GIFT OF Ul, v/yycLL \ A DIFFERENTIAL AND INTEGRAL CALCULUS y&i&fe THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO A FIRST COURSE IN THE DIFFERENTIAL AND INTEGRAL CALCULUS • BY WILLIAM F. OSGOOD, PH.D. PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY REVISED EDITION Neu) gorfc THE MACMILLAN COMPANY 1909 All rights reserved Oft c Q r C. € ^p^ "K, Copyright, 1907, 1909, By THE MACMILLAN COMPANY. Set up and electrotyped. Published October, 1907. Reprinted April, June, 1908. New edition with additions, February, 1909. Norijjooto $regs J. S. Cushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE The treatment of the calculus that here follows is based on the courses which I have given in this subject in Harvard Col- lege for a number of years and corresponds in its main outlines to the course as given by Professor B. 0. Peirce in the early eighties. The introduction of the integral as the limit of a sum at an early stage is due to Professor Byerly, who made this important change more than a dozen years ago. Professor Byerly, moreover, was a pioneer in this country in teaching the calculus by means of problems, his work in this direction dat- ing from the seventies. The chief characteristics of the treatment are the close touch between the calculus and those problems of physics, including geometry, to which it owed its origin ; and the simplicity and directness with which the principles of the calculus are set forth. It is important that the formal side of the calculus should be thoroughly taught in a first course, and great stress has been laid on this side. But nowhere do the ideas that underlie the calculus come out more clearly than in its appli- cations to curve tracing and the study of curves and surfaces, in definite integrals with their varied applications to physics and geometry, and in mechanics. For this reason these sub- jects have been taken up at an early stage and illustrated by many examples not usually found in American text-books* It is exceedingly difficult to cover in a first course in the cal- culus all the subjects that claim a place there. Some teachers will wish to see a fuller treatment of the geometry of special * Professor Campbell's book : The Elements of the Differential and Integral Calculus, Macmillan, 1904, in its excellent treatment of the inte- gral as the limit of a sum, is a notable exception. v 38351 vi PREFACE curves than I have found room for. But I beg to call attention to the importance of the subject of functions of several vari- ables and the elements of the geometry of surfaces and twisted curves for all students of the calculus. This subject ought not to be set completely aside, to be taken up in the second course in the calculus, to which, unfortunately, too few of those who take the first course proceed. Only a slight knowledge of par- tial differentiation is here necessary. . It has been my practice to take up in four or five lectures near the end of the first year as much about the latter subject as is contained in §§ 3-9 of Chap. XIV, omitting the proofs in §§ 7-9, but laying stress on the theorems of these paragraphs and illustrating them by such examples as those given in the text ; and to proceed then to the simpler applications of Chap. XV. Thus the way is pre- pared for a thorough treatment of partial differentiation, a sub- ject important alike for the student of pure and of applied mathematics. This subject was given in the older English text-books in such a manner that the student who worked through their exercises was able to deal with the problems that arise in practice. But modern text-books in the English lan- guage are inferior to their predecessors in this respect.* Multiple integrals are usually postponed for a second course, and when they are taken up, some of the things that it is most important to say about them are omitted in the text-books. It is the conception of the double and triple integral in its relation to the formulation of such physical ideas as the moment of iner- tia and the area of a surface that needs to be set in the fore- front of the course in the calculus. And the theorem that such an integral can be computed by a succession of simple integra- tions (the iterated integrals) should appear as a tool, as a de- vice for accomplishing a material end. The conception, then, of the double integral, its application to the formulation of physical concepts, and its evaluation are the things with which * I have here to except Goursat-Hedrick, A Course in Mathematical Analysis, vol. I ; Ginn & Co., Boston, 1904. PREFACE vii Chap. XVIII deals. In Chap. XIX the triple integral is ex- plained by analogy and computed, the analytical justification being left for those who are going to specialize in analysis. The solution of numerical equations by successive approxi- mations and other methods, illustrated geometrically, and the computation of areas by Simpson's Rule and Amsler's planim- eter are taken up in Chap. XX. In an appendix the ordinary definition of the logarithm is justified and it is shown that this function and its inverse, the exponential function, are con- tinuous. The great majority of problems in the calculus have come down to us from former generations, the Tripos Examinations and the older English text-books having contributed an im- portant share.* For the newer problems I am indebted in great measure to old examination papers set by Professor Byerly and by Professor B. 0. Peirce,f and to recent Ameri- can text-books. It is not possible to acknowledge each time the author, even in the case of the more recent problems, but I wish to cite at least a few of the sources in detail. I am in- debted to Campbell $ for Ex. 4, p. 181; to Granville § for Ex. 45, p. 108 ; to Greenhill || for Ex. 16, p. 188 ; and to Osborne IF for Ex. 43 on p. 107. * In particular, Williamson, An Elementary Treatise on the Differen- tial Calculus, University Press, Dublin, and Todhunter, A Treatise on the Integral Calculus, Macmillan. t Many of these problems have been collected and published, with others, by Professor Byerly in his Problems in Differential Calculus; Ginn & Co., Boston, 1895. With the kind permission of the author I have drawn freely from this source. I Campbell, I.e., Chaps. XXXVI and XXXVII. § Granville, Elements of the Differential and Integral Calculus, p. 129, Ex. 47; Ginn & Co., Boston, 1904. II Greenhill, A Treatise on Hydrostatics, p. 318 ; Macmillan, 1894. H Osborne, Differential and Integral Calculus, p. 129, Ex. 33 ; D. C. Heath & Co., Boston, revised edition, 1906. This book contains an espe- cially large collection of exercises. viii PREFACE In choosing illustrative examples to be worked in the text I have taken so far as possible the same examples that my prede- cessors have used, in order not to reduce further the fund of good examples for class-room purposes by publishing solutions of the same. It is in the interest of good instruction that writers of text-books observe this principle. As regards the time required to cover the course here pre- sented, I would say that without the aid of text-books which I could follow at all closely, I have repeatedly taken up what is here given in about one hundred and thirty-five lectures, ex- tending over a year and a half, three lectures a week. The time thus corresponds roughly to a five-hour course extending throughout one year. To Mr. H. D. Gaylord, who has given me much assistance in reading the proof, and to Dr. W. H. Eoever and Mr. Dunham Jackson, who have aided me with the figures, I wish to ex- press my appreciation of their kindness. Cambridge, September 12, 1907. CONTENTS CHAPTER I Introduction ART. PAGE 1. Functions 1 2. Slope of a Curve 5 CHAPTER II Differentiation of Algebraic Functions. General Theorems 1. Definition of the Derivative . . 9 2. Differentiation of x n 11 3. Derivative of a Constant 13 4. General Formulas of Differentiation 13 6. Three Theorems about Limits 15 6. General Formulas of Differentiation, Concluded .... 21 7. Differentiation of Radicals 25 8. Continuation ; x n , n Fractional 27 9. Differentiation of Algebraic Functions 34 CHAPTER III Applications 1. Tangents and Normals .37 2. Maxima and Minima 39 3. Continuation ; Auxiliary Variables 43 4. Velocity 46 5. Increasing and Decreasing Functions 49 6. Curve Tracing 51 7. Relative Maxima and Minima. Points of Inflection . . .53 8. On the Roots of Equations 57 ix X CONTENTS CHAPTER IV Differentiation of Transcendental Functions ART. PAGE 1. Differentiation of sin a; 62 2. Differentiation of cos a;, tanx, etc 67 3. Inverse Functions .68 4. The Inverse Trigonometric Functions 69 5. Logarithms and Exponentials . . . . . . . .74 6. Differentiation of log x 78 7. The Compound Interest Law 82 8. Differentiation of e x , a x 83 CHAPTER V Infinitesimals and Differentials 1. Infinitesimals 85 2. Fundamental Theorem 89 3. Tangents in Polar Coordinates .90 4. Differentials 91 5. Technique of Differentiation 94 6. Differential of Arc 99 7. Rates and Velocities 101 CHAPTER VI Integration 1. The Area under a Curve Ill 2. The Integral 114 3. Special Formulas of Integration . . . . , . .118 4. Integration by Substitution .. 119 5. Integration by Ingenious Devices 122 6. Integration by Parts 125 7. Use of the Tables 126 8. Length of the Arc of a Curve . . . . . . 129 CHAPTER VII Curvature. Evolutes 1. Curvature 134 2. The Osculating Circle 138 CONTENTS xi ART. PAGE 8. TheEvolute 139 4. Properties of the Evolute 143 CHAPTER VIII The Cycloid 1. The Equations of the Cycloid 146 2. Properties of the Cycloid 147 3. The Epicycloid and the Hypocycloid 149 CHAPTER IX Definite Integrals 1. A New Expression for the Area under a Curve . lj 2. The Fundamental Theorem of the Integral Calculus . 3. ( Volume of a Solid of Revolution 157 4. Other Volumes 159 5. Fluid Pressure 161 6. Duhamel's Theorem 164 7. Length of a Curve 166 8. Area of a Surface of Revolution ...... 167 9. Centre of Gravity 169 10. Centre of Gravity of Solids and Surfaces of Revolution . . 170 11. Centre of Gravity of Plane Areas 172 12. General Formulation 174 13. Centre of Fluid Pressure 175 14. Moment of Inertia 176 15. A General Theorem 179 16. The Attraction of Gravitation .181 17. Proof of Formula (3). Variable Limits of Integration . 184 CHAPTER X * Mechanics 1. The Laws of Motion 190 2. Absolute Units of Force 194 3. Simple Harmonic Motion 201 4. Motion under the Attraction of Gravitation .... 206 5. Constrained Motion 209 xii CONTENTS ART. PAGE 6. Motion in a Resisting Medium 212 7. Graph of the Resistance 216 8. Motion under an Attractive Force with Damping . . . 218 9. Motion of a Projectile 224 CHAPTER XI The Law of the Mean. Indeterminate Forms 1. Rolle's Theorem . . 229 2. The Law of the Mean 230 3. Application 231 4. Indeterminate Forms. The Limit § 231 5. A More General Form of the Law of the Mean .... 234 6. The Limit #, Concluded 235 7. The Limit = . . . .236 8. The Limit • oo 238 9. The Limits 0°, 1", oo°, and oo - oo 239 CHAPTER XII Convergence op Infinite Series 1. The Geometric Series 243 2. Definition of an Infinite Series ....:.. 244 3. Tests for Convergence 245 4. Divergent Series 248 5. The Test-Ratio Test 250 6. Alternating Series 252 7. Series of Positive and Negative Terms ; General Case . . 254 8. Power Series 257 9. Operations with Infinite Series . 258 CHAPTER XIII Taylor's Theorem 1. Maclaurin's Series . 262 2. Taylor's Series 264 3. Proof of Taylor's Theorem 266 4. A Second Form for the Remainder 269 5. Development of e x , sin x, cos x 269 CONTENTS xiii ART. PAGE 6. The Binomial Theorem 272 7. Development of sin -1 x 274 8. Development of tanx 274 9. Applications 275 CHAPTER XIV Partial Differentiation 1. Functions of Several Variables. Limits and Continuity . . 282 2. Formulas of Solid Analytic Geometry 283 3. Partial Derivatives 287 4. Geometric Interpretation 288 5. Derivatives of Higher Order 290 6. The Total Differential 292 7. Continuation. Change of Variable 295 8. Conclusion 298 9. Euler's Theorem for Homogeneous Functions .... 300 10. Differentiation of Implicit Functions 302 11. A Question of Notation . . 306 12. Small Errors 306 13. Directional Derivatives 308 14. Exact Differentials 309 CHAPTER XV Applications to the Geometry of Space 1. Tangent Plane and Normal Line of a Surface .... 316 2. Tangent Line and Normal Plane of a Space Curve . . . 317 3. The Osculating Plane 323 4. Confocal Quadrics 326 5. Curves on the Sphere, Cylinder, and Cone .... 329 6. Mercator's Chart 331 CHAPTER XVI Taylor's Theorem for Functions of Several Variables 1. The Law of the Mean 334 2. Taylor's Theorem 335 Maxima and Minima 336 4. Test by the Derivatives of the Second Order .... 340 xiv CONTENTS CHAPTER XVII Envelopes ART. PAGK 1. Envelope of a Family of Curves 344 2. Envelope of Tangents and Normals 348 3. Caustics 350 CHAPTER XVIII Double Integrals 1. Volume of Any Solid 351 2. Two Expressions for the Volume under a Surface ; First Method 353 3. Continuation ; Second Method 356 4. The Fundamental Theorem of the Integral Calculus . . . 357 5. Moments of Inertia 359 6. Theorems of Pappus 362 7. Polar Coordinates 363 8. Areas of Surfaces 367 9. Cylindrical Surfaces 370 10. Analytical Proof of the Fundamental Theorem ; Cartesian Co- ordinates ... 371 11. Continuation ; Polar Coordinates 373 12. Surface Integrals 374 CHAPTER XIX Triple Integrals 1. Definition of the Triple Integral 380 2. The Iterated Integral '. 382 3. Continuation ; Polar Coordinates 386 4. Line Integrals 390 CHAPTER XX Approximate Computations. Hyperbolic Functions 1. The Problem of Numerical Computation . . . . . 398 2. Solution of Equations. Known Graphs 398 3. Newton's Method 399 4. Direct Use of the Tables 402 5. Successive Approximations 403 CONTENTS xv ART. PAGE 6. Definite Integrals. Simpson's Rule 406 7. Amsler's Planimeter ., . 409 8. The Hyperbolic Functions 412 APPENDIX A. The Exponential Function 417 B. Functions without Derivatives 422 Supplementary Exercises 424 Index ............ 460 CALCULUS CHAPTER I INTRODUCTION 1. Functions. The student has already met the idea of the function in the graphs he has plotted and used in Algebra and Analytic Geometry. For example, if the graph is a straight line ; if y 2 = 2 mx, y = ± V2mx, it is a parabola, and if ?/ = sin x, we get a succession of arches that recur periodically. Thus a function was thought of originally as an expression involving x and having a definite value when any special value is given to x: f(x)=2x + 3, or f(x) = ± V2ma?, or f(x) = sin x. Other letters used to denote a function are (x), \f/(x) } F(x), etc. We read /(a;) as "/of xP Further examples of functions are the following: (a) the volume of a sphere, V, as a function of the length of the radius, r : r-.4«f«i 2 CALCULUS , (&) the distance s that a stone falls when dropped from rest, as a function of the time t that it has been falling : (c) the sum of the first n terms of a geometric progression : s n = a + ar -f- ar 2 H -f ar n ~ l , as a function of n : a — ar n s„ = 1-r In higher mathematics the conception of the function is enlarged so as to include not merely the case that y is actually expressed in terms of x by a mathematical formula, but also the case of any law whatever by which, when x is given, y is determined. We will state this conception as a formal definition. Definition of a Function, y is said to be a function of x if when x is given, y is determined. As an example of this broader notion of the function con- sider the curve traced out by the pen of a self -registering ther- mometer. Here, a sheet of paper is wound round a drum which turns slowly and at uniform speed, its axis being ver- tical. A pen, pressing against this drum, is attached to a thermometer and can move vertically up and down. The height of the pen above the lower edge of the paper depends on the temperature and is proportional to the height of the temperature above a given degree, say the freezing point. Thus if the drum makes one revolution a day, the curve will show the temperature at any time of the day in question. The sheet of paper, unwound and spread out flat, exhibits, then, the temperature y as a function of the time x* * Objection has been raised to such illustrations as the above on the ground that the ink mark does not define y accurately for a given x, since the material graph has appreciable breadth. True ; but we may proceed here as in geometry, when we idealize the right line. What we see with our eyes is a taut string or a line drawn with a ruler or a portion of the INTRODUCTION 3 Other examples of this broader conception of the function are : the pressure per square inch of a gas enclosed in a vessel, regarded as a function of the temperature ; and the resistance of the atmosphere to the motion of a rifle bullet, regarded as a function of the velocity. A function may involve one or more constants, as, for example : f(x)=ax-\-b, (x) = tan ax. Here, a and b are any two numbers, which, however,- once chosen, are held fast and do not vary with x. If y is a function of x, y =f(x), then x is called the indepen- dent variable and y the dependent variable. The independent variable is the one which we think of as chosen arbitrarily, i.e. we assign to it at pleasure any values which it can take on under the conditions of the problem. The other variable or variables are then determined. Thus when we write: s=16f, we think of t as the independent, s as the dependent variable. But if we solve for t and write : l ~ 4' then we think of s, which is here necessarily restricted to posi- tive values, as the independent, t as the dependent variable. In general, if two variables are connected by a single equation, as for example ~ pv = C, where C is a constant, either may be chosen as the indepen- dent variable, the other thus becoming the function. path of a stone thro ?n hard. These are not straight lines ; but they sug- gest a concept obtained by thinking of finer and ever finer threads and narrower and ever narrower lines, and thus we get at the straight lines of geometry. So here, we may think of the actual temperature at each instant as having a single definite value and thus being a function of the time, the ideal graph of this function, then, being a geometric curve that lies within the material belt of ink traced out by the pen. 4 CALCULUS A function may depend on more than one independent vari- able. Thus the area of a rectangle is equal to the product of two adjacent sides. Further examples: /0,2/) = ilog 2 + 2/ 2 )> (x, y, z) = ax 2 + by 2 -f- cz 2 — d. Again, it may happen that to one value of x correspond several values of y, as when x 2 + y 2 = a 2 , y=z±va 2 — x 2 . y is then said to be a multiple-valued function of x. But usually, when this is the case, it is natural to group the values so as to form a number of single-valued functions. In the above example these will be* y = Va 2 — x 2 and V — — Va 2 — x 2 . In this book a function will be understood to be single- valued unless the contrary is explicitly stated. A function is said to be continuous if a small change in the variable gives rise to a small change in the function. Thus the function 1 y = -> X whose graph is a hyperbola, is in general continuous ; but when x approaches 0, y increases numerically without limit, and so at the point x — the function is discontinuous. * The sign y/ means the positive square root of the radicand, not, either the positive or the negative square root at pleasure. Thus, V4 is 2, and not — 2. This does not mean that 4 has only one square root. It means that the notation VI calls for the positive, and not for the negative, of these two roots. Again, Vz 2 = x, if a; is positive ; VaJ 2 = _ x, if x is negative. A similar remark applies to the symbol 2 ^/, which likewise is used to mean the positive 2 wth root. INTRODUCTION It is frequently desirable to use merely the numerical or absolute value of a quantity, and to have a notation for the same. The notation is : \x\, read " the absolute value of x." Thus, | — 3 | = 3 ; 1 3 | = 3. Again, whether a be posi- tive or negative, we always have Va 2 = | a\. 2. Slope of a Curve. We have learned in Analytic Geometry how to find the slope of some of the simpler curves. The method is of fundamental importance in the Calculus, and so we begin by recalling it. Consider, for example, the parabola : (1) y X 2 . Let P, with the coordinates (x , 2/ ), be an arbitrary point on this curve, and let P': (x', y'), be a second point. Pass a secant through P and F. Let x' = x + h, 2/o + &• Then the slope of the secant will be : tanr' = O Fig. 1 .'/o a% When-P f approaches P as its limit, the secant rotates about P and approaches the tangent as its limit, its slope approaching the slope of the tangent : * lim taiiT' = taiiT. p=p We wish to evaluate this limit. * The sign.= means that limit, without, how- ever, ever being allowed to reach this limit. For, if i* were to coincide with P, we should no longer have a determinate secant, one point not to determine a straight line. 6 CALCULUS Suppose the point Pis the point (1, 1). Let us compute k and tan r' = k/h for a few values of h. Here, x = 1, y Q = 1. Ifh=.l, I , then x 1 — x + h = 1.1, y 1 =zy + k = 1.21 and k = y'-y = .21; hence tanr' = - = 2.1. The following table shows further sets of values of h, k, and tan r' that belong together : ft A". tan r = j .1 .01 .001 .21 .0201 .002001 2.1 2.01 2.001 It is the last column that we are chiefly interested in, and its numbers appear to be approaching nearer and nearer to 2 as their limit. Let us prove that this is really so. As the proof is just as simple for an arbitrary point P, we will return to the general case. Since P and P' lie on the curve (1), we have : (2) y = x 2 , (3) y Q + k = (x + hy = x* + 2x Q h + h\ Hence, subtracting (2) from (3), we get : (4) k = 2x h + h 2 , k it and finally : tanr' ixo + h. Now let P' approach P as its limit. Then h approaches and we have : k lim tan r' = lim- = lim(2a? 4- 7i), P'=P 7i=0 Jl h±Q (5) tanr = 2ic . If, in particular, sc =l, then tan t=2, as we set oux to prove. INTRODUCTION 7 EXERCISES 1. If f(x) = x*-3x + 2, show that /(l) == 0, and compute /(0), /(- 1), /(1£). find/(V2) correct to three significant figures. Ans. —.0204. 3. If F(x) = (x — x 3 )smx, find all the values of x for which F(x) = 0. 4. If a )=»-Va*-a>", find "/(a) and /(0). 5-7T 6. If in the preceding question a = cos — -, compute /(0) to 6 three' significant figures. 7. If ij/(x) = x$ — x~%, find if/(S). 8. If f(x) = x\og 10 (12-x*), find/(-2)and/(3i). 9. Solve the equation X s — xy -\- S = 5y for y, thus expressing y as a function of x, 10. If A*) 838 *! show that /(a?) /(?/) « A* + 2/)- 11. Continue the table of § 2 two lines further, using the values h = .0001 and h = .00001. 8 CALCULUS 12. Find the slope of the curve Sy = Sx s at the point (2, 3), first preparing a table similar to that of § 2 and then proving that the apparent limit is actually the limit. 13. Find the slope of the curve y = X s — x 2 at any point (x , y Q ). 14. Find the slope of the curve y = ax 2 + bx + c at the point (x , y ). 15. Find the slope of the curve X at Oo, 2/o). CHAPTER II DIFFERENTIATION OF ALGEBRAIC FUNCTIONS GENERAL THEOREMS 1. Definition of the Derivative. The Calculus deals with varying quantity. If y is a function of x, then x is thought of, not as having one or another special value, but as flowing or growing, just as we think of time or of the expanding cir- cular ripples made by a stone dropped into a placid pond. And y varies with x, sometimes increasing, sometimes de- creasing. Now if we consider the change in x for a short interval, say from x = x to x = x', the corresponding change in y, as y goes from y to y\ will be in general almost pro- portional to the change in x, as we see by looking at the graph of the function ; for X O/Q and tanr' approaches the limit tanr, i.e. comes nearer and nearer to the fixed value tan t, the slope of the tangent. The determination of this limit : lim y^^ = ttmT, z'±xo X — Xq is a problem of first importance, and we shall devote the next few chapters to solving it for the functions we are already familiar with and to giving various applications of the results. 9 10 CALCULUS We will first formulate the idea we have just explained as a definition. Let (i) y=m be a given function of x. Form the function for an arbitrary- value x of x : (2) 2/o =/(*«), and then give to x an increment, Ax-, i.e. consider a second value x' of x and denote the difference x' — x by the symbol * Ax: x' — x = Ax ; x' = x + Ax. The function ?/ will thereby have changed from the value y a to the value (3) V' =/(*») and hence have received an increment y'-y ==Ay; \j = y + Ay. From (3), written in the form : (4) y + Ay=f(x + Ax), and (2), we obtain by subtraction : Ay =f(x + Ax) -f(x ), L hence and hence A .y = /Oo + Aa -ZOO Ax Ax Definition of a Derivative. The limit which the ratio (5) approaches ivhen Ax approaches : * The student must not think of this symbol as meaning A times x. We might have used a single letter, as h, to represent the difference in question : x' = x + h ; but h would not have reminded us that it is the increment of #, and not of y, with which we are concerned. The nota- tion is read " delta aj." DIFFERENTIATION' OF ALGEBRAIC FUNCTIONS 11 (6) Km £* or lim /(*• + **)-/(*.) Ax = 0A.X" Ax = AX is called the derivative of y with respect to x and is denoted by D x y, (read: "Dxofy"): (?) As=o Ax D x y. y pJ / / m J^L ~'\ Ay ^-~-_____— ~ = cu , y + Ay = c(u + Au) i hence Ay = cAu, Ax Ax' and lira *M= lira (c^\ Az=oA# Ax=oV Ax) 14 CALCULUS The limit of the left-hand side is D x y. On the right. An/ 'Ax approaches D x u as its limit, hence the limit of the right-hand side is* cD x u, and we have D x (cu) — cD x u, q. e. d. Theorem II. The derivative of the sum of two functions is equal to the sum of their derivatives : (II) D x (u + v) = D x u + D x v. For, let y = u + v. Then y = u + v 0} y + Ay = Uq + Au + % + Av, hence Ay = Au + Av, and Ay _ Au Av Ax Ax Ax When Ax approaches 0, the first term on the right approaches D x u and the second D x v. Hence the whole right-hand side approaches * D x u -+- D x v, and we have t Aw v [Au , Av\ t Au . v &v lim — £ == lira [ 1 \ = hm f- hm — Ax=oAa? Ax=o^Aa7 Ax) Ax=o Ax Ax=oAx or D x y = D x u + D x v, q. e. d. Corollary. The derivative of the sum of any number of functions is equal to the sum of their derivatives. If we have the sum of three functions, we can write U-\-V + W=:U+(v-\- W). Hence D x (u + v + w) = D x u + D x (v -f w) = D x u + D x v + D x w. * For a careful proof of this point cf. § 5. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 15 Next, we can consider the snm of four functions, and so on. Or we can extend the proof of Theorem II immediately to the sum of n functions. Polynomials. We are now in a position to differentiate any polynomial. For example : D„(7a?*-5 + • • • + lim X n . ' Theorem B. TJie limit of a product is equal to the product of the limits: lim (IF)=(lim X)(lim Y). * Here XY = (A + e) (B + v ) = AB + Arj + Be + erj. By Theorem A, Corollary, the limit of the right-hand side is AB + lim (A v ) + lim (Be) + lim (erj). The last limit is obviously 0. As regards the first two, it is easy to see that * if a variable (as rj) approaches 0, then the product of any constant (as A) times this variable must also approach 0. An unfavorable case would be that in which the constant is very large, say 10,000,000. But even then the variable, as it decreases, will finally become and remain numerically less than 10" 7 = 100 J )(X)0 , and so the product becomes DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 17 less than 1. As the variable decreases still further, it becomes and remains numerically less than 10~ 8 , then less than 10~ 9 , and so on ; the product thus becoming and remaining numeri- cally less than y 1 ^, y-J-^, and so on. Hence the limit of the product is 0. Thus each of the limits in the above expression is seen to be 0, and hence lim(XY)=AB, q.e.d. In particular, we see that the limit of a constant times a vari- able is equal to the product of the constant into the limit of the variable: lim((7X) = Clim(X). For, a constant is a special case of a variable. Corollary. The limit of the product of n variables is equal to the product of their limits, n being any fixed number : lim (X, X 2 • • • X n ) = (lim X x ) (lim X 2 ) ■ . . (lim X n ) . Theorem C. The limit of the quotient of two variables is equal to the quotient of their limits, provided that the limit of the variable that forms the denominator is not : Hffi?=-iE-|L if limF^O. Y lim Y For X A - A + * A = Be-A v Y B B + n B B(B + V )' X_A Be-A v 1 Y~B+ & £ B The limit of the first fraction in the last term is 0, by Theo- rems A and B. The second fraction ultimately becomes posi- tive and remains less than 2, even if rj is negative. For, since lim 77 = 0, /B will finally become and remain algebraically greater than - 1 . V ^-1 . V 1+J 18 CALCULUS Hence the last term becomes and remains numerically less than twice the first factor, and consequently its limit is 0. •••limf = |, 1-e.d. In particular, we see that, if a variable approaches unity as its limit, its reciprocal also approaches unity : If limX=l, then limi = l. Remark. If the denominator Y approaches as its limit, no general inference about the limit of the fraction can be drawn, as the following examples show. Let Y have the values : Y= i_ j_ __i_ _i_ 10' 100' 1000' ' "' 10"' " (1) If the corresponding values of X are : x= l_ _i L_ J_ 10 2 ' 100 2 ' 1000 2 ' '"' 10 2n ' '" , then lim — = lim — = 0. (2) If X = Y 10» 111 1 Vio' yioo' yiooo' ' 1Q « then X/Y=10 n/2 approaches no limit, but increases beyond all limit. /Q\ Tf Y" — c c c . . . G V) ±- 1Q > 100 > 1000 > . 1Qn » where c is any arbitrarily chosen fixed number, then lim |= ft DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 19 111 1 (4) If X = 10' 100' 1000' 10,000' then X/Y assumes alternately the values +1 and —1, and hence, although remaining finite, approaches no limit. To sum up, then, we see that when X and Y both approach as their limit, their ratio may approach any limit whatever, or it may increase beyond all limit, or finally, although remain- ing finite, i.e. always lying between two fixed numbers, no mat- ter how widely the latter may differ from each other in value, — it may jump about and so fail to approach a limit. Infinity. If lim X=A=f=0 and lim Y=0, then X/Y in- creases beyond all limit, or becomes infinite. A variable Z is said to become infinite when it ultimately becomes and remains greater numerically than any preassigned quantity, however large. If it takes on only positive values, it becomes positively infinite; if only negative values, it becomes negatively infinite. We express its behavior by the notation : limZ=oo or limZ=-fco or \\mZ = — oo . But this notation does not imply that infinity is a limit ; the variable in this case approaches no limit. And so the nota- tion should not be read " Z approaches infinity " or " Z equals infinity ; " but " Z becomes infinite." Thus if the graph of a function has its tangent at a certain point parallel to the axis of ordinates, we shall have for that point : lim — * = oo ; Ax = AX read: " Ay /Ax becomes infinite when Ax approaches 0." Some writers find it convenient to use the expression "a variable approaches a limit" to include the case that the vari- able becomes infinite. We shall not adopt this mode of ex- pression, but shall understand the words " approaches a limit " in their strict sense. 20 CALCULUS If a function f(x) becomes infinite when x approaches a cer- tain value a, as for example f(x) = - for a = 0, x we denote this by writing /(a) =00 (or /(a) = -f- co or = — co , if this happens to be the case and we wish to call attention to the fact). Definition of a Continuous Function. We can now make more explicit the definition given in Chapter I by saying : f(x) is continuous at the point x = a if lim f(x)=f (a). From Exercises 1-3 below it follows that the polynomials are continuous for all values of x, and that the fractional rational functions are continuous except when the denominator vanishes. EXERCISES 1. Show that, if n is any positive integer, lim(X w ) = (limX) rt . 2. If O (x) = c + c x x + c 2 x 2 H \- c n x n , then lim O (x) = G(a) = c 4 c^a 4- c 2 a? -f ••• 4- c n a n : x=a 3. If G (x) and F(x) are any two polynomials and if F(a) =£ 0, then lim ^M = ^M. «*• F(x) F(a) 4. If X remains finite and Y approaches as its limit, show that lim(XY) = 0. 5. Show that ,. x 2 + l 1 lim = - • *=oo3x 2 + 2a;-l 3 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 21 Suggestion : Begin by dividing the numerator and the denominator by x 2 . 6. Evaluate the following limits : / \ v x + 1 / jn v ax -f- bx~ x (a) lim-— — -; (d) hm — — -; (b) hm — - — — - — ■ — -; (e) lim — — — — ; ( C) lim flag+6flr ) ; (/) lim ^=* w x=:oo ex -f dec" 1 ^Vl + a 4 *. 6. General Formulas of Differentiation, Concluded. Theorem III. The derivative of a product is given by the formula : (III) D x (uv) = uD x v + uD x ?*. Let 2/ = uv. Then y = «o% y + Ay = (uq + Aw) O + Ay), Ay = w Av -f- v Aw -f- Au Av, Aw Av . Aw , A Av ^ = M + v + Am-—, Ax Ax Ax Ax and, by Theorem A, § 5 : lim ^= lira L*?) + lim (V-") + lim (Au^Y a*=oAx Ax=o\^ Aay Axio\^ A.ty ax^o \ Aay By Theorem B, § 5, the last limit has the value 0, since lim Au = and lim (Av/ Ax) = D x v. The first two limits have the values u D x v and v D x u respectively.* Hence, dropping the subscripts, we have : D x y= uD x v + vD x u, q. e. d. * More strictly, the notation should read here, before the subscripts are dropped: [D x v'] x=x , etc. Similarly in the proofs of Theorems I, II, and V. 22 CALCULUS By a repeated application of this theorem the product of any number of functions can be differentiated. When more than two factors are present, the formula is conveniently written in the form : (15) D x (uvw) = D x u [ D x v j D x w UVW U V w For a reason that will appear later, this is called the loga- rithmic derivative of uvw. Theorem IV. The derivative of a quotient is given by the formula:* ,. vD D (x), then (V) ft/(y) = ft/(y)ftjr or - (V) D x u = D y u.D x y. Here y = <£ (x ), u Q =/(y ), y + Ay = <£ (xq + Ax), u + Au =/(y + Ay), Att=/Gfo + %)-/(y ), A^ _ f(y + Ay) — /(y ) < Ay # Ax Ay Ax When Ax approaches 0, Ay also approaches 0, and hence the limit of the right-hand side is ( lim fbh+m) -/<*)) (n m *i) = D f( y) DxV . \a*=o Ay /\Ax=oAxy The limit of the left-hand side is D x u, and hence D x u = D y u-D x y, q. e. d. The truth of the theorem does not depend on the particular letters by which the variables are denoted. We may replace, for example, x by t and y by x. Dividing through by the sec- ond factor on the right, we thus obtain the formula : tv<) ft.,, fis. 24 CALCULUS Example. Let u = (ax -f- 6) n , where n is a positive integer. Set y = ax-\-b. Then u=f(y) = y n and D x « = D x 2/ n — D y y n - B x y — ny n ~ x • a = wa(ax -f 6)"- 1 . EXERCISES Differentiate the following functions : 1. y=-^—- Ans. D x y= 1 + a?2 - . 1 t 3 2 - IT-TTtr 4 - 2/ = l + « 2 1 — a? „ 1-1 x 2 3. s = - 5. y = 1 + t " (1 - a) 2 6. ?/ = aj(a + 6a?)\ ^4rcs. D x y = [a + (n -f 1) to] (a + foe)* -1 . 7. y = — , where m is a positive integer. x m l-\-x-\-x 2 n S — x^ + Sx* 8 . y== t -r — 9> x a; ar 10. Show that Formula (12) holds when n is a negative integer. 15. *±£?. 2 + a 16. (a + bx 4- ca 2 )' 1 Dif ferentiate further 11. (*+3)(2-*). 12. (a — x) s . 13. l-3x-x* ^ 9 14. s 17 (1-ary 18 . ,-?t=l (2s + 3) ! 19. Find the slope of the curve y = - at the point (1, 1). DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 25 20. Find the slope of the curve 240?/ = (1 - x) (2 - x) (3 - x) (4 -x) at the point x = 0, y = fa. Ans. — - -^ . 7. Differentiation of Radicals. Let us differentiate y = Va;. Here, y = V# , y + Ay = V# 4- Aa;, Ay __ V^o + Ax — V % n . Ax Ax We cannot as yet see what limit the right-hand side approaches when Ax approaches 0, for both numerator and denominator approach 0, and - has no meaning, cf. § 5. We can, however, transform the fraction by multiplying numerator and denomi- nator by the sum of the radicals and recalling the formula of Elementary Algebra : a 2 - b 2 = (a - b) (a + b). „ A?/ V# + Ax — ' V#„ V# + Ax -}- V# Thus t~ r * -— === — Aa Ax Va + Ax + V# _. 1 , (x + Ax)—x 1 A# ^/ Xq _|_ Ax + V^ V# 4- A# 4- W and hence lim — ^ = lim — ■ — = — • a*=o Ao; ax= o -yJ Xo -|- Aa? 4- Va? 2 V» Dropping the subscript, we have : (16) D x ^x = 1 2Vx We can now differentiate a variety of functions. 26 CALCULUS Example. To differentiate y = Va 2 — x 2 . Introduce a new variable z^a 2 -^ (I / and then apply Theorem Y : * 2/ = Va, A2/ = AV^=AV^.Z>^=-^r-(-2a ; ) 2Vz Va 2 — a 2 Hence Z> x Va 2 -a 2 = ~" g Va 2 — x 2 EXERCISES Differentiate the following functions : 1. y = Va 2 + x 2 . 6. y = —-> Ans. j Vx 2x? 2. u = Va + x 4- Va — #. 7 — . Va 2 — a? 2 3. y = a?Vl— a?. 8. y = ^/'2mx. x-\-\ v a? 5. , Ans. — • vr^? (Vi-z 2 ) 3 ' vi+ x * The student should observe that Theorem V is not dependent on the special letters used to designate the variables. Thus if, as here, y=f(z) and z = \J/ (x) we have D x y = D z y • D z z. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 27 8. Continuation : x n , n Fractional. • The Laivs of Fractional Exponents. Let n=p/q, where p and q are positive integers prime to each other, and consider the function p (17) y = x n = x q for positive values of x. If q is odd, the function is single- valued ; but if q is even, there are two ^th roots of x, and we might define the function of (17) to be double-valued, namely, as ±(-\/x) p . This is. however, inexpedient, and usage has determined that the notation (17) shall be defined to mean the positive root : p x * = (yx) p . If n is a negative fraction, n = — m, then x n = — Moreover, a = 1. x m In Elementary Algebra the following laws of exponents are established : (A) These laws hold without exception when a and b are both positive and m and n are any positive or negative integers or fractions, inclusive of 0. Graph of the Function x n . When n is an integer, n = 1, 2, 3, •••, 10, •••, the graphs are as indicated in Fig. 3; for the slope of the curve (17) : tanr — D x y — nx n ~ Y is positive and increases steadily as x increases if n > 1. I. a m -a n = a m+n ; II. (a m ) n = a mn ; III. a n • b n = (ab) n . 28 CALCULUS Consider next the case that p = 1, q > 1. Here (18) y = x< or Ax Hm Ax y D v x" 30 CALCULUS Keplacing y by its value, a^ /? , we have : i l l- 9 1 y l ~«=:(x q ) 9 = x~=oc? , and thus 1 1 §-' (19) D x x q =-x q • Q This shows that Formula (12) holds even when n = 1/q. Turning now to the general case : p y = x«, i let z = x q ; ?/ = z p . Then by Theorem V, § 5, and (19) : 1 ~i D x y = D x z p = D z z p >D x z = pz p ~ x • -x q , z p - i =(x^y Hence D x y=*x q - x q =£-x q , or (20) ZJ.afssnaj- 1 , when n is any positive integer or fraction. If n is a negative integer or fraction : n = — m, then aj w = — , and hence £c w can be differentiated by the aid of Theorem IV, a; m xr m or -D x a? n = waj n-1 . Consequently Formula (12) holds for all commensurable values of n. We shall show later that it holds for incommensurable values, too, and thus is true for any fixed value of n. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 31 i If n = i, we have 2> X 2Vx ' which agrees with (16). . Example. To differentiate ^ = Va 2 — a^. Let z = a 2 — x 2 . Then 2/ = z% D x y = D x z? = D z z^D x z = \z-*(-2x) = 3(Va 2 -^) 2 Inequalities for x n . If n is a positive integer, then r x n > 1 when # > 1 ; 1 x n 1. Then, by I, 2/* < 1- But y q — x, and #<1 is contrary to hypothesis. Similarly when x <1. Finally, relations I. hold when n=p/q is any positive frac- tion. For x q = (iC 9 ) , and if a?>l, then 1 / \p x q >l and hence (x 9 ) >1. Similarly for a? < 1. When, however, n < 0, the first inequality sign in each line of I. must be reversed and the value x = excluded. The function x n is nevertheless always positive when x > 0. 32 CALCULUS Theorem. If n' > n, then (a) x n ' > x n when x > 1 ; (b) x n ' 0, we see from the relations I. that when a?>l, this last expression is positive ; when x <1, it is negative. Hence the theorem. Graph of the Function x n ; Conclusion. From the theorem just established it follows that the graphs of x n for different positive fractional values of n lie as suggested in Fig. 3, namely : they all pass through the origin and the point (1, 1), and they have no other point of intersection. Their slope is always positive. Of two graphs corresponding to n and n' > n, the latter lies below the former when x < 1 ; above it when a?>l* The student is requested to write out similar statements for the case that n < 0, and to draw the graphs. It is better, how- ever, not to complicate Fig. 3 with these latter graphs, but to deduce them when needed from Fig. 3. Thus confusion will be avoided. Fig. 3 should be permanently visualized. The student should construct such a figure for himself accurately on coordinate paper, using the tables of squares and cubes. * These curves penetrate every part (a) of the square, the coordinates of whose points (x, y) satisfy the relations : 0- s-1 21. If y^x = l + x, ftnd D x y. -4w*. 2a;VaJ 34 CALCULUS 9. Differentiation of Algebraic Functions. When x and y are connected by such a relation as x 2 -f- y 2 = a 2 or X s — xy -f y 5 = or xysmy = x + ylogx, i.e. if y is given as a function of x by the equation or its equivalent, ®(x, y) = ty(x, y), where neither nor ^ re- duces to y, then y is said to be an implicit function of x. If we solve the equation for y, thus obtaining : y =/(«), 2/ thereby becomes an explicit function of a?. It is often difficult or impossible to effect the solution ; but even when it is possible, it is usually easier to differentiate the function in the implicit form. Thus in the case of the first example we have, on dif- ferentiating the equation as it stands with respect to x: D x x 2 + D x y 2 = D x a 2 or 2x + 2yD x y = 0. Hence D x y = — - • y If we differentiate the second equation in a similar manner, we get : 3 x 2 — x D x y — y + 5 y K D x y = 0. Solving for D x y, we obtain : r, 3 x 2 — y by 4 — x DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 35 When F(x, y) is a polynomial in x and y, the function y, denned by the equation F(x,y) = is called an algebraic function. Thus all polynomials and fractional rational functions are algebraic. Moreover, all functions expressed by radicals, as 2/ =Va 2 -x 2 or y = ^ljz£?_^4-V*, * 1 -f-X are algebraic, for the radicals can be eliminated and the result- ing equation brought into the above form. But the converse is not true: not all algebraic equations can be solved by means of radicals. It can be shown that an algebraic function in general is continuous. In case the function is multiple-valued it can be considered as made up of a number of branches, each of which is single-valued and continuous. Assuming this theorem we can find the derivative of an algebraic function in the manner illustrated in the foregoing differentiations. On the assumption just mentioned a short proof of Formula (12), § 2, can be given for the case that n—p/q. Since p y = x q , we have: y 9 = x p . Differentiate each side. with respect to x: D x y q = D x x p = pxr-\ Now D m tf = D y y* • D x y = q X? The proof of this formula which we gave in § 8 does not depend on the above assumption, but is a complete proof. 36 CALCULUS EXERCISES 1. Differentiate y in both, ways, where xy + 4:y = 3x, and show that the results agree. 2. The same for y 2 — 2mx. 3. Find the slope of the curve x 4 -2xy 2 + y 5 =13 at the point (2, 1). Ans. 10. 4. Show that the curves Sy = 2x + x 4 fy 2y + 3x + y 5 = x 3 y intersect at right angles at the origin. CHAPTER III APPLICATIONS 1. Tangents and Normals. The equation of a line passing through a point (# , y ) and having the slope a is y-yo = H x ~ x o), and the equation 'of its perpendicular through the same point is y-yo=-~(x-x ) or x — x + \(y — y ) = 0. A Since the slope of a curve y=f(x) or F(x,y)=0. in the point (x , y ) is [D x y~\ x=x , the equation of the tangent line in that point is (1) 2/ — 2/o= [D x y~] x=:XQ (x - x ). Similarly, the equation of the normal is seen to be : (2) 2/-2/o=-jt— — (x-xo) or x- x + [2>»y],. s (y-3fii)-»0. Example 1. To find the equation of the tangent to the curve y = x? in the point x = J, y = J. Here D x2 / = 3^ [Z> x2 /] x=x =[3^] x = i = |. Hence the equation of the tangent is 2/ — | = f(cc — i) or 3a;— 4?/ — 1 = 0. 37 38 CALCULUS Example 2. Let the curve be an ellipse : a 2 ^b 2 Differentiating the equation as it stands, we get : 2x . 2y n a ^ b 2 x — + -rf D x y = 0, J),?/ - ~ -T- * a 2 b 1 a-y Hence the equation of the tangent is Xt\ / \ «l/o This can be transformed as follows : a 2 y y — a 2 y 2 = — b 2 x x + b 2 x 2 , b 2 x x + a 2 y y = a 2 y 2 + b 2 x x P=0. Let us compute, then, D X P and set it equal to 0: D X P= -^ + 3 = 0, ^100 L^=5.77. 3 ' ys Consequently the best length for the lever is 2x = 11.4 ft., the corresponding value of P being 34.6 lbs. Example. A box is to.be made out of a square piece of card- board 4 in. on a side, by cutting out equal squares from the corners and turning up the sides. Find the dimensions of the largest box that can be made in this way. First express the volume V oi the box in terms of the length x of a side of one of the squares cut out, and plot the graph of V, thus determining approximately the best value for x. Then solve by the Calculus. The foregoing examples suggest a simple test for a maximum or a minimum. Test for a Maximum. If the f miction is continuous ivithin the interval a<.x0, y increases ; Qj) if D x y < 0, y decreases. As an application consider the condition that a curve y =f(x) have its concave side turned upward, as in Fig. 8. The slope of the curve is a function of a?: y tanr = (x). sider the tangent line at a vari- able point P. If w think of 7' as ^ tracing out the r tA - ve and carrying FlG< 8 50 CALCULUS the tangent along with it, the tangent will turn in the counter clock-wise sense, the slope thus increasing algebraically as x increases, whenever the curve is concave upward. And con- versely, if the slope increases as x increases, the tangent will turn in the counter clock-wise sense and the curve will be con- cave upward. Now by the above theorem, when D x tan t > 0, tanr increases as x increases. Hence the curve is concave upward, when D x tan t is positive. The derivative D x tan t is the derivative of the derivative of y. This is called the second derivative of y : (read : " D x second of y "). # The^test for the curve's being concave downward is obtained in a similar manner, and thus we are led to the following important theorem. Test for a Curve's being Concave Upward, etc. The curve y =f(x) is concave upward ivhen D x 2 y>0; concave downward when D x 2 y<0. Fig. 9 * The derivative of the second derivative, D x v /* 2 y) , is called the third derivative and is written D x z y, and so on. APPLICATIONS 51 A point at which the curve changes from being concave up- ward and becomes concave downward (or vice versa) is called a point of inflection. Since D*y changes sign at such a point, this function will necessarily, if continuous, vanish there. Hence : A necessary condition for a point of inflection is that D x 2 y = 0. 6. Curve Tracing. In the early work of plotting curves from their equations the only way we had of finding out what the graph of a function looked like was by computing a large number of its points. We are now in possession of powerful methods for determining the character of the graph with scarcely any computation. For, first, we can find the slope of the curve at any point ; and secondly we can determine in what intervals it is concave upward, in what concave down- ward. As an example let us plot the graph of the function : (1) y = -. x 3 + px + q. Consider first the case q = = 0: (2) y = : X s -J- pX. Here D t y = >&*+i* DJy- = 6x. From the last equation it follows that the curve is concave upward for all positive values of x. Moreover, when x be- comes positively infinite, D x y also becomes positively infinite, no matter what value p may have. The curve goes through the origin and its slope there is 52 CALCULUS Hence the graph is, in character, for positive values of x as shown in Fig. 10. 2»0 * Fig. 10 To obtain the graph for negative values of x we need only observe that the curve is symmetric with respect to the origin. For, if (x, y) be any point of the curve, then x 1 — — x, y' = — y is also a point of the curve. When, therefore, we have once plotted the curve for positive values of x, we need only to rotate the graph through 180° about the origin in order to get the remainder of the curve. Finally we can get the graph of (1) from that of (2) by merely shifting the axis of x to a parallel axis. The formulas for this transformation are : x = x', y = y'-q. Thus (2) becomes : y' — q == x' s + px'. Transposing q and dropping the accents, we get equation (1). This curve is symmetric with respect to the point x = 0,y = $. EXERCISES Use coordinate paper in working these exercises. 1. Show that the curve ? = 3 V 3 + x 2 is concave downward in the interval — 1 < x < 1 and concave upward for all other values of x. Find its slope in its points of APPLICATIONS X""53 inflection and draw the tangent line in each of these points. Hence plot the curve. 2. Plot the curve 4y = x* - 6x? + 8, determining ^^* (a) its intersections with the coordinate axes ; (b) the intervals in which it is concave upward and those in which it is concave downward ; (c) its points of inflection ; (d) the points where its tangent is parallel to the axis of x. Plot the points (a), (c), (d) accurately, using a scale of 2 cm. as the unit, and draw accurately the tangent in each of these points. Hence construct the curve. Plot the following curves : 3. 10y = x*-12x + 9. Note that the curve cuts the axis of x in the point x = 3. 4. y = a? + 2x 2 -13x + 10. 9. y 2 = x 2 + x^. 5. y = x — x*. 10. y 2 =x(x — l)(x — 2). T 1 6. y= - ^11. y = z s' • 1 -fiC 2 1 —x 1 „ T 2 8. 2/ = -^-- 13. y 2 = ~ 1-x ' l+x 2 7. Relative Maxima and Minima. Points of Inflection. A function (i) y=f{x) is said to have a maximum at a point x = x if its value at .^ is larger than at any other point in the neighborhood of x . But such a maximum need not represent the largest value of the 54 CALCULUS function in the complete interval a<±x0, it has a minimum. This condition is sufficient, but not necessary. Thus the function (2) y = a? evidently has a minimum at the point x = 0. But here D x 2 y is not positive, but = 0. Still, the above test is adequate for the great majority of cases that arise in practice. We shall obtain a more general test later. Example. Let y = o?-3a? + l. APPLICATIONS 55 Here D x y = 6x* - 6x = 6x(x> - 1) (x 2 + 1), and hence D x y = for a; = — 1, 0, 1. Furthermore, D x 2 y = 30 a! 4 - 6. Thus [A, 2 #]x=-i= 24 >0, .-. x = — 1 gives a minimum; [D x 2 y~\ x=0 =— 6<0, .*. a? = " " maximum; \_D x y~] x = x = 24 >0, .-. & = 1 " " minimum. As a further application of the test just found let us obtain a sufficient condition for a point of inflection. We have seen that a necessary condition is that D x y = 0; but this is not sufficient, as the example of the function (2) above shows. A geometric feature characteristic of a point of inflection is that the tangent ceases rotating in one direction and, turning back, begins to rotate in the opposite direction. Hence the slope of the curve, tanT, has either a maximum or a minimum at a point of inflection. Conversely, if tan t has a maximum or a minimum, the curve will have a point of inflection. For, suppose tan r is at a maximum when x = x . Then as x, starting with the value x , increases, tan t, i.e. the slope of the curve, decreases alge- braically, and so the curve is concave downward to the right of Xq. On the other hand, as x decreases, tanr also decreases, and so the curve is concave upward to the left of x . Now, by the above theorem, tan t will have a maximum or a minimum if D x tanr = 0, D x 2 tanr ^ 0. Hence, remembering that t2LTLT = D x yj we obtain the following Test for a Point of Inflection. If [Z> x 2 2/],„=0, [D'yl^+0, the curve has a point of inflection at x = x . 56 CALCULUS This test, like the foregoing for a maximum or a minimum, is sufficient, but not necessary. Then Example. Let 12y = x 4 + 2x i -12x 2 + Ux-l. 12 D x y = 4a 3 + 6x 2 - 24a + 14, 12 D 2 y = 12 x 2 + 12* - 24 = 12 (x - 1) (a? + 2), 12Z) x 3 2/ = 12(2a + l). Setting ZVy = 0, we get the points x = 1 and x = — 2. And since 12[A 3 2/],=. = 36^0, l2[2>.'y]„_ 1 = a -36*0 ) we see that both of these points are points of inflection. The slope of the curve in these points is given by the equations : 12[A2/]x,i = 0, 12[D„y]„_ i = 64. Hence the curve is parallel to the axis of x at the first of these points ; at the second its slope is 4i. EXERCISES Test the following curves for maxima, minima, and points of inflection, and determine the slope of the curve in each point of inflection. 1. y = 4x? -15ar + 12a? + l. 4. y = (x- l) 3 (a + 2) 2 . x 2. y = x 3 + x 4 + x 5 . 3. 6y = x 6 -3x 4 + 3x 2 -l. 5 ' y = 2 + 3x* 6. y = (l-x*) s . 7. Deduce a test for dis- tinguishing between two such points of inflection as those indicated in Fig. 12. Fig. 12 APPLICATIONS 57 8. On the Roots of Equations. The problem of solving the equation can be formulated geometrically as follows : To find the points of intersection of the curve with the axis of x : 2/ = 0. Hence we see that we can approximate to the roots as closely as we please by plotting the curve with greater and greater accuracy near the points where it meets the axis of x. It is often a matter of importance to know how many roots there are in a given interval ; for example, the number of posi- tive roots that an equation possesses. One means of answer- ing this question is by the methods of curve tracing above set forth. Consider, for example, the equation : x 6 - 3^ + 1 = 0. The function y=x«-3x 2 + 1 is positive for values of x that are numerically large. For («-!+!> Here the parenthesis apprr Alt hes a positive limit when x in- creases without limit ; the f~ % factor increases without limit, and so the product increases *vi thou t limit. Again, the function has a maximum when x = (see §7), its value there being positive, namely 1 ; and at x — 1 it has a minimum, its value there being negative. Consequently the curve must have crossed the positive axis of x between x = and x = 1, and again when x > 1, and so the above equation has at least two positive real roots. Has it more ? 58 CALCULUS In the interval < x < 1 D x y = 6x(x*-I)(x> + 1) is negative and hence the function is steadily decreasing. The graph, therefore, can have crossed the axis of x but once. Again, when x > 1, D x y is positive, and so the function is always increasing. Hence x the graph can have crossed the axis h of x beyond this point but once. Thus we see that the equation has v!y just two positive roots, and since to FlG 13 each root x = a corresponds a second root x = — a, it has just two nega- tive roots, and so in all just four real roots. A general principle is illustrated in this example, which may be stated as follows. Theorem. If a continuous function f(x) changes sign in an interval a ; nple. The points in which this curve is cut by the line (3) 2/=-? evidently have for their abscissas the roots of the cubic (1).* Now if p ^> 0, the graph will correspond to the first of the two * Another geometric formulation of the problem of finding the roots of the cubic (1) is to consider the intersections of the curves y = x s , y--px-q. APPLICATIONS 59 figures in Fig. 10. Thus the line (3) will cut the curve (2) in just one point, and so equation (1) will have just one real root.* But if p < 0, then the graph corresponds to the second figure in Fig. 10, and we see that it depends on the relative magnitudes of p and q as to whether there are three points of intersection or fewer. The maxima and minima of the function (2) are obtained by setting , D x y = 3a*+p = 0, x=±yJ-£, and it turns out that a minimum occurs at the point f f-V-f' y= 3\-!' a maximum at 2 „ = _?I?J_£. 3' y 3\ 3 Hence if q is numerically greater than these equal and oppo- site values of y, i.e. if . s y Z 27' the cubic (1) will have one real root. If q, however, is numeri- cally smaller : - <, * 27' it will have three real roots ; and if q 2 = 2 _ ±P S it will have two real roots, one of which counts twice, except in the case above mentioned, p = q = 0, where it has one triple root. The first is a special cubic, which can be plotted accurately from the tables once for all ; and then the straight line can be drawn as soon as p and q are assigned special values. Thus we get a graphical solution for any cubic of the above type. * In the special case : p — 0, q = 0, the cubic (1) reduces to x z — 0. It is customary to say that this equation has three coincident roots. t Observe thatp<0, so that fpV— p/S is negative and, further down, — 4p 3 /27 is positive. 60 CALCULUS We can collect all cases under the following Theorem. The cubic equation has X s +_pa ' + 9 = (a) 1 real root when t+ 27 >0; 0) 3 « a a a <0; (C) a a a f\ \ P and Q »< (<%) 1 « a « a « \ p = q=0. In case (cj) one root counts twice; and in (c 2 ) the root counts three times. EXERCISES 1. How many real roots has each of the following equations ? (a) x 5 ~ 5x- 1 = 0. (c) 3.t 4 -4x 3 + 12.t 2 + 7 = 0. (6) 3a? 4 + 4^ + 6a 2 - 1 = 0. (d) x 2n+l +px + a = 0. -4ws. (a) three; (6) two; (c) none. 2., How many real roots have the cubics : (a) a3 3 + 7a-l = 0; (c) a 3 -3a-2 = 0; (6) ^-4^ + 1=0; "(d) a! 3 -a + 3 = 0? 3. How many positive roots has the equation 6x i + Sx s -12x 2 -24:X-l = 0? Ans. One. 4. Show that the function /( * ) = (ar-l) 8 + (^2)» has just one root in the interval 1 < x < 2. 5. How many real roots has the equation Ix 3 - 15 x 2 + 12a + 1 = 0? ^Ins. Three. APPLICATIONS 61 6. Show that, by a suitable transformation of the coordinate axes to parallel axes, the new origin being on the axis of x, namely : x = x' + h, y=y', the equation : y = x?+p 1 x 2 +p 2 x+p 3 can be carried over into the equation : y = x*+px + q. Hence obtain the condition that the cubic: x 3 +p x x 2 +p a x +p s = have three real roots. CHAPTER IV DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS 1. Differentiation of sin a;. First Method. The graph of the function (1) y = sin x can be constructed geometrically by drawing a circle of unit radius and measuring the ordinates corresponding to different angles ; the angle itself, measured in radians, giving the abscissa and being computed arithmetically. y In order to differentiate sin x we have to give to x an arbi- trary value x and compute the corresponding value of y : y = smx . Then give x an increment, Ax, and compute again the corre- sponding value of y : y + Ay = sin (x -f Ax). TRANSCENDENTAL FUNCTIONS 63 Hence (2) Ay = sin (x + Aa;) — sin # , Ay _ sin (# -f- Aa?) — sin x Ax Ax Let us follow these steps geometrically by constructing the successive magnitudes. Fig. 16 explains itself. The radius of the circle is unity, and so MP = sin x Q , M'F = sin (x + Ax). QP' = sin (x -h Ax) — sin x = Ay, PP' = Ax. Hence (3) ^=«Z\ Aa; ppi Now it is easy to see what limit this last ratio ap- proaches when P' approaches P. Suppose first we had in the denominator the chord PP'. Then S^=sin = ZQPT=?-x , pr=p 2 lim -3£— = cos x . p'= ppp* The chord of a small angle differs, however, from the arc only by a small percentage of either. We readily see that W r PP' 1 lim — — = 1. p f =p ppt The student is requested to draw an accurate figure represent- ing the arc and the chord of an angle of 30° and also of 15°. 64 CALCULUS Eeturning now to the ratio (3) and writing it in the form : QP^^QT PP PP PP' pp'' we have, when P' approaches P as its limit : Ax=oAa? \p'=pppij\p'=pppij or dropping the subscripts : (5) D x sin x = cos x. EXERCISE Prove in a similar manner that (6) D x cos x = — sin x. Second Method. The foregoing method has the advantage of being easily remembered. Each analytic step is mirrored in a simple geometric construction. It has the disadvantage, however, of incompleteness. For, first, we have allowed Ax, in approaching 0, to pass through only positive values ; and secondly we have assumed x {) to lie between and ir/2. Hence there are in all seven more cases to consider. An analytic method that is simple and at the same time general is the following. Recall the Addition Theorem for the sine: sin (a -j- b) = sin a cos b + cos a sin b, sin (a — b) = sin a cos b — cos a sin b, whence sin (a + b) — sin (a — b) = 2 cos a sin b } and let a + b = x Q + A#, a — b = x* . Solving these last equations for a and b, we get : . Ax , A& a = Xq H -, b = — • TRANSCENDENTAL FUNCTIONS 65 Thus sin (x -f Ax) — sin x ^rV n ^> Ax" 2; and the difference-quotient (2) becomes : Ay Ax cos . Ax ( ^\ "2" The limit of the first fac- tor on the right is cos# . The limit of the second is of the form : a = a and here, again, we have to do with the limit of the ratio of the arc to the chord. Fig. 17: ^ PP' = 2sina, PP' = 2a. For, in We have seen that the limit (4) has the value 1 by direct inspection of the figure. We can give a formal proof based on the axioms of geometry as follows. First of all, a straight line is the shortest distance between two points, and so PP'(y). Then (y) is called the inverse of the function / (x) . The graph of the former function serves as the graph of the latter, provided in the latter case we take y as the independent, x as the dependent variable. In order to obtain the graph of the * The versed sine and the coversed sine are defined as follows : vers x — 1 — cos x, covers x = 1 — sin x. <2. sin 2x. ,3. cos 2 a;. *4. , X tan-. 2 5. sin a; a + b cos x 14. Prove that TRANSCENDENTAL FUNCTIONS 69 inverse function with x as the independent variable, transform the x, y plane, and with it the above graph, as follows : let a = 2/', V = x'. This is equivalent to rotat- ing the plane through 180° about the bisector of the angle made by the positive coordinate axes. In other words, it amounts to a reflection of the plane in that bisector. We have met an example of inverse functions in the radical, x l '«, Chap. II, § 8. If, as x increases, y steadily increases (or if y steadily decreases), the inverse function will obviously be single-valued. In this case the derivative of the inverse function is obtained from the definition: , y = (x) if x=f(y), and the relation : Ax Ax 1 Ay Fig. 21 »*-& hence lim^/ = lim_, Ax=oAa; Ay=oAa? Ay provided D y x =^0. 4. The Inverse Trigonometric Functions, (a) sin -1 a. The inverse of the function (1) y = sin x is obtained as explained in § 3 by solving this equation for x as a function of y, and is written : (1') x = sin- 1 y, 70 CALCULUS read "the antisine of y." In order to obtain the graph of the function (2) y = sin -1 x we have, then, to reflect the graph of (1) in the bisector of the angle made by the positive coordinate axes. We are thus led to a multiple-valued function, since the line x = x'(—l 0, we have and so finally : * (4) cos?/= Vl — X 2 , 1 D x sm~ 1 x = Vl-x 2 (6) cos l x. The treatment here is precisely similar. We define : (5) y — cos -1 x if # = cos?/, and we make the inverse function single-valued by choosing that value of y which satisfies the relation : (6) 0l. The ordinate of any one of the points of intersection : (1) y = a n , is determined as soon as the value of n has been assigned, and is thus a function of n. From the theorem of p. 32 we know that, when n increases, y increases. Moreover, y is always posi- tive ; it increases without limit when n = -f- oo , and it approaches when n = — oo . As yet the function (1) has been defined only for commen- surable values of What value shall it have when, for ex- ample, n = V2? If we allow n, passing through rational values, to approach V2 as V = e its limit, it turns out that a n approaches a definite limit. We define a^ 2 as this limit : lim a n = a^ 2 . n=V2 And similarly for every other incom- mensurable value of the exponent. The function thus ob- tained: (2) y=a-, is continuous. Its graph for the special TRANSCENDENTAL FUNCTIONS 75 value a = 2.72 is shown in Fig. 25. For a proof of the fore- going statements cf. the Appendix. The chief properties of the exponential function thus denned are expressed by the equations (I) a u+v = a u a v , called the Addition TJieorem; and (II) (a u ) v = a™. The inverse of the exponential function is the logarithm : (3) y = \og a x if x — a y . It is single-valued and continuous for all positive values of x. Moreover, log a l = 0, log a a = l, log a O = -oo, log a (+co) = + oo. V y = \ogx Fig. 26 The graph is obtained in the usual manner from that of (2) by reflecting in the bisector of the positive coordinate axes, and is shown in Fig. 26. The chief properties of logarithms follow from (I) and (II) : (A) (B) log„a;" = nlog a a;. 76 CALCULUS The proof of (A) is as follows. Let u = log a x, whence x = a u -, v = log a y, " y = a v . Then (I) becomes: a u+v = xy, whence u-\-v = \og a xy, or log a a; + log y = log 8 ajy, q. e. d. To prove (B) write (II) in the form (a u f = a nw and substitute for a u its value a; : x n = a WM , whence tiw = log a # n , or log a a?" = n log a #, q. e. d. A third relation is of importance when we have to change from one base to another. It is : (C) log a * = j^. It is easily remembered because of its formal analogy with the formula (V") of Chap. II, which, when the variables are denoted by a, 6, and x, becomes : A « The proof is as follows. Let (a) w = log a sc, whence x = a u ; (/?) v = log 6 a, « x = b v ; (y) <7=log 6 a, " a = b G . We wish to prove that v From (y) follows that b = a d . TRANSCENDENTAL FUNCTIONS 77 Substituting this value of b in (/?) we get : x = a c . Substituting this value of x in (a) we get : a u = a d . And now it merely remains to take the logarithm of each side. In particular, if we set x = b, (C) becomes : (4) log^^-L.. log* a The following identity, which is often useful, is obtained by replacing y in the second equation of (3) by its value from the first equation : (5) x = a Thus (6) x n — a We have assumed hitherto that a > 1. If < a < 1, the graph of (2), Fig. 25, must be reflected in the axis of ordinates, and the graph of (3), Fig. 26, in the axis of abscissas. EXERCISES 1. Show that (a) log.-=-log.aj. x (b) log a £ = log a P-lo go Q. (c) lo&VT+a^lo&CL + a 2 ). (d) log a (a? - y 2 ) = log a (x + y) + log a (x-y). (e) log a (x + h)- log a x = log A + ^ • 2. Simplify the following expressions : vi, V,, {2 >)>, vpz, i, £i;. 78 CALCULUS 3. Solve the equation : a x — a~ x = V a * 4. a -x for x in terms of y. 4. Show that l0 *- <%+*> - log, [(i +*)!]. 5. Are (a) xX and (a x ) x the same thing ? 6. If b x = c, show that 3log 6 = log a c. 6. Differentiation of log a. To differentiate the function y = log a x we have to form the difference-quotient : (7) Ay = lo g a (x + Ax)-log a x == \ xj Ax Ax A3. and see what limit it approaches when Ax approaches 0. If we set Ax we can write this last expression in the form : 3q A3 \ X J~X Q fl \ flj" 3 Da |_V fij J When A3 approaches 0, fx becomes infinite, and the question is : What is the value of the limit : limfl+i^ IV. TRANSCENDENTAL FUNCTIONS 79 First, let /x become infinite passing through only positive integral values : fji = 7l, ri = l, 2, 3, •••. If we write •M»)-(l+jjT> we get by direct computation : *(1) =2, $(2) =2.25, $(3) =2.37, $(10) =2.59, $(100) =2.70, $(1000) =2.72, $(10,000) =2.72. Hence we see that, as n increases, (ri) increases, but does not appear to mount above a number somewhat less than 3. We can show this to be the case no matter how large n. By the Binomial Theorem : (a + b) n = a n + na n ~ l b + n ( n ~ 1 ) a n - 2 b 2 + ... . JL • £ we have : n(n-l)(n-2) /l V 1-2.3 Uy - Mf 1 --) n \ 7iy \ 71/ H to n + 1 terms 1- = 1 + 1 + T-| + V 1^2. 3 ' V + These terms are all positive and each increases when n in- creases. Moreover, when n increases, additional positive terms present themselves. And so, for both reasons, $(71) increases: $(tt + l)>$(7l). 80 CALCULUS Secondly, <{>(n) is always less than 3. For, the above terms after the first two are less than the terms of the series: 1 + 1 + 1 + 1+ ... +_!_. ^ T 2 2 2 2 n ~ x Kecalling the formula for the sum of the first n terms of a geometric progression, Chap. I, § 1, and setting a = 1, r = J, we get: 1 + 1+1+ ...+_L = lz^i)! = 2— 1, and so: (/i) <3— -— -< 3. r v ' 2 n Now if we have a function of the positive integer n which always increases as n increases, but never exceeds a certain -h 1 " i ?r\ ml i 1 2 ' 3 Fig. 27 fixed number, A, it must approach a limit not greater than A, when n = oo (Fundamental Principle for the existence of a limit, Chap. XII, § 2) . Hence (n) approaches a limit whose value, e, is not greater than 3 : (9) lin/l+iY=e. „=» ^ nj We shall see later that e = 2.718.-.. /u. irrational and negative. We can now show that, when fx becomes positively infinite, varying continuously, i.e. passing through all positive values, <£(/u.) still approaches e as its limit. Let n (ji) = e. M = + «> Finally, let p be negative, p = — r. Then K)-=K)"H^)H-^H 1+ ^J When fi = — oo , r = +00 , and lim <£(/*) = e. /A = — 00 Returning now to equations (7) and (8) and remembering that log a # is a continuous function, we see that &y lim^= lim rilog/l + iYl = ilogriim fl + -Y' = -log a e, or, dropping the subscript, we have : D x \og a x = 1 -^. x 82 CALCULUS The base a of the system of logarithms which we will use is at our disposal. We may choose it so that the constant factor log„e = 1, namely, by taking e as the base : a = e. Thus (10) becomes : (11) D.log>=i. This base, e, is called the natural base, and logarithms taken with e as the base are called natural or Napierian logarithms, in distinction from denary or Briggs's logarithms, for which the base is 10. Natural logarithms are used in the Calculus be- cause of the gain in simplicity in the formulas of differentia- tion and integration, — a gain precisely analogous to that in the differentiation of the trigonometric functions when the angle is measured in radians. It is customary in the Calculus not to write the subscript e and to understand by log a; the natural logarithm of x, the denary logarithm being expressed as log 10 #. 7. The Compound Interest Law. The limit (9) of § 6 pre- sents itself in a variety of problems, typical for which is that of finding how much interest a given sum of money would bear if the interest were compounded continuously, so that there is no loss whatever. For example, $1000, put at interest at 6 %, amounts in a year to $1060, if the interest is not com- pounded at all. If it is compounded every six months, we have $1000 ( 1 + tt) as the amount at the end of the first six months, and this must be multiplied by (1 + '-— - J to yield the amount at the end of the second six months, the final amount thus being $1000 (>+3)f TRANSCENDENTAL FUNCTIONS 83 It is readily seen that if the interest is compounded n times in a year, the principal and interest at the end of the year will amount to 1000(1 +^)* dollars, and we wish to find the limit of this expression when n = 00 . To do so, write it in the form : 1000 (i+^ n J and set n/.06 = fi. The bracket thus becomes ♦w-(i + iy: and its limit is e. Hence the desired result is 1000 e 06 = 1061.84* EXERCISE If $ 1000 is put at interest at 4 % compare the amounts of principal and interest at the end of 10 years, (a) when the interest is compounded semi-annually, and (&) when it is com- pounded continuously. Ans. A difference of -$5.88. 8. Differentiation of e*, a x . Since a x and log a x are inverse functions, we have : (12) y=a x if x = \og a y. In particular : (13) y=e* if x = \ogy. Differentiating this last equation with respect to x, we obtain : D x x = D x \ogy = D y logyD x y, 1 = \ DxVy (14) .-. D x e x = (f. *The actual computation here is expeditiously done by means of series ; see the chapter on Taylor's Theorem. 84 CALCULUS If we proceed with (12) in a similar manner, we get as our first result : y By (4) in § 5 : log a e = log e a (15) .*. D x a x = a* log a. Differentiation ofx n , n irrational. Formula (12) of Chap. II can now be shown to hold when n is irrational. Since by §5,(6): if we set z — n\ogx, we have D x x n = D x e* = D*e" • D x z=e'n- = nx n ~ 1 , q. e. d. x We are now in a position to differentiate any of the elemen- tary functions without evaluating new limits, for any such differentiations can be reduced, by the aid of Theorems I-V of Chap. II, to special formulas already in our possession. An important aid, however, in the technique of differentiation is furnished by the method of differentials, which we will con- sider in the next chapter, and so we shall postpone the drill work on this chapter till that method has been taken up. EXERCISES Differentiate the following functions : 1. y = log 10 oj. Ans. D x p = x 2. y = 10 x . Ans. D x y = 2.303 x l(f 3. 2/ = log sin x. 4. y = e C08x . 5. 2/ = k)g tan (^+2^ Ans. D 9 y= — \J2 4/- cos x CHAPTER V INFINITESIMALS AND DIFFERENTIALS 1. Infinitesimals. An infinitesimal is a variable which it is usually desirable to consider only for values numerically small and which, when the formulation of the problem in hand has progressed to a certain stage, is allowed to approach as its limit. Thus in the problem of differentiation Ax and Ay are infinitesimals ; for we allow Aa; to approach as its limit and then Ay in general also approaches 0. Again, in Chap. IV, we had to do with lim- — ^- Here a and sin a are infinitesimals. a=0 a Further examples of infinitesimals are furnished by the fol- lowing magnitudes of Fig. 28 : (1) a = AP, ft=AQ = tsma, t y = MA = l-cosa, 8 = AJST, ^< K e=PQ, Z = AQ+QP, etc. Fig. 28. That infinitesimal which is chosen as the independent vari- able is called the principal infinitesimal. Two infinitesimals, a and ft, are said to be of the same order if their ratio approaches a limit not when the principal infini- tesimal approaches : lim £= K^O. a If their ratio approaches as its limit, ft is said to be of higher order than a, and if it becomes infinite, ft is of lower order. 85 86 CALCULUS Thus if /3 = tana, ft __ tan a _ 1 since a a cos a a Ld lim£ = l=£0. a = a ence tan a is of the same order as a. Again, if y = 1 — cos a, 2sin 2 £ y 1 — cos a 2 sin£ • a 2 sin-. 2 a 2 ence lim 1 = a=0C6 and 1 — cos a is of higher order than a. An example of an infinitesimal of lower order is V«. For V« 1 A r 1 = — - and lim — = = oo a V^ a=0 ^/ a (read : "1/Va becomes infinite when a approaches 0"). It is readily seen that, if (3 and y are infinitesima s of the same order, or if y is of higher order than j3, and if further- more p is of higher order than a, then y is also of higher order than a. For, since lim 1=K. we have: 1 = k+€, y=#/3 + eft where c is infinitesimal. Hence a ct J and since lim p/a = by hypothesis, we see that lim y/a = 0. INFINITESIMALS AND DIFFERENTIALS 87 Similarly, if ft and y are of the same order, or if y is of lower order than /?, and if furthermore /? is of lower order than «, then y is also of lower order than a. An infinitesimal y3 is said to be of the n-th order if p/a n approaches a limit not 0, a being the principal infinitesimal : lim £ = K^ 0. a=oa n For example, if a is the principal infinitesimal, sin a and tan a are of the first order, Va is of order J-, a n is of order n, and 1 — cos a can be shown to be of order 2. For 2 sin 2 ? COSrt sin- and lim a=b0 cos a Theorem. If two infinitesimals a and fi differ from each other by an infinitesimal of higher order than either, then P lim tL a I And conversely: If limp /a = 1, then a and /? differ from each other by an infinitesimal of higher order than either : /?-«= lim 0. First, our hypothesis is that, if we write ^8 — a = c, then Dividing through by a we have : lim 1 = 0. a Hence lim |8_ lim H-. q. e. d. 88 CALCULUS To prove the converse, write £=! + „. a Then rj is infinitesimal. Multiplying up we have : P = a 4- ya> The difference, ft— a = r)a = e, is evidently an infinitesimal of higher order than a and hence also than /?. Definition. If a is the principal infinitesimal and if lim fi _ jT- a=0 so that, when we write £=ir +£) a we have (3 = Ka + ea, then the term Ka is called the principal part of p. EXERCISES ^ 1. Show that a — 2 a 2 and 3 a -f a? are infinitesimals of the same order. 1 2. Show that ot— sin a is of higher order than a. « 3. Show that asin« is an infinitesimal of the second order. ' 4. In Fig. 28 show that PQ and MA are infinitesimals of the same order. 5. Determine the order of AR, referred to a. ) 6. Show that AN is of higher order than RQ. 7. Show that AQ and MP are of the same order. 8. Show that PQ is of the second order, referred to a. 9. Determine the order of each of the following infini- tesimals : (a) a + sin a; (b) Vsina; (c) VI — cos a- INFINITESIMALS AND DIFFERENTIALS 89 10. Show that the sum of two positive infinitesimals, each of the first order, is always an infinitesimal of the first order, and that the difference is never of lower order. Cite an example to show that the difference may be of higher order. 11. Determine the principal part of each of the infinitesi- mals in the text numbered (1). 12. If two infinitesimals have the same principal part, show that they differ from each other by a small percentage of the value of either, and that this percentage is infinitesimal, pro- vided that their principal part is not 0. # 2. Fundamental Theorem. There are two theorems that are fundamental relating to the replacement of infinitesimals by other infinitesimals that differ from them respectively by infinitesimals of higher order. One is the theorem of this paragraph ; the other is Duhamel's Theorem of Chap. IX, § 6. Theorem. In taking the limit of the ratio of tivo infinitesi- mals, each infinitesimal may be replaced by another one which differs from it by an infinitesimal of higher order : lim£ = lim£' y 7 if lim^ = l and lim^ = l. P y For, since |' = l + € or £' = 0(1 + c) P and -= =1 + v or y' = y( 1 + >?)> where c and rj are infinitesimals, we have : y' y! + v Hence lim # = Aim £Ylim ±±±) = lim &, q. e. d. y' \ yA i + W y 90 CALCULUS 3. Tangents in Polar Coordinates. Let be the equation of a curve in polar coordinates. We wish to find the direction of its tangent. The direction will be known if we can determine the angle \jj between the radius vector produced and the tangent. Let P, with the coordinates (r , ), be an arbitrary point of the curve and P':(r + Ar, + A0) a neighboring point. Draw the chord PP and denote theZOP'P by«//. Then obviously lim «//= xjjq. Fig. 29 To determine \f/ 0j drop a perpendicular PM from P on the radius vector OP and draw an arc PN of a circle with as centre. The right triangle MP'P is a tri- angle of reference for the angle if/' and tani// MP PM Hence tan \p Q = lim tan «//= lim MP p±pP'M In the latter ratio we can, by virtue of the Fundamental Theorem of § 2, replace MP and P'M by more convenient infinitesimals. We observe that MP=r o sinA0, Furthermore, PN= Ar, hence so that ,. MP lim A9=or f ,A0 1. lim P'M 1. A0=o Ar INFINITESIMALS AND DIFFERENTIALS 91 Hence we have : \ /> lim#£=lim^ = [ri^_,, p±pPM m*o Ar L J0, is a spiral, except when A. = 0, which coils round the origin infinitely often. Here, D e r = a\e K », D r = -?- = — -, tan^ = ~, or cotxj/ = \. D 9 r a\e K9 \ Hence the tangent always makes the same angle, cot -1 A., with the radius vector produced. For this reason the curve is called the equiangular spiral. EXERCISES 1. Plot the curve r = 6, and determine the angle at which it crosses the prime vector when r = 2tt. Ans. \f/ = 81°, nearly. 2. The equation of a parabola referred to its focus as pole is r (1 + cos 0) = m. Find the value of if/ when = and when = tt/2. 3. The equation of a cardioid is r=a(l — cos0). Determine \f/. 4. Differentials. Let be a function of a; and let D x y be its derivative : Km *H = D x y. Ax=oAa; ~) 92 CALCULUS Then Ay Ax D x y + t, where c is infinitesimal, and Ay = D x y Ax + e Ax. Since x is the independent variable, we may take Ax as the principal infinitesimal, and this last relation represents Ay as the sum of its principal part, D x yAx, and an infinitesimal of higher order, e Ax, Definition. The principal part of the increment Ay of the function (1) is called the differential of y and is denoted by dy : (2) dy = D x yAx. We may, in particular, choose f(x) as the function x. Then (2) becomes : (3) dx=D x xAx = Ax. Thus we see that the differential of the independent variable, x, is equal to the increment of that variable. But this is not in general true of the dependent variable, since e does not in general vanish. By means of (3) equation (2) can be written in the form : dy = D x ydx. D*y. x' Fig. 30 Geometrically, the in- crement Ay of the func- tion is represented by the line MP' in Fig. 30, while the differential, dy, is equal to MQ. It is obvious geometrically that the dif- ference between Ay and dy, namely the line QP 1 , is of higher order than INFINITESIMALS AND DIFFERENTIALS 93 Ax — PM. The triangle PMQ is a triangle of reference for t, and . dy tan t = -^ • dx In the above definition x has been taken as the independent variable, Ax as the principal infinitesimal. The following theorem is fundamental in the theory of differentials. Theorem. The relation (4) : dy = D x ydx, is true, even when x and y are both dependent on a third vari- able, t. Suppose, namely, that x and y come to us as functions of a third variable, t : (6) x = (t), y = t(t), and that, when we eliminate t between these two equations, we obtain the function (1). Then dx and dy have the following values, in accordance with the above definition, since t, not x, is now the independent variable, At the principal infinitesimal : dy = D t y At, dx = D t xAt. We wish to prove that dy = D x ydx. Now by Theorem V of Chap. II : D t y = D x yD t x. Hence, multiplying through by A£, we get : D t yAt = D x y.D t xAt, or dy = D x ydx, q. e. d. With this theorem the explicit use of Theorem V in Chap. II disappears, Formula V of that theorem now taking on the form of an algebraic identity : du _ du dy dx dy dx To this fact is due the chief advantage of differentials in the technique of differentiation. 94 CALCULUS Differentials of Higher Order. It is possible to introduce differentials of higher order by a similar definition : d 2 y = DJyAx 2 , d 3 y = D^yAx 3 , etc. But inasmuch as a theorem analogous to the above for differen- tials of the first order does not hold true here, the chief advan- tage of the differentials of the first order is lost. We shall, therefore, refrain from introducing differentials of higher order and regard the expressions : % % etc, dor dx 3 not as ratios, but merely as another notation for the deriva- tives D 2 y, D x s y, etc.* Remark. The operator D x is written in differential form as dx Thus t-^t 1 - means ^Vr^-' dx*l — x * 1 — x and similarly for higher derivatives. 5. Technique of Differentiation. Theorems I-IV of Chap. II, written in terms of differentials, are as follows. General Formulas of Differentiation I. d (cu) = c du. II. d (u + v) = du + dv. III. d (uv) =udv+ v du. v , / 'a\ _vdu — udv vj v 2 * Differentials of higher order are still used in some branches of mathe- matics, notably in differential geometry. For a treatment of such differ- entials cf. Goursat-Hedrick, Mathematical Analysis, vol. I, § 14. INFINITESIMALS AND DIFFERENTIALS 95 Consider, for example, the first : rw \ ,-> . d(cu) du D x (cu) = cD x u, i.e. — * — t- = c — , dx dx and it remains only to multiply through by dx. — We have already noted the disappearance of Theorem V. To these are to be added the special formulas of Chapters II and IV. Beside the derivatives that were there worked out ab initio it is useful to include in this list a few others. Special Formulas of Differentiation 1. dc = 0. 2. dx n = nx n ~ 1 dx. 3. dsinx = cos x dx. 4. dcosx = — sin x dx. 5. d tan x = sec 2 xdx. 6. dlogx _dx X 7. de x = e x dx. 8. c?sin _1 a; dx 9. dtsai^x VI -x 2 dx 1+a*' This list is somewhat elastic. Some students will prefer to include the formulas : 10. dcos~ l x t= 11. dvers -1 # = Vl-x* dx V2X-X 2 12. da* =a x logadx. But it is better to err on the side of brevity. 96 CALCULUS All other functions that occur as combinations of the above, — the so-called elementary functions, — can be differentiated by the aid of these two sets of formulas. We will illustrate the use of differentials by some examples. Example 1. To differentiate y— Va^ — x 2 . Let z — a 2 — x 2 . Then y = z\ dy= dz* = ±z~*dz =\z~% (— 2xdx), dy _ 2 x ax 3(x)logf(x). INFINITESIMALS AND DIFFERENTIALS 99 Or, what amounts to the same thing, write f(x) = e log/(x) , \_fi x )'V >{x) = eW x)l0 *' (x) . Thus, to differentiate y = x x , write log y — x logic or y — e xloex . Hence -^=d(x log x) — etc. or dy — e x log x d (x log x) = etc. y g=af(l+log*). Differentiate each of the following functions : i 26. y = x x . 28. 2/ = (cos#) tan *. 27. y = x* [nx . 29. ?/ = (sin x)' inx . 6. Differential of Arc. Let s denote the length of the arc of the curve y =/(#), measured from a fixed point A, and let As denote the length of the arc PP. Then (see Fig. 30) PP' 2 = Aa 2 + Ay 2 , hence lim (*£)* = 1 + lim /^Y. Ax=o \ A# / Ax=o\^.xJ In taking the first limit we can replace the chord PP' by the arc As, since * lim—— = 1. p f =p PP * A formal proof of this statement can be given as follows. We have : (A) PP' + MP>, lim (&)*- lim (£*Y + ] im (ML\ '. A/-=o\ Ar J Ar=o\ Ar J A/-=o^ Ar J (3) (D r s) 2 = 1 + i*(D r ff)* or ds 2 = dr 2 + rW. As P' approaches P, the ratio * dx* PQ _ Vdx* + dy 2 PP~ VAx* -f Ay 2 ^ L Ay 2 Ax 2 V obviously approaches 1. ^P', which is equal numerically to Ay — dy, is an infinitesimal of higher order than Ax and hence than the chord PP ; hence QP/PP approaches and thus the limit of the right-hand mem- ber is 1. Hence the middle term approaches 1, q. e. d. INFINITESIMALS AND PIFFF.KENTi ALS 101 Furthermore : ,a\ , rdO , dr (4) sin *=^. cos *=s? the tangent PT being drawn in the direction of the increasing s. Beside these there is the formula of § 2 : (5 ) tan *=fr 7. R ates and Velo cities. The principles of velocities and rates were treated in Chap. II. We are now in a position to deal with a larger class of problems. Example 1. A railroad train is running at 30 miles an hour along a curve in the form of a parabola : (A) y 2 = 500x, the axis of the parabola being east and west and the foot being taken as the unit of length. The sun is just rising in the east. Find how fast the shadow of the locomotive is mov- ing along the wall of the station, which is north and south. Since 30 m. an h. is equivalent to 44 ft. a sec, the problem is : Given ^ = 44; to find ft. dt dt From (A) : 2ydy = 500 dx, dx = ^. V O * Substituting this value of dx in (1), § 6, we FlG> 31 get: ds* = da? + dtf = ^ + dy>, dy- 250 2 V250 2 +2/ 2 Hence, dividing through by dt and writing for ds/ dt its value, we get : dy_ 250x44 dt V250 2 + 2/ 2 ' In particular : — = — ft. a sec, or 21.2 m. an h. L^_u =250 v'2 102 CALCULUS Example 2. A man standing on a wharf is drawing in the painter of a boat at the rate of 2 ft. a sec. His hands are 6 ft. above the bow of the boat. How fast is the boat moving through the water when there are still 10 ft. of painter out ? 4 Let r be the number of feet of painter out at any instant. Then dr = -2. dt Fig. 32 j? or dr / dt gives the rate at which r is in- creasing with the time, and since r is decreasing, the rate is negative. We wish to find the rate at which P is moving. Let s denote the horizontal distance PB of P from the wharf. Then ds /dt gives this rate numerically, but algebraically ds/dt is negative. We desire, then, the value of —ds/dt. Since s and r are connected by the relation : we have 2 s ds = 2 r dr, _ds _ _ r dr _ 2r dt s dt y r 2_36 Hence, finally : ■[-—'] =f 2r 1 =24 ft. a sec. The student will note that the method of solution consists in determining first the velocity at an arbitrary instant, and then substituting the particular value r = 10 into the result thus obtained. EXERCISES 1. A lamp-post is distant 10 ft. from a street-crossing and 60 ft. from the houses on the opposite side of the street. A man crosses the street, walking on the crossing at the rate of 4 m. an h. in the direction toward the lamp-post. How fast is his shadow moving along the walls of the houses when he INFINITESIMALS AND DIFFERENTIALS 103 is two-thirds of the way over ? When he is 55 ft. from the houses? Ans. 6 m. an h. and 96 m. an h., respectively. 2. A kite is 150 ft. high and there are 250 ft. of cord out. If the kite moves horizontally at the rate of 4 m. an h. directly away from the person who is flying it, how fast is the cord being paid out ? Ans. 3^ m. an h. 3. A point describes a circle with constant velocity. Show that the velocity with which its projection moves along a given diameter is proportional to the distance of the point from this diameter. 4. A revolving light sends out a bundle of rays that are approximately parallel, its distance from the shore, which is a straight beach, being half a mile, and it makes one revolution in a minute. Find how fast the light is travelling along the beach when at the distance of a mile from the nearest point of the beach. Ans. 15.7 m. a m. 5. The sun is just setting as a base ball is thrown vertically upward so that its shadow mounts to the highest point of the dome of an observatory. The dome is 50 ft. in diameter. Find how fast the shadow of the ball is moving along the dome one second after it begins to fall, and also how fast it is moving just after it begins to fall. 104 CALCULUS EXERCISES Determine the maxima and minima of the following functions : 1. x log X. 2. X COS X. 3. xe~*. 4. x n e~ x . Ans. A minimum when x = .3679. Ans • ii 5. x 2 log A minimum 9. sin 2 x — x. 10. sin x cos 3 a;, cos a; maximum when x =cot x, < x < l n ; -£ir A block of stone is to be drawn along the floor by a rope. Find the angle which the rope should make with the horizontal in order that the tension may be as small as possible. Ans. The angle of friction. 31. Into a full conical wine-glass whose depth is a and generating angle a there is carefully dropped a spherical ball of such a size as to cause the greatest overflow. Show that the radius of the ball is a sin a sin a -f- cos 2 a 32. The illumination of a small plane surface by a luminous point is proportional to the cosine of the angle between the rays of light and the normal to the surface, and inversely pro- portional to the square of the distance of the luminous point from the surface. At what height on the wall should a gas- burner be placed in order to light most brightly a portion of the floor a ft. distant from the wall ? Ans. About -J-^a ft. above the floor. 106 CALCULUS 33. A gutter whose cross-section is an arc of a circle is to be made by bending into shape a strip of copper. If the width of the strip is a, find the radius of the cross-section when the carrying capacity of the gutter is a maximum. Ans. a/ir. 34. If, in the preceding problem, the cross-section of the gutter is to be a broken line made up of three pieces each 4 in. long, the middle piece being horizontal, how wide should the gutter be at the top ? ' Ans. 8 in. 35. A wall 27 ft. high is 64 ft. from a house. Find the length of the shortest ladder that will reach the house if one end rests on the ground outside the wall. 36. A long strip of paper 8 in. wide is cut off square at one end. A corner of this end is folded over onto the opposite side, thus forming a triangle. Find the area of the smallest triangle that can thus be formed. 37. In the preceding question, when will the length of the crease be a minimum? 38. The captain of a man-of-war saw, one dark night, a privateersman crossing his path at right angles and at a distance ahead of c miles. The privateersman was making a miles an hour, while the man-of-war could make only b miles in the same time. The captain's only hope was to cross the track of the privateersman at as short a distance as possible under his stern, and to disable him by one or two well-directed shots ; so the ship's lights were put out and her course altered in accordance with this plan. Show that the man-of-war crossed the privateersman' s track - V) v. Fig. 33 52. Find the velocity of the piston of a locomotive when the speed of the axle of the drivers is given. 53. A draw-bridge 30 ft. long is being slowly raised by chains passing over a windlass and being drawn in at the rate of 8 ft. a minute. A distant electric light sends out hori- zontal rays and the bridge thus casts a shadow on a vertical wall, consisting of the other half of the bridge, which has been already raised. Find how fast the shadow is creeping up the wall when half the chain has been drawn in. Fig. 34 110 CALCULUS 54. The sun is just setting in the west as a horse is running around an elliptical track at the rate of m miles an hour. The axis of the ellipse lies in the meridian. Find the rate at which the horse's shadow moves on a fence beyond the track and parallel to the axis. 55. Differentiate y when 2x sin y = 3y sin x. Ans. gg = 3ycosa»-2siny. dx 2x cosy — 3 sin x 56. Differentiate y when y = x\og(x-y). 57. Plot the curve r = acos30, determining where the tangent is parallel to the axis of a lobe. 58. Plot the following curves : (a) y = x-\-sinx. (c) r = asin30. (b) y = xe~\ (d) r = i. d 59. Locate the roots of the equation x = cot x and hence discuss completely the maxima and minima of the function in Question 2. 60. The equation 0(l + cos0)=2sin0 has one root in the interval — ir/2 < 6 < v/2, namely = 0. Has it others ? 61. Find all the roots of the equation (l+& 2 )tan- 1 6 = 6. CHAPTER VI INTEGRATION 1. The Area under a Curve. Let it be required to compute the area bounded by the curve (i) y=m, the axis of x, and two ordinates whose abscissas are x = a and x = b, (a< b). We can proceed as follows. Consider first the variable area, A, bounded by the first three lines just mentioned and an ordinate whose abscissa x is variable. Then A is a function of x. For, when we assign to x any value between the limits a and b in question, the corresponding value of the area is thereby determined and could actually be computed by plotting the figure on squared paper and counting the squares, or by cutting the figure out of a sheet of paper or tin and weighing the piece. If, then, we can obtain an analytic expression for this function of x, holding for all values of x from a to b, we can then set x = b in this formula and thus solve the above problem. To do this, begin by giving to x an arbitrary value, x = x , and denoting the corre- y sponding value of A by A . Next, give to x an incre- ment, Ax, and denote the corresponding increment in A by A A. We can ap- proximate to the area AA by means of two rectangles, as shown in the figure, and thus we get: y Ax < A^l < (y + Ay) Ax. Ill Fig. 35 112 CALCULUS Hence y < — -< y + Ay. Ax (Iif(x) decreases as x increases, the inequality signs will be reversed.) Allowing now Ax to approach as its limit, we see that the variable AA/Ax always lies between the fixed quan- tity y and the variable y + Ay, whose limit is y Q . Hence ,. AA Ax=0 AX or, dropping the subscripts, we have : (2) D x A = y. For example, let the curve be y = x* and let a = 1, b = 4. Then D x A = x 2 and the question is : What function must we differentiate in order to get x 2 ? We readily see that X s /3 is such a function. But this is not the only one. For, if we add any constant, a?/ 3 + C will also differentiate into x 2 . We shall see later that this is the most general function whose derivative is x 2 , and hence A must be of the form : (3) A= i +C - This formula is not wholly definite, for C may be any con- stant. On the other hand we have not as yet brought all our data into play, for we have as yet said nothing about the fact that the left-hand ordinate shall correspond to the abscissa x = l. Now the variable area A will be small when x is only a little greater than 1, and it will approach as its limit when x ap- proaches 1. If, then, (3) is to be a true formula, it must give as the value of A when x = 1, or (4) .0 = i + C, 0=-J. INTEGRATION 113 (5) ,: ^=f-|. Having thus found the variable area, we can now obtain the area we set out to compute by putting x = 4 in (5) : The process of finding the area under a curve is thus seen to be as follows. First find a function which when differen- tiated will give the ordinate y=f(x) of the curve (1) before us ; and add an undetermined constant to this function. Next, determine this constant by requiring that A shall = when x = a. Thus the variable area is completely expressed as a function of x. Lastly, substitute x — b in this formula. EXERCISES 1. Show that, if the area in the foregoing example had been measured from the ordinate x = 2, the value of the constant C would have been — 21 : X s A = --2%: 3 3 ' and if it had been measured from the origin, then C would have been = : 2. If, in (1), y=f(x) =x, the curve is a straight line; and if a = 6, b = 20, the figure is a trapezoid. Compute its area by the above method and check your result by elementary geometry. 3. Find the area under the curve V = A lying between the ordinates whose abscissas are a = 10 and a; = 20. Ans. 620,000. 114 CALCULUS 4. Find the area of one arch of the curve y = sin x. Ans. 2 5. Find the area under that portion of the curve 2/ = l — JB 2 which lies above the axis of x. 6. A river bends around a meadow, making a curve that is approximately a parabola : y — x — 4sc 2 , referred to a straight road that crosses the river, as axis of x ; the mile is taken as the unit. How many acres of meadow are there between the road and the river? Ans. 6 J, nearly. 2. The Integral. In the preceding chapters we have treated the problem : Given a function ; to find its derivative. The examples of the last paragraph are typical for the inverse problem: Given the derivative of a function; to find the function. Stated in equations, the problem is this. If D x U= u, or dXJ—u cfcc, where u is given, to find U. The function U is called the integral of u with respect to x and is denoted as follows : U= I u dx. Thus we have the following Definition op an Integral. The function U is said to be the integral of u : JJ= I udx y if D x U=u, or dU=udx. The given function u is called the integrand. INTEGRATION 115 More precisely, we should say that U is an integral of u\ for U-\- G is evidently an integral, too, C being any constant. For example : (6) C x *dx = -?^- + C i n*-l. J n + 1 For, if we differentiate the function: /y.M+1 U=^— +c n + 1 with respect to x, we get : D x U=x n , and x n is the integrand u of the integral in question. The following theorem is fundamental in the theory of integration. Theorem A. If two functions have the same derivative : (A) D x f(x) = D x 9i Find the area above the positive a,xis of x bounded by the curve : 10. Find the area enclosed between the two parabolas : y = x 2 , y*=x. 3. Special Formulas of Integration. Corresponding to the Special Formulas of Differentiation, of Chap. IV, we can write down a list of special formulas of integration, by means o"f which, together with the general methods discussed in this chapter, all the simpler integrals can be evaluated. Each formula can be proven by differentiating each side of the equation. Special Formulas of Integration /■ x- n+l 1. I x n dx = -, n^P—l. n + l I- smxdx = —cos a;. 3. I cos xdx = sin x. /dx . --log*- 5. iedx — e x . 6. T-^-o = tan" 1 * Jl + x* INTEGRATION 119 dx ■Vl-x 2 = sin _1 a;, = — cos _1 a; sec 2 xdx = tana. 9. I osc 2 xdx = —cot a;. /• To these may be added the formulas : 10. C dx = vers" 1 a;. J ^2x-x> 11. Ca'dx = r-^-- J log a We have omitted the constant of integration each time for the sake of simplicity. But the student must not forget to insert it in applying these formulas in a given example. Moreover, we have not included the formula: /• Odx=C. 4. Integration by Substitution. Many integrals can be ob- tained from the special formulas of § 3 by introducing a new- variable of integration. The following examples will illus- trate the method. Example 1. To find I V« + bx dx. Let a + bx = y. Then b dx — dy and -y/a + bx dx — -y^ dy^^\s Integrating each side of this equation, we get : C^a~+bxdx=:j- fy*dy = *y*+C. 120 CALCULUS Hence fvG+ta dx = 2V <" + te )' + G. Example 2. To find I cos ax dx. Let ax = y. Then adx = dy, cos axdx = - cos ?/ cfa/, a and lcosaxdx = - I cosy dy = - sin y-f (7= -sin ax+ C. J aj a a Example 3. To find / x ^/tf + x 2 etc. Let x 2 — y. Then 2xdx = dy, x-VaFTx^ dx = xVtf^f^=\^a T +~y dy, and I x^/a 2 + x 2 dx = ± I Va 2 + y dy. This last integral is a special case of the integral of Ex- ample 1. For, if the a of that formula is replaced by a 2 , the b by 1, and the x by y, we have the present integral. Hence jx VaFTx 2 dx = \{a 2 + x 2 ) * + G. We might have set a 2 -{-x 2 = y. Example 4. To find I tan x dx. Here funxdx = /**"»*« f ~ ** cos * = - log cos s + C. J J COS X J COS£ In substance, we have introduced a new variable, cos x = #. But in. practice it is often simpler, as here, to refrain from actually writing a new letter. INTEGRATION 121 In the above examples we have tacitly assumed that if x and y are functions one of the other, and if f(x) and (y)dy, then jf(x) dx = j (y) dy. We can justify this assumption without difficulty. For D x ff(x)dx=f(x), A JV (y) dy = D y j (y) dy -D x y = (y) D x y ; and since /(#) = <£ (y) -%- , ctx it follows that the above integrals differ from each other at most by a constant, k. Hence, if the constant of integration in one of these integrals is chosen at pleasure, the constant of integration in the other can be so determined that k = 0. This theorem in integration corresponds to Theorem V of Chap. II in differentiation. And, as in the case of that theorem, the use of differentials, — and it is to this fact that their importance is due, — reduces the theorem in form to an algebraic identity : /•*-/[*S]j EXERCISES Evaluate the following integrals : 1. / Vl - x dx. Ans. _ |(1 _«>)*•+ C. 2. l^/l + 2xdx. Ans. f(l + 2aj)*+C. 122 CALCULUS 4 - / 4/ * 6. I sin ax dx. 8. / sin (ttx + v) dx. J Va + bx J J 7J J a 2 -far* a a == • J.ws. sin -1 - x 2 a 10. C dx . ^n«. sin-^+O. J Va 2 ii. r gda? . ^rcs. _v^=^+o. 12. I x 2 y/a 3 + x i dx. 14. / %-- 16. I a; sin x 2 dx. J J a + bx J /' , _ . _ (* xdx , _, f* dx xe x dx. 15. I . • 17. /- -• 18. I cot # cte. J.«s. log sin x -f- O. 5. Integration by Ingenious Devices. Example 1. To find I cos 2 Odd. Set cos 2 = i (1 rhjcojL2 6). Then /cos 2 cW = } f(l + cos 2 ff) dO = 1 0+ J sin 2 + C, (8) .-. j cos 2 dO = ±(0 + smO cos 0) + C. We can now evaluate an important integral, namely : , / Va 2 — x 2 dx. Let x = a sin ; c?sc = a cos d$, ^a 2 -x 2 dx = a 2 cos 2 $d6 } INTEGRATION 123 / Va 2 -x 2 dx = a 2 J cos 2 d0=^($ + sin ens 0) + C, (9) .-. / Vo 2 "^ 2 " tte = i r*VS^a?+ a 2 sin" 1 ^ J+ G Example 2. To find T-/ a The integrand can be written in the form : _j_=j_r^ L.T. a 2 —x 2 2a \_x -f- a x — aj Hence f-i*-. = 1- CJSL. _ J_ fj?L Jas — x 2 2aJ x-\-a 2aJ x — a = ~riog(aJ + a)-log(«-a)l+C7; Example 3. To find f-^ J s First Method. J sin 6 J , sini c?0 # 2 sin - cos - 2 2 *In case — a _ J* sec 2 d $__ /Vtaj C dQ . . # / x— 7;= log tan- +( Second Method. f # f sec 2 d Anan<£ i . J _ J sin * cos S J tan sin 2 ■ r^cos^ = _l log l + cos^ Jl-cos 2 2 g l-^o7^ +C# The fraction 2 l + cosfl_ CQS 2 1 i-cos0~ • 2 e ~7~70 sm 2 ^ tan 2 - (11) .'. f-^=logtan^+G sin EXERCISES 1. Jsm 2 6d0. Ans. i(0-smOcosO) + G. 2. C de . 3 /"_*_ J 1 + cos 0_\^ ' J 1 - COS 4 '/cS' ^logtaag + g+C, 5. or llog ^ + Sm ^ +fi or log (sec 0-f tan 0)+ C. -L sin u r dx J V¥+tf' log(^ + V^+a 2 )-ha Suggestion : Let x = a tan 0. 6 * J V^= 2 ' ^ log^-fV^^H-a INTEGRATION 125 6. Integration by Parts. The formula of d "erentiation : d (uv) = udv-\-v du, leads to a formula of integration : (III) I udv = uv — I v du. Integration by means of this formula is known as integration by parts. Example 1. To find I xe x dx. Let u = x, dv =e x dx; then du = dx, and v= I e x dxz=e x , I xe x dx = xe x — I e x dx = (x — 1) e x + C. / / Example 2. To find I log x dx. Let u = log x, dv — dx\ then du = — , v=x, x and j log x dx = x log a; — /# — = #(log# — 1)+ C. Example 3. To find / Va 2 + x?dx. Let w = Va 2 + x 2 , dv = dx; xdx then du = xdx , v = x fVa ir +tfdx = x^/a 2 + x 2 - f-^ J J Va 2 4-x 2 Again, Va 2 4- x* = — , VV + iC 2 126 CALCULUS ■dx fVtf + ^dx = a 2 C-**—± f-£ J J -Va 2 + x? J Va 2 + X 2 Adding these two equations and recalling Ex. 5 in the preced- ing Exercises, we have : (12) / Va 2 + x 2 dx = i[xVa 2 + x 2 + a 2 log(x + ^a 2 + x 2 )]+C. EXERCISES Evaluate the following integrals : 1. I xe ax dx. 5. Ixcosaxdx. 9. I x tan" 1 x dx. 2. I xPe^dx. 6. I sin- l xdx. 10. I xlogxdx. 3. / x*e ax dx. 7. /tan -1 x dx. 11. I e ax smxdx. 4. Ixsmxdx. 8. I xsin^xdx. 12. I e ax cosxdx. 13. I^a? — a 2 dx. Ans. %[x Vx 2 — a 2 — a 2 log (# -f- V&* 2 — a 2 )] + O. 7. Use of the Tables. The integrals that ordinarily arise in practice and which can be evaluated in terms of the elemen- tary functions can be found in such a table of integrals as Professor Peirce's,* and for this reason it is not necessary for us to go further into the theory of integration in this course. We have learned how to differentiate all the elementary functions, but not all these functions can be integrated in terms of the elementary functions. Thus the integral : dx (* d(b or ' f d4> (0<& 2 <1), J Vl-& 2 sin 2 <£ * B. O. Peirce, A Short Table of Integrals, Revised Edition, 1899 or later, Ginn & Co., Boston. INTEGRATION 127 leads to a new class of transcendental functions, the Elliptic Integrals, and cannot be evaluated in terms of algebraic functions, sines and cosines, etc. There are, however, large classes of functions that can be integrated,* and the classes that are important in practice have been tabulated. The student is requested to examine with care the classification in the Tables above referred to. Example 1. To find by aid of the Tables C ® dx . The integrand is a rational function of x, and so we look under "II. Rational Algebraic Functions," p. 5. There we find "A. — Expressions Involving a -\-bx." Formula 31 gives us the integral we want : / Example 2. To find / - xdx 1 . 1 . „ (l- x y~~ l-x 2(l-af) 2 * dx + X + X 2 Here the integrand involves rationally an expression of the form X = a -f bx -f ex 2 , and so we look under C, p. 10. Two formulas, 67 and 68, give this integral. But since # = 4ac — b 2 = 3 is positive, the second formula would introduce imagi- naries. The first gives : /: ^_ = _l_tan-?£±l + a 1 + x + x 2 V3 V3 Example 3. To find / - J ■ VI + x + x 2 Here the integrand involves VX, and so we look under " III. Irrational Algebraic Functions," and find under D, p. 23, Formulas 160, 161. Since c = 1 > 0, we choose No. 160 : * When we say, a function can be integrated, we mean, can be inte- grated in terms of the elementary functions. Every continuous function has an integral, for the area under its graph is an integral. 128 CALCULUS / dx log/Vl+a + ^ + a + ^Wo. VI + x + x 2 \ Example 4. To find / sin 6 a; dx. The integrand is a transcendental function. Turning to V, p. 35, and looking down the list we come to No. 263 : /\ • n , sin n-1 # cos x , n — 1 {* • n * j sm n x dx = ■ -\ I sm n 2 x dx. n n J If we set here n = 6, we reduce the given integral to an . ex- pression involving l sin 4 sc dx, and this integral can in turn be reduced by the same formula, written for n = 4. Thus we get finally : /■ sin 6 # dx sin 5 a; cos cc 5 sin 3 a; cos x 5sin#cos x 5x ^ 6~ ~W "16 16 "* Example 5. To find f- — dx 4 cos x Formula 300 gives : r^^ = |tan-[3tan|]+a J 5— 4 cos x 3 2 J EXERCISES Evaluate the following integrals with the aid of the Tables. + a Ans - h log (4 — 5x) + /xdx (4,-Bxf , r ax " .Mi-*) / dx . f * xdx (ar-2)(a>-3)" ' Ja- 2 -5z + 6 5x Ans. h log h C. x 1 — x INTEGRATION 129 j5 + 3^ V15 1^5;^ / * da? J » + ^ + *» ^na. i log ^ — L tan" 1 ? ^— + 0. 9. f^-=^dx. Ans. 2Vr^ + log A /l ~ a; ~ 1 -K7. «/ * VT^oJ + 1 10. r-^=. 12 . /v^e?* 11. f— p=r 13. r vf +^^ t/ a;VlH-ar t/ a; 14. / V— 1-f 4a — a 2 (to. \2 7 2 V3 15. r * — r is. r d * . J (7 -9X + 2X 2 )* J xVtf+px + q 17. f dx 18 . f S m 2 Ocos 2 $dO. J(l-^)Vl + ^ J 8. Length of the Arc of a Curve. We have seen in Chap. V, § 6, that the differential of the arc of a curve is given by the formula : ds= Veto 2 + dy 2 = \/l +^dx. * dx 2 Hence the length of the arc can be obtained by integration, and we have : < 13 > s =f\k+% dx - 130 V CALCULUS Example 1. Let us find the length of the arc of the parabola : y = x\ Here dy — 2x dx, -Vdx 2 -f- dy 2 = Vl 4- 4 x 2 dx, s = I Vl + 4:X 2 dx, and this integral can be reduced at once to (12) in § 6, or to Formula 124 of the Tables: s = 2J Vi + x 2 dx =WiT? + ilog(>+ y/±+¥)+ a If we measure the arc from the vertex, s = when x = 0, and we have for the determination of C : = ±log4+C, '(7=1 log 2. Hence s = \x V1 + 4& 2 + \ log (2x + V 1 + 4a; 2 ). In particular, the length of the arc to the point (1, 1) is M* = i = iV5 + ilog(2-fV5). On p. Ill of the Tables we find a table of natural logarithms, from which we see that log (2 + V5) = log 4.24 = 1.45. Hence [s] x=1 = 1.48. As a check on this result we note that the length of the chord is V2 = 1.41; on the other hand, the length of the broken line consisting of the abscissa and the ordinate of the point (1, 1) is 2. Consequently the length of the arc in question must lie between 1.41 and 2. Example 2. To find the length of the arc of the equiangular spiral : r = ae ke < X = cot a. Here ds = Vdr 2 + rW, dr = aXe^dO, INTEGRATION 131 ds = VT+X 2 ae^ dO = , + X dr = dr sec a, A = sec a I dr = r sec a + k. If we measure the arc from the point = 0, r = a, then s = when r = a and = a sec « 4- k. .-. s = (r — a) sec a. When = — cc , the spiral coils round the pole r = infinitely often, and r approaches as its limit. The value of s, taken numerically, when r < a, is : | s | = — s = (a — r) sec a. Thus we see that the length of the spiral does not increase beyond all limit when = — go, but lim | s | = a sec a. 0=-co EXERCISES 1. Find the length of the cardioid : r = a (1 — cos 6). Ans. 8 a. 2. Find the length of the spiral r = from the pole to the point where it crosses the prime vector for the first time, = 2tt. Ans. 21.3. 3. Find the length of the arc of the curve 27 y 2 = a? included between the origin and the point whose abscissa is 15. Ans. 19. "4. Find the length of the arc of the spiral r = 1/6, measured from the point = 1, r = 1. 5. Prove that the length of the arc of the catenary -J measured from the vertex, x = 0, is : s= -le a — e a J- 132 CALCULUS 6. Assuming the equation of a parabola, referred to the focus as pole, in the form : m 1 — cos<£' find the perimeter of the segment cut off by the latus rectum. Check your answer. EXERCISES Obtain the following integrals without the aid of the Tables. sin x dx + b cos x dx e x + e~ x ' 'sin OdO cos 2 6 1. fV2^xdx. 9. A**"**. 17. C-™ J J x N J a + r dx /v 4-1 r ' 2. I 10. / — HJi(tt. 18. I e cosx sin a; do;. 3. f(a-xfdx. 11. /*— — • 19. Csm 3 xdx. 4. frcfi-xtydx. *12. C^p^ldx. 20. f- 5 - JrS- i3 - J 1 ™*- 2l - /- J l + a> J v ; Jl + sma 7. T^icte. 15. C ^~~ 1 ^ dr. '23. fsetfxdx. 8. I 16. I xcosafdx. 24. I cos 3 xdx. J a + bx* J J 25. Let J[ denote the area bounded by the curve a fixed radius vector 6 = , and a variable radius vector = $, see Fig. 29. Show that and thus obtain the theorem : INTEGRATION 133 (14) i Cr*dO. 26. Find the area of the cardioid : r = a (1 — cos ) . Ans. f ira?. 27. Determine the area cut out of the first quadrant by the arc of the equiangular spiral r = ae Ke corresponding to values of 6 between and %tt. 28. Obtain the area of one lobe of the lemniscate : r 2 = a 2 cos 2 0. 29. The same for r = a sin 36. 30. The same for r = a cos nO. 31. Find the area of the ellipse : — + ^ = 1. Ans. nab. a 2 b 2 32. Prove that the length of the arc of the curve a 2 a 2 — or measured from the origin, is t a-\-x s = a log — ■ x. a — x 33. Prove that the length of the arc of the curve Sa 2 y 2 =x 2 (a 2 -x 2 ) is s = y + — sin" 1 - . '34. Prove that the area of the curve [y J =a 2 — x 2 is ira 2 . J 35. Determine the area of the loop of the curve y 2 = x 2 + x 3 ' Ans. T 8 7 CHAPTER VTT CURVATURE. EVOLUTES 1. Curvature. We speak of a sharp curve on a railroad and thus express a qualitative characteristic of the curve. Let us see if we cannot get a quantitative determination of the degree of sharpness or flatness of curves in general. If we consider the angle <£ by which the tangent of a curve has changed direction when a point that traces out the curve has moved from P to P, then this angle will depend, not only on the sharpness of the curve, but also on the distance from P to P'. We can nearly eliminate this latter element Fig. 86 when P' is near P by taking the average change of angle per unit of arc, /PP\ This ratio we define as the average curvature : -£- = average curvature for arc PP. PP' The limit approached by this average curvature is what we understand by the curvature at P; it is denoted by k : (1) k = lim -^- — actual curvature at P. p'=p PP Thus for a circle of radius a, PP'=od>, -£--., lim-i^l PP> (i p'=p ppi a 134 CURVATURE. EVOLUTES 135 and the average curvature does not change with P. The curvature of a circle is the same at all points and is equal to the reciprocal of the radius. Again, the curvature of a straight line is 0. To evaluate the limit (1) for any curve, y =f(x), we observe that, if we write PP'=AS, = ±T, then k = lim — = D s r, a«=o As where t denotes as usual the angle which the tangent of thB curve makes with the axis of x. More precisely, it is the numerical value of D a r which we want, for k is an essentially positive quantity (or 0). Hence (2) k=±— , or better: K = \ — w ds Ids From the foregoing definition we see that the curvature is the rate at which the tangent turns when a point describes the curve wifh unit velocity. To compute dr/ds we have (3) tanr = ^ or r = tan" 1 ^. dx dx It will be convenient to introduce a shorter notation for deriva- tives and we shall adopt Lagrange's, which employs accents : y *=/(?), It follows, then, that dy' = ( &dx = ( ^ ) dx = y"dx dx dx 2 and ds = Vdx 2 -f dy 2 = VI -\-y' 2 dx. Returning to (3) and differentiating we have: 136 CALCULUS r-taa-V, ^=«^ = fJE dr ds i+y' 2 i+y' 2) .'/ (l+y ,2 P (4) I y" I (1+2/' 2 ) [ ♦*r The reciprocal of the curvature is called the radius of curva- ture and is usually denoted by p : * l__ (l+y") f L dtf \y"\ " (?) /> = K = dx 2 ; The radius of curvature of a circle is its radius. The curva- ture of a curve at a point of inflection is in general ; for y" = at such a point if y" is continuous there. Example. To find the curvature of the parabola f 2mx. Here 2ydy = 2mdx, y' y dy' = -- 2 dy, y __ ™?\y m" = (m 2 + y*) ? (m 2 + * The student can always recall which of these two ratios is the curva- ture, which the radius of curvature, by the check of dimensions. If we re- gard x and y each as of the first degree in length, then y' = dy/dx is of the 0-th and y" — dy' /dx of the — 1st degree. Hence the bracket is of the 0-th degree and | y" | of the — 1st, and the ratio must therefore be written so as to yield p of the 1st, k of the — 1st degree in length. CURVATURE. EVOLUTES 137 EXERCISES Find the curvature of each of the following curves. 1. 1 2. L 3. A. 5. ^6. y=x*. y=3?, at the origin. y = log esc x. The ellipse: 2? + £«.l. a 2 6^ The hyperbola: — — ^- = 1. a 2 b 2 The equilateral hyperbola : xy Arts, k = • (1 + 4^)2 Ans. K = |sinic]. 4— « 4&4 (6V + aV) f a 2 2* (y + 2/ 2 ) 2 7. Show that the radius of curvature of the curve y — x^ approaches as its limit when the point P approaches the cusp, (0, 0). 8. Find the radius of curvature of the curve 54 y = 10a 5 - 19x 4 + 11jb 8 + x 2 - 12x at the origin. Ans. p = 125. 9. Find the radius of curvature of the catenary at the vertex. Ans. p = a. 10. At what points of the curve y=x? is the curvature greatest ? 11. Find the locus of the points in which the curvature of the curves of Fig. 4 : y =x n , x > 0, n> 0, is greatest. 138 CALCULUS 2. The Osculating Circle. At an arbitrary point P of a curve let the normal be drawn toward the concave side of the curve and let a distance be laid off on this normal equal to the radius of curvature, p. The point Q thus obtained is called the centre of curvature. The circle constructed with Q as centre and with radius p stands in an important relation to the curve. It is called the osculating circle and has the property that it repre- sents the curve more accurately near P than any other circle does. Consider the family of circles drawn tangent to the curve at P and with their centres on the concave side. Those whose radii are very short are curved too sharply, — more sharply than the given curve. Now let the circles grow. If we pass to the other extreme of circles with very large radii, these will be too flat. Evidently, then, certain intermediate circles come nearer to the shape of the curve at P than these extreme ones do. It is not difficult to find a criterion by means of which one circle is characterized as better than all the others. Draw the tangent at P and drop a perpendicu- lar from P' on it meeting it in M and cutting an arbitrary one of the circles in R. Then, as we shall show later, MP will in general be an infinitesimal of the second order referred to the arc PP as principal infinitesimal, and PR will also be of the second order for a circle taken at random. We can, however, in general find one circle for Fig. 37 which PR will be an infinitesimal of the third order, and it turns out that this circle is precisely the oscu- lating circle. We shall give the proof later (Chap. XIII, § 9). The osculating circle cuts the curve in general at the point of tangency ; but there may be certain exceptional points at which this is not the case. Near such a point P'R is an infini- tesimal of even higher order than the third, in general, of the fourth. EXERCISE Construct carefully the parabola y = x 2 for values of x : — | < x < f , taking 10 cm. as the unit. Draw the osculating CURVATURE. EVOLUTES 139 circle at the point x = £, y = \, and also at the vertex. Ink in the parabola in a fine black line, the first osculating circle in red, and the second in a different colored ink or in pencil. 3. The Evolute. When a point P traces out a curve, the centre of curvature Q traces out a second curve. This latter curve — the locus of Q — is called the evolute of the given curve. We proceed to deduce its equation and to discuss its properties. The point Q can be found analytically by writing down the equation of the normal at P and determining the intersection of this line with a circle of radius p, hav- ing its centre at P. The equation of the normal is (6) X-x + y'(Y-y) = 0, where (X, Y) are the running coordinates, i.e. the coordinates of a variable point on the normal, and (a?, y) the coordinates of P, — the latter being held fast during the FlG * 3b following investigation. The equation of the circle is (?) (x-xy+(Y- y y= P * a+y ,z ) To find where (6) and (7) intersect, eliminate X : (l+y>*)(Y-yy a + y' 2 ) y" 2 ' Y-y i+y 2 y" Which sign must we take? Notice that when the curve is concave upward, as in the figure, Y-y>0 and a _ 4 — 2Z 3 . We can solve these equations respectively for x 2 and y 2 and substitute the values thus ob- tained in (12) : / ax, \t / by Y \ \a 2 —b 2 ) ^{tf-b 2 ) = 1. Fig. 40 Dropping the accents we have as the final equation of the e volute of the ellipse : (13) (ooj)* + (6y)* = (a 2 - 6 2 )l EXERCISES Find the equation of the evolute of each of the following curves. x 2 ?/ 2 1. The hyperbola: = 1. Ans. (as)* - (byy = (a 2 + b 2 ) K 2. The hyperbola : 2xy — a 2 . Ans. (x + y)% — (x— y)% = 2c$. 3. The catenary : y = ^(e x + e~ x ). Ans. x x = x — \ (e 2x — e~ 2x ), y l = 2y ; 2\4 -*(i±v5 4. ic^ + y* = a*. Ans. (x + #) $ + (a; — y)» = 2cA CURVATURE. EVOLUTES 143 Ans. x 2 + ?/ 2 5. a; = a (cos + # sin#), y = a(sinO — $ cos0). 6. x = acos 3 0, y = asin 3 6. Ans. (x + y) 1 * -\- (x — y)* 4. Properties of the Evolute. The property of the evolute to which the curve owes its name is the following. Suppose a material cylinder to be constructed on the concave side of the evolute and a string to be wound on the cylinder, Fig. 41. Let a pencil be fastened to the end of the string, the point being placed at a point P of the given curve and the string drawn taut and fastened at a point A of the evolute so that it cannot slip. If now the pencil is moved along the paper so that the string unwinds from the evolute or winds up, the pencil will describe the given curve. To prove this, let P be an arbitrary point of the given curve, Q the corresponding point of the evolute, and P' the position of the pencil \g^r when the string leaves the evolute at Q. We wish to prove that P' coincides with P. To do this it is sufficient to show (a) that QP is tangent to the evolute, so that P' lies on QP; and (b) that QP' = QP= P . ad (a) Writing equations (9) in the form : y y and differentiating with respect to x, we have ; * _ (1 + y-) ^_ - _, r // * The student may find it more convenient in working out these differ- entiations to retain the form (9). Lagrange's form is more compact. 144 CALCULUS dy l _ 1 _ dx x ~ y'~ 1 dy J dx and thus the slope of the evolute at Q is seen to be the ne?* ■ tive reciprocal of the slope of the given curve at P. Hei. QP is tangent to the evolute, q. e. d. ad (b) If we denote by s ± the length of the arc Q Q of the evolute, then ( = s 1 + p , and we wish to show that this quantity is eqr to p : «i + po = p- It is evidently sufficient to show that d 9 i _ dp dx dx Now dsf = dx x 2 -f dyi\ +2/i f2 =^ 2 (i+j!T)=2/i' 2 (i+y 2 ). And again : ft= ± 3 y y-(u.f).v"V rT7'= ±*m+F. dx y" 2 Hence P=±£, dx dx and since we have taken s x so that it increases when p in- creases, the upper sign holds : ds 1 = dp, s ± = p+ C. At Qo, s 1 = and p = p , hence = p + C y and P = s 1 + Po , q.e. d. We have shown incidentally that the normals to the given curve are tangent to the evolute. Thus it appears that the evolute is the envelope of the normals of the given curve. This property can be used as the definition of the evolute and its equation is then readily deduced by the method of Chap. XVII. CURVATURE. EVOLUTES 145 EXERCISES 1. If the equation of the curve is given in po ar coordinates, r=/(0), then (see Fig. 29) At = A^ + A0 , , dr d\b . d& and hence — = — + — • ds ds ds Remembering that d$ r tan^ = r— = -, dr r 1 where r' = dr/dd, obtain the formula, drfl* [_ dO 2 } (14) P=± ^ ^r 9 ^ d0 2 ^ d0 2 Find the radius of curvature of each of the following curves at any point. (r 2 + a 2 )* 2. The spiral of Archimedes r = ad. Arts, p = v _ . \ • r r 2 + 2a 2 I. — 3. The cardioid r = 2 a (1 — cos <£). Ans. p = f V ar. a 2 4. The lemniscate 7^ = a 2 cos 2 0. ^ws. p = — . or r 3 5. The equilateral hyperbola r 2 cos 2 = a 2 . ^Lns. f> = -j- 6. The equiangular spiral r = ae K0 . 7 . The trisectrix r = 2 a cos — a. . a(5-4cos0)* S ^ ' = 9-6cos0 ' CHAPTER VIII THE CYCLOID 1. The Equations of the Cycloid. The cycloid is the path traced out by a point in the riui of a wheel as it rolls, i.e. by a point in the circumference of a circle which rolls without slip- ping on a straight line, always remaining in the same plane. Let the given line be taken as the axis of x and let 6 be the angle through which the circle has turned since the point P was last in contact with the line at 0. The coordinates of P, Fig. 42 x = OM and y — MP, can be expressed as follows in terms of 6. We notice that the arc NP=aO of the circle and the seg- ment ON of the line are of equal length, since the circle rolls without slipping. Hence OM= ON- Also, and so we have: (1) MP=NS {x =a(6 y = a(l as the equations of the cycloid. 146 MN^aO — asmS. LS = a — a cos ; sin 0), cos 0), THE CYCLOID 147 It is possible to eliminate 6 between these equations and thus obtain a single equation between x and y. But the func- tions thus introduced are less simple than those of equations (1) and it is more convenient to discuss the properties of the curve directly by means of these equations. EXERCISES 1. The equations of the cycloid referred to parallel axes with the new origin at the vertex, i.e. the highest point, are : J x = ad + a sin 0, ^ ' \y=—a + a cos 0, the angle 6 now being the angle through which the circle has turned since the point P was at the vertex. Obtain these equations geometrically, drawing first the requisite figure, and verify the result analytically by transforming the equa- tions (1) : x = x' + -n-a, y = y'-\-2a, 6 = 0' -f ir. 2. Show that the equations of an inverted cycloid referred to the vertex as origin can be written in the form : x = aO -f a sin 6, a — a cos 6. x Draw the figure and interpret $ geometrically. 2. Properties of the Cycloid. The slope of the curve is dy _ a sin Odd sinfl _ 2 sin \B cos \0 _ , x ~ dx adO — acosOdO 1 — cos0 2sm 2 ±0 or tanr = cot^0. From this, result we infer that the tangent at P is perpen- dicular to the chord PN, Fig. 43. For the latter makes an angle of \6 with the negative axis of x and hence its slope is — tani0, i.e. the negative reciprocal of the slope of the tan- gent. Thus we see that the normal at P goes through the 148 CALCULUS 7 V p )i ^/ 2 S / » f fcsSN. Fig. 43 and (6) aV d dy dx dx lowest point of the generating circie and hence the tangent at P goes through the highest point. The equation of the tangent at the point (x , y ), $ = , is (4) y-yo = w>HOo(F-Xo), and of the normal : (5) x-x + cot \B Q (y - 2/ ) = 0. The Evolute. We have seen that ^=coU0. Hence l+^- 2 =csc 2 i-0 dx dx 2 l ± csc 2 ±0d0 add — a cos OdQ 4asin 4 i0 4asin 4 i0' sin 3 i0 4a sin \B. It is now easy to construct the centre of curvature and thus find the evolute. We have merely to lay off on the normal PN a distance PQ = 4asin|0, i.e. double the distance PN. The locus of the point Q is thus seen to be an equal cycloid having its vertex at the origin O. We leave the proof, which is simple, to the student, referring him to Fig. 43. Fig. 44 The Arc. We have : ds 2 = dx 2 + dy 2 = a 2 [(l - cos 0) 2 + sin 2 0]d0 2 = 4a 2 sin 2 i0d0 2 , s = 2a /sini0d0 = 4a / sini0d(i0) = -4acos|-0 + C. THE CYCLOID 149 If we measure the arc from the origin, 0=-4a + C, C=4a, (7) .-. s = 4 -(Z-o\\, 2/=iT^-i>S' = (a + 6)sin0-6cosr^-/'|-0^1. 150 CALCULUS Furthermore, the arc AN= a0 and the arc NP=b are equal a0 = b. Hence we have as the equations of the epicycloid : x = (a + b) cos - b cos ^±^0, (9) y = (a + b) sin 0—b sin b a + b, If the variable circle rolls on the inside of the fixed circle, the path of the point P is a hypocycloid. Its equations are obtained in a similar manner and are: (10) x = (a — b) cos 6 -\-b cos 6, y = (a — b) sin — & sin 6. The following special cases are of interest. (1) If a = 26, the hypocycloid reduces to a segment of a straight line, namely, the diameter of the circle, y = 0. Thus a journal on the rim of a toothed wheel which meshes inter- nally with another wheel of twice the diameter describes a right line, so that circular motion is thereby converted into rectilinear motion. (2) When a = 4&, the equations of the hypocycloid reduce to the following (cf. Tables, Formulas 580 and 585) : r x = 3b cosO +bcos30 = acos 3 0, \ y = 36 sin 6 — b sin 3 = a sin 3 6. Hence x s + y 1 Fig 46 This is the equation of the four-cusped hypocycloid. The cycloids play an important part in Applied Mechanics, in the theory of the shape in which the teeth of gears should be cut. For a more extensive discussion of the subject of this chap- ter see Williamson, Differential Calculus, Chap. XIX, Roulettes. THE CYCLOID 151 EXERCISES 1. Show by means of the equation of the normal of the cycloid, (5), that the normal goes through the lowest point of the generating circle. 2. Obtain the equations of the path of the journal of the driver of a locomotive and plot the curve. 3. Obtain the equations of the path of a point on the outer edge of the flange of a driver. 4. Obtain the equations of the path of the pedal of a bicycle. 5. Obtain the equation of the path of an arbitrary point in the wheels of a sidewheel steamboat. The curves of Exs. 2-5 are called trochoids. 6. Find the velocity, v, of the point that generates a cycloid. Ans. v = 2awsin^0 = 2Fsini0, where wis the angular velocity of the wheel and V the linear velocity of the hub. At the vertex v= 2V, i.e. the velocity of the highest point of the wheel is twice that of the hub. 7. Find the area included between an arch of the cycloid and its evolute. Ans. 4:ira 2 . 8. Show that the length of the arc of an inverted cycloid (3), measured from the vertex is s = 4asin t. 9. Obtain the equations of the evolute of the cycloid ana- lytically, by means of equations (9) in Chap. VII. 10. At what points is the trochoid x=a$ — b sin 6, y = a—bcos6, (b), the cusp being taken as the pole. Obtain this result from equations (9). 13. Obtain the result in question 12 directly geometrically. 14. Prove by elementary geometry that the hypocycloid for which b — \a is a straight line. 15. Show that the equation of the normal of the hypocycloid is: (sin O + sin £ZL_ $ )(x — x ) = (cos O — cos ^=— O ) (y — y ). 16. Prove that the normal of the hypocycloid passes through the point of contact of the rolling circle. 17. Work out questions 15 and 16 for the epicycloid. 18. Show that the hypocycloid for which b = \a and that for which b = j a are the same curve. 19. Show that the length of the four-cusped hypocycloid is three times the diameter of the fixed circle. 20. Find the area of the four-cusped hypocycloid. Ans. W 8 21. Find the area enclosed between one arch of a four-cusped epicycloid and the fixed circle. Ans. 22. Obtain the equations of the epitrochoid. 23. Obtain the equations of the hypotrochoid. 24. How many revolutions does the rolling circle make in tracing out a cardioid ? a four-cusped hypocycloid ? How many revolutions does the moon make in a lunar month ? 25. How many cusps does an epicycloid have when a and b are commensurable: a/b=p/q? What can you say about this curve when a and b are incommensurable ? CHAPTER IX DEFINITE INTEGRALS 1. A New Expression for the Area under a Curve. In Chap. VI we learned how to compute the area A under a continuous curve, y=f(x), by integration. We found tnat D x A = y and hence finally : A— lydx+C, We will now consider a new method of computing the same area. Let the interval (a, b) of the axis of x : a A* + ... +/(a>n-i) A*, and thus, allowing n to increase without limit, we obtain the result : (3) A=* lim t/(a*,)Aa>+/(aJi) Aa> + ••• +/ and compute its value. Determine the limit of the sum (2) for this function by means of the indefinite integral. 2. Taking as the interval (x) is the equation of the generating curve. For example, let it be required to find the volume of a segment of a sphere. Here the generating curve is a circle, . at-hy^T*, and, h denoting the altitude of the segment, the abscissas of the bases are r — h and r. Hence 158 CALCULUS F = tt ffr 2 -aAdx o x irx 3 = |M If, in particular, /i——r, we have the complete sphere and obtain the familiar result f tt?* 3 . EXERCISES 1. Show that the volume of an ellipsoid of revolution is |7ra6 2 , where a denotes the half-length of the axis. 2. A spindle is formed by the rotation of an arch of the curve y = sin x about its base. Find its volume. Ans. 4.93480. 3. Show that the volume of a cone is ^irr 2 h, and that the vol- ume of a frustum is 4. Show that the volume of a segment of a paraboloid of revolution, of arbitrary altitude, is one-half that of the circum- scribing cylinder. .5. A cycloid revolves about its base. Show that the vol- ume of the solid generated is 57r 2 a 3 . 6. The four-cusped hypocycloid x% + y$ = a* rotates about the axis of x. Find the volume of the solid generated. 4 32-rra 3 Ans. _. 7. Find the volume of a segment of the solid of revolution generated by the catenary : when it rotates about the axis of x, the plane x = forming one of the bases. Am w(£ _ g -» + a DEFINITE INTEGRALS 159 4. Other Volumes. We will begin with the following ex- ample. A wood-cutter starts to fell a tree 4 ft. in diameter and cuts half way through. One face of the cut is horizontal, and the other face is inclined to the horizontal at an angle of 45°. How much of the wood is lost in chips ? Since the solid whose volume we wish to compute is sym- metric, we may confine ourselves to the portion OABC. Divide the edge OA into n equal parts and pass planes through these points of division per- pendicular to OA. The solid is thus divided into slabs that are nearly prisms ; only the face QRB'Q' is not a plane. Let us meet this difficulty by constructing a right prism on PQR as base and with PP' as altitude. Then its volume will be a little greater than that of the actual slab. The solid formed by the n prisms thus constructed differs in volume but slightly from the actual solid.* We will next formulate analytically the volume of the prisms. The base PQR is a 45° right triangle. Let OP=x k and PQ = y k . Then, by the Pythagorean Theorem, Hence the volume of this prism is i yi ?Ax = ±(4-x*)Ax, * Let the student not proceed further till this point is perfectly clear to him. Let him make a model of the actual solid out of cardboard or a piece of wood and draw neatly the lines in which the plane sections through P and P' perpendicular to OA cut the solid. He will then be able to visualize the auxiliary prisms without difficulty and to perceive that the sum of their volumes approaches as its limit the volume to be computed. Fig. 49 160 CALCULUS and the volume of the solid we wish to compute is (8) lim [i (4 - V) *b + i (4 - x*) A* + . • . + J (4 - x n _ 1 2 ) Ax]. n=oo The problem before us is thus reduced to that of computing the limit (8). Now inspection of this limit shows that it is of the same type as the limit (3) of § 1 ; in fact, the two vari- ables become identical if we put Hence the limit (8) can be computed by integrating the function J (4 — x 2 ) and taking the integral between the limits x = a = and x = b — 2 : / 4(4-^)fe = 2x-J, The total volume is twice this amount, and thus it appears that there were 5J cu. ft. of chips hewn out. EXERCISES 1. A banister cap is bounded by two equal cylinders of revolution whose axes intersect at right angles in the plane of the base of the cap. Find the volume of the cap. Ans. -fa 3 . 2. A Rugby foot-ball is 16 in. long, and a plane section con- taining a seam of the cover is an ellipse 8 in. broad. Find the volume of the ball, assuming that the leather is so stiff that every plane cross-section is a square. Ans. 341-J cu. in. 3. Do the preceding problem on the assumption that the leather is so soft that every plane cross-section is a circle. Ans. 536 cu. in. 4. A solid is generated by a variable hexagon which moves so that its plane is always perpendicular to a given diameter of a fixed circle, the centre of the hexagon lying in this diam- DEFINITE INTEGRALS 161 eter, and its size varying so that two of its vertices always lie on the circle. Find the volume of the solid. Ans. 2V3a 3 . 5. A conoid is a wedge-shaped solid whose lateral surface is generated by a straight line which moves so as always to keep par- allel to a fixed plane and to pass through a fixed circle and a fixed straight line; both the line and the plane of the circle being per- pendicular to the fixed plane. Find the volume of the solid. Ans. %ira 2 h. Fig. 50 6. Find the superficial area of two of the solids considered above. 7. Show that the volume of an ellipsoid whose semi-axes are of lengths a, b, c is ^-n-abc. 5. Fluid Pressure. We will next consider the problem of finding the pressure of a liquid on a vertical wall. Let the surface be bounded as indicated in the figure and let it be divided into n strips by ordinates that are equally spaced. Denote the pressure on the ft-th strip by AP k . Then we can approximate to &P k as follows. Consider the rectangle cut out of this strip by a parallel to the axis of x through the point (x k , y k ). The pressure on this rectangle is less than that on the given strip ; but we do not yet know how great it is. Still, if we turn the rectangle through 90° about its upper side, the ordinate y k , we shall obviously have decreased the pressure further. Now the pressure on the rectangle in this new posi- Xk*l Fig. 51 162 CALCULUS tion is readily computed. It is precisely the weight of a col- umn of the liquid standing on this rectangle as base. The volume of such a column is (x k -\- c)y k Ax, and if we denote by w the weight of a cubic unit of the liquid, then the weight of the column in question is w(x k + c)y k Ax. This is less than AP k . In like manner we can find a major approximation by con- sidering the rectangle that circumscribes the given strip and whose altitude is y k+ i, and then turning it over on its lower base. The pressure on it in its new position is and this is larger than AP k . We thus obtain : (9) w(x k + c)y k Ax < AP k dx = -|(9 - a 2 )* | = 9, and the total pressure is 2 x 62 J x 9 = 1120 lbs. EXERCISES 1. A vertical masonry dam in the form of a trapezoid is 200 ft. long at the surface of the water, 150 ft. long at the bottom, and is 60 ft. high. What pressure must it withstand ? Ans. 9300 tons. 2. A cross-section of a trough is a parabola with vertex downward, the latus rectum lying in the surface and being 4 ft. long. Find the pressure on the end of the trough when it is full of water. Ans. 66 lbs. 3. One end of an unfinished watermain 4 ft. in diameter is closed by a temporary bulkhead and the water is let in from the reservoir. Find the pressure on the bulkhead if its centre is 40 ft. below the surface of the water in the reservoir. Ans. Nearly 16 tons. 6. Duhamel's Theorem. Let «i + «2 + ... • + a n be a sum of positive infinitesimals which approaches a limit when n becomes infinite; and let Pi + P* + -+Pn be a second sum such that fi k differs from a k by an infinitesimal of higher order : where lim^ = l, A.1+, % h, Pk = = a k -f **«*, : k is infinitesimal. Then Iim[ft + ft4- ••• + ftj = lim [«! + «*+ • "+<]. DEFINITE INTEGRALS 165 In accordance with the hypothesis of the theorem we have 4- — V ^ € ^ — Vf y^tn^y. — yct 1 <€ l a l ^r]a u < — 7ja n ^ e M « ;l ^ ry«. < Hence — (ai + «2+ h« w )^S £ i a i+ hc»« B g(a l +a 2 + |-«n)i/- But 17 approaches and a x + « 2 + ••• + «„ remains finite. This completes the proof. Application. As a typical appli- cation of Duhamel's Theorem we will give the completion of the proof of formula (12). Let y[ be the minimum ordinate in the A:-th strip, and let y' k ' be the maximum ordinate. Then we have by like reasoning to that of § 5 : Vu Fig. 52 w (x k + c)y' k Ax < AP k < w (x k+l + c) y' k ' Ax, and hence P lies between the two variables (13) w (xq + c) yl Ax + w (x x + c) y[ Ax -\ \-w (aj B _i + c)yi_i Aa? and (14) !«(#! + c)^'Aa? + w(a 2 + c)2/i'A£H r-w(»»4-c)yi'_iAaj. Neither of these variables is of the type to which the Funda- mental Theorem of § 2 applies ; but each suggests the variable (15) w(pc + c)y Ax-{-w(x 1 -{-c)y l Ax-\ \-w(x n _ 1 + c)y n . 1 £iX 9 whose limit is the definite integral (12), and each approaches the value of this integral as its limit, as we will now show. 166 CALCULUS Let a k = w (x k + c) y k Aaj, P k = w (x k + c) y[ Az. Then lim(a 1 4-« 2 + \- o=/< w(x-\-c)ydx. Furthermore, & = t£&±£}#jA* = |J i im .^ = 1 . a* wfe + c^Aa ?/ A n=»2/, Hence /?i-f- A> + ••• + /?«> i.e. the variable (13), approaches the value of the above integral as its limit. In like manner it is shown that (14) approaches this same limit. Hence P is equal to this limit and (12) holds in all cases. 7. Length of a Curve. In Chap. VI § 8 we found the length of a curve by means of the indefinite integral, =jV'+(i: dx. We can also evaluate the required length by con- sidering it as the limit of a sum. Divide the inter- val from x = a to x = b into n equal parts, erect ordinates at the points of division, and inscribe a broken line in the arc to be measured. The length of this line is Fig. 53 X VAa^ + Aft* = VAx 2 + A?/ 2 + VAz 2 + A?/f H f- vAz 2 + A?/ n _i 2 and the limit of its length is the length s to be determined. Now this latter variable is not of a type whose limit is a definite integral, but it suggests a new sum which is and DEFINITE INTEGRALS 167 whose limit, moreover, can be identified with the above limit by DuhamePs Theorem ; namely, since lim Ay k / Ax =f'(x k ): VI +f(x Q fAx + VTTfW 2 Ax + • • • + Vl+/'K_ 1 ) 2 Aa;. For, letting «* = vr+T 7 ^? ax, p k =yji + ^ a*, we see that lim & = 1. n=oo 0^ Hence 5 (16) #= fvi+f(xydx, a and this agrees with the earlier result above referred to. 8. Area of a Surface of Revolution. To find the lateral area of a surface of revolution we proceed in a manner similar to that employed in finding the volume, § 3. Divide the interval from x = a to x = b into n equal parts, erect ordinates, and inscribe a broken line in the arc of the generating curve, as in the preceding paragraph. This broken line, when it rotates about the axis of revolution, generates the lateral surfaces of a series of frusta of cones. Let us compute the lateral area of the ft-th frustum. The lateral area of a cone is half the product of its slant height by the perimeter of its base, -n-rl. The corre- sponding formula for the frustum is the product of the slant height by the circumference of the circular cross-section made by a plane passed midway between the bases, 7r(r + M)l Hence the lateral area in question is (17) v (y k + Jfc+i) VAz 2 + Ay k % and the area of the surface S that we wish to compute is thus seen to be : (18) 8 = lim V * (y k + a* i) Va^+a^7- n = oo -f> Ans. 2tt(& 2 + ^sin^e DEFINITE INTEGRALS 169 3. The curve # J H- y — q> rotates about the axis of x. Find the total superficial area of the surface generated. . 12 -n-a 2 5 4. An arch of a cycloid rotates about its base. Determine the superficial area of the surface generated. , 647rcr o * 5. Show that in polar coordinates £ (20) ^2Wfritt^r»+ ££d§, a < 6. Find the area of the surface generated by the rotation of the cardioid r = 2a(l — cos0) about its axis. Ans. +— • 5 7. Show that the area of a surface of revolution is given by the formula (21) 8 = 2tt I yds, where the coordinates x, y of a point of the generating curve are expressed as functions of the length of the arc, s. 9. Centre of Gravity. A Law of Statics. Let n particles, of masses m lf m 2 , •••, rn in , be situated on a straight line, which we will take as the axis of x, and let their coordinates be x l} x 2 , -•-,x n . Then the coordinate x of their centre of gravity is given by the formula : * (22) X — m i ,T i ~f~ m z x 2 4- • • • -f m m 1 + m< i + ••- +m n * A proof of this formula may be found in any work on Mechanics, for example, Jeans, Theoretical Mechanics, Chap. VI. 170 CALCULUS If the particles are situated in a plane at the points (x 1 , y^, •••, 0„, y„)_or in space at the points (x 1} y u z,), ..., (x n , y n , z n ), and if (x, y) or (x, y, z) is their centre of gravity, then x is given by the same formula (22), and y and 2 are given by similar formulas, in which the cc's are all replaced by 2/'s or z's. Example. A granite column 6 ft. high and 1% ft. in diam- eter is capped by a ball of the same substance 2 ft. in diameter and stands on a cylindrical granite pedestal 9 in. high and 2 ft. in diameter. How high above the ground is the centre of gravity of the whole post ? Ans. 4.26 ft. 10. Centre of Gravity of Solids and Surfaces of Revolution. By the aid of the Calculus we can compute the centre of gravity of bodies not made up of a finite number of particles or of bodies whose centres of gravity are known. We will begin with homogeneous solids of revolution. Their centre of gravity always lies somewhere in the axis of symmetry. Divide the body into slabs as in § 3, Fig. 48, and denote the abscissa of the centre of gravity of the k-th slab by x£. Then, if p denote the density of the substance, the mass of the k-th slab is pAV k , and /3 AF + /3 AF 1 +-"+ /0 AF n _ 1 Here b V—tt I y 2 dx, and it remains to compute the value of the numerator. Since this sum has the same value for all values of n, namely x V, we may allow n to increase without limit, and we shall have (23) Km [x' A F + x[ A V x + • • • + a£_i A V n -{\ =xV. n=oo Now this bracket readily suggests a sum whose limit can be computed by integration. Since DEFINITE INTEGRALS 171 and ^yl 2 Ax^ A V k ^ iry' k ' 2 Ax, where y[ and yH are respectively the smallest and the largest radii of any cross-section of the &-th slab, we see that irx k y' h 2 Ax =*■ l!o = T- and K = l^r^ = |A, i.e. the centre of gravity is three-fourths of the distance from 172 CALCULUS EXERCISES 1. How far is the centre of gravity of a hemisphere from the centre of the sphere ? Ans. § r. * 2. Find the centre of gravity of a segment of a paraboloid of revolution. 3. Find the centre of gravity of a segment of a sphere. * 4. Find the centre of gravity of a frustum of a cone. Ans. Distance from smaller base, - • _^ "*" _ — . 4 R 2 + Rr + r 2 ' 5. The curve y = sin x, ^ a? ^ - > rotates about the axis of x. Find the centre of gravity of the solid generated. ^ - = * 1 = ua 4 7T «> 6. Show that the centre of gravity of a surface of revolution bounded by two planes perpendicular to the axis is giiien by the formula •w I xyds 2tt I ^2/A/l-f -^dx H 7. Prove that the centre of gravity of any zone of a sphere lies midway between the bases of the zone. " 8. Find the centre of gravity of the lateral surface of a cone of revolution. Ans. |7i. 9. Find the centre of gravity of the lateral surface of a segment of a paraboloid. 11. Centre of Gravity of Plane Areas. To find the abscissa of the centre of gravity of the area under a curve, y =f(x), § 1, Fig. 47, divide the area into n >.*n-ips of equal breadth as there ■ . li I eonsid" i ' ., >■■■-->■. ■ , : ' U strip be denotet ; ..■..;.. , t^e^uper- i DEFINITE INTEGRALS 173 ficial density, supposed constant (i.e. the mass of one square unit of the slab), by p, then the mass of the A:-th strip will be pAA k . Let the abscissa of its centre of gravity be x' k . Then we shall have - = P AA ■ a^ + pAA • x[+ ••• +pA^ n _ 1 . x f n _ x P AA + p&A 1 + ••'+pAA n _ 1 _ x' AA + x l *A l + • • • + <-i AA-i A b Here jydx, and it remains to compute the value of the numerator. The reasoning is precisely similar to that of the preceding para- graph. We allow n to become infinite and thus obtain lim [4 A Aq + x[ &A X ■{ 4- a?i_j A A n _{\ = x A. Now x k (x) is the equation of the lower boundary, and y" =f(x) that of the upper one. 5. Show that the centre of gravity of a triangle lies at the intersection of the medians. 6. Show that the centre of gravity of a uniform wire of length I is given by the formula: / x ds j #-%/ 1+Xdx dx 2 x I I 7. Find the centre of gravity of a uniform semi-circular wire. A 2r nnrr Ans. x — — = .637r. 7T 12. General Formulation. The foregoing examples may all be brought under one general formulation, which applies furthermore to bodies of variable density and wholly arbitrary shape. Let the body be divided into small pieces, and denote the mass of any piece by £±M k , the abscissa of its centre of gravity by x[. Then n-1 M DEFINITE INTEGRALS 175 and hence (26) a_— a M The latter limit can always be computed by means of integrals, but it may be necessary to employ double or triple integrals, cf. the later chapters. Formula (26) is sometimes written in the form : CxdM (27) »-^r? 13. Centre of Fluid Pressure. Let us determine at what height a horizontal brace should be applied to hold the pressure of the liquid computed in § 5, without there being any tendency of the surface to rotate about the brace. Divide the surface into strips, as in § 5, Fig. 51, and consider the pressures on the successive strips. As in the problem of the centre of gravity of n particles, we have here again to do with the ab- scissa of the resultant of a system of parallel forces. Let the abscissa of the point at which the pressure &P k on the A;-th strip acts be denoted by x[, the abscissa of the brace by x. Then g = gjAP + ariAP 1 + - +<-i±P n -i APo + AA+.-.+AP^ The value of the denominator is b P=w I (x-\-c)ydx. Allowing n in the numerator to increase without limit, we obtain by reasoning now familiar to us the result : 6 ? I (x-\-c)xydx 176 CALCULUS For example, to find the centre of pressure for the rectangle considered in § 5. Here b h h s 7c X = fA, and the brace should be applied to the end of the tank two- thirds of the way down. EXERCISE Find the depth of the centre of pressure in the case of the dam described in § 5, Ex. 1. 14. Moment of Inertia. By the moment of inertia of a system of particles about a straight line in space, called the axis, is meant the quantity n (28) ^ m k r k 2 = m v r? + m 2 r 2 2 -\ \- m n r n 2 , where r k denotes the perpendicular distance of the k-th. particle, whose mass is m k , from the axis. If a body consists of a continuous distribution of matter, like a wire or a plate or a solid body, its moment of inertia is defined as follows. Let the body be divided up into small pieces, of mass m k , and let the mass of each piece be concen- trated at one of its points, whose distance from the axis shall be denoted by r k . Form the sum (28) for all the pieces. Then the limit of this sum as the pieces grow smaller and smaller is the moment of inertia of the body : (29) / = limV m fc r fc 2 . The physical meaning of the moment of inertia is the measure of the resistance which the body opposes, through DEFINITE INTEGRALS 177 its inertia, to being rotated about the axis. The moment of inertia also has an important application in the theory of the strength of materials. By means of the Calculus we can compute the moment of inertia of any body. Let us begin with a circular wire, of radius r and mass m, the axis being perpendicular to the plane of the circle and passing through its centre. Here every point in the matter in question is at the same distance r from the axis, and so the moment of inertia is I=mr 2 . Next, consider a uniform circular disc. Divide its radius into n equal parts: r = 0, r u r 2 , •••, r n = a, and cut the disc up into rings by concentric circles of radii r lf •••, r n _ x . The mo- ment of inertia of the whole disc is equal to the sum of the moments of inertia of these rings. Now the moment of inertia of the Axth ring, A7" x ., evidently is greater than what it would be if its mass were concentrated along its inner boundary, but less than if its mass were concentrated along its outer boundary. Hence (30) r, 2 A Jf, < AJ, < r, +1 2 AM k . Furthermore, AM k = pAA k} where p denotes the density and AA k the area : A.4, = 7rr, +1 2 - Trr, 2 = *(r k + Ar) 2 - Trr, 2 = 2irr*Ar + tt Ar 2 , (31) AM k = 2tt P r k Ar + irp Ar 2 . We are now in a position to apply Duhamel's Theorem. We have I=AJ -hAJi+... -f-Ak., = lim[A/e4-AJ 1 +- 4-Al^]. n=oo On the other hand, formulas (30) and (31) suggest a simpler infinitesimal by which to replace A/ A ., namely a k =.27rpr k 3 Ar. 178 CALCULUS In fact, if we divide (30) through by a k : - k 2 AM k Al k ^WAJf, 2 irpfj? Ar 2 Trpr^. 3 Ar 2 irpr k s Ar ,we see that the limit of either extreme is 1, and so the limit of the middle expression must also be 1. Putting, then, f$ k = AI k , we get : n=oc a k Hence a I = lim V 2*pr k **r = 2 vp fr>dr = ^ . — 3 ^ 2 The mass of the disc is M = irpa 2 , and consequently /may be written in the form : (32) I = M°L. Definition. If the moment of inertia of a body be written in the form: I=Mk 2 , k is called the radius of gyration. The radius of gyration is defined, then, as s/I J M. It may be interpreted as follows : if all the mass were spread out uniformly along a circular wire of radius k, the axis passing through the centre of the ring at right angles to its plane, the moment of inertia would still be the same : / = M k 2 . The radius of gyration of the above cir- cular plate is a/ ^/2. EXERCISES Determine the following moments of inertia. 1. A uniform rod, of length 2 a, about a perpendicular bisector. An ^ Ma 2 2. A square whose sides are of length 2 a, about a parallel to a side through the centre. a Ma 2 ^ • o DEFINITE INTEGRALS 179 3. A uniform rod of length I about a perpendicular through one end. * Ml 2 3 njr 2 v 4. A circular disc about a diameter. Ans. — — • 4 /&. An isosceles triangle about the median through the vertex. Ma 2 Ans. — - f where a is half the length of the base. 6. A scalene triangle about a median. Mh 2 Ans. , where h is the distance of either vertex from 6 the median. J 7. A circular wire about a diameter. s 8. A cone of revolution about its axis. 9. A sphere about a diameter. 15. A General Theorem. When the moment of inertia of a body about an axis is once known, its moment of inertia about any parallel axis can be found without performing a new inte- gration. The theorem is as follows. Theorem. If the moment of inertia of a body about an arbitrary axis be denoted by I , that about a parallel axis through the centre of gravity by i, then (33) I = I+Mh\ where h denotes the distance between the axes. We will prove the theorem first for a system of particles. Assume a set of cartesian coordinates (x, y, z), the axis of z being taken as the first axis of the theorem, and then take a second set of cartesian coordinates (x', y', z') parallel to the first, the origin being at the centre of gravity. Then we have : Ans. Ma 2 2 Ans. 33fr 2 10 Ans. 2Ma 2 180 CALCULUS •4=2) wfS = 2) m (^ 2 + 2/ 2 )> Furthermore, x=x' + x, y = y' + y, z = z' + z, where (x, y, z) is the centre of gravity referred to the (x, y, z) axes. Hence 2) m(x* + y 2 ) = 2) m (x' 2 + 2/' 2 ) +2x2) m *' + 2 ? 2) «? Now 2) m ^' = 0> 2) Wl ^' = ^ For, recall formula (22) in § 9. Applying that formula to the present system of particles, referred to the (V, y', 2')-axes, we see that the abscissa of the centre of gravity, x\ is : 7 mx' M But the centre of gravity is at the new origin of coordinates, and so x' = 0, hence 2) mx> = 0- Similarly, 2j m . ? / = 0- It remains only to interpret the terms that are left, and thus the theorem is proven for a system of particles. If we have a body consisting of a continuous distribution of matter, we divide it up into small pieces, concentrate the mass of each piece at its centre of gravity, form the above sums, and take their limits. We shall have as before 2)mx'=0, V my ] = 0, and hence 2) m(x 2 + 2/ 2 ) = 2) m(aj' 2 + y' 2 ) + Mli\ lim 2) m(x 2 +if)= lim ^m(x' 2 + y' 2 ) + Mh 2 , q. e. d. DEFINITE INTEGRALS 181 EXERCISES Determine the following moments of inertia. 1. A circular disc about a point in its circumference.* Ans. SMa 2 2. A uniform rod, of length 2 a, about a point in its perpen- dicular bisector. Ans. Jlff^ + h* 3. A rectangle, of sides 2 a and 26, about its centre of gravity. A M^ 2 ±^). o 4. The following figures about the axis through the centre of gravity parallel to the lines of the page : [^ in X D Fig. 54 16. The Attraction of Gravitation. Sir Isaac Newton dis- covered the law of universal gravitation. This law asserts that any two particles in the universe attract each other with a force proportional to their masses and inversely proportional to the square of the distance between them: (34) /* mm' /-.*■ mm' where K is a physical constant.f By means of the Calculus we can compute the force with which bodies consisting of a continuous distribution of matter attract one another. Let us determine the force which a uni- form rod of mass M exerts on a particle of mass m situated * By the moment of inertia of any distribution of matter in a plane about a point in that plane is meant the moment of inertia about an axis through the point perpendicular to the plane. t Called the gravitational constant. Its value is 6.5 x 10 8 cm 3 sec~ 2 gr- 1 . 182 CALCULUS in its own line. Divide the rod up into n equal parts and denote the attraction of the k-th segment by AA k . The mass 6 of this segment is pAx, ;------ — - I ' ■ "";. ■ ' ' ' ' H*. . where p denotes the den- x* a x o x \ x k «*+i x n r sity of the rod. Now if FlG * 55 this whole mass pAx were concentrated at the nearer end, its attraction would be greater than AA k ; and similarly, if it were concentrated at the further end, its attraction would be less. Hence jr«e£f<.A4 t k+1 < AF k < K^cos k , and hence we infer by the usual method of rea- soning that „ *» Fig. 56 Here r 2 = h 2 + x 2 , cos = — , ■Vh' + x 2 and so we have F=2Km P h f dx J V(ft 2 + ar>) 3 From Peirce's Tables, Formula 138, dx x f. V(^ 2 +^) 3 hWh 2 + x 2 and consequently 2Kmpa _ ^ mM hVh 2 +a 2 h^/h 2 + a 2 F= 2Emp VW+x 2 EXERCISES Compute the following attractions. 1. A rod whose density varies as the distance from one of its ends, on a particle in its own line. An8 . ?^M"i 0g i±^ +7 A_-.r i 2 & h i+h * This relation holds for the part of the rod we are considering, namely, when 0. For the other half a modification is necessary. 184 CALCULUS t 2. A semicircular wire, on a particle at its centre. Ans. iS^f. ^ 3. The same wire, on a particle in the circumference situated symmetrically as regards the wire. * KmM ^ . 3?r '*(**• 8 ' ^ 4. A rod AB, on a particle situated at in a perpendicular O-B at one end. Ans. A force of 2KmM sin \AOB, making an angle of hi \AOB with OB. \l 5. A circular disc, on a particle in the perpendicular to the disc at its centre. \ 2KmM 1- 2 4- a 2 J 6. A rectangle, on a particle in a parallel to two of the sides through the centre. For further simple problems in attraction cf. Peirce, New- tonian Potential Function. 17. Proof of Formula (3). We can give a proof of (3) as follows. Suppose that y increases with x, as in Fig. 57. Then the above rectangles are all inscribed in the curve and their sum is less than the area A : (35) /(a^Aaj+Z^Aa; + - +/(av_ I )Aa;(x'). We may denote the variable of in- tegration, x, by t, and at the same time change x' to x. Thus we have : (39) *(*) = J/(0 dt. The integral on the right-hand side represents the area under the curve, bounded by a variable ordinate whose abscissa is x. Hence its derivative has the value (Chap. VI, § 1) f(x), and thus we see that X (40) *'(*)=/(*) or *.ff(f)dt=f(x). a Finally, the variable of integration, t, is often denoted by the same letter as the variable upper limit, (40) thus being written : X (41) (x)=Jf(x)dx. DEFINITE INTEGRALS 187 EXERCISES 1. A cycloid revolves about the tangent at its vertex. Show that the volume of the solid generated is 7r 2 a 3 . 2. Show that the volume of the solid generated by a curve that revolves about the axis of y is given by the formula : V = 7T I xPdy. 3. A cycloid revolves about its axis, i.e. the line through the vertex perpendicular to the base. Show that the volume of the solid generated is iro?l-^ )• 4. If the curve y 2 (x — 4 a) = ax (x — 3 a) revolve about the axis of x, show that the volume of the solid generated by the loop is ?r a 3 (7| — 8 log 2). Compute this vol- ume correct to three significant figures when a — 1. Ans. 6.12. 5. The curve y 2 — x (x — 1) (x — 2) revolves about the axis of x. Show that the volume of the solid generated by the oval is \tt. 6. Find the volume of the solid generated by the catenary y = i(e* + e- x ) when it rotates about the axis of y. Ans. £[(r 2 -2r + 2)e r +(?* 2 + 2r + 2)e- r -4], where r de- notes the radius of the base. 7. Find the volume of a torus (anchor ring). Ans. 2ir i o 2 b. 8. Find the area of the surface of a torus. 9. Find the area of the loop of the curve X s — 3axy + y s = 0. . Ans. fa 2 . 188 CALCULUS 10. Find the area of the loop of the curve r cos = a cos 20. Ans. (2--)a 2 . 11. Obtain the area of the surface of a segment of the solid generated by the rotation of a catenary (a) about the axis of x ; (b) about the axis of y. Ans. (a) 7r(ys-\-ax)', (b) 2 7r(a 2 -\-xs — ay), where s denotes the length of the arc measured from the origin. 12. The kinetic energy of a system of particles, moving in any manner, is the sum of the kinetic energies of the individ- ual particles, 2^^m k v k 2 . Show that the kinetic energy of a uniform rod of length 2 a, which is rotating about its perpendicular bisector with angular velocity o>, is i Jfa 2 o> 2 . 13. A pendulum consisting of a rectangular lamina oscillates about an axis perpendicular to its plane and passing through the middle point of one of its sides. Compute its kinetic energy. . Ans. M(±a? + \b 2 )oi 2 . 14. A homogeneous cylinder rotates about its axis. Find its kinetic energy. Ans. \Md 2 o> 2 . 15. Show that the kinetic energy of a rigid body, rotating with angular velocity s = igt* + K, iT=0, and we have the law for freely falling bodies deduced directly from Newton's Second Law of Motion, the hypothesis being merely that the force of gravity is constant. Substituting in the last equation the observed values s = S, t = T, we get : 9 fit if jt2 If we use English units for mass, space, and time, g has, to two significant figures, the value 32, i.e. the absolute unit of force in this system, a pounded, is equal nearly to half an ounce. If we use c.g.s. units, g ranges from 978 to 983 at different parts of the earth, and has in Cambridge the value 980. The absolute unit of force in this system is called the dyne. Since g is equal to the acceleration with which a body falls Again, in differentiating the logarithm, we found D x \og a x=(\og a e)-. x This multiplier reduces to unity when we take a = e. Hence the defini- tion : The ' absolute (natural) base of logarithms is that base which makes the multiplier log a e in the above equation equal to unity. 196 CALCULUS freely under the attraction of gravity, g is called the accelera- tion of gravity. But this is not our definition of g ; it is a theorem about g that follows from Newton's Second Law of Motion. The student can now readily prove the following theorem, which is often taken as the definition of the absolute unit of force in elementary physics: The absolute unit of force is that force which, acting on the unit of mass for the unit of time, generates the unit of velocity. Example 1. A body is projected down an inclined plane with an initial velocity of v feet per second. Determine the motion completely. The forces which act are : the component of gravity, mg sin y absolute units, down the plane, and the force of friction, fj,R = fxmg cos y up the plane. Hence ma = mg sin y — /xmg cos y dv or — s= g sin y — fig cos y. at Integrating this equation, we get V as g (sin y — fx COS y) t + C, v = + C, (A) v= cti = 9 ^ sin y-i MG0S y) t + v <>- v t, A second integration gives (B) 8 = \ g (sin y — fi cos y) f the constant of integration here being 0. To find v in terms of s we may eliminate t between (A) and (B). Or we can begin by using formula (3) for the acceleration : dv . . v ' U ^" :=9,( ^ Siny "" /xC0S ^' ■|-'y 2 = gf(siny — ^,COSy)s + K, W= +K, v 2 = 2g (sin y — /x, cos y) s -f v 2 . MECHANICS 197 Example 2. An Atwood's machine has equal weights, M and M , attached to the cord, and a rider of mass m is added to one of the weights. Determine the motion. We apply Newton's Second Law to each of the weights M and M+m individually. The forces are indicated in the diagram, the tension in the string, whose weight is negligible, being the same at all points. Moreover, since the space traversed by both weights is the same, s, their velocities and accelerations are also equal. Thus M a = T - Mg, (M + m)a= (M + m)g-T, mg 2M+m T = (2M+27n)M t 2M+m " Fig. 61 From the last formula it appears that the tension is constant and that it lies between the values Mg and (M+m)g. The student can work out for himself the formulas that give v and s in terms of t, and v in terms of s. EXERCISES* 1. A weightless cord passes over a smooth pulley and car- ries weights of 8 and 9 pounds at its ends. The system starts from rest. Find how far the 9 pound weight will descend before it has acquired a velocity of one foot a second. What is the tension in the cord ? Arts. \\ ft. ; 8.4 lbs. 2. Obtain the usual formulas for the motion of a body pro- jected vertically : or =-2gs+v 2 -, = -gt + v Q \ = -igt 2 +v t. 3. On the surface of the moon a pound weighs only one sixth as much as on the surface of the earth. If a mouse can jump v 2 = 2gs + v 2 gt + v Q or or See foot note on p. 193. 198 CALCULUS up 1 foot on the surface of the earth, how high could it jump on the surface of the moon ? Compare the time it is in the air in the two cases. 4. A bullet fired from a revolver penetrates a block of wood to a distance of 6 inches. How much greater would its velocity have to be to make it go in 12 inches ? Assume the resistance to be the same at all points, for all velocities. Ans. About 40 percent. 5. Kegarding the big locomotive exhibited at the World's Fair in 1904 by the Baltimore and Ohio Railroad the Scientific American says : " Previous to sending the engine to St. Louis, the engine was tested at Schenectady, where she took a 63-car train weighing 3,150 tons up a one-per-cent. grade." Find how long it would take the engine to develop a speed of 15 m. per h. in the same train on the level, starting from rest, the draw-bar pull being assumed to be the same as on the grade. 6. A block of iron weighing 100 pounds rests on a smooth table. A cord, attached to the iron, runs over a smooth pulley at the edge of the table and carries a weight of 15 pounds, which hangs vertically. The system is released with the iron 10 feet from the pulley. How long will it be before the iron reaches the pulley, and how fast will it be moving ? Ans. 2.19 sec. ; 9.1 ft. a sec. 7. Solve the same problem on the assumption that the table is rough, /a =2*0, an( ^ tnat tne P u ^ e y exerts a constant re- tarding force of 4 ounces. 8. If Sir Isaac Newton registered 170 pounds on a spring balance in an elevator at rest, and if, when the elevator was moving, he weighed only 169 pounds, what inference would he draw about the motion of the elevator ? 9. What does a man whose weight is 180 pounds weigh in an elevator that is descending with an acceleration of 2 feet per second per second ? MECHANICS 199 (1 10. A chest-weight consists of a movable pulley and a fixed pulley, as shown in the diagram. If a 16 pound weight is attached to the movable pulley and if the cord carries a 9 pound weight at its free end, how far will the 9 pound weight descend before it has acquired a velocity of one foot a second ? What is the tension in the cord ? Ans. J| ft. ; 8.3 lbs. Fl «- 62 11. In a system of pulleys like that of question 10 a 4 pound weight is attached to the movable pulley, and to the free end of the cord is fastened a weight of 1 pound and 15 ounces, and in addition a rider weighing 2 ounces is laid on. The sys- tem starts from rest, and after the rider has descended 8 feet it is removed. Determine the motion. 12. A bucket of water, at the bottom of which there rests a stone, forms one weight of an Atwood's machine. The bucket with its contents weighs 16 pounds, and the other weight is 18 pounds. If the stone weighs 12 pounds and its specific gravity is 3, find how hard it presses on the bottom of the bucket when the system is released. 13. In the bucket described in the preceding question there is a cork, of specific gravity \, submerged and held under by a thread tied to the bottom of the bucket. Will the tension in the thread be increased or diminished after the system is released ? 14. What is the mechanical effect on one's stomach when one is in an elevator which, starting from rest, is allowed suddenly to descend? 15. A block of ice is resting on a sled, the coefficient of fric- tion between the ice and the sled being ■£-$. The sled is drawn along, starting from rest. Find the shortest possible time in which the ice can be moved 10 ft. 16. A man weighing 180 pounds is at the top of a building 60 feet above the ground. He has a rope which just reaches 200 CALCULUS to the ground and which can bear a strain of only 170 pounds. Can he slide down the rope to the ground in safety ? Interpret the velocity with which he reaches the ground by finding the height from which he would have to drop in order to acquire the same velocity. 17. Find the shortest time in which a bale weighing 160 pounds can be raised from the ground to a window 25 feet high (coming to rest at the window) by means of the rope of the preceding question, if the rope passes over a fixed pulley just above the window and is drawn in over the drum of a dummy engine. 18. If the speed of a train is being uniformly retarded by the brakes, prove that a plumb line will hang at rest relatively to the train at a certain angle, and determine this angle. 19. In the train described in the preceding exercises, ques- tion 6, there is a bucket of water. Find the angle which the surface of the water makes with the plane of the tracks after the water has ceased to surge. 20. At what angle ought a man to stand in a car that is starting with an acceleration of 3 feet per second per second ? 21. The drivers of a locomotive are keyed to the axle and are being transported on a platform car. The axle is perpen- dicular to the track, the diameter of the wheels is 6 ft., and they are blocked by pieces of joist 3 in. thick. The brakes being put on hard, so that the train loses 3. J- miles an hour of speed every second, find whether the drivers will jump the cleats. 22. A body slides down a smooth inclined plane. Show that the velocity with which it reaches the foot of the plane is the same that the body would have acquired in falling freely through the same difference in level. 23. Chords are drawn from the highest point O of a vertical circle. Show that the time of descent of a bead from rest at MECHANICS 201 0, down a smooth wire coinciding with any one of these chords, is constant. 24. A point O is distant 10 feet from an inclined plane, whose angle of inclination is a. Find the shortest time in which a bead can reach the plane if it starts from rest at and slides down a smooth straight wire. 25. The draw bar of the locomotive in Example 5 weighs 50 pounds. How much harder does the engine pull on the draw bar than the draw bar pulls on the train ? 3. Simple Harmonic Motion. Problem. One end of an elas- tic string is made fast at a point A and to the other end is fastened a weight. The weight is carefully brought to rest and then is given a slight vertical displacement. Determine the motion. Let AB be the natural length of the string, the point of equilibrium of the weight, and let P be the , position of the weight at any instant after it is released; C, the point from which it is released. The forces that act on it are : the force of gravity, 7ng, downward and the tension T of the string upward, — we neglect the damping due to the atmosphere. Hence we have from Newton's Second Law of Motion (5) m -L = T-mg. From Hooke's law, which says that the tension in a o stretched elastic string is proportional to the stretch- ing, it follows that p -- TIP ( 6 ) t =^> d where X is Young's modulus,* provided the cross-section of the * The physical constant X is sometimes interpreted as that force which would be required to double the length of the string, provided this could be done without exceeding the elastic limit. 202 CALCULUS string is unity. Since at the tension is just equal to the force of gravity, we have furthermore (7) mg = \22. Hence from (6) and (7) : and thus (5) becomes (8) m M = x T The variables s and x are connected by the relation : s + x=OC=h, where h denotes the original displacement. Thus ^£ . ^ _ a ds _ dx ~dt ~dt~ ° r dt" dt and '^— *?• dt 2 dt 2 Substituting in (8) we get d 2 x = X dt 2 ml or, setting X/ml = n 2 : This differential equation is characteristic for Simple Harmonic Motion. To integrate (I) multiply through by 2dx/dt and note that* d dx 2 _r>dxd 2 x dt dt 2 ~ ~dt dt 2 ' have, then : ^dx d 2 x _ 2n 2 x dx dt dt 2 dt * This method is evidently applicable to any differential equation of the form : MECHANICS 203 Integrating each side with respect to t we get dx* dt 2 C-2n 2 x — dt = -2n 2 Cxdx = -n 2 x>+C. To determine C observe that initially x m h, while the velocity, equal numerically to dx/dt, is : = -n 2 h 2 + C. Hence (ii) g.,*<»_rf). From this result we infer (a) that the maximum velocity is attained when x = and is ixh ; (6) that the height to which the body rises, determined by putting dx/dt = in (II), corre- sponds to x = — h. The latter inference, however, is legitimate only on the assumption that the point C: x = — h, is not higher than B, i.e. that OC = OB. For otherwise the body will rise above B, and since the string cannot push, a new law of force becomes operative, the force now being simply that of gravity, and so (I) is no longer true. We return to equation (II) and write it in the form dx ^^-nVW-x 2 , dt the minus sign holding so long as the body is rising, since x decreases as t increases. To integrate this equation write it as follows: , dx Hence nt J VV-x 2 h Initially t = and x = h, therefore 0=0 + and we have 204 CALCULUS (9) nt = cos" 1 -, hence (III) x = h cos nt. We have deduced this result merely for the interval that the body is rising. When the body begins to descend, dx/dt becomes positive and we have nt = + I _ • J Vh 2 -x 2 This integral can, however, still be expressed by the formula (10) w^cos-^+O, provided that, contrary to our usual agreement, we choose that determination of the multiple-valued function which lies be- tween 7r and 2tt. To determine Owe have from (9) that when x — —h, t = 7r/n. Substituting these values in (10) we get 7r = cos - 1 (-l)+0, (7 = 0, and thus (9) and (III) hold throughout the descent. From this point on the motion repeats itself, — a fact that is mirrored analytically in equation (III) by the periodicity of the function cos nt. Thus formula (III) holds without restriction. Turning now to a detailed discussion of these results we see that the time from C to is ^ = ir/2n. The same time is also required from to C, then from C back to 0, and lastly from to C. Thus the total time from C back to C is T= —> n In descending, the velocity is the same in magnitude as when the body was going up, only reversed in sense ; and the time required to descend from C to an arbitrary point P is the same as that required to rise from P to C. The time T is called the period of the oscillation. If we consider the body at an arbitrary point P and time t, then at MECHANICS 205 the instant T seconds later, the body will be at the same point and moving with the same velocity, both in magnitude and sense, — this fact is expressed by saying that the phase is the same, — for x = hcosn( H — — ) = h cos nt, \ n J — = -hnsmn(t+ — ^ = - hn sin nt. dt \ n) Finally, we observe that the amplitude 2 h of the oscillation has no effect on the period. EXERCISES 1. One end of an elastic string is fastened at a point A, and to the other end is attached a weight that would just double the length of the string. The weight being dropped from A, find how far it will descend. Assume the string to be 3 feet long and the mass of the weight to be 2 pounds. Ana. 11.2 ft. 2. If the weight in the preceding question is brought to a point 9 feet below A and released, how high will it rise ? How long will it take for it to return to the starting point ? 3. A slender rod is clamped at one end so as to be horizontal when not loaded. A ball of lead is then fastened to the free end and brought carefully to the position of equilibrium, the ball dropping by less than 3 % of the length of the rod. The ball being given a slight vertical displacement, show that the oscillation will be approximately simple harmonic motion and determine the period. Neglect the deviation of the path of the ball from a vertical straight line, and assume that the force that the rod exerts is proportional to the distance which the free end has been dis- placed from equilibrium. 4. A steel wire of one square millimeter cross-section is hung up in Bunker Hill Monument, and a weight of 25 kilo- 206 CALCULUS grammes is fastened to its lower end and carefully brought to rest. The weight is then given a slight vertical displacement. Determine the period of the oscillation. Given that the force required to double the length of the wire is 21,000 kilogrammes, and that the length of the wire is 210 feet. Ans. A little over half a second. 4. Motion under the Attraction of Gravitation. Problem. To find the velocity which a stone acquires in falling to the earth from interstellar space. Assume the earth to be at rest and consider only the force which the earth exerts. Let r be the distance of the stone from the centre O of the earth, and s, the distance it has travelled from the starting point A. Then the force acting on it is and since f—mg when r=R, the radius of the earth: mg = — and / = — ^- — R 2 T Hence, from Newton's Second Law of Motion, (11) m d 2 s _ mgR 2 dt 2 r 2 Furthermore, s + r = l, where I denotes the initial distance OA, and consequently ds ,dr_r\ d 2 s d 2 r_Q Fig^64 dt dt~ ' dt 2 dt 2 Equation (11) tnus becomes : / x d 2 r gR 2 To integrate this equation, we employ the method of § 3 and multiply by 2dr/dt: 2 dr (18) s=^°(l-e~™). From this result is also evident that the boat will never cover a distance of 8 ft. while the above approximation lasts. EXERCISE If the man and the boat together weigh 300 lbs. and if a steady force of 3 lbs. is just sufficient to maintain a speed of 3 miles an hour in still water, show that when the boat has gone 20 ft., the speed has fallen off by a little less than a mile an hour. Problem 2. A drop of rain falls from a cloud with an initial velocity of v ft. a sec. Determine the motion. We assume that the drop is already of its final size, — not MECHANICS 215 gathering further moisture as it proceeds, — and take as the law of resistance : R = cv 2 . Hence dv m— = dt = mg — cv 2 , ds mg — cv 2 i m ds = mvdv mg — cv 2 ' ^-^log^-c^ + C, = -^log(mg-cv 2 ) + C, and thus finally (19) » = £log m g- CT 5'- 2 c mg — cv 2 Solving for v we have ^ _ mg - c V m# — ctr (20) o. This last condition would not be fulfilled in the case of § 3 if the " string " were a steel wire and the weight a piece of lead, for k would then be very small. It could be realized, however, if the "string" is a spiral spring and the weight is provided with a collar, to act like an inverted parachute and increase the damping. To determine A and B in this case we have that initially £ = 0, x=h-, hence (25) h = A + B. Furthermore, from (24), — = - mxAe-^x* - m^Be-™*, dt 2 , and initially dx/dt = ; hence 220 CALCULUS (26) = m 1 A-{-m 2 B. From (25) and (26) A and B can at once be determined : a __ m 2 h p _ — m x h ra 2 — -m/ m 2 — m 1 and hence (27) — =■ - — - — (e "v — e W V). dt m 2 — w*i The motion is now completely determined. The particle starts from rest and moves upward with increasing velocity for a time, then slows up and approaches the point « = 0as its limit, when t = oo , — practically, of course, reaching this point after a comparatively short time. All this we read off from (24) and (27) : lim x = lim (Ae-** + Be-™*' ) = ; lim^ = 0; t=oo dt 0 + V :r T sin . Hence (24) becomes : x=Ae 2 (cosVn 2 -iK 2 i+ V-lsinvV-iK 2 *) + J3e 2 (cos -Vn 2 -^ K H-V~1 sin Vn 2 -i/< 2 0, and this result can be written in the form MECHANICS 221 (29) x = e 2 (a cos V^ 2 - i V « + 6 sin Vn 2 - ^ * 2 *), where a and 6 are constants, to which arbitrary real values can be assigned. The foregoing explanation, by means of imaginaries, is in no wise essential to the validity of the final formula (29). The student can prove directly that the function (29) really is a solution, no matter what values a and b may have, by actually substituting it in (51). Another form in which the solution (29) may be written is the following : (30) x = Ce 2 sin (Vn 2 - \k* t + y), where C and y are now the constants of integration. Instead of the sine in the last formula the cosine may equally well be written. Keturning to the special problem before us, we have, for the determination of C and y in (30), initially : x = h, t = 0: (31) ft=<7siny. Furthermore, ^ = Ce"~ s % V»* -J* 1 cos (Vw* - ± #c» 1 4- y) _!sin(Vn 2 -iK 2 * + y)], and initially dx/dt = : (32) = C\ Vri 2 -i* 2 cos y - ^ sin y 1 From (32) it follows that COty: 2Vn 2 -i* 2 If we take the solution that lies in the first quadrant 00. It can, however, also be regarded as a limiting case under n 2 — Jk 2 >0, the very first point of intersection of the curve with the axis of abscissas having receded to infinity. 9. Motion of a Projectile. Problem. To find the path of a projectile acted on only by the force of gravity. The degree of accuracy of the approximation to the true motion obtained in the following solution depends on the projectile and on the velocity with which it moves. For a cannon ball it is crude, whereas for the 16 lb. shot used in putting the shot it is decidedly good. Hitherto we have known the path of the body ; here we do not. We may state Newton's Second Law of Motion for a plane path as follows : * d 2 x ^r m^=Y, dt 2 where X, Fare the components of the resultant force along the axes. In the present case X=0, Y= —mg, and we have d 2 x A dt 2 ' d?y m — £ = — mg. dt 2 y Fig. 72 If we suppose the body projected from with velocity v at an angle a with the horizontal, the integration of these equa- tions gives : •— = G = v cos a, x = v t cos a ; etc * The form of Newton's Second Law that covers all cases, both in the plane and in space, be the motion constrained or free, is that the product of the mass by the vector acceleration is equal to the vector force. MECHANICS 225 -1 = v sin a — gt, y = v Q t sin a — %gt*. (XL Eliminating t we get : (35) y = x tana - 2v 2 cos 2 a The curve has a maximum at the point A : v Q 2 sin a cos a v 2 sin 2 a *>=^-l — > *=v- Transforming to a set of parallel axes through A : a; = aj' + aj 1 , y = y' + y lf we find: y< = -— -i* — • 2 V cos « This curve is a parabola with its vertex at A. The height of its directrix above A is v 2 cos 2 a/2g, and hence the height of the directrix of (35) above is v 2 sm 2 a v 2 cos 2 a _v 2 2g + 2g ~2g' This result is independent of the angle of elevation a, and so it appears that all the paths traced out by projectiles leaving with the same velocity have their directrices at the same level, the distance of this level above being the height to which the projectile would rise if shot perpendicularly upward. EXERCISES 1. Show that the range on the horizontal is and that the maximum range R is attained when a = 45°: 9 The height of the directrix above is half this latter range. 226 CALCULUS 2. A projectile is launched with a velocity of v ft. a sec. and is to hit a mark at the same level and within range. Show that there are two possible angles of elevation and that one is as much greater than 45° as the other is less. 3. Find the range on a plane inclined at an angle /? to the horizon and show that the maximum range is R a = g 1 + sin/3 4. A small boy can throw a stone 100 ft. on the level. He is on top of a house 40 ft. high. Show that he can throw the stone 134 ft. from the house. Neglect the height of his hand above the levels in question. 5. The best collegiate record for putting the shot is 46 ft. (F. Beck, Yale, 1903) ; the amateur and world's record is 49 ft. 6 in. (W. W. Coe, Portland, Ore., 1905). If a man puts the shot 46 ft. and the shot leaves his hand at a height of 6 ft. 3 in. above the ground, find the velocity with which he launches it, assuming that the angle of elevation a is the most advantageous one. Ans. v Q = 35.87. 6. How much better record can the man of the preceding question make than a shorter man of equal strength and skill, the shot leaving the latter's hand at a height of 5 ft. 3 in. ? 7. Show that it is possible to hit a mark B : (x b ,y b ), provided Vb 4- vV + y b - 8. A revolver can give a bullet a muzzle velocity of 200 ft. a sec. Is it possible to hit the vane on a church spire a quarter of a mile away, the height of the spire being 100 ft. ? MECHANICS 227 EXERCISES 1. A cylindrical spar buoy (specific gravity %) is anchored so that it is just submerged at high water. If the cable should break at high tide, show that the spar would jump entirely out of the water. 2. A number of iron weights are attached to one end of a long round wooden spar, so that, when left to itself, the spar floats vertically in water. A ten-kilogramme weight having be- come accidentally detached, the spar is seen to oscillate with a period of 4 seconds. The radius of the spar is 10 centi- metres. Find the sum of the weights of the spar and attached iron. Through what distance does the spar oscillate ? Ans. (a) About 125 kilogrammes ; (b) 0.64 metre. 3. A chain rests partly on a smooth table, a piece of the chain hanging over the edge of the table. The chain being re- leased, find the velocity with which it will leave the table. 4. Solve the same problem for a rough table, the chain passing over a smooth pulley at the edge of the table. 5. A particle of mass 2 lbs. lies on a rough horizontal table, and is fastened to a post by an elastic band whose unstretched length is 10 inches. The coefficient of friction is \, and the band is doubled in length by hanging it vertically with the weight at its lower end. If the particle be drawn out to a distance of 15 inches from the post and then projected directly away from the post with an initial velocity of 5 ft. a sec, find where it will stop for good. 6. Show that if two spheres, each one foot in diameter and of density equal to the earth's mean density (specific gravity 5.6) were placed with their surfaces \ of an inch apart and were acted on by no other forces than their mutual attractions, they would come together in about five minutes and a half. Given that the spheres attract as if all their mass were con- centrated at their centres. 228 CALCULUS 7. A particle is projected horizontally along the mner sur- face of a smooth vertical tube. Determine its motion. 8. A man and a parachute weigh 150 pounds. How large must the parachute be that the man may trust himself to it at any height, if 25 ft. a sec. is a safe velocity with which to reach the ground ? Given that the resistance of the air is as the square of the velocity and is equal to 2 pounds per square foot of opposing surface for a velocity of 30 ft. a sec. Ans. About 12 ft. in diameter. 9. A toboggan slide of constant slope is a quarter of a mile long and has a fall of 200 ft. Assuming that the coefficient of friction is T f -$, that the resistance of the air is proportional to the square of the velocity and is equal to 2 pounds per square foot of opposing surface for a velocity of 30 ft. a sec, that a loaded toboggan weighs 300 pounds and presents a surface of 3 sq. ft. to the resistance of the air; find the velocity acquired during the descent and the time required to reach the bottom. Find the limit of the velocity that could be acquired by a toboggan under the given conditions if the hill were of infinite length. Ans. (a) 68 ft. a sec. ; (6) 30 sees. ; (c) 74 ft. a sec. 10. The ropes of an elevator break and the elevator falls without obstruction till it enters an air chamber at the bottom of the shaft. The elevator weighs 2 tons and it falls from a height of 50 ft. The cross section of the well is 6 x 6 ft. and its depth is 12 ft. If no air escaped from the well, how far would the elevator sink in? What would be the maximum weight of a man of 170 pounds ? Given that the pressure and the volume of air when compressed without gain or loss of heat follow the law : pv 1A1 = const., and that the atmospheric pressure is 14 pounds to the square inch. CHAPTER XI THE LAW OF THE MEAN. INDETERMINATE FORMS 1. Rolle's Theorem. A theorem which lies at the founda- tion of the theoretical development of the Calculus is that of Rolle, from which follows the Law of the Mean. Rolle's Theorem. If (x) is a function of x, continuous throughout the interval a^x*£.b and vanishing at its extremities : 4>(a) = 0, '(x), at every interior point ax of the interval, then ' (x) must vanish for at least one point within the interval : <£'(X)=0, a (x) is always = 0, for which case the y theorem is obviously true. Sup- pose, then, that (x) is positive in a part of the interval. Then (x) will have a maximum at some point x = X within the in- terval, and at this point the derivative, '(x) = tan t, will van- ish, cf. Chap. Ill, §7: <£'(X)=0, a(x) is negative, it will have a minimum, and thus the theorem is proven. 229 a X *\ Fig. 73 230 CALCULUS 2. The Law of the Mean. Let the function /(x) be contin- uous throughout the interval a ^ x <^ b and let it have a deriv- ative, df(x)/dx=f , (x) } at every interior point of the interval. Draw the graph and let L M be the secant connecting its ex- tremities. Then there will be at least one point X within the interval at which the tangent is parallel to the secant LM. For, consider the distance from a point P of the curve to the secant, measured along an ordinate, PQ. This distance (taken algebraically) will have a maximum or a minimum value, and at such a point the tangent is evidently parallel to the secant. Now the slope of the secant is tan ZJg£Jf= A 6 } -/<«>, b — a A )-K a ) an( j ^ s i p e f the curve at x = X is /'(X). Hence b—a (A) /(&)-/<<*) = Q> - a) f(X), a)-/(<») _/(X) =0 , a(x) here employed is pre- cisely the distance PQ. 3. Application. As a first application of the Law of the Mean we will give the proof of Theorem A in Chap. VI, § 2. In that theorem 3>' (x) = by hypothesis for all values of x, or at least for all in a certain interval. If, then, a and b = x 1 are two points of this interval, we have from the Law of the Mean (A) : (x 1 )-&(a) = 0, i.e. ®(x 1 ) = $(a) for all points x x in question. Hence &(x) is a constant. Exercise. Show that, if f(x) satisfies the conditions of § 2 and if furthermore /' (x) > at all points within the interval, then 4. Indeterminate Forms. The Limit J. If both the numer- ator and the denominator of a fraction (i) /M vanish for a particular value of x, x = a: /(a)=0, 2P(a)=0, 232 CALCULUS the fraction takes on the form # and thus ceases to have any meaning. The fraction will, however, in general approach a limit when x approaches a, and we proceed to determine this limit. Sometimes this can be done by a simple transformation. Thus if f(x) _ x — a F(x)~ x 2 -a 2 ' we need only divide numerator and denominator by x — a and we have : v x — a v 1 1 lim — = lim sd — . x =ax 2 — a 2 x±ax + a 2a Again, if /M = *anx F(x) x and a = 0, we have lim tan f = Hm J__ sinx = 1 _ x=o x x=o cos a; x When, however', such simple devices as the foregoing are not available, we can apply the Law of the Mean. Let b=x be any point near a. Then, remembering that f(a) = and F (a) = 0, we have : f(x) = (x-a)f'(X), F{x) = (x-a)F'(X'), where X and X both lie between a and x, and hence f(*y = r(X) F(x) F'(X) When x approaches a, X and X both approach a, too, and so, if f'(x) and F' (x) are continuous, as is usually the case in practice, lim/'(X) =/'(a), lim F' (X) = F'(o). If, then, F* (a) ^ 0, we have : (2) i im /©=/M LAW OF THE MEAN. INDETERMINATE FORMS 233 The limit of such a fraction as the one above considered is referred to for brevity as the limit $.* t?„„™^i„ rp„ i±„ r i •!• logo; Here ue. J. O I1I1U 11 x = Ul . = 1 1 —X /(*)= = \ogx, /' *=o x 2Va x=o sin 2 x 2 13. Obtain the limits in Exs. 2, 4, 7,9 by differentiation. * This limit is also called the "true value" of the "indeterminate form" f(x)/F(x) for x = a. Both terms are based on a false conception. In the early days of the Calculus mathematicians thought of the fraction as really having a value when x = a, only the value cannot be computed because the form of the fraction eludes us. This is wrong. Division by is not a process which we define in Algebra. It is convenient, how- ever, to retain the term indeterminate form as applying to such expres- sions as the above and others considered in this chapter, which for a cer- tain value of the independent variable cease to have a meaning, but which approach a limit when the independent variable converges toward the exceptional value. 234 CALCULUS Obtain the following limits by differentiation. 3=0 X TV 15. lim- = loga. •< —4 21. Hm 1 - V2sip,ra; = -1, 16. li m ^^! = log«. -il-V2oos^x «*o x b 5 3 22 . lim ^-l+(x-l^ = _3, o« t Va; — Va 3 1 -.« r e* — e~ x n 23. lira- — = -a ? . 18. km — ; = 2. *** F(b)-F(a) F'(X)' LAW OF THE MEAN. INDETERMINATE FORMS 235 The proof is as follows. Form the function : This function satisfies all the conditions of Rolle's Theorem, and hence its derivative, must vanish for a value of x within the interval. Hence F^F% F,( - X) - f{X) = ' a(y), F(x) =F(a + ^ = '(y) = f(*) *'(y) F'{x) ■ If', then, f'(x)/F'(x) approaches a limit when x approaches a, ' (y) / ^' (y) will approach the same limit when y=cc. 238 < CALCULUS Hence (y)/& (y) will approach this limit, too. But <£ (y) /<£ (y) =f(x)/F(x). This completes the proof. * Example. To find lim—. *=« e x We have : lim - == lim - = 0. 8. The Limit • oo . If we have the product of two func- tions : f(x) .<£(>), one of which approaches as x approaches a, while the other becomes infinite, we can determine the limit of this product by throwing the latter into one of the forms : m or m, i.e. the form 0/0 or oo/oo, and then applying the foregoing methods. Example. To find lim x logic. Here it is better to choose the form 1/x for then the logarithm will disappear on differentiation : limlES* lim-l^L m i im (_ x) = o. x =o x~ L x=o — 1/ar x±o * The theorem contained in (2) goes back to 1' Hospital, 1696. The theorem of this paragraph is due to Cauchy, 1823 and 1829, who proved it, however, only on the assumption that f(x) / F(x) approaches a limit. Stolz extended it in 1879 as in the text, showing that, if /' (x) / F' (x) approaches a limit, then /(a:) / F(x) will also approach a limit, and this will be the same limit. LAW OF THE MEAN. INDETERMINATE FORMS 239 EXERCISES Determine the following limits. 2 1. lim-- Ans. 0. 6. lim x log sin x. Ans. 0. *=« e x *=° 2. lim?". Ans. 0. 7. lim-^iL ^tw. 3. 3=00 e x i=o cot 3x 3. lim a; cot trx. Ans.— 8. lim£c a logx, a>0. ^4ws. 0. x = 7T x=0 4. ii m ?2gi?. ^4rts. 0. 9. lim-^-. .4ws. oo. i = oo X 2 = oo l0gi(7 B. l im l°££,»>0. Aw. 0. 10. iim lo g sin2a; . J,*.! x=oo a w x=o log sin a 9. The Limits 0°, l 00 , go , and oo — oo. The expression (5) /(s)*<*> ceases to have a meaning when/(#) and <£(#) take on certain pairs of values. If we write (cf. Formula (5) on p. 77) f(x) = e log/ ( x) fix)^^ = e* (x) los/( - x ^> we see that the expression (5) becomes indeterminate when the exponent of e takes on the form 0-oo. We are thus led to consider new limits of the types : (a) /(a) = 0, *(a)=0; 0°. (b) /(a) = l, *(o) = oo; 1". (c) /(a) = », *(a)=0; oo . The limiting value of the exponent of e. can be obtained by the method of § 8, and hence the limit of (5) determined. Example. To find lim (cos x)* 3 . I log COS X (cos x) x3 = e * , lim log C osx = lim -sinx. x =o a? x=o3arcosa? 240 CALCULUS This last limit can be obtained immediately by a simple trans- formation : — sin x _ 1 sin x o x 2 cos x 3x cos x x Hence we see that the exponent of c becomes negatively infinite if x approaches from the positive side, and so 1 3 = V ' If, however, x approaches from the negative side, the ex- ponent of e becomes positively infinite, and j_ lim (cos#)* 3 =oo. z=0 A convenient notation for distinguishing between these two cases is the following : _L J_ lim (cos x) x3 = 0, lim (cos x) x3 — oo. x=0+ » = 0- TJie Limit co — oo . If we have the difference of two functions, each of which is becoming infinite, as log (x + 1) — log x when x = oc, it is sometimes possible to evaluate the limit by a simple transformation. For example : log (x + 1) - log x = log (l + 1\ lim log (* + -Y= 0. More often, however, the simplest method is that of infinite series, cf. Chap. XIII. EXERCISES Determine the following limits. -i- 1 1. lim of . Arts. 1. 3. lim# 1-a: # Ans. -■ x = x=l 6 2. lim(l + sina;) cotz . Ans. e. 4. lim(Vl + X 2 — a;). ^4ns. 0. LAW OF THE MEAN. INDETERMINATE FORMS 241 5. Km (cot x) x . Ans. 1. ft ,. / , fa V •*f^ 6. limf - + 1 ] • Ans. e a . *=«\x J JL 1 7. lim (cos a;)* 2 . Ans. — . 1Q H ^4?is. — 1. 8. iiai[2-.-Y"«- log a? log a? y ^4ns. — 1. EXERCISES Determine the following limits. 1. lim 2-3x + 4 : x 5 * = * Ix + tf + lx 5 o r 3 4-" a? 2. lim — -• x =* 4 — 9 a? 4- a; 2 3 . Iim V9W, 11. lim 12. lim 13. lim sin a; (log a;) 2 . x=« (x — a)'' 14. lim" 5. lim I , . 15. lim csc 2 /3x log cos aa;. a — x a 4- a? J *-° 6. limvY?^ 2 cot^J^. 16. lim **'" 1 ' 7. lim cos -1 a; »*» Vl-a; 2 7ra; a?sina? 8. lim 17. lim (1 — a?) tan z = l ii 18. lima~ x loga?. cos a; x 19. lim a? 2 — a? 9. lim n sin - n = » 71 10. lim—. x = « x 3 *=»1 — a; 4- log a; »/3— — 20. lim l_a + 2Vl4-a- 2 4-a; 4 242 2i. lim}*— * -J*- 22. lim csc # sin (tan as). CALCULUS 23. lim x a (logxy, a>0, £>0. 24. lira Qog^y m > 0, n > 0. 25. lim G(x)e~ x , where G(x) is a polynomial. 26. Show that p X* lim— = 0, n being any constant whatever. CHAPTER XII CONVERGENCE OF INFINITE SERIES* 1. The Geometric Series. We have met in Algebra the Geometric Progression : a-t-ar + ar 2 -\ , the snm of the first n terms of which is given by the formula: a —ar n i 1 — r Suppose, for example, that a = 1, r = 1. Then Sl = l =1 etc. If we plot on a line the points which represent s r , s 2 , s 3 , •••, it is easy to see how to obtain s n from its predecessor, s n _ u Si Sn So . —\ 1 — I ^-A4 Fig. 75 namely : s n lies half way between s n _j and the point 2. Hence it appears that, when n grows larger and larger without limit, s n approaches 2 as its limit. * This chapter is in substance a reproduction of Chapter I of the author's Introduction to Infinite Series, published by Harvard University. ^ 243 244 CALCULUS In general, if r is numerically less than 1, |'r|n; but which (2) never exceeds some definite fixed number, A: no matter what value n has, then s n approaches a limit, U: lim *„ = U. M= 00, The limit U is not greater than A : U^ A. u a Si So Sa ii r i in Fig. 76 CONVERGENCE OF INFINITE SERIES 247 EXERCISE State the Principle for a variable which is always decreasing, but which remains greater than a certain fixed quantity, and draw the corresponding diagram. By means of the foregoing principle we can state a simple test for the convergence of an infinite series of positive terms. Direct Comparison Test. Let u + u l -\- ••• be a series of positive terms which is to be tested for conver- gence. If a seco?id series of positive terms already known to be convergent : can be found whose terms are greater than or at most equal to the corresponding terms of the series to be tested : u n 1 ; cf. Infinite Series, § 6. CONVERGENCE OF INFINITE SERIES 249 This condition, however, is only sufficient, not necessary, as the following example shows : ^2^3^4^ If we strike in anywhere in this series and add as many more terms as the number that have preceded : 1 +-k+- ■ * n + 1 n-\-2 n + n we get a sum >^. For each term just written down is > l/2w, and there are n of them. If, then, we can get a sum greater than \ out of the series as often as we like, we can get a sum that exceeds a billion, or any other number you choose to name, by adding a sufficient number of terms together. Hence the series diverges in spite of the fact that its terms are growing smaller and smaller. This series is known as the harmonic series. A further test for divergence corresponding to the test of § 3 for convergence is as follows. Direct Comparison Test. Let «o + «H be a series of positive terms which is to be tested for divergence. If a second series of positive terms already known to be divergent : a<> + aH can be found whose terms are less than or at most equal to the corresponding terms of the series to be tested : then that series diverges. The proof is similar to that of the test of § 3 for conver- gence and is left to the student as an exercise. 250 CALCULUS EXERCISES Prove the following series to be divergent. i. i+Jt+A :+ J =+ .. % V2 V3 V4 5. The Test-Ratio Test. The most useful test for the conver- gence or the divergence of a series is the following, which holds regardless of whether the terms are positive or negative. It makes use of the ratio of the general term to its predecessor, ^n+i/ w n> — the test-ratio, as we shall call it. The Test-Ratio Test. Let tto + «i+ •'•■ be an infinite series and let the limit approached by its test-ratio be denoted byt: lim n=oo U n+l _ ^ u n Then if l«!i, " " diverges; « \t\ = i, the test fails. We shall prove the theorem in this paragraph, so far as it relates to convergence, only for the case that the terms are all positive. Then £ J> and \t\=t. Suppose t < 1. Let y be chosen between t and 1 : t < y < 1. Since the variable u n+l /u n approaches t as its limit, the points representing this variable cluster about the point t and hence CONVERGENCE OF INFINITE SERIES 251 ultimately, — i.e. from a definite value of n on: ft^m, — lie to the left of the point y : ^m. u n t y H 1 1 1 Fig. 77 Now give to n successively the values m, m + 1, etc. : — W m+1 <^ ro y; ft = m + l, -=±* 1, the terms now being either positive or negative. Then, when n ^ m, K±lJ>l or K +1 |>|« n |, I U n | i.e. all later terms are numerically greater than the constant ft m , and so they do not approach as their limit. Hence the series diverges. * The student should notice that it is not enough, in order to insure convergence, that the test-ratio remain less than unity when n^m. Thus for the harmonic series u n+ \/u n = n/{n + 1)<1 for all values of n, and yet the series diverges. But the limit of the test-ratio is not less than 1. What is needed for the proof is that the test-ratio should ulti- mately become and remain less than some constant quantity, y, itself less than 1. 252 CALCULUS Lastly, if 1 1 1 = 1, we can draw no inference about the con- vergence of the series, for both convergent and divergent series may have the limit of their test-ratio equal to unity. Thus for the harmonic series, known to be divergent : g»±? = — ?L^ = 1 , lim^l^l; n while for the convergent series of § 3, Ex. 6 : ^■^/-JL-V, and lim^±i = l. u n w + iy »=» u n EXERCISES Test the following series for convergence or divergence. 1^34 1. 2 + f2 + 2~3 + 2 4+ "" AUS ' Convergent 1.2 1.2.3 1.2.3-4 A ~. 2 4- 4- -I . Ans. Divergent. 1.1-2 L2.3 2^ 3100 4100 ' 3 3-5 3-5-7 ' 2 2 2 2 3 4. *+* + *+.;<, 6. ^ + -^ + ^4-.... 25 r 2 10 2 15 5 3 10 3 15 3 For what values of x are the following series convergent, for what values divergent ? 7. 1 + ^4^+.'.., 9. 1 +- + - + -+ .... 8. .T 3 + a 5 + a: 7 +.... 10. l-fz 2 +— + — +.... 2! 3! 6. Alternating Series. Theorem. Let the terms of an in- finite series be alternately positive and negative : u — u x -\- a 2 — ••'. CONVERGENCE OF INFINITE SERIES 253 If (1) each u is less than or equal to its predecessor : u n+1 n = («« — «*l) + ( U 2 ~ U 3 ) H h (Wom-2 — MjJm-l)- Thus s 2 /« always increases or remains unchanged when m increases. If n is odd, n = 2m + l, and we see that s 2m+1 steadily decreases or remains unchanged when m increases. Furthermore, s 2m does not exceed the fixed value 8 V For S 2m = ,S 2»i+l U 2m = S 2m+1 = S l ■ Hence, by the Fundamental Principle of § 3, s 2l)l approaches a limit. In like manner it is shown that s 2m+1 is never less than s 2 . For S 2m+l =z S 2m ~f" U 2m ^ S 2m ^ S 2 ■ Hence s 2m+l also approaches a limit. Finally, these limits are equal. For, since s 2m +i = s 2m +«**, lim s 2w+1 = lim s 2m + lim w 2m , and, by hypothesis, lim u n = 0. Hence s n approaches a limit when n becomes infinite passing through both odd and even values, and the series converges, q. e. d. It is easily seen that the error made by breaking an alter- nating series off at any given term does not exceed numeri- cally the value of the last term retained. 254 CALCULUS 7. Series of Positive and Negative Terms ; General Case. Let then s'n = (T m + V Hence the series of absolute values converges if both the v-series and the ly-series converge. Conversely, if the series of absolute values converges, then both the v-series and the w-series converge and we have Case 1. For both of the latter series are series of positive terms, and no matter how many terms be added in either series, the sum cannot exceed the value U' of the series of absolute values. Hence by the Principle of § 3 each of these series converges. Series whose absolute value series converge are said to be absolutely or unconditionally convergent; other convergent series are conditionally convergent. We can now complete the proof of the theorem of § 5, namely, for the case that lim^±!=*, |*|<1. «=*> u n Here the series of absolute values converges, for and hence lim \^n±ll = \t I < 1. ... \u n \ Consequently the w-series converges absolutely. Example 1. To test the convergence of the series x 2 . x 3 x 2 3 Here ^»±i= —x, lim£s±!=-^ u n n + 1 n=w u n and hence the series converges when — 1 < x < 1 and diverges outside of this interval. Divergent — 1 1 Divergent Convergen t 256 CALCULUS At the extremities of the interval the test fails. But we see directly that for x = l the series is a convergent alternating series; for x = — 1, the negative of the harmonic series, and hence divergent. Example 2. The series ^1-2 T 1-2.3 ^ has for its general term, u k : _ n(w-l) ••• ( n -fc-f 1) „* If n is a positive integer, the later terms are all and the series reduces to a polynomial, namely the binomial expansion of (1 + x) H . When n is not a positive integer, the value of the test-ratio is 1 h±l = r Lnl X) and \\m v ^=-x. U k K + 1 fc = Q0 U k Hence the series converges when — 1 < x < 1 and diverges when | x | > 1. For the determination of whether the series is convergent or divergent at the extremities of the interval of convergence more elaborate tests are necessary. EXERCISES Eor what values of x are the following series convergent ? Indicate the interval of convergence each time by a figure. / 1. l + a: + 2(x) = a b + (a b 1 + a 1 b )x + (a b 2 + aA + a 2 b )x 2 + .... CONVERGENCE OF INFINITE SERIES 259 The resulting series thus obtained will converge at least for all values of x lying within the smaller of the two intervals of convergence of the given series. It is even possible to divide one power series by another as if they were both polynomials. We shall make use of this property in the next chapter when we come to develop tan x. An especially important operation with power series is that of differentiating or integrating the series term by term, i.e. as if it were a polynomial. For example, take the geometric progression : — = l + « + ^ + ^+ •-. 1 — x Differentiating each side with respect to x, we have (1 — xy a result that can easily be verified by multiplying the first series by itself as explained above. Again, integrating each side of the equation — — = l-x + x 2 -x* + ... 1 + x between the limits and h, we get, since dx ! % 1 + x = lo g (l + X ) = log(l + A), the important series : log(l + 70 = A-| + |--. By means of this series and others immediately deduced from it natural and denary logarithms are computed. In like manner we get from the series = l-x 2 + 3t i 1 + x* a series for tan -1 ft: 260 CALCULUS h 1-f-z 2 3 5 By means of this series the value of ir can be expeditiously computed with great accuracy. It is of value for the student at this stage, before proceeding to the further study of series, to see how the simpler series are actually used in practice as a means of computation. He is referred for a treatment of this subject to the Infinite Series, Chap. II : " Series as a Means of Computation," see the foot- note at the beginning of this chapter. The processes with infinite series, of which we have given a brief account in this paragraph, are also taken up and estab- lished in the Infinite /Series, Chap. IV : " Algebraic Transforma- tions of Series," and Chap. V : " Continuity, Integration, and Differentiation of Series." In the latter chapter will also be found a proof of the theorem that a power series always repre- sents a continuous function throughout its whole interval of convergence. EXERCISES 1. If ao + Oi + --- is any absolutely convergent series and p > Pi>**' any set of numbers, positive or negative, that merely remain finite as n increases: \p n \ W .... Obtain three terms in each of the following developments. 5. tan# = #-f Ja5 3 + T ? 5« 5 + •••• 6. secx = l+%x 2 + -%\x 4 '■-. 7. c §iHX = l-f a + ^o 2 — £a 4 + •-•. 264 CALCULUS 2. Taylor's Series. It may, however, happen that no devel- opment according to powers of x is possible. Thus if /(aj)=loga>, /(0) = — oo. But a power series represents a continuous func- tion and so no power series in x can be expected to represent log x. It is evident generally that, whenever the function or any one of its derivatives becomes discontinuous for x = 0, the function cannot be developed in a Maclaurin's Series. A power series is most useful for computation if the values we have to assign to its argument (i.e. the independent vari- able) are small. Now it may happen that we know the value of the function and of all its derivatives at a single point, x = x , or at least can easily compute them. In such a case we can find the value of the function at points x=x + h near by if we develop /(a?), not according to powers of x, but according to powers of h. Setting, then, x = x + h, h = x — x , we shall have, if a development be possible : fix) =f(x + h) = c +Ci/i + c 2 h 2 H . We can determine the coefficients here as in the case of Mac- laurin's Series. Thus, setting h = 0, we find, f(x ) = c . Differentiating with respect to h and remembering that x is a constant, we obtain : df(x) r== df(x) dx = ,„. = dh dx dh K J f'(x + h)=:c l + 2c 2 h + 3c 3 h 2 + -., /' f (aj +7i) = 2.1c 2 + 3.2c 3 /i4-4.3c 4 ^ 2 H- ••., /"(a )=2.1c 2 , c 2 =^f^ TAYLOR'S THEOREM 265 A / (n) 0<>) and so on : c n = J - — ^- yz n\ If, then, f(x) can be developed in powers of h, the series will have the form : (3) fix, + h) =/O ) +/'(z ) h+ f -^£ tf + . ... When h is replaced by x, (3) becomes : (3') /(») =/Oo) +/ , (%)(*-ab) +-^f)(*-*„) 2 + •- These series are known as Taylor's Series. For example, let /(a?) = logo?, x = l. Then ™-1 /'(i) = i; rwr>& /"(i) = -i; /"(-)=—, /"'(l) = 2!; f™(x\-( W+i< n - -1)! f<">m = r-n» +1 ('«- and the series will have the form : log(l+A) = A-|- + |— -. This agrees with the result obtained by integration in the pre- ceding chapter. The series converges for values of h numer- ically less than 1. Maclaurin's Series is a special case jrf Taylor's Series, ob- tained by letting x = 0. But conversely, Taylor's Series can be obtained from Maclaurin's by replacing x by h as above and developing f(x + h) in a Maclaurin's Series. 266 CALCULUS EXERCISES Assuming that the function can be developed in a Taylor's Series, obtain the following developments. 1. e a+h =e a + e a h + —Ji 2 + .... 21 2. sin (x + h) = sin x -f h cos x — — sin x — —- cos x + • • .. & • o ! V* ; V2L 2!^3!^ J 4. x n = (a + fc>" = a!' + j*a*- l ft + ^^f^— a n 2 h 2 ^ 1.2-3 a fl "^ * ^6 7 2 T 2 22! 2 3! 6. log* = log2 + ^_ife^ + ^fc^)- 3 --. 2 2 2 2 3 2 3 Obtain three terms in the development of each of the fol- lowing functions. 7. logCl+z 2 ), x = 3. Ans. 2.303 + .6 (x - 3) - .08 (a - 3) 2 + • . -. 8. tanx, x = -' 10. , x = — 1. 4 ic 9. log(e x + e" x ), a = 0. 11. 10 x , x = 0. 3. Proof of Taylor's Theorem. Let the function f(x) be continuous throughout the interval a < x < b and let it have continuous derivatives of all orders throughout this interval. Let x be an arbitrary point of the interval, which, once chosen, shall be held fast, and let x -f h be any second point of the interval. We will see if we can approximate to the value of the function by means of the first n + 1 terms of the corresponding Taylor's Series : TAYLOR'S THEOREM 267 (4) / (3% + h) =/O ) + f(x ) *+'-+ "CifiSJ *" + 5, n where R denotes the error, i.e. the difference between the value of the function and the value of the approximation. In order to see how good this approximation is, we must have an expression for R that will throw light on the numerical value of this quantity. Such an expression can be found as follows. Let us write R in the form : 7»«+l 7,n+l R = — P, i.e. let P=R+ + 1)! + 1)! Then (4) becomes, on transposing terms : (5) f(x, + h) -/(*„) - hf (*) £/">(*.) - j^-P=0. n\ (ti+1)! We now proceed to form arbitrarily the following function of z : * (s) =/(X) _/(X) is obviously =0, and if we compare (a? ) with the left-hand side of (5), we see that (x ) vanishes, too. Hence the derivative of (z) must vanish at some point within the interval. Now, on computing the derivative we find that the terms cancel each other to a large extent : * <*>' (?) = ~f (*) + /' (•) - (X- z)f" (z) + (X - z)f» (z) - (- X ' — g ) n f(n+l)/g\ _|_ (X—Z) n p^ so that there remain finally only two terms : * The student is requested to write out the terms in this differentiation for n = 1, 2, and 3. 268 CALCULUS n\ n\ Consequently the conclusion of Rolle's Theorem: 4>'(Z) = 0, x (x ) *L-f»\ (x ) -hP=0, ni and we form the function of z, $ (z) =/(X) -/(*) - (X - z)f (z) - (X^f"(z) - where X—x -{-h. This function satisfies the conditions of Rolle's Theorem in the interval x fj z < X, and so its deriva- tive, *' (z) - - (X ~ Z) V (W+1) (*) + P, must vanish at some point Z = x + 0h within the interval. Hence, (9) B = 0--°y hn+1 f(n + » (^ + eh y 5. Development of e x , sin x, cos x. The function e* can be developed by Taylor's Theorem about the point x = 0. Here /(V)=6*, /'<*)»«", • • • /»(»)-*, /(0) = 1, /'(0) = 1, • • • /™(0) = 1, and the remainder R as given by (6) has the form : fcn+l i£ = — e°\ (n + 1)! If h<0, e 0h 0, e 0h m. n 2 If we denote, then, the product of the first m factors, taken numerically, by C, and replace each of the subsequent factors by !-, we shall have : 7,71+1 I /1\n-m+l (n + 1) ! J \2, The limit of this last expression is when n = , and conse- quently * lim h n+1 / (n + 1) ! = 0. We have, then, lim R = and hence, replacing h by x : n=oo (io) ,-i+.+j£+!;+.... The series converges and represents the function for all values of x. To develop sin x we observe that f(x)=smx, ,/(0)=0, /' (a^cosz, /(0) = 1, f"(x) = -smx, /"(0)=0, />"(X)=-C08X, /'"(0)=-l, and from this point on these values repeat themselves. It is not difficult to get a general expression for the n-th. derivative, namely : * We might have given a short proof of this relation by observing that fe n+1 / (n + 1) 1 is the general term of a convergent series : 1 + 7* + — +£-+.... 21 3! TAYLOR'S THEOREM 271 /(«>(*) = sin^ + ^\ This formula obviously holds for n = 1, 2, 3, 4, and from that point on the right-hand member repeats itself, as it should. Thus we find : R= hn+ ] S m( Oh + 1)! V + W7r\ 2/ The second factor is never greater than 1 numerically, and the first factor, as we have just seen, approaches as its limit. Hence lim R = and we have, on replacing h by x : n=xo (11) sino^-^ + fl-.... In a similar manner it is shown that (12) cosa = l-^ + — . V J 2r 4! EXERCISES 1. Compute the value of e 06 (cf. Chap. IV, § 7) to six signifi- cant figures. 2. Show that e x can be developed by Taylor's Theorem about any point x . 3. Obtain a general expression for the n-th derivative of cos x and hence prove the development (12). 4. Show that sin x and cos x can be developed by Taylor's Theorem about any point x . 5. Remembering that 1° is equal to tt/180 radians, compute sin 1° correct to six significant figures. By about what percent- age of either does sin 1° differ from its arc in the unit circle ? 272 CALCULUS 6. The Binomial Theorem. Let f(x)=x», where n is any constant, integral, fractional, or incommensur- able, positive or negative ; and let x Q = 1. Then /(1) = land f(x) = nx*-\ /'(l) = n, /' ' (a?) = n (n - 1) x n ~\ f" (1) = n (n - 1), fW(x) = n(n - 1) ... (n - k + l)x n ~ k , f»(l) = n(n-l).>.(n-k + l). For the remainder R it is better here to employ the second form, (9). Thus B = 0--°y hk+1 . »( n - 1) ... ( n - k)(l + ^)"~*- 1 1 k ! The last factor remains finite, whatever the value of 0, pro- vided | h | < 1. For, since < 6 < 1, i-|ft| lj (i + ehy-*<(i-\h\y-\ n and D «* + :... This is the Binomial Theorem for negative and fractional exponents. When n is or a positive integer, the series breaks off of itself with a finite number of terms and we have a polynomial, namely : (1 + x) n . In all other cases the series converges when x is numerically less than 1 and represents the function (1 -f- x) n ; and it diverges when x is numerically greater than 1. The following developments obtained from especially useful. ■VI -x 9 1 * L3 (15) yi-^ = i-^-^-^*« EXERCISES 1. Show that, when | a | > | b | : 1 • 2i 1 • £ > o ^2. Compute V3 correct to seven significant figures by means of the series (13). Suggestion: Eegin by 'writing 3 =(f) 2 (ff). Here £ is one of the convergents in the development of V3 by continued fractions. 3. Compute V30 to five significant figures. 4. Obtain from (13) the development : l_^_2z + 3^-4* 3 +-". (1 + x) 2 274 CALCULUS 5. Obtain the development : log(l + A) = A-| + |-- by the method of this paragraph. 7. Development of sin -1 a?. We can now obtain the develop- ment of sin -1 a; in a manner similar to that employed for tan -1 x. Integrating each side of (14) gives : ' , lh* , l-.SVv r dx J vr= The value of the left-hand side is sin -1 ft. h by x, we have : 1 a 3 , 1 • 3 ar 5 Hence, replacing (16) sin *o? X + 23+f^L 5 + The series converges and represents the function when | x | < 1. 8. Development of tan x. We have : , sin a; x — 4-a^4- -A-naP— ••• tan x = = o \ , • cos a? 1 — -g-ar-f ^ar — ... Now it can be shown that one power series can be divided by another just as if both were polynomials, the resulting series converging throughout a certain interval, cf. Infinite Series, § 36. Hence 1 ^2 l 1 /»4 . _ _ <\ 1' A *+A )x a; — lx? + J- ar* ^--^ + * ^ + "3 i* 3 We can obtain in this way as many terms in the development of tana; as we wish, although the law of the series does not become obvions. (17) tdLnx = x + ^x 3 + -&x 5 -i TAYLOR'S THEOREM 275 9. Applications. We shall consider here only two or three applications of Taylor's Theorem, referring the student for further applications to the Infinite Series, Chaps. II, III, and IV. (1) Test for Maxima, Minima, and Points of Inflection. We can now state wider sufficient conditions for maxima, minima, and points of inflection than those given in Chap. III. Suppose that the function f(x), together with its first n de- rivatives, is continuous in the neighborhood of the point x = x and that /'(*o) = 0, /"(z ) = 0, . . . /(.-i) (a; )=0, but that f n \x )=t=0. Then we shall have, by Taylor's Theorem with the Remainder, Formula (7) : (18) / (a* + h) -f(x ) = h«f^ (xo + Oh)/n !. If n is even, h n will be positive on both sides of the point h = 0, x = x ; and since f (n) (x) is continuous, it will preserve the sign it has at x throughout a certain interval about this point : x — a < a; <# -f a, — a(z ) = 0, f^(x )^0, the function f(x) will have a maximum at x = x if / (2m) (x ) < ; a minimum " " " / (2m) (x ) > 0. If, on the other hand, n is odd, the right-hand side of (18) will change sign with h and we shall have a point of inflection parallel to the axis of x. More generally, since the condition for a point of inflection, be it parallel to the axis of x or not, is that tan t =/'(#) be at a maximum or a minimum, we deduce 276 CALCULUS from the test just obtained, applied, not to/(#), but to /'(#) = tanr, the following Test for a Point of Inflection. If /"O*o)=0, f"'(x )=0, . . . /»(ai) a 0, / (2OT+1) ^o)^0, the curve y=f(x) has a point of infection in the point (x Q , y Q ). (2) Order of Contact of Two Curves. Let two curves, C l and (7 2 , be tangent to each other at an ordinary point P of either curve, and draw their common tangent PT. At a point M of PT infinitely near to P (by this is meant that M is taken con- veniently near to P and is later going to be made to approach P as its limit) erect a perpendicular cutting C Y in P x and C 2 in P 2 . PM and the arcs PPj , PP 2 are obviously all infinitesimals of the same order. It will be convenient to take PM as the principal infinitesimal. Denote by n the order of the infini- tesimal PiP 2 . Then the curves C x and C 2 are said to have contact of the n — lst order. For example, the parabola c » ft: y = x> has contact of the first order with its tan- gent at its vertex : 2 : 2/ = 0. But the curve y —x^ has contact of the second order with its tangent at the origin ; this point being a point of inflection for the latter curve. And the curves y = x s f y = X s — x* have contact of the third order. Since we can always transform our coordinate axes so that the tangent PT will be parallel to the axis of x — such a trans- formation evidently has no influence on the order of contact of the curves — we may without loss of generality assume the equations of the curves in the form * C,: y =/(*), 2 : y = (x), TAYLOR'S THEOREM 277 where y =/(«b) =<£ 0»o) and f'(x )=0, '(x ) = 0. Hence, by Taylor's Theorem with the Remainder, (7) : ci: y-y.=^/"K)+- +^f l " ) (*o+M), On y-y^ ^" (^)+- + ^ n) (^ + e'h). The infinitesimal P X P 2 on which the order of contact of these curves depends is numerically equal to the difference between the ordinate y of C x and the ordinate y of C 2 , ie. to (i9) fr/'^o)- (a* + M)~L Now the curvature of these curves at the point (x Q , y ) is, since /' (x ) = and ' (x ) = : *i=|/"te>) |, « s =|+"0*>)|. Hence the curves will have contact of the first order if theyl have different curvatures at P, or if they have the same curva- ture (#=()), one curve being concave upward and the other con- cave downward. But if they have the same curvature and (in case the curvature of both is =#= 0) if they both present their concave side in the same direction, then they will have contact of at least the second order. Thus at an ordinary point a curve has contact of the first order with its tangent. In particular, let C 2 be the osculating circle of C x at P. Then C 2 has the same curvature as C x and is concave toward the same side of the tangent. Hence it has in general contact of the second order with d ; but at special points it may have contact of higher order. At an ordinary point of inflection the tangent line has con- tact of the second order with the curve. For here, if we take C 2 as the tangent line, (x) = for all values of x, and hence the derivatives <£"(a\,), '"(xo), etc. all vanish. On the other hand, /" (x ) = 0, /'" (x ) =£ 0. Consequently (19) becomes 278 CALCULUS ||/»'(ai + «) and lim ££ = $/»' (**,)*<). (3) Evaluation of the Limits -, oo — oo, etc. The limit of the fraction lim^M, x=« jF(a!) when /(a) = and F(a) =0, can be obtained without the labor of differentiating whenever the numerator and the denomina- tor can be expressed as power series in terms of x — a = h. For example, to find v x - sin x lim x ±ox— tana; By the aid of the series for sin x and tan x, we have x — sin x _ ^ x 3 + higher powers of x x — tan x — ^ x 3 + higher powers of x Hence, cancelling x 3 from the numerator and the denominator, we see that the value of the limit is — |. The method of series is often of service in evaluating the limit oo — oo . For example, to find lim ( Vl + v? — x). Here we can take out # as a factor : (aK-> and then express the radical, since x > 1, as a series in 1/x by means of the Binomial Theorem : v .1-1+1. 1+3.I+ Hence x(^l + J-l)-| • £ + f • J + TAYLOR'S THEOREM 279 When x = , the terms of this power series in 1 /x approach as their limit, and since a power series represents a continuous function, the value of the limit in question is seen to be 0. EXERCISES 1. Show that the function y = 2 cos x + x sin x has a maximum when x — 0. 2. Have the following functions maxima, minima, or points of inflection when x = ? (a) 5 sin x — 4 sin 2 x 4- sin 3 #. (6) 2ar 5 -3e* + 6sina + - x . (c) locosx — 6 cos 2 a; + cos 3 a;. 3. Determine all the maxima, minima, and points of inflec- tion of the function y — \x — \ sin x + y 1 ^ sin 2x, and hence plot the graph. £.. Show that the curve y — cos x has contact of the fifth order at the point (0, 1) with the curve y = l-$x? + ^x\ 5. Show that the curve y = sinx has contact of the sixth order at the origin with the curve y = x-lx 3 + T %- u x 5 . 6. Determine the parabola y = a + bx-\-cx 2 so that it shall have contact of the second order with the curve y = e x , when x = 0. 7. The same when x = 1. Ans. y = %e + ^ex 2 . 280 CALCULUS 8. Show that, when the function f(x) is represented by a Taylor's Series, the n-th. approximation curve : y = s n (x)=f(x ) +f(xo) (x-x ) + ... + f ^^ (x-x y, has contact of at least the n-th. order with the curve y =f(x) at the point (x , y ). When will it have contact of higher order ? 9. Show that the curve ax + P y yx + 8 can in general be so determined as to have contact of the second order with the curve y=f(x) at the point (x , y ). For simplicity, assume x = and y = 0. What cases are exceptions ? 10. Show that o i (M f^!-dx = ~ L- + -JL (a>0). W J 1 + x* a a + b^a + 2b ' V ; (0j e~**dx — x — - -f- 3 5 21 73! 11. Evaluate to three significant figures IT ! since , dx. Evaluate the following limits : 12. limfcota; )• Ans. 0. 13. lim(Vl +x — x). Ans. — oo. as = \^ XJ * = oo TAYLOR'S THEOREM 281 i 1 14 . limfcY. Ans. 1. 16. lim fcY. Ans. 0. x=Q \ X J * = \ X J 15. limf- 1 — ^) • Ans. -^. 17. lim(cosx)-. 1 *±o\ x J y e x^o ^ ng> J_ # Ve 18. Show that, when two curves have contact of even order, they cross each other ; when they have contact of odd order, they do not cross. 19. If f(x)<(x) is .^/(s) -^- L dx dx and f(x ) = tj>(x ) i is f(x + h)^(x + h), h>0 ? 21. Show that sin a — a is an iufinitesimal of the third order, referred to a as principal infinitesimal. _a2 22. Determine the order of the infinitesimal cos a — e 2 * 23. Show that the equation sin cf> = 1 has one and only one root lying between and 7r/2. r CHAPTER XIV PARTIAL DIFFERENTIATION 1. Functions of Several Variables. Limits and Continuity. We shall consider in this chapter functions that depend on more than one variable. Thus the area z of a rectangle is the product of its two sides, x and y : z = xy; and the volume u of a rectangular parallelopiped is the product of its three edges x, y, and z : u — xyz. If the number of independent variables is two, we can rep- resent the function (1) *=/(*, V) geometrically as a surface. Such a function is said to be continuous at the point (x , y , z ) if a small change in the values of x and y gives rise only to a small change in the value of the function. And the function is said to approach a limit, z , when the point (x, y) approaches (x , y ), if the point (x, y, z) of the surface (1) approaches a limiting point (x , y , z Q ) in space, no matter how the point (x, y) in the plane may approach the point (a? , y ) as its limit. To formulate this latter definition in a more precise manner and at the same time in a way that is applicable to functions of more than two variables, let c be an arbitrarily small positive quantity. If a positive 8 can be found such that 282 PARTIAL DIFFERENTIATION 283 \f(x,y)-z \ ^o are constants, find 4. Geometric Interpretation. Geometrically the meaning of the partial derivatives in case there are but two independent variables is as follows. Holding y fast is equivalent to cut- ting the surface (1) by the plane y = y Q . The section is a plane curve : */ '\ and — - is the slope of this curve. Similarly —■ is the slope of the curve ~ t N * =/Oo, y)- We thus have the slopes of two tangent lines to the surface (1) at the point (x , y , z ), and hence we can readily determine the equation of the tangent plane through this point. For the tangent plane at a point contains all the tangent lines at the point and is determined by any two of them. If, therefore, we write the equation of the tangent plane with undetermined coefficients in the form : z - z = A (x - x ) + B (y — y ), we have only to require that the slope of the line in which this plane is cut by the plane y = y , i.e. PARTIAL DIFFERENTIATION 289 z — z = A (x — x ) be dz/dx, formed for the point (sc , 2/ ), — we will denote this quantity by (dz/dx) Q , — and similarly that the slope of the line in which the plane is cut by the plane x == x : z-z = B(y-y ), be (dz/dy) Q . Hence dxja and we obtain as the equation of the tangent plane : (31) Z - 2 » = (l) (a; -^ + (|) ^-^ From (28) it follows that the equations of the normal line (or simply the normal) to the surface (1) at the point P: 0»o> Vo, 3o) are: (32) (-) (-) \dxj \dyj The direction cosines of the normal are given by the relations (33) cos«:cos0:cos 7 = (!) o :(!) ; -l. EXERCISES Find the equations of the tangent plane and the normal to the following surfaces : 1. z = tan -1 ?. x Ans. y x — x y + (x 2 + y 2 )(z — z Q ) = 0; Vo ~ —Xq «o 2 + 2/o 2 ' 290 CALCULUS 2. z = ax 2 -f by 2 . Ans. For the tangent plane : z = 2ax x + 2 6^^ — 2o» 3. « 2 4- y 2 + a; 2 = a 2 . 4. Show that the surface z = xy is tangent to the x, y plane at the origin. , 5. The sphere: ^y+>.ii and the ellipsoid: 3 ^ + ^ + ^ = 2 intersect in the point (—1, — 2, 3). Find the angle at which they cut each other there. Ans. 23° 33'. 6. What angle does the tangent plane of the ellipsoid in the preceding question make with the x, y plane ? Ans. 59° 2'. 7. At what angle is the surface z as Sxy 2 — 5x 2 y — 7x + 3y cut by the axis of x at the origin ? Ans. 65° 41 '. ' 5. Derivatives of Higher Order. The first partial deriva- tives of the function are themselves functions of sc and #, and can in turn be differentiated : d 2 u * , n /. / \ /dw\ d 2 it ^ / N , It can be shown that the order of differentiation does not matter, provided merely that the derivatives concerned are continuous functions : d 2 u d 2 u (34) cxdy dydx PARTIAL DIFFERENTIATION 291 The theorem holds for functions of any number of variables.* Let us verify the theorem in some special cases. (a) u — e x cos y ; du r • d fdu = -e*smy, dy (P) u d fdu\ „ • Hfe)— "-»■ a? log 2 # y d 2 u 1 dzdx yz d 2 u _1^ du__\ogz dx y du _ x - fo 2/z' dscdz yz EXERCISES / 1. Verify the theorem for the other two pairs of cross deriv- atives in (6). 2. Verify the theorem in each of the following cases: (a) u = z sin xy ; (b) u — log (xy 2 ) ; (c) w = y s . d 3 u d*u ffu 3. Prove that dx 2 dy dydx 2 dxdydx f 4. If w = logV# 2 -f y 2 , then ** + *•** dx 2 dy 2 * The proofs of this theorem formerly given are not rigorous. For a critique of these see Gibson, Elementary Treatise on the Calculus, § 93, where a correct proof, due to Schwarz, is to be found ; or Goursat- Hedrick, Mathematical Analysis, vol. 1, § 11. A simple proof can be given by integration ; cf. Whittemore, Bulletin Amer. Math, fioc, ser. 2, vol. 4 (1898), p. 389. 292 CALCULUS 1 ' 5. If u = V x 2 -j- y 2 -\-z 2 then ^ + ^ + ^ = 0. dx 2 ^ dy 2 T dz 2 1 6 If — = *tH and — = — — 0sb. % dy dx then ^ + ^ = 0. v 6. The Total Differential. Let us form the increment of the function „, N Au =f(x + A#, y + Ay) -/(# , 2/ )- If we subtract and add the quantity f(x , y 4- Ay), we shall have: Aw =f(x + Ax, y + Ay) -f(x , y + Ay) +/Oo, 2/o + Ay) -/(«<,, 2/ ). Applying the law of the mean to these two differences gives : (35) Au = Axf x (x + 6 Ax, y + Ay) + Ayf y (x ,y + 0' Ay). Now if f x (x, y) and i /y(a;, y) are continuous functions of x, y, f x (x + 6Ax, y + Ay) will approach f x (x , y ) as its limit when Ax and Ay both approach zero, and hence will differ but slightly fromXC^o? 2/o) when Ax and Ay are numerically small: / x (z + OAx, y + Ay) =f x (z , y ) + e, where c is infinitesimal : lim e = 0. Ax = 0, Ay=0 Similarly, the limit of/ y (# , y o + 0'Ay) isf y (x , y ) and / y 0» o , 2/o + 0'Ay) =/ y Oo, y ) + iy, where rj is infinitesimal. Hence (35) may be written in the form : PARTIAL DIFFERENTIATION 293 (36) Aw = ^ Az + |* Ay + eAz + ,7 Ay, where we have dropped the subscripts and replaced f x (x, y), f y (x, y) by the alternative notation. Formula (36) is analogous to the second formula on p. 92, and so it is natural to describe the linear terms : du . , du A — &x + —Ay ex cy as the principal part of An. The remaining terms form an infinitesimal of higher order.* Definition. We define the total differential of u as the principal part of Aw : (37) *-g A . + g A * Since this definition holds for all functions u, we may in particular set u = x. From (37) follows then that (38) dx = Ax. Similarly, setting u = y, we get : (39) dy = Ay. Substituting these values in (37) gives * If £ is an infinitesimal depending on several, let us say two, inde- pendent variables, a and |8, and if we take these variables as the princi- pal infinitesimals, then f is said to be an infinitesimal of higher order than a and /3 if lim f = 0. £ is said to be of the same order if K<—L—(r, s), y = ^(r, s). 296 CALCULUS If we hold s fast and allow r alone to vary, we have the case just treated, the independent variable now being r instead of t. Hence (44) is still valid, the derivatives with respect to r now being partial : /to du _dudx du dy dr dx dr dy dr In like manner : du _dudx du dy ds dx ds dy ds ' Let us state this result in the form of a theorem. It is ap- plicable to functions of any number of variables. Theorem 1. If and if each of the arguments x, y z, • • • is made to depend, on r, 8, • • • ; x = cf>(r, s, • • •), y = «/'(r, «,•••)> z = (r, s), y=t( r > s )> then, by the definition (37), du du . , du . Also Hence dx= i Ar+8 i As > d y=i Ar+d i As - du 7 , du , 0a> ty (dud* ,dudy\ * r ,(?ufo + d J:M\± s \fo dr dy drj [dx^ds B^ds) du . , du . , q. e. d. PARTIAL DIFFERENTIATION 299 We will state the result as Theorem 2. If u=f(?,y,z, • • •), and if each of the arguments x, y, z, • • • is made to depend on r, s, • • • : *= $(?>*> • • •)> y—^( r t s > • - •)> * = w (*j s > • • •)> then, if all the first partial derivatives are continuous, we shall have : tfa; (72/ ^ no matter whether the independent variables are x, y, z, • • • or r, s, • • -. The number of variables in each class, (x, y, z, • • •) and (r, s, • ' •), is arbitrary. It is readily shown that the general theorems relating to the differentials of functions of a single variable : d(cu) = cdu, d (u + v) = du + dv, d(uv) = udv + vdu, 'u\ vdu—udv d V V hold for functions of several variables. Moreover, the differ- ential of a constant, considered as a function of several variables, is 0: dc = 0. Example. Let us work the example of § 7 by means of the above theorem. du = ye xy dx + xe xv dy, r & dx = . dr + r 2 + s 2 ■ ^ + ^ > g y * r 2 -f s 2 r -f s 2 300 CALCULUS Hence du = ry ~ 8 f e>*dr + sy + ™e**ds r 2 + s 2 r^ + s 2 du 7 , du, = —dr-\-—-ds. dr ds Now dr = Ar and ds = As are independent variables, and consequently we can equate their coefficients on the two sides of the last equation : * du _ ry — sx xy du _ sy + rx d r ~ ? + s 2 ' ds" r* + s 2 EXERCISES 1. Work the first four exercises at the end of § 7 by the method just explained. 2. If u=f(x + a,y + b), , .* du du , du du show that —- = — and —- = —-. dx da 7 dy db 3. If u=f(x) and x = 3r + 2s + 7t, show that — = 2 — . cs dx 9. Euler's Theorem for Homogeneous Functions. A function u is said to be homogeneous if, when each of the arguments is multiplied by one and the same quantity, the function is merely multiplied by a power of this quantity. For definite- ness we will assume three arguments : u=f(®, y, z), (46) / O, Xy, Xz) m \»f (x, y, z). * The reasoning here, given at greater length, is as follows. Since dr and ds are both arbitrary, we may set ds = 0, dr =£■ 0, and then cancel dr. Thus the coefficients of dr on both sides of the equation are seen to be equal. Similarly, setting dr = 0, ds ^t 0, we infer the equality of, the coefficients of ds. PARTIAL DIFFERENTIATION 301 The exponent n of A. is called the order of the function. Thus the functions u = aa?+bxy + cy 2 , u = J log (ar 2 -f- # 2 ) - log a, ax -{-by z , _,w ca + cty -^2 _j_ y 2 x are homogeneous of order 2, 0, 0, \, 1, resoectively. If in particular we set A. = -, we have x (47) /+/,(*', y', z')y+f(x', y', z')z = n\^f(x,y y %% where f x (x', y', z f ) denotes as usual the partial derivative of f(x, y, z) with respect to x, the arguments being subsequently replaced by x', y', z' respectively. If we now put A = l, (49) assumes the form (C), and the theorem is proven. We have stated and proved the theorem for a function of three variables. But theorem and proof hold for a function of any number of variables. 302 CALCULUS EXERCISE Verify Euler's Theorem for each of the above examples. 10. Differentiation of Implicit Functions. Let y be defined implicitly as a function of x by the equation (cf. Chap. II, § 9) : (50) F(x,y)=0. To differentiate y we begin by setting u = F(x,y) and forming the total differential of u : du = ?fdx + d -fdy. Cx Cy This relation is true, no matter what the independent variables are, § 8, Theorem 2. We may, therefore, in particular choose y so that the equation (50) is satisfied. Then du = 0, and we have: dF (51) ^+M:^ = o or 4-_*L K } dx^dydx dx dF dy In like manner, if z is defined by the equation : (52) F(x,y,z) = 0, we can differentiate z partially by setting u = F(x, y, z) and taking the total differential of each side : , dF , . dF, , dF, du = — - dx -f — - dy 4- — - dz. cx cy cz This equation is true, no matter what the independent variables are. If in particular z be so chosen that the equation (52) is satisfied, then du = 0, and PARTIAL DIFFERENTIATION 303 <7a? 02/ (72 But dz now has the value : dz = — d» + — - cfa/. da; dy Hence, eliminating dz, we have {W + W^fa,{£FdFdz\ d =Q \0a? dzdx) \dy dzdy) Here cto = A# and dy = Ay are independent variables. We may, therefore, set dy = 0, dx =£ 0, and divide through by cfcc : ^qx dF.dFdz ft (53) to + &fcT°' £? £2 A similar equation holds for dz/dy, x being replaced through- out in (53) by y. The student should notice carefully what the independent variables are in each differentiation. Thus dF /dx is the derivative of a function of three independent variables, x, y, z, and the values of these variables are not in general such as to satisfy the equation (52). At this stage of the work (52) is irrelevant, does not exist for us, has not as yet come into play. The same is true of dF/dy and dF/dz. When we come to dz/dx, however, this 2 is a function of the two independent variables, x and y, — and such a function that (52) is satisfied. The generalization to a function u of any number of variables is now obvious : F(u,x,y, z, ■ • .) = 0, dF dFdu dF_ du_ dx ( 54 ) du dz+~dx~°> dx~~~dF' du etc. 304 CALCULUS Example. Differentiate z partially, where a 2 ^6 2 c 2 ' Here and we have: a 2 ^~&Yx~ * dx c 2 x m dz dy c 2 y b 2 z b 2 c 2 dy ' Several Implicit Functions. We may have two implicit functions, u and v, of any number of variables, x, y } • • •, defined implicitly by two equations : 1 <&(u, v, x,y, - - -) = 0. For definiteness, let the number of variables x, y, • • • be two. Setting U = F(u, v, x, y), V=&(u, v, x, y), and taking differentials, we have : i (56) , TT dF , , dF , , dF, . dF , d(7= — - c?m -f — - dv + — - dx + — dy, cu cv ex cy d F = — du + — - d v + -x- daj + — dy, cu cv ox cv no matter what the independent variables are. If we now require that u and v be so determined that the equations (55) be satisfied, we get: d{7=0, dV=0- ? and furthermore : du = -^ dx + -^ dy, dv = — - dec + — - dy. ex cy Substituting these values of du and dv in the right-hand sides PARTIAL DIFFERENTIATION 305 of (56), we see that the coefficients of dx and dy are equal to 0, and hence we get the two equations : dFBu M^+^-aO du dx dv dx dx ' f^^ + ^^ + ^-0 du dy dv dy dy and two similar equations, in which F is replaced by S>. These latter equations the student should write out for him- self. From the first and third of these four equations we can solve for du/dx and dv/dx, and from the second and fourth, for du/dy and dv/dy. Thus (57) du dx' F. K * x ** F K *. with similar formulas for dv/dx, du/dy, dv/dy. The student should also write these out clearly and neatly. The generalization is now obvious. Thus if F(u, v, w,x,y, . . .) = 0, (58) J $ (u, v, w, x, y, • • •) = 0.. 1 *(u, v,w, x,y,.- .)=0, we shall have F F F cb cb ^*x ^v ^*w (59-) du - \If \b \b * X T « ^w -• W dx— F F F $« ®v ^«, *u % *«, The determinant that appears in the denominators : dF dF dF dF dF du dv dw du dv d& d$> d<& (60) d& ao du dv y du dv dw d* a* d* i du dv dw 306 CALCULUS is called the Jacobian of the functions F, <$, or F, 3>, #. In the foregoing it has been tacitly assumed that all the partial derivatives are continuous and that the Jacobian does not vanish. 11. A Question of Notation. Problem. Suppose u=f(x,y), y = cf>(x,z), to find f*. ex Before beginning a partial differentiation the first question which we must ask ourselves is: Wliat are the independent variables f Hitherto the notation has always been such as to suggest readily what the independent variables are. In the present case they may be : (a) x and y ; or (b) x and z ; or (c) y and z. We can indicate which case is meant by writing the independ- ent variables as subscripts, thus : In case (c) — has no meaning. ex : ' Another notation sometimes employed is to mark the vari- able or variables that are held fast, thus : dx \: w dx Let the student compute -— in cases (a) and (b). ox 12. Small Errors. In the case of functions of a single vari- able we have seen that the linear term in the expansion of Taylor's Theorem : fix) =f(x ) +f'(x )(x -x ) + • • ., can frequently be used to express with sufficient accuracy the effect of a small error of observation on the final result, cf. PARTIAL DIFFERENTIATION 307 Infinite Series, § 27. This term, /' (x ) (x — x ), is precisely the differential of the function, df, for x = x Q . The differential of a function of several variables can be used for a similar purpose. If x, y, • • • are the observed quan- tities and u the magnitude to be computed, then the precise error in u due to errors of observation Ax = dx, Ay = dy, etc. is Au. But *,„!**+£*+. .. ex cy will frequently differ from Au by a quantity so small that either is as accurate as the observations will warrant, — and du is more easily computed. Example. The period of a simple pendulum is 4 T=2^!-. 9 To find the error caused by errors in measuring I and g, or in the variation of I due to temperature and of g due to the loca- tion on the earth's surface. # Here dT=-±=dl- £\fl dg sllg 9^9 or . Ans. The percentage error is given by the formula : dc_ (a — b cos , du (62) gi=^ cos " + ^ co ^+fe co ^' where a, /?, y are the angles that C makes at P with the axes. EXERCISES 1. If a normal be drawn to a curve at any point P and if r denote the distance of a variable point of the plane from a fixed point 0; y, the angle between PO and the direction of the normal, show that (63) ' £ = -<*>sy. 2. Explain the meaning of -^ and show that or (64) dn^dr K J dr dn 14. Exact Differentials. If in the expression (65) Pdx+Qdy P and Q are functions of x and y subject to no restriction ex- 310 CALCULUS cept that, along with whatever derivatives we wish to use, they be continuous, there may or may not be a function u =f(x, y) whose total differential : ■, du , . du , du = —dx + —dy ex cy coincides with (65) . If there is such a function, then Now since dx d_fdu\_ dy\dx)~ du _ dy = Q- we see that P and Q are subject to the restriction: dP = dQ v ' dy dx It can be shown that conversely, when P and Q do satisfy (66), there always does exist a function u, of which (65) is the total differential. # In this case the expression (65) is said to be an exact differential. Example. Consider the expression : (2ax + by + l)dx + (bx + 2cy + m) dy. tt dP , dQ , Here — = b, Tr = h > cy ex and hence we have an exact differential before us. To inte- grate it, begin with ^ = P=2ax + by + l, ex and integrate each side with respect to x, regarding y as constant : u = ax 2 + bxy + lx + (y), the constant of integration depending, of course, on y. Now differentiate this expression for u with respect to y : *Cf. Goursat-Hedrick, Mathematical Analysis, vol. 1, §§ 151, 152. PARTIAL DIFFERENTIATION 311 Comparing this last expression with Q = bx + 2 ct/ -f- m, we see that '(y) = 2cy-\-m, (y) = cy 2 + my+C. Hence w = ax 2 + &#?/ + cy 2 + Ix + m?/ + O. If we have three independent variables and the expression Pdx+Qdy + Rdz, the necessary and sufficient condition that it be an exact differ- ential is that (67) d X= d 3 dQ = dR dR = dP dy dx' dz dy ' dx dz ' It is assumed that the partial derivatives are continuous. EXERCISES Determine which of the following expressions are exact dif- ferentials and integrate such as are : 1. ( e x cos y . ) dx — (e x sin y -f 7 sec 2 y) dy. \ VI — x 2 / 2. (x + y)dx + (x-y)dy. 3 . yz e xyz dx + zx e xy * dy •+- xy e yz dz. 312 CALCULUS EXERCISES 1. If pv 1Al =C, find ^. dp 2. If u = ^l, X or ■x = r i -s, ,<1/ = e '> 3. If u = e* amv + xlog(x-\-y), x = pqr, y — r sin -1 (qr) ; findf^. dq 4. If u = 2xy and 2# + 32/-t-5z = l, explain all the meanings which •— may have, and evaluate ex this derivative in each case. 5. If find*?. ox {u 5 + v 5 4- x 5 = 3#, W 3 + v 3_ | _ 2/ 3 = _3 a . > 6. If F=2mv dV and j ■*■+* + *- %> U S + V S + yS == _ Sx> dx ue v -\-vx = ysinu 9 u cos u = a? 2 -+- y% find- ax findf^ dy PARTIAL DIFFERENTIATION 313 8. From the equations it follows that I _ dx du dx dv du dx dv dx ' q__ dy du , dydv du dx dv dx Explain the meaning of each of the partial derivatives. Com- {x SB u 4- vu v , y = v — uv u , du j du pute _ and — dx dy /9. If find §*. dx ' 10. If u = x 2 -f y 2 + z 2 and z = xyt, explain all the meanings of — • dx I (x,y) = 0, dz d dz d(f> show that * = ^"frfe . dx d dy 12. If u=f(x + at,y + pt) y show that h.= a *L+B^ 9 dt dx^ P dy f and obtain the general formula for -^ • dt n 314 CALCULUS 13. If u =/(y + ax) -f <\> (y — ax), show that d 2 u _ 2 d 2 u dx 2 dy 2 14. If »-'(§: show that du du _ r. dx. dy 15. If u=f(x + u, y-u), ex 16. If u=f(xu,y), find^. dx 17. Use the method of differentials to find ~ t — and -^, dx cy c)t in terms of / f (£, >?),/„(£ *?), if w =f(x + ut, y — ut). 18. If w is a function merely of the differences of the argu- ments x u x 2 , • • •, x n show that du du_ , t t , du _ ~ dx x dx 2 dx n 19. If u and v are two functions of x and y satisfying the relations : du __dv du _ _ #y dx dy dy dx' show that, on introducing polar coordinates : x = r cos , y = r sin <£, we have du __ldv ldu _ dv _ dr r d dr PARTIAL DIFFERENTIATION 315 20. If f(x, y)=0 and (a?, z) = 0, show that *A%(P>V, = °> show that -£ — _H -s _. 1. <7£ 0V (7/) Explain the meaning of each of the partial derivatives. 22. Under the hypotheses of question 19, show that d 2 u ldu 1 d 2 u _ ~ dt 2, r dr r 2 dcf> 2 23. If u=f(x, y) is homogeneous of order n, show that <>d 2 u . o d 2 u , 9 d 2 u t -, x ar — - -f- 2 xy 7 —-~ -f V tt-s = n ( n — 1 ) w. &» 2 J dxdy * dy 2 y J 24. If w is a function of x, y, z and x, y, z are connected by a single relation, is it true that dy dz dy 25. If dU=6dS-pdv is an exact differential, and if S and v can be expressed as functions of the independent variables 6, p, show that (t), z=«K0; (6) as the intersection of two cylinders : (9) y = *(«), * = The form (b) may be regarded as a special case under (a), namely that in which x = t. On the other hand, it is a special case under (c). The Direction Cosines. To find the direction cosines of the tangent to a space curve at a point P: (x Q , y , z ), pass a secant through P and a neighboring point P : (x -f Ax, y +Ay, z + Az). The direction cosines of the secant are : cosa' = -^:, cos£' = -^-, cosy' = =, PP PP> PP and hence, for the tangent, APPLICATIONS TO THE GEOMETRY OF SPACE 319 v Ax ,. (Ax cos a — lim = hm with similar formulas for cos /?, cos y. Hence PP'J (14) cos a dx cos /» = &, COSy dz ds ds' , ' ds Here the tangent is thought of as drawn in the direction in which s is increasing. If it is drawn in the opposite direction, the minus sign must precede each derivative. From (14) it follows at once that (15) ds 2 = dx 2 + dy 2 +dz 2 . This important formula can be proven directly from the rela- tion ppi 2 = Ax 2 -f Ay 2 + Az 2 . If we assume the form (a), ds 2 = [f(t) 2 -hcf>'(ty + ^(t) 2 ^dt 2 Fig. 83 and (16) (17) cos a = f(t) VfW+fffl+rW = Jy/f (}f + ' (ff+V (ty at. cos/} = COSv = (18) Applying these results to (9), we get 1 cos a Af ^dx'^dx 2 etc., 320 CALCULUS (19) -/^♦s+s* The Equations of the Tangent Line and the Normal Plane. For the tangent line we have, in case (a) : (20) and in (6) : (21) y-yo = «z-_#o = y — y o = « - «o . f(t ) *'(f ) ffr)' (x-x ) 9 - z ^(fX {x ~ xo) - \dxj The normal plane is given by (22) f (?.) (x - x ) + *' (t ) (y - y ) + ^ ft) (i -z ) = in (a) ; and in (6) by f dy y (23) a-#o + c?rc ,^-*>- + (S>-*>- a On the other hand, the tangent line in case (c) may be obtained most simply as the intersection of the tangent planes to the surfaces at the point in question : (24) 1 ©.<-* >♦(&<»-*>+ These equations may be thrown into the equivalent form x — ^o _ y — .Vo _ g — z q . F y F z ^ ^x ^ *; % *. <*>* x *x % (25) Hence we see that the direction cosines of the tangent line to the curve of intersection of the surfaces (10) are given at (x, y, z) by the proportion : (26) cos « . cos /? : cos y = The equations of the normal plane can now be written down at once. 1 F F F z F x K F s 1 y z ^ f z ^x ** *, APPLICATIONS TO THE GEOMETRY OF SPACE 321 EXERCISES Find the equations of the tangent line and the normal plane to the following space curves : 1. The helix (11) and the curve (12). 2. The curve : y 2 = 2 mx, z 2 = m — x. 3. The curve: 2x 2 + 3y 2 + z 2 = 9, ^ = 3^ + ^, at the point (1, — 1, 2). 4. Find the angle that the tangent line in the preceding question makes with the axis of x. 5. Compute the length of the arc of the helix : a? = cos0, y = sir\0, 5z = 0, when it has made one complete turn around the cylinder. 6. How steep is the helix in the preceding question ? 7. Show that the condition that the surfaces (10) cut orthogonally is that i (27) dFd,y,2)=0, *(x,y,z)=0, intersecting at the point (x , y , z ), be tangent to one and the same line there is that, in this point, dF dF dF dx dy dz d® d® d® dx dy dz d* d* ?W dx dy dz It is assumed that in no row do all the elements vanish. (28) = 0. 322 CALCULUS 9. The surfaces a? + y 2 + z 2 = 3, xyz = l, z — xy, all go through the point (1, 1, 1). Find the angles at which they intersect there. 10. Obtain the condition that the surface (4) and the curve (8) meet at right angles. 11. Find the direction of the curve x = t 2 , y — f, z = t* in the point (1, 1, 1). 12. Find the direction of the curve xyz = 1, y 2 = x in the point (1, 1, 1). 13. Find all the points in which the curve x = t 2 , y = P, z = t* meets the surface z 2 = x + 2y-2, and show that, when it meets the surface, it is tangent to it. 14. Show that the surfaces cr b* & in general never cut orthogonally ; but that, if i+i-i-o a 2 + 6 2 c* - > they cut orthogonally along their whole line of intersection. 15. When will the spheres ^ + ^ + ^ = 1, (x-a) 2 + {y-bf+(z-cy = l cut orthogonally ? 16. Two space curves have their equations written in the form (13). They intersect at a point P. Show that the angle e between them at P is given by the equation : APPLICATIONS TO THE GEOMETRY OF SPACE 323 cos c = x[x[ + y[y' 2 + afjaj, where as{=^p, etc. 17. The ellipsoid: cc 2 -|- 3 2/ 2 -+- 2z 2 = 9 and the sphere: 252 _|_ y2 _|_ ^ _ g mtersect in the point (2, 1, 1). Find the angle between their tangent planes at this point. 3. The Osculating Plane. Let P : (x ,y , z ) be an arbitrary- point of a space curve (8), and pass a plane (29) A(x-x ) + B(y- y ) + C(z-z ) = through P. Then the distance D of a point P'l x = f(t + h), y = cf>(t + h), z = if;(t + h) of the curve from this plane will be in general an infinitesimal of the first order with reference to PP' as principal infini- tesimal. For ±D = A(x-x ) + B(y-y () ) + C(z-z ) ^A 2 + B 2 +C 2 where x, y, z are the coordinates of P'. Hence ±D = A ^ & + *> ~/ft)1 + * C*ft> + *) ~ * W1 + etc ' . V^ + Jff+C* Applying Taylor's Theorem with the Eemainder to each bracket : f(to + h)-f(t ) = hf>(t )+ 1 £f<'(t (i + eh), etc., and setting V-4 2 4- 5 2 + O 2 = A, we obtain ± D = A r^/'ft) + B +' (O + «A ' (' (t ) if/' (/ ) all vanish. "We exclude this case. On the other hand, PP' = As and h = M are infinitesimals of the same order, since As lim ^ = D t s = V/' (t y+ ' (t y + ^ (t y * o. Thus the above statement is proven. If, however, A, B, and C are so chosen that (30) Af'(t ) + B'(t ) + Of ft) - 0, then lim ± D/h = and ,. ±D Af"(t ) + B"(t ) + C"(t ) + Cy'fo) = 0, then even lim ± D/h 2 = and D becomes an infinitesimal of still higher order ; — of the third order, as is readily shown, if Equations (30) and (31) serve in general to define the ratios of the coefficients A, B, C uniquely. The latter may, therefore, be eliminated from (29), (30), and (31), and thus we obtain the equation of the osculating plane : x-x y-y z—z (32) f'(t ) 4>'(to) «A'(*o) =0. f"(t ) 4>»(t ) f(to) The osculating plane as thus defined is a tangent plane having contact of higher order than one of the tangent planes APPLICATIONS TO THE GEOMETRY OF SPACE 325 taken at random. There is in general only one osculating plane at a given point. But in the case of a straight line all tangent planes osculate. Again, if f"(t ) = "(t ) = ) = ^> the same is true. The osculating plane cuts the curve in general at the point of tangency; for the numerator of the expression for ± D changes sign when h passes through the value 0. It is easy to make a simple model that will show the oscu- lating plane approximately. Wind a piece of soft iron wire round a broom handle, thus making a helix, and then cut out an inch of the wire and lay it down on a table. The piece will look almost like a plane curve in the plane of the table, and the latter will be approximately the osculating plane. The normal line to a space curve, drawn in the osculating plane, is called the principal normal. The centre of curvature lies on this line, the radius of curvature being obtained by pro- jecting the curve orthogonally on the osculating plane and taking the radius of curvature of this projection. If a body move under the action of any forces, the vector acceleration of its centre of gravity always lies in the osculat- ing plane of the path. When the equation of the curve is given in the form (9), the equation (32) becomes: EXERCISES 1. Find the equation of the osculating plane of the curve (12) at the point t = w. 2. Find the equation of the osculating plane of the curve of intersection of the cylinders : sc 2 + y 2 = a 2 , x* 4- z 2 = a 2 , and interpret the result. 326 CALCULUS Suggestion. Express x, y, z in terms of t : x = a cos t, y = a sin £, 2 = a sin £. 3. Show that the centre of curvature of a helix lies on the radius of the cylinder produced. 4. Show that the osculating plane of the curve V = A z 2 = l-y at the point (0, 0, 1) has contact of higher order than the second. 4. Conf ocal Quadrics. * Consider the family of surfaces : where X is a parameter taking on different values. Each sur- face of the family is symmetric with regard to each of the co- ordinate planes. We may, therefore, confine ourselves to the first octant. If A. > — c 2 , we have an ellipsoid, which for large positive values of X resembles a huge sphere. As X decreases, the sur- face contracts, and as X approaches — c 2 , the ellipsoid, whose equation can be thrown into the form : v T \ a 2 + X b 2 + X/ * No further knowledge of quadric surfaces is here involved than their mere classification when their equation is written in the normal form a 2 X b 2 X C 2 See Bailey and Woods, Analytic Geometry, p. 316. It is desirable that the student have access to models of the three types here involved. The student should work out for himself, after a first reading of this paragraph, the corresponding treatment of the confocal conies in the plane : r'2 + =^-=1 a 2 + X 6 2 + X APPLICATIONS TO THE GEOMETRY OF SPACE 327 flattens down toward the plane z = as its limit, — more pre- cisely, toward the surface of the ellipse In so doing, it sweeps out the whole first octant just once, as we shall presently show analytically. Let A continue to decrease. We then get the family : ( 35 ) ~iT- + ^T 7JV-\ = 1 > -& 2 0<-c 2 . These are hyperboloids of one nappe, and they rise from coin- cidence with the plane z = for values of /a just under — c 2 , sweep out the whole octant, and flatten out again toward the plane y = as their limit when /a approaches — b 2 . Finally, let A trace out the interval from — 6 2 to — a 2 . We then get the hyperboloids of two nappes : (36) S- 7^ r A r-i -a 2 0, y > 0, z > 0. Hold x, y, z fast and consider the function of A: The function is continuous except when A = — c 2 , — b 2 , or —a*. In the interval — c 2 < A < -f- oo we have * /(+»)= -1, lim /(A) =+oo. A=-c*+ * The notation lim /(x), lim f(x) is explained in Chap. XI, § 9. x=a+ x=a— 328 CALCULUS Hence the curve crosses the axis of abscissas at least once in this interval. On the other hand ft m ^ V z ^ q J K J (a 2 + A) 2 (& 2 + A) 2 C^ + A) 2 * Hence /(A.) always increases as X decreases, and so the curve cuts the axis only once in this interval. We see, therefore, that one and only one ellipsoid passes through the point P. Similar reasoning applied to the intervals (— b 2 , — c 2 ) and (—a 2 , — b 2 ) shows that one and only one hyperbola of one nappe, and one and only one hyperbola of two nappes pass through P. Theorem 2. The three quadrics through P intersect at right angles there. The condition that two surfaces intersect at right angles is given by (27). Applying this theorem to (34), and (35) we wish to show that 2x 2x 2y 2y 2z 2z =Q a? + ka 2 + (j. b 2 + \b 2 + f*. c 2 +kc 2 + n Now subtract (35) from (34) : ( A * . y 2 i ^ 2 L q ^ ; L(a 2 + A)(a 2 + / x)" h (6 2 +A)(6 2 + / .)" 1 "(c 2 + A)(c 2 + /A )J ' and since /* — A =£ 0, this proves the theorem. The three systems of surfaces that we have here investigated are analogous to the three families of planes in cartesian coor- dinates, to the spheres, planes, and cones in spherical polar coordinates, and to the planes, cylinders, and planes in cylindri- cal polar coordinates. They form what is called an orthogonal system of surfaces, and enable us to assign to the points of the first octant the coordinates (A, /*, v), where -c 2 ) = 0. To determine the angle w between this curve and a parallel of latitude, draw the meridians and the parallels of latitude through an arbitrary point P : (0 O > <£o) and a neighboring point P* > (#o + A0, <£ + A<£) of this curve. We thus obtain a small curvilinear rectangle, of which the arc PP' is the diagonal. We wish to determine the angle ' = ZM 1 PP' of this triangle evidently approaches w as its limit when P approaches P. Fig. 84 We have tan M X P PMi Now PM l differs from PM= a cos <£ A0 by an infinitesimal of higher order and likewise M X P differs from MP' = aA by an infinitesimal of higher order. Hence, by the theorem of Chap. V, § 2, we obtain : lim tan w' = lim —J — = hm £ — -, P'±f p'±p PM X A0=o a cos A0 tan cos<£ ( De, 330 CALCULUS or, dropping the subscript : (38) tano> = -J—^. \ " ' cos <\> dO In order to obtain the differential of the arc of the curve (37) we write down the Pythagorean Theorem for the triangle PM^: PP' 2 = PM l 2 -{-M 1 P'\ divide through by A0 2 and then let A0 approach as its limit. Since the chord PP' differs from the arc As by an infinitesimal of higher order, we have : lim f^] 2 = lim f — Y= a 2 cos 2 4> + a 2 lim f^Y p>±p\&0J p'±p\A0j ^ p>±p\A0J , (D d sy = a 2 cos 2 + a 2 (D e ) 2 , (39) ds* = a 2 [cos 2 <£d0 2 + d 2 ]. Rhumb Lines. A rhumb line or loxodrome is the path of a ship that sails without altering her course, i.e. a curve that cuts the meridians always at one and the same angle. If we denote the complement of this angle by o>, then we have from (38) for the determination of the curve : — £- = c!0tanG>, cos (40) <9tano>= fJ±- = \ogten(± + Z) + a v ' J cos<£ \2 4y This is the equation of an equiangular spiral on the sphere, which winds round each of the poles an infinite number of times. EXERCISES 1. Show that the total length of a rhumb line on the sphere is finite. APPLICATIONS TO THE GEOMETRY OF SPACE 331 2. The cartesian coordinates of a point on the surface of a sphere are given by the equations : x = a cos <£ cos 0, y = a cos sin 0, z = a sin <£ . Deduce (39) from these relations and the equation : ds 2 = dx> + dy 2 + dz 2 . 3. Taking as the coordinates of a point on the surface of a cone (p,0), where p is the distance from the vertex and is the longitude, show that (41) tan-- 3&— . pdOsina 4. Obtain the equation and the length of a rhumb line on the cone. 5. The preceding two questions for a cylinder. 6. Mercator's Chart. In mapping the earth on a sheet of paper it is not possible to preserve the shapes of the countries and the islands, the lakes and the peninsulas represented. Some distortion is inevitable, and the problem of cartography is to render its disturbing effect as slight as possible. This demand will be met satisfactorily if we can make the angle at which two curves intersect on the earth's surface go over into the same angle on the map. For then a small triangle on the surface of the earth, made by arcs of great circles, will appear in the map as a small curvilinear triangle having the same angles and almost straight sides, and so it will look very similar to the original triangle. What is true of triangles is true of other small figures, and thus we should get a map hi which Cuba will look like Cuba and Iceland like Iceland, though the scale for Cuba and the scale for Iceland may be quite different. A map meeting the above requirement may be made as fol- lows. Regarding the earth as a perfect sphere, construct a cylinder tangent to the earth along the equator. Then the 332 CALCULUS meridians shall go over into the elements of the cylinder and the parallels of latitude into its circular cross-sections as fol- lows : Let P be an arbitrary point on the earth, Q> its image on the cylinder. (a) Q shall have the same longitude, 0, as P. (b) To the latitude of P shall correspond a distance z of Q from the equator such that the angle w which an arbitrary curve C through P makes with the parallel of latitude through Pand the angle On the other hand, dz tan (D l = adO Fia. 85 Hence, setting a for convenience = 1, we get d * =^ or dz = M- dO cos cf> dO cos $' J cos 4> + 1). t l0S ""\2 the constant of integration vanishing because z = corresponds to<£ = 0. Thus a point in latitude 60° N. goes over into a point distant 1.32 units from the equator. The cylinder can now be cut along an element, rolled out on a plane, and the map thus obtained reduced to the desired scale. This map is known as Mercator's Chart.* It has the property that the meridians and the parallels of latitude go over into two orthogonal families of parallel straight lines. Furthermore, a rhumb line on the earth is represented by a straight line on the map. *G. Kremer, the latinized form of whose name was Mercator, com- pleted a map of the world on the plan here set forth in 1569. APPLICATIONS TO THE GEOMETRY OF SPACE 333 We call attention to the fact that the above map cannot be obtained by projecting the points of the sphere on the cylinder along a bundle of rays from the centre. EXERCISE Turn to an atlas and test the Mercator's charts there found by actual measurement and computation. CHAPTER XVI TAYLOR'S THEOREM FOR FUNCTIONS OF SEVERAL VARIABLES 1. The Law of the Mean. Let f(x, y) be a continuous func* tion of the two independent variables x and y, having continuous first partial derivatives. We wish to obtain an expression for f(x + h,y + k) analogous to the Law of the Mean for functions of a single variable, Chap. XI, § 2. One such expression has been found in Chap. XIV, § 6; but there is a simpler one. Form the function : *(*) =/(**> + th, 2/0 + tic), < * < 1, where x , y , h, k are constants and t alone varies. Notice that (t), setting a = 0, 6 = 1, we get : *(l)=fc(O)+l.*'(0), O<0<1. Now *'(*) = hf x (x + th, 2/ + th) + kf y (x + th, y + th). Hence f(%o + h, y + k) = (1) /(%* Sfe) + hf x (x + M, 2/o + Oh) + A;/, (a*, + 6h, y + 0fc)> where < < 1, and this is the form we sought for the Law of the Mean for functions of two independent variables. The extension to functions of n > 2 variables is obvious. 334 FUNCTIONS OF SEVERAL VARIABLES 835 2. Taylor's Theorem. We obtain Taylor's Theorem with the Remainder if we write the corresponding theorem for $(£): *(i)=*(0)+*'(<>) + ... +^ (w) (0)+ 1 lv * ln+1) (0), and then substitute for <£ and its derivatives their values. Thus when w = lwe get (2) f(x + h, 2/0 + &) =f(x , 2/ ) + hf m (a%, 2/o) 4- &/„ (a*>, 2/o) +|[tf/>(X, F) + 2M-/ xy (X, F)+*«/>(X, F)], where X = # + 07*, F= y -+■ 0&, and < 6 < 1. The student should write out the formula for the next ease, 71 = 2. The general term, 3> (n) (0)/w !, can be expressed symbolically as nl\_ dx dy_\ x=x iy=v and the remainder as (W + 1)!| £# 0v] x = z + *A |y = y o + 0* The extension to functions of n > 2 variables is immediate. If the remainder converges toward zero when n becomes infinite, we obtain an infinite series whose terms are homo- geneous polynomials and which converges toward the value of the function. If furthermore the series whose terms consist of the monomials that make up the terms of the latter series con- verges for all values of h and k within certain limits : \h\ y , u ). Hence we see that at (a? , y ) A similar statement holds for a minimum. The necessary condition contained in (4) can -be extended at once to functions of n > 2 variables. For, if any one of the first partial derivatives, du/dx, for example, were ^0 at (# , y , z , •••), then the function f(x, y , z , •••), a function of x alone, would be increasing as x passes through the value x , or else it would be decreasing, according to the sign of du/dx. The conditions (4) are frequently sufficient to determine a maximum or a minimum. Example 1. Given three particles of masses m lf m 2 , ra 3 , situated at the points (x l} y^, (x 29 y 2 ), (« 3 , 2/ 3 ). To find the point about which the moment of inertia of these particles will be a minimum. Here it is clear that for all distant points of the plane the moment of inertia is large, becoming infinite in the infinite region of the plane. Furthermore, the moment of inertia is a positive continuous function. Hence the surface u = I=m 1 [(« -x i y + (y-y i y]+m 2 [(x- x 2 ) 2 + (y - y 2 ) 2 ~\ + ra 3 [(a - x 3 ) 2 + (y - 2/ 3 ) 2 ] must have at least one minimum, and at such a point Y = 2 [>! (oj — x Y ) + ra 2 (x — x 2 ) + m 3 (x — a$] = 0, y = 2[m 1 (y-y 1 )+m 2 (y--y 2 ) + m ;i (y-y 3 )'] = 0. FUNCTIONS OF SEVERAL VARIABLES 337 But these equations determine the centre of gravity of the particles and are satisfied by no other point. Hence the centre of gravity is the point about which the moment of inertia is least. The result is in accordance with the general theorem of Chap. IX, § 15, and it holds for any system of particles whatever. Auxiliary Variables. As in the case of functions of a single variable, so here it frequently happens that it is best to express the quantity to be made a maximum or a minimum in terms of more variables than are necessary, one or more relations existing between these variables. The student must, therefore, in all cases begin by considering how many independent varia- bles there are, and then write down all the relations between the letters that enter; and he must make up his mind as to what letters he will take as independent variables before he begins to differentiate. Example 2. What is the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid : We assume that the faces are to be parallel to the coordinate planes and thus obtain for the volume : V= 8 xyz. But x, y, z cannot all be chosen at pleasure. They are con- nected by the relation (5). So the number of independent variables is here two, and we may take them as x and y. We have, then : £-«»(■+■©-*• From (5) we obtai n : 338 CALCULUS dz __ _ c 2 x dz _ c 2 y dx a?z' dy b 2 z Now neither x = nor y = can lead to a solution, and the only remaining possibility is that ^ _ y 2 _ z 2 a 2 ~6 2 ~c 2 ' Thus the parallelopiped whose vertices lie at the intersections of these lines with the ellipsoid, i.e. on the diagonals of the circumscribed parallelopiped x = ± a, y—±b, z = ± c, is the one required,* and its volume is pr_ Sabc^ _ 3V3* EXERCISES 1. Required the parallelopiped of given volume and mini- mum surface. Ans. A cube. 2. Required the parallelopiped of given surface and maxi- mum volume. A?is. A cube. 3. A tank in the form of a rectangular parallelopiped, open at the top, is to be built, and it is to hold a given amount of water. Find what proportions it should have, in order that the cost of lining it may be as small as possible. How many independent variables are there in this problem ? Ans. Length and breadth each double the depth. * The reasoning, given at length, is as follows. V is a continuous positive function of x and y at all such points of the quadrant of the ellipse «2 x ft* ' for which x > 0, y > 0, and it vanishes on the boundary of this region. Hence it must have at least one maximum inside. But we find only one point, as = a/V3, y = b/y/B at which V can possibly be a maximum. Hence, etc. FUNCTIONS OF SEVERAL VARIABLES 339 4. Find the shortest distance between the lines y = 2x, | y = 3x + 7 f z=5x, | z — x. 5. Show without using the calculus that the function x 4 + y* + 4:X — 32 2/ — 7 has a minimum. Suggestion. Use polar coordinates. 6. Find the minimum in the preceding problem. 7. A hundred tenement houses of given cubical content are to be built in a factory town. They are to have a rectangular ground plan and a gable roof. Find the dimensions for which the area of walls and roof will be least.* 8. A torpedo in the form of a cylinder with equal conical ends is to be made out of boiler plates and is just to float when loaded. The displacement of the torpedo being given, what must be its proportions, that it may carry the greatest weight of dynamite ? Ans. The length of the torpedo must be three times the length of the cylindrical portion, and the diameter must be V5 times the length of the cylindrical portion. 9. Find the point so situated that the sum of its distances from the three vertices of an acute-angled triangle is a mini- mum. Ans. The lines joining the point with the vertices make angles of 120° with one another, f 10. Find the most economical dimensions for a powder house of given cubical content, if it is built in the form of a cylinder and the roof is a cone. * The problem is identical with that of finding the best shape for a wall-tent. t For a complete discussion of the problem for any triangle see Goursat- Hedrick, Mathematical Analysis, vol. 1, § 62. 340 CALCULUS 11. Find approximately the most economical dimensions for a two-gallon milk can. Assume the upper part of the can to be a complete cone. 4. Test by the Derivatives of the Second Order. We proceed to deduce a sufficient condition for a maximum or a minimum in terms of the derivatives of the second order. Suppose the necessary conditions (4) are fulfilled at (x , y ). Then from (2) we get : (6) f(x + h, y -f- k) -f(x , y ) = i (Ah 2 + 2Bhk + Ck 2 ), where A =f x * (x + Oh, y + Bk), B =f xy (x + Oh, y + $k), C=f y *(x + 6h,y + dk), and for a minimum the difference (6) must be positive for all points x=.x + h, y = y -{-k near (x , y ) except for this one point, where it vanishes. Definite Quadratic Forms. A homogeneous polynomial of the second degree in any number of variables is called a quad- ratic form,* and is said to be definite if it vanishes only when all the variables vanish. Thus 7i 2 + fc 2 , 2h 2 + 3k 2 + 5l 2 are examples of definite quadratic forms in two and three variables respectively ; h 2 , Sh 2 + Ihk + 2k 2 = (3h + k)(h + 2k), regarded as quadratic forms in two variables, are not definite. A definite quadratic form never changes sign. Theorem. In order that U=Ah 2 + 2Bhk+Clc i , *For some purposes it is desirable to define an algebraic form merely as a polynomial. But we are concerned here only with homogeneous poly- nomials. Moreover, we exclude the case that all the coefficients vanish. FUNCTIONS OF SEVERAL VARIABLES 341 where A, B, C are independent of h and k, be a definite form, it is necessary and sufficient that (7) B 2 -AC<0. That this condition is sufficient is at once evident. For, if it is fulfilled, surely neither A nor C can vanish, and we can write : U = -[(Ah + Bk) 2 +(AC- £ 2 )& 2 ]. A. Hence U can vanish only when Ah + Bk=0 and fc = 0, i.e. only when h = k = 0, q. e. d. We leave the proof that the condition is necessary to the student. When the condition (7) is fulfilled, A and C necessarily have the same sign, and this is the sign of U. Corollary. If A, B, C depend on h and k in any manner whatever, and if, for a pair of values (h, k) not both zero, the con- dition (7) is fulfilled, then for these values U has the same sign as A and C. Application to Maxima and Minima. Returning now to equa- tions (6), let us suppose that ^ 8) \dx~d^)~dx^df <0 at (x , y ) and that these derivatives are continuous in the vicinity of this point. Then the relation (8) will hold for all points near (x , y ) and furthermore, for such points, both — and — -^ will preserve the sign they have at (# , y )- Hence dx 2 dy 2 the right-hand side of (6) will vanish only at (x , y ), and at other points in the neighborhood will have the sign common to these latter derivatives. We are thus led to the following : 342 CALCULUS Sufficient Condition for a Maximum or a Minimum. If at the point (x , y ) -v tiu A du (a) — = — = (&> dx d 2 u\ dy /j^Y- — — <0 \dxdy) dx 2 dy 2 and if the derivatives of the second order are continuous near (x , y ), then u will have a maximum at (x , y ) if d 2 u and a minimum there if dx 2 2 variables, the method of procedure is similar. First, the algebraic theorem about quadratic forms has to be generalized. Thus for three variables, (9) U— a u x^ + a 22 x 2 2 + a^x z 2 + 2a 12 x 1 x 2 + 2a 13 x x x 3 + 2a 23 x 2 x 3 , and the necessary and sufficient condition that U be a positive definite quadratic form is that a n a 12 a 13 (10) On>0, a n a^ a 21 a 22 >o, G&21 ^22 ^23 a 31 a 32 a.33 >o, where a v = a jt . This form of statement suggests the general- ization for n = n. If U is to be a negative definite quadratic form, the first, third, fifth, etc. inequality signs in (10) must be reversed. For a proof by Gibbs, arranged by Saurel, cf . the Annals of Mathe- matics, ser. 2, vol. 4 (1902-03), p. 62. FUNCTIONS OF SEVERAL VARIABLES 343 The case of implicit functions, treated by Lagrange's multi- pliers, is given in Goursat-Hedrick, Mathematical Analysis, vol. 1, § 61. EXERCISES 1. Show that the surface z = xy has neither a maximum nor a minimum at the origin. 2. Test the function x? + 3^ _ 2x y -f 5y 2 - 4^ for maxima and minima. 3. Determine the maxima and minima of the surface x? + 2 y 2 + 3 z 2 - 2 xy - 2yz = 2. CHAPTER XVII ENVELOPES 1. Envelope of a Family of Curves. Consider a family of circles, of equal radii, whose centres all lie on a right line : (1) (x-ay + tf^l, where the parameter a runs through all values. The lines (2) 2/ = l and y — ~ 1 are touched by all the curves of this family. Again, let a rod slide with one end on the floor and the other touching a vertical wall, the rod always remaining in the same vertical plane. It is clear that the rod in its successive positions is always tangent to a certain curve. This curve, like the lines (2) in the preceding Fig. 86 example, is called the envelope of the family of curves. Turning now to the general case, we see that the family of curves (3) f(x,y, a ) = may have one or more curves to which, as a varies, the succes- sive members of the family are tangent. When this is so, two curves of the family corresponding to values of a differing but slightly from each other : (4) fix, y, « ) = 0, fix, y, a + A«) = 0, 344 ENVELOPES 345 will usually intersect near the points of contact of these curves with the envelope, as is illustrated in the above examples. So if we determine the limiting position of this point P of inter- section of the curves (4), we shall obtain a point of the enve- lope. Now a third curve through P is the following : (5) =f(x, y,ao + Aa) —f(x, y, «b) = Aa/ a (x 9 y> Oq + 0Aa). For, the coordinates of P satisfy the equation of this curve. Hence, allowing Aa to approach 0, we get* (6) /* 0,2/, 0=0. Thus the coordinates of a point of the envelope, when one exists, are seen to satisfy the simultaneous equations : f /(®>y>a)=o, Conversely, the locus (7) will be tangent to each curve (3) provided that df/dx, df/dy do not both vanish along this locus. To prove this, observe that the slope of a curve of the family (3) is given by the equation : (8) ^ + ^ = . dx dy dx In order to find the slope of the envelope, we may think of equations (7) as solved for x and y : (9) x=(a), 2/ = a) =f a (x, y, «o) = 0. 346 CALCULUS Then the slope of the envelope is dy = ^'(a) i dx \a) Now take the total differential of / (x, y, a) : df= d J-dx + d 4-dy + d if-da. ox cy Ca If x and y satisfy (9), then df=0, dx = ' (a) da, dy = if/' (a) da f and -^ = 0. Hence da (10) 0= d l dx + lf d y ox cy or dx cy '(a) 0. Thns (10) gives the same slope that (8) does, and the envelope is tangent to the family. Example 1. Applying the formulas (7) to the family of circles (1) we get : |£ == _2(.t-«) = 0. ca The elimination of a between this equation and (1) gives f or 2/ = l and y — — l. Example 2. To find the envelope of the family of ellipses whose axes coincide and whose areas are constant. Here, Fig. 87 (a) (P) "1 "t" Ti ~" > a 2 6 2 Trab = k. It is more convenient to retain both param- eters, rather than to eliminate, but we must be careful to remember that only one is inde- pendent. If we choose a as that one, a = a, and differentiate with respect to a, we have ; ENVELOPES 347 a 3 b 3 da \ da) and hence Between (a), (6), and (c) we can eliminate a and 6 and thus get a single equation in # and y, which will be the equation of the envelope. To do this, solve (a) and (c) for a 2 and 6 2 , thus getting a 2 = 2aj*, & 2 = 22/ 2 , and then substitute the values of a and 6 from these equations in (6): ± 2-rrXy = K, a pair of equilateral hyperbolas. The equations x= ± a^/2, y = ± bV2, combined with (b), give the coordinates of the points of the envelope in which the particular ellipse corresponding to that pair of values of a and b is tangent to it. This remark applies generally whenever the coordinates x and y of a point of the envelope are obtained as functions of a. EXERCISES In each of the following questions draw a rough figure to indi- cate the curves of the family and the envelope. 1. Find the envelope of the family of parabolas : y 2 = Sax — a 3 . 2. Circles are drawn on the double ordinates of a pa- rabola as diameters. Show that their envelope is an equal parabola. 3. Show that the envelope of all ellipses having coincident axes, the straight line joining the extremities of the axes being of constant length, is a square. 348 CALCULUS 4. Find the envelope of straight lines drawn perpendicular to the normals of a parabola at the points where they cut the axis. 5. Show that the envelope of the lines in the second exam- ple of § 1, p. 344, is an arc of a four-cusped hypocycloid. 6. The legs of a variable right triangle lie along two fixed lines. If the area of the triangle remains constant, find the envelope of the hypothenuse. 7. Find the envelope of a circle which is always tangent to the axis of x and always has its centre on the parabola y = x 2 . 8. What is the envelope of all the chords of a circle which are of a given length ? 9. Find the envelope of the family of circles which pass through the origin and have their centres on the hyperbola xy = l. 10. A straight line moves in such a way that the sum of its intercepts on two rectangular axes is constant. Find its envelope. Draw an accurate figure. 11. The streams of water in a fountain issue from the nozzle, which is small, in all directions, but with the same velocity, v Q . Show that the form of the fountain is approxi- mately a paraboloid of revolution. 2. Envelope of Tangents and Normals. Any curve may be regarded as the envelope of its tangents. Thus the equation of the tangent to the parabola (ii) at the point (x , y ) is y-yo="-(?-x ) or (12) y 2 — 2ma y = — (x- mx 2/0 ' 2/o 2 ENVELOPES 349 Hence the envelope of the lines (12), where y is regarded as a parameter, must be the parabola (11), and the student can readily assure himself that this is the case. The evolute of a curve was defined as the locus of the centres of curvature, and it was shown that the normal to the curve is tangent to the evolute. Hence the evolute is the envelope of the normals, and thus we have a new method for determining the evolute. For example, the equation of the normal to the parabola at the point (x , y ) is x-x + 2x (y-y )=0 or x-\-2x y — x — 2# 3 = 0, and we get at once as the envelope of this family of lines : y = Sx Q 2 + J, x = — 4a? 3 , or &-«?-«*- EXERCISES 1. Obtain the equation of the evolute of the ellipse : x == a cos <£, y = b sin , as the envelope of its normals. 2. Obtain the evolute of the cycloid : x = a(0 — sin 0), y = a(l — cos0). 3. Obtain the coordinates (a^, ?/i) of any point on the en- velope of the normals to the curve y =f(x): v — a*, + f («b) (y — y ) = 0, and show that the result agrees with the formulas of Chap. VII, § 3. 350 CALCULUS Fig. 88 3. Caustics. When rays of light that are nearly parallel fall on the concave side of a napkin ring or a water glass, a portion of the table cloth is illuminated. Let us determine the equation of the boundary. Suppose we have a narrow semicircular band, on the polished concave side of which a bundle of parallel rays fall. The rays are reflected at the same angle with the normal as the angle of inci- dence, and so we wish to find the envelope of the reflected rays. Take the radius of the band as 1. Then the equa- tion of the reflected ray is (13) y - sin = tan 2 (x - cos 0). To get the envelope of the family, we differentiate with respect to : — cos = 2sec 2 2 (x — cos 0) + tan 2 sin 0, 2(y) =f(x', y) (x' } constant) between the ordinates corresponding to the abscissas y=Y and y = Y x . Hence -/ f(x', y) dy. Dropping the accent, which has now served its purpose, we have : (3) A(x)=Jf(x,y)dy, where we must remember that x is constant, y being the vari- able of integration, and that F and Y x are functions of x. It remains only to integrate A with respect to x between the limits x = a and x = b, where a is the smallest abscissa 354 CALCULUS that any point in S has, and b is the largest. We thus obtain : h -J A (x) dx. This last integral is commonly written in either of the forms : * b Yi b Y t I dx I f(x, y) dy or / / f(x, y) dy dx. a to a Yq It is called the iterated integral of f(x, y) (not the double integral; the latter will be explained later), since it is the result of two ordinary integrations performed in succession. Instead of integrating first with regard to y and then with regard to x, we might have reversed the order, integrating first with regard to x. We should thus obtain the formula : X t V = fdyjf(x, y)dx. For example, let us compute the volume cut off from the paraboloid : ^ x 2 y 2 z = l 2- 4 9 by the (x, 2/)-plane. Since the surface is obviously symmetric with respect both to the (x, z) and the (y, z) planes, it is suffi- cient to compute the part of the volume that lies in the first octant, and then multiply the result by 4. To get A we have * Another form sometimes employed is to be avoided, namely : b Y t §{[f(x,y)dxdy. viation The second form given in the text is to be thought of as an abbre for b r, J|j/(z, 2/)#}-!!>{'-t"--S}>-. = i(4-z' 2 )V4-o;' 2 . Hence, dropping the accent, we get : A = $(±-x 2 )i Finally, integrating A from the smallest x in S to the largest, we have (see Tables, No. 137) : 2 1 C(±-x 2 )§dx = iL(4-^ + 6«V4=^+24sin- 1 |~| 2 = ^, 16|_ 2J 4 and so the total volume is 37r = 9.42. 356 CALCULUS EXERCISES 1. A round hole of radius unity is bored through the solid just considered, the axis of the hole being the axis of z. Find the volume removed. 2. Compute the volume of a cylindrical column standing on the area common to the two parabolas x = y 2 , y = x> as base and cut off by the surface z = 12 + y-x 2 . 3. Work each of the foregoing examples, integrating first with regard to x and then with regard to y. 3. Continuation. Second Method. Another way of finding the above volume is as follows. Divide the region S up into small pieces, called elements of area, of arbitrary shape, and denote the area of any one of them by b.S k . Let (x k , y k ) be an arbitrary point of the kth. element. Construct a cylinder on this element as base and of height f(x k , y k ) ; see Fig. 102. The volume of this column is Consider now the totality of such columns. They form a solid whose volume, (4) 2/(*»y,)AS„ differs only slightly from the volume T^we wish to compute. As n grows larger and larger, the maximum diameter of each of the elementary areas approaching as its limit, it is clear that the limit of (4) is V: (5) V=limV f(x k ,y k )AS k . This is the second expression for the volume we set out to obtain. DOUBLE INTEGRALS 357 Z I Fig. 95 We remark that it is not important that the elementary- areas just fill out the region S. Thus we might divide the plane by parallels to the coordinate axes into rectangles whose sides are of length Ax and Ay, and then take as the elementary areas (a) all the rectangles that lie wholly within S j or (6) all those just mentioned and in addition such as contain at least one point of the boundary of S in their interior or on their boundary ; or (c) any set intermediate between (a) and (&). In each case the sum (4) would clearly have as its limit the volume V. 4. The Fundamental Theorem of the Integral Calculus. Just as in Chap. IX, § 2, we equated the two expressions for the area under a curve to each other and thus obtained an analyti- cal theorem regarding limits, so here we equate the two expres- sions just found for the volume under a surface and thereby deduce a corresponding theorem for functions of two inde- pendent variables. Fundamental Theorem of the Integral Calculus. Let f(x, y) be a continuous function of x and y throughout a region S of the (x, y)-plane. Divide this region up into n pieces of area ASq, ASu •••, A$ n _! and form the sum: f(x„ y ) ASo+ZO-!, ft) A£i+ '•• +/ + V 2 + VV' +1"*). EXERCISES 1. Determine by double integration the moment of inertia of a right triangle of constant density about the vertex of the right angle. A M(a? + b 2 ) 6 2. Compute the moment of inertia about the focus of the segment of a parabola cut off by the latus rectum. 3. Show that the moment of inertia of a lamina about the axis of y is „ „ 1= J I p x 2 dS. 4. Find the moment of inertia about the axis of y of a uniform lamina bounded by the parabola y 2 = 4ax, the line x + y =Sa, and the axis of x. Work the problem both ways, integrating first with regard to x, then with regard to y\ and then in the opposite order. . » 46 pa 4 362 CALCULUS 6. Theorems of Pappus. Theorem I. If a closed curve rotate about an external axis lying in its plane, the volume of the ring thus generated is the same as that of a cylinder whose base is the region S enclosed by the curve and whose altitude is the distance through which the centre of gravity of S has travelled: (10) . V=27rh-A, where h denotes the distance of the centre of gravity of S from the axis, and A, the area of S. We will confine ourselves to the case that the boundary curve is met at most in two points by a parallel to the axis of rotation, which we will take as the axis of ordinates. Divide the area into strips of breadth Ax by parallels to the axis of y, and approximate to the volume generated by the fcth strip by means of the volume generated by a rectangle with the left-hand boundary of this strip for one of its sides and with base Ax* This latter volume can be computed at once as the difference between two cylinders of revolution, and is *4 + i(y'!e-y f k) - WW-y'k) =27rxM-yd ^ +»(j£-i© ^% y'z=(x) being the equation of the lower boundary, and y" =f(x) that of the upper one. Hence F=limV \27rx k (y:-y' k )Ax + ' n '(y:-y' k )Ax i l This last expression can be simplified by DuhameFs Theorem, and thus n-l h F=lim y\2 7r x k (yZ-y k )Ax = 2TT fx(y"-y')dx. a ■ Recalling the result of Ex. 4, p. 174, we see that the value of this integral is xA = hA, and this completes the proof. If the curve rotates only through an angle © instead of completely round the axis, we have merely to replace 2ir by ©. * The student should draw the requisite figure. DOUBLE INTEGRALS 363 Finally, the form of the proof is somewhat simplified by- means of double integrals, the above restriction on the boundary, as well as the use of Duhamel's Theorem, being then unneces- sary. We have at once : C CxdS. F=lim^27rxA^=27r C CxdS, ^ = ~ § ~A Theorem II. If a plane curve, closed or not closed, rotate about an axis not cutting it and lying in its plane, the area of the surface thus generated is the same as that part of the cylindrical surface having the given curve as generatrix, which lies between two parallel planes whose distance apart is the distance traversed by the centre of gravity of the given curve : S = 2tt7i'1 or ®h-L The proof is similar to that of the first theorem, and is left as an exercise for the student. 7. Polar Coordinates. We have computed the volume V under the surface z=f(x, y) by iterated integration, using cartesian coordinates. Let us now compute the same volume, using polar coordinates. To do this we ^ divide the solid up into thin wedge- shaped slabs (the slab not extending in general clear to the edge of the wedge) by means of n equally spaced planes through the axis of z : 6 = = a, B x , •••, n = /?, and approximate to the volume of the A;-th slab, AV k , as follows. Let A k be the area of the section of the plane == k with the solid, and let this section rotate about the axis of z through the angle A0. Then, by the first theorem of Pappus, § 6, the volume generated is A0 • h k A k , and the sum of such volumes, k=0 364 CALCULUS is a good approximation for V. In fact, when we visualize the totality of these pieces, we see that the volume of the solid thus obtained approaches Fas its limit, when 71 = 00. Hence 8 (11) V= Hm V h k A k A0= ChAdB. — JTo s 7 Furthermore, let us consider the product hA corresponding to the cross-section made by an arbitrary plane 6 = 6'. Writing the equation of the surface in the form z=f(;x,y) = F(r,B) and recalling the general formula for the centre of gravity : b I xydx we have here to set x—r, x — h, y = z = F (r, &), a = r' } b = r", and we thus obtain : r" hA= frF(r, 6') dr. r' Substituting this last expression in (11), we get the final formula : i F= d6 rF(r, 6)dr, and hence the Theorem : 6 r" (12) C Cf(t, 6)dS= jd$ frF(r, 6) dr. DOUBLE INTEGRALS 365 The first integration is performed on the supposition that 6 is held fast and that r varies from the smallest value r', which it has in JS corresponding to the given value of to the largest value, r". Fig. 100 TJie Inverse Order of Integration. If instead of using the planes $ = , $ lt --,0 n we had divided the solid up by the cylin- ders r-. result : r« = a, r. * n = b, we should have been led to the U V (13) f fF(r,8)dS= fdr frF(r, 0)d$. \&*~-._ / Here, the first integration is performed on the supposition that r is held fast and that varies from the smallest value, $', which it has in S correspond- ing to the given value of r to the largest value, 0". Fig. 101 ' Example. To find the moment of inertia of a uniform circu- lar disc about its centre. Here 2ir a = P f fr'dS^p fdO fr 3 dr = P - and hence /= Ma 2 /2. This problem we have solved before by single integration. The solution by double integration is simpler in form, though in substance the two solutions are closely related. EXERCISES 1. The density of a circular disc is proportional to the dis- tance from the centre. Find the radius of gyration of the disc about its centre. Ans. aV|. 366 CALCULUS 2. Determine the moment of inertia about the focus of the segment of the parabola : r= w 1 — cos bounded by the latus rectum. 3. The density of a square lamina is proportional to the distance from one corner. Find its moment of inertia about this corner. 4. Find the moment of inertia about the origin of the part of the first quadrant bounded by two successive coils of the equiangular spiral r — e 9 , the inner boundary going through the point 6 = 0, r = 1. 5. Find the moment of inertia of the lemniscate : r 2 = a 2 cos 2 6, about the point r = 0. 6. Show that the abscissa of the centre or* gravity of any plane area is given by the formula: SJ> pxdS M 7. Find the centre of gravity of the lemniscate of question 5. 8. Show that the area of any plane region S is expressed by the integrals : A= I fdxdy= f IrdrdB. s s 9. Find the area bounded by the curve = sin r and the portion of the axis of x between the origin and the point x = 7r. Ans. it. DOUBLE INTEGRALS 367 8. Areas of Surfaces. We have determined the area under a plane curve and the lateral area of a surface of revolution by means of simple integrals. The general problem of finding the area of any curved surface is solved by double integration. Let the equation of the surface be ***/(*, y) and let the projection on the x, y plane of the part @ of this surface whose area A is to be computed, be the region S. Divide 8 up into elementary areas and erect on the perimeter of each as generatrix a cylindrical surface. By means of these cylinders the surface @ is divided into elementary pieces, of area &A k , (k = 0, 1, •••, Ti — 1), and we next consider how we may approximate to these partial areas. Evidently this may be done by constructing the tangent plane at a point (x k , y k , z k ) of the Avfch elementary area and computing the area cut out of this plane by the cylinder in question. Now the orthogonal cross- section of this cylinder is of area b£ ki and hence the oblique section will have the area A/^secy,, where y k is the angle between the planes, or between their normals. The desired approximation is thus seen to be n — i ^ &S k sec y k , Fig. 102 and consequently A is equal to the limit of this sum, or * It is a fundamental principle of elementary geometry to refer all geo- metrical truth back directly to the definitions and axioms. What are the axioms on which this formula depends ? The answer is : The formula itself is an axiom. The justification for this axiom is the same as for any other physical law, namely, that the physical science, here geometry, built on it is in accord with experience. 368 CALCULUS (14) A = I j sec y dS. The angle y is the angle between the normal to the surface and the axis of z. Hence by Chap. XV, § 1 : (15) dx 2 dy 2 If the equation of the surface is written in the form F(x,y,z) = 0, we have (16) sec ^IH!)* + © dF cz Example. Two equal cylinders of revolution are tangent to each other externally along a diameter of a sphere, whose radius is double that of the cylinders. Find the area of the surface of the sphere interior to the cylinders. It is sufficient to compute the area in the first octant and multiply the result by 8. We have to extend the integral (14) over the region S indicated in Fig. 104. Here, x 2 -+- y 2 + z 2 — a 2 , and by (16) sec v = - = Va 2 — r 2 r 2 = x 2 + y 2 . Fig. 103 Since the integrand, sec y, depends in £ * a simple way on r, it will probably be well to use polar coordinates in the iterated integral. We have, then: M-I/ 1 -'-- /-/^S DOUBLE INTEGRALS 369 J a cos 9 * ardr = — a Va 2 — r 2 ^/cf — r 2 a cos & = a 2 (l-sin0), 2 '• I ^ = a " T(l - sin 0) d£ = a 2 (| -1 V yl = 4 7ra 2 -8a 2 . Fia. 104 Objection may be raised to the foregoing solution on the ground that the integrand, sec y = a/ Va 2 — r 2 , does not remain finite throughout S, but becomes infinite at the point = 0, r = a. We may avoid this difficulty by computing first only so much of the area as lies over the angle a ;< and its limit, when a approaches 0, is EXERCISES 1. A cylinder is constructed on a single loop of the curve r = a cos n6 as generatrix, its elements being perpendicular to the plane of this curve. Determine the area of the portion of the surface of the sphere a? + y 2 + z 2 = 2 az which the cylinder intercepts. j 2 (tt — 2) a 2 2. Compute the moment of inertia about the axis of z of the surface whose area was determined above in the text. 3. A square hole is cut through a sphere, the axis of the hole 2b 370 CALCULUS coinciding with a diameter of the sphere. Find the area of the .surface removed. . Aa , . _* b Q 2 . . b 2 Ans. loaosm l — — 8 or sin 1 — -. Va 2 -6 2 a - & 2 4. Determine the area of the surface z = xy included within the cylinder X 2 -f- y 2 = a 2 . 5. M cylindrical surface is erected on the curve r = as generatrix, the elements being perpendicular to the plane of this curv«. Find the area of the portion of the surface z — xy which is bounded by the y, z plane and so much of the cylindri- cal surface as corresponds to ^ 6 <^ tt/2. 9. Cylindrical Surfaces. If the surface @ is a cylinder, the area can be expressed explicitly as a simple integral. Let the elements of the cylinder be parallel to the axis of y. The equa- tion of the surface then becomes : Hence A = f f secydS = I dx I VI +f(x) 2 dy, S ay 1 b (i7) a = fvr+TW (y" - y') dx - EXERCISES 1. Two cylinders of revolution, of equal radii, intersect, their axes cutting each other at right angles. Show that the total area of the surface of the solid included within these cylinders is 16 a 2 . 2. Obtain formula (17) directly, without the use of double integrals. DOUBLE INTEGRALS 371 3. Write out formula (17) when the elements of the cylinder are perpendicular (a) to the x, y plane ; (b) to the y, z plane. 4. Show that the lateral area of that part of either of the cylinders discussed in the example of § 8 which is contained In the sphere is 4 a 2 . 5. The area of a region S of the x, y plane may be written in the form : A= f CdS = f(y"-y')dx = f(x"-x')dy. S a a By means of the last formula compute the area of the region common to the circle and the parabola : ar J + 2/ 2 = 16a 2 , 2/ 2 = 6 ax. 6. Deduce from formula (14) the formula of Chap. IX, § 8, for the area of a surface of revolution : A = 2tt Cy^/l+f'ixydx. 10. Analytical Proof of the Fundamental Theorem. Carte- sian Coordinates. In the sum : (is) SJ/to.y-)^*, *=0 whose limit is the double integral (19) ff fd8 > :g: x t x in Fig. 105 we may choose as elementary areas rectangles with sides Ax, Ay, thus making AS k = AxAy, and then add all those terms together which correspond to rectangles lying in a column parallel to the axis of y. This partial sum can be represented as follows : 372 CALCULUS where we have assigned new indices, i and j, to the coordinates of the point (x k} y k ), and where furthermore we have chosen the points (x k , y k ) of this column so that they all have the same abscissa, x ( . If, now, holding x i and Ax fast, we allow q to increase with- out limit, Ay approaching as its limit, we have n H (20) Ax lim § f(x t> y 3 ) Ay = Ax Cf(x i9 y)dy. ?=oo i=b J, Next, we add all the limits of these columns together: -i ^ ^Ax J f(x i ,y)dy, y'i and allow p to increase without limit, Aaj approaching 0. This gives y'i b >/" lim^ Ax j f(x i ,y)dy= f dx jf(x,y)dy, Pmm i= ° y'i a y ' i.e. the iterated integral of the Fundamental Theorem. This method of deduction is less rigorous than the former one, for we have not proven that we get the same result when we take the limit by columns and then take the limit of the sum of the columns, as when we allow all the AS k s to approach simultaneously in the manner prescribed in the definition of the double integral.* It is nevertheless useful as giving us * For a complete analytical treatment of the subject of this paragraph along the lines here indicated, which in point of elegance and rigor leaves nothing to be desired, see Goursat-Hedrick, Mathematical Analysis, Chap. VI. DOUBLE INTEGRALS 373 additional insight into the structure of the iterated integral, for it enables us to think of the first integration as correspond- ing to a summation of the elements in (18) by columns, and of ^che second integration as corresponding to the summation of these columns. Moreover, when we come to polar coordinates in the next paragraph, it helps to explain and make evident the limits of integration. 11. Continuation; Polar Coordinates. Let the region S be divided up into elementary areas by the circles r = r if r »+i — ?*» = Ar, and the straight lines = j} 6 j+1 — i = A0. Then AS k = r k kr A0 + \ Ar 2 A0, and hence, in taking the limit of the sum (18), &S k may, by DuhamePs Theorem, be replaced by r*ArA0. Writing f(x,y) = F(r,0) we have, therefore, f ffdS = lim V F(r k , k ) r k Ar Ad. In order to evaluate this latter limit, we may replace (r k , 6 k ) by (r t , Oj) and, holding 6 i fast, add together those terms that correspond to elementary areas lying in the angle between the rays 9 =0j and = J+1 , thus getting : AO^Ffa, Oj)r { Ar. The limit of this sum, as p = oo, is Ad I F(r, 6j)rdr. 5 Next, add all the limits thus obtained for the successive elementary angles together and take the limit of this sum. We thus get Fig. 106 374 CALCULUS lim ^ AS Cf(v, 6j)rdr= CdO /V(r, 0)rdr, i.e. the first iterated integral, (12), of § 7. If on the other hand we hold r t fast and add the terms that correspond to elementary areas lying in the circular ring bounded by the radii r ■— r< and r = r i+1) we get and the limit of this sum, when q = oo, is Ar /V(r 0) rd<5, where r denotes the distance of a variable point P of the sur- face from a fixed point of space and is the angle that the outer normal of the surface at P makes with the line OP produced. EXERCISES 1. Find the volume cut out of the first octant by the cylinders z = 1 — x 2 , x = l — y 2 . 2. Compute the value of the integral : e x2+ **'dS, Ans. |f. //• s extended over the interior of the circle x* + y 2 = l. Ans. 5.40. 3. Evaluate //' (x*-3ay)dS, where S is a square with its vertices on the coordinate axes, the length of its diagonal being 2a. Ans. \a A . 4. Express as an iterated integral in polar coordinates the double integral ' r fdS, > extended over a right triangle having an acute angle in the pole. Give both orders of integration. 376 CALCULUS 5. Express the iterated integral 2 2a cos 9 fdd ifrdr ~ 2 as a double integral, and state over what region the latter is extended. 6. The same for n 2 b esc 9 (a) jdoCfrdr; 2a V^ay (P) idy jfdx. 7. Change the order of integration in the following in- tegrals : i i (a) jdx Jf(x,y)dy, a y+a (b) Idy jf(x,y)dx. 8. The density of a square lamina is proportional io the distance from one corner. Determine the mass of the lamina. Arts. .765 Aa 3 . 9. Find the centre of gravity of the lamina in the pre- ceding question. A ^ -_-_ ar7V2-2+31og(l+_V2)l 8[V2 + log(l + V2)] 10. Two circles are tangent to each other internally. De- termine the moment of inertia of the region between them about the point of tangency. DOUBLE INTEGRALS 377 li. Find the attraction of a uniform circular diso on a particle situated in a line perpendicular to the plane of the disc at its centre. 12. Solve the same problem for a rectangular disc. Ans. K^tan-i ab . 13. Determine the attraction of a uniform rectangle on an exterior particle situated in a parallel to two of its sides, pass- ing through its centre. Ans. K^log lab 'h±a t b + Vjli - a) 2 + b 2 h-a & + V(7i + a) 2 +b 2 - 14. The intensity of light issuing from a point source is inversely proportional to the square of the distance from the source. Formulate as an integral the total illumination of a plane region by an arc light exterior to the plane. 15. Compute the illumination in the foregoing question on the interior of the curve r 2 = l-0\ the light being situated in the perpendicular to the plane of the curve at r = 0. Ans. 2\(l — h cot -1 h). 16. One loop of the curve r^ a 3 cos 30 is immersed in a liquid, the pole being at the surface and the initial line vertical and directed downward. Find the pressure on the surface. * wa 3 V3 8 ' 17. One loop of the lemniscate r 2 = a 2 cos20 is immersed as the loop of the curve in the preceding question. Find the centre of pressure. Ans. Distance below the surface = a V2 ( h - V \37r 4y 378 CALCULUS x 18. Formulate the volume of a solid of revolution as a double integral. 19. The curve cos0 = 3-3r + r 2 rotates about the initial line. Find the volume of the solid generated. . 23 x ~30" 20. Find the volume cut from a circular cylinder whose axis is parallel to the axis of z 9 by the x, y plane and the surface xy an az. Assume that the cylinder does not cut the coordinate axes. Ans. =**£. a 21. A cone of revolution has its vertex in the surface of a sphere, its axis coinciding with a diameter. Find the volume common to the two surfaces. Ans. f 7ra 3 (l — cos 4 a). 22. Determine the volume of an anchor ring. 23. Determine the area of the surface of an anchor ring. 24. Find the moment of inertia of an anchor ring about its axis - Ans. M@£ + »\ 25. Find the area of that part of the surface v 2 a* tann- ic which lies in the first octant below the plane z = ir/2 and within the cylinder x 2 4- y 1 — 1. 26. Obtain a formula for the centre of gravity of a curved surface of variable density. 27. Obtain a formula for the components of the attraction which a surface of constant or of variable density exerts on a particle of matter not lying in the surface. DOUBLE INTEGRALS 379 Hence show that the force with which a homogeneous piece of the surface of a sphere lying wholly in one hemisphere and symmetrical with reference to the diameter perpendicular to the base of the hemisphere attracts a particle situated at the centre of the sphere is proportional to the projection of the piece on the base. 28. Find the moment of inertia about the origin of the portion of the first quadrant bounded by the curve (x + l)(2/ + l)=4, correct to three significant figures. 29. Find the volume of a column capped by the surface z = xy, the base of the column being the portion of the first quadrant in the x, y plane which lies between two successive coils of the logarithmic spiral : r = ae . Ans. ~(e^-l) (e 2 M-l). 30. Find the abscissa of the centre of gravity of the above .column. 31. A square hole 2b on a side is bored through a cylinder of radius a, the axis of the hole intersecting the axis of the cylinder at right angles. Find the volume of the chips cut out. Ans. 4 b 2 V a 2 — ¥ -f 4 a 2 b sin" 1 -- Of- 32. A square hole 26 on a side is bored through a sphere of radius a, the axis of the hole going through the centre of the sphere. Find the volume of the chips cut out. Ans. 2-, ■a 2 - Sa^sm- 1 a -* tan- 1 , b -1 L V2(a 2 -6 2 ) « Va 2 -2& 2 J CHAPTER XIX TRIPLE INTEGRALS 1. Definition of the Triple Integral. Let a function of three independent variables,/ (x,y, z), be given, continuous throughout a region V of three dimensional space. Let this region be divided in any manner into small pieces, of vohrYne AV k , and let (x kf y k , z k ) be an arbitrary point of the ft-th piece. Form the product f(x k , y k ,z k ) AV k and add all these products together: (1) §/(*,*, %)AF* fcsmO When n is made to grow larger and larger without limit, the greatest diameter of each of the elementary volumes approaching as its limit, the sum (1) approaches a limit, and this limit is defined as the triple or volume integral of the function /throughout the region V: (2) lim^f(x k , y k , z k )AV k = JJJfdV. It is not essential that the totality of the elementary volumes should just fill out the region V. We might, for example, divide space up into small rectangular parallelopipeds, the lengths of whose edges are A#, Ay, Az, and consider such as are interior to V, or such as have at least one point of Fin their interior or on their boundary. The integral is also written as follows : 380 TRIPLE INTEGRALS 381 / / //0> y,z)dxdydz. V The proof involved in the above definition, that the sum (1) actually approaches a limit, has to be given along different lines for triple integrals, from what was possible in the case of double integrals. There, we were able to represent the sum n — 1 A = by a variable volume which obviously approached a fixed volume as its limit. Here, we should need a four dimensional space in which to represent geometrically the sum (1). It is necessary, therefore, to fall back on an analytical proof. Such a proof will be found in Goursat-Hedrick, Mathematical Analysis, Vol. 1, Chap. VII. The proofs of this and the later theorems of this chapter belong properly to a later stage of analysis. The theorems themselves, however, are easily in- telligible from their analogy with the corresponding theorems for double integrals, and it is our purpose here to state them and to explain their uses. EXERCISES 1. Show that the mass of a body, of variable density p, is P dV, and that f-///' , JII pXdV fff pXdV Jjfav m I= ffP dV > where r denotes the distance of a variable point from the axis. 382 CALCULUS 2. Formulate as a triple integral the attraction of a body- on a particle exterior to it. 2. The Iterated Integral. In order to compute the value of the volume integral denned in § 1 we introduce an iterated integral The method is that of Chap. XVIII, §§ 10, 11. Let the region V be divided up by planes parallel to the coordinate planes into rectangular parallelopipeds whose edges are of lengths Ax, Ay, Az, and let us take as our elements of volume these little solids. Then AV k = Ax Ay Az, and the sum (1) becomes n— 1 We will select from this sum the terms that correspond to elements situated in a column parallel to the axis of z and add them together, see Fig. 108 : AxAy^f(x { ,y J9 z { )Az 9 1=0 where we have assigned new indices, i, j, and I, to the co- ordinates of the point (x k , y k , z k ) and where furthermore we have chosen the points (x k , y k , z k ) of this column so thSt they all lie in the line x = x i9 y = y j . If, now, still holding x t9 y j9 Ax, and A.v fast, we allow s to increase without limit, Az approach- ing 0, we have z" «— i /» Ax Aylim Vf(x if y jf z t ) Az = Ax Ay I /(«„ y s , z)dz, z where z' is the smallest ordinate of the points of V on the line a5 = a7 »> y = Vj) and z" is the largest, — we assume for simplicity that the surface of V is met by a parallel to any one of the coordinate axes which traverses the interior of V in two points. Next, we add all the limits of these columns together : TRIPLE INTEGRALS 383 where we have set s> f(*,y>*)dz = Q(x,y), and take the limit of this sum. The region JS of the x, y plane over which this summation is extended con- sists of the projections of V/ the points of V on that Fig. 108 plane, and hence the limit of this sum is the double integral of (x, y), extended over S : b y" lim V $ ( x i > Vj) &x&y= I I <&dS = l dx I ®(x, y) dy. S ay' We are thus led to the final result : Fundamental Theorem of the Integral Calculus: w fff fdV=1 ff dS f f(x ' y ' z)dz V 8 z' b y" z" = jdx \dy jf(x,y,z)dz. Another notation for the iterated integral is as follows : f f lf(x,y,z)dzdydx. a y' z' Any other choice of the orders of integration is equally- allowable. An example or two will serve to illustrate the process. Example. Find the moment of inertia of a tetrahedron whose face angles at a vertex O are all right angles, about an edge adjacent to 0. 384 CALCULUS Take as the origin of coordinates and the three adjacent edges as the axes. Then /= p fff(*? + f) dV= pfdxfdyf(x> + y*) dz, where the limits of integration are as follows. First, the limit z' = and the limit z" = Z is the maximum ordinate in V cor- responding to an arbitrary pair of values x, y ; i.e. the ordinate of a point in the oblique face of the tetrahedron : a b c F». 109 Hen0e Z=C ' and the result of the first integration is : a by <$> (x, y) = /V + f) dz = (x 2 + 2/ 2 ) 8 [*H y+ i Next, this latter function must be integrated over the sur- face S consisting of a triangle bounded by the positive axes of x and y, and the line a o This double integral may be computed by iterated integration, the limits of integration for y being y' = and and those for x, and a. The remainder of the computation is, therefore, as follows : TRIPLE INTEGRALS 385 a r z J dx H { + y*)dz = c ^( a * + b 2 ); M(a 2 + b 2 ) 10 The student can verify the answer by slicing the tetrahe- dron up by planes parallel to the x, y plane and employing the result of Ex. 1 at the end of § 5 in Chap. XVIII. EXERCISES 1. Find the centre of gravity of the above tetrahedron. 2. Determine the moment of inertia of a rectangular paral- lelopiped about an axis passing through its centre and parallel to four of its edges. 3. A square column has for its upper base a plane inclined to the horizon at an angle of 45° and cutting off equal inter- cepts on two opposite edges. How far is the centre of gravity of the column from the axis ? * a?_ 3h' 3. Continuation ; Polar Coordinates. In space there are two systems of polar coordinates in common use, namely, spherical coordinates and cylindrical coordinates. Spherical Coordinates. Let P, with the cartesian coordinates x, y, z, be any point of space. Its spherical coordinates are defined as indicated in the figure. If we think of P as a point of a sphere with its centre at and of radius r, then is the longitude and is the colatitude of P. z We have ic = rsin <£cos#, . y — rsin sin0, z — r cos <£. Fig. 110 We propose the problem of computing the volume integral ( 5 ) |*£<| /(**< «m **) ^ Vk = jJJ fdV 2c 386 CALCULUS by means of iterated integration in spherical coordinates. For this purpose we will divide the region Tup into elementary volumes as follows. Construct (a) a set of spheres with as their common centre, r = r ii their radii increasing by Ar; (b) a set of half-planes = jf the angle between two successive planes being A0; and lastly (c) a set of cones = iy their semi-vertical angle increasing by A<£ : l+1 — t — A<£. The element of volume thus obtained is indicated in Fig. 111. The lengths of the three edges that meet at right angles at P are Ar, rA, r sin <£A0, and hence this volume A V differs from the volume of a rectangular parallel- opiped with the edges just named : (6) r 2 sin<£ArA0A<£ by an infinitesimal of higher order : AF lim = 1. r 2 sin Ar A0 A<£ It follows, then, from DuhamePs Theorem that in the limit of the sum (5) we may replace AT* by the infinitesimal (6). If we set Fig. ill f(x,y,z) = F(r,6,), we have J J j fdV= lim 2J F(r„ 6 k , k )r k 2 sin k Ar A0 A<£. Can we evaluate this last limit by iterated integration ? It is easy to see that we can. For the sum is of the type of the sum (3), and hence the method of § 2 is applicable. Fol- lowing that method, let us select, for example, those terms for which 6 and <£ have a constant value, and add them together : p-i A0A*2JF(r o $ J> *i)tf simfcAr, where $j and <£, are constant. They correspond to elementary volumes lying in a row bounded by the planes = 8j and TRIPLE INTEGRALS 387 = $ j+lf and by the cones cf> = fa and = l+1 . Now allow p to increase without limit, Ar approaching 0. This gives, as the limit of the above sum, A0A<£sin t )dr, where r ' is the distance of the nearest point of V to on the line = 0j, = l , and r", that of the farthest. We assume for simplicity that the surface of V is met by any one of the lines : 6 = const., = const., r = const., = const., r — const., — const., which traverses the interior of V, in two points. Next, we add all the limits thus obtained together where we have set r' and take the limit of this sum. If we interpret 8 and as the coordinates of a point on the surface of a sphere r = const, (say, r = 1), then the region S over which the above sum is to be extended consists of those points in which radii vectores drawn to points of V pierce the surface of this sphere. Hence the limit of this sum is the double integral of ^> (6, ), extended over S : lim2j*(^,^)AdA*= f f*(6> ) d S s /5 " r" = I dO f sin d f F(r, 6, )r>dr. We are thus led to the following result: CALCUIXS :: :lf ::rf- to#; (6) with respect I7«r ili JJJp*fim+drmd+ and /"/'/*-- w * From Duhamel's Theorem it follows, ' "J then, that in taking the limit of the sum (1), A7i may be replaced by r t Ar A0Az, and so, setting S (x,y,z) = F(r,0,z), we obtain: / / I fdV= lim ^ F(r k , k , z k )r k ArA$\z. This last limit can be computed by iterated integration in a manner precisely similar to that set forth in the case of spherical coordinates. We thus obtain : (8) fff /dV =f d2 f d6 f /rdr > together with similar formulas yielded by adopting a different order of integration. The above volume integral and the iterated integral are also written in the forms: J J ffrdrdOdz and C j ffrdrdBdz. r Example. To find the attraction of a cylindrical bar on a particle of unit mass situated in its axis. The magnitude of the attraction is evidently* * The unit of force is here taken as the gravitational unit. 390 CALCULUS SSf^r- Here r 2 = r 2 + ^ 2 , Hence cos^=- = 2tt h+l o ;* o a ^ vV + z 2 o Fig. 114 = 1- h+l a h+l fdzf *** =i- r * d * Va 2 + z 2 ' - / _ Va 2 + (ft+ If + Va 2 + ft 5 ; .'. A = 2tt P [1 + VaF+W - Va 2 + (ft + r) 2 ] . EXERCISES 1. Determine the attraction of a straight pipe on a particle situated in its axis. 2. Find the force with which a cone of revolution attracts a particle at its vertex. Ans. 2irph (1 — cos a). 3. Show that the force with which a piece of a spherical shell cut out by a cone of revolution with its vertex at the centre attracts a particle at depends, for a given cone, only on the thickness of the shell. 4. Prove the preceding theorem for any cone. 4. Line Integrals. Line integrals present themselves in such physical problems as that of finding the work done by a variable force when the point of application describes a curve. TRIPLE INTEGRALS 391 Let a plane curve C: y=f(x) or F(x,y) = 0, be given. Its coordinates can always be expressed as func- tions of the arc s, measured from an arbitrary point. Thus in the case of the circle x 2 + y 2 = a 2 we can write s . s x = a cos - , y = a sin-, a' 9 a' where s is measured from the point (a, 0). We will think of the equation of the carve C, therefore, as expressed in the form : (1) x=(s), y = $(8). Consider next a function F(s) defined at each point of the curve. It may be given as a function both of the coordinates x, y of a variable point P of the plane and of the arc s: f(x, y, s). But in the latter case P is to lie on C, and so x and y have the values given by (1), / (x, y, s) thus becoming a function of s alone : fix, y, s)=f[(s), $(8), s] = F(s). We will now divide the arc up into n equal parts by the points s = 0, s lf • • •, s n _ 1} s n = I and form the sum : fc = The limit of this sum as n becomes infinite is i /■ F(s)ds, and is called the line integral of the function F(s) or f(x, y, s) taken along C. Other notations for this integral are I fix, y, s) ds and / f(x, y, s) ds, 392 CALCULUS where (x Q , y ) and (x l} y x ) are the coordinates of the extremities of the arc C. Geometrically the line integral admits of a simple interpreta- tion. Let a cylinder be constructed on C as generatrix, its elements being perpendicular to the x, y plane, and let the values of the function F(s) be laid off along the elements of this cylinder. Then the area of the cylinder bounded by this curve and the generatrix represents the line integral in question. As an example of a line integral, suppose a point moves in a field of force. Let the magnitude of the force be g and let the force make an angle 6 with the tangent to C drawn in the direc- tion of the motion. Then the compo- nent of the force along the curve is gcos#, and the work done by the force is / (2) W= I g cos dds Fig. 115 / f A Second Form of the Line Integral. A second form in which line integrals appear is the following : I Pdx+Qdy, the meaning of the integral being this. Two functions P(x, y), Q (x, y) of the independent variables x, y are given, the curve is divided as before, and the sum n-l (3) . 2 l p ( x *> 2/*) Aa * + Q ( x k, Vk) A^.] fc = is formed, Ax k denoting the difference x k . +1 — x k , and similarly for Ay*. The limit of this sum is the limit in question. To evaluate the limit, we may write the summand in the form: (P(x>, y.) ^+ Q(x„, y k ) £»W TRIPLE INTEGRALS 393 Now v Ax lim — = cost, a«=o As lim — * = sin t, As=0 AS and hence by Duhamel's Theorem the limit of (3) and the limit of the sum n— 1 2} [P(a?*, yk)co8T k +-Q(x k , y k )sm Tk '] As 4=0 are the same. But the limit of the latter sum is f [P(a>, y) cost + Q(x 9 y) sin r] ds = /Ti^ + ^ffW where the x and 2/ in the integrands are given by (1). As an example of the second form of line integral consider again a field of force, the components of the force along the axes being denoted at each point by X, Y. Then the work done by the force when the point of application describes the curve C is (4) W= fxdx+Ydy. The relation between formulas (2) and (4) for the work be- comes clear when we consider the special case that the point of application P moves in a right line, the force not changing in magnitude or direction. One expression for the work, — that corresponding to (2), — is W=(gcos0)Z. On the other hand, the work done by the component X is (Xcos t) I = X (xj — x ), and that done by Y f Fig. 116 (Fsinr) J =^-2/0). Hence we ought to have : $lcos6=X(x 1 -x )+ Y(y l -y ). 394 CALCULUS That this is in fact a true relation is readily seen. For the component of gf along the line P describes, namely PM=% cos e, is equal to the sum of the components of X and Y, namely, PN=Xcosr and NM = Fsinr. But cost = ^— ^2, s inr = ^^. ( i Hence gcos0 = X^^+F 2l=-&, i I and thus the above relation is seen to be true. When the force changes and the path is a curve, we still have gcos0 = XcosT+rsinT=X— + F^, ds ds and hence g cos Qds = Xdx 4- Fdy. &pace Curves. Both line integrals admit of immediate ex- tension to space curves C : X=(s), y = $(s), « = «(*), the first integral giving i 1 1 //(*> y> *$ s)ds= //[>(«), if,(s) y »(«), s]ds = I F(s)ds, and the second, Prite + Qdy + Rdz. Thus in the case of a field of force we should have for the work : TT= jXdx+Ydy + Zdz. TRIPLE INTEGRALS 395 Example. Let a particle move along a given path in inter- planetary space. To find the work done on it by the earth, supposed stationary. Assume a system of cartesian axes with the origin at the centre of the earth. Then the magnitude of the attraction will be \ *=v and we shall have X = gcos« = ^ \ x r 2 r Z = gc0Sy=A.? ; r r J r 8 J r 2 \r rj Thus we see that the work done depends only on the positions of the extremities of O, not on the particular path joining the points, i.e. we have a conservative field of force. In connection with this subject we will mention the follow- ing definition. Hitherto we have defined the definite integral : j f{x) dx only for the case that a < b. If a > b, the definition is, how- ever, still valid, Ax=(b — a)/n now being negative. Hence in all cases Jf(x)dx = -Jj f(x) dx. Furthermore we agree that 396 CALCULUS a ff(x)dx = 0. From these relations we infer that b c b Cf(x)dx = Jf(x)dx + jf(x)dx, a a c no matter how a, b, and c are related to each other. We can also write : b c a Cf(x)dx+ lf(x)dx + ff(x)dx = 0. a b c EXERCISES 1. The density of a rectangular parallelopiped is propor- tional to the square of the distance from one vertex. Find its mass. Am Xabc ^ + &2 + ^ > o 2. Determine accurately the volume of the element in spherical polar coordinates, Fig. 111. 3. Find the centre of gravity of the volume in the preceding question. 4. Express the iterated integral -Va 2 -x* 2+4X+&U dx I dy I fdz x+y as a volume integral, and state throughout what region of space the latter is to be extended. 5. The same for a 4 2 a cos (f> I cos 6 dO j sin d I dr. _v 2 b cob TRIPLE INTEGRALS 397 6. Write down the five equivalent forms of the integral ;, y, z) dz, jdy jdx jf(x, obtained by changing the order of the integrations. 7. Two spheres are tangent to each other internally, and also to the x, y plane at the origin. Denoting the space included between the spheres by V, express the volume integral ///' fdV by means of iterated integrals in cartesian coordinates. 8. The temperature within a spherical shell is inversely proportional to the distance from the centre, and has the value T on the inner surface. Given that the quantity of heat required to raise any piece of the shell from one uniform temperature to another is proportional jointly to the volume of the piece and the rise in temperature, and that C units of heat are required to raise the temperature of a cubic unit of the shell by one degree, find how much heat the shell will give out in cooling to the temperature 0°. Ans. 2irCT a(b 2 — a 2 ). 9. The interior of an iron pipe is kept at 100° C. and the exterior at 15°. The length of the inner radius of the pipe is 2 cm., that of the outer radius, 3 cm. The temperature at any interior point is given by the formula : !T=alogr + A where r is the distance from the axis and the constants a, fi are to be determined from the above data. Taking the specific heat of iron as .11, and its specific gravity as 7.8, how much heat will a segment of the pipe 30 cm. long give out in cooling to 0° ? Ans. 21,000 calories. 10. Determine the attraction of a bar, of rectangular cross- section, on an exterior particle situated in its axis. CHAPTER XX APPROXIMATE COMPUTATIONS. HYPERBOLIC FUNCTIONS 1. The Problem of Numerical Computation. It frequently happens in practice that we wish to know the value of a func- tion for a special value of the independent variable or that we wish to compute a definite integral. In all such cases only a limited number of decimal places or of significant figures, as the case may be, are of interest in the result, for the data of the problem are accompanied by errors of observation or are otherwise inexact, and as soon as these errors begin to make themselves felt, we have obviously reached the limit of accu- racy for the result in hand. Hence any method that will enable us to obtain the result with the degree of accuracy above indicated yields a solution of our problem. On the other hand, rough approximate solutions of the kind we are about to take up serve as useful checks for solutions obtained by other methods. 2. Solution of Equations. Known Graphs. Example* Let it be required to solve the equation (1) cos x + \x = 0, 00, x>0, and hence the graph is always concave upward. Finally, /(l) = e-e-'-2 T % = .2ir>0, and so the equation has one and only one positive root and this root lies between and 1. It will probably be better to locate the root with somewhat greater accuracy before beginning to apply the above method. Let us compute, therefore, /(J). By the aid of the Tables, p. 121, we find : /(.5) = 1.6487 - .6065 - 1.0667 = - .0245 <0. Comparing these two values of the function : /(.5)=-.02, /(1)=.22, and remembering that the curve is concave upward, so that the root is somewhat larger than the value obtained by direct interpolation (this value corresponding to the intersection of the chord with the axis of x) we are led to choose as our first approximation a x = .6 : /(.6) = 1.8221 - .5488 - 1.2800 = - .0067, /(.6) =r 1.8221 + .5488 - 2.1333 = .2376, a 2 = .6- ~ - 0067 = .6 + .0282 = .628. 2 .2376 To get the next approximation, a 3 , we compute /(.628) = 1.8739 - .5337 - 1.3397 = .0005. 2d 402 CALCULUS Hence the value of the root to three significant figures is .628 with a possible error of a unit or two in the last place, and the value of a we set out to compute is, therefore, 15/.628 = 23.9. 4. Direct Use of the Tables. While explaining methods of solution more or less obvious geometrically, we must not over- look an immediate solution of the problem in certain cases by mere inspection of the tables. For example, the equation cos x = x has one and only one root, as we see by inspection of the graphs of y = cos x and y = x. To find this root, turn to the Tables, p. 134. There we find : RADIANS DEGREES COSINES .7418 42° 20' 42° 30' .7392 19 .7373 Hence x = .7391, corresponding to an angle of 42° 21'. The interpolation by which x was found is a neat problem in ele- mentary algebra. It is left to the student. The example of § 3 is nearly a case in point. The hyper- bolic sine and cosine are defined by the equations (cf. § 8) : sh# = e x — e~ ch# = e x + e~ x Tables of values of these functions are given on pp. 120-123 of the Tables. The problem in hand thus becomes the fol- lowing : to solve the equation shsc = -j-f cc. From the tables on p. 121 we find : APPROXIMATE COMPUTATIONS 403 X H* sha; .62 .63 .6613 .6720 107 .6605 n „ n „ 120 .6725 Hence the value of x given by direct interpolation is .626. Solve the equation : EXERCISE cot X = X. 5. Successive Approximations. The method of successive approximations is most easily understood by inspection of the graphs of the functions. There are two cases, both illustrated in the accompanying figures. If the slopes of the curves both have the same sign, let Q: F(x,y)=0 be the one that is less steep, C 2 : or 3> (x, y) = or V =f(x) x = (y) Cd /Ci 11 Jf\ ^ \ A j ' X '• Cj x 2 y \ r"^ 1 4 3 X x 2 .; c l Fig. 118 the other. Then, making the best guess we can to start with, x = x 1 , compute 2/i =/Oi) and substitute this value in the equation of C 2 , thus getting the second approximation: a 2 = + 1 cos = 15, cos <£ -f- 1 sin = 10. Hence cos 2 <£ — sin 2 <£=10 cos <£ — 15 sin . Now an expression of the form a cos cji — b sin Fig. 119 can always be written as VaT+tf( cos - Wa 2 + & 2 where cos a — — Va 2 +6 2 sin <£ ) = Va 2 + b' 2 cos (<£ -h a), sma = Va 2 + & 2 Va 2 + 6 2 In the present case, then : cos 2 = V325 cos (<£ -f a), where cos a — 10 sma = 15 V325 V325 Thus a is an angle of the first quadrant and tan a = 4, a = 56° 16'. APPROXIMATE COMPUTATIONS 405 Our problem may be formulated, then, as follows : To find the abscissa of the point of intersection of the curves : y — cos 2 , y = V325 cos ( 4- a). We know a good approximation to start with, namely : tan <£ = £, <£ = 33°44'. For this value of the slopes are given by the equations : ^.^ = -2sin2d>=-2sin67°28 f =-1.8, ■k d ' 1??.-^=- V325sin(d,-f a) =- V325 = -18. ir d Hence we have : C x : y — cos 2 ; (L : y = V325 cos (d> + a) or = cos * — % == — a. V325 Beginning with the approximation when its 2/ = 2/i: y x = V325 cos (<£ 2 + a), 2 = 32° 31'. We now repeat the process, beginning with <£ 2 = 32° 31' and find: 2/ 2 = cos 65° 02', y 2 = V325 cos (fc + a), 3 = 32° 23'. A further repetition gives <£ 4 = 32° 22', and this is the value of the root we set out to determine. 406 CALCULUS EXERCISES 1. Solve the same problem for a beam 2 ft. thick. 2. A cord 1 ft. long has one end fastened at a point 2 ft. above a rough table, and the other end is tied to a rod 2 ft. long. How far can the rod be displaced from the vertical through and still remain in equilibrium when released ? The equations on which the solution de- pends are : 2cot0 + - = cot = 2. If the coefficient of friction /x = ^, find the value of cf>. 3. A heavy ring can slide on a smooth vertical rod. To the ring is fastened a weightless cord of length 2 a, carrying an equal ring knotted at its middle point and having its further end made fast at a distance a from the rod. Find the position of equilibrium of the system. 4. Solve the example worked out in § 3 by the method of successive approximations. 5. In the example worked in the text replace cos by its value in terms of sin<£, reduce the resulting equation to the form of an algebraic equation in sin and solve the latter by Horner's Method. 6. Definite Integrals. Simpson's Rule. If we wish actually to compute the area under a curve numerically, we can make an obvious improvement on the method of inscribed rectangles by using trapezoids, as shown in Fig. 53. We begin as before by dividing the interval (a, b) into n equal parts, and we denote the length of each part by h. The area of the k-th trapezoid is i(y*+-y»+i)* and hence the approximation thus obtained is APPROXIMATE COMPUTATIONS 407 This formula is known as the Trapezoidal Rule. If the curve is concave downward, as in Fig. 53, A x is too small. Again, if we take n as an even integer and draw tangents at the points (x u y x ), (x 3 , y s ), ••• (#„_! y«_i), we get some trapezoids as shown in the figure, the area of any one being 2y k h, where k is odd. Hence ^r^r A 2 = 2h[y l + y 3 -\ hy—i] 'vlx i Vm is an approximation which is too large, and A X x = x k + h, y k+1 = a + bh + ch?', x = x k — h, y k _ x = a — bh + ch?. Hence a = y k , 2ch 2 = y k _ 1 -2 y k -f y k+1 . Thus the area under the parabolic arc is seen to have the value 408 CALCULUS iftfot-i + ^t + ft+i). Adding these areas for k = l 9 3, ••• n — 1, we get a new ap- proximation : ^3 = P[2/o + 2/ n + 2G/ 2 + 2/4+ •••2/„- 2 ) + 4(2/ 1 + 2/ 3 + — +|f«-i)]. This formula is known as Simpson's Rule. If we set u = y + y n , v = yi + Vs+ — 4-2/n-i, ™ = 2/ 2 + 2/H by— s, we have: A 1 = ^h(u + 2v + 2w), A 2 — 2hv f A 3 = ±h(u+4:V + 2w). It turns out that A 3 = %A 1 + ±A 2 . 2 Example.* Consider / — , and let n = 10. Then h = . land it = 1.5, v = 3.459 539 4, to = 2.728 174 6. Hence ^ = .693 771, A 2 = .691 908, A 3 = .693 150. The value of the integral is ( Tables, p. 109) : log 2 = .693 147. Thus A x differs from the true value by less than 7 parts in about 7000, or one tenth of one percent. A 2 differs by about 12 parts in 7000 ; while A 3 is in error by less than 3 parts in 600,000, or 1 part in 200,000. l EXERCISES 1. Compute I e x dx; taking w = 10, and compare the result with that obtained by integration. Note the tables on pp. 120, 121 of the Tables. * These figures are taken from Gibson's Elementary Treatise on the Calculus, p. 331, to which the student is referred for further examples. A more extended treatment of the subject of this paragraph will be found in Goursat-Hedrick, Mathematical Analysis, vol. 1, § 100. APPROXIMATE COMPUTATIONS 409 / dx ' 1 3. Obtain an approximate formula for the content of a cask whose bung diameter is a, head diameter, b, and length, I. Ans. ^[8a 2 + 4a&+.3& 2 ]. 4. If in the preceding question a is only slightly greater than b, the formula may be replaced by the simpler one : iral(a + 2b). 7. Amsler's Planimeter. A curve may be given graphically, as in naval architecture, when the plans of a ship are made by drawing to scale successive cross-sections. Again, take the indicator diagrams of a steam engine. A pencil or stylus is carried over a sheet of paper, tracing a curve as shown in Fig. 122. The height of the pencil above the axis of abscissas represents the pressure p of the steam on the piston, and the abscissa is proportional to the distance the piston has travelled. Hence the work done p ( \* in the direct stroke is proportional to * pdx, f> the ordinate p being given by the upper part of the curve. When the piston returns, negative work is done, and the amount is - I pdx or I pdx, Since x is proportional to the volume of steam behind the piston, we may also write the work as pdv. J 410 CALCULUS the ordinate now being given by the lower part of the curve. Hence the total work done is proportional to the algebraic sum of these two integrals, namely, the line integral I pdx or I pdv, taken round the complete boundary, i.e. the work is propor- tional to the area enclosed by the curve. In order to compute such areas one method is that of § 6, and this is the one employed in naval architecture. Another method is by means of integrating machines, integraphs, or planimeters, as they are called, and this is the one employed for measuring indicator diagrams. There are several such machines in use, one of which, Amsler's Planimeter, we will now describe. It consists of two arms, OP and PQ, jointed at P. One arm is pivoted at ; the other has a point at its end Q, and Q is made to trace out the curve whose area is sought. The theory is as follows. Consider the area swept out by the arm PQ> Give to this arm an infinitesimal displacement, its new position being P'Q'. The corresponding infinitesimal z-^ , increment of area, AA, is seen to ( yt s differ from the area PQSQ'P'P, sC^s® where SQ is congruent to the arc y^^h\C J PP' &&& makes the same angle with jjj* ays PQ, by an infinitesimal of higher •/^\/($L order. But this latter area is ob- ^^cP viously equal to FlGl23 Ih + ± I 2 Acf>, where h denotes the perpendicular distance from P'S to PQ and I is the length of PQ. Hence AA = lh + %l 2 A + c, where c is an infinitesimal of higher order. In order to measure h, a disc is attached to the arm PQ at R, the axis of the disc coinciding with that arm.* The disc can * As a matter of fact, B lies in the line QP produced. This alters nothing in the theory, the distance PR = a merely being taken negative. APPROXIMATE COMPUTATIONS 411 turn freely on its axis and the rim of the disc rests on the paper. Now suppose that the arm PQ were brought into its new position P'Q' as follows: (a) PQ is moved in its own line till P reaches the foot of the perpendicular dropped from P on its line ; (b) PQ is moved perpendicular to itself till it comes into the position PS ; (c) PQ is rotated about P' as a pivot till it comes into the final position P'Q'. It is now easy to compute the angle through which the disc has turned. During the movement (a) it does not turn at all. During (b) it turns through an angle proportional to h, h/r, where r is the radius of the disc ; and during (c) through an angle aA/r, where a denotes the length PP. The total angle thus obtained, (/i + aA)/r, will differ from the angle Aw due to the actual displacement at most by an infinitesimal of higher order, rj : « A h-\-aAd> . Aa> sx — £ £ -f- ri. r This assumption is an axiom or physical law, borne out by experience, on which the whole theory of this machine rests. If we eliminate h between the equation for AA and that for Aw, we get : A A = ZrAw + (1Z 2 - al) A<£ - lr v + c. Dividing by A<£ and allowing A<£ to approach as its limit, we obtain : D^A^lrD^+^-al) and hence dA = Ir d. The simplest case is that in which, as Q describes the closed curve in question, <£ steadily increases for one arc from <£ to <£i and steadily decreases for the remaining arc from fa to <£ . The total area swept out for the first arc is A x = ^-(wj — w ) -h (-J-Z 2 — al) (fa — ). 412 CALCULUS For the second arc, <£ is decreasing, and the area will be negative : A 2 = lr(Q - cuO + (il 2 - oZ)(^o- ^)- The area of the curve is the algebraic sum of these two areas : A = A X + A 2 = Ir (O — o>o) and hence is proportional to the angle O — o> through which the disc has turned. This angle is read off on the vernier, and the constant multiplier is known or determined for the par- ticular machine that is being used. It can be shown generally that the area of any closed curve is given by the same formula, provided comes back to its initial value, the method being merely to divide the area enclosed by the curve up into pieces, for each of which the above determination is applicable. But if the bar PQ makes a complete rotation, so that cj> changes by 2-rr, the integral of the last term in the expression for dA will not vanish, but will contribute (-J- I 2 — at) • 2 it to the result. 8. The Hyperbolic Functions. Certain functions analogous to the trigonometric functions, called the hyperbolic functions, have recently come into general use. They go back, however, to Riccati (1757) and are defined as follows : . , e x -e~ x sinn x = 2 e * + e -* cosh x = — l - — m sinha; tanh x coshx' etc. (read " hyperbolic sine of x," etc.). An abbreviated nota- tion for sinh x, cosh x, tanh x, is sh x, ch #, th x. The graphs of these functions are shown in Fig. 124. The functions satisfy the following functional relations, sh# and tho; being odd functions, ch x an even function : HYPERBOLIC FUNCTIONS 413 sh(— x)= — sh^, ch(— x) = ch#, th(— £)= — th#. Moreover: shO = 0, chO = l, thO = 0. Also : ch 2 x — sh 2 x — 1, 1 — th 2 x = sech 2 #, coth 2 a; — 1 = csch 2 a;. ?/ = th.r ttL Fig. 124 The Addition Theorems are as follows : sh (cc + ?/) = sh ic ch ?/ -f- ch ^ sh y ; ch (x + y) = ch x ch ?/ + sh x sh # ; 1 1 / , x th x 4- th y th (x + w) = — f— • / V ; 1 + thajthy From these relations follow at once : sh2# = 2sh£chx, ch 2x = ch 2 x + sh 2 a = 2 ch 2 a -1 = 1 + 2sh 2 a,\ Derivatives of the Hyperbolic Functions. The derivatives have the values: d sh x , d ch x , = ch #, = sh x, dx dthx dx = sech 2 aj, dx d coth x dx = — csch 2 #, etc. Tlie Inverse Functions. The inverse of the hyperbolic sine is called the antihyperbolic sine : y = ah~ 1 x if x = shy. Hence ««-.}(* — *-*). 414 CALCULUS Solving for e p , we get : e v = x ± vT+a?. Since e y > for all values of y, the upper sign alone is possible, and y=sh- 1 x = \og(x-\-^/l-{-x 2 ). The antihyperbolic cosine, however, is multiple-valued, as appears from a glance at its graph, obtained as usual in the case of an inverse function by rotating the graph of the direct function about the bisector of the angle made by the positive coordinate axes : ch _1 a; = log (a; ± s/x 2 — 1), x^l. The upper sign corresponds to positive values of ch" 1 **;. Also: th" 1 a; = -ilogi±^, -l C dx , _,x — = :sin 1 -. f — =sh l -; Vtf-x 2 a JVa 2 + « 2 a /* dx 1 , _,x C dx 1 .v_i» — — ^-tan" 1 -, I- — 3 = -th »-. cr + ar a a J a 2 — or a a A collection of formulas relating to the hyperbolic functions will be found in Peirce's Tables, pp. 81-83, and tables for shx and ch# are given there on pp. 119-123. HYPERBOLIC FUNCTIONS 415 Relation to the Equilateral Hyperbola. The formula: Vl — 3?dx == \ x Vl — x 2 + 1 sin" 1 expresses the area OQPA under a circle in terms of the function sin -1 # and enables us, on subtracting the area of the triangle OQP from each side of the equation, to interpret sin -1 a; as twice the area of the circular sector OP A. y y \ \ x Q Fig. 125 There is a similar interpretation for sh _1 a; with reference to the equilateral hyperbola y 2 = l + x?. J Vl + x 2 dx = %x VT+aT 2 + i log (x + VT+xV i^Vl+^ + Jsh- 1 ^. Thus we see that sh _1 # is represented by twice the area of the hyperbolic sector, OPA. To the formulas for the circle : & + y 2 = l, x = sin u, y = cos u, correspond the following formulas for the hyperbola: x — sh u, y = ch u, the parameter u being represented geometrically in each case by twice the area of one of the above sectors. 416 CALCULUS The analogy of the hyperbolic functions to the trigonometric functions is but one phase of the fact that in the domain of complex quantities the trigonometric and the exponential functions and their inverse functions, the ant itrigonome trie functions and the logarithms are closely related. We have already had occasion to mention the formula: e*' = cos + i sin , i = V — • 1. Thus sin z = 2i zi i_ a —zi cos z = — sin -1 z = - log (zi ± Vl — z 2 ), i tan i* = _log— , where 2! = # + y£ is any complex quantity . 2%e Gudermannian. Let <£ be defined as a function of x by the relation : sh# = tan<£, <£ = tan -1 sh#, —-< is called the Gudermannian of # and is denoted as follows : * <£=gdx. We have : sh# = = tan , chx = : sec <£, thx = sin , csch x = = COt , sech x = : COS <£, cotha? = csc 4>; and since e x = ch a; + sh a;, e x -. = tanfc +i) a; = log tan f - + V4 !)■ * Also called the hyperbolic amplitude and denoted by amhse. APPENDIX A. — THE EXPONENTIAL FUNCTION In Chap. II, § 8, it was shown that, when x = a > 1, (1) a n '>a n if ri>n, where n and n' are two positive or negative rational numbers. Moreover (2) a">0 for all rational values of n ; and (3) lim a M = + oo , lim a n = 0. n=+» n=— oo One further relation, which we will now prove, is important, namely: (4) lim a n == 1. M = When 0< n < 1, the curve y = x n is concave downward, for D x 2 y = n(n — l)af~ 2 <0, and so it lies below its tangent. The equation of the latter in the point (1, 1) is : y = n(x-l)+l. Hence for such values of n, the ordinate of the curve, a n , is less than the ordinate of the tangent, n (a — 1) -f 1 : 1 < a n < n (a - 1) + 1, < n< 1. 2e 417 418 CALCULUS Thus (4) is seen, at least, to be true whe« n approaches from the positive side. Similarly it is shown that, when n<0, the curve is concave upward, and l>a n >n(a — 1)4-1, n<0. Hence (4) is true when n approaches from the negative side, too, and the relation is thus established generally. We can now prove the following theorem. Theorem 1. Ifvbe any irrational number and n be allowed to approach v, passing only through rational values, then a n approaches a limit. First, let n approach v from below, n v. n=v— Here, V is any rational number greater than v. Again, let n approach v from above, n = n' > v. Then, by similar reasoning, a nf approaches a limit A 1 > a 1 , where I is any rational number less than v. (6) lim a n ' = A' > a 1 , l A' and a 1 < A, we infer that a v -a l >A'-A. Setting V = 1 4- h, we get : Q^A'-A x . No- where now V is any number >x . Similarly, lim a* = A' > a 1 , J 1. It is shown without difficulty that Theorems 1 and 2 hold when < a ?g 1. Theorem 3. 27ie relations (A) q/* (7/iap. II, § 8, ^o/d w/ien m awe? w are one or both irrational. Consider, for example, the second relation : (a m ) n = a mn . Let m approach an irrational value,, fi, as its limit. Then, since x n is a continuous function of x when n is rational, we have: lim (a w ) w = (lim a m ) tt = (a*) n . »»==/* m = /x ' On the right-hand side, lima mn = a' An , m = jA and hence (a») n = a» n . If here we allow n to approach an irrational number v as its limit, we see by Theorems 1 and 2 that («")" a^ v . THE EXPONENTIAL FUNCTION 421 The proof that (a™)" = a mv depends on Theorem 1 alone. The other relations of (A) are proven in a similar manner. We have now established rigorously all that was assumed in Chap. IV for the purpose of defining the logarithm and of differentiating the logarithm and the exponential functions. Hence we are entitled to the conclusion of that chapter that x n is continuous and has a derivative when n is irrational. We have also the material for proving the final statements of Chap. II, § 8, respecting the graph of x n . If x = a, (0 < a < 1 or a > 1) and y = b>0 are chosen arbitrarily, one and only one value of n can be found for which the curve y = x n will go through the point (a, 6), namely : b = a% w = log a 6 = ^. log a The whole subject of logarithms, exponentials, and fractional exponents can be treated with great simplicity by basing all of these functions on the logarithm, defined as the definite integral : X dx x x / Cf . a paper by Bradshaw, Annals of Mathematics, ser. 2, vol. 4 (1903), p. 51 ; or Osgood, Lehrbuch der Funktionentheorie, vol. 1, p. 487. 422 CALCULUS B.- FUNCTIONS WITHOUT DERIVATIVES In recent years much attention has been paid to discontinu- ous functions and to functions which, though continuous, still do not have a derivative. Consider, for example, the function . 1 y = sm - • x When x approaches as its limit, y oscillates between the values + 1 and — 1, and thus the function, while remaining finite, approaches no limit. It does not even approach one limit when x approaches from the positive side and another limit when x approaches from the negative side. The reader can easily plot the graph roughly. Let us now form the following function : f(x) = x sin-, x^O; x /(0) = 0. This function is continuous for all values of x, and its graph is comprised between the lines y = x and y = — x. At the point x — Oj however, it has no derivative. For, form, the difference-quotient : /(0+A*)-/(0) _ dn l Ax Ax This variable — the slope of a secant through the origin and a variable point P with the coordinates Ax and Ay = Ax sin-- Ax oscillates between + 1 and — 1, i.e. the secant OP turns to and fro, and approaches no limit whatever. Again, a function may have a first derivative, but no second derivative, as for example: (x) = x 2 sm-, a?=£0; x *(<)) = o. FUNCTIONS WITHOUT DERIVATIVES 423 The foregoing functions have a derivative, to be sure, in general ; only for a single point is there trouble. But exam- ples can be adduced of functions that, though continuous for all values of x, do not for one single value of x have a deriva- tive. In the light of these facts it might seem as if a thorough- going revision of all we have said in the early chapters were necessary. The revision, however, is simple. So far as our theorems about derivatives are applied to special functions we have fortified ourselves by showing that the elementary func- tions actually possess derivatives unless possibly at exceptional points easily recognized. In the statement of the general theorems of Chap. II, § 4, however, it is true that we need to add the requirement that the functions u and v shall possess a derivative. With this supplementary condition Theorems I-V are true in all cases. The proof of Theorem V, however, requires a modification, of which we will speak presently. Curves. A further restriction on the functions we have treated, which is essential for some of the proofs, is this, that the curve y —f{x) shall have at most a finite number of max- ima and minima in a finite interval. The function sf(x) and (ic) of the above examples do not have this property. In the neighborhood of the point x = 0, they both have an infinite number of maxima and minima. We can impose this restric- tion, however, throughout the Calculus and still the functions will be general enough for most purposes. With this restriction the proof of Theorem V is valid. Without it, the theorem can still be proven by the aid of the Law of the Mean. The proof of convergence required to justify the definition of the definite integral, Chap. IX, § 17, rests on this assumption. SUPPLEMENTARY EXERCISES A. — Introduction Find the slope of each of the following curves. 1. y = x — x 2 , x = l. Ans. — 1. 2. y = 4 -+- x 2 , x = 0. Ans. 0. x = i. ^4ns. — 9. Xq — Xq . 2a J 3. 1-X y = _, X 4. 2x y (3 - xf y 5. x 2 + a 2 y = — 7—, * 6. a; " 3-2a 8. a; 0. a; ar , a? = sc . ^4ws. 1 — In- 6 + 2x ^"* (3- *o) 3 x 2 + 2ax Ans. *=-£. 59 *L=i -2a 59 ' 7-a 3 SUPPLEMENTARY EXERCISES 427 as 2 -2bs + c aH t 2 -l e 62. 1 (l-*) 2 64. 2x (5 - xy 66. 2 1 (4 -art* 4-a; 67. a 2 -* 2 a + t 69. 14 3 2-x l+2a? 2-3a; 71. 12 4 61. t 63 2 (4 -3a;) 3 65. X 2 (a + bx) n Ans. 68. 2s 1-s 2 ' (4,-xf 70. a a-\-x a — x 72. — £ — + 1. t ' 1-t 2 2 + 3* 2 1-s s-s 2 ^ g»(7 + 2aQ« 14^(24 a? H-35)(7 + 2a ; ) 3 7 (14- 3a:) 9 ' (14 -3a;) 10 75. 3 * 74. 6x x 2 — x -f- 1 76. ar*-8 tf + 2x + ± 77. a; 4 — a + 1 #2 + 0? + 1 79. a? Va 2 — af . 81. a; v 2 +l Ans. 1. 78. 2a? + 1 + 3s. (l-2a;) 2 l- 2a; 80. a; -y/x 2 + a 2 . 82 X?±^ 2 - x + a a + x 83. J a + a? - 84. \ a. — cf. a — x V2 ax 85. (a 2 + a^)Va-a;. 86. (* 2 - 1) VI - 4*. 3— 3 /r -r^ 87 "& 88. Jla± A/l + r ^c + c/^ 428 CALCULUS 89 . _A£±£_. 90 . « + ft* Va — /to -V2ax — x 2 91. arVA-ua; 2 . 92. (/3 2 - x 2 )Va - fix. 93. V4-5a;+ar J . 94. V^6-5a;-a; 2 . 95. V-a;. 96. aV4-5a? 2 . 97. r s V-l+3r-2r». 98. *V^I 99. t-t i_ . 100. Sx V^T? (* + l)V4-3s» 101. a? — Vl+7a£ 102. Va 2 -^-Va 2 -t-£ 103. * 104. VI-*-* . Va - a-— Va + a; Vl — a + a 105. . p + ^ + 1. 106 _ J a ;2_ a ._|_^ \ 2 7- cos a 107. (2a-36a;)V(a + 6a;) 3 . ^4n«. — -• 108. (8a 2 -12a^ + 15 6V)VFT^.^ iQSWV^+ to : ' 2 109. (2 a — 6a;) Va + 6a;. 110. (8 a 2 - 4 a&x + 3 6W) Va + 6oj. 111. VC^-s 2 ) 3 . 112. V(a 2 + ^) 3 . 113. V(m 2 — y 2 ) 5 . 115. 117 Vv 2 — 1' 2 V 4a; + 17 11Q Vl7-8ic-9a; 2 5— a; ^5 + 10a; 114. V(p 2 +c 2 ) 5 . 116. V^?. 118. ^^ V25 + 6a;-12a; 2 V-5 + 10aj-aj"^-10aj+^J' 6 00 (_5 + 10a;-a; 2 )^ SUPPLEMENTARY EXERCISES 429 120 . * + 1 ( n + iX+2 ^ - Ans. 2 121. ( P x-l)V P (l + x 2 )-2x 2 ? n 2 1 1 Ans. V/o(l + x 2 ) -2a; pi l *P y/p(l+x 2 )-2x 122. 4a^ — 6x~^+x$* 123. 12aj*-20»* + 8af*. 124. -\/x 125. 1-* ■Vt 126. 127. 1 + Vx y/x 128. #?-Va?-f 2(^a>)«. An*. 2 +3Vi. 3^a> 2 129. (V») 130. V2 131. l + x+x 2 y/x 132. x — x~ l ViC 133. 135. (7 -9x)\/2x. 1-r 2 Vmr 134. 136. Vax — » .2 |-3 (a» — a;*) . 137. ■y/x 138. 140. 0-a y/sd 139. V4-a 2 -5 V2-a; 141. m n naj n _ ma ."» _|_ w _|_ ?l# 142. cv 1Al . 143. (ZiC* 4- ma 6 ) 2 . 144. »(l-af)* 145. (a + x)(a 2 -x 2 )K 146. (a + &af) F . 147. (x*-xfi /4tt#, 2 or* (a? •+.!)* (a? + 1)^-1)* 430 CALCULUS 148. [V^=^-Vo+^] n . 149 [pii + x^-sxy. 150. x n — x~ n 151. — x n n m 2 152. af+i + af-K !53. (x a + x «) 2 . 154. X 1 1 KK * 155. i + yi-a 2 x-Vi+x 2 157. Vi_vi + a/ % 159. y(x-y)-x 2 = a. 161. x + xy = a + b. 163. 6-V + «y = a 2 6 2 . 165. aj 2 y 2 _ ^ a 2 6 2 .167. o sc — y 2/"= -• z + y XB6. (_£_Y. 158. 2« 2 + 52/ 2 =7. * 160. xy~ax-\-by. 162. x 3 y = x 3 y . 164. ar 5 4-2/ 3 -3aa^/ = 0. 166. y* = a 2 (x — y). 168. 2^ — 3^ — y 2 = l. 169. a 3 — xy + y 4 = 0. 170. aPy — y^ + atf — aby^O. 171. y 2 — 2ax—2by -\-x*. 172. a 4 + 2/ 4 + l — 6a&y = 0. 173. ^-3^ + 4a# 2 +lly 7 =l. C. — Differentiation of Transcendental Functions The further special formulas of p. 95 are required, which may also be written as follows. The student should notice, however, that this form is no more and no less general than that of p. 95. 3. dsinw = cos udu. 4. dcosu = — sin udu. 5. d tan u = sec 2 udu. 6. dlogu __du u 7. de u = e u du. SUPPLEMENTARY EXERCISES 431 du 8. dsin -1 w VI -u 2 9. a*tan _1 w = 1 + u 2 To these may be added, if desired : du 10. dcos~ x u = 11. c?vers _1 w = du ■V2u-u 2 12. da M = a M logadw. The student should note further the trigonometrical for- mulas : * (A) cos -1 u = - — sin -1 u. (B) cot^u = tan~ x 1 = - - tan" 1 !*. w u 2 (C) csc -1 w = sin- 1 -. u (D) sec -1 m = cos -1 - • u Differentiate the following functions.f 174. sin ax. 175. 1 — cos irx. 176. sin 5 a;. 177. cos-- 178. csc 2 x. 179. ^?- a x 180. I + sme 181 . S ec(0-a). 182. cotf. l-sm0 v } 2 183. tan-^-. ' 184. cos ^^ "^ - 185. csc^- 1 — x 2 n *For further formulas relating to the trigonometric functions, cf. Peirce's Tables, pp. 73-80. f Cf . also the examples on pp. 96-99. 432 CALCULUS 186 . siD< ^ ■ 187. 1 -° 0S ^. 188. 1 + C0S ^ 1 — cos sin <£ 189. cos n 0. 190. sin w irx. 191. sinaxcos&x. 192. sin(n — m)6, 193. cos (n + ra) <£. 194. x 2 cos ax. s . (2n + l)x . f2n-l)x sin V ! 1— COS x - sin mx 2 o 195. »"*•'"*. 196. = 197. - a? ■ x • x • X sin- sm- 198. vers ax. 199. xversx. 200. vers 2 ax. 201. Vl-A^sin^. 202. xH -»' 203. -45 204. sin + vers 0. 2Q5 J versa; sin d — revs * 2 — rers COS _™„ X cos y 1 -J- cos X 206. sin — vers * 2 — vers x Vvers . 207. tang - 1\ 208. cot (Z - g. % 209. • + COtT*. V .210. cot a; — csc a. 211. cot-- .212. cot x + esc x. £ «„„ o cos3x , • o A «j r i sin7x — cos9x 213. sm2x hsm3x. 214. cos5xH 3 63 215. sin2x = 2sinxcosx. 216. cos2x= cos 2 x — sin 2 x. » 217. sin" 1 ^^. ■ 218. cos" 1 -- 219. tan- ll ~ 2x . 2 n 3 220. tan -1 (2 tan x). L 221. cos -1 (n sin x). 222. sin _1 ax. 223. cot" 1 ^— -• 224. tan-^sinJ— a + x V ^2a ii) ( — )• 226. vers- 1 ** a 227. sin- 1 ^/-* 228. vers" 1 a 6 a SUPPLEMENTARY EXERCISES 433 229. 232. tan" sin" , 1 l-2a; 5a; x b + a cos x a + b cos x 230. sin- 1 ^^ 231. cos -il An$. 233. sin 235. ,236. 239. _, f Va 2 — b 2 sin x~ \ a + b cos a; ^-. rva'-ysin g-i 6 + a- cos a; log (2 — 3a). 237. log (a— a), log (1 + x 2 ). « 240. log ^—^- • 234. tan - - Va 2 -& 2 a -f & cos a? '[>te>*"} ^4ns. Va 2 -6 2 .242. log VI + cos 6. ♦ 243. log a + b cos x 238. logo 2 . 241. log (a + 2a 2 ). Va — Va — x Va ♦244. 246. 248. 251. 254. 257. 260. 263. 266. 268. 270. 272. 273. log (a -a;) 2 . »+& .245. e~ x . ^247. e l ~ x \ e*. log 249. e xl °e a . 252. Va^. 255. e mx — e~ 258. cP"*. e _nt cos (at — /3). 261. e _x logaa. a x — q~* a z sin log a?. log(a + VoM^)- ^ein -1 x 264. a log< *». 250. a- x . 253. -Z/&. 256. log log a;. 259. (10 1+ ') 2 . 262. e z8ina: . 265. lO"* 3 . 267. logcos n a. 269. a x log a. 271. VHK J_ log Va + ^-Va t a>a Va Va -f 6a + Va -**e*J*±E t a<0. V-a * -« 2f -4ws. ^4ns. a; Va + 6a; 1 a; Va -f- bx 434 CALCULUS 27 4. lo g <* + Vtt 2 ±* 2 275. togy V# + a 4- V # — a V# + a — V& — a 276. log(V21-8x + x 2 + a -4). — a scVa 2 ± or 2 1 .4ns 2Var>-a 2 1 A 7is V21 . — 8 a; -f a,* 2 1 277. log(V-12 + 16a + 4a>» + 2a> + 4). Ans. "V — O "T"4 X-\-X l 278. (sinaz)*. 279. (l + x)\ 280. (log#) x \ ~\tana; i 281. (cos- 1 - ) . 282. (cos a)* 8 . 283. (4-tana;) x . a J 284. (a 2 + x i ) x . 285. (tan- 1 a) 8in_lx . 286. r r2 . i 287. (log cos x) x . 288. (log tan x) x . 289. x logx . 290. x = Ue l + e~~ c ). Ans. ^== — , or e c_ e c V a; c 291. e x cos^y —(1 — x) log (1 + y) = 1. Ans. dy = ** CQS " ly + log (1 + y) /I, Wl^~ ' *» e *Vl+2/+(l-a)Vr3^ + -^ ^ 292. r 2 = a 2 cos20. 293. x y — y x = Q. 294. = sin(0 + <£). 295. x = ysmy. 296. 2/ — A: sin y — a + a;. 297. tan _1 a?— tan" 1 ?/ = &# + &?/. 298. #e y — ?/ log x = 1. 299. asin(r#) = &0 + r. 300. re r = 0sin0. 301. a£ + ^ = ai 302. V^+V^=Va. 3 ° 3 ' (a) ~~(f) 3 = 1 ' 304 " ( 1 °g £C ) 2/ -2/ COS7raj = :I - {# = acos 5 0, (# = tan0 — 0, . • k» 306. { y = osm^. [y = cos6. SUPPLEMENTARY EXERCISES 435 (x = a -f- bt + ct* 308. y = a ' + b't + c't 2 . 309. i 1-A 2 x = -a. 1+A 2 310. 2a v = a. * 1 + A 2 1 + * 3 x=- -, 2/ = 311. Find the slopes of the curves in Exs. 308-310 where these curves cross the axis of x. 312. Remembering that i r «+i l + r + r 2 + ... +r w = r , 1 — r show that l + 2r + 3r 2 + ... +nr^ = 1 ~ ( w + ^ + < " (1-r) 2 313. Using the result of the preceding question, obtain a formula for the sum : l+2 2 r + 3V+ — 4-nV- 1 . 314. In the formula ^ w-a T(> + &) 2 ' i2, a, 6, and c are constants and p, v, and T denote pressure, volume, and temperature, respectively. (a) Considering T as constant, find the derivative of p with respect to v. (b) Considering v as constant, find the derivative of p with respect to T. (c) Considering p as constant, find the derivative of T with respect to v. D. — Applications Find the value of x for which y is a maximum or a minimum in each of the following cases. 315. 12y = x s -6x\ x>0. 316. ay = x 7 - 35 x 5 , x < 0. 436 CALCULUS 317. y = x B - 15a 3 + 25, x >0. 318. y = a? - 2a?- %x, x>l. 319. 2/ = r -^— • 320. y = 3 + a; 2 * 4 + 7(2-z) 2 321. ,-*+!. 322. r -£±* x x 323. J.- 13 -** + * , ->3. 324. » = |±*£, *>3. cc — 2 lx — 6 325. Divide 12 into two such parts that the product of one of these parts by the square of the other is as large as possible. 326. Divide a into two such parts that the product of one of them by the cube of the other is as large as possible. 327. Divide 30 into two such parts that twice the cube of one of them increased by three times the square of the other may be as large as possible. 328. If the strength of a beam is proportional to its breadth and the square of its depth, find the dimensions of the strongest beam that can be cut from a log of circular cross-section. 329. What is the shortest distance from the point x = 12, y = to the curve y 2 = 4:X? 330. What point of the curve y = x% is nearest to the point 08 = 1, y = 0? 331. Tangents are drawn to the parabola y 2 = a 2 — 2 ax. What one of them cuts off the smallest triangle from the first quadrant ? 332. In the foregoing problem, what tangent has the short- est segment in the first quadrant ? 333. A trough is to be made of a long rectangular-shaped piece of copper by bending up the edges so as to give a rect- angular cross-section. How deep should it be made, in order that its carrying capacity may be as great as possible ? SUPPLEMENTARY EXERCISES 437 334. A block of stone is to be drawn up an inclined plane by a rope. Find the angle that the rope should make with the plane in order that the tension may be as small as possible. 335. Show that the abscissa of that point of the curve y = log x which is nearest to the origin is given by the equa- tion: iogi=*>. X 336. Find the value of x from the foregoing equation correct to two significant figures. Suggestion. Tabulate for some trial values of x the values of x 2 , 1/x, log 1/x, read off directly from convenient tables, such as Huntington's Four Place Tables. 337. Assuming the density of water to be given from 0° to 30° C. by the formula p= Po (l + at + l3t 2 + yt 3 ), where p Q denotes the density at freezing, t the temperature, and a = 5.30 x 10- 5 , y8 = - 6.53 x 10" 6 , y = 1.4 x 10" 8 , show that the maximum density occurs at t = 4.08°. Determine by inspection whether the following functions are increasing algebraically or decreasing as the independent variable increases. 338. a — x. 339. t(t — c), t> c. 340. -i-, x>-\. 341. -1—, o?>0. 1 + x at + x 2 342. x(x-l)(x-2),x>2. 343. x(l - x 2 ), x> 1. 344. _L-, *<0. 345. l±i, 0<*<1. 2 2 346. a ~ X . 0-l. 369. r = 3P + 50-ll. 370. y = 9 X . 371. y = a; + sina;. a 2 + a,* 2 372. y = sin a; -f- cos x. 373. y = e~ x sin x. 374. a? = A sin (n« — y), A > 0, n > 0. 375. aj=(7cos(w« + €), C > 0, n > 0. 376. x = e _a ' sin (n£ — e), n > 0. In what intervals are the following curves concave upward, and in what downward ? 377. y = x?-9x 2 + 7x-3. 378. y = 2ar ? + 3^-1. 379. y = 3x 5 + 5x 4 -2x + 7. 380. y = x 11 - x M . 381. y = ax 3 +bx 2 + cx + d. 382. y = ax 2 +bx + c. 383. y = 2z 6 -9ar i + 10a: 4 -a;. 384. y = — -^ + — • B y 56 42^30 385. y = x-|-cosa;. 386. y = sina? — a;. SUPPLEMENTARY EXERCISES 439 387. y = x\ogx. 388. y = xe~ x . 389. y = ™%£. 390. y = x x . x 391. y = e~ x \ 392. y = logcosx. 393. Plot the curves which represent the functions in Exs. 384-392. 394. A point moves along the hyperbola xy = 100 and its ordinate increases uniformly at the rate of 20 ft. a sec. Find how fast the abscissa is decreasing when y = 15. 395. A point moves along the curve r=l/0 at the rate of 6 ft. a sec. How fast is radius vector turning when = 2tt ? 396. If a drop of rain, as it moves through moist air, receives accretions so that its radius is increasing at the rate of c cm. a sec, at what rate is its volume increasing when its radius is a cm.? 397. A point describes the cardioid r = 2a(l — cos#) with uniform velocity c. How fast is its distance from the pole r = changing when 9 = \-k ? 398. A man in a train that is running at full speed looks out of the window in a direction perpendicular to the track. If he fixes his attention successively for short intervals of time on objects at different distances from the train, show that the rate at which he has to turn his eyes to follow a given object is inversely proportional to its distance from him. 399. A point describes the circle x 2 -f- y 2 = 25 with a velocity of 12 ft. a sec. Find the component velocity along the axis of x when x = 4. 400. At what rate is the ordinate of the curve y — x—x^ changing when x= 1, if the abscissa is increasing at the rate of 10 metres a second ? Is the ordinate increasing or decreasing? 401. A man walks across the floor of a semicircular rotunda 100 ft. in diameter, his speed being 4 ft. a sec, and his path the radius perpendicular to the diameter joining the extremi- 440 . CALCULUS ties of the semicircle. There is a light at one of the latter points. Find how fast the man's shadow is moving along the wall of the rotunda when he is half way across. 402. Water is flowing out of a hemispherical bowl from an opening at the lowest point. If the rate of efflux is c cu. cm. a sec. when the level of the water is half way between the hole and the centre of the bowl, how fast is the level falling ? Suggestion. Compare the volume of water that flows out in At seconds with the fall in level, Ax, and thus compute directly lim Ax/ At. E. — Errors of Observation* Let x be an observed quantity and y a second quantity, dependent on x, to be computed: y =/(*)• An error of Ax in observing x, the true value being x , will give rise to an error Ay in the computed value, where ty =/(^o + A^) —f(x ). Now we are concerned only with an approximate value of Ay and hence any other quantity that differs from Ay by less than the error in Ay which we are willing to admit is an equally faithful representative of the error in y. Such a representative is found for most of the cases that arise in practice in dy, since, if the derivative f'(x) is finite for x = x and =£0, Ay and dy will differ from each other only by a small percentage of either, when Ax is small, that is : l im Ay- 1 is a constant of the apparatus. If log k (called the " logarithmic decrement ") be denoted by A., then A = log op - log a m s m Assuming that a is correct, but that every subsequent ampli- tude a m is subject to an absolute error 8 independent of m, find the absolute error in X in terms of a , k, and m, and show that it is a minimum when m satisfies the equation k m — e = 2.718, so that under the conditions of this problem one should use, in determining X, that a m which is nearest a /2.718 to get the best results. SUPPLEMENTARY EXERCISES 445 423. Assuming that in the preceding problem both a and a m are subject to the absolute error 8, and that the resulting error in A is the sum of the absolute values of the errors which would be produced by the errors in a and a m acting separately, show that the best value of m is given by the equation 424. Solve the foregoing equation by approximate methods and show that k m or a /a m should have the value 3.59. 425. In Ex. 423 assume that the error in A has its "most probable value," which is the square root of the sum of the squares of the errors which would be produced by the errors in a and in a m acting separately. Show that the best value for a Q /a m is now 3.03. 426. The period of a pendulum varies inversely as the square root of the force of gravity, g, at the place of observa- tion. With what percentage accuracy must the period be observed if g is to be determined to one part in ten thousand ? 427. * If the period of such a pendulum is very nearly one second, the period can be compared with that of the pendulum of a clock beating seconds. Suppose that the two are beating exactly together at the time t lf that the unknown pendulum gains on the other, and that they beat together again at the time t 2 . Then in exactly n = t 2 — t x sec. the unknown pendulum has made exactly n -\- 1 swings, and its period is n/(n + 1) sec. How large must n be to make an error of 1 sec. in determining n allowable under the conditions of the preceding question ? 428. If the n of the last question is 45, what percentage error in the period would result from an error of 2 sec. in determining n? 429. If the allowable error in the period is a tenth of one per cent, how great must n be to make an error in it of 10 sec. allowable under the conditions of Ex. 427 ? * The solutions of Exs. 427-429 belong in a course in physics, for the mathematical processes involved are exceedingly simple. They are inserted merely to indicate more clearly the bearing of such work as is here given in errors of observation, on the actual problems that arise in practice. 446 CALCULUS F. — Integration Integrate each of the following functions with respect to x. 430. 12x 5 -10x i -16x? + 2x. 431. a/-5a> 6 + a? + l. M 432. ax h — bx a + ab. 433. (m + n)x n + m — n. 434. 1- 4a -or 5 + H # 10 . 435. - 3 + x + £«*- \x*. 436. 4a? 9 -13x 6 -5ar 3 -l. 437. 1^-1^ + 2^ + 1. 438. 9 - 3 " + " 2 . 439. *-g* + *-l. 12 3x 440. ic 2 +--l. 441. 3-- + 3 . a? a; 442. a£ + af£ + l. 443. a^ + af* — 2. 444. VaJ — -— + -• 445. af* — a" 1 + x~K ■y/x x 446. -* ?-- i + a 447. £L±^±£^. 448. 9** -4 -4" 449. iri£. 450. 1 + ^ g -(l + a?)Vg. 451. x(a+&V2a?). a? 452. (5 — VaJ)(a> + 3Va>). 453. (a + bx)^/cx-. 454. (Vc-V2x) 2 . 455. [-{/a5 — -g-. \ wax. 456. (a 2 - x 2 ) VTax. 457. (Va-V3to) 3 . Evaluate the following integrals. 458 - J(r^-ir->- 459 - fiihi-^*- 460 . /•-*_. 461. rj^p i=L J Vl-ay J \ Va L + £ * a. SUPPLEMENTARY EXERCISES 447 462. f**L. 463. ffjLL ?«_U J a-x J \a + t a-3tj 464. /*±-?fe 465 . fr-W-«-l fc 7466. f (aam2x-bcos3x)dx. 467. j e - a2x dx. 4e8 - /(^I- 1 )^- 469 - f(j=i—J* 470. rfe tC0Sa -~\dt. 471. i*—^—dx. Jtf2. Ccot-dx. S ^473. Csm — dx. 474. I sec ax dx. 475. fcsc^dx. 476. fsec 2 2xdx. 477. /7l + cot 2 |^d «. 478. j cos (nt-e)dt. 479. /V*~ /' o o i /iai (*CSC 2 xdx tan s xsec 2 xdx. M 45i * I — , * J Vcota? "tan" a 2 -}-^ 2 r -i da 48^/ / 482. I cos J aj — « / J vr^ 2 " ^ 484. v 7e x cose*cta. J 485. J e~ x (l- e~ x ) 3 dx. 486.1JVF+2** 48V^ J^pT 488. Pt{$F* -■*■"*)&. V^ 89 ' f ^ a *J r ^ dx. 448 CALCULUS 490. f(a - 3 bx)%dx. 491. / V"^ dx, {x < 0). 492. C ^ ads . 493. f(k-r) m ~ n dr. J Vc + 3as J 494. C(Z±^dr. 495. C(a 2 -2ax)^dx. [(axy^-iaxy^dx. 497. j ^7=' r^tan — da?. 499. AinS^Sdft J 3 3 J Va 498. 500. I a cos 502 504 r a cos _2g-3a ote. 501 . T-Va sin ( Va *) dt. /kao /"2 cos aa;— sin flsc, (2 e - - c^«) dr. 503 - J g + /8 daj - . C(a + b-axfdx. ■ 505. C(a + t)*to. • J v (ife.- 4 )^ 507 - /(rf-«- 4 cto. V 508. fx(a 2 -x>)*dx. 509. j?V~a + b-afidt. 510 . /•_**=. 511. A^* • 512. fx(l+aa?)idx. 513. ix(a + bx 2 ) 514. A** 515. /" * Jl + ct 6 J 3 «- dic. 516 46a- 5 SUPPLEMENTARY EXERCISES 449 518. Ilosx — ' V619. / x tan irx 2 dx. 520 522 524 526. • Jlogx^' Uld. J a . jxe-^dx. Vfel. flog^/1—xdx. . jx\ogVa 2 -x 2 dx. 523. Cxf—^-^-b^dx. . C ** + * *>. 525. r ( b -^ dx . J x 2 +px + q J ( a + 3bx- X s ) 2 fap\pl _{]-%. 527. f 2atdt r_l^_. 529. f-3* J5-2* 2 J5 + 2 540 542 2 a; 2 ' 530. f J* -• 531. f_*- J (x + a)* + (x-a)* Jl + 7i 532. f-*L. 533. f- J ?-i> 2 J 63 + 27 s 2 534. /*- -^ -. 535. f J (x + a) 2 -(x-af J i 2ds + 2 dq m 2 -j- n 2 q 2 dx 'x 536. A™*^. 537. A 1 ^ ,/ sin 2 a; ^z cos 5 : 538. f-^-v 539. f * ■ J p + qu 2 J b + (x — a) 2 f cfa 541 f (34-5^)0 5V* . f -***-• 543. f-^JL. J 1 + x* J m-y* 544. f X ' dx . 545. fj^*. J5 + 13* 16 J 1-x 2 2g 450 CALCULUS 546 . f *x l dx 547< jtB,n(nx-2)dx. J Vlog* J 548 . /V-^dr. 549. JV~ -(n + 2*), cos a? VI — & 2 sin 2 a? dec. 551. I -« ^ cos 2 - i /C sin OdO e-» in9 cos Odd. 553. / ,- =' J V4 — costf e?as as 554. f 'sin (9 log cos OdO. 555. /cos log a; ^ 556. n + cosj^ 557# f C0S 2 esindde . J 1-COS0 J 558. fsmxco$L2xdx. 559. / cos * sin 2 * da*. 560. /sin 2 * cos a- d*. 561. I sin 2 a; cos 2 * d*. 562. / sin w *cos*d*. 563. I cos n *sin*d*. 564. I cos 2 mxdx. 565. I sm 2 mxdx. V 566. /* d* ^ 1 * _i/&tan* ^4?is. — ■ tan' cos 2 *-f-& 2 sin 2 * ct& 567 f sinftcosftda; . ^ s . 1 log (a cos 2 * + 6 sin 2 *). J a cos 2 * + 5 sin 2 * 2(&-a) C cos*d* 568. / J Vl — ft 2 sin* a; /, sin (m — w) * sin (m + fl) * smm*smn*d*. <4n*. 2 \ m _ J ! ) 2 (m + w) SUPPLEMENTARY EXERCISES „_-. C . 1 a cos (m — n)x cos (m-f n)x 570. I smmxcosnxax. Ans. — ■? f — ) — ! — f-- J 2(m — ri) 2 (m -f n) .».. C j a sin(m — n)x , sin(m-fw)a; 571. I cos mx cos na; dx. ins. ——) f- -\ — ) ' — f- . J 2 (ra — ti) 2(m + %) Evaluate the following ; grals by the aid of the Tables. 572. C ^ . 573. T— — — J3~7x + 2x i j3-2z + 7(* /* _ 7 _ / * asdar " J 3s f-Saj 8 " ' j3-2x + 7a*' 576 C——^—~ 577 C——^—— ' Jl3-7x + 2x 2 y J 3aj 9 -2ar , + 7q!* 578 . r ^ 579. r — *» — J V3-7a> + 2a* J V3-2a; + 7ar J 580. f- ' =- 581. /^ * J xV8-7o; + 2it* 2 J #V3-2o; + 7ar J 582. f ** 583. f *» -• J (3_7^ + 2.t 2 )^ J (3- 2a; + 7^* f— ^ 585. f- J 4 — 5 cos a; ^7 5 584. 586. 588. 590. 592. 5--4cosa; dx r Mx r J 3 + Ttanx' ' J 11 + f * 589. f J (5 — 4 cos a;) 2 ^7 < r * 59i. r J sin x (5 — 4 cos a?) j 1 f fe 593. f-J^ J 10 — cos a; + 2 sin a; J1 + cc 13 sin x dx cos #(5 — 4cos#) dx 9 — 7 cos 2 ic dx COSiC 452 CALCULUS G. — Definite Integrals 594. The hyperbola xy — lOOl rotates about the axis of x. Find the volume of that part of the solid thus generated which is contained between the planes ^rpendicular to the axis and corresponding to x = 5 and x = 2 595. The curve y = sec x revolvt '"♦out the axis of x. Find the volume of the solid whose bases ; ,rrespond to a? = \v and 596. The curve y = x — x A rotates about the axis of x. Find the volume of the solid generated bytthat part of it which lies above the axis of x. 597. The hyperbola ^_?f « 2 i> 2 = 1 revolves about the axis of x. Find the volume cut off from one of the two # solids thus obtained by a plane perpendicular to the axis and distant h from the vertex. Ans>?¥¥(3a + h). 3d 2 J 598. So much of that arc of the curve y = eosx — icos2# which cuts the axis of ordinates and lies above the axis of x rotates about the latter axis. Find the volume of the solid generated. 599. The curve y = cos _1 ic rotates about the axis of x. Find the volume generated by that part of the curve for which <^ y < ir, the base being a plane perpendicular to the a: | = -1. 600. The parabola x 1 + y 1 = a' £ rotates about the axis of x. Find the volume of the solid bounded by the arc which is tan- gent to the coordinate axes at its extremities, the base being formed by a plane through the origin perpendicular to the axis. 601. Determine the volume of the following solid. Think of the axis of y as vertical and consider the cylinder whose SUPPLEMENTARY EXERCISES 453 elements are perpendicular to the curve y = 12 — 12 a£ Next, turn this cylinder about the axis of y through 90°. The two cylinders and a horizontal plane through the origin bound the solid in question. 602. If the base of the conoid of Ex. 5, p. 161, is an ellipse whose plane is parallel to the fixed line, show that the volume is ^irabh, where h denotes the distance from the line to the plane. 603. The solid of p. 159, Fig. 49, is cut by a plane through O, perpendicular to the plane of the base AOB and making an angle of 45° with OA. Determine the volume of the part with the vertex A. 604. A horn is generated by a variable circle whose plane turns about a fixed line. The point of the circle nearest the line describes a quadrant AB of a circle of radius a, and the radius of the variable circle is cO, where denotes the angle between the variable plane and its initial position, when it passes through A. Show that the volume of the horn is j^v^iSa + Sirc). 605. Find the area of the lateral surface of the solid described in Ex. 603. 606. Find the area of the lateral surface of the solid described in Ex. 601. Ans. 64, nearly. 607. An arbitrary closed curve is drawn on the surface of a sphere, catting out a region S from that surface. Show that the volume of the cone whose vertex is at the centre of the sphere and whose base is S is where A denotes the area of S and R the radius of the sphere. 608. The curve r=f($) rotates about the axis = 0. As- suming f(6) to be single-valued and continuous for a _ _ /3, where _ a < /3^v, obtain a formula for the volume of the solid generated by the rotation of the plane region bounded by the curve and the two radii vectores drawn to its extremities. 454 CALCULUS 609. Hence determine the volume of the solid generated by the rotation (a) of the curve r = a cos 20; (b) of the lemniscate ?* 2 = a 2 cos 2 6 j (c) of the curve r = 1 — 6 2 . 610. Show that, if two solids are so related to each other that, when cut by any plane parallel to a certain fixed plane, the areas of the two cross-sections are equal, then the volumes of the solids are equal. (Cavalieri's Theorem.) 611. Find the areas of the surfaces in (a) Ex. 594; (b) Ex. 597. 612. Find the fluid pressure on the vertical plane area bounded by the curve a-, a 4 + ar and the double ordinate x = h, the axis of y lying in the sur- face of the liquid. 613. Assuming that the density of water at a distance of x ft. below the surface is p = p (l + . 000 001 3 x), find how much greater the pressure is on a vertical rectangle 10 ft. broad and a mile deep, with one side in the surface, than what it would be if water were incompressible. 614. Find the pressure on the end of the trough described in Ex. 2, p. 164, if the density is a linear function of the dis- tance below the surface. 615. If the density p of any curve is variable, show that the mass of the curve is t M— I pds. /< 616. The density of a rod is proportional to the distance from one end. Find its mass. SUPPLEMENTARY EXERCISES 455 617. The density of a semicircular wire is proportional to the perpendicular distance from the diameter joining its ends. Find its mass. 618. If in the preceding problem the density is proportional to the square of the perpendicular distance from the radius drawn perpendicular to the above diameter, what is the mass ? 619. Find the centre of gravity of the rod in Ex. 616. 620. Find the centre of gravity of a quadrant of the wire in Ex. 617 and in Ex. 618. 621. The density of a spherical surface at any point is pro- portional to the distance of the point from a fixed diameter. Eequired the mass. 622. The same problem for a cone of revolution, the density being proportional to the distance from the axis. 623. The density at each point of a sphere is proportional to the distance of the point from the centre. Find its mass. 624. If the density is a -f 6r, where r denotes the distance from the centre, required the mass. Determine the following moments of inertia and radii of gyration : 625. A rod whose density is proportional to the distance from one end, about a perpendicular at that end. 626. The same rod, about a perpendicular bisector. 627. The circular wire of Ex. 617 about the diameter. 628. The same, about the radius perpendicular to the diameter. 629. The circular wire of Ex. 618 about the diameter. 630. The same, about the radius perpendicular to the diameter. 631. A circular disk whose density is proportional to the distance from the centre, about the centre. 456 CALCULUS 632. The same, about a diameter. 633. A circular disk whose density is proportional to the dis- tance from the circumference, about the centre. 634. The same, about a diameter. 635. A circular disk whose density at any point is \vb 2 — r 2 , r denoting the distance from the centre, about the centre. 636. The same, about a diameter. 637. A conical surface of revolution, about the axis of the cone. 638. A spherical surface, about a diameter. 639. A conical surface of revolution whose density is pro- portional to the distance from the axis, about the axis. 640. A spherical surface whose density is proportional to the distance from a diameter, about that diameter. 641. A sphere whose density is proportional to the distance from the centre, about a diameter. 642. A sphere whose density is proportional to the distance from a diameter, about that diameter. 643. A sphere whose density is any linear function of the distance from the centre, about a diameter. 644. A triangle whose density is proportional to the dis- tance from one side, about that side. 645. A semicircle whose density is proportional to the dis- tance from the bounding diameter, about that diameter. Determine the following centres of gravity : 646. A uniform circular segment. 647. A uniform circular sector. Check your answer. 648. A segment of the equilateral hyperbola x 2 — y 2 — a 2 , cut off by the double ordinate x = a + h. SUPPLEMENTARY EXERCISES 457 649. The corresponding segment for any hyperbola. 650. A triangle whose density is proportional to the dis- tance from one side. 651. A uniform parabolic wire, extending equal distances to each side of the vertex. 652. A semicircular wire whose density is proportional to the length of the arc measured from its middle point. 653. The same, when the density is proportional to the dis- tance from the diameter through its extremities. 654. A semicircle whose density is proportional to the dis- tance from the bounding diameter. 655. A semicircle whose density is proportional to the dis- tance from the centre. (Suggestion. First obtain a formula for the centre of gravity of a semicircle whose density is an arbitrary function of the distance from the centre.) 656. Show that the ordinate of the centre of gravity of the uniform plane area of § 1, p. 153, is given by the formula : * 6 I y 2 dx 657. Find the attraction of a quadrant of a circle on a particle at the centre of the circle. 658. Find the attraction of so much of a cylindrical surface of revolution as lies between two planes normal to the axis, on a particle situated in the axis. 659. The same, when the density of the surface is pro- portional to the distance from one of the planes. 660. Find the attraction of a homogeneous hemispherical sur- face on a particle situated at the centre of the sphere. * This formula was given to me by Mr. Rogers Sherman Hoar, at that time a student in the first course in the Calculus. 458 CALCULUS 661. The same question, only that the density of the surface is proportional to the distance from the axis. 662. The same, when the density is proportional to the distance from the base measured along the arc of a great circle meeting the base at right angles. 663. Find the attraction of the semicircular wire (a) in Ex. 617 ; (6) in Ex. 618, on a particle at the centre of the circle. 664. The same for a particle on the circumference extended, at the point situated symmetrically with respect to the wire. 665. Find the attraction of a homogeneous surface in the form of a right cone of revolution on a particle at the centre of the base. 666. Find the attraction of the surface of a frustum of a cone of revolution on a particle at the centre of the smaller base. 667. Evaluate the double integral // xydS, where S is the rectangle whose vertices lie at the points (1, 2), (1, 5), (3, 2), (3, 5). Ans. 42. 668. The same integral, extended over the triangle cut off from the first quadrant by the line joining the points (0, 3) and (3, 0). Ans. 3f . 669. Compute C C(±o+ x +f)ds, the region S being the piece of the plane bounded by the parabola y = x 2 — x and the right line y — x. Ans. 55^. 670. Extend the integral //WHO" SUPPLEMENTARY EXERCISES 459 over the same region, and check your answer by inverting the order of integration. 671. Compute the value of // 675. Find the moment of inertia of a homogeneous right cone of revolution about an axis through the vertex perpen- dicular to the axis of figure. INDEX Absolute value, 5. Acceleration, 190. Addition theorem, 75. Arasler's plani meter, 409. Approximate computations, Chap. XX. Area under a curve, 111, 117, 153, 407 ; — in polar coordinates, 132 ; — of a surface of revolution, 167 ; — of any surface, 367 ; — of cylindrical surface, 370; further problems in areas, 453. Attraction of gravitation, 181, 457. Cardioid, 152. Catenary, 131,400. Caustics, 350. Centre of fluid pressure, 175. Centre of gravity, 169 ; general formulation for — , 174, 381 ; further problems in — , 456. Change of variable, 295. Chart, A Mercator's, 331. Compound interest law, 82. Computation, Numerical, Chap. XX. Concave upward, Test for, 50; further examples for — , 438. Confocal quadrics, 326. Conservative field of force, 395. Contact of curves, 276. Continuity, 4, 20, 283. Convergence, Fundamental principle for, 246. Cubic equation, 58, 61. Curvature, Definition of, 134; radius of — , 136 ; formula for — in polar coordinates, 145. Curve tracing, 51 ; further problems in — , 110, 439. Curves on the sphere, cylinder, and cone, 329; — without tangents, 422. Cycloid, 146. Definite integral, 156, 358, 380; further problems relating to — , 452. Density, Examples of variable, 183, 188, 360, 376, 381, 396, 454. Derivative, 9; logarithmic — , 22 ; partial — , 287 ; directional — , 308; normal — , 309. Developments in series, Chap. XIII. Differentials, Definition of, 91 ; — of higher order, 94; — of arc, 99 ; — of functions of several variables, 293; total — , 293 ; exact — , 309. Differentiation, General formulas of, 13,21,94; special formulas of — , 95 ; — of implicit functions, 34, 302. Direction cosines, 283, 289, 319. Directional derivatives, 308. Double integral, 358, 458. Duhamel's theorem, 164. Elementary functions, 96. Envelopes, 344. Epicycloid, 149. Equiangular spiral, 91, 130. Errors of observation, etc., 306, 440. Euler's theorem for homogeneous functions, 300. Evolnte, 139, 349. Exact differentials, 309. 460 INDEX 461 Fluid pressure, 161. Four-cusped hypocycloid, 150. Fractional exponents, 27. Function, Definition of a, 2; continuous — , 4, 20, 58 ; implicit — , 34, 302 ; algebraic —,35; test for increasing or decreasing -49; further examples of latter, 281, 437 ; inverse — , 68 ; elementary —,96; — of several variables, 282 ; hyperbolic — , 412 ; exponential — , 417 ; — without a derivative, 422. Gudermannian, 416. Guldin, c/. Pappus. Helix, 318. Homogeneous functions, 300. Hyperbolic functions, 412. Hypocycloid, 150. Implicit functions, 34, 302. Indefinite integral, 156. Indeterminate forms, 231, 278, 280. Indicator diagrams, Area of, 409. , Infinitesimals, 85; — with several variables, 293. Infinity, 19. Inflection, Point of, 51, 53, 276. Integral, Definition of (indefinite), 114; tables of — ', 126; — as the limit of a sum, 155, 358, 380 ; definite — , 156, 358, 380 ; iterated — , 354, 382 ; double — , 358 ; surface — , 374 ; triple — , 380 ; line — , 390 ; numerical computation of definite —,406; further problems in double and triple — , 458. Integration, Special formulas of, 118; further problems in — , 446 ; variable limits of — , 186 ; mechanical — , 409. Iterated integral, 354, 382. Jacobian determinant, 305. Law of the mean, 230, 334; generalized — , 234. Laws of motion, 190. Lemniscate, 133. Length of a curve, 129, 166. Limits, Three theorems about, 15; the — %, etc., 231, 278, 280; — for functions of several variables, 282. Line integrals, 390. Logarithmic spiral, cf. Equiangular spiral. Loxodrome, 330. Mass in terms of density, 360, 381. Maxima and minima, 39, 53, 275, 336; further problems in — , 104, 279, 435. Mercator's chart, 331. Minima, cf. Maxima. Moment of inertia, 176, 359, 381 ; further problems in — , 455. Napierian logarithm, Reason for, 82. Natural logarithm, cf. Napierian — . Newton's ba/ws of motion, 1'K); — method for numerical equations, 399. Normal, Equation of, 37; — lines and planes, 286, 289, 316, 317 ; — derivative, 309; principal — , 325. Numerical computation, Chap. XX. Osculating circle, 138, 277 ; — plane, 323. Pappus, Theorems of, 362. Peirce's Tables, 126. Planimeter, Amsler's, 409. Principal normal, 325. Quadratic forms, 340. Quadric surfaces, Confocal, 326. Radian measure, Reason for, Radius of gyration, 178. Rates, cf. Velocity. Rhumb "line, 330. Rolle's theorem, 229. 66. 462 INDEX Roots of equations, 57, 398; further problems in — , 110, 281 , 322. Roulettes, 150. Semi-cubical parabola, 141. Series, Definition of infinite, 244; power - - , 257 ; Maclaurin's — , 202 ; Taylor's — , 264, 266, 335 ; binomial — , 272. Simple harmonic motion, 201. Simpson's rule, 406. Slope of a curve, 5. Spirals, 131 ; — of Archimedes, 145 ; also Equiangular spiral. Square root sign, 4. Successive approximations, 403. Surface integral, 374. Tables of integrals, 126. Tangent, Slope «tf, 5, 11 ; Tangent, Equation of, 37 ; — in polar coordinates, 90 ; — plane, 289, 316 ; — line to space curve, 317. Taylor's series, 264 ; — theorem, 266, 3:35. Test-ratio, 250. Total differential, 292. Triple integral, .380, 459. Trisectrix, 145. Trochoids, 151. Velocity, Definition of, 46 ; rates and — , 101 ; further problems in — , 108, 109, 439. Volume of a solid of revolution, 157; volumes of other solids, 159, 352; further problems relating to such volumes, 452. Work done by a variable force, 393, 409. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. EC 16 25F fi h ' B 7 T S RECD LU FEB il 19b/ j I * w£C?P ^ fl$ ' io¥ '58CS ftEC'D 1 O m iH^ 3Fe'59ESl WZCT5 05 M h W 133ec6lL2 OtG \ 5 ^ -; 4^r'62TDf REC'D LD : L AUG 23 1972 91 'D CIRC DEI'T FEB » 74 LD 21-100m-6,'56 (B9311sl0)476 Genera! Library University of California Berkeley m msjiy UNIVERSITY OF CALIFORNIA LIBRARY