Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/differentialinteOOIoverich DIFFERENTIAL AND INTEGRAL CALCULUS THE MACIMILLAN COMPANY NEW YORK • BOSTON ■ CHICAGO • DALLAS ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO DIFFERENTIAL AND INTEGRAL CALCULUS BY CLYDE E. LOVE, Ph.D. ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN 'j ^„o J i i> l^efe gnrit THE MACMILLAN COMPANY 1917 All rights reserved ^ hlol Copyright, 1916, By the MACMILLAN COMPANY. Set up and electrotyped. Published September, 1916. Reprinted March, 1917. 3 NoriDDDli iPrfgg J. S. dishing Co. — Berwick & Smith Co, Norwood, Mass., U.S.A. w^ ^A PREFACE This book presents a first course in the calculus sub- stantially as the author has taught it at the University of Michigan for a number of years. The following points may be mentioned as more or less prominent features of the book. * In the treatment of each topic, the text is intended to contain a precise statement of the fundamental principle involved, and to insure the student's clear understanding of this principle, without distracting his attention by the discussion of a multitude of details. The accompanying exercises are intended to present the problem in hand in a great variety of forms and guises, and to train the stu- dent in adapting the g eneral methods of the text to fit these various forms. The constant aim is to prevent the work from degenerating into mere mechanical routine, as it so often tends to do. Wherever possible, except in the purely formal parts of the course, the summarizing of the theory into rules or formulas which can be applied blindly has been avoided. For instance, in the chapter on geo- metric applications of the definite integral, stress is laid on the fact that the basic formulas are those of elemen- tary geometry, and special formulas involving a coordinate system are omitted. Where the passage from theory to practice would be too difficult for the average student, worked examples are inserted. It seems clear that so-called applications in which the student is made to use a formula without explanation of •laa^Q'T vi PREFACE its meaning and derivation, are of little value. In the present text the non-geometric applications are taken sys- tematically from one subject, mechanics, and the theory is developed as fully as in the calculus proper. A feature of the book is its insistence on the importance of checking the results of exercises, either directly or by solving in more than one way. The latter method is largely used in the integral calculus, on account of the variety of elementary transformations possible with defi- nite integrals. The answers to many of the exercises are given, but seldom where a knowledge of the answer would help in the solution, or where a simple means of checking the answer exists. Topics of minor importance are presented in such a way that they may be omitted if it is desired to give a short course. The chapter on curve tracing is introduced as early as possible, so that the results are available for use through- out the course. Some instructors will wish to begin the use of integral tables immediately after the chapters on formal integra- tion. This of course can easily be done. In spite of obvious difficulties, a chapter embodying a first treatment of centroids and moments of inertia is introduced before multiple integrals have been defined. By this arrangement the student is brought to realize the fact that in most cases of practical importance mass- moments of the first and second orders can be found by simple integration, whereas from the usual treatment he gets exactly the opposite idea. In the chapters on differential equations, emphasis is laid on those types most likely to be met by the student of engineering or the mathematical sciences. In the last chapter the average student will doubtless require con- PREFACE vii siderable help from the instructor, but it is hoped that, if properly presented, the chapter may give the student some facility in writing and solving the simpler differen- tial equations of mechanics and in interpreting the results. To Professor Alexander Ziwet, who has read the entire manuscript, the author makes grateful acknowledgment, not only for valuable advice and criticism, but for his unfailing encouragement and support. Thanks are also due to Professor T. H. Hildebrandt, who has kindly assisted in reading the proofs, and has made a number of useful suggestions. CLYDE E. LOVE. Ann Arbor, August, 1916. CONTENTS CHAPTER I FUNCTIONS. LIMITS. CONTINUITY ART, PAGK 1. Functions 1 2. Geometric representation ....... 2 3. Independent variable ........ 3 4. Kinds of functions 3 5. One-valued and many-valued functions .... 3 6. Rate of change ; slope 5 7. Limits 6 8. Theorems on limits 7 9. Limit of a function ........ 8 10. Infinitesimals .......... 8 11. Limit of the ratio of two infinitesimals .... 8 12. Continuity 10 13. Infinity 11 14. Function with infinite argument 12 CHAPTER II THE DERIVATIVE 15. The derivative 14 16. Higher derivatives . 18 CHAPTER III DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 17. Introduction 19 18. Derivative of a constant ....... 19 19. Derivative of a sum ; a product ; a quotient ... 19 20. Derivative of a function of a function 21 ix CONTENTS 21. Derivative of x^, n a positive integer 22. Derivative of x", n fractional 23. The general power formula . 24. Implicit functions .... 25. Differentiation of implicit functions 26. Inverse functions . ... PAGE 22 24 24 26 26 27 CHAPTER IV GEOMETRIC APPLICATIONS 27. Tangents and normals to curves -.29 28. Length of tangent, subtangent, normal, and subnormal . 30 29. Increasing and decreasing functions ..... 32 30. Maxima and minima 33 31. Concavity 33 32. Points of inflection 35 '33. Summary of tests for maxima and minima, etc. ... 35 34. Applications of maxima and minima 37 35. Derived curves 42 CHAPTER V DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS I. Trigonometric and Inverse Trigonometric Functions 36. Trigonometric functions 37. Differentiation of sin x , 38. Limit of sin a/ a as ct approaches 39. Differentiation of cos x, tan x, etc. 40. Inverse trigonometric functions 41. Restriction to a single branch 42. Differentiation of the inverse trigonometric functions II. Exponential and Logarithmic Functions 43. Exponentials and logarithms . - . 44. Properties of logarithms 45. The derivative of the logarithm 46. The limit e 47. Differentiation of the exponential function 48. Hyperbolic functions 45 46 47 48 51 51 53 55 57 58 60 62 64 CONTENTS xi CHAPTER VI THE DIFFERENTIAL ART. PAGE r 49. Order of infinitesimals 68 \ 50. The differential 69 \51. Parametric equations ; implicit functions .... 72 CHAPTER VII CURVATURE 52. Differential of arc 75 53. Curvature 76 54. Radius of curvature 78 CHAPTER VIIT APPLICATIONS OF THE DERIVATIVE IN MECHANICS //^ . .... jf55. Velocity and acceleration in rectilinear motion ... 80 56. Vectors ... 82 57. Velocity in curvilinear motion 82 58. Rotation ........... 83 59. Acceleration in curvilinear motion ..... 85 60. Time-rates 88 ■/S- *v \ CHAPTER IX t CURVE TRACING IN CARTESIAN COORDINATES I. Algebraic Curves 61. Introduction 92 62. Singular points ......... 92 63. Determination of tangents by inspection . . . . 93 64. Kinds of singular points ....... 95 65. Asymptotes 97 66. Exceptional cases ......... 101 67. General directions for tracing algebraic curves . . . 101 xu CONTENTS II. Transcendental Curves ART 68. 69. 70. 71. Tracing of transcendental curves . Curve tracing by composition of ordinates Graphic solution of equations The cycloid 72. The epicycloid 73. The hypocycloid PAGE 105 106 107 108 109 110 CHAPTER X CURVE TRACING IN POLAR, COORDINATES 74. Slope of a curve in polar coordinates .... 75. Maxima and minima 76. Curve tracing 112 114 114 CHAPTER XI THE INDEFINITE INTEGRAL 77. Integration 78. Integration an indirect process 79. Constant of integration . . . . 80. Functions having the same derivative . 81. Geometric interpretation of an integral 82. Variable of integration . . . , 83. Change of the variable of integration . 84. Integration by substitution . 116 118 118 119 120 122 122 123 CHAPTER XII STANDARD FORMULAS OF INTEGRATION 85. Standard formulas 86. Formulas (l)-(3) 87. Formula (4) : Powers . 88. Formulas (5) -(6) : Logarithms and exponentials 89. Formulas (7)-(9) : Trigonometric functions 90. Formulas (lO)-(ll) : Inverse trigonometric functions 91. Formula (12) : Integration by parts .... 92. Integration by substitution ...... 126 127 127 129 131 132 132 134 CONTENTS xiil CHAPTER XIII INTEGRATION OF RATIONAL FRACTIONS ART. PAGE 93. Preliminary step 137 94. Partial fractions 137 95. Distinct linear factors 137 96. Repeated linear factors 139 97. Quadratic factors 140 CHAPTER XIV THE DEFINITE INTEGRAL 98. The definite integral 143 99. Geometric interpretation of a definite integral . . . 144 100. Interchanging limits 145 101. Change of limits corresponding to a change of variable . 145 CHAPTER XV THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM 102. Area under a curve 148 * 108. Evaluation of the limit 150 • 104. The fundamental theorem 150 •• 105. Plane areas in cartesian coordinates ..... 151 • 106. Plane areas in polar coordinates . . . . . .154 107. Volumes of revolution 156 108. Volumes of revolution : second method .... 157 109. A theorem on infinitesimals 158 110. Other volumes 161 111. Line integrals 163 112. Geometric interpretation of the line integral . . . 164 113. Fundamental theorem for line integrals .... 165 114. Evaluation of line integrals . ...... 165 115. Length of a curvilinear arc ....... 167 116. Surfaces of revolution ........ 168 117. Cylindrical surfaces 170 xiv ' CONTENTS CHAPTER XVI INTEGRAL TABLES ART. PAGE 118. Use of tables , . . 172 CHAPTER XVII IMPROPER INTEGRALS 119. Definitions 175 120. Geometric interpretation 177 CHAPTER XVIII CENTROIDS. MOMENTS OF INERTIA I. Centroids 121. Mass; density . . . ... . . . .179 122. Moment of mass 180 123. Centroid • . 180 124. Centroids of geometrical figures . . . . . . 182 125. Determination of centroids by integration .... 183 126. Centroids of plane areas 185 127. Centroids of volumes 187 128. Centroids of lines 188 129. Centroids of curved surfaces 189 II. Moments of Inertia 130. Moment of inertia 190 131. Radius of gyration 190 132. Determination of moment of inertia by integration . . 190 133. Moment of inertia with respect to a plane .... 194 134. General theorems on moments of inertia . . . ., 195 135. Kinetic energy of a rotating body . . . » . 198 CHAPTER XIX LAW OF THE MEAN. EVALUATION OF LIMITS 136. Rolle's theorem 200 137. The law of the mean . . . . '. . . .200 138. Other forms of the law of the mean 201 CONTENTS XV ART. 139. The indeterminate forms -, — . 00 140. The indeterminate forms • go, co — oo 141. General remarks on evaluation of limits PAGE 202 204 206 CHAPTER XX INFINITE SERIES. TAYLOR'S THEOREM I. Series of Constant Terms 142. Series of n terms 209 143. Infinite series . 209 144. Sum of an infinite series 210 145. Convergence and divergence . ' 211 146. Tests for convergence . . . . . . . .211 147. Cauchy's integral test 212 148. Comparison test 214 149. Ratio test 216 150. Alternating series 218 151. Absolute convergence 219 II. Power Series 152. Power series . 153. Maclanrin's series . 154. Taylor's series X|(^55. Taylor's theorem . "^^56. Approximate computation by seiies 157. Operations with power series 158. Computation of logarithms . 220 222 223 226 228 230 234 CHAPTER XXI FUNCTIONS OF SEVERAL VARIABLES I. Partial Differentiation 159. Functions of several variables 160. Limits; continuity ....... 161. Partial derivatives 236 236 237 XVI CONTENTS 162. Geometric interpretation of partial derivatives 163. Higher derivatives ..... 164. Total differentials 165. Differentiation of implicit functions . PAGE 238 238 240 241 II. Applications to Solid Analytic Geometry 166. Tangent plane to a surface 244 167. Normal line to a surface 245 168. Angle between two surfaces ; between a line and a surface 246 169. Space curves 248 170. Tangent line and normal plane to a space curve . . 248 171. Direction cosines of the tangent 250 172. Length of a space curve 250 CHAPTER XXII ENVELOPES. EVOLUTES 173. Envelope of a family of plane curves . 174. Determination of the envelope . 175. Envelope of tangents ..... 176. The evolute 252 252 254 256 CHAPTER XXIII MULTIPLE INTEGRALS 177. Volume under a surface 178. Volume under a surface : second method 179. Interpretation of the given function . 180. The double integral .... 181. The double integral in polar coordinates 182. Transformation of double integrals 183. Area of a surface 184. Triple integrals 185. Heterogeneous masses .... 186. Centroids and moments of inertia : the general case 258 262 263 264 265 266 268 270 273 276 CONTENTS xvii ART. 187. 188. 189. 190. CHAPTER XXIV FLUID PRESSURE Force Force distributed over an area Fluid pressure Resultant of parallel forces . 191. Center of pressure PAGE 279 280 281 283 284 CHAPTER XXY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER I. General Introduction 192. Differential equations . 193. Order of a differential equation . 194. Solutions of a differential equation 286 287 287 II. Equations of the First Order 195. The general solution 287 196. Particular solutions ........ 289 197. Geometrical interpretation 290 198. Separation of variables 292 199. Coefficients homogeneous of the same degree . . . 294 200. Exact differentials .295 201. Exact differential equations 296 202. Integrating factors 297 203. The linear equation 298 204. Equations linear in/ (j/) 300 205. Geometric applications , 301 CHAPTER XXVI DIFFERENTIAL EQUATIONS OF HIGHER ORDER I. Introduction 206. General and particular solutions . . . . . . 304 207. Geometric interpretation 305 XVlll CONTENTS 11. The Linear Equation with Constant Coefficients 208. The linear equation .... 209. The homogeneous linear equation 210. The characteristic equation . . ' . 211. Distinct roots 212. Repeated roots .....' 213. Complex roots ..... 214. Extension to equations of higher order 215. The non -homogeneous linear equation 307 307 308 309 309 310 312 313 III. Miscellaneous Equations of the Second Order 216. The equation ?/"= /(^) 316 217. The equation y" = /{y) 317 218. Dependent variable absent 318 219. Independent variable absent 319 CHAPTER XXVII APPLICATIONS OF DIFFERENTIAL EQUATIONS IN MECHANICS I. Rectilinear Motion 220. Rectilinear motion .... 221. Motion of a particle under given forces 222. The equation of motion 223. Uniformly accelerated motion 224. Momentum ; impulse .... 225. The principle of impulse and momentum 226. Work 227. The principle of kinetic energy and work 228. Constrained motion .... 229. Simple harmonic motion 230. Attraction proportional to the distance 231. Hooke's law 321 322 323 324 326 326 327 328 329 330 332 333 11. Plane Curvilinear Motion 232. Rotation 335 233. The simple pendulum ........ 336 234. The equations of motion . . . . . . . 337 235. Projectiles . . 338 DIFFERENTIAL AND INTEGRAL CALCULUS CALCULUS CHAPTER I FUNCTIONS. LIMITS. CONTINUITY 1. Functions. If a variable y depends upon a variable X so that to every value of x there corresponds a value of y, then y is said to be ?, function of x. For example, (a) the area of a circle is determined by the radius and is therefore a function of the radius ; (S) the attraction (or repulsion) between two magnetic poles is a function of the distance between them ; (c) the volume of a given mass of gas at a constant temperature is a function of the pressure upon the gas. A complete study of the properties of a function is possible in general only when the function is given by a definite mathematical expression. For this reason we shall be concerned almost entirely with functions defined in this way. Thus, in the examples above, we have (a) A = 7rr2, (c) for a "perfect gas," h v = -. V But the existence of a functional relation between two quantities does not imply the possibility of giving this relation a mathematical formulation. If by any means whatever a value of y is determined corresponding to B 1 2 CALCULUS everj ^^a«.ue of x und'3^ consideration, then y is a function of X. For example, the temperature of the air at any point of the earth's surface is a function of the time at which the thermometer is read, although no mathematical law connecting the two variables is known. We often wish to express merely the fact that «/ is a function of x^ without assigning the particular form of the function. This is done by writing y =/(^) (read y equals / of a;). Other letters may of course be used in the functional symbol, as F{x)^ (^)i '^C^)? ^tc. The value of f(x) when x = ai^ denoted by the symbol /(a). Thus, if f(x)=x^-Zx-l, then /(a) = a2 - 3 a - 1, /(2) = - 3, f(^x + ii) = {x + hy - s(x + A)- 1. Except where the contrary is explicitly stated, the vari- ables and functions with which we shall have to deal are restricted to real values. This restriction is introduced for the sake of simplicity, and also because in the elemen- tary applications only real quantities are of importance. 2. Geometric representation. The student is already familiar with the geometric representation of a function as the ordinate of a plane curve. Thus in (a) of § 1 the graph is a parabola ; in (5) it is a certain cubic curve; in (c) it is an equilateral hyperbola. Even though no mathematical expression for the func- tion is known, it may still be represented graphically. For instance, to represent the temperature at a point of the earth's surface as a function of the time, let a large number of readings be taken, the corresponding points be plotted on coordinate paper with time as abscissa and temperature as FUNCTIONS. LIMITS. CONTINUITY 3 ordinate, and a smooth curve be drawn through the points. This curve will represent approximately the variation of temperature throughout the time-interval in question. 3. Independent variable. We usually think of x as varying arbitrarily — i.e. we assign values to x at pleasure, and compute the corresponding values of y. The variable X is then called the independent variable^ or argument. But it is clear that if y is a function of x^ x is likewise a func- tion of y, and in general either one may be chosen as the independent variable. The values assigned to x must of course be compatible with the conditions of the problem in hand. In most cases X is restricted to a definite range or interval ; for instance, if the function we are dealing with is «/ = V2;, we restrict X to positive values. 4. Kinds of functions. We shall have to deal with both algebraic and transcendental functions. The algebraic functions are rational integral functions^ or polynomials ; rational fractions^ or quotients of polynomials ; and irra- tional functions, of which the simplest are those formed from rational functions by the extraction of roots. The elementary transcendental functions are trigonometric and inverse trigonometric functions ; exponential functions., in which the variable occurs as an exponent; and logarithms. Function algebraic transcendental rational irrational elementary higher integral fractional trigonometric exponential inverse trig'c logarithmic 5. One-valued and many-valued functions. A function y =f(x') is said to be one-valued, if to every value of x corresponds a single value of y ; two-valued, if to every value of X correspond two values of y, etc. In the case of a many-valued function it is usual to 4 CALCULUS group the values in such a way as to form a number of one- valued functions, called the branches of the original func- tion. Thus the equation defines a two-valued function whose branches are y = — '^x. In dealing with many- valued functions, we shall in gen- eral confine our attention to a particular branch. / EXERCISES , 1. Express the surface and volume of a sphere as functions of the radius ; the radius as a function of the surface and of the volume. 2. Express the surface and volume of a cube as functions of the length of its edge. 3. Represent geometrically each of the functions of Ex. 2. 4. rind/(0,/(3),/(-l),/(0),/(:r + A),if (a) f{x) = 2 a: + 5 ; {h) f{x) = x^ - 3 x + 3 ; (c)/(x) = sin7rx; {d) f{x) = 2\ 5. Exhibit graphically each of the functions of Ex. 4. 6. Plot the graph of each of the functions (a), (6), (c) of § 1. 7. Restate the examples (a), (h), (^) of § 1 both in words and by an equation, with the independent and the dependent variable inter- changed. 8. Plot the graph of each of the following functions: l-\- x^ (0 y = 7-^2' ("^^ y = r-^' 1 + X^ 1 — X f 9. Show that (a) the graph of a one-valued function is met by any parallel to the y-axis in not more than one point ; (b) the graph of a many-valued function consists of a number of branches (not necessarily disconnected), each of which has this same property. Give examples. 10. Show that the equation i/^ = x'^ — a^ defines y as a two-valued function of x, and draw the graph. FUNCTIONS. LIMITS. CONTINUITY 11. The freezing point of water is 32° Fahrenheit, 0° Centigrade ; the boiling point, 212^ F., 100° C. Express temperature in degrees F. as a function of temperature in degrees C, both analytically and graphically. 12. A sum of money is placed at simple interest. Express the amount at any time as a function of the time, and draw the graph. , 6. Rate of change ; slope. A fundamental problem in studying the nature of a function is the determination of its rate of change. Let P : (x^ y) be a point on the graph of the function Assign to X an arbitrary change, or increment^ Ax (read delta ic, not delta times a:), usually taken positive, and denote by Ay the corresponding change in ?/, so that the point P' : (x -{- Aa:, y + Ay) is a second point on the curve. The ratio Fig. 1 — ^ is the average rate of change Ax of g with respect to x in the in- terval Ax; geometrically this ratio is the slope of the chord PP', If now we let Ax approach 0, the ratio — ^ in gen- Ax eral approaches a definite limiting value, which is de- fined as the rate of change of g with respect to x at the point P. The geometric interpretation is obvious : when Ax is taken smaller and smaller, P' approaches P along the curve, the chord PP' approaches the tangent at P as its limiting position, and —^ approaches as its limit the slope of the tangent. Hence the rate of change of a function is the slope of its graph. 6 CALCULUS 7. Limits. From what has just been said, it appears that the determination of the rate of change of a function, or the slope of a curve, requires the evaluation of a certain limit. It will therefore be well to introduce at this point a brief discussion of the subject of limits. When the successive values of a variable x approach nearer and nearer a fixed number a, in such a way that the difference a — x becomes and remains numerically less than any preassigned positive number however small, the con- stant a is called the limit of a:, and x is said to approach the limit a — in symbols, lim X = a. Examples are easily found in elementary work : (a) If a regular polygon be inscribed in a circle, the difference between the area Ap of the polygon and the area Ac oi the circle becomes arbitrarily small (less than any preassigned number) as the number of sides increases in- definitely. Hence Irm Ap = Ac (6) We know from elementary algebra that the sum Sn of the geometric series IS + 1 2n-l 2- 1 2n-l is 1 1-i The difference between 2 and S^ is 2n-l This difference becomes arbitrarily small as the number of terms increases indefinitely ; hence lim S,, = 2. ((?) If a steel spring of length I suspended vertically be stretched to a length I -{- a and then released, tlie end of FUNCTIONS. LIMITS. CONTINUITY 7 the spring will oscillate about its original position. The length x of the spring will be alternately greater and less than the original length Z, but as the oscillations be- come smaller the difference between x and I will become and remain arbitrarily small. Thus lim x — l. In this example, the variable actually reaches its limit, since the spring soon ceases to oscillate at all. In many cases, however, the variable never reaches its limit. This is true in (a) above, since no matter how many sides the polygon may have, its area is always less than that of the circle. 8. Theorems on limits. We shall have occasion to use the following theorems on limits, which we assunae without formal proof. Theorem I* : The limit of the sum of two variables is equal to the sum of their limits. Theorem II : The limit of the product of two variables is equal to the product of their limits. Theorem III: The limit of the quotient of two variables is equal to the quotient of their limits^ provided the limit of the denominator is not 0. f increases i Theorem IV : If a variable steadily \ , \ but I decreases j never becomes . \ than some fixed number A., the vari- able approaches a limit which is not \ \ than A. Theorems I and II may evidently be extended to the case of any number of variables. * In theorems I, II, III it is of course implied that the limits of the two variables exist. We shall see later (§§ 139, 140) that the sum of two vari- ables, for instance, may approach a limit when neither of the two variables taken by itself approaches a limit. 8 CALCULUS 9. Limit of a function. We have frequently to observe the behayior of a function f(x) as the argument x ap- proaches a limit. If, as x approaches a, the difference be- tween f(x) and some fixed number I ultimately becomes and remains numerically less than any preassigned constant however small, the function f(x) is said to approach the limit Z, and we write Unless otherwise specified it is supposed that the same limit is approached whether x comes up to a from the positive or the negative direction. If we wish to consider what happens when x approaches a from the positive side only, we write ^^\ /(^) ; from the negative side only, '"" fix). 10. Infinitesimals. An infinitesimal is a variable whose limit is 0. Thus a constant, however small, is not an in- finitesimal. An infinitesimal is not necessarily small at all stages of its variation ; the only thing necessary is that ultimately it must become and remain numerically less than any assignable constant however small. If one infinitesimal is a function of another, the inde- pendent variable is called the principal infinitesimal. In the problem of § 6, both Aa; and A^ are infinitesimals, with ^x as the principal infinitesimal. 11. Limit of the ratio of two infinitesimals. We return to the exceptional case of theorem III, § 8, in which the denominator is infinitesimal. Given any fraction - in which u approaches 0, two cases are to be distinguished : (a) V also approaches ; (J) V does not approach 0. It is clear that in case (6) the fraction — may be made u to assume values greater than any assignable constant by FUNCTIONS. LIMITS. CONTINUITY 9 taking u sufficiently small ; hence the fraction can ap- proach no limit. But consider case (a), in which both u and V are infinitesimal. Theorem III does not apply; the ratio of the limits is -, which is quite meaningless ; never- theless the limit of the ratio may exist, as we shall find in many cases in the next few chaj)ters. The determination of the limit of the ratio of two infinitesimals is a problem of the greatest importance ; in fact, it is clear from the discussion of § 6 that this problem always arises in finding the rate of change of a function, or the slope of a curve. EXERCISES 1. Determine (a) 1"^ (x^ - 3 x^ - 5 a: - 5) ; "- ^ x>— 1 /j\ lim ^ — X — \ Which of the theorems of § 8 are needed? 2. Determine (a) ^i^] x-^ - 3 a; + 2 (b) l^^l (sin X + cos x) ; (c) lira ^. Ans. (c) -2 2x Which of the theorems of § 8 are needed ? 3. Determine lim x'^ - 3 .r + 2 ^y^ich of the theorems of § 8 •^■>l x -1 are used? Ans. — 1. 4. Evaluate ^i"^ ^ ~ ^' • 5. Evaluate lim Vl - x\ ^^^^ 1 Vg 6. Evaluate J^^^^^- ^ns. 1. ■^^■^tan X 7. Evaluate ^i^ sin_2£. Ans. 2. •^■^0 tan X 10 CALCULUS 8. Evaluate ^"^ ^^^Ix •^■^0 sin X 9. Show that, if n is a positive integer, lim x"" = (lini xy. 10. Show that, if P{x) is a polynomial in x, 11. Show that, if P^ix) and P2(x) are polynomials, lim^iM = ^PiOO ^>« PaCa:) P2(a)' provided PgC*^) =^ ^^ 12. Under what circumstances may the limit in Ex. 11 exist when P2(a) = ? Give an example. 13. Does the limit in Ex. 11 always exist when Pi(a) = P2(a) = 0? Give examples. 12. Continuity. An important idea in the study of functions is that of continuity. A function /(re) is said to be continuous at the point a; = a if This means, first, that the function is jigfined when x = a^ and second, that the difference between /(a:) and /(a) be- comes and remains arbitrarily small (numerically less than any assignable constant) as x approaches a. The curve y =f(x') passes through the point x = a without a gap or break. A function is said to be continuous in an interval of values of the argument if it is continuous at all points of the interval. ^ In the discussion of § 6, it is ta_citly assumed that the function is continuous in an interval including the point P\ this assumption is an essential part of the argument. All the functions treated in this hook are continuous., ex- cept perhaps for certain particular values of the variable, and such values are either excluded or subjected to special investigation. FUNCTIONS. LIMITS. CONTINUITY 11 13. Infinity. The most important type of discontinuity is that in which the function increases numerically without limit, or, as we say, becomes infinite^ as x approaches a. In this case we write lim fix)=cc. But it must be noted that this equation is merely symbolic, for the reason that the symbol oo does not represeyit a num- ber. The symbolic equation tells us, not that f(x) ap- proaches some vague, indefinite, very large limiting value, but that it increases beyond any limit whatever. Graphically the occurrence of such a discontinuity means that the curve y = f(x) approaches nearer and nearer the line x= a^ usually without ever reaching it, at the same time receding indefinitely from the a:-axis. Examples: (a) As x approaches 0, the function 1 a;* becomes infinite (Fig. 2) linii =00. a>>0 x^ Fig. 2 (5) The function Fig. 3 y = X 12 CALCULUS becomes positively or negatively infinite according as x ap- proaches 2 from the right or the left (Fig. 3) : lim ^ =+Q0, lira _l_ = _co. x-^'i^ a; — 2 .r->2"' 2; — 2 14. Function with infinite argument. We have fre- quently to investigate the behavior of a function as the argument becomes infinite. If when X increases indefinitely the difference between f(x^ and some fixed number I ultimately becomes and re- mains numerically less than any preassigned constant how- ever small, we write Graphically this means that the curve y = f(x) ap- proaches nearer and nearer the line y = U usually without ever reaching it, at the same time receding indefinitely from the ?/-axis. Examples : (^oC) As x increases indefinitely in either direction, the function y = —^ approaches (Fig. 2): lim 1 = 0. 1 + 1 (5) lim ^_+l_ lim ^ = 1. 1- x EXERCISES 1. Show that a polynomial is continuous for all values of x (see Ex. 10, p. 10). 2. For what values of a: is a rational fraction discontinuous?- 3. For what values of x is the function discontinuous? a;2-4 4. Evaluate lira -^ — - ■ Trace the curve y = — — r« 6. Evaluate («) ^''\ ^+i ; (b) ^^"^ ^^^1. x->0+ x x->0- X FUNCTIONS. LIMITS. CONTINUITY 13 6. Evaluate l^n -r^^ll^. 7. Evaluate 4 X — (a) lim 3^2+ 5x . j-^ao a;2 — 3 a; — 1 (&) lim ^ a:->oo 3 x- - 4 (c) lim 10^ ; (^) Urn 10-; (e) lim a;2 + 3z+l. x^^ a: - 5 ' (/) lim tana:. Ans. (a) 3; (c) 0; (/) non-existent. 8. Does sin x approach any limit as x becomes infinite ? Does sin X q Dogg tan x ^ X X 9. Show that as x approaches 0, the function sin - oscillates be- X tween — 1 and 1, without approaching any limit. 10. Discuss the behavior of tan - near the origin. 1 11. Discuss the behavior of 10* near the origin. 12. Evaluate lim 2- sin-. 13. Is the function x-^ — 4 continuous at x = 2 ? Can/(2) be so defined as to make/(a:) contin- uous? 14. If f(x) is continuous, is its square continuous? Is its reciprocal ? 15. Given two continuous functions, what can be said of the con- tinuity of theu' sum ? Their product? Their quotient? 16. Are the trigonometric functions continuous for all values of the argument ? Discuss fully. CHAPTER II THE DERIVATIVE 15. The derivative. We return now to the problem (§6) of finding the rate of change of a function, or the slope of a curve. Given a function continuous at the point P : (x^ «/), let us assign to x an ar- bitrary increment A2;, and compute the corresponding in- crement ^y of y. We have y + ^y^f(x + Lx), so that ^y=Ax + Lx)-f(x). Now form the ratio Fig. 1 dkX Ax The limit of the ratio —^ as Ax approaches is called the Ax derivative of y with respect to x. The derivative is designated by the symbol -^ : ^= lim ^= lim fCx-hAx)-f(x) Other commonly used symbols for the derivative are y'^ The operation of finding the derivative is called differentiation . 14 THE DERIVATIVE 15 It follows from § 6 that the derivative of a function is ' identical with its rate of change. Geometrically the deriva- tive of a function is the slope of its graph. Only differ eyitiable functions (i.e. those having a deriva- tive) are considered in this hook. In some cases the de- rivative may fail to exist for particular values of the argument, but such values are either excluded or subjected to special investigation. To hnd th6 derivative of a given function, we have merely to huild up the " difference-quotient " — ^ and then pass to the limit as Ax approaches 0. It will be remembered that this is essentially the method used in analytic geometry to find the slope of a curve. Since Ax and Ay approach together, our problem is to find the limit of the ratio of two infinitesimals (cf. §11). In general, this limit cannot be evaluated until some suitable transformation, algebraic or otherwise, has been applied to the quotient _^ • Ax The process of finding the derivative is illustrated by the following Examples : (a) Find the slope of the parabola y = 2x^-6x-[-4: at the point (ir, 3/) ; at the point (1, 0). If y=f(x) = 2x^-6x + 4:, then y -\- Ay = f(x -\- Ax} = 2(x -h Axy-6(x -h Ax} -h 4, A^ = 4 xAx + 2 A? — 6 Ax^ ^ = 4.x+2Ax-6, Ax y'= lini ^ = 4 a; _ 6. ^ Ax^O Ax Fig. 4 16 CALCULUS Hence the slope at any point (x, y) is 4 a^ — 6 ; in par- ticular, the slope at the point (1, 0) is — 2. (5) Given t dt We have 1 * + ^'-« + A«' A„_ 1 1 « + A« t Reducing the fractions in the right member to a common denominator, we find ^^ ^ ^ - (^ + AQ ^ -A/^ Whence Ag^ -i A^ " (^ + AO*' ^= lim 4^=_i. Geometrically this means that the slope of the hyper- bola s = - at the point (^, s) is — -• Zl Li (c) Find the rate of change of the function y = V^ at the point (rr, y); at the point (4, 2). If y=^x, then y -\- l^y = Va: + Aa;, A?/ = Vo:; H- Aa; — Va: = ( Va; 4- b^x — Va;) _ (a: + Aa;) — a; Va; + Aa:; 4- Va; A^ ■ Va; + Aa: + Va; Va: + Aa: 4- Va; Va: + Aa; + Va; THE DERIVATIVE 17 ^= liiii %^ 1 . At the point (4, 2), the rate of change is EXERCISES ^X 1^=4 ■! Find the slopes of the following curves at the points indicated. 1. y = X — x^ 2ht (x, y) ; at :c = 2. Trace the curve. 2. 3/ = a:^ + 1 at (z, ?/). Trace the curve. 3. y = x^ — x^ at the points where the curve crosses the a:-axis. Trace the curve. 4. 3/ = at X = 2. ' Ans. X -\-\ 6. y = — at a: = 2. x^ 6. y = a:^ - 3 a;2 + 2 at (a;, 3/). 7. 3/ = X + - at a; = 2 . 8. If 3/ = -, find '^. a:^ 6/a; 9. li y = V3 — a:, find y' . Ans. - 1 2\/3-x 10. If/(:c)= ^^ -,find/'(a:). Ans. ^ (1 - xy (1- xy 11. If s is measured in feet and t in seconds, find the rate at which s is changing at the end of 2 seconds when (a) s =- ^; (^) s =Vt -\- 1. Ans. (a) | ft. per second. X 1 . . 12. At what points does the curve y = have the slope -? a: + 1 4 Am. (1, i), (-3,1). 13. Differentiate 3/ = — -• Ans. ^^-■. Va; 2xi ^14. Find ^ if r = $1 Ans. ^oK du c 18 CALCULUS 15. Differentiate y = ^ (x-l)2 16. If f{x) = Vrt2 _ a;2^ find f'(x). Ans. — X Va2 17. Find the angle between the curve y = • and the line a; + 1 y = X at each point of intersection. 16. Higher derivatives. The derivative of ^ with re- spect to X is itself a function of x. The derivative of the first derivative is called the second derivative^ and is written — ^ (read d second 7/ over dx square); the derivative of (JiX the second derivative is called the third derivative, written ^- etc Other symbols for the higher derivatives are y", i/'^', • . • ; D.%D.%...;f"(x-),f"'Cx%.... Example: In example (a), § 15, we found . y' = 4:X — 6. Hence ■ y + A^' = 4:(x -h Ax^ — 6, Ay' = 4 Aa;, Ax y" = lim — ^ = 4. Aa:->0 Aa^ In this case all the higher derivatives are 0. EXERCISES 1. Find y" and y'" in Exs. 2, 3, 5, p. 17. 2. In example (?>), § 15, find 3. In Ex. 10, p. 17, find/"(x). 4. In Ex. 11, p. 17, find how fast -y- is changing when t = 2 seconds. CHAPTER III DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 17. Introduction. In this and a later chapter (V) we develop certain standard formulas by means of which any elementary function may be differentiated. The use of these formulas effects a great saving of time, and obviates the necessity of evaluating a special limit in every problem. The formulas of §§ 19-20 are direct consequences of the definition of the derivative, and are valid for all functions {i.e. all functions that are continuous., one-valued.^ and differentiahle ; see §§ 12, 5, 15). 18. Derivative of a constant. We note first that the derivative of a constant is 0: (1) ^=0. ^ ^ dx For, if y = c, then no matter what the values of x and Aa: may be, y will remain unchanged, and hence Ay = : ^ = 0,^= lim ^=0. Ax dx Ax->o Aa; The line ?/ = o Aa; dx dx Proof of (3): y = uv, y -\- Ay = (^u + Aw)(v + A?;), Ay = uAv H- vAii + AuAv^ Ay Av , Au , . Av Ax Ax Ax Ax ^^ lim ^ = u—-{-v~ dx Ax->oA2; dx dx Proof of C^): u y = -'' V . u-\- Au y + ^y= T. ^ V + Av . u-\- Au u uv + vAu — uv — uAv Ay = • -— = — — — -— , V -\- Av V {y -\- Av)v DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 21 Au Av V u — A^ Ax Ax Ax~ Qv + Av)v du dv V- u — dg _ i[^^ A^ _ dx dx dx~ A-^-^oAa: v^ Formulas (2) and (3) can be extended to the case where n functions are involved. For three functions, (3) becomes d du , dv , dw —uvw = vw—~ +WU \- uv dx dx dx dx In the special case when t* = (?, a constant, (3) and (4) become (3') —cv=c--. dx dx dv r4'^ -lf-__^ ^ ^ dxv~ v'' ' 20. Derivative of a function of a function. A function is sometimes expressed in terms of an auxiliary variable which in turn is a function of the independent variable ; for example, ?/ = 5 w^ + 2 i*, where u = x^ -\- ox-{-l. The variable u in the first equation may of course be replaced by its value in terms of a:, and — ^ can then be dx determined directly ; but it is desirable to have a formula by which — ^ can be found without eliminating u. dX Let y =f(u)^ where u = 4>(_x). Assign to X an increment Ax, and denote by Au and Ai/ the corresponding changes in u and ?/. Then A?/ _ A?/ Au Ax Au Ax and, passing to the limit, we find 22 CALCULUS T Ay T Ay -,. Au lim — ^ = lim — ^ • lim — , Ax->^Ax aj:->o A-w Ax->oAa; or ^ ^ dx du dx This very important formula is easily remembered from the fact that inform it is a mere identity. 21. Derivative of jc**, n a positive integer. If ?/ = a;", where w is a positive- integer, then (1) 4^ 1=^^^'^" For, y +Ay=(x + Axy = 3;"H-na;"-^A2;+ ^^^^^ ~ ^^ x^'-'^Ax + ••• + A^", 2 I Ay = na^"~iAa; + ^^^7 — ^^""^A^;^ + • • • + Ax"", 2 ! ^ = nx'^-^ H- ^(^- ^) a:^-2Aa; + -. + A^"-^ Ax 2! — ^ = lim — ^ = na;'^"^ 6?a: A^->.o Ax In particular, if ?i = 1, i.e. ii y = x., . dx __ H dx which is obvious geometrically. By means of (4'), it can be shown that (1) is true when w is a negative integer. Examples: (a) Find the derivative of y = ?>7^ + l x^+1. dx dx dx = 9 a:2 + 14 a:. (5) Differentiate y = x^ a;-f 3 I DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 23 y ^ ax ax ^ {x+-6f _ {x + ^')2 X - x^ ^ x^ + Q X (2; +3)2 ix+2>y EXERCISES Differentiate the following functions. 1. (a) y = bx^-2x; (b) s = t -'St^ + t^. 2. (a) ?/ = x4 - 3 x3 - 2 a:2 _ 1 ; (6) y ^ x\o x^ + 3). ,.3. (a) y = l-2x-3a;5; (6) y ={x^ - l)(x2 + 3x + 2). 1 +a:2 ^ a;2-l (x2-l)2 - 5. y = L±2^LZL^. 6. 2/=(xB-l)(:.3+i). x^ + 6 ,7. v = ^^- ■ 8. y = (l+^)a-2x). 1 - a: X 9. If ^^= -^'-^ , find^. Am. -\ X dx^ x' 10. Find — (5 a:3 + 7 a:2 + 8 a:)(x2 + 3 x + 4). dx 11. Differentiate ?/ =(a: + l)(a: + 2)(x + 3). 12. If a: = -, find — • t^ dfi _ /v^+i (1-0 13. If F{t) = {-!—Y, find F'(0. ^m. 9 3 14. Find the rate of change of s = t —- -\ t t^ 15. In the proof of (1), § 21, why is n assumed to be a positive integer? 16. Show directly from the definition of the derivative that for- mula (1) of § 21 holds when n = ± I. 17. Find ^ if ^/ = 2 w^ _ 4, m = 3a;2 + 1. dx 18. Find the slope of the curve y = x (x + l)(x + 2) at the points where it crosses the x-axis. Trace the curve. 24 CALCULUS 19. At what points is the tangent to the curve y =:(x — 3) '-(a: — 2) parallel to 0X1 Trace the curve. 20. Prove formula (l)of § 21 when n is a negative integer. 21. Given a polynomial of the n-th degree, prove that all the deriva- tives after the n-th are identically 0. 22. Derivative of x^, n fractionaL By means of formula (5), the power formula (1) of § 21 can be extended at once to the case when n is a rational fraction. If 2 where p and q are positive integers, then y^ = xP. Differentiating each member of this equation with respect to x^ we find, by (5), dx q y'^~^ q y'^ q x^ q This shows that formula (1) of § 21 holds even when n is a positive rational fraction. By using (4'), the formula can be shown to hold when ?i is a negative rational fraction. In more advanced texts it is proved that the power formula holds when n is irrational, and hence is valid for all values of n. 23. The general power formula. Suppose y = u^^ where u = . 15. ?/ = — — • vlns. ?/' = V(a2_x2)3 - (a2-x2)l 16. Find the slope of the hyperbola x'^ — y^ = 12 at (4, — 2). Ans. -2. *47. If <^(y) =^ ^ , find <^'(tO, <^"(0. <^"'(^') 8(1 -t;)2 18. Find 1 Vl - s • Vl + s^. 19. If y =2 V^, find v". 20. Find — (at - xl)i 26 CALCULUS 21. Differentiate ^ =( ) . Ans. y' = ^ \^1 + Vl - x'^J x-\/l - a;2 22. Find the slope of the curve y = (x^ — l)^ at each of the points where y" = 0. Trace the curve. 23. Draw the graph of the function y = a:" for n = i, 1, |, 2, 3. 24. Implicit functions. Up to this point we have been concerned with functions defined explicitly by an equation of the form i/=f(x}. It may happen, however, that x and i/ are connected by an equation not solved for i/ ; for example, x^ -^ y^ =. aP'. In such a case y is called an implicit function of x^ and the relation is expressed by writing The definition becomes explicit if we solve for y ; in the above example, y = ± Va^ — x'^. 25. Differentiation of implicit functions. To find the derivative of an implicit function, we proceed as follows : Differentiate each term of the equation Fix, y) = 0, hearing in mind that^ owing to the equation^ y is a fu7iction of X. Example : Find the slope of the ellipse x^ — xy + y"^ + x^l at the point (a:, ?/). We have d {x^ — xyi-y^-^-x^^Oy ax or by (6) and (5) 2x-x^-y-h2y^ + l = 0, dx dx dy _ 2x — y -\-l dx X — 2y DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 27 26. Inverse functions. The equation (1) y=/(^) defines x implicitly as a function of y ; when solved for ^, it takes the form (2) ^ = <^(y> The function (^(^) is called the inverse of the direct func- tion f{x). For example, (a) if ^ = ^^ then x=^±. Vy ; (J) \i y = ± Va^ — a:^, then a: = ± Va^ (c) if ?/ = «•% then x = log^-grT""' In each case the second func- tion is the inverse of the first. To construct the graph of the inverse function from that of the direct function, we have only to interchange x and y^ which amounts to a reflection in the line y = x. This is shown in the figure for example (a) above. If f(x) and (f>(^y) are in- verse functions, then 1 y (3) For, since 6.x J_ Ay Fig. 5 (*'(2/)^0). we have, passing to the limit, dy^l_ dx dx'' dy which by (1) and (2) is the desired formula. 28 CALCULUS EXERCISES 1. Express y explicitly as a function of x, when (a) x^ — y^ — 1\ (6) 2 a;?/ + 2/2 — 4 I (c) sin (x + 2/) = 1 ; {d) xa^ = 1. Find the slopes of the following curves at the points indicated. 2. a;2 + 7/2 = 25 at (x, y) ; at (3, 4). Ans, - -; - -• y 4 - 3. a:2 + x?/ + ?/2 = 3 at (1, 1). ^m. - 1. 4. 2 a:2 + 2 2/^ - 9 a:?/ = at (1, 2). - 5. xy^ = 3 at (3, 1). Do this in two ways. Find ^ in the following cases. dx ^ 6. 1-^^=1. Ans. — a^ h^ a^y 7. {x — y) (x- + yy = a^. 8. x^ -\- y^ — 3 axy = 0. ,9. x-\- V(2x - 'dyy = 0. 10. x^y^ + 2 x^y - xy^ + 2 = 0. 11. a; = (1 — 3 ?/)2. Solve in two ways. 12. If i/2 = 4 ax, find 3/". Cf. Ex. 19, p. 25. Ans. - ^. yZ ~" 13. Prove that a tangent to a circle is perpendicular to the radius drawn to the point of contact. 14. If a:2 + 2/^ = a", find y" . 15. Show that d'^y d^x dx'^ f dv \ df \dxl 16. Find the inverse of each of the following functions : (a)?/ = 3a:-4; (&)2/ = £z^; X — 2 (c) y = (x2-l)2; (f7) 2/=logioa:. 17. In Ex. 16, are the direct functions one-valued? Are the in- verse functions? 18. In Ex. 16 (a), (c), verify formula (3), §26. 19. In Ex. 16 (a), (c), verify that the graph of the inverse function may be found from that of the direct function by reflection in the line y = x. 20. In Ex. 16 (&), (c), discuss the continuity of the direct and in- verse functions. CHAPTER IV GEOMETRIC APPLICATIONS 27. Tangents and normals to curves. It is known from analytic geometry that the equation of a line through the point (a^Q, y^) with slope m i% (1) y-y^ = m(ix-XQ). Let P : (a-Q, y^) be a point on the curve F(x. y-) = 0, and denote by ^q' the value of the derivative at the point (a^Q, ?/q). The equation of the tangent to the curve at P c^ then be written, by (1), in the form The equation of the normal — i.e. the line perpendicular to the tangent at the point of contact — can be found at once from that of the tangent, by recalling that if two lines are perpendicular, the slope of one is the negative reciprocal of the slope of the other. Examples: (a) Find the tangent and normal to the ellipse 42:2 + 9^/2 = 40 at the point (1, — 2). , t We have Sx-hlSyy'=0, hence '^ = 2/0 = - 4:X' (1, -2) Therefore the equation of^the tangent is 2/ + 2=f(a:-l), that of the normal is Fig. 6 y + 2 = -ICx-l). 29 30 CALCULUS Fig. 7 (6) Find the equation of a tangent to the curve y = 7^ parallel to the line «/ = 3 :r + 1. The slope of the required tangent is 3. But the slope at any point (a;, ?/) of the curve y = 7?\'& y=3a:2. Hence the coordinates of the point- of contact are found by solving the simul- taneous equations 3 a;^ = 3, y =^Qi^. This gives the points (1, 1), (— 1, — 1), and the required tangents are ^ _ 1 = 3(^ _ 1), ^ + 1 = 3(a; 4- 1). 28. Length of tangent, subtangent, normal, and sub- normal. Let P : {x, y} be a point on the curve FCx, 2/) = 0. The segment TP of the tangent intercepted between the point of tangency and the 2:-axis is called the length of the tangent; its projection TQ on OX is called the length of the subtangent. The segment iVP of the normal inter- cepted between P and the a:-axis is called the length of the normal; its pro- jection $iV on the a:-axis is called the length of the subnormal. It is customary to consider all these lengths as essentially positive. They are evidently de- termined by the coordinates (a;, y} of the point P and the slope at P. EXERCISES Find the tangent and normal to each of the following curves at the points indicated. 1. (a) y = 1 — X — x^ at (1, — 1) ; (c) x2 + y2 = 25at(-3, 4); (b) xy = 2 a-2 at (a, 2 a) ; {d) y = a: + - at (1, 2). Ans. (a) ^ + 1 = - 3(a; - 1), i/ + 1 = \{x - 1). GEOMETRIC APPLICATIONS 31 "^ x^ — 2 xy -\- 2 y^ — X = at the points where x = 1. Ans. At (1, 0); 2y = x-l, i/-\-2x = 2; at (1, l):2?/ = a: + l, ?/ + 2a:=3. "St y = — — - at x = 2a. Ans. x-\-2u = ia, y = 2x — ^a. 4. Find the equation of the tangent to (a) y^ = iax at (x^, y^) ; (&) ^^ ± |^ zz: 1 at (xq, y^) . Ans. (a) y,y = 2a(x + x,); (b) ^ ± M ^ i. 5. Find the subtangent, subnormal, tangent, and normal lengths in each of the cases of Ex. 1. Draw a figure in each case. 6. Find the angle between the parabolas y^ = x, y = x^ at each of their points of intersection. 7. Find the tangent and normal to the curve y"^ = 2 x^ — x^ at the points X = 1. Ans. At (1, 1) : ^ 2y = x-{-l,y + 2x=3', at (1, - 1) : x + 2y + 1 = 0, y = 2x - 3. _ 8. Show that the subtangent to the parabola y^ = 4: ax is bisected at the vertex, and that the subnormal is constant. Hence give a geometric construction for drawing the tangent and normal; also show how to find the focus of a parabola if the axis is given. 9. Find a tangent to the parabola y^ = 4ax making an angle of 45"^ with the a:-axis. Ans. y = x + a. 10. Find the tangents to the hyperbola '^x^ — 9y- -{■ 36 = per- pendicular to the line 2 y ■\- 5x = 10. Ans. 2x— 5y=±8. —11. Find a tangent to the curve y = 1 — x^ parallel to the line X — 2 y = d. 12. Find a normal to the parabola y = x'^ perpendicular to the line 3 X — 2 y = 1. -13. Show that the portion of the tangent to the hypocycloid 2 2 2 x3^y'5 = «3 intercepted between the axes is constant. 14. Find the tangents to the circle x^ + y'^ = 5 which are parallel to the line x -r 3 y = 0. Draw the figure. Ans. y±iV2 = -l(x±iV2). 15. Find the tangents to the curve y = \ x^ — x'^ -{- 5 X which make an angle of 45° with the x-axis. Plot the curve. 16. Find the angle between the line y = — 2 x and the curve y = x2(l — x) at each point of intersection. 32 CALCULUS 17. Find the equation of a tangent to the curve y^ = 1 — x parallel to the ?/-axis. Trace the curve. ^18. Show that the area of the triangle formed by the coordinate axes and the tangent to the hyperbola 2 xij = a^ is constant. 19. Show that the length of the normal is constant (equal to a) in the circle (x — c)^ + 7j'^ = a^, where c is arbitrary, and explain geometrically. 20. Show that the sum of the intercepts on the axes of the tangent 1 1 1 . to the parabola x^ + y^ = a^ is constant. 21. Show that, in the curve y"^ = ax, the subtangent is n times the abscissa of the point of contact. Hence show how to draw the tan- gent at any point of the curve y — ax". 22. Find the length of the tangent, subtangent, normal, and sub- normal to the curve y = f(x) at the point {x, ?/). Ans. Tangent, ■^ Vl + y''^\, subtangent, ^; normal, ^/Vl + y''^; subnormal, yy' . y' y' 29. Increasing and decreasing functions. In studying the properties of a function it is usually of great assistance to represent the function graphically. In tracing a curve, it is well to begin by locating several points, e.g. the intersections with the axes, and finding the slope at those points ; it is also useful to note the behavior of y for large positive and negative values of x. In addition to giving the slope at any point, the differ- ential calculus is of assistance in a variety of other ways, as will be shown in the next few articles. We shall assume as usual that the function in question is one-valued, continuous, and differentiable. We note first that, as x increases, the curve rises if the slope is positive, as on the arc AB (Fig. 9); it falls if the slope is negative., as along BD : li y' > 0, ?/ increases ; If y^ < 0., y decreases. GEOMETRIC APPLICATIONS 33 At a point such as B (Fig. 9), y F Fig. 9 Of course this also appears at once from the fact that y' is the rate of change of y. 30. Maxima and minima. where the function is al- gebraically greater than at any neighboring point, the function is said to have a maximum^ value ^ and the point is called a maximum point. Simi- larly, at a point such as D the function has a minimum value. It is evident that at such a point the tangent is parallel to OX ; i.e. / = 0. But the vanishing of the derivative does not mean that the function is necessarily a maximum or a minimum ; the tangent is parallel to OX at ^, yet the function is neither a maximum nor a minimum there. It appears from the figure that the test is as follows : At a point where ?/' = 0, if y' changes from positive to negative (as x increases), y is a maximum ; if y' changes from negative to positive, y is a minimum ; If y' does not change sign, y is neither a maximum nor a minimum. Since the function is continuous, the maxima and minima must alternate : between two maxima there must be a minimum, and vice versa. The points at which «/' = are called critical points, and the corresponding values of x are the critical values of X. 31. Concavity. The second derivative is the rate of change of the first derivative. It follows that wKen ^ is positive, y is increasing : as x increases the tangent turns in counterclockwise sense and the curve is concave upward. 34 CALCULUS When «/'' is negative, y^ decreases : the^curve is concave downward. At a maximum point the curve is concave downward, and hence «/'^, if it is not 0, must be negative. At a minimum «/", if not 0, must be positive. If the second derivative is easily obtained and if it does not happen to be at the critical point in question, it is usually more convenient to determine whether we have a maximum or a minimum by finding the sign of «/'' ; but the test of § 30 has the ad- vantage of being perfectly general. However, in practice other considerations usually enable us to distinguish be- tween maxima and minima without the application of either of these tests. Example: Find the maximum and minimum values of the function y = x^ — '^ X., and trace the curve. This curve crosses the a:-axis at a: = 0, ± VS. Since ^' = 3 2:2 - 3, the slope at (0, 0) is — 3, at ( ± V3, 0) it is 6. Setting / = 0, we find the critical points (—1, 2), (1, —2). When x is large and negative, y is large and negative ; when x is large and positive, y is large and positive. It is therefore clear that the curve must rise to a maximum at ( — 1, 2), fall to a mini- mum at (1, — 2), and then rise in- definitely. These conclusions may be verified as follows. The second de- rivative, y'^ = 6 a:, is negative at a; = — 1, so that \ the Fig. 10 point ( — 1, 2) is a maximum ; y\ is GEOMETRIC APPLICATIONS 35 positive at x=l, hence (1, — 2), is a minimum. The curve is shown in the figure. EXERCISES Examine the following functions for maxima and minima, and trace the curves. 1. y = x{x + 5). 2. y = x^ - 2x'^ + x. ■^3. y = x^x'^ - 8). 4. ?/ = (x2 - 4)2. .. b. y ■= x^ +1. 6. ?/ = a;3. 9. ^ = ^,- -r 10. a: =0 + 1)3. 32. Points of inflection. A point at which the curve changes from concave upward to concave downward, or vice versa, is called a point of inflection. At a point of inflection the tangent reverses the sense in which it turns, so that y" changes sign. Hence at such a point y" ^ if it is continuous, must vanish. In Fig. 9 the points (7, U, F are points of inflection. Since y^' — i.e. the rate of change of the slope — vanishes at a point of inflection, the tangent is sometimes said to be stationary for an instant at such a point, and in the neighborhood of the point it turns very slowly. Hence the inflectional tangent agrees more closely with the curve near its point of contact than does an ordinary tangent ; it is therefore especially useful in tracing the curve to draw the tangent at each point of inflection. 33. Summary of tests for maxima and minima, etc. The results of §§ 29-32 may be summarized as follows : Let the function y =/ (x) and its first and second derivatives be one- valued and continuous. (a) In an interval where y' > 0, the curve rises; where y' < 0, the curve falls. (J) A point where y' = is a maximum or a minimum point., unless at the same point y" = 0, in ivhich case see (e) 36 CALCULUS below. If y' > at the left* of this point and y' <0 at the right., y is a maximum; if y^ < at the left atid y' "> at the right., y is a minimum. Or., if y" 0, ?/ is a minimum. ^c) Iri an hiterval where ^jy" > Q, the curve is concave upward; where y" < 0, the curve is concave downwc^rd.. (c?) A point at which^['_==3^is a^^^ inflection., pro- vided, y" changes sign as the curve ^(^sses through the point. (g) A point at which both y' =^ and y" = is a maxi- mum or a minimum if y' changes sign as the curve passes through the point; it is a point of inflection with a horizontal tangent if y' does not change sign. Example: Trace the curve y = x(x — 1)^. This curve crosses the a;-axis at (0, 0), (1, 0). For large positive or negative values of x, y is large and positive. The derivatives are y z= (^ _ 1)3 j^^^x(x- 1)2= (x - 1)2(4 X - 1), y" = %x - 1)(4 a: - 1) + 4(a; - 1)2= 6(:r - 1)(2 x - 1). The slope at (0, 0) is — 1 ; at (1, 0) it is 0. The critical values are a; = 1, \. When a:=|, y" is positive ; hence (^, — 2V6 ) ^^ ^ minimum point. When 2; = 1, y" = ' and the test by the second de- rivative fails. Since y' does not change sign as -X ^ passes through 1, the function has neither a Pjq jj maximum nor a mini- mum at that point. The second derivative vanishes at (J, — y^g), (1, 0) ; these are points of inflection. The slope at (^, — y^g) is ^. The curve is shown in the figure. V * That is, immediately at the left. GEOMETRIC APPLICATIONS 37 EXERCISES Trace the following curves. Where possible, find the points of intersection with the axes, determine the behavior of y for large values of x, find the maxima and minima and points of inflection, and draw the tangent at each point of inflection. 1. ^/ = a:3 - 6 a;2 + 9 a; + 3. 2. ?/ = 4 + 3 x - a;^. /3. ?/ = x3 - 3 x^ + 6 a: + 10. 4. ^/ = (x - 3)2(x - 2). 5. y = (l- ^2)3. 6. ?/ = (4 - x^y\ /7. y={x- \y{x + 2)2. ^. y = a;3 - 3x2 - 9 x + 5. 9. y = x^. 10. y = x^. 11. y =x(x -l)(x-2). 12.?/= ^^^ 14. y = x2 + 4 a2 1 (1 +X2')2 15. Show that the curve y = has three points of inflection x2 + a'^ lying on a straight line. Trace the curve. 16. Show that, for the curve y = x", where n = 2, 3, 4, •••, the origin is a minimum or a point of inflection according as n is even or odd. 34. Applications of maxima and minima. The theory of maxima and minima finds application in a great variety of problems. In the applications it is rarely necessary to use either of the tests of ,§ 33 to distinguish between maxima and minima ; the critical value that gives the desired result can usually be selected by inspection of the conditions of tlie problem. It frequently happens that the function to be teste4 for maxima or minima can be most simply expressed in ter ms of two variables. When this is done, a relation between the t wo variable s must be found from the condi - ti ons of the problem - By means of this relation one of the variables can be eliminated, after which the maxima 'and minima can be found in the usual way. However, it is often more convenient not to eliminate, but to proceed as in example (5) below. 38 CALCULUS Examples: (a) A box is to be made of a piece of card- board 4 in. square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in this way. Let X be the length of the side of each of the squares cut out. Then the volume of the box is (1) r=<4V-^^.2. The derivative is ^={(4 _ 2 a;)2- 4 a;(4 - 2 z) = (4-2:r)(4-6x). Setting ^y dx we find — 9 ^T- 2 x^z ov 3 Since V vanishes when a; = and again when a: = 2, it must reach a maximum at some intermediate point; it therefore follows without the application of further tests that the critical value x — ^ gives the required maximum volume : F„»x. = |(4-|)^ = Wcu. in. The graph of equation (1) is shown in the figure, the F^-scale being one fourth as large as the a?-scale. Since by the condi- tions of the problem x is restricted to values between and 2, the dotted portions of the curve have no meaning" in the present case. Fig. 12 & r (5) Find the dimensions of the largest rectangle that can be inscribed in a given circle. Take the coordinate axes parallel to the sides of the rectangle. The area of the rectangle is (2) • A = 4.xy. GEOMETRIC APPLICATIONS 39 This can be expressed in terms of x (or y) by means of the relation (3) x'^y'^a^ which gives y = Va^ — a;^, ^ = 4 x^ 0^ — 2:^. From this point the method is the same as in {a). Fig, 13 The problem can be solved without eliminating x or y, as follows: Differentiating equation (2) with respect to x and setting the derivative equal tu 0, we have, since ?/ is a function of x^ ^ = 4f^^+.Vo, or dx dy dx dx y. X Differentiating equation (3), we get ax or dy _ x dx y Equating values of -^, we find dx _ ^ = _ ^, X y whence y = x'. the maximum rectangle is a square. EXERCISES 1. What is the largest rectangular area that can be inclosed by 800 yd. of fencing? 40 CALCULUS 2. For a rectangle of given area, what shape has the minimum perimeter ? •f^ 3. Find the most economical proportions for a cylindrical tin cup of given volume. Ans, Radius = height. 4. A rectangular field is to be fenced off along the bank of a straight river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing? 5. The equal sides of an isosceles triangle are 10 in. long. Find the length of the base if the area is a maximum. 6. Find the rectangle of maximum perimeter inscribed in a given circle. ^'7. Find the most economical proportions for a box with an open top and a square base. + 8. Find the most economical proportions for a covered box whose base is a rectangle with one side twice the other. A ns. Altitude = | x shorter side. •f- 9. Find the dimensions of the largest right circular cylinder that can be inscribed in a given sphere. Ans. Diameter = V2 x height. 10. In Ex. 9, find the form of the cylinder if its convex surface is a maximum. •^11. Find the dimensions of the largest rectangle that can be in- scribed in a given right triangle. Ans. x = ^ a. -fl2. Find the most economical proportions for a conical tent of given capacity. Ans. h =V2 r. J^. A man in a rowboat 6 miles from shore desires to reach a point on the shore at a distance of 10 miles from his present position. If he can walk 4 miles per hour and row 3 miles per hour, where should he land in order to reach his destination in the shortest possible time ? Ans. 1.2 miles from his destination. 14. A rectangular field of given area is to be inclosed, and divided into two lots by a parallel to one of the sides. What must be the shape of the field if the amount of fencing is to be a minimum ? 15. A Norman window consists of a rectangle surmounted by a semicircle. What shape gives the most light for a given perime'ter? A ns. Breadth = height. •f-16. Find the most economical proportions for a quart can. Ans. Diameter = length. GEOMETRIC APPLICATIONS 41 "/---i7. The strength of a rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the strongest beam that can be cut from a log of given diameter. A ns. Depth = V2 x breadth. 18. Find the volume of the largest box that can be made by cutting equal squares out of the corners of a piece of cardboard 6 x 16 in. and turning up the sides. Ans. if^^ cu. in. AS. A gutter is to be made of a strip of tin 12 in. \ / wide, the cross section having the form shown in the V y figure. What depth gives a maximum carrying \ 4 / capacity? ' Fig. 14 20. Find the most economical proportions for a cylindrical cup of given capacity, if the bottom is to be three times as thick as the sides. 21. Find the most economical proportions for an A-tent of given volume, whose sides slope at 45° to the horizontal. 22. Find the dimensions of the largest right circular cylinder that can be inscribed in a given right circular cone. Ans. Altitude = \h. 23. Solve Ex. 22 if the convex surface of the cylinder is to be a maximum. 24. Find the right circular cone of maximum volume inscribed in a given sphere. 25. Find the cone of minimum volume circumscribed about a given sphere. 26. The base of a box is a rectangle with one side twice the other. The top and front are to be made of oak, the remainder of pine. If oak is twice as valuable as pine, find the most economical proportions. 27. A wall tent 12 x 16 ft. is to contain a given volume. Find the most economical proportions. Ans. Height above eaves = 2 a/8 ft. 28. An oil can is made in the shape of a cylinder surmounted by a cone. If the radius of the cone is three fourths of its height, find the most economical proportions. Ans. Altitude of cylinder = altitude of cone. 29. A cupboard 5 ft. high and having shelves 1 ft. apart is to be made from a given amount of material. If a front, but no back, is required, what shape gives the greatest amount of shelf room ? Ans. Width =: twice depth. 30. A silo is made in the form of a cylinder, with a hemispherical roof; there is a floor of the same thickness as the wall and roof. Find the most economical shape. Ans. Diameter = total height. 42 CALCULUS 31. Solve Ex. 30 if the floor is twice as thick as the wall and roof. Ans. Height of cylinder = diameter. 32. The cost of erecting an office building is $ 100,000 for the first floor, $ 105,000 for the second, 1 110,000 for the third, etc. ; other ex- penses (lot, plans, excavation, etc.) are $700,000. The net annual income is 1 10,000 for each story. How high should the building be, to return the maximum rate of interest on the investment? . Ans. 17 stories. 35. Derived curves. The curves y z=zf'(^x)^ y =f"(x), y =f^'^(jc)^ ... are called th.Q first., second., third., ... derived curves^ corresponding ^ to the curve y =f(x)* The relations be- tween the given func- tion and its first and second derivatives, which have been formulated analytic- ally in § 33, are well brought out graphi- cally by drawing the original curve and its first and second de- rived curves. This is shown in Fig. 15 for the curve y = X'^ x^ The work should be arranged with the several axes of ordi- nates lying in the same vertical line. It often happens in practical work that a function is defined in such a way — for instance by experimental data Fig. 15 GEOMETRIC APPLICATIONS 43 — that no mathematical expression for it is known. If then we wish to examine tlie behavior of one of the deriv- atives, as is frequently the case, we plot the graph of the original function from the given data, and construct the required derived curve graphically. The process is ob- vious. We can measure the slope at any point of the original curve ; the number thus obtained is the ordinate of the corresponding point on the first derived curve. In this way as many points as desired may be plotted on the first derived curve and a smooth curve drawn through them, after which the second derived curve may be con- structed in a similar way; etc. EXERCISES 1. What can be said of the first derived curve (a) in an interval where the original curve is < „ ,,. >? ° i falling J (b) in an interval where the original curve is concave \ ^ ^ ? [downward] (c) at a point where the original curve has a < . . >? [ minimum J (d) at a point where the original curve has a point of inflection ? 2. What can be said of the second derived curve (a) in an interval where the original curve is concave J upward | r, \ downward j (b) at a point where the original curve has a point of inflection? (c) in an interval where the first derived curve is i ^^® I v ■\ falling J * (d) at a point where the first derived curve has a | ^^[^^cimum 1 i minimum J ' 3. What can be said of the original curve at a point where the second derived curve touches the a:-axis without crossing it? 4. Plot the curve ij = x^ + x'^ and its first, second, and third derived curves. 5. Plot the curve y = sin x, and construct the first derived curve. What well-known curve does the latter resemble ? 44 CALCULUS 6. Draw a smooth curve, on a large scale, through the points a- _4 _2 2 4 6 8 10 12 14 16 18 20 22 24 y 0-1-106 10 3 0-1-1-1 1 9 20 35 and construct the first and second derived curves. 7. The national debt of the United States at the indicated dates is given in the accompanying table, the unit being $100,000,000. Con- struct the curve showing the rate at which the debt has increased or decreased. Date Debt Date Debt 1850 0.6 75 20.9 ^55 0.4 '80 19.2 '60 0.6 '85 13.8 '61 0.9 '90 8.9 '62 5.0 '95 9.0 '63 11.1 '99 11.6 '64 17.1 1900 11.1 '68 24.8 I '10 9.9 10.5 '65 26.8 '05 '70 23.3 CHAPTER V DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS I. Trigonometric and Inverse Trigonometric Functions 36. Trigonometric functions. The student is already familiar with the elementary properties of the trigono- metric functions. They are one-valued and continuous for all values of the argument a:, except that the tangent and secant become infinite when a:=±(2?i + l) — , the co- tangent and cosecant become infinite when x = ± wtt, where n is a positive integer. The sine and cosine, and their re- ciprocals, the cosecant and secant, are periodic with the period 2 tt ^the tangent and cotangent are periodic with the period tt. y =. sin 9CX Fig. 16 The properties just mentioned are well exhibited by the graphs of the various functions. The graphs of the sine, y = cos X Fig. 17 cosine, and tangent are shown in Figs. 16-18 ; the student should draw the graphs of the other functions. 45 46 CALCULUS function by the general method others can be obtained. The derivative is an important aid in the fur- ther study of these functions. Since all the func- tions can be ex- pressed in terms of the sincj it will be sufficient to find the deriva- tive of this one from this result all the 37. Differentiation of sin x. The derivative of sin x may be obtained directly from the definition of the deriva- tive (§ 15). We have y = sin X, y + Ay = sin {x -\- Aa;), A?/ = sin (x + A:r) — sin x^ t^y _ sin {x + A^;) — sin x Ax Ax Expanding sin (^x + Ax^ by the addition formula of trigo- nometry, we get Ay _ sin x cos Ax ■+■ cos x sin Ax — sin x Ax Ax By trigonometry, so that (1) cos Aa: = 1 — 2 sin^ 1 Ax^ Ay _ cos X sin Ax — 2 sin x sin^ ^ Ax Ax Ax sin Ax . sin A- Ax . . . = cos X • — sin X ' — r-2 . sin ^ Ax. Ax iAx TRANSCENDENTAL FUNCTIONS 47 It will be shown in the next article that lim?i!l^=l. Assuming this result for the moment, we see that lim sin i Ax T sin A:r -, iim — = 1, 2 I Ax = 1. Ajr->0 Ax Hence, passing to the limit in equation (1), we find dy _ d dx dx In some applications, it is convenient to write this formula in the form sm X = cos X. dx sin X = sm x-\ — 2 If u is any function of x^ it follows from formula (5) of Chapter III that du or (7) d ' d . —- sin u = —~ sin u • -— , dx du dx d . du . , , — sm z/ = cos w — = sin i/ + TT\du 2Jdx dx dx Example: Differentiate sin 5 x^. By (7), with u = 5x^ — • sin 5 a:^ = 10 a; cos 5 x^. dx 38. Limit of sin a/a as a approaches 0. In the differen- tiation of sin X we had to make use of the fact that lim5iE^=l. a-^0 a This result may be obtained as follows. Let P, Q be two points on a circle such that the chord PQ subtends an angle 2 a < tt. As a approaches 0, the ratio of the chord to the arc approaches unity : a->o SLTcPQ Fig. 19 48 CALCULUS But chord PQ=2r sin oe, and, if a is measured in radians, arcP§ = 2ra. Hence T chord PO T 2 r sin « t sin a ^ lim — ^ = iim = iim = 1; a->.o arc PQ a->o 2 ra a->o a When a is in degrees, the length of the arc is arc PQ =2r ' a, ^ 180 and the formula for — sin x is much less simple than dx when radians are used (see Ex. 26, p. 50). For this reason angles in the calculus are always measured in radians unless the contrary is stated. 39. Differentiation of cosj:, tanjr, etc. The derivatives of the other trigonometric functions can also be obtained directly from the definition of the derivative, but they are more easily found from (7). To differentiate cos x^ we write cos X = sm [ a; + — ], d d ' ( , nT\ f , TT — cos x = — sin [x -\- ~]= cos [x -{ — dx dx \ 2j V 2 = — sin X. If u is any function of rr, we find by formula (5) of Chapter III, ^o\ d . du f . 'TT\dU (8) — cos w = — sin w — = cos M + - — ^ ^ dx dx \ 2jdx The remaining trigonometric functions may be differen- tiated by expressing them in terms of the sine and cosine. The results are as follows : ■'""^^-^ (1) — tan X = sec^ a;, dx TRANSCENDENTAL FUNCTIONS 49 (2) — cot x = — cosec^ x^ dx (3) — sec X = sec x tan x, dx (4) — - cosec x = — cosec x cot x. dx If u is aii}^ function of x, we find by formula (5) of Chapter III, (9) — tan u = sec^ u — , dx dx d . n du — cot u = — cosec^ u — , dx dx d . du — - sec u = sec u tan w— -, dx dx d , du —- cosec u = — cosec u cot u — • dx dx EXERCISES 1. Trace the curve // = sin x, finding maxima and minima and points of inflection, and drawing the inflectional tangents. 2. Proceed as in Ex. 1 with the curves (a) y = cos X ; j^ y = tan x ; (c) y = sec x. Differentiate the following functions. 3. (a) sin 2 x ; j^ cos - ; j^) tan (tt + a:) ; (d) X sec X ; (e) a;^ cot a:; (^) (3 ^ + 1) cos 3 6; (g) ^; (A) sin^o;; • (0 cos3 2^. Ans. (h) —sin-; (K) 2 sin a; cos a- ; (/) — -6 cos^ 2 ^sin2 ^. x^ X 4t. y = X tan 2 x + Vl + x'^. I 4 X 5. ?/ = vl + sin a,-. * 6. ?/ = -; • sm X 7. ?/ = cot^ 4 a:. vl ns. — 12 cot^ 4 a: cosec^ 4 x. 8. ?/ = sin3 3 e. 9. r ::= sec (2 ^ + 1). Find ~ in the following cases. dx 10. Q.O'&ly — .r- + 4. ,4rlr 3/ sin x = cos 2 x. E 50 CALCULUS 12 y sin 2 x 13. y ^ Vl + tan^x. ^ * "^ 1 + cos 2 a; 14. 3/2 ^ sin 2 X. ^5- 3/' -2/ = tan |- 16. If ^ = sin a;, find ?/",?/'",•••, 2^('*). clx dP'x 17. If a; = cos a>/, find — -, — — -• (1 X 18. If a: = ^ sin ^^ + 5 cos kt, show that — -r = — k'^x. dt'^ 19. Obtain each of the formulas (l)-(4), § 39. 20. From the trigonometric formula for sin {x + a), deduce by differentiation the formula for cos (x + a). 21. Find the tangent and normal to the curve y = sin a: at x = — • 22. Find tangents to the curve y = tan x parallel to the line ^ = 2x4-5. 23. If /(x) = cos2 X, find/"(x),/'"(x), •••,/(")(a;). d^y 24. If V = ^ sin X, find —^ • ^ dx^ d^x 25. If X = f cos kt, find — —- • ' dfi 26. Show that if x is measured in degrees, the formula for the derivative of sin x becomes £sina; = j|5Cosx. 27. Differentiate cos x directly from the definition of the derivative. Sill X 28. Writing tanx in the form tan x = , obtain the derivative cos X of tan X directly from the definition of the derivative. 29. Find the maximum rectangle inscribed in a circle, using trigo- nometric functions. 30. Find the rectangle of maximum perimeter inscribed in a circle. 31. Find the right circular cylinder of maximum volume inscribed in a sphere. 32. Find the largest right circular cone that can be inscribed in a given sphere. Ans. V=^ira^. 33. A steel girder 30 ft. long is carried along a passage 10 ft. wide and into a corridor at right angles to tlie passage. The thickness of the girder being neglected, how wide must the corridor be in order that the girder may go round the corner? TRANSCENDENTAL FUNCTIONS 51 34. A wall 8 ft. high is 27 ft. from a house. Find the length of the shortest ladder that will reach the house when one end rests on the ground outside the wall. 40. Inverse trigonometric functions. The symbol arc- sin x^ or sin~i a;, denotes the angle ivliose sine is x: y = arcsin x ii x= sin y. That is, the function arcsin x is the inverse (§ 26) of the function sin x. The graph of 1/ = arcsin x is as shown in Fig. 20. It is of course the same as that of sin a;, with the coordinate axes interchanged; i.e. it is the reflection of the sine curve in the line i/ = x. The functions i/ = arccosa;, ?/ = arctan2:, etc., are de- fined in a similar way. In §§ 41-42 we consider only the three principal func- tions arcsin x, arccos x, arctan x. The other three func- tions may be treated similarly. 41. Restriction to a single branch. The trigonometric functions are one-valued : to a given value of the argu- ment there corresponds one and but one value of the func- tion. The inverse trigono- metric functions, on the other hand, are infinitely mani/- val- ued : corresponding to a given value of the variable there are infinitely many values of the function. Geometrically this means that a line x = Xq, if it meets the curve at all, meets it in infinitely many points ; the truth of this state- ment is evident from a glance at Figs. 20-22. Following the rule of S 5, 7 ,, ,, •^. ?/ = arcsin X we snail confine our attention -pio '>o -10 52 CALCULUS y = arctan x Fig. 22 to a single branch of each of these functions ; the branch chosen is the one drawn full in each figure. Thus in our future work the function arcsin x, for example, is restricted to the interval IT . . . TT — < arcsin x < — > 2= =2 This means that arcsin (— 1) = IT 2' IT not - — ; etc. Similarly, — < arctan x < — 2= -2 therefore arctan (— 1)= — — , arctan (— oo) = — — etc. TRANSCENDENTAL FUNCTIONS EXERCISES \ In the following, the restrictions laid down in § 41 are assumed to hold. 1. Find (a) arcsin |, (b) arcsin (— l), (c) arctan (— V.S), (d) arc- tan oo, (e) arccos (— |), (/) arccos (-1). 2. Show that arcsin x + arcsin (— x) = 0. 3. Show that arccos x + arccos (— x)= tt. 4. Show that (a) arccos x = - — arcsin x ; Ch) arccot x = arctan x = arctan - ; 2 X / N 1 TT . 1 (c) arcsec x = arccos- = arcsm - ; (a) arccosec x = arcsni -• X 42. Differentiation of the inverse trigonometric functions. To differentiate the function ?/ = arcsin a:, let us pass to the direct form sin y = X, Differentiating by the rule for finding the derivative of an implicit function (§ 25), we find dii -, cos y^ =1, dx dy ^ 1^ 1 ^ 1 dx cos^ Vl-sin2«/ VH^' hence or d . 1 arcsin x = dx ^i _ X- It should be noticed that cos y is put equal to Vl — sin^ y rather than — Vl — sin^ y. This is correct because, as Fig. 20 shows, the slope of the curve y — arcsin x is posi- tive at all points of the branch that we are considering. 54 CALCULUS In a similar way we find d -1 ■arccos x = dx VI - a;2' d , 1 — arctan x — r, • dx 1 + x^ By formula (5) of Chapter III we find that if u is any function of a:, du rM\\ d dx (10) — arcsin u — dx Vl - 1/2 du d dx arccos w= — dx Vl-i^2 da (11) ^ arctan w = ^ dr 1 + 1/2 While in the above discussion we confine our attention to a single branch of the function, it appears from Figs. 20-22 that if we know the slope at every point of one branch, we can at once find the slope at every point of any other branch. EXERCISES Find the derivatives of the following functions. 1. y = arcsin 2 x. 2. 1 ?/ = arccos - • ^ X 3. y = arctan (1 + 2 a:). 4. y = arcsin Vx. 5. y = arccot (2 x + 5) 2. 8. 1 V = arccosec -— • -^ 2x 7. s = t arcsin 3 t. p = Vl — arcsin v. 9. 2/ = (arcsin x) 2. 10. V- ^ 11. Ans. 2 arcsin X y = arctan i: . arctan x TRANSCENDENTAL FUNCTIONS 55 ^2. 5 = Vl - 2t arccos V2T. 13. y = z arctan 4 r. 14. V = arcsin — • Ans. - ■ — -, 15. y = t"^ SLTcsin-- 16. y = 2 V arcsin 2 x 17. ?/ = arcsin x + arccos x. Explain the meaning of the result. 18. If y^ sin x + y = arctan x, find y' . 19. Find tangents to the curve y = arctan x perpendicular to the line 4 a: + 3/ = 0. 20. Obtain — • arccos x from the relation dx arccos x = arcsin x. 2 21. Show that arctan x = arcsin and obtain — arctan x from this fact. dx 22. If 2/ = arcsin x, find ^-^, ^— ^« dx^ dx^ 23. If ?/ = arctan x, find — ^, — ^ . dx'^ dx^ 24. Show that d . - 1 ' arccot x = Vl + ; rf 1 — arcsec x — dx x^x'^ — 1 d ^ - 1 — arccosec x = dx xVx"^ - 1 25. Trace the curve y = arccot x. 26. Trace the curve y = arcsec x. 27. Trace the curve y = arccosec x. - II. Exponential and Logarithmic Functions 43. Exponentials and logarithms. The number a"(a > 0) is defined in algebra for all rational values of n. In the calcu- lus it becomes necessary to attach a meaning to the function ?/ = a^ (a>0) as X varies continuously. 56 CALCULUS Let Xq be any given irrational number. It can be shown that when x approaches Xq passing through rational values, the function a-^ approaches a definite limit. This limit is denoted by a^» : lim a^=^a-">. X-^Xo The function a-^ thus becomes defined for all values of x. This function is one-valued and continuous, and obeys the ordinary laws of exponents, viz. : (1) a^ ' a^ = a x+t («')' a xt The inverse of the exponential function is the logarithm^ defined by the statement that y = \og„x '\{ x=^ dJf (a>l*). This function is one-valued and^eonttnuous^for all positive values of x. The number a is called the hase of the sys- tem of logarithms. The graph of the function where g = 2.718..- (see §46), is shown in Fig. 23; the graph of its inverse y = log^a? is shown in Fig. 24. y = log X Fig. 24 * The assumption a > 1 is introduced for simplicity ; this condition is satisfied in all cases of practical importance. TRANSCENDENTAL FUNCTIONS 57 44. Properties of logarithms. For convenient reference we recall the fundamental properties of logarithms : ( 1 ) log^a:// = log^x + log„?/, (^) loga - = log«2: - log„^, (3) logaX'' = it log„2;, (4) log^a^ = X, (5) ' a}''%''=x, (6) log^a; = log^a; • log„^, To prove (1), let then we must show that p = m -{- n. Passing to the direct form, we have x^ = a^, X = a"*, y = a", so that Hence, by (1) of § 43, p z=z m + n. Formulas (2) and (3) may be proved in a similar way. Formula (4) is merely a restatement of the definition of the logarithm ; formula (5) is the converse. To prove (5), set and take logarithms to the base a on each side : log^a: = log„^, whence t = X. To prove (6), let m = lo g^a: and n = log^ a; ; then . x = a"^ — J". If we take logarithms to the base a on each side of' the equation 58 CALCULUS it appears that As a special case take x= a: the formula gives (7), 1 logft^ = loga^ EXERCISES 1. Find X, if (a) log^^x = 2, (b) log^^x = - h (c) log^x = 4, (d) logi„a;3= 4, (e) loga^; = 0, (/) \ogaX = 1. 2. Simplify (a) a^os^ (6) a'^'^ss^ (c) a^i^s^ (rf) aSiogx^ (•g) a'+iogx^ (/) «*"5^°^*, the logarithms being taken to the base a in each case. Ans. (b) i. 3. Prove formulas (2) and (3) of § 44. 4. Show that negative numbers have no (real) logarithms, 5. Show that numbers between and 1 have negative logarithms ; numbers greater than 1, positive logarithms. 6. Show that lim logoa: = — oo . 7. Find the inverse of the function a^ — a~* y = • a^ 4- a-^ 8. For what two values of a: is a^ = (a*) ? 45. The derivative of the logarithm. To obtain the de- rivative of the logarithm we proceed by the general method of § 15 : 7/ = log„a:, y + A?/ = log„(2; + Arc), X -\- Ax Ay = log„(2; + Ax) - \og^x = log„ by property (2) of § 44. Hence, Ax Ax \ X J X TRANSCENDENTAL FUNCTIONS 59 Let us multiply and divide by x and then make use of (3), §44: ^ = l.flog/l + ^ C^x X l\x \ X =iiogii+^r. X \ X J Hence, (1) f^ = lim ^ = 1 lim log/l + ^f (2) =- log„ X Aa^V^ lim 1 + Ax->0 V X J It will appear in the next article that the limit lim ( 1 4- - ) exists and is a number lying between 2 and 3. This number is denoted by the letter e ; we shall find later (Ex. 4, p. 230) that e= lim Tl + -Y= 2.71828 •••. n->- 2. Hence e lies between 2 and 3. As already stated, we shall see later that e = 2.71828 .... 47. Differentiation of the exponential function. The de- rivative of the exponential function «^ may be found as follows. If y = a^, then (1) log„ y = x. Differentiating (1) by the rule for implicit functions (§ 25), we find ^ ^ y ax 7 = z-^ — = y loge a, dx logo e by (7), § 44 ; hence d —- a^ = a^ logg a. dx For the case a = e, this formula becomes simply dx If w is a function of x^ we have — a" = a" loge a ' dx dx This formula, too, becomes simpler when a= e : (13) -^6^* = ^**^". dx dx TRANSCENDENTAL FUNCTIONS 63 EXERCISES 1. Show that common logarithms are transformed into natural logarithms by the formula log'io X = logio « • loge X = 0.4343 logex. 2. Show that logea; = 2.3026 l.ogiox. 3. By means of a table of common logarithms, show that log2 = 0.693, log 3 = 1.099, log 5 = 1.609. 4. Using the results of Ex. 3, find log ^, log VS, log 6, log 0.1, log ^9. Find the derivatives of the following functions. ^5. y = log 2 X. 6. y = log(\ +2:2). 8. 10. 1 (l-xy ^-^"^1 + 2.- '7. y = log y/D — X. 9. ,.logV^. y = logio 2 X. 11. y = logio (X2 - 1). 12. y = log« a;2. 13. y = log sin x. 14. y = X log X. 15. logx. ^ X 16. ?/ = (1 - x^) log a;. 17. y = log^ X. 18. ^ = log log x. 19. y = log log (1 - x). 20. y = e2x. 21. y = ex\ 22. y = xe-". 23. y = e'' log X. 24. y = 10-. 25. y = 2< 26. y = ee'. 27. r = eo^. 28. r = e^ cos 2 ^. 29. y = log cos 2 X. 30. ?/ = arcsin log x. 31. y = ecoss*^ 32. y = log 33. y = sin^ e". 34. 36. tan ? 35. y = Vl + log X. y 3= arctan c*. 37. y=(il-e^r. 1 X- 38. y = log e^*. 39. liy = e - , find y. 40. If / (6) = log log sin 2 (9, find /' (0) . 2 64 CALCULUS (ly 41. Given log (^x + y)= x^ + y"^^ find —-- 42. liy = log 2-, find ^/", t/'", •••, y("). ^ 43. li y = e% find ?/", y'", •-■, ?/<^"^. .4ns. ?/("■) = a"e°^. 44. Find the inverse of the function y = e^'°^. 45. Find the inverse of the function y = log cos 2 x. 46. Find the tangent and normal to the curve y = log x at (a) y = 0; (b) y=-^] (c) x = e\ 47. Show that the curve y =. e" has a constant subtangent. Hence devise a simple geometric construction for drawing the tangent to y — e'' dX any point. 48. Show how to draw the tangent to the curve y = log x. 49. Find the maximum and minimum points on the curve y = X log X. Trace the curve. 50. Trace the curve y = e 51. li y = xe"", find y", y'", •••, ?/''\ 52. Trace the curve y = xe^. 53. Find the equation of a tangent to the curve y =x log x parallel to the line 3 x — 2y = 5. 64. In passing from (1) to (2), § 45, we make use of the principle that lim (log x) = log (lim x). From which one of our assumptions concerning the logarithm does this principle follow ? 48. Hyperbolic functions. A class of exponential func- tions of frequent occurrence in some applications are known as hyperbolic functions. They are denoted by' the symbols sinh x (read hyperbolic sine of a;), cosh rr, and tanh x^ and are defined as follows : sinh X = , 1-1 cosh x= , 2 , 1 g^ — e~^ sinh X ' tanh X = = — gx _j_ g X cosh X TRANSCENDENTAL FUNCTIONS 65 The reciprocals of these are cosech x^ sech x^ and coth x respectively. Tables of hyperbolic functions have been computed ; see, for example, Peirce's Short Table of In- tegrals (Ginn and Co.). The inverses of the hyperbolic functions are called anti- Tiyperholie functions : y — sinli~^a; \i x=^ sinh ?/, etc. The fundamental properties of the hyperbolic functions are easily obtained from the definitions ; their derivation is left to the student in the exercises below. EXERCISES 1. Show that cosh^ X — sinh- a; = 1, 1 — tanh^ X = sech^ x, sinh 2 a; = 2 sinh x cosh x, cosh 2 X = cosh- x -h sinh'^x. 2. Show that — - sinh X = cosh x, -z- cosh x — sinh x ax fix 3. Show that sinh-i X = log (a; + Vl + x-). Fundamental Differentiation Formulas (1) 1=0. ^«N tf , dv du ^ ^ dx dx dx ^«/x d dv du dv y— — M— /A\ d_u_ dx dx ^^^ dxv 7^ ' 66 CALCULUS ^ ^ dxv~ v^dx' ^^ dx du' dx' ^ ^ dx dx' du (6') |^V^ = -^, (7) |sin.= cos.|.sin(.,-|)|, (8) |cos. = -sin.|=cos(. + |)|, (9) ^tanw=sec2z/^, OAT dr du (10) ^ arcsin w = ^ fir Vl - 1/2 (11) -T-arctanw = = -^ ^ ^ dx 1 + Z/2 du (12)^log„=f, (13) ^e^^e**^. MISCELLANEOUS EXERCISES Find the derivatives of the following functions. 1. sin3 - . 2. log tan 3 x. 3. aretana:3. 4. (1 - xy-(2 x + Sy. 5. e'=°»=^. 6. Vl - c6t X. 7. X arcsin -- 8. ^ 2 VI - a,-2 9. log^sin^. 10. arctan (1 - ^2). TRANSCENDENTAL FUNCTIONS 67 11. qos^2 X. 13. arcsin - -- V3-4a: 10. • 17. log log cos X. 19. 22^. 21. X log Vl — X. 23. sin a: cos 2 x. -.0 i 25. xVx^ + 1 27. (a: - 1)^ 12 (3x2-4)2 x^+1 14. tan2(l - x). X X 16. €- —e ^ 2 18. Cos2 (--x\ \4 1 20. Vsin x'^.. 22. (e'' - 1)4. 24. arccos log x. 26. 2 2 3 (a3 _ j:^)2. 28 sin2 2 e (1- cos 2^)2 30. logtan(| + |). (a:2 + 3 a: + 3)^ 29. log(e2x+ 1). Find y' in the following cases. 31. sin (x + y) = cos (x - y). 32. e^^/ = a; -f ?/. 33. ^^ + ^ = 3. 34. a:-?/ = tan(a:-?/). 35. Find ?/", if ay^ = x^ 36. Find y'", if a:2 - ?/2 ^ ^2. Find the slope of each of the following curves at the point indicated. 37. (a: - 2/)2 = 3 a: + 4 ?/ - 14 at (2, 2). 38. y = log X at the point where y = — 2. 39. 3^ = e^ (a) at the point where y = 2 ; (b) at the point where X = log 3. 40. arcsin a: + a;^ = 0, at the point a; = — 1. CHAPTER VI THE DIFFERENTIAL 49. Order of infinitesimals. We have found tliat in the problem of differentiation the increments Ax and A?/ are in- finitesimals, with Ax as the principal infinitesimal (§ 10). An idea of fundamental importance in the study of in- finitesimals is that of order. Let an infinitesimal v be de- fined as a function of a principal infinitesimal u. If where k=^0^ then u and v are said to be infinitesimal of the same order ; if —"- ^^ - ».-'■— —^„ .. lim !1 = 0, «->o u V is said to be of higher order than u. More precisely, if a number n can be found such that lim iL =. y^, where A; =?^ 0, v is said to be infinitesimal of the n-th order with respect to'u. If u and V are of the same order, we may write V = ku -\- eu^ where e is an infinitesimal. It is clear that when u ap- proaches 0, the term eu approaches more rapidly than does ku^ so that for small values of u the term ku is numerically the larger. For this reason the term ku is called the priyieipal part of v. Example : When the side Z of a square increases by an amount Al, the area increases by an amount AA = {1 + Aiy- P =2lAl+ 'Af. 68 IM M' J THE DIFFERENTIAL 69 If A^ approaches 0, AA does also. The two infinitesimals are of the same order, since lim ^A = lim (2 I + Al) = 2 I. ^i-X) Al AZ->o The principal part of AA is 2 lAl. The figure illustrates the fact that AA consists of a term of the first Fig. 2o and a term of the second order. EXERCISES 1. What is the increase AV in the volume of a cube of edge / when the side increases by an amount Al ? Show that if Al is infinitesimal, •AFis infinitesimal of the same order, and find the principal^part^f A V. Illustrate by a figure. Ans. A T'' =: 3 r^Al + 3 ZAI'^ + Al^- 2. Of the functions sin ^. sec 0, tan ^, 1 - cos ^, which are infini- tesimal with respect to as the principal infinitesimal? 3. As the radius of a right circular cylinder of given altitude approaches 0, the volume and the total surface do likewise. Show that the volume is infinitesimal of higher order than the total surface. 4. Given a right circular cylinder and a right circular cone of the same base and altitude, show that (a) if the altitude is infinitesimal, the lateral surface of the cylin- der is infinitesimal of a higher order than that of the cone; (h) if the radius is infinitesimal, the lateral surf aces. of the cylinder and cone are of the same order and (for small values of the radius) the former is approximately twice the latter. 5. Is the sum of two infinitesimals itself infinitesimal? Is the product? Is the quotient? 50. The differential. It follows from the above defini- tion that Al/ and Ax are in general infinitesimals of the same order. For, the limit of their ratio i^_^J/'^3'nd this in general exists and.Js dii|erentJrom 0. Further, the principal part of A^ is evidently y'Ax. It is easily seen from Fig. 26 that the principal part of A?/ = QP' is QB, 70 CALCULUS Fig. 26 the segment of A?/ cut off by the tangent at P. For, the slope at P is , _ QR _ QR ^ PQ Ax' so that QR = ^'Ax. The principal part of Ay (the length QR in Fig. 26) i% called the differential'^ of y and is writ- -X ten dy : dy = y' Ax. Hence the increment Ay consists in general of the differ- ential dy plus an infinitesimal of higher order. This is illustrated by the example of § 49. In particular, let y = a: ; then y' = 1, and dy = dx = Ax ; i.e. the differential of the independent variable is the incre- ment of the variable. We may therefore write dy^y'dx. Thus the differential of any function is equal to its derivative multiplied by the differential of the independent variable. The derivative of ^ with respect to x may now be thought of as a quotient — the differential of y divided by the differential of x. This is the reason for using the symbol -^ to denote the derivative. The symbol -^ may thus be dx dx considered as representing an actual division — the ratio dy -i- dx. It must be kept clearly in mind, however, that the derivative is a certain limit, viz. ^= lim ^. dx A^-^-o Ax If y = f (x)^ instead of writing dx * In case y' ^ 0. If y' = 0, then dii = 0. THE DIFFERENTIAL 71 we may, and often do, write dy=f'(x)dx. Thus the fundamental formulas of differentiation are often written in this so-called differential notation ; e.g. d(x'^') = nx^~'^dx^ d(\og u) = — , etc. u Examples: (a) If ^ = sin 2 0, then dy=2cos2edd, (5) Find an approximate formula for the area of a narrow circular ring. The area of a circle of radius r is A = 7rr2. If the radius be increased by an amount Ar, the area is increased by an amount AA whose principal part is dA = 2 7rr dr. Hence the area A^ of a narrow circular ring is approxi- mately the product of the circumference * by the width w : Af=2 irrw. EXERCISES Find the differential of each of the following functions. 1. (a) x2; (6) cos^; (c) ^^ _ i ; (^) log a;; (e) arcsm?/; (/) tan 2 a ; {g) ^^— ^; (^) sin2y. Ans. (a) 2xdx\ (h) -sin 6 dO. 2. (a) (l-3a:2)2; (6) log (1 - cos2 ^) ; (c) ue^; (d) arctane'; (e) xVa-i-bx; (/) — • Vx vT Z. y = x(l — x2)3. 4. y ■. 4: X 5. V = u sin2 u. Q. X = y log y. 7. y =—• " 8. s = arcsin (1 — f). * Either the inner or the outer circumference. 72 CALCULUS ft cos 6 ^ - . , 9. r = — -^ — 10. y=e~'^^u\kx. 11. F = I irr^. 12. X = t sin aL 13. y =(1 + OL^) arctan «. 14. // = cos^2x. 15. Find the difference between dy and A?/, if ?/ = a:^. Draw the figure. 16. Proceed as in Ex. 15 for the function y = x'^ — x'^. 17. If y = ^x, find A?/ and dy and show geometrically w^hy they are equal. 18. If s = 16 /2 + 25 ^, find the difference between As and ds when f = 12 and A« = .02. 19. Draw figures to show that dy may be equal to, greater than, or less than A?/. 20. Show that the error committed in using the approximate formula of example (6), § 50, is irw^. When r = 10 ft., what is the greatest allowable value of ic if accuracy to within 5% is required? Ans. About 1 ft. 21. If A is the area of a rectangle one of whose sides is twice the other, draw a figure showing the difference between dA and A^ when the length of the side changes (cf. Fig. 25). 22. If V is the volume of a cube, draw a figure showing the differ- ence between dV and AF when the length of the edge of the cube changes. 23. Find an approximate formula for the volume of a thin cylin- drical shell of thickness /. Ans. 2irrht. 24. Find an approximate formula for the volume of a thin spheri- cal shell. What is the greatest allowable thickness for a radius of 5 ft. if accuracy to 1 % is required? Ans. About 0.6 in. 25. Find approximately the volume of wood required to make a covered cubical box of edge 3 ft., using half -inch boards. Ans. 2\ cu. ft. 26. Work Ex. 25 if the dimensions of the box are 6, 4, and 2 ft. 51. Parametric equations ; implicit functions. A curve is frequently not determined by an equation between x and y^ but by two equations giving x and y in terms of a third variable, or parameter. These equations are called parametric equations of the curve. THE DIFFERENTIAL 73 For instance, the coordinates of a point moving in a plane are functions of the time : These two equations may be considered as parametric equations of the path. Again, the equations of an ellipse in terms of the eccentric angle (f) are x= a cos (f)^ ^ = b sin cf). While it may be possible to eliminate the parameter, thus obtaining the ordinary cartesian equation of the curve, it is often more convenient not to do so. When dealing with parametric equations, it is conven- ient to use differentials in finding derivatives, particu- larly the derivatives of higher order. The method is illustrated by example (a) below. Differentials can also be used conveniently in finding derivatives when the relation between the variables is an implicit one. Examples : (a) Find — ^ and — ^ when dx dx^ x=Zt^ y = t^ — ■^. We have ,_dgp = S dt^ dy = 2 t dt^ — ^ = ^^ • dx 3 To find — ^, put (for convenience) -^=y'. Then da^ dx hence dy' = I dt, d^y _ dy' __ ^dt _ ^ dx^ dx Sdt ^' (6) Find y' and y^' when x^ -^ y"^ = a^. Differentiating both sides of the equation, we get 2 X dx + 2 y dy = 0, y' = — -; y 74 CALCULUS J ! _ y dx — X dy dy x^ — y + x-^ ~ y yti ^§¥ ^ ^ ^ y_ dx y'^ y'^ _ — y*^ — x^ _ _a^ - ^3 yZ EXERCISES Find -^ and — ^ in the following cases. dx dx^ 1. (a) x=t'^, y = t -^] (b) x= fi + 1, y = t^', (c) X = cos 2 0, y = sin 2 6 ; (d) x = a cos^ 6, y = a sin^ 6 ; (e) X = e"^', ?/ = e« + 1 ; (/) a: = a (cos ^ + sin ^), y = a (sin ^ — ^ cos ^) . Ans. (a) ^ = -±- ^ ^ dx^ 4^3 2. (a) 2/2 = 4 ax; (b) x^ - y^ = 1 ; (c) a:^ + y~^ = a^; (^) xt + ?/^ = a^ ; (e) x-if = y^\ (/) x^ + ?/8 = 3 ^(a.^,. . , . r/2y 4a2 Ans. (a) — ^ = • dx'^ 7/3 Find -^, using differentials. 3. 3x33/2 — x?/ + ^2 — ?/ — 5 = 0. 4. 7/ = cos (x - 2/). 6. e^'+y' = xy. 6. ^^^ + ?/2 = 5; ■ X + y 7. log V^MT' = X. Ans. ^ = ^' + y'-^ . dx y 8. xy — x^y'^ -\- b y = 5. 9. x^ — 3 x^ + xy^ —2/2 = 0. CHAPTER VII CURVATURE 52. Differential of arc. Let s denote the length of the arc of the plane curve counted from some initial point Pq up to the point P : (x^ ^), and suppose for detiniteness that s increases as X increases. The arc s can be regarded as a function of ds X. Its derivative — may be found as follows : Fig. 27 As i^x ^__As_ ^ Va2:' + A^ As pp i^x pp' ^x where As is the length of the arc, PP the length of the chord, from P : (x, y) to P' : (x + Aa:, y + A?/). Since, as pointed out in § 38, lim >« , = 1' we have (1) Ax-^^ PP ^ =3 lim f^ = V' Hr If s increases as x decreases, then Ax \AxJ 75 76 CALCULUS and (2) ^=-Jl+(&f. dx ^ \dxj After squaring and clearing of fractions, equation (1) (or (2)), becomes i.e. ds is the hypotenuse of the right triangle whose sides are dx and c?«/. If the tangent to the curve at P makes an angle a with OX^ then dx . dy cos a = — , sm a = -^ > ds ds 53. Curvature. We say in ordinary language that a curve whose direction changes rapidly has great curvature^ or is sharply curved. Thus a circular arc is said to have greater curvature when the radius is small than when it is large. This somewhat vague idea may be made precise as follows. Consider, first, two points P, P' on a circle, and denote the arc PP' by As, the angle between the tangents at P, P' by Aa. The quotient — is evidently the change in the direction of the curve, per unit of arc* ; it is called the curvature of the circle. If now the curve in question is not a circle, the direction of the curve no longer changes uniformly, and the quotient — represents As Fig. 28 merely the average curvature of the arc As. But if P' be made to ap- proach P along the curve, so that As and Aa approach 0, the quantity — in general approaches a limit — , which * It is easily seen that, in the case of the circle, this quotient is constant. CURVATURE )?7 is called the curvature at the point P : K= lim Aa^rfa^ As-^o As ds The definition is of course independent of the particular coordinate system used ; the angle a is the angle made by the tangent at P with a7i7/ fixed line in the plane of the curve. When the equation of the curve is given in car- tesian coordinates, it is convenient to take a as the slope- angle of the tangent — i.e. the angle between the tangent and the a:-axis. The curvature fc is then easily expressed in terms of the coordinates. For, tan a = — ^ = ?/', ax da a = arctan y', Also, by § 52, Hence (1) ds = Vl -f y'^ dx. da y" ds (1 + y'2y It is customary to consider fc as essentially positive, so that, strictly speaking, we should write K = da 1-'^ ds (1 + y'^y where the symbol | a \ means the absolute or numerical value of a. It should be noted that when ^' = 0, formula (1) re- duces to fC=l/" Thus the value of the second derivative at any point is equal to the curvature at that point when the coordinate axes are so chosen that the first derivative is 0. 78 CALCULUS 54. Radius of curvature. The reciprocal of the curva- ture is called the radius of curvature^ and is denoted by p : _l _ds (1+ y'^y ^ ^~K~da~ y^ * This quantity is also to be considered as essentially positive. If a length equal to the radius of curvature p at the point P be laid off on the normal from P toward the con- cave side of the curve, the extremity Q of this segment is called the center of curvature. It can be shown that the circle with radius p and center Q represents the curve near P more closely than any other circle. This circle is called the osculating cir- cle., or circle of curvature. In general, the circle of curvature crosses the curve at P, as is the case in Fig. 29. EXERCISES 1. Show that the curvature of a straight line is everywhere 0. 2. Show that the radius of curvature of a circle is the radius of the circle. Find the radius of curvature of the following curves. 3. y = x'^ (a) at any point ; (b) at the vertex. 4. y'^ = 4: ax. Fig. 29 4a2 5. The equilateral hyperbola 2 xy = a^ at (a, I a). 6. y = x^ + 5 x^+ Qx at (0, 0). Ans. fVSa. Ans. 22.51. 7. The ellipse ^4-^=1 a^ b^ (aV + Mx^) CURVATURE 79 8. The hyperbola ^ - ^ = 1. 1 9. The hypocycloid a:^ + ?/^ = a^. Ans. ^(axy)^. 10. The ellipse x = a cos cf), y = b sin <^. 11. The curve x = t'^, y — \ — t^. 12. The curve x = ^t\ y = iM - t^. Ans. |(1 + t'^)'^. 13. The catenary y = ^\e'' -i- e V at the point (0, a). Ans. a. 14. Show that the curvature at a point of inflection is 0. 16. Find the point of maximum curvature on the curve y = e'. Ans. (-0.347,0.707). 16. At what points of the curve y = x^ \s the curvature greatest? 17. Plot the parabola a:^ = 4 y accurately, on a large scale, in the interval from a; = — | to a; = f , and draw the osculating circles at the points a: = 0, X = I, a; = 1. CHAPTER VIII APPLICATIONS OF THE DERIVATIVE IN MECHANICS 55. Velocity and acceleration in rectilinear motion. If a point P moving in a straight line describes equal spaces in equal times, its motion is said to be uniform. Its dis- tance X from the starting point is evidently proportional to the time : X = v^t. The constant factor v^ is called the velocity of the moving point ; it is equal to the space passed over per unit time. If the motion is not uniform, we introduce the idea of velocity at a jpoint or instant. Suppose that a distance ^x including the point P is described O P in time M : then the quotient — Fig. 30 ^' . -; - ■ IS the average velocity during that interval of time. If now A^ approaches in such a way that P always remains in Aa:, the quotient — - approaches a limit which is called the velocity at the point P. This limit is of course the derivative of x with respect to t : y^ lim ^^dx Thus the velocity is the time-rate of change of space described (cf. § 6). The rate of change of the velocity is called the acceleration : j= ~ dt 80 d'^x dt^ APPLICATIONS OF THE DERIVATIVE 81 If the acceleration y is constant, the motion is said to be nn'^ormly accelerated. An important case of uniformly ac^celerated motion is that of a body falling toward the earth from a point near the earth's surface, all resistances being neglected. The attraction of the earth gives the body an acceleration ^, called the acceleration of gravity^ equal to 32 ft. per second per second'approximately. Tn any problem, it is instructive to draw the graphs of ^ V;j^ and^^^as functions of ^^^^ In doing this, it should be remembered that the graph of v is the first derived curve (§35), the graph ofy is the second derived curve, cor- responding to the graph of x. EXERCISES * 1. A stone is thrown upward with a velocity of 64 ft. per second. The distance from the starting point at the time t (in seconds) is y = 16 r^ - 64 U the positive sense being downward. Find the velocity and the acceleration. How high will the stone rise and for how long a time ? Where is the stone and what is its velocity after 5 seconds of motion? What distance is covered in the sixth second V 2. In Ex. 1, draw the graphs of _?/, i\ and /. 3. A particle slides down an inclined plane. The distance from the starting point at any time t is a— 4 f2 _ 20 t. Discuss the motion. 4. A point moves according to the law a; = 5 cos 2 t. Discuss the motion. Draw the graphs of x, \\ and j. 5. A point moves according to the law ^ = 32 (1 — e-*). Discuss the motion. 6. A point moves according to the law x — log (1 + 2 t). Discuss the motion. Draw the graphs of x, i\ and/. 7. A point moves according to the law x — e~' sin 2 t. Discuss the motion. * The types of motion considered here will be discussed more in detail in Chapter XXVII. G 82 CALCULUS 8. The positions of a point at the ends of successive seconds are observed as follows : t\0 1 3 4 5 6 7 X\0 -1 -A 32 '3 Fig. 31 -1 I 4 7 Draw the graphs of v and J, and find an approximate expression for V andy in terms of t. ■ , ' 9fi^ Vectors. A right line segment of definite length, direction, and sense is called a vector. Vectors are of great importance in physics because they can be used to repre- sent velocities, accelerations, forces, and other fundamental quantities. The resultant of two vectors AB, ^C^ is the diagonal AD of the parallelogram having AB, AC SiS adjacent sides. Two forces acting on the same par- ticle are equivalent to a single force, their resultant ; similarly for other vectors. This is the parallelogram law. The operation of finding the resultant by the parallelogram law is called geometric addition. The original vectors AB, AC are called components of AD. It is evident that any vector may be resolved into cetpappnents in an infinite number of ways. sV'i., Velocity in curvilinear motion. If a moving point describes a plane curve, its coordinates are functions of the time : x={t), 7/ = ylr(t). The distance s passed over along the curve is also a function of the time. The velocity at any point P is defined as the vector, laid off on the tangent to the path from P, of magnitude y^ lim As^Js At^O ^f fit Fig. 32 APPLICATIONS OF THE DERIVATIVE §3 ^'\ The components of the velocity parallel to the coordi- nate axes are Vj. = v cos a, Vy= V sin a, where a is the angle between OJTand the tangent at P. By § 52, ds dx dx • ds dv dy V cos « = — • — = — , V sin a = — • -^ = -^ , dt ds dt dt ds dt so that , dx dy dt ^ dt Thus the rectangular components of the velocity of P are the velocities of the projections P^ and Py of P on the coordinate axes. By § ^Q^ the total velocity v is inclined to the ic-axis at an angle V * a = arctan -^ • The equations a: = 0 A^ laid off in the limiting direc- tion of Av, is called the ac- celeration of P at the time t. It is the so-called geometric derivative^ or vector deriva- tive, of v with respect to t. To find an expression for j in terms of the coordinates of P, we may resolve j into components jj. and jy parallel ,< CALCULUS to the coordinate axes. Denoting by <^' the angle between Av and the 2;-axis, let us project the triangle OFF' on OX : Av cos ', is the angle between j and the a:-axis. Similarly . . . dv^ d?x 1 sin ,2 + t,„2 V Hence, . dv Again. ) Jn = Jv COS a - j^ sin a _ dVy dx dv^ dy dt ds dt ds By § 54, the radius of curvature of the path is ds P=da' so that p\dt da dt da J 1 f dVy dv^ p\ ^ da ^ da Now, differentiating both sides of the equation (§ 57) a = arctan -^ 88 CALCULUS with respect to a, we find dv„ dv 1 1 = V 1 da da so that Hence, V 2 dv^ dv^ V — V da da dv^ dv^ n V — V — - = v^. da aa i;2 Jn — P Thus the acceleration j is equal to — 07dy in the case of rectilinear motion; in curvilinear motion — represents ■ dt the tangential component oi the acceleration. EXERCISES 1. In Ex. 4, p. 84, find y^, y^, y^, y„. Find y, (a) as the resultant of /j. andy^^, (&) as the resultant oi jt andy,i. 2. In Ex. 6, p. 84, find the total acceleration in magnitude and direction when t = 2. 3. Show that in uniform circular motion the acceleration is directed toward the center and -is proportional to the radius of the circle. 4. In Ex. 8, p. 85, find ji and y„, if the radius of the circle is 10 ft. 5. A point describes the parabola ^/^ = 4 x with a constant velocity of 6 ft. per second. Find v^, Vy, j\, andy^ at the point (1, 2). / 60. Time-rates. The question of determining time- rates arises in a variety of problems beside those that have been considered. If in any problem the quantity whose rate gf cjiange is to be found can be expressed directry as a function of the time, the result can of course be obtained at once by APPLICATIONS OF THE DERIVATIVE 89 differentiating with respect to the time. Frequently, however, the problem is solved by expressing the quantity in question Jii tgiuns of anotlier quantity whose rate of change is known, and then differentiating the equation connecting them. The methodTis best explained by an JExample : Water is flowing into a conical reservoir 20 ft. deep and 10 ft. across the top, at the rate of 15 cu. ft. per minute. Find how fast the surface is rising when the water is 8 ft. deep. ' The volume of water is r= i irr^h. By similar triangles. r 5 A~20' *■" v- Hence 7rA3 48' dV='^}Uh, 16 dV irh^dh dt 16 dt But we have given that ^^=15, dt so that ttA^ dh _^r dh 16 dt~ ' dt 240 7rA2 When A = 8, dh 15 , -i f\ £j^ = — = 1.19 ft. per minute. dt 4:77 ^ EXERCISE 1. Water is flowing into a cylindrical tank of radius 5 ft. at the^ rate of 20 gallons per second. Find how fast the surface is rising. 2. In the example of § 60, find how fast the water is flowing in if, when the water is 5 ft. deep, the surface is rising 2 ft. per minute. 90 CALCULUS 3. Water is flowing into an inverted conical tank 32 ft. deep and 12 ft. across at the bottom, at the rate of 4 cu. ft. per second. How fast is the surface rising ? 4. Two trains start from the same point at the same time, one going due east at the rate of 40 mi. per hour, the other north 60 mi. per hour. At what rate do they separate ? A7is. 72.1 mi. per hour. 5. Two railroad tracks intersect at right angles. At noon there is a train on each track approaching the crossing at 40 mi. per hour, one being 100 mi., the other 200 mi. distant. Find (a) how fast they are approaching each other, (b) when they will be the nearest together, and (c) what will be their minimum distance apart. Ans. (b) 3 : 4.5 p.m. ; (c) 70.7 mi. 6. A ladder 20 ft. long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of 2 ft. per second, how fast is the top descending when the lower end is 12 ft. from the wall ? 7. A man 6 ft. tall walks away from a lamp-post 10 ft. high at the rate of 4 mi. per hour, (a) How fast is the further end of his shadow moving? (b) How fast is the shadow lengthening? 8. A man on a wharf 20 ft. above the water pulls in a rope, to which a boat is attached, at the rate of 4 ft. per second. At what rate is the boat approaching the shore when there is 25 ft. of rope out? 9. A kite is 120 ft. high, with 130 ft. of cord out. If the kite moves horizontally 4 mi. per hour directly away from the boy flying it, how fast is the cord being paid out ? 10. A stone dropped into a pond sends out a series of concentric ripples. If the radius of the outer ripple increases steadily at the rate of 6 ft. per second, how fast is the disturbed area increasing at the end of 2 seconds? Ans. 452 sq. ft. per sec. 11. The path traced by a moving point is the parabola y = x^ + 2x4-3. If y^ = 3 f t. per second, find Vj^ and the total velocity v. Ans. Uj, = 6 a: + 6. 12. A point moves on the hyperbola x^ — y^ = 144 with v^ = 15 ft. per second. Find v at the point (13, 5). -f» 13. As a man walks across a bridge at the rate of 5 ft. per second, a boat passes directly beneath him at 10 ft. per second. If the bridge is 30 ft. above the water, how fast are the man and the boat separat- ing 3 seconds later? Ans. 8i ft. per sec APPLICATIONS OF THE DERIVATIVE 91 14. A light is placed on the ground 30 ft. from a building. A man 6 ft. tall walks from the light toward the building, at the rate of 5 ft. per second. Find the rate at which his shadow on the wall is shortening when he is 15 ft. from the building. Ans. 4 ft. per sec. -15. Solve Ex. 14 if the light is 8 ft. above the ground. 16. An elevated train on a track 30 ft. above the ground crosses a street at the rate of 20 ft. per second, at the instant that an auto- mobile, approaching at the rate of 30 ft. per second, is 40 ft. up the street. Find how fast the train and the automobile are separating 2 seconds later. 17. In Ex. 16, find when the train and the automobile are nearest together. Ans. ^f sec. 18. A light stands 60 ft. from a building. A man walks along a path 20 ft. from the building, at the rate of 5 ft. per second. How fast does his shadow move on the building? 19. An arc light hangs at a height of 30 ft. above the center of a street 60 ft. wide. A man 6 ft. tall walks along the sidewalk at the rate of 4 ft. per second. How fast is his shadow lengthening when he is 40 ft. up the street? Ajis. 0.8 ft. per sec. 20. In Ex. 19, how fast is the tip of the shadow moving? 21. A light stands 30 ft. from a house, and 20 ft. from the path leading from the house to the street. A man walks along the path at 5 ft. per second. How fast does his shadow move on the waU when he is 20 ft. from the house ? Xi CHAPTER IX CURVE TRACING IN CARTESIAN COORDINATES I. Algebraic Curves 61. Introduction. In Chapter IV we learned how to trace simple curves whose equations are given in the ex- plicit form and for which y, y\ and ^" are one-valued and continuous. In the present chapter we shall attempt a more general treatment of the subject of curve tracing. In §§ 62-67 we confine our attention to algebraic curves — i.e. curves for which the ordinate y is an algebraic func- tion of X. 62. Singular points. If y is defined implicitly as a function of x by the equation F(x, y) = 0, the derivative in general takes the form of a fraction whose numerator and denominator are functions of both X and y : say , ^ A(x, y) ^ Bix, y) If A(x., y') and B(x^ y^ both vanish at the point P : (x^ y) on the curve, tlie slope at that point assumes the inde- terminate form -. A point at which the derivative takes the form - is called a singular point. To find the singular points of a curve we must therefore j 92 CURVE TRACING IN CARTESIAN COORDINATES ^^S find the values of x and y that satisf}^ the three equations Fix, y) = 0, A(x, y)=0, B(x, y)=0. As we have but two unknowns x and y to satisfy three equations, it appears that a curve will have singular points only under certain conditions. It will be sufficient to consider an algebraic curve having a singular point at the origin. If a singularity occurs at any other point (A, ^), the origin may be trans- ferred to that point by the substitutions X = x^ -\- h^ y = y^ + k. 63. Determination of tangents by inspection. Let the equation of the curve be written in the form F(ix, y)= Uq + V + ^1^+ V^ + V^ + (^2^+ ••• + ^n^"= ^■ Differentiating, we find {bQ-\-2cQX-{- c^y + •..)c?2: + (^i + c^x -\- 2c\^y -i, )dy = 0, dy^_ 6 q+ 2cqX + ^1^ + '" dx 5j + c-^x + 2 c^y + '" The origin is on the curve only if a^ = 0. In that case the equation of the tangent at (0, 0) is found by the usual method (§ 27) to be h^x + h^y = 0, provided b^ and b^ are not both ; i.e. the equation of the tangent at tlie origin may be found by simply equating to the group of terms of the first degree. In case a^, 6^ and b-^ are all 0, the origin is on the curve and the derivative is indeterminate at that point; hence the origin is a singular point. In this case the equation of the curve evidently contains no terms of lower degree than the second. 94 CALCULUS For convenience let us put^ = ^2(^ - m^x)iy -m^x). Then The abscissas of the points of intersection of the line y = mx with this curve are given by the equation Two roots of this equation are : every line y = mx inter- sects the curve in two coincident points at the origin. But the above equation in x also shows that if we let m approach either m^ or ^2, the coefficient of oc^ approaches ; i.e. a third point of intersection of the curve with the line y = mx approaches the origin, and the lines y = m^x^ y = m^ are both tangent to the curve at the singular point. These lines may of course be real and distinct, real and coincident, or imaginary. Since H^y - ^1^) (y - ^2^) = v^ + H^y + ^2^^ we see that the equations of the two tangents are obtained by equating the second degree terms to 0, and factoring the left member of the resulting equation. The argument we have used can be extended to show that if F{x^ ?/) has no terms of degree lower than the A;th, any line through the origin meets the curve there in k points, and the k tangents to the curve at the origin are obtained by equating the group of terms of lowest degree to 0. * The argument is readily modified to take care of the case c^ — 0. CURVE TRACING IN CARTESIAN COORDINATES ^5 64. Kinds of singular points. A point at which there are two tangents (whether distinct, coincident, or imagi- nary) is called a double point; one at which there are three tangents is a triple point ; etc. It follows from § 63 that the origin is a double point if the equation F(x^ y) = has terms of the second, but none of lower, degree; a triple point if the equation has terms of the third but none of lower degree ; etc. If the tangents at a double point are real and different, the point is called a node : two branches of the curve cross each other, as in Fisf. 39. Ti- ^1 . . . , Fig. 39 It the tangents are imaginary, the point is called an isolated or conjugate point : there are no other points of the curve in its vicinity. Such a point is F in Fig. 40. If the tangents are real and coincident, there are several possibilities. The simplest singularity in this case is the cusp of the first kind : two branches of the curve touch each other, coming up on opposite sides of Fig. 40 ' ^^6 cuspidal tangent, as in Fig. 41. At a cusp of the second kind the two branches lie on the same side of the tangent, as in Fig. 42. Fre- quently the point is a double cusp^ or point of osculation, Fig. 41 Fig. 42 Fig. 43 the commonest form of which is shown in Fig. 43. And in some cases the point may be an isolated point. 96 CALCULUS Example : Examine the curve y^ = 7? — x^ for singular points. Since there are no terms of lower than the second degree, the curve has a singular point at the origin. The tangents at that point are given by the equation 'if' = — 7?' \ i.e. they are the lines y = ix, y — — ix. These lines are imaginary, and the origin is an isolated point. There are no other singular points. For, , Zx^ — ^x and the coordinates of the singular point must satisfy the three equations Sx^-2x = 0, 2y = 0, y^ z= a^ — x^. The first two equations are satisfied by the coordinates (0, 0), (|, 0), but the second pair do not satisfy the last equation. (See also Ex. 20 below.) EXERCISES Show that the origin is a singular point for each of the following curves, write the equations of the tangents there, and determine the nature of the singularity. 1. The folium x^ + ^^ _ 3 f^j.^^ ^^^j Xode. 2. x'^y^ = a^(x'^ +11'^)- 3. y x+ y 4. The eissoid ?/ — ^^j^. Cusp of the first kind. 2 a — x 6. y^ =2 ax'^ — x^. 6. y^{x^ + y-) = a^x'^. Ans. Double cus]). 7. y\x- — a^) = x^. Ans. Isolated point. CURVE TRACING IN CARTESIAN COORDINATES 8. {y — x^Y = x^. Ans. Cusp of the second kind. 9. y'^ = x^ + x^. 10. x^ — xy'^ = y^. Ans. Triple point. 11. The leniniseate Qx~ -\-y^)'^z= a^(x^ — y^). 12. Show that the conchoid x^y'^ = (a + yy~(b'^ — y"^) has a node at (0, - a) iib>a. 13. Show that the curve a(y — xy-=(x — ay has a cusp at (a, a). Find the singular points of the following curves. 14. if =x(x - ly. 15. 2/2 ^ a:(2 x +1)2. 16. y^ = x(x^ — !)• ^^5- None. 17. ay^ = x\x - ay. Ans. Cusps at (0, 0), (a, 0). 18. y^ + y^= (x^ - 1)2. 19. xy^ + x^y = a^. 20. Prove that a cubic curve cannot have more than one singular point. 21. Prove that the graph of a one-valued algebraic function cannot have any singular points. 65, Asymptotes. As the point of contact of a tangent to a curve recedes indefinitelj;.^from the origin, the tan- gent may or may not approach a limiting position. If it does, the line approached is called an asymptote. For example, the hyperbola has the lines a as asymptotes. On the other hand, the parabola has no asymptotes, since as the point of tangency recedes the tangent does not approach any limiting position. Although general methods for finding asymptotes exist, they are frequently difficult to apply. The following .tests are sufficient in ordinary cases. 98 CALCULUS (a) To find the conditions that must be satisfied in or- der that the line y — mx 4- k shall be an asymptote to the algebraic curve (1) F{x, y) = a^x^ -}- a^x'^-'^y + ... +«„?/» + h^^-'^ + V"~^«/+ ••• = ^' let us substitute mx + k for y in the equation of the curve. This gives an equation of the nth. degree in x whose roots are the abscissas of the n (real or imaginary) points of in- tersection of the line with the curve. It is shown in algebra that one root of the equation A^x"" + A^x""-^ -f ^2^"~^ + . . . H- J.^ = becomes infinite if Aq approaches ; two roots become in- finite if both Aq and A^ approach ; etc. Hence if we equate to the coefficients of x^ and 2;"~\ we shall in gen- eral determine values of m and k such that the line y = mx + k will intersect the curve in two infinitely distant points. Such a line is in general an asymptote. Of course if the coefficients of x"^ and x^~^ cannot both vanish, there are no asymptotes (except such as may be given by (6) below). Example : («) Find the asymptotes of the hyperbola x^-y'^-2x-2y-[-l==0. Substituting y = mx -f-A:, we get x^ - (mx + k)^- 2x - 2(mx + A:) -|- 1 = 0, or (1 - m2)2:2H-(- 2mk-2m-2^x+ ... = 0. Equating to the coefficients of the two highest powers of a;, we find 1 - w2 = 0, Whence w = 1, ^ = — 2, m = — 1, A; = 0, CURVE TRACL^TG IN CARTESIAN COORDLVATES 99 Fig. 44 and the asymptotes are y = x-2, y = -x. The curve is shown in Fig. 44. (5) Asymptotes parallel to the ?/-axis are not given by test (a), since their equations cannot be writ- ten in the slope form. Let us arrange the equation of the curve in descending powers of y: F{x, y) = a^- + (a^x + ;5i)?/-i + (a^x^ + ^^x + 72)r"^ + ... =0. If the term in y^ is present, every value of x gives n finite (real or imaginary) values of ?/, and no line x= k can in- tersect the curve in infinitely distant points. But if a^ = 0, then every line x = k intersects the curve in one infinitely distant point. If now the coefficient of y'^~^ involves x (i.e. if a^ ^ 0), that coefficient equated to gives us the equation of a line parallel to the ?/-axis which intersects the curve in tivo infinitely distant points, and is an asymptote. This result can be extended to the case when the co- efficient of the highest power of ?/ is a polynomial of higher degree in x. By equating this polynomial to we find an asymptote parallel to 01^ corresponding to each real linear factor of the polynomial. Similarly, asymptotes parallel to OX may be found by equating to the coefficient of the highest power of x. Such asymptotes are detected in general by test (a), but it may be easier to find them by the present method. Example : (5) Test the curve (x^ — V)y'^ = 2^ for •asymptotes. Equating to the coefficient of the highest power of ?/, 100 CALCULUS we find the lines x^ — 1 = 0, i.e. X — ±\ as asymptotes parallel to OY. The coefficient of the highest power of x cannot be equated to 0, so there are no asymptotes parallel to OX. To test for asymptotes oblique to the axes, put y = mx + k : {x^ — V){if)^7p' -f 2 mhx + A;2) = 7?. Equating to the coefficients of the two highest powers of x^ we find m^ = 0, 2 mk — 1. These equations are incompatible ; hence there are no oblique asymptotes. EXERCISES Test the following curves for asymptotes . 1. xy -\- X = b. ' 2. x^ -{- y^ = \. Ans. X ■\- y = Q. x^ 3. y^ = ax^ + x^. 4. The cissoid y'^ = 2 rt — X 5. x^ + 7 xy + 12 y^ + X -\- 4 y - 16 = 0. Ajis. a; + 3 ?/ + 1 = 0, x -\- "1 y = 0. 6. x"^ — xy + y^ -\- 5 X = 0. Ans. None. 7. x'^y'^ = n^(x'^ + ?/-). 8. The folium x^ + y^ — 3 axy. 9. x'^y'^ + lx'^+ ?/- + X = 0. 10. x~y + xy'^ = a^. 11. x3 - 4 xy-^ - 3 x^ + 12 xy - 12 ;/2 + 8 x + 2 ?/ + 4 = 0. A71S. a; + 3 = 0, X - 2 ?/ = 0, r + 2 ?/ = 6. 12. ay^ — ay'^ — x^ -\- ax^ + a^. .4ns. x = a, x ± y -\- a = 0. 13. a?/2 = x^ + xy^. 14. Prove that a parabola has no asymptotes, but that every line parallel to the axis meets the curve in one infinitely distant point. 15. Prove that every line parallel to an asymptote meets the curve in one infinitely distant point. 16. In example (h), § 65, prove that every line parallel to the* ar-axis meets the curve in one infinitely distant point. CURVE TRACING IN CARTESai^,N ,COQ;RD.INAT£S IQl 17. Show that a curve of the n-th degree cannot have more than n asymptotes. 18. Show that the curve y = P(^), where P(x) is any polynomial in x, has no asymptotes. 66. Exceptional cases. Exceptions may arise to the theory of § 65. For instance, it may happen that the co- efficient of x'^~^ vanishes identically for some value of m for which the term in x"" disappears, so that all lines having this value of m as their slope meet the curve in two points at infinity. In this case there are in general two or more parallel asymptotes having the given slope, and the values of k are determined by equating to the coefficient of the highest power of x that does not disappear identically. The exceptional cases are rare and unimportant in elementary work, and a fuller discussion of them is un- necessary. EXERCISES 1. Show that the curve (a; + y)'^(x^ + xy -\- y'^) = a^y- + cfi{x — y) has the pair of parallel asymptotes x + y = ± a. 2. Show that every line parallel to the x-axis meets the curve y^ = x^ -{- X in two infinitely distant points, but that the curve has no asymptotes. 67. General directions for tracing algebraic curves. The following questions should be answered as fully as possible before trying to trace an algebraic curve. (1) Is the curve symmetric with respect to the coordinate axes? (It is symmetric with respect to 01^ if the equation is unchanged when x is changed to — rr ; etc.) (2) Where does it intersect the axes ? (3) Has it any asymptotes? If so, locate each of the points where the curve intersects its asymptotes. (4) Is it possible to determine certain regions of the plane within which the curve must lie? 102 CALCULUS (5) Has the curve any singular points ? If so, determine the tangents at each point, and the nature of the singu- larity ; draw the tangents if they are reaL (6) Has it any maximum and minimum points ? The above is only a general outline of the process to be followed ; other steps will often suggest themselves. In many cases the points of inflection should be found and the inflectional tangents drawn, but this is not worth while if the second derivative is complicated. Translation or rotation of axes is occasionally useful. The elementary method of tracing the curve by plotting points is too la- borious to be used extensively, but it is frequently advis- able to plot a few points, merely as a check on the analysis. Examples : (a) Trace the curve ?/^ = 3 ax^ — x?. (1) The curve is not symmetric with respect to either axis. (2) When a: = 0, ^ = ; when ?/ = 0, a; = or 3 a. (3) By § ^b, the line y ^ a — x is an asymptote. Substituting a — x for y in the equa- tion of the curve, we find a^ — 3 c^x + 3 a:i^ — rr^ _ 3 ^^2_ ^3^ The highest powers of x drop out, as they should, and we find that the curve crosses its asymptote at a; = - . o (4) Writing the equation in the form ?/3 = x^{Z a — a:), we see that 1/ > if a: < 3 «, ^ < if a: > 3 a. Hence the curve is above the a:-axis at the left of a; = 3 a, below at the right of that point. (5) Since there are no terms of lower than the second degree, the origin is a singular point. The tangents are given by the equation 3 aa;2 = : CURVE TRACING IN CARTESIAN COORDINATES 103 they coincide in the ?/-axis. Since, by (4), the curve near the origin cannot go below OX^ the point is a cusp. For small values of x on either side of 0, y is real, hence there is a branch on each side of OY^ and the origin is a cusp of the first kind. Since the curve is a cubic, there can be no other singu- lar points (Ex. 20, p. 97). (6) The derivative is dy _Q ax — Sx^ dx 3 ?/2 The numerator vanishes when x = or 2 a. Rejecting the value a; = 0, which gives the singular point, we have a; = 2 a as the only critical value. It will appear presently that the point (2 a, V4 a) is a maximum point. To trace the curve, let us begin at the extreme left. In that region, the curve must be just below its asymp- tote, since it has to pass through the origin and can cross a the asymptote only at rr=-. It comes down to the a 2 a\ T, . -, — ). It IS now origin tangent to the j/-axis, turns back on the other side, and crosses the asymptote at clear that the critical value x=2 a corresponds to a maximum point. Return- ing from the maximum, the curve crosses OX at (3 a, 0) and again approaches the asymptote. The curve is shown in the figure. (5) Trace the curve ap'y'^ = x^^cp' — dtp'). (1) The curve is symmet- ric with respect to both axes. Fig. 45 104 CALCULUS (2) When a; = 0, ?/ = ; when 7/ = 0^ x = 0,'± a. (3) There are no asymptotes. (4) When x is numerically greater than a, 1/^ is negative and ?/ is imaginary. Hence, the curve lies entirely be- tween the lines x = ± a. (5) The origin is a singular point. The tangents are real and different, hence the point is a node. There are no other singular points. (6) The first derivative is dy _ 2 a^x — 4a^ _ x(^a^ — 2 x^^ dx 2 ^ y This vanishes when «2 _ 2 a;2 = 0, a: = ± -^ . V2 Corresponding to each of these values the curve has, on account of its symmetry, a maximum y =■- and a mini- a mum %i = ^ 2 The student may draw the curve. EXERCISES Trace the following curves. \-x 1 + a; a'-x A ,„ _ a.-^ + «' 1. ^ = rT-r2- 2. y = x{x - 1)2. {x — a)'-^ 5. ?/2 = 4- .r ^ yi ^ x^Cl + a,-) , ^ 1 -a: 7. ?/ = 2 x3 - 9 a-2 + 12 a; - 3. 8. ?/ = a:^ - 3 a:^ + a:. 9. ?/2(a;2 + «2) ^ ^2^2. 10. ?/(a:2 _ ^2) ^ ^ _^ X. 11. a;3 + y3 3, 3 aa-y. 12. 3/2 ^ 3,(3,2 _ i). 13. 3/ = -/^- 14. 3/^ = a: a;2 — a^ 16. ,_x^(x'^-n^) ■^ x^ + a2 18. (a:^ — a'^)y'^ = ax^. 20. xy"^ + a;2?/ = a^. 22. xhf = (a; + a)2(4 a2 _ -x^). 24. (^ — x^)^ = x^. 26. ay^ =(a;2 - a2)2. 28. V- ^^ . CURVE TRACING IN CARTESIAN COORDINATES 105 15. 3/2(a;2 + r/2) = arx'^. 17. (^ - a;)2 = a:3. 19. y= i^--y . 21. a;2/ = o:^{x'^ + y^). 23. (a:2 - ?/2) (a: - 3 ?/) = x. 25. y2 ^ 1 - a: . ^ 1 + a;2 27. :.3^ = «2(^ + «)2. ,^ ^^^^^^ II. Transcendental Curves 68, Tracing of transcendental curves. In tracing tran- scendental curves, we follow much the same procedure as m § 67, except in the matter of as3"mptotes and singular points. While the definitions of §§ 62-65 hold for all curves, the tests there given apply only to algebraic curves. We shall in this article confine our attention to tran- scendental curves having no singular points. To find asj^mptotes, the following rule may be used : In general^ if y becomes infinite as x approaches a definite limit a, the line x= a is an asymptote ; if x becomes infinite as y approaches 5, the line y — b is an asymptote. In rare instances, the derivative may behave in such a way that although the conditions of the rule are satisfied, the tangent does not approach any limiting position, and hence there is no asymptote ; but the rule holds in all cases that are apt to arise in practice. Example : Trace the curve y = xe^. (1) There is no symmetry. (2) The curve crosses the axes at (0, 0). (3) As X becomes large and negative, y approaches * ; * This statement is easily made plausible ; a strict proof will be given later (§ 140). CALCULUS hence the negative x-axis is an asymptote. When x is large and positive, y is large and positive. (4) Since e^" is always positive, y has always the same sign as x : the curve lies in the first and third quadrants. (5) Since y' = xe"" + e% the only critical point is [ — 1, ]. The slope at (0, 0) is 1. ^ ^^ (6) y''= xe^ -\- 2e^. There is a point of inflection at f — 2, j ; the slope of the inflectional tangent is -e-2 = -0.14. The curve is shown in Fig. 46. Fig. 46 EXERCISES Trace the following curves. 1. y = e-^\ 2. y = xe-^"". 3. y = tan x. ^. y = sec X. - logx X 6. y = e'\ 7. y = sin^ x. 10. 2/2 = ^°§'^. X 8. y'^ = sin x. 11. 2/2 = log X. 9. 12. X y = logx y = -' X 13. y=,-^' logx 16. Show that 14. y = g-^ sin x. Urn 2 + sin.^^O; 16. y = X log X. then show that the curve 2 + sin x^ furnishes an exception to the rule of § 68. ^' 69. Curve tracing by composition of ordinates. The curve can be traced very readil)^ U f (x) has the form CURVE TRACING IN CARTESIAN COORDINATES 3w^ where are curves whose form is easily obtained. We have only to add the ordinates of the two latter curves to obtain the required curve. Less frequently curves may be conveniently traced by multiplying or dividing ordinates in a similar way. W^ EXERCISES Trace the following curves. 1. y = X + \ogx. 2. y = X -\ X Z. y — e^ — X. ^. y ■= sin x + cos x. 6. y = X -\- sin X. Q. y = sinh x = • „ sin X o cos X 7. y = 8. y = X X 9. The catenary is the curve in which a homogeneous cord or chain hangs when suspended from two of its points under its own weight. The equation is X a I ^ ^\ y = acosh-=-f ea_,.g-aj. Ttace the curve. 70. Graphic solution of equations. The roots of the equation are the abscissas of the points where the curve y=f{x) crosses the 2:-axis. Hence if we trace the curve y=zf{x) and measure its intercepts on OX^ we have a graphic solution of the equation /(a;) = 0. It is usually best to get the general form of the curve by the methods of the preceding articles, and then plot it carefully, on a large scale, in the neighborhood of each of its a;-intersections. The roots of the equation (1) f {X-) = (^)- 108 i:/ CALCULUS In case these two curves are easily traced, we thus ob- tain with little labor a graphic solution of (1). This method is frequently preferable to the first one mentioned above. Such methods may be useful in various ways. If no high degree of approximation is required, the graphical result may be sufficient in itself; it may be used as a rough check on a more accurate result obtained in some other way ; it gives a first approximation that may be needed as a starting point for more elaborate methods, or it may suggest some value of the variable which by substitution is found to satisfy the equation exactly. EXERCISES Solve the following equations graphically. 1. a;4- 3a-3 + 3=0. 2. Sa:* - 2^3 - 21 x^ - 4x + 11 = 0. 3. X + 10"^ = 0. 4. a: + 2 cos x = 0. 5. X -\- logio X = 0. Q. X + cos X = I. 7. Trace the curve y = x sin x, locating maxima and minima graphically. 8. Solve the equation x log x = 1. 9. A gutter whose cross-section is an arc of a circle is to be made by bending into shape a strip of tin of width 8 inches. Find the radius of the cross-section when the carrying capacity of the gutter is a maximum. Ans. 2.55 in. ,71. The cycloid. In the remainder of this chapter we consider several special tran- scendental curves. The path traced by any point A on the rim of a wheel that rolls without slipping along a straight track is called a cycloid. Let the circle of radius a roll along the a:-axis, and take the initial jiosition of A as origin. Then, if (a:, ^) are the coordinates of A^ CURVE TRACING IN CARTESIAN COORDINATES "^l^ Y Fig. 48 OB = arc AB = aO, x= OB- AC=ae -a^m =^ a{6 - sin 6), y = BO' - CO' = a-aGosd^ a{l - cos 6). These are the parametric equations of the cycloid in terms of the angle 6 through which the circle has rolled. The coordinates of the center of the rolling circle are (a^, a). The rate at which the center is advancing is d r n\ du at at where o) is the angular velocity. The curve is shown in Fig. 48. 72. The epicycloid. If a circle rolls without slipping on the outside of a fixed circle, a point on the circumference of the roll- ing circle generates an epicycloid. Let a and h be the radius of the fixed circle and that of the rolling circle respectively, and suppose the point A was originally at E. Then arc AL = a»rc UL, or hcj) = ad. I Fig. 49 The equations of the path of A in terms of the parara- ^ CALCULUS eter 6 may be obtained as follows : X = 0M= OF+FM= 0F+ DA = (a + 5) cos 6 -\-h ^\n *-(!-. jj = (a + 6) COS ^ — 5 cos {6 + <^) = (a + 6) cos 6> - 5 cos (<9 + 1^) = (a 4- 5) cos 6 — h cos , ^ ; y = MA = FI) = FO' - = (a + 6) sin ^ — 6 cos h DO' (9 = (a +5) sin (9 -5 sin ^4^(9. "~73. The hypocycloid. A point on the circumference of a circle that rolls on the inside of a fixed circle gener- ates a hypocycloid. Its equations are obtained in the same manner as those of the epicycloid. They are x = (a — h^ cos 6 -\-h cos — - a — h 0, y = (^a — J) sin ^ — J sin '^^ — 6. EXERCISES 1. Show that the tangent to the cycloid passes through the highest point of the rolling circle. 2. A wheel of radius 2 ft. rolls on a straight track with a velocity of 6 ft. per second. Find jx, jy, j, jti and _/„ at the points ^ = 0, ^ = ^, ^ = TT. 3. The highest point on an arch of the cycloid js called its vertex. Show that by taking the origin at the vertex and replacing 6 by 01 — — TT, the equations of the cycloid become x' = a(^' + sin^'), y =- a(l-cos^'), CURVE TRACING IN CARTESIAN COORDINATES Hi or, if we change the sense of the ?/-axis and drop subscripts, X — a(d -\- sin^), y = a(l — cosO). 4. Sketch the epicycloid for which the rolling circle and the fixed circle have the same radius. If -— = — radians per second, find v and at 2 J, and also find between what limits these quantities will vary. 5. Show that the hypocycloid for which b = -r is a diameter of the fixed circle. 6. Show that the equations of the hypocycloid of four cusps, for which b = -, may be written X = a cos^ 6, y = a sin^ 0. Hence find its cartesian equation. Trace the curve. 7. Give the cartesian equation of the cycloid. 8. Find the radius of curvature of the cycloid. Ans. 4 a sin -. 9. Trace the epicycloid for which & = - . CHAPTER X CURVE TRACING IN POLAR COORDINATES 74. Slope of a curve in polar coordinates. We have seen that in sketching a curve it is helpful to know the direction of the curve at any point. In cartesian coordinates the direc- tion at the point P^ : (x^^ y^ is most easily determined by giving the in- clination of the curve to the straight line y = y^ — i.e. the "slope" of the curve — since this is found by a mere differentiation. Similarly, given the equation of a curve in polar coordinates, the direction at the point Pq : (r^, 0^ is best found Fig. 50 by means of the inclination to the curve r = r^, of course a circle through P^ with center at 0. reason the quantity tan (/>, where <^ is the angle between the curve and the circle just mentioned, will be called the polar slope. To find the polar slope we proceed as follows.' Consider a fixed point P : (r, ^) and a neighboring point P' : (r 4- Ar, 6 + A^) on the curve (Fig. 51), and drop a perpendicu- lar PJSr from P upon OP'. Let ' = ANPP'. Then NP' OP' - which is For this Fig. 51 tan (^' = ON pjsr pjsr 112 CURVE TRACING IN POLAR COORDINATES^^ But OP' =r + Ar, 0N= r cos A6>, PN= r sin A(9. Hence tan ' r + Ar — r cos A^ r sin A^ Ar + r(l — cos A^) r sin A^ Ar + 2 r sin^ \ A(9 r sin A^ Ar , r sin i A^ . i * zi A6> J A6> 2 sin A(9 I When P' approaches P along the curve, ' approaches (/). By § 38, sin a Hence lim a->o a = 1. tan ' = dr A0->O or 4. JL ^1" tan 9 = — -. The formula for tan cj) may be remembered as follows. Strike through P a circular arc PM with center at (Fig. 52). Then arc PM = rdO, MP' = dr (approximately), and MP' tan (f) = (approximately ) . arc PM This at once suggests the formula. I Fig. 52 114 CALCULUS cliT 75. Maxima and minima. When -— = 0, <^ = 0, and in du general r is a maximum or a minimum, as at A and B^ Fig. 53. Just as in cartesian coordinates, there is a pos- sible exception : the curve may have the form shown at C. However, the exceptional case is rare in the simpler curves. 76. Curve tracing. Before sketch- ing a curve whose equation is given in polar coordinates, the following questions should be considered* : (V) Is the curve symmetric with respect to the initial line ? (It is, if the equation is unchanged when 6 is replaced by — ^; other tests may frequently be used.) (2) Is it possible to determifie any particular regions of the plane within which the curve must lie ? (3) At what poiiits is the polar slope ? Is the radius vector a maximum or a minimum at each of these points f The above discussion is frequently insufficient to deter- mine the general form of the curve, in which case addi- tional points must be plotted. Example : Trace the lemniscate r^ = cl^ cos 2 6, (1) The curve is symmetric about the initial line. (2) cos 2 ^ is negative, and r is imaginary, when also when — < At) < -—, I.e. -T < c/ < — -; 2 2 4 4 A 2 4 4 Hence the curve lies entirely within the sectors AOB^ COD (Fig. 54). * For brevity, the discussion of singular points and asymptotes is omitted. CURVE TRACING IN POLAR COORDINATES 115 (3) Since r dr = — a? sin 2 6 dO^ - a^ sin 2 e tan (^ = 0^ cos 2 ^ = - tan 2 e. From this the direction of the curve at any point may be found. When tan (^ = - tan 2 ^ = 0, 2 ^ = 0, TT, 2 TT, 3 TT, and 2 2 Only the values and it give real values of r ; ' at each of these points r is a maximum, viz. r = a. The curve passes through the origin whenever r = 0; i.e. when cos 2 ^ = 0, or ^ = — , — — , 4 4 etc. The curve is shown in Fig. 54. EXERCISES Trace the following curves. I. r = 2 a cos d. 2. The spiral of Archimedes r = aB. 3. r = a sec ^. 4. r = a cos 2 6. 5. r^ = a^ sin 6. 6. r = a cos 3 ^. 7. r2 sin 2 ^ = a^. 8. The linia^on r=ft — acos^, (a) when h = 2a\ (h) when J = a ; (c) when h = \a. 9. The conic r = 7;, where e is the eccentricity, (a) if 1 — e cos^ el. 10. The logarithmic spiral r = e*^. Show that tan <^ is constant. II. What is the form of the curve r = a cos nO, (a) when n is even; (6) when n is odd? CHAPTER XI THE INDEFINITE INTEGRAL 77. Integration. We have been occupied up to this point with the problem : Given a function, to find Its de- rivative. Many of the most important applications of the calculus lead to the inverse problem : Given the derivative of a function, to find the function. The required function is called an integral of the given derivative, or integrand^ and the process of finding it is called integration. If f(x) is a given function and F(x') is a function whose derivative is f(x)^ the relation between them is ex- pressed by writing F(x) = j' f{x)dx. where the " integral sign " j indicates that we are to per- form the operation of integration upon f(x)dx. For reasons that will appear later, it is customary to write after the integral sign the differential f(x) dx, rather than the derivative /(a:). Examples: (a) Find the equation of a curve whose slope at every point is equal to twice the abscissa of the point. We have to find a function y such that -^ = 2 x^ or dy — 2 X dx ; dx hence 7/ = \ 2 X dx. 116 THE INDEFINITE INTEGRAL 117 It appears at once that 2 a; is the derivative of ^. Thus a curve having the desired property is the parabola y = x^. But it is clear that if (1) y = :r2 + 0, • where is any constant whatever, we still have dy = 2 X dx^ and our data are satisfied by any one of the family of parabolas rep- resented by (1). In order to obtain a unique answer to our problem, we must have some additional informa- tion about the curve. Thus, if it is to pass through the point P : (1, |), we substitute these coordinates in (1) = and the answer is y = x^-\-\. (b) Find the velocity of a body falling freely under gravity at the end of 5 seconds, if the initial velocity is 20 ft. per second upward. Taking motion downward as positive, we have to find v from the relation (see § 55) Fig. 55 dv Hence V = i g dt = gt -\- C. Making use of the fact that v= — 20 when ^ = 0, we get - 20 = + C, and the velocity at the time t is i;=^^-20. 118 CALCULUS At the end of 5 seconds we have, taking g = 32, V = 140 ft. per second. (f?) Find the space covered in the fifth second of the motion in (5). Here v=—=S2t-20, dt X = r(32 t - 20}dt =16t^-20ti-O. No data are given for determining O. But if we denote by Xt the space covered in t seconds, the space described in the fifth second is ■ 2^5 - 2:4 = (16 . 25 - 20 . 5 + (7) - (16 . 16 - 20 . 4 + (7) = 124 ft., the unknown constant having disappeared. In fact, is merely Xq, the distance of tlie starting point from some arbitrarily chosen origin, so that the distance passed over between any two instants must necessarily be independent of a 78. Integration an indirect process. Differentiation is a direct process ; by means of the fundamental formulas the derivative of any elementary function may be found. On the other hand, to find an integral of a given function, we must be able to discover a function whose derivative is the given integrand, and this is always in the last analysis a matter of trial. The problem can by no means always be solved ; in fact, there are many comparatively simple functions whose integrals cannot be expressed in terms of elementary functions*. 79. Constant of integration. In each of the examples of § 77, an arbitrary constant presented itself. It is clear that this will be the case in general ; i.e. a function whose * It will be shown in § 81 that for every continuous function an integral exists^ although it may not be an elementary function. THE INDEFINITE INTEGRAL 119 derivative is given is not completely determined, since it contains an arbitrary additive constant, the constant of integration. On account of the presence of this undetermined con- stant, the function if(x)dx is called the indefinite integral oifix). 80. Functions having the same derivative. In § 79 it was tacitly assumed that if the derivative of a function is given, the function is determined aside from an additive constant. That this is true follows from the Theorem : Two functions having the same derivative differ only hy a constant. The theorem is almost self-evident. Let (x) and ^(x) be the two functions, and place y = ix)-^^r{x). By hypothesis, dx The rate of change of y with respect to x is everywhere 0, hence y is constant. EXERCISES Evaluate the following integrals, checking the answer in each case by differentiation. dx. 1. (rt)^^;^^; {h)^(2x-x'^)dx] (c) f (1 - 4/4)r7f ; (r/)|(l + 3/)%; (6)0; (/)|(^^--^) 2. J^. 3. ^^inOdd. 4. rsin2^^^. 5. ( Vx + 1 dx. 6. j Vl - X dx. 7. i e'^dx. 8. ({l-\-2xydx. 9. JxVa^ + x^ dx. Ans. ^(a'^ + x^y + C. 10. Find the equation of the family of curves whose slope at every point is equal to the square of the abscissa of the point. Exhibit graphically. 120 CALCULUS 11. Find the equation of that one of the curves of Ex. 10 that Ans. y = — — 14. ^ 3 passes through the point (8, — 5). 12. A body falls from rest under gravity. Find the velocity at the end of 3 seconds, and the distance traveled in that time. 13. Find the equation of the curve for which y" = 4 at every point, if the curve touches the line ^/ = 3 x at (2, 6). Ans. y = 2 x'^ — 5 X -\- 8. 14. A body moves under an acceleration numerically equal to the time. If the initial velocity is 10 ft. per second in the direction of the acceleration, find v and x at the end of 4 seconds, x being measured from the starting point. 15. In Ex. 14, find the initial velocity if the body moves 10 ft. in the first second. 16. Find the equation of the curve for which y" = , if the x'^ curve makes an angle of 45° with OX at the point (1, 0). 17. Find the equation of the curve through (1, 2) and (2, 3) («) if y" = 0; (b) if y" = Q x ; (c) if y" =—. Trace the curve in x^ each case. .4/^s". (6) y = x^ — 6x + 7. 81. Geometric interpretation of an integral. Consider the area A bounded by the curve ^ ==y(a:), the a:-axis, the fixed ordinate x = a^ and a vari- able ordinate x = x. This area is evidently a function of x. We proceed to find the deriva- f{x-\-Ax) ^i^® ^f -^ with respect to x. When X is increased by an ^.^ 2: ^ amount Ax^ A assumes an in- crement A^, the area KLRN in Fig. 56. It appears from the figure that AA is greater than the area f(x)Ax of the inscribed rectangle KLMN^ and less than the area f(x-\- Ax^Ax of the circumscribed rectangle*: f(x)Ax< AA0 Ax Since the derivative of A is /(a:), it follows by the defi- nition of the integral that = jf(x)dx. In case the position of the fixed ordinate x = a is given, the constant of integration may be determined by the fact that ^ = when x = a. We have thus proved the following result : • ^]i£^irUf/n,ifji integral ifCx^dx represents the area hounded hy the curve_]j =zJls)'> the Xj^JOlxis, a fixed_ordinate, and a variabl£~M^dinate. It is evident that if /(a:) is continuous, this area always exists ; hence every contirmous functioyi has an integral (cf. § 78). " Since the formula for the area uiider the curve is frequently written r/r{y \ ly A= \y^ ^ Example: Find the area bounded by the parabola y = x^^ the a^-axis, and the lines x = 1^ x = 4. The area from a; = 1 to any variable ordinate is A= \ ydx= \ x^dx= — -{- O. 122 CALCULUS Since A = when a: = 1, we have = 3^4-C, C = — 3, or "^-3 3 In particular, the area from a: = 1 to a; = 4 is EXERCISES In the following, find the area bounded by the x-axis, the given curve, and the indicated ordinates. Check roughly by drawing the figure on coordinate paper and estimating the area. 1. y = x^, X = 0, X =4: . Ans. 64. 2. The parabola y^ = 4i x and its latus rectum. 3. The hyperbola y =-, x = 1, x = 3. Ans. 1.099. X 4. Find the area of one arch of the sine curve. Ans. 2. 5. Find the area bounded by the parabola y = 1 — x^ and the x-axis. 82. Variable of integration. In the last article we had occasion to use the symbol j ?/ dx. In order that such a symbol shall have any meaning, j/ must be directly or indirectly a function of x. The variable whose differential occurs is called the variable of integration^ any other variables appearing under the integral sign must be functions of the variable of integration, and their values in terms of that variable must be introduced before the integral can be evaluated. Xhaiact that the differential occurring tells us which variable is the variable of integration is one of the reasons for using the notation i f(x)dx rather than the notation J/(^). 83. Change of the variable of integration. If x is so re- lated to y that (1) (f>(x)dx = y\r(^y)dy, THE INDEFINITE INTEGRAL 123 we may replace j (j>(x)dx by j "^(^y^dy. For, let ^(x)= C(f)(x)dx, ^(^) = ffOj)dy. Now ^ r:;?2; c?^ c^a; dx The two functions therefore have the same derivative, by (1), and hence differ only by a constant. Since each function contains a constant of integration, these constants inay_be„so_ chosen that The device of replacing a given integral by an equiva- ^ 1 lent integral in a different variable is very useful in many problems. 84. Integration by substitution. A change of variable is usually brought about by means of an explicit substitu- ^^"' tion j^^^^dx = (^' (%i)du. The process is called integration hy substitution^ and is "^ighly'' important. It is tu be remembered that not merely x^ hut dx as well, must be replaced by the proper value in terms of the new variable. Example : Evaluate \ x^l — x dx. Let us put 1 — a; = w, so that a: = 1 — u, dx = — du. 124 CALCULUS Then j xVl — xdx=— 1(1 — u)u^ du= — j (u^ — u^'ydu = - 1(1 -2:)^ + 1(1 -2:)^+ a EXERCISES 1. Work the above example by placing I — x = u\ 2. Evaluate i x(3 — 2 x^)dx by substituting S — 2 x^ = u. Check by expanding the given integrand and integrating directly. Evaluate the following integrals by means of the indicated sub- stitution, and check the results by differentiation. 3. I Va + bx dx, a + hx = u. 4. I sin^ 6 cos dO, sin 6 = u. 5. \z -: — ;, 2 X = U. h + 4a:;2 6. CJ^A^, 1-x^ = u. n ~Z — Z' l + tan^ = •^ 1 + tan 6 8, J 7 J 1 u. Vl - x^ 9. If the velocity of a point moving in a straight line is given as a function of the time, show that the distance covered may be found by the formula =J^ dt. Given v = 10 t + 20, find x in terms of t by substituting the value of V in the above integral ; also find x in terms of v by substituting for dt. Show that the two values of x are equivalent. 10. Given x = t^, y = dt, find A = \ y dx as a function of t by substituting for y and dx their values in terms of ^ Ans. A =:2t^ + C. THE INDEFINITE INTEGRAL 125 11. In Ex. 10, find y in terms of x by eliminating t. Then express yl as a function of x by substituting for y and integrating with re- 3 spect to X. Ans. A = 2 x^ + C. 12. In Ex. 11, find A as a function of y by substituting for dx. Ans. A = ijy^ + C. 13. Show that the three values of A found in Exs. 10-12 are equivalent. 14. By the formula A = j ?/ dx, find the area under the curve y'^ = x from x = to x = 4, («) by sub- stituting for y in terras of x ; (&) by substituting for dx in terms of y and dy. 15. Proceed as in Ex. 14 for the area under the curve y = e^ from a; = to a: = 1. 16. Proceed as in Ex. 14 for the area of half an arch of the curve y = \ sin 4 x. Ans. x\. CHAPTER XII STANDARD FORMULAS OF INTEGRATION 85. Standard formulas. If we were to try at present to solve any but the simplest applied problems involving integration, we must fail through inability to evaluate the indefinite integrals involved. We shall therefore devote this and the following chapter to the technique of integration — the formal evaluation of indefinite integrals — after which the question of applications will be treated at some length. As the first step toward facility in integration, the student must become thoroughly familiar with the fol- lowing Fundamental Integration Formulas (1) fdu=u + C. (2) f(du -\- dv) =fdu + Cdv, (3) Ccdu = cCdu, /i,n+l n-hl (5) f^ = logu-\-Cr\ (6) fe^dw = e" + C, (7) J cos w i/z/ = sin M + C = cos f M — ^ j + C, 126 STANDARD FORMULAS OF INTEGRATION 127 (8) \ sin u du = — cos u -\- C = sin I u — -]-{- C, (9) j sec^ udu = tan w + C, (10) /. du . u , ^ = arcsin - + C, Va^ - u^ a <-'"> S^ du 1 4.« w , ^ = - arctan - + C, + w^ a (12) j udv=uv— i vdu. It is strongly recommended that each of the formulas be written out by the student in words, and memorized in that form. The test of the correctness of an iiitegral is that its derivative must Jje Jhe^^iven^ jntegrand. The above formulas may be verified at once by differentiation. 86. Formulas (l)-(3). Formula (1) merely embodies 'the definition of an integral. Formula (2) is readily extended to the case of any number of terms. The formula shows that if the inte- grand consists of a sum of terms each term may be inte- grated separately. Formula (3) says that if the integrand contains a con- stant factor, that factor may be written before the integral sign. As a corollary, we may introduce a constant facto?' into the integrand^ provided we place its reciprocal before the integral sign. But the student must beware of intro- ducing variable factors by this rule. 87. Formula (4) : Powers. This formula evidently fails when n = — l. The exceptional case I — is taken care of by formula (5). 128 CALCULUS Exaynples : (a) CU 2;3 4- 1 4- -^^dx = 8j"a;3 dx -\-Jdx + If 3^''^ dx = — — -A-x — -X ^ + C' 4 2 4 2x (b) Jx(l + :r2)2 dx =f(x +2x^-\- a^)dx /VIM ry*^ /y»0 A better method : Introducing a factor 2 after the integral sign and its reciprocal in front, we have fx(l + x^)^dx =1/(1 + x'^y -2xdx. Since 2 x dx is the differential of 1 -\- x^^ formula (4) applies with u=l -\- x^^ n — 2. Hence i fci + x^y .2xdx = ^il±^ + 6^= 1(1 + x'^y 4- C. 2^ 2 3 t) In substance we have introduced a new variable u =4 i^^ as in § 84. But with a little practice one is able to think of the quantity 1 -\- x^ directly as the variable of integra- tion, without writing out a formal substitution, thus effecting a great saving of time. The student should compare the two answers that have been obtained in this example. Va -j- bx dx = - 1 Va -\-hx - h dx _ 1 {a-\- hx) 2 ^ O Here the quantity a -{- hx is taken as the variable of inte- gration, the factor h is introduced to give the proper differential, and (4) then applies with n= |. STANDARD FORMULAS OF INTEGRATION 129 EXERCISES Evaluate the following integrals; check the results by differen- tiation. 1- 1(1+-,)''^- 2. j"(v';+2V^ + -i^)rfx. ' 3. j'-£i+ji±-lrf,, 4. j4(:.+ l)Vix. 6. ^(1 - 2 xyiJx. 6. ({X + \){x + 2) dx. 9. f x\/«2 + a--^/.r. 10. f- - ^'^^-^ • c/ ^ Vl - ^3 11. r(«2_a;2)2f/x. 12. f V(I+17)=^r/^ 13. fe^vTT^'/y. 14. f(x- + l)(x2 + 2a:+ 6)r/a;. J 2 2 3 -/,. 2 2 5 16. rsin2 6lcos^^/^. 17. r ^^"^-^ • 18. r(rt' -xh^f/x. 19. f — ^^^ • *^ -^ V;^ - 4 X 20. (aVflj-vix. 21. I — ~ V "^ (1 - 8&N.,Formulas (5)-(6) : Logarithms and exponentials. — • a^ -f ar If the factor 2 be inserted under the integral sign, the numerator becomes the differential of the denominator, and (5) applies : — dx. x + 1 K 130 CALCULUS By division we find = X— z -] x+ 1 x-[-l Whence f^^dx= f(x-2-^^-)dx ^ x-\-l ^ \ x + lj x^ (^. 27. f— ^-i^. 28. f.fsec^x^/x. ^ f 1 - x^^ J 29. farctanx^/x. 30. (^^J^^^^ dx. 31. je^^rfx. 32. j*//(6-^)V rf^2^^' 34. p-'(l +e-)3rfx. 35. p(l +^'\ /x-. 36. fsec2^tan3^^l9. 37. Jci - x3)2^/x. 38. J(a - x^ dx. 39. Jx(«-x)t.x. ^O-L-^ + 10 41. J-^. 42. J.^ log(l + .^)rfx. 43. f-^^^. 44. J-^ ^ 1 + X8 -^ 1 - d e^ dx 3e* 45 f ^^—. 46. f^^ + ^^^'^'cZx. 47. flog^^^^-i'- 48. f _^^'^ -^ -^ Vx(l + x) 49. /.=.-", we may assign any two other values to x^ say x = \ and x=% thus obtaining two simultaneous equa- tions to solve for (7 and D ; or we may equate the coefficients of like powers of x in the two members of the identity. Equate coefficients oi x^ -. A + D = 1, D = 2. Equate coefficients oi x^ : S A -i- + 21) = 0, O = -1. Whence = -log^- -^ +-l--+21og(.r+l)+C-'. {X + 1)^ X -\-l 140 CALCULUS EXERCISES Evaluate the following integrals. 1 r dx 2 r dx ^ x^ — x^ •' (1 — x)^ o C xdx ^ C{3fi_j-_Y)dx ■ J (1 - .r^5* • J .rr.x- 2^2 5 (1 - :c)S J x{x - 2)2 J 13 1 ^ns. loq:a; log (x-\- 1) logCx — 1) h C. 4 * V -r y ^ s V y 2(2' - 1) a;4 + 12 x-3 + 52 a;2 + 96 a: + 64 A -5a:— 12 , oi a:4-4, ^ Arts. + 2 log — — -\- C. a:2 + 6a; + 8 ^a.-+2 7. C^^ Ans. i e-* - - + i log (e^ - 2) + C. 97. Quadratic factors. Corresponding to a factor in the denominator of the form * ax^ + hx + c where ^^ — 4 ac < 0, we assume the terms — ; ' ^H — ax'^ + ox -\- c ax^ -{•bx-{- c The case in which the denominator has repeated quad- i-atic factors is of less importance, and will be omitted. Example: Evaluate I —— — dx. Assume x^Jr^x + lQ ^A BC2X+2) C a^ -^ 2x^ -^ 5x X x'^ -\- 2 X -\- f) x^ -\- 2 x -^ 5 2^2 -I- 4 a; + 10 = A(2;2 + 2 a: + 5) + Bx(2 x-{-2)-{- Cx. Put 2^ = 0: 5^ = 10, ^ = 2. Equate coefficients oi x'^ : A -j- 2 B = 1, B = — ^. Equate coefficients oi x : 2 A-\- 2 B -^ C = i, = 1. * Of course such a factor might be broken up into complex linear fac- tors, after which the process of § 95 would apply. The present method has the advantage of avoiding imaginaries. INTEGRATION OF RATIONAL FRACTIONS 141 Whence J x^ + 2 2 + 43;+ 10 dx -\-zx^-\- b X = c(^-i. ^"+2 + I V- -' \x 2 x'^-\-2x-\- b x^-[- 2 x-\- 0/ = 2 log x — ~ log (a;2 + 2 2: -h 5) + - arctan ^-^ — h C. EXERCISES Evaluate the following integrals. 1. f ^'^^ ylns. x + 21og(a:2-4a:+5)+3arctan(a:-2) + C. -^ a:- — 4 x + 5 ' J 1 + ^4* ' J a:3 + 4 a;2 + 8 x' 4. \ A ns. — log — ■ arctan - + C . ^ o^ - 2-4 4 a a - a; 2 a a J.r dx g r X r/x 1 - a:^ ■ J a;"^ + X + 1 •1^ 6 xdx x^-\-x'^ + '^x + 4 ,4 „.5. J^ log (a:2 + 4) - i log (x + 1) + I arctan | + C. Q r x^f/a: q C X dx J (1 + x'-^)^' ■ J x-^ - 2 a: + 2' ^ x3 -^ x%l + x^) 12. C—Jl A?is. 1^+ ^log(sin(9+cos(9)+ C. *^ 1 + tan 6 MISCELLANEOUS EXERCISES Evaluate the following integrals. dx 1 r dx 2 r^ *^ (a + bx)'^ ' ^ a + hx g r xdx ^ C X- dx -^ (« + x)2" * -^ (^/ + x)2 _ r xdx ■ g r x dx J (a + xy(h + x)' J (a- + x^yi' 142 CALCULUS 7. ('-^^^d.. 8. (• ^" . ^ «4 - a;4 ^ 2 a;2 + 8 a; + 1 1 - x^' ' J {2-x) 13 15 tan 6 — cot ^ f_^. 16. f 17. fi^i^iy.. 18 r^^ + i rf^. CHAPTER XIV THE DEFINITE INTEGRAL 98. The definite integral. Let /(a:) be a given function, F(x) an integral of f(^x')^ andjr== a and x=h two given values of x. The change in the value of the integral F(x) as X changes from a to 6 is called the definite integral of /(a;) between the "- limits " a and 5, or simply the definite integral from a to Z>, and is denoted by the symbol Jf(jc)dx. Its value is evidently F{h^ — F(a^. a This change in the value of the integral between two values of the variable is required in many important problems. It is called the definite integral because its value is independent of the constant of integration. To evaluate a definite integral, we have merely to find the indefinite integral, and then subtract its value at the " lower limit 11, « if rom its value at the '> ujipe^-limit " h. It is custoniary to use the symbol F(x) F(h)-F (a), „ .Thus ^-^--.^ fy(x)dx = F(x) '=F(b}-F(a). Since the constant of integration disappears, there is no object in writing it at all. Examples : (o^) In ((z^, we must either return to the original variahle before substi- tuting the limits, or change the limits to correspond with the change of variahle. The latter method is usually preferable. The new limits are found, of course, from the equation of substitution X = (^(2), 146 CALCULUS as in the following Example : Evaluate j x^\ — x dx. Let 1— x = z^x=l— z^ dx= — dz. When a:=-l, 2 = 2; when x=l, 2=0. Hence j xVl — xdx= — j (1 — 2)2^ 6?2 = fee* - ^4)^^ = I ^^ - I ^']"= - t\V2. EXERCISES 1. Work the above example, putting 1 — x = z^. Evaluate the following integrals. 6. f' ^^-^ . .4m. J^. 7. T-^^.^ns. (|V2-2)a. 8. J2 1 _x4' ®' Jo "I ^ f/x a:'^ + 1 10. The area bounded by the parabola 3/^ = 4 ax, the x-axis, and the latus rectum is, by § 99, A = £ydx. Evaluate this integral (a) by substituting for y ; (b) by substituting for dx and changing limits. 11. Find the area under the curve y — e'' from x — to a: = 1 by the two methods of Ex. 10. 12. Find the area of half an arch of the cosine curve by the two methods of Ex. 10. THE DEFINITE .INTEGRAL 147 13. Given y = sin x, evaluate i xdy in. two ways. 14. The velocity of a point moving in a straight line is y = 4 cos - • 2 Find in two ways the distance from the starting point at the end of - seconds, o CHAPTER XV THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM 102. Area under a curve. We have seen in § 99 that the area ABQD bounded by the plane curve y =f(x), the a:-axis, and the lines x= a^ x = b is given by the definite inte- gral I f(x)dx. We will now obtain a new expression for the same area. In what follows, the functionj^(a:j is assumed to be one- valued and continuous, and to have only a finite number^ of maxima and minima in the interval from x = a_to x==J>; in fact, we may suppose for definiteness that the curve rises throughout the interval. The argument is readily modified to fit the case when the curve steadily falls, or rises and falls alternately. We may evidently get an approximate expression for the area A by dividing the base AB into n equal inter- vals Ax, erecting the ordi- nates at the points of di- vision, and taking the sum of the inscribed rectangles AEFD, etc. The areas of these rectangles are respec- tively Fig. 58 where f(x^Ax, f(x^)Ax,'", f(xJAx, X-, = a. x^ = a -\- Ax, Xn = a -\- 2 Ax, x^= a -{- (ii — 1)A2: 148 b — Ax. DEFINITE INTEGRAL AS LIMIT OF A SUM 149 Hence, approximately^ A =f{x^}Ax -^f(x^}Ax 4- i=l Now it is geometrically evident that this sum of rect- angles may be made to represent the area A with an error less than any preassigned constant however small, by taking n sufficiently large. Hence, by the definition of § 14, we have exactly —^-^.^ (1) A= lim ^f(Xi)Ax. Fig. 59 A formal proof of this state- ment may be given as follows. Let J.J denote the sum of the inscribed rectangles, A^ the sum of the circumscribed rectangles (Fig. 59). Then for all values of n A^< A< A^. Now the difference between A^ and A^ is the sum of the shaded rectangles. Sliding all these across into the last column, we see that A^-A^ = IO.Ax = [f(b) -/(«)]A;r. As n increases indefinitely Ax approaches 0, so that A2 — A^ also approaches 0. Since A always lies between A^ and ^.g, it follows that A^ and Ac^ approach A as their common limit. Hence we have formula (1).* The rectangles f(^x^Ax are called elements of area. As n increases indefinitely, each of the elements approaches : i.e. they are infinitesimals. * The n parts into which AB is divided need not be taken all equal; the same limit is obtained provided the width of each rectangle approaches as the number of divisions is indefinitely increased. Further, it is clear that we may take the limit of the sum of either the inscribed or the cir- cumscribed rectangles, or of any set intermediate between these t^v^o. 150 CALCULUS Questions like that of the present article, in which we have to deal with the limit of a sum of infinitesimal ele- ments, will arise many times in this and later chapters. As in every case the existence of the limit will be evident by geometric intuition, we shall in future omit formal proofs. 103. Evaluation of the limit. Equating the values of A found in §§ 99 and 102, we find A = lim y/(a:,)Aa: = f(x)dx. Thus the limit occurring in § 102 can always be evaluated by a definite integration.* The fact that the quantity f(x)dx appearing under the integral sign represents the area of a rectangle of altitude f(x) and base dx = Aa:, and thus suggests the sum from which the integral was derived, is the chief reason for using the notation j f(x)dx (see § 77). In fact, the inte- gral sign j is historically a somewhat conventionalized S^ meaning sum. 104. The fundamental theorem. In § 102 we have ex- pressed the area under a plane curve as the limit of a sum of rectangles ; in § 99 we have found the same area as a definite integral. It is clear that the arguments used will hold no matter ivliat may he the geometric or physical meaning of the given function^ for any function whatever may be interpreted as the ordinate of a point on a plane curve. We therefore have at once the following Fundamental Theorem for Definite Integrals : Griven a function /(a;), continuous in the interval from x = a to X = h, divide this interval into n equal parts A2:, * More precisely, the limit can always be expressed as a definite inte- gral; the actual evaluation of the integral is often impossible (see § 78). DEFINITE INTEGRAL AS LIMIT OF A S\jM 153 n and form the sum ^f(^Xi)Ax^ where x^ = a, x^ = a - . .., Xn=' a +(n— l)Aa;. The7i Hence^ if in any problem an arbitrarily close approxima- tion * to the required quantity can be found by adding up terms of the typef(x')Ax from x = a to x = b^ that quantity is given exactly by the definite integral I f{x)dx. We have now presented the ^efiiiite integral in two distinct aspects : first, as the change in the value of tHe indefi nite integral between two values of the variable ; second, as the limit of a sum of infinitesimal elements. The great advantage of this latter point of view will become apparent as we proceed. It may be remarked that if the function f(x^ has a finite number of finite discontinuities in the interval from a to 6, as in Fig. 60, it can still be inte- grated. For it will be con- tinuous in a number of sub- ^^^' ^ intervals such as AO^ QD^ DB^ to each of which the fundamental theorem can be applied and the results added. 105. Plane areas in cartesian coordinates. Not only the area considered in § 102, but any plane area bounded by curves whose equations are given in cartesian coordinates, can be found as follows. Imagine inscribed in the area a set of n elementary rectangles of altitude h^ and width AZ, n in such a way that the sum V ^^A^ may be made to repre- <=i * That is, an approximation in which the error may be made less than any preassigned constant. 150 CALCULUS Quer^e area to any desired degree of approximation by hav" rising n. Then at once, by the fundamental theorem, A= lim V7z,A/= ChdU the limits being chosen in such a way as to extend the integration over the whole area. Of course in any par- ticular problem li and dl must be ex- pressed in terms of the coordinates. In every problem the student should make a sketch of the area to be found, draw an element in a general position, and obtain the area of the element directly from the figure, as in the following Examples: (a) Find the area in the first quadrant bounded by the parabola ?/2 = 4:ax, the a^-axis, and the line x— a. An arbitrarily close approximation to this area can be found by forming a sum of rectangles of altitude y, base dx^ and area y dx^ as shown in Fig. 61. Hence, we have exactly ra /»« 2 2 i' A= \ y dx = 2 \ \:ax dx = ~ • - (ax^ ^ Jo "^ Jo [ ^a 3^ ^ (K) Find the area in the first quadrant between the parabolas (1) y^ = -^ ^^, (2) ?/2 = 8a2:- 4^2. Let us take as the element a rectangle parallel * to OX. The area of the rect- angle is evidently (x^ — x^dy, where x^ and x^ are the abscissas of the points on the curves (1) and (2) respectively. The fig. 62 Fig. 61 = -a^ *That is, with its finite side parallel to OX. DEFINITE INTEGRAL AS LIMIT OF A SUM 153 curves are found to intersect at (a, 2 a). Hence A =r(r,^ - X,) dy =r(f + 5 -Pjdy *^o ^0 V8 a 2 4 aj '£ \:^ SaJ ^ 12 24a 2 = - a^. 3 EXERCISES 1. Find the area bounded by the curve y = x^ and (a) the lines y = 0, X = 2, (h) the lines x = 0, y = 1. 2. Solve example (a), § 105, taking the element parallel to OX. Evaluate the integral in two ways. 3. Find the area bounded by the curve ay'^ = x^ and the line X = '^a. Ans. i|^ a"^. 4. Find the area of a circle (see Ex. 52, p. 136). 5. Find the area of an ellipse, using the cartesian equation ; check by using the equations x = a cos (f>, y — b sin cf) (see Ex. 51, p. 136). A ns. irab. 6. Find the area of half an arch of the curve y = I cos 2 x. 7. Find the area between the parabolas y^ = 4:ax and x^ = 4 ay. ■ A71S. 1/ a-. 8. Show that the area bounded by a parabola and any chord perpendicular to the axis is two thirds of the circumscribing rectangle. 9. Find in two ways the area bounded by the parabola y = x'^, the ?/-axis, and the lines y = 1, y = 4. 10. Find the area bounded by the curve y — log x, the axes, and the line ?/ + 1 = 0. 11. Find the area bounded by the curve y = log x, the x-axis, and the line x =2. Ans. 0.386. a I - — ^\ 12. Find the area under the catenary ?/=-(ga+e aj, from X — — a \iO X = a. An^. aP-\ e e 13. Find the area between the curve x^ = 4 a^ — ay and the x-axis, taking the element («) parallel to 0Y\ {h) parallel to OX. Ill 14. Find the area bounded by the parabola X'^ -\- y'^ = a'^ and the coordinate axes. Ans. ^a^. 154 CALCULUS 15. Find the area of a circular segment of height h. Check by- putting A = 2 r. 16. Find the area of one arch of the cycloid x = a(6 — sin 6), y = a(l — cos 6). Ans. 3 na^. 17. Find the area bounded by the curves y = x, y = 2 x, y = x^. Ans. |. 18. Find in two ways the area in the first quadrant bounded by the curves y = x^, x'^ = 2 — y, y = 0. 19. Find in two ways the area bounded by the curve y=(l — x^y and the a;-axis. 20. Find the area bounded by the curve y =(x — Sy^(x — 2) and the a'-axis. 21. Find the area bounded by the curve y = —^ — -, its asymptote, and the maximum ordinate. Ans. 0.698. 22. Trace the curve y^(x'^ + a"^) = aP'x'^, and find the area bounded by the curve and the line x = a. Ans. 0.83 a^. 23. Trace the curve ay- = ax^ + x^, and find the area of the loop. 24. Find the area bounded by the curve y^ = x^(l + x) and the ar-axis. Why is the answer negative ? 25. Trace the curve y = ;- , and find the area under the curve from X = - 2 to X :ir 0. ^ + ^^ ^^,^ j 37_ 26. Find in two ways the area bounded by the coordinate axes and the curve y^ = 1 — 2 x — xy. 27. Find the area of the circle x = a sin 2 6, y = a cos 2 6. 28. Find the area bounded by the curve y = — ^—, the a:-axis, and the maximum ordinate. Ans. ^. 29. Find the area in the first quadrant under the curve y^ = , X — I between the minimum ordinate and the ordinate at a; = 3. Ans. 2.05. 30. Trace the curve y^ = x'^(x + 4:), and find the area inclosed by it. 212 Ans. — — . 105 106. Plane areas in polar coordinates. Given the equa- tion of a plane curve in polar coordinates, let us try to find the area bounded by the curve and the radii vectores corre- DEFINITE INTEGRAL AS LIMIT OF A SUM 155 spending to 6 = a, 6 = /3. We can obtain an arbitrarily close approximation to the area by inscribing in it n circu- lar sectors of radius r^ and angle A^, hence of area* | r,^A^, and forming n the sum V J r,^A6. Hence, by the fundamental theorem, i=l By means of a theorem to be ^^^- ^^ proved in § 109, it can be shown that this result may also be obtained by choosing as the element a triangle of altitude r^, base rjA^, and area J r'jAB. EXERCISES 1. Find the area swept out by the radius vector of the spiral of Archimedes r=a6, in the interval from ^ == to ^ = 2 tt. 2. Solve Ex. 1 for the logarithmic spiral log r = a6. 3. Find by integration the area of the triangle bounded by the lines r = a sec 0, B = (), 6 = — • 4 4. Find the area inside the lemniscate r^ = a^ cos 2 6. Ans. a^. 5. Find the area of the curve r^ = a^ cos $. 6. Find the entire area of the cardioid r = a(l+ cos 0) . Ans. f ira^. n 7. Find the area between the parabola r = a sec'-^- and its latus rectum. " Ans.^a^. 8. Find the area of the curve r = a sin 2 6. Ans. ^wa^. 9. Show that the area of one loop of the curve r = a cos nO is 2 , hence the total area inside the curve is one fourth or one half the 4 71 area of the circumscribed circle, according as 7i is odd or even. * By elementary geometry, the area of a circular sector of radius r and angle a is A = l r^a. 156 CALCULUS 107. Volumes of revolution. The volume of a solid of revolution may be found very readily by the fundamental theorem. Suppose the volume is generated by revolv- ing the area ABCD about the line AB. If we inscribe in the re- volving area a set of n rectangles of alti- tude r» and- base AA, each rectangle will gen- erate in its rotation a circular disk^ or cyl- inder, of radius r^, alti- tude Ah, and volume irr^Ah. Further, as n increases the sum of these cylindrical vol- umes approaches as its limit the required vol- FiG. 64 b ume. Hence, by the fundamental theorem, i=l the limits being chosen so as to include the whole volume. Of course in any problem both rand dh must be expressed in terms of the coordinates. When the axis of revolution does not form part of the boundary of the revolving area, we may choose as elements a set of circular rings, as in example (6) below. Examples : (a) The area in example (a), § 105, revolves about OX. Find the volume generated. Dividing the area into elements as in Fig. 61, we see DEFINITE INTEGRAL AS LIMIT OF A SUM 157 that each rectangle generates a cylindrical volume-element of radius ?/, altitude dx^ and volume iry'^dx. Hence V ="X' y^ dx — \iTa\ xdx= 27ra^. (6) The above area rotates about OY. Find the volume generated. If we divide the area into elements as in the figure, each element generates a circular ring of outer radius a, inner radius x^ thickness dy^ and volume ir^a^ — ^^dy. Further, the limit of the sum of these volumes is the required volume. Hence F= 'rr£\a'^ - x^)dy = ttJ'Y^^ - Fig. 65 y' 16 a^ dy r_ 80 a^ = -Tra*- This result could have been obtained equally well by finding the volume generated by rotating the area OBQ about Oy, and subtracting this from the volume of the cjdinder formed by revolving the rectangle OABQ. But in case it is possible to simplify the integral before integrating, as often happens, the first method is to be preferred. 108. Volumes of revolution : second method. The fol- lowing method for finding volumes of solids of revolution is often preferable to that of § 107. Let us take as an element of the area ABO (Fig. (^6^ a rectangle of length hi parallel to the axis of revolution AB^ and of width Ar. This rectangle generates by its rotation a cylindrical shell of inner radius r^, altitude A^, and thick- ness Ar. The volume of the shell is evidently 7r(ri -t- Ar')Vii — irrfhi = 2 TrrihiAr + Tr/^^Ar^, and the limit of the sum of these elementary shells is evi- 158 CALCULUS dently the required volume. Now it will be shown in the next article that, in passing to the limit, we may neglect the infinitesimal of higher order* 7rhiAr\ Hence, by the fundamental theorem, Fig. 66 II V= Urn ^ 2 TTTihiAr = 2Tr Crh dr, the integration being ex- tended through the whole region. This result is easily re- membered from the fact that the integrand is the differential of volume of a right circular cylinder, the altitude being constant: F=7rr2A, dV=2 7rrhdr. Example : Solve example (6), § 107, by the present method. Divide the rotating area into rectangles parallel to OF, as in Fig. 61. Each rectangle generates a cylindrical shell of radius x, altitude g, and thickness dx. Hence V= 2 TT j xg dx = 4: ira^ I x^ dx = f ira^. 109. A theorem on infinitesimals. It often happens, as in the preceding article, that in applying the fundamental n * This is easily shown directly. The quantity neglected is ^ vhi^r^. n i=l This may be written irAr ^ hi^r. When Ar approaches 0, the sum n 1=1 ^ hiAr approaches a finite limit, viz. the generating area, so that the 1=1 whole quantity approaches 0. DEFINITE INTEGRAL AS LIMIT OF A SUM 159 theorem we have to replace the element as originally chosen by another element differing from the first one by an infinitesimal of higher order. That this is allowable appears from the following Theorem; The limit of a sum of positive infinitesimals is unchanged when each infinitesimal is replaced hy another that differs from it hy an infinitesimal of higher order. It follows that, in taking the limit of such a sum, all in- finitesimals of higher order may he neglected^ as was done in §108. In this connection it should perhaps be mentioned ex- plicitly that two infinitesimals differ from each other by an infinitesimal of higher order whenever the limit of their ratio is 1, and conversely. To prove the theorem, let Wj, 2^2' ••' ^n be a set of posi- tive infinitesimals such that lim ^ ^» exists; and let Vj, ^2, •••, v„ be another set such that Vi differs from u^ by an infinitesimal of higher order : ^t = "^i + «^t?/t, where w^ is infinitesimal. Then n n n lim 2 Vi= lim ^Ui-^ lim ^ ?/\Wj. Denoting by w the absolute value of the largest of the t^'s, we have w n n -W^Ui<^ Willi ^W^Ui. t=l i—\ t=l Since the first and third of these quantities both approach 0, the second must do likewise. Hence n n lim ^ Vi — lim ^ w^, and the theorem is proved. 160 CALCULUS EXERCISES 1. Find the volume of a sphere. 2. Find the volume of a right circular cone. 3. The hyperbola x^ — y'^ = a^ revolves about its transverse axis. Find the volume of a segment of height a of the hyperboloid gen- erated. Ans. |7ra3. 4. Find the volume generated by revolving the four-cusped hypo- cycloid a;3 + 3/3 _ ^3 about OX. Ans. j%% ira^ 5. Find the volume generated by revolving the area under the curve ?/ = e^ from x = to x = 1 (a) about OX; {b) about OY; (c) about the line x = 1. Ans. (c) 2 7r(e — 2). 6. The area under one arch of the sine-curve revolves (a) about OX; (b) about OF. Find the volume generated. Ans. (a) ^i (b) 2 7r2. 7. Find the volume obtained by revolving about OX the area X X under the catenary ?/ = -(ga + e «), from x = — a to x = a. Ans. — 7- 4 V^'-i) 8. The area OBC of Fig. 65 revolves (a) about the line CB ; (b) about AB. Find the volume generated; check by solving in two ways. 9. Find the volume of an oblate spheroid, using («) the ordinary equation of the ellipse; (6) the parametric equations a; = acos<^, y = bsm^. Solve each part in two ways. A71S. f Tra^ft. 10. The area in example (b), § 105, revolves about OX. Find the volume generated. 11. Find the volume of a spherical segment of height h. Check by putting h = 2 r. 12. Trace the curve a'^y'^ = x^(2 n — x), and find the volume gen- erated by revolving the curve about the a:-axis. 13. Find the volume generated by revolving (a) about OX, (b) about OY, the area between the curves 2 y =^ x^, y = x'^. Check the results by solving in two ways. 14. Trace the curve (x — 4 a)y'^ = ax(x — 3 a), and find the closed volume generated by revolving it about the a:-axis. Ans. 6.12 a3. DEFINITE INTEGRAL AS LIMIT OF A SUM 161 15. The curve y^ = x(x — l)(x Find the closed volume oeuerated. 2) rotates about the a;-axis. A ns. — • 4 16. Find the volume generated by revolving one arch of the cy- cloid X = a(0 — sin 0), ij = a(l — cos 0) about the ar-axis. Ans. 5 ir^a^. 17. Trace the curve y(^x^ + y^) = a(^"^ — y'^), and find the volume generated by revolving the loop about OY. Ans. 0.053 ira^. 18. Find the volume of a torus. Solve in two ways. Ans. 2 Tr^a%. 19. The area bounded by the curve y = (1 — x-)^ and the a:-axis revolves about the ?/-axis. Find the volume generated, (a) by the method of § 107 ; (h) by the method of § 108. In (a) evaluate the integral in two ways, first by substituting for x^, next by substituting for dy. 20. Trace the curve a^y'^ = a^x'*^ — x^, and find the volume gener- ated by revolving one loop about OY. Ans. ^^ ira^. 21. A round hole of radius a is bored through the center of a sphere of radius 2 a. Find the volume cut out. 22. Find the closed volume generated by revolving the curve X^Cx^ Q, ^ y~ = — ^, . — T- about the x-axis. Trace the curve. Ans. 0.072 ttci^. 23. Find the volume inside the cylinder x- + ?/- = 2 a^ and outside the hyperboloid x^ + y'^ — z^ — a'^. A7is. | ira^. 24. Find the closed volume generated by revolving the curve y- = x-^(x + 4), (a) about the x-axis, (b) about the ^/-axis. 25. Find in two ways the volume generated by revolv- ing about the ?/-axis the area bounded by the curve y = X and the coordinate axes. 110. Other volumes. The volume of any solid can be expressed as a defi- nite integral, provided we know the area of every plane section parallel to some fixed plane. Let us divide the volume into slices of thickness Ah by M Fig. 67 162 CALCULUS means of n planes parallel to this fixed plane. If on each of these plane sections we erect a cylinder of altitude Ah and base A^ where A is the area of the section, the sum of the n cylindrical volumes thus formed will be approxi- mately the required volume, and the limit of this sum will be exactly the volume. Hence V= limXAi^h = fAdh, n-^QC t=l *^ with properly chosen limits. Example : A woodsman chops halfway through a tree 4 ft. in diameter, one face of the cut being horizontal, the other in- clined at 45°. Find the volume of wood cut out. The figure shows one half of the required volume. If we slice up the volume by planes parallel to the ^2-plane, the element of volume is a triangular plate of width 1/, altitude 2, and thickness dx. Hence V= 2 Pj ^z dx. But Fig. 68 so that 2 = y, and 2/ = V4 — X' V= r(4:-x^}dx=5^cu. Jo ft. EXERCISES 1. Solve the above example in a different way. 2. Find the volume of a tetrahedron with three mutually perpen- dicular faces. Ans. ^abc. 3. Find the volume sliced off from a right circular cylinder by a plane through a diameter of one base and tangent to the other base. Ans. I rt^A. DEFINITE INTEGRAL AS LIMIT OF A SUM 163 4. Find the volume of a right pyramid with a square base. 5. Find the vohime of an ellipsoid, using the answer to Ex. 5, p. 153. Alls, ^irabc. 6. Find the volume of an elliptic cone. Ans. ^ Trabh. 7. Find the volume of a spherical wedge. Ans. f aa^. 8. Find the volume of a wedge cut from a right circular cone by two planes through the axis. Ans. i aa%. 9. Obtain a formula for the volume of a wedge cut from any solid of revolution by two planes through the axis. 10. A carpenter chisels a square hole of side 2 in. through a round post of radius 2 in., the axis of the hole intersecting that of the post at right angles. Find the volume of wood cut out. Ans. 15.3 cu. in. 11. Find the volume cut from the cylinder x'^ + t/'^ = a^ by the planes z = ?nx, z = nx. Solve in two ways. 12. A right circular conoid is generated by a straight line which moves always parallel to the a:y-plane and passes through the line y = h in the ?/2-plane and the circle x'^ -{- z^ = a^ in the xs-plane. Find the volume of the conoid. Ans. | ira^. 13. Find the volume in the first octant bounded by the hyperbolic paraboloid generated by a straight line moving always parallel to the z?/-plane and passing through the lines y + z = a in the yz-plane and X = b in. the a^s-plane. Ans. ^ a%. 14. A banister cap is bounded by two equal cylinders of revolution 8 in. in diameter, whose axes intersect at right angles in the plane of the base of the cap. Find the volume of the cap in two ways. 15. Find the volume of a right pyramid whose base is a regular hexagon. 16. Find the volume in the first octant under the plane z = x and inside a cylinder standing on the parabola ?/ = 4 — a;^ as a base. Solve in two ways. 17. Solve Ex. 12 if the line ?/ = A is replaced by the line y -\- z = h (/i>a). Ans. ^TTO^h. 18. Solve Ex. 13 if the line x = b is replaced by the line x = z. 19. Find the volume in the first octant bounded by the planes X ■\- z = a, X + 2^ + 22 = 2 a. 111. Line integrals. The ordinary definite integral depends on all the values of a given function /(a:) along a straight line segment — the segment of the a;-axis from 164 CALCULUS Fig. 69 x= a to X = h. It happens frequently that we have to compute a quantity that depends in a similar way on the values of a function F along a curvilinear arc 0. The function F is in general dependent on both coordinates of the point on the curve : F= Fix, y). But since y is given as a function of x by the equation of the curve, the function F(x^ y) reduces at once to a func- tion of one variable. Given a function F(x, y) de- fined at all points of a plane curve (7, let us inscribe in C a broken line of n segments As/ having equal projections ^x on the a:-axis, multiply each seg- ment by the value of F(x^ ^/) at the corresponding point of division* P^ (Fig. 69), and form n the sum of these products, 2j -^fe^ Vi) ^^l ' The limit of this sum, as the number of divisions becomes infinite, is called the line integral of F(^x^ y) along the arc (7, and is denoted by the symbol j F{x^ y) ds : lis. % ^(^i' y^^'l =£^(^^ y) <^'- 112. Geometric interpretation of the line integral. The existence of the limit last written may be made evident geometrically. Let us interpret the function F(^x^ y) as the 2-coordinate of a point on a surface in space : (1) z = F(ix,y). * It is of course merely for convenience that the broken-line segments are drawn with equal projections on OX. Tlie division may be made in any manner provided in the limit every segment approaches 0. Further, Asj' may be multiplied by the value of i^(x, y) at either end-point of ASj' or at any point on the subtended arc. DEFINITE INTEGRAL AS LIMIT OF A SUM 165 On the curve C as directrix erect a cylindrical sur- face with generators perpendicular to the rr^z-plane. Each of the quan- tities F(Xi,yi) As/ represents the area of a rectangle in- scribed in this cyl- inder, having a base As/ and an al- titude F(x^^ 1/i), and the sum of these rectangles evi- dently approaches as its limit that part of the cylin- drical surface ly- ing between the a;^-plane and the surface (1). 113. Fundamental theorem for line integrals. The theorem of § 104 takes the following form for line in- tegrals : Given a function FQx^ ?/) defined at all points of an arc C, iyiscrihe in the arc a broken line of n segments As/ having n equal projections on OX, and form the sum ^ F{Xi, t/i) ASi, where (xi, y^ is the i-th point of division on C. Then Urn V Fix^. y^)As^ = (f(x, y)ds. Fig. 70 n-^cc t=l Hence, if in any problem an arbitrarily close approxima- tion to the required quantity can be found by addi?ig up terms of the type F(^x, y')As', the quantity is given exactly 114. Evaluation of line integrals. To evaluate a line integral, we have in general to express both F(^x, ?/) and ds in terms of x, y, or some other suitable variable, and 166 CALCULUS then integrate between limits in such a way as to extend the integration over the given arc. Thus, if X is chosen as the variable of integration, we replace ds by its value (§ 52) ^^=Vi+(tT^-' \dx) and obtain JFix, y-)d»=£Fix, y)^ll +(^)V where a and b are the abscissas of the end points of C,* and where y must be replaced by its value in terms of x from the equation of the curve, f - * It is assumed that no parallel to the y-Rxis can meet C in more than one point. If this condition is not satisfied, C must consist of several portions for each of which the condition holds, and each portion may be considered separately. t This transformation of the line integral into an ordinary integral may be justified by the theorem of § 109. We have evidently ASi' = Va^' + A^' = \ 1 + ^Ax. ^ Ax Now the limit of the sum ^ F (Xi, Pi)^! + ^^ Ax cannot be ex- !=i Ax^ pressed directly as a definite integral by the theorem of § 104, since the summand depends not only on Xi but on Ax as well. But the infinitesi- mals F(Xi, ?/i)\/l -\--=2^^ ^^^ F{Xi, yi)y/l + y''^ Ax differ from each Ax other by an infinitesimal of higher order, since the limit of their ratio is evidently 1, and hence the latter may be substituted for the former, by § 109. Therefore \ F{x, y)ds = lim V F(x,, ?/,)\/l + ^ Ax JC n->oo ^ ^ Ax^ ■ n = lim V F(Xi, yi)Vl+yi'2Ax = Cf(x, 7j)y/l + y'-^dx, Ja by the fundamental theorem of § 104. DEFINITE INTEGRAL AS LIMIT OF A SUM 167 To integrate with respect to ^, we put and express the entire integrand in terms of y. If X and y are given in terms of a parameter t^ we use In the discussion of line integrals, we have spoken in terms of cartesian coordinates, but the argument is evi- dently independent of the particular coordinate system used. Some simple types of line integrals are considered in the next three sections ; other examples will be met with later. 115. Length of a curvilinear arc. To find the length of an arc of a plane curve (7, we proceed as follows : Inscribe in Q a broken line of n segments As/ as in § 111, and n form the sum V As/. This sum is of course the length of the broken line, and its limit is the length s of the arc. It is evidently the line integral I c?s, the given function in this case being ^{x^ y')-—!: s= lim VAsJ=fds. The process of finding the length of a curve is some- times called rectification of the curve. Example : Find the circumference of the circle x^ -\- y^= a^. Here dy _ _x dx y^ 168 so that CALCULUS i C dx = 4a I — r^:= X = 4 a arcsin - Va2 - x^ «. = 2 7ra. EXERCISES 1. Find the circumference of the circle x = a cos 9, y = a sin ^. 2. Rectify the semicubical parabola ay^ = x^ from a:.= to a: = 5 a. Ans. -^T^j^-a. 3. Trace the curve 9 y"^ ■= 4:(1 + x-y, and find its length from i: = to a; = 2. 4. Rectify the catenary 3/ = - [e« + e~o J ylm. -232. e <» 2 2 from a: = to a: = a: 2\ 2. L ^ 5. Find the length of the four-cusped hypocycloid x^ -^ y^ = a^ Ans. 6 a. 6. Rectify the curve x = t% y = fi from ^ = to ^ = VS. 5 7. Find the length of the curve y = ^x^ between the origin and the point a; = 4. Ans. ^^^(l+ev^). 8. Find the length of one arch of the cycloid. Ans. 8 a. 9. Trace the curve 9 ay'^ = x{x — 3 a)^, and find the circumference of the loop. Ans. 4 a V3. 10. Find the length of the curve y — e^ from a: = to ar = f log 2. 116. Surfaces of revolution. The surface generated by the rotation of a plane curve about a line AB in its plane is easily expressed as a line integral. Fig. 71 Let US inscribe in the curve DEFINITE INTEGRAL AS LIMIT OF A SUM 169 Q a broken line of n segments As/ . Each of the segments As/ generates in the rotation the frustum of a right cir- cular cone, the radii of whose bases may be called r^ and r^ -\- Ar^-. The surface of this conical frustum is, by ele- mentary geometry, the circumference of the middle sec- tion multiplied by the slant height, or ^ir^r^ -h |^Ar^)As/. n The sum of these surfaces, V 2 7r(rj + \ Ar^) As/, ap- 1=1 proaches as its limit the required surface of revolution. Hence, by the theorems of § 113 and § 109, we have S= lim V2'rrnAs/ =2-17 r ds. Example: Find the surface of a paraboloid of revolu- tion bounded by a right section through the focus. Given the equation of the generating parabola ?/^ = 4 ax^ we have = 2 TT r%\/l + ^dx =1iT f V?/2 + 4 ^2 dx = 2 TT I V4 ax -\- ■iaP'dx = (^ax + a^) ^ Jo a 3 . = |7r«2(2V5-l). EXERCISES 1. Work the above example, using y as the variable of integration. 2. Find the surface of a sphere, using polar coordinates. 3. Find the surface of a sphere, using cartesian coordinates. Evaluate the integral in various ways (cf. Ex. 10, p. 146). 4. Find the surface generated by revolving the cubical parabola a^y = x^ about OX, from x = to x = a. Ans. ^ (10 VIO - 1). 27 5. Find the surface generated by revolving (r/) about OX, (h) about OY, the arc of the carve y = ' between the minimum point 6 X . ^ and the point x = 2. A ns. (a) i^; {h) ^(15 + 4 log 2). 16 4 170 CALCULUS 1 x^ 6. Trace the curve ?/ = - log x — ~ (cf. § 69), and obtain the sur- face generated by revolving the curve about OY from x = 1 to x = 2. Arts. 10.47. 7. Find the surface generated by revolving the catenary _a '- _£ y — 2 (e** + e "), («) about OZ, (b) about OF, from a; = to a; = a. Ans. (b) 27ra2^1--]. 8. Find the surface generated by the revolution of an arch of the cycloid about its base. Ans. %* Tra^. 9. Find the surface of a torus. Ans. Aiir'-ah. 10. Find the entire surface generated by revolving the curve 8 a V = a:2(a2 _ a;2) about OZ. Ans.^ira^ 11. Find the surface formed by revolving the four-cusped hypo- 1 2. 2 cycloid x^ + y^ = a^ about OX. Ans. '^^ira^. 12. Find the surface cut from a sphere by a circular cone of half- angle a with its vertex at the center of the sphere. 117. Cylindrical surfaces. Given a cylinder whose di- rectrix is a plane curve (7, the area of any portion of the cylinder may be found as follows : Inscribe in (7 a broken line of sesf- ments As/, •••, As„', and inscribe in the required area a set of rectangles of alti- tude A< and base AsJ. The limit of the sum of these rect- angular areas is the area on the cylinder : n ^ S= lim VhiLSi' = \ hds. Fig. 70 ^ DEFINITE INTEGRAL AS LIMIT OF A SUM 171 Example : Find the area on the cylinder oi^ -\- z^ = a^ included between the planes y = 0^ y = mx. Denoting by O the circular arc APB, we have aS' = 4J yds = -i\ mx^l = 4m| x\l -] — dx i C^xdx = 4 ma I Jo z = — 4 ma I (?2 = 4 ma'^. Fig. 72 EXERCISES 1. In the above example, find the area of the section cut by the plane y = mx. 2. Find the surface of the cap in Ex. 14, p. 163. Ans. 128 sq. in. 3. Find the surface of the cylinder x^ + y^ =a^ included between the planes z = x, z = 3 x. 4. Find the surface cut off on the cylinder x"^ + y^ = a^ by the paraboloid of revolution x'^ -\- y^ = hx Ans. 2 7r 5. Find the area, in the first octant, of the section of the cone x^ — 2/^ + 2'^ =1 by the plane x -\- y = a. 6. The center of a sphere of radius 2 a is on the surface of a cyl- inder of radius a. Find the surface of the cylinder intercepted by the sphere. Ans. 16 a^. 7. Find the surface on the cylinder z^ = Alqx inside the cylinder y2 = 4 ax, from x = to x = "d a. Ans. ^^ or. 8. Work the example of § 117, using polar coordinates. CHAPTER XVI INTEGRAL TABLES 118. Use of tables. In the solution of problems in- volving integration, the work may frequently be much shortened by the use of a table of integrals. Many such tables have been prepared ; the references below are to B. O. Peirce's Short Table of Integrals (Ginn & Co.). The chief object in using a table is to save time. The student is therefore not making the best use of the table unless he is so familiar with its contents and arrange- ment that he can tell at a glance whether the desired formula is likely to be given and where it is to be found. Further, it should be remembered that in many cases the result may be found by the methods of Chap- ter XII in less time than would be required if the table were used. — x{\ — x^ ;(1 — aP") Let us use formula bS of the table, with a = 1, s, dx 1 1 x^ lop- J ^2 2 I a^e^ dx. This integral is not given explicitly in the table, but it resembles formula 402. Making the substitution a:2 = 2, 2xdx= dz^ we find j A'^'^dx = \ j ze''dz = ^e'(z — l) ' 172 4 — 3. ~ 2 +h INTEGRAL TABLES 173 EXERCISES Evaluate the following integrals, using a table whenever a saving of time may be effected by so doing. 1. C^ — 'I^ ., Ans. \ arctan " ^ + ^ + C. 2. j - " "^ z r. vln.s. log ( Vl + a; + x"-^ + a: + |) + C Vl + a: + a;2 . c fix g r^_^_^/^^ J (1 +x-)'^* ■ -^ (1 -2x)2' 6. f xVl + ^-r/x. 7. C ^^^^ . J *^ (1 + 3 xy 8. ixy/\ — xdx. 9. I sin'* a- (/x. 0. I 11. Xxcosx^dx. •^ \ + cos X ♦^ 2. fx'V'^a:. 13. ^ xh-^ dx. 4. i arcsin x f/ar. 15. \\o^^xdx. 6. i arcsin x f/ar. 15. j log"^ C" __^dx__^ yj p I »/o ,/.,2 I ^2 * ^0 ,/;;2 //y Vft^ + x'^ *^° V«^ + //'^ ^ j^ V2; + t'^dt: Ans. Va - i log (2 + V3). flog 2 . 71- >• J Ve^ - 1 dx. Ans.2--^- 20. Find the area bounded by the hyperbola x^ — if- = o?- and the line X = 2a. 21. Find the length of the arc of a parabola from the vertex to the end of the latus rectum. Ans. 2.29 a. 22. Find the surface generated by rev^olving the curve y — e"" about OX from a: = to :c = 1. 23. Find the area inside the four-cusped hy^Docycloid x = a cos^ 6, y = a sin^O. Ans. f Tra^. 24. Find the area of the ellipse r= ;:, where e is the ^ . .^ 1 + e cos^ eccentricity. 174 CALCULUS 25. Find the volume generated by revolving one arch of the cycloid about its base. Ans. 5 Tr^a*. 26. Find the surface generated by revolving the curve 3 a^x + y^ = about OF from y = to y = a. 27. Find the surface generated by revolving one arch of the sine curve about the a:-axis. Ans. 2 7r[V'2 + log (1 + V2)]. CHAPTER XVII IMPROPER INTEGRALS 119. Definitions. Definite integrals in which either or both of the limits of integration are infinite, and also those in which the integrand becomes infinite within the inter- val of integration, are called improper integrals. Such in- tegrals have no meaning under the definitions so far laid down (see §§ 98, 102) ; we proceed to show how they may arise, and to find under what conditions a meaning can be assigned to them. Examples : (a) The area under the curve y = —^ from a; = 1 to a: = 6 is evidently •^1 X^ X x'- When h becomes infinite, the area approaches the limit 1. This limit we define as the area " bounded " by the curve, the a:-axis, and the line a: = 1, although it is not properly a bounded area in the usual sense of the term. Symboli- cally we write r dx^_X *^\ x^ x_ 175 = 1. 176 CALCULUS The first thought might be that the area in the figure would increase indefinitely as the right-hand boundary recedes. Our result shows that this is not the case — the area is always less than 1. . (5) The area under the curve x'lp' = 1 from x= a (a> O') to a; = 1 is A = r^ = 2Vx ^" ^x = 2-2V^. When a approaches 0, the initial ordi- nate becomes infinite ; the area ap- proaches the limit 2. This limit we define as the area in the first quadrant " bounded " by the curve, the axes, and the line x== 1. For brevity, we write merely X = 2 Fig. 74 '0 ^x but it must be borne in mind that the geometric interpretation is quite different from that of the ordinary integral. (a+, fQc)dx=\\m I f{x)dx^ provided the limits exist. 3. If /(a:) becomes infinite as a;->c, where a < c < b, Jrb re' . rb I f(x)dx= lim I f\x^dx-{- lim I /{x^dx^ provided the limits exist. 120. Geometric interpretation. It is obvious that an in- tegral with an infinite limit may be interpreted in general as the area under a curve which is asymptotic to the 2:-axis ; an integral whose integrand becomes infinite may usually be thought of as the area between a curve and a vertical asymptote. Of course, as in example (c) of § 119, these integrals may not have any meaning in a given case. EXERCISES Evaluate the following integrals. 1- («) C~'^ Q^) T— 5 (0 Tcosxr/x; (^0 C e^dx. Ans. (a) i; (Jj) meaningless; (c) meaningless; (d) ^. 2. \ — ' Ans. TT. 3. i ^- Ans. Meaningless. ^-Wl-x^ ^0 X 4. r — Ans. 2. 5. C" ^^ • Ans. 2\/2^. ^ ^ Vx ^^ V2a-t N 178 CALCULUS 6. Trace the curve y = — — , and find the area between the curve and its asymptote. Ans. 4:7ra^. 7. Find the volume generated by revolving the area of Ex. 6 about the asymptote. Ans. 47r%^. 8. Find the area in the fourth quadrant bounded by the curve xy^ = (x — 1)'-^ and the coordinate axes. 9. Find the area in the second quadrant under the curve y = e''. 10. Find the volume generated by revolving (a) about OX, (b) about OY, the. area in the second quadrant under the curve y = e'. Ans. (a) |; (6) 2 7r. 11. The area in example (6), § 119, revolves about the ?/-axis. Find the volume generated. 12. Find the surface generated by revolving about OX that portion of the curve y = e"^ which lies to the left of the ^-axis. Ans. 2.29 tt. 13. Trace the curve x(x — yY = a^, and find the area bounded by the curve, the ?/-axis, and the line a; = 4 a. Ans. 2 a^. 14. Find the volume generated by revolving about OY the area under the curve y = e~2=^\ Check by solving in tvv^o ways. CHAPTER XVIII CENTROIDS. MOMENTS OF INERTIA I. Centroids 121. Mass; density. The student is assumed to be familiar with the idea of mass as introduced in physics. A mass is said to be homogeneous if the masses contained in any two equal volumes are equal. In all other cases the mass is heterogeneous. In the present chapter we con- fine our attention to homogeneous masses. The density S of a homogeneous mass is the ratio of the mass Mio the volume l^that it occupies : V That is, the density is the mass per unit volume. Although every physical mass occupies a certain volume or three-dimensional portion of space, nevertheless it is frequently desirable to introduce the idea of the material particle^ or geometric point endowed with mass.* The mass-point may be imagined as the limiting form ap- proached by a body whose dimensions approach 0, while the density increases in such a way that the mass remains finite. Similarly we may think of masses of one dimension and of two dimensions — i.e. of material curves and surfaces. Such masses are represented approximately, for example, by slender wires and thin sheets of metal. In these cases we define the density as "linear density," or mass per unit length, S * This notion is fundamental in studying the motion of a rigid body. 179 180 CALCULUS and "surface density," or mass per unit area, respectively. 122. Moment of mass. The product of a mass m, con- centrated at a point P, by the distance Z of P from a given point, line, or plane, is called the moment* of m with re- spect to the point, line, or plane. Denoting this moment by (x, we have Q — ml. If a system of points P^ P^^ •••, P„, having masses Wj, jTZg, •••, m^ respectively, be referred to cartesian coordinate axes, the moments of the system with respect to the three coordinate planes are respectively n ^1 n In case the particles all lie in one of the coordinate planes, the moments with respect to coordinate planes reduce to moments with respect to coordinate axes. The idea of mass-moment may be extended to the case of a continuous mass by thinking of 'the mass as composed of an indefinitely large number of particles. A precise definition will be laid down in § 187. The actual compu- tation of such a moment is usually effected by means of definite integrals; we return to this question presently. 123. Centroid. Given any mass if, let G-y^^ G-^^ G-^y de- note the moments of the mass with respect to the coordi- nate planes. The point C whose coordinates ^, y, ^, are given by the formulas Mx = a^,. My = a,,, M-z=a,y * More precisely, the simple moment^ or moment of first order. CENTROIDS. MOMENTS OF INERTIA 181 clearly has the property that the moment of the mass with respect to each of the coordinate planes is the same as if the whole mass were concentrated at that point. It is easily shown that this property holds for moments with respect to any other plane. The proof for the gen- eral case requires the use of multiple integrals (Chapter XXIII); for a system of mass-particles the proof is as follows. Let (1) ax -\- hy -\- cz = p be the equation of any plane in the normal form ; let ^, ^j, p^, '"'> Pn be the distances of the points O, P^ P^^ •••, P^ from this plane. Now p^ = ax^ -f %i + cz^ — p, p^= ax^ + hyr, + cz^ - p, so that n n n n n t=l J=l i=l ?=1 1=1 = aMx -h hMy + c3Iz - Mp = M(^ax -{- by -\- ez — p} = Mp. That is, the moment of the system with respect to the plane (1) is the same as if the whole mass were concen- trated at O, The point O is called the center of mass^ or centroid * ; The centroid of a mass is a point such that the moment of the mass ivith respect to any plane is the same as if the ivhole mass ivere concentrated at that point. By § 122, the coordinates of the centroid of a system of particles are given by the formulas n 71 . n i=\ i=l i=l * The centroid coincides with the center of gravity^ and is frequently so-called ; but the term centroid is in some respects preferable. 182 CALCULUS In the actual determination of centroids, the following considerations are often useful (the first two apply only to homogeneous masses): (a) If the body has a geometrical center, that point is the centroid. (5) Any plane or line of symmetry naust contain the centroid. (c) If the body consists of several portions for each of which the centroid can be found, each portion may be imagined concentrated at its centroid: the problem thus reduces to the consideration of a set of particles. 124. Centroids of geometrical figures. It is clear that, for a homogeneous body of given size and shape, both the mass and its moment with respect to any plane are pro- portional to the density h. Hence, in the formulas for ^, y^ i, 8 cancels out from both members, leaving the co- ordinates of the centroid independent of the density. We may therefore without loss of generality take S=l, and are thus led to speak of centroids of geometrical figures — volumes, areas, and lines — without reference to the idea of mass. EXERCISES 1. Find the centroid of the following plane systems of particles : (a) Equal particles at (0, 0), (4, 2), (3, - 5), (- 2, - 3). (h) A mass of 2 units at (0, 1), one of 3 units at (3, — 3), one of 6 units at (4, 1). 2. Four particles of mass 2, 4, 6, 8 units are placed at the points (0, 0, 0), (0, 2, 2), (4, 1, 5), (- 3, 2, - 1) respectively. Find the centroid. 3. Show that the centroid of two particles divides the line joining them into segments inversely proportional to the masses. 4. Show that the centroid of three equal particles lies at the in- tersection of the medians of the triangle having the three points as vertices. 5. Equal particles are placed at five of the six vertices of a regular hexagon. Find the centroid. CENTROIDS. MOMENTS OF INERTIA 183 6. Particles of mass 1, 2, •••, 8 units are placed at the successive vertices of a regular octagon. Find the centroid. 7. Find the centroid of the cross section of an angle-iron, the sides being 5 in. and 8 in., and the thickness of each flange 1 in. Ans. (-V, f). 8. Find the centroid of the T-iron section (a) of Fig. 75, (b) of Fig. 76. I 9. Find the centroid of a wire frame in the shape of the perimeter (a) of Fig. 75, (ft) of Fig. 76. \' <-2'-U 1" n" ^ / — I ^3^ _^ <-3'^ <-2^ 4" — 6'' Fig. 75 -8^ Fig. 76 10. From a circular disk a round hole is punched out, the two circles being tangent internally. Find the centroid of the remaining figure. 11. From a circular plate of radius 4 in. a hole 2 in. square is cut out, one corner of the square being at the center of the plate. Find the centroid of the remainder. 12. Find the centroid of a cylindrical basin of radius 4 in. and depth 3 in., if the bottom is twice as thick as the sides. 13. A monument is composed of a block of stone of base 4 ft. by 3 ft. and height 2 ft. 6 in., surmounted by a cube of side 2 ft., this in turn supporting a sphere of radius 1 ft. Find the centroid of the whole figure. 125. Determination of centroids by integration. To find the centroid of a continuous mass, we must in gen- eral resort to integration. In the most general case mul- tiple integrals (see Chapter XXIII) must be used, but in most cases of practical importance the result may be obtained by a single integration. In the following discussion we restrict ourselves to one-, two-, or three-dimensional bodies of the forms considered 184 CALCULUS in Chapter JXV.* Let us choose, as in that chapter, a suit- able geometrical element (of volume, area, or length), and denote the mass contained in this element by Am^. Let x^^ yi, Zi be the coordinates of the centroid-f of Aw,. Then the n sum ^XiArrii represents approximately the moment of the 1=1 mass with respect to the ^2-plane (or the «/-axis, in the case of a plane mass in the a^^-plane), and the limit of this sum as n becomes infinitJ^ is exactly the moment in question. In this way we obtain the following formulas for the coordinates of the centroid : M'x = lim Va^Awf, n M'^ = lim ^i/Anii, n Mz = lim ^zAnii. Now upon recalling the meaning of Aw^, we see that in any given case the above limits may be expressed as definite integrals by the fundamental theorem of § 104. The result may be written in the following form : (1) Mx= ( X dm, My= iydm, Mz= (zdm, where x, «/, z are the coordinates J of the centroid of the mass-element. In any problem each integrand must be expressed in terms of a single variable, and limits are to be assigned in such a way as to extend the integration over the whole mass. In the next few articles we explain more in detail tlie application of the above formulas. * Nevertheless the formulas obtained are applicable, with proper interpretation, to all masses, with no restriction whatever. t It follows from the theorem of § 109 that in the expressions for these coordinates, all infinitesimals may be neglected. I Apart from infinitesimals. » CENTROIDS. MOMENTS OF INERTIA 185 126. Centroids of plane areas.* To find the centroid of a plane area (thin sheet or plate) we choose an element of area as in § 105 (or § 106, if polar coor- dinates are used), and find the centroid . from the formulas Ax = Cx dA, Ay = i y dA, *' where x and y are the coordinates of the centroid of the element. Uxample : Find the centroid of the area in the first quadrant bounded by the parabola y^= -iax and its latus rectum. It With the element of area chosen as in Fig. 61, we have ^ = I y dx = ^ a^ ; Fig. 61 Ax— I xydx-=^^a\ . Jo Jo Ay == j U ' ydx — 2ai x^ dx= i a^ dx a^. Hence ^ = f«' «/ = !«• We may also find y very easily by taking the element parallel to OX. Thus -^y — I y(.^ ~ ^^dy = 2 a j (a — x^dx — aP^ etc. Jo Jo EXERCISES Find the centroid of the following areas. In each case, draw a figure and estimate the coordinates of the centroid, thus obtaining a rough check on the result. 1. An isosceles triangle. 2. A semicircular area. Evaluate the integral in two ways. * The problem of this article is of particular importance in the theory of the flexure or bending of beams. 186 CALCULUS 3. One quadrant of an ellipse, using (a) the equation — + ^ = 1 (6) the equations x = a cos cf>, y = b sin . 11. The area bounded by the parabola 3/^ = 4 ax, the ?/-axis, and the line y = 2a, with respect to each coordinate axis. 12. The area in Fig. 75, p. 183, (a) with respect to the base, (h) with respect to the line of symmetry. 13. The area in Fig. 76, with respect to the base. 14. A sphere with respect to a diameter. Ans. f Ma^. 15. A cylinder of revolution, with respect to its axis. Ans. | Ma^. 16. A right circular cone with respect to its axis. Ajis. ^q Ma^. 17. A paraboloid of revolution bounded by the right section through the focus, with respect to the axis. 18. An ellipsoid generated by revolving an ellipse about its major axis, with respect to the axis of revolution. Use («) the cartesian equation of the ellipse, (h) the equations x = a cos <^, y = h sin ^ where pi is the distance of nii from the plane through I perpendicular to the plane of I and I. Hence Fig. 77 II II n n ^ rriidi- = ^ viid^^ + d^^mi — 2d^ miPi. i—i i=\ • i—i i=i n But the quantity ^ ^^^ipt ^^ the mass-moment of first i=i order of the system with respect to the plane through I perpendicular to the plane determined by I and I, and since this perpendicular plane contains the centroid, the moment in question is 0. Hence 1,=: !-,-{. Md\ The proof of theorem II for two parallel planes is still simpler. It is left to the student. EXERCISES Find the following moments of inertia. 1. A right circular cylinder with respect to (a) a plane through the axis ; (b) a generator ; (c) a diameter of the middle section ; (o X This fraction takes the form - when ri; = 0, so that the theorem applies : 1 . tan X 1 . sec^ x ^ lim = lim — — - = 1. a:->0 X x-^0 1 140. The indeterminate forms • go , oo — oo. Given the product of two functions /(a;) • FQx), suppose that as X approaches a one function approaches while the other increases indefinitely. The product is then said to take the indeterminate form • oo . If we write fix) . Fix) = ^, Tix) it appears that the quotient last written assumes the form or ^, and the theorem of § 139 may be applied. If, as X approaches a, each of two functions /(a;), F(x) increases indefinitely, their difference fXx')— F(x) is said to assume the indeterminate form oo— oo. Here also LAW OF THE MEAN 205 we express /(^) — ^(2:) as a fraction which takes the form - or ^ , and then apply the theorem. Example : Evaluate lim x log x. This takes the form • go. If we write it in the form — 5^, the theorem of § 139 becomes applicable : ~x 1 lim X log X = lim — ^ — = lim = lim ( — 2:) = 0. X X^ EXERCISES Evahiate the following limits, when they exist. 1. 1. . cos X lini • ^ TT IT 2^ 2 3. lim x->4 x'^ + X — 20 5. lim sin 2 x a— >0 X 7. lim a; log sin a;. 9. lim 3x^-4^ , x^^ 2^2- 3a; + 1 11. lim ^'-4:^3 13. lim ^ — arcsin ^ e^-o sin^ 6 Trace the following curves. 14. y = X log a:. 16. lOQ' X V = ■ — 2. lim — • 4. x->-i a: + 1 6. lim a:e~*. a— ^cc 8. lim log cos a: x->o a; 10. T a: — 1 lim x->x a:-'^ + 1 12. lim (sec x— tan x). ^n5. --. 6 15. ?/ = are~*. 17. y = X' log x. 18. Find the area in the fourth quadrant bounded by the curve y = log X and the coordinate axes. 19. Find the centroid of the area in the second quadrant under the curve y — e^. Obtain each coordinate in two ways. 206 CALCULUS 20. Find the moment of inertia of the area in Ex. 19, about each coordinate axis. 21. Find the area bounded by the curve y = xlog x and the x-axis. 141. General remarks on evaluation of limits. While the methods of §§ 189-140 are frequently very useful in investigating the limit of a function at a point where the function ceases to be defined, they are by no means always applicable. In the first place, the function may lose its meaning in some other way than by taking the form TT or ^ (or a form reducible to one of these), so that the theorem of § 139 cannot be brought into play, yet it may be possible to show the existence of a limit by other methods. Even when the function '^j^ ^ does take ^^""^ the form - or ^ the theorem may fail to apply because '^—^ — - approaches no limit, yet the limit of the original quotient may exist. Again, the function ^_^ ^ may take /' (x) the form i^ ov —^ and at the same time -^,, ; may ap- proach a limit, yet it may be impossible to obtain any result by the use of the theorem. Finally, there is always the possibility that a function undefined at a; = a may fail to approach any limit as x approaches a. Each of these cases is illustrated by the following sin X Examples: (a) Evaluate lim • Here the denominator increases indefinitely, while the numerator oscillates between — 1 and 1, without ap- sm X ' proaching any fixed value. Nevertheless, since is X never numerically greater than -, it clearly approaches 0. 37 LAW OF THE MEAN 207 1 a:^ cos - (b) Evaluate lim — : x-^o Sin X Since cos - lies always between — 1 and 1, the numer- ator approaches and the fraction takes the form -x • Differentiating numerator and denominator separately, z X cos - + sm - X X we obtain a new fraction Tliis fraction cos X approaches no limit, since sin - oscillates between — 1 and 1, and the theorem is therefore inapplicable. But the original fraction has a limit, which may be found directly : X'^ COS - X . X ' 1 lim — : = lim — • lim x cos - = 0, x^o ^^^ ^ -r->o sin X x-^o X X 1 since lim -^— = 1 and lim x cos - = 0. x->o sm X x->o X Ox (c) Evaluate lim — • X-^co O This fraction takes the form —- By § 139, y 2^ ,. 2- log 2 T 2^1og2 2 lim — = lim & — = lim ^ — = ••-. a>->oo 3^ a>->oo 3^ log 3 ^-».cc 3^ log^ 3 No matter how many times we differentiate, we cannot 2^ get rid of the quotient — • Yet if the function be writ- ten in the form (f)^ it is seen to approach the limit 0. (c?) Evaluate lim x-^co X In this case no limit is approached. For no matter how large x be taken, as x varies from /^tt to (n -|- l)7r 208 CALCULUS the function tan x^ and hence , ranges through all possible values from — go to +20. EXERCISES a. Evaluate the following limits, when they exist. 1- 1^"^^/ ^ris. 0. 2. lim sin <^ cot <^. Ans. -1. n M->00 2' cos X x-^-o X 5. lim ^°S<^1 + ^') a— ^00 .y 7. ,. tan X — X lim ^ a:_>0 ^* — Sni X 9. lim '^^^^ . j;_>oo cos 2 X 1 ^•,^_^ sin a; — sin a ^ns. 0. 4. lim (a:^ - x). 6. lim ' 8. lim X sin - • x->0 X tan ^ 10. Hm «_. flrtan 3 B 12. Imi x_^a X- a x->i a;3 - a-2 _ a; + 1 1 a; cos - 13. lim — : 14. lim (cot a: — cosec a:) . x-^Q sm a; a-^-o 15. lim^/?^±i. 16. lim —. Ans. 0. Trace the following curves. ^„ sin a; ^o tana; 17. y = 18. y = X " X 19. y = x^e-''. 20. y =—- X lo£r a; 21. Find the area bounded by the curve y = , the a:-axis, and the maximum ordinate. Trace the curve. Aris. |. 22. Trace the curve y = — ^ , and find the area under the curve X log X from X = 2 to a: = e. Ans. 0.367. 23. Trace the curve y = xe ^ , and find the area under the curve in the first quadrant. 24. Find the moment of inertia of the area in Ex. 23, with respect to each coordinate axis. CHAPTER XX INFINITE SERIES. TAYLOR'S THEOREM I. Series of Constant Terms 142. Series of n terms. A series of n terras is an expres- sion of the form ^1 + ^2 + ^3+ 1" ^»' where each term is formed from the preceding one by some definite law. Examples are the arithmetic series a -{-[a -h d] -\- [a -h 2 d~\ + ••• -{-[a + ^n- l)d], in which each term is formed from the preceding by the addition of a constant cZ, and the geometric series a -{- ar -\- ar^ + • • • + ar""^, in which each term is equal to r times the one before it. 143. Infinite series. When the number of terms in- creases indefinitely, a series of n terms becomes an infinite series^ denoted by the symbol aj -f «2 + ^3 + •••• The series is defined by the law of formation of successive terms, or, what amounts to the same thing, by the n-th or general term. The .general term may frequently be written down by inspection of the first few terms, as in the following Examples : (a) In the series the general term is -• n (J) In the geometric series l_l-}-l_i-i- ... 2 "^ 4 8 ' ' the general term is (— J)"~^. p " 209 210 CALCULUS (^„ = ^[2a+(^-l)(^]. Similarly the sum of any series of a finite number of terms can be found. On the other hand, an infinite series has no sum in the ordinary sense of the term, since no matter how many terms we might add up, there would always be an infinite number left over. We may, however, give a meaning to the term " sum " even in this case by laying down the fol- lowing definition : The sum of an infinite series is defined as the limits as n increases iiidefinitely., of the sum of the first n terms : provided the limit exists. Thus the " sum " of an infinite series is the limit of an ordinary sum. Example : By (1), the sum of the first n terms of the infinite geometric series a -\- ar + ar'^ + • • • + a?'"~^ + • • • is aSL = a — ar 1-r INFINITE SERIES. TAYLOR'S THEOREM 211 Hence the sum of the series, if the sum exists, is ^__ lim a—_ar^ When r is numerically less than 1, the quantity ar" ap- proaches as 72 increases, and 1 — r When r is numerically greater than 1, the quantity ar" in- creases indefinitely, and the above limit does not exist; the series has no sum. 145. Convergence and divergence. If the series has a sum aS', i.e. if S^ approaches a limit when n increases, the series is said to be convergent., or to converge to the value S; if the limit does not exist, the series is divergent. It follows from the above example that a geometric series converges to the value if |r|< 1; it diverges 1 — 7' if |r| >1. A series may diverge, as in the case of a geometric series for which r > 1, because aS'„ increases indefinitely as n increases ; or it may diverge because S^ increases and decreases alternately, or oscillates, without approaching any limit. In the latter case the series is called oscillatory. EXERCISES 1. Show that every infinite arithmetic series is divergent. 2. Find the sum of a geometric series of n terms for which r = 1, (a) directly, (6) by applying the theorem of § 189 to formula (1), § 144. 3. Show that the infinite geometric series for which r = 1 is divergent. 4. Show that the infinite geometric series for which r = — 1, viz. a — a-\-a— a-{- •••, is oscillatory. 146. Tests for convergence. In the elementary applica- tions divergent series are of no importance. Before being 212 CALCULUS » able to use a given series we must determine whether it converges or diverges. If S^ can be expressed explicitly as a function of n, as in the case of the arithmetic and geometric series, we can in general determine the con- vergence or divergence of the series directly, and find the sum if it exists ; but S^ cannot be so expressed in most cases. A necessary condition for convergence is that the gen- eral term approach as its limit : lim a^ = 0. n— >>oo For, when this condition is not satisfied, each term that is added changes S^ by an amount that does not approach 0, so that the difference between S^ and a fixed number S obviously cannot become and remain arbitrarily small. This condition, though necessary, is not sufficient; i.e. if the condition is not satisfied, the series diverges, but if it is satisfied, the series still may diverge. This is illus- trated by the harmonic series 14-14-14-14- ... which will be shown in the next article to be divergent, although lim a„ = lim - = 0. Many special tests for convergence have been devised, applicable to more or less broad classes of series. Several of the simplest are considered in the next few articles. 147. Cauchy's integral test. We will begin with an JExample: Prove that the harmonic series (1) i+i + j+i+- is divergent. Here the general term is n INFINITE SERIES. TAYLOR'S THEOREM 213 Let us draw the curve 1 y=fi^~) = X erect the ordinates at 2: = 1, 2, 3, ••., 71, and complete the circumscribed rectangles as shown in the figure. Then the areas of the rectangles are, respectively, 1 1 1 ' 2' 3' Fig. 82 1 n so that the sum of these areas is the sum of the first n terms of the series (1) : A 6 n But the sum of the rectangles is clearly greater than the area under the curve from a; = 1 to x=n'. y dx= \ — = log 92. 1 *^l X When n increases, the area log n under the curve becomes infinite, hence S^ does likewise and the series diverges. This example illustrates CaucTiyB integral test : Given a series of positive terms (2) ai + «2 + ^3+ *••» P^t an = fin). If the function f(x) is defined not only for positive integral values, but for all positive values of x, and if fix) never in- creases with X, then the series (2) converges or diverges ac- fQx) dx does or does not exist. The proof of this test is easily written out by drawing the curve y =fQc) and following the process suggested by the above example. The details are left to the student. * For the definition of this improper integral, see § 119. 214 CALCULUS EXERCISES 1. Write out the proof of Cauchy's integral test. 2. In the statement of the integral test, why is it assumed that /(x) never increases with xl Show that it would be sufficient to assume that/(x) never increases with x after some fixed point x — x^. 3. Prove that the series 1.2 2.3 3-4 4.5 is convergent. 4. Prove that the series 2P 3^ 4^ converges if 7? > 1, diverges if p ^ 1. Test the following series for convergence. 1.2 3.4 5.6 6. 1 + — i — + -*- + -^ + '•'. 1+22 1+32 1+4^ 7. i_2+3-4+ .-.. 8. 1-f +i-f+ •••• 9. 1+i + i + f +-. 2 3 4 10, 1 + -A — + _2_ + ^ + .... ^ 1 + 22 1 + 32 1 + 42 11. Test the geometric series for convergence by the integral test. 148. Comparison test. Let w J -h ^2 + ^3 + * * • be a series of positive terms to be tested. (a) If a series ^ ^ ^ a^-\- a^-\- a^-\- ... of positive terms, known to be convergent, can he found such that ^ then the series to be tested is convergent. (5) If a series 1,1,7, INFINITE SERIES. TAYLOR'S THEOREM 215 of positive terms^ known to be divergent^ can be found such that -^ 1 Un ^ bn-> then the series to be tested is divergent. To prove (a), let S^Cu) be the sum of the first n terms of the 2^-series, Sn^a} the sum of the first n terms of the a-series, and S(^a~) the sum of the a-series. Since all the terms w„ are positive, S^Cu') always increases with n. On the other hand, we have S^Cu') < S^a') < S(a'). Since S^^u) always increases, but never exceeds the fixed number >S'(a), it approaches a limit, by theorem IV of § 8, which is not greater than SQa'). The proof of (b) is left to the student. The success of the test depends on our ability to find a convergent series whose terms are greater than the corresponding terms of the series to be tested, or a diver- gent series whose terms are less than those of the series to be tested. To show that the terms of the w-series are greater than those of some convergent series, or less than those of some divergent series, proves nothing. It is clear that the convergence of a series is not affected by discarding any finite number of terms from the series. Hence the conditions of the test do not need to be satisfied from the very beginning of the series, but only after a certain pointy all the terms up to that point being neglected. Example : Test the series The series ^1.2 2.3 3.4 '22 32 42 is known to converge (Ex. 4, p. 214). Discarding the 216 CALCULUS first term of the series to be tested, we have 1 u„ = n(n -h 1) The general term of the known series is 1 a„ = n^ Since the series in question converges. EXERCISES Test the following series as to convergence or divergence. 2. 1. 1+ — + — + 2 ! 3 ! 1 4-A + A + 1,1.1, v'2 V3 V4 K 1 , 1 , 1 , 1 , 2 2-22 3-23 4-2* 1-2 34 5-6 4. 1 + 1 + U.... 3 5 7 6. 1 +1 + 1+.... 3-^ 52 149. Ratio test. There are many ''ratio tests"; the simplest is the following : Given the series to be tested for convergence, form the ratio — ^^ of a general term * to the one preceding it. U, (a) If lim (J) If lim < 1, the series converges. u. the series diverges. >1, or if u n+l Ur increases indefinitely/. («) If 1'™ }J_^00 u n+l Ur = 1, the test fails. * We may divide the (n + l)-th term by the w-th, the (71 + 10)-th by the (n + 9)-th — a7iy general term by the one before it, since the question of convergence is not affected by dropping any finite number of terms. INFINITE SERIES. TAYLOR'S THEOREM 217 This test holds for any series whatever, not merely for series of positive terras. Suppose we have case (aj : ^^^'^ u n+l = L<1. At present we shall consider only the case in which all the terms are positive, and show later ' how the proof may be completed. r-- 4- 1 — ;- Let us choose some number r be- -p^ go tween L and 1. By the definition of limit, the difference between the ratio _!^ and its limit L ultimately becomes and remains as small as we please ; therefore a number m can be found such that for all values oi n^m we have '^n-f-l ^ < r. Hence '^m+3 • • • Discarding the first m terms of our series, we see that the remaining terms are less than the corresponding terms of the series u^r -{- u^r^ + u,y -\ . But this latter series, being a geometric series with ratio r < 1, is convergent ; hence the given series is conver- gent, by the comparison test. Case (5) may be proved by showing that the general term w„ does not approach when n increases indefinitely. The details are left to the student. The test may be shown to fail in case (c) by the follow- ing example : For the series 2p 3^ 218 CALCULUS • the test ratio is T n _n 4- tT= 1+1 and 11 111 1 ■ % >'0O 1+1 L ^J = 1. But, by Ex. 4, p. 214, when jt? < 1, this series diverges ; when jo > 1, the series converges. Hence there are both convergent and divergent series for which the limit of the test ratio is 1. 150. Alternating series. A series whose terms are alternately positive and negative is called an alternating series. Such series are of frequent occurrence. Their most important properties are contained in the following theorems. Theorem I : If after a certain point the terms of an alternating series never increase numerically^ and if the limit of the n-th term is 0, the series is convergent. Theorem II : In a convergent alternating series^ the difference between the sum of the series and the sum of the first n terms is not greater numerically than the (n-\-V)-th term : l^~^n| ^ l^n+ll- To prove theorem I, let us write the series in the form Wj — 1^2 + ^3 — ^4+ rt-^ where all the u'^ are positive. When n is even, say II = 2 m, we may write S^ in the two forms (1) ^2m = (^1 - ^2) + (i^3 - W4) + • • • + (^2m-l - '^2m)'> (2) Sor^i = '^^1 — (^2 — Wg) — • • . — (U2,a~2 " '^2m-l) " %m- Equation (1) shows that 82^ always increases, since each INFINITE SERIES. TAYLOR'S THEOREM 219 of the parentheses is positive. Equation (2) shows that S2jn is always less than u^ Since ^2ff» always increases, but never exceeds the fixed number u^^ it approaches a limit S not greater than i^j, by theorem IV of § 8. Further, the sum of an odd number of terms S2m+i ^-p- proaches the same limit, since lim (S.,^+^- jS^,,,) = lim W2m+i = 0- Theorem II follows at once. For, if n is even, the dif- ference S — Sn is the alternating series and we have just shown that the sum of an alternating series is not greater than the first term. Similarly if n -^is odd. 151. Absolute convergence. A series is said to be abso- lutely coyivergent if the series formed from it by replacing all its terms by their absolute values is convergent. It can be shown that a series always converges if the series of absolute values converges. From this fact the proof in § 149 is easily completed for the case when the terms are not all positive. EXERCISES Determine whether the following series are convergent or divergent. 1. 2 2^ 2^ 3. fi'l /p3 ^4 - + - + -+ —. 2 3 4 5. 1 ! 2 ! 3! 10 102 103 7. l-i + i-i+- 9. l_l+l_i+ 92 93 94 2. l-l+i--.... 2! 3! 4. 4 V4/ V4/ 6. 1 1.3 1.3.5 3 3-6 3.6.9 8. 1_1+1_1+.... 32 ^52 72 ^ LO. l-i + i-f + .". 11. Are the series in Exs. 7-10 absolutely convergent ? 12. Carry out the proof of case (h), § 149. 220 CALCULUS y\\. Power Series 152. Power series. Up to this point we have con- sidered only series whose terms are constants. The case of greatest practical importance, however, is that in which the terms are functions of a variable. In what follows, we shall be chiefly concerned with the class known as power series. A series of the form a^-\- a-^ ■\- a^x^ -{- •••, where a; is a variable and a^, a^, a^^ ••• are constants, is called a power series. Such series are of especial impor- tance in practice. A power series may converge for all values of the variable x^ or for no values except ; but usually it will converge for all values in some finite interval, and diverge for all values outside that interval. The interval of convergence always extends equal distances on each side of the point 2; = 0. The interval of convergence can usually be determined by the ratio test. We illustrate the process by an Example : Find the interval of convergence of the series Here 1 + 0; + ^+- + 2 3 x^ n n+l X + — + n lim u n+l Ur = lim X 71 + 1 X"^ n = lim - n-^ao n-j- 1 x\ = \x\ by the theorem* of § 139. * Objection may be raised to the use of this theorem in the present instance, on the ground that n is not a continuous variable. The objec- tion, liowever, is easily disposed of. For, if we can prove by the theorem that a given function of n approaches a certain limit when n varies con- tinuously, it is certain that the same limit will be approached when n varies through positive integral values. INFINITE SERIES. TAYLOR'S THEOREM 221 (a) The series converges when \x\ < 1, i.e. — 1< x <, 1. (5) The series diverges when \x\ > 1. (c) The test fails when x= ±1. But when x = l the series is the harmonic series and therefore diverges ; when x = — 1^ the series is which converges by § 150. Hence the interval of convergence is — 1 < a; < 1. EXERCISES Find the interval of convergence of the following series. 1. 1 + X + x^ i- x^ + •••. 2. 1- 2a; -h'dx-^ - i x^ -\- -. 3. i_? + ^_^+ .... 3 9 27 4. .r — — H •••. ^n5. All values of x. 3 ! 5 ! 5. 1 + lOar + 2 . .100a:2 + 3 • 1000 x^ + •••. 6. 1 + a: + 2!a;2+ 3!x3+ .... 7. 1 + nx + ^-(iLlll) x2 4- K^-l)(>^-2) ^3 + .... 2! 3! 2 3 8. 1 +a: + — + — + •••• ^ns. All values of a:. 2 ! 3 ! 9. If a^ + Og + 03 + ••• is an absolutely convergent series and &p &2» ^jp'''^; set of numbers that remain finite as n increases: I ft„ I < M, where M is a constant, show that the series a A + «2^2 + «3^3 + ••• converges absolutely. 10. Prove that the series sin 3 X , sin 5 x sm X 1 • — . • • 32 5-^ converges absolutely for all values of x. 11. If a^ + flo + (Is + ••• is an absolutely convergent series and if Mj + u^ + M„ + ••• is a series such that ^ approaches a limit when n «n increases, show that the w-series converges absolutely. 222 CALCULUS 12. State and prove a theorem for divergent series analogous to that of Ex. IL 153. Maclaurin's series. It is shown in algebra that the quantity (1 + x')^, where m is not a positive integer, may be developed into an infinite series in powers of x by the binomial theorem : (1 -\-x)"^=l-\-mx-\ ^-— — ^x^-\ ^^ ^ -or + •••, the expansion being valid for all values of x numerically less than 1. Consider now the problem of developing any given function f(^x) in powers of x. We will assume for the present that such a development is possible, and write (1) fXx) =Cq + c^x + G^x^ 4- ••• + c,,x'' + ..., where the coefficients Cq, c-^, Cg, ••• are constants to be de^ termined. Letting 2j=0, we get/(0) = c?Q; i.e. c^ is the value of the given function at a: = 0. Differentiating (1), f\x^ = Cj H- 2 ^2^ -h 3 ^32:2 ^ ...^ and setting 2: = 0, we find /'(0)=.i. Proceeding in this way, we find /"(0) = 2.1.„ /"_(0)=8_.2.1.3, /<»>(0) = m!e,„ • • • Hence (1) takes the following form, called Maclaurins series : (2) /(^) = /(0) + /'(0):r+-M^+CT):.3+.... It must be remembered that as yet we have not proved the validity of this result ; we have merely shown that, . if a development in powers of x is possible, it must have the form (2). Evidently a necessary condition for the INFINITE SERIES. TAYLOR'S THEOREM 223 existence of Maclaurin's series is that the function and its successive derivatives be defined at x = 0. Example : Expand e^ in Maclaurin's series. Here /(^)=6^ hence /(0) = 1, f"ix)=e^ /"(0)=1, • • • • • • • • • Therefore the development is 154. Taylor's series. More generally, let it be required to develop a function f(^x') in powers of x — a^ where a is a given number. Assuming /(rr) = (?()+ c^{x — a) -\- c^(x — ay -\- c^(x — ay + •••, and setting x = a^ we find /(^)=^o- Proceeding as in § 158, we obtain finally Taylor s series: (1) /W=/(a)+/'(oo U, = lim n->-oc 7^4-l (x— ly n = la;-l|. INFINITE SERIES. TAYLOR'S THEOREM 225 Thus the series converges when |a: — 1|<1, i.e. when < 2; < 2. When 2; = 0, the series is — 1 — 2 — 3 — •"-, which is divergent (§ 1-47). When x=2, the series is 1 _ 1 4_l which is convergent (§ 150). Hence, finally, the interval of convergence is < 2: <2. EXERCISES 1. In the Maclaurin series for e* (§ 153), show that the series converges for all values of x. In each of the following, determine the interval of convergence of the series. 2. Expand sin z in powers of x. Ans. sin x =x 1- — — •••.all values of x. 3 ! 5 ! ' 3. Expand cos x about the origin. x^ X* Ans. cos X = 1 1 •••, all values of x. 2! 4! 4. Expand e^ in Taylor's series about the point x = 2. 6. By replacing a: by 1 + a: in the example of § 154, obtain the development of log (1 + x) in powers of x. Ans. log(l + a:)=a:-^'+|-^+..., -l^ ^x^ H ^^ ^ ^x^ + •••. 2! 3! 8. Expand sin x about the point a: = -• 9. Show that log x cannot be expanded in powers of x. 10. Expand arctan x about the origin. Ans. arctan x = x 1 .... 3 5 11. Show that, if P(x) is a polynomial of the n-th degree in x, F(x) = P(a) + P'{a){x - a) + ^^ {x - a)^ + ... + ^^^-li^(x - a)", 2 ! n\ whatever may be the values of a and x. Q 226 CALCULUS 155. Taylor's theorem. Let the function f(x) and its first n -{-1 derivatives be continuous in an interval includ- ing the point a; = a, and let a; = a + A be a second point of the interval. Let R^ denote the difference between f(a + K) and the sum of the first n-{-l terms of the corresponding Taylor's series (2), § 15J: ; i.e. set (1) fQa + h')=fia) + f(a-)h+l^B+- n ! For convenience, write R^ in the form (n + 1) I so that (2) /(a + A)=/(a)+/'(a)A+ - +f^-^h'^+ J^P„. Consider now the auxiliary function (x) =fia + K) -/(x-) -Qa + h- a:)/'(.r) {a + h-xY j,,,.^-^ ^^^ (a + h-xy j,^^)^^^ 2! n\ (a-{-h-xy+'^ p (^ + 1)1 This function evidently vanishes when x = a -\- h, and, by (2), it also vanishes when x = a. Further, it results from our hypotheses that (f>C^^ has a derivative ' (x) in the in- terval from x=a to x=a-\-h. Hence (i>(x) satisfies all the conditions of Rolle's theorem (§ 136) in that interval, and its derivative must vanish at some point x^ of that interval. Differentiating (x)^ we find after simplifying that n\ ■ By Rolle's theorem, f(a;i)=0, hence INFINITE SERIES. TAYLOR'S THEOREM 227 Substituting this value of P„ in (2), we get (3) /(a + A)=/(a)+/'(a)A + -^^^A2+ ... +/!::^^n or, writing x— a for A, (4) f{x)=f{a) + f{d){x- a) +^{x-af-{- ... n! ^ ^ (n+1)! -^ ^ '^' where a^j lies between a and a::. P^ormula (4), or its equivalent (3), is called Taylor s theorem tuith a remainder. The last term is called the re- mainder after n -{- 1 terms : (n H- 1) 1 For 71=0, Taylor's theorem reduces to the law of the mean (§ 137): f{x-)=fia-) + (x-a)f'i^{)- If n increases indefinitely, the right member of (4) be- comes an infinite series, the Taylor's series iov f(x). The necessary and sufficient condition that the series shall con- verge to the value /(a:), and hence that the function shall be developable in Taylor's series, is that ^im R^ = 0. Example : Prove that the function e* can be developed in powers of x for all values of x. Here a = and /(a:) = e-, f'Qx-) = 6^ ••-, /^"^^>(a:) = 6% so that the remainder in Taylor's theorem has the form n+l Rn = . . e-^^ (/l+l)I where x^ is between and x. 228 CALCULUS ^n+l By Ex. 8, p. 221, the quantity ■- — — is the general (w + 1) ! term of a series which converges for all values of x^ so that, by § 146, this quantity converges to when n becomes infinite. Hence, for all values of x, and the proof is complete. EXERCISES 1. In the Maclaurin series for sin x, prove that the remainder con- verges to for all values of x. 2. Prove that cos x can be expanded in powers of x for all values of X. 156. Approximate computation by series. We have found in the preceding article that any function whatever, provided certain conditions regarding continuity are satisfied, can be represented by Taylor's theorem as a polynomial of arbitrary degree, with a certain remaiyider Rn- It is clear that i2„ is the error committed if we re- place the function by the polynomial. This suggests a method for computing approximately the numerical value, for any given value of the argument, of functions such as the sine, cosine, logarithm, etc., whose value cannot be found directly. We have only to build up the Taylor polynomial for the function in question, and show that the error R^ is less than the allowable limit of error for the problem in hand. If now our function can be developed in Taylor's series, we know at once that its value to any desired degree of ac- curacy can be found by merely adding up a sufficient num- ber of terms at the beginning of the series. For, by § 155, the remainder, or error, R^ converges to as ti increases indefinitely, and hence, by the definition of § 14, can be made as small as we please by taking n sufficiently large. An upper limit for the error committed by stopping at INFINITE SERIES. TAYLOR'S THEOREM 229 any point may frequently be found from the general prop- erties of series. Thus in the case of an alternating aeries the error is less than the first term neglected, by theorem II of § 150. Example : Compute sin 3° correct to five decimal places. Since ^ ^ sin:r=2:-— +--••., it follows that sin3° = sin^ = — -i — + , , 60 601 6V60y 120V60y = 0.052365-0.000024+ •••. Since this is an alternating series, the error committed by stopping with any term is less than the next term. With- out computing the third term, we see that it is much too small to affect the fifth decimal place, hence we need keep only two terms : sin 3° =0.05234. To be of practical use in computation, a series should converge rapidly, as in the above example, so that a few terms are enough to give the desired degree of accuracy. In this connection the following point should be noted. Our choice of a in Taylor's theorem is governed only by the necessity of knowing at that point the value of /C^:) and its derivatives. Since the remainder Mn contains the factor (x— a)"+\ it is clear that, in general, the smaller the difference x— a^ the faster the remainder will approach 0. Hence in general, of all possible values for a, we should choose that one lying nearest to the value of x in question. Thus to compute sin 3°, we took a = ; if we had to com- pute sin 47°, we would take * a = — , i.e. we would expand •i sin x in powers of 2; ; etc. 4 * Assuming of course that we know the value of the sine and cosine only for the "principal angles '0, — , -, etc. 6 4 230 CALCULUS EXERCISES 1. Draw on the same axes, on a large scale, the curve y = sinx and the first and second " approximation curves " y = x, y = x — — ^ in 6 the interval < a; < tt. Estimate the interval within which each of the approximating polynomials is correct to one decimal place. 2. Proceed as in Ex. 1 with the curve ?/ = e^ and the successive approximation curves y = 1 -hx, y = l + x + — ,y = l + x-\ 1 — ,in the interval - 2 < a: < 2. 3. Compute (a) sin 1° to five places; (h) sin 9° to three places; (c) cos 3° to four places. 4. Find the value of e to five decimal places. Ans. e = 2.71828 •••. 5. Compute (a) sin 47°, (b) cos 31°, each to four places. 6. Find the tenth root of e to five decimal places. 7. Find the value of e^-^^ to five decimal places. 8. Show that an arc of a great circle of the earth 2| miles long recedes 1 ft. from its chord. 9. Taking the circumference of the earth as 40,000,000 meters, show that the difference between the circumference and the perimeter of a regular inscribed polygon of 1,000,000 sides is less than one fif- teenth of a millimeter. 10. Within what interval can sin 6 be replaced by 0, if accuracy to three decimal places is required ? 157. Operations with power series. Operations that can always be performed upon series of a finite number of terms, such as rearrangement of terms, multiplication of one series by another, term-by-term differentiation or integration, etc., cannot be assumed offhand to be allow- able with infinite series, and in fact it is easily shown that they are not allowable in all cases. In dealing with developments in Taylor's series, it is frequently desirable to know just when such elementary operations are permissible. We therefore state, without proof, the following theorems regarding power series. Theorem I : Addition. Two poiver series may he added together for all values of x for which both series are convergent. INFINITE SERIES. TAYLOR'S THEOREM 231 That is, if the series (a;) = (Iq + a^x + a^x^ +•••■, are both convergent, the series obtained by adding these together will converge to the value <^(a:) + '^(x) : (j)(x) H- yjr^x) = a^ -f 6q + (aj + ^i):^^ + (^2 + ^2)^ + •*•• Theorem II : Multiplication. Two power series may he multiplied together for all values of x for which both series are absolutely convergent. That is, if the series (f>(x) = ^0 + a^x H- a<^x^ + •••, -^(x) = ^0 + ^1^ + M^ + ••• are both absolutely convergent, then (f>(x) -^(x) = a^b^ H- (^i^o + «o^i)a: + («2^o + ^1^1 + <^oh)^^ + • • • • Theorem III: Division. One convergent power series may be divided by another., provided the constant term in the denominator is not 0. The result holds within a certain interval, the determination of which is beyond the scope of this discussion. Theorem IV : Substitution. If the series z = a^^-\- a^y -{■ a^y'^ + ••• converges for all values of y, aiid the series y = bQ-}-b-^x + b^x^ + ••• converges for all values of x, the seizes for y may be sub- stituted in the series for z and the result arranged in poivers of X. This result holds for all values of x. Theorem V : Differentiation. A poiver series may he differentiated term hy term for all values of x within its interval of convergence.* Theorem VI : Integration. A poiver series may he integrated term by term between any limits lying within the interval of convergence.* * The endpoints of the interval are excluded. 232 CALCULUS These theorems enable us to obtain many Taylor ex- pansions in which the evaluation of the successive deriva- tives would be very tedious, and in which the law of formation of the coefficients is so complicated that the interval of convergence could not be determined directly. Example : Expand e^^^^ in powers of a;, to x^ inclusive. By Exs. 1, 2, p. 225, we have sin^ -, , . . sin^a^ , sin^a: , sin*a; , 21 3! 4! sin ic = 2^ — — + Since both these series converge for all values of a:, the series for sin x may, by theorem IV, be substituted in the series for e^^^^: .-=i + ^,_|i + ...)+(|_,_|! + ...J ^1/ 3;8 N3 1 / a? , Y_^ + 3l("-3T+-) +r! I" -3! + •••;■*"•••• Expanding the parentheses by theorem II and collecting terms, we find By theorem IV, this series converges for all values of x. EXERCISES Expand the following functions in powers of x, and determine the interval of convergence in each case. 1. sin^a:. Ans. x^ - | a:* + 53 x^ + •••, all values. 2. cos^o:. Compare this result with that of Ex. 1. •3 5 3. e'^sin x. Ans. x +x^ -{-—- — + ••• , all values. 3 30 4. log i^il£. 6. e=«<=08*. 1 — X 6. By integrating the series =1— X+X^ — X^ + --' 1 + X between the limits Oandx, obtain the Maclaurin series for log (1 + x). INFINITE SERIES. TAYLOR'S THEORExM 233 7. Expand arcsin x in powers of x by integrating the binomial f 1 expansion of . /8. Expand by division, and integrate the resulting series ^ 1 + x"-^ term by term. Cf. Ex. 10, p. 225. 9. By differentiating the Maclaurin series for log (1 — x) (Ex. 6, p. 225), prove the formula of elementary algebra for the sum of an infinite geometric series. 10. By means of series, prove the formula sin 2 x = 2 sin x cos x. Expand each of the following in powers of a:; the interval of con- vergence is as indicated in each case. 11. -^. -l /y»*t log (\ — X) = -~ X each series holding for values of x numerically less than 1. From these vre may deduce a series that is better adapted to numerical computation than either of the above series. Subtracting the second equation from the first (by theorem I, § 157), we find 1 l-\-x ^( , x^ . x^ , for values of x numerically less than unity. Let us put 1 + ^_ "^ + 1 1 — X m or 1 x = 2m-f l' where m may have any positive value. Then (1) log (1 + m) = log m-\-2 1 + 1 2 m -hi 3(2 m + 1)3 + L_. .. 5(2^ + 1)5 This series converges rapidly, and is therefore well adapted to computation. It is easily shown that for values of m > 1 the error committed by stopping at any point is only slightly greater than the first term neglected. INFINITE SERIES. TAYLOR'S THEOREM 235 Example : Taking m = 1, we have log2 = 2g 3.33^5.35^7.3' = 2[0.3333 + 0.0123 + 0.0008 + .••] = 0.693. From this the logarithms of 4, 8, ... may be found directly. With m = 2, we find log 3 ; from this and the previous result we may obtain the logarithms of all numbers whose only prime factors are 2 and 3. In fact, it is clear that only the logarithms of prime numbers need be computed by the series. EXERCISES 1. Compute to three decimal places the natural logarithms of all integers from 3 to 10 inclusive. 2. After finding log 10, obtam logj^ 2, logjQ 3. 3. By comparison with the geometric series 1 1 Ti , 1 , 1 2n + l (2 7n + l) fl I 1 1 1 , ...1 ^«+iL (2 m + 1)2 (2 7n + 1)4 J' prove that the error committed by stopping with the n-th term of the 1 1 (2 w + 1)2 series (1) is less than times the first term neglected CHAPTER XXI FUNCTIONS OF SEVERAL VARIABLES I. Partial Differentiation 159. Functions of several variables. Up to this point we have been concerned with functions of a single argu- ment. A function may, however, depend upon several independent variables. For example, the volume of a cir- cular cylinder is a function of its radius and altitude ; the acceleration of a moving particle is a function of all the forces acting on it ; the strength of a rectangular beam is a function of its breadth and depth. If 2 is a function of two variables x and y, we write with a similar notation for functions of more than two variables. Geometrically a function of two variables may be rep- resented as the ordinate of a surface in space. Thus the equation z = ax -\- hy -\- c represents a plane ; the equation z^Q? — if' represents a hyperbolic paraboloid, etc. A thorough study of functions of several variables is beyond the scope of a first course in the calculus. In the present chapter we set forth a few of the most important definitions and theorems, confining our attention chiefly to functions of two arguments. 160. Limits ; continuity. Suppose we have given a function of two variables (1) 2 = /(^, y) 236 FUNCTIONS OF SEVERAL VARIABLES 237 representing a surface in space. When x and y approach the respective values x^^ ?/q, the function z is said to approach a limit Zq if the point (2;, y, 2) of the surface (1) approaches a definite limiting point (a^^, y^, Zq). In other words, if when x is sufficiently near Xq and y is sufficiently near y^ the difference between z and Zq becomes and re- mains numerically less than any preassigned quantity however small, then z is said to approach the limit Zq : in symbols, lim f(^x, y) = z^. A function /(a;, y) is said to be continuous at the point \\mf(x, y) = f(xQ, y^). X—^Xq Similar definitions are laid down for functions of more than two variables. In what follows, it is supposed that all functions occur- ring are continuous at all points under consideration. 161. Partial derivatives. If y be kept jixed^ the func- tion ^, , becomes a function of x alone, and its derivative may be found by the ordinary rules. This derivative is called the partial derivative of z with respect to a;, and is denoted by any one of the symbols ax dx The partial derivative with respect to y has a similar meaning. The idea of partial differentiation may be extended at once to functions of any number of variables. We have only to remember that in differentiating with respect to any one variable, all the other variables are treated as constants. 238 CALCULUS Fig, 84 162. Geometric interpretation of partial derivatives. To keep y constant, say y = y^^ in the equation means geometrically that we cut the surface by the plane y = ^q. . ■ dz . The partial derivative — is tliere- fore the slope of the curve of in- tersection of the surface and the plane, i.e. of the curve whose equa- tions are The partial derivative — may be interpreted similarly. ^ 3z dz 163. Higher derivatives. The derivatives —-, — - are * ydxdy themselves functions of x and y, and their partial deriva- tives can in turn be found. They are denoted by the following symbols : by \dxj by ox dx\dyj dxdy ""^ , . dy\dyj dy^ The process can of course be repeated to find still higher derivatives. It can be shown that the two "cross-derivatives" 5% dh dydx dxdy are identical : d^z dydx dxdy FUNCTIONS OF SEVERAL VARIABLES 239 That is, the order of differentiation is immaterial. This is true for derivatives of all orders, and for functions of any number of variables. EXERCISES •J <% Find -T- , ^— for the following functions. ox dy 1. z = x^ -\- xy — S X -{- .5. Ans. — = 2 x + y — S ', ■— = x. dx oy 2. 2 = a:^ + 3 x'^y — xy -\- 2 y — 3. 3. 2 =(x2 - 2 xyy. ^. z= ^^^~^y^ . 5. 2 = ecos^cx-y)^ 6. 2 = logVa:2 + 3^2. « . y A dz — y 7. z = arctan --^ • ^ "^s. — = — — ^ . a; 5a; a;^ + ?/^ 8. Given /(a:, y, 2) = xyz + 3 .r^?/ + z^, find/,, /j^,/,. -4 ns. /, = y(z + 6 a:). 9. It u = x^ + y^ — z^ + X + 2 y, find — ■ , — , — • dx oy oz 10. Find the slope of the curve cut from the hyperbolic paraboloid 2 = a:2 — 2^2 by the plane ?/ = 3, at the point (4, 3, — 2). Ans. 8. 11. Find tlie equations of the tangent to the parabola 2 = 3 x2 + 4 ?/2, a; = 2 at the point (2, 1, 16). Ans. x = 2,z=Sy^^. 12. If M = x^y- - 2 xy^ + 3 x^y^, show that x-:^ + y~-bu. 13. If w = (?/ - z){z - x)(^x - y), show that ^ + -^ + -^ = 0. ox oy oz (92^ fi-2z S-" 52- 14. Given 2 = a: + a;S^2 + 2 xY, find ^, -^, -^, ^ • oa;2 ayaa; aa;a_y oy^ ^•2-y Q'2-, A ns. — - = 6 a;?/2 + 4:y*; '^ = 6 a-2?/ + 16 xy^. dx^ oyox d'-~ 52- 15. Given 2 = a:2^2 _j_ 3 ^^a^^s _ ^2^^ verify that dydx dxdy d^z 5^2 d^z 16. If 2 = cos (a: — y), show^ that = = - — — • ayox^ oxoyox ox'-oy 17. If M = ei-2/-2^, verify that 52^ _ 52^ . d'^u _ d'^u dxdy dydx dydz dzdy 240 CALCULUS 18. If ^ = i log {x^^-y^), show that |^ + ^ = 0. 19. li z = x'^y, show that x—- + ^/-^ = 02. ^ dx dy 20. If M = x^+ y^+ y^, show that x^ + y^ + zp: = 2u. dx dy dz 21. lif(x, y, z) = ^ show that/^2 +/^2 +/,, = 0. Vx^ + ^/'-^ + z'^ 22. Prove that if two functions u and y are so related that du _ dv du _ dv dx dy dy dx then ' d^^d^ = o. dx^ dy'^ 164. Total differentials. When x and ?/ change by amounts Ax and A^, the function z=f(x,^} changes by an amount Az. It can be shown that Az may be expressed in the form dz dz Az = —Ax-\ Ay + eAx + riAy, dx dy where € and 7] are infinitesimals. The quantity — Ax-\ Ay is called the principal part dx dy (cf . § 49) of the infinitesimal Az. The total differential of z is defined as the principal part of Az : J dz . . dz . dz = — Ax-\ Ay. dx dy In particular, ii z = x, — =1 and — = 0, so that dx dy dx = Ax. Similarly dy = Ay. Hence we may write (1) dz = ^dx + ^^dy. dx dy FUNCTIONS OF SEVERAL VARIABLES 241 For functions of more than two arguments a similar formula holds. Thus, if (2) du = ^—dx -\ dy -\ dz. ox dy dz If x^ y, z are functions of a fourth variable f, then u be- comes a function of t alone, and its differential has been defined in § 50. It can be shown that the value of du as given by (2) agrees with the earlier definition, so that (2) still holds even when x, y, z are functions of a single variable. Example: Find approximately the increase in the area of a rectangle if each of its dimensions increases by a small amount. We have A = ah^ hence dA = hda -|- adh. The actual increase in the area is ^A=(a-\- da)(h-\-dh)—ah = hda -f adh + dadh. If da and dh are so small that their product can be neglected in comparison with the other terms occurring, the total differential dA represents the actual change AA with sufficient accuracy. 165. Differentiation of implicit functions. Let y be de- fined as a function of x by the equation K^, 2/) = 0. Let us for an instant put then by (1), § 164, dz = ^-dx -[-^- dy, dx dy da bda (5 a Si- I db df dy dx dx df 9y 242 CALCULUS But in the present instance 2 = 0, hence df f dz = -^ dx -\- ^^ d^ = 0, dx dy or (1) -£=-"^ (1^0). The value of — ^ as g^iven by this formula is of course dx ^ -^ identical with that given by the method of § 25. Again, let z be defined implicitly as a function of the two independent variables x and y by the equation F(:x,y,z)=0. Put u = F(x, y, z) ; then , dF. , dF. , dF. du = — -dx+ -—-ay -\ dz, dx dy dz But since t^ = 0, du= likewise, and — dx-\ dy -\ dz — 0. dx dy dz Further, since s is a function of x and y, we may write dz dz dz = — dx -\ dy. dx dy Eliminating dz between these two equations, we find dx dz dx] \dy dz dy) To find — , keep y fixed, so that dy = 0. Then dx dF dFdz^^Q dx dz dx FUNCTIONS OF SEVERAL VARIABLES 243 or dF dz dx dx dF dz dF dz dy dy dF dz Similarly (3) £=-.^ {~*0 EXERCISES Find the total differential of each of the following functions. 1. z = x^ - 'S xy + y- + 2 y. 2. z = cos^ (x - y). 3. u = X -\- y + z. ^. u = log tan ^' X 5. Let V be the volume, S the total surface, of a right circular cylinder. If r and h change by an amount Ar and A^ respectively, find d V, A V, dS, A5. Draw a figure. 6. In Ex. 5, if r = 5 ft., A = 10 ft., Ar = AA = 2 in., compute the percentage of error made by using dV in place of AF and dS in place of A5. 7. The dimensions x, y, z oi a rectangular parallelepiped change by amounts Aa:, Ay, Az. Find dV, AV. Also obtain dV and AF directly by inspection of a figure. Find -^ in the following cases, using formula (1) of § 165. 8. 3x-4ty + 2xy = \. 9. (2x2- 3?/2)2+ 1 _ a;^^o. 10. Arctan ^ = x. X 11. Given x^ + y^ -\- z^ = 1, find ^, ^ by formulas (2) and (3) dx dy ■ of § 165. 12. Find the equations of the tangent to the circle x'2-i-y^ + z^=d6, y = 4'at the point (2, 4, 4). 13. Find the equations of the tangent to the ellipse ar^ + 3 3/2 = z^, s = 4 at the point (2, 2, 4). 244 CALCULUS II. Applications to Solid Analytic Geometry 166. Tangent plane to a surface. It can be shown that all the lines tangent to a surface z = f{x, y) at a point P : (a^Q, ^q, Zq) lie in a plane,* the tangent plane to the surface at that point. This plane is of course determined by any two of the tangent lines. We have already learned (§ 162) how to find the equations of the tangent lines lying in the planes x = Xq^ y = y^. Let us assume the equation of the tangent plane in the form 2 - 2^0 = %(^' - ^o) + ^2(^ - ^o)' where m^ and m^ are to be determined. Now the line of intersection of this plane with the plane y — y^ has the slope my But this line is the tangent lying in the plane bz y = ^Q, and, by § 162, its slope is the value of ^^ at P, Bz which value we shall denote by the symbol ^ dx dxjp dz Hence Similarly we find ^z-^ ^y\p Thus the equation of the plane tangent to the surface dz (1) ^~^o = ^ (^-^o) + i5 p iy-yo)' dyl More generally, let the equation of the surface be given in the implicit form (2) FCx,y,z) = 0, * Provided s, — , and ^ are continuous at P. dx dy FUNCTIONS OF SEVERAL VARIABLES 245 where the partial derivatives^ — , — , ^— do not all vanish dx dy dz at P : (xq, t/q, Zq). Suppose for definiteness that -— ^0. dz _Jp We may imagine equation (2) solved for z, and may then write the equation of the tangent plane by formula (1). But, by (2) and (3) of § 165, dF dF dz___^ i£^_^. ex~ dF' By dF dz dz Substituting these values in (1), we find BF; dx 'dz dF dy _ jp (x — Xq) Q-p ~dz iy - Vo)^ or ^ ^ fix p dy_ (j^-j/„)+g"' p dz (z - z.) = 0. 167. Normal line to a surface. The normal to a surface at a point P is the line through P perpendicular to the tangent plane. It will be recalled from solid analytic geometry that the direction cosines of any line perpendicular to the plane Ax + By + Cz + D=0 are proportional to the coefficients A^ B, C. Hence, since the normal is perpendicular to the tangent plane (3) of § 166, we have at once the following Theorem: The direction cosines of the normal to the sur- face F(x,y.z)=0 at any point are proportional to the values of — , — , — at ,7 , . ^ dx dy dz that point. ^ This theorem is fundamental in the geometry of surfaces. 246 CALCULUS By analytic geometry, the equations of a line through % ^0' ^0 with direction cosines proportional to a, 5, c are ^ ~ ^0 _ ^ ~ l/o _ ^ ~ ^0 . From this the equations of the normal at any point may be written down at once. 168. Angle between two surfaces ; between a line and a surface. The angle between two surfaces at a point of in- tersection is defined as the angle between the tangent planes at that point, and this in turn is equal to the angle be- tween the normals. This angle may be found by the theorem of analytic geometry that, if two lines have direc- tion cosines Zp m^, n^ and l^, m^, n^ respectively, the angle between them is given by the formula cos 6 = l^l^ + m^rric^ + n^n^- The angle at which a line pierces a surface is defined as the angle between the line and the tangent plane at the piercing-point. This is evidently the complement of the angle between the line and the normal. Example : Find the angle between the cylinder ?/2 = 4 a; and the ellipsoid 2x^ -\- g^ + z^ = 1 ?Lt the point (1, 2, 1). For the ellipsoid, the partial derivatives are dF . dF ^ dF ,-, — =4 a;, — = ^J/^ — = 2 z, dx By dz hence the direction cosines of the normal at (1, 2, 1) are proportional to 4, 4, 2, and their actual values are J, |, J. For the cylinder, dX dg dz hence the direction cosines of the normal are n, — -<> 0. Therefore "^2 V2 cos(9 = - — .? + — .?=0: V2 3 V2 B the surfaces intersect at right angles. FUNCTIONS OF SEVERAL VARIABLES 247 EXERCISES Find the equations of the tangent plane and normal line to each of the following surfaces at the point indicated. 1. The cone x- + 'd if = z^ dit (2, 2, 4) ; draw the figure. Ans. x+^y -2z = 0; -^-^ = ^Ln^ ^ ^jui . ^ '13-2 2. The paraboloid z — x^ — ?/'^ at (1, 1, 0). 3. The cylinder y^ = 4: ax at (a, 2 a, «) ; draw the figure. 4. The paraboloid x = yz a,t the origin. 5. The sphere x^ -\- y^ -{- z^ — a^ at {xq, y^, z^). Ans. x^x + y^y + zqz = a^. 6. The surface — ±f-±-=lat (a^o, yo, Zq). 7. Find the equations of the tangent to the circle x'^ -\- y'^ -\- z^ = ^, x + y-\-z=^^t the point (1, 2, 2) ; draw the figure. 8. Find the angle between the sphere x^ + y^ + s;^ = 14 and the ellipsoid 3 a;^ + 2 ^2 ^ s^ ^ 20 at the point (-1, - 2, 3). Ans. 23° 33'. 9. Show that at any point on the 2-axis there are two tangent planes to the surface a^y'^ = x^(b'^ — z^) . 10. Show that the sum of the squares of the intercepts on the axes 2 2 2 2 made by a tangent plane to the surface x^ + y'^ -{■ z'^ = a^ is constant. Sketch this surface. 11. Prove that the tetrahedron formed by the coordinate planes and a tangent plane to the surface xyz = a^ is of constant volume. 12. Find the angle at which the normal to the hyperboloid y^ — x^ -{- 4: z^ = IQ at the point (2, 2, 2) intersects the a:^-plane. Draw the figure. 13. Find the equations of the projections on the coordinate planes of the normal to the cylinder x = y + z~ Sit (2, 1, 1). 14. Find the equations of the normal to the surface x-y -\- y^-\- z^ = S at the point (1, 1, 1). 15. Show that the sphere x- + y^ + z"^ = 2 a'^ and the hyperbolic cylinder xy = a^ are tangent to each other at the point (a, a, 0). 16. Determine a and b so that the ellipsoid x-+2y^ + z^ = 7 and the paraboloid z = ax^ + hy^ may intersect at right angles at (1, 1, 2). Ans. a =: 3, & = — 1. 17. Find the angle between the normal to the oblate spheroid 3.2 _|_ ^2 ^ 2 5;2 — 10 at (2, 2, 1) and the line joining the origin to that point. Ans, Arccos | V3. 248 CALCULUS 18. In Ex. 17, find the shortest distance from the origin to the normal in question. Ans. ^V6. 19. Find the angle at which the line - = ^ = - pierces the ellipsoid 2 a;2 + i ?/2 + s2 ^ 2.5. Ans. 67° 48'. 20. Prove that every line through the center of a sphere intersects the sphere at right angles. 169. Space curves. Two surfaces (1) ^(x, y, z) = 0, ^{x, y, 2) = intersect in general in a curve in space. The curve is de- termined by the equations of the two surfaces considered as simultaneous. Since there are an infinite number of surfaces through a given curve, and since the equations of any two of these surfaces in general determine the curve, it follows that the equations of the curve may be given in an infinite number of ways. A particularly simple way is to give the equa- tions of two of the " projecting cylinders " — i.e. the cylin- ders through the curve with generators perpendicular to the coordinate planes. Eliminating y and z in turn be- tween the equations (1), we find two equations of the form (2) ct>(ix,z}=0,ylr(x,y} = 0. These equations represent cylinders through the curve (1) with generators perpendicular to the a:2-plane and the 2;^-plane respectively. We have seen that the coordinates of a point on a plane curve are frequently given in terms of a parameter t. The same device is often employed with curves in space : the curve is given by the three parametric equations (3) X =/(«), 2/ = KO.^ = KO- By eliminating the parameter between two different pairs of these equations, we obtain equations of the form (2). 170. Tangent line and normal plane to a space curve. The tangent to the curve (1) of § 169 at the point P : (a:Q, «/q, Zq) is the intersection of the tangent planes to FUNCTIONS OF SEVERAL VARIABLES 249 the two surfaces Hence its equations can be written down at once. The normal plane is the plane through P perpendicular to the tangent. To find its equation, we have only to transform the equations of the tangent line to the " symmetric form " a h c after which the equation of the normal plane can be written directly. Example : Find the equations of the tangent line and the normal plane to the curve ^2 _^ ^2 + 2;2 = 3, at the point (1, 1, 1). The equations of the tangent are found by formula (3), § 166, to be a; + ?/ + 2 = 3, 32; + ^ — 2^=2. To put these equations in the symmetric form, let us eliminate y and z in turn, thus representing the line by two of its projecting planes : 22:-32 = -l, 5a^ + 3i/ = 8. Equating the values of x from these two equations, we find _ -3j-h8_3^-l or 3~' -5 ~ 2 ' * The equation of the normal plane is therefore ^ 3(:r-l)-5(y-l)4-2(2-l)=0. 250 CALCULUS 171. Direction cosines of the tangent. Let us draw a secant (Fig. 86) through P : (xq, y^, Zq) and a second point P' : (^Xq + Ax, 7/q + A^, Zq + A^;), and denote by s the length of the arc from a fixed point to P, by As the length of the arc PP' . The direction cosines of the secant are , Ax ot ^y I Az cos a' = , cos p = ^ , cos 7' = . pp! ppi ppi those of the tangent are T Aa; T Aa; As c?:c cosa= lim = iim ■ = — , pp'^qPP' pp-^o^s pp> ds _dy ds dz ds (1) cos/3 = cos 7 = Hence the direction cosines of the tangent are propor- tional to dx, dy, dz. From this fact we obtain at once the equations of the tangent to the curve (3) of § 169. They are ^~ ^o -_ y ~ Vq _ ^ ~ ^0 /'Co) ^''(^o) ^'(. Ans. (ax)^ + (byy = (a^ - &2)f. 2 2 1 4. Find the evolute of the hypocycloid x'^ + y^ = a^, the equation of whose normal is y cos a — X sin a = a cos 2 «. 2. 2. 2. Ans. (x + y)^ + {x — y)^ = 2 a^. CHAPTER XXIII MULTIPLE INTEGRALS 177. Volume under a surface. Let us try to find the volume V bounded by a portion T of the surface the area S into which T projects in the 2:?/-plane, and the cylindrical surface through the boundaries of S and T. Z Fig. 90 We can get an approximate expression for the required volume as follows. Draw in S a set of n lines parallel to 258 MULTIPLE INTEGRALS .259 the «/-axis and a set of m lines parallel to the a;~axis, as in Fig. 90, thus dividing S into rectangles of area A«/ A2;, together with a number of irregular portions around the boundary. By passing through each line of the two sets a plane perpendicular to the a;?/-plane we divide V into vertical rectangular columns, together with smaller ir- regular columns. The upper boundary of each rectangu- lar column is a portion of the surface T. Through that point of the upper boundary of each column which is nearest the :ri/-plane, pass a horizontal plane, thus form- ing a set of rectangular prisms lying wholly within V. The sum of the volumes of these prisms is evidently an approximation to the required volume, the error com- mitted being the sum of the irregular columns around the outside, together with the portions lying above the upper bases of the rectangular prisms. That is, approxi- mately^ where /(a;<, ^,) is the altitude of the prism. It is obvious that the error in this approximation may be made arbitrarily small by taking both b^x and A?/ suf- ficiently small. Hence the required volume is exactly n m (1) r= lim V VKx,, y^l^y Ax. The "double limit" (1) may be evaluated by two successive applications of the fundamental theorem of § 104, as follows. Let us fix our attention on the rectangle P-Pl'Ql'QI in S (Fig. 90). The volume AF/ whose base is this rec- tangle may be found approximately by adding the volumes of all the included elementary prisms. Hence, by the fundamental theorem of § 104, AF"/ is given exactly 260. CALCULUS by the formula m AF/ = lim ^f(Xi, y>}^yi^x Xi and ^x remaining constant as we pass to the limit. Now if we add all the volumes of this type, we have approximately the required volume. It is to be noticed that in the expression for A Vl the coefficient of ^x is a function of Xi alone, since the limits yl and yl' are func- tions of Xi alone. Thus we may apply again the theorem of § 104, and find that the required volume under the sur- face z =f(x, y) is dx^ F= lim 5 P f{Xi, y)dy Ax = f \ P f(x, y)dy where a and h are the extreme values of x on the bound- ary of S. The quantity just found is usually written without the brackets, thus : (2) V = j I f(ix,y)dydx. It is called a double integral^ or more properly an iterated integral^ being merely an integral of an integral. It is to be noted that the inner integral sign belongs with the inner differential, and that during the integration with respect to ?/, x remains constant. Further, the first or inner limits of integration are in general variables, but the outer limits are always constants. Of course we might integrate first with respect to x^ then with respect to y. The same argument as before would lead to the formula (3) V= rr"Ax,y)dxdy, y remaining constant during the first integration. MULTIPLE INTEGRALS 261 In the foregoing argument, we have assumed our vol- ume to be divided into rectangular columns perpendicular to the a^?/-plane. Frequently, however, it is more conven- ient to erect columns perpendicular to one of the other coordinate planes (cf. example (5) below). Such varia- tions offer no difficulty provided the geometric meaning of the successive integrations be kept clearly in mind. In every problem, a sketch of the required volume should be made, and the required double integral built up by inspec- tion of the figure. Examples: («) Find the volume in the first octant bounded by the plane z = x-\- y and the cylinder y = 1 — x'^. Integrating in the order ?/, a;, we have •l-J-2 = i i (x + y)dydx =Xi^^+% r 1— X2 dx -£{-^-'-^^> — 31 ~ 60* Fig. 91 (5) Find the volume common to the circular cylinder ?/2 _|_ ^2 — ay and the sphere x"^ -{- y^ -\- z^ = a^. Let us divide the volume into columns perpendicular to the ^z-plane : X dzdy Va*^ — y'^ -~ z^ dz dy^ etc. 262 CALCULUS 178. Volume under a surface : second method. The re- sult of § 177 may be obtained by a somewhat different method. The area of the section by a particular one of the planes parallel to the i/z-plaue is evidently Hence, by § 110, the volume is A(x)dx= I I f(x^y)dydx, . a %/a*/y' The actual work of obtaining the volume in any particular case is therefore the same by the two methods — the only difference is in the geometric interpretation of the succes- sive steps. The great advantage of the method of § 177 is that it lends itself readily to the discussion of a great variety of other problems besides the computing of vol- umes, as we shall see in the next few articles. Of course when ^(rr) is known to start with, the volume may be found by a single integration as in § 110. This is the case in several of the exercises below. EXERCISES In each of the following exercises, the limits of integration should be obtained directly from a figure. 1. Find the volume in the first octant bounded by the planes x = l, z = x-\-y and the cylinder y'^ = x. Ans. y . 2. Find the volume in the first octant bounded by the cylinder a:^ + ?/2 = a^ and the plane z = x + y. Ans. f a^. 3. Find the volume of a cylindrical column standing on the area common to the two parabolas x = y'^, y = x'^ as base and cut off by the surface 2 = 1 + ?/ — x^. Check the result by integrating in two ways : first in the order y, x ; next in the order x, y. 4. Find the volume in the first octant bounded by the plane y + z = 1 and the surface x = i — z — y^. Check as in Ex. 3. 5. Find the volume cut off from the paraboloid y = 1 ~ y ~ 77 by the xz-^lane. MULTIPLE INTEGRALS 263 6. Find the volume in the first octant under the surface z — xy bounded by the cylinder ij — x^ and the plane ?/ = 1. Solve in two ways, 7. Find the volume bounded by the surface z = xy, the cylinder 2/2 = ax, and tlie planes x-\-y = 2a, y — 0, z = 0. 8. Find the volume sliced off from the paraboloid az = a^ — x^ — y^ by the plane y -\- z = a. 9. Find the volume cut out of the first octant by the cylinders z — \ — x'^, X = \ — y'^. Ans. \%. 10. Find the volume of. a segment of an elliptic paraboloid bounded by a right section. 11. Find the volume bounded by the surfaces 4 y'^ -\- 4: z^ = x^, X = 4: y, X = 2 a, z = 0. Ans. sVC-Itt — 3V3)a^. 12. Find in two ways the volume in the first octant bounded by the paraboloid y = xz and the planes z = x, z= 2 — x. 13. Find the volume of a segment of a hyperboloid of two nappes bounded by a right section. 2 2^2 14. Find the volume bounded by the surface x^ -\- y^ -{■ z^ = a^. Sketch this surface. Ans. /^ tt a^. 15. Find the volume in the first octant inside the cylinder x^ + y''^ = 2 ax and outside the paraboloid x^ + y'^ = az. A71S. | tt a^. 16. Find the volume in the first octant bounded by the surface 111 a J \hj \cJ 90 17. Show that the volume of any cone or pyramid is one third the area of the base times the altitude. 18. Write out six different double integrals for the volume in the first octant bounded by the cylinders y = x-, x^ -{■ z'^ = 1. 19. Find the volume in Ex. 18 by simple integration. 179. Interpretation of the given function. Any function /(a;, z/) of two independent variables may be interpreted as the 2-coordinate of a point on a surface in space. If, then, in any problem, we can express the required quantity as -a double limit of tlie form (1), § 177, no matter ivliat may he the geometric or physical meaning of the given function f(x^ ?/), the limit may be evaluated by an 264 CALCULUS iterated integration as in § 177. Thus the result of that article is by no means confined to the determination of volumes — we shall, as was mentioned in § 178, apply it to the study of a variety of problems. 180. The double integraL In the argument of § 177 it is not necessary that the function / be expressed in terms of cartesian coordinates x and y ; further, the area S need not be divided into elements in the particular way there adopted. The essential points are, first, that we have a function / of two independent variables de- fined at all points of the region S\ second, that we divide S into n elements AaS' which are infinitesimal of the second order.^ When S is divided in this way, the double limit Fig. 92 lim V/,A^ is called the double integral of the function / over the region aS', and is denoted by the symbol j j f dS : s As noted in § 177, the integral I I ^ f(x^ V^^y dx is often called a double integral, and it is evidently equal to j \ f dS \ but it is clear that the latter integral is the * That is, such that the maximum distance between two points on the boundary of A 6^ approaches 0. MULTIPLE INTEGRALS 265 more general, since it does not tie us down to a particular coordinate system, or to a particular mode of division of S. The integrals (2) and (3) of § 177 are merely two special forms of the double integral. 181. The double integral in polar coordinates. Given a function /(>, 6) of the polar coordinates r, ^, the double integral I ( / dS may be evaluated as follows. Divide jS s into elements by a set of circles with center at the origin and a set of lines radiating from the origin, as in Fig. 93. Then AS is the difference between two circular sectors of angle A^ and radius r and r + Ar respec- tively ; i.e. AS = l(r + Ar)2A6' - ~i- r^AO = (rAr 4- J aP)AO. We may now repeat the argu- ment of § 177, integrating first with respect to r, and noting that, by § 109, the infinitesimal of higher order ^aPaO may be neglected. This leads to the result fffdS=lim^^f(r, e)rArAe =f^X^'fir, Q}rdrdB. s EXERCISES Fig. 93 1. A round hole is bored through the center of a sphere. Find the volume cut out, using polar coordinates. 2. A cyhnder is erected on the circle ?- = a cos ^ as a base. Find the volume of the cylinder inside a sphere of radius a with center at the origin. 3. Find the volume above the a;y-plane common to the paraboloid 2 = 4 — x^ — r/2 and the cylinder x^ ■}- ij- = 1, using polar coordinates. 266 CALCULUS 4. A square hole of side 2 whose axis is the 2-axis is cut through the paraboloid of Ex. 8. Find the volume cut out. 5. Find the volume of a spherical wedge by double integTation. 6. Prove that when a curve r =f(0) revolves about the initial line, the volume of revolution generated is given by the formula F = 2 TT P f ' r sin 6-rdr d6. Solve the following by the method of Ex. 6. 7. Find the volume of a sphere. 8. Find the volume generated by revolving the cardioid r =a(l — sin ^) about its line of symmetry, Ans. |7ra^. 9,. The curve r^ = a^ sin 6 revolves about the y-axis. Find the volume generated. 10. Find the volume generated by revolving one loop of the curve r = a cos 2 $ about its line of symmetry. 11. Find the volume generated by revolving a circle about one of its tangents. 12. Find the volume cut from a sphere by a cone of half-angle ^ o with its vertex at the center of the sphere. Check by using cartesian coordinates. 13. Find the volume of the prolate spheroid generated by revolv- ing about its major axis the ellipse I r — 1 — e cos A 7S where e is the eccentricity. Ans. — 3(l_e2)2 14. Find the volume of a paraboloid of revolution bounded by a right section through the focus, taking the equation of the generating n parabola in the form r = — -• Ans. 2 7ra^. 1 — cos 6 182. Transformation of double integrals. We have seen that the integrals (2) and (3) of § 177 are merely dif- ferent forms of the double integral j I fdS. It may s happen that an integral given in the form (2) is difficult or impossible to evaluate, but that when transformed to the form (3), it becomes simple. Or sometimes after evaluating the form (2) we change to the form (3) and MULTIPLE INTEGRALS 267 evaluate again, merely as a check on the result. The process of changing the form of an integral from (2) to (3), or vice versa^ is called inverting the order of integration. Another transformation of importance is the change from one coordinate system to another — for instance, from cartesian to polar coordinates. Example : Evaluate I \ — cIt/ dx. This integral cannot be evaluated di- rectly, since the function — is not inte- y grable in terms of elementary functions. But a study of the limits shows that the field of integration is the triangle bounded by the lines 2: = 0, y — x^ y = a. Hence Fig. 94 I I —ay dx c/Q *Jx y *^o •^o y ■s Jo eydy = e° — 1. y X dy EXERCISES 1. Check the result in example (a), § 177, by inverting the order of integration. 2. Invert the order of integration in Exs. 1 and 2, p. 262. 3. Find the volume bounded by the cylinder x'^ = 4 ay and the planes x + y + z = a, z = 0, a: = 0, integrating in two different ways. 4. Express the volume of Ex. 3 as a double integral in two other ways. 5. Interpret the integral \ " i " V4 a^ — y^ dy dx as a volume, and write out five other double integrals (all in cartesian coordi- nates) for this same volume. 6. Evaluate i " ( " ^(.^ + y^dy dx, and check by inverting the order of integration. Interpret geometrically. 7. Evaluate C ^^ e'''' dy dx. Ans. 0.859. 268 CALCULUS 8. Evaluate \ \ Jo Ji dmates. 1 /'V^i-a 'O 9. Evaluate sin 1 r2 2 6^2+2/2 (ffj fix by transforming to polar coor- Ans. 1.35. TTiJ Jo J-z r ■ dy dx. , 4 Ans. —• ir 10. Express \ i/(/5 (a) in cartesian coordinates, (&) in polar coor- s dinates, where >S' is the triangle bounded by tlie lines x = a, y = 0, y = x. /• 2 /• o cos S 11. Transform 4 "^ I e''*^"<> r^/r//^ to cartesian coordinates. Jo Jo 12. Compute the value of I | cos (x^ + y'^)dS extended over the interior of the circle x^ + ?/'-= 1, • Ans. 2.644. 13. Find the area in the first quadrant under the curve y = e'2'^ by noting that '^^'dxyU^e-'-y'dyy^^^^Je-'-^^^y'-\lydx Ans. ^^ Fig. 95 14. Find the centroid of the area in Ex. 13. 183. Area of a sur- face. Let us try to find the area cr of a portion of the surface Y Suppose that the xy-^vo- jection of a is the region aS'. Let us divide S into elements AaS' in any suit- able way, and fix our attention on a particular one of these elements. This element is the hori- zontal projection of tlie MULTIPLE INTEGRALS 269 portion Act of a. If we draw the tangent plane at some point P of Act, then AaS' will be the horizontal projection of a certain area Act' on the tangent plane. Hence AaS' = Ao-' cos 7, where 7 is the angle between the 2-axis and the normal PiV^to o- at P, and Ao-' = cos 7 If now we form the sum of the quantities Act' and pass to the limit, we have ^s^o^'^cosY ^^-^ cos 7 In case it is more convenient to project the area on the xz- or the ^5;-plane, the corresponding formula is readily developed. Example : Find the area of that part of the surface z= y -\-x'^ whose projection on the :c?/-j)lane is the triangle bounded by the lines y = ()^ y — x, x = \. Writing the equation of the surface in the form 2 — ^ — 2:^ = 0, we have for the partial derivatives the values dF o ^F . OF . — = — zx^ — = — 1, — =1- dx dy dz Hence by the theorem of § 167, the direction cosines of the normal are proportional to — 2 2:, — 1 , 1, and 1 cos 7 = — • V4 2:2 4- 1 + 1 Therefore the required area is o- =f^f^ V4 x^ + 2 dy dx = f xV4: x^ H- 2 dx Jo = l.f(4a-2+2)- = ^5(6^- 20- 270 CALCULUS EXERCISES 1. Find the area cut out of the plane x-\-y + 2z = 2a by the cylinder a;2 + ?/- = rt-. Ans. ^VQ-n-a'^. 3 2. Find the area of that part of the surface z = y + ^x^ whosfe projection in the ^-^/-plane is the triangle bounded by the lines y = 0, Ans. if (2 +\/2). The center of a sphere of radius a is on the surface of a cylinder y = X, X = 2 Find the surface of the cylinder intercepted by the Ans. 4 rt^. of diameter a. sphere. 4. In Ex. 3, find the surface of the sphere intercepted by the cylinder. Ans. 2(7r — 2)a^. 5. How much of the conical surface z^ = x^ -\- y^ lies above a square of side 2 a. in the x?/-plaue whose center is the origin ? 6. How much of the surface az = xy lies within the cylinder 3.2 ^ y2 — f^2 9 (Use polar coordinates.) 7. A square hole is cut through a sphere, the axis of the hole coin- ciding with a diameter of the sphere. Find the area cut from the surface of the sphere. Ans. 16rt&arcsin 8 a^ arcsin &2 62 184. Triple integrals. z Fig. 9G We have seen that the integral of a function of one variable, extended over a given inter- val, may be inter- preted as the area under a plane curve. Again, the integral of a func- tion of two varia- bles extended over a plane region may be interpreted as the volume under a surface. If now we have a function of three variables MULTIPLE INTEGRALS 271 defined at all points of a portion of space, no similar geomet- ric interpretation for the integral of the function over the given region is possible, since geometric intuition fails in space of four dimensions. Nevertheless the meaning of such an integral may be made plain by analogy with the earlier cases. Suppose we have given a function /(a:, ?/, z) defined at all points of a three-dimensional region V. Let us pass through V three sets of planes parallel to the coordinate planes, thus dividing V into elementary rectangular parallelepipeds of volume Aa: A?/ A2, together with smaller irregular portions around the boundary. Now multiply the volume of each element by the value of the function at some point within the element, say at its center, and form the sum of these products. The triple limit Aa;->0 ^^ ^^ ^^ is defined as the value of the trijjle integral of /(a:, ?/, z) over the region V. This limit may be evaluated by three successive integrations (cf. §177) : r = lim y V y fix. y, z)Ax Ay Az Ax->0 ^^ "^ '^ I I fix,y,z)dzdydx. a %Jy ^ z' The first integration extends over a vertical column of base Ay Ax\ the limits z\ z" are the extreme values of z in this column, and are in general functions of both x and y. The integration with respect to y is extended over a slice parallel to the ?/2-plane ; the limits y' and ^" are the extreme values of y in this slice, and are functions of X alone. In the final integration the limits are of course the extreme values of x in the whole region. 272 CALCULUS More generally, the function / may be given in terms of any system of coordinates, and the region V may be divided into elements in any suitable way.* We write in general, for the value of the triple integral of / over the region F^ V It is hardly necessary to say that such transformations as inversion of order and change from one coordinate system to another are allowable and useful with triple integrals, just as with double integrals. It may be well to observe at this point that applica- tions of triple integration are comparatively rare in ele- mentary work. In the problems treated in the next two articles, triple integrals are sometimes required. The volume 1^ itself may be expressed as a triple inte- gral, the given function/ being taken equal to unity : V It is true that the volume may be found more directly by methods previously studied ; nevertheless it may be worth while to solve a few exercises by the present method for the sake of practice in determining the limits in triple integration. Example : Find the volume cut off from the paraboloid z = l — x^ — ^ by the a;j/-plane. ^ In this case * The element must of course be infinitesimal of the third order. MULTIPLE INTEGRALS 273 3*8*2 3' 12 2\/l-x2 , ax EXERCISES Finci the following volumes by triple integration, drawing a figure in each case. 1. The tetrahedron bounded by the coordinate planes and the , X y z ^ plane -■{■•} + -= I. a c 2. The volume bounded by the paraboloid x^ -\- y"^ = az, the cylin- der a:- + y- = 2 ax, and the plane 2 = 0. ' Ans. ^ na^. 3. Interpret the triple integral \ \ \ ^'~^ dz dy dx geometri- cally, and express the same volume as a triple integral in several other ways, drawing a figure for each case. 185. Heterogeneous masses. The density of a homo- geneous mass has been defined in § 121 as the ratio of the mass to the volume it occupies : V For a, heterogeneous mass, i.e. one whose density varies from point to point, we must introduce the idea of density at a point. Consider an element of volume A F^ including a point P, and let Ail[f denote the mass contained in AF". Then the ratio — - is the average density in A Fi If A I^^ approaches in such a way that P is always included, the ratio in general approaches a limit * S, called the density at the * In general this is true only if AT" is infinitesimal of the third order, as in § 184. T 274 CALCULUS point P : 5 ,. AM dM 6 = lim - — = AF->oA7 dV The mass of a heterogeneous body whose density at any point is given as a function of the coordinates of the point can be found by integration. We have only to choose a suitable mass-element and integrate over the whole body. The great point to be noted is that in general the element itself must be homogeneous^* since otherwise the mass of the element cannot be computed and hence the integral cannot be built up. In many cases it is possible to choose an element as in Chapter XV and obtain the result by a simple integra- tion ; in more complicated problems double or triple in- tegration may be necessary. We give the argument in full only for the general case where triple integrals are employed. Given a mass M occupying a volume FI divide I^into elements as in § 184, and multiply each element A J^ by the density B at one of its points. Then the sum ^^^^SAFis an arbitrarily close approximation to the mass Mii AP^be taken suffi- ciently small, and the mass is therefore given exactly by the formula r For a mass distributed over a surface S, the idea of "surface density" must be introduced: g^ lim Aiff ^ dM whence, by argument now familiar, M = ffBdS. * By this is meant that the density at different points of the element varies only by infinitesimal amounts ; cf. example («) below. By the theorem of § 109, the infinitesimal variations may be neglected. MULTIPLE INTEGRALS 275 Similarly, for a mass distributed along a curve (7, the "linear density" is g^ Urn AM^dM^ As">.o As ds ' and M =x Sds. Fig. 97 Examples : (a) Find the mass of a circular cone whose density q varies as * the distance from the axis. Let us take the vertex of the cone at the origin and its axis along OX. If we divide the mass into cylindrical shells about the axis, each element will be "homogeneous" of density B = kr = ky. We have dV = 2'7ry{h — x)di/^ M=fdM=j8dV=2 7rkf"y\h-x)dy = 2 7rk \ y^lh y \dy = -irka^h. (5) The density at any point of a cube is proportional to the sum of the distances from three adjacent faces. Find the mass of the cube. Taking the three faces mentioned as coordinate planes, and choosing the element as in § 184, we have M= -^ P P p(3^ + ^ H- z')dz dy dx. * To say that a varies as b, or a is proportional to 6, means that a = kb, where k is constant. 276 CALCULUS EXERCISES Determine the following masses. 1. A straight rod whose density is proportional to the n-th power of the distance from one end. 2. A semicircular wire whose density varies as the distance from the diameter joining the ends. Ans. 2 ka^. 3. A circular plate whose density varies (a) as the distance from the center; (b) as the distance from a fixed diameter. Ans. (b) f ka^. 4. A spherical surface whose density varies as the distance (a) from a fixed diameter ; (^>) from a diametral plane. Ans. (a) kir'^a^. 5. A sphere whose density is proportional to the distance from the center. Ans. kira^. 6. A rectangle whose density is proportional to the sum of the distances from two adjacent sides. 7. A circular plate whose density varies as the distance from a point on the circumference. 8. A square whose density is proportional to the distance from one corner. A ns. .765 ka^. 9. The tetrahedron bounded by the coordinate planes and the plane x + y + z = a, if the density is proportional to the sum of the distances from the coordinate planes. 186. Centroids and moments of inertia: the general case. We are now in position to lay down precise definitions of the moment of first order, and the moment of inertia, of any mass. Divide the mass into elements AF'as in § 184, and multiply each element by the density 8 at a point P : (^x, I/, z) of the element. Then the moment of the first order with respect to the ^z-plane is defined as lim yy:VxSAv= fffxSdv, A K->o ^^ '^ '^ *^ y^ with a similar formula for the moment with respect to any other plane. The centroid is defined as the point (i, ^, z) whose coordinates are given by the formulas Mx = fffxhdV,Mi/=fffyhdV,Mz=fffzhdV. MULTIPLE INTEGRALS 277 Similarly, the moment of inertia with respect to any axis is defined as V where r is the distance of a point of the element from the axis. While the above formulas are important from the theoretical standpoint on account of their generality, it must not be foi'gotten that in the actual computation of moments of the first order and moments of inertia multiple integrals are very rarely needed, at least for homogeneous masses. It will be remembered that the theorems of § 134 have been proved only for a set of particles. The reader will now have no difficulty in extending the proof to the general case. EXERCISES 1. Find the centroid of the volume in the first octant bounded by the paraboloid az — x^ + y'^ and the planes ij — x^ x — a. Ans. (I a, ^9^-a, I'jn). 2. Find the moment of inertia of the volume in Ex. 1, with respect to the 2:-axis. 3. Find the centroid of the volume in Ex. 3, p. 262. 4. Find in two ways the centroid of the volume in Ex. 9, p. 263. 5. Find the moment of inertia, with respect to the x-axis, of the volume in Ex. 4, p. 262. Check by inverting the order of integration. In Exs. 6-10, use polar coordinates. 6. Find the moment of inertia, with respect to the 2-axis, of the volume in Ex. 2, p. 265. 7. Find the centroid of a hemisphere (cf. Ex. 6, p. 266). 8. Find the moment of inertia of a sphere about a diameter. 9. Find the centroid of a spherical wedge of half -angle a. Check by putting ct = 7- 10. For the wire of Ex. 2, p. 276, find (a) the centroid ; also the 278 CALCULUS moment of inertia with respect to (b) the diameter joining the ends, (c) the radius perpendicular to this diameter. Ans. (a) x = \Tra; (b) ^Ma'^; (c) I Ma^. 11. Find the moment of inertia of a circular disk whose density varies as the distance from the center, (a) about the axis of the disk, (b) about a diameter. Ans. (a) ^Ma^. 12. Find the centroid of a rectangle whose density is proportional to the sum of the distances from two adjacent sides. 13. Find the moment of inertia with respect to (a) the a:^-plane, (6) the a;-axis, of the volume bounded by the planes z = x -\-y, x+y = a, and the coordinate planes. Ans. (a) \ Ma^. 14. Find the moment of inertia, with respect to the yz-plane, of the volume bounded by the planes x = 0, y = 0, z ■= a, z = x -^ y, integrat- ing in the order z., y, x\ check by integrating in the order x, y, z. 16. Find the centroid of a straight rod whose density is propor- tional to the n-th power of the distance from one end. . _ n -t- 1 , Ans. x= — ■ — I. n-h 2 16. By dividing a triangle into strips parallel to the base and con- centrating the mass of each strip at its center, show that in finding the centroid of the triangle we may replace the triangle by a straight line lying along the median and having a density proportional to the distance from the vertex. Hence find the centroid of any triangle by the result of Ex. 1.5. 17. By a method analogous to that of Ex. 16, find the centroid of any cone or pyramid. 18. Prove theorems I and IT of § 134 for the general case of any continuous mass. CHAPTER XXIV FLUID PRESSURE 187. Force. If a particle of mass m moves with an ac- celeration y, the product of the mass by the acceleration is called force : F=mj\ and the motion is said to be due to the action of the force. Since force is a mere numerical multiple of acceleration, it follows that force is a vector (§ 5(j'). If several forces act on the same particle, their combined effects are equiva- lent to that of a single force, their resultant. Usually the resultant is most easily found analytically by resolving each force into components parallel to the coordinate axes and summing in each direction to get the rectangular com- ponents of the resultant, after which the resultant is found by compounding these rectangular components (see the example below). If there is no force acting on the particle, or, what is the same thing, if the resultant of all the forces is 0, the par- ticle is said to be in equilibrium. It follows from § 59 that a particle in equilibrium is either at rest or moving uniformly in a straight line. If several forces act at various points of a body, it is not always possible to compound them into a single result- ant. In what follows, we shall consider only cases in which this is possible. Example : Find the resultant of a plane system of forces i^i=10 1bs., ^2 =20 lbs., i^3 = 8 1bs., 7^^=15 lbs. acting as in the figure, where a = arctan |. 279 280 CALCULUS Fig. 98 Hence The components parallel to OX are, of F^, 10; of F^, 20 cos a =16; of F^, 0; of F^^ — 15 sin a = — 9. "-^ Hence the a;-component Rj. of the resultant is i2^ = 10 + 16 + - 9 = 17. Similarly, i^^ = + 12 - 8 - 12 = - 8. B = VI72 + 82 = V353 = 18.8 lbs., inclined to the a^-axis at an angle arctan(-y\)=-25°12'. 188. Force distributed over an area. We have frequently to consider a force not acting at a single point, but dis- tributed over an area. Examples are the pressure of a body of water upon a dam, that of a carload of sand against the sides of the car, the attraction of an electric point-charge upon an electrified plate, etc. If the mass upon which the force acts be thought of as composed ulti- mately of particles, such a distributed force may be regarded as comprising the totality of forces acting on the separate particles. We shall consider only the case in which all these separate forces taken together are equiva- lent to a single resultant ; the resultant is the total force acting on the body. Consider a force acting in the same direction at all points of a plane surface /S', and suppose for concreteness that the force is normal to the surface. If we denote by Ai^ the total force acting on an element of area AaS' chosen A Ti^ as in § 180, then the ratio — - is called the average pres- sure on ^S. Now, if AaS' approaches in such a way that a FLUID PRESSURE 281 A IT certain point Q is always included, the ratio — - in general approaches a limit, called the pressure at the point Q: T AF dF V = hm ■ = ^ A,s'->o AS dS When the pressure at every point is given as a func- tion of the coordinates, the total force F can be found by integration. In the most general case the force ap- pears as a double integral, by § 180 : s but in most cases of practical importance the element of area can be so chosen that a single integration is sufficient. 189. Fluid pressure. An example of force acting normally to a surface is furnished by the pressure of a fluid against a retaining wall. The pressure, at any point of an incompressible fluid, due to the weight of the fluid is equal to the weight per unit volume times the depth h of the point below the surface of the fluid : p = wh. We will assume the retaining area to be plane and vertical. Let us divide this area into horizontal rectangular elements of area Z^AA as in the figure. If we denote by Pi the pressure at the depth A^, the force acting on the rectangle liAh is approximately * pJiAh = whiliAh. n Then the sum ^ whiliAh is approximately the total force, * The actual force on the rectangle evidently differs from the quantity PiliAh by an infinitesimal of higher order, which may be neglected. Surface Fig. 99 282 CALCULUS or total pressure * P, on the whole area, and the limit of this sum is exactly P. Hence, by the fundamental theorem of § 104, where limits of integration are to be assigned in such a way as to extend the integration over the whole area. Example : A trough, whose cross-section is an equilateral tri- angle of side 2 ft., is full of water. Find the total pressure on one end. Let us take the origin at the lower vertex of the triangle. Then the equation of the line OA is ^ = VS X. The total pressure on the triangle is dx P = 2wf "( V3 - ^)x d^ = 2 wJ\VS - V§ x)x^S = 6 w \ (1 — x^x dx= Q tv\ = w = 62 lbs., nearly. EXERCISES 1. A particle is acted on by two forces F^, F^ lying in the same vertical plane and inclined to the horizon at angles ctj, a^. Find their resultant in magnitude and direction, if F^ = 527 lbs., F^ = 272 lbs., «! = 127° 52', a2 = 32° 13'. Ans. 569 lbs., inclined to the horizon at 99° 26'. 2. Six forces, of 1, 2, 3, 4, 5, 6 lbs. respectively, act at the same point, making angles of 60° with each other. Find their resultant. Ans. 6 lbs., acting along the line of the 5-lb. force. * Care must be taken not to confuse the total pressure^ P, with the pressure at a point., p. The former is a force, the latter, a force per unit area. FLUID PRESSURE 283 3. Work the example of § 189 with the origin at B. 4. Find the total pressure on one side of a plank 2 x 8 ft. sub- merged vertically with its upper end (a) in the surface, (h) 4 ft. below the surface. 5. A horizontal cylindrical boiler 4 ft. in diameter is half full of water. Find the total pressure on one end. Ans. 330 lbs. 6. Work Ex. 5 if the boiler is full of water. 7. What force must be withstood by a vertical dam 100 ft. long and 20 ft. deep? 8. Work Ex. 7 if the dam is a trapezoid 100 ft. long at the top and 80 ft. long at the bottom, taking the origin at an upper corner. Check by solving again with the origin in a different position. 9. Find the total pressure on one side of a right triangle of sides AB = 3 ft., AC = 4 ft., submerged with A C vertical and (a) A in the surface, (h) A 2 ft. deep, (c) C 2 ft. deep. In each case check as in Ex. 8. 10. Find the total pressure on one face of a square 2 ft. on a side, submerged with one diagonal vertical and one corner in the surface. 11. Find the force on one end of a parabolic trough full of water, if the depth is 2 ft. and the width across the top 2 ft. Ans. ff w. 12. A trough 4 ft. deep and 6 ft. wide has semi-elliptical ends. If the trough is full of water, find the pressure on one end. 13. Find the force that must be withstood by a bulkhead closing a watermain 4 ft. in diameter, if the surface of the water in the reservoir is 40 ft. above the center of the bulkhead. Ans. 16 tons, 14. Show that the problem of § 189 is analytically equivalent to the following : To find the mass of a thin plate, if the density is propor- tional to the distance from a line in the plane of the plate. 190. Resultant of parallel forces. Suppose we have given a set of parallel forces /j, f^, •■•-,fn-, whose resultant (algebraic sum) F is not 0. The problem of finding the line of action of the resultant is analogous to that of find- ing the centroid of a set of mass particles. Let us take the a;«/-plane perpendicular to the given forces, and let (rr.-, «/,) be the point where the line of action of /< pierces this plane. The moment of the result- ant about each coordinate axis must equal the sum of the 284 CALCULUS moments of the forces about the same axis. Hence the line of action of ^pierces the a;?/-plane at the point whose' coordinates ^, 'y are given by the formulas n n Fx = ^fiXi, Fy = ^fiVi. 191. Center of pressure. More generally, consider again the case of a force acting normally at all points of a plane area. Take the given plane as rr^z-plane, and divide the surface into elements A/Sas in § 180 ; then the force on AaS* is approximately p^S^ where p is the pressure at a point (a:, y) of A/S^, and the moment of this force about the «/-axis is xpAS. The sum of these moments is approxi- mately the moment of the whole force, and the limit of the sum is exactly that moment. Similarly, we can find the moment about the 2:-axis. Hence the resultant acts at the point whose coordinates ^, ^ are given by the formulas Fx=ffxpdS, Fy^ffypdS, * s s where F is the total force. The point (x^ ^) is called the center of pressure. As usual, it happens in many problems that the double integrals reduce to simple integrals, if the element be properly chosen. In particular, in the problem of fluid pressure it is easily seen that the depth of the center of pressure below the surface is given by the formula Ph = tv^hH dh, where P is the total pressure. EXERCISES 1. A straight beam AB 50 ft. long bears loads as follows: 100 lbs. at A, 100 lbs. at C, 200 lbs. at A 50 lbs. at B; AC = 10 ft., AD = 20 ft. Find the point of application of the resultant. • 2. Work Ex. 1 if the segment AD bears a uniformly distributed load of 5 lbs. per foot. FLUID PRESSURE 285 3. Work Ex. 2 if the segment DB bears a distributed load which increases uniformly from 5 lbs. per foot at D to 15 lbs. per foot at B. 4. A platform ABCD 20 ft. square bears a single concentrated load. The reactions are, at A, 50 lbs.; at B^ 80 lbs.; at C, 100 lbs. ; at D, 70 lbs. Where is the load ? 5. Find the most advantageous length for a lever to lift a weight of 100 lbs., if the distance from the weight to the fulcrum is 4 ft. and the lever weighs 4 lbs. per foot. Find the depth of the center of pressure in the following cases. 6. A rectangle submerged vertically (a) with one edge in the surface, (h) with its upper edge at a depth c. Ans. («) f a. 7. An isosceles triangle submerged with the line of symmetry vertical and (a) the vertex, (6) the base, in the surface. Ans. (a) I A; (6) \ h. 8. Any triangle submerged with one side in the surface. 9. One end of the parabolic trough of Ex. 11, p. 283. 10. A semicircle submerged with its bounding diameter in the surface. 11. In each case of Ex. 9, p. 283, if the pressure is removed from one side of the triangle, at what point must a brace be applied in order to hold the triangle in position ? .4ns. (a) With AB, AC as axes, (|, 2). 12. Show that the problem of § 191 is analytically equivalent to that of finding the centroid of a plane mass of variable density p. CHAPTER XXV DIFFERENTIAL EQUATIONS OF THE FIRST ORDER I. General Introduction 192. Differential equations. A differential equation is an equation that involves derivatives or differentials. Various examples have arisen in our previous work, of which the following may be mentioned : (§15) (§16) (§50) (Ex. 43, p. 64) (§51) (Ex. 18, p. 50) (Ex. 1, p. 81) (Ex. 2, p. 78) (Ex. 1, p. 2.39) (10) ?1 + f^ = 0- (Ex. 22, p. 240) Equations containing partial derivatives, such as ex- amples (9) and (10), are called partial differential equa- tions. Such equations are of great importance, but a study of them is beyond the limits of this book. 286 (1) ds 1 dt t^ (2) y"=4. (3) dy = 2 cos 2 6 d6. W y{n) ^ ^n^ax^ (5) xdx + y dy = ^, (6) (7) (8) (i+y¥ = „. y (9) f = 2^ + 2/-.3. ox DIFFERENTIAL EQUATIONS OF FIRST ORDER 287 193. Order of a differential equation. The order of a differential equation is the order of the highest derivative that occurs in it. Tlius, in § 192, examples (1), (3), (5) are of the first order, (2), (6), (7), (8) are of the second order, (4) is of the n-th order. In the applications, equations of the first and second orders are of predominant importance, and we shall be chiefly concerned with these two types. 194. Solutions of a differential equation. A solution of a differential equation is any relation between x and y by virtue of which the differential equation is satisfied. Thus equation (1) of § 192 is true if 1^ where c is arbitrary ; hence this relation is a solution of the equation. A solution of (2) is easily seen to be It appears from these examples that a solution of a differential equation may involve one or more arbitrary constants ; we shall find this to be true in general. It follows that each equation has an infinity of solutions, obtained by assigning different values to the arbitrary constants. By analogy with the integral calculus, a solution of a differential equation is often called an integral of the equation, and the arbitrary constants are called constants of integration. II. Equations of the First Order 195. The general solution. Suppose there is given a relation (free of derivatives) between x, y and an arbi- trary constant : (1) Fix, y, C-) = 0. 288 CALCULUS Geometrically this equation represents a family of curves^ whose individual members are obtained by assigning par- ticular values to c. If we differentiate (1) with respect to x^ the arbitrary constant c may be eliminated from the equation thus formed and the original equation. The result of this elimination is evidently an equation involving x^ y^ and y^ ; i,e. it is a differential equation of the first order : (2) ■ *Cr^,y,j/')=0. As this equation does not contain c, it represents a prop- erty common to all the curves of the above-mentioned family. Since equation (2) is true by virtue of equation (1), it follows that (1) is a solution of (2). If a solution of a differential equation of the first order contains an arbitrary constant, it is called the general solution: hence (1) is the general solution of (2). It can be shown that, in general, corresponding to every differential equation of the form (2) there exists a gen- eral solution (1) ; methods of finding this solution in various cases will be considered presently. It may be worth while to point out that, if the differ- ential equation has the simple form ax the integral calculus gives us the general solution at once y=jf(x)dx + c. It should also be noted that while in the integral cal- culus the constant of integration always appears as an additive constant, this is not true in general in the solu- tion of a differential equation ; the constant often enters in other ways. DIFFERENTIAL EQUATIONS OF FIRST ORDER 289 Examples : (a) Find the differential equation whose general solution is Differentiating, we find di/ = 2 ce^ dx ; eliminating c by division, we get ^ = 2dx. y This example illustrates the fact that the arbitrary con- stant is not always additive. (5) Find by inspection the general solution of the equation X dy •\- y dx=(). The answer is seen at once to be xy = e. 196. Particular solutions. A solution obtained from the general solution by assigning a particular value to the arbitrary constant is called a particular solution of the differential equation. Thus in example (^), § 195, the equations xy =0, xy = 5, etc., are particular solutions. In applied problems involving differential equations we are often concerned with a particular solution. Nevertheless the determination of the general solution is usually a necessary preliminary step, after which the required particular solution is found by determining the arbitrary constant from given initial conditions. The process is illustrated by the examples of § 77, which should be reviewed at this point. Differential equations involving y' to a degree higher than the first may in some cases have a so-called singular solution., which cannot be obtained from the general solu- tion by assigning a particular value to the arbitrary con- stant. As such solutions are of little importance in most of the elementary applications, we shall omit a discussion of them. 290 CALCULUS EXERCISES In the following cases, find the differential equation whose general solution is the given equation. 1. y = x^ -{- c. 2. y — ex. Z. y = ce^. A. y = ex -\- c^. 6. log r = kO. 6. xy -\- cy = \. 7. s = sin t -f c cos t. 8. c^ + 2 ey = x^. Find by inspection the general solution of each of the following differential equations. 9. dy — sin x dx = 0. 10. x dx -\- y dy = d. 11. ^=xdx. 12. ^i^ = ^'. y y X 13. xdy + y dx -\- 2 dy = 0. 14. Find the equation of a curve M'hose slope at any point is equal* to the abscissa of the point. How many such curves are there? Draw several of them. 15. In Ex. 14, find the curve that passes through (4, — 3). 16. Solve Ex. 14, reading " ordinate " instead of " abscissa." 17. A point, starting with a velocity of 10 ft. per second, moves under a constant acceleration of 8 ft. per second per second. Find (a) the velocity, (/>) the distance from the starting point, after t seconds of motion. 18. A point moves under an acceleration dv 4 o * — = — 4 cos 2 t. dt If 17 = and X = 1 when ^ = 0, find v and x in terms of t. 197. Geometrical interpretation. In analytic geometry we find that the locus of a point whose coordinates x^ y are connected by an equation is a certain curve, the graph of the equation. In general, any value whatever may be assigned to x^ and the corre- sponding value of y determined. * That is, the number representing the slope is the same as that rep- resenting the abscissa. It is only in this sense that a ratio, such as the slope of a curve, can be equal to a length, such as the abscissa of a point. DIFFERENTIAL EQUATIONS OF FIRST ORDER 291 If now we have given a differential equation of the first order, and of the first degree in y\ i.e. a relation between a;, y^ and y' of the form (1) y' = Fix, y-), it is clear that, in general, any values whatever may be assigned to x and y provided we associate with them the value of «/' given by the equation. Thus, equation (1) is satisfied by the coordinates of any point (x^ y) provided the point is moving in the proper direction. Starting with any assumed initial position, and moving always in the direction required by the given equation, the point de- scribes a curve ; the values of a;, y^ y' at any point of the curve satisfy the differential equation. Further, since the initial position is entirely arbitrary, it is clear that the point may be made to describe any one of a family of curves, the so-called integral curves. The equation of this family is, of course, the general solution of the differ- ential equation ; it contains, as it should, an arbitrary parameter, viz., the constant of integration. The graph of any particular solution is merely one of the family of integral curves. Example ; Interpret geometrically the differential equa- tion X dx -i-y dy = 0. Writing the equation in the form dy _ _x dx y we see that the point (a:, y') must always be moving in a direction perpendicular to the line joining it to the origin. Its path is therefore any one of the family of circles with center at the origin. This may be verified by observing that the general solution of the differential equation is 2^ -{- y^ = c. 292 CALCULUS EXERCISES In each of the following cases find the equation of the family of integral curves and draw several curves of the family. 1. ^f = 0. 2. y< = 5. ^ ax X 5. ^ = -y. 6. y' = y. dx X 7. Find the differential equation of the family of circles through the origin with centers on the x-axis. A ns. 2 xyy' = y"^ — x^. 8. Find the differential equation of the family of parabolas with foci at the origin and axes coinciding with the x-axis. 9. Interpret geometrically the equations in Exs. 9, 12, and 14, p. 290. 198. Separation of variables. In the remainder of this chapter we show how to find the general solution of a differential equation of the first order in some of the simpler cases. Every differential equation of the first order, and of the first degree in «/', can evidently be written in the form Mdx + iVc?y = 0, where in general il!f and iVare functions of both x and t/. It is often possible to transform the equation so that Jf is a function of x alone and iVis a function of ^ alone ; this transformation is called separation of variables. When the variables have been separated, the differential equa- tion may be solved by a simple integration, as in the fol- lowing Example : Solve the equation XT/ dx -j- (a;^ + 1) c?^ = 0. After division by y(x^ + 1) the equation takes the form xdx ,dy_r. DIFFERENTIAL EQUATIONS OF FIRST ORDER 293 Integrating, we get ll0g(2;2+ 1) + log?/ = (7, or log y^x^ -f- 1 = c, «/Va;2 -f 1 = gc^ y\x^ + 1) = c\ where ^' = e^c^ EXERCISES Solve the following differential equations. 1. (1 + x)y dx + (1 — y)x dy = 0. Ans. log {xy) + x — y = c. 2. y' = axy'^. Ans. ax~y + cy + 2 = 0. 3. &mx cosy dx ■= co^ X s,\n y dy. Ans. cosy— c cos x. 4. -^ + /zza2. Ans. ?L±^ = ce^a., dx y — Cb 1 + y 1 - x 6. (1 + x)y^dx — x^ dy = 0. 7. Vl - 2/2c^x + Vl -x2f/2/ = 0. J^ns. xVl - y^ -^ yVl x"- = c. 8. '^=-kv^. dt 9. ^ = _ cos 2 t. dt 10. Show that the function y = ce* is the only function that is unchanged by differentiation. 11. Find a function whose first derivative is equal to the square of the original function.* Interpret geometrically. 12. Determine the family of curves whose slope at any point is equal to the product of the coordinates of the point. Find the curve of this family that passes through the point (0, 1), and trace it. 13. A particle falls under gravity, the resistance of the air being neglected. If the initial velocity is y^, find y and a; in terms of t. * Cf . footnote, p. 290. 294 CALCULUS 14. Determine the family of curves represented by the equation dx 15. In Ex. 14, find the curve (a) that passes through (0, 0) ; (&) that crosses the line a; = 1 at an angle of 45°. Trace these curves. 199. Coefficients homogeneous of the same degree. A polynomial in x and y is said to be homogeneous if all the terms are of the same degree in x and y. More generally, an}^ function of x and y is said to be homogeneous of the n-th degree if, when x and y are replaced by kx and ky respec- tively, the result is the original function multiplied by k"^. Thus the function X + Va;^ — y'^ -\- y log ^ X is homogeneous of the first degree. If, in the equation Mdx + Ndy = 0, the coefficients iHfand iVare homogeneous functions of the same degree^ it is easily seen that the equation when solved for y^ takes the form i.e. y' is a function of ^ alone. This suggests the substitu- X tion of a new variable v for the ratio ^; i.e. the substitu- .. X tion y = vx., dy = V dx -\- X dv. This substitution always produces a differential equation in V and x in which the variables are separable. Example : Solve the equation (rr -\- y)dx — x dy = ^. Substituting y= vx^ dy = V dx + X dv., we find (ic -H vx)dx — x(y dx -\- x dv) = 0, or dx — xdv = ^. DIFFERENTIAL EQUATIONS OF FIRST ORDER 295 The variables can ] now be separated : dx - dv=0, X log X — V -\- c = ^^ or, since V =^^ X y =^ X log X -\- ex. EXERCISES Solve the following differential equations. 1. {x + y)y' -}- X — y= 0. Ans. arctan ^ + - log (x^ + y^) = c. X 2 2. (^x'^ + y^)dx — 2 xy dy = 0. Ans. x- — y^ = ex. du *- 3. (xy — x^)— = y'^. Ans. y = ce'. dx 4. x^ dy + y^ dx = 0. 6. x^ dx + y^ dy = 0. 6. u dv — V du — Vu'^ 4- v'^ du = 0. Ans. u^ = c^ + 2 cv. 7. x dx + Va;"^ + 1 ^^ = 0. 8. 2 uv du + (^2 - 3 u^)dv = 0. Ans. v^ = c(u^ - v^). 9. V — = — X — V. dx 10. Show that, if AI and N are homogeneous of the same degree, the equation M dx + N dy =0 can always be put in the form 11. Give a general proof of the fact that, in the problem of § 199, the substitution y = vx always leads to an equation in which the variables are separable. 200. Exact differentials. The differential of a function u of two variables x and y is given by formula (1) of §164: (1) du = — dx -\ dy. dx dy 296 CALCULUS The quantity (2) Mdx + Ndy is called an exact differential if it is precisely the dif- ferential of some function u. Thus, the quantity X dy •\- y dx is an exact differential, viz. d(xy^ ; on the other hand, the quantity x dy — y dx is not an exact differential. If the quantity (2) is an exact differential, it appears by comparison with (1) that there must exist a function u such that (3) ^ = M, dx (4) ^^ = K dy Differentiating (3) with respect to y and (4) with respect to x^ we find J2^ ^ QM BH _ dN dy dx dy dxdy dx c)lJI filJf Equating values of - — — and -, by § 163, we get the dy dx dx dy relation dM^dJsr dy dx as a necessary condition that (2) be an exact differential. It can be shown that this condition is not only necessary but sufficient: i.e. the quantity M dx -\- N dy is an exact differential if and only if dM ^dN dy ~ dx' 201. Exact differential equations. The equation (1) Mdx-\-Ndy=Q is called an exact differential equation if its left member is an exact differential. DIFFERENTIAL EQUATIONS OF FIRST ORDER 297 Since equation (1), when exact, has the form du = 0, its general solution is evidently u = c. While a general method can be given for finding the function u^ we shall consider only cases in which this function is readily found by inspection. 202. Integrating factors. If the equation (1) Mdx-\-Ndi/ = is not exact, its solution can still be put in the form (2) u = c by merely solving for the arbitrary constant. By differ- entiating (2) we obtain an equation of the first order that is satisfied whenever (1) is satisfied: this equation must therefore have the form v(Mdx-\-]Srdi/}=0, where v is in general a function of both x and y. Thus for every differential equation* (1) there exists a function V, called an integratiiig factor^ whose introduction renders the equation exact. It can be shown that every differential equation has not merely one, but infinitely many, integrating factors ; nevertheless it is frequently impossible to find one of them. In various cases, some of which will be considered presently, an integrating factor can be found by direct processes ; in other cases it is best found by inspection. It should be noticed that in separating variables, as in § 198, we are really introducing an integrating factor. Thus, in the example of that article, the integrating factor is 1 K^ + 1)' * Assuming the existence of the general solution. Cf. § 195. 298 ' CALCULUS Example : Solve the differential equation X dy — y dx =^ 0. If we note thalt the differential of ^ is ^ ^V - V ^^ ^ it X x^ appears that — is an integrating factor in the present x^ instance : X dy — y dx _ r. x^ x y = ex. Other integrating factors are — (which merely sepa- rates the variables), — , — -• y^ x^ ± y^ EXERCISES 1. Solve the above example by using each of the integrating factors there mentioned, and compare the results. 2. Solve Ex. 1, p. 295, by means of an integrating factor. Solve the following equations. 3. xdy —(x -\- y) dx = 0. 4. (2 a: + 2 ?/) dx + (2x + f)dy = 0. 5. (x — y^) dx + 2 xy dy = 0. 6. xdy — y dx = (x^ + y'^) dx. 7. (a: + y + 1) dx -\- (x - y) dy = 0. 8. xdx -^ y dy + x dy — y dx = 0. 9. xy' = y -\- Vx^ — y'^. 10. u(u + 2 v) du + (m2 - v^) dv = 0. 11. ^dv__ 1 ^ds s^' 12. (sin y + 2 x) dx -\- X cos y dy = 0. 203. The linear equation. A differential equation of the tirst order is said to be linear if it is of the first degree in y and y' , Every such equation may evidently be DIFFERENTIAL EQUATIONS OF FIRST ORDER 299 written in the form (1) y' +Py=- Q, where P and Q are functions of x alone. We shall find that the linear equation is of especial importance in the applications. Before undertaking to solve equation (1), let us con- sider the special case (2) / + P2/=0. Here the variables are separable, and the solution may be obtained at once : ^-{-Pdx = 0, y whence log y + \ P dx = c^ (3) ye^^'^-" = c'. Now, differentiating (3), we get elPdx(^^y _|_ py ^^^ ^ Q^ which shows that e^^'^'^ is an integrating factor for equa- tion (2). But since Q is a function of x alone, it follows that e^^^"^ is likewise an integrating factor for equation (1). Examples : (a) Solve the equation dy -\-^y dx= x dx. Here P=2, CPdx = 2x, e^^'^'^ = e^. Introducing the integrating factor e^^^ and integrating, we find ye^ = I xe^"" dx == I- xe^"" — je^"" -\- (?, whence y = \x — \-\- ce-2^. (6) Solve the equation xy' — x^ — y = 0. Writing this in the form (4) dy — ^dx = x^dx, X 300 CALCULUS we have P = , ] P dx= — logx^ X ^ whence X by formula (5) of § 44. Hence, dividing equation (4) by X and integrating, we get ^ = i xdx = — \- c, X *^ 2 2^ = a^ -\- c'x. 204. Equations linear in/(i/). The equation (1) /w+^/w=e, where P and Q are functions of x alone, is evidently linear in /(«/), and may be solved by the method of the preceding article. An equation not given directly in the form (1) may sometimes be reduced to that form by a simple trans- formation. In particular, this is always possible with the equation. The process is as shown in the following Example: Solve the equation X y^ Let us write the equation in the form y^ dy -\-'^dx= dx. X If we multiply through by 3, so that the first term be- comes c?(^^), this equation is seen to be linear in y^ : 3 v^ Sy^dy -i — ^ dx=S dx. X Here P =z- e^^ ^^ = g31og X _ ^ X DIFFERENTIAL EQUATIONS OF FIRST ORDER 301 whence the solution of the equation is ^3^3 _ 3 i 2'^ dx = I x'^ -{- c, ^3 _ I ^ _j_ cx~^. EXERCISES Solve the following equations. 1. -^-\- y = X. Ans. y = x — 1 + cc*. dx 2. {x + \) dy - 2y dx =(x + \)^dx. Ans. 2y =(x + ly -\- c(x + l)^. 3. y' - xy = X. 4. X— + (1 -\- x)y = e^. Atis. 2 x?/e^ = e'^ + c. dx 5. (x - 2y + 5) dx -j- (2 x + 4:)dy = 0. 6. y' sin y + sin x cos y = sin x. 7. dy + 1/(1 — xy^) dx = 0. Ans. — = a: H h ce^. 8. ^y^y' — 2//^ = X -f 1. ^ns. ?/^ = ce^ — 2 ^ — !• 9. -— = g — kv. Solve in two ways. 10. — ^ cos X -\- y sin x = 1. c/a: 11. —+ y cos a: = sin 2 x. 12. (1 + a;2) -^ -I- ?/ = arctan x. dx 13. — z= — V + cos ^ 14. (xy"^ + ?/)f/a: — x dy = 0. 15. dydx + (x + a:?/2) f//y = 0. 16. ydy + (xy^ — x) dx = 0. Solve in two ways. 17. X dy + {xey -\)dx — ^. 205. Geometric applications. Many of the properties of a curve depend not only on the coordinates x^ y^ but on the slope y^ as well. When a curve is defined by such 302 CALCULUS properties, the analytic expression of the given data leads to a relation between x^ y, and y' — in other words, to a differential equation of the first order. The general solution of this equation represents the family of ''inte- gral curves," as seen in § 197 ; in many cases additional data are given that enable us to determine the constant of integration. Example : Find the equation of the curves whose normal always passes through a fixed point. Let us take the fixed point as origin of coordinates. The slope of the normal at (2;, ^) is j ; but since the normal passes through the origin, its slope is ^- Hence the differential equation of the required curves is or xdx-[- y dy = 0. Solving, we get x^ + y^ = c. The only curves having the given property are circles with center at the given fixed point. EXERCISES 1. Find the equation of the curves whose subnormal is constant. Draw the figure. (See Ex. 22, p. 32 ; cf . also Ex. 8, p. 31.) 2. Find the equation of the curves whose subtangent is constant and equal to a. Draw the figure. ^^^ _ ^^^^ 3. Determine the curves in which the normal at any point is perpendicular to the radius vector (i.e. the line joining the point to the origin). 4. Determine tlie curves in which the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact. DIFFERENTIAL EQUATIONS OF FIRST ORDER 303 5. Determine the curves in which the area inclosed between the tangent and the coordinate axes is, equal to n'^. 6. Determine the curves such that the area included between the curve, the coordinate axes, and any ordinate is proportional to the ordinate. . ' Ans. y = ce". 7. Find the curve of Ex. 6 that crosses the ^-axis (a) at (0, 2 a) ; (6) at an angle of 45°. MISCELLANEOUS EXERCISES Solve the following equations. 1. x^dy — (1 + xhj)dx = 0. 2. V — = 1 — v'^. Solve in two ways. dx 3. dy — sin x dx = 2y dx. 4. y dx -f dy = y'^ dx. Solve in two ways. 5. (x — y)dx + (1 — X — 2 y)dy = 0. 6. dy + x^y dx = 0. Solve in two ways. 7. — = a — cos kt. dt 8. ^ + (log V — \)dx = 0. Solve in tw^o ways. y 9. y = 1 - y + sin t. 10. (a:2 - 4 xy)dx + y"^ dy = 0. 11. x^ -y = xVx^ + y\ 12. (1 + x^)dy - (1 + xy)dx = 0. 13. 'I^ = a^- kH\ 14. v — =y-v. dt dy CHAPTER XXVI DIFFERENTIAL EQUATIONS OF HIGHER ORDER I. Introduction 206. General and particular solutions. Being given a relation between x^ y^ and n arbitrary constants, say (1) ^(^. y, G^ •••,0=^' let us differentiate this relation n times in succession. The equations thus obtained form with the original equation a set of n + 1 equations from which the n con- stants may be eliminated. The result is a differential equation of the n-\\\ order, (2) ^(:r, ^, y, ...,y">) = 0. Conversely, corresponding to a differential equation of the form (2), there exists in general a relation of the form (1) which satisfies the differential equation. Equation (1) is called the general solution of equation (2). Thus the general solution of a differential equation of the n-th order involves n arbitrary constants. It is understood that the general solution contains n essential constants : i.e. that it cannot be replaced by an equally general form containing a smaller number of con- stants. Thus the equation y = c^e^'^^^ appears at first sight to contain two constants, but there is really only one. For, writing the equation in the form y z= c^e^ • e*^2 304 DIFFERENTIAL EQUATIONS OF HIGHER ORDER 305 and setting c^e''^ = (7, we see that the equation ^ = Ce"" is equally general. A particular solution is one that is obtained from the general solution by assigning particular values to one or more of the arbitrary constants. Thus a particular solution may contain any number of constants less than the maximum number, n. For example, it follows from Ex. 18, p. 50, that the equation dt^ is satisfied by the equation x = A cos kt -\- B sin kt, where A and B are arbitrary. Since the differential equation is of the second order, the solution here given, containing two constants, is the general solution. Par- ticular solutions are x= A cos kU X = A(cos kt + sin A;f), x=^ sin kt^ rc = 0, etc. 207. Geometric interpretation. Given a differential equation of the second order, and of the first degree in y"^ y" =f(x,y, y), we may in general assign values at pleasure to x^ ?/, and y\ and compute the corresponding value of y". The equation is satisfied by the coordinates of any point (2:, y') moving in any direction, provided its direction is changing at the proper rate. Or, since the value of y", together with the assumed value of y\ determines the curvature of the path, we may also say that the differential equation is satisfied 306 CALCULUS by the coordinates of any point moving in any direction, provided its path has always the proper curvature. The paths of the point (re, y) moving in the manner just described are called, as in § 197, the integral curves of the given differential equation. The ordinary equation of the family of integral curves is of course the general solution of the differential equation ; since this solution contains two arbitrary constants, or parameters, it follows that the integral curves form a doubly -iyijinite system. The point (x^ ?/) may start from any assumed initial position in any direction ; hence through any point in the plane there pass infinitely many integral curves. The above discussion is readily extended to differential equations of the third and higher orders. EXERCISES Find the differential equation whose general solution is as follows. 1. y = c^ + c^e'^='. Ans. y" — 2 y' = 0. 2. y = c\e^ + c^e'^. 3. y = Cjfi^ + c^xe^. 4. y = Ci sin x + c^ cos x, Ans. y" ■\- y — 0. 5. y — c\-\- CgX + a;2. 6. ^ = Cl(l + Xy^ + Cg. Solve the following differential equations, and discuss the nature of the integral curves. 7. y" = 0. 8. y" = 1. 9. y" = 6 ar. 10. y" = y' . 11. ^^ + -^'"3" = a. (Cf . Ex. 2, p. 78.) 2/ . 12. Solve Ex. 8, (a) if the curve touches the line y —2 x 2X (1,2); (i) if the curve passes through the points (1, 2), (3, 3) ; (c) if the curve passes through (1, 1) ; {d) if the curve intersects the ly-axis at right angles. Draw the curve (or several of the curves) in each case. 13. Solve Ex. 9 for each of the cases of Ex. 12. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 307 II. The Linear Equation with Constant Coefficients 208. The linear equation. We have already (§ 203) defined the linear equation of the first order as an equa- tion that is of the first degree in y and y^ . More gener- ally, a differential equation of the n-t\\ order is said to be linear if it is of the first degree in ?/, ?/', •••, y^^^\ Thus every linear differential equation of the n-th order can be written in the form where the coefficients /?i, •"•, Pn ^^^^ the right member X are functions of x. In what follows, we shall be concerned entirely with the important special case in which the functions pi, •••, p„ are constants : (1) y^^^ + ay-^^ + ... +a,y = X. 209. The homogeneous linear equation. A linear differ- ential equation whose right-hand member is is said to be homogeneous.^ Thus the general form of the homo- geneous linear equation with constant coefficients is (1) ?/<'^) + ay-^^ + -• +a^y = 0. This equation is important not only in itself but because its solution must be determined before that of the non- homogeneous equation (1) of § 208 can be found. If 1/ = ?/j is a particular solution of equation (1), then y = ^1^1, where e^ is arbitrary, is also a solution, as ap- pears at once by substitution in (1). Further, if y = y^ is a second particular solution,! then not only y = c^y,^ but also y = e^y^ + c^y^ * That is, it is homogeneous in y and its derivatives. See § 199. t That is, a solution not of the form y = Ciyi. 308 CALCULUS is a solution. Finally, if y = HVv y = ^2^2' y = CnVn are n distinct particular solutions, then y = ^1^1 + ^22/2+ ••• +^n«/n is a solution, and since it contains n arbitrary constants, it is the general solution. We proceed to show that the general solution of equa- tion (1) can always be written down, provided a certain algebraic equation of the n-\h degree can be solved. The theory will be developed in detail only for the equation of the second order. 210. The characteristic equation. The homogeneous linear equation of the first order, viz., y' + HV = ^' is evidently satisfied by y = e~^'^^ . This suggests the possibility of determining m so that y _ gmx will be a solution of the equation (1) y'^ + a^y' + a^y = 0. Substituting in (1) the values y = e"»^, y^ zzzme^""^ y" = m^e'^^ and bracketing out the factor e"*% we find that the differ- ential equation is satisfied, provided (2) m? H- a^m -\- a^z= 0. Equation (2) is called the characteristic equatio7i* cor- responding to (1). Thus y z= e'^^ is a solution of equation (1) if and 07dy if m is a root of the characteristic equation. * Also called the auxiliary equation. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 309 211. Distinct roots. If the roots w^, m^ of the char- acteristic equation are distinct, we obtain at once two distinct particular solutions of the differential equation, viz., y = e^i"", y = e'^i''. Hence, by § 209, the general solution is (1) y= Cie»^^i*+ c^e^^"". Example: Solve the differential equation y" -y'~'2y = (i. The characteristic equation is m^ — m — 2=0, whence w = 2 or — 1. Thus the general solution of the given equation is 212. Repeated roots. When the charactexistic equation has equal roots, the method of the previous article does not give the general solution. For, if m-^ = m^^ equation (1) above becomes y = {jj^"*!^ -f (?2e"'i^ hence the solution contains only a single constant, and is a particular solution. To find a second particular solution, let us try y = xe^^^^ whence y^ = e'^^^'i^m^x + 1), yif _ e'^^''(m^x -\- 2 Wj). Substituting in the differential equation, we find that y = xe^^^ will be a solution, provided (1) {m^^ + a^m^ + a^yx + 2 w^ + «i = 0. Now the coefficient of x vanishes because m^ is a root of the characteristic equation. Further, since m-^ = Wg, it 310 CALCULUS follows that «! %=-^, or 2m^-\- a^ = 0. Thus (1) holds, and y —xe'^^'^ is a second particular solution. Therefore the general solution of the differential equa- tion is 213. Complex roots. If the characteristic equation has complex roots a ± z/9, the general solution takes the form (1) = e^^Qc^e^^'^ -\- c^e-'^^^. Up to this point the exponential function has not been defined for imaginary values of the exponent. If, however, we expand e'''' formally in Maclaurin's series, and compare with the series for sin x and cos x^ we obtain the relation (2) e»^ = cos X + i sin x. In the theory of functions of a complex variable, this formula is taken as the definition of the imaginary expo- nential function. By means of (2), the right member of (1) may be sim- plified. For, gi^x _ (3Qg ^^ + z sin ^x^ ^-iPx — Qos ^x — i sin fix. Whence (1) becomes y = g''^[((Tj + Cg) cos fix + z(^i — C2) sin fix^^ or, if we place Cj -f- y = 0. 9. y" - 4?/' + 4i/ = 0. ^ns. ?/ = qe2x ^ C2a;e2*. 10. ^.= 0. . 11. 4^+4^' + r = 0. 12. 9/' + 12/ + 4// = 0. 13. y" -oy'^by. 14. y" + 2ij' + by —0. Ans. y = e~''{c^co&2 x + c^sm^x). 15. ?/" - 4 ?/' + 6 ?/ = 0. 16. ^ = - k'^x. 17. y' + 93/ = 0. 18. y" + 2y' + y = 0. 19. 8/'+ 16/+ Qy = 0. 20. Find the equation of a curve for which y" = y, if it crosses the 3/-axis at right angles at (0, 1). 21. Find the equation of a curve for which y" = — y, ii it touches the line ?/ = x + 1 at (0, 1). 22. Determine the curves for which the rate of change of the slope is equal to the slope. 23. In Ex. 22, find the curve that touches the line y = 2 a: at the origin. Ans. y = 2e^ — 2. 24. In Ex. 22, find the curves that cross the y-3,xh at 45°. 25. In Ex. 22, find the curve that passes through (0, 1) and ap- . proaches the negative a:-axis asymptotically. 26. Show that e '^ = i, e^' = - 1, e-""' = 1. 27. Derive formula (2) of § 213 by comparison of the Maclaurin series for e'^, sin x, and cos x. 28. Show that, if the characteristic equation has equal roots m^, the equation /' + «i/y' + a^y = can be reduced to the form z" = by the substitution y = ze^^i", and derive the result of § 212 from this fact. 312 CALCULUS 214. Extension to equations of higher order. The the- ory of §§ 210-213 is readily extended to equations of higher than the second order. We give the results with- out proof : Let there be given a differential equation (1) ?/(«> + ay-^^ ^ ... j^a„y=0. (a) If the roots Wj, mg, •••, m^ of the characteristic equation m^ -\- a^m^'^ -\- ••• + a,^ = are all distinct, the general solution of (1) is y = c^e'"^^ + c^e"^^-'' + • • • + c^e"^n^. (6) Corresponding to a double root m-^^ the terms in the general solution are y = c-^e^-^^ 4- c^xe^^^ ; corresponding to a triple root, y = c^e"H^ -|- c^xe"^^^ -f- c^x^e"^-^"^ ; etc. (c) A pair of complex roots a ± ^/3 give rise to the terms y = e«^((?j cos ^x + ^2 sin ^x} ; a pair of double roots a ± iff give rise to the terms y = e'^(^c^ cos I3x + ^2 sin fix + c^x cos fix 4- o^x sin fix^ ; etc. EXERCISES Solve the following equations. 1. y'" _ 7 y' + 6 ?/ = 0. 2. y"' = 4 ?/. 3. ?/'" = y" + 6 ;/. 4. y^'^^ - 12 y" + 27 y = 0. A71S. y — c^e^ + c^e ^ + c^xe * + c^x^e~''. 6. y"' = 0. 7. ?/(''^ -2y" + y = 0. 10. ?/" - 6 .?/" + 13 ^' = 0. Ans. y=c^^ e^''{c..2 cos 2x + c^ sin 2 a:). 11. '^ = x. 12. ^^ + 4^ = 0. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 313 13. 2/(*) - 4 y'" + 14 y" - 2Q y' + 25 ^ = 0. 14. y'" + 3 /' + 3 ^' + ^ =: 0. 15. y'" - 2 y" - y' + 2 ^ =: 0. 16. Prove the results of § 214 for the equation of the third order. 215. The non-homogeneous linear equation. Let us con- sider now the non-homogeneous linear equation (1) y«) + ^y-i)+ ... +a^y = X In solving this equation, the first step is to write down the general solution of the homogeneous equation obtained from (1) by making the right member 0. The quantity Y is called the comple- mentary function. The next step is to obtain, by any means whatever, a particular integral of (1), Then the equation y = Y + y is a solution of (1), as appears at once by substitution, and since it contains n arbitrary constants, it is the gen- eral solution. Various methods are known for finding the particular solution — y = y- The method given below, though not entirely general, is usually the best method when it applies, and it is avail- able in nearly all cases that arise in the simpler appli- cations. We begin with an Example : Solve the equation (2) y" —by' -\-Qy = x -\- e^"^. The complementary function, i.e. the solution of the IS Y = c^e^^ -\- c^e^^. 314 CALCULUS To obtain a particular integral of (2), proceed as follows; Differentiating twice, we obtain (3) ?/^4) _ ^yin J^Qy" = 4e2x^ Differentiating again, we get (4) ?/«> - 5 ?/^4) 4_ 6 y"f = 8 e^^. Multiplying equation (3) by 2 and subtracting from (4), we get the homogeneous equation (5) yib) _ 7 ya) ^ 16 y'" _ 12 7j" = 0. It is easily seen that the complementary function Y forms part of the solution of this equation ; hence two of the roots of the characteristic equation m^ — 7 m* + 16 m.^ — 12 w^ = are m = 2, 3. The other roots are 2, 0, 0, Thus the general solution of (5) is (6) ^ = c^e^"" 4- o^e^"" + ^3 4- o^x + e^ xe^"^. Let us substitute y in the original equation as a trial solution, noting, however, that the terms arising from the complementary function must disappear identically after the substitution, so that it is sufficient to substitute* (7) y = c^-{- c^x -{- c^xe^"". We have , , o 2^ . 2t ?/' = dt:^ 8. y" — ^ !i' + ^y = cos X — 9. ^^^ + 4 X :--_ sin 3 t + fi. df^ ?^. Ans. y = c^e^^ + c^e^ + j^ocosa: — x^o sin X + xe^"". A ns. X — c■^^ cos 2 f + c^ sin 2t — I sin 3 ^ + 1 i^ — ^. 10. y" — 2y' + y = xe"^. Ans. y = e^(c^ + c^x + Ix^). 11. ?/" + // = 1 + 2 cos^. 12. dhj , — -I + ?/ = x sin x. yl«s. y = Cj^C 13. d^u = u. dir" 14. v= 0. dv^ 15. y'" _ ?jij" + 2y = 3a:-4 16. /"-2/' + y =e-. 17. a; 18. ^ ns. u = \v^ + c-^ + Co. J.n,s. 2/ = ci + e^(c^ + c^x + \ x^). 19. Prove the statement that the complementary function corre- sponding to the original equation is always a part of the solution of the derived homogeneous equation. III. Miscellaneous Equations of the Second Order 216. The equation y" =/(a:). In this section we con- sider various classes of equations of the second order which can be solved by special devices. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 317 The simplest case is that in which the second derivative is a function of the independent variable : y" =/(^)- This equation can be solved directly by two successive integrations. In fact, it is obvious that the equation can be solved by n successive integrations. 217. The equation y" =f(y)- An equation in which the second derivative is a function of the dependent variable, can always be rendered exact by introducing the integrat- ing factor 2 y' dx in the left member, and its equivalent 2 d^ in the right member : 2y'y"dx = 2f(y)dy. Integrating, we find y'^ = '^JKy^dy + c,. After extracting the square root of both sides, we have a differential equation of the first order, and of the first degree in y\ in which the variables can be separated. Example : Solve the equation yZ Multiplying through by 2 y' dx, we get Id 2 y' y" dx = ^ , whence y2 = _ + ^ 1 ?/2 '^c^y'^ — 1 y = ± y 318 CALCULUS Separating variables, we have ± ^ c^y^ — 1 whence ^ — ^ — dx^ ± '^c^y^ — 1 = c^a: 4- 0, ^ / — arctan — ^ — = a: -f- c C 2 This may be simplified by writing -j- in place of c^ — arctan ^ = a: 4- ^2* 320 CALCULUS EXERCISES Solve the following equations. 1. ^'^ = X8. dx^ 3. ..'" = «Y 4. y"= '• Vy dh 1 5. dc^ fi 7. x^y" = y'. 9. y" = 1 - y'- 11. y" + yy' = o. 12. d-^y ^ 1 <^3/ _ dx'^ X dx d^x -, Ans. ax = log(y + Vy'^ + c^) + Cg. 6 — =:-! 8. /y^//" = 7/. 10. ?/" = a:e*. Ans. y = c^ log X + c 2' X (IX 13. !^= t-^^y. dt^ ^ U, t—+^=l. Ans. X = t + c,]ogt + c^. dt^ dt 1 & - 16. y^ = - . Ans. (x - ciY +{y - c^y = a\ (1 + y"^)l « 16. ^ = - k^x. df^ 17. d^ = _kH. dfi 18. Of the above exercises, which ones can be solved by the methods of section II? 19. Solve Exs. 16, 17 by the methods of section II. 20. Show how to solve the equation y" + Py' + Qy'^ = 0, where P and Q are functions of x alone. CHAPTER XXVII APPLICATIONS OF DIFFERENTIAL EQUATIONS IN MECHANICS I. Rectilinear Motion 220. Rectilinear motion. Consider a point P moving in a straight line : for instance, the centroid of a falling body, of the piston of a steam en- q p gine, or of a train running on a '* x- >' ^ straight track. The position of the ^^^- ^^^ point at any instant is determined by its abscissa OP = x, counted from an arbitrarily chosen origin on the line, a definite sense along the line being selected as positive. As the point moves, its abscissa 2; is a function of the time : X=(f>(t). If this function is known, the motion of the point is com- pletely determined. The velocity v is found as the first dx derivative — , and the acceleration / as the second deriv- d X ative — — , of the abscissa x with respect to the time (see §55). In most applications, however, it is the converse prob- lem that presents itself. Thus, the velocity may be given as a function of ^ or a: or both, say so that in order to determine the position of the point at any time it is necessary to solve this differential equation Y 321 322 CALCULUS of the first order. Or, and this is the most common case, the acceleration may be given as a function of t^ x^ and v (or of any one or two of these), say (1) S^-^^*'"''"^- The abscissa x is found in terms of t by solving this differential equation of the second order. It should be noted that when the acceleration (or the velocity) is given, the motion is not completely determined unless " initial conditions " are also given by means of which the constants of integration can be determined. 221. Motion of a particle under given forces. Suppose the '' point " whose motion was discussed in the preceding article is a material particle moving under given forces. If the particle is free to move in any direction, the motion will be rectilinear only if the resultant F of all the applied forces lies in the same straight line with the initial velocity. The product of the mass by the acceleration is equal to the resultant force, by § 187. If we multiply both members of equation (1) above by m, and write F(t^ x^ v) in place of m/(f, x^ v}, that equation takes the form drX m-^ = F(t,x,v-). This equation and equation (1) of § 220 are mathematically equivalent, since one is a mere constant multiple of the other. The difference lies in the physical meaning of the quantities involved. It should be noted that the term " particle " as here used does not mean necessarily a mere mass-point. The '•'• particle " may be a body of any size or shape, provided that all the forces acting may be regarded as applied at a single point, and that the motion of one point determines the motion of the whole mass, as in the case of a rigid body moving without rotation. APPLICATIONS OF DIFFERENTIA^. EQUATIONS 323 222. The equation of motion. The equation (1) mg = F(f, X, K), or its equivalent (2) S=/(^^-'')' is called the equation of motion. It follows from what has been said that the rectilinear motion of a particle is determined by the equation of motion together with the initial conditions. In each problem there are in general three steps : first, to write the equation of motion ; second, to solve this equation, determining the constants of integration in accordance with given initial conditions ; third, to in- terpret the results. When the forces acting are given, the equation of motion can be written at once : we have only to equate ox w^— — to the sum of the components of the forces in the di- (X/b rection of motion. In the most general case, the equation of motion may be expressed as a differential equation of the second order in X and t by substituting dx dt Special cases, however, are common. If the force is a function of t only, the method of § 216 evidently applies. If # is a function of t and v, we may use the method of § 218, writing ^ON d'^x _ dv ^ ^ 'd^~~dt' If jP is a function of x and v, the method of § 219 applies : in this case, since d^x _ dv _dv dx dt^ dt dx dt 324 CALCULUS we substitute ^ ^ dt^ dx We shall find that in many cases a variety of methods may be used. In any problem we may desire to know the position of the particle at any time, the velocity at any time, and the velocity at any position. We should therefore try to obtain three equations, giving * x in terms of <, v in terms of f, and V in terms of x^ respectively. The {x^ f)-equation is of course obtained by solving the equation of motion (1) (or (2)) as an equation in x and ^, and determining the constants. The (v, ^) -equation may be found by differentiation of the (a:, Q -equation, after which the (v, a:) -equation may be obtained (theoretically at least) by eliminating t between the other two. If it is possible to introduce (3) and apply the method of § 218, the (v, ^) -equation results directly from the first integration ; if formula (4) and § 219 can be used, the (v, 2:)-equation is obtained directly. 223. Uniformly accelerated motion. A motion is said to be uniformly accelerated if the applied force, and hence the acceleration, is constant (cf. § ^b). If the constant ac- celeration be denoted by h^ the equation of motion is simply di^x _ 7 EXERCISES 1. Write the differential equation of uniform rectilinear motion (§ 55), and find x in terms of t, v in terms of t, and v in terms of x, if a: = 2 and v = 4 when t — 0. Solve the equation of motion in three ways, by the methods of §§ 212, 216, and 219, and obtain the (y, t)- equation and the {v, a:)-equatiou in each of the ways suggested in § 222. Draw the graph of each equation. * Explicitly if possible. APPLICATIONS OF DIFFERENTIAL EQUATIONS 325 2. Solve Ex. 1 if X = 10 when t = 5 and a; = 22 when t = 9. Find the values of x and v when ( = 0. 3. The velocity of a particle at the time t is u = 6 ^ — 5. Find (a) the acceleration ; (b) the space covered in 4 seconds ; (c) the velocity when x = 6 (x being measured from the starting point). Describe the motion in words. 4. The velocity of a particle at the distance x from the starting point is V =Vx + 10. Find X in terms of /; also find the acceleration. 5. A particle falls under gravity, all resistances being neglected. Write the equation of motion, taking motion downward as positive, and solve it by three methods. Explain the meaning of the con- stants of integration. 6. Determine the constants of integration in Ex. 6 if the particle falls from rest, the starting point being taken as origin. Draw the graph of the (x, t)- and (v, ^)-equations, noting that the latter is the first derived curve of the former (§ 3.5). 7. (a) Solve Ex. 5 if the initial velocity is 10 ft. per second up- ward, (b) How far and how long does the particle rise ? (c) Find V and t when the particle is 20 ft. below the starting point. Ai^s. (c) V = 37.1 ft. per second. 8. Solve Ex. 5 if a: = 10 when t = 1 and x = 100 when t = 3. Does the particle at first move upward or downward ? Find the ve- locity at the end of 1 second. Ans. 13 ft. per second. 9. If a stone dropped from a balloon while ascending at the rate of 20 ft. per second reaches the ground in 10 seconds, what was the height of the balloon when the stone was dropped ? With what ve- locity does the stone strike the ground ? 10. Solve Ex. 5 if the velocity 2 ft. below the starting point is 23 ft. per second. If the starting point is 500 ft. above the earth's surface, when and with what velocity does the particle reach the earth ? Ans. t = 5 or 6i seconds. 11. Show that the velocity acquired by a body falling from rest through a height h is V = V2 gh. Derive the formula in two ways. 12. A body falls 50 ft. in the third second of its motion. Find the initial velocity. 326 ' CALCULUS 13. A body falls under gravity. Find the distance covered in 6 seconds if at the end of 2 seconds the distance below the starting point is 84 ft. 14. The motion of a railroad train is uniformly accelerated. If when the train is 250 ft. from a station the velocity is 30 ft. per second, when 600 ft. from the station it is 40 ft. per second, find the acceleration, and the velocity when passing the station. Ans. Vq = 20 ft. per second. 15. A stone is thrown vertically upward from the top of a tower. At the end of 2 seconds it is 400 ft. above the ground, and is still rising, with a velocity of 10 ft. per second. Find the height of the tower. Ans. 316 ft. 16. A stone thrown upward from the top of a tower with a velocity of 100 ft. per second reaches the ground with a velocity of 140 ft. per second. Discuss the motion. What is the height of the tower? Ans. 150 ft. 224. Momentum ; impulse. When a particle of mass m is moving with a velocity v^ the product mv of the mass by the velocity is called the momentum of the particle. When a particle moves under a constant force F from the time Iq to the time fj, the product F(^t^ — to) of the force by the time during which it acts is called the impulse of the force for that time-interval. More generally, if F varies from instant to instant, let us divide the time from ^Q to t^ into n equal intervals A^, multiply each A^ by the value of F at the beginning (or any other instant) of the interval, and form the sum of the products thus obtained. The limit of this sum, as A^ approaches 0, is the impulse of the variable force F during the interval from Iq to t-^ : /= lim y FAt = f'Fdt. 225. The principle of impulse and momentum. Let us write the equation APPLICATIONS OF DIFFERENTIAL EQUATIONS 327 in the form dV XT dt Multiplying by dt and integrating from the time ^q, when the velocity is v^^ to the time t^ when the velocity is v, we find (1) mv — tuVq =1 Fdt By § 224, the left member of (1) is the change of momentum in the time-interval from tf^ to t^, the right member is the impulse of the force F. Henoe we have the Theorem : If a particle moves in a straight line^ the change of momentum in any time-interval is equal to the impulse of the force during that interval. This theorem will be referred to as the principle of impulse and momentum. It should be observed that what we have really done here is to find a first integral of the equation of motion by the method of § 218. Since the force F is always either directly or indirectly a function of ^, the above theorem is true in general ; but in order actually to com- pute the impulse directly in a given case, the force must of course be given explicitly as a function of t : F=F(t). If the force F is constant, equation (1) becomes simply mv — 'MVq = Ft— FtQ. 226. Work. When a particle moves in a straight line under the action of a constant force F, the work done is defined as the product of the force by the distance passed over : W = Fx. When the force is variable, we proceed as follows: Take the line of motion as a:-axis, and suppose the body moves from x— a to x=h. Divide the interval into ' i 328 CALCULUS segments Ax, and multiply each segment A2: by the value of F at some point of Ax. The limit of the sum of the products thus obtained is defined as the work of the vari- able force during the motion : 227. The principle of kinetic energy and work. Let us write the equation d^x TTT m — - = J^ in the form dV jy dx Multiplying by dx and integrating between the a;-limits Xq and X and the corresponding v-limits Vq and v, we find (1) lmv^-lmv,^= Crdx. By § 135, the quantity -| mv^ is the kinetic energy of the particle, hence the left member of (1) is the change in kinetic energy from Xq to x. By § 226, the right member is the work done- during the motion. Hence we have the Theorem : If a 'particle moves in a straight line, the change of kinetic energy in any space-interval is equal to the work do7ie hy the force in that interval. This is the principle of kinetic energy and work. Here we have merely applied to the equation of motion the method of § 219. In order to compute the work directly, the force must of course be given explicitly as a function of a; : F=F(x). If the force is constant, equation (1) reduces to J mv^ — }^ mv^ = Fx — Fxq. APPLICATIONS OF DIFFERENTIAL EQUATIONS 329 EXERCISES 1. Verify the principle of impulse and momentum in Exs. 7, 8, p. 325. 2. Verify the principle of kinetic energy and work in Exs. 7, 8, p. 325. 3. Solve Ex. 14, p. 326, by the principle of kinetic energy and work. 4. Solve Ex. 15, p. 326, by the principles of §§ 225, 227. 5. A ball of mass 5^ oz. strikes a bat with a velocity of 12^ ft. per second, and returns in the same line with a velocity of 32 ft. per second. If the blow lasts -^^ second, what force is exerted by the batter? Ans. 9 lbs. 6. A ball of mass 5^ oz. moving at 50 ft. per second is caught and brought to rest in a distance of 6 in. AVhat is the average pressure on the hand? Ans. 26 lbs. 228. Constrained motion. The motion of a body some- times depends on other conditions than the given forces. Thus, the piston of a steam engine can move only along the cylinder, a body sliding down an inclined plane can- not fall through the plane, etc. The motion in such cases is said to be constrained. In the case of constrained motion, let the applied force be resolved into components along, and at right angles to, the path. The component in the direction of motion is the "effective force" ; the motion is due entirely to this com- ponent, and hence it is only this component that appears in the equation of motion. For example, when a particle slides down a smooth inclined plane, the effective force is the component of gravity parallel to the plane.* Further, it is evident that, in the definitions and theorems of §§ 224-227, the force F must be taken as merely the effective component. The component normal to the path cannot do work, or contribute to a change of momentum. * The motion is supposed to take place alone; a ''line of greatest slope" — i.e. a line at right angles to a horizontal line in the plane. 330 CALCULUS EXERCISES 1. Write the equation of motion down an inclined plane, and solve it in a variety of ways. Explain the meaning of the constants. 2. Determine the constants in Ex. 1 if the angle of inclination to the horizon is 30°, and the initial velocity is (a) 0; (b) 10 ft. per second up the plane. In (b), how far and how long will the body move up the plane ? A7is. (b) 3} ft. 3. A bead is strung on a smooth straight wire inclined at 45° to the horizontal. What initial velocity must the bead be given to raise it to a vertical height of 10 ft.? 4. A railroad train is running up a grade of 1 in 200 at the rate of 20 miles per hour when the coupling of the last car breaks. Fric- tion being neglected, (a) how far will the car have gone after 2 min- utes from the point where the break occurred ? (b) When will it be- gin moving down the grade? (c) How far will it be behind the train at that moment ? (c?) If the grade extends 1 500 ft. below the point where the break occurred, with what velocity will it arrive at the foot of the grade? Ajis. (a) 2368 ft.; (b) 3 minutes 3 seconds ; (c) 2689 ft. ; (d) 25 miles per hour. 6. Show that it takes a body twice as long to slide down a plane of 30° inclination as it would take to fall through the " height " of the plane. 6. Show that in sliding down a smooth inclined plane a body ac- quires the same velocity as in falling vertically through the height of the plane. 7. A mass of 12 lbs. rests on a smooth horizontal table. A cord at- tached to this mass runs over a pulley on the edge of the table ; from the cord a mass of 4 lbs. is suspended. Discuss the motion. If the 12 lb. mass is originally 5 ft. from the edge of the table, find when and with what velocity it reaches the edge. Check by the principles of §§ 225, 227. 8. A cord hangs over a vertical pulley and carries equal weights of 10 lbs. at each end. If a 1-lb. weight be added at one end, discuss the motion of the system. Find v when the system has moved 6 ft. 229. Simple harmonic motion. If a point P moves in a circle with constant angular velocity ©, the motion of the projection P^ of P on a diameter of the circle is called simple harmonic motion. As P moves in the circle APPLICATIONS OF DIFFERENTIAL EQUATIONS 331 uniformly, P^ oscillates from A through to B and back again. p Suppose Pj. is at A at the time y^ f = 0. Then in time t the angle / ^/ AOP swept out by the radius / ^ vector of P is equal to ojf, hence -^T P^ lA the distance x of P^ from is V / (1) x= a cos (ot^ ^ -^ where a is the radius of the circle. ^^' ^^^ If when t= the point P is not at ^, but at some point P' such that the angle AOP' is equal to e, the equation (1) is evidently replaced by (2) x = a cos (^Q)t-\- e'). The abscissa x is called the displacement of P^.; the maximum displacement a is the amplitude of the motion. The time of completing one whole oscillation from A to B and back is called the period; it is evidently equal to the time required for P to make one complete revolu- tion, and is therefore CD The number of oscillations per unit time is called the frequency ; it is obviously the reciprocal of the period : n = - = -^ T lir' The angle o)^ + e is called the phase-angle^ or simply the phase, of the motion. Differentiating (2), we get the velocity dx v= —— — aw sin (&)^ + e), and the acceleration (3) j = —-=- aoi^ cos ((ot + e). 332 CALCULUS Combining (2) and (3), we may write the acceleration in the form — - = — (o^x : i.e. the acceleration is proportional to the displacement., and is always directed opposite to it. 230. Attraction proportional to the distance. If a particle moves in a straight line under the action of a force directed toward a fixed point in the line of motion, and propor- tional to the distance x from that point, the equation of motion can evidently be written in the form (1) m — — = — mk'^x., where A; is a constant, the minus sign being chosen be- cause the force is always directed opposite to the displace- ment X. The fixed point toward which the force is directed is called the center offeree. Integrating equation (1) by the method of § 213, we get x= c-^ cos kt H- (?2 sin kt^ whence Cm 'V V = — = — ^(?isin kt-{- kccf cos kt. dt ^ ^ Then 2' Take v = and X = a when ^ = 0. a = (?^, = kc. whence c^ = a, (?2 = ^» and finally (2) x= a cos kt, V = — ak sin kt. Since x has here the same form as in equation (1) of § 229, it follows that a particle moving under the con- ditions of this article performs simple harmonic oscilla- tions about the center of force 0. This is a fact of great importance, as forces directed toward a fixed point and proportional to the distance from that point are of frequent occurrence in nature. APPLICATIONS OF DIFFERENTIAL EQUATIONS 333 231. Hooke's law. When a spiral steel spring of length AO=^l is stretched to a length AP = I + x, the tension in the spring, or the force tending to restore it to its natural length, is proportional to the extension x. This law, known as Hookes law^ is obeyed very closely (pro- vided the extension is not too great) by all so-called elastic materials. Suppose a steel spring of negligible mass is placed on a smooth horizontal table with one end fast at A. Let the natural lens^th of the spring be A0= ?. A par- j^i I 9 ^ Q m tide 01 mass m attached to the free end is drawn out to the position P and then released. The only force acting is the tension in the spring, which by Hooke's law is directed toward the position of equilibrium and is proportional to the distance from 0. If the spring offers the same resistance to compression as to extension, it fol- lows from § 230 that the particle performs simple harmonic oscillations about 0. The equation of motion is Ct X 7 9 7n — - = — mfc^x. Of course if the resistance to compression is not the same as to extension, a different equation comes into play as soon as the particle passes through 0. EXERCISES 1. In the problem of simple harmonic motion, trace the curves showing X, v, and j as functions of t, remembering that the graph of v is the first derived curve, the graph of J the second derived curve, of the graph of x. Take rt = l,(o = 2, e = 0. 2. Show that, if x performs periodic oscillations as in § 229, v and / do likewise. Prove the following from the equations of § 229, and verify by the curves of Ex. 1 : the periods of all three are the same ; the amplitude of v is o> times that of x, the amplitude of / is w times that of y ; in phase, v differs from a: by ^ and J differs from v by ^ • 334 CALCULUS 3. In the problem of § 230, obtain the (y, i)-equation by two methods. Ans. v = ±k Va^ — x^, 4. A particle has simple harmonic motion. Proceeding from equation (1) of § 230, find x in terms of <, v in terms of i, and v in terms of a:, if v = v^ and a: = when t — 0. 5. Show directly from equation (2) of § 230 that the particle performs periodic vibrations about the center, and find the amplitude and the period. Find when and where the velocity is a maximum, and find the magnitude of the maximum velocity. 6. A steel spring offering the same resistance to compression as to extension is placed on a smooth horizontal table with one end fixed. The spring is stretched to a length 6 in. greater than the natural length and then released. Discuss the subsequent motion of a mass attached to the free end. Take Ic^ = 4. Find the period. Ans. T" = TT seconds. 7. In Ex. 6, find the work done by the force in a quarter-oscilla- tion. Check by the theorem of § 227. 8. Work Ex. 6 if the steel spring is replaced by a rubber band of natural length 1 ft. Ans, T = 7.14 seconds. 9. In Exs. 6 and 8, discuss the effect of increasing the con- stant k^. 10. Work Ex. 8 if F = 512. Ans. T = 0.6 second. jij 11. A rubber band of natural length AB = l is suspended vertically with a weight attached. The effect of the weight is to stretch the band to a length AO — I 4- h. The weight is given a displacement OP = a and then released. Write the equation of motion and solve it completely. Show that the particle performs simple harmonic oscillations about O, provided a < A. 12. Solve Ex. 11 \ia>h. 13. In Ex. 11, find in two ways the work done by the forces as the particle moves from P to 0. 14. In Ex. 11, the weight is let fall from a height h above B. Determine the greatest extension of the rubber band. 15. A bead is strung on a smooth straight wire, and is '^" attached by a rubber band of very short natural length to a point in the perpendicular bisector of the wire. Taking the wire as axis of 2/, show that, if gravity can be neglected, the equation of / J5-I- h 0- a APPLICATIONS OF DIFFERENTIAL EQUATIONS 335 motion of the bead is approximately Discuss the motion completely. 16. A particle is acted upon by a force of repulsion from a point O proportional to the distance from 0. Neglecting gravity, write the equation of motion and solve it completely, taking x = and v =Vq when t = 0. Discuss the solution. 17. In Ex. 16, find the work done in the first 10 ft. of the motion. Check by the principle of kinetic energy and work. 18. In Ex. 16, find the impulse of the force during the first second. Check by the principle of impulse and momentum. 19. It is shown in the theory of attraction that the attraction of a spherical mass on a particle within the mass is directed toward the center of the sphere and is proportional to the distance from the center. Discuss the motion of a particle moving in a straight tube through the center of the earth, if the velocity at the surface is 0. Determine the proportionality constant k^^irom. the fact that the force at the surface is — mg. 20. In Ex. 19, how long does it take the particle to pass through the earth? .4ws. 42^ luinutes. 21. A straight tube is bored through the earth connecting two points of its surface. Show that the equation of motion of a particle sliding in this tube is d^x mq m — = - -^2. X, df^ R where R is the radius of the earth and x is the distance of the particle from the midpoint of the tube. Discuss the motion. Show that the time of passing through such a tube is independent of the posi- tion of the endpoints. II. Plane Curvilinear Motion 232. Rotation. In discussing circular motion, it is usually convenient to take as dependent variable the angle 6 swept out in the time t. The problem of uniformly accelerated circular motion (§ 58) is closely analogous to that of uniformly accelerated 336 CALCULUS rectilinear motion. The equation of motion is evidently where k is the constant angular acceleration. 233. The simple pendulum. A simple pendulum is a point swinging in a vertical circle under the acceleration of gravity. Let P be a particle of mass m connected to the point by a cord or rod of length ?, and denote by 6 the angle between OP and the vertical, by s the length of the arc AP. The effective force acting on P is the component of gravity tan- gent to the circle ; since this is directed opposite to s, it must be given the minus sign. The equation of motion of P is therefore ci) '^'' Fig. 105 m — - = — mg sin 6, dv" But s = ie, so that (1) may be written (2) ml — - = — mg sm v. A first integration of (2) can be performed by the method of § 219 ; the general solution, however, cannot be expressed in terms of elementary functions. We shall therefore consider only the case in which the oscillations are so small that sin 6 may be replaced by (see § 156), and (2) written in the form (3) ml — = — mqO. dfi ^ This equation shows that for small oscillations the motion is approximately simple harmonic. The remainder of the discussion is left to the student. APPLICATIONS OF DIFFERENTIAL EQUATIONS 337 EXERCISES 1. Write the equation of uniform circular motion, and solve it in a variety of ways, explaining the meaning of the constants. Exhibit the results graphically. 2. Proceed as in Ex. 1 for uniformly accelerated circular motion. 3. A wheel is making 400 R.P. M. when a resistance begins to retard its motion at the rate of 10 radians per second. When will it come to rest? How many revolutions will it make before stopping ? 4. Solve equation (3), § 233, taking = Oq and v = when t = 0. For convenience, put ^ = k'\ 6. Show that, in the problem of the simple pendulum, the time of one swing or beat is 9 6. Find the length of the "seconds pendulum" — i.e. a pendulum making one swing per second — at a place where g = 32.17. Ans. 3.2595 ft. 7. Find the angular velocity o in terms of 6 if the oscillations are so large that (2), § 233, must be used. 8. Study the motion of a pendulum making small oscillations, if the resistance of the air is proportional to the velocity. 234. The equations of motion. In the general case of motion in a plane curve, it is convenient to resolve all the applied forces into components parallel to the coord i- nate axes. The product m — - of the mass by the a^com- ponent of the acceleration (see § 59) is equal to the sum Fx of the 2;-components of all the forces; similarly for the y-components. We thus have the two equations of motion : ■F:,, -F„. 338 CALCULUS In the most general case, both F^ and Fy are functions of x^ y^ t, and the velocity-components i;^ = — , ?; = -^. dt dt We shall, however, confine our attention to the case in which F^ is a function only of x^ v^., and ^, and Fy is a function of 7/, Vy^ and t. In this case the two equations of motion may be integrated separately. We thus obtain two equa- tions giving respectively x and y in terms of t ; these are parametric equations of the path of the moving point. By the same methods as those already used we find equa- tions giving v^ and Vy in terms of ^, and in terms of x and y respectively. The total velocity v may be found by § 57. 235. Projectiles. A simple example of curvilinear motion is furnished by a projectile moving under gravity alone — i.e. in a medium whose resistance can be neglected. Let a particle be projected with an initial velocity v^ inclined at an angle a to the horizontal. With the start- ing point as origin and the ^-axis positive upward, the initial conditions are 2: = 0, ?/ = 0, Vx = VqCOs a, Vy = Vq sin a when f = 0. The force of gravity acts vertically downward ; there is no horizontal force. Hence the equations of motion are d^x f. d^y These may be integrated and the constants determined precisely as in our earlier work. EXERCISES 1. Solve the problem of §235 completely, finding a:, y, v^, and Vy in terms of t, v^ in terms of x, and Vy in terms of y. 2. In Ex. 1, by eliminating t from the (x, t)- and (y, ^) -equations, show that the path is a parabola. APPLICATIONS OF DIFFERENTIAL EQUATIONS 339 3. Show that, in the ideal case of § 235, where all resistances are negligible, a projectile whose initial velocity is horizontal will strike the ground in the same time as a body let fall from rest from the same height. 4. The range of a projectile is the distance from the starting point to the point where it strikes the ground. Show that the range on a horizontal plane is 2 i2 = ^ sin 2 a. 9 5. What elevation gives the greatest range on a horizontal plane ? 6. The time of flight is the time from the starting point until the projectile strikes the ground. Show that on a horizontal plane the time of flight is 9 7. A stone is thrown horizontally from the top of a tower 400 ft. high, with a velocity of 20 ft. per second. (a) AVhen, (h) where, and (c) with what velocity does it strike the ground ? Ans. (a) 5 seconds ; (c) 161.2 ft. per second, at 7° 8' to the vertical. 8. Find the work done by gravity in Ex. 7. 9. A stone slides down a roof sloping 30° to the horizon, through a distance of 12 ft. If the lower edge of the roof is 50 ft. high, (a) when, (h) w^here, (c) with what velocity does the stone strike the ground? Ans. (6) 25.1 ft. from the building ; (c) 59.7 ft. per second, at 16" 30' to the vertical. 10. A pitcher throws a ball with a speed of 100 ft. per second, the ball leaving his hand horizontally at a height of 5 ft. Show that under the assumptions of § 235 the ball would strike the ground before reaching the batter 60 ft. away. 11. A particle slides on a smooth roof inclined at 45° to the hori- zontal. If the initial velocity is 10 ft. per second parallel to the edge of the roof and the starting point is 20 ft. above the edge, find when, where, and with what velocity the particle leaves the roof. 12. A particle moves under the action of a force directed toward the origin and proportional to the distance from (cf. §230). If the initial conditions are a: = 10, y = 0, v^ = 0, Vy = 20 ft. per second, discuss the motion completely. Take k = 1. 13. Find the cartesian equation of the path in Ex. 12. 14. In Ex. 12, find the work done in one quarter of the period. Check by the principle of kinetic energy and work. INDEX (The references are to pages.) Acceleration, 80, 279, 322, 323 angular — , 84, 335 components of — , 86-88 in curvilinear motion, 85, 337 of gravity, 81 Algebraic functions, 3 differentiation of — , 19-25 Arc centroid of — , 188 of a plane curve, 167 of a space curve, 250 Area (plane), 121, 148, 175; 176 centroid of — , 185 in cartesian coordinates, 151 in polar coordinates, 154 Asymptotes, 97 tests for — , 98, 99, 105 Catenary, 107, 153, 160, 168, 170 Center of mass, 181 Centroid, 180, 276 of arcs, 188 of areas, 185 of surfaces, 189 of volumes, 187 Characteristic equation, 210 Composition of ordinates, 106 Concavity, 33-36 Conjugate point, 95 Constant of integration, 118, 143, 287 determination of — , 117-118, 322 Continuity, 10, 202 of a function of two variables, 237 Critical point, 33 Critical value, 33 Curvature, 76, 306 center of — , 78 circle of — , 78 radius of — , 78 Curve tracing, 32, 101, 105 by composition of ordinates, 106 in polar coordinates, 114 Cusp, 95 Cycloid, 108, 111, 154, 161, 168, 170 Definite integral, 143 as limit of a sum, 150 change of variable in — , 145 fundamental theorem for — , 150 geometric meaning of — , 144 Density, 179, 274 linear — , 179, 275 surface — , 180, 274 Derivative, 14 as quotient of differentials, 70 geometric meaning of — , 15 higher — , 18, 73 of a function of a function, 21 Derived curves, 42, 81, 325 Difference-quotient, 15 Differential, 70 exact — , 296 of arc, 76 of independent variable, 70 total — , 240 Differential equation, 286 exact — , 296 general solution of — , 287, 305 geometric meaning of — , 290, 304 linear — , 298, 300, 307 order of — , 287 partial — , 286 particular solution of — , 289, 305 singular solution of — , 289 Differentiation, 14 of algebraic functions, 19-25 of exponential functions, 62 of implicit functions, 26, 73, 241- 243 of inverse trigonometric functions, 53 of logarithms, 58 of trigonometric functions, 46-49 standard formulas of — , 65-66 Double integrals, 260, 262, 264 in polar coordinates, 265 transformation of — , 266 Double point, 95 341 342 INDEX Envelope, 252 of normals, 256 of tangents, 254 Epicycloid, 109 Equation of motion, 323 in curvilinear motion, 337 Evolute, 256 Exponential function, 56 differentiation of — , 62 graph of — , 56 imaginary — , 310 Force, 279, 322, 323, 327 distributed — , 280 parallel — , 283 Function, 1 anti-hyperbolic — , 65 branches of — , 4, 51 continuous — , 10, 19, 32, 121, 237 differentiate — , 15, 19, 32 homogeneous — , 294 hyperbolic — , 64 impUcit — , 26, 73, 92, 241-243 increasing and decreasing — , 32 integrable — , 121, 151 inverse — , 27, 51, 56 kinds of — , 3 limit of — , 8, 11, 12, 237 of several variables, 236 one-valued — , 3, 19, 32 rate of change of — , 5, 15 (See also algebraic — , exponential — , etc.) Geometric addition, 82 Geometric derivative, 85 Graphic solution of equations, 107 Heterogeneous masses, 273 Hooke's law, 333 Hypocycloid, 110 of four cusps, 111, 160, 168, 170, 173 Implicit functions, 26, 92 differentiation of — , 26, 73, 241- 243 of several variables, 242-243 Improper integrals, 175-177 geometric meaning of — , 177 Impulse, 326, 327 Indefinite integral, 116-118 geometric meaning of — , 121 Indeterminate forms, 202-205 Infinitesimals, 8, 240 a theorem on — , 159 limit of ratio of — , 8, 15, 70 order of — , 68 principal — , 8 principal part of — , 68, 70, 240 Infinity, 11, 12 Integral curves, 291, 306 Integral tables, 172 Integrating factor, 297, 299, 317 Integration, 116 by parts, 132 by substitution, 123, 134 change of variable in — , 122, 145 of rational fractions, 137-141 standard formulas of — , 126-127 Inverse trigonometric functions, 51 differentiation of — , 53 graphs of — , 51-52 Isolated point, 95 Kinetic energy, 198, 328 of a rotating body, 198 Law of the mean, 201-202 Limit, 6 evaluation of — , 202-208 of function, 8, 11, 12 of function of two variables, 237 of ratio of infinitesimals, 8 of sin a/a, 47 the — e, 60, 230 theorems on — , 7, 159 Line integrals, 163, 167, 168. 170, 188, 189 evaluation of — , 165-167 fundamental theorem for — , 165 geometric meaning of — , 164 Logarithms, 56 common, 63 computation of — , 234 differentiation of — , 58 graph of — , 56 Napierian — , 60 natural — , 60 properties of — , 57-58 Maxima and minima, 33, 102 applications of — , 37-42 in polar coordinates, 114 tests for — , 33-36 Moment of force, 283-284 of inertia, 190-198, 277 of mass, 180, 276 INDEX 343 Momentum, 326, 327 Motion circular — , 83, 335 constrained — , 329 curvHinear — , 82, 85-88, 335-339 equation of — , 323, 337 plane — , 80-88, 321-339 rectilinear — , 80-81, 321-335 simple harmonic — , 330-333, 336 uniform — , 55, 324 uniformly accelerated — , 56, 324 Node, 95 Normal, 29 length of — , 30, 32 to a surface, 245 Normal plane, 249 Osculating circle, 78 Pappus, propositions of, 186, 189 Parametric equations, 72, 83, 167, 253 Partial derivatives, 237 geometric meaning of — , 238 higher — , 238 Partial fractions, 137 Pendulum, simple, 336 Point of inflection, 35, 36, 102 tangent at — , 35 Point of osculation, 95 Pressure, 280-281, 284 center of — , 284 fluid — , 281 Projectiles, 338 Radius of gyration, 190, 191 Radius of inertia, 190 Rate of change, 5, 15, 33, 80, 84, 88 of derivative, 33 Rectification of curves, 167, 250 RoUe's theorem, 200, 226 Rotation, 83, 335 Series, 209 absolute convergence of — , 219 alternating — , 218, 229 computation by — , 228, 234 convergence and divergence of — , 211, 220 geometric — , 6, 209, 210, 211 Maclaurin's — , 222 of n terms, 209 power — , 220 sum of — , 210 Taylor's — , 223 tests for convergence of — , 211- 218, 220 transformation of — , 230-232 Singular points, 92 kinds of — , 95 Slope of curve, 5, 15, 32 polar—, 112 Subnormal, 30, 32 Subtangent, 30, 32 Surfaces angle between lines and — , 240 angle between — , 246 centroids of — , 189 cylindrical — , 170 general — , 268 normal to — , 245 of revolution, 168 tangent plane to — , 244 Symmetry, 101, 114 Tangent, 29 determination of — by inspection, 93 inflectional — , 35, 102 length of — , 30, 32 stationary — , 35 to space curve, 248, 250 Tangent plane, 244 Taylor's theorem, 227 Time-rates, 88-91 Trigonometric functions, 45 differentiation of — , 46-49 elementary properties of — , 45 graphs of — , 45-46 Triple integrals, 270-272 Triple point, 95 Vector, 82 components of — , 82 derivative, 85 resultant of — , 82 Velocity, 80, 84, 320, 324 angular — , 83, 84 components of — , 83 in curvilinear motion, 82, 338 Volumes centroids of — , 187, 276 general — , 161, 258-262, 265, 272 of revolution, 156-158, 266 under a surface, 258-262, 265 Work, 327, 328 Printed in the United States of America. W^'' < DAY USE 14 DAY USE LOAN DEPT. Renewed books are sub^ectM "» "t ^P6iz«R« LD 2lA-60m-10.'65 (F7763sl0)476B neneral Library '4 . ,