy=0 
 
 A ' dy (5) 
 
 The above discussion applies to a particular case, namely, to a 
 quarter of a hemispherical hill; but the volume of any hill what- 
 
 * This value of ha is the value of x (for the given constant value of y) 
 when 2 = 0, as found from equation (1). 
 
102 
 
 CALCULUS. 
 
 ever can be found in the same way. Find by a first integration, 
 the area ^4 of a plane section of the hill parallel to the xz plane 
 and at a distance y therefrom, and then integrate A • dy to 
 get the required volume; or find by a first integration the area 
 A' of a plane section of the hill parallel to the yz plane and at a 
 distance x therefrom, and then integrate A' * dx to get the 
 required volume. 
 
 The prismoid formula. — A prismoid is a solid with parallel top 
 and bottom faces with sides generated by straight lines. Thus a 
 cylinder is a prismoid, any wedge even if its base is a polygon or 
 curve is a prismoid, Fig. 47a represents a prismoid. 
 
 The volume of a prismoid is given by the formula 
 
 T + ^M + B 
 6 
 
 Xh 
 
 (6) 
 
 where T is the area of the top face, B is the area of the bottom 
 face, M is the area of the middle section (which is parallel to 
 top and bottom faces) and h is the altitude of the prismoid. 
 
 Equation (6) also gives the volume of any figure bounded by a 
 second degree surface ss in Fig. 476 between parallel top and 
 
 Fig. 47a. 
 
 Fig. 476. 
 
 bottom faces, and equation (6) can be used to give a very close 
 approximation to the volume of any solid between parallel ends 
 with sides smoothly curved from end to end. 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 103 
 
 PROBLEMS. 
 
 1. Integrate equations (4) and (5), above, and find the volume 
 of one quarter of a hemispherical hill as shown in Fig. 41a. Verify 
 the result from the formula: volume of a sphere equals ^tt times 
 its diameter cubed. 
 
 2. Find the volume of the entire wedge (20 inches long) in Fig. 
 p2, the wedge being 8 inches wide in a direction perpendicular to 
 the plane of the paper. 
 
 Ans. 800 cubic inches. 
 
 Note. — The volume of a wedge is, of course, easily found by using the prin- 
 ciples of elementary geometry. It is 
 intended, however, that this problem be 
 solved by integration as follows: Let v 
 be the volume of the shaded portion of 
 the wedge as shown in Fig, p2. Then 
 t; is a function of x, and the increment 
 of V due to an increment of a: is the 
 volume of the thin slab shown in Fig. ^2. 
 
 The area of the face of the slab is — 
 
 X 10 inches X 8 inches = 4a: square inches. 
 Therefore the volume of the slab is 4a;-dx 
 cubic inches. That is, 
 
 dv = Ax'dx 
 
 ^ x-axU 
 
 Fig. 2)2. 
 
 and the desired volume is found by integrating this expression from ic = to 
 aj = 20 inches. 
 
 3. Find the volume of the shaded portion of the wedge in 
 
 30 inches 
 
 Fig. p3. 
 
 jj i0_inche8 
 
 Fig. p4. 
 
104 
 
 CALCULUS. 
 
 Fig. p3, the wedge being 6 inches wide in a direction perpendicular 
 to the plane of the paper. 
 Ans. 92 L6 cubic inches. 
 
 4. Find the volume of the frustum of a cone which is shown in 
 Fig. p4:. 
 
 Ans. 229.2 cubic inches. 
 
 5. Find the volume of the paraboloid of revolution which is 
 shown in Fig. p5. 
 
 Ans. 1060.7 cubic inches. 
 
 Note. — ^The equation of the paraboloid is y = px', where the axis of 
 revolution is the ^/-axis of reference; and the value of p is determined from 
 the condition that x = 7}4 inches when y = 12 inches. 
 
 ' 7 inches 
 f 
 
 5 inches 
 
 Fig. p5. 
 
 Fig. p6. 
 
 6. Find the volume of the shaded portion of the paraboloid in 
 Fig. p6. 
 
 Ans. 699.8 cubic inches. 
 
 7. Find the volume of the spherical segment which is shown in 
 Fig. p7. 
 
 Ans. 276.79 cubic inches. 
 
 8. Find the volume of the segment of a sphere which is shown 
 in Fig. pS. 
 
 Ans. 119.71 cubic inches. 
 ' 9. Find the volume between the xy plane and the plane 
 z = a -{-hx + cy and inside of the cylinder {x — ey-\-{y—fY = g^ 
 when a = 10 inches, 6 = 3, c = 2, e = 8 inches, / = 9 inches 
 and gr = 5 inches. 
 
 Ans. 4085.7 cubic inches. 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 105 
 
 Note. — Let A-dy be the expression which must be integrated with respect 
 to y to give the desired volume. Then A is obtained by integrating z • dx 
 between Umits which are the two values of x found from (x — e)^ + (?/ — ff 
 = g^ for the given value of y, and then A-dy is integrated between the 
 limits y ^ f -\- g and y = f — g. 
 
 I 
 
 Is 
 
 Fig. p7. 
 
 finches 
 Pinches^ ^ 
 
 '^ 
 
 Fig. p8. 
 
 I 
 I 
 
 II 
 
 
 li 
 1^. 
 
 -^^ 
 
 10. Find the total volume which is common to the two cylinders: 
 1/2 _|_ 2-2 _ 25 and x^ + y^ = 25, everything being expressed in 
 inches. 
 
 Ans. 666.7 cubic inches. 
 
 Note. — One should be able to determine the proper limits for each integra- 
 tion in this problem if one understands problem 9. 
 
 11. The inside dimensions of a barrel are 20 inches in diameter 
 at each end, 23 inches in diameter at the middle and 31 inches long. 
 What is the capacity of the barrel in gallons? 
 
 Ans. 51.3 gallons. 
 
 69. Area of the surface of a hill. Another example of partial 
 integration. — It is desired to find the area of the portion caCA 
 of the surface of the hill shown in Fig. 48 when the equation of 
 the surface of the hill is given; this equation, of course, expresses 
 the height z of the hill at a point as a function of x and y as 
 explained in Art. 68. Let S be the area of caCA. Then the 
 
106 
 
 CALCULUS. 
 
 normal to dt 
 
 increment of S due to an infinitesimal increment of y {= Bb 
 
 in the figure) is the area dS of 
 the narrow strip ca. To find the 
 area of this narrow strip let s 
 be the area of the portion from 
 c to ds in the figure, then s is a 
 function of x, and the increment 
 of s due to an infinitesimal in- 
 crement of X is the small ele- 
 ment of area ds which caps the 
 prism dx . dy. This small ele- 
 ment of area is sensibly coinci- 
 dent with the tangent plane at 
 ds, and therefore the normal to 
 Fig. 48. ds makes an angle 7 with the 
 
 z-axis whose tangent is equal to 
 
 ^/(i^(i)' 
 
 which is the resultant slope of the hill at ds, 
 
 according to Art. 64. 
 
 Now the area of the base of the prism, dx 
 of the area ds, and consequently we have ; 
 
 dx ' dy — cos 7 • ds 
 dx • dy 
 
 dy, is the projection 
 
 or 
 
 ds = 
 
 But if 
 we find: 
 
 cos 7 
 
 (1) 
 
 <-^'4(MFW)' 
 
 COS 7 = 
 
 ^Rl)■-(sJ 
 
 so that equation (1) becomes: 
 
 (2) 
 
 '^ = ^' + {'iy+{fJ-'^-'y (2) 
 
 .Jjl^ 
 
PARTIAL DIFFERENTIATION AND INTEGRATION. 107 
 
 10 inches 
 
 Now the expression under the radical is a function of x and y 
 which may be found by differentiating the equation of the surface 
 of the hill (which is given); and the total area of the strip ca 
 (which is equal to dS) is the integral of (3) from x = to x = ha* 
 In this integration y and dy are constants. The value of dS 
 so found is the required infinitesimal increment of the area caCA 
 due to an infinitesimal increment of y, and the value of the area 
 caCA is found by integrating this expression for dS between the 
 limits y = and y = y (meaning any value of y). If the 
 entire area of the hill in Fig. 48 is to be found, the expression for 
 dS must be integrated from 2/ = to y = the intercept of the 
 hill on the ^/-axis.f 
 
 PROBLEMS. 
 
 1 . The hill shown in Fig. 48 is a quarter of a hemisphere. Find 
 the area of its surface and verify 
 your result from the formula: 
 area of a sphere = 47r times its 
 radius squared. 
 
 2. Find the area of the curved 
 surface of the cone shown in 
 Fig. p2, 
 
 Ans. 135.8 square inches. 
 
 Note. — Let a be the area of the i jp ly-i 
 
 shaded portion of th cone in Fig. p2. " *" 
 
 Then a is a function of x; and the in- Yig. v2. 
 crement of a due to an increment of x 
 
 is the area of the narrow strip shown in the figure. The width of this strip 
 
 (parallel to the slant height of the cone) is ^^^ "^ ^^ X dx, and the length 
 
 of the strip (cu-cumference of the cone) is ~.XS X tt inches. Therefore we 
 
 get: 
 
 da = OM^irX'dx 
 
 * The value of 6a is found from the given equation of the surface of the hill. 
 It is the value of x for the given constant value of y when 2 = 0. 
 
 t This intercept is found from the equation of the surface of the hill by plac- 
 ing X and z both equal to zero and solving for y. 
 
108 
 
 CALCULUS. 
 
 and the required area is found by integrating this expression from a; = to 
 X = 10 inches. 
 
 3. Find the area of the curved surface of the frustum of the 
 cone in Fig. p3. 
 
 Ans. 144.38 square inches. 
 
 4. Find the area of the part of the surface of the cylinder 
 
 .^^' 
 
 ^>.. 
 
 j^ JUf.Jnche8 ^ 
 
 Ilg.p3. 
 
 ^1 
 
 4 inches 
 Pinches, 
 
 m 
 
 
 1^ 
 
 I 
 
 Fig. p5. 
 
 a;2 -f ^2 _ 100 which is inside of the cylinder x^ -\- y^ = 100, the 
 radii of the cylinders being measured in inches. 
 Ans. 800 square inches. 
 
 5. Find the area of the convex surface of the segment of a 
 sphere which is shown in Fig. p5. 
 
 Ans. 989.5 square inches. 
 
 6. Find the area of a zone on the earth's surface between 40° 
 and 80° north latitude, taking the earth as a sphere 4,000 miles 
 in radius. 
 
 Ans. 34,500,000 square miles. 
 
CHAPTER V. 
 MISCELLANEOUS APPLICATIONS. 
 
 70. Influence of errors of observation upon a result. When 
 one measures a thing with great care repeatedly the successive 
 measurements never agree with each other, and it is easy to 
 determine what is called the probable error* of the measurement 
 by considering the discrepancies in a set of observed values. 
 Thus the length of a bar might be found by repeated measure- 
 ment to be 103.625 centimeters with a probable error of, say, 
 db 0.0036 centimeter; the mass of a body, as determined by re- 
 peated weighing on a balance, might be 67.2382 grams with a 
 probable error of, say, ± 0.0006 gram. 
 
 In many cases, however, it is not the directly measured 
 quantities that are important but a result which is calculated 
 therefrom; and it is often desirable to calculate the probable 
 error in such a result when the probable errors in the individual 
 measurements are known. For the sake of simplicity we will 
 discuss a special case, as follows: 
 
 A steel ball is weighed on a balance and its mass is found to be 
 M grams with a probable error of ± m grams, and the diameter 
 of the ball is measured and found to be L centimeters with a 
 probable error of ± I centimeters. The density of the steel 
 as calculated from the measured values of M and L is: 
 
 ^ = & (^) 
 
 and it is required to find the probable error in D. The probable 
 
 *See Franklin, Crawford and MacNutt's Praclicol Physics, Vol. I, pages 
 1-9 for a very simple discussion of errors of observation. For more complete 
 discussion see Merriman's Least Squares, John Wiley & Sons, New York. See 
 also Palmer's Theory of Measurements, McGraw-Hill Co., New York, 1912. 
 
 109 
 
110 CALCULUS. 
 
 errors m and I are usually very small so that it is sufficiently 
 accurate to calculate ADm (the probable error in D due to 
 =t m) and ADi (the probable error in D due to ± I) by the 
 formulas: 
 
 AI>« = l^-m (2) 
 
 and 
 
 AD, = gj (3) 
 
 Let AD be the probable error in the result due to m and I 
 together. Then AD is not equal to the sum AD;;^ + AD^, but 
 according to the theory of probability we have: 
 
 AD = V(ADJ2 + (ADO^ (4) 
 
 PROBLEMS. 
 
 1 . The scale of an alternating current ammeter reads in degrees, 
 and the value of the current is i = k^, where fc is a constant 
 and is the deflection in degrees due to a current of i amperes. 
 The error of reading is probably ± i^ degree. What is the 
 probable error in the result, t, in fractions of an ampere (a) when 
 the deflection is in the neighborhood of 200° and (6) when the 
 deflection is in the neighborhood of 20°? Take the value of k 
 to be such that 10 amperes gives 100° deflection. Ans: (a) 
 =t 0.0089 ampere, (6) ± 0.028 ampere. 
 
 2. What is the percentage error (probable) in the value of i 
 under conditions (a) and (6) in problem 1? Ans. (a) =t 0.06 per 
 cent., (6) ± 0.6 per cent. 
 
 Note. The percentage error is — X 100. 
 
 3. The diameter of a steel sphere is repeatedly measured and 
 found to be 0.8762 inch with a probable error of ± 0.0016 inch. 
 Calculate the volume of the sphere and calculate the probable 
 error in the calculated volume. Ans. 0.3525 cubic inch 
 =b 0.0019 cubic inch. 
 
 
MISCELLANEOUS APPLICATIONS. 
 
 Ill 
 
 Note. It is interesting to note that the percentage error in the volume i3 
 three times as great as the percentage error in the measured diameter, that is 
 
 0.0019 . 
 
 0.3525 
 
 is three times as great as 
 
 0.0016 
 0.8762 ' 
 
 4. Show that the percentage error of x^ is twice as great as 
 the percentage error of x, when a; is an observed quantity and 
 x^ is a calculated result. 
 
 5. By measurement the length of a rectangle is 25 inches 
 with a probable error of =^ 0.02 inch, and the width of the rec- 
 tangle is 10 inches with a probable error of ± 0.015 inch. What 
 is the probable error in the calculated area of 250 square inches? 
 Ans. ± 0.421 square inch. 
 
 6. The diameter of a steel ball is 2.542 centimeters ± 0.001 
 centimeter, and the mass is 116.25 grams ± 0.02 gram. What is 
 the density and what is the probable error in the density? Ans. 
 7.781 grams per cubic centimeter =t 0.0168 gram per cubic 
 centimeter. 
 
 71. Maximum and minimum values. It is evident that a 
 growing thing reaches its maximum size when it stops growing; 
 but a thing may stop growing for a while and then go on growing 
 
 Fig. 49. 
 
 . . dv . ... 
 
 A maximum; -^ is positive when x is 
 
 less than a, and negative when x is 
 greater than a. 
 
 Fig. 50. 
 Not a maximum; -^ is positive when 
 
 X is less than a, 
 greater than a. 
 
 also when x is 
 
 again. Also it is evident that a decreasing thing reaches its 
 minimum size when it stops decreasing; but a thing may stop 
 decreasing for a while and then go on decreasing again. Thus 
 
112 
 
 CALCULUS. 
 
 ■^ is equal to zero when x = a in all four of the following 
 
 figures, 49, 50, 51 and 52, but 6 is a maximum value of y in 
 Fig. 49 only, and 6 is a minimum value of y in Fig. 51 only. 
 
 y-ojdB 
 
 I x-axi8 
 . — a ->l 
 
 Fig. 51. 
 
 dti 
 A minimum; -^ is negative when x 
 
 is less tlffen a, and positive when x 
 is greater than a. 
 
 y-€ixi8 
 
 Fig. 52. 
 
 dv 
 Not a minimum; -^ is negative 
 
 when X is less than a, also when 
 X is greater than a. 
 
 To determine the values of x for which a given function of x 
 is a maximum or a minimum, differentiate the function, place its 
 derivative "equal- to zero,-and 'solve for x. Then for each value a 
 of X so found, determine the algebraic sign of the derivative for a 
 value of X slightly less than a and also for a value of x slightly 
 greater than a. Interpret the results by referring to Figs. 49 to 52* 
 
 PROBLEMS. 
 
 In the first five problems, following, find for what values of x 
 is 2/ a maximum or minimum and compute the maximum or 
 minimum value in^each case. • 
 
 1. 2/ = a:'' — 6x2 + 9x -f 3. ^^s. x = 1) y = 7, a maxi- 
 mum: X = 3; 2/ = 3, a minimum. 
 
 ^ x^ + 4x + 44 
 
 2. 2/ = 
 
 Ans. 
 
 a^ = 8; 2/ = I, 
 
 a mimmum: 
 
 x2 + 8x + 40 
 X = — 6; 2/ = 2, a maximum. 
 
 * A slightly easier method is to consider the values of the second derivative 
 of the given function but the above method is sufficient. 
 
MISCELLANEOUS APPLICATIONS. 113 
 
 3. y = (x — 4)3 {x + ly. Ans. x = — 1; y = Oj a maxi- 
 mum : X = 1; y = — 108, a minimum. 
 
 4. y = 9e2=^ + 25e~2^. Ans. a: = log Vf; 2/ = 30, a minimum 
 
 5. 1/ = 3 sin a; + 4 cos a;. Ans. a; = tan~^f in the first quad- 
 rant; y = 5, a maximum: x = tan-^J in the third quadrant; 
 y = — 6, a minimum. 
 
 6. Divide the number 20 into two parts such that the product 
 of one part by the square of the other part shall be a maximum. 
 Ans. 13.33 and 6.67. 
 
 7. What number added to one half the square of its reciprocal 
 gives the least possible sum. Ans. 1. 
 
 8. The cost of an electrical power transmission line is approxi- 
 mately proportional to the weight of copper used because the 
 cost of poles and the cost of erection can be expressed with a fair 
 degree of accuracy as an addition of so many cents per pound on 
 the cost of copper. Therefore the annual money cost represented 
 by interest on investment and depreciation, being reckoned as 
 so many per cent, of first cost, is proportional to the weight of 
 copper used, or it is equal to kw where A; is a constant and w 
 is the weight of copper used. 
 
 The amount of power lost in a transmission line is halved if 
 the amount of copper is doubled. Therefore the annual money 
 loss due to loss of energy in the line is universely proportional 
 
 to the weight of copper, or it is equal to — where m is a con- 
 stant. 
 
 Therefore the total annual money loss is kw H . Find the 
 
 w 
 
 value of w (expressed in terms of k and w) which will make 
 this total annual money loss a minimum. Ans. w = ^llH 
 
 Note. This matter is discussed in electrical engineering books under the 
 head of Kelvin's law. See Franklin' s Electric Light ing^ Art. 26, pages 66-70. 
 The Macmillan Co., New York, 1912.' 
 
 8. It is desired to construct of sheet metal a cylindrical gallon 
 
114 CALCULUS. 
 
 measure without a cover. Find what its depth must be in 
 order that the amount of sheet metal used shall be a minimum. 
 
 Ans. y = jj?^ (= 4,19) inches. 
 
 Note. The amount of sheet metal used will be a minimum when the 
 surface area S of the vessel is a minimum; but S = 2irxy + ttx^, where x 
 is the radius of the base of the cylinder and y is its depth. Since the measure 
 is to contain 1 gallon (=231 cubic inches), irx^y =231; and, substituting for 
 z its value in terms of y, we have: 
 
 ^ = 2>/23i^ + ??i 
 
 The value of y which makes this expreasion a minimum is to be obtained. 
 
 10. Determine the altitude of the cylinder of greatest convex 
 surface that can be inscribed in a sphere whose radius is 10 inches. 
 Ans. 14.142 inches. 
 
 11. Determine the altitude of the cylinder of greatest volume 
 that can be inscribed in a sphere whose radius is 10 inches. 
 Ans. 11.547 inches. 
 
 12. Determine the volume of the smallest cone which can be 
 circumscribed about a sphere whose radius is 10 inches. Ans. 
 
 — ^— (= 838.1) cubic inches. 
 o 
 
 13. A man who can row at a speed of 4 miles per hour and 
 
 run at a speed of 6 miles per 
 hour wishes to reach the 
 
 gi \ point p from a boat at h 
 
 %^->.^ ^\ as shown in Fig. pl3 in the 
 
 ^1 \ \^^ least possible time. Find the 
 
 T^ch — ^ ^ V. ' beach distance ap that the man 
 
 t;,. '"'-t* must run on the beach. Ans. 
 
 rig. pl3. 
 
 1.056 miles. 
 
 14. The strength of a rectangular beam is proportional to its 
 breadth and to the square of its depth. Find the breadth and 
 depth of the strongest beam which can be cut from a cylindrical 
 
 'N 
 
MISCELLANEOUS APPLICATIONS. 115 
 
 log whose radius is 18 inches. Ans. Breadth, 20.8 inches; depth, 
 29.4 inches. 
 
 15. For what angle of deflection does the old-fashioned tangent 
 galvanometer give the greatest percentage accuracy in the value 
 of the current? Ans. 45°. 
 
 Note. The equation of the tangent galvanometer is: 
 
 i = k tan 6 
 
 where i is the current in amperes, 6 is the observed angle of deflection, and 
 fc is a constant. Let dd represent the probable error of the observed value 
 of 0, then the probable error in the value of i is: 
 
 ,. k.de 
 di = 
 
 COS2 d 
 
 and the percentage error in the value of i is: 
 
 _di _ dd 
 ^ ~ i "" sin COS 9 
 
 in which dd being the probable error of 6 is to be treated as a constant. 
 
 A general discussion of this important matter of precision as dependent 
 upon the choice of magnitudes of the quantities to be observed is given in 
 Chapter XII of Palmer's Theory of MeasurementSy McGraw-Hill Co., New 
 York, 1912. 
 
 72. The problem of the bent beam. The formulas used in 
 
 connection with a bent beam are based on Hookers law, and the 
 
 F sectional area a p p sectional area 8 p 
 
 original length L m! original length L y 
 
 increase of length I --^ decrease of length l-^ 
 
 Fig. 53. Fig. 54. 
 
 derivation of two of the simpler formulas furnishes a good 
 
 example of the use of calculus. 
 
 Hookers law. Figure 53 represents a rod subjected to a 
 
 tp 
 
 stretching force F. Experiment shows that — is proportional 
 
 s 
 
116 CALCULUS. 
 
 to y (Hookers law). Therefore we may write 
 
 L 
 
 or 
 
 a L 
 
 F = E-^ (1) 
 
 in which ^ is a proportionality factor and it is called the stretch 
 modulus of the substance of which the rod is made. Equation 
 (1) also gives the compressing force F required to produce a 
 slight shortening of the rod as represented in Fig. 54. 
 
 When a beam is bent the filaments on one side of the beam are 
 lengthened, and the filaments on the other side are shortened. 
 If we wish to determine the degree of lengthening or shortening 
 it is necessary to consider a very short portion of the beam. 
 But a very short portion of a bent beam has the same curvature 
 as the osculating circle as shown in Fig. 55, and the easiest way 
 
 ^^ mediaa Une 
 
 very short portion 
 beam 
 
 Fig. 55. Fig. 66. 
 
 to think of the very short portion of the bent beam is to think of it 
 as a portion of a long beam which is actually bent into the arc of a 
 circle. Thus Fig. 56 represents a beam bent into the arc of a 
 circle. 
 
 A certain filament in the bent beam is unchanged in length. 
 
MISCELLANEOUS APPLICATIONS. 
 
 117 
 
 pulU 
 
 This filament is located in what is called the median line (or 
 plane) of the beam. The median plane of a beam of rectangular 
 section is the middle plane of the beam. 
 
 Let us consider the forces which act across a section qq of a 
 bent beam as shown in Fig. 57. Considering these forces as 
 acting on A5, they are a set of 
 pulls on A and a set of pushes 
 on B, and together they con- 
 stitute a turning force or torque, 
 T, about an axis perpen- 
 dicular to the plane of the paper. 
 It is the object of this discussion 
 to derive an expression for T 
 in terms of 6, D, R and Ej 
 where h is the breadth of the 
 
 beam, D is the depth of the beam, R is the radius of curvature 
 of the median line of the beam, and E is the stretch modulus 
 of the material of which the beam is made. 
 
 Let us consider the portion of the beam which lies between 
 the radii RR in Fig. 56. This portion of the beam is shown to 
 
 I 
 
 I median line 
 
 jIL. — .TP . 
 
 \x 
 
 length MR+x)^ 
 length s=Re 
 
 length =(R-jc;d 
 
 Fig. 58. 
 
 a larger scale in Fig. 58. The original length of every filament 
 of this portion of the beam is Rd. Consider the upper filament 
 
118 
 
 CALCULUS. 
 
 Jf of the beam. The stretched length of this filament is 
 {R + x)dj because // is a circular arc of radius (R + x) and 
 it subtends the angle 6. Therefore the filament }} has been 
 increased in length by the amount xB by the bending. Similarly 
 
 it may be shown that the lower 
 
 stretched 
 
 _ 5 /'Stretch 
 
 eompresaed- 
 
 •ectional viewjof Jtean 
 
 Fig. 59. 
 
 filament /'/' in Fig. 58 has been 
 shortened by the amount xd by 
 the bending. 
 
 Figure 59 is a sectional view 
 of the beam with its depth arbi- 
 trarily increased by added layers 
 of material on top and bottom, 
 and we wish to find the incre- 
 ment of T due to this assumed 
 increase of depth of the beam. 
 The stretching force in the top 
 
 layer can be found from equation (1) by substituting xd for Z, 
 
 h . dx for s and RO for L, which gives 
 
 F = -r^'X'dx 
 
 this stretching force is a pull at A in Fig. 57, and its torque 
 action about the axis 00 in Fig. 59 is counter-clock wise as 
 seen in Fig. 57, and it is equal to 
 
 Fx = ^'xHx 
 
 The compressing force in the bottom layer is a push at B 
 in Fig. 57, and its torque action about the axis 00 is also counter- 
 clock wise as seen in Fig. 57, and it is equal to 
 
 Fx = 
 
 Eh 
 R 
 
 x^'dx 
 
 Therefore the total torque action due to the two added layers is: 
 dT = ?|^-x2.daj (2) 
 
MISCELLANEOUS APPLICATIONS. 
 
 119 
 
 Whence, by integrating, we have: 
 2 Ehx^ 
 
 T = 
 
 3 R 
 
 + a constant 
 
 (3) 
 
 But it is evident that T = when x = 0, because a beam of 
 zero depth would of course have no stiffness at aU. Therefore 
 the constant of integration in equation (3) is zero, and equation 
 (3) becomes: 
 
 
 (4) 
 
 It is somewhat simpler to express T in terms of the depth D 
 of the beam {= 2x). Therefore substituting x = ^ inequation 
 (4), we have: 
 
 12 B 
 
 (5) 
 
 This equation applies to a beam bent in any way whatever, T 
 
 being the torque action 
 
 across a particular section 
 
 of the beam, and R the 
 
 radius of curvature of the 
 
 median line at that point. 
 
 Thus Fig. 60 shows a beam 
 
 with one end fixed in a wall 
 
 and carrying a weight W 
 
 at the other end. In this 
 
 case the bending torque at 
 
 p is equal to W(a — x)j 
 
 and the torque action across 
 
 the section of the beam at 
 
 p is equal and opposite to 
 
 W{a — x) Therefore, ignoring algebraic signs, we have 
 
 Fig. 60. 
 
 T = W(a - x) 
 
 (6) 
 
120 CALCULUS. 
 
 Substituting this value for T in equation (5) and solving for R 
 
 we have: 
 
 _ 1 EhD^ 
 
 ^~T2W(a-x) ^^^ 
 
 or 
 
 R = -^ (8) 
 
 a — X 
 
 where 
 
 c = -i.^^ (9) 
 
 Let it be required to find the equation of the curve formed by 
 the median line of the beam in Fig. 60. According to Art. 49 
 the radius of curvature of any curve at a point is: 
 
 n.hM 
 
 If the bending of the beam in Fig. 60 is very slight the value of 
 -p will be everywhere very small so that, as a first approxima- 
 tion, we may neglect (~f) in the numerator of equation (10) where 
 
 it is added to the much larger constant quantity 1. Therefore 
 from equations (8) and (10) we have 
 
 i = ^ 
 
 (Py a — X 
 
 dx" 
 or 
 
 d^y _ a x 
 
 dx^ c c 
 whence, by integrating,* we have 
 
 (11) 
 
 (12) 
 
 dy ax x^ , . . 
 
 ;p = — + Si constant 
 
 uX C JiiC 
 
 (13) 
 
 * By recognizing the function of which is the derivative. 
 
MISCELLANEOUS APPLICATIONS. 
 
 121 
 
 But ~ is actually* equal to zero when a; = in Fig. 60, 
 that the constant of integration in equation (13) is zero, or: 
 
 so 
 
 dy 
 dx 
 
 ax 
 c 
 
 2c 
 
 Integrating this expression, we have: 
 
 y = 
 
 ax2 
 
 2c ~ 6c 
 
 + a constant 
 
 (14) 
 
 (15) 
 
 But y = when x = in Fig. 60, so that the constant of 
 integration in equation (15) is zero, and equation (15) becomes: 
 
 y 
 
 ax" 
 2c 
 
 6c 
 
 (16) 
 
 in which c the constant defined by equation (9). 
 
 73. Average value of a function. Let y he sl function of x 
 as represented by the curve 
 cc in Fig. 61. The average 
 value of y between x = a 
 and X = h is the shaded area 
 in Fig. 61 divided by the base 
 {b — a). This gives the height 
 of a rectangle of base {h — a) 
 the area of the rectangle being 
 equal to the shaded area in 
 the figure. From this definition of average value we have: 
 
 {average value of y 
 between x = a and x 
 
 Fig. 61. 
 
 =i]^^a£y- 
 
 dx (1) 
 
 PROBLEMS. 
 
 1. The velocity of a falling body is v = gt. Find the average 
 value of V from i = to t = 10. Ans. 5g. 
 * Not approximately. 
 
122 
 
 CALCULUS. 
 
 2. The velocity of a falling body, starting with an initial 
 velocity w, is v = u -\- gt. Find the average value of v from 
 t = 5 to < = 10. Ans. u + 7.5 g. 
 
 3. Find the average value of y — sinx from a; = to a; = tt. 
 The answer is shown in Fig. p3. 
 
 y=8inx 
 
 Fig. p3. 
 
 4. Find the average value of e = E sin (at during a whole 
 cycle from cot = to w^ = 27r. Ans. Zero. 
 
 5. Find the average value oi y = sin^ x during a half cycle 
 (x = to X = tt) and during a whole cycle (x = to x = 27r). 
 The answer is shown in Fig. p5. 
 
 Fig. p5. 
 
 6. Find the average value oi y = sin a; sin {x — 0) during a 
 whole cycle {x — to x — 2t). Ans. J^ cos d. 
 
MISCELLANEOUS APPLICATIONS. 123 
 
 CENTER OF GRAVITY. 
 
 74. Resultant of a set of parallel forces. The single force R 
 in Fig. 62 is equivalent to the three parallel forces A, B and C 
 
 combined, and it is called the ^ ^ 
 
 resultant of A, B and C. k X >j 
 
 The value of R is equal to it__a— ^ ! j 
 
 A -{- B -\- C, and the point of i^ j j^. ji 
 
 apphcation of R is determined ^ M |b 1^1^ 
 
 by the condition that the } 
 
 torque action of R about any Ijij 
 
 arbitrarily chosen axis X 
 
 (perpendicular to the plane of p- g2. 
 
 the paper) must be equal to 
 
 the combined torque action of the given forces A, 5, and C 
 
 about that axis. That is: 
 
 R^ A + B + C (1) 
 
 (2) 
 
 A + B + C ^^^ 
 
 Definition of the center of gravity of a body. Every portion 
 of a body is pulled downwards by gravity, and the forces which 
 thus act upon the various parts of a body are together equivalent 
 to a single force (their resultant) the point of application of which 
 is called the center of gravity of the body. Thus the short arrows 
 /// in Fig. 63 represent the forces with which gravity pulls on a 
 body, these forces are together equivalent to the single force F, 
 and the point of application of F is the center of gravity, C, 
 of the body. 
 
 The force with which gravity pulls on a body is proportional 
 to the mass, m, of the body. Therefore, if we use the pull 
 of gravity on a one-pound body as our unit of force, the pull 
 
 and 
 
 
 
 RX = Aa + Bh + Cc 
 
 so that 
 
 ■*, 
 
 
 V _Aa + Bh-{-Cc 
 
124 
 
 CALCULUS. 
 
 of gravity on any body is numerically equal to the mass of the 
 body in pounds. Thus a 10-pound body would be pulled by 10 
 
 units of force. This unit 
 of force is called a pound. 
 It is evident therefore 
 that the word pound has 
 two meanings; it signifies 
 a unit of mass when we 
 speak of so many pounds 
 of sugar or coal, and it 
 signifies a unit of force 
 when we speak of so 
 many pounds of force. 
 Consider a small particle of the body in Fig. 63 of which the 
 mass is dm pounds and of which the abscissa is x as shown. 
 The pull of gravity on the particle is dF = dm, the torque 
 action of this force about the axis (perpendicular to the plane 
 of the paper) is x-dm, the combined torque action about 
 of all the forces fff may be expressed as the integral J x-dm, 
 and this total torque action is equal to the torque action about 
 of the resultant F. Therefore we have: 
 
 Fig. 63. 
 
 FX = Jx'dm 
 
 (4) 
 
 Now the force F is equal to the total mass M of the body as 
 above explained. Therefore from equation (4) we have: 
 
 X = 
 
 fx'dm 
 
 M 
 
 (5) 
 
 Similarly, the y and z coordinates of the center of gravity are 
 given by the equations: 
 
 F = 
 
 fy-dm 
 
 M 
 
 (6) 
 
MISCELLANEOUS APPLICATIONS 
 
 125 
 
 and 
 
 Z = 
 
 f Z'dm 
 
 M 
 
 (7) 
 
 Equations (4) and (5) have exactly the same significance as 
 equations (2) and (3); and when the value of fx-dm is found 
 for the whole body, then the distance X of the center of gravity 
 from the origin can be calculated if the total mass of the body 
 is known. 
 
 75. Center of gravity of a bent rod. It is desired to find the 
 coordinates X and Y of the center of gravity of a rod which is 
 represented by the heavy black portion of the parabola which is, 
 shown in Fig. 64 and whose equation is y = kx^. Let h be 
 
 
 y-dxis 
 
 1 
 
 1 
 
 
 I.. V i 
 
 \ 
 
 \ 
 \ 
 
 / 
 
 r ( 
 
 1 
 
 •- 1 
 
 '^x. 
 
 --■»< 
 
 x-axifi 
 
 
 t-"!-,- 
 
 
 
 t 
 1 
 
 1 
 / 
 
 
 / 1 
 
 \ 
 \ 
 
 J / i 
 
 
 /l>i 
 
 N. 
 
 ,X \ Ijc-ojcis 
 
 
 ^ JC -^ 
 
 Fig. 64. Fig. 65. 
 
 the mass of the rod per unit of length (pounds per foot). Then 
 
 }i'd^ = hA{dxY-\- {dyf 
 
 is the mass in pounds of the portion dfs of the rod, as may be 
 understood with the help of Fig. 65. But from the equation of 
 the parabola , y — fcx^, we have 
 
 dy — 2kx'dx. 
 
126 CALCULUS. 
 
 Therefore h ^{dxY + {dyY is equal to ^ Vl + ^k'^x'^ • dx and 
 this expression is to be used instead of dm in equation (5) of 
 the previous article so that: 
 
 £6 
 
 Now the mass of the rod in Fig. 64 is the length of the rod 
 multiplied by ^, and the length is given by the integral of 
 Vl + Wx^ • dx from x = a to x = h. That is 
 
 /'»x=6 
 
 M = h \ Vl + Wx'^'dx (2) 
 
 When M is determined by this equation its value can be used 
 in equation (1) to give the value X. 
 
 In a similar manner equation (6) of Art. 74 gives, for the bent 
 rod shown in Fig. 64: 
 
 hh 
 
 £=6 
 xHl -^Wx^'dx 
 a 
 
 Y — M (3) 
 
 from which Y can be calculated when M is known Jsee equation ^ 
 (2)]. 
 
 76. Center of gravity of a solid cone. Consider the solid 
 cone shown in Fig. 66. To determine the location of the center 
 of gravity of this cone equations (5), (6) and (7) of Art. 74 can 
 be used as in the case of a bent rod, but it is worth while to give 
 a slightly different argument as follows: Let T be the torque 
 action of the pull of gravity on the cone, T being reckoned about 
 the axis (perpendicular to the plane of the paper). Then T 
 is evidently a function of the length x of the cone {x is shown in 
 Fig. 67), and the increment of T due to an increment of x is: 
 
 dT = X'dF (1) 
 
 J^ 
 
MISCELLANEOUS APPLICATIONS. 
 
 127 
 
 where dF is the force with which gravity pulls on the thin slab 
 of material in Fig. 66, the radius of the slab being y and its 
 thickness dx. The volume of this slab is iry^ • dx and its mass is 
 
 Fig. 66. 
 
 X'OxU 
 
 Fig. 67. 
 
 irDy^'dx (= dm) where D 
 pounds per cubic foot. But 
 of the half-angle of the cone, 
 so that equation (1) becomes 
 
 is the density of the material in 
 y = kx, where k is the tangent 
 Therefore dm = dF = irkWx^'dx, 
 
 whence 
 
 dT = TrkWx^'dx 
 T = — -i — x^-\- a. constant 
 
 (2) 
 (3) 
 
 but it is evident that T is zero when x = because when x = 
 there is no cone at all. Therefore the constant of integration in 
 equation (3) is equal to zero, and equation (3) becomes 
 
 T=^-m.^ 
 
 (4) 
 
 Now the volume of the cone in Fig. 66 is a function of x whose 
 increment dV is: 
 
 dV = Try^'dx 
 
 (5) 
 
128 CALCULUS, 
 
 or, using y = A;a; as before, we have 
 
 dV = wk^x^dx (6) 
 
 so that 
 
 V = iirk^x^ (7) 
 
 the constant of integration being zero because V = when 
 X = 0. The mass, Af, of the cone is equal to D7 pounds, and 
 the total pull of gravity on the cone is therefore F = DV pounds. 
 The torque action of this force about the axis is 
 
 FX = DVX = DX times JttAjV 
 
 and this torque action is equal to T. Therefore, substituting 
 DX times iirk^x^ for T in equation (4) and solving for X, 
 we have: 
 
 X = ix (8) 
 
 77. Average distance of the particles of a body from a plane. 
 A new view of center of gravity. Imagine a body to be made 
 up of minute particles of equal mass. Then the number of 
 particles in a piece of the body would be proportional to the 
 mass of the piece, that is, equal to A^ X mass of piece, where 
 A^ is a constant. Let dm be the mass of a small piece of a body; 
 then N'dm is the number of particles in dm. If x is the 
 abscissa of dm, then x X N'dm is the sum of the abscissas 
 of all the particles in dm, and the integral J Nx • dm or Nj x • dm 
 (extended so as to include the entire body under discussion) is 
 the sum of the abscissas of all the particles of the body. Further- 
 more NM is the total number of particles in the body, M being 
 the mass of the body. Therefore, dividing Nfx-dm by NM 
 we have the sum of the abscissas of all the particles divided by the 
 number of particles, which is the average abscissa of the entire body. 
 Representing this average abscissa by X we have: 
 
 NjX'dm fx-dm 
 ^ ^ NM ^ ~~M~~ 
 
MISCELLANEOUS APPLICATIONS. 
 
 129 
 
 Therefore, comparing this with equation (5) of Art. 74 we see 
 
 that the abscissa of the center of gravity of a body is the average 
 
 abscissa of all the particles of the body, and similar statements can 
 
 be made with reference to Y and Z as given by equations (6) 
 
 and (7) of Art. 74. From the point of view of averages the 
 
 center of gravity of a body is usually called the center of mass 
 
 of the body. 
 
 PROBLEMS. 
 
 1. Four downward forces of 100 pounds, 125 pounds, 200 
 pounds and 50 pounds act upon a bar at distances of 10 inches, 
 16 inches, 18 inches and 24 inches from the end of the bar, 
 respectively. Find the value of the resultant of the four forces 
 and the distance from the end of the bar to the point of applica- 
 tion of the resultant. Ans. 475 pounds, 16.4 inches. 
 
 2. Find the center of gravity of a straight bar 15 inches long 
 and of uniform sectional area on the assumption that the density 
 of the material is given by the equation D = kx^; where D is 
 the density of the material at a point, x is the distance of the 
 point from the end A of the rod, and fc is a constant. Ans. 12 
 inches from end A. 
 
 B-\ i ^-. 
 
 O 
 
 1f-axi8 
 
 circulitr rod 
 
 \ 
 
 
 Fig. pSa. 
 
 Fig. p36. 
 
 3. Find the location of the center of gravity of a rod bent 
 into the arc of a circle as shown in Fig. p3a. Ans. 8.27 inches 
 from the center of the circle in the line AB, 
 10 
 
130 
 
 CALCULUS. 
 
 Note, Figure p36 shows how this problem may be formulated. The length 
 
 dx 
 
 of the portion ds of the rod is where 6 is the angle between ds and 
 
 ^ cos » * 
 
 the X-axis and it is equal to the complement of the angle between r and the 
 X-axis. That is cos ^ = -. The total added mass corresponding to dx is 
 the mass of the two short portions ds. Therefore 
 
 where h is the mass of the rod per unit length. With this start it is easy to 
 carry the problem through to a conclusion. 
 
 4. Find the distance from the 16 inch base to the center of 
 gravity of the entire wedge (30 inches long) which is shown in 
 Fig. p4. Ans. 10 inches. 
 
 5. Find the distance from the 16 inch base to the center of 
 gravity of the frustum of a wedge (12 inches long) which is shown 
 in Fig. p4. Ans. 5.5 inches. 
 
 6. Find the distance from the apex to* the center of gravity 
 of the entire cone (10 inches long) shown in Fig. pQ. Ans. 7.5 
 inches. 
 
 7. Find the distance from the apex to the center of gravity 
 of tlie frustum of a cone (5 inches long) which is shown in Fig. p6. 
 Ans. 8.04 inches. 
 
 8. Find the position of the center of gravity of a solid hemi- 
 
 ^^- 
 
MISCELLANEOUS APPLICATIONS. 
 
 131 
 
 sphere whose radius is 16 inches. Ans. 6 inches from the 
 center of the sphere on the axis of symmetry. 
 
 Note. This problem can be formulated in a manner very similar to the 
 formulation of the problem of the cone which is discussed in Art. 76; and Fig. 
 p8 will suggest the method of formulation. 
 
 9. Figure p9 represents a segment of a solid sphere. Locate 
 its center of gravity. Ans. 14.32 inches from the center of 
 the sphere on the axis of symmetry of the segment. 
 
 y'^^^^y^segment of sphere 
 
 (r-x) 
 
 x-axis 
 
 \ 
 
 T 
 ■*—8lab of I 
 
 thickness dx / 
 / 
 / 
 / 
 
 Fig. p8. 
 
 I 
 
 11 
 
 Fig. p9. 
 
 10. Locate the center of gravity of a thin hemispherical shell 
 whose radius is 16 inches. Ans. 8 inches from the center of 
 the sphere on the axis of symmetry of the shell. 
 
 iVote. Let Fig. p36 represent a thin spherical shell. Then 2-Ky.ds is the 
 area of the portion of the shell corresponding to dx. The mass dm of this 
 portion of the shell is 2-Ky .ds X a, where a is the mass of the shell in pounds 
 per square foot. With this start it is easy to carry the problem through to 
 a conclusion. 
 
 1 1 . Locate the center of gravity of the solid paraboloid which 
 is shown in Fig. pll. Ans. 8 inches from the vertex of the 
 paraboloid on the axis of the paraboloid. 
 
 12. Locate the center of gravity of the frustum of a solid 
 
132 
 
 CALCULUS. 
 
 paraboloid which is shown in Fig. pl2. Ans. 9.91 inches from 
 the vertex of the paraboloid on the axis. 
 
 13. Locate the center of gravity of the thin paraboloidal shell 
 
 / 
 
 L7inch±8_ 
 
 \ 
 
 5 inches 
 
 Fig. pU. 
 
 Fig. pl2. 
 
 which is shown in Fig. pll. Ans. 6.73 inches from the vertex 
 of the paraboloid on the axis. 
 
 Note. In the solution of this problem it is necessary to integrate an ex- 
 pression of the form J a: Va + 6aJ • dx. This expression can be put into a 
 form whose integral is easily recognized by substituting * 
 
 z^ := a -\- bx 
 
 80 that 
 
 X = 
 
 dx = 
 
 z" - c 
 h 
 
 2zdz 
 
 14. Figure pl4 represents a flat metal plate cut with a parabolic 
 edge. Locate the center of gravity of the whole plate. Ans. 
 5.4 inches from the vertex of the parabola on the axis. 
 
 15. Locate the center of gravity of the portion AB of the 
 plate shown in Fig. pl4. Ans. X = 6.66 inches, 7 = 0. 
 
 16. Locate the center of gravity of the portion A of the plate 
 Bhown in Fig. pl4. Ans. X — 6.66 inches, Y = 3.42 inches. 
 
 Note. Consider a small element of the plate as represented by the shaded 
 square in Fig. pl6. The area of this element is dx.dy and its mass is dm = 
 k'dx'dy, where & is a confitant. Therefore the integrals fx-dm and 
 
 ■au.^: 
 
MISCELLANEOUS APPLICATIONS. 
 
 133 
 
 J y-dm become kjjx'dx-dy and kjfy-dxdy, respectively. The double 
 integral signs indicate that two integrations are necessary in each case, one 
 with respect to x and the other with respect to y. The limits of these integra- 
 tions are most easily assigned when the integration with respect to y is made 
 
 first. In this case the integration with respect to y is between the Umits 
 y = and y = Vax (where y^ = ax is the equation of the parabola). In 
 this integration x and dx are treated as constants. Then the integration 
 with respect to a; is made between the limits x = 4 inches and x = 9 inches. 
 
 17. Locate the center of gravity of a fiat plate which is in the 
 shape of a sector of a circle as shown in Fig. pl7. Ans. In the 
 line AB at a distance 6.36 inches from A. 
 
 y-axi8 
 
 hemiapfiere 
 thin slab 
 
 x^axia 
 
 Fig. pl7. 
 
 Fig. pl8. 
 
134 CALCULUS. 
 
 18. Suppose that the density of the material of which a sphere 
 is made is kr where A; is a constant and r is any distance from 
 the center of the sphere. Locate the center of gravity of half 
 of the sphere, its radius being 2 feet. Ans. X = 0.8 foot. 
 
 Note. Consider a ring-shaped element of material lying in the thin slab in 
 Fig. pl8; the radius of the ring being y and the thickness and breadth of the 
 ring being dx and dy respectively. The volume of this ring is ^irydx-dy 
 and its mass is equal to kr X 2iry'dx'dy, where r = Vx* + y^ as shown 
 in the figure. Therefore the pull of g ravity (parallel to the y-axis, say) on the 
 ring-shaped element is 2irky Vx^ + y^ -dx-dy and the torque action of this pul 
 about the axis (perpendicular to the plane of the paper) is: 
 
 dT = 2Trkxy\x^ + y^' dx-dy 
 
 With this start the problem can be formulated without difficulty. The 
 total mass of the hemisphere can be found by integrating, between proper 
 limits, the expression: 
 
 dM = 2irky^x^ -{- y^ 'dx-dy 
 
 19. What is the average distance of the points on a semi-circle 
 from the diameter which bounds the semi-circle? Ans. 0.64 
 of the radius. 
 
 Note. Let N be the nimiber of points per unit length of a line. Then 
 N-ds \a the number of points in the line element ds, and iVTrr is the number 
 of points on the semi-circle. 
 
 20. What is the average distance of all the points in a semi- 
 circular area from the diameter which bounds the semi-circle? 
 Ans. 0.42 of the radius. 
 
 Note. Let N represent the number of points in one unit of area. Then 
 
 Ttr^ 
 N-dA is the number of points in an element of area dA and A^* -7^- is the 
 
 number of points in the semi-circle. 
 
 CENTER OF PRESSURE. 
 78. Force exerted on a water gate. Figure 68 shows a water 
 gate covering an opening in a dam. The short arrows fff repre- 
 sent the forces with which the water pushes against the gate. 
 These forces are together equivalent to the single force F, their 
 
 .^ 
 
MISCELLANEOUS APPLICATIONS. 
 
 135 
 
 water level r rrrr 
 
 dam 
 
 Fig. 68. 
 
 resultant, the point of application of which is called the center 
 of pressure on the gate. The force F is equal to the sum of all 
 the forces ///, and the torque action of F 
 about an arbitrarily chosen axis (see 
 Fig. 69) is equal to the sum of the torque 
 actions about of the forces ///. 
 
 Consider a horizontal strip of the gate 
 of which the width is dx as shown in Fig. 
 69, and of which the length is w. The area 
 of this strip is w-dx and the force dF ex- 
 erted on the strip is:* 
 
 dF = Dwx-dx (1) 
 
 Therefore 
 
 I X'dx (2) 
 
 x=a Fig. 69. 
 
 * The pressure in pounds per square foot at a place x feet beneath the 
 surface of water is p = Dx, where D is the density of water in pounds per 
 cubic foot (= 62K). Furthermore, the force exerted on the strip is found 
 by multiplying the area of the strip in square feet by the pressure in pounds 
 per square foot. 
 
136 CALCULUS. 
 
 which gives 
 
 F = iDw{¥ - a^) (3) 
 
 The torque action of the force dF about the axis is x-dF 
 or Dwx^'dXf and the total torque action about of all the 
 forces /// is 
 
 '~\^'dx (4) 
 
 T = DwT 
 
 *Jx=a , 
 
 which gives 
 
 T = \DwQy' - a^) (5) 
 
 This torque action must be equal to XF, Therefore using the 
 value of F from equation (3) we have 
 
 \Dw{h'' - a2)X = \Dw{W ~ a^) (6) 
 
 whence 
 
 79. Force exerted on a curved surface by a stationary fluid 
 under pressure. The short arrows /// in Fig. 70 represent 
 the forces with which the water pushes on the curved surface of 
 a dam. All the forces /// are together equivalent to a single 
 force F (see Fig. 73) which is called their resultant. Let Fx 
 and Fy be the x and y components of F as shown in Fig. 73. 
 Then Fx is the sum of the a:-components of the forces fff 
 in Fig. 70, and Fy is the sum of the ^/-components of the 
 forces ///. Let X and Y be the coordinates of the point of 
 appHcationof the resultant force F (see Fig. 73). Then we have 
 the following four conditions which determine Fxy Fy, X and Y: 
 
 (a) Fx is the sum of the aj-components of the forces fff. 
 
 (h) Fy is the sum of the ^/-components of the forces fff. 
 
 (c) The torque action of Fx about the arbitrarily chosen 
 axis is FxY, as may be seen from Fig. 73, ahd this torque 
 action is equal to the sum of the torque actions about of 
 the a;-components of the forces ///. 
 
MISCELLANEOUS APPLICATIONS. 
 
 137 
 
 (d) The torque action of Fy about is FyX, and this 
 torque action is equal to sum of the torque actions about of 
 the ^/-components of the forces ///. 
 
 In order to formulate these four conditions it is necessary to 
 derive expressions for the x and y components of the force dF 
 which is exerted by a fluid on an element of a curved surface. 
 Consider the surface element AB, Fig. 71, the width of the 
 element being ds and its length I (perpendicular to the plane 
 of the paper). The area of AB is l.ds square feet, and the force 
 in pounds exerted on AB is dF = pl.dSf where p is the pressure 
 
 ,umter level 
 
 Fig. 70. 
 
 of the fluid in pounds per square foot. The x and y components 
 of dF are, respectively: 
 
 and 
 
 dF:, = sin0-dF 
 dFy = cos (f)'dF 
 
 dx 
 
 Therefore, using pl-ds for dF, using -p for cos 0, and using 
 
 -r for sin 6, we have 
 ds 
 
 and 
 
 dF. ==pl-dy (1) 
 
 dFy = phdx (2) 
 
 To determine the total force exerted on the face of the dam in 
 
138 
 
 CALCULUS. 
 
 Fig. 70, that is, to determine the two components Fx and Fy 
 of the total force and the coordinates X and Y of its point of 
 appHcation as shown in Fig. 73, consider the element of area 
 hds as shown in Fig. 72, where I is the length of the dam per- 
 
 pendicular to the plane of the paper. The pressure of the water 
 at this element is Dy pounds per square foot as explained in 
 the footnote to Art. 78. Therefore, using Dy for p in equations 
 (1) and (2), we have for the x and y components of the force 
 dF in Fig. 72: 
 
 dFx = Dlydy (3) 
 
 and 
 
 dFy = Dlydx (4) 
 
 From equation (3) we find the x-component of the total force 
 exerted on the dam by integrating between the limits y = Q 
 to y = hj which gives: 
 
 F, = iDlh^ (5) 
 
 The equation of the parabola in Fig. 72 is 
 
 ^ 2 
 
 y = p-^^ 
 
 Therefore, using this value of y in equation (4) and integrating 
 
MISCELLANEOUS APPLICATIONS. 139 
 
 between the limits x = to x = h, we find the 2/-component 
 of the total force exerted on the dam, namely: 
 
 Fy = \Dlhh (6) 
 
 To find the distance X in Fig. 73 it is necessary to find the 
 total torque action about of all the forces dFy and place the 
 result equal to FyX. The torque action of dFy (the i/-com- 
 ponent of dF in Fig. 73) about is a; . dFyj which by equation 
 (4) becomes Dlxy-dx. But 
 
 so that the torque action of dFy about is DlroOc^-dx, and the 
 torque action of all the forces dFy about is 
 
 Dlrh I x^-dx 
 
 O^Jx=0 
 
 which is equal to \I)lhh^. Therefore we have 
 
 XFy = \DlhW (7) 
 
 or, using the value of Fy from (6), we have: 
 
 X = 16 (8) 
 
 The torque action of dFx (the x-component of dF in Fig. 73) 
 about is y-dFx, which by equation (3) becomes Dly^-dy; 
 and the total torque action of all the forces dFx is 
 
 Jf*V=n 
 
 which is equal to iDl¥. Therefore we have: 
 
 YFx = iDW (9) 
 
 or, using the value of Fx from (5), we have 
 
 Y = ih (10) 
 
140 
 
 CALCULUS. 
 
 y-axis 
 
 PROBLEMS. 
 
 1. Find the total force F acting on the gate in Fig. 68 and 
 find the distance X, when a = 4 feet, 6 = 8 feet and ly = 4 
 feet. Ans. 6000 pounds, 6.22 feet. 
 
 2. A dam has a circular hole through 
 it 6 feet in diameter. The center of 
 the hole is 7 feet beneath the surface of 
 the water. The hole is covered with 
 a gate. Find the total force with 
 _ which the water pushes on the gate 
 
 4---(vr/77r^ 1 ^^^ fiiid the distance from the surface 
 
 I V///////1 "V r I 
 
 of the water to point of application 
 of this force. Ans. 12,370 pounds, 7.32 
 feet. 
 
 X-axis Note. Consider the element of area which is 
 
 » represented by the small shaded square m Fig. p2. 
 
 ^^' ^ • The pressure at this area is kx where A; is a 
 
 constant. Therefore the force acting on the element of area is kx-dx'dy, 
 
 and the total force on the gate is jj kx'dx'dy. To specify the limits of this 
 
 integration let y = f(x) be the equation of the circle in Fig. p2 (the gate). 
 Then if we integrate first with respect to y the limits are y = — fix) to 
 y = +/(x). We may then integrate with respect to x between the Umits 
 X = h — r to x = 6 + r. 
 
 The torque action of the force kx-dx'dy about the axis (perpendicular 
 to the plane of the paper) is kx^-dx-dy, and the total torque action of the 
 
 force acting on the entire gate is given by the integral J J kx^'dx-dy. The 
 limits are the same as stated above. 
 
 3. Find the abscissa X and the ordinate Y of the center of 
 pressure of the shaded half of the circular water gate in Fig. p2, 
 where 6 = 7 feet and r = 6 feet. Ans. X = 7.32 feet, Y = 
 1.27 feet. 
 
 MOMENT OF INERTIA. 
 
 80. Kinetic energy of a rotating body. Definition of moment 
 of inertia. Consider a wheel which is rotating n revolutions 
 
MISCELLANEOUS APPLICATIONS. 141 
 
 per second. Its speed in radians per second is 
 
 CO = 2irn (1) 
 
 because there are 2% radians in one revolution. The speed of 
 a rotating body in radians per second is called the angular 
 velocity or the spin velocity of the body. 
 
 Consider a particle of the wheel at a distance r from the axis 
 of rotation. As the wheel rotates this particle travels in a circle 
 of which the circumference is 27rr and it travels n times round 
 this circle per second so that the velocity v of the particle is 
 
 V = 2Trnr (2) 
 
 or using w for 27rn according to equation (1), we have: 
 
 v = ojr (3) 
 
 If the spin velocity of the wheel is doubled it is evident from 
 this equation that the velocity of every particle in the wheel 
 will be doubled so that the kinetic energy of every particle in the 
 wheel will be quadrupled. Therefore the total kinetic energy, 
 W, of a rotating wheel is quadrupled if the spin velocity co 
 of the wheel is doubled; that is, for a given wheel, W is pro- 
 portional to £0^, and for the given wheel there is a definite 
 constant by which co^ may be multiplied to give W. That is: 
 
 W = iKo)'' (4) 
 
 where (^K) is the proportionality factor for the given wheel. 
 The constant K is called the moment of inertia of the wheel,* and 
 it depends upon the size, shape, and mass of the wheel. 
 
 * The kinetic energy of a particle is equal to ^mv^, where m is the mass 
 of the particle in pounds, v is its velocity in feet per second, and kinetic energy 
 is expressed not in foot-pounds, but in foot-poundals. Throughout this dis- 
 cussion of moment of inertia distance is expressed in feet, velocity in feet per 
 second, mass in pounds, force in poundals, torque in poundal-feet, work or 
 energy in foot-poundals, and moment of inertia in pound-feet.^ 
 
142 CALCULUS. 
 
 81. General integral expression for moment of inertia. Imag- 
 ine a small particle of mass dm to be added to the spinning 
 body at a distance r from the axis of spin. Then the velocity 
 V of the added particle is cor, and the kinetic energy of the added 
 particle is ^dm X coV which is, of course, the infinitesimal 
 increment of the kinetic energy of the spinning body due to the 
 added particle. Therefore 
 
 dW = ioy^rHm 
 and by integration* we have 
 
 W = Wfr^-dm (5) 
 
 Comparing this equation with equation (4) it is evident that 
 
 K^fr^'dm (6) 
 
 82. Average value of the square of the distances of all the 
 particles of a body from an axis. Definition of radius of gyration. 
 
 Using the ideas of Art. 77, N-dm is the number of particles in a 
 small piece dm of a body. If r is the distance of the small 
 piece dm from a chosen axis, then Nr^-dm is the sum of the 
 squares of the distances of all the particles in dm from the axis, 
 and the integral jNr^.dm or Nfr^-dm extended so as to 
 include an entire body is the sum of the squares of the distances 
 of all the particles of the body from the axis. But NM is 
 the number of particles in the body. Therefore 
 
 N f r^ ' dm fr^ • dm 
 
 — ^ = ^ (I) 
 
 is the average value of the squares of the distances of all the 
 particles of a body from the axis. This average may be repre- 
 
 * This integration can only be indicated. Before the actual value of the 
 integral can be found r and dm must be expressed in terms of one or more 
 independent variables, and the limits of the integration must be such as to 
 include the entire spinning body. 
 
MISCELLANEOUS APPLICATIONS. 
 
 143 
 
 sented by p^ so that 
 
 f r^ • dm 
 
 M 
 
 or 
 
 p2M = fr^-dm 
 
 (2) 
 
 (3) 
 
 but J 7-2 • dm is the moment of inertia K of the body with respect 
 to the chosen axis, according to equation (6) of Art. 81. There- 
 fore we have: 
 
 K = pm (4) 
 
 The distance p, which is the square-root-of-thc-aVerage-value- 
 of-the-squares-of-the-distances-of-all-the-particles-of-a-body-from- 
 a-chosen-axis, is called the radius of gyration of the body with 
 respect to the chosen axis. 
 
 83. Moment of inertia of a circular saw about its axis of 
 spin. The moment of inertia of a circular saw could be easily 
 derived from equation (6) of Art. 81, but it is instructive to give 
 the complete argument again as follows: Let W be the kinetic 
 energy of a circular disk r feet in radius rotating at a constant 
 spin velocity of co radians per second, as shown in Fig. 74, and 
 
 n— -.^- 
 
 added , 
 material 
 
 Fig. 74. 
 
 let it be required to find the infinitesimal increment of W due 
 to an arbitrary infinitesimal increment of r. Let t be the thick- 
 ness of the disk and let D pounds per cubic foot be the density 
 of the material of which the disk is made. Imagine the radius 
 
144 CALCULUS. 
 
 of the spinning disk to be increased by the addition of material 
 as indicated in Fig. 74. The volume of the added material is 
 2irr X t X dr, the mass of the added material is 2-Ktr'dr X D 
 pounds, the velocity of the added material is or, the kinetic 
 energy of the added material is equal to one half the product of 
 its mass times the square of its velocity, and the kinetic energy 
 of the added material is the desired infinitesimal increment dW. 
 Therefore : 
 
 dW = rDUa'^r^'dr (1) 
 
 W and r being the only variables. Therefore, by integration 
 we get: 
 
 W = JttD^coV^ + a constant (2) 
 
 But W must evidently be zero when r is zero. Therefore the 
 constant of integration must be equal to zero, so that equation 
 (2) becomes 
 
 W = iirDtJ'r' (3) 
 
 Now irrHD is equal to the mass m of the disk in pounds so that 
 equation (3) becomes: 
 
 W = iw^mr^ (4) 
 
 But according to Art. 80 the kinetic energy of any rotating body 
 can be expressed as ^Ku^, where K is the moment of inertia 
 of the body. Therefore we have 
 
 ^v W = i«2mr2 = ^Kc^ (5) 
 
 from which we have: 
 
 K = imr^ (6) 
 
 That is, the moment of inertia of a circular saw about its axis 
 of spin is equal to one half the mass of the saw in pounds multi- 
 plied by the square of the radius of the saw. 
 
 Remark. The thickness i in the above discussion may be 
 anything whatever. Therefore equation (6) expresses the mo- 
 ment of inertia of a circular cylinder of any length rotating about 
 its axis of figure. 
 
MISCELLANEOUS APPLICATIONS. 
 
 145 
 
 84. Moment of inertia of a rectangular bar. To determine 
 the moment of inertia of the rectangular bar shown in Fig. 75 
 the general equation (6) of 
 Art. 81 will be used. Figure 
 76 represents a top view of 
 the bar, being the axis of 
 rotation. Consider the ele- 
 ment of the barwhich is rep- 
 resented by the small black 
 square in Fig. 76. This ele- 
 ment is understood to ex- 
 tend entirely through the 
 bar parallel to the axis 
 paper in Fig. 76). 
 
 Fig. 75. 
 
 (at right angles to the plane of the 
 Therefore the volume of the element is 
 t'dx'dy and its mass is tD-dx' dy. The distance of the element 
 from the axis is r = Vx^ + y^, so that 
 
 r^ = x^ + y^ 
 
 Therefore the general equation (6) of Art. 81 becomes: 
 
 K = tnffix^ + 2/2) 'dx'dy (1) 
 
 in which the double integral sign is used because two integrations, 
 one with respect to x and one with respect to y are necessary, 
 as explained in the note to problem 16 on page 132. 
 
 If we integrate with respect to y (treating x and dx as con- 
 stants) between the limits 
 L 
 
 y = 
 
 to 
 
 t/ = + - we 
 
 Fig. 76. 
 
 get the moment of in- 
 ertia with respect to the 
 axis of the thin slab be- 
 tween the dotted lines in 
 Fig. 76. We may then inte- 
 l . . I 
 
 grate with respect to x between the limits x = —-^ to x 
 
 11 
 
 = + 2 
 
146 CALCULUS. 
 
 and we get the desired expression for K, namely: 
 
 K = ^\V + ^) (2) 
 
 But Dtwl is the mass m of the bar in pounds, so that equation 
 (2) becomes: 
 
 X=^(P + «,»)« (3) 
 
 Remark. The same final result is obtained if we integrate 
 first with respect to x (treating y and dy as constants), and 
 then with respect to y. What is the meaning of the result of 
 the first integration in this case? 
 
 85. Torque required to increase the spin- velocity of a body. 
 
 The velocity of a particle in a rotating wheel is t; = rw, according 
 to equation (3) of Art. 80. Therefore when « increases, v 
 increases r times as fast as w. That is: 
 
 dv du) ,^v 
 
 This -r. is the acceleration of the particle in the direction at 
 right angles to r*, and 
 
 dv J / d(a J \ 
 
 is the sidewise force (at right angles to r) which must act on the 
 particle to produce the sidewise acceleration. The torque action 
 of this sidewise force about the axis of rotation is: 
 
 dT = r^'dmXr 
 dt 
 
 or 
 
 dT = ~ -r'-dm (2) 
 
 * We are not here concerned with the radial acceleration of the particle 
 which is discussed in Art. 50. 
 
MISCELLANEOUS APPLICATIONS. 147 
 
 so that the total torque action required to increase the spin 
 velocity of the body at the rate -t: is : 
 
 ^''^r'^-dm (3) 
 
 T = — C 
 dtj 
 
 The integral is of course understood to be extended over the 
 whole body and it is equal to the moment of inertia of the body 
 according to equation (6) of Art. 81. Therefore equation (3) 
 may be written 
 
 r = Kf (4) 
 
 That is, the spin acceleration -^ of a body multiplied by the 
 
 moment of inertia of the body is equal to the torque which must 
 be exerted on the body to produce the spin acceleration. 
 
 86. Moments of inertia about par- ^dsf 
 
 allel axes. Let K be the moment ^y"^ 
 
 of inertia of a body about an axis ^^^ A 
 
 (perpendicular to the plane of the ^^ / 1 
 
 paper in Fig. 77) which passes through *i-.^^2 ^-. 
 
 the center of gravity of the body, pjg 77^ 
 
 and let K' be the moment of inertia 
 
 of the body about an axis P (perpendicular to the plane of the 
 
 paper) which is at a distance a from 0. Then 
 
 K' = Z + aW (1) 
 
 where M is the total mass of the body in grams or pounds as the 
 case may be. 
 
 Proof. From the triangle in Fig. 77 we have: 
 
 gf2 = (j2 _j_ j,2 _j_ 2ar cos B 
 or 
 
 52 = (j2 4. y.2 _|_ 2ax (2) 
 
 where x is the abscissa of the element of material dm referred 
 
148 CALCULUS. 
 
 to the center of gravity as an origin. Now according to 
 equation (6) of Art. 81 we have 
 
 K' = fq'-dm (3) 
 
 Therefore, using the value of q^ from equation (2), we have: 
 
 K' = fa^-dm + fr^-dm + f2ax-dm 
 
 K' = a'fdm + Jr^-dm + 2afx'dm (4) 
 
 But J dm = the total mass M of the body, jr^-dm is the 
 
 moment of inertia K referred to the axis 0, and jx-dm is 
 zero according to equation (5) of Art. 74, because the center of 
 gravity in Fig. 77 is taken as the origin. Therefore equation 
 (4) becomes: 
 
 K' = am + K (1) 
 
 87. Moment of section of a beam. Consider equation (2) of 
 Art. 72, namely: 
 
 dT = ^^x^'dx (1) 
 
 The product 2b -dx is the area of the shaded strips in Fig. 59, and 
 X is the distance of this area from the axis 00, Let us use dA 
 for 2b 'dx, then equation (1) becomes: 
 
 dT = ^x^'dA (2) 
 
 80 that 
 
 ^ "* :^'dA (3) 
 
 -f/' 
 
 The integral fx* • dA is called the moment of section of the beam; 
 and according to equation (3) the torque T which bends any 
 
 beam is equal to -h- » where E is the stretch modulus of the 
 K 
 
 material of the beam, R is the radius of curvature of the median 
 line of the beam, and S[ =fx* • dA) is the moment of section of 
 the beam. 
 
MISCELLANEOUS APPLICATIONS. 
 
 149 
 
 The integral fx^-dA is similar in form to the integral fr^-dm 
 in equation (6) of Art. 81, and because of this similarity engineers 
 call J x^ ' dA the "moment of inertia " of the section of the beam. 
 The term wmnent of section is however the correct term. 
 
 PROBLEMS. 
 
 1. Find the moment of inertia of a circular saw six feet in 
 diameter and of which the mass is 125 pounds. Ans. 562.5 Ib.-ft.^ 
 
 2. Find the amount of energy in foot-pounds stored in the 
 saw of problem 1 at a speed of 600 revolutions per minute. Ans. 
 34482 foot-pounds. 
 
 Note. In the formula W = }4Ku^, K is expressed in pound-feet-squared, 
 w is expressed in radians per second, and W is expressed in foot-poundals 
 (not in foot-pounds). There are 32.2 foot-poundals in one foot-pound. 
 
 3. Find the moment of inertia of a cylinder 2 feet in diameter 
 and 3 feet long with respect to its axis, the density of the material 
 being 420 pounds per cubic foot. Ans. 1980 Ib.-ft.^ 
 
 4. Find the moment of inertia of a hollow-cylinder, the external 
 dimensions being the same as in problem 3, inside diameter being 
 1 foot, the axis of revolution being the axis of the cylinder, and 
 the density of the material being 420 pounds per cubic foot. 
 Ans. 1114 lb.-ft.2 
 
 5. Find the moment of inertia of 
 the rectangular bar in Figs. 75 and 
 76. The bar is 5 feet long, 1 foot 
 wide and 0.5 foot thick and it has 
 a mass of 1000 pounds. Ans. 
 2167 lb.-ft.2 
 
 6. Find the moment of inertia of 
 the bar in problem 5 when the axis 
 of rotation coincides with the edge 
 t in Fig. 75. Ans. 8667 Ib.-ft.^ 
 
 7. Find the moment of inertia 
 of a very thin circular disk when 
 
 If-axis 
 ,^axi8pf^ revolution 
 
 x-axi8 
 
 Fig. p7. 
 
150 
 
 CALCULUS. 
 
 the axis of rotation is a diameter of the disk; the mass of the disk 
 being 120 pounds and its diameter being 6 feet. Ans. 270 Ib.-ft.^ 
 
 Note. Let the circle in Fig. p7 represent the disk. Take the narrow 
 vertical black strip as dm. Then dm = k X2y-dx where A; is pounds per 
 square foot of disk area. 
 
 8. Find the moment of inertia of the thin circular disk with 
 respect to the axis as shown in Fig. p8, the density of the 
 
 front view 
 
 Fig. p8. 
 
 / 7?2 \ 
 material being D. Ans. K = tR^D l-j-^x^j-dx. 
 
 Note. The mass of the narrow strip in Fig. p8 is 
 
 2D a//? _ y^'dx'dy = dm 
 and the distance of the strip from the axis is 
 
 Va:» + 2/2 = r 
 Of course x and dx are constants in this problem. 
 
 30 inches 
 
 Fig. p9. 
 
 hoop 3 feet 
 in diameter 
 
 Fig. plO. 
 
MISCELLANEOUS APPLICATIONS. 
 
 151 
 
 9. Find the moment of inertia of the soUd cylinder shown in 
 Fig. p9; the density of the material being 0.28 pound per cubic 
 inch. Ans. 1833 Ib.-ft.^ 
 
 Note. The method of formulating this problem is suggested by problem 8. 
 
 10. A slender circular ring or hoop has a mass of 10 pounds 
 and it is 3 feet in diameter, (a) Find its moment of inertia 
 about the axis C (perpendicular to the plane of the paper in Fig. 
 plO), and (6) find its moment of inertia about the axis ab. 
 Ans. (a) 22.5 Ib.-ft.^ (6) 11.25 
 lb.-ft.2 
 
 11. Find the moment of in- 
 ertia of a solid steel sphere 10 
 inches in diameter, the axis of 
 revolution being a diameter of 
 the sphere. The density of steel 
 is 0.28 pound per cubic inch. 
 Ans. 22.4 Ib.-ft.^ 
 
 cuds of revolution 
 solid sphere 
 
 cylindrical 
 shell 
 
 Note. Let the circle m Fig. pll 
 
 represent the sphere. Take for dm 
 
 a thin cylindrical shell of radius r and 
 thickness dr. 
 
 Fig. pll. 
 
 12. Find the moment of inertia of a steel governor ball, 4 
 inches in diameter, with respect to the axis of revolution of the 
 governor, the center of the ball being 6 inches from the axis. 
 Ans. 2.45 Ib.-ft.^ 
 
 TABLE OF MOMENTS OF INERTIA. 
 
 Axis through center of mass in each case, 
 m = mass of body in grams or pounds. 
 K = moment of inertia. 
 
 I and having uniform section, 
 
 1. Thin straight bar of length 
 axis at right angles to bar. K ■■ 
 
 2. Rectangular parallelopiped, axis parallel to one edge, 
 and b being lengths of other edges. K = j^ia^ + b'^)m 
 
 a 
 
152 CALCULUS. 
 
 3. Cylinder or disk of radius r, referred to axis of cylinder. 
 K = ir^m. 
 Referred to axis at right angles to axis of cylinder 
 
 --(5+0 
 
 m 
 
 where I is the length of the cylinder. 
 
 4. Hollow cylinder oirsidu R and r, referred to axis of cylinder. 
 K = ^{R^ + r^)m. 
 
 5. Sphere of radius r referred to a diameter. K — \r^m. 
 
 Note. If the axis of rotation does not pass through the center of mass, use 
 equation (1) of Art. 86. 
 
CHAPTER VI. 
 
 EXPANSIONS IN SERIES. USE OF COMPLEX QUANTITY. 
 
 88. Maclaurin's theorem. — Let y be any function whatever 
 of X which is finite and continuous and of which all of the 
 derivatives with respect to x are finite and continuous. Then: 
 
 y = A4-Bx+^Cx2 + -ilDx«+|^Ex^+... (1) 
 
 where A, B, C, D, etc., are constants as follows: 
 
 A 
 
 is the value of y 
 
 when X 
 
 = 
 
 B 
 
 is the value of y^ 
 dx 
 
 when X 
 
 = 
 
 C 
 
 is the value of -r^ 
 
 when X 
 
 = 
 
 D 
 
 is the value of , , 
 
 when X 
 
 = 
 
 E 
 
 is the value of ^ 
 
 when X 
 
 = 
 
 
 etc., 
 
 etc. 
 
 
 Equation (1) expresses what is known as Maclaurin^s theorem. 
 
 Proof. — Let the curve cc in Fig. 78 represent the given 
 function. Then the value of y which is to be expressed by 
 equation (1) is the ordinate of the point p. To establish equa- 
 tion (1) we will make a series of approximations and consider 
 the limit towards which this series of approximations trends, 
 as follows: 
 
 First approximation. — To get a first approximation let us 
 
 dv dv 
 
 assume that -^ is equal to the constant B (the value of -^ 
 
 153 
 
154 
 
 CALCULUS. 
 
 when X = 0) everywhere between the points p and q in Fig. 
 78. That is by assumption we have: 
 
 dx 
 
 ^ B 
 
 (2) 
 
 Integrating this differential equation we have: 
 
 y = Bx -\- a. constant 
 but y = A when x = 0, so that the constant of integration is 
 
 Fig. 78. 
 
 evidently equal to A. Therefore as a first approximation we 
 have: 
 
 y = A + Bx (3)* 
 
 dv 
 It is interesting to note that to assume -^ = B everywhere is 
 
 the same thing as to take the inclined dotted line in Fig. 78 as 
 an approximation to the curve cc; and the ordinate of this 
 inclined straight line is A + Bx. 
 
 * Figure 79 shows a curve pq for which 
 
 dy 
 
 dx 
 
 B everywhere, but the 
 
 ordinate of p is not equal to A -{■ Bx because of the discontinuity or jump 
 at d. The function y and all of its derivatives must be finite and continuous 
 everywhere between p and q, as stated at the beginning. Any discontinuity 
 outside of the region between p and q does not vitiate equation (1) for the 
 region between p and q. 
 
EXPANSIONS IN SERIES. 155 
 
 Second approximation. — To get a second approximation let 
 us assume that -j-^ is equal to the constant C (the value of 
 
 -T^ when a; = 0) everywhere between p and q in Fig. 78. 
 That is, by assumption, we have: 
 
 d^y 
 
 Integrating once we have: 
 
 d.^ = ^ w 
 
 ^ = Cx + a constant 
 dx 
 
 dii 
 But -f — B when a; = 0, so that the constant of integration 
 
 is equal to 5, giving: 
 
 % = B + Cx (5) 
 
 Integrating again we have : 
 
 y = Bx + -x Cx^ + a constant 
 
 But y = A when x = 0, so that the constant of integration 
 is equal to A, giving as a second approximation: 
 
 y = A + Bx + ^Cx2 (6) 
 
 Third approximation. — To get a third approximation let us 
 
 assume that -r^ is equal to the constant D (the value of -~ 
 
 when X = 0) everywhere between p and q in Fig. 78. That is, 
 by assumption, we have: 
 
 By three successive integrations (the constant of each Integra- 
 
156 CALCULUS. 
 
 tion being determined as above) we get, as a third approximation: 
 
 y = A + Bx4-|cx2 + iDx3 (8) 
 
 nth approximation. — To get an nth approximation let us 
 
 assume that 7-^ is equal to the constant N (the value of t^ 
 
 when a; = 0) everywhere between p and q in Fig. 78. That is, 
 by assumption, we have: 
 
 g = ^ (9) 
 
 By n successive integrations this differential equation gives, 
 as the nth approximation: 
 
 y= A + Bx +^Cx2 + ||Dx3 + . . . + j^Nx" (10) 
 
 To show that the nth approximation approaches the true value of 
 y as a limit as n approaches infinity. In the first place it is 
 evident that equation (10) gives equation (1) when n is in- 
 creased more and more, but it remains to be shown, that y in 
 equation (10) approaches the correct value of ?/ as a limit as n 
 is increased. 
 
 d^ti 
 The nth derivative -t\ is assumed to be everywhere finite 
 
 and it must have therefore a definite largest value V and a 
 definite smallest value v between p and q. If the largest value 
 V is used instead of N in equation (10) we get too large a value, 
 a, for y. If the smallest value v is used instead of N in equa- 
 tion (10) we get too small a value, b, for y.* That is: 
 
 a = A + Bx+~Cx^+"'n'Vx'' (11) 
 
 * This statement happens to be plausible and therefore the reader is apt 
 to accept it as true without actually perceiving its truth, which is indeed 
 evident if one takes the trouble to think about it. 
 
EXPANSIONS IN SERIES. 157 
 
 and 
 
 h = A -i- Rv 4- ^^ , 
 
 2 ' 1^ 
 
 A + Bx+lcx" + . . . ^. t;x" (12) 
 
 and the true value of y lies between a and b. But, subtracting 
 equation (12) from equation (11) member by member we get: 
 
 and this difference approaches zero as n approaches infinity, 
 because when n is increased by 1 the numerator is multiplied 
 by the finite quantity x, whereas the denominator is multipUed 
 by the quantity n + 1, which becomes as large as you please. 
 It is evident therefore that the value of a as given by equation 
 (11) becomes more and more nearly equal to the value of 6 
 as given by equation (12). But the true value of y Hes between 
 a and h. Therefore equations (11) and (12) both approach the 
 true value oi" y as a limit as n is increased; and equations (11) 
 and (12) reduce to equation (1) when n is indefinitely great. 
 
 Taylor's series. — Heretofore any algebraic expression contain- 
 ing X has been spoken of as a function of x. Thus ex -\- e is 
 a function of x, c and e being costants. Also such expressions 
 as e^a;+/i)^ sin(x + h), tan (a; -f h) are functions of x. If, however, 
 a: is a constant and h a variable we would think of these expres- 
 sions as functions of h. 
 
 Let y be any function whatever of {y + h), let a; be a 
 constant and h a variable, and let it be understood that y and 
 all of its derivatives are finite and continuous. Then y may 
 be expanded by Maclaurin's theorem, giving: 
 
 y [any function of (x + h)] -=A+Bh-\- ~Ch^ + ^D¥ + • • • (14) 
 
 where A, B, C, D, etc., are constants, as follows: 
 
 A is the value of y when h = 
 
158 CALCULUS. 
 
 dv 
 B is the value of -4 when ^ = 
 an 
 
 C is the value of -r^ when h = 
 etc., etc. 
 
 Equation (14) is sometimes called Taylor^s series, but it is 
 identical to Maclaurin's series and to give it a separate name is 
 to create a false distinction. 
 
 89. Examples, (a) Expansion of e*. — The successive deriva- 
 tives of e* are as follows: 
 
 y = e' which is equal to 1 when x = 
 
 dxi 
 
 -^ = e^ which is equal to 1 when a; = 
 
 j\ = e* which is equal to 1 when a; = 
 
 etc., etc. 
 
 Therefore A = B = C = D = etc. = 1, and equation (1) of 
 Art. 88 gives: 
 
 e' = 1 + X -^-^x^ + j^ + ^x' -^ etc. (1) 
 
 (6) Expansion of sin x. — The successive derivatives of sin x 
 are as follows: 
 
 2/ = sin a; which is equal to when re = 
 
 3^ = cos a; which is equal to 1 when a; = 
 ax 
 
 -t5 = — sin a; which is equal to when a; = 
 
 -M = — cos a; which is equal to — 1 when a; = 
 
 t4 = sin a; which is equal to when x = 
 ax* 
 
 and so on in endless repetition. 
 
EXPANSIONS IN SERIES. 159 
 
 Therefore A = 0, 5 = 1, C = 0, D = ~ 1, ^ = 0, etc., and 
 equation (1) of Art. 88 gives: 
 
 />t3 /v»5 /v»7 /v»v 
 
 Binx = x-j- + j--j- + -^ etc. (2) 
 
 (c) Expansion of cos a;. — The successive derivatives of cos x 
 are as follows: 
 
 y = cos X which is equal to 1 when x = 
 
 -^ = — sin X which is equal to when x = 
 
 — I = — cos X which is equal to — 1 when x = 
 
 -^ = sin ic which is equal to when a; = 
 
 -T-^ = cos X which is equal to 1 when x = 
 
 and so on in endless repetition. 
 
 Therefore A = 1, 5 = 0, C = - 1, D = 0, ^ = 1, etc., and 
 equation (1) of Art. 88 gives: 
 
 /ji»2 /v«4 /J.6 /»8 
 
 COS X = 1 -^ + ^ - y + -|g- etc. (3) 
 
 90. Maclaurin*s theorem applied to a function of two inde- 
 pendent variables. — If u is a function of x and y which is 
 itself finite and continuous, and if all of its derivatives (partial 
 derivatives) are finite and continuous, then: 
 
 u = A 
 + B.X + Byy 
 
 + i(C..x2 + 2C.yXy + Cyyy') 
 
 + ||-(i>xXXa^ + SD,,yX^y + 3D,yyXy^ + Dyyyf) (1) 
 
 etc. etc. etc. 
 
160 CALCULUS. 
 
 Where A is the value of u when x and y are both zero. 
 
 Bx is the value of -j- when x and y are both zero. 
 
 d/li 
 
 By is the value of -r- when x and y are both zero. 
 
 Cxx is the value of ^J when a: and ?/ are both zero. 
 
 Cxy is the value of , , or -r-^ when x and 2/ are both 
 
 zero. 
 
 Cyy is the value of -r-y when a; and y are both zero. 
 
 etc. etc. etc. 
 
 The proof of Maclaurin's theorem as applied to a function of two or more 
 variables is essentially identical to the proof of the theorem as applied to a 
 function of one variable. To enable one to appreciate the modified character 
 of argument and especially as an interesting example of partial integration, 
 let us consider what we may call the second approximation, that is the expres- 
 sion we get for u on the assumption that the respective second derivatives 
 are everywhere constant and equal to Cxx, Cxy and Cyy (their respective 
 values when a; = and j/ = 0) as follows: 
 
 dhi ^ 
 
 Cxy (3) 
 
 dhi 
 dxdy 
 
 |S = C. (4, 
 
 Integrating equation (2) with respect to x and equation (3) with respect to y 
 we have 
 
 T- = CxxX + any function of y (5) 
 
 and 
 
 T- = CxyV + any function of x (6) 
 
 Now -T- is equal to Bx when x and y are both equal to zero. Therefore 
 
 the constant term in "any function of y" in (5) is equal to Bx, and the constant 
 term in "any function of a;" in (6) is equal to Bx. Therefore, using the 
 
EXPANSIONS IN SERIES. 161 
 
 expression "vanishing function of ?/" for a function of y which is equal to 
 zero when y = 0, equations (5) and (6) become: 
 
 -3- = CxxX + (a vanishing function of y) + Bx (7) 
 
 and 
 
 ;t- '= Cxyy -\- {Q' vanishing function of x) -\- Bx (8) 
 
 But these two expressions must be identical, consequently the "vanishing 
 function of ?/" in (7) must be dyy and the "vanishing function of x^' in 
 (8) must be CxxX. Therefore both of these equations reduce to 
 
 ^ = C.x + C.,7/ + jB. (9) 
 
 In a similar manner we may integrate equation (4) with respect to y 
 and equation (3) with respect to x, and get the equation: 
 
 ^=Cyyy+CxyX-\-By (10) 
 
 Now equations (9) and (10) may be integrated, giving: 
 
 u = \CxxX^ + Cxyxy + BxX + any function of y (11) 
 
 and 
 
 w = ICyyy^ + CxyXy + Byy + any function of x (12) 
 
 But u = A when x and y are both equal to zero, therefore " any function of 
 ?/" in (11) must be ("a vanishing function of ^") + -4, and likewise, "any 
 function of x" in (12) must be ("a vanishing function of re") -{-A. Further- 
 more, equations (11) and (12) must be identical so that the "vanishing func- 
 tion of 1/" in (11) must be }4Cyyy^ + Byy, and the "vanishing function of 
 x" in (12) must be }4CxxX^ + BxX. Therefore equations (11) and (12) both 
 reduce to: 
 
 u = A 
 
 + BxX + Byy 
 + liCxxX^ + 2Cxyxy + Cyyy^) (13) 
 
 which is the approximate value of u as derived from equations (2), (3) and (4). 
 
 PROBLEMS. 
 1. Expand log(l + x) by Maclaurin^s theorem. Ans. 
 
 />»2 ^3 /v»4 /v*5 
 
 log(l+a;) = x-2+3--4 + 5 
 
 12 
 
162 CALCULUS. 
 
 Note. — The function log x and all of its derivatives become infinite for 
 a; — 0, and therefore log x cannot be expanded in powers of x. 
 
 2 . Using the answer to problem 1 make an attempt to calculate 
 the logarithm of by placing x = — 1 and adding together a 
 number of terms of the series. 
 
 Note. — The series obtained by Maclaurin's theorem for log (1 + x) cannot 
 be used for values of x which lead up to or beyond a point where log (1 + x) 
 or any of its derivatives become discontinuous or infinite. 
 
 3. Expand cos(a; + h) in a series of ascending powers of h, 
 Ans. 
 
 cos (a; + ^) = cos a; — /i sin x — i^cos x + "to" sin a; + • • • 
 
 4. Expand y = tan x by Maclaurin's theorem. Ans. 
 
 , , x^ , 2x^ , 
 
 tana: = a;+3-+-j5-+ ••• 
 
 Note. — ^When x = - the given function and its derivatives become infinite. 
 Therefore the series found in answer to this problem does not give the value of 
 tan X for values of x equal to or greater than ^. 
 
 91. Demoivre*s Theorem. — An important algebraic identity, 
 which was discovered by Demoivre, is expressed by the equation: 
 
 e'* = cos a; + j sin X (1) 
 
 where e is the Napierian base and j = V— 1.* This relation 
 is known as Demoivre^s theorem, and it is very useful in certain 
 transformations for purposes of integration, and it is also useful 
 in the theory of alternating currents. 
 
 To establish equation (1) write jx for x in equation (1) of 
 Art. 89, remembering that f = — 1, f — — jj j^ = + 1, etc., 
 and we have: 
 
 -I x^ . x* x^ . . , . / a^ x^ ^' I i. \ /o\ 
 
 * It is customary in the th eory o f alternating currents to use i for electric 
 current, and j is used for V — 1* 
 
EXPANSIONS IN SERIES. 163 
 
 But the real terms in this series give a series identical to equation 
 (3) of Art. 89, and the imaginary terms give a series identical to 
 equation (2) of Art. 89. Therefore from equation (2) we get 
 g/^ = cos X + j sin X. 
 
 Definition of complex quantity. Geometric representation of 
 complex quantity. — Any expression like a-\-h V — 1, which is 
 part real and part imaginary is called a complex quantity. Thus 
 the right-hand member of equation (1) is a complex quantity; 
 and of course e^'* is a complex 
 quantity because equation (1) 
 shows that e^'* is part real (cos x) 
 and part imaginary (j sin x). 
 
 The accepted method of rep- 
 resenting a complex quantity geo- 
 metrically is shown in Fig. 80. ' pjg go. 
 The vector E is the complex quan- 
 tity, its a:-component is thought of as a real quantity a, 
 its ^/-component is thought of as the imaginary quantity jh, 
 and the vector is the sum of its two components. That is 
 
 E = a+jh* 
 
 * This algebraic expression of a vector as a complex quantity is one aspect 
 of the important use of complex quantity in the theory of alternating currents. 
 See Frank lin and Esty's Elements of Electrical Engineering^ Vol. II, Chapter V; 
 The Macmillan Co., New York, 1908. 
 
 Another aspect of the use of complex quantity in the theory of alternating 
 currents is exhibited in Chapter VII of this treatise where the fundamental 
 differential equations of alternating currents are integrated with the help of 
 transformations involving the use of complex quantity. 
 
 The practical use of complex quantity in alternating-current theory is due 
 chiefly to C. P. Steinmetz; but the use of complex quantity in the solution of 
 linear differential equations as explained in Chapter VII contains everything 
 that is now known of the use of complex quantity in alternating-current 
 theory in a manner which is self-evident, and the use of complex quantity as 
 exemplified in Chapter VII is much older than electrical engineering. 
 
164 CALCULUS. 
 
 92. Euler*s expressions for sin x and cos x. — Many differential 
 expressions can be reduced to simple recognizable forms for 
 purposes of integration with the help of Demoivre's theorem, 
 and it is in some cases convenient to modify equation (1) of 
 Art. 91 so as to express sin x and cos x in terms of exponentials 
 Such expressions are due to Euler and they are; 
 
 sm X = — ^ — (1) 
 
 and 
 
 cos X = » (2) 
 
 The use of these equations is exemplified below. They are 
 derived as follows: From Demoivre's theorem we have: 
 
 g;* = cos a; + 3 sin x (3) 
 
 Write — a; for a; in this expression, remembering that 
 
 cos(— x) = cos a; 
 and that 
 
 sin(— x) = — sin x 
 and we have: 
 
 g-yx = cos X — j sin x (4) 
 
 Subtracting equation (3) from equation (4) member from member 
 we get equation (1), and adding equations (3) and (4) member 
 to member we get equation (2). 
 
 93. Example showing use of Euler's equations. — Consider the 
 function : 
 
 z = sin mx cos nx (1) 
 
 To find the average value of this function between x = and 
 a; = 27r it is necessary to integrate z-dx between the Hmits 
 x = and a; = 27r as explained in Art. 73. Now the function 
 given in equation (1) is a function whose average value is of 
 fundamental importance in connection with Fourier's theorem 
 (see Chapter VIII) and therefore it is important to be able to 
 
EXPANSIONS IN SERIES. 165 
 
 reduce the differential expression sin mx cos nx -dx to a combina- 
 tion of fundamental forms which can be found in Class A of the 
 table of integrals in appendix B. This reduction may be easily 
 made with the help of Euler's equations as follows: 
 It is required to integrate the differential equation: 
 
 dy = sin mx cos nx • dx (2) 
 
 Writing mx for x in Euler's expression for sin x we get an 
 expression for sin mx, and writing nx for x in Euler's expression 
 for cos X we get an expression for cos nx. Substituting these 
 expressions for sin mx and cos nx in equation (2), we get: 
 
 dy = ^. e^"("'+">^-rfa; + ^.e''^"'-''^'-"dx 
 
 - }^ e-i(»»+")=^ 'dx-^. e-J («-">^ • dx (3) 
 
 Each term in the second member of this equation is of the form 
 ae^^-dx of which the integral (ignoring constant of integration) 
 
 is r • &^. Thus, in case of the first term a = —. and 
 
 5 = 2{^ + n) 
 
 so that the integral of the first term is 
 
 1 
 
 4(m + n) 
 
 e' 
 
 {m-\-n)x 
 
 Proceeding in a similar manner with each term and arranging 
 the result systematically we get: 
 
 ^ " 4(m + n) ^^ ^ ^ 
 
 4(m — n) 
 
 Fg— y(m-n)x _ g;(m— n)il /^\ 
 
 And this expression can be easily reduced to form 24 in the 
 table of integrals by using Euler's equations. 
 
166 CALCULUS. 
 
 PROBLEMS. 
 
 1. Reduce dy = sm^x-dx to a combination of standard 
 forms as given in the table of integrals. Ans. 
 
 dy ^ h-. (e~'"'* - e^' + Z&' - Se"'') *dx 
 
 2. Reduce dy = coB>^X'dx to a combination of standard forms. 
 Ans. 
 
 dy = -hie^"^ + e-''^ + ^&^ + ^e-^^ -\-^)'dx 
 
 94. Hjrperbolic sines and cosines.* — In the solution of the 
 differential equation of the alternating-current transmission linef 
 the expressions (e* — e~^) and {e' + e"') occur, and trans- 
 mission line calculations are facilitated by the use of tables giving 
 the values of (e* — e~^) and (e* + e~^) for various values of x, 
 and of course the discussion of such calculations is simplified 
 by having names for {e" — e~*) and {e' + e~'). Indeed the 
 
 expressions ^ and ^ are related to the equilateral 
 
 hyperbola in the same way that ^. ( = sin x) and 
 
 ^ ( = COS x) are related to the circle. Therefore these 
 
 expressions are called the hyperbolic sine of x (sinh a;) and the 
 * Tables giving values of a, h, c and d in the expressions 
 
 cosh(aj + jy) = a -^-jb 
 and 
 
 sinh(x -^jy) = c -\-jd 
 
 for various values of x and y are published in a supplement to the General 
 Electric Review for May, 1910. 
 
 t The simplest discussion of this subject is that which is given in Franklin's 
 Electric Waves, pages 141-153, The Macmillan Co., New York, 1909. This 
 discussion is essentially complete, although no mention is made of hyperbolic 
 sines and cosines. 
 
EXPANSIONS IN SERIES. 167 
 
 hyperbolic cosine of x (cosh x) respectively. That is: 
 
 and 
 
 sinh X = ^ (1) 
 
 cosha; = ^^i^ (2) 
 
 PROBLEMS. 
 
 1 . Differentiate y = sinh x» 
 
 2. Differentiate y = cosh x, 
 
 3 . Integrate dy = sinh x • dx. 
 
 4. Integrate dy = coshx-dx. 
 
 For answers see forms 29 and 30 of table of integrals in 
 Appendix B. 
 
CHAPTER VII. 
 SOME ORDINARY DIFFERENTIAL EQUATIONS. 
 
 95. Degree and order of a differential equation. — The simple 
 equation az -{- h = is said to be linear with respect to z 
 because it contains no power of z higher than the first power. 
 A differential equation of the form: 
 
 is called a first degree or linear differential equation because it 
 
 contains no products of y, ~ and -—, and no powers of 2/, ^, 
 
 etc., higher than the first. The coefficients A, B and C in a 
 linear differential equation may contain the independent vari- 
 able X, but we shall confine our attention in this chapter 
 chiefly to linear differential equations with constant coefficients. 
 
 If the first derivative,* only, appears in a differential equation, 
 the differential equation is said to be of the first order. If the 
 second derivative occurs (with or without the first derivative) 
 the differential equation is said to be of the second order; and 
 so on. 
 
 96. Ordinary and partial differential equations. — A differential 
 equation expressing the law of growth of a function of one inde- 
 pendent variable is called an ordinary differential equation. 
 Many examples of ordinary differential equations are given in 
 Chapters I and V. See Art. 24 in particular. 
 
 * The terms degree and order apply to partial differential equations as well 
 as to ordinary differential equations. A function of two variables, however, 
 has two first derivatives, three second derivatives, and so on as explained in 
 Art. 59. Therefore it is somewhat misleading to speak of the first derivative, 
 or the second derivative in explaining what is meant by a differential equation 
 of the first or second order. 
 
 168 
 
 i 
 
ORDINARY DIFFERENTIAL EQUATIONS. 169 
 
 A differential equation which expresses the law of growth of a 
 function of two or more independent variables is called a partial 
 differential equation. Some examples of partial differential equa- 
 tions are given in Chapter V. See Arts. 60 and 90 in particular. 
 
 This chapter is devoted to the discussion of a few important 
 ordinary differential equations, and several important partial 
 differential equations are discussed in Chapters VIII and IX. 
 
 97. Pure and mixed differential equations.* — A pure differ- 
 ential equation contains but one derivative (the first or any- 
 higher derivative) and does not contain the dependent variable y. 
 
 Thus dy = Qx- dx, -^ = sin x, ^\ = log x are pure differential 
 
 equations. 
 
 A mixed differential equation contains more than one derivative, 
 or it contains one or more derivatives and the dependent variable 
 y. Thus 
 
 are mixed differential equations. 
 
 Solution of pure differential equations. — A pure differential 
 equation can be integrated by looking up the appropriate form in 
 the table of integrals. This is exemplified by every problem in 
 integration heretofore given in this treatise. In the case of a 
 pure differential equation of the second or third order, successive 
 integrations are of course necessary. For example, consider the 
 third order pure differential equation: 
 
 Let q represent ~, then equation (1) becomes: 
 
 1 = -' (2) 
 
 *This and the following articles refer primarily to ordinary differential 
 equations. 
 
170 CALCULUS. 
 
 This equation can be integrated, giving; 
 
 9 = 3 = 4<^ + C (3) 
 
 Let p represent -p, then equation (3) becomes: 
 
 %-ia^ + C (4) 
 
 This equation can be integrated, giving: 
 
 P = ^ = i\ax' + CX + D (5) 
 
 This equation can be integrated, giving 
 
 y = -hax^ -f ^Cx^ + Dx + E (6) 
 
 Solution of mixed differential equations. Separation of 
 variables. — The linear differential equation (1) of Art. 95 is of 
 course a mixed equation, and the general solution of this equation 
 with constant coefficients is given in a subsequent article. We 
 will here consider one or two examples of a simple transformation, 
 called the separation of variables, which can sometimes be used 
 to bring a mixed differential equation into a form which can be 
 integrated by looking up appropriate forms in the table of 
 integrals. 
 
 As a first example consider: 
 
 This equation reduces to: 
 
 1 = ^ (^) 
 
 ^ = x'^'dx (8) 
 
 y 
 
 and this equation can be integrated with the help of forms 1 and 
 2 of the table of integrals, giving 
 
 log2/ = |x3 + C (9) 
 
ORDINARY DIFFERENTIAL EQUATIONS. 171 
 
 As a second example consider 
 
 dij 
 Let p represent -^ and this equation becomes 
 
 g = x«p (12) 
 
 and the integral of this equation, according to equation (9), is: 
 
 \ogv = \:x? -\- C 
 or 
 
 P = I = ei^'^" (13) 
 
 This is a pure differential equation and its integral can be found 
 by looking up the appropriate forms in the table of integrals. 
 
 PROBLEMS. 
 Solve the following differential equations: 
 
 1. x3 ^ = 2. Ans. 2/ = log a; + Cx" ■\- Dx -^ E, 
 
 2. ^ = xe', Ans. y = {x - 2)e' + Cx + D. 
 
 3. ^ = 27 sin^x. Ans. y = 21 cos x- cos^ x -\-Cx^ +Dx -{-E, 
 
 4. 1 - 2/ + (1 + a;) ^ = 0. Ans. log \-^ = C. 
 
 ax 1 ~~ 2/ 
 
 5. 1 - 21/ = 3^. Ans. C(l - 2y) = e"?. 
 
 6. sin 2/-cix + a; cos i/-(f?/ = 0. Ans. xsiny = C. 
 
 7. sin a; cos y-dx — cos x sin y - dy = 0. Ans. cos y = C cos a;. 
 
 8. fa + a;i/ )rfa; + (a; — xy) dy = 0. Ans. logxy = C — x + y. 
 
 9. Vl — 2/^ • (ia; + V 1 — a;^ • (^2/ = 0. Ans. sin""^ x-\- sin"^ y=C. 
 
 ^°-S + .-| = "- Ans.Clogx = . + D. 
 
172 CALCULUS. 
 
 11. ^ + (^f^y + 1=0. Ans. y = log cos {x - C) + D. 
 
 12.(l+.^)g + .| + ax = 0. 
 
 Ans. y = D — ax -\- C log [x -\- Vl + x^]. 
 
 98. The general solution and particular solutions of a differ- 
 ential equation. — The general solution of a differential equation 
 is an expression for y (the dependent variable) in terms of x 
 (the independent variable) which includes every possible* func- 
 tion which satisfies the differential equation. Thus equation (6) 
 of Art. 97 is the general solution of equation (1) of that article. 
 Concerning the three constants of integration it is evident that 
 E is the value of y when a; = 0. Also it is evident from equa- 
 tion (5) of Art. 97 that D is the value of -^ when x = 0, and 
 
 it is evident from equation (3) that C is the value of — g when 
 
 X = 0. 
 
 The general solution of a differential equation of the nth. order 
 contains n undetermined constants of integration. 
 
 When one or more of the constants of integration in the general 
 
 solution of a differential equation have particular values assigned 
 
 to them we have what is called a particular solution of the 
 
 differential equation. For example, let C = 2, let D = and 
 
 let E = in equation (6) of Art. 97, then the equation becomes 
 
 ax^ 
 y = j^ -\- x^ which is a particular solution of equation (1) of 
 
 Art. 97. 
 
 99. Discussion of the first order linear differential equation 
 with constant coefficients. — Consider the differential equation: 
 
 y + Ag = o (1) 
 
 * This statement is subject to some qualification because of what is called 
 the singular solution. See Johnson's Ordinary and Partial Differential Equa- 
 tions, page 43, John Wiley & Sons, New York, 1890. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 173 
 
 The most obvious method for solving this differential equation 
 is to "separate the variables" as explained in Art. 97. Thus 
 equation (1) is easily reduced to: 
 
 A^=-dx (2) 
 
 y ^ ^ 
 
 and this equation can be integrated by looking up appropriate 
 forms in the table of integrals, giving: 
 
 Alogy = -x-VC (3) 
 
 It is desirable, however, to solve equation (1) by the following 
 method because the method applies to a linear differential equa- 
 tion of any order when the coefficients are constants. 
 Let 
 
 y = Ce*^ (4) 
 
 where C and k are constants, and e is the Napierian base. 
 Then: 
 
 Substituting the values of y and ~- from (4) and (5) in equation 
 
 (1), we have: 
 
 Ce^^ + AkCe^' = (6) 
 
 whence by cancellation we have: 
 
 1 + AA; = 
 
 or 
 
 t = — 
 A 
 
 k = -1 (7) 
 
 Therefore if h in equation (4) has the value — -j then equation 
 (4) satisfies equation (1); that is, equation (4) is a solution of 
 (1) if k = — -T-', indeed equation (4) is the general solution of 
 
174 CALCULUS. 
 
 (1) because it contains one undetermined constant C, the 
 constant of integration. 
 
 100. The principle of superposition. — A principle of extremely 
 wide application in physics is the so-called principle of super-posi- 
 tion. From the physical point 
 window No, 1 window No. 2 of view a general statement of 
 
 this principle is scarcely possi- 
 ble and therefore the following 
 examples must suffice: (a) A 
 person at A (Fig. 81) can see 
 window No. 1 and another per- 
 son at B can see window No. 2 
 Fig. 8L at the same time. This means 
 
 that two beams of light a and h 
 can travel through the same region at the same time and not 
 get tangled up together as it were, each beam behaving as if it 
 were traveling through the region alone. (6) Two sounds can 
 travel through the same body of air simultaneously, each sound 
 behaving as if it were traveling through the body of air alone, 
 (c) Two systems of water waves can travel over the same part 
 of a pond simultaneously, each system behaving as if the other 
 were not present, {d) Two messages* can travel over a tele- 
 graph wire at the same time and not get mixed up. (e) Two 
 forces F and G exerted simultaneously upon an elastic structure 
 produce an effect which is the sum of the effects which would be 
 produced by the forces separately, provided the sum of the forces 
 does not exceed the elastic limit of the structure; therefore each 
 force may be thought of as producing the same effect that it 
 would produce if it were acting alone 
 
 All of the effects in physics which are superposable — and this 
 
 * Indeed any number of messages can travel over a telegraph wire in 
 either direction or in both directions simultaneously. The only limiting 
 feature in multiplex telegraphy, when line-loss is negligible, is in the design of 
 the sending and receiving apparatus; and the same is true in wireless teleg- 
 raphy. In each of the above examples the word two means iwo or more. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 175 
 
 includes the greater part of the effects in mechanics, heat, 
 electricity and magnetism, light and sound, and a great many- 
 effects in chemistry — are expressible in terms of linear differential 
 equations with constant coefficients, and the principle of super- 
 position may be thought of as a property of such a differential 
 equation as follows . If y is a function of x which satisfies a 
 linear differential equation with constant coefficients, and if z is 
 another function of x which satisfies the same differential equation, 
 then (y + z) is a function of x which satisfies the equation* 
 
 Proof. — Let the given linear differential equation be: 
 
 , A du . „ d?u . f. /^v 
 
 If a function y satisfies this equation, then: 
 
 If another function z satisfies equation (1), then: 
 
 Now ^^y + ^'> = ^ + ^ and ' ^^^ + ^> _ ^ . ^ There- 
 ^""^ dx dx^ dx *""* dx^ dx' + dx»- "^"^^ 
 
 fore, adding equations (2) and (3), we get: 
 
 (, + .)+A'?(^ + B^J-i)+...=0 (4) 
 
 But this is the same form as equation (1) which shows that the 
 function {y + z) satisfies (1). 
 
 * This proposition is true for a partial linear differential equation also. 
 Indeed most of the superposable effects in physics are expressible in terms of 
 partial linear differential equations with constant coefl&cients. Examples 
 are given in Chapters VIII and IX. 
 
176 CALCULUS. 
 
 101. Discussion of the second order linear differential equa- 
 tion with constant coefficients. — Consider the differential equa- 
 tion: 
 
 It is not possible to "separate the variables" in this differential 
 equation and therefore the second method of Art. 99 must be 
 used. Therefore let: 
 
 y = Ce** (2) 
 
 then 
 
 | = iCe*. (3) 
 
 and 
 
 Substituting these values of y, -^ and -r-^ in equation (1) and 
 
 cancelling the common factor Ce**, we have : 
 
 1 4- A/b + M2 = f\ ^ (5) 
 
 whence 
 
 Therefore using a for one of these values of k and using j8 for 
 the other value of A;, we get two solutions of (1), namely: 
 
 w = Ce'^ (7) 
 
 and 
 
 z = De^' (8) 
 
 But according to Art. 100 the sum (w + z) is also a solution. 
 Therefore using y for (w -{- z), we have as a solution of (1): 
 
 2/ = C6- + i)e^- iW^-v-ii--^ (9) 
 
 and this is the general solution of (1) because n contains the two 
 undetermined constants C and D. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 177 
 
 102. The starting of a boat. — At a certain instant (t = 0) a 
 
 constant force E begins to act on a boat, and it is desired to 
 
 find an expression for the increasing velocity i of the boat. 
 
 At very low speeds the backward drag or friction of the water on 
 
 a boat is proportional to the velocity of the boat. Let us assume 
 
 that this proportional relation is exact, then the frictional drag 
 
 of the water on a boat is equal to Ri, where i is the velocity of 
 
 the boat and R is sl constant which depends on the shape and 
 
 size of the boat. Therefore the net accelerating force acting on 
 
 the boat is E — Ri. This force is equal* to the product of 
 
 di 
 the mass L of the boat and the acceleration -r;. Therefore we 
 
 at 
 
 have: 
 
 L%^E-Ri (1) 
 
 To get this equation into the standard form of a linear differential 
 equation, let 
 
 y = E-Ri (2) 
 
 Then 
 
 ^ = _ p^ 
 dt ^dt 
 
 and equation (1) becomes: 
 
 R dt ^ 
 or 
 
 ^ + 1-1 = (^) 
 
 The general solution of this equation is given in Art. 99, but 
 it is worth while to work it out anew, as follows: Let 
 
 y = Ce^« (4) 
 
 * If force is expressed in poundals (or dynes), mass in pounds (or grams), and 
 acceleration in feet per second per second (or centimeters per second per 
 second). 
 13 
 
178 CALCULUS, 
 
 then 
 
 dt 
 
 ^^ = A;Ce« (5) 
 
 dv 
 Substituting these values of y and -^ in equation (3) and 
 
 cancelling out the factor Ce*', we have: 
 
 80 that 
 
 " L 
 
 Therefore equation (4) becomes: 
 
 y = Ce"^"' 
 or, substituting E — Ri for yj we have: 
 
 (6) 
 
 E- Ri = Ce ^ 
 
 (7) 
 
 Now i — Q when t = 0. Therefore placing this pair of values 
 in equation (7) we have: 
 
 E ^C (8) 
 
 so that the constant of integration is determined, and equation 
 (7) becomes: 
 
 E - Ri ^ Ee ^ 
 or 
 
 R r" 
 
 |-|e-^- (9) 
 
 This equation expresses the value at each instant of the increasing 
 velocity of the boat as a function of the elapsed time t. 
 
 103. The stopping of a boat. — A boat is moving at velocity I 
 when the propelling force ceases to act, and it is required to find 
 an expression for the decreasing velocity, i, of the boat as it 
 
ORDINARY DIFFERENTIAL EQUATIONS. 
 
 179 
 
 gradually comes to rest. In this case the only force acting on 
 the boat is the retarding force Ri, and in this case the accelera- 
 
 di 
 tion T7 is negative so that: 
 
 or 
 
 ^ + §•1 = 
 
 The general solution of this differential equation is: 
 
 (1) 
 (2) 
 
 (3) 
 
 But i = / when t = 0. Therefore C = /, and equation (3) 
 becomes: 
 
 i = le 
 
 (4) 
 
 Remark. — The notation used in this discussion of the starting 
 and stopping of a boat is the standard notation used in electrical 
 theory. This notation is used here and in the following articles 
 
 35 
 30 
 
 25 
 «20 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ■ 
 
 -5 
 
 p- 
 
 . 
 
 
 
 
 
 
 
 ^ 
 
 
 -^ 
 
 ' 
 
 
 
 
 
 
 
 ^ 
 
 >^ 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 . 
 
 
 
 
 / 
 
 
 grc 
 R= 
 
 wing 
 '3 oh 
 
 current 
 
 ms 1 
 
 R 
 
 
 
 
 / 
 
 
 L = 
 
 0.04 
 110 
 
 hem 
 polts 
 
 V 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 J_ 
 
 
 
 3 4 5 6 7 8 9 
 
 hundredths of a second 
 
 Fig. 82. 
 
180 
 
 CALCULUS. 
 
 because the mechanical problems discussed in this and the 
 following articles are strictly analogous to certain electrical 
 problems which constitute the foundation of the theory of 
 alternating currents. Thus the curve in Fig. 82 is a graphical 
 representation of equation (9) of Art. 102; the ordinates of this 
 curve represent the growing values of the current i in a circuit 
 
 
 \- 
 
 
 
 
 
 
 
 
 
 
 35 
 
 K 
 
 
 
 
 
 
 
 
 
 
 J^ 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 •^b 
 
 
 \ 
 
 
 dei 
 R = 
 
 :ayin 
 
 -3 oh 
 
 0.04 
 
 367 
 
 g current 
 img 
 
 
 
 
 ^20 
 
 
 \ 
 
 \ 
 
 
 henry 
 amperes 
 
 
 
 
 
 
 
 \ 
 
 \ 
 
 
 
 
 
 
 
 lO 
 
 
 
 
 \ 
 
 \. 
 
 
 
 
 
 
 S 
 
 
 
 
 
 
 
 
 
 
 
 , 
 
 
 
 • 
 
 2 
 
 3 
 
 mndr 
 
 edth 
 
 5 ( 
 
 8 of < 
 
 3 
 
 I sec 
 
 1 
 ond 
 
 3 < 
 
 ? 
 
 Fig. 83. 
 
 of resistance R and inductance L after a battery of electro- 
 motive force E is connected to the circuit. The curve in Fig. 
 83 is a graphical representation of equation (4) of Art. 103; the 
 ordinates of this curve represent the decaying values of the 
 current i when an initial current of / amperes is left to die 
 away in a circuit of resistance R and inductance L. 
 
 PROBLEMS. 
 Solve the following differential equations. 
 
 Ans. y = Ce-' + De^'. 
 
 -3-2-^^ = 0- 
 
 ^- da? 
 
 4y. Ans. y = Ce^' + De-^', 
 
 "- -^ ^^ 
 
ORDINARY DIFFERENTIAL EQUATIONS. 
 
 181 
 
 ^-4^ + 3)/ = 0. Ans. y = Ce' + D^'. 
 
 <Py 
 
 •n^^+^ 
 
 ) = 
 
 10 
 
 dy 
 dx 
 
 '■%-4i^'t = '- 
 
 Ans. y = Ce^* + Be^'. 
 Ans. y = Ce^^ + De^^ + E. 
 
 104. Undamped oscillations. — Figure 84 shows a weight sus- 
 pended by a spring, and the weight stands in 
 equilibrium in a certain position. If the weight 
 happens to be q feet above or below its equilib- 
 rium position an unbalanced force proportional 
 
 to q, and therefore equal to y^* times q acts 
 
 upon the weight. This force is downwards when 
 q is upwards, and upwards when q is down- 
 wards. Therefore the force is equal to — p -q- 
 
 This force accelerates (or retards) the weight, 
 and we have 
 
 Y di _ 
 
 (1) 
 
 spring 
 
 Fig. 84. 
 
 where L is the mass in pounds of the attached 
 
 di 
 weight in Fig. 84, and -^ is the acceleration (it is 
 
 a retardation when it is negative) of the attached 
 
 weight. But the velocity i of the weight is the 
 
 rate of change of its distance q from the equilibrium position 
 
 That is 
 
 * dt 
 
 so that 
 
 (U _^ 
 
 dt ~ dt^ ^"^^ 
 
 * This is simply a proportionality factor. It is written in this form so a 
 to conform to the standard notation of electrical theory. 
 
 (2) 
 
182 CALCULUS. 
 
 Therefore equation (1) becomes 
 
 or 
 
 T ^ _ _q 
 dt^ C 
 
 . + §•9 = (4) 
 
 The general solution of this second order linear differential equa- 
 tion is given in Art. 101, but it is worth while to work out the 
 solution anew as follows: 
 Let 
 
 q = Me''* (5) 
 
 then 
 
 ^ = /bM6** (6) 
 
 and 
 
 kme"* (7) 
 
 Substituting these values of q and ■— in equation (4) and 
 cancelling the common factor ilfe**, we have: 
 
 1 + ^'k'' = (8) 
 
 so that 
 
 k = yj 
 
 -c («) 
 
 It is evident that k is imaginary because L and C are positive 
 quantities. Therefore let us write j<a for k where j is V — 1,* 
 then: _ 
 
 IL 
 
 \C 
 
 (9) 
 
 * It is customary in electrical theory to use j for V — I because the letter 
 t is used for electric current. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 183 
 
 Now of course w may be either positive or negative, that is, k 
 may be + jco or — jw. Therefore 
 
 V = Me+i'^* (10) 
 
 and 
 
 z = Ne-i^' (11) 
 
 are two particular solutions of equation (4), where M and N 
 are undetermined constants. Therefore, according to Art. 100, 
 (v + z) is also a solution of (4). Consequently, writing u for 
 {v ■{• z), we have as a solution of (4) : 
 
 u = Me+'"' + Ne-^"** (12) 
 
 and this is the general solution of (4) because it contains the 
 two undetermined constants M and N. 
 
 Now from Demoivre's theorem it is evident that u as given 
 by equation (12) is complex (part real and part imaginary), and 
 for the sake of generahty M and N may be thought of as 
 complex also. That is, for M and N we may write: 
 
 M = M' -{- jM'' (13) 
 
 and 
 
 iV = iV' + jN" (14) 
 
 Now any complex equation is equivalent to two simple equations, 
 namely the equation which expresses the equaUty of the real 
 parts of the members of the complex equation, and the equation 
 which expresses the equality of the imaginary parts of the com- 
 plex equation; and this latter equation becomes real when j 
 is cancelled from its members. Thus equation (12) may be 
 broken up into two simple equations each of which is a solution 
 of equation (4). 
 
 To spHt equation (12) up into two equations use the values 
 for M and N from equations (13) and (14), and reduce e+^"'*" 
 and e~''^' by means of Demoivre^s theorem as explained in Art. 
 91. We thus get: 
 
 r -Vjs = {W + jM") (cos (at + j sin coO /. ^x 
 
 + {N' + jN") (cos co< - 3 sin coQ ^ ^ 
 
184 
 
 CALCULUS. 
 
 where r + js has been written for q. Separating real and 
 imaginary terms in equation (15) we get: 
 
 and 
 
 r = (ilf ' + N') cos a)t - {M" - N") sin co< (16) 
 
 8 = (M" + N") cos oit + {M' - N') sin cot (17) 
 
 Now in either of these equations the coefficient of cos o)t is any 
 constant, and the coefficient of sin cat is amj constant. There- 
 fore, q for r or s, both of these equations reduce to: 
 
 q = G cos cot + H sin cot 
 
 (18) 
 
 which is the final general solution of equation (4), G and H 
 being undetermined constants. 
 
 In alternating current theory this equation is used in a slightly 
 modified form as follows: 
 
 Let 
 
 G = - sin (19) 
 
 and 
 
 H = Qcosd 
 
 Then equation (18) becomes: 
 
 g = Q sin (co< — d) 
 
 (20) 
 
 (21) 
 
 This equation is represented by Fig. 85, in which the line Q 
 
 rotates at angular velocity co 
 radians per second, and the 
 value of q at each instant is 
 represented by the projection 
 of Q on the fixed line ah. 
 
 The two integration con- 
 stants Q and d are deter- 
 mined by the initial condi- 
 tions as follows: For example 
 let the weight be started 
 oscillating by pulling it up- 
 wards 2 feet (q = 2 feet), 
 
 axis of angle. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 185 
 
 holding it still ( 37 = 1 and releasing it at the instant t = 0. 
 The values of Q and 6 are then determined by substituting the 
 values 37 = and i = in the equation 
 
 ~ = ojQ cos (Q)t — 6) 
 
 and by substituting the values q = 2 and t = inequation (21). 
 
 105. Damped oscillations. — The weight in Fig. 84 is acted 
 
 upon by the force — ^ when it is at a distance q above or 
 
 below its equiHbrium position as explained in Art. 104; and, 
 if the weight moves up and down through a resisting fluid, the 
 backward drag due to friction will be approximately proportional 
 
 to the velocity -^ of the weight. Assuming this proportional 
 
 relationship to be exact, the backward drag of friction may be 
 
 written R'-^, and since it is always opposed to the velocity 
 
 -^ it must be written — R~, Therefore the total force acting 
 
 on the weight is — ^ — 72 -^, and this force is equal to the 
 
 (Pa 
 product of the mass L of the weight and the acceleration — 
 
 of the weight. Therefore we have: 
 
 T^ - - 1 -. J?^ 
 ^dt^ ~ C ^dt 
 or 
 
 3 + Cijf + Lcg = (1) 
 
 The general solution of this equation may be found by the 
 method used in Art. 104. Reduced to simple form the general 
 
186 
 
 CALCULUS. 
 
 solution is; 
 
 q = Qe-^ sin {(at - B) 
 
 (2) 
 
 where Q and B are undetermined constants of integration and 
 where a and w are written for the following: 
 
 a — 
 
 and 
 
 2L 
 
 \Z^~ 
 
 4L2 
 
 (3) 
 
 (4) 
 
 Equation (2) completely expresses the motion of the weight in 
 Fig. 84 when the motion of the weight is opposed by friction as 
 above explained. Indeed equation (2) expresses a kind of 
 harmonic motion in which the amplitude (Qe~*0 decreases 
 continually. 
 
 q-axU 
 
 '"^^^^^^^xponentiai curve 
 
 Fig. 86. 
 
 The varying value of q as expressed by equation (2) is repre- 
 sented by the ordinates of the wavy curve in Fig. 86 for the case 
 
 in which B = — 
 
 The dotted curves are exponential curves 
 
ORDINARY DIFFERENTIAL EQUATIONS. 
 
 187 
 
 of which the equations are g = ± Qe~^, Another graphical 
 representation of equation (2) is shown in Fig. 87 where q is 
 the projection on the fixed line ah of a line Qe"** which rotates 
 
 01 
 
 I ^logarithmic spiral 
 
 Fig. 87. 
 
 at a constant angular velocity of w radians per second and 
 grows continually shorter and shorter. The end of the Une 
 Qe~** describes the dotted curve which is called an exponential 
 spiral. 
 
 PROBLEMS. 
 
 Solve the following differential equations: 
 
 
 + 42/ =- 0. Ans. 2/ = C sin 2a; + D cos 2x. 
 
 2. ^ - 2 ^ + 5i/ = 0. Ans. y = e-(C sin 2x + D cos 2x). 
 
 3. ^ - 8 ^ + 25?/ = 0. Ans. y = e^(C sin Zx + D cos Sx). 
 
 4. ^ - 24 ^ + 1692/ = 0. Ans. y = e^^x^e gi^ 5x-{-D cos 5a;). 
 
188 
 
 CALCULUS. 
 
 5. ^H-^ + 2/ = 0. Ans. y = e-^'(c 8m"fx+D co8^x\ 
 
 106. Forced or maintained oscillations. — The weight in Fig. 
 84 is acted upon by the force — ^ due to the altered stretch of 
 
 the spring, and by the force ~ ^ jT due to the frictional drag 
 
 of the fluid in which the weight moves up and down. Let us 
 suppose that an outside periodic force, E sin pt, acts upon the 
 weight. Then the total force acting on the weight will be 
 
 and this will be equal to the product 
 
 l-Bf^+E^npt, 
 
 of mass times acceleration 
 
 m 
 
 Therefore we will have: 
 
 q + CR^ + LC^ = CEsmpt 
 
 (1) 
 
 circuit 
 
 This is the differential equation of motion of the weight under 
 
 the assumed conditions, and 
 this identical differential equa- 
 tion expresses the mode of vari- 
 ation of the charge q in the 
 condenser C in Fig. 88 when 
 an alternator of which the 
 electromotive force is E sin pt 
 delivers electric current to a cir- 
 n cuit of resistance R and induc- 
 tance L, the circuit containing 
 a condenser of capacity C. 
 The charge q in the condenser 
 is not so familiar a thing as 
 
 I alterwUor 
 
 circuit 
 
 Fig. 88. 
 
 the current i 
 
 (■§) 
 
 which flows in the circuit, but it is con- 
 venient to use q as our variable instead of i. 
 
ORDINARY DIFFERENTIAL EQUATIONS. 189 
 
 Everyone who has had any experience with alternating-current 
 phenomena knows that in general there are two recognizable 
 alternating currents in the circuit shown in Fig. 88, namely, (a) 
 The alternating current which is maintained hy the alternator. 
 This alternating current is of the same frequency as the electro- 
 motive force E sin yt of the alternator, and (6) A decaying 
 oscillatory current, as if the condenser had been initially charged 
 and allowed to discharge through the circuit. This decaying 
 alternating current is independent of the alternator, it is expressed 
 by equation (2) of Art. 105, and its frequency is determined by 
 equation (4) of Art. 105. 
 
 The finding of the general solution of equation (1) is accom- 
 plished in two steps. The first step is the finding of a so-called 
 singular solution of (1), a solution which expresses the maintained 
 alternating current above mentioned, and the second step is to 
 add to this singular solution the general solution of equation (1) 
 of Art. 105 as given by equation (2) of Art. 105. This addition 
 is permissible inasmuch as it is very easy to show that v + z 
 is a solution of equation (1) of this article if t; is a solution, and 
 if z is a solution of equation (1) of Art. 105. 
 
 This procedure involves elaborate transformations which are 
 unintelligible to the beginner, and it is therefore useless to carry 
 it out. Fortunately, however, a much simpler method is avail- 
 able for the establishment of the desired result, a result which is 
 the foundation of the more important part of the theory of 
 alternating currents.* 
 
 * See pages 66-68 and 73-74 of Franklin and Esty's Elements of Electrical 
 Engineering, Vol. II. 
 
CHAPTER VIII. 
 
 THE PARTIAL DIFFERENTIAL EQUATION OF WAVE MOTION. 
 
 107. Differential equation of travel. Equation of a traveling 
 curve. — ^When a point is stationary its abscissa x is constant, 
 and when a point travels at constant velocity along the x-axis of 
 reference its abscissa increases (or decreases) at a constant rate 
 so that X = kt -\- any constant. This is all very simple but the 
 equation of a traveling curve which is so extensively used in the 
 theory of wave motion is not so simple, and therefore some dis- 
 cussion of it is necessary. 
 
 The curve cc in Fig. 89 is stationary with respect to the origin 
 O'j and the equation of the curve referred to the origin 0' is: 
 
 y = Fix') 
 
 (1) 
 
 stationary 
 y-axia 
 
 ^mooing 
 ^ y-axia 
 
 1 
 
 j x-axU 
 
 vt * X 
 
 d X — — — - 
 
 Fig. 89. 
 
 But the origin 0' and the curve cc are both assumed to be 
 traveling to the right at velocity v, so that the abscissa of the 
 moving origin 0' referred to the stationary origin is vt. There- 
 fore the abscissa x of any point on the curve cc referred to the 
 stationary origin is equal to x' + vt, or x' =^ x — vt. Therefore, 
 substituting x — vt for x' in equation (1) we have 
 
 y - F{x — vt) 
 190 
 
 (2) 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 191 
 
 which is the equation of the traveling curve cc referred to the 
 stationary origin 0. 
 Similarly it may be shown that 
 
 y=f{x + vt) (3) 
 
 is the equation of a curve which is traveling to the left at velocity v. 
 The two equations (2) and (3) satisfy one of the most important 
 of the differential equations of physics, namely, the differential 
 equation of wave motion. 
 
 Let the expression x — vt be represented by the single letter z. 
 That is 
 
 z = X — vt (4) 
 
 so that 
 
 dz 
 
 and 
 
 
 Then equation (2) becomes 
 
 y = F(z) ^ ^ (7) 
 
 Let ^ be represented by F'(z), and let ^ [ = ^^^^'l be 
 
 represented by F"{z). Then, according to the rule for differ- 
 entiating a function of a function (see Art. 34), we have (using 
 
 the above values of — and — ) : 
 dx dt I 
 
 dy _dy dz _ ,. . . . 
 
 Fx-Tz'd-x-^^'^ (^> 
 
 differentiating again we have : 
 
 S=«5^-l= '"•(•) ») 
 
 Similarly we have: 
 
 S4M=--'(^) ao) 
 
192 CALCULUS, 
 
 and 
 
 Therefore, comparing equations (9) and (11), we have 
 
 and it may be shown, as above, that equation (3) leads to this 
 same differential equation. 
 
 A partial differential equation may be recognized as one which 
 contains more than one independent variable. Therefore it is 
 unnecessary to use the symbol d instead of the symbol d. Here- 
 after the symbol d will be used exclusively. 
 
 General solution of eqvxition {12). — Equations (2) and (3) both 
 satisfy equation (12), and therefore equations (2) and (3) are both 
 solutions of (12). Therefore, according to the principle of super- 
 position as explained in Art. 100, the sum of F{x — vt) and 
 fix + vt) or 
 
 y = F(x-vt)+f{x + vt) (13) 
 
 is a solution of equation (12). Indeed this is the general solution 
 of (12) because it contains two undetermined fimctions.* 
 
 PROBLEMS. 
 l.Findthevaluesof |,3,Jandgwhen 
 
 y = A sin (x — vt); 
 also when 
 
 y = A sin (x -\- vt). 
 
 Show that both of these functions satisfy equation (12) of Art. 107. 
 
 * The general solution of an ordinary differential equation of the second 
 order contains two undetermined constants, and the general solution of a 
 partial differential equation of the second order contains two undetermined 
 functions. 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 193 
 
 2. Show that 
 
 A sin {x — yQ + A sin (x + vi) = 2A sin x cos vi^ 
 
 and plot the curve y = 2A sin x cos vi for the following values 
 of vtj namely, 
 
 0, ^, 2' ^» 'T^ -4 » -2^ -^ and ^tt. 
 
 iVote. — The curves thus plotted show the successive configurations of a 
 string which is vibrating in what is called a simple mode. See Franklin and 
 MacNutt, Light and Sound, page 243, The Macmillan Co., New York, 1909 
 
 108. Equation of motion of a stretched string. — When a 
 stretched string is in equilibrium it is of course straight. Let 
 us choose this equilibrium position of the string as the a:-axis of 
 reference as shown in Fig. 90, and let us set up the differential 
 
 equation which expresses the mode of motion of the string while 
 the string is vibrating or while a bend is traveling along the string 
 as a wave. We will assume that each part of the string moves 
 only up and down (at right angles to the a;-axis in Fig. 90) and we 
 will assume that the string is perfectly flexible.* Under these 
 conditions the a;-component of the tension of the string is always 
 and everywhere equal to a constant T. 
 
 Let the curve ccc, Fig. 90 be the configuration of the string 
 at a given instant t, that is, ccc is what a photographer would 
 call a snapshot of the moving string. The shape of the curve ccc 
 
 * This means that the only thing that keeps the string straight is its tension. 
 14 
 
194 CALCULUS. 
 
 defines y as a. function of x, and the steepness of the curve at a 
 
 dv 
 point is the value of -^ at that point. 
 
 Consider the very short portion ab of the string. The length 
 of this portion of the string when the string lies along the rc-axis 
 (in equilibrium) is dx, and the mass of the portion is m - dx 
 pounds or grams as the case may be, where m is the mass per unit 
 length of the string. An enlarged view of the very short portion 
 of the string is shown in Fig. 91. The adjacent parts of the 
 string pull on the portion ab in the direction of the string at a 
 and at 6, and the forces R and R' are thus exerted on the 
 portion ah. The a;-component of R is the force T towards the 
 left, and the a;-component of R' is an equal force T towards the 
 right. Therefore the downward force D on the end a of the 
 portion ah \s D = T tan B, and the upward force U acting on 
 the end h is U = T tan 6'. Therefore the net upward force 
 exerted on ah is 
 
 dF = T tan 6' - T tan (1) 
 
 dv 
 But tan 6 is equal to the value of -^ at a, and tan 0' is equal 
 
 dv 
 to the value of -p at h. Therefore the difference, tan 0'— tan 0, 
 
 dv 
 is the increase of -^ from a to h, and this increase is equal to 
 
 cPv d^v 
 
 -t4 • dx. This is evident when we consider that the value of -j^ 
 
 dv 
 means the rate of increase of -f- with respect to x. Therefore, 
 
 d^v 
 substituting j^ • dx for tan 6' — tan d in equation (1), we 
 
 have: 
 
 dF =T^^'dx (2) 
 
 Now according to Newton's laws of motion the net upward 
 force dF acting on the portion ah of the string is equal to the 
 
 ^<^- 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 195 
 
 mass m ' dx oi the portion multiplied by the upward acceleration 
 -z^ of the portion. The 
 in equation (2), we have: 
 
 -z^ of the portion. Therefore, substituting m -^ ' dx for dF 
 
 d^ d^ 
 
 ^ dt' ^ dx' 
 or 
 
 dp- m dx^ ^^ 
 
 This differential equation is here derived for the case of wave 
 motion on a string, but identically the same form of equation 
 applies to sound waves in air, to electric waves, to waves of light 
 (which indeed are electric waves), and to certain kinds of water 
 waves.* Equation (3) is therefore very important. Its general 
 solution as found in Art. 107 is: 
 
 y = F{x-vt)+f(x + vt) (4) 
 
 where _ 
 
 V = a/- (5) 
 
 The term F(x — vt) represents a wave (a bend of any shape) 
 traveling to the right at velocity v^ and , the term f{x + vt) 
 represents a wave (a bend of any shape) traveling to the left at 
 velocity v. 
 
 109. Idea of wave motion established without the help of 
 equation (3) of Art. 108. — It is evident from Art. 107 that travelj 
 purely and simply, is about the only thing that is estabUshed by 
 the solution of equation (3) of Art. 108. Therefore one might 
 expect to obtain equations (4) and (5) without the help of equation 
 (3) by introducing the idea of travel at the beginning. With this 
 end in view let us consider a bend of any shape and let us imagine 
 that this bend is traveling along the string to the right at any 
 
 * We are here considering only the simple case in which there are no appreci- 
 able energy losses as the wave travels along. 
 
196 CALCULUS. 
 
 velocity v. One could make a bend of definite shape travel 
 along a stretched string by threading the string through a bent 
 tube and moving the bent tube along, and this state of affairs 
 would be entirely unchanged if we imagine the tube to he stationary 
 and the stretched string to he drawn through it as indicated in Fig. 92. 
 
 Fig. 92. 
 
 It is assumed that the string slides through the tube without 
 friction. At any point p the string pushes against the side d 
 of the tube because of its tension, and the string pushes outwards 
 against the side c of the tube because of centrifugal action. 
 Let r be the radius of curvature of the tube at p. Then the 
 particles of the string near p may be considered to travel along 
 a circular path of radius r. Therefore, according to Art. 50, the 
 
 radial* acceleration of a particle of the string at p is - . Consider 
 
 r 
 
 an infinitely short piece of the string whose length is ds and 
 
 whose mass is m • ds. Then m ' ds X - is the radial force 
 
 r 
 
 which must act upon the short piece of string to produce the 
 
 specified acceleration — . If the string has no tension, then all of 
 
 this radial force is exerted by the side c of the tube. 
 
 If the string is under tension T, then the tension produces a 
 
 T 
 radial force equal to - • ds on the portion ds of the string, 
 
 according to Art. 51; and if the string is not moving this force is 
 exerted against the side d of the tube. 
 
 * In the direction of r and towards the center of the osculating circle. 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 197 
 If the string is moving at a velocity which satisfies the equation: 
 
 or 
 
 V' T 
 m - = - 
 r r 
 
 
 (1) 
 
 then the radial force due to the tension of the string is just sufficient 
 to produce the necessary radial acceleration, and no force at all 
 is exerted on the string by the guiding tube. Under these con- 
 ditions the guiding tube might be removed and the bend would 
 stand in a fixed position on the traveling string; or if the string 
 were standing still the bend would travel along the string at 
 velocity v. 
 
 In this discussion the bend can be of any shape and it can 
 travel at velocity v in either direction. 
 
 110. The vibration of a plucked string.* — A string is pulled 
 to one side as shown in Fig. 93 and released. The string is thus 
 
 Fig. 93. 
 
 * The vibration of a string which is struck with a hammer as in a piano is 
 easy to formulate. The vibration of a string which is set vibrating by a bow 
 as in a violin is somewhat more difficult to formulate. 
 
 A good discussion of this subject is given in Byerly's Fourier's Series and 
 Spherical Harmonics, pages 30-145, Ginn and Co., Boston, 1893. 
 
 A discussion of the motion of piano strings and of violin strings is given 
 in Appendices V and VI of Helmholtz's Tonempfindungen. English trans- 
 lation by Alexander J. Ellis is entitled Sensations of tone. Published by 
 Longmans, Green and Co., 1885. 
 
198 CALCULUS. 
 
 set vibrating. The vibrating string has at each instant a definite 
 shape, that is, it forms a definite curve. It is desired to find an 
 equation expressing y in terms of x and elapsed time t, which 
 equation will be at each instant the equation of the curve formed 
 by the moving string. 
 
 There are two conditions which must be satisfied by the ex- 
 pression for ?/, namely: (a) y must be zero at all times when 
 x = and when x = tt, and (6) at the instant t = the ex- 
 pression for y must be the equation of the curve formed by the 
 plucked string at the moment of release. 
 
 Also, of course, the expression for y must satisfy the differential 
 equation (3) of Art. 108, namely, 
 
 (Py^T^ 
 
 dt^ rndx" ^^ 
 
 Now it can be shown (see problem 2 on page 193) that the following 
 expressions satisfy equation (1) 
 
 y = A sin nx sin nvt (2) 
 
 2/ = A sin nx cos nvt (3) 
 
 y = A cos nx sin nvt (4) 
 
 y = A cos nx cos nvt (5) 
 
 where A and n have any values whatever. But only (2) and 
 (3) give ?/ = when a; = 0, and n must he an integer to give 
 2/ = when x = w. But (2) cannot be used because it gives 
 2/ = everywhere when ^ = 0. Therefore our problem is to 
 take 
 
 2/ = A sin nx cos nvt (3) 
 
 which satisfies condition (a), and add a large number of such solu- 
 tions together (using different values of A and n in each) to 
 get an expression for y which satisfies condition (6), above. The 
 possibility of doing this was discovered by Fourier* and is embodied 
 in what is called Fourier's theorem. 
 
 * Fourier's original discussion is very simple and interesting. See Fourier's 
 Theory of Heal, translated by Alexander Freeman, Cambridge, 1878. 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 199 
 
 111. Fourier's theorem. — ^Let y be any given function of x. 
 Consider a certain portion AB of the curve which represents y 
 as shown in Fig. 94, and let the distance AB be 27r (this is 
 
 AB 
 equivalent to choosing -^r— as our unit of length). Then, ac- 
 cording to Fourier we may express the function y within the 
 region AB as follows: 
 
 y = Ao -{- Ai sin X + Az sin 2x + ^3 sin 3x + 
 + Bi cos X + B2 cos 2x + B3 cos Sx 
 
 ::::) 
 
 (1) 
 
 where the coefficients Ao, Ai, A2, • • • and Bi, B2, B3, • • • are 
 constants whose values are : 
 
 1 /»a;=2rr 
 
 Ao = ^ J ydx (2) 
 
 -j r*x=2ir 
 
 An=' ~ \ y minx ' dx (3) 
 
 '^ Jx=0 
 -t nx=Z7r 
 
 — I y cosnx ' dx (4) 
 
 '^ Jx=Q 
 
 =277 
 
 B. I 
 
 '" lx=Q 
 
 A complete proof of Fourier's theorem would involve a proof 
 that the right-hand member of equation (1) is a convergent 
 
 
 \^^' 
 
 series.* This, however, will here be taken for granted. The 
 proof then reduces to a derivation of equations (2), (3) and (4). 
 
 * This matter is discussed in Chapter III of Byerly's Fourier's Series and 
 Spherical Harmonics, Ginn and Co., Boston, 1893. 
 
200 CALCULUS. 
 
 In this proof a consideration of average values is most important, 
 
 and the student should therefore look over Art. 73 again. 
 
 Another matter of import- 
 ance is to understand clearly 
 'V^A^uin fix ^^ ^YiQ meaning of the equation 
 y = sin nx (and y = cos nx) 
 when n is an integer. Now 
 y — sin nx is the equation 
 of a sine curve. If n = 1 
 this sine curve makes a posi- 
 tive "arch" between x = 
 
 and a; = TT, and a negative "arch'* between x = t and x = 27r. 
 
 In general the sine curve y = sin nx makes a positive arch 
 
 between x = and x = -, and a negative arch between x = - 
 
 n' ^ n 
 
 and X = — as shown in Fig. 95. Therefore the sine curve 
 
 n ^ 
 
 y = sinnx makes n pairs of positive and negative arches between 
 X = and x = 2t, 
 
 The derivation of equations (2), (3) and (4) depends upon the 
 following propositions, n and m being integers. 
 
 Proposition I. — The average value of sin nx (or of cos nx) between 
 X — and x = ^tt is zero. This is evident when we consider 
 that sin nx (or cos nx) passes through exactly similar sets of 
 positive and negative values which exactly offset each other 
 between a; = and x = 27r. 
 
 Proposition n. — The average value of sin^ nx (or of cos^ nx) 
 between x = and x = 2x is equal to \. This may be shown 
 
 -I /*Z=ilt 
 
 by finding the value of ^ I m^nx-dx (or of the correspond- 
 
 ing expression for cos^ nx) as explained in Art. 73. 
 
 Proposition m. — The average value of sin nx sin mx between 
 X = and x = ^ir is zero when n and m are different integers. 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 201 
 This may be shown by finding the value of the expression 
 
 'oZ I sin nx sin mx • dXy 
 
 which may be easily done with the help of form 24 in the table 
 of integrals. 
 
 Proposition IV. — The average value of sin nx cos mx between 
 X = and x = 2t is zero whether the integers n and m he the 
 same or not. This may be shown with the help of form 25 in the 
 table of integrals. 
 
 Proposition V. — The average value of cos nx cos mx from x = 
 to X = 2t is zero when m and n are different integers. This can 
 be shown with the help of form 26 in the table of integrals. 
 
 Derivation of equation (2). — Consider the average value of each 
 member of equation (1) between x — and x = 27r. The 
 average value of the first member is, by definition, equal to 
 
 — I y-dx, and the average value of the second member is Ao 
 
 according to proposition / above. Therefore ^ I y'dx = Ao 
 
 which is equation (2). 
 
 Derivation of equation (3). — Multiply both members of equa- 
 tion (1) by sin nx and we have: 
 
 y sin nx = Ao sin nx + Ai sin nx sin x -{- • • • -{-An sin^ nx 
 
 + • • • + -Bi sin na; cos X + • • • . 
 
 Consider the average value of each member of this equation 
 between a; = and x = 27r. The average value of the first 
 
 member is ;r- ( y sin nx-dx. The average value of every term 
 
 ^T^ Jx=0 
 
 of the second member is zero according to the above propositions 
 except the term An sin^ nx, and the average value of this term 
 
202 
 
 CALCULUS. 
 
 is }An according to proposition IL Therefore 
 ^1 y sinnx ' dx = iAn 
 
 which is equation (3) 
 
 Derivation of equation (4). — ^Multiply both members of equa- 
 tion (1) by cos nx, consider the average value of each member 
 of the resulting equation between a; = and x = 27r, and 
 equation (4) is obtained. 
 
 Simplification of equations (3) and (4) due to symmetry of given 
 curve. — Let ccc, Fig. 96a, be the curve which is to be expressed 
 by equation (1). It is permissible to take the base of ccc as tt. 
 
 Fig. 96a. 
 
 Under these conditions there are two important methods for 
 choosing the second half of the complete curve, namely, the 
 portion between tt and 27r, as follows: 
 
 Curves having sine terms only. — The second half of the complete 
 curve may be chosen like ccc turned upside down and turned 
 end for end, as shown by the dotted curve c'c' in Fig. 96a. In 
 this case y = — y' as shown in Fig. 96a. Now sin nx has equal 
 and opposite values at p and at p', therefore y sin nx • dx has the 
 same value and sign at p and at p', and therefore 
 
 ry %\Ti. nx • dx = 2 \ y ^\ 
 
 sin nx • dx 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 203 
 Consequently equation (3) becomes: 
 
 An = - I y sinnx ' dx (5) 
 
 TT Jo 
 
 Furthermore cos nx has the same value and the same sign at p 
 and at p', therefore y cos nx • dx has the same value but has 
 
 /»27r 
 
 opposite signs at p and at p\ and therefore I y cos nx • dx is 
 
 zero. Consequently all of the cosine terms drop out of equation 
 (1). Furthermore it is evident from the symmetry of the com- 
 
 plete curve cccc^c' in Fig. 96a that 1 y-dx = so that 
 
 Jo 
 Ao = 0. Therefore the complete curve in Fig. 96a is given by: 
 
 y = Ai sin X + Az sin 2x + A3 sin 3x + • • • (6) 
 
 Curves having cosine terms only. — The second half of the complete 
 curve may be chosen like ccc turned end for end but not turned 
 upside down as shown by the dotted curve c'c'c' in Fig. 966. 
 
 y'tixia 
 
 K 
 
 x-axis 
 
 ^ 0' 
 
 1 
 
 — X -ii ! K (x) 
 
 ^ — . — ~-;r-^ ^ 
 
 Fig. 966. 
 
 Under these conditions it may be shown, by an argument some- 
 what similar to the above, that equations (1) and (4), respectively, 
 become 
 
 y = Ao + Bi cos X -\- B2 cos 2x + B3 cos 3a; + • • • (7) 
 
 and 
 
 2 f' 
 
 Bn = ~ \ y COS nx ' dx (8) 
 
204 
 
 CALCULUS. 
 
 112. The vibration of a plucked string completely formulated. — 
 In Art. 110 the problem of the plucked string was reduced to the 
 problem of adding together a number of terms like A sin nx cos nvt 
 so as to get an expression for y which is the equation of the curve 
 formed by the string when t = 0. This is evidently the same 
 thing as expanding the given function y '\n o. series of sines as 
 explained in Art. HI. That is, we must find the coefficients in 
 the series: 
 
 2/ = ill sin a; + Ai sin 2x + ^la sin 3a; + • • • (1) 
 
 and the general expression for the coefficient An is: 
 An = - \ 2/ sin nx • dx 
 
 (2) 
 
 according to equation (5) of Art. Ill, the use of which means 
 that we have assumed the length of our string as tt units and 
 
 middle of string 
 -string 
 
 Fig. 97o. 
 
 have completed the curve as indicated by the heavy dotted line 
 in Fig. 97a. 
 
 Now the equation of the curve formed by the string at the 
 beginning {t = 0), as shown in Fig. 97a, is: 
 
 y 
 
 2c 
 
 from X = to ^ = -^ 
 
 2c 
 
 y = 2c X from x = ^ to x = 
 
 IT Z 
 
 (3) 
 (4) 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 205 
 Therefore the integral (2) must be evaluated in two parts, namely: 
 
 An = - t — xsmnx ' ax + - I [2c x ] smnx-dx (5) 
 
 which gives: 
 
 . Scl . nx 
 
 ir^w- 2 
 
 Therefore the equation of the curve formed by the string at the 
 beginning is: 
 
 y = -^ (sin x — ^ sin 3x + ^V sin 5a; — • • • ) (6) 
 
 and the equation of the curve formed by the string at an instant 
 t seconds after the beginning is: 
 
 y = —^ (sin X cos vt — ^ sin Sx cos 3vt ■}- • • • ) (7) 
 
 Description of the motion of a plucked string. — It would seem 
 from the elaborate difficulties of the above problem [and perhaps 
 
 >qTTTTTTTpr 
 TTTTTlXTTrt\^ ^ 
 
 "T^iXnTTTTTTTTTTpf 
 
 Fig. 976. 
 
206 
 
 CALCULUS. 
 
 the complexity of the result as given in equation (6) would suggest] 
 that the motion of a plucked string is very complicated. As a 
 matter of fact, however, the motion of a plucked string is extremely 
 simple and easy to describe. Thus Fig. 976 shows six successive 
 snapshots of a string which has been plucked in the middle and 
 released. The moving part of the string is straight and parallel to 
 the equilibrium position of the string and the motion is indicated 
 by the short arrows. 
 
 PROBLEMS. 
 
 The most important practical problem involving Fourier's theorem is, 
 perhaps, the finding of the coefficients in equation (1) of Art. Ill when the 
 function y is given not as an algebraic function of x but as an experimentally 
 determined curve. 
 
 A very good set of directions for making the necessary calculations is 
 given in Bedell and Pierce's Direct and Alternating Current Manual, 2nd edition, 
 pages 331-338, D. Van Nostrand, New York, 1913; and a discussion of the 
 origin and proof of the method is given on pages 339-344. 
 
 The harmonic analyzer is a machine for finding the coefficients in equation 
 (1) of Art. Ill when the function y is given as an experimentally determined 
 curve. The eariiest harmonic analyzer is that of Lord Kelvin. A brief 
 description of this machine is given in Franklin's Electric Waves, pages 340- 
 342, The Macmillan Co., New York, 1909. See also section 37 of the article 
 on Tides in the ninth edition of the Encyclopedia Britannica. See also articles 
 by James Thomson and by Sir Wm. Thomson (Lord Kelvin) in Proceedings of 
 the Royal Society, Vol. XXIV, 1876, page 262 and pages 269 and 271. The 
 harmonic analyzer of G. U. Yule is described in an article by J. N. Le Conte, 
 Physical Review, Vol. VII, pages 27-34, 1898. An especially interesting de- 
 scription of a harmonic analyzer is given on pages 68-74 of A. A. Michelson's 
 Light waves and their uses, University of Chicago Press, Chicago, 1903. 
 
 1. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p\. 
 
 Fig. pi. 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 207 
 
 Ans: y = (sin x + J sin 2a; + J sin 3a; + • • • ) 
 
 TT 
 
 2. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p2. 
 
 Ans; y = -^ (sin a; + J sin 3a; + J sin 5a; + • • •) 
 
 TT 
 
 
 y-axis 
 
 
 
 
 
 
 c 
 
 
 
 
 
 
 T 
 
 r 
 
 1 
 
 
 c \a 
 
 c 
 
 
 
 1 
 
 1 
 * 
 
 
 ! x-ojcw 
 
 
 1 
 
 1 
 
 
 1 
 
 c \a 
 
 c 
 
 
 
 ' 
 
 
 i c 
 
 
 
 
 h 
 
 K -^ -J(— ^- — ^ 
 
 Fig. p2. 
 
 3. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p3. 
 
 Ans: y = —2 (cos a; — | cos 3a; + i cos 5a; — | cos 7a; + • • •) 
 
 
 y-axis 
 e 
 
 
 c 
 
 
 n r — 
 
 f 
 X 
 
 c c 
 x-curis 
 
 
 1 
 
 I 
 1 
 
 1 
 
 L J 
 
 c 
 
 1 
 la 
 
 c 
 
 1 
 
 
 i 7( *-^ 
 
 , 
 
 Fig. p3. 
 
 4. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p4. 
 
 Ans: y = — (sin a; — J sin 3a; + -jV sin 5a; — :jV sin 7a; + • • •) 
 
208 
 
 CALCULUS. 
 
 5. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p5. 
 
 y-€ixi8 
 
 Sa 
 
 Ads: y = -^ (cos a; + i cos 3a; + ^ cos 5a; + • • • ) 
 
 
 y-axis 
 
 ./ 
 
 " ^^ »-axw ^ 1 \. 
 
 ■■--.--'• 
 
 ^^\i>^ 1 "^^-^^- 
 
 
 < — n ^—-n i 
 
 Fig. p5. 
 
 6. Determine the coefficients in Fourier's series to give the 
 curve cc which is shown in Fig. p6. 
 
 Ans: y = — (sin a; — J sin 2a; + J sin 3a; — i sin 4a; + • • • ) 
 
 t 
 
DIFFERENTIAL EQUATION OF WAVE MOTION. 209 
 
 7. Derive the answer to problem 4 by integrating the answer 
 to problem 3. 
 
 Note. — Any Fourier series gives a convergent series when integrated term 
 by term. 
 
 8. Plot the curve which represents the integral of the curve 
 given in Fig. p2, the constant of integration being such as to 
 make the integral curve pass through the origin; and determine 
 the coefficients in a Fourier's series to give the integral curve. 
 
 Note. — The constant of integration is zero in problem 7 and therefore 
 easily overlooked. 
 
 9. Derive the answer to problem 3 from the answer to problem 2 
 by shifting the origin ^ to the right. 
 
 10. Differentiate the answer to problem 4 and interpret the 
 result. 
 
 Note- — A Fourier series gives a divergent series by differentiation unless 
 the function ij is continuous. Thus y in Fig. p4 is continuous, whereas y 
 in Figs, pi, p2, p3 and p6 is discontinuous. 
 
 15 
 
CHAPTER IX 
 
 VECTOR ANALYSIS. 
 
 SCALAR AND VECTOR FIELDS. 
 
 113. Space analysis. — A system of space analysis, commonly 
 called vector analysis, is developed in this chapter, and it is ex- 
 tremely useful and important in every branch of physics where 
 variations in space are involved. Thus the theory of electricity 
 and magnetism (aside from the electron theory) is a simple appli- 
 cation of vector analysis. Also the theory of heat flow, the 
 theory of fluid motion, and a great part of the theory of elasticity 
 and wave motion are applications of vector analysis. It is almost 
 impossible for a student to make a beginning in any branch of 
 theoretical physics without some understanding of vector analysis. 
 
 Vector analysis as outlined in this chapter is a very different 
 thing from the familiar use of vectors and of complex quantity 
 in the theory of alternating currents. In the one case we have a 
 comprehensive system of space analysis, and in the other case 
 we have a narrow scheme for representing the solution of a linear 
 partial differential equation, as explained in chapter VII. Indeed 
 the kind of "vector analysis'' which is used in the theory of 
 alternating currents cannot possibly be extended to three dimen- 
 sions as a consistent system of space analysis. 
 
 Vector analysis originated in Sir William Hamilton's theory of 
 quaternions.* The theory of quaternions, however, contains 
 much that is of doubtful value in theoretical physics. Indeed 
 Maxwell in his great treatise on Electricity and Magnetism (Oxford, 
 1873) used only a few of Hamilton's ideas. The study of Max- 
 
 * See Sir William Rowan Hamilton's Lectures on Quaternions, and Elements 
 of Quaternions. These books are now out of print but they may be found 
 in most good libraries. An Elementa ry Treatise on OuMernions by P. G. Tait, 
 second edition, Oxford, 1873, is the standard treatise on the subject. 
 
 210 
 
VECTOR ANALYSIS. 211 
 
 well's treatise has kept the subject of space analysis before the 
 serious student of physics for more than a generation, and two 
 significant attempts have been made to formulate the more useful 
 of Hamilton's ideas in what is now called vector analysis. The 
 first of these was that of Willard Gibbs, whose view of vector 
 analysis was outlined in a very condensed form in a pamphlet 
 printed (for private circulation, only) in New Haven in 1883.* 
 The second attempt to formulate the more useful of Hamilton's 
 ideas was that of Oliver Heaviside.f 
 
 The first attempt to place vector analysis before the student of 
 physics in simple form was that of E. L. Nichols and W. S. Frank- 
 lin.J The best discussion of vector analysis for the student is 
 that of Abraham and Foppl in their Theorie der Electridtdt, Vol. I, 
 pages 1-125, Leipsig, 1907. 
 
 114. Scalar and vector quantities. — A scalar quantity is a 
 quantity which has magnitude only. Thus every one recog- 
 nizes at once that to specify 10 cubic yards of sand, 25 pounds of 
 sugar, 5 hours of time is in each case to make a complete speci- 
 fication. Such quantities as volume, mass, time and energy are 
 scalar quantities. 
 
 A vector quantity is a quantity which has both magnitude and 
 direction, and to completely specify a vector quantity one must 
 give both its magnitude and its direction. This necessity of 
 specifying both the magnitude and the direction of a vector is 
 especially evident when one is concerned with the relationship of 
 two or more vectors. Thus if a man travels a stretch of 10 miles 
 and then a stretch of 5 miles more, he is by no means necessarily 
 15 miles from home. If T)ne man pulls on a post with a force of 
 
 * Professor Gibbs' point of view is set forth in Vector Analysis by E. B. 
 "V^ilson, New Haven, Yale University Press, 1901. 
 
 t See Heaviside 's Electromagneti c Theory ^ Vol. I, pages 132-305, The 
 Electrician Publishing Co., London, 1893. This discussion of Heaviside'a 
 is unusually interesting, but we cannot agree with Heaviside in his statement 
 that vector analysis is independent of the quaternion. 
 
 X See Nichols and F rankUn's Elements of Physics, Vol. II, first edition, The 
 Macmillan Co., New York, 1895. 
 
212 CALCULUS. 
 
 200 units and another man with a force of 100 units, the total 
 force acting on the post is by no means necessarily equal to 300 
 units. A New Yorker traveling steadily at a speed of 60 miles 
 an hour would by no means necessarily reach Boston in 5 hours, 
 because he might be traveling in some other direction. 
 
 Many cases arise in physics where it is necessary to consider 
 the single force which is equivalent to the combined action of 
 several given forces; where it is necessary to consider the actual 
 velocity of a body due to the combined action of several causes, 
 each of which alone would produce a certain amount of velocity 
 in a given direction; and so on. The single force or single velocity 
 is in each case called the vector sum or the resultant of the given 
 forces or given velocities. 
 
 The addition of vector quantities is exemplified by the addition 
 of forces as follows: Two given forces are represented by the 
 lines a and 6, in Fig. 98, and the sum or resultant of the forces 
 is represented by the diagonal r of the parallelogram constructed 
 on a and h as sides. It is evident that the geometrical relation 
 
 between a, b and r is completely shown by the triangle in Fig. 
 99, in which the line which represents force h is drawn from the 
 extremity of the line which represents force a, and r is the 
 closing line of the triangle. 
 
 The geometrical construction of Fig. 99 gives a method of 
 adding any number of forces with the least amount of compli- 
 cation, as shown in Fig. 100. Thus to add four given forces 
 a, h, c and d: draw a Hne representing force a; from the end of 
 this line draw a line representing force b; from the end of this 
 
VECTOR ANALYSIS. 213 
 
 line draw a line representing force c; and so on. Then the closing 
 line r of the polygon abed represents the sum of the forces in 
 magnitude and in direction. 
 
 To represent a vector in an alge- 
 braic discussion it is usually most 
 convenient to represent the vec- 
 tor in terms of its components in 
 the directions of three chosen 
 rectangular axes of reference. 
 Thus if X, F, Z are the compo- 
 nents of a given force F (or any p. .^ 
 vector whatever) then the force may 
 be represented as 
 
 F = X+Y + Z. 
 
 In this expression vector addition is of course understood 
 because X, Y and Z are vectors. 
 
 ■ 115. Vector products. Case I. Parallel vectors. — The product 
 
 of two parallel vectors is a scalar. This fact is exemphfied in 
 
 every branch of physics. Thus to multiply a force by the distance 
 
 a body has moved in the direction of the force gives the work done 
 
 by the force, and work is a scalar quantity. To multiply a force 
 
 by the velocity with which a body moves in the direction of the 
 
 force gives the power developed by the force, and power is a 
 
 scalar quantity. The square of the velocity of a body determines 
 
 its kinetic energy, and energy is a scalar quantity. A plane area 
 
 is a vector quantity and its vector direction is the direction of the 
 
 normal to the area. To multiply the area of the base of a prism 
 
 by the altitude of a prism gives the volume of the prism, and 
 
 volume is a scalar quantity. 
 
 The quotient of two parallel vectors is a scalar quantity. Thus 
 
 let F be the force exerted by a fluid on a flat surface of area a. 
 
 F 
 Then the quotient - is the hydrostatic pressure of the fluid, and 
 a 
 
 hydrostatic pressure is a scalar quantity. 
 
214 CALCULUS. 
 
 Case II. Orthogonal vectors. — Vectors which are at right angles 
 to each other are said to be orthogonal. The product of two 
 orthogonal vectors is a third vector at right angles to both of the 
 given vectors. Thus to multiply the length by the breadth of a 
 rectangle gives the area of the rectangle, and area is a vector as 
 above explained. To multiply a force F by the perpendicular 
 distance I from the Une of action of the force to a chosen axis 
 gives the torque action IF of the force about the axis, and torque 
 action is a vector quantity. 
 
 The quotient of two orthogonal vectors is a third vector at right 
 angles to both of the given vectors. Thus to divide the area of a 
 rectangle by the length of one side gives the length of the other 
 side. 
 
 Case III. Oblique vectors. — The product of two oblique vectors 
 consists of two parts; one part is a scalar and the other part is 
 a vector. This proposition is a necessary result of the above 
 statements concerning the product of parallel vectors and the 
 product of orthogonal vectors. Thus Fig. 101 shows two obUque 
 vectors U and V. The vector V can be resolved into the 
 
 ">^ components 7' and V" parallel to U 
 
 vy\ j and perpendicular to U respectively, 
 
 y^ ^\ 1 ^ and then 7' + V" can be substi- 
 
 v ^ ^^ ^ ^1 tuted for V in the product UV, giv- 
 
 — ^— ' ing UV + VV". But UV is the 
 
 pj jQj product of two parallel vectors and it 
 
 is a scalar, and UV" is the product 
 of two orthogonal vectors and it is a vector. 
 
 From Fig. 101 it is evident that V = V cos 6 and that 
 7" = 7 sin 6. Therefore we have the following important 
 propositions: 
 
 (a) The scalar part of the product of two oblique vectors is 
 equal to the product of the numerical values of the respective 
 vectors and the cosine of the angle between them. 
 
 (6) The numerical value of the vector part of the product of 
 two oblique vectors is equal to the product of the numerical values 
 
VECTOR ANALYSIS. 
 
 215 
 
 of the respective vectors and the sine of the angle between them; 
 and of course the direction of the vector part of the product is 
 at right angles to the plane of the given oblique vectors. 
 
 There are but few cases in physics where the scalar part and the 
 vector part of the product of two oblique vectors are both important, 
 although there are many cases where the scalar part of a vector 
 product is important, and many other cases where the vector part 
 of a vector product is important. Thus a force F acts on a car 
 as shown in Fig. 102. Imagine the car to move a distance I in 
 
 Fig. 102. 
 
 the direction of the track, then the work done by the force is the 
 scalar part of the product of the two obhque vectors F and l^ 
 or: the work done is equal to numerical value of I X numerical 
 value of F X cos 6. The vector part of the product IF in this 
 case has no very important meaning. 
 
 A force F acts on a crank as shown in Fig. 103. The torque 
 action of the force about the axis of the crankshaft is the vector 
 part of the product of the two oblique vectors F and I; the 
 numerical value of the torque is equal to numerical value of I X 
 numerical value of F X sin 6, and the direction of the torque, 
 considered as a vector, is parallel to the axis of the shaft. 
 
 Exact direction of the vector part of a vector product. — ^The 
 product UV in Fig. 101 is a vector at right angles to the plane 
 of the paper, but is it towards the reader or away from the reader? 
 Consistency* requires us to admit that if the product UV" is 
 
 * This is shown by the following discussion of Fig. 104 and by the inter- 
 pretation of equation (1). Sir William R. Hamilton gives a very full discus- 
 sion of this in his Lectures and in his Elements of Quaternions. 
 
216 
 
 CALCULUS. 
 
 towards the reader in Fig. 101, then V'U must be away from the 
 
 reader. That is, we must admit that UV" = — V"U where 
 
 U and V" are orthogonal vectors. 
 This matter is exemphfied as follows: A force F acts upon a 
 
 crank-arm I as shown in Fig. 104. The vector part of the product 
 
 IF is the torque action of the force 
 about the axis (perpendicular to the 
 plane of the paper). Now the torque 
 action of F in Fig. 104 can be ex- 
 pressed in terms of the components of 
 F, namely, X and F, and the com- 
 ponents of I, namely, x and y. 
 Thus xF is a counter-clockwise torque 
 about 0, and yX is a clockwise torque 
 
 about in Fig. 104. 
 as positive, we have*. 
 
 Therefore, taking counter-clockwise torques 
 
 Total torque action of F 
 about in Fig. 104, or the 
 vector part of the product IF 
 
 I =a;y- 
 
 yx 
 
 (1) 
 
 The vector direction of a torque may be thought of as the direc- 
 tion (along the axis of the torque) in which a right-handed screw 
 would travel if turned by the torque. Adopting this convention 
 and choosing the positive direction of the 2-axis of reference towards 
 the reader in Fig. 104 we have from equation (1) : 
 
 (a) xY is an x-vector multiplied by a ^/-vector, and it is a 
 vector in the positive direction of the 2-axis. 
 
 (6) yX is a ?/-vector multiplied by an a;-vector, and it is a vector 
 in the negative direction of the 2-axis. 
 
 Expressing F and I in terms of their components as shown in 
 Fig. 104, we have: 
 
 F = X + y (2) 
 
 and 
 
 l = x-\-y (3) 
 
 Multiplying equations (2) and (3) member by member, using the 
 
VECTOR ANALYSIS. 217 
 
 ordinary rules of algebra, we get : 
 
 IF = xX -\- yY -{- xY -{- yX (4) 
 
 But it is shown above that xY and yX are opposite in sign. 
 Now the opposite signs oi xY and yX are shown in equation (4), 
 according to the above discussion, hy the fact that in one case we 
 have an x-vector multiplied by a y-vector and in the other case we have 
 a y-vector multiplied hy an x-vector. But it is very inconvenient 
 to have algebraic signs so indirectly indicated; it is better to 
 indicate the sign explicitly and take xY — yX as the vector part 
 of the product IF. When this is done we pay no attention to the 
 order of the factors, because the inverse order in the two terms is 
 taken account of in the negative sign of the second term. With this 
 understanding, therefore, we have the following important propo- 
 sitions: 
 
 (a) The scalar part of the product IF in Fig. 104 is equal to 
 xX -h yY. This is evident when we consider that xX and yY 
 are the only scalar terms in equation (4). 
 
 (&) The vector part of the product of IF in Fig. 104 is equal to 
 xY — yX, and it is in the direction towards the reader in Fig. 104. 
 
 116. Scalar fields. — It is sometimes important to consider the 
 temperature at various points in a body, the pressure at various 
 points in a fluid, the density at various points of a substance or 
 the electric charge per unit volume at various points in a region. 
 Such a distribution of temperature, hydrostatic pressure, or density 
 is called a scalar field because the quantity under consideration is 
 a scalar and it has a definite value at each point in a region or 
 field of space. The distribution is said to be homogeneous or 
 uniform when the scalar quantity has the same value at every 
 point in the field; otherwise the distribution is said to be non- 
 homogeneous. An example of a non-homogeneous scalar field is the 
 temperature of an iron rod one end of which is red hot and the 
 other end of which is cold. Atmospheric pressure is also non- 
 homogeneously distributed because it is different at different 
 places and different at different altitudes. 
 
218 CALCULUS. 
 
 A scalar field is sometimes called a distrihvied scalar. Thus 
 when the scalar nature of temperature is to be emphasized it is 
 helpful to call temperature (throughout a hot iron rod, for example) 
 a distributed scalar. 
 
 A scalar field is sometimes called a scalar function, A distri- 
 buted scalar has a definite value at each point of space; that is, 
 the value of a distributed scalar at a point is a function of the 
 coordinates of the point. When this fact is to be emphasized it 
 is helpful to call temperature (or any distributed scalar) a scalar 
 function. 
 
 117. Vector fields. — It is sometimes important to consider the 
 velocity at different points of a fluid, the direction and intensity 
 of heat flow at different points of a substance, or the direction and 
 intensity of an electric or magnetic field at different points in 
 space. Such a distribution of fluid velocity, or other vector, is 
 called a vector field because the quantity imder consideration is a 
 vector, and it has a definite value and direction at each point in 
 a region or field of space. The distribution is said to be homo- 
 geneoiLs or uniform when the vector has the same value and is in 
 the same direction at every point in the field; otherwise the distri- 
 bution is said to be non-homogeneous. The water in a rotating 
 bowl is an example of a non-homogeneous vector field because the 
 water moves in different directions and at different velocities at 
 different points in the bowl. The magnetic field around an 
 electric wire is a non-homogeneous vector field because the mag- 
 netic field does not have the same intensity and the same direction 
 everywhere. 
 
 A vector field is sometimes called a distributed vector. Thus 
 when the vector nature of fluid velocity is to be emphasized it is 
 helpful to call fluid velocity a distributed vector. 
 
 A vector field is sometimes called a vector function. Each 
 component (the a;-component, the ^/-component and the 2-com- 
 ponent) of the velocity of a fluid has a definite value at each point 
 of space; that is, the values of the components at a point p are 
 
VECTOR ANALYSIS. 219 
 
 functions of the coordinates of p. When this fact is to be empha- 
 sized it is helpful to call fluid velocity (or any distributed vector) 
 a vector function. 
 
 118. Volume integral of a distributed scalar. — For the sake of 
 simplicity let us consider a special case, namely, the density of a 
 substance, and let us assume that the density varies from point to 
 point. Let ^ be the density at a given point, then \l/ ' dr is 
 the mass of material in the volume element dr at the point, and 
 the total mass M of the body is: 
 
 M = frP ' dr (1) 
 
 This integral is called the volume integral of the distributed scalar ^. 
 The physical significance of volume integral is not in every case 
 so simple as in the case of density. If \l/ is the volume density of 
 electric ch?irge, then equation (1) gives the total electric charge 
 in the region throughout which the integration is extended. 
 If ^ is the energy density in an electric or magnetic field, or in a 
 strained solid, or in a moving fluid, then equation (1) gives the 
 total energy in the region throughout which the integration is 
 extended. 
 
 119. Gradient of a distributed scalar. — Any distributed scalar 
 like temperature, or density, or hydrostatic pressure has a definite 
 value at each point of a field and therefore the value of any distri- 
 buted scalar at a point may be thought of as a function of the 
 coordinates x, y and z of the point. Indeed, this matter has 
 already been discussed in Arts. 62 to 66, where it is explained 
 that a distributed scalar has a definite gradient at each point. 
 Thus if ^ is the temperature at a point in a body, then 
 
 (1) 
 
 (2) 
 (3) 
 
 X ■ 
 
 dx 
 
 Y 
 
 dyp 
 
 dy 
 
 Z 
 
 drP 
 
 
 dz 
 
220 CALCULUS. 
 
 where X, Y and Z are the component gradients of \f/. The 
 gradient of a distributed scalar has a definite value and a definite 
 direction at each point and it is, therefore a distributed vector. 
 
 PROBLEMS. 
 
 1. A vessel one meter wide, one meter long, and one meter 
 deep, contains a fluid of which the density is one gram per cubic 
 centimeter at the top, increasing uniformly to two grams per 
 cubic centimeter at the bottom. Find the volume integral of 
 the density of the fluid. Ans. 1,500,000 grams. 
 
 2 . The gradient of the density in problem 1 is uniform through- 
 out the vessel and equal to one gram per cubic centimeter per 
 meter, and it is directed vertically downwards. Suppose the 
 downward gradient of the density to be a linear function of the 
 distance from the top of the vessel, changing from zero at the top 
 to two grams per cubic centimeter per meter at the bottom. 
 Find the volume integral of the density of the fluid, the density 
 being one gram per cubic centimeter at the top. Ans. 1,333,333 
 grams. 
 
 3. The value of a distributed scalar at any point p is \p = — 
 
 where Q is a constant and r the distance of p from the origin 
 
 of coordinates. Find the a:, y and z components of the gradient 
 
 Qx 
 of \l/. Ans. The x-component is — -r-r-; — t"; — 5Ti' 
 
 {x^ + 2/2 H- z^)* 
 
 4. Find the gradient of ^ ( = ~ ) in the direction of r. Ans. 
 
 5. The a;-component of the gradient of any distributed scalar 
 may be thought of as a distributed scalar. Find the ^/-component 
 
 of the gradient of the a;-component of the gradient of -• Ans. 
 
 SQxy 
 
 (3^-\-y^ + 2^)1 
 
VECTOR ANALYSIS. 221 
 
 120. Permanent and varying states of scalar distribution. — 
 
 When the temperature at each point of a body remains unchanged, 
 the distribution of temperature throughout the body is said to be 
 permanent, although, of course, the temperature of the body 
 may not be everywhere the same. When the temperature at each 
 point of a body is changing we have what is called a varying state 
 of distribution. Thus the density in a gas which is being com- 
 pressed, and the temperature throughout a body which is being 
 heated or cooled are examples of varying scalar distributions. 
 
 121. Stream lines of a distributed vector. — A line drawn through 
 a moving fluid so as to be at each point in the direction in which 
 the fluid at the point is moving is called a stream line. The 
 geometrical idea of a stream line applies to any vector field what- 
 ever, and the manner of distribution of a vector is clearly repre- 
 sented by the use in imagination of such lines. In electric and 
 magnetic fields these lines are called lines of force, but the term 
 stream line will be used in general statements. 
 
 122. Permanent and varying states of vector distribution. — 
 
 When the velocity of a moving fluid remains unchanged in magni- 
 tude and direction at every point, we have what is called a perma- 
 nent state of fluid motion. When the velocity of a fluid is changing 
 at each point, we have what is called a varying state of fluid motion. 
 Thus, when an orifice in a large tank of water is suddenly opened, 
 a perceptible time elapses before the jet of water becomes estab- 
 lished. During this time the velocity of the water is changing 
 rapidly at each point in the jet. After the jet becomes steady, 
 however, the velocity of the water at each point remains constant 
 in magnitude and in direction. The magnetic field in the neigh- 
 borhood of a moving magnet or in the neighborhood of a moving 
 or changing electric current is an example of a varying vector 
 distribution. 
 
 Rate of change of a distributed vector at a point. — Let the line a, 
 Fig. 105, represent the value at a given instant, of the velocity of 
 a fluid at the point p, and let the line a + Aa represent the 
 
222 CALCULUS. 
 
 velocity of the fluid at the same point after a time-interval A^ has 
 
 elapsed. The Hmiting value oi -r- as At approaches zero is the 
 
 rate of change of a at the point p.* This rate of change of a 
 has a definite value and is in a definite direction at each point of 
 space and it is therefore a distributed vector. 
 
 123. Line integral of a distributed vector. — Consider a line or 
 path pp' in a vector field as shown in Fig. 106. Let As be an 
 element of this line or path, let R be the value of the distributed 
 
 ^direction o/j^atp N\ 
 
 ""\^. ^^^ >P' 
 
 Fig. 105. Fig. 106. 
 
 vector at the element As, and let e be the angle between U and 
 As. Then R cos € is the resolved part of R parallel to As, 
 and R cos € • As is the scalar part of the product of R and As; 
 and: 
 
 E = f Rcose- ds (1) 
 
 is called the line integral of the distributed vector R along the line 
 or path over which this summation is extended. The angle e is 
 reckoned between R and the positive direction of As, the positive 
 
 * In this illustration the velocity under consideration is the changing 
 velocity of the successive particles of the fluid as they pass the point p, not 
 the changing velocity of a given particle while it is traveling along near p. 
 The latter velocity may be changing from instant to instant even though the 
 former is invariable. 
 
VECTOR ANALYSIS. 223 
 
 direction of As being the direction in which As would be passed 
 over in traveling along the line pp' in a chosen direction. If the 
 chosen direction be changed, cos e will change sign at each ele- 
 ment. Therefore the line integral from p to p' is equal to the 
 line integral from p' to p in Fig. 106, but opposite in sign. 
 
 Examples. — The line integral of electric field along a path is 
 called the electromotive force along the path. The Une integral of 
 a magnetic field along a path is called the magnetomotive force 
 along the path. The line integral of fluid velocity along a path 
 is called the circulation of the fluid along the path. 
 
 Cartesian expression for line integral. — Let X, Y and Z be 
 
 the components of the vector R, and let dx, dy and dz be the 
 components of the line element ds. Then we have: 
 
 R = X +Y + Z (a vector equation) (2) 
 
 and 
 
 ds = dx + dy + dz (a vector equation) (3) 
 
 The product of the two vectors R and ds is part scalar and 
 part vector as explained in Arts. 114 and 115, and the scalar part 
 of the product is R cos e - ds or X - dx + Y - dy + Z • dz. 
 Therefore equation (1) may be written: 
 
 E = J{X 'dx+Y 'dyi-Z -dz) (4) 
 
 Line integral of the gradient of a distributed scalar.— Consider 
 a distributed scalar rf/. Let us call it temperature for the sake 
 of inteUigibility. Let R in Fig. 106 be the gradient of \p, that 
 is R is the temperature gradient. Then the line integral of R 
 along the path pp' is equal to \J/' — ^, where ^' is the tempera- 
 ture at p' and \f/ is the temperature at p. This is evident 
 from the following considerations. The product R cos € in Fig. 
 106 is the resolved part of the temperature gradient in the direction 
 of the line element As, and i^ cos € • As is the change of tempera- 
 ture along As. Therefore 2/2 cos e • As is the sum of the changes 
 of temperature along all parts of the path pp' or the total change 
 of temperature from p to p'. 
 
224 
 
 CALCULUS. 
 
 Line integral of a gradient along co-terminous paths. — The 
 line integral of R along any path pp' is the temperature difference 
 between p and p'. Therefore the line integral of R is the same 
 for all paths from p to p\ There is an important exception to 
 this proposition as follows: 
 
 Imagine an ordinary auger to be placed with its axis at right 
 angles to the plane of the paper in Fig. 107, the plane of the 
 
 Fig. 107. 
 Slope-lines on an auger-hill. 
 
 Fig. 108. 
 
 Magnetic lines of force around an 
 electric wire. 
 
 paper being the base plane and the winding-stair-like surface of 
 the auger being looked upon as the surface of a hill raised above the 
 base plane. This hill we will call an auger-hill, and its slope 
 lines are shown by the fine-line circles in Fig. 107. The height 
 of this auger-hill above any point p has many values differing 
 from each other by, say, one inch, where one inch is the ** pitch" 
 of the auger. 
 
 Now the line integral of the slope of a hill along any path is the 
 difference of level of the ends of the path. A path which returns 
 to its starting point (a closed path) comes back to its initial level 
 on an ordinary hill, but a path from p back to p in Fig. 107 does 
 not come back to its initial level if the path circles round the axis 
 of the auger. Therefore the line integral of slope along the 
 path 1 from p to p' in Fig. 107 is not the same as the line integral 
 of slope along path 2 from p to p'. 
 
VECTOR ANALYSIS. 225 
 
 The fine-line circles in Fig. 108 are the magnetic lines of force 
 (slope lines of the magnetic-potential hill) in the neighborhood 
 of an electric wire perpendicular to the plane of the paper. The 
 line integral of the magnetic-potential slope along path 1 from 
 p to p' in Fig. 108 is not the same as the line integral of the 
 magnetic-potential slope along path 2 from p to p'. 
 
 124. Potential of a vector field. — Consider any distributed vec- 
 tor R, Then the '^height" at each point of space of an imagined 
 "hill'' whose slope or gradient is everywhere equal to R is called 
 the potential of R. Thus the temperature at each point of a 
 body is the "height" at that point of a hill whose slope is every- 
 where equal to the temperature gradient, therefore temperature 
 is the potential of temperature gradient. The "height" in volts 
 at each point of space of an imagined "hill" whose slope is every- 
 where equal to the electric field intensity (in volts per centimeter) 
 is called electric potential. The "height" at each point of space 
 of an imagined "hill" whose slope is everywhere equal to the 
 velocity of a moving fluid is called the velocity potential of the fluid. 
 
 The temperature at a point of a body can of course be deter- 
 mined by placing a thermometer at that point. That is, tempera- 
 ture is an actual physical condition. The electric potential at 
 a point, however, is not an actual measurable physical condition 
 at that point; indeed electric potential exists only in the imagi- 
 nation as a sort of mathematical fiction whose usefulness grows 
 out of the fact that it is often very helpful to think of a given 
 vector field as a gradient. An example showing the usefulness 
 of the idea of potential is given in Art. 133. 
 
 Theorem as to the existence of potential. — Let ^ be a distri- 
 buted scalar, like temperature. Then the component gradients 
 of yp are given by equation (1), (2) and (3) of Art. 119. Differ- 
 entiating the first of these equations with respect to y and the 
 second with respect to x we have: 
 
 d^yp dX 
 
 dy ' dx dy 
 
 16 
 
226 CALCULUS, 
 
 and 
 
 ^ dY (2^ 
 
 dx ' dy dx 
 
 But J , and , , are always identical as stated in 
 dy ' dx dx ' dy 
 
 Art. 59 and as demonstrated in Art. 134. Therefore from equa- 
 tions (1) and (2) we have: 
 
 ^ ^dY 
 
 dy dx 
 
 Similarly from equations (2) and (3) of Art. 119, we obtain: 
 
 f-^ = (4) 
 
 dz dy 
 
 and frona equations (1) and (3), we obtain: 
 
 dZ dX , V 
 
 In these equations (3), (4) and (5) X, Y and Z are the com- 
 ponent gradients of any scalar function \J/, and any distributed 
 vector whose components satisfy equations (3), (4) and (5) can have 
 a potential, or, in other words, any distributed vector whose com- 
 ponents satisfy (3), (4) and (5) can be looked upon as the gradient 
 of an imagined "hilV 
 
 Multivalued potential. — The auger-hill which is represented in 
 Fig. 107 has a multiplicity of heights above any point p. Similarly 
 the magnetic-potential in Fig. 108 has a multiplicity of values at 
 any point p. Of course the "height" of the potential "hill" 
 in Fig. 108 may be thought of as an actual height measured 
 upward from the plane of the paper; but one must not forget that 
 the magnetic field under consideration fills all space, and that 
 there is a multiplicity of values of magnetic potential at each 
 point in space. See Art. 66. 
 
VECTOR ANALYSIS. 
 
 227 
 
 125. Surface integral of a distributed vector. Flux. — 
 Let AOOB, Fig. 109, be a diaphragm stretching across a closed 
 loop of wire, the dots A and B being where the wire passes 
 through the plane of the paper. Let AS be the area of an ele- 
 
 diaphram 
 
 Fig. 109. 
 
 Fig. 110. 
 
 ment of the diaphragm, let R be the value at AS of a distributed 
 vector, and let e be the angle between R and the normal to AS 
 this normal being always drawn outward from the same side 00 
 of the diaphragm. Then R cos e is the resolved part of R normal 
 to A>S, and R cos e ' AS is the scalar part of R - AS; and: 
 
 ^ = f Rcose ' dS 
 
 (1) 
 
 is called the surface integral of R over the portion of the diaphragm 
 over which the integration is extended. 
 
 ^ If the normal to AS in Fig. 109 is reversed, as shown in Fig 
 110, then cos e will be reversed in sign, and the surface integral, 
 retaining its numerical value, will be reversed in sign. 
 
 In the integration over a closed surface like a box or sphere, 
 the normal is understood, throughout the following discussion, 
 to be drawn outwards. 
 
 Examples. — If R in Fig. 109 is the velocity of a fluid at AS, 
 then R cos e is the resolved part of the velocity normal to AS, 
 
228 CALCULUS. 
 
 R cos c • A*S is the volume of fluid per second flowing across 
 AS, and JR cos e • dS is the total volume of fluid per second 
 flowing through the loop of wire AB. The volume of fluid per 
 second passing through a loop is called the fliix of the fluid through 
 the loop, and the surface integral of any distributed vector over a 
 surface is called the fiux of the vector across the surface. Thus 
 magnetic flux is the surface integral of magnetic field and electric 
 flux is the surface integral of electric fleld. 
 
 The surface integral of a distributed vector over a closed surface 
 like a box or sphere is called the flux into or oiU of the region 
 enclosed by the box or sphere. 
 
 Cartesian expression for surface integral. — Let X, Y and Z 
 
 be the componenfs of the vector R in Fig. 109, and let da, db 
 and dc be the areas of the projections of dS on the yz, on the 
 xz, and on the xy planes, respectively. Then: 
 
 R = X + Y + Z (a vector equation) (2) 
 
 and 
 
 dS = da -{- dh + dc (a vector equation) (3) 
 
 The scalar part of the product R - dS is 
 
 X ' da + Y ' dh -\- Z ' dc i= R cos € ' dS), 
 
 But the shapes of the surface elements da, db and dc may be 
 anything whatever. Therefore we may write dy • dz for da, 
 dx • dz for dh, and dx • dy for dc. Hence equation (1) becomes: 
 
 ^ = ff (X ' dy ' dz -{- Y ' dx ' dz + Z ' dx ' dy) (4) 
 
 126. Divergence of a distributed vector. — In some cases a dis- 
 tributed vector "flows" outwards or emanates from a region of 
 space. Thus the liquid in a tank flows outwards from the end 
 of a supply pipe. When a gas is expanding each portion of the 
 gas is growing less dense and there is an outward flow from every 
 small part of the region occupied by the expanding gas. What is 
 called electric field emanates from electrically charged bodies, 
 
VECTOR ANALYSIS. 229 
 
 and when electric charge is spread throughout a region, electric 
 
 field emanates from every small part of the region occupied by 
 
 the charge. 
 
 Consider a small volume element Ar in the neighborhood of a 
 
 point p in a vector field. Let A$ be the flux of the distributed 
 
 vector R out of the volume element Ar. It can be shown,* 
 
 when i2 is a continuous function of the coordinates x, y and 2;, 
 
 A$ 
 that the ratio 7- approaches a definite limiting value as the 
 
 At 
 
 volume element Ar grows smaller and smaller. This limiting 
 
 value, -J- is called the divergence of the vector field at the point p. 
 
 Therefore, representing the divergence of a vector field at a point 
 by p, we have: 
 
 d^ = P' dr (1) 
 
 in which d^ is the flux of R coming out of the volume element dr. 
 
 When a vector field flows into each small part of a region, the 
 divergence is negative. Negative divergence is sometimes called 
 convergence. 
 
 The flux $ of a vector field across a surface is a scalar quantity, 
 as is evident from the discussion of equation (1) of Art. 125. 
 
 (A<l>\ 
 — 1 
 
 is a scalar quantity; and since a vector field has a definite diver- 
 gence at each point (of course the divergence may be zero) of 
 space, it is evident that the divergence of a vector function (a 
 distributed vector) is a scalar function (a distributed scalar). 
 
 Cartesian expression for divergence. — Consider a small cube of 
 which the edges are dx, dy and dz as shown in Fig. 111. Let 
 Xj Y and Z be the components of the given distributed vector R 
 
 A€» 
 * The actual proof that — has a definite limiting value may be estab- 
 lished without great difficulty by considering the portion v" of the linear 
 vector field in Art. 133. 
 
230 
 
 CALCULUS. 
 
 at the point p. Then the flux of R into the cube across the 
 face a is X * dy ' dzj because X is the component of R normal 
 to the face a and dy • dz is the area of face a. If X is the 
 
 ic-component of R at any point of the face a, then f X + -7- * dx\ 
 
 
 z-axis 
 
 ^y^r- 
 
 p^ 
 
 
 
 JL. 
 
 
 ^^#^ 
 
 
 
 1 
 
 1 
 \' 
 
 1 
 
 X-axis 
 
 yy-axi 
 
 B 
 
 X 
 
 1 
 
 / 
 
 dx 
 
 Fig. 111. 
 
 is the a;-component of R at the corresponding point of the face 6, 
 and if X - dy - dz is the flux into the cube across face a then 
 
 X + ^ ' dx) ' dy ' dz is the flux out of the cube across face &.* 
 
 Therefore the flux of R out of the cube across face h exceeds the 
 
 jy 
 
 flux of R m^o the cube across face a by the amount -7- -dx'dy'dz. 
 
 In the same manner it may be shown that the net flux of R out of 
 the cube across the two faces which are perpendicular to the y-axis 
 
 is -T~ ' dx • dy ' dz; and that the net flux oui of the cube across 
 
 the two faces which are perpendicular to the 2-axis is -r 'dx-dy- dz. 
 Therefore the total flux out of the cube is: 
 
 ,^ /dX,dY^dZ\ , , , 
 ^^=Kdx-^dy^-dz)' <^^'dy'dz 
 
 (2) 
 
 *The argument here given is not entirely rigorous, and the peculiar 
 wording of this sentence is intended to suggest a form of argument which is 
 rigorous. 
 
VECTOR ANALYSIS. 231 
 
 But dx • dy ' dz is the volume dr oi the cube. Therefore, using 
 equation (1) we have: 
 
 . dX.dY.dZ .^. 
 
 PROBLEMS. 
 
 1. All space is filled with a fluid moving parallel to the a;-axis 
 of reference at a uniform velocity of 10 centimeters per second. 
 Find an expression for the velocity potential of the fluid. Ans. 
 xp = 10a; + any constant. 
 
 2. The velocity components of a moving fluid parallel to the 
 axes of reference are everywhere equal to a, h and c respectively. 
 Find the velocity potential. Ans. rj/ = ax + hy -\- cz + any con- 
 stant. 
 
 3. The velocity components of a fluid parallel to the axes of 
 reference are ax, hy and cz, respectively. Find the velocity 
 potential. Ans. \J/ = \ax^ + \hy'^ + \cz^ + any constant. 
 
 4. A viscous fluid flowing over a plane has a velocity which is 
 everywhere given by the equation X = ay as shown in Fig. p4. 
 Show that no velocity potential exists. 
 
 y-axi8 
 
 *■ i ,> X a* gj/ 
 
 Jub:. 
 
 x-axi8 
 
 Fig. p4. 
 
 5. A vessel of water rotates uniformly at a speed of two revo- 
 lutions per second about a vertical axis (the 2-axis). Derive 
 expressions for the velocity components of the moving water and 
 show that the velocity has no potential. 
 
 6. Liquid flows over an infinite plane towards a circular spot 
 
232 CALCULUS. 
 
 20 centimeters in radius where the liquid leaks through the plane 
 at the rate of two cubic centimeters per second for each square 
 centimeter of the leaky portion of the plane. The depth of the 
 liquid is everywhere 10 centimeters. Choose the x and 2/-axes 
 of reference in the plane with the origin at the center of the leaky 
 portion, and derive expressions for the x and 2/-components of 
 the fluid velocity in the region over the leaky portion. Ans. x- 
 
 component = r^:. 
 
 Note. — In this and the following problems consider only the horizontal 
 part of the motion and assume the velocity to be the same from top to bottom 
 of the liquid. 
 
 7. Find an expression for the velocity potential in the region 
 
 x^ -I— ifi 
 over the leaky portion. Ans. — ^ — h any constant. 
 
 8. Derive expressions for the x and y components of the 
 
 fluid velocity in the region outside of the leaky portion. Ans. x- 
 
 40x 
 component = -^x~2- 
 
 9. Find an expression for the velocity potential in the region 
 outside of the leaky portion. Ans. 20 logc (x^ -f y"^) + any constant. 
 
 10. Determine the constants of integration involved in the 
 answers to problems 7 and 9 on the assumption that the velocity 
 potential is zero at the edge of the leaky spot. Ans. (a) — 20, 
 (6) - 20 log. 400. 
 
 11. A straight "fence" 20 centimeters long and 10 centimeters 
 high is placed in the layer of moving liquid on the plane in 
 problem 6. Find the surface integral of the fluid velocity over 
 the fence (a) when the ends of the fence are at a distance of Vl25 
 centimeters from the ce nter of the leaky spot, and (6) when the 
 ends of the fence are V 1,000 centimeters from the center of the 
 leaky spot. Ans. (a) 100 cubic centimeters per second, (6) 800 
 tan~i 0.333 cubic centimeters per second. 
 
 12. Find the divergence of the fluid velocity in problem 6 (a) 
 over the leaky spot and (6) in the region outside of the leaky spot. 
 
VECTOR ANALYSIS. 
 
 233 
 
 Ans. (a) 0.2 cubic centimeter per second per cubic centimeter, (h) 
 zero. 
 
 13. Find the divergence of the fluid velocity specified in prob- 
 lem 3. Ans. a -f- 6 + c; 
 
 14. One end of a rubber band is fixed, and the band is stretched 
 by moving the other end of the band at a velocity of 2 centimeters 
 per second. The band is 24 centimeters long at a given instant. 
 Find the divergence of v where v is the distributed velocity of 
 the band. Assume that the band does not contract laterally. Ans. 
 0.0833 cubic centimeters per second per cubic centimeter. 
 
 15. Find the distribution of pressure in a tank of water rotating 
 at a speed of one revolution per second (w = 27r radians per 
 second), the density of water being 62.5 pounds per cubic foot. 
 Ans. p = 123.3 r^ where p is the pressure in poundals per square 
 foot at a point r feet from the axis of rotation. 
 
 Note. — Consider an element of the rotating liquid as indicated by the 
 shaded area in Fig. pl5, the dimen- 
 sion perpendicular to the plane of the 
 paper being I. The area of side a is 
 Ir . do and the outward push on a is 
 plr . do. The area of the side h is 
 Z(r 4- dr) . do, and the pressure at this 
 face is (p + dp) ; therefore, dropping 
 infinitesimals of the third order, the 
 force pushing inwards on face h is plr 
 . do -{- pl-dr .do +lr .pd.dd. The 
 area of faces c and e is I . dr and 
 the pressure over these faces may be 
 considered as equal to p; therefore the 
 normal forces on faces c and e are 
 pl.dr as shown. The outward compo- 
 nent of these two forces, according to 
 
 Fig. pl5. 
 
 Art. 51, is 
 
 pi . dr 
 
 Xr 
 
 (omitting infinitesimals of the third order). 
 
 Therefore the net inward force due to pressure acting on the element of liquid 
 is Ir . dp . de, and this must be equal to the product of the mass Dlr . dd . dr 
 of the element of liquid and its radial acceleration coV. 
 
 16. Find the expressions for the pressure gradients in the 
 
234 CALCULUS. 
 
 rotating liquid of problem 15 in the directions of x and y axes 
 lying in a plane at right angles to the axis of rotation. Ans. 
 X = 246.6X, Y = 246.6y. 
 
 17. Find the divergence of the pressure gradient in the rotating 
 fluid of problem 15. Ans. 493.2. 
 
 127. Reduction of a surface integral over a closed surface to a 
 volume integral extended throughout the enclosed region. — 
 A very important transformation in the theory of electricity and 
 magnetism is the reduction of a surface integral to a volume 
 integral or vice versa. Let J? be a distributed vector, and let 
 us consider its surface integral over a closed surface (normal 
 directed outwards). Imagine the entire enclosed region to be 
 broken up into small cells Uke the individual bubbles in a mass of 
 foam. Then the integral of R over the hounding surface of the region 
 is equal to the sum of the integrals of R over the hounding surfaces of 
 the individuxil cells (normal directed outwards in each case). This 
 proposition is evident when we consider that every wall which 
 separates two cells is integrated over twice with directions of 
 normal opposite (see Art. 125), so that the surface integrals over 
 dividing walls are thus cancelled. The only surface integrals 
 which are not thus cancelled are the integrals over the parts of 
 the surface which bounds the region. 
 
 Let d$ be the surface integral over one of the cells (the flux 
 of R out of the cell), and let Jr cos e - dS be the surface 
 mtegral of R over the bounding surface of the enclosed region. 
 Then from the above proposition we have; 
 
 fR cos €' dS = fd^ (1) 
 
 But d^ = p ' dr according to equation (1) of Art. 126, so that 
 equation (1) becomes: 
 
 Jr cos € • dS = fp ' dr (2) 
 
 This equation may be expressed in words as follows : The integral 
 of a distributed vector R over a closed surface is equal to the integral 
 
VECTOR ANALYSIS. 235 
 
 of the divergence of R throughout the enclosed region. The first 
 member of (2) is of course a surface integral and the second 
 member is a volume integral. 
 
 128. Solenoidal vector fields. — A vector field is said to be 
 solenoidal when its divergence is everywhere zero. The surface 
 integral of such a vector field over any closed surface is zero 
 according to equation (2) of Art. 127. 
 
 Tube of flow. — Imagine stream lines to be drawn in a vector 
 field from each point of the periphery of a closed curve or loop. 
 These stream lines form a tubular surface which is called a tube 
 of flow of the given distributed vector. Consider a number of 
 diaphragms which stretch across a tube of flow. The flux is the 
 same across each diaphragm. This is evident when we consider 
 that any two diaphragms, together with the walls of the tube, 
 constitute a closed surface out of which the total flux must be 
 zero because the distributed vector under consideration is assumed 
 to be solenoidal; but the flux across the walls of the tube is zero 
 because the vector field is everywhere parallel to the walls. There- 
 fore the flux into the enclosed space across one diaphragm must 
 be equal to the flux out of the enclosed space across the other 
 diaphragm. 
 
 Unit tube. — A tube of flow is called a unit tube when the flux 
 through the tube is unity. Thus, for example, a unit tube has a 
 sectional area of 0.1 of a square inch at a point in a moving fluid 
 where the velocity of the fluid is 10 inches per second. A unit 
 tube has a sectional area of 0.1 of a square centimeter at a point 
 in a magnetic fleld where the intensity of the field is 10 gausses. 
 Imagine the solenoidal region of a distributed vector to be divided 
 up into unit tubes. Then the flux across any surface anywhere 
 in the region will be equal to the number of these unit tubes which 
 pass through the surface. Each unit tube may be conveniently 
 represented in imagination by the single stream line along the axis 
 of the tube. When stream lines are drawn in this way, so that 
 each stream line represents a unit tube, then the flux across any 
 
236 CALCULUS. 
 
 surface in the vector field is equal to the number of stream Unes 
 which pass through the surface. The lines of force in a magnetic 
 or electric field are always thought of as being drawn so that each 
 line represents a unit tube, and the quantity of magnetic flux or 
 electric flux through a surface is expressed by the number of lines 
 of force which pass through the surface. 
 
 129. Curl of a distributed vector. — In Art. 126 examples were 
 given of vector fields which emanate or "flow out" from a region 
 or from every small part of a region. There are important cases 
 in which a vector field curls round a region or round every small 
 part of a region. Thus the stream lines in a rotating bowl of water 
 curl round the central portions of the bowl. The magnetic field 
 due to an electric wire curls round the wire. The electric field 
 induced by an iron rod while the rod is being magnetized curls 
 round the rod. 
 
 Consider a small plane area AS at a point p in a. vector field. 
 Let AL be the fine integral of the distributed vector R round the 
 boundary of this element of area. It can be shown,* when R is 
 a continuous function of the coordinates Xy y and z, that the 
 
 ratio -r-s approaches a definite limiting value as the element 
 
 of area AS grows smaller and smaller. This limiting value, -r^ 
 
 is the component of a new vectorf C in the direction of the normal 
 to dSf and this new vector C is called the curl of the given 
 vector R at the point p. From this definition we have: 
 
 dL = C cose- dS (1) 
 
 where dL is the line integral of R round the boundary of a 
 
 * The actual proof that — „ approaches a definite limiting value may be 
 
 established without great difficulty by considering the part v' of the linear 
 vector field in Art. 133; see Art. 135. 
 t See Art. 135. 
 
VECTOR ANALYSIS. 237 
 
 small plane element of area dS, C is the curl of R at the element 
 of area, and e is the angle between C and the normal to dS. 
 
 Example of curl. — Consider a uniformly rotating bowl of water, 
 the speed of rotation being oi radians per second. Consider the 
 character of the fluid motion in the immediate neighborhood of 
 the point p distant D from the axis of rotation of the bowl as 
 shown in Fig. 112. The motion of a small portion of the water 
 near p may be thought of as a combination of (1) A motion of 
 translation at velocity Deo, and (h) A simple motion of rotation 
 at a speed of w radians per second about an axis through the 
 point p.* Now in considering the line integral of the fluid 
 velocity around the circle cc, the motion of translation evidently 
 need not be considered. Let r be the radius of the circle cc. 
 The velocity of the fluid at every point of the circumference of cc 
 (rotatory motion only being considered) is cor and it is everywhere 
 in the direction of the circumference. There- 
 fore the angle € is zero and the expression 
 for the line integral around the circle reduces 
 to circumference X (>ir. Therefore AL = 
 27rr2aj, where AL is the line integral of the 
 fluid velocity around the circle cc. But the 
 area of the circle is dS = ttt^. Therefore from 
 equation (1) we get: C = 2aj. That is, the 
 curl of the fluid velocity in the rotating bowl is Yig. 112. 
 
 equal to two times the speed of the bowl in radi- 
 ans per second; and since the expression for C does not contain D 
 it is evident that C has the same value everywhere in the bowl. 
 
 Now the angular velocity co of the bowl is a vector and its 
 vector direction is parallel to the axis of rotation and towards the 
 reader in Fig. 112. Therefore C is a vector, and it is towards the 
 reader at every point in Fig. 112. The "stream lines" of the 
 vector C are straight lines parallel to the axis of rotation of the 
 bowl. 
 
 * See Franklin and MacNutt's Mechanics and Heat, pages 174-175, The 
 Macmillan Co., New York, 1910. 
 
238 
 
 CALCULUS. 
 
 Cartesian expression for curl. — Consider a small rectangular 
 plane area of which the edges are dy and dz, as shown in Figs. 
 113 and 114. Let X, Y and Z be the components at p of the 
 
 y-axia 
 
 z-axU 
 
 y-axi$ 
 
 Q 
 
 Z•^^'f 
 
 z ^ 
 
 z-axia 
 
 Fig. 113. 
 
 Fig. 114. 
 
 given distributed vector R. To find the line integral of R around 
 the small rectangle in the direction of the small curved arrow, let 
 us consider first th^ two edges a and 6 in Fig. 113. The line 
 integral along the edge a is — Y • dy because the component Y 
 is counter to the direction of the small curved arrow. If Y is 
 the i/-component of R at any point of the edge a, then the 
 2/-component of R at the corresponding point of the edge h 
 
 dz; and if the line 
 
 being -3- • dz greater, is equal to F + , 
 az dz 
 
 integral of R along edge a is — Y ' dy, then the line integral 
 of R along the edge h is + (y -{- -^-dzj-dy* Therefore the 
 net value of the line integral along the two edges a and b is 
 -f ' dy » dz. In the same way it may be shown from Fig. 114 
 that the net line integral (in the direction of the small curved 
 arrow) along the edges e and /is — -r- ' dy - dz. Therefore 
 
 dZ , 
 
 -Ty'^y 
 
 * The argument here given is not entirely rigorous, and the peculiar wording 
 of this sentence is intended to suggest a slightly modified argument which is 
 rigorous. 
 
VECTOR ANALYSIS. 239 
 
 the total line integral dL of R around the small rectangle is: 
 
 But dy ' dz is the area dS of the rectangle. Therefore using 
 equation (1) and remembering that the x-component of the curl 
 is at right angles to the rectangle in Figs. 113 and 114, we have: 
 
 ^^- dz~ dy ^^^ 
 
 In a similar manner we may get: 
 
 ^'~ dx dz ^^^ 
 
 and 
 
 ^-g-S («) 
 
 where d, Cy and C z are respectively the x^ y and z com- 
 ponents of the curl of the distributed vector R whose x, y and z 
 components are X, Y and Z respectively. 
 
 The curl of a gradient is necessarily equal to zero. — This is 
 evident when we consider that equations (3), (4) and (5) of Art. 
 124 are satisfied in a vector field when the vector is a gradient. 
 
 130. Reduction of a line integral around a closed loop to a 
 surface integral over a diaphragm bounded by the loop. — A very 
 important transformation in the theory of electricity and mag- 
 netism is the transformation of a line integral to a surface integral 
 or vice versa. Let E be a distributed vector, and let us consider 
 its fine integral around a closed curve or loop AB, Fig. 115, the 
 heavy arrow showing the direction in which the line integral is 
 taken. Imagine a diaphragm of any shape whatever stretched 
 across the loop AB, and imagine this diaphragm to be divided 
 up into small meshes as if the diaphragm were made of wire 
 
240 
 
 CALCULUS. 
 
 gauze. Then the integral of R round the loop AB is equal to 
 the sum of the integrals of R round the individuul meshes of the 
 diaphragm, the integrals round the meshes being taken in the 
 
 same direction as the integral round 
 the loop as indicated by the small 
 curled arrows. This proposition is 
 evident when we consider that every 
 line dividing two meshes is inte- 
 grated over twice in opposite direc- 
 tions (see Art. 123), and when we 
 consider that in integrating round 
 the various meshes we eventually 
 integrate along every portion of the 
 loop AB once in the direction of the heavy arrow in Fig. 115. 
 
 Let dL be the line integral of R round one of the meshes, 
 and let jR cos e - ds be the Hne integral of R roujid the loop 
 AB, Then from the above proposition we have: 
 
 Fig. 115. 
 
 J R cos € ' ds = f dL 
 
 (1) 
 
 But dL = C cos € ' dS, according to equation (1) of Art. 129, so | 
 that equation (1) becomes: ■ 
 
 J R cos e ' ds = J C cos e • dS 
 
 (2) 
 
 This equation may be expressed in words as follows: The integral 
 of a distributed vector R round a closed loop is eqvxil to the integral 
 of the curl of R over any diaphragm stretched across the loop. The 
 first member of (2) is of course a line integral and the second 
 member is a surface integral. 
 
 The divergence of the curl of a distributed vector is always and 
 everjnvhere equal to zero. — Consider two diaphragms stretched 
 across a closed loop. The integral of C is the same over both of 
 these diaphragms because each surface integral is equal to the 
 line integral of R round the loop. If the direction of the normal 
 
VECTOR ANALYSIS. 241 
 
 be reversed in the integral over one of the diaphragms the integral 
 will be reversed in sign (see Art. 125). Therefore the surface 
 integral of C over a closed surface, namely, the two diaphragms, 
 with normal directed outwards, is equal to zero. That is C is a 
 vector whose flux out of or into a closed surface is always equal to 
 zero. Therefore C is a solenoidal vector and its divergence is 
 everywhere zero. See Art. 127. 
 
 131. Vector potential. — Let us imagine a distributed vector of 
 which a given distributed vector is the curl. Then the imagined 
 distributed vector is called the vector potential of the given distri- 
 buted vector. The divergence of a curl is always and everywhere 
 necessarily equal to zero, as explained in the previous article. 
 Therefore, to have a vector potential, a given distributed vector 
 
 r. . dX , dY , dZ . _^, ,. 
 must have no divergence. But ^ — \- -^ — \- -f~ is the divergence 
 
 of a distributed vector whose components at a point are X, Y 
 and Z, according to equation (3) of Art. 126. Therefore the 
 existence of a vector potential (of a given distributed vector) is 
 determined by the condition; 
 
 dx^ dy^ dz " ^ ^ 
 
 where X, Y and Z are the components at a point of the given 
 distributed vector. 
 
 Example. — In certain cases of fluid motion each particle of the 
 fluid is rotating at a definite speed about a definite axis. This 
 spinning motion of the particles of a fluid is a distributed vector* 
 and it is called vortex motion. Now it was shown in Art. 129, 
 that the curl of the velocity of the water at a point p in a rotating 
 bowl is equal to two times the spin velocity w of the particle of 
 water at p. That is, ignoring the factor 2, the curl of fluid velocity 
 
 * Spin is a vector quantity, its vector direction being along the axis of 
 spin in the direction in which a right-handed screw would travel if turned in 
 the direction of the spin. 
 17 
 
242 CALCULUS. 
 
 IS the vortex motion; therefore the vortex motion of a fluid is a 
 distributed vector whose vector potential is the fluid velocity. 
 In this case the vector potential (of vortex motion) is physically 
 existent; in general, however, the vector potential of a given 
 distributed vector is merely imagined to exist. The idea of vector 
 potential is useful because it is sometimes helpful to think of a 
 given distributed vector as a curl. 
 
 132. Rotational and irrotational vector fields. Vector potential 
 and scalar potential. — ^When the velocity of a moving fluid has 
 curl the particles of the fluid are in rotation. In consequence of 
 this fact any vector field which has curl is called a rotational vector ( ^ 
 field. A vector field which has no curl is called an irrotational 
 vector fiM. 
 
 In an irrotational vector field equations (3), (4) and (5) of 
 Art. 129 reduce to equations (3), (4) and (5) of Art. 124. There- 
 fore an irrotational vector field can be thought of as a gradient, 
 or, in other words, an irrotational vector field has a potential (a 
 scalar potential). 
 
 In a rotational vector field equations (3), (4) and (5) of Art. 124 
 are not satisfied, and therefore a rotational vector field cannot be 
 thought of as a gradient, or, in other words, a rotational vector 
 field has no potential (no scalar potential). 
 
 A vector field which has no divergence is called a solenoidal vector 
 fiM, and a solenoidal vector field can be thought of as a curl, or, 
 in other words, a solenoidal vector field has a vector potential. 
 
 A vector field which has divergence is called a non-solenoidal 
 vector field, and such a vector field cannot be thought of as a 
 curl, or, in other words, a non-solenoidal vector field has no vector 
 potential. 
 
 PROBLEMS. 
 
 1 . A viscous liquid moves over the xy plane so that X = az. 
 Find the curl. Ans: — a. 
 
 2. Find the vector potential of fluid velocity when the fluid 
 is everywhere moving in the direction of the 2-axis at a velocity 
 
VECTOR ANALYSIS. 243 
 
 of + 100 centimeters per second. Ans. IOO2! centimeters squared 
 per second squared. 
 
 Note. — A distributed vector which has neither divergence nor curl in a 
 given region may have a scalar potential and it may have a vector potential. 
 
 133. Character of a very small portion of any vector field. — 
 
 Let X, Y and Z be the components of any distributed vector v 
 at a point p whose coordinates are x, y and z. Then X, Y 
 and Z are functions of x, y and 2, and indeed we will assume 
 them to be continuous functions. If one is to reach a clear 
 understanding of divergence and curl it is necessary to examine 
 carefully into the manner of distribution of v in a very small 
 region near a chosen point. It is most convenient to locate the 
 origin of coordinates at the chosen point. Then the coordinates 
 X, y and z of any point in the very small region are infinitesimals, 
 and the squares and products and higher powers of x, y and z 
 are negligible. 
 
 Now the components X, Y and Z being continuous functions 
 of X, y and z can be expanded by Maclaurin's theorem as ex- 
 plained in Art. 90, and all terms in these expansions which contain 
 squares and products and higher powers of x, y and z may be 
 discarded, giving: 
 
 X = Xo + aix + a2y + a^z (1) 
 
 F = Fo + 61X + h,y + hz (2) 
 
 and 
 
 Z = Zo + CiX + C22/ + Cziz (3) 
 
 in which the constant coefficients are the values of the derivatives 
 at the origin, as shown in the following schedule : 
 
 (I) 
 
 dX 
 dx 
 
 dX 
 
 "'-dy 
 
 dX 
 
 "'- dz 
 
 ^'- dx 
 
 ^' = dy 
 
 dY 
 ^'= dz 
 
 dZ 
 
 dZ 
 
 '^-dy 
 
 dZ 
 
244 CALCULUS. 
 
 The " linear " vector field. — The simplest type of vector field 
 is the homogeneous field; in such a field X has the same value 
 throughout the field, Y has the same value throughout the field, 
 and Z has the same value throughout the field. When X, Y 
 and Z are linear functions of x, y and z we have what is called 
 a linear vector field. Thus equations (1), (2) and (3) represent 
 a linear vector field. 
 
 Resolution of a linear vector field into simple parts. — The dis- 
 cussion of divergence and curl in Arts. 126 and 129 is not rigorous, 
 and, as is always the case in a merely plausible discussion, the 
 harder one tries to understand it the more vague and unintelligible 
 it becomes. It is now proposed, therefore, to establish rigorously 
 the ideas of divergence and curl by considering a linear vector 
 field as expressed by equations (1), (2) and (3). For this purpose 
 it is necessary to resolve the given linear vector field into three 
 parts, and to be able to speak of these parts intelligibly let us 
 think of the given linear vector field as the velocity t; of a moving 
 fluid, the components of v being X, Y and Z as given in equa- 
 tions (1), (2) and (3). The three parts into which v is to be 
 resolved are: 
 
 1. A uniform translatory motion of the entire body of fluid. 
 This part of v will be represented by Vo and its components are 
 Xo, Yq and Zq. 
 
 2. A simple motion of rotation of the entire body of fluid. 
 This part of v will be represented by v' and its components will 
 be represented by X', Y' and Z', 
 
 3. A continuous stretching of the entire body of fluid in three 
 mutually perpendicular directions. This part of v will be repre- 
 sented by v" and its components will be represented by X", 
 Y" and Z", 
 
 Discussion of v'. — Consider a rotating body of which the axis 
 of rotation passes through the origin of coordinates, and let cox, 
 03 y and 0)2 be the components of the angular velocity of the 
 body around the x, y and z axes of reference, respectively. 
 
VECTOR ANALYSIS. 
 
 245 
 
 Consider a point p of the body of which the coordinates are 
 X, y and z as indicated in Figs. 116, 117 and 118. The velocity 
 
 Fig. 116. 
 
 Fig. 117. 
 
 Fig. 118. 
 
 of p due to CO a; consists of two components yo)x and — zccx as 
 indicated in Fig. 116, and similar statements may be made con- 
 cerning Figs. 117 and 118. Therefore, picking out the x-compo- 
 nents of the velocity of p in Figs. 117 and 118, we get the value 
 of X', and in a similar manner we get expressions for Y' and Z' 
 as follows: 
 
 X' = " ' — oi^y + ojyZ (4) 
 
 y = + 0)eX • • • — OiixZ (5) 
 
 Z' = — 03yX + £0x2/ 
 
 (6) 
 
 The values of cox, coj, and co^, expressed in terms of the coefficients 
 in equations (1), (2) and (3), can best be determined after the 
 expressions for v" have been formulated. 
 
 Discussion of v''. — As in case of v' it is necessary to find the 
 forms of expressions which give X'', F'' and Z'\ For this 
 purpose consider three mutually perpendicular axes of reference 
 Xi, yi and Zi. Then a continuous stretch of the portion of fluid 
 parallel to the Xi-axis is represented by the equation Xi = axi, 
 where Xi is a velocity parallel to the rci-axis and a is a constant. 
 Analogous expressions using j8 and y as constants represent 
 continuous stretches parallel to the 2/1 and Zi axes. Therefore 
 
246 CALCULUS. 
 
 continuous stretches in three mutually perpendicular directions 
 (which directions are for the moment chosen as the axes of refer- 
 ence) are expressed by the equations: 
 
 Xi = axi (7) 
 
 1^1 = Pyi (8) 
 
 and 
 
 Zi = 721 (9) 
 
 It is desired to transform these equations so as to express 
 exactly the same fluid motion, but to express it by giving the com- 
 ponents of v" parallel to new axes a;, y and z, and as functions 
 of the new coordinates x,y and z. This transformation is enorm- 
 ously simplified by expressing the given fluid motion [equations 
 (7), (8) and (9)] in terms of its velocity potential P, which is a 
 
 distributed scalar whose Xi-gradient is 3— = axu whose 2/1-gradi- 
 
 axi 
 
 ent is ^ = Pyi and whose 2i-gradient is -p = yzi. Integrating 
 
 these three differential equations and ignoring the constant* of 
 integration we get: 
 
 P = IciXr'^ + \W + \W (10) 
 
 Now to transform equations (7), (8) and (9) as stated above we 
 need only to substitute in equation (10) the following values 
 for Xif 2/1 and 2i [see equations (11), (12) and (13)], and then 
 find the gradients of P in the directions of the new axes. In 
 this way we get equations (14), (15) and (16). 
 
 Xi = hx + m\y -\- niZ (11) 
 
 2/1 = Ux + rn^y -f n2.z (12) 
 
 Zi = I2PC -{- mzy + n^ (13) 
 
 where U mi ni, U W2 ^2 and h rriz riz are the direction cosines of 
 
 * The three equations are partial differential equations, but the functions 
 of integration reduce to a single imdetennined constant. 
 
 I 
 
VECTOR ANALYSIS. 247 
 
 the new x, y and z axes referred to the old Xi, 2/1 and Zi axes. 
 
 X" =fx + py + qz (14) 
 
 F' = px + gy + rz (15) 
 
 Z" = qx + ry-\-hz (16) 
 
 The coefficients in these equations are of course expressions involv- 
 ing a, jS and 7 and the direction cosines in equations (11), (12) 
 and (13), but they are represented by the single letters /, g, h, 
 p, q and r for the sake of clearness. Anyway, the only im- 
 portant feature about equations (14), (15), and (16) is the sym- 
 metry of the coefficien s which means that X'\ Y" and Z" are the 
 components of a fluid motion consisting of three continuous 
 mutually perpendicular stretches, because this degree of symmetry 
 is the only necessary consequence of equations (7), (8) and (9). 
 
 Determination of coefficients in equations (4), (5) and (6) and 
 in equations (14), (15) and (16) in terms of the coefficients in 
 equations (1), (2) and (3). — Equations (4), (5) and (6) and (14), 
 (15) and (16) show only the degree of symmetry that the co- 
 efficients must have in order that equations (4), (5) and (6) may 
 represent a simple rotation and in order that equations (14), 
 (15) and (16) may represent three continuous mutually perpendicu- 
 lar stretches, and the coefficients in equations (4), (5) and (6) 
 and (14), (15) and (16) may be thought of as undetermined. 
 It remains to determine these coefficients cox, co^, co^, /, g, h, p, q 
 and r, so that Vo, v' and v" may be component parts of the 
 given fluid velocity v which is represented by equations (1), 
 (2) and (3). To do this add equations (4) and (14) and place the 
 coefficients of x, y and z in the resulting equations equal to the 
 coefficients of x, y and 2, respectively, in equation (1); and 
 proceed in a similar manner with the other pairs of equations. 
 In this way we get the following schedule of equations: 
 
 f = ai p — coz = az q + o)y = as^ 
 
 P + 03z = hi g = ^2 r - COx = 63 p (II) 
 
 q — 03y = Ci r + Wx = C2 h = C3J 
 
248 CALCULUS. 
 
 from which we get the following values: 
 
 w, = i(c2 — hz) 
 o)y — \{az — ci) 
 
 Or, using the values of ai, a^, 03, 61, 62, etc., from the schedule I 
 
 of equations, we have: 
 
 
 dZ dY 
 
 2-^ = dy dz 
 
 (17) 
 
 dX dZ 
 
 2co« = -3 3- 
 
 ^ dz dx 
 
 (18) 
 
 _ dY dX 
 2"^ - dx dy 
 
 (19) 
 
 and from equations II we get also: 
 
 
 dX 
 ■' dx 
 
 (20) 
 
 dY 
 ^~ dy 
 
 (21) 
 
 dz 
 
 (22) 
 
 P = ^{dx + dy) 
 
 (23) 
 
 JdX , dZ\ 
 ^ = Kdz+dx) 
 
 (24) 
 
 ,/dZ , dY\ 
 
 (25) 
 
 PROBLEMS. 
 
 1. Find the flux of v" out of a rectangular parallelopiped 
 which is I feet long in the direction of the a:i-axis, w feet wide 
 in the direction of the 2/1-axis, and t feet thick in the direction 
 of the 2i-axis, the constants in equations (7), (8) and (9) being 
 expressed in reciprocal seconds. Ans. lwt(a + 18 + 7) cubic feet 
 per second. 
 
VECTOR ANALYSIS. 249 
 
 2. Find the divergence of v" , Ans. a + /S + 7- 
 
 "Note. — It is here intended that the student find the flux $ which comes 
 out of a given volume t, and then determine the limit of - as t approaches 
 zero. It is simplest to take for r the rectangular parallelopiped of problem 1. 
 
 3. Show that the integral of v" around any closed curve what- 
 ever is zero. 
 
 ^ote. — Choose the axes of reference as in equations (7), (8) and (9). Let 
 ds be an element of the closed path or curve. Then the scalar part of the 
 product v" . ds can easily be found, and it is easy to show that the integral 
 of this scalar product aroimd the closed curve is zero. 
 
 4. It is evident that the velocity v\ which is a simple motion 
 of rotation, has zero flux into or out of any closed region whatever. 
 Therefore the divergence of v' is everywhere equal to zero. Find 
 the line integral of v' around the ellipse of Fig. 122, the radius 
 of the cylinder being r, the axis of the cylinder being parallel to 
 the axis of rotation of the fluid, and the angular velocity of rotation 
 being co. Ans. l-KV^iti. 
 
 134. Proof that ^ -j- and -i, -r- are identical when z is 
 
 dy • dx dx • dy 
 
 a continuous function of x and y. — This proposition may be 
 
 stated so as to appeal to one's geometric sense as follows: Let z 
 
 be a function of x and y and let this function be represented 
 
 by a hill built upon the xy plane. Then z is the height of this 
 
 hill above the point p' in the base plane, x and y being the 
 
 coordinates of p' as shown in Figs. 41a and 416 in Art. 62. Let 
 
 x(^-p\ be the a;-component of the slope of the hill at the point p 
 
 (see Figs. 41a and 416), and let ^( = ^) be the ^/-component 
 
 of the slope of the hill at p. 
 
 Figs. 119 and 120 represent a top view of the hill. Let dz be 
 the difference of level of the points p and q on the hill in Figs. 
 
250 
 
 CALCULUS. 
 
 119 and 120. Let us travel from p to g along the sides a and h 
 
 x^axU 
 
 dx 
 
 />>. 
 
 <^-»e 
 
 r X 
 
 e 9 A X-^^'dy 
 
 IHveu 
 
 
 ^X-axis 
 
 AS 
 
 dx^ 
 
 < 
 
 y-axl 
 
 "a 
 
 c b 
 
 
 Y 
 
 8 
 
 ^^gM 
 
 Fig. 119. 
 
 Fisc. 120. 
 
 of the infinitesimal square, and the rise will be; 
 
 dz = X • rfa; + F + 
 
 dY 
 
 y ' dx 
 
 dx] ' dy 
 
 Let us travel from p to q along the sides c and e of the in- 
 finitesimal square, and the rise will be: 
 
 dz = Y ' dy ■{- (x -\-^ ' d^ ' dx 
 
 Now if z is a continuous function of x and y these two expressions 
 for dz must be identical, that is, one must rise by the same amount 
 in going along a and 6 as in going along c and e in Figs. 119 
 
 { 
 
 and 120. Therefore 
 dz 
 
 dY ^ , 1 . ^^ 
 
 -J- must be equal to -y- . 
 dx ^ dy 
 
 But 
 
 X = — 
 dx 
 
 and 7 = 
 
 dy' 
 
 Therefore 
 
 ^z ., , ^ d^z 
 
 -3 -J- must be equal to -; 
 
 dy ' dx dx 
 
 dy 
 
 The above argument is not entirely rigorous but it brings out very cleariy 
 the geometrical significance of equations (3), (4) and (5) of Art. 124. To 
 make the argument rigorous a slightly altered point of view suffices, as follows: 
 The gradient of the hill at any point in the side e of the infinitesimal square 
 exceeds the gradient at the corresponding point in the side a by the amount 
 dX 
 
 dy 
 
 dy, so that the rise along e exceeds the rise along a by the amount 
 
VECTOR ANALYSIS. 
 
 251 
 
 ( 7P • % ) • ^^- Similarly the rise along side b exceeds the rise along side c 
 
 by the amount (-j- » dx) ' dy. Therefore the rise along sides c and e 
 
 exceeds the rise along a and b by the amoimt l^ ^ J • dx • dy. But 
 
 the rise along c and d must be the same as the rise along a and 6. There- 
 
 fore 
 
 dX dY 
 
 -r- must be equal to zero. 
 
 dL 
 dS 
 
 135. To show that -^ is the resolved part of a vector (the 
 
 curl of R) in a direction normal to dS, where dL is the line integral 
 of R around the boundary of a surface element dS. — It can be 
 
 shown without difficulty that the two parts Vo and v" of the 
 Hnear vector field v of Art. 133 have zero Hne integrals around 
 any closed curve. Furthermore any distributed vector R may- 
 be looked upon as having a linear distribution throughout a 
 very small region. Therefore to establish the above proposition 
 it is sufficient to consider the portion v' of a linear vector field. 
 
 Now v' is a simple motion of rotation about a definite axis. 
 Let CO be the angular velocity of this rotation. The line integral 
 of v' around the circle cc in Fig. 121 is equal to 27rr^aj, so that 
 
 cylinder 
 
 Fig. 121. 
 
 normal to ^-^ ^ >^ 
 Fig. 122. 
 
 the line integral divided by the area of the circle is equal to 2a>, 
 as explained in Art. 129. Consider a cylindrical surface of which 
 the end view is the circle cc in Fig. 121. A side view of this 
 
262 CALCULUS. 
 
 cylinder is shown in Fig. 122. Let dS be the area of an oblique 
 plane section of the cylinder. Let dL be the Une integral of 
 v' around the boundary of dS. It can be shown without difl&culty 
 
 that dL is equal to 2Trr^o). But the area of dS is ;:. There- 
 cos 6 
 
 fore -75 is equal to 2co cos 6. That is, 2w is a vector whose 
 component in the direction of the normal to dS is equal to -75-. 
 
 136. The distortion of a small portion a body which has been 
 twisted or bent in any manner whatever. — Consider a particle p 
 of an unbent beam. Imagine the beam to be bent or twisted in 
 any manner whatever, and let 1; be a line drawn from p to p', 
 where p' is the position of the particle after the beam is bent. 
 Then the line v is a vector, it is called the displacement vector of 
 the particle p, and it may be thought of as being at p because 
 it refers to the point p. Now v has a definite value and a definite 
 direction at each point of the unbent beam, therefore it is a distri- 
 buted vector. That is the unbent beam may be thought of as a 
 vector field in its relation to the bent or twisted beam. 
 
 Consider a very small portion of the unbent beam. The 
 displacement vector v may be thought of as having a linear distri- 
 bution throughout this small portion of the beam as explained 
 in Art. 133. Therefore the displacement-vector field v may be 
 resolved into three parts as explained in Art. 133. One part Vq 
 is a simple translatory displacement of the entire small portion 
 of the beam; another part v' is a simple rotation of the small 
 portion of the beam about a definite axis; and a third part v" 
 consists of three mutually perpendicular stretches.* This is a 
 fundamental proposition in the theory of elasticity.! 
 
 * The expression cordinuom stretches as used in Art. 133 referred to what 
 takes place in a rubber band while it is being stretched, and the word stretch 
 as here used refers to what has taken place in a rubber band after it has been 
 stretched. 
 
 t See the chapter on elasticity in Franklin and MacNutt's Mechanics and 
 Heat, The Macmillan Co., New York, 1910. 
 
 I 
 I 
 
VECTOR ANALYSIS. 253 
 
 PROBLEM. 
 
 1. One end of a rubber band is fixed and the band is stretched 
 by moving the other end of the band. The initial length of the 
 band is 10 inches, the increase of length is 0.5 inch. Express the 
 displacement of each particle of the stretched band in terms of 
 the initial distance x of the particle from the fixed end of the 
 band, ignoring the lateral contraction of the band. Ans. Displace- 
 ment = O.OSic. 
 
APPENDIX A. 
 PROBLEMS GROUP 1. 
 
 (After Art. 34.) 
 Differentiate the following: 
 
 1. 2/ = 
 
 n+ 1' 
 
 dx 
 
 g = 20.3(. + l). 
 
 3. y = 6a;8 - 5x^ + 4^^ - Sx^ 
 dy 
 
 dx 
 
 4,y = x^x + 7), 
 
 5. t/ = 3a;'(a;2 - 5a;), 
 
 6. e = ar*(6r3 - cr + d), 7^b<*-^ 
 
 ^ = ar3(76r3 - 5cr + 4d). 
 ar 
 
 7. = (r + 3)(2r + 5), 
 
 ^^ ^ I 11 
 -=- = 4r + 11. 
 
 ar 
 
 8. e = (2r2 + 3r)(3r2 + 2r), 
 
 ^ = 3r(8r2 + 13r + 4). 
 
CALCULUS. 
 
 dr 
 
 ^ = (r2 - r + l)(5r2 - 3r + 1). 
 ar 
 
 37 = 2act -{- ae + be, 
 at 
 
 9. = (2 - r)(3 - r), 
 
 dr '' 
 
 10. = (r^ - r2 + r)(r2 - r + 1), 
 
 de ^ 
 
 dr 
 
 11. s = iat + h)(ct-\-e), 
 
 ds ^ 
 dt 
 
 12. « = (a - 60(c - eO, 
 
 ds _ 
 dt ~ 
 
 13. s = (3^ - 2< - 1)(1 - 3^2 4- 5^3)^ 
 
 14. 8 = (2 H- 0^ 
 
 15. 5 = (1 + 2x)^ 
 
 16. 8 = ^(2< 4- 3)«, 
 
 5? =26e<- (ae + bc). 
 at 
 
 5^ = 75<* - 76^3 + 3^ + 12^ - 2. 
 at 
 
 ^ = 3(2 + t)\ 
 
 ^ = 6(1 + 2x)\ 
 ax 
 
 I =2«(7< + 3)(2< + 3)^ 
 
 17. s = a«H6^ + c« + e), 
 
 ^ = a^(56^ + 4d + 3e). 
 
 18. 2/ = (1+ 2x)H2x2 + 1)\ 
 
 ^ = 2(1 + 2a;)2(2x2 + 1)3(3 + Sx + 22x2). 
 ax 
 
 19. 2/ = (ax + by{a + 6x)3, 
 
 ^ = (ax + 6) (a + 6x)2(5a6x + 2a2 + 36^). 
 ax 
 

 APPENDIX A 
 
 L. 
 
 2°-^ = x + 2' 
 
 
 A^: 
 
 
 dy 2 
 
 v- 
 
 
 da; {X + 2Y' 
 
 
 ^t CI — X 
 
 dy a — h 
 dx~ (b-xy 
 
 
 ^^- ^" (2x + 5)4' 
 
 dy 2(a; + 3)(2a; + 7) 
 
 
 dx {2x 
 
 + 5)» 
 
 23 .-(^^ + ^)' 
 
 t 
 
 
 z' 1 + a; \ 
 
 d^ _ {ax + bYiSae - 76c - 4aca;) 
 dx ~ (ex + ey 
 
 25. 
 
 26. 
 
 /2x- 1\ 
 
 /ax^ -hx + cY 
 
 dy _ 
 dx 
 
 3(1 + xy 
 
 {2x + 3)* 
 
 • 
 
 dy 
 
 12(2x - 
 
 -D' 
 
 dx- 
 
 (X- 
 
 2)' • 
 
 dy __ 5{ax^ — c)(ax'^ — hx -\- c) 
 dx e^x^ 
 
 27. Find the slope of the curve y = x^ — 2x^ + Z at (a) 
 x = 0, (6) X = 2, and (c) x = - 2. Ans. (a) 0, (6) 4, (c) - 20. 
 
 28. At what angles does the curve y = x{x — 2){x — 3) cut 
 the a;-axis? Ans. (a) At a; = 0, slope is 6, (6) at x = 2, slope is 
 — 2, (c) at a; = 3, slope is 3. 
 
 18 
 
4 CALCULUS. 
 
 2a; 4- 3 
 
 29. Is -. — r-^ an increasing or a decreasing function of x? 
 
 4x + o 
 
 Ans. Decreasing. 
 
 30. Is o _t A ^^ increasing or a decreasing function of «? 
 
 oX -p 4 
 
 Ans. Increasing. 
 
 31. When y = 2x^ — Qx + 5, for what values of x is t/ (o) 
 an increasing function, and (6) a decreasing function? Ans. (a) 
 All values of x less than — 1 or greater than + 1, (6) all values of 
 X between — 1 and + 1. 
 
 PROBLEMS GROUP 2. 
 
 (After Art. 89.) 
 
 Differentiate the following functions 
 
 Note. — The forms referred to in the answers to the following 
 problems are found in the table of integrals, see Appendix B. Thus 
 the answer to problem 4 is given by form 45, as 
 dy 1 
 
 dx 
 
 V^ 
 
 + hx 
 
 
 the answer to problem 5 is 
 
 found 
 
 to be 
 
 
 dy _ 
 
 x 
 
 =, etc. 
 
 
 l.V = ^-3x + l-l„ 
 
 
 dy _ 
 dx 
 
 2x-3-%-^. 
 
 2. J/ = 3V?-2V?, 
 
 
 dy 
 dx 
 
 
 2Vx' 5 
 
 
 dy _ 
 dx 
 
 3 , , 20 , 
 
 b 
 
 
 
 Form 45. 
 
 S. y= Va;2 + a^, 
 
 Form 61. 
 
 6. 2/ = - Va^ — x^y 
 
 
 
 Form 53. 
 
APPENDIX A. 5 
 
 7. e = (r^- a^y^, ^ = 5r%r' - a')K 
 
 8. y = ib V(aT6i)^ Form 42. 
 
 9. y = i V(rc2 + a2)3, Form 58. 
 
 10. 2/ = - J V(a2 - a;2)3, Form 49, 
 
 11. 1/ = (a. - x.)t, ^ = ^. 
 
 d^ r3(7 - 8r) 
 
 12. d == r^-ylr - r^, 
 
 rfr 2 -^7- _ 7.2 
 
 2(2a - hx) ^a + hx ^ 
 
 14. 2/ = — 25^ , Form 46. 
 
 _ 2(2a - Zhx) V(a + M^ ^ 
 
 15. 2/ = -^ ^^ '-y Form 43. 
 
 2(8a2 - 4:ahx + 36V) Va + 6rc 
 
 16. 2/ = ~- — ^553 > Form 47. 
 
 17. 2/ = -^ 1Q553 ^> Form 44. 
 
 18. 2/ = 7,/ ■ I. x > Form 34. 
 
 o{a + oaj) 
 
 19 ^ = _ ^^l-~^\ Form 64. 
 
 20. y --= , ,1 , Form 66. 
 
 aNa;2 + a2' 
 
 22. „ = - ^'^^ - ^, Form 73. 
 ^ ax 
 
 - (3^' - Q')^ ^ = (a;^ - a^)K7a;^ + 9a') 
 
 ^^' ^ ~ a;5 ' dx . 6a;§ 
 
 |a2"- aj2 <^^ - 4a2a; 
 
 " NoH^T^' S ^ 3(a2 - a;2)§(a2"+^t 
 
CALCULUS. 
 
 25., = log„.n, 1 = ^ 
 
 26. u = log (3t;2 + 4t;3)», 
 
 du ^ 18(1 + 2t;) 
 dv v{3 + 4t;) ' 
 
 27. 2/ = ^ log (a + 6a;), Form 33. 
 
 28. 2/ = ^ log ^a;2 + ^V Form 39. 
 
 29. 2/ = log (x + Vl + x^)y Form 10. 
 
 30. y = log (a; + V^~^), Form 11. 
 
 31. 2/ = log (x + Va;2 + a^), Form 60. 
 
 32. 2/ = - log(^ +iJJ+l), From 16. 
 
 33. 2/ = - log (i + ij^2 - l)» Form 15. 
 
 35. s - ti log (a^ - 60* I = %faT-^ W "^ ^* ^""^ ^""^ " ^^^' 
 
 ^^•^ = ^^^S^' ^^^^^• 
 
 37. 2/ = i log i-±^, Form 13. 
 
 1 X 
 
 38. 2/ = - log — 1 — P==, Form 63. 
 
 39. 2/ = - log "^^' + ^' - ^ Form 63. 
 "a X 
 
 __J___, 2cx + 6 - Ate2 _ ^ac ^ .. 
 
 40. 2/ = h, . log X r-i—F — ■ f Form 41. 
 
 ^ V62 - 4ac *^ 2cx + 6 + a/^^ - 4ac 
 
 41. 2/ = p a + 6x — a log (a + 6a;) , Form 35. 
 
 42. 2/ = J [log (a + 6a;) + ^^^], Form 36. 
 
APPENDIX A. 7 
 
 43. 2/ = ^a^ - x^ — a log — ' , Form 51. 
 
 X 
 
 44. 1/ - I -VSM^ + |- log (a; + Va;2 + a^), Form 57. 
 
 45. 2/ = I Va;2 + a^ - ^ log (a; + V^^ + ^2)^ Form 62. 
 
 46. 2/ = |(2a;2 + Sa^) Va:^ + a^ + ~ log {x + Vx^ + a^), Form 65. 
 
 47. y = |(2a;2 + a^) ^2 _|_ ^2 _ |' i^g (^^ ^^2-^:^2)^ Form 59. 
 
 48. 2/ = log. (30. + 2), J = 3^-2 
 
 49. ^ = ^ % = _ L±J2g^ 
 
 a; log x* dx {x log xY ' 
 
 50. 2/ = a:i°«=", 1^ = loga-x*°««-i 
 
 51. 2/ = a^, ^ = 2a;a-Moga. 
 
 ^"^^ ^ "^ » da: ~ ic • 
 
 53. ^ = c=^e^ 3- = c^e^(l + log c). 
 
 54. 2/ = e^'x^ -^ = e=^a;^-K^ + ^)' 
 
 55. 2/ = ^^x\ ^ = 4-a:3(4 + x log 4). 
 
 _ ?!±? dy /- 2\ ?!±2 
 
 56.2/ = e«, ^=(^l-_je-^. 
 
 58„-5£_+3^^ # 24X-W 
 
 5'- ^=(^''-2)'. , %-W^r 
 
8 CALCULUS. 
 
 11 1 dy 1 + log X 
 
 60. y = loglog^ - j^, ^ = -^(bi^- 
 
 61. y = ^{e^ - e-2-), ^ = c{e'^^ + 6-2-). 
 
 62. s = (2e2* - 1)(1 - 3e-30, ^^ = 4e2< + Ge"* - Qe-^'. 
 /:o -.. i^« log ^ ^2/ I 
 
 MO, 
 
 y 
 
 -^^^l + logx' 
 
 64. 
 
 y 
 
 = glog„x _ a;log««^ 
 
 65. 
 
 y 
 
 e' + x- 
 
 dx a; log X (1 + log x) 
 
 ^= loga_e/ ,,^,_^,,^A 
 
 dx iC \ / 
 
 ^ _ 2e'xf'-\a — x) 
 
 dx " (e* - x")2 
 
 66. 2/ = x[(log xY - 2 log X + 2], ^ = (log x)2. 
 
 2/ - log gnx _^ 1 _^ g-nx » ^a- - g2«x + 1 + e'^"* * 
 
 riogi(e- + l) __ log (6"- -1) 1 
 
 = log (e"'' - 1) %^ L e"^- 1 e"^ + 1 J 
 
 ^^* ^ log (€"- +1)' da; [log (e-- + 1)]2 
 
 PROBLEMS GROUP 3. 
 {After Art. 42.) 
 Differentiate the following: 
 
 1. 2/ = sin hx, 
 
 2. 2/ = cos nXy 
 
 3. 2/ = sin 3a;2, 
 
 4. 2/ = 2 + 4 Sin 2a;, 
 
 a; 1 
 
 5. 2/ = 2 "" 4 sin 2a;, 
 
 d2/ 
 
 5 cos 5a;. 
 
 dx~ 
 
 — n sin na;. 
 
 dy_ 
 dx 
 
 6a; cos Sa;^. 
 
 
 Form 21. 
 
 
 Form 20. 
 
APPENDIX A. y 
 
 dii 
 6. y = sin Zx cos 2x + cos 3a; sin 2x, 3^ = 5 cos 5x. 
 
 sin (m - n)x sin (m + n)a; 
 ^- ^ 2(m - n) ^ 2(m + n) ' '''' 
 
 ^ «m (m - n)x _ sin (m + n)a; ^^^^ 24. 
 
 ^ 2(m - n) 2(m + n) ' 
 
 ^ cos (m - n)x cos(m + n)a: ^^^^ 25. 
 
 ^ 2(m - ri) ^ 2(m + n) ' 
 
 10. y = log sin x, 
 
 11. 2/ = — log cos a;, 
 
 12. s = sin(<2 -st + 2), 
 
 13. s = log cos (1 - f), 
 
 14. y =log(a sin2 x dy ^ 2(a — h) tan x 
 
 + & cos^ x), dx a tan^ x -{- h ' 
 
 15. w = 2x2 sin 2a; dy . ^ ^ 
 
 , o o . ^ / = 4a:2 cos 2a;. 
 + 2a; cos 2a; — sm 2a;, dx 
 
 16. 2/ = 6=^ sin (2 - a;2), ^ = e^ [sin (2-a;2) -2a; cos (2-a;2)]. 
 
 ^o re -«^ ^^ 0[2 cos (e« - 6"^) - ^e^ + 6-«) 
 
 17. r - 02 cos (e^ — e "), t^ = • / « _flM 
 
 ^ ^' dd sin (e^ — e '')]. 
 
 , cos a; dy sin c 
 
 18. i/ = log 
 
 
 Form 8. 
 
 
 Form 7. 
 
 J^ = (2^ - 3)cos(i2 _ 
 
 3« + 2). 
 
 1 = 2aan (1 - i2). 
 
 
 cos (a; + c) ' da; cos x cos (a; + c) 
 
 19. V = sin' 2x cos^ 3a;, i^ = 6 sin^ 2a; cos 3a; cos 5a; . 
 ^ aa; 
 
 20. v = e^^'^^^ -^ = 2a;cosa;2.e^^^^ 
 ^ dx 
 
 * ^ ^ sin a; + cos a; ' da; cos 2a; * 
 
 , sin J(a; — a) d?/ sin a 
 
 22 1/ = log — - — ^ = 
 
 ' ^ sin |(a; + a)' dx 2 sin J(a; — a) sin |(a; + a) 
 
10 
 
 CALCULUS. 
 
 23. The lengths of crank radius and connecting rod of a steam 
 engine are 2 feet and 10 feet respectively, as shown in Fig. p23, 
 and the crank revolves at the uniform speed of 2 revolutions per 
 second. Find the velocity of the piston for the following values 
 of the angle 6. (a) 6 = 0, (h) 6 = 45°, (c) 6 = 90° and (d) 6 = 
 135°. Ans. (a) zero, (h) 20.32 feet per second, (c) 25.13 feet per 
 second, (d) 15.24 feet per second. Fig. p2Z is 1 J" high X 4" long. 
 
 Fia. p23. 
 
 PROBLEMS GROUP 4. 
 {After Art. 43.) 
 Differentiate the following: 
 
 1. 2/ = cot X, 
 
 2. y = sec x, 
 
 3. 2/ = cosec Xf 
 
 4. 2/ = vers x, 
 
 5. 2/ = tan {x^ — 2a:), 
 
 6. 2/ = sec" nx, 
 
 7. 2/ = log tan-, 
 
 J = — cosec^ X. 
 ax 
 
 dy . 
 
 -J- = sec X tan x, 
 
 ax 
 
 dy . 
 
 :r — — cosec x cot ». 
 aa; 
 
 = smx. 
 
 dy 
 dx 
 
 ^ = 2(x - 1) sec2 (a;2 - 2x). 
 
 dy „ 
 
 ■f^ = rv- sec" na; tan nx. 
 
 ox 
 
 Form la 
 
APPENDIX A. 
 
 11 
 
 S, y = log tan ^ + I j, 
 
 9. y = log (sec x + tan x), 
 
 10. y = log (cosec x — cot x), 
 
 11. 2/ = e^* cosec 2x, 
 
 Form 18. 
 
 Form 18. 
 Form 19. 
 
 4y _ 
 
 12. 1/ = (tan w— Scot w) Vtan^.-^p = r— — v— . 
 
 ' dw 2 tan5 w 
 
 da; 
 dy 
 
 = 2e2' cosec 2x(l - cot 2x), 
 
 13. s = <" cosec"* nt, 
 2taiid 
 
 15. 1/ = log [tan (e^ + S)*]^, 
 
 16. e = log 
 
 tan ^ — 2 
 
 17. r = 
 
 2 tan ^ - 1 
 a sin ^ + 6 vers ^ 
 
 -T. =ni"~^ cosec*" n<(l — mi cosnQ. 
 
 di/ _ 2 sec^ e 
 Te~{l - tan2 ^)2* 
 
 dx sin [2(e^ + 3)i]* 
 
 dg^ 3 
 
 dr 4 — 5 sin r 
 
 dr 
 
 2ab vers 6 
 
 a sin — 6 vers d^ (a sin — 6 vers ^)2' 
 
 PROBLEMS GROUP 5. 
 (After Art. U-) 
 Differentiate the following: 
 
 dy 1 
 
 1.1/ = COS~^ X, 
 
 2. y = tan~^ a;, 
 
 3. 2/ = cot~^ x, 
 
 4. y = sec~^ a;, 
 
 5. 2/ = cosec"^ a;, 
 
 6. 2/ = vers~^ a;, 
 
 da; 
 
 Vl- a;2 
 
 
 
 
 Form 12. 
 
 dy 
 dx 
 
 1 
 
 l^x^' 
 
 
 Form 14. 
 
 dy _ 
 
 1 
 
 
 
 dx 
 
 x^lx'' - 
 
 1 
 
 
 Form 17. 
 
12 CALCULUS. 
 
 X 
 
 7. 2/ = sin-i -, Form 52. 
 
 8. y = -tan-i-, Form 37. 
 
 c c 
 
 2 , 2cx + h 
 
 Form 40. 
 
 1 X 
 
 10. J/ = - sec-i-, Form 67. 
 
 o a 
 
 11. y = vers-i -, Form 70. 
 
 12. 2/ = x sin-i X + Vl - a^, Form 27. 
 
 13. 2/ = x cos-i a: — V 1 — x^ Form 28. 
 
 14. 2/ = I Va2 - x2 + ^sin-i-, Form 48. 
 
 15. 2/ = - I Vo^"^^ + ^sin-i -, Form 54. 
 
 16. 2/ = 1(2x2 - a^) Va2-x2 + |^sin-i -, Form 50. 
 
 17. 2/ = J(5a2 - 2a;2) Va^-rc^ + ^' sin-^-, Form 55. 
 
 o o O 
 
 18. to = COS"* Vvers t -jj= — h Vl + sec i. 
 
 ^ ,3^ - 2 , ^ , 3(9 - 12 dr ^ 
 
 19. , = tan-^- + cot- -^^^, ^ = 0. 
 
 20. r = cos-i-3- - 2^-^, rf", = aJ^ - r 
 
 21. s = t^ sec-i I - 2 Vi2-4, ^=2^ sec-i|. 
 
 22. 0) = V2[sec-* (sec i + tan 01, 
 
 dt V (sec t + tan tan t 
 «. ^ ,4 + 5 tan x dy S 
 
 23. 2/ = tan 3 d^ = 5 + 4sin2x 
 
 OA - -l!i±£l' ^ - - 2 
 
 Z4. s - tan ^, _ ^_,, ^^ - ^2, ^ g-2f 
 
APPENDIX A 
 
 25. y = tan-' (|tan x) - tan-'^, g = - 
 
 26. y = tan 
 
 13 
 
 ab 
 
 sin^ X + ¥ cos^ x' 
 J g^ tan a; - 6^ dy a6 
 
 a?)(l + tan xY 
 
 27. 1/ = V2ax — x^ + a 
 
 vers" 
 
 28. 2/ = — V 2aa; — x^+ a 
 
 vers"* — . 
 
 X — a ITT 5 , a^ , X 
 
 29. 2/ = — ^ — V 2ax — a;^ + — vers~i -, 
 
 30, y = ' — ^ V 2ax - a;2 + — vers-^ -, 
 
 dx o? sin^ X ■\-\P- cos^ x 
 
 Form 72. 
 
 Form 71. 
 
 Form 68. 
 
 Form 69. 
 
 PROBLEMS GROUP 6 
 
 {Ajter Art. ^.) 
 
 Differentiate the following: 
 1. 2/ = ^'^'j 
 
 2. = (1 - ty-', 
 
 3. s = Zt'\ 
 
 4. 2/ = x^^''^*^", 
 
 5. 2/= (iogxy°«- 
 
 6. 2/ = (3x2 + 1)2^^-2), 
 
 ^ = nx"*(l + log x). 
 ax 
 
 I = - (1 - 0^-'[i + log(i - 01 
 
 | = 3<^'ni + 31og<]. 
 ^= (n+l)(logx)«x^'"«*>"-i 
 
 ^'[n-log(logx)]. 
 
 dy ^ (log x) 
 dx 
 
 dy 
 dx 
 
 (3x= + !)«-«' {2 log(3a;» + 1) + ^3^^^ -^} • 
 
14 APPENDIX A. 
 
 PROBLEMS GROUP 7. 
 {After rule III of Art. 67.) 
 Integrate the following differential equations. 
 
 1. dy = (3x'-7^+ 1 - I'jdx, y = ^-1^+ log^ +^ + C. 
 
 2. dy = (2 - ^)V ■ dx, j,=^_^ + |'+C. 
 
 3. dy = 3(2j? + S)3? • dx, y = ^2?f-±-^+C. 
 
 dx 
 
 - , dx 
 
 ^ , X ' dx 
 
 7. dy = Va + hx' • dx, Form 42. 
 
 dx 
 
 8. di/ = ■ , , Form 45. 
 
 VaH- hx 
 
 9. dy = X Va2 - a;^ • c?x Form 49. 
 
 3/ * dx 
 
 10. di/ = ■ , Form 53. 
 
 Sa^ — x^ 
 
 dx 
 
 11. dy =—-==, Form 73. 
 
 ^<2ax-x^ 
 
 12. dy = a' ' dx, Form 4, 
 
 13.%=(e- + a- + 36-)dx, , = ^^+^_^^+C. 
 
 15. dy — sin X cos x ' dx y = — ^ — + C 
 
 16. dy = tan a; • dx, Form 7. 
 JVote. — ^Take tan x = . 
 
 C06 X 
 
APPENDIX A. 15 
 
 17. dy = cot X • dXj Form 8. 
 
 -o , . . , , 2qo^^x cos^x , ^ 
 
 18. a?/ = sm^ xdx, y = — cos x H ^ ^ f- C. 
 
 iVote. — Take sin" x = iX — cos^ x)" sin x. 
 
 ,-, J -1 K J sin^a; 2siD^a; , sin® a; , ^ 
 
 19. dy = sm* a; cos^ x - dXy y = —z = 1 ^ h C. 
 
 20. dy = sin^ a; • c^a;, Form 20. 
 1 — cos 2a; 
 
 iVote. — Put sin^ x = 
 
 2 
 
 21. c?2/ = cos^ X ' dx, Form 21. 
 - - , fi 7 5a; , sin 2a; sin^ 2a; , 3 sin 4a; , ^ 
 
 22. dy = cos' X'dx, 2/ = jg + -j is" "*" "~64~ "^ ^• 
 
 23 . (??/ = sin mx sin na; • da;, Form 24. 
 iVote. — Put sin mx sin nx = }^ cos(m — n)a; — 3^ cos (m + n)a;. 
 
 - . , , J. , tan^ a; tan^ a; ' , ^ 
 
 24. ai/ = tan^ x - dx, y = — -. ^r log cos x -^ C. 
 
 4: ^ 
 
 Note. — Put tan" x = tan' x (sec^ x — I). 
 
 cot^ X 
 
 25. dy = cot^ X ' dx, y = x -\- cot x ^ h C. 
 
 o^ J fi , ^ X 2 cot^ a; cot^ x 
 
 26. ai/ = cosec^ x - dx, y = C— cot x i^— ^ — . 
 
 o 5 
 
 iVofe. — Put cosec* x = (1 + cot^ x)''. 
 
 ofT J J. 1 fi J tan'' x , tan^ a; , tan^ x , ^ 
 
 27. dy = tan' a; sec^ x - dx, y = — 1 1 ^ f-C 
 
 4 o o 
 
 iVo^e.—Put sec* x = (1 + tan^ a:)2. 
 
 v,o J , . , J sec^ X 2 sec^ a; , sec' x , ^ 
 
 28. dy = tan^ a; sec' x - dx, y = —^ = 1 ^r — |- C. 
 
 7 o o 
 
 Note. — Put tan^ x = (sec^ x — 1)^ and write tan" x sec' x = (sec* x — 1) 
 sec^x-sec xtanx-dx. 
 
 29. dy = -p=^=, 2/ = i sin-i ~ + C, 
 
 ^- , dx 1 , 
 
 * " VliP'^l' 2^ = 5 log (5x+V25x2-4) + C. 
 
16 APPENDIX A. 
 
 ^^' '^'' = 4T25:^' J/ = 2 tan-' 2- + C. 
 
 ,~ J dx 1 , 2 + 6a; , - 
 
 35. ., = ^^ . dx, , = I log (4x-6)+^-?^ '°^ £|+^- 
 
 36. Find the length of the arc of the parabola in Fig. 13 on page 
 24 from x = to x = 10 inches, the value of A; being 2 when 
 X and y are both expressed in inches. 
 
 Ans. 201 square inches. 
 
 PROBLEMS GROUP 8. 
 (After rule IV of Art. 67.) 
 Verify the following: 
 
 1. Jx smx • dx = — a; cos a; + sin a; + C. 
 
 Note. — Use rule IV. Try u = em x, dv = x-dx] also try u = x, dv = 
 mnx-dx. 
 
 2. fx COS a; • da; = a; sin a; + COS a; + C 
 
 /x^ 
 x^logx ' dx = -J (log X — I) + C. 
 
 4. fx{^' + e-'')dx = I (e^x _ g-sx) _ i (^sx + g-3x) + q. 
 
 5. J* log (ax + 6) • cte = - (ax + h) log (ax + &) — a; + C. 
 
 6. fa^smx'dx= — x^ cos x + Sx^ sin x + 6x cos x — 6 sin x + C. 
 
 Note. — Applying rule IV, taking u = x^ and dv = smx'dx, we get an 
 expression involving Jx^ cos x-dx. This expression can again be reduced by 
 
APPENDIX A 17 
 
 applying rule IV (to the part to be integrated), making w = a;^ and dv » 
 cosx'dx] and so on. 
 
 7. fx^ sin 2x ' dx = §(2a;2 - 1) sin 2x - iC^s^ - 3x)cos 2x + C, 
 
 „ /• „^ . , J e°^(a sin 6x — 6 cos 6a;) , ^ 
 
 8. I e«^ sm hx • (ix = — ^^ . , ,, ^ + C. 
 
 •^ a^ -jr 0^ 
 
 Note. — Integrate by parts, taking u = e<**, then: 
 
 /, • I. J e"" C08 hx , a r ^ , , ,^ . 
 
 e'^'^embx-dx = ^^ ^ hj ^ ^°^ oa;*aa; (1) 
 
 Integrate by parts, taking u = sin bx, then: 
 
 /e"* sin bx'dx = / e"* cos bx'dx (2) 
 
 a a«/ 
 
 Take the sum of equation (1) multiplied by ¥ and equation (2) multiplied 
 by a^ in order to eliminate the term containing J e°* cos 6a: • dx and the result 
 given above is obtained. By subtracting equation (2) from equation (1), 
 J e"* sin bx'dx can be eliminated and the value of I e"* cos bx-dx obtained. 
 
 9 . J e^^ cos 5x ' dx = -^{5 sin 5x + 3 cos 5x) + C. 
 
 e~2* sin a; • (ia; = ^ (2 sin x + cos ic) + C 
 
 X 
 
 11. fe~^ COS ^' dx = -TY" (2 sin r- — 3 cos ^j + C. 
 
 PROBLEMS GROUP 9. 
 
 (Special methods a of Art. 67.) 
 
 Verify the following: 
 
 r 5a; + 4 _ /" 3 - c?a; /•2j^ 
 
 ^•Ja;2 + x-2*'^~Ja;-l"*"Ja; + 2 
 
 = 3 log (a; - 1) + 2 log (a; + 2) + C. 
 
 Note. — A full discussion of the integration of rational fractions is given in 
 Byerly's Integral Calculus. See note on page 86. 
 
 In this particular case the denominator is equal to (x — 1) (x + 2) and 
 
18 APPENDIX A. 
 
 5x4-4 
 
 «« + a;-2 
 
 may be broken up into what are called partial fractions as follows: 
 &C + 4 _ A B 
 
 a:* + x-2 X - 1 x-{-2 
 
 Clear of fractions and equate coefficients of like powers of x and we find 
 A = 3 and 5=2. 
 
 2. f ^^^^l^ -dx = \og(x + i) + \og(x+l) + C. 
 
 + \ log (x - 2) + C. 
 
 = ^' + 15x - 6 log (x - 1) + 41 log (x - 2) + C. 
 
 Note. — \Mienever the degree of the numerator of a rational fraction is 
 gre^ater than or equal to the degree of the denominator, the fraction should be 
 reduced to a mixed quantity. 
 
 5-/^^^-'i- = - + glog(--3)-|log(. + 2) 
 
 + \ log (x + 1) + C. 
 
 '^ • / (X - IRx' + 1 ) • "^ ° ^°g ^^ ~ ^^ " W^) 
 
 - I log (x^ + 1) + C. 
 
 X* 
 
 Note. — __ ^ - ^ is broken into partial fractions by putting it equal 
 
 .^ A B , Cx-\-D 
 
 ^ x-l'^ ix-1)*'^ x« + 1 * 
 
 «• J i(^^-:^4? • ^^ = - i^^4 + ^^S^^^ + ^- 
 
APPENDIX A. 19 
 
 PROBLEMS GROUP 10. 
 (Special methods b of Art. 67.) 
 
 Integrate the following and for answers see specined forms in the 
 table of integrals of Appendix B. 
 
 1. /Vo^-^^ • dx Form 48. 
 
 Note. — Let a: = a sin ^, then dx — a cos d-dd; and we get: 
 
 (^CL^ — x^-dx = C<a* — a' sin* d-a cos 6-6$ 
 = a^ fcoa^d'dd 
 
 = a^ 
 
 X, Ix^a^ - X* 
 
 » - xn 
 
 o» J 
 
 a* . . X . 1 ^r-r -r- 
 
 2. Jx^^a" - x^ ' dx, 
 
 , rVa2 - x^ , 
 
 3. J— ^— -da:, 
 
 4. f{a^ - x^)i • dx, 
 
 6. fV^M- a^ • dx, 
 
 7. faj^ Va;2 -{- a^ • dic, 
 8. 
 
 
 9. 
 
 10 
 
 \x^ + a2 
 dx 
 
 dx 
 
 / rix 
 (x2-a2)r 
 
 ^ a^x^ ' 
 X 
 
 Form 50. 
 Form 51. 
 Form 55. 
 Form 56. 
 Form 57. 
 Form 59. 
 Form 62. 
 
 Form 64. 
 Form 66. 
 
 19 
 
20 APPENDIX A. 
 
 PROBLEMS GROUP 11. 
 (Special methods c of Art. 67.) 
 Verify the following: 
 
 J r x^ ' dx ^ {x-\- ay-^ {x + ay-^ 
 
 * J (a; + a)"* 4 — m ^ 3 — m 
 
 2 — m 1 — m 
 
 i\rote. — ^Let X -^ a = y, dx = dy. 
 X ' dx 3 , . , .. 3 
 
 2- /(STS-r|.(« + ^*)'- i (« + ''-)' + ^- 
 
 JVote. — Let a -\-hx = y, h'dx = dy, 
 1. r.f" , = tan-i e- + C. 
 
 J 6^ + e-* 
 
 A^ofe. — Let e* = y, x = log w, dx = — . 
 
 2/ 
 
 4. r(„,-xi)sd.=£(5Li^'-5H?Lr:4^^LiM) 
 
 J oa^ 16as 
 
 
 iVote. — Let X ^^ a sin' 9. 
 
APPENDIX B. 
 
 TABLE OF INTEGRALS. 
 
 A fairly extensive table of integrals is A table of integrals, arranged by 
 B. O. Peirce and published by Ginn & Co., Boston, 1890; revised edition 1899. 
 
Explanatory Notes. 
 
 To say that j2x'dx is x^, is exactly the same thing as to say that the 
 
 differential of x^ is 2x-dx. 
 
 Constants of integration are omitted from this table. 
 
 Any of the integrals given in this table can be verified by differentia- 
 
 — r-Tj -\- C) according to Art. 30, 
 
 — jjy + a constant J is the integral of 
 
 x^'dx. 
 
 sin X 
 One has frequent occasion to use the following formulas: tan x = , 
 
 cos a; 1 1 J 1 
 
 cot X = - — ■, sec x = , cosec x = — — , and vers x = 1 — cos x. 
 
 sm x' cos X sm x 
 
 The definitions of the hyperbolic functions sinh re, cosh x, etc., are given 
 
 m Chapter VI, Art. 94. 
 
 The letter e stands for the base of the Napierian logarithms ( = 2.7182818) ; 
 
 log X stands for Napierian logarithm of x. 
 
APPENDIX B. 
 A. Fundamental forms. — 
 
 x"^ ' dx = — j^ when n is no%=-equal to — 1. 
 
 3. J e* • dx = e^ 
 
 a' - dx — i . 
 
 log a 
 
 5. j sin re • c^x = — cos x. 
 
 6. I cos X ' dx = sin a;. 
 
 7. J tan X ' dx = — log cos x. 
 
 8. I cot X • da; = log sin x. 
 
 / dx 
 
 0. C - = sinh~^ re = log (re + Vl -f a^O* 
 J Vl + x^ 
 
 1. C = cosh-1 X = log (x + Vx^ — 1). 
 
 J ^lx^ — I 
 
 dx 
 
 4^x2 
 
 dx . ^ , 1., 1+ X 
 
 2. J T o = tan~i X, 
 
 ^- Jr^2 = ^"^^"^ = 2^°^i-x 
 
 J xa/x^"^^ 
 
 dx , 
 
 4. I — , = sec~^ X 
 
24 APPENDIX B. 
 
 17. I — = = vers~i a;. 
 
 B. Integrals involving trigonometric functions.— 
 
 18. Jsccx • dx = log (sec a: + tan x) = log tan f ^ -f | j. 
 
 19. Jcosec X ^ dx = log (cosec x ~ cot x) = log tan|. 
 
 /x 1 
 sin^ .T • da; = - - ^ sin 2a;. 
 
 /x 1 
 cos^ a; • c^a; = 2 + T sin 2a;. 
 
 22. j sec^ a; • rfa; = tan a;. 
 
 23. J cosec^ X - dx = — cot a;. 
 
 24. fsin 7.U; sin n^ . dx = ^^^/^ " ^^^ - ^^^/^^^ + ^,)^ . 
 J 2(w - n) 2(m + n) 
 
 o< To; .^ J cos (m — n)x cos (m + w)a; 
 
 25. I sin ma; cos na; • oa; = ^ / ^ — -- — ~. 
 
 J 2(m — n) 2(m + n) 
 
 OA /^^^^ J sin (m — n)a; , sin (m + n)a; 
 
 26. I cos wa; cos iia; • da; = —p-r ^ + —7^7^ — , / . 
 
 •/ 2(w - n) 2(w + n) 
 
 27. J sin-i a; • da; = a; sin-^ x + Vl - x^. 
 
 28. J cos~i a; • da; = a; cos"^ a; — Vl — a;^. 
 
 C. Integrals involving hyperbolic functions. — 
 
 29. J sinh x • dx = cosh a;. 
 
APPENDIX B. 25 
 
 30. j cosh X ' dx = sinh x. 
 
 31. j tanh x ' dx = log cosh x. 
 
 32. j coth X ' dx — log sinh re. 
 
 D. Integrals involving (a + 6a;). — 
 
 J (a 
 
 + 6a; 
 c?x 1 
 
 (a + 6x)2 6(a + hx) ' 
 
 35.- j~j^ = p [a + 6a; - a log (a + 6.t)]. 
 
 E. Integrals involving (a -f 6a;2). — 
 
 ^^ C dx 1 , , ic 
 
 37. J — -r — „ = - tan-i - 
 
 38. 
 
 c^ -\- x'^ c c 
 
 dx 1 , c -\- X 
 
 r dx 1 , 
 
 & — x^ 2c c — X 
 
 -,« r ^ ' dx 1 , / , , a\ 
 
 F. Integrals involving (a + 6a; + ca;^). — For brevity let X 
 stand for (a + 6a; + cx^) and let g stand for (4ac — 6^). 
 
 ,^ rdx 2 . ,2cx + h , u.' 
 
 40. I v^ = — /^ tan~i ^ — when q is positive. 
 
 4.1 C~— , loe ; — - when o is negative. 
 
 ^^' J X aT^ ^2ca; + 6+A/-3 
 
 G. Integrals involving ^la + 6a;. — 
 42. J Va + 6x • rfa; = 25 '^(^ + ^^)^* 
 
26 
 
 APPENDIX B. 
 
 43. fx^^+Tx ■ & = - ^(^^ - ^^g/° + ^^)'. 
 dx 
 
 44 
 
 1562 
 
 _ 2(8a2 - 12ahx + ISb^x^) V(^TW' 
 1056^ 
 
 2Va + 6i 
 
 * *^ Va + 6a; 6 
 
 r a; ' da; ^ ^ 2(2a - 6x) A/aT6x 
 -^ Va + 6x 362 
 
 /^ x^ ' dx ^ 2(8a^ - 4fl6a; + ^h^x^) Va + 6a; 
 
 H. Integrals involving Vo^^^^"^ — 
 
 48. C-ylaT^^^ ' dx = ~-<a^^^^ + ^ sin-'-. 
 J 2 2 a 
 
 49. fx-ylaF^"^^ • dx = - ^^i{a' - x^y . 
 
 50. fx^ Va2 - x2 • (ia; = f (2a;2 - a^) V^IT^a 4, |' gin-i ? 
 »/ o o a 
 
 52. r-p^= 
 
 •^ A/a2--x2 
 
 Va2 - a;2 , ^-^ -_ , a + Va^ - a;^ 
 
 ' dx = -sa^ — x^ — a log 
 
 sm" 
 
 Xj_^ 
 
 Va2 - a;2 
 a;2 . <?a; 
 
 S3, f-^1^- = _ V^ 
 
 X" 
 
 ^. r x^ ' dx X /-; ^ , a^ . , a; 
 
 54. I —. = - Pi Va2 - a;2 + - sm-i - . 
 
 •^ Va2^^2 2 2 a 
 
 55. J'Co^ - a;2)* * ^^ = | (5«' " ^a;^) Va^ - x" + 
 
 3a4 . ,x 
 
 <^ 
 
 «/<.?^ 
 
 x^)^ a^^a'-x^' 
 
APPENDIX B. 27 
 
 I'. Integrals involving Va;^ + a^. — 
 
 57. J'Vx2~+^2 ' dx = ^ A/iH-~a2 + I' log {x + ViM^"S2). 
 
 SS, fx^¥^~a^ ' dx = ~^I'(^~+^K 
 
 59. J'a;^ Vx2~+~a2 • da; = | {2x^ + a^) Vi2Tp^2 
 
 - I log (x + V^+ a2). 
 
 60.. f , ^^ = log (a; + Va;^ + a^). 
 •-^ Va;^ + a^ 
 
 61. r_Ai^= V^+^2. 
 
 62 
 
 + 
 x^ ' dx X i-TT-; — i. a^ 
 
 . f^^= = \ V:.^ + a^-|- log(x + V^H^a^). 
 63. r--^==i log ^— =i log ^^^±^ 
 
 ' -^ a;2 V^M^^ a2^ 
 
 65. r(a;2 + a2)? • c?x = |(2a;2 + 5a^) ^x^ + a^ 
 
 + — log (a; + Va;2 + a^). 
 
 66 f ^^ ^ ^ 
 
 • J (a;2 + a2)^ ^2 ^^.2 _|_ ^2" 
 
 r'. Integrals involving Vx^ - a^.—AU of the forms 57 to 66 
 apply in this case by simply changing the sign of a^ except form 
 63 which becomes: 
 
 ^^ r dx I .X 
 
 67. I — , = - sec-^ -. 
 
 -^ x^^x^ - a^ « a 
 
28 APPENDIX B. 
 
 68. 
 
 
 69. rx>55irr^.<te=-3^±^z2^^g^^^r^^^^^^.5 
 
 •/ D J a 
 
 70. f-, ^ - =:vere-i^. 
 
 71. C-^LM==. = - V2ax - x2 + a vers-i-*-. 
 J •V2aa; - a:^ a 
 
 72. r^^-^> d:c = V2ax-x2 + a vers-^?. 
 J X o, 
 
 73. f ^ = - V2aa; - 7^ 
 J X'^2ax — a? ox 
 
APPENDIX C 
 
 A SELECTED LIST OF TREATISES ON MATHEMATICS AND ON 
 THE VARIOUS BRANCHES OF MATHEMATICAL PHYSICS. 
 
 HISTORY. 
 
 See A History of Mathemati cs by Florian Cajori. The Macmillan 
 Co., New York, 1894 
 
 A good sketch of the history of calculus is given in the article 
 Infinitedmal Calculus in the 9th edition of the Encyclopedia 
 Britannica. 
 
 Sir David Brewster's Life of Newton j Thomas Constable & Co., 
 Edinburgh, 1855, is well worth reading by anyone who is inter- 
 ested in the history of mathematics. 
 
 The theory of infinitesimals and the theory of limits have 
 occasioned a very great deal of discussion in the past. These 
 matters are discussed somewhat at length in the Britannica article 
 above mentioned, but the most edifying discussion of this subject 
 is that which is given in the introductory chapter to Augustus 
 De Morgan's Differential and Integral Calculus, London, 1842. 
 
 GENERAL TREATISES. 
 
 An extremely interesting and facinating book for the student of 
 mathematics is W. K. Cliff ord's* Common Sense _oi the Ejcact 
 Sciences, The International Scientific Series, D. Appleton & Co., 
 New York, 1888. 
 
 One of the best general treatises is Edouard_Gou.rsat's Cot^rs 
 d^ Analyse M athematique , two volumes, Gauthier-Villars, Paris, 
 1905. This valuable work has been translated into English by 
 
 *W. K. Clifford's philosophical essays (two volumes, Macmillan & Co., 
 London, 1879) and especially his small book entitled Seeing and Thinking 
 (Macmillan & Co., London, 1880) are extremely readable and interesting. 
 
 29 
 
30 APPENDIX C. 
 
 E. R. Hedrick and it is published as Goursat-Hedrick Mathe- 
 matical Analysis, Ginn & Co., Boston. 
 
 The various mathematical articles in the ninth and eleventh 
 editions of the Encyclopedia Britannica are a great help to the 
 student of mathematics. 
 
 The great reference work in pure and applied mathematics is 
 the Encyclopddie der mathematischen Wissenschaften, B. G. Teub- 
 ner, Leipsig. The publication of this encyclopedia was begun in 
 1905 and it is not yet finished. In the meantime the work is 
 being republished in a revised and enlarged form in a French 
 translation. 
 
 The most extensive general treatise on mathematical physics 
 is Winkelman^s Handbuch der Physik in six volumes, second 
 edition, J. A. Earth, Leipsig, 1909. 
 
 THEORY OF FUNCTIONS. 
 
 The theory of functions is one of the most facinating branches 
 of pure mathematics. 
 
 A good discussion of this subject is given in the second volume 
 of Goursat's Cours d^ Analyse Mathematique (Goursat-Hedrick 
 Mathematical Analysis). One of the best books on the subject 
 is H. Durdge's Elements of the Theory of Functions (English trans- 
 lation by Fisher and Schwatt), Fisher and Schwatt, Philadelphia, 
 1896. 
 
 Two very extensive treatises on the function theory are A. R. 
 Forsythe's Theory of Functions of a Complex Variable, Cambridge 
 University Press, 1893; and Harkness and Morley^s Theory of 
 Functions^ Macmillan & Co., London, 1893. 
 
 DIFFERENTIAL EQUATIONS 
 
 A very good elementary treatise is D. A. Mur ray ^s IntrodmAory 
 Cours&JtL^DiJlerential Equ ations , Longmans Green & Co., New 
 York, 1897. See also W. W. Johnson ' s Differential Eq uations^ 
 John Wiley & Sons, New York, 1890. The second volume of 
 
APPENDIX C. 31 
 
 Goursat-Hedrick's Mathematical Analysis contains a good dis- 
 cussion of differential equations. 
 
 A very good treatise on partial differential equations is W. E. 
 Byerly^s treatise entitled Fou rier's Series and Spherical, C ylindrical 
 and Ellipsoidal Harmonics, Ginn & Co., Boston, 1893. 
 
 An important general method of solving linear differential 
 equations is given by P. A. Lambert in the Annals of Mathematics. 
 First paper in Vol. XI, pages 185-192, July, 1910; second paper, 
 Vol. XIII, pages 1-10, September, 1911. 
 
 QUATERNIONS AND VECTOR ANALYSIS. 
 (See page 211.) 
 
 HYPER-GEOMETRY AND RELATIVITY. 
 
 A very interesting development of mathematics is what is 
 called non-Euclidean geometry or hyper-geometry. Perhaps the 
 most interesting thing foi a beginner to read on this subject is 
 Helmholtz's popular lecture entitled **Ueber den Ursprung und 
 die Bedeutung der geometrischen Axiome" Vortrdge und Reden, 
 Vol. II, F. Vieweg & Son, Braunschweig, 1884. These very inter- 
 esting lectures of Helmholtz have been translated into English, 
 first series by Dr. Pye-Smith, second series by E. Atkinson, and 
 both series are published by Longmans, Green & Co. 
 
 See section VI of article Geometry in the 11th edition of the 
 Encyclopedia Britannica. A good discussion of non-Euclidean 
 geometry is given in A. N. Whitehe ad's Universal Alg ebra, Cam- 
 bridge University Press, 1898. See also Stackel und Engel. 
 TMm:ie_jderParallellinim von Eu klid his auf Gauss, Leipsig. 1895; 
 F. Klein, Nicht Euklidi sche Geometric, Gottingen, 1893; and P. 
 Barbarin, La Geometric non-Euclidienne, Paris, 1902. 
 
 The subject of non-Euclidean geometry is closely related to a 
 very recent development in theoretical physics called the theory 
 or relativity. A very simple discussion of this subject is given 
 by W. S. Franklin in an article entitled The Principle of Relativity, 
 Journal of the Franklin Institute, July, 1911. A very interesting 
 article on this subject is the eighth lecture in Max Planck's Acht 
 
32 APPENDIX C. 
 
 Volemngen uher theoretische Physik, S. Hirzel, Leipsig, 1910. The 
 following articles by Q, N. Lewis^ R. C. Tolm an and E. B. Wilson^ 
 are interesting, and all but the last one are easy to read. Philo- 
 sophical Magazine, Vol. 16, pages 705-717; American Academy Pro- 
 ceedings, Vol. 44, pages 711-724; American Academy Proceedings, 
 Vol. 48, pages 389-507. A general treatise is M. Laue's Das Relativi- 
 tdtsprimip, Braunschweig, F. Vieweg und Sohn, second edition, 1913. 
 
 ASTRONOMY.* 
 
 Most of the recent researches in astronomy have been in the 
 line of what is called astro-physics, and but few additions have 
 been made to the older theoretical astronomy with its beautifully 
 complete mathematical theory. 
 
 Good books on theoretical astronomy for the beginner are C. A . 
 Young's General Astronomy, Ginn & Co., Boston, 1898; and W. W » 
 Camp bei rs Elements of Practical Astronomy , The Macmillan Co., 
 New York, 1899. 
 
 Perhaps the best treatise on astronomical measurements is 
 Wm.. Qhauvenet's Syherical and Prac tical Astronomy , two volumes, 
 J. B. Lippincott, Philadelphia, 1863. For a treatise on astro- 
 nomical calculations see Jas. C. W atson's The oretical Astronomy , 
 J. B. Lippincott, Philadelphia, 1867. See also ^ahnbes timmung 
 der_ Komet en und Pla neten, T. R. v. Oppolzer, two volumes, Wm. 
 Engelmann, Leipsig, 1882. 
 
 Perhaps the greatest treatise on theoretical astronomy is Traiti. 
 d&.JLechanique Cileste, P. S. Laplace, five volumes, Paris, 1799. 
 English translation by Nathaniel Bowditch, Boston, 1829. 
 
 Theoretical astronomy goes far beyond every other branch of 
 
 * Although we are here concerned chiefly with books on mathematical 
 theory it is worth while, perhaps, to mention some of the extremely interesting 
 descriptive books on astronomy: ^imon New co mb^s Popular Astronomy f 
 Harper Bros., New York, 1882. J. Norm an Lockyer' s Star gazin g, Macmillan 
 & Co., London, 1878. Richard A. Proctor's Other Worlds tha n Ours , Longmans, 
 Green & Co., London, 1878. Also C. A. Young's General Astronomy is ar- 
 ranged so that the general reader can easily omit the more difficult parts 
 and the book then becomes a very good descriptive treatise. 
 
APPENDIX C. 33 
 
 applied mathematics in the legitimate use of elaborate formulas 
 and in the tremendously laborious numerical calculations that 
 are involved. The student can get an idea of the former and a 
 faint suggestion of the latter by looking over the four large volumes 
 of The Collected Mathematical Works of George W. Hill which have 
 been recently pubhshed (1905) by The Carnegie Institution of 
 Washington and placed in every important library in the United 
 States. Dr. Hill's greatest contribution to theoretical astronomy 
 was a new method for calculating the motion of the moon. 
 
 PROBABILITY. 
 
 In many respects the theory of probability is the most important 
 branch of mathematics for the experimental scientist. 
 
 The article Prdbahility in the eleventh edition of the Encyclopedia 
 Britannica is a very good outline of the theory of probability. 
 See also De Morgan's treatise on probabiHty in the Cabinet 
 Cyclopedia, London, 1838. Laplace's Theorie analytiq ue des 
 Prohahilites^Vsins, 1820; is one of the great treatises on the subject. 
 
 The application of the theory of probability to measurement is 
 treated in Mansfi eld Mer riman's Least Squares, John Wiley & 
 Sons, New York, eighth edition, 1903; in R. S. Woodward's 
 Prdbahility and Theory of Errors, John Wiley & Sons, New York, 
 1906; and in A. de F. Palmer's Theory of Measurements, McGraw- 
 Hill Book Co., New York, 1912. 
 
 The application of the theory of probability to the study of 
 statistics of all kinds and especially to biometrics (the quantitative 
 study of variation of plants and animals) is of great importance. 
 A simple discussion of this subject is given in the article Variation 
 and Selection in the eleventh edition of the Encyclopedia Britannica. 
 
 The kinetic theory of gases, one of the most important branches 
 of theoretical physics is a phase of the theory of probability. See 
 Watson's Kinetic Theory of Gases, The Clarendon Press, Oxford, 
 1893. See also the epoch-making work, T^dwig "Rn1t7mfl.Tm\q 
 Ymksungm^uh^.GaMMQrie^iT^ two parts, J. A. Barth, Leipsig, 
 1895 and 1898. The kinetic theory of gases is treated in a very 
 
34 APPENDIX C. 
 
 general way by J. Willard Gi bbs in his Stat istical MechanicSf 
 Charles Scribner's Sons, New York, 1902. 
 
 A very good discussion of the kinetic theory of gases is given in 
 Nernst's Theoretical Chemistry, Chapter II, translated by Lehfeldt, 
 Macmillan & Co., London, 1904. 
 
 The ideas of the kinetic theory of gases are extensively used in 
 the recent theories of the discharge of electricity through gases 
 and in the theories of radioactivity. See especially J. J. Thomson's 
 Conduction of Electricity through Gases, Cambridge University 
 Press, 1906. See also E. Rutherford's Radioactive Transforma- 
 tions, Charles Scribner's Sons, New York, 1906. 
 
 The modem statistical theory of radiant heat is also a branch 
 of the theory of probability. Some idea of this subject can be 
 obtained from lecture 6 of Acht Vorlesungen iiher theoretische 
 Physik, Max Planck, S. Hirgel, Leipsig, 1910. 
 
 THERMODYNAMICS. 
 
 There is a widespread notion that theoretical thermodynamics 
 makes a very severe demand upon the methods of higher mathe- 
 matics, whereas, as a matter of fact its demands are less perhaps 
 than any other branch of mathematical physics. 
 
 An extremely simple development of the mathematical theory 
 of thermodynamics is given in Franklin and MacNutt's Mechanics 
 and Heat, pages 350-397, The Macmillan Co., 1910. 
 
 A very good advanced treatise is Max Planck's Treatise on 
 Ther modynamics (English translation by Alexander Ogg), Long- 
 mans Green & Co., London, 1903. Another important treatise is 
 Edgar Buckin gham's Theo ry o f Th ermodynamics ^ The Macmillan 
 Co., New York, 1900. 
 
 R. Clau sius' Mec hanische Wdrmetheorie , third edition, F. 
 Vieweg & Sons, Braunschweig, 1887, is a very important work. 
 
 The student interested in theoretical thermodynamics will find 
 it well worth while to read the celebrated paper of J. Van't Hoff 
 entitled "The function of osmotic pressure in the analogy between 
 solutions and gases," Philosophical Magazine, Vol. XXVI, pages 
 81-105, August, 1888. 
 
APPENDIX C. 35 
 
 The best treatment of thermodynamics for the experimental 
 chemist is that which is given in a more or less disconnected 
 manner in W. Nernst's Theoretical Chemistry , English translation 
 by R. A. Lehfeldt, Macmillan & Co., London, 1904. 
 
 A good treatment of thermodynamics for the steam engineer 
 is that which is given in chapters II and III (pages 37-159) of 
 JL_A,_-Ewing' s The Steam Engin e _ and othe r Heat Engines ^ third 
 edition, Cambridge University Press, 1910. 
 
 THEORETICAL MECHANICS. 
 
 . There are many good beginner's treatises on theoretical me- 
 chanics. A good advanced treatise is Alexander Ziwet's Ele- 
 mentary Treatise on Theoretical Mechanics, in two parts, The Mac- 
 millan Co., New York, 1893. 
 
 Every student of mathematical physics should read a portion, 
 at least, of Sir Is aac Newton's Pri ndp ia. 
 
 An extremely interesting and readable book is Poinso t's Theorie 
 Nouvelle de la Rotation des Corps , second edition, Paris, 1851. 
 Harold Crabtree's Spinning Tops, Longmans Green & Co., 
 London, 1909 (about), is a good mathematical treatment of the 
 theory of the rotation of a rigid body. 
 
 The most exhaustive treatises on theoretical mechanics are; 
 Geo. M. Minchin's Treatise on Statics, third edition, two volumes. 
 The Clarendon Press, Oxford, 1886; Ed ward J. Rout h's Ele- 
 mentary Rigid Dynamics, Macmillan & Co., London, 1860; and 
 Edward J. Routh's Advanced Rigid Dynamics, Macmillan & Co., 
 London, 1860. 
 
 See lecture 7 in Acht Vorlesungen uber theoretische Physik by 
 Max Planck, S. Hirzel, Leipsig, 1910. This lecture deals with the 
 most fundamental principle of physics, the so-called principle of 
 least action. 
 
 THEORY OF SOUND. ♦ 
 
 One of the best books on this subject for the beginner is J. H. 
 Poynting and J. J. Thomson's Sound, Charles Griffin & Co., 
 London, 1899. See also the article Acoustics in the 9th edition 
 
36 APPENDIX C. 
 
 of the Encyclopedia Britannica. Helmholtz^s great work The 
 Sensations of Tone (English translation by Alexander J. Ellis), 
 contains a series of appendices on mathematical theory. 
 
 The most comprehensive treatise is L ord R ayleigh^s Theory of 
 Sound in two volumes, Macmillan & Co., London, 1877. Second 
 edition revised and enlarged 1894. 
 
 THEORY OF ELECTRICITY AND MAGNETISM. 
 
 An understanding of vector analysis, as developed in chapter IX 
 of this text, is absolutely necessary before one can begin the 
 study of Maxwell's theory of electricity and magnetism. 
 
 A very simple discussion of Maxwell's theory is given in Frank - 
 lin's Electric Waves, pages 186-196, The Macmillan Co., New 
 York, 1909. The beginner will be greatly helped by E. Atkinson's 
 translation of Mascart and Joubert's Treatise on Electricity and 
 Magneti sm, Vol. I, Thos. de la Rue & Co., London, 1883. 
 
 The great treatise on this subject is Maxwell' s original treatise 
 in two volumes entitled Electricity and Magnetism. The first 
 edition of this work appeared in 1873. Third, edition very 
 slightly altered, The Clarendon Press, 1891. 
 
 The study of Maxwell's treatise is greatly facilitated by Mascart 
 and Joubert's treatise above mentioned ; by the study of Heinrich 
 Hertz's Ele ctric Waves , translated by D. E. Jones, Macmillan & 
 Co., London, 1893; and by the study of A. G. Webster's Electricity 
 and Magnetism , Macmillan & Co., London, 1897. 
 
 A most excellent treatise for the student is Abr aham and 
 Fop pl's Theor ie de r Ehctrizitdt; Vol. I, Einfuhrung in die Max- 
 wellsche Theorie, 3d edition, Leipsig, 1907; Vol. II, Electromag- 
 neschie Theorie der Strahlung (Electronentheorie) , Leipsig, 1905. 
 
 Recen t Researches Electricity and Magnetism by J. J. Thomson, 
 The Clarendon Press, Oxford, 1893, contains a great deal of 
 interest especially on the subject of electric waves. See also J. J . 
 Thomson's _ Cowc^t^c^ion qf_ Electricity thr ough Gases, Cambridge 
 University Press, 1906. This important book deals principally 
 with the electron theory, although, of course, the book deals 
 almost entirely with experimental researches. 
 
APPENDIX C. 37 
 
 THE THEORY OF LIGHT. 
 
 One of the best books for the student is Thomas P rest on's 
 Theory of Lig hts third edition, Macmillan & Co., London, 1901. 
 
 An extremely interesting book, partly theoretical and partly 
 descriptive, is A. A. Michelson's Ligh t Waves and th eir U seSf Uni- 
 versity of Chicago Press, 1903. R. W . Wood's PhydcaLQpiic&f 
 second edition. The Macmillan Co., New York, 1910, contains a 
 great deal of interesting and important theory. 
 
 The theory of lenses and optical instruments is treated in a 
 very exhaustive manner by Czapski, von Rohr and Eppfenstein 
 in Winkelmann's Handbuch der Phydk. 
 
 One of the most interesting series of original memoirs is that 
 of Augustin Fresnel, see Fresnel's Oevres Completes^ Vol. I, pages 
 1-382, Paris, 1866. 
 
 A very complete treatise on the theory of light (electro- 
 magnetic) is that of P. Prude ; Enghsh translation by C. R. Mann 
 and R. A. MiUikan, Longmans Green & Co., New York, 1902. 
 
 HYDROMECHANICS. 
 
 One of the best treatises on this subject for the student is the 
 article Hydromechanics in the ninth edition of the Encyclopedia 
 Britannica. Part I of this article is devoted to Hydrostatics. In 
 this part the problem of the figure of the earth is discussed and 
 also the important practical problem of the stability of floating 
 bodies. Part II of this article is devoted to Hydrodynamics, the 
 highly mathematical theory of the motion of a frictionless fluid. 
 Part III of this article is devoted to Hydraulics from the point of 
 view of the experimental physicist and the engineer. Parts 
 I and II were written by A. G. Greenhill and Part III was written 
 by W. C. Unwin. Professor Unwin's article has been pubUshed 
 as a separate treatise, Macmillan & Co., London, 1907. 
 
 See also G. M. Minchin's Treatise on Hydrostatics, two volumes, 
 
 second edition, Clarendon Press, Oxford, 1912; Horace Lamb's 
 
 Hydrodynamics I Cambridge University Press, 1879; third edition 
 
 1906; and A. B. Bassett's Hydrodynamics, two volumes, Deighton 
 
 Bell & Co., Cambridge, 1888. 
 20 
 
38 APPENDIX C. 
 
 THEORY OF ELASTICITY. 
 
 An understanding of vector analysis as developed in chapter IX 
 of this text is very helpful in the study of the theory of elasticity. 
 Thus the ideas which are established in Art. 136 are the foundation 
 of the elementary theory of elasticity as given ii; Franklin and 
 MacNutt's Mechanics and Heat, pages 182-218, The Macmillan 
 Co., New York, 1910. This is perhaps the simplest existing 
 elementary treatise on the theory of elasticity. 
 
 The matter which is discussed in Art. 136, namely, the theory 
 of elastic strains, is discussed in W. K. Clifford's Kinematic, 
 part I, Macmillan & Co., London, 1878, pages 158-221. 
 
 One of the best treatises on the theory of elasticity is W. J. 
 Ibbetson's Mathematical Theory of Elasticity, Macmillan & Co., 
 London, 1887; see also A. E. H. L ove's Mathematical Theory of 
 Elasticity, two volumes, Cambridge University Press, 1892. 
 
 The greatest reference work in the theory of elasticity is Karl 
 Pearson's History. 
 
 CRYSTALLOGRAPHY. 
 
 It is not generally known that one of the most interesting and 
 remarkable branches of mathematical physics is the theory of 
 crystallography. Thus the purely mathematical theory of regu- 
 lar-point-systems in space is in complete accord with experi- 
 mental studies of crystal forms. A regular-point-system in space 
 is called a space lattice. The purely mathematical theory of the 
 space lattice is treated in Sohncke's Theorie der Krystallstruktur, 
 B. G. Tuebner, Leipsig, 1879. Very complete treatises on 
 crystallography are Groth's Physikalische Krystallographie, Wm. 
 Englemann, Leipsig, 1895; and Liebisch's Grundriss der Physi- 
 kalischen Krystallographie, Veit & Co., Leipsig, 1896. In both of 
 these books the geometrical theory of crystallography is fully 
 treated. 
 
INDEX 
 
 Acceleration and velocity, 55 
 
 in circular motion, 70 
 Algebraic integration, 82 
 Angular acceleration and torque, 146 
 Arbitrary variations, 9 
 Area under a curve, 28 
 Artificial functions and natural func- 
 tions, 15 
 Average, discussion of, 121 
 
 value of a function, 121 
 
 Barrel hoop, discussion of, 72 
 Beam, the problem of the bent, 116 
 Bent beam, problem of, 115 
 Boat, starting of a, 177 
 stopping of a, 178 
 
 Center of gravity, 123 
 
 of mass. See center of gravity, 
 of pressure, 134 
 Change, rate of, 2 
 Circle, the osculating, 66 
 Complex quantity, definition of, 163 
 geometrical representation 
 
 of, 163 
 use of, 153 
 Component slopes and resultant 
 
 slopes, 96 
 Constant of integration, 33 
 Constant quantities and variable 
 
 quantities, 1 
 Continuous variables and discon- 
 tinuous variables, 4 
 Convergence of a vector field, 229 
 Convergent series, 43 
 Cosines, hyperbolic, 166 
 Curl, cartesian expression for, 238 
 divergence of, 240 
 example of, 237 
 of a gradient, 239 
 of a vector field, 236 
 Curvature, 61 
 
 radius of, 65 
 Cyclometer, the, 74 
 
 Dam, force exerted upon, 136 
 
 Damped oscillations, 185 
 Decrements and increments, 2 
 Demoivre's theorem, 162 
 Dependent variables, 13 
 Derivative, Graphical representation 
 of a, 18 
 of a function, 17 
 Derivatives, successive, 54 
 Definite integrals and indefinite in- 
 tegrals, 34 
 Differential equation, 31 and 168 
 Degree and order of, 168 
 general solutions and par- 
 ticular solutions of, 172 
 linear, 172 
 of wave motion, 190 
 ordinary and partial, 168 
 partial, 89 
 pure and mixed, 169 
 Differentiation and integration, 32 
 formulas for, 38 
 by development, 38 
 by rule, 38 
 geometric, 70 
 graphical, 38 
 
 of an exponential function, 48 
 of a function of a function, 42 
 of an implicit function, 91 
 of a logarithm, 46 
 of a product, 40 
 of a quotient, 41 
 of a sum, 39 
 
 of trigonometric functions, 50 
 partial, 87 
 
 successive, 88 
 physical, 38 
 rules for, 38-54 
 successive, 54 
 Discontinuous variables and con- 
 tinuous variables, 4 
 Distributed scalar, 218 
 
 vector, 218 
 Divergence, Cartesian expression for, 
 229 
 of a curl, 240 
 of a vector field, 228 
 
40 
 
 INDEX. 
 
 Equation, differential, partial, 89 
 Errors of observation, influence upon 
 
 a result, 109 
 Euler's expression for sine and cosine, 
 
 164 
 Expansions in series, 153 
 
 Fields, scalar and vector, 217, 218 
 Forced oscillations, 188 
 Fourier's theorem, 199 
 
 application of, 204 
 Function, definition of, 13 
 
 derivative of a, 17 
 
 graphical representation of a, 16 
 
 implicit differentiation of, 91 
 
 law of growth of, 31 
 
 tabulation of, 15 
 Functions, natural and artificial, 15 
 
 which have the same derivative, 
 25 
 
 Gravity, center of, 123 
 Geometric differentiation, 70 
 Grade, definition of, 6 
 Gradient, curl of, 239 
 
 definition of, 6 
 
 line integral of, 223 
 
 of a distributed scalar, 219 
 
 of a hill, 93 
 
 temperature, discussion of, 7 
 Gradients, examples of, 98 
 Graphical representation of a deriva- 
 tive, 18 
 of a function, 16 
 Gyration, radius of, 142 
 
 Harmonic motion, 58 
 Hill, gradient of a, 93 
 
 slope of a, 93 
 
 surface, area of, 105 
 
 volume of a, lOO 
 Hooke's law, 115 
 Hoop, discussion of, 72 
 HyperboUc sines and cosines, 166 
 
 Implicit function, differentiation of, 
 
 91 
 Increments and decrements, 2 
 Indefinite integrals and definite inte- 
 grals, 34 
 Independent variables, 13 
 Inertia, moment of. 140 
 
 moments of, about parallel axes, 
 147 
 
 Inertia, moments of, table of, 151 
 Infinitesimals, method of, 22 
 Integrals, definite and indefinite, 34 
 
 table of. See Appendix B. 
 Integrating machine, 74 
 Integration, 74 
 
 algebraic, 82 
 
 and differentiation, 32 
 formulas for, 38 
 
 by rules. See footnote, page 83. 
 
 by steps, 79 
 
 constant of, 33 
 
 mechanical, 74 
 
 multiple. See Integration, Par- 
 tial. 
 
 partial, 87 and 99 
 
 rules for, 84 
 Irrotational vector fields, 242 
 
 Kelvm's law, 113 
 
 Line integral, Cartesian expression 
 for, 223 
 of a gradient, 223 
 of a vector field, 222 
 reduction of to a surface in- 
 tegral, 239 
 Linear vector field, the, 243 
 
 Maclaurin's theorem, 153 
 
 appHed to the function of 
 two variables, 159 
 Mass, center of. See Center of 
 
 gravity. 
 Maximum and minimum values, 111 
 Mechanical integration, 74 
 Median line of beam, definition of, 117 
 Method of infinitesimal, the, 22 
 Minimum and maximum values. 111 
 Moment of inertia, 140 
 
 of section of a beam, 148 
 of inertia about parallel axes, 147 
 Table of, 151 
 Motion, harmonic, 58 
 
 in a circle, 70 
 Multiple integration. See Partial 
 Integration. 
 
 Natural functions and artificial func- 
 tions, 15 
 
 Oscillations, damped, 185 
 
 forced, 188 
 
 undamped, 181 
 Osculating circle, 66 
 
INDEX. 
 
 41 
 
 Partial differential equation, 89 
 
 differentiation, 87 
 successive, 88 
 
 integration, 87 and 99 
 Planimeter, the, 74 
 Plucked string, the vibration of, 197 
 Potential, existence theorem of, 225 
 
 multivalued, 226 
 
 of a vector field, 225 
 
 scalar, 225 
 
 and vector, 242 
 Pressure, center of, 134 
 Principle of superposition, 174 
 Prismoid formula, the, 102 
 
 Quantities, constant and variable, 1 
 
 Radius of curvature, 65 
 
 of gyration, 142 
 Rate of change, 2 
 Resultant slopes and component 
 
 slopes, 96 
 Rotational vector fields, 242 
 
 Scalar and vector quantities, 211 
 field, 217 
 
 gradient of, 219 
 volume integral of, 219 
 function, 218 
 potential, 225 
 
 and vector potential, 242 
 Series, convergent, 43 
 Sines, hyperbolic, 166 
 Slope of a hill, 93 
 
 Slopes, component and resultant, 96 
 Solenoidal vector fields, 235 
 Space analysis, 210 
 Speed-time curve, the, 79 
 Spring, work required to stretch a, 26 
 Starting of a boat, 177 
 Stopping of a boat, 178 
 Stream lines, 221 
 Stretched string, equation of motion 
 
 of, 193 
 String, equation of motion of, 193 
 
 plucked, vibration of, 197 
 Successive derivatives, 54 
 differentiation, 54 
 partial differentiation, 88 
 Superposition, the principle of, 174 
 Surface integral, Cartesian expression 
 for, 228 
 of distributed vector, 227 
 
 Surface integral, reduction of to line 
 mtegral, 239 
 of to volume integral, 
 234 
 of a hill, area of, 105 
 
 Table of moments of inertia, 151 
 Tabulation of a function, 15 
 Tangent plane, equation of, 94 
 Taylor's theorem, 157 
 Temperature gradient, discussion of, 7 
 Torque and angular acceleration, 146 
 Tube of flow in a vector field, 235 
 
 Undamped oscillations, 181 
 Unit tube in a vector field, 235 
 
 Variable and constant quantities, 1 
 Variables, continuous and discon- 
 tinuous, 4 
 dependent and independent, 13 
 Variations, arbitrary, 9 
 Vibrations, damped, 185 
 forced, 188 
 undamped, 181 
 Vector analysis, 210 
 
 and scalar quantities, 211 
 field, 218 
 
 convergence of, 229 
 curl of, 236 
 divergence of, 228 
 line integral of, 222 
 potential of, 225 
 rotational and irrotational, 
 
 242 
 solenoidal, 235 
 surface integral of, 227 
 the linear, 243 
 function, 218 
 potential, 241 
 
 and scalar potential, 242 
 products, 213 
 Velocity and acceleration, 55 
 Volume integral of scalar field, 219 
 
 reduction of to surface in- 
 tegral, 234 
 of a hill, 100 
 Vortex motion, 241 
 
 Water gate, force exerted upon, 134 
 Watt-hour-meter, the, 74 
 Wave motion, differential equation of, 
 190 
 
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