w n 4 + + * * I >■ * >$ »| 4- V ^ 4 ¥ *\ ■ 4- % Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/elementarytextboOOsnydrich ELEMENTARY TEXTBOOK ON THE CALCULUS BY VIRGIL SNYDER, Ph.D. AND JOHN IRWIN HUTCHINSON, Ph.D Of Cornell University NEW YORK •:• CINCINNATI •:• CHICAGO AMERICAN BOOK COMPANY THE MODERN MATHEMATICAL SERIES. Lucien Augustus Wait, {Senior Professor of Mathematics in Cornell University ,) General Editor. This series includes the following works : BRIEF ANALYTIC GEOMETRY. By J. H. Tanner and Joseph Allen. ELEMENTARY ANALYTIC GEOMETRY. By J. H. Tanner and Joseph Allen. DIFFERENTIAL CALCULUS. By James McMahon and Virgil Snyder. INTEGRAL CALCULUS. By D. A. Murray. DIFFERENTIAL AND INTEGRAL CALCULUS. By Virgil Snyder and J. I. Hutchinson. ELEMENTARY TEXTBOOK ON THE CALCULUS. By Virgil Snyder and J. I. Hutchinson. HIGH SCHOOL ALGEBRA. By J. H. Tanner. ELEMENTARY ALGEBRA. By J. H. Tanner. ELEMENTARY GEOMETRY. By James McMahon. COPYRIGttl', 1912, BY AMERICAN BOOK COMPANY EL. CALCULUS. W. P. 1 Q fV3 03 53/ PREFACE The present volume is the outgrowth of the requirements for students in engineering and science in Cornell University, for whom a somewhat brief but adequate introduction to the Calculus is prescribed. The guiding principle in the selection and presentation of the topics in the following pages has been the ever increasing pressure on the present-day curriculum, especially in applied science, to limit the study of mathematics to a minimum of time and to the topics that are deemed of most immediate use to the professional course for which it is preparatory. To what extent it is wise and justifiable to yield to this pressure it is not our purpose to discuss. But the constantly accumulating details in every pure and applied science makes this attitude a very natural one towards mathematics, as well as towards several other subjects which are subsidiary to the main object of the given course. This desire to curtail mathematical training is strikingly evidenced by the numerous recent books treating of Calculus for engineers, for chemists, or for various other professional students. Such books have no doubt served a useful purpose in various ways. But we are of the opinion that, in spite of the unquestioned advantages of learning a new method by means of its application to a specific field, a student would ordinarily acquire too vague and inaccurate a command of the fundamental ideas of the Calculus by this one-sided presenta- tion. While a suitable illustration may clear up the difficulties 3 262792 4/ ( • ;/f ;'l :*•/;• \ * ., 'breface of an abstract theory, too constant a dwelling among applica- tions alone, especially from one point of view, is quite as likely to prevent the learner from grasping the real significance of a vital principle. In recognition of the demand just referred to, we have made special effort to present the Calculus in as simple and direct a form as possible consistent with accuracy and thoroughness. Among the different features of our treatment, we may single out the following for notice. The derivative is presented rigorously as a limit. This does not seem to be a difficult idea for the student to grasp, espe- cially when introduced by its geometrical interpretation as the slope of the line tangent to the graph of the given func- tion. For the student has already become familiar with this notion in Analytic Geometry, and will easily see that the analytic method is virtually equivalent to a particular case of the process of differentiation employed in the Calculus. In order to stimulate the student's interest, easy applications of the Differential Calculus to maxima and minima, tangents and normals, inflexions, asymptotes, and curve tracing have been introduced as soon as the formal processes of differentia- tion have been developed. These are followed by a discussion of functions of two or more independent variables, before the more difficult subject of infinite series is introduced. In the chapter on expansion, no previous knowledge of series is assumed, but conditions for convergence are discussed, and the criteria for determining the interval of convergence of those series that are usually met with in practice are derived. A chapter on the evaluation of indeterminate forms and three chapters on geometric applications furnish ample illus- PREFACE 5 tration of the uses of infinite series in a wide range of problems. By reason of its significance in applications, it does not seem advisable to omit the important principle of rates. Arising out of the familiar notion of velocity, it affords an early glimpse into applications of the Calculus to Mechanics and Physics. We do not propose to make the Calculus a treatise on Mechanics, as seems to be the tendency with some writers; but a final chapter on applications to such topics of Mechanics as are easy to comprehend at this stage is thought advisable and sufficient. Especially in treating of center of gravity, the formulas have been derived in detail, first for n particles, and then, by a limit- ing process, for a continuous mass. This was considered the more desirable, as textbooks in applied mathematics frequently lack in rigor in discussing the transition from discrete particles to a continuous mass. Besides, the derivation of these formu- las affords a very good application of the idea of the definite integral as the limit of a sum. This idea has been freely and consistently used in the derivation of all applied formulas in the Integral Calculus. However, as the formula for the length of arc in polar coordinates is especially difficult of derivation by this method, we have deduced it from the corresponding formula for rectangular coordinates by a transformation of the variable of integration. In-order to make the number of new ideas as few as possible, the notions of infinitesimals and orders of infinitesimals have been postponed to the last article on Duhamel's principle. This principle seems to flow naturally and easily from the need of completing the proof of the formulas for center of gravity. The teacher may omit this article, but its presence should at 6 PREFACE least serve the important end of calling the attention of the student to the fact that there is something yet to be done in order to make the derivations complete. Some teachers will undoubtedly prefer to do a minimum amount of work in formal integration and use integral tables in the chapters on the applications. For such the first chapter of the Integral Calculus might suffice for drill in pure integration. The problems in this chapter are numerous, and, for the most part, quite easy, and should furnish the student a ready insight into the essential principles of integration. The characteristic features of the books on the Calculus previously published in this series have been retained. The extensive use of these books by others, and a searching yearly test in our own classroom experience convince us that any far- reaching change could not be undertaken without endangering the merits of the book. The changes that have been made are either in the nature of a slight rearrangement, or of the addi- tion of new illustrative material, particularly in the applications. We wish to acknowledge our indebtedness to our colleagues, who have added many helpful suggestions ; to Professor I. P. Church, of the College of Civil Engineering, for a number of very useful problems in applications of integration (See Exs. 14-18, pp. 318-320, and Exs. 6-7, pp. 323-324), and particu- larly to Professor James McMahon, who has carefully read all the manuscript, assisted throughout in the proof reading, and made many improvements in the text. CONTENTS DIFFERENTIAL CALCULUS CHAPTER I Fundamental Principles ARTICLE PAGE 1. Elementary definitions 15 2. Illustration : Slope of a tangent to a curve . . . .16 3. Fundamental theorems concerning limits 17 4. Continuity of functions 19 5. Comparison of simultaneous increments of two related variables 20 6. Definition of a derivative 21 7. Process of differentiation 22 8. Differentiation of a function of a function .... 23 CHAPTER II Differentiation of the Elementary Forms 9. Differentiation of the product of a constant and a variable 25 10. Differentiation of a sum . 26 11. Differentiation of a product 27 12. Differentiation of a quotient 28 13. Differentiation of a commensurable power of a function . 29 14. Differentiation of implicit functions 33 15. Elementary transcendental functions .... 34 16. Differentiation of log a x and log a u 34 17. Differentiation of the simple exponential function 36 18. Differentiation of an incommensurable power . 37 19. Limit of ^— as 6 approaches 38 20. Differentiation of sin u 39 21. Differentiation of cos u 40 22. Differentiation of tan u 40 23. Differentiation of sin -1 u 42 24. Table of fundamental forms 44 CONTENTS CHAPTER III Successive Differentiation ARTICLE PAGE 25. Definition of the nth derivative 47 26. Expression for the nth derivative in certain cases ... 49 CHAPTER IV Maxima and Minima 27. Increasing and decreasing functions 51 28. Test for determining intervals of increasing and decreasing . 51 29. Turning values of a function 53 30. Critical values of the variable 55 31. Method of determining whether <f>'(x) changes its sign in pass- ing through zero or infinity ....... 55 32. Second method of determining whether <p'(x) changes its sign in passing through zero ... ..... 57 33. The maxima and minima of any continuous function occur alternately .59 34. Simplifications that do not alter critical values .... 59 35. Geometric problems in maxima and minima .... 60 CHAPTER V Rates and Differentials 36. Rates. Time as independent variable 68 37. Abbreviated notation for rates 72 38. Differentials often substituted for rates 74 39. Theorem of mean value 74 CHAPTER VI Differential of an Area, Arc, Volume, and Surface of Revolution 40. Differential of an area 78 41. Differential of an arc 79 42. Trigonometric meaning of — , — .80 dx dy 43. Differential of the volume of a surface of revolution ... 81 CONTENTS ARTICLE 44. Differential of a surface of revolution 45. Differential of arc in polar coordinates 46. Differential of area in polar coordinates 81 82 83 CHAPTER VII Applications to Curve Tracing 47. Equation of tangent and normal .... 48. Length of tangent, normal, subtangent, and subnormal 49. Concavity upward and downward 60. Algebraic test for positive and negative bending 51. Concavity and convexity toward the axis . 52. Hyperbolic and parabolic branches . 53. Definition of a rectilinear asymptote . 85 85 89 90 94 95 96 Determination of Asymptotes 54. Method of limiting intercepts 96 55. Method of inspection. Infinite ordinates, asymptotes parallel to axes 97 56. Method of substitution. Oblique asymptotes .... 99 57. Number of asymptotes 102 Polar Coordinates 58. Meaning of p — 104 dp 59. Relation between ^ and p— 105 dx dp 60. Length of tangent, normal, polar subtangent, and polar sub- normal 105 CHAPTER VIII Differentiation of Function 61. Definition of continuity 62. Partial differentiation 63. Total differential 64. Total derivative . • 65. Differentiation of implicit functions 66. Geometric interpretation . 67. Successive partial differentiation 68. Order of differentiation indifferent s of Two Variables 109 110 112 115 116 118 122 122 10 CONTENTS CHAPTER IX Change of Variable article page 69. Interchange of dependent and independent variables . . 124 70. Change of the dependent variable 125 71. Change of the independent variable 126 72. Simultaneous changes of dependent and of independent variables 126 CHAPTER X Expansion of Functions 73. Convergence and divergence of series 132 74. General test for convergence , 133 75. Interval of convergence 138 70. Remainder after n terms , . 140 77. Maclaurin's expansion of a function in a power series . .141 78. Taylor's series 148 79. Rolle's theorem 150 80. Form of remainder in Maclaurin's series 150 81. Another expression for the remainder 153 CHAPTER XI t Indeterminate Forms 82. Definition of an indeterminate form » 157 83. Indeterminate forms may have determinate values . . . 158 84. Evaluation by development 160 85. Evaluation by differentiation ....... 161 86. Evaluation of the indeterminate form g- 165 CHAPTER XII Contact and Curvature 87. Order of contact 167 88. Number of conditions implied by contact 168 89. Contact of odd and of even order 169 90. Circle of curvature 172 91 . Length of radius of curvature ; coordinates of center of curvature 1 72 92. Limiting intersection of normals 174 93. Direction of radius of curvature . . . ' • • 175 CONTENTS 11 ARTICLE PAGE 94. Total curvature of a given arc ; average curvature . . . 176 95. Measure of curvature at a given point 177 . 96. Curvature of an arc of a circle 178 97. Curvature of osculating circle 178 98. Direct derivation of the expressions for k and B in polar co- ordinates 180 EvOLUTES AND INVOLUTES 99. Definition of an evolute 182 100. Properties of the evolute 184 CHAPTER XIII Singular Points 101. Definition of a singular point 102. Determination of singular points of algebraic curves 103. Multiple points 104. Cusps ......... 105. Conjugate points 191 191 193 194 197 CHAPTER XIV Envelopes 106. Family of curves 107. Envelope of a family of curves 108. The envelope touches every curve of the family 109. Envelope of normals of a given curve 110. Two parameters, one equation of condition 200 201 202 203 204 12 CONTENTS INTEGRAL CALCULUS CHAPTER I General Principled oe Integration ARTICLE PAGE 111. The fundamental problem 209 112. Integration by inspection 210 113. The fundamental formulas of integration .... 211 114. Certain general principles 212 115. Integration by parts 216 116. Integration by substitution 219 117. Additional standard forms 222 118. Integrals of the forms C(^x + B)dx and C ( Ax + B)dx # »M J ax 2 + bx + c J ^/ ax -2 + bx + c 119. Integrals of the forms f — and J (Ax + B) Vax 2 + bx + c dx 5 (Ax + B) 2 Vax 2 + bx + c 121. Decomposition of rational fractions 122. Case I. Factors of the first degree, none repeated 123. Case II. Factors of the first degree, some repeated 124. Case II J. Occurrence of quadratic factors, none repeated 125. Case IV. Occurrence of quadratic factors, some repeated 126. General theorem 226 CHAPTER II 120. Reduction Formulas 229 CHAPTER III Integration of Rational Fractions 238 240 241 243 245 247 CHAPTER IV Integration by Rationalization 127. Integration of functions containing the irrationality y/ax + b 248 128. Integration of expressions containing Vax 2 + bx + c . . 249 129. The substitution V± t 1 ± k 2 -z 253 CONTENTS 13 CHAPTER V Integration of Trigonometric Functions ARTICLE 130. Integration by substitution 181. Integration of ( sec 2n a; dx, \ csc 2 »x dx 132. Integration of I sec TO x tan' 2n+1 x dx, j csc m x coV^+h; dx 133. Integration of i tan"x dx, \ cot n x dx 134. Integration of ( sin'"x cos n x dx 135. Integration of f — , f- J a + b cos nx J a k dx b sin nx dx + b sin nx -f c cos nx 136. Integration of I e ax sin wxcJx, je^cos nxdx PAGE 255 255 257 258 260 264 266 CHAPTER VI Integration as a Summation. Areas 137. Areas 268 138. Expression of area as a definite integral 270 139. Generalization of the area formula 273 140. Certain properties of definite integrals 274 141. Maclaurin's formula 276 142. Remarks on the area formula - 277 143. Precautions to be observed in evaluating definite integrals . 283 144. Calculation of area when x and y are expressible in terms of a third variable 289 145. Areas in polar coordinates 291 146. Approximate integration. The trapezoidal rule . . . 292 147. Simpson's rule 294 148. The limit of error in approximate integration .... 295 14 CONTENTS CHAPTER VII Geometrical Applications ARTICLE PAGE 149. Volumes by single integration 298 150. Volume of solid of revolution 302 151. Lengths of curves. Rectangular coordinates .... 306 152. Lengths of curves. Polar coordinates 309 153. Measurement of arcs by the aid of parametric representation . 310 154. Area of surface of revolution 312 155. Various geometrical problems leading to integration . 315 CHAPTER VIII Successive Integration 156. Functions of a single variable . 157. Integration of functions of several variables 158. Integration of a total differential 159. Multiple integrals . . . 160. Definite multiple integrals 161. Plane areas by double integration . 162. Volumes 321 324 326 328 329 330 334 CHAPTER IX Some Applications of Integral Calculus to Problems of Mechanics 163. Liquid pressure on a plane vertical wall . . . . 338 164. Center of gravity . 340 165. Moment of inertia . . 346 166. Duhamel's theorem . . 348 Trigonometric Formulas . 352 Logarithmic Formulas . 353 DIFFERENTIAL CALCULUS ^XKc CHAPTER I FUNDAMENTAL PRINCIPLES 1. Elementary definitions. A constant number is one that retains the same value throughout an investigation in which it occurs. A variable number is one that changes from one value to another during an investigation. If the variation of a number can be assigned at will, the variable is called independent; if the value of one number is determined when that of another is known, the former is called a dependent variable. The depend- ent variable is called also a function of the independent variable. E.g., 3 x 2 , 4vx — 1, cos x, are all functions of x. Functions of one variable x will be denoted by the symbols /(#), <f>(x), • ••, which are read as "/of x" " <f> of x" etc. ; simi- larly, functions of two variables, x, y, will be denoted by such expressions as f(?,y),F(x,y), •••• When a variable approaches a constant in such a way that the difference between the variable and the constant may be- come and remain smaller than any fixed number, previously assigned, the constant is called the limit of the variable. 15 16 DIFFERENTIAL CALCULUS 2. Illustration : Slope of a tangent to a curve. To obtain the slope of the tangent to a curve at a point P upon it, first take the slope of the line joining P = (x ly y x ) to another point (x 2 , y 2 ) upon the curve, then determine the limiting value of the slope m as the second point approaches to coincidence with the first, always remaining on the curve. Ex. 1. Determine the slope of the tangent to the curve 2-FM + fc at the point (2, 4) upon it. Here, x\ = 2, y\ = 4. Let x 2 = 2 + h, yi = 4 + k, where h, k are so related that the point (x 2 , y*) lies on the curve. Thus 4 + k = (2 + h)\ or h = 4 A + A 2 - (1) The slope m = y* - x 2 - -Xi becomes 4+ Tc 2 + h -4 _ 2 k Fig. 1 which from (1) may be written in the form k = 4 + h. (2) The ratio k : h measures the slope of the line joining (x h y x ) to (ar 2 , ys) • When the second point approaches the first as a limiting position, the first member of equation (2) assumes the indeterminate form -, but the second member approaches the definite limit 4. When the two points approach coincidence, a definite slope 4 is obtained, which is that of the tangent to the curve y = x 2 at the point (2, 4). It may happen that h, k appear in both members of the equation which defines the slope, as in the next example. FUNDAMENTAL PRINCIPLES 17 Fig. 2 Ex. 2. If x 2 + y 1 — « 2 ? find the slope of the tangent at the point Oi> yd- Since Xl * + i/! 2 = a 2 , (asi + ny+ (t/1 + ky = a\ hence 2 hx 1 + A 2 + 2 /fr/i + fc 2 = 0, from which - = — - — — — h 2 ?/i + k k To obtain the limit of -, put h, k h each equal to zevo in tlie second member. lim * = _*!. h±o h ?/i This step is more fully justified in the next article. This result agrees with that obtained by elementary geome- try. The slope of the radius to the circle a 2 + y 2 = a 2 through the point (x lf y x ) is — , and the slope of the tangent is the nega- tive reciprocal of that of the radius to the point of tangency, since the two lines are perpendicular. 3. Fundamental theorems concerning limits. The following theorems are useful in the processes of the Calculus. Theorem 1. If a variable a approaches zero as a limit, then lea will also approach zero, k being any finite constant. That is, if a = 0, then Jca = 0. For, let c be any assigned number. By hypothesis, a can be- come less than -, hence ka can become less than c, the arbi- k ■ ' * For convenience, the symbol = will be used to indicate that a variable approaches a constant as a limit; thus the symbolic form x = a is to be read " the variable x approaches the constant a as a limit." el. calc — 2 18 DIFFERENTIAL CALCULUS trary, assigned number, hence ka approaches zero as a limit. (Definition of a limit.) Theorem 2. Given any finite number n of variables a, (3, y, •••, each of which approaches zero as a limit, then their sum will approach zero as a limit. For the sum of the n variables does not at any stage numerically exceed n times the largest of them, which by Theorem 1 approaches zero. Theorem 3. If each of two variables approaches zero as a limit, their product will approach zero as a limit. More gen- erally, if one variable approaches zero as a limit, then its product with any other variable having a finite limit will have the limit zero, by Theorem 1. Theorem 4. If the sum of a finite number of variables is variable, then the limit of their sum is equal to the sum of their limits ; i.e., lim (x + y + • • •) = lim x + lim y + For, if x = a, y = b, • • •, then x = a + a, y = b -\- (3, •-•, wherein a = 0, fi = 0, •• • ; (Def . of limit) hence x + y+ ••• = (o+ &+ •••) + (« + fi+ •••)> but a + p+->- =0, (Th. 2) hence, from the definition of a limit, lim (x + y + •••) = a-\-b-\- ••• = lim x -f- lim y + •••. Theorem 5. If the product of a finite number of variables is .variable, then the limit of their product is equal to the product of their limits. For, let x = a + a, y = b+($, wherein a = 0, (3 = 0, so that lim x = a, lim y = b. FUNDAMENTAL PRINCIPLES 19 Form the product xy = (a + a)(b + fi) = ab + «6 -f /3a + «0. Then lim sc?/ = lim (ab + ab + (5a + a(5) = ab + lim ab + lim 0a + lim a/3 (Th. 2) = ab. (Th. 1) Hence lim xy = lim a; • lim y. In the case of a product of three variables x, y, z, we have lim xyz = lim xy • lim z (Th. 5) = lim x lim y lim 3, and so on, for any finite number of variables. Theorem 6. If the quotient of two variables as, y is vari- able, then the limit of their quotient is equal to the quotient of their limits, provided these limits are not both infinite or not both zero. (Th. 5) y lim y 4. Continuity of functions. When an independent variable x, in passing from a to b, passes through every intermediate value, it is called a continuous variable. A function f(x) of an independent variable x is said to be continuous at any value x l} or in the vicinity of x Xi when f(x^) is real, finite, and determi- nate, and such that in whatever way x approaches a^, From the definition of a limit it follows that corresponding to a small increment of the variable, the increment of the For, since X x = -y, y lim x = lim - lim y y y and hence ,. x lim x lim - = 20 DIFFERENTIAL CALCULUS function is also small, and that corresponding to any number c, previously assigned, another number 8 can be determined, such that when h remains numerically less than 8, the differ- is numerically less than c. 2/,+e a>.+5 Fig. 3 Thus, the function of Fig. 3 is continuous between the values x x and x Y -f- 8, while the functions of Fig. 4 and Fig. 5 are dis- continuous. In the former of these two the function becomes infinite at x = c, while in the latter the difference between the value of the function at c + h and c — h does not approach zero with h, but approaches the finite value AB as h ap- proaches zero. When a function is continuous for every value of x between a and b, it is said to be continuous within the interval from a to b. 5. Comparison of simultaneous increments of two related vari- ables. The illustrations of Art. 2 suggest the following general procedure for comparing the changes of two related variables. Starting from any fixed pair of values x 1} y^ represented graph- ically by the abscissa and ordinate of a chosen point P on a given curve whose equation is given, we change the values of FUNDAMENTAL PRINCIPLES 21 x and y by the addition of small amounts h and k respectively, so chosen that the new values x L + h and y l + k shall be the coordinates of a point P 2 on the curve. The amount h added to x x is called the increment of x and is entirely arbitrary. Likewise, k is called the increment of y ; it is not arbitrary but depends upon the value of h ; its value can be calcu- lated when the equation of the curve is given, as is shown by equation (1). These increments are not necessarily positive. In the case of continuous functions, h may always be taken positive. The sign of k will then depend upon the function under consideration. The slope of the line PjP 2 is then - and the slope of the tangent line at Pj is the limit of - as h and consequently k approach zero. The determination of the limit of the ratio of k to h as h and k approach zero is the fundamental problem of the Differential Calculus. The process is systematized in the following ar- ticles. While the related variables are here represented by ordinate and abscissa of a curve, they may be any two related magnitudes, such as space and time, or volume and pressure of a gas, etc. 6. Definition of a derivative. If to a variable a small incre- ment is given, and if the corresponding increment of a contin- uous function of the variable is determined, then the limit of the ratio of the increment of the function to the increment of the variable, when the latter increment approaches the limit zero, is called the derivative of the function as to the variable. 22 DIFFERENTIAL CALCULUS k That is, the derivative is the limit of - as h approaches zero, or liin (k For the purpose of obtaining a derivative in a given case it is convenient to express the process in terms of the following steps: 1. Give a small increment to the variable. 2. Compute the resulting increment of the function. 3. Divide the increment of the function by the increment of the variable. 4. Obtain the limit of this quotient as the increment of the variable approaches zero. 7. Process of differentiation. In the preceding illustrations, the fixed values of x and of y have been written with sub- scripts to show that only the increments h, k vary during the algebraic process of finding the derivative, also to emphasize the fact that the limit of the ratio of the simultaneous incre- ments h, k depends upon the particular values which the variables x, y have, when they are supposed to take these in- crements. With this understanding the subscripts will hence- forth be omitted. Moreover, the increments h, k will, for greater distinctness, be denoted by the symbols Ax, Ay, read " increment of x," " increment of y." If the four steps of Art. 6 are applied to the function y = <£ (x), the results become y + £fy=<f>(x + \x), Ay = <j>(x + Ax) — <f>(x) = A<f> (x), Ay _<f>(x + Ax) — <j> (x) _ A<£ (x) Ax Ax Ax . lira - A J = Km { +(* + **)- »(*) | =Um Aj>^ A# Ax Ax ' FUNDAMENTAL PRINCIPLES 23 The operation here indicated is for brevity denoted by the symbol — , and the resulting derivative function by <f>'(x); thus dx dy _d<f>(x) _ lim f <f>(x + Ax)-<j>(x) ' dx dx Aa; = Ax = +-(*). The new symbol -^ is not (at the present stage at least) to ax be looked upon as a quotient of two numbers dy, dx, but rather as a single symbol used for the sake of brevity in place of the expression " derivative of y with regard to x." The process of performing this indicated operation is called the differentiation of <f> (x) with regard to x. EXERCISES Find the derivatives of the following functions with regard to x. 5. I. X 3 6. x n , n being a positive integer. 7-2 7. 1. x 2 - 2x-, 2x\ 3; x. 2. 3x*-4:x + 3. 3. 1 4*' 4. **-2 + i. X 2 9. y = Vx. 10. y = x~$. 8. ar+1 x f 1 [Put # 2 = x, and apply the rules.] 8. Differentiation of a function of a function. Suppose that y, instead of being given directly as a function of x, is expressed as a function of another variable u, which is itself expressed as a function of x. Let it be required to find the derivative of y with regard to the independent variable x. Let y =f(u), in which u is a function of x. When x changes to the value au + Aaj, let u and y, under the given relations, 24 DIFFERENTIAL CALCULUS change to the values u + Aw, y + A?/. Then A?/ _ Aw Aw , Aa; — Aw Ax hence, equating limits (Th. 5, Art. 3), dy _dy da _ df(u) du dx ~~ du dx~ du dx This result may be stated as follows : The derivative of a function of u with regard to x is equal to the product of the derivative of the function with regard to w, and the derivative of u with regard to x. EXERCISES 1. Given v = 3u 2 -l, M = 3x 2 + 4; find ^- dy du du dx dx du dx K ' 2. Given ?/ =3m 2 -4u+ 5,« = 2x 3 -5; find ^ • 3. Given y = -,w = 5a; 2 -2x + 4; find ^ • 1 -r 3 3 _ , rfv 3 m 2 3 a: 3 aar CHAPTER II DIFFERENTIATION OF THE ELEMENTARY FORMS dv In recent articles, the meaning of the symbol -f was ex- ctx plained and illustrated ; and a method of expressing its value, as a function of x, was exemplified, in cases in which y was a simple algebraic function of x, by direct use of the definition. This method is not always the most convenient one in the dif- ferentiation of more complicated functions. The present chapter will be devoted to the establishment of some general rules of differentiation which will, in many cases, save the trouble of going back to the definition. The next five articles treat of the differentiation of algebraic functions and of algebraic combinations of other differentiable functions. 9. Differentiation of the product of a constant and a variable. Let y = cu, Then y + Ay = c(u + Au), A?/ = c(m + Au) — cu = cAu, Ay Au Ax Ax' therefore dy du dx~~ dx Thus d(cu) _ du dx dx 25 (1^ 26 DIFFERENTIAL CALCULUS The derivative of the product of a coyistant and a variable is equal to the constant multiplied by the derivative of the variable. 10. Differentiation of a sum. Let 2/ == M_[_<y_ W ;_j_ ... in which u. v, w, ••> are functions of x. Then y + Ay = u + Au + v + Av — w — Aw + • • •, Ay = Au + A?; — Aiv + • • •, Ay _ Au ,Av_ Aw Ax Ax Ax Ax ' dy _du dv dw dx dx dx dx Hence -f(u + v- w+ . ..)=f^+^-^+ .. (2) doc doc doc doc The derivative of the sum of a finite number of fractions is equal to the sum of their derivatives. Cor. If y = u + c, c being a constant, then y + Ay = u + Au + c, hence Ay = Au, and dy = du dx dx This last equation asserts that all functions which differ from each other only by an additive constant have the same derivative. Geometrically, the addition of a constant has the effect of moving the curve y = u(x) parallel to the y-axis ; this opera- tion will obviously not change the slope at points that have the same x. -c /rtN dy du , dc From (2), -f- = — + — ; dx dx dx DIFFERENTIATION OF THE ELEMENTARY FORMS 27 but from the fourth equation above, dy _du % dx dx' dc hence, it follows that — = 0. dx The derivative of a constant is zero. If the number of functions is infinite, Theorem 4 of Art. 3 may not apply; that is, the limit of the sum may not be equal to the sum of the limits, and hence the derivative of the sum may not be equal to the sum of the derivatives. Thus the derivative of an infinite series cannot always be found by differentiating it term by term. 11. Differentiation of a product. Let y = uv, wherein u, v are both functions of x. Then ^=( U + * U W + * V )- UV = u ^ + v^ + ^ . to,. Ax Ax Ax Ax Ax Now let Aa; approach zero, using Art. 3, Theorems 4, 5, and noting that if — has a finite limit, then the limit of Avf—) Ax \AxJ is zero. ' The result may be written in the form d(uv) = u dv + v du (3) doc dx doc The derivative of the product of two functions is equal to the sum of the products of the first factor by the derivative of the sec- ond, and the second factor by the derivative of the first. This rule for differentiating a product of two functions may be stated thus : Differentiate the product, treating the first factor as constant, then treating the second factor as constant, and add the two results. 28 DIFFERENTIAL CALCULUS Cor. To find the derivative of the product of three functions uvw. Let y = uvw. By (3), *y = w ±(uv)+uv^ dx dx dx = w(u dv du\ dw dx dx ) dx The result may be written in the form d(uvw) =uv dw + vw du + wu dv (4 doc dx dm dx By induction the following rule is at once derived : The derivative of the product of any finite number of factors is equal to the sum of the products obtained by midtiplying the de- rivative of each factor by cdl the other factors. 12. Differentiation of a quotient. Let y = - , u, v both being functions of x. Then u -f Au u Au Av ! V U A?/ v -\- Av v _ Ax Ax Ax ~~ Ax v(y + Av) Passing to the limit, we obtain the result v du- u dv d (u\- dx dx (5) dx\v J v 1 TJie derivative of a fraction, the quotient of two functions, is equal to the denominator multiplied by the derivative of the nu- merator minus the numerator multiplied by the derivative of the denominator, divided by the square of the denominator. DIFFERENTIATION OF THE ELEMENTARY FORMS 29 13. Differentiation of a commensurable power of a function. Let y = u n , in which it is a function of x. Then there are three cases to consider : 1. n a positive integer. 2. n a negative integer. 3. n a commensurable fraction. 1. n a positive integer. This is a particular case of (4), the factors u, v, w, ••• all being equal. Thus dy n _ x du dx dx 2. n a negative integer. Let n = — m, in which m is a positive integer. Then y = u n = u~ m = — , «* d l = ^- d l by (5), and Case (1) hence dx u 2m dx — mu~ m ~ idu, dx' dy _ dx n-l dtt wit" — • dx 3. 7i a commensurable fraction. Let n=*-, where p, g are both integers, which may be either • q positive or nega tive. p Then y = u n = u 9 ; hence if = fir, and i.e. dec cte 30 DIFFERENTIAL CALCULUS Solving for the required derivative, we obtain dx V dx J hence -*- U n = nu n - 1 4". (6) dx, doc The derivative of any commensurable power of a function is equal to the exponent of the power multiplied by the power with its exponent diminished by unity, multiplied by the derivative of the function. It should be noticed that Vu = it 2 , u hence ±^=±*« > ±(1\ = = 1*». dx 2^/ u dx dx\u) u 2 dx These theorems will be found sufficient for the differentia- tion of any function that involves only the operations of addi- tion, subtraction, multiplication, division, and involution in which the exponent is an integer or commensurable fraction. The following examples will serve to illustrate the theo- rems, and will show the combined application of the general forms (1) to (6). ILLUSTRATIVE EXAMPLES t 3 x 1 - 2 ~ , dy 1. y = ; find -*-• J x + 1 dx (x + l)-f(3 *»- 2) - (3x3 - 2) L (x-r 1) d / = ~ 7—TO ^ ^^ dx (x + l) 2 ± (3^-2)=^ (3^)--f(2) (by 2) dx dx dx = 6x. (by 1,6) f(si-l)=£? = l. (by 2) ax dx DIFFERENTIATION OF THE ELEMENTARY FORMS 31 Substitute these results in the expression for -f- . Then dx dy = (x + 1)6 x - (3 x 2 - 2) = 3 x* + 6 x + 2 . <fc (*+l) 2 (x + 1) 2 2. u = (3s 2 + 2)Vl + 5s 2 ; find ^?. ^ = (3 s 2 + 2) - VI + 5 s 2 + Vl + 5s 2 • — (3 s 2 + 2) . (by 3) ds ds ds — vTToT 2 = — (1 + 5 s 2 )* </* ds . = l(l+5s 2 )-^(l + 5s 2 ) (byG) 2 as 5s Vl + 5 s 2 -^(3s 2 + 2>=6s. (by 6) ds Substitute these values in the expression for — Then ds du = 5j jS* + 2) , 6gVrTT72 = 45*»+.16» rf* Vl + 5s 2 Vl~+5s 2 3 v= vr+^ + vr^g ;find $. Vi + x 2 - vT^ 2 d* First, as a quotient, (VT+T 2 - VI - x *)^(VTTx* + VI^ 2 ) </a; (Vl + z 2 -Vl-x 2 ) 2 (Vi + x 2 + vi - x 2 ) — ( vrr^ 2 - vi - x 2 ) , ^ , (by 5) (Vl + z 2 - Vl-z 2 ) 2 — ( vr+T 2 + vr^) = -^ vrr^ 2 + — vn^ 2 . (by 2) dx dx dx •-f VIT^ = # (1 + * 2 )* = 1 (1 + * 2 ) _i -7- (1 + * 2 )- (by 6) dx dx 2 dx lL(l+x*)=2x. (by 2 and 6) dx 32 DIFFERENTIAL CALCULUS Similarly for the other terms. Combining the results, we have dx x 3 { VI - X*J Ex. 3 may also be worked by first rationalizing the denominator. EXERCISES Find the ^-derivatives of the following functions : 1. y = x xo . 2. y — x~ s . 3. y = cVx. 4. 2 Vi '' Vx 3 5. y = v / ^'~ 5 - 6. y = (x + a) n . 7. y — x n -\- a n . 8. X y = . 14. y = (2 a\ + x2) V a i + ^i. 15 y-i * r |l + Vl-**J 16. 1.7- :r n -f 1 v — — - — S x n - 1 18 1 1 (a + x) m (b + x) w 19. 3x 3 + 2 v = — - Va 2 - x* X ( x s + !)| 9 y== «jL?. 20. y = 3(x 2 + l)t(4x 2 -3). x 2 + 2 21. y = 3 m" - 7. 10. 2/=(* + l)vW2. 22> y = 4w 3_ 6w2+1 o M _ 3 . 11. Va + a 23. y =(1-3 u 2 + 6u 4 )(l + u 2 ) 3 . v « + V * 24. 7/ = us. i2. y = JT±J 25 - $ =■ M * + 3 * w2 + * 4 - 1 ~ X " 26. y = "" . 13.,= *==. (« + *>• x + Vl — x 2 27. y = t* 2 j? 8 io. 28. Given (a + a:) 5 = a 5 + 5 a 4 x + 10 « 3 -r 2 + 10 a 2 x 3 + 5 a* 4 + x* ; find (a + x) A by differentiation. 29. Show that the slope of the tangent to the curve y = x 3 is never negative. Show where the slope increases or decreases. DIFFERENTIATION OF THE ELEMENTARY FORMS 33 30. Given b 2 x 2 + a 2 y 2 = d 2 b 2 , find d JL : (1) by differentiating as to dx x ; (2) by differentiating as to y ; (3) by solving for y and differen- tiating as to x. Compare the results of the three methods. 31. Show that form (1), p. 25, is a special case of (3), p. 27. 32. At what point of the curve y 2 = ax 3 is the slope ? — 1 ? + 1 ? 33. Trace the curve y = x 3 + 3x 2 + x — 1. 34. y - - 3 ^ 2 + 7 and u = 5 x 2 - 1 ; find (J l . V7 u 2 + 5 tf* 35. At what angle do the curves y 2 = \2x and y 2 + x 2 + ;r — G3 = intersect ? 14. Differentiation of implicit functions. When a functional relation between x and y cannot be readily solved for y, the preceding rules may be applied directly to the implicit function. The derivative will usually contain both x and ?/. Thus the derivative of an algebraic function, defined by equating a poly- nomial in x and y to zero, may be obtained by the process illus- trated in the following examples : Ex. 1. Given the function y of x, defined by the equation x 5 4- y r ° — 5 xy + 1 = 0, find^. dx Since dx - 5 xy + 1) = 0, hence 5x 4 + 5</^-5 . dx y - r d - 5 x ,- = dx o, G>: f2, 3) Solving for -*. , we obtain dx <]y _ dx •'■ 4 x - -7/ Ex. 2. xy 2 -\- x 2 y = 1. Find dy dx Ex. 3. *+ y + (x-y) 2 +(2 x — 3 2/) 3 = 0. Find EL. CALC. — 3 34 DIFFERENTIAL CALCULUS 15. Elementary transcendental functions. The following func- tions are called transcendental functions : Simple exponential functions, consisting of a constant number raised to a power whose exponent is variable, as 4 X , a* ; the logarithmic functions, as log a x, log 6 u ; the incommensurable powers of a variable, as x^ 2 , u n ; the trigonometric functions, as sin u, cos u ; the inverse trigonometric functions, as sin -1 u, tan -1 x. There are still other transcendental functions, but they will not be considered in this book. The next four articles treat of the logarithmic, the two ex- ponential functions, and the incommensurable power. 16. Differentiation of log a od and log a u. Let y = log a x. Phen y-\-Ay = log a (x + Ax), Ay log a (x + Ax) — log a x Ax Ax Lio g /*_±M iX \ X J Ax For convenience writing h for Ax, and rearranging, we obtain Ay Ax _l.«Wl+») x h \ xj H-;»»K'+;)i- -) DIFFERENTIATION OF THE ELEMENTARY FORMS 35 SB f h\* To evaluate the expression ( 1 + - ) when h = 0, expand it by the binomial theorem, supposing - to be a positive inte- ger m. The expansion may be written \ mj m 1-2 m 2 1-2-3 m 3 which can be put in the form \ mj 1 2 ^1 2 3 1 2 Now as m becomes very large, the terms — , — , ••• become in m very small and m increases without limit as h approaches zero. As m = cc the series approaches the limit 1 + 1+— + — + — + •••, 2 ! 3 ! 4 ! which will be discussed later. The numerical value of this limit can be readily calculated to any desired approximation. This number is an important constant, which is denoted by the letter e, and is equal to 2.7182818-..; thus lim A + IV = e = 2.7182818-...* * This method of obtaining e is rather too brief to be rigorous ; it assumes that — is a positive integer, but that is equivalent to restricting Ax to ap- Ax proach zero in a particular way. It also applies the theorems of limits to the sum and product of an infinite number of terms. The proof is completed on p. 315 of McMahon and Snyder's " Differential Calculus," 36 DIFFERENTIAL CALCULUS The number e is known as the natural or Naperian base; and logarithms to this base are called natural or Naperian log- arithms. Natural logarithms will be written without a sub- script, as logcc; for other bases a subscript, as in \og a x, will generally be used to designate the base. The logarithm of e to any base a is called the modulus of the system whose base is a. X li / 7A* If the value , ™ f 1 + - ) = e is substituted in the expres- sion for _?, the result is dx dy 1 , / = - • log, e. ax x More generally, by Art. 8, d , logffle du f ms s log„ M = - s -^. (7) In the particular case in which a = e, d . _ 1 du /ox doc ~~ u dx The derivative of the logarithm of a function is the product of the derivative of the function and the modulus of the system of logarithms, divided by the function. 17. Differentiation of the simple exponential function. Let / y = a u . Then log y = u log a. Differentiating both members of this identity as to x, we obtain 1 dy , du ,, oX dy , du therefore -j- <*> u = l«g a • a ' 1 " ^ ■ ^ DIFFERENTIATION OF THE ELEMENTARY FORMS 37 In the particular case in which a = e, d du , + n x as •"=•"• a*- (10) The derivative of an exponential function with a constant base » is equal to the product of the function, the natural logarithm of the base, and the derivative of the exponent. 18. Differentiation of an incommensurable power. Let y = u n , in which n is an incommensurable constant. Then log y = n log u, 1 dy _ n du ydx u dx' dy_ y du dx u dx' d _ __, du dx dx This has the same form as (6), so that the qualifying word " commensurable " of Art. 13 can now be omitted. EXERCISES Find the x derivatives of the following functions 1. y- = log(x+ a). 2. y- = log (ax + b). 3. y = = log (4 x 2 - 7 x 4-2) 4. y = = logj + *. 1 — X 5. 2/ = 1 — X 2 6. y = = x log X. 7. y = - x n log X. 8. y = - x n log x m . N 9. y = log Vl — X 2 . 10. y = Vx — log ( y/x + 1) . 11. y = \og a (3x 2 -V2 + x). 12. y = log 10 (x 2 + 7 x). 13. y = log x a. 14. y = e xa . 154 y = e 4 *+ 5 . l 16. y=-e l+x . 38 DIFFERENTIAL CALCULUS 17. y = — - — . 18. y = e x (l -z«). -X . — X 19. y 23. 24. log Vq -f Vx e* + e~ x 20. y = log (e* - c-*). 21. y = log (x + e x ). 22. v = x n a*. Va 1 logx' 25. 3/= (log*)*. 26. y = log (log a). 27. y 28. y = a l °s x log - . x The following functions can be easily differentiated by first taking the logarithms of both members of the equations. 31. ¥ =*&+*!. 32. y = x\a -f 3 x)\a - 2 x) 2 . 33. y = V (*+ g )'. Vx — a 29. (*-*)■ (a;-2)*(a;-3)* 30. y = xVl - x(l + a:). 19. Limit of sin as 9 approaches 0. Before proceeding to determine the derivatives of the trigonometric functions it is necessary to prove the following lemma : lim sin = 1. = With as a center and OA = r as radius, describe the circular arc AB. Let the tangent at A -4 meet OB produced in D; draw Fig. 7 BC perpendicular to OA, cutting OA in C. Let the angle OAB = in radian measure, then arc AB = r#, (75 < arc AB < .4Z), by geometry i.e. r sin < r6 < r tan 0, . sin 6 < < tan 0. DIFFERENTIATION OF THE ELEMENTARY FORMS 39 By dividing each member of these inequalities by sin 0, 1 <^-<sec0; sin but sec = 1, when = sin , lim ., A lim sin -, hence > 0=o^ = 1 > and »=o-vr = 1 20. Differentiation of sin ?/. Let y = sin u. ,p. Ay _ sin (u 4- A?/) — sin w Ait A./ 1 Aw A/ To evaluate the expression sin (u + Aw) — sin w, we make use of the formulas for the sine of the sum and the sine of the difference of two angles. Since sin (a + b) = sin a cos b 4- cos a sin b, sin (a — b) = sin a cos 6 — cos a sin 6, hence, by subtracting the second equation from the first, sin (a + b) — sin (a — b) = 2 cos a sin b. This equation is true for all values of a and of b. In particu- lar, then, putting ' ' F 5 a + 6 = w + Aw, and a — b = u, that is, a = u + — , and 6 = — , we obtain sin (w + Aw) — sin w = 2 cos (u-\ — - ) sin — . 40 DIFFERENTIAL CALCULUS The expression for — J. may now be written in the form Ax Aw sm — A?/ / Au\ 2 A?* — ^ = cos [u + — — - , Ax V 2 J Au Ax hence dy .. -f- = cos u • li m dx Aw = sm 2 Aw' Aw 2~ c?w dx~' hence, by Art. 19, ^- sin w = cos u — rfx doc (ID 77>e derivative of the sine of a function is equal to the product of the cosine of the function and the derivative of the function. 21. Differentiation of cos u. Let ?/ = cos u — sinf - — u \ bl dx dx du -w)-^( -u\ dx\2 d — i doc u = — sinu doc (12) The derivative of the cosine of a function is equal to minus the product of the sine of the function and the derivative of the function. 22. Differentiation of tan u. Let Then y = tan u sin?* cos?* d . d cos u • — sin u — sin u • — cos u dy dx dx - (bv 5) dx ~~ cos- u • o du . • Q . du cos J u h sm- U — dx dx du dx (by 11, 12) cos 2 u cos 2 u that is, ^tant* = sec 2 t*^ doc doc (13) DIFFERENTIATION OF THE ELEMENTARY FORMS 41 The derivative of the tangent of a function is equal to the product of the square of the secant of the function and the deriva- tive of the function. Since the remaining elementary trigonometric functions can be expressed as rational functions of those already con- sidered, their derivatives can be obtained by means of the preceding rules. The results are 4-wtu= -csc 2 t*^. (14) dx dx ~ sec u = sec u tan u — • (15) dx dx lL esc u = - esc w cot w — (16) dx dx EXERCISES Find the ^-derivatives of the following functions : 1. y — sin 7 x. 2. y — cos 5 x 3. y = sin x 2 . 4. y = sin 2 x cos x. 5. y = sin 8 x. 6. y = sin 5 x 2 . 7. y = sin 2 7 x. 8. y — \ tan 3 x — tan x. 9. y = sin 3 x cos x. 10. ?/ = tan x + sec x. 11. y = sin 2 (1-2 x 2 ) 2 . 12. # = tan (3 -5 a; 2 ) 2 . 13. y — tan 2 a: — log (sec 2 a;). 14. y =log tan ($# + Jir). ■ 15. y = logsinVx. 29. ?/ — tan (x -f y) . 16. * i y = tan a*. 17. y = sin narsin n a;. 18. y = sin (w + b) cos (w — J). 19. sin m nx y = . cos n mx 20. y = x + log cos ( x - j J. 21. ?/ = sin (sin t/). 22. ?/ = sin 2 ^. 23. y = sin e x • log a\ 24. ?/ = Vsin x 2 . 25. y = esc 2 4 x. 26. y — sec(4 x — 3) 2 . 27. r/ = cot x 2 -f sec Var. 28. y ~ sin a:?/. 42 DIFFERENTIAL CALCULUS 30. Find — (cos u) directly from the definition of the derivative. dx Also — (tan u). dx 31. Find — (cos u) from the relation sin 2 u + cos 2 u = 1. dx 23. Differentiation of sin- 1 u- Let y = sin~ 1 u. Then sin y = u, and, by differentiating both members of this identity, dy ,du cos y — = — ; d# dx hence dy _ 1 fZw 1 d?t. da; cos ?/ d.r _j_ Vl— sin 2 y dx ' d • _i 1 dw i.e. — - sm -1 it = ± d# VI _ tf d^ The ambiguity of sign accords with the fact that sin -1 u is a many- valued function of u, since, for any value of u between — 1 and 1, there is a»series of angles whose sine is u : and, when u receives an increase, some of these angles increase and some decrease; hence, for some of them, sm — - is positive, and du for some negative. It will be seen that, when sin -1 u lies in the first or fourth quarter, it increases with u, and, when in the second or third, it decreases as u increases. Hence, for the angles of the first and fourth quarters, jL Sin- 1 !* = --f- COS" 1 !*:= + ■— 1— gg. (17) €loc doc VI — «*2 dor In the other quarters the minus sign is to be used before the radical. DIFFERENTIATION OF THE ELEMENTARY FORMS 43 The derivatives of the other inverse trigonometric functions can be easily obtained by the method employed in the present article. The most important of the remaining ones are tan -1 u, sec -1 u ; 4- tan" 1 u = - 4- cot 1 u = -L-^. doc ddc l + u 2 doc x sin -1 a;. 2. y = cos -1 Vl — x' 2 . 17 # y 3. y = sin-i (3 x - 1). 18. y = e**" 1 *. 4. y = sin- 1 (3x -4 x 3 ). 1Q _., 1 1 — x 2 5. y = sm-i- — -• 1 + r 6. y = vsin -1 x. 1. y — tan -1 ?*. 8. y = cos -1 log x. 9. y = sin -1 (tan a-). 10. y = sec -1 — VI -x 2 11. # = CSC -1 -• a: 12. y^tan^f 1 V Va: 2 - 1 13. y = tan- 1 -* 14. y = sin" 1 Vsin x. 15. y = tan- 1 V^^ 1 -f cos (18) -^-sec ~ 1 u = --^-csc- 1 m = !__*•. (19) das dx u^u 2 —l (lQC EXERCISES Find the ar-derivatives of each of the following functions : 1. y = sin -1 2 x 2 . 16. y = tan a; • tan -1 x. 2 a; 2 - 1 20. y = sec" 1 *L±i. ^ a; 2 - 1 21. y-tan- lV ^+ V «. 1 — V«ar 22. y = cos x 23. y = tan -1 (n tan a). 24. y = cos -1 (cos 2 a-). 25. y = cos -1 (2 cos a:). 26. y = tan -1 (Vl + x 2 - X). 27. y = 2 tan" 1 aL— ' M + a: 28. y = tan -1 — |- tan iV3 .^b-x a:V3 44 DIFFERENTIAL CALCULUS 24. Table of fundamental forms. d(cu) = c du . (1) dx dx Jt {u + V -. w)= du+dv_dw. 2 dx dx dx dx *&»1 = u^ + v^ (3) dx dx dx d -(uvw) = uv m + vw m + wu <M. (4) dx dx dx dx v du_ u dv d u _ dx dx dx v »,2 (5) ^a w =loga-a u ^ (9) da? da? A e « =e ud ^ (10) <ia? da? -f-sinw =costi~> (11) <?a? da? ^-cosi* =-smu dw . (12) c?a? da? ^tanw =sec*tt^. (13) da? da? -£- cot ft = - esc 2 u d ^ (14) da? da? DIFFERENTIATION OF THE ELEMENTARY FORMS 45 4- sec u = sec u tan u ^- (15) doc doc ^cscu = -cscMcotw~ ( 16 ) doc doc JL sin l u=-4- COS" 1 M = 1 *L (17) «a? «a? Vi _ ^2 «a? 4- tan" 1 t* = - # cot 1 w = —1— *!. (18) dx doc i + w 2 da? 4- sec 1 u=-~- esc 1 t* = -J ^* (19) doc doc u y/u* -i d & EXERCISES ON CHAPTER II Find the ^-derivatives of the following functions : 1. y = 3 Z 2 + 5 a,- 3 - 7. o 3 5 1 3. y = (x + 5) Vx - 3. 4. ?/ = xVa' 2 — x 2 . 5. ?/ = x log sinx. a 10. y = log-- . a* 11. 1-x 2 y = Vl + x 2 12. y = e x cos x. 13. , = 00^(1). 14. . , 4 sin x ?/ = tan -1 3 + 5 cos x 15 v — ( x 4- rtH.an- 1 -!/- 6 - y=-V«2_ x 2. ( r 2 ? > 7. y= c - ec- x 8. J ,= ta„2z,z = t.n-i(2»-l). ^ . 16. w = cot x — — — 2 9. y = e Vu , u = log sin x. a; 17. ?/ = tan 4 x — 2 tan 2 x +.log(sec* x). 18. y = £l2«£+log(l- a; ). 19. y = C0S-i 3 + 5c0S:C . 5 + 3 cos x 46 DIFFERENTIAL CALCULUS 20. y.= log (l±£^-|tan-i*. 21. # = log (x + Vx 2 - a 2 ) + sec" 1 -- a 22. y = e u , u — log x. 25. x 2 ?/ 2 + x 8 + y 3 = 0. 23. y = log s 2 + e", s = sec x. 26. x 3 -f x = y + y 8 . 24. x s + y s — 3 axy = 0. 27. xy 2 + x 2 # — x-\-y. 28. y = sin (2 u - 7), u = log x 2 . 29. By means of differentiation eliminate the constant p from the equation y =px 2 . 30. At what points is the tangent to the curve y = cos x parallel to the x-axis ? 31. Show that the x-derivative of tan -1 \ ~~ cos x is not a func- * 1 + cos X tion of x. x 2 ?/ 2 32. Find at what points of the ellipse h *— — 1 the tangents cut _, . a 2 Z> 2 off equal intercepts on the axes. 33. Find the points at which the slope of the curve y = tan x is twice that of the line y = x. 34. Find the angle which the curves y = sin x and y = cos x make with each other at their point of intersection. CHAPTER III SUCCESSIVE DIFFERENTIATION 25. Definition of nth derivative. When a given function y~<f>(x) is differentiated with regard to x by the rules of Chapter I, then the result d y is a new function of x which may itself be differentiated by the same rules. Thus, 7 , , . , dx\dxj dx The left-hand member is usually abbreviated to _J?, and the dx 2 right-hand member to <f>"(x) ; that is, Differentiating again and using a similar notation, we obtain ±(cPy\_d*i_ dx\dx 2 J dx 3 and so on for any number of differentiations. Thus the sym- d 2 y bol -r^ expresses that y is to be differentiated with regard to x, and that the resulting derivative is then to be differentiated. d s y . . d Similarly, -^ indicates the performance of the operation — three times, j-( j~(;p ))• ^ n general, the symbol -~ means that y is to be differentiated n times in succession with regard to x. 47 48 DIFFERENTIAL CALCULUS Ex. 1. If y = x 4 + sin 2 x, -£ = 4 x 8 + 2 cos 2 x, ufx g = 12x 2 -4sin2x, ?^ = 24x-8cos2x, «x 3 ^ = 24 + 16sin2x. ■ dx 4 If an implicit equation between x and y is given and the derivatives of y with regard to x are required, it is not neces- sary to solve the equation for either variable before perform- ing the differentiation. Ex. 2. Given x 4 + y 4 + 4 a 2 xy = ; find v^ . dx' 2 ^-(x* + y*+±a 2 xy)=0, 4 x 8 + 4 ?/ 3 ^ + 4 a 2 * ^ + 4 a 2 # = 0. The last equation is now to be solved for -j-, dy x s + a 2 y dx y s + a 2 x Differentiating again, we obtain (i) d 2 y _ d fx 8 + a 2 y\ dx 2 ~ dx\y 8 + a 2 x] w (y 8 + a 2 x) — (x 3 + a 2 y) - (x 8 + a*y) JL (y 8 + a 2 x) dx ' dx (y T T~a 2 7y 2 (y 8 + a 2 x) ('dx 2 + a 2 '- 1 !') - (x 8 + a 2 ?/) fsv 2 ^ + a") V '/x/ " V ax / (*/ 3 + d 2 x) 2 SUCCESSIVE DIFFERENTIATION 49 The value of JL from (1) is now to be substituted in the last dx equation, and the resulting expression simplified. The final form may be written : dhj 2 a*xy - 1 a 2 x s y s - a 4 O 4 + y 4 ) - 3 x 2 y 2 O 4 + # 4 ) dx 2 ~ (y 8 + a 2 x) 8 In like manner higher derivatives may be found. 26. Expression for the nth derivative in certain cases. For cer- tain functions, a general expression for the nth derivative can be readily obtained in terms of n. where n is any positive integer. If y = e ax , ~- = a n e ax . Ex. 2. If y '= sin x, dy 1 tt\ ^|=cos:r = sin^+^j, g=cos(x + |) = sin(x+^), 6/ n v . / n7r\ If y~- - sin ax. d n y 1 mr\ d^ = aSln [ ax+ 2-)- EXERCISES ON CHAPTER III 1. y=3x*+5x2+3x-9; find^. 5. ij = tana:; find ^l. dx 3 J dx 8 2. y = 2x 2 + Zx + 5; find v|. 6. y = e*logz; find ^X 3 ' ^ = * ; fmd S' 7."y = *«loga:j find g. 4. y = ^ _I 2 ; find g. 8. y = sec**; find g. EL. CALC. — 4 50 DIFFERENTIAL CALCULUS 9. y = logsina:; find g. 18. 3^^-p find J*. 10. y = sin x cos a; ; find -r\> 19. y z= cos ma: ; find ~^- U. , = ^L ; flnd % 20. y = , X . ; find £». 12. y = ** logs'; find ^. 21. y = log (a + *)"; find J. , 13. y = sin,; find g. 22. y 2 = 2^; find g- 14. y = log (* + *-) ; find g • 23. J + £ = 1 ; find g. 15. y = (^- 3^ + 3)^; find g. 24. a* + y 3 = 3axy; find g- t 16. 3/ = a; 4 log a;; find g. 25. «■+» = zy; find g. 17. y = e ax I find g- 26. y = 1 + xe* ; find ^- d u (in 27. y = «*sina;; prove ^J,-2-^+2y = 0. 28. y = aa;sina;; prove x 2 ~ - 2a; -£ + 2 + 2)y = 0. d 2 y 29. y = ax n + 1 + bx~ n \ prove a; 2 ^J> = n(n + l)y. 30. y =(sin-^)2; prove (1 - z 2 ) ^ 2 - X-+ = 2. »■» = ££ P^g- 1 -* _ , d n y 32. y = x n_1 loga:; find -=-£• 33. „ = *.; prove ^ = 2^-^ + 9«-. 34. y = cos 2 a:; find -^. CHAPTER IV MAXIMA AND MINIMA 27. Increasing and decreasing functions. A function is said to be increasing if it increases as the variable increases and decreases as the variable decreases. A function is said to be decreasing if it decreases as the variable increases and increases as the variable decreases. When the graph of the function is known it will indicate whether the function is increasing or decreasing for an assigned value of x ; conversely, a knowledge of the fact whether a function is increasing or decreasing is of great assistance in drawing the graph. Usually a function is increasing for certain values of x and decreasing for others. 28. Test for determining intervals of increasing and decreasing. Let y — 4>(x) be a continuous function having a derivative for all values of x from a to b. By the above definition y is in- creasing or decreasing at a point x l , according as k = Qfa + h) - <£(zi) has or has not the same sign as h, where h is a sufficiently small number. Hence <f>(x) is an increasing or a decreasing function at the value x x according as is positive or negative. 51 52 DIFFERENTIAL CALCULUS Thus, the function y = <f>(x) is increasing, if <f>'(x) is positive ; if 4>'(x) is negative, the function is decreasing. In order that a function shall change from an increasing function to a decreasing function or vice versa, it is necessary and sufficient that its derivative shall change sign. If the derivative is continuous, this can happen only when the deriva- tive passes through the value zero. The derivative may also change sign when it becomes infinite, and, notwithstanding this discontinuity of the derivative, the original function may still be continuous. In the graph of the function this requires that at such a point the tangent to the locus shall be parallel to the y-axis. The process will be illustrated by a few examples. Ex. Find the intervals in which the function <f>(x) = 2 a; 8 - 9a; 2 + 12a; - 6 is increasing or decreasing. The derivative is <f>'(x) = 6a; 2 - 18 a; + 12 = Q(x - \){x - 2); hence, as x passes from -co to 1, the derived function <f>'(x) is posi- tive and <f>(x) increases from <£( — oo) to <£(1), i>e. from <f>=—ccto<f>= — 1; as x passes from 1 to 2, <f>'(x) is nega- tive, and <f>(x) decreases from <£(1) to <£(2), i.e. from — 1 to - 2; and as x passes from 2 to + co, <£'(*) is posi- tive, and <f>(x) increases from <£(2) to <£(ao), i.e. from — 2 to + oo . The locus of the equation y = <£(a;) is shown in Fig. 8. At points where <f>'(x) = 0, the function <f>(x) is neither increas- ing nor decreasing. At such points the tangent is parallel to the axis of x. Thus in this illustration, at x = 1, x = 2, the tangent is parallel to the x-axis. Fig. 8 MAXIMA AND MINIMA 53 EXERCISES 1. Find the intervals of increasing and decreasing for the function <f>(x) = x s + 2x 2 + x - 4. Here <£'(*) = 3x 2 +4.r+l = (3z+ l)(x + 1). The function increases from x = — coto x = — 1 ; decreases from a: = — ltoar=-— |; increases from a: = — \ to a: = go . 2. Find the intervals of increasing and decreasing for the function y = X s - 2 x 2 + x - 4, and show where the curve is parallel to the a>axis. 3. At how many points can the slope of the tangent to the curve # = 2x 3 -3x 2 + l be 1 ? - 1 ? Find the points. 4. Compute the angle at which the following curves intersect ; y = 3 x 2 - 1, y = 2 x 2 + 3. 29. Turning values of a function. It follows that the values of x at which <f> (x) ceases to increase and begins to decrease are those at which <j>'(x) changes sign from positive to nega- tive ; and that the values of x at which <f> (x) ceases to decrease and begins to increase are those at which <f>'(x) changes its sign from negative to positive. In the former case, <£(#) is said to pass through a maximum, in the latter, a minimum, value. Ex. 1. Find the turning values of the function <£0) = 2 x s - 3 x 2 - 12 x + 4, and exhibit the mode of variation of the function by sketching the curve , , x V = <K X )- Here <f>'(x)=6x 2 - Qx- 12 = 6(z + 1)0 - 2), 54 DIFFERENTIAL CALCULUS Fig. 9 9, - 16, - 5, 36, oo. hence <j>'(x) is negative when x lies between - 1 and + 2, and positive for all other values of x. Thus <f)(x) increases from x = - oo to x = - 1 ; decreases from { x =— 1 to a: = 2; and increases from x = 2 to x = oo. Hence </>(- 1) is a maximum value of <f>(x), and <f>(2) a minimum. The general form of the curve y = <f>(x) (Fig. 9) may be inferred from the last statement, and from the following simultaneous values of x and y : x = - oo, - 2, - 1, 0, 1, 2, 3, 4, oo. y = -co, 0, 11, 4, Ex. 2. Exhibit the variation of the function <f>(x) = (x—l)~$+ 2, especially its turning values. Since <f>'(x) = - -, 6 (x - 1)* hence <f>'(x) changes sign at x = l, being negative when x < 1, infinite if x = 1, and positive if z> 1. Thus <f>(l) = 2 is a minimum turning value of tf>(x). The graph of the function is as shown in Fig. 10, with a vertical tangent at the point (1, 2). Ex. 3. Examine for maxima and minima the function Fig. 10 <f>(x) = (x 1 iy + 1, Here <f>'(x) = 3 0-l)§ Fie. 11 hence cf>'(x) never changes sign, but is always positive. There is accordingly no turning- value. The curve y =' <f>(x) has a verti- cal tangent at the point (1, 1), since -^ dx is infinite when x = l. (Fig. 11.) MAXIMA AND MINIMA 55 30. Critical values of the variable. It has been shown that the necessary and sufficient condition for a turning value of <f>(x) is that <f>'(x) shall change its sign. Now a function can change its sign only when it passes through zero, as in Ex. 1 (Art. 29), or when its reciprocal passes through zero, as in Ex. 2. In the latter case it is usual to say that the function passes through infinity. It is not true, conversely, that a function always changes its sign in passing through zero or infinity, e.g. x 2 and x~ 2 . Nevertheless all the values of x, at which <j>'(x) passes through zero or infinity, are called critical values of x, be- cause they are to be further examined to determine whether <f>'(x) actually changes sign as x passes through each such value ; and whether, in consequence, <f> (x) passes through a turning value. For instance, in Ex. 1, the derivative <f)'(x) vanishes when x = — 1, and when x = 2, and it does not become infinite for any finite value of x. Thus the critical values are — 1, 2, both of which give turning values to <f> (x). Again, in Exs. 2, 3, the critical value is x = 1, since it makes <f>'(x) infinite ; it gives a turning value to <£ (x) in Ex. 2, but not in Ex. 3. 31. Method of determining whether <f>'(&) changes its sign in passing through zero or infinity. Let a be a critical value of x; in other words, let <£'(a) be either zero or infinite, and let h be a very small positive number, so that a — h and a + h are two numbers very close to a, and on opposite sides of it. In order to determine whether <f>'(x) changes sign as x increases through the value a, it is necessary only to compare the signs of <f>'(a -f- h) and <f>'(a — h). If it is possible to take h so 56 DIFFERENTIAL CALCULUS small that cf>'(a —h) is positive and <f>'(a + h) negative, then cf>'(x) changes sign as x passes through the value a, and <f>(x) passes through a maximum value <f>(a). Similarly, if <j>\a — h) is negative and <f>'(a + h) positive, then <f>(x) passes through a minimum value <£ (a). If <f>'(a — h) and <j>'(a + h) have the same sign, however small h may be, then <f> (a) is not a turning value of <£ (x). Ex. Find the turning values of the function 4>0) = 0-i) 2 + i) 8 . Here <f>'(x) = 2(x- 1)0 + 1) 8 + S(x - l) 2 (x + l) 2 = (a:-l)(x+ 1) 2 (5 x- 1). Hence <j>'(x) becomes zero at x = — 1, \, and 1 ; it does not become infinite for any finite value of x. Thus, the critical values are — 1, |, 1. Fia. 12 When x has any value less than — 1, the three factors of <f>'(x) take the signs — + — , hence <f>'(x) is +, and when x has a value MAXIMA AND MINIMA 57 between — 1 and .J they become — + — , and <f>'(x) is still + ; hence <f>(— 1) = is not a turning value of <f>(x). When x has any value between \ and 1, the signs are — + + and <f>'(x) is — ; hence <f>($) is a maximum. Finally, if x has any value greater than 1, the signs are + + +; hence <f>'(x) changes sign from — to + as x increases through 1, and <£(1) = is a minimum value of <j>(x). The general march of the function may be exhibited graphically by tracing the curve y = <{>(x) (Fig. 12), using the foregoing results and observing the following simultaneous values of x and y : x^-co, - 2, - 1, 0, \, 1, 2, go. y = - oo, - 9, 0, 1, 1 • 1 ••-, 0, 27, oo. 32. Second method of determining whether <!>'(») changes sign in passing through zero. The following method may be employed when the function and its derivatives are continuous in the vicinity of the critical value x = a. Suppose, when x increases through the value a, that <j>'( x ) changes sign from positive through zero to negative. Its change from positive to zero is a decrease, and so is the change from zero to negative ; thus <f>'(x) is a decreasing function at x = a, and hence its derivative <£%e) is negative at x= a. On the other hand, if <f>'( x ) changes sign from negative through zero to positive, it is an increasing function and <f>"(x) is positive at x = a ; hence : TJie function cf>(x) has a maximum value <f>(a), when <f>'(a) = and <f>"(a) is negative; $(x) has a minimum value <f>(a), when cf>'(a) = and <f>"(a) is positive. It may happen, however, that <f>"(a) is also zero. . To determine in this case whether <£(#) has a turning value, it is necessary to proceed to the higher derivatives. If <f>(x) is 4 58 DIFFERENTIAL CALCULUS a maximum, <j>"(x) is negative just before vanishing, and negative just after, for the reason given above ; but the change from negative to zero is an increase, and the change from zero to negative is a decrease ; thus <f>"(x) changes from increasing to decreasing as x passes through a. Hence (f>'"(x) changes sign from positive through zero to negative, and it follows, as before, that its derivative <f> IV (x) is negative. Thus <f>(a) is a maximum value of <f>(x) if <t>'(a) = 0, <f>"(a) = 0, <f>'"(a) = 0, <j> JV (a) negative. Similarly, <f>(a) is a minimum value of <f>(x) if *'(«) = 0, <f>"(a) = 0, <£'"(a) = 0, and <£ lv (a) positive. If it happens that <f> lv (a) = 0, it is necessary to proceed to still higher derivatives to test for turning values. The result may then be generalized as follows : The function <£(#) has a maximum (or minimum) value at x = a if one or more of the derivatives <£'(a), <f>"(a), <f>'"(a) vanish and if the first one that does not vanish is of even order, and negative (or positive). Ex. Find the critical values in the example of Art. 31 by the second method. <f>"(x) = (x+iy(5x-l)+2(x-l)(x+l)(ox-l) + 5(x-l)(x + l)' i = 4(5a: 8 + 3a: 2 -3x-l), <£"(1) = 16, hence <£(1) i s a minimum value of <f>(x)', <f>"(— 1) = 0, hence it is necessary to find <£'"( — 1) ; <£'"(V) = 12(5 a: 2 + 2 a; - 1), <£'"( — i.) — 24, hence <£(— 1) is neither a maximum nor a minimum value of <j>(x). Again, <£"(£) = 5(J - l)(i 4- l) 2 is negative, hence <£(£) is a maxi- mum value of <£(V). MAXIMA AND MINIMA 59 33. The maxima and minima of any continuous function occur alternately. It has been seen that the maximum and minimum values of a rational polynomial occur alternately when the variable is continually increased, or diminished. This principle is true also in the case of every continuous function of a single variable. For, let <£(a), <f>(b) be two maximum values of <f>(x), iu which a is supposed less than b. Then, when x = a + h, the function is decreasing ; when x = b — h, the function is increasing, h being taken sufficiently small and positive. But in passing from a decreasing to an increasing state, a continuous function must, at some inter- mediate value of x, change from decreasing to increasing, that is, must pass through a minimum. Hence, between two maxima there must be at least one minimum. It can be similarly proved that between two minima there must be at least one maximum. 34. Simplifications that do not alter critical values. The work of finding the critical values of the variable, in the case of any given function, may often be simplified by means of the follow- ing self-evident principles. 1. When c is independent of x, any value of x that gives a turning value to c<jy(x), gives a turning value to <f>(x) also; and conversely. These two turning values are of the same or opposite kind according as c is positive or negative. 2. Any value of x that gives a turning value to c + <f>(x) gives a turning value of the same kind to <f>(x) also ; and conversely. 3. When n is independent of x, any value of x that gives a turning value to [<£(aj)] B gives a turning value to <f>(x) also; and conversely. These turning values are of the same or opposite kind according as ?i[<£(a,*)] n_1 is positive or negative. 60 DIFFERENTIAL CALCULUS EXERCISES Find the critical values of x in the following functions, determine the nature of the function at each, and obtain the graph of the function. 1. M = X(x 2 — 1). 2. u = 2 x 3 - 15 x 2 + 36 x - 4. 3. u = (x - l) 3 (x - 2) 2 . 4. u = sin x + cos x. 5. u =Q-*) 3 . a - 2x 7. u = 5 + 12 x - 3 x 1 - 2 x 8 . 8. logx X 9. m = sin 2 x cos 3 x. 10. ^ _ x 2 - X + 1 X 2 + X - 1 11. w _(x + 3)(x+l) (x - l)(x-2) 6. w = x(x + l) 2 - 5. 12. Show that a quadratic integral function always has one maxi- mum, or one minimum, but never both. 13. Show that a cubic integral function has in general both a maximum and a minimum value, but may have neither. 14. Show that the function (x — b)* has neither a maximum nor a minimum value. 35. Geometric problems in maxima and minima. The theory of the turning values of a function has important applications in solving problems concerning geometric maxima or minima, i.e. the determination of the largest or the smallest value a magnitude may have while satisfying certain stated geometric conditions. The first step is to express the magnitude in question algebraically. Jf the resulting expression contains more than one variable, the stated conditions will furnish enough relations between these variables, so that all the others may be expressed in terms of one. The expression to be maximized or minimized, being thus made a function of a single variable, can be treated by the preceding rules. MAXIMA AND MINIMA 61 Ex. 1. Find the largest rectangle whose perimeter is 100. Let x, y denote the dimensions of any of the rectangles whose perimeter is 100. The expression to be maximized is the area u = xy, (1) in which the variables x, y are subject to the stated condition 2x + 2r/ = 100, i.e. y =50 -x; (2) hence the function to be maximized, expressed in terms of the single variable x, is u _ <£(V) = ^50 -x)= 50 x - x 2 . (3) The critical value of x is found from the equation <P(x) = 50-2x = to be x = 25. When x increases through this value, <f>'(x) changes sign from positive to negative, and hence <f>(x) is a maximum when x = 25. Equation (2) shows that the corresponding value of y is 25. Hence the maximum rectangle whose perimeter is 100 is the square whose side is 25. Ex. 2. If, from a square piece of tin whose side is a, a square be cut out at each corner, find the side of the latter square in order that the remainder may form a box of maximum capacity, with open top. Let x be a side of each square cut out. Then the bottom of the box will be a square whose side is a — 2 x, and the depth of the box will be x. Hence the volume is v = x(a - 2 x) 2 , which is to be made a maximum by varying x. Here — = (a-2i) 2 ^i(a-2z) dx = (a-2x)(a -6.i). This derivative vanishes when x — -, and when x — - • It will be 2 6 found, by applying the usual test, that x = ^ gives v the minimum 62 DIFFERENTIAL CALCULUS value zero, and that x = - gives it the maximum value " a • Hence 6 27 the side of the square to be cut out is one sixth the side of the given square. Ex. 3. Find the area of the greatest rectangle that can be inscribed in a given ellipse. An inscribed rec- tangle will evidently be symmetric with regard to the principal axes of the ellipse. Let a, b denote the lengths of the semi- axes OA, OB (Fig. 14) ; let 2 x, 2 y be the dimen- Fig. 14 sions of an inscribed rectangle. Then the area is u = 4 x?j, (1) in which the variables x, y may be regarded as the coordinates of the vertex P, and are therefore subject to the equation of the ellipse 2 b l (2) It is geometrically evident that there is some position of P for which the inscribed rectangle is a maximum. The elimination of y from (1), by means of (2), gives the function of x to be maximized, 4 j } u = — x a V^ (3) By Art. 34, the critical values of x are not altered if this function is divided by the constant — , and then squared. Hence, the values a of x which render u a maximum, give also a maximum value to the function +(x) = x*(a* - &) = aW - x*. Here <f>'( x ) = 2 a 2 x - 4 x z = 2 x(a* - 2 x°-), cf>"(x) =2a 2 -12r 2 ; MAXIMA AND MINIMA 63 hence, by the usual tests, the critical values ± ■ — render cb(x), V2 and therefore the area w, a maximum. The corresponding values of y are given by (2), and the vertex P may be at any of the four points denoted by , a , b ±— :, y = ±— , V2 V2 giving in each case the same maximum inscribed rectangle, whose dimensions are a V2, b V2, and whose area is 2 ab, or half that of the circumscribed rectangle. Ex. 4. Find the greatest cylinder that can be cut from a given right cone, whose height is k, and the radius of whose base is a. Let the cone be generated by the revolution of the triangle 0^45 (Fig. 15), and the inscribed cylin- der be generated by the revolution of the rectangle A P. Let A = h, AB =a, and let the coordinates of P be (x, ?j) . Then the function to be maximized is Try' 2 (h r- x) subject to the relation ^ = - x h This expression becomes v = ^.x*a- /i 2 v Fig. 15 The critical value of x is f h, and V x). ijraVi 27 EXERCISES ON CHAPTER IV 1. What is the width of the rectangle of maximum area that can be inscribed in a given right segment of a parabola? .. 2. Divide 10 into two parts such that the sum of their squares is a minimum. 3. Find the number that exceeds its square by the greatest pos- sible quantity. 64 DIFFERENTIAL CALCULUS >^4. What number added to its reciprocal gives the least possible sum? 5. Given the slant height of a right cone; find its* altitude when the volume is a maximum. 6. A rectangular piece of pasteboard 30 in. long and 14 in. wide has a square cut out at each corner. Find the side of this square so that the remainder may form a box of maximum contents. 7. Find the altitude of the right cylinder of greatest volume in- scribed in a sphere of radius r. 8. Determine the greatest rectangle that can be inscribed in a given triangle whose base is 2 b, and whose altitude is 2 a. 9. A rectangular court is to be built so as to contain a given area c 2 , and a wall already constructed is available for one of its sides. Find its dimensions so that the expense incurred in building the walls for the other sides may be the least possible. 10. The volume of a cylinder of revolution being constant, find the relation between its altitude and the radius of its base when the entire surface is a minimum. 11. Assuming that the stiffness of a beam of rectangular cross section varies directly as the breadth and as the cube of the depth, what must be the breadth of the stiffest beam that can be cut from a log 16 in. in diameter? 12. A man who can row 4 mi. per hour, and can walk 5 mi. per hour, is in a boat 3 mi. from the nearest point on a straight beach, and wishes to reach in the shortest time a place on the shore 5 mi. from this point. Where must he land ? 13. If the cost per hour for the fuel required to run a given steamer is proportional to the cube of her speed and is $20 an hour for a speed of 10 knots, and if other expenses amount to $135 an hour, find the most economical rate at which to run her over a course s. MAXIMA AND MINIMA 65 14. If the cost per hour of running a boat in still water is propor- tional to the cube of the velocity, find the most economical rate at which to run the steamer upstream against a current of a miles per hour. 15. A Norman window consists of a rectangle surmounted by a semicircle. If the perimeter of the window is given, what must be its proportions in order to admit as much light as possible? 16. Find the most economical proportions for a cylindrical dipper which is to hold a pint. 17. The gate in front of a man's house is 20 yd. from the car track. If the man walks at the rate of 4 mi. an hour and the car on which he is coming home is running at the rate of 12 mi. an hour, where ought he to get off in order to reach home as early as possible? 18. How much water should be poured into a cylindrical tin dip- per in order to bring the center of gravity as low down as possible ? [Omit until after reading Art. 164.] 19. A statue 10 ft. high stands on a pedestal that is 50 ft. high. How far ought a man whose eyes are 5 ft. above the ground to stand from the pedestal in order that the statue may subtend the greatest possible angle? 20. The sum of the surfaces of a sphere and a cube is given. How do their dimensions compare when the sum of their volumes is a minimum ? 21. An electric light is to be placed directly over the center of a circular plot of grass 100 ft. in diameter. Assuming that the inten- sity of light varies directly as the sine of the angle under which it strikes an illuminated surface and inversely as the square of its dis- tance from the surface, how high should the light be hung in order that the most light possible shall fall on a walk along the circumfer- ence of the plot ? 22. Find the relation between length of circular arc and radius, in order that the area of a circular sector of a given perimeter shall be a maximum. el. calc. — 5 66 DIFFERENTIAL CALCULUS 23. On the line joining the centers of two mutually external spheres of radii r, R, find the distance of the point from the center of the first sphere from which the maximum of spherical surface is visible. 24. The radius of a circular piece of paper is r. Find the arc of the sector which must be cut from it so that the remaining sector may form the convex surface of a cone of maximum volume. 25. Describe a circle with its center on a given circle so that the length of the arc intercepted within the given circle shall be a maxi- mum. 26. Through a given point within an angle draw a straight line which shall cut off a minimum triangle. ,27. What is the length of the axis, and the area, of the maximum parabola which can be cut from a given right circular cone, given that the area of the parabola is equal to two thirds of the product of its base and altitude? A parabola is cut from the cone by a plane parallel to an element. 28. Through the point (a, b) a line is drawn such that the part intercepted between the rectangular coordinate axes is a minimum. Find its length. 29. The lower corner of a leaf, whose edge is a, is folded over so as just to reach the inner edge of the page. Find the width of the part folded over when the length of the crease is a minimum. 30. What is the length of the shortest line that can be drawn tan- gent to the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 and having its ends on the co- ordinate axes ? 31. Given a point on the axis of the parabola y 2 = 2 px at a dis- tance a from the vertex. Find the abscissa of the point of the curve nearest to it. 32. A wall 6 ft. high is parallel to the front of a house and 8 ft. from it. Find the length of the shortest ladder that will reach the house if one end rests on the ground outside the wall. MAXIMA AND MINIMA 67 33. It is required to construct from two circular iron plates of radius a a buoy, composed of two equal cones having a common base, which shall have the greatest possible volume. Find the radius of the base. 34. A weight W is to be raised by means of a lever with force F at one end and the point of support at the other. If the weight is suspended from a point at a distance a from the point of support, and the weight of the beam is w pounds per linear foot, what should be the length of the lever in order that the force required to lift the weight shall be a minimum? 35. A load is hauled up an inclined plane by a horizontal force ; it is required to find the inclination of the plane so that the mechanical efficiency may be greatest, assuming that the efficiency 77 is defined by the formula tan Q ' tan(0+4>)' where <f> is the angle of friction; i.e. tan <f> — fi, the coefficient of fric- tion between the load and the plane. 36. If the plane is of cast iron and the load is steel, and if the coefficient of friction between these substances is fi = 0.347, at what angle is the efficiency of the inclined plane a maximum? 37. Prove that a conical tent of given capacity will require the least amount of canvas when the height is v2 times the radius of the base. 38. If given currents c and c' produce deflections a and a' in a tangent galvanometer, so that tan «/tan «' = c/c', show that a — a' is a maximum when a + a' = -. CHAPTER V RATES AND DIFFERENTIALS 36. Rates. Time as independent variable. Suppose a particle P is moving in any path, straight or curved, and let s be the number of space units passed over in t seconds. Then s may be taken as the dependent variable, and t as the independent variable. The motion of P is said to be uniform when equal spaces are passed over in equal times. The number of space units passed over in one second is called the velocity of P, The velocity v is thus connected with the space s and the time t by the formula s v — -. t The motion of P is said to be non-uniform when equal spaces are not passed over in equal times. If s is the number of space units passed over in t seconds, then the average velocity during these t seconds is denned as -. If during the time A£ the num- ber of space units As are described, then the average velocity As during the time A£ is — . The actual velocity of P at any in- stant of time t is the limit which the average velocity approaches as At is made to approach zero as a limit. rr, lim As ds Thus v = A , . n — = — A * = °A* dt is the actual velocity of P at the time denoted by t. It is evidently the number of space units that would be passed over 68 RATES AND DIFFERENTIALS 69 in the next second if the velocity remained uniform from the time t to the time t -f 1. It may be observed that if the more general term, " rate of change,'' is substituted for the word " velocity," the above statements will apply to any quantity that varies with the time, whether it be length, volume, strength of current, or any other function of the time. For instance, let the quantity of an electric current be C at the time t, and C -\- AC at the time t + At. Then the average rate of change of current in the in- terval At is — ; this is the average increase in current-units At per second. And the actual rate of change at the instant de- noted by t is k rt jn J lim AC dC te±° At~ dt' This is the number of current-units that would be gained in the next second if the rate of gain were uniform from the time t to the time t + 1. Since, by Art. 8, dy_dy. dx dx dt dt hence -^ measures the ratio of the rates of change of y and dx oi x. It follows that the result of differentiating 2/=/(*) (1) may be written in either of the forms *-m (2) *-/«!■ (3) 70 DIFFERENTIAL CALCULUS The latter form is often convenient, and may also be obtained directly from (1) by differentiating both sides with regard to t. It may be read: the rate of change of y is f(x) times the rate of change of x. Ketnrning to the illustration of a moving point P, let its dx coordinates at time t be x and y. Then measures the rate dt of change of the ^-coordinate. Since velocity has been defined as the rate at which a point is moving, the rate —may be called the velocity which the dt point P has in the direction of the a>axis, or, more briefly, the ^-component of the velocity of P. It was shown on p. 68 that the actual velocity at any instant t is equal to the space that woiild be passed over in a unit of time, provided the velocity were *lrt\B / c^r&- uniform during that unit. Ac- cordingly, the ^-component of dt velocity — may be represented by the distance PA (Fig. 16) which P would pass over in the direction of the a>axis during a unit of time if the velocity remained uniform. Similarly -# is the ^/-component of the velocity of P, and may be represented by the distance PB. The velocity — of P along the curve can be represented by Civ the distance PC, measured on the tangent line to the curve at P. It is evident from the parallelogram of velocities that PC is the diagonal of the rectangle PA, PB. ^.dx Fig. 16 RATES AND DIFFERENTIALS 71 Since PC 2 = PA 2 + PB 2 , it follows that S)'=(f)" + (*)' « Ex. 1. If a point describes the straight Hue 3 x + 4y = 5, and if x increases h units per second, find the rates of increase of y and of s. Since y = £ - £ x, hence -7-— — 7-r" rf< 4 eft When — = A, dt it follows that ^ = - f A, ^ = VW^TJ^i - 5 A< eft </£ Ex. Si. A point describes the parabola # 2 = 12 r in such a way that when x = 3 the abscissa is increasing at the rate of 2 ft. per second ; at what rate is y then increasing? Find also the rate of increase of s. Since y*=12*, then "(It '" dt dy Qdx 6. dx , dt y dt Vl2x dt ' hence when x = 3 and — = 2, it follows that ^ = ± 2. A ^- (D 2 = (fT + B) ' •-- |=^ft. per second. Ex. 3. A person is walking toward the foot of a tower on a hori- zontal plane at the rate of 5 mi. per hour. At what rate is he ap- proaching the top, which is 60 ft. high, when he is 80 ft. from the bottom ? 72 DIFFERENTIAL CALCULUS Let x be the distance from the foot of the tower at time t, and y the distance from the top at the same time. Then x 2 + 60 2 = if, and *** = «& dt -dt When x is 80 ft., y is 100 ft.; hence if — is 5 mi. per hour, dy ■ A • , dt -^ is 4 mi. per hour. dt k 37. Abbreviated notation for rates. When, as in the above examples, a time derivative is a factor of each member of an equation, it is usually convenient to write, instead of the symbols — , -^, the abbreviations dx and dy, for the rates of J dt dt J ' change of the variables x and y. Thus the result of differen- tiatillg y=m (l) may be written in either of the forms *-/«, (2) dt ' } dt' w dy=f'(x)dx. (4) It is to be observed that the last form is not to be regarded as derived from equation (2) by separation of the symbols, dy, dx\ for the derivative -M- has been defined as the result of dx performing upon y an indicated operation represented by the symbol — , and thus the dy and dx of the symbol -M- have dx dx been given no separate meaning. The dy and dx of equation (4) stand for the rates, or time derivatives, ~ and •-— occur- dt - dt RATES AND DIFFERENTIALS 73 ring in (3), while the latter equation is itself obtained from (1) by differentiation with regard to t, by Art. 8. In case the dependence of y upon x is not indicated by a functional operation/, equations (3), (4) take the form dy dy dx dt dx dt ' dy = -fax. dx In the abbreviated notation, equation (4) of the last article is written (ds) 2 — (dx) 2 -f (dy) 2 or ds 2 = dx 2 + dy 2 . Ex. 1. A point describing the parabola y 2 — 2 px is moving at the time t with a velocity of v ft. per second. Find the rate of increase of the coordinates x and y at the same instant. Differentiating the given equation with regard to t, we obtain ydy = pdx. But dx, dy also satisfy the relation dx 2 + dy 2 = v 2 ; hence, by solving these simultaneous equations, we obtain dx = — ' v, dy = ^ v, in feet per second. v V + p 2 Vy* 2 + p 2 Ex. 2. A vertical wheel of radius 10 ft. is making 5 revolutions per second about a fixed axis. Find the horizontal and vertical velocities of a point on the circumference situated 30° from the horizontal. Since x = 10 cos 6, y = 10 sin 6, then dx = -10 sin OdO, dy = 10 cos Odd. But dO = 10 7T = 31.416 radians per second, hence dx = — 314.16 sin $ = — 157.08 ft. per second, and dy = 314.16 cos = 272.06 ft. per second. Ex. 3. Trace the changes in the horizontal and vertical velocity in a complete revolution. 74 DIFFERENTIAL CALCULUS 38. Differentials often substituted for rates. The symbols dx, dy have been denned above as the rates of change of x and y per second. Sometimes, however, they may conveniently be allowed to stand for any two numbers, large or small, that are propor- tional to- these rates ; the equations, being homogeneous in them, will not be affected. It is usual in such cases to speak of the numbers dx and dy by the more general name of differ- entials ; they may then be either the rates themselves, or any two numbers in the same ratio. This will be especially convenient in problems in which the time variable is not explicitly mentioned. 39. Theorem of mean value. Let f(x) be a continuous func- tion of x which has a derivative. It can then be represented by the ordinates of a curve whose equation is y —f(x). In Pig. 17, let x = ON, x + h = OR, f(x) = NH, f(x + h) = RK. Then f(x + li)-f(x) = MK, and f(x+iC\-f(xS MK Fig. 17 = ^±± = tan MHK. h HM But at some point 8 between H and A" the tangent to the curve is parallel to the secant IIK. Since the abscissa of S is greater than x and less than x + h it may be represented by x-\-0h, in which 6 is a positive number less than unity. The slope of the tangent at S is then expressed by/'(.x* + 6h), hence f(x + h) -f(x) = / {x + m)t from which f(x + h)=f(x)+hf'(x+Oh). RATES AND DIFFERENTIALS 75 The theorem expressed by this formula is known as the theorem of mean value. If in this equation we put f(x + K) — f (a?) = dy, h = dx, in which h is an arbitrary increment, then the relation between the increment of the variable and the actual increment of the function will be expressed by the equation dy=f'(x + 6dx)dx, whereas if dy, dx are regarded as differentials (dy not an actual but a virtual increment), then the relation becomes dy=f\x)dx. This more clearly illustrates that the differential dy is de- fined as the change that would take place in the function y, corresponding to the actual change dx in the independent vari- able x } provided the rate of change remained constant. EXERCISES 1. When x increases from 45° to 45° 15', find the increase of logio sin x, assuming that the ratio of the rates of change of the func- tion and the variable remains constant throughout the short interval. Here dy = logio e • cot xdx = .4343 cot xdx = .4343 dx. Let dx— .004163 (the number of radians in 15'). Then dy = .001895, which is the approximate increment of logio sin x. But log 10 sin 45° = - \ log 2 = - .150515, therefore log 10 sin 45° 15' = - .148620. 2. Show that log, ft .r increases more slowly than x, when x > log 10 e, that is, x > 0.4343. 76 DIFFERENTIAL CALCULUS 3. A man is walking at the rate of 5 mi. per hour towards the foot of a tower 60 ft. high standing on a horizontal plane. At what rate is the angle of elevation of the top changing when he is 80 ft. from the foot of the tower ? 4. An arc light is hung 12 ft. directly above a straight horizontal walk on which a man 5 ft. in height is walking. How fast is the man's shadow lengthening when he is walking aw T ay from the light at the rate of 168 ft. per minute? 5. At what point on the ellipse 16 x 2 + 9 y 2 — 400 does y decrease at the same rate that x increases? 6. A vessel is sailing northwest at the rate of 10 mi. per hour. At what rate is she making north latitude? 7. In the parabola y 2 = 12 x, find the point at which the ordinate and abscissa are increasing equally. 8. At what part of the first quadrant does the angle increase twice as fast as its sine ? 9. Find the rate of change in the area of a square when the side b is increasing at a ft. per second. 10. In the function y = 2 x z + 6, what is the value of x at the point where y increases 24 times as fast as x ? 11. A circular plate of metal expands by heat so that its diam- eter increases uniformly at the rate of 2 in. per second. At what rate is the surface increasing when the diameter is 5 in.? 12. What is the value of x at the point at which x 8 — 5 x' 2 + 17 x and x s — 3 x change at the same rate ? 13. Find the points at which the rate of change of the ordinate y = x 3 - 6 x 2 + 3 x + 5 is equal to the rate of change of the slope of the tangent to the curve. 14. The relation between s, the space through which a body falls, rates and differentials 77 and t, the time of falling, is s = 16 t 2 . Show that the velocity is equal to 32 t. The rate of change of velocity is called acceleration and is denoted by a. TT dv d 2 s Hence a = — = -— • dt dt 2 Show that the acceleration of the falling body is a constant. 15. A body moves according to the law s — cos (nt + e). Show that its acceleration is proportional to the space through which it has moved. 16. If a body is projected upwards in a vacuum with an initial velocity v , to what height will it rise, and what will be the time of ascent? 17. A body is projected upwards with a velocity of a ft. per second. After what time will it return? 18. If A is the area of a circle of radius x, show that the circum- dA ference is — . Interpret this fact geometrically. 19. A point describing the circle x 2 + y 2 = 25 passes through (3,4) with a velocity of 20 ft. per second. Find its component velocities parallel to the axes. 20. Let a point P move with uniform velocity on a circle of radius a with center ; let AB be any diameter, and Q the orthogonal projec- tion of P on AB. Find an expression for the velocity of Q in terms of the angular velocity of P, and show how this velocity varies during a revolution of P. The motion of the point Q along AB is called harmonic. 21. A point P moves along the curve y = x z at the rate of 3 ft. per second. At what rate is the angle <£, which the tangent to the curve makes with the x-axis, increasing when P is passing through the point (1, 1)? V CHAPTER VI DIFFERENTIAL OF AN AREA, ARC, VOLUME, AND SURFACE OF REVOLUTION 40. Differential of an area. If the coordinates of P are (x, y) and those of Q (x 4- Ax, y-\- Ay), then MN= PR = Ax, and PS = RQ=Ay. If the area OAPM is denoted by A, then A is evidently some function of the abscissae; also if area OAQN is denoted by A 4- AA then the area MNQP is AA ; it is the incre- ment taken by the function A, when x takes the increment Ax. But MNQP lies between the rectangles MR, MQ ; hence i/Aa; < A.4 < (y + Ay)Aa;, and A. 4 y<—-<y + ^y- Ax Therefore, when Ax, Ay, AA all approach zero, lim A A dA Ax (lx = ?/ Hence, if the ordinate and the area are expressed each as a function of the abscissa, the derivative of the area function with regard to the abscissa is equal to the ordinate function. 78 AREA, ARC, VOLUME, AND SURFACE 79 In the notation of differentials we may say : The differential of the area between a curve and the axis of x is measured by the product of the ordinate and the differential of x. dA = ydx. Ex. If the area included between a curve, the axis of x, and the ordinate whose abscissa is a:, is given by the equation A = x s , find the equation of the curve. Here =M = 3z2. dx 41. Differential of an arc. A segment of a straight line is measured by applying the unit of measure successively to the segment to be measured. In the case of a curve this is gen- erally impossible. We define the length of a given curve between two points upon it as the limit of the sum of the chords joining points on the curve when the lengths of these chords approach the limit zero. We shall then assume that the ratio of the arc to the chord approaches the limit 1 when the length of the chord approaches the limit zero. [Compare § 19.] Let PQ be two points on the curve (Fig. 19) ; let x, y be the coordinates of P ; x + Ax, y + Ay Y those of Q ; s the length of the arc AP; s -+- As that of the arc AQ. Draw the ordinates MP, NQ ; and draw PR parallel to MN. Then PR = Ax, RQ = Ay ; arc PQ = As. - Hence chord PQ = V(A xf+ (Ay) 2 , M N Fig. 19 PQ Ax Therefore ' = Ai Ax As 1>Q AXy PQ = As Ax PQ A?/Y v i+ is- 80 DIFFERENTIAL CALCULUS Taking the limit of both members as Ax approaches zero lim As and putting ™ Q -~ = 1, we obtain %-Mif (1) Si ,u„„l y , |-V' + (*/. (2) Moreover, from Art. 36, (ds\*/dx\* (dy\* [dtj \dtj \dtj' (3) or in the differential notation, ds 2 = dx 2 -f dy 2 . W 42. Trigonometric meaning of — , — . doc dy cs- A.r A.r PQ »r>/^ Since — = • — -3S = cos PPQ As PQ As As' it follows by taking the limit that dx , — = cos <£, ds wherein <f>, being the limit of the angle RPQ, is the angle which the tangent at the point (x, y) makes with the o-axis. Similarly, -* = sin </> ; whence — = sec <£, — = esc <£. ds dx dy By using the idea of a rate or differential, all these relations may be conveniently exhibited by Fig. 20. These results may also be de- rived from equations (1), (2) of Art. 41, by putting -'^tan d>. Fig. 20 J F dx *■ ■ .» QL^dx AREA, ARC, VOLUME, AND SURFACE 81 43. Differential of the volume of a solid of revolution. Let the curve APQ (Fig. 21) revolve about the x-axis, and thus generate a surface of revo- lution ; let V be the volume in- cluded between this surface, the plane generated by the fixed ordinate at A, and the plane generated by any ordinate MP. Let A V be the volume gener- ated by the area PMNQ. Then AV lies between the vol- umes of the cylinders generated by the rectangles PMNR and SMNQ) that is, Try 9 Ax < A V < 7T (y -f Ay) 2 Ax. Y s A / R X M A J Fig. 21 Dividing by Ax and taking limits, we obtain dV dx Try 2 , dV = 7ry 2 dx. 44. Differential of a surface of revolution. Let S be the area of the surface generated by the arc AP (Fig. 22), and AS that generated by the arc PQ, whose length is As. Draw PQ', QP parallel to OX and equal in length to the arc PQ. Then it may be assumed as an axiom that the area generated by PQ lies between the areas gen- iL erated by PQ' and P'Q; i.e. 2 iry As < AS < 2 tv (y + Ay) As. I M N Fig. 22 EL. CALC. — DIFFERENTIAL CALCULUS Dividing by As and passing to the limit, f = 2^ (1) dx ds dx \ \dx) 45. Differential of arc in polar coordinates. Let p, 6 be the coordinates of P (Fig. 23) ; p ■+ Ap, 6 + A6 those of $ ; s the length of the arc KP ; As that of the L arcPQ; draw PM perpendicular to OQ. Then PM = p sin A0, p MQ = OQ-OM=p + A P -p cos A<9 = p (1 — cos A0) + A/a Fig. 23 = 2 P sin2 i A # + V Hence PQ 2 = (p sin A0) 2 + (2 p sin 2 1 A0 + Ap) 2 , /PQV = ^sinA^V + ( p sin J A0 . S Ai^ + V)\ Replacing the first member by f — - %■ • -^ ] , passing to the \ .AN -Af7 / limit when A0 = 0, and putting lira— ^ = 1, lira — — = 1, As A9 lira sin * A6 = 1, we obtain i^0 ^V = , 2 + ^Y. that is, ^ = V^ +/ d<9 ' Vd0 y AREA, ARC, VOLUME, AND SURFACE 83 In the rate or differential notation this formula may be conveniently written ^ = rf * + 2 ^ 46. Differential of area in polar coordinates. Let A be the area of OKP (Fig. 24) measured from a fixed radius vector OK to any other radius vec- tor OP ; let A^4 be the area of OPQ. Draw arcs PM, QN, with as a center. Then the area POQ lies between the areas of the sectors OPM and ONQ; i.e. Fig. 24 1 p 2 A0 < A.-l < J (p + Ap) 2 A0. Dividing by A0 and passing to the limit, when A0 = 0, we obtain dA d$ = ip 2 - Hence, in the differential notation we may write the formula dA = \ phlO. EXERCISES ON CHAPTER VI 1. In the parabola y 2 = 4 ax, find — , — , — - , dx dx dx dx 2. Find — and — for the circle x 2 + y 2 = a 2 . dx dy ds 3. Find — for the curve e v cos x = 1. dx 4. Find the x-derivative of the volume of the cone generated by revolving the line y = ax about the axis of x. 5. Find the it-derivative of the volume of the ellipsoid of revolu- tion, formed bv revolving — + •*- = 1 about its major axis. a 2 b 2 84 DIFFERENTIAL CALCULUS ds 6. In the curve p = a 9 find — du 7. Given p = a (1 + cos 6) ; find ^ 8. In p 2 cos 2 0, find ^. 9. The parabolic arc y 2 = 9 x measured from the vertex to a variable point P = (x, y) is revolving about the a>axis. If P moves along the curve at the rate of 2 in. per second, what is the rate of increase of the surface of revolution when P is passing through the point (4, 6)? What is the rate of increase of the volume of revolution? 10. The radius vector to the cardioid p = 2 (1 — cos $) is rotating about the origin with an angular velocity of 18° per second. Find the rate at which the extremity P of the radius vector is moving along the curve, taking the inch as unit of length. At what points of the curve will P be moving fastest? slowest? Find the velocities at these points. CHAPTER VII APPLICATIONS TO CURVE TRACING 47. Equation of tangent and normal. The function y=f(x) may be represented by a plane ciu-ve. It will now be shown how to obtain several of the properties of this curve by means of the principles already established. The tangent line at a point (x-t, y{) on the curve passes through the point and has the slope — , the symbol meaning that the coordinates x x , y x are substituted in the first derivative after the differentiation has been performed. Its equation may be written in the form y- 2/1 = ^' (*-*.)• (1) The normal to the curve at the point (x u y Y ) is the straight line through this point, perpendicular to the tangent. Since the slope of the normal is the negative reciprocal of that of the tangent, its equation may be written in the form *-*i + p(*- *)-0. (2) 48. Length of tangent, normal, subtangent, subnormal. The segments of the tangent and normal intercepted between the point of tangency and the axis OX are called, respectively, the tangent length and the normal length, and their projections on OX are called the subtangent and the subnormal. 85 DIFFERENTIAL CALCULUS C Fig. 25 a Fig. 25 6 Thus, in Fig. 25, a, b let the tangent and normal to the curve PC at P meet the axis OX in T and N, and let i/P be the ordi- nate of P. Then TP is the tangent length, PNthe normal length, TM the subtangent, MN the subnormal. These will be denoted, respectively, by t, n, r, v. Let the angle XTP be denoted by cf>, and write tan <f> = -^*. " Then hence d^x 2/i 1 + (for, y H§, fdjh \dx x The subtangent is measured from the intersection of the tangent to the foot of the ordinate ; it is therefore positive when the foot of the ordinate is to the right of the intersec- tion of tangent. The subnormal is measured from the foot of the ordinate to the intersection of normal, and is positive when the normal cuts OX to the right of the foot of the ordi- APPLICATIONS TO CURVE TRACING 87 nate. Both are therefore positive or negative, according as <£ is acute or obtuse. The expressions for t, v may be obtained also by finding from equations (1), (2), Art. 47, the intercepts made by the tangent and normal on the axis OX. The intercept of the tangent subtracted from x l gives t, and x x subtracted from the intercept of the normal gives v. Ex. Find the intercepts made upon the axes by the tangent at the point (x v y x ) on the curve Va: + Vy = Va, and show that their sum is constant. Differentiating the equation of the curve, we obtain 2 V* 2V~ydx Hence the equation of the tangent is x i The x intercept is x l + y/x l y v and the y intercept is y x 4- y/x l y v hence their sum is , ,— ,— N0 If a series of lines is drawn such that the sum of the intercepts of each is the same constant, account being taken of the signs, the form of the parabola to which they are all tangent can be readily seen. EXERCISES 1. Find the equations of the tangent and the normal to the ellipse X 2 V 2 (- V- — 1 at the point (x v ?/,). Compare the process with that era- a' 2 b 2 i ployed irr-analytic geometry to obtain the same results. 2. Find the equation of the tangent to the curve x\x +y)= a 2 (x -y) at the origin. 88 DIFFERENTIAL CALCULUS 3. Find the equations of the tangent and normal at tho point (1, 3) on the curve y' 2 = 9 x 3 . 4. Find the equations of the tangent and normal to each of the following curves at the point indicated : (a) y = — , at the point for which x = 2 a. 4 a 2 + x 2 (jg) y 2 = 2 x 2 — x 3 , at the points for which x = 1. (y) y 2 ~ ipx, at the point (p, 2p). 5. Find the value of the subtangent of y 2 = 3 x 2 — 12 at x = 4. Compare the process with that given in analytic geometry. 6. Find the length of the tangent to the curve y 2 = 2 x at x = 8. 7. Find the points at which the tangent is parallel to the axis of x, and at which it is perpendicular to that axis for each of the fol- lowing curves : (ft) a,? + 2 hxy + bf = 1. (/J) , =?^iL\ ax (■y) y 3 = x 2 (2 a — x). 8. Find the condition that the conies ax 2 4- by 2 = 1, a'x 2 -f b'y 2 — 1 shall cut at right angles. 9. Find the angle at which x 2 = y 2 + 5 intersects 8 x 2 -f 18 y 2 = 144. Compare with Ex. 8. 10. Show that in the equilateral hyperbola 2 xy — a 2 the area of the triangle formed by a variable tangent and the coordinate axes is constant and equal to a 2 . 11. At what angle does y 2 = 8 x intersect 4 x 2 + 2 y 2 =48 ? 12. Determine the subnormal to the curve y n — aP-^x. 13. Find the values of x for which the tangent to the curve 7/3 = ( x - a ) 2 (x - c) is parallel to the axis of x. APPLICATIONS TO CURVE TRACING 89 14. Show that the subtangent of the hyperbola xy = a 2 is equal to the abscissa of the point of tangency, but opposite in sign. 15. Prove that the parabola y 2 = 4 ax has a constant subnormal. 16. Show analytically that in the curve x 2 + y 2 = a 2 the length of the normal is constant. 17. Show that in the tractrix, the length of the tangent is con- stant, the equation of the tractrix being x = Vc*-Tj* + £ log c-V c 2 -y\ - c + vV 2 — y 2 X 18. Show that the exponential curve y = ae c has a constant sub- tangent. 19. Find the point on the parabola y 2 = 4 px at which the angle between the tangent and the line joining the point to the vertex shall be a maximum. 49. Concavity upward and downward. A curve is said to be concave downward in the vicinity of a point P when, for a finite distance on each side of P, the curve is situated below the tangent drawn at that point, as in the arcs AD, FTL It is concave upward when the curve lies above the tangent, as in the arcs DF, HK. 90 DIFFERENTIAL CALCULUS By drawing successive tangents to the curve, as in the fig- ure, we easily see that if the point of contact advances to the right, the tangent swings in the positive direction of rotation when the concavity is upward, and in the negative direction when the concavity is downward. Hence upward concavity may be called a positive bending of the curve, and downward concavity, a negative bending. A point at which the direction of bending changes con- tinuously from positive to negative, or vice versa, as at F or at D, is called a point of inflexion, and the tangent at such a point is called a stationary tangent. The points of the curve that are situated just before and just after the point of inflexion are thus on opposite sides of the stationary tangent, and hence the tangent crosses the curve, as at D, F, H. 50. Algebraic test for positive and negative bending. Let the inclination of the tangent line, measured from the positive end of the ^axis toward the forward end of the tangent, be denoted by <£. Then <£ is an increasing or decreasing function of the abscissa according as the bending is positive or negative ; for instance, in the arc AD, the angle <f> diminishes from 4- — Z through zero to — - ; in the arc DF, cfy increases from — - 4 4 through zero to - ; in the arc FIT, <£ decreases from -f - through o o zero to — y ; and in the arc HK, <f> increases from — y through zero to + -. At a point of inflexion $ has evidently a turning value which is a maximum or a minimum, according as the concavity changes from upward to downward, or conversely. APPLICATIONS TO CURVE TRACING 91 Thus in Fig. 26, cf> is a maximum at F, and a minimum at D and at II. Instead of recording the variation of the angle <f>, it is gen- erally convenient to consider the variation of the slope, tan <£, which is easily expressed as a function of x by the equation tan <£ = -^. dx Since tan <f> is always an increasing function of <f>, it follows that the slope function -^ is an increasing or a decreasing dx function of x, according as the concavity is upward or down- ward, and hence that its x-derivative is positive or negative. Thus the bending of the curve is in the positive or negative d-y direction of rotation, according as the function — ^ is positive dx- or negative. At a point of inflexion the slope — is a maximum or a dx d 2 v minimum, and therefore its derivative — - changes sign from dx 2 positive to negative or from negative to positive. This latter condition is evidently both necessary and sufficient in order that the point (x, y) may be a point of inflexion on the given curve. Hence, the coordinates of the points of inflexion on the curve may be found by solving the equations /"O) = o, /"(*)=«,, and then testing whether f"(x) changes its sign as x passes through the critical values thus obtained. To any critical value a that satisfies the test corresponds the point of inflexion 92 DIFFERENTIAL CALCULUS Ex. 1. For the curve y = (x 2 — l) 2 find the points of inflexion, and show the mode of variation of the slope and of the ordinate. Here ^/ = 4a-(> 2 -l), dx (Py_ dx 2 = 4(3:r 2 -l), 1 hence the critical values for inflexions are x = ± — -. It will be seen i vs that as x increases through = , the second derivative changes sign V3 from positive to negative, hence there is an inflexion at which the concavity changes from upward to downward. Similarly, at x = ^ — =3 the concavitv changes from downward to upward. The V3 following numerical table will help to show the mode of variation of the ordinate and of the slope, and the direction of bending. As x increases from — oo to = the bending is positive, and Va the slope continually increases from — go through zero to a maximum I Q value — — , which is the slope of 3V3 the stationary tangent drawn at the point f , - ). V V3 9/ As x continues to increase from L_ to H = , the bending is neg- V3 V3 Q ative, and the slope decreases from + — - through zero to a minimum oV3 o value _, which is the slope of the stationary tangent at 3V<3 X y dy dx d*y dx 2 — 00 + co — GO + _ 2 + 9 -24 + -1 + V3 -1 8 3V3 1 _ V3 H 8 3V3 1 + + CO + 00 + GO + ^■tl APPLICATIONS TO CURVE TRACING 93 Finally, as x increases from -\ to + oo, the bending is positive V3 and the slope increases from the o value through zero to + oo. 3V3 The values x = — 1, 0, +1, at which the slope passes through zero, correspond to turning values of the ordinate. Ex. 2. Examine for inflexions the curve x + 4 = (y - 2) 3 . V J Fig. 27 In this case y = 2+(x + 4)\ dy 1 dx d 2 y (x + 4)" dx 2 + 4) dy Fig. 28 Hence, at the point ( - 4, 2), ^ dx and — ¥ are infinite. When x< - 4, dx' 2 d 2 u d 2 v —± is positive, and when x > — 4, — ^ is negative. dx 2 dx 2 Thus there is a point of inflexion at (— 4, 2), at which the slope is infinite, and the bending changes from the positive to the negative direction. Ex. 3. Consider the curve dy _ dx 4 a: 8 , y = ar d 2 y dx 2 12 x 2 . d 2 y At (0, 0), ^ is zero, but the dx 2 curve has no inflexion, for — & dx 2 never changes sign (Fig. 29). Fig. 29 94 DIFFERENTIAL CALCULUS 51. Concavity and convexity toward the axis. A curve is said to be convex or concave toward a line, in the vicinity of a given point on the curve, according as the tangent at the point does or does not lie between the curve and the line, for a finite distance on each side of the point of contact. Fig. 30 a Fig. 30 6 First, let the curve be convex toward the #-axis, as in the left- hand figure. Then if y is positive, the bending is positive cl 2 v and — " is positive : but if y is negative, the bending is nega- dx 2 tive and —^ is negative. Hence in either case the product dx 2 sin dx 2 y—^- is positive. Next, let the curve be concave toward the a>axis, as in the right-hand figure. Then if y is positive, the bending is nega- tive and — ^- is negative ; but if y is negative, the bending is dx 2 positive and — -^ is positive. Thus in either case the product dx 2 d 2 v y — -- is negative. Hence : dx 2 In the vicinity of a given point (x, y) the curve is convex or concave to the x-axis. according as the product y —^ is positive or dx 2 APPLICATIONS TO CURVE TRACING 95 EXERCISES 1. Examine the curve y — 2 — 3(x — 2) 5 for points of inflexion. 2. Show that the curve a 2 y = x(a 2 — x 2 ) has a point of inflexion at the origin. 3. Find the points of inflexion on the curve y x 2 + 4 a 2 ■m 4. In the curve ay = x«, prove that the origin is a point of in- flexion if m and n are positive odd integers. 5. Show that the curve y = c sin - has an infinite number of a points of inflexion lying on a straight line. 6. Show that the curve y(x 2 -f- a 2 ) = x has three points of inflexion lying on a straight line ; find the equation of the line. 7. If y 2 =f(x s ) is the equation of a curve, prove that the abscissas of its points of inflexion satisfy the equation [/'(*)] 2 = 2/(z) •/''(*). 8. Draw the part of the curve a' 2 y = ~ — ax 2 -f 2 a 3 near its point o of inflexion, and find the equation of the stationary tangent. 9. Show that the curve y = x 2n has no points of inflexion, n being any positive integer. Sketch the curve. 10. Show that the curve (1 + x 2 )y =1 — x has three points of in- flexion, and that they lie in a straight line. 52. Hyperbolic and parabolic branches. When a curve has a branch extending to infinity, the tangents drawn at successive points of this branch may tend to coincide with a definite fixed line, as in the familiar case of the hyperbola. On the other hand, the successive tangents may move farther and farther out of the field, as in the case of the parabola. These two kinds of infinite branches may be called hyperbolic and parabolic. The character of each of the infinite branches of a curve can always be determined when the equation of the curve is known. 96 DIFFERENTIAL CALCULUS 53. Definition of a rectilinear asymptote. If the tangents at successive points of a curve approach a fixed straight line as a limiting position when the point of contact moves farther and farther along any infinite branch of the given curve, then the fixed line is called an asymptote of the curve. This definition may be stated more briefly but less precisely as follows : An asymptote to a curve is a tangent whose point of contact is at infinity, but which is not itself entirely at infinity. DETERMINATION OF ASYMPTOTES 54. Method of limiting intercepts. The equation of the tan- gent at any point (x 1} y x ) being y-yi = ~(^-^i), ClXi the intercepts made by this line on the coordinate axes are dx l Xq — x x — y x — dy x J (1) Suppose the curve has a branch on which x == oo and y = oo . Then from (1) the limits can be found to which the intercepts x , y approach as the coordinates x lf y x of the point of contact tend to become infinite. If these limits are denoted by a, b, the equation of the corresponding asymptote is a b Except in special cases this method is usually too compli-' cated to be of practical use in determining the equations of the asymptotes of a given curve. There are two other methods, which together will always suffice to determine the APPLICATIONS TO CURVE TRACING 97 asymptotes of curves whose equations involve only algebraic functions. These may be called the methods of inspection and of substitution. 55. Method of inspection. Infinite ordinates, asymptotes parallel to axes. When an algebraic equation in two coordinates x and y is rationalized, cleared of fractions, and arranged according to powers of one of the coordinates, say y, it takes the form a^+(6aj + c)r- 1 + (^ + ^+/)r" 2 + - +«»-# + «» = <>, in which u n is a polynomial of the degree n in terms of the other coordinate x, and u n . l is of degree n — 1. When any value is given to x, the equation determines n values for y. Let it be required to find for what value of x the correspond- ing ordinate y has an infinite value. For this purpose the following theorem from algebra will be recalled : Given an algebraic equation of degree n, ay n + (3y n - 1 + yy n - 2 + ••• =0; if « = 0,one root y becomes infinite; if a = and /? = 0, two roots y become infinite; and in general if the coefficients of each of the k highest powers of y vanish, the equation will have k infinite roots. Suppose at first that the term in y n is present ; in other words, that the coefficient a is not zero. Then, when any finite value is given to x, all of the n values of y are finite, and there are accordingly no infinite ordinates for finite values of the abscissa. Next suppose that a is zero, and b, c, not zero. In this case one value of y is infinite for every finite value of x, and el. calc. — 7 98 DIFFERENTIAL CALCULUS hence the curve passes through the point at infinity on the y axis. There is one particular value of x, namely, x = ^—. for b which an additional root of the equation in y becomes infinite. For, when x has this value, the coefficient bx -f- c of the high- est power of y remaining in the equation vanishes. Geometrically, every line parallel to the y axis has one point of intersection with the curve at infinity, but the line bx-\-c = has two points of intersection with the curve at infinity. A line having two coincident points of intersection with a curve is a tangent to the curve ; and when the coinci- dent points are at infinity, but the line itself not altogether at infinity, the tangent is an asymptote. Hence, an ordinate that becomes infinite for a definite value of x is an asymptote. Again, if not only a, but also b and c are zero, there are two values of x that make y infinite ; namely, those values of x that make dx 2 + ex +f= 0. The equations of the infinite ordinates are found by factoring this last equation; and so on. Similarly, by arranging the equation of the curve according to powers of x, we can easily find what values of y give an infinite value to x. Ex. 1. In the curve 2 x s + z 2 y + zy* = x 2 — y 2 — 5, find the equation of the infinite ordinate, and determine the finite point in which this line meets the curve. This is a cubic equation in which the coefficient of y 8 is zero. Arranged in powers of y it is y 2( x + 1) + yx 2 + (2 x 8 t- z 2 + 5) =0. APPLICATIONS TO CURVE TRACING 99 When x = — 1, the equation for y becomes • y 2 + y + 2 = 0, the two roots of which are y = oo , y = — 2 ; hence the equation of the infinite ordinate is x + 1 = 0. The infinite ordinate meets the curve again in the finite point (— 1, — 2). Since the tern^in x 8 is present, there are no infinite values of x for finite values of y. Ex. 2. Show that the lines x = a, and y = are asymptotes to the curve a 2 x = y(x — a) 2 (Fig. 31). Fig. 31 Ex. 3. Find the asymptotes of the curve x 2 (y — a) + xy 2 = a 3 . 56. Method of substitution. Oblique asymptotes. The as- ymptotes that are not parallel to either axis can be found by the method of substitution, which is applicable to all algebraic curves, and is of especial value when the equation is given in the implicit form /(*»y) = o. (i) Consider the straight line y = mx + b, (2) 100 DIFFERENTIAL CALCULUS and let it be required to determine m and b so that this line shall be an asymptote to the curve /(a*, y) = 0. Since an asymptote is the limiting position of a line that meets the curve in two points that tend to coincide at infinity, then, by making (1) and (2) simultaneous, the resulting equa- f(x, mx + b) = 0, is to have two of its roots infinite. This requires that the coefficients of the two highest powers of x shall vanish. These coefficients, equated to zero, furnish two equations from which the required values of m and b can be determined. These values, substituted in (2), will give the equation of an asymptote. Ex. 4. Find the asymptotes to the curve y 3 = x 2 (2 a — x). In the first place, there are evidently no asymptotes parallel to either of the coordinate axes. To determine the oblique asymptotes, make the equation of the curve simultaneous with y = mx + b, and eliminate y. Then (mx + b) 3 = x 2 (2a-x), or, arranged in powers of x, (1 + m 3 )x 3 + (3 m 2 b - 2 a)x 2 + 3 b 2 mx + b s = 0. Let m 3 + 1 = and 3 m 2 b - 2 a = 0. Then m = — 1, /> = =—; o hence y = - x + — o is the equation of an asymptote. The third intersection of this line with the given curve is found 2 a from the equation 3 mb 2 x -f b 8 = 0, whence x = —• APPLICATIONS TO CURVE'. TRACING: 161 Fig. 32 This is the only oblique asymptote, as the other roots of the equation for m are imaginary. Ex. 5. Find the asymptotes to the curve y(a 2 + x 2 ) = a 2 (a - x). Y Fig. 33 Here the line y = is a horizontal asymptote by Art. 55. To find the oblique asymptotes, put y = mx + b. Then (mx + b) (a 2 4- x 2 ) = a 2 (a - x), i.e. mx* + bx 2 + (ma 2 + a 2 )x + (a 2 b - a 8 ) = ; hence m = 0, b = 0, for an asymptote. Thus the only asymptote is the line y = already found. 1#2 DIFFERENTIAL CALCULUS 57. Number of asymptotes. The illustrations of the last article show that if all the terms are present in the general equation of an nth degree curve, then the equation for deter- mining m is of the nth degree and there are accordingly n values of ra, real or imaginary. The equation for finding b is usually of the first degree, but for certain curves one or more values of m may cause the coefficients of x n and x n ~ x both to vanish, irrespective of b. In such cases any line whose equa- tion is of the form y = m y x + c will have two points at infinity on the curve independent of c ; but by equating the coefficient of x n ~ 2 to zero, two values of b can be found such that the re- sulting lines have three points at infinity in common with the curve. These two lines are parallel ; and it will be seen that in each case in which this happens the equation defining m has a double root, so that the total number of asymptotes is not increased. Hence the total number of asymptotes, real and imaginary, is in general equal to the degree of the equation of the curve. This number must be reduced whenever a curve has a para- bolic branch, since in this case a value of m which makes the coefficient of x n vanish does not correspond to any finite value of b. Ex. 6. Find the asymptotes of the curve (x — y) 3 = 2 x. The equation in m is (m — l) 3 = 0. The coefficient of z 3 vanishes identi- cally when m = 1 ; that of x is 3(?n — l)b 2 - 2 which cannot be made to vanish for any finite value of b when m = 1* The curve has no asymptotes. Ex. 7. Find the asymptotes of the curve 0r-l)( 2-^ * ' x - 3 and trace the curve. (Fig. 34.) APPLICATIONS TO CURVE TRACING 103 Fig. U EXERCISES Find the asymptotes of each of the following curves: 1. y(a 2 - x 2 ) = b(2z + c). 9 x 2 -2ax 3. x 2 y 2 = a\x 2 - y 2 ). b* 4. y = a + 5. y z = x 2 (a — x). 6. y\x - l) = x 2 . 15. x 8 + 2 x 2 y - xy 2 - 2 y s + 4 y 2 + 2 xy + y = 1 7. (* + a)y«=(y+6)*«. 8. x 2 y 2 = a: 3 -f # + y. 9. xy 2 + x 2 y = a 3 . 10. y(x 2 + 3a 2 )=a* 11. x 3 - 3 axy + y 8 = 0. 12. j; 3 + y 3 = a 3 - 13. x 4 - x 2 y 2 + a 2 x 2 + b* = 0. 14. x 4 - y 4 - a 2 *^. 104 DIFFERENTIAL CALCULUS POLAR COORDINATES 58. When the equation of a curve is expressed in polar coordinates, the vectorial angle is usually regarded as the independent variable. To determine the direction of the curve at any point, it is most convenient to make use of the angle between the tangent and the radius vector to the point of tangency. Let P, Q be two points on the curve (Fig. 35). Join P, Q with the pole 0, and drop a perpendic- ular PM from P on OQ. Let p, 6 be the coordinates of P ; p + Ap, + A6 those of Q. Then the angle POQ = A0', PM=p sin A0; and Fig. 35 MQ = OQ- OM=p + Ap-p cos A0. Hence tan MQP p sin A6 p -f Ap — p cos A0 When Q moves to coincidence with P, the angle MQP approaches as a limit the angle between the radius vector and the tangent line at the point P. This angle will be designated by if/. Thus tan if/ lim * A0 = O p sin A0 p + Ap — p cos A0 But p (1 - cos A0) = 2 P sin 2 i A0, hence tan i/> = A0 l ™ p sin A0 1 A/i sin i A0 . Ap sm i A 0--r-hr- + ^ ±A0 APPLICATIONS TO CURVE TRACING 105 Since A l lr ? ft sm - == 1, the preceding equation reduces to i±0 tan if/ = dp dB dO dp (3) Ex. 1. A point describes a circle of radius p. Prove that at any instant the arc velocity is p times the angle velocity, ds dt ' dt ' dt' Ex. 2. When a point describes a given curve, prove that at any instant the velocity — has a radius component l -&- and a com- dt dt ponent perpendicular to the radius vector p — , and hence that r dt Fig. 37 cos \p ds i dO , . r< , sin \p = p — , tan \p = p - ds This furnishes a dynamical proof of equation (3). If 59. Relation between -£- and p V doc dp the initial line is taken as the axis of x, the tangent line at P makes an angle $ with this line. Hence -f \p = <£ ; dd\ .. _Udy y i.e., + tan" p — i = tan" d Pj dx Fig. 38 60. Length of tangent, normal, polar subtangent, and polar sub- normal. The portions of the tangent and normal intercepted between the point of tangency Pand the line through the pole perpendicular to the radius vector OP, are called the polar 106 DIFFERENTIAL CALCULUS tangent length and the polar normal length; their projections on this perpendicular are called the polar subtangent and polar subnormal. Fig. 39 a Fig. 39 b Thus, let the tangent and normal at P (Figs. 39 a, b) meet the perpendicular to OP in the points jVand M. Then PN is the polar tangent length, PM is the polar normal length, ON is the polar subtangent, OM is the polar subnormal. They are all seen to be independent of the direction of the initial line. The lengths of these lines will now be determined. Since PN=OP- sec OPN= p sec ^ flO 4 •\|j +1 hence polar tangent length = p dp (16 d P \P ' + D v '<$)'■ Again, ON= OP tan OPN= p tan if/ = p hence polar subtangent = p' 2 — dp 2 d$ dp' APPLICATIONS TO CURVE TRACING 107 PM = OP • esc OPN= p esc ^ = yj/ + (&\ , hence polar normal length = -v/p 2 -h ( — ^* ) • OM= OP cot OPN = *£, dO hence polar subnormal = -£ • The signs of the polar tangent length and polar normal length are ambiguous on account of the radical. The direc- d6 tion of the subtangent is determined by the sign of p 2 — • 09. dp When — is positive, the distance ON should be measured to dp the right, and when negative, to the left of an observer placed at and looking along OP; for when increases with d6 p, — is positive (Art. 28), and \p is an acute angle (as in dp d6 Fig. 39 b) ; when 6 decreases as p increases, _ is negative, and if/ is obtuse (Fig. 39 a). EXERCISES 1. In the curve p = a sin 0, find \J/. 2. In the spiral of Archimedes p = aO, show that tan xp = 6 and find the polar subtangent, polar normal, and polar subnormal. Trace the curve. 3. Find for the curve p 2 = a 2 cos 2 the values of all the expres- sions treated in this article. 4. Show that in the curve p6 = a the polar subtangent is of con- stant length. Trace the curve. 5. In the curve p = a(l — cos 0), find if/ and the polar subtangent. 108 DIFFERENTIAL CALCULUS 6. Show that in the curve p = b • e 9cota the tangent makes a con- stant angle a with the radius vector. For this reason, this curve is called the equiangular spiral. 7. Find the angle of intersection of the curves p = a{l+ cos 0), p = h(l — cos 6). a 8. In the parabola p = a sec 2 -, show that cp + $ = tt. EXERCISES ON CHAPTER VII Trace the following curves. Find asymptotes, intervals of in- creasing and decreasing ordinate and direction of bending, as well as intercepts on the axes. 1. y = x s + 2 x 2 - 7 x + 1. 5. y 2 = x s . 2. y 2 = x 3 + 2 x 2 -7 x + 1. 6. ay 2 = x 3 - hx 2 . 3. y = 2 -l) 2 . 7. z 4 -7/ 4 = 2x. 4. x 3 + ?/ 3 = 1. In the following curves find \p, determine whether p can become infinite, and obtain the (angular) intervals of increasing and decreas- ing p. 8. p = a cos 2 6. 10. p = «(1 - cos 0). a 9. p = a sin 3 0. "■• p = «sec 2 -. CHAPTER VIII DIFFERENTIATION OF FUNCTIONS OF TWO VARIABLES Thus far only functions of a single variable have been con- sidered. The present chapter will be devoted to the study of functions of two independent variables x, y. They will be represented by the symbol 2 =/(?, y)- If the simultaneous values of the three variables x, ?/, z are represented as the rectangular coordinates of a point in space, the locus of all such points is a surface having the equation 61. Definition of continuity. A function z of x and y, z=f(x, ?/), is said to be continuous in the vicinity of any point (a, b) when /(a, b) is real, finite, and determinate, and such h Y Tof(a + h,b + k)=f(a,b), k = however h and k approach zero. When a pair of values a, b exists at which any one of these properties does not hold, the function is said to be discontinu- ous at the point (a, b). E.g., let z= x -+JL. x - y When x = 0, then z — — 1 for every value of y ; when y = then 2=4-1 for every value of x. In general, if y — mx, - - 1 + m 1 — m 109 110 DIFFERENTIAL CALCULUS and z may be made to have any value whatever at (0, 0) by giving an appropriate value to m. Geometrically speaking, when the point (x, y) moves up to (G, 0), the limiting value of the ordinate z depends upon the direction of approach. 62. Partial differentiation. If in the function * =/to y) a fixed value y x is given to y, then is a function of x only, and the rate of change in z caused by a change in x is expressed by dz = — dx, (1) dx dz in which — is obtained on the supposition that y is constant. dx To indicate this fact without the qualifying verbal state- ment, equation (1) will be written in the form d^Ax. (2) OX dz The symbol — represents the result obtained by differentiat- ed ing z with regard to x, the variable y being treated as a con- stant; it is called the partial derivative of z with regard to x. From the definition of differentiation, Art. 6, the partial derivative is the result of the indicated operation dz = lim f(x + Ax, y)-f(x,y) dx Ax ~° Ax Similarly, the symbol — represents the result obtained by differentiating z with regard to y, the variable x being treated FUNCTIONS OF TWO VARIABLES 111 as a constant; it is called the partial derivative of z with regard to y. The partial derivative of z with regard to y is accordingly the result of the indicated operation dz = lim f(x, y + &y)-f(x,y) dy A y = ° Ay dz d x z = — die is called the partial x-differential of z, and dx dz cLz = — dy is called the partial y-differential of z. by EXERCISES 1. Given u = x* + 3 x*y* - 7 xy s , prove that x ~-hy^=iu. ox By 2. Given u = tan -1 ^, show that x— + y — = 0. x dx dy 3. u = log (e* + e») ; find ^ + & dx dy 4. n = sin ary; find ^ + ^. dx dy 5. u = log (x + V^T^) ; find x~ + y|^ 6. m = log (tan x + tan y + tan 2) ; show that sin 2x^ + sin 2 y^ + sin 2 *<?H = 2. dx dy dz 7. II = log (a; + y) ; show that f? + ^ = 1. 5a: dy e w 8. ti = -3L; show that *^ + «£!?=«. *+3/ dx *dy 9. u = (y-«)( 2 -x)(x-y); show that ^ + f^ + ^ = 0. dx dy dz 10 - » = ^ show that g+g = (» + »-D«. 11. u = log (x 3 + y 8 + z 8 - 3 xyz) ; show that du , du .Qu 3 dx dy dz x + y-\- z 112 DIFFERENTIAL CALCULUS 63. Total differential. If both x and y are allowed to vary in the function z=f(x, y), the first question that naturally arises is with regard to the meaning of the differential of z. Let z x = f(x x ,%), and z x + Az = f(x l -\- Ax, y x -\- Ay) be two values of the function corresponding to the two pairs of values of the variables a^, y x and x x + Ax, y x + Ay. The difference Az = f(x x + Ax, y x + Ay) - f(x x , y x ) may be regarded as composed of two parts, the first part being the increment which z takes when x changes from x x to x x + Ax, while y remains constant (yz=y x ), and the second part being the additional increment which z takes when y changes from 2/i to y x -f- Ay, while x remains constant (x = x x + Ax). The increment Az may then be written Az = f{x x + Ax, ^ + Ay) - ffa + Ax, ^) 4-/(«i + Ax, 2/0 -f(x x , y x ) - /fa + Aa? > 2/1 + A -y) ^ /fa + A:g > yQ a?/ Ay + f(x x + Ax,y x )-f(x x ,y l ) ^ x Ax From the theorem of mean value, Art. 39, the last equation may be written Az = A/fa + 6 Ax, y x ) Ax + A/fa + Ax, y, + $ x Ay) Ay. (3) dx dy It represents the actual increment Az which the dependent variable z takes when the independent variables x and y take the increments Ax and Ay. FUNCTIONS OF TWO VARIABLES 113 To illustrate, let z = f(x, y) be the equation of a surface (Fig. 40). Let Ai = (xi, y\), A 2 = (xi + Ax, y\), A%= (xi + Ax, y\ + Ay), so that AiPi=f(x h y x ), AiPi~f{xi + Ax, y x ), A z Pi=f{x l + Ax, y x + Ay), Q 2 P 2 =/(2i + Ax, yi) -f(x h yi) = Aiz, Q z Pz=f(xi + Ax, yi + Ay) -f(xi + Ax, y{) = A 2 z, RzP&=f{xi + Ax, y x + Ay) -/(a*, yi) = A x z + A 2 z = Az. As the moving point P passes from P to P-2 along the plane curve P\P 2 , the ordinate takes the increment T t n i? f-' Q 3 ' Q* / i?a A ' V A 2 A 3 Fig. 40 where the derivative is taken at the inter- mediate point x = x\ + 6 Ax, y — y\ (Art. 39). Similarly, as P passes from P 2 to Pz along / the plane curve P 2 Ps, the ordinate takes the further increment where the derivative is taken at the intermediate point y = y\ 4- diAy, x = xi -f Ax. The sum of these two partial increments gives the total increment Az. In the preceding equation (3) let Ax, Ay, Az be replaced by c • dx, € • dy, e • dz respectively, in which dx, dy are entirely arbitrary. After removing the common factor e, let e approach zero. The result is (4) dx dy The differential dz denned by this equation is called the total differential of z. It is not an actual increment of z, but the increment which z would take if its change continued uniform while x changed from x x to x x + dx and y changed from y x to Vi + dy. EL. CALC — 8 114 DIFFERENTIAL CALCULUS In other words, dz is the rate of change of the variable 2 when the independent variables x and y change simultane- ously at the rates of dx, dy respectively. Equation (4) may be written in the form dz = ^dx + -^dy = d x z + d y z, dx dy from which the following theorem can be stated ; the total dif- ferent icd of a function of two variables is equal to the sum of iU partial differentials taken with regard to the separate variables, or the total rate of change of z is equal to the sum of its par- tial rates. The same method can be applied directly to functions of three or more variables. Thus, if u is a function of the vari- ables x, y, z, u=^(x,y J z) ) then du = ~dx-\-^- dy + — * dz. ox dy dz Ex.1. Given z = axy' 2 -\-bx 2 y + ex 3 + cy, then dz = (ay 2 + 2 bxy + 3 cx 2 )dx + (2 axy + bx' 2 + c)dy. Ex. 2. Given u = tan -1 --, show that du — ^ ~ " — x x' 1 + y' 2 Ex. 3. Assuming the characteristic equation of a perfect gas, vp = Rt, in which v is volume, p pressure, t ahsolute temperature, and 11 a constant, express each of the differentials dv, dp, dt, in terms of the other two. Ex. 4. A particle moves on the spherical surface x 2 + y 2 + z 2 = a 2 in a vertical meridian plane inclined at an angle of 60° to the em- plane. If the a; -component of its velocity is ^- feet per second when x — - , find the y-component and the ^-component velocities. Since z = Va 2 — x 2 — y 2 , then dz = ^ ■ y - V Va' 2 — x 2 — y' 2 vV- — x 2 — y' 2 FUNCTIONS OF TWO VARIABLES 115 But since dx = — , and the equation of the given meridian plane is V = x tan 60°, hence dy — VS dx = a — , and y = ^—— • Therefore u j 10 4 dz = - — - & = - — in feet per second. 2V3 2 15 Ex. 5. A triangle has a base of 10 units and an altitude of 6 units. The base is made to increase at the rate of 2 units and the altitude to decrease at the rate of \ unit. At what rate does the area change? Ex. 6. A point on the hvperboloid x 2 — &- = 1 in the position 4 5 x = 2, y — 2 moves so that x increases at the rate of 2 units per sec- ond, while y decreases at the rate of 3 units per second. Find the rate of change of z. Ex. 7. If the area of a rectangle A = xy is incorrectly measured owing to a small error dx, dy in the length of each side, how close is dA = xdy + ydx to the actual error in the area? 64. Total derivative. If in the relation z=f(x, y), the vari- ables x, y are not independent, but both are functions of another variable s, the process of the preceding article can still be applied. The variable z is now a function of s, and its derivative as to s may be expressed in the form dz _ dz dx dz dy ds dx ds dy ds In particular, if y is not independent, but is a function of x, then s may be chosen as x itself, and the preceding equation becomes ' dz _ dz dz dy dx dx dy dx If the functional relation between x and y is given, y = <*>0)> 116 DIFFERENTIAL CALCULUS dz then the same result will be obtained, whether — is deter- dx mined by the present method, or y is first eliminated from the relation *=f(®, y), and the resulting equation is differentiated as to x. The method of this article frequently shortens the process. It is here well to note the difference between — and — • dx dx The former is the partial derivative of the functional expres- sion for z with regard to x, on the supposition that y is con- stant. The latter is the total derivative of z with regard to x, when account is taken of the fact that y is itself a func- tion of x. In the former case the differentiation with regard to x is merely explicit ; in the latter it is both explicit and implicit. dz Ex. 1. Given z = Vx 2 -f y 2 , y = log x; find dx dz _ x y dy dx y/ x 2 _f. yi \Ac 2 + y 2 dx d]L-_ dx _1 x dz X 2 + y dx x yJx 2 + yl hence Ex. 2. If z = tan-i ^- and 4 x 2 + y 2 = 1, show that ^ = — . 2x dx y 65. Differentiation of implicit functions. If, in the relation z =f(x, y), z is assumed to be constant, then dz = : FUNCTIONS OF TWO VARIABLES 117 hence ¥-dx + d fdy = 0, (1) dx By from which ^ = -37- ( 2 ) dy_ dx df dx dy In all such cases either variable is an implicit function of the other, and thus the last equation furnishes a rule for finding the derivative of an implicit function. Ex. 1. Given x 3 + y 3 + 3 axy = c, find -^. dx Since (3 x 2 + 3 ay) + (3 y 2 + 3 ax) ^ = 0, ^ = - x * + ay . dx dx y 2 + ax Ex. 2. /(a* + ty) = c ; J£ = a/'(ax + by) ; |£ = &/'(«* + by) ; <7y _ _ a </x~ &' Ex. 3. If ax 2 + 2 fary 4- % 2 + 2 gx + 2fy + c = 0, find ^. </x Ex. 4. Given x 4 - y4 = c, find^. </x Ex. 5. If x increases at the rate of 2 inches per second as it passes through the value x = 3 inches, at what rate must y change when y = 1 inch in order that the function 2 xy 2 — 3 x 2 y shall remain constant ? If u = 2 xy 2 - 3 x 2 y, then du n i a du a — = 2y 2 -Qxy, ^-=± X y- dx dy du dy dy _ dx _ 2y 2 -6xy _dt dx ~ du ~ 4 xy — 3 x 2 ~~ </x dy dt Since x = 3, y — 1, -y = 2, hence, -y = - 2 T 2 5 inches per second. 118 DIFFERENTIAL CALCULUS Ex. 6. u — v 2 -f vy, v = log s, y = e 8 . Find — Ex. 7. u = sin" 1 (r - a), r = 3 7, s = 4 * 3 . Fin Ex. 8. e » _ c * + a^ = o. Find ^. Ex. 9. sin (*#) - e** - xhj = 0. Find ^. du dt It is to be noticed that the result of differentiating any implicit function of x, y by the method of the present article will agree with the result of differentiation according to the rules of Chapter II. 66. Geometric interpretation. Geometrically, the equation z =f(x, y) represents a surface. The equation y = y x defines a plane parallel to the ^-coordinate plane. The two equations treated simultaneously therefore define the plane section made on the surface z = f(x, y) by the plane y = y x . The derivative dz — - defines the slope of the tangent line to this curve at the ax x point (x l} y u z x ). Similarly, the plane x = x l cuts the surface in a section parallel to the yz-coordinate plane. The slope of the tangent dz line to this second curve is defined by — -. The equations of these two lines are V = 2/i, z - z 1 = — i (x - x x ), dx x dz x = x x , z-z x = -± (y - ?/,). They have the point (x x , y x , z x ) in common ; hence the two lines will define a plane. The equation of any plane through the first line will be of the form dXi + *(//-*/,)= o, FUNCTIONS OF TWO VARIABLES 119 and similarly, the equation of any plane through the second line will be of the form oyi + k'(x-x 1 )=0. These two equations will be identical when dz, , dz-, K = ~ ^H~> k = — — -j tyi Mi hence the equation of the plane containing both lines is 2 - *i = -p- (x - a^H ^-{y - vi). dx x dy x It is called the tangent plane to the surface z = f(x, y) at the point («!, y v z x ). From the equation dM = %Ldx+pLdg, (3) ax l dy 1 it is seen that if x, y receive the arbitrary increments dx, dy, then the increment dz is defined by the sums of the products of these increments by the corresponding partial derivatives. Thus, if dx = x — X], dy = y — y l9 dz = z — z l9 it is seen that the point (x, y, z) always lies in the tangent plane to the sur- face z =f(x, y), however the increments dx and dy approach zero. Moreover, the equations of the line joining (x lf y l9 z x ) to Xj -f A#, ?/! + Ay, z y + Az on the surface will be of the form x - :> \ __ y - y l _ z - z x Ax A?/ Az 120 DIFFERENTIAL CALCULUS Now as Ax, Ay approach zero, the point always remaining on the surface, the line becomes a tangent in the limit, and its equations are x~x 1 = y-y 1 = z-z x (i) dx dy dz ' W wherein dx, dy depend upon the direction of approach, and dz is defined by (3). But a tangent line to the surface is also tangent to any plane section passing through the line, and the line (4) is seen to lie in the tangent plane, hence : TJie tangent lines to all the plane sections of the surface z =f(x, y) passing through the point (x x , y x , Zj) lie in the tangent plane at that point. The line through (x Y , y ly z ± ) perpendicular to the tangent plane dz dz t-^^-^+dy 1 ^-^ is called the normal to the surface at the point (x^ y x , z^). Its equations are X - Xl y-y x . z-z x dz x dz x — 1 dx x dy x If the equation of the surface is given in the implicit form F(x, y, z) = 0, then since dF, , dF , , dF, ft -—dx-\ rt?y -| dz = 0, ox dy ' dz the equation of the tangent plane becomes, if F(a; 1 , y lf z^) = F 1? dxt dt/t dz l FUNCTIONS OF TWO VARIABLES 121 and those of the normal are 3-3* 1 = y—V\ _ g-g i bFy dF\ BF X ' dx x dy 1 dzj EXERCISES 1. Show that the plane z = touches the surface z = xy at (0, 0, 0). 2. Find the equation of the tangent plane to the paraboloid z = 2 x 2 + 4 y 2 at the point (2, 1, 12). 3. Find the equations of the normal to the hyperboloid x 2_4y 2 +2z a =6at (2,2,3). 4. Show that the normal at any point (x h yi, z{) on the sphere x 2 + y 2 + z 2 = 16 will pass through the center. 5. Find the equation of-the tangent plane at any point (x\, y h z\) of the surface x 1 + y* + z 5 = a* and show that the sum of the squares of the intercepts which it makes on the coordinate axes is constant. 6. Show that the volume of the tetrahedron cut from the coor- dinate planes by any tangent plane to the surface xyz = a 3 is constant. 7. The sphere x 2 + y 2 + z 2 = 14 and the ellipsoid 3 x 2 +2 y 2 + z 2 = 20 pass through the point (—1,-2,-3). Determine the angle at which their tangent planes at this point intersect. 8. How far distant from the origin is the tangent plane to the ellipsoid x 2 + 3 y 2 + 2 z 2 = 9 at the point (2, -1, 1) ? 9. Find the equation of the tangent plane and of the normal to the cone z 2 = 2 x 2 + y 2 at (x ly yi, z\) on the surface. Show that the plane will always pass through the vertex of the cone. 10. Find the equations of the tangent line to the circle x 2 + y 2 + z 2 = 25, x + z = 5, at the point (2, 2V3, 3). 122 DIFFERENTIAL CALCULUS 67. Successive partial differentiation. The expressions dz dz — , — which were denned in Art. 62 are functions of both ox oy x and y. dz If — is differentiated partially as to x, the result is written ox d_fdz\ == dh m dx\dxj dx 2 This expression is called the second partial derivative of z as to x. Similarly, the results of the operations indicated by JL{<>?\ JL(te\ JL(te\ dy\dxf dx\dyf dy\dyj dh dh d 2 z are written -— -, — — , -— respectively. oy ox ox oy oy z Beginning with the left, we call these expressions the second partial derivative of z as to x and y, the second partial derivative of z as to y and x, and the second partial derivative of z as to y. 68. Order of differentiation indifferent. Theorem. The successive partial derivatives d 2 z dh By dx dx dy are equal for any values of x and y in the vicinity of which z and its first and second partial x- and ?/-derivatives are continuous. The truth of this theorem will be assumed. Tt should be verified for special cases as in the following examples. FUNCTIONS OF TWO VARIABLES 123 Cor. It follows directly that under corresponding conditions the order of differentiation in the higher partial derivatives is indifferent. dh d s z . ffhs E.g., dx dy dx dx 2 dy dy dx 2 EXERCISES 1. Verify that -^_ = J£}L. % w hen u = xh/\ dx dy dy dx 2. Verify that -_^H_ = J^, when u = xhj + xy 9 . dx dy 2 dy' 2 dx 3. Verify that -^L = -^-, when u = y log (1 + xy). dx dy dy dx 4. In Ex. 3 are there any exceptional values of x, y for which the relation is not true? 5. Given u = (x 2 + y 2 )l, verify the formula * 2 ^ + 2*^ + ^ = 0. x 2 dx dy dy 2 6. Given u = (x s -f t/ s ) I, show that the expression in the left member of the differential equation in Ex. 5 is equal to — . 7. Given u = (x* + y* + z *yh ■ prove that B ~ + ^ + = 0. 8. Given u = sec (y+ax) + tan (?/ - ax) ; prove that — = a^' u - dx 2 dy 2 9. Given M=sinx cosy; verify that d u = ^ = ^ u . J dy 2 dx 2 dxdydxdy dx 2 dy 2 d 2 u d 2 u 10. Given «=(4a&- c 2 )-| ; prove that :rr = r - ^!* 11. If «=*&L, show that o;^ + „-i^-=2^. * + # dx* J dxdy dx 12. Given u = log (a: 2 + y 2 ), prove 5 ^ + — = 0. a< 2 ■ ay 13. If u = (x n + y n y, show that the equation of Ex. 5 is satisfied. 14. Given u = (x* + y* + z 2 + w 2 )" 1 , prove ^ + <*% + ^f + S^M. = d* 2 9^ a* 2 Qw* CHAPTER IX CHANGE OF VARIABLE 69. Interchange of dependent and independent variables. If y is a continuous function of x, defined by the equation f(x, y) = 0, the symbol — represents the derivative of y with regard to x, when one exists. If x is regarded as a function of y, defined by the same equation, the symbol — represents the dy derivative of x with regard to y, when one exists. It is re- quired to find the relation between -^ and — dx dy Let x, y change from the initial values x ly y x to the values x 1 + Ax, y x + Ay, subject to the relation f(x, y) = 0. Then, since A Ay _ 1 Ax Ax Ay it follows, by taking the limit, that dx dx dy (1) Hence, if y and x are connected by a functional relation, the derivative of y with regard to x is the reciprocal of the derivative of x with regard to y. This process is known as changing the independent variable from x to y. The corresponding relations for the -higher de- 124 CHANGE OF VARIABLE 125 rivatives are less simple. They are obtained in the following manner : d y dx d x To express — 4 in terms of — , — - differentiate (1) as to x ; dx 2 dy dy 2 d 2 y _ d [ 1 d [ 1 dy _ d \ 1 1 dx 2 dx 1 dx dy dx dx dy dx idy dy\ idyl dx dy But hence In a similar manner, — l 1 }- dy dx 1 idyl d 2 x dy 2 , /dx\ 2 [dyj d?x d 2 y dx 2 dy 2 fdx\ 3 ' \dy) ner, d 3 y d 3 x dx o fd 2 x\i d?dy WJ dx 3 fdx\ 5 \dy) (2) (3) 70. Change of the dependent variable. If y is a function of z, let it be required to express — — , •••in terms of — , — -, •••. dx dx- dx dx-' Suppose y = <f>(z). Then dy ^dycte = ^u\dz_ dx dz dx dx dx 2 dxV KJ dx dz d .i^\,,i/^d 2 z 126 DIFFERENTIAL CALCULUS But ^-<l>'(z) = -cp'(z)- = <p"(z) — ■ dz* K ; dz v ; <to v v "eta S-^S'+^S- (4) The higher ^derivatives of y can be similarly expressed in terms of ^-derivatives of z. 71. Change of the independent variable. Let y be a function of x, and let both x and y be functions of a new variable I. It is required to express — in terms of -3L. and — in terms of 1 , cto eft' dx 2 -■- and — a- eft eft 2 By Art. 8, e% % = d* ? (1) die do? hence d 2 ?/ C?£C d 2 ^ e% ^y_ ~dfdi~~d^di (9) fdx\* dt In practical examples it is usually better to work by the methods here illustrated than to use the resulting formulas. 72. Simultaneous changes of dependent and of independent vari- ables. Suppose, for example, that an equation involving x, y, — , . . . is given, and it is required to transform the equation into polar coordinates by means of the formulas x = p cos 0, y = p sin 6. Since the variables x and y are connected by some equation (y being a function of x), we may regard x, y, p as CHANGE OF VARIABLE functions of 6. E.g., consider the function 12' 11 = HtJf dry dx 2 rom Art. 71, dy dy _d0 f dx dx dO dx d-y dy drx d 2 y dO ' d6- dO ' d0\ dx- /dxX 3 tie. By substituting these values in the expression for E, it becomes R dx\ 2 , (dy dO \d0 dx d~y dy ^ d 2 x dO ' ~d&- ~ dd ' d¥ This is in terms of a new independent variable 0. We have now to express these ^-derivatives of x and y in terms of p and 6. From the relations x = p cos 0, y = p sin we have g = - p cos*-2sin*| + cos*g, dhj d$- ' = - p sin + 2 cos =c 4- sin dp d$ de 2 128 DIFFERENTIAL CALCULUS Upon substituting these values in the last expression for B, we obtain B = W] f + 2 («!>)*- p *E (16 EXERCISES 1. Change the independent variable from x to z in the equation x 2 — \ + x — '- + y = 0, when x = e*. dx 2 dx a dx dz dx 2 dz 2 dz Hence x 2 ^- + x^- + y = becomes ^ + y = 0. 2. Interchange the function and the variable in the equation dx 2 \dxj 3. Interchange x and y in the equation </x 2 4. Change the independent variable from ar to y in the equation JdY\ 2 _dy d*«_d 2 y (djY = o V^ 2 / dx dx* dx 2 \dx) 5. Change the dependent variable from y to z in the equation !^= 1 +2(1 +.V)fM', when y = tans. </x2 1 + y* \dxl * CHANGE OF VARIABLE 129 6. Change the independent variable from x to y in the equation x 2 h x \- u = 0, when y = log x. dx 2 dx 7. If y is a function of x, and x a function of the time t, express the ^/-acceleration in terms of the ar-acceleration, and the x-velocity. Since dy = dydx dt dx dt hence d 2 y _ dy (Px .dx d fdy\ dt 2 ~ dx dt 2 dt <it\dx)' c ]l( ( 111\ = ^_((J][\flx = d 2 y(lx it\dxl ~ dxKdxj dt dx 2 dt' But dt d 2 y _ dy d 2 x d 2 y ldx\ 2 hence ^-s^ + sUi' In the abbreviated notation for /-derivatives, dx dx 2 8. Change the independent variable from x to u in the equation dry 2 x dy y dx 2 1 + x 2 dx (1 + x 2 ) 2 ] y + y = 0, when x = tan u. 9. Change the independent variable from x to t in the equation (1 _ X 2) A _ x ( Il = o, when x = cos t. dx 2 d-x 10. Show that the equation d 2 y , dy . A dx 2 dx J remains unchanged in form by the substitution x = . EL. CALC 9 130 DIFFERENTIAL CALCULUS 11. Interchange the variable and the function in the equation dx 2 \dxl y \dx) 12. Change the dependent variable from y to z in the equation |j*+ a - y)% + y 2 = °' when y = z ' 2 - Change the independent variable from x to t in the equations : 13. CI — x 2 ) — — x— + y = 0, given a: = cos t. K clx 2 dx y ' h 14. x 3 — - + 3 x 2 \- x 1- v = 0, given x — e l . dx 3 dx' 1 dx 15. x 4^ +fl 2^ = 0, x = -. dx 2 t dy X dx~ V 16. Transform by assuming x = p cos 6, y = p sin 0. 17. Given z = 7 + * 2 , ?/ = 3 + * 2 - 3 l\ Find — . CHAPTER X EXPANSION OF FUNCTIONS It is sometimes necessary to expand a given function in a series of powers of the independent variable. For instance, in order to compute and tabulate the successive numerical values of sin x for different values of x, it is convenient to have sin x developed in a series of powers of x with coeffi- cients independent of x. Simple cases of such development have been met with in algebra. For example, by the binomial theorem, (a + x) n = a n + na n ~ l x + n l - 1 ) a — v + • . . ; (1) J. • A and again, by ordinary division, j-L-=l + s + 3» + aj »+.... (2) It is to be observed, however, that the series is a proper representative of the function only for values of x within a certain interval. For instance, the identity in (1) holds only for values of x between — a and -f a when n is not a positive integer ; and the identity in (2) holds only for values of x between — 1 and -f 1. In each of these examples, if a finite value outside of the stated limits is given to x, the sum of an infinite number of terms of the series will be infinite, while the function in the first member will be finite. 131 132 DIFFERENTIAL CALCULUS 73. Convergence and divergence of series.* An infinite series is said to be convergent or divergent according as the sum of the first n terms of the series does or does not approach a finite limit when n is increased without limit. Those values of x for which a series of powers of x is con- vergent constitute the interval of convergence of the series. For example, the sum of the first n terms of the geometric series , , 2 , s , a -f- ax 4- ax z + ax? -\ ail — x n ) is s n = -1 >-. 1 — x First let x be numerically less than unity. Then when n is taken sufficiently large, the term x n approaches zero ; hence Hm s n = -^-- W = co l_ x Next let x be numerically greater than unity. Then x n be- comes infinite when n is infinite ; hence, in this case lim ,,-od *» = «>• Thus the given series is convergent or divergent according as x is numerically less or greater than unity. The condition for convergence may then be written . -l<a<i, and the interval of convergence is between — 1 and -|- 1. Similarly the geometric series l-3z + 9z 2 -27^+..., > * For an elementary, yet comprehensive and rigorous, treatment of this subject, see Professor Osgood's " Introduction to Infinite Series" (Harvard University Press, 1897). EXPANSION OF FUNCTIONS 133 whose common ratio is — 3 x, is convergent or divergent accord- ing as 3 a; is numerically less or greater than unity. The condition for convergence is — 1 < 3 x < 1, and hence the interval of convergence is between — ^ and -f- ^. 74. General test for convergence. Let £ = ?<! + u 2 + u 3 -\ hM n + M B+1 H be a series of positive terms having the property that w+1 < r (r a fixed proper fraction) for all values of n that exceed a def- inite integer k that can be assigned. We wish to prove that under these conditions S is convergent. This is called the ratio test for convergence. According to hypothesis we have the inequalities ^±l<r, ^*±*<r, ^±- 3 <r, etc. U k U k+\ U k+2 By multiplying the first two equalities together we obtain ^^ < r 2 ; then, multiplying this result by the third of the u k given inequalities we deduce further -^ ±? <r 3 ; and so on. These results may be written in the form %+i < ru k , u k+2 < r*u k , w 4+3 < rhi k , .-• , u k+p < r'u k . Hence we have the inequality S < u x -f u 2 H (- u k -f ru k -f r 2 ^ + rX -\ But the series in the right member, which may be denoted by S'j can be put in the form S' = ^ + W2 -f — + m*_i + u h (1 -f r + r 2 + r 3 + •••) = w x + w 2 H f- ?^_! + -^*- • 1— r 134 DIFFERENTIAL CALCULUS The terms u x , u 2 , ••• , u k being assumed finite, it follows that S' is finite and hence S, which is less than S', also is finite. Since S is formed by the successive addition of positive terms, it follows that the series S converges towards a definite finite limit. If the series S contains an infinite number of negative, as well as of positive, terms, it converges whenever the series formed by the positive, or absolute, values of its terms eon- verges. The series is then said to be absolutely convergent. In order to prove the preceding theorem, we obeerve that the positive terms of S taken alone form a converging series, whose limit will be denoted by P, and the negative terms taken alone will form a converging series whose limit will be denoted by — J¥. t Let S m denote the sum of the first m terms of S and suppose that these consist of p positive terms whose sum is denoted by P p and of n negative terms whose sum is — N n . Then we have S m = P p — N n . Now when m becomes infinite, p and n also become infinite, and hence a lim - e . . Km p _ lira jy __ p _ *r m i oo m p = oo p n = oo » Therefore, S is convergent. When a series is convergent, but the series formed with the absolute values of its terms is not convergent, the given series is said to be conditionally convergent* The absolute value of a real number u is its numerical value taken positively, and is written | u |. If a series consists of terms that are alternately positive * The appropriateness of this terminology is due to the fact that the terms of an absolutely convergent series can be rearranged in any way, "without altering the limit of the sum of the series ; and that this is not true of a con- ditionally convergent series. For a simple proof, see Osgood, pp. 43, 44. EXPANSION OF FUNCTIONS 135 and negative, and if, after any definite term of the series, each, succeeding term is numerically less than the preceding one, then the series is convergent. For, suppose that beginning with the term n k , the series is &' = V k ~ V k + l + %+2 — U k+3 + U k+i ~ '" > in which u k , v k+l , etc. represent positive numbers and n k+l < u k , u k+2 < Uk+u ••' j n m+i < l 'mi f° r every value of m greater than 7c. By grouping the terms in pairs, (u k — ii k+l ), (u k+2 —u k+ ^), ••• , each of which is positive, it is seen that S' has a positive value, which .may be finite or infinite. But S' may also be written in the form & = u k - [(%+i - *<*+ 2 ) + (m*+8 - u k+i) + •••], wherein the terms in brackets are all positive, hence S' has a value less than u k . It therefore converges towards a definite finite limit. It now follows that the approximate value of S' obtained by algebraically adding u k , u k+l , ••• , u m differs from the true value of the series by a number less than u m . This fact can be shown in precisely the same way as that by which S' has just been shown to have a value less than u k . Ex. 1. Is the series 1 1 \- ... + (— l)' 1-1 - + ••• con- 2 3 4 n vergent i Since the terms are alternately positive and negative and their numerical values are always decreasing, it follows at once from the preceding paragraph that this series is convergent. It will be found later that its value is log 2. Ex. 2. Prove the convergence of the series met with in Art. 16, a+ i. + l + i + ...+J- + ...-. •2! 31 4! T n\ 136 DIFFERENTIAL CALCULUS In this case u n = 1 , w n+1 = 1— . Hence ^±1 = — 1— . This n ! (w + 1) ! m„ n + 1 ratio is less than \ for all values of n greater than 2, and the ratio condition for convergence is satisfied. Ex. 3. Prove the divergence of the harmonic series The ratio u n+ i : u n becomes greater than r when n is sufficiently large. By grouping the terms it may be written in the form 1 + *+(* + *) + (* + * + * + »+-, the succeeding groups having 2 3 , 2 4 , ••• , 2 n , ••• consecutive terms re- spectively. The sum of the terms in any group is greater than \. For, in the nth group the last term — has the least value, and as there are 2 n_1 terms in the group their sum is greater than 2 n_1 — = -• As there is an infinity of such groups, their sum is infinite. Ex. 4. The series £=1 + 1 + 1-+ ... +1+... is convergent for p > 1. Let the terms of S be grouped in the following manner : the nth group beginning with and containing 2 n_1 terms. The nth group is accordingly less than its first term multiplied by the number of terms in the group, that is, < 2 n_1 • — = — — . & r ' ' (2" _1 ) p (2 n— 1 ) p_1 Hence we deduce the inequality the right member of which is a geometric series having — — as the common ratio. It is therefore convergent, and hence S is convergent, EXPANSION OF FUNCTIONS 137 if < 1. This inequality is satisfied for every value of ^greater than unity. Moreover, it was shown in Ex. 3 that ior p = 1 the series 5 is divergent. When p < 1, S is divergent. For in that case — ->-, n is any positive integer (except 1), and therefore the n p n terms of S are greater than the corresponding terms of the harmonic series. Hence : The necessary and sufficient condition that the series 1 -\ — + - \- ••• may converge is p > 1. Ex. 5. Show that the- series 1 = — \- 12 2-3 3.4+ n(n+l) is convergent. This may fc>e proved by comparison with the series in Ex. 4 for the particular case p = 2. 1 .-, 1 .1 1 .1 1 .1 Since l-2 < ' 2-3 < 2 2 ' 3-4 < 3 2 ' '" ' n(n + 1) < n*' "' it follows that the value of the given series is less than that of which is known to be convergent on account of the theorem deduced in the preceding example. Ex. 6. Examine for convergence the series whose nth term is — — ■ n 2 +l r n = * ^ - 1 Hi nt. n 2 + 1 1 ^ n + 1 n -\ n Ex. 7. Examine for convergence the series l_ 2 f i _ . , (~l) w ~^ 2 5 10 n 2 + 1 Ex. 8. Determine whether the series whose nth term is n 2 + 1 convergent or not; the series whose general term is — — . n 3 + l 138 DIFFERENTIAL CALCULUS 75. Interval of convergence. If the terms u 1} u 2 , ••• of a given series are functions of a variable x, then the series will usually converge for .some values of x and diverge for all others. In such a case the problem is to determine the interval of conver- gence, that is, the range of values of x for which the series is convergent The following examples will illustrate the method of procedure. Ex. 1. Determine the interval of convergence of the series 1 + x -f 2 x 2 + 3 x s + ••• + nx n + ... . In this case u n = (n — l)^"" -1 and u n+1 = nx n .. Hence, ^ = "*" = -JL- x. u n (n-l)x n - 1 n-1 According to the ratio condition for convergence, it is necessary that this ratio shall be numerically less than 1 for all values of n exceeding a fixed number k. As n increases, the fraction 71—1 approaches unity. Hence if [ x \ has any fixed value less than 1, the given series is absolutely convergent. The interval of convergence is defined by the inequalities — 1 < x < 1. It is evident from the preceding example that the ratio con- dition for the absolute convergence of a series ma}^ be written lim l ±n±l < 1, (3) which is especially convenient for application. Ex. 2. Find the interval of convergence of the series 1 + 2 • 2 x + 3 • 4 x 2 + 4 • 8 x* + 5 • 16 x* +.- . EXPANSION OF FUNCTIONS 139 Here the nth term u n is n 2 n_1 x n_1 , and the (n -f l)th term u n+1 is (n + \)2 n x n ; i ^n+i (n 4- l)2"z* (n + 1) 2 x lience -J!±±=*-— — ^ = i — — i w n w2 n - 1 x" » n therefore when n = co, -^ = 2 x. It follows by (3) that the series is absolutely convergent when — 1 <2#<1, and that the interval of convergence is between — \ and + \. The series is evidently not convergent when x has either of the extreme values. Ex. 3. Find the interval of convergence of the series x 1-3 3 x 3 x 5 x" > (_l)n-i^2 M -i ^ • 3 3 5-3 5 7-3 7 "" (2n-l)3 2 "- 1 Here t/ H 4.i I _ 2n - 1 3 2 ""* a; 2 "* 1 = 2n-l z 2 . w ft 2 n + 1 ' b 2 "* 1 ' x 2 "-i 2 n + 1 ' 3 2 ' v Inn hence n = co 3 2 ' a: 2 thus the series is absolutely convergent when — < 1, i.e., when b — 3 < x < 3, and the interval of convergence is from — 3 to + 3. The extreme values of x, in the present case, render the series con- ditionally convergent. Ex. 4. Determine the interval of convergence of the series 2 ! 4 ! 6 ! . ' (2 n - 2) ! Since even powers of x are positive, the terms of this series are alternately positive and negative. The term w n+1 is derived from u n by multiplying it by . For all values of n such that (2 n — 1)2 n this fraction is less than 1, we shall have the condition [ u n+1 | < | n n \ and the series is convergent on account of the property of series with alternately positive and negative terms. 140 DIFFERENTIAL CALCULUS Ex. 5. Prove the convergence of the series Inthiscase | u n |=z 2 - f 1 +-+••• + — Y Notice that h . m I u J \9 4 2"- 1 / n=<x> ' n[ is not zero. The series is nevertheless convergent, but not absolutely convergent. Ex. 6. Determine the interval of convergence for the series 16 n Ex. 7. Determine the interval of convergence for the series -A_ + 2 3 + ... + _n _ + .... x-1 (x-1) 2 (ar-l)* (ar-l) w Ex. 8. Find the interval of convergence for the binomial series -, , , a(a — 1) 9 , a(a — Y)(a — 2) „ , 1 4-rtxH — i — *- x 2 H — ^ ^ '- x 3 + in which a is any constant. Ex. 9. Show that the series has the same interval of convergence as that of Ex. 3 ; but that the extreme values of x render the series absolutely convergent. 76. Remainder after n terms. The last article treated of the interval of convergence of a given series without reference to the question whether or not it was the development of any known function. On the- other hand, the series that present themselves in this chapter are the developments of given func- tions, and the first question that arises is concerning those values of x for which the function is equivalent to its development. EXPANSION OF FUNCTIONS 141 When a series 1ms such a generating function, the difference between the value of the function and the sum of the first n terms of its development is called the remainder after n terms. Accord- ingly, if f(x) is the function, S n (x) the sum of the first n terms of the series, and li n (x) the remainder obtained by subtracting S„(x) from fix), then in which S n (x), R n (x) are functions of n as well as of x. If JToO R -^) = > tllei1 Hi" «-(*) =/(*) i thus the limit of the series S n (x) is the generating function when the limit of the remainder is zero. Frequently this is a sufficient test for the convergence of a series. If a series is expressed in integral powers of x — a, the pre- ceding conditions are to be modified by substituting x — a for x ; in other respects each criterion is to be applied as before. 77. Maclaurin's expansion of a function in a power-series.* It will now be shown that all the developments of functions in power-series given in algebra and trigonometry are but special cases of one general formula of expansion. It is proposed to find a formula for the expansion, in ascending positive integral powers of x — a, of any assigned function which, with its successive derivatives, is continuous in the vicinity of the value x = a. The preliminary investigation will proceed on the hypothe- sis that the assigned function f(x) has such a development, ♦Named after Colin Maclaurin (1608-1746), who published it in his " Treatise on Fluxions " (1742) ; but he distinctly says it was known by James Stirling (1692-1770), who also published it in his " Methodus Differ- entialis "(1730), and by Taylor (see Art. 78). 142 DIFFERENTIAL CALCULUS and that the latter can be treated as identical with the former for all values of x within a certain interval of equivalence that includes the value x = a. From this hypothesis the coeffi- cients of the different powers of x — a will be determined. It will then remain to test the validity of the result by finding the conditions that must be fulfilled, in order that the series so obtained may be a proper representation of the generating function. Let the assumed identity be fix) = A+B(x-a) + C(x- a) 2 + D(x- a) 3 + E(x-ay+-, (1) in which A, B, C, ••• are undetermined coefficients, indepen- dent of x. Successive differentiation with regard to x supplies the following additional identities, on the hypothesis that the derivative of each series can be obtained by differentiating it term by term, and that it has some interval of equivalence with its corresponding function : f(x) = B + 2C(x-a)+ 3D(x-a) 2 + ±E(x-af+ ••• f"(x)= 2(7 +3-2D(x-a) + 4 . 3E(x-af+ •-. f"(x)= 3.2Z> +4.3.2^(»-o)+-.- If, now, the special value a is given to x, the following equations will be obtained: f(a)=A,f(a) = B, /"(a) = 2 C,f"(a) = 3-2D, -. Hence, A =/(«), B=/(a), C= f -^f, D= f -^, ■- EXPANSION OF FUNCTIONS 143 The coefficients in (1) are now determined, and the required development is /(a5)=/(a) + /(a)(a5-a)+^^(aj-a)«+£^(a5-a)» (2) 2! This is known as Maclaurin's series, and the theorem ex- pressed in the formula is called Maclaurin's theorem. Ex. 1. Expand log x in powers of x — a, a being positive 1 x Here /(*) = log x,f(x) = £,/"(*) 1 1.0 V'"(*) =:L T Hence, /(a) = log a,f'(a) = If" (a) = - I, /"'(a) = ^ 1-2 /«(a) and, by (2), the required development is (-l)-i( w -l)| log x = log a + 1 (* - a) - J- O - a) 2 + -L (a; - a) J a 2 a 2 o a 3 The condition for the convergence of this series is lira I (x — a)^ 1 . (x - a)> (n + l)a n + 1 x — a na n <1, <1, \x - a \<a, 0<x<2a. • This series may be called the development of log x in the vicinity of x — a. Its development in the vicinity of x = 1 has the simpler form log X = X - 1 - 1(X - I)* + \(X - l) 3 - .., which holds for values of x between and 2. 144 DIFFERENTIAL CALCULUS In using this series for the computation of a table of logarithms we may put for a any number whose logarithm is already known, and for x any number near a in magnitude. It is a great advantage to keep x — a so small that the power-series in x — a may be not merely convergent, but may converge to its limit so rapidly that all powers of x — a above the fourth or fifth may be neglected without affecting the desired degree of accuracy. E.g., being given log 10 = 2.302585, suppose it is required to com- pute log 11, log 12, •••, log 20. Put a = 10, and x = 11. Then logi^iogio+^-Kr^HKi^^K^+Ki^-Ki^HKiV) 7 --- The numerical work may be tabulated in the following form : + 2.30258509 .10000000 - .00500000 .00033333 .00002500 .00000200 .00000017 .00000001 - .00502517 2.40292043 - .00502517 2.39789526 Hence log 11 = 2.397895-., correct to six places of decimals. To make sure of the sixth figure it is well to carry the work to seven or eight figures. The remaining terms of the series after KrV) 7 cannot affect this result, because their sum is less than an infinite decreasing geometric progression whose first term is IGV) 8 and whose ratio is fo. From the formula 1 - r i it follows that the remainder is less than 72 • 10 9 To calculate log 12, log 13, ••• we could now keep a = 10 and let x = 12, 13, ••• successively, but in order to secure rapid convergence it is better to change the value of a, choosing for a the nearest number EXPANSION OF FUNCTIONS 145 whose logarithm has been found. Thus, in computing log 12 we can use either of the two series log 12 = log io + a - KA) 2 + K&Y - -. log 12 = log li + A - KA) 2 + Kt 1 !) 3 - - ; but it will be found that five terms of the second furnish as close an approximation as nine terms of the first. The practical advantage of the step-by-step process will depend on how many of the intermedi- ate values we actually require. If we are given log 10 and wish to compute log 15, it may be easier to compute the latter directly with- out determining the intermediate values. Ex. 2. Develop f(x) = x s — 2 x 1 + 5 x — 7 in powers of x — 1 and use the result to compute/(1.02),/(1.01),/(.99),/(.98). Ex. 3. Develop 3 y 2 — 14 y + 7 in powers of y — 2. Ex. 4. Expand sin x in powers of x — a and use the result to com- pute sin 31°. Let a = 30°, x = 31°. In the formula sin x = sin a + cos a(x - a)-^^(x - a) 2 - ^^ (x - a) 3 •», the difference x — a becomes 1° or .001745 radians, and the coefficients of its various powers are all known ; since sin a = .5, cos a = .866025 the work is now reduced to numerical calculation in which three terms are sufficient to obtain the result correct to six places of decimals. In general, to calculate sin x or cos x, take for a the nearest value for which sin a, cos a are known. The expansion of a function f(x) in a series of ascending powers of x can be obtained at once from formula (2) by giv- ing a the particular value zero. The series then becomes A*)=A0) + /-(0)x + q|i^ + ... + ^^ + .... (3) EL. CALC. — 10 146 DIFFERENTIAL CALCULUS Ex. 5. Expand sin x in powers of x, and find the interval of con- vergence of the series. Efeve /0) = sin x, /(0) = 0, f(x) =cosx, /'(()) =1, f"(x) = - sin a;, /"(0) = 0, /'"(*) = - cos*, /"'(())= -1 f"(x) = sinx, /iv(0)=0, /v(ar) = cos x, / v (0) = 1, Hence, by (3), + 1 • x + • x 2 - i. x- 3 + • x 4 + o ! thus the required development is sin a: = a; x s H x 5 x 7 + ••• + -^ — 2!! — a: 2 "- 1 + .... 3! 5! 7! (2w-l)! To find the interval of convergence of the series, use the method of Art. 74. The ratio of u n+1 to u n is u n +\_ a> +1 Z 2 "" 1 x 2 « n (2n + l)! (2n-l)l (2n + l)2n This ratio approaches the limit zero, when n becomes infinite, how- ever large may be the fixed value assigned to x. This limit being less than unity, the series is convergent for any finite value of x, and hence the interval of convergence is from — oo to -f oo. The preceding series may be used to compute the numerical value of sin x for any given value of x. It is rapidly convergent when x is small. Take, for example, x = .5 radians. Then sin(.5)_.o-— +2>3>4>5 2.3.^.5.6.7+ -> EXPANSION OF FUNCTIONS 147 = .5000000 - .0208333 + .0002604 - .0000015 + .0000000 sin (.5) = .4791256 •• Show that the ratio of u 5 to u A is 2 \-g ; and hence that the error in stopping at u 4 is numerically less than u t [^i? + Ciis)^ ••*]» ^ na ^ is, < ?fa u 4 . When x is not small, it is better to use the more general series in powers of x — a. Ex. 6. Show that the development of cos x is cosx = l - J L + ±--±- + ... +i ii ± + .... 2! 4! 6! (2n-2)! ' and that the interval of convergence is from — oo to + oo. Ex. 7. Develop the exponential functions a x , e x . Here /(*) = «*, /'(*)=a*loga, /"(*) = a*(log «) 2 , .», /*■>(*) =a-(loga)»; hence /(0)=l; /'(0)=log a,/"(0) = (loga)2, ...,/W(0) = (log a)», and a-H (log a)* + ^ a )V + ... + ( lo g q )V + .... •2! n ! As a special case, put a = e. Then log a = log e = 1, and ex^i + ^ + rL + rL-f ... +£!+ .... 2.! 3! n! These series are convergent for every finite value of x. Ex. 8. Compute 10* when x = .1. Ex. 9. Compute 10* when r = 2.01. 148 DIFFERENTIAL CALCULUS Ex. 10. Defining the hyperbolic cosine and the hyperbolic sine by the equations cosh x = \(e x + e x ), sinh x = \{e x — e~ x ), prove — cosh x = sinh x, — sinh x = cosh x, dx dx cosh 0=1, sinh 0=0; and hence expand cosh x and sinh x in powers of x. Verify that cosh x + sinh x = e x , and cosh x — sinh x = e~ x . Compute cosh 2 and sinh 2 to four decimal places. Show that the error made in stopping the series at any term is much less than the last term used. 78. Taylor's series. If a function of the sum of two num- bers a and x is given, f(a + x), it is frequently desir- able to expand the function in powers of one of them, say x. In the function /(a + x), a is to be regarded as constant, so that, considered as a function of x, it may be expanded by formula (3) of the preceding article. In that formula, the constant term in the expansion is the value which the func- tion has when x is made equal to zero, hence the first term in the expansion of f(a 4- x) may be written /(a). In the same manner the coefficients of the successive powers of x are the corresponding derivatives of f(a + x) as to x, in which x is put equal to zero after the differentiation has been per- formed. The expansion may therefore be written f(a + x) = f(a) + f(a)x +£^r-x 2 + ••• + f< [ a) oc n + •••• 9. ' m T This series, from the name of its discoverer, is known as Taylor } s series, and the theorem expressed by the formula is known as Taylor's theorem. EXPANSION OF FUNCTIONS 149 Ex. 1. Expand sin (a + x) in powers of x. Here f(a + x) — sin (a + x), hence f(a) = sin a, and f'( a ) — cos a, Hence sin (a + x) = sin a + cos a • x - s i^x2 _ cos_a 3 v ' 2! 3! Ex. 2. Compute sin 61°, by putting a = 60°. EXERCISES 1. Expand tan x in powers of x. Obtain three terms. 2. Compare the expansion of tan x with the quotient derived by dividing the series for sin x by that for cos x. 3. Find a limit for the error which occurs in replacing cos x by the first three terms of its expansion in powers of x when x = \ of a radian. 4. Prove that log (x + Vl + x 2 ) = x - -±— + -^- • 2-32-4 5. Provelogc'os^-^-^- 1 -^ 6 -?!^!.... 8 2 4! 6! 8! 6. Compute sin 1° cdrrect to six places of decimals. 7. Expand Vl — x 2 in powers of x, and compare with the expan- sion by the binomial theorem. 8. Expand cos x in powers of x — -. 9. Expand e x+h in powers of h. 10. Arrange 3 x 3 — 5 x 2 + 8 x — 5 in powers of x — 2. 11. Expand log (x + K) in powers of h. 12. Arrange x 4 — 1 in powers of x + 1. X"~N w ! Cl n ~ r X r 13. Prove the binomial theorem (a + x) n = a n + ••• = x, - 7~, \T- r=0 v J Find the form of the series when n is not an integer, and determine the interval of convergence. 150 DIFFERENTIAL CALCULUS 14. Find V12G = V125 + 1 = 5 vl + T ^ 3 to three places of deci- mals by the binomial theorem. 15. Find S^Lm 16. Calculate log 31. 17. What is the greatest value of m that will permit the approxi- mation (1 + m) 4 = 1 + 4 m with an error not exceeding .001 ? 18. Expand - in powers of x — 1 and find the interval of con- x vergence. 79. Rolle's theorem. By Art. 76 a series can be the correct representation of its generating function only when the re- mainder after n terms can be made as small as desired by taking n large enough. Before obtaining the form of this remainder it is necessary to introduce the following lemma. Rolle's theorem. If f(x) and its first derivative are continu- ous for all values of x between a and b, and if f(a), f(b) both vanish, then f'(x) will vanish for some value of x between a and b. * The proof follows immediately from the theorem of mean value (Art. 39). See Figure 41. 80. Form of remainder in Maclaurin's series. Let the re- mainder after n terms be de- noted by B n (x, a), which is a function of x and a as well as of n. Since each of the succeeding terms is divisible by (x — a) n , R n may be con- Fig. 41 veniently written in the form R n (x, a) = ^ — pi- <f> (x, a). EXPANSION OF FUNCTIONS 151 The problem is now to determine <£(#, a) so that the relation f(x)=f(a)+f(a)(x-a)+^(x- af+ .'.. + 7 \f, (x - a) H J + yv ' ' (z - ci) n (1) (n — 1)1 n\ may be an algebraic identity, in which the right-hand mem- ber contains only the first n terms of the series, with the remainder after n terms. Thus, by transposing, f(x) _/(«)_/(«)(» - a) - f J£p(x-ay - - _/->) (a! _ B w_ »fe«) (a _ o) .= o. (2) (h-1)! V ; n! ^ ; w Let a new function, F(z), be defined as follows: F(z) =f(x) -/(*) -f\z){x - z) -£& (x - zf - -. -O^**-'^ 1 -*^*-*- (3) This function F(z) vanishes when z—x, as is seen by inspection, and it also vanishes when z = a, since it then becomes identical with the left-hand member of (2) ; hence, by Rolle's theorem, its derivative F'(z) vanishes for some value of z between x and a, say z v But F\z) = -/(«) -|-/(2) -/»(»)(* - »)+/"(*)(» -«)-... (n-l)! V ' ^(n-l)! 1 - ' 152 DIFFERENTIAL CALCULUS These terms cancel each other in pairs except the last two ; hence _ x Since F'(z) vanishes when z = z 1} it follows that 4>(x, a) =/<»>&). (4) In this expression z x lies between x and a, and may thus be represented by J Zl = a + 0(x- a), where 6 is a positive proper fraction. Hence from (4) 4>(x, a)=f»[a + d(x-a)], and R n {x, a) = / w l> + fl(*-«) 1 {x _ o) - .• The complete form of the expansion of f(x) is then /(a>) =/(«) + /»(«) fas- a) +^f^ (oc-a) 2 + - +f"^ (*-a)»-* + / (W We(g--«)) («-«)". (5) (n — 1); w! in which n is any positive integer. The series may be carried to any desired number of terms by increasing n, and the last term in (5) gives the remainder (or error) after the first n terms of the series. The symbol / (n) (« 4- (x — a)) indicates that f(x) is to be differentiated n times with regard to x, and that x is then to be replaced by a + 0(x — a). * This form of the remainder was found by Lagrange (1736-1813), who pub- lished it in the Memoires de l'Academie des Sciences a Berlin, 1772. EXPANSION OF FUNCTIONS 153 81. Another expression for the remainder. Instead of putting RJx, a) in the form ( x _ a \» r-^fo a), n\ it is sometimes convenient to write it R n (x, a) = (x — a) ^(#, a). Proceeding as before, we find the expression for F'(z), (w-1)! In order for this to vanish when z = z lf it is necessary that (n-l)l in which z t = a + 0(x — a), x — z l = (x — a)(l — 0). Hence *fe a) = / ( ">(a + <?(*- a)) (1 _ e) .- 1(a . _ a) »-. ; (n-1)! and !}„(*, «)=(!- g)-' /'"y^- a ) >(s - a)".* An example of the use of this form of remainder is fur- nished by the series for log x in powers of x— a, when x—a is negative, and also in the expansion of (a -+- #) m - Ex. 1. Find the interval of equivalence for the development of log x in powers of x — a, when a is a positive number. Here, from Art. 77, Ex. 1, X n hence /<»>(a f 6(x -a)) = (-l)n-i (" >- 1) » (a + 0(x — a))'* and, by Art. 80, ^ fl) = ^--'c*-*)- = (-D- 1 r *-« r. V ' n(a+0{x-a)) n n la + 6(x-a)\ * This form of the remainder was found by Cauchy (1789-1857), and first published in his "Le<,:ons sur le calcul infinitesimal," 1826. 154 DIFFERENTIAL CALCULUS First let x — a be positive. Then when it lies between and a, it is numerically less than a + 0(x - a), since is a positive proper fraction ; hence when n = oo x ~ a T = 0, and RJx, a) = 0. Again, when x — a is negative and numerically less than a, the second form of the remainder must be employed. As before hence R n (x, a) = (1 - 0)«-i . f" 1 ^"^ = (l_0)»-i. - («-')' [a-(9(a-.r)]» t (a — x) — 0{a — x) l "- 1 a — x a -0(a - x) J ' a - ${a - x)' The factor within the brackets is numerically less than 1, hence the (n — l)th power can be made less than any given number, by tak- ing n large enough. This is true for all values of x between and a. Therefore, log x and its development in powers of x — a are equiva- lent within the interval of convergence of the series, that is, for all values of x between and 2 a. _i Ex. 2. Show that the development of x ? in positive powers of x — a holds for all values of x that make the series convergent ; that is, when, a; lies between and 2 a. If the function is expanded in powers of x, the complete form will be ffljj. , /'-"(0)^-i /(*)=/(0)+/'(0)*+-^^+ ... +^~y;*" + /^M*. (1) EXPANSION OF FUNCTIONS 155 for the first form of remainder, and /(*)=/(<>) +/'(0)* + /^-'+ ... +M^-i £W(1 _ gy-l . x , ( 2 ) for the second form of remainder. Similarly, the complete form of Taylor's series (Art. 78) becomes 0| (n-l)l /»>(q + flaQ , for the first form of remainder, and /(a + x) =f(a) +f(a)x + -^z 2 + • •• + /•"«„ 2 , , /'-"(a) „„-, <> | (»-l) (4) for the second form of remainder. These forms are of no value for numerical computation unless f (n) (x) can be determined, but may sometimes be used to advantage to obtain a maximum error, corresponding to small values of n. It should be observed that when n = 1, the theorem of mean value results. (Art. 39.) Ex. 3. Obtain the limit of error in retaining but two non-vanish- ing terms in the expansion of log (x + Vl + x' 1 ) when x = \. log (x + viTi" 2 ) = x- -*. + P- ?/ - n - ?/ ; r wherein y = 6x. LV J J ■ 24 156 DIFFERENTIAL CALCULUS The next step is to obtain the largest and the smallest value which the expression in brackets assumes for values of y within the interval to \. For this purpose, consider the function „- 3.v(3-2.y*) (1 + y*)! ^ We find that — is positive for all values of y between y = and y = \; dy hence u has its largest value when y = J, and the corresponding value of the last term is .000284. Ex. 4. How many terms should be used in the expansion of e x in powers of x to insure a result correct to four places of decimals when x = |? Ex. 5. In the expansion of logio (1 + x) in powers of x how many terms should be used in order to obtain the value of logio (1 • 8) cor- rect to 5 decimals ? CHAPTER XI INDETERMINATE FORMS 82. Hitherto the values of a given function f(x), corresponding to assigned values of the variable x, have been obtained by direct substitution. The function may, however, involve the variable in such a way that for certain values of the latter the correspond- ing values of the function cannot be found by mere substitution. For example, the function sin x for the value x = 0, assumes the form -, and the correspond- ing value of the function is thus not directly determined. In such a case the expression for the function is said to assume an indeterminate form for the assigned value of the variable. The example just given illustrates the indeterminateness of most frequent occurrence; namely, that in which the given function is the quotient of two other functions that vanish for the same value of the variable. Thus if f(x) = *M, and if, when x takes the special value a, the functions <f>(x) and \p(x) both vanish, then J K ' *(a) is indeterminate in form, and cannot be rendered determinate without further transformation. 157 158 DIFFERENTIAL CALCULUS 83. Indeterminate forms may have determinate values. A case has already been noticed (Art. 2) in which an expression that assumes the form - for a certain value of its variable takes a definite value, dependent upon the law of variation of the function in the vicinity of the assigned value of the variable. As another example, consider the function x 2 — a 2 y = x— a If this relation between x and y is written in the forms y(x — a)=x 2 — a 2 , (x — a)(y — x — a) = 0, it will be seen that it can be represented graphically, as in Fig. 42, by the pair of lines x — a = 0, y — x — a = 0. Hence when x has the value of a, there is an indefinite number of corresponding points on the locus, all situated on the line x = a; and accordingly for this value of x the function y may have any value whatever, and is therefore indeterminate. When x has any value different from a, the corresponding value of y is determined from the equation y = x-\-a. Now, of the infinite number of different values of y corresponding to x = a, there is one particular value AP which is continuous with the series of values taken by y when x takes successive values in the vicinity of x = a. This may be called the de- terminate value of y when x — a. It is obtained by putting x = a in the equation y = x + a, and is therefore y = 2 a. Fig. 42 INDETERMINATE FORMS 159 This result may be stated without reference to a locus as follows. When x = a, the function x— a is indeterminate, and has an infinite number of different values ; but among these values there is one determinate value which is continuous with the series of values taken by the function as x increases through the value a ; this determinate or singular value may then be defined by liui xr— a 2 ^ x=a x — a In evaluating this limit the common factor x— a may be re- moved from numerator and denominator, since this factor is not zero while x is different from a; hence the determinate value of the function is lim a? + « _o„ Ex. 1. Find the determinate value, when x = 1, of the function x 3 + 2 x 2 -3x 3 x b _ 3 x a _ x + i ' which, at the limit, takes the form — This expression may be written in the form (x* + Zx)(z- 1) (3s 3 - 1)0- 1)' which reduces to x + ' x • When x = 1, this becomes - = 2. 3 x 2 - 1 2 Ex. 2. Evaluate the expression x 8 + ax 2 + a 2 x 4- a 3 , x 8 4- b 2 x + az 2 + a& 2 when x = — a. 160 DIFFERENTIAL CALCULUS Ex. 3. Determine the value of x 8 - 7 x 2 + 3 x + 14 x 3 + 3 x 2 - 17 x + 14 when x = 2. Ex. 4. Evaluate - — — — when x = 0. (Multiply both numerator and denominator by a -f Va 2 — x 2 .) 84. Evaluation by development. In some cases the common vanishing factor can be best removed after expansion in series. Ex. 1. Consider the function mentioned in Art. 82, e* - e~* sin x When numerator and denominator are developed in powers of x, the expression becomes 21 31^ V 21 31 / *-£+... 3! 2, + ^+... 2+f+... x 3 , - X 2 . — ST+- 1 -« + - which has the determinate value 2, when x takes the value zero. Ex. 2. As another example, evaluate, when x= 0, the function x — sin- 1 a: sin 8 x By development it becomes « + X J + Removing the common factor, and then putting x = 0, we obtain INDETERMINATE FORMS 161 In these two examples the assigned value of x, for which the indeterininateness occurs, is zero, and the developments are made in powers of x. If the assigned value of x is some other number, as a, then the development should be made in powers of x — a. Ex. 3. Evaluate, when x = -, the function 2 COS X 1 — sin x IT By putting x — - = h, x = - + h, the expression becomes eo.(=+») . -»+£-... _l+£ \ 2 / — sm 7i 6 6 . . hr , ,\ 1 -cos A A 2 A 4 , /i A 8 , 1 — sin — \- h) — h ••• h ••• \2 / 2 24 2 24^ which becomes infinite when 7* = ; that is, when x = -• 2 TJ lim cos x Hence . m - = ± oo, x=fl-siuo: ' according as h approaches zero from the negative or positive side. 85. Evaluation by Differentiation. Let the given function be of the form ^~, and suppose that /(a) = 0, cf>(a) = 0. It is required to find ^ ^-L-i . 4 x = a <f>(x) We assume that f(x), <j>(x) can be developed in the vicinity of x = a, by expanding them in powers of x — a. Then m /(«)+/'(«)(* - «)+-^ ( f (»- «) 2 + - *(*) <l>(a) + 4>'(a)(x - a) + *!M (x - uy+ - /•(«)( a( _ a ) + ^2l( ( ,_«)i + ... ^«)(*-a) + ^(z-a) 2 +... EL. CALC. 11 162 DIFFERENTIAL CALCULUS By dividing by x — a and then letting x = a, we obtain lim f(x)_f(a) «-« + («) +'(o) By hypothesis the functions f'(a), <£'(a) will both be finite. If /'(a) = 0, <£'(a) =£ 0, then £i& = 0. <j>(a) lf/'(a)*0, *'(a) =0, then ^ = oo. <£(a) If /'(a) and <£'( a ) are both zero, the limiting value of J-^-l is to be obtained by carrying Taylor's development one term farther, removing the common factor (x — a) 2 , and then letting f"(a) x approach a. The result is J —~ • Similarly, if /(a), /'(a), /"(a); </>(a), <^)'(a), <^>" (a) all vanish, it is proved in the same manner that lim f(x) = f'"(a) x = a '<j>(x) <£'"(«)' and so on, until a result is obtained that is not indeterminate in form. Hence the rule : To evaluate an expression of the form -, differentiate numer- ator and denominator separately ; substitute the critical value of x in their derivatives, and equate the quotient of the derivatives to the indeterminate form. Ex. 1. Evaluate * ~ T ^ when = 0. Put f(0) = 1 - cos 6, <f>(0) = e*. Then f(0) = sin 0, <f>'(0)=2$ t and /'(0) - 0, <£'(0) = 0. INDETERMINATE FORMS 163 Thus, the function \ * J is also indeterminate at = 0. It is there- 9 (0) f"({y\ fore necessary to obtain </ ,,„ ' • Accordingly, f"{6) = cos 0, p"(0) = 2, /'(0)=1, 9"(0)=2, hence lim 1 — cos _ 1 J = 02 - 2 * Ex.2. Find x= , lim i* + e~* + 2 cos a: — 4 lim e x — e~ x — 2 sin a: x = 4 a: 3 lim e* + e * — 2 COS X x=0 12 a: 2 lim e x — e -* + 2 sin x x = 24 x lim e z + e -* + 2 cosx x = 24 1 6' Ex. 3. Find lim a; - siD3rcosa: . x = x 8 Ex.4. Find M" « , -2»'i**+»«-*. * = 1 a: 4 - 2 a: 3 + 2 a: - 1 In this example, show that z — 1 is a factor of both numerator and denominator. ^ _ „. , lim 3 tan x - 3 a: - a: 8 Ex. 5. Find „ . n In applying this process to particular problems, the work can often be shortened by evaluating a non-vanishing factor in either numerator or denominator before performing the differentiation. 164 DIFFERENTIAL CALCULUS Ex. 6. Find ";■> (»-*)' tan* . X == " X The given expression may be written lim . tanx _ lim lim tans = 16-1 = 16. In general, if f(x) = *P(%)x( x )> an( ^ ^ «M°0 = 0> x( a )^0, (h (a) = 0, then ,. ., N ,,, * llm /M = v r^^M. * = «*(*) XU ^>'(a) For lim iKx)x(x) = lim , v . lim </<») a) . ^!M x = a $(x) x = a^ K > x = a^{x) * ; <f>'(a) lim sin x cos 2 x Ex. 7. Find x = $ (2 x - tt) 2 Ex.8. Find lim (.-3) 2 log(2 x } > x = 1 sin(x - 1) EXERCISES Evaluate the following expressions : 1. 1 - CO8ar whenx = 0. 7. e * + e ~' ~ 2 when x = 0. 2. e * . e when a; = 0. 8 tan a;- sin x cos a: when ^_ a 3. * 3 - 1 U 1 when x = 1. x- 1 4. when x = 0. &* - 1 5. sin ax , „ when x = 0. sin frx l 9. when x = 0. tan -1 a; 10. g*B"i*-*-* when a- = 0. x 2 + a;log (1-a:) >. (1 4- arV» — 1 , n -.-. tan a; — sin a; , n 6. V x ^ x ' 1 when x = 0. 11. - Q when x = 0. x X s There are other indeterminate forms than -• They are £,00-00,0°, r, 00°. INDETERMINATE FORMS 165 86. Evaluation of the indeterminate form ® . GO fix) Let the function J \ ' become 5°- when x = a. It is re- <j>(x) <*> quired to find lim &&• This function can be written <K*) _1_' which takes the form - when x = a, and can therefore be evaluated by the preceding rule. When x = a, — vv, lim /(») _ lim x = a <£(#) £ = a 1 4,\x) *(*)_ lim [*(*)]* 1 ic = a f'(x) lim ~" " x — a -/W T+'(*). (1) If both members are divided by im -£-^J when this limit is not nor co, the equation becomes 1= lim f(x) +\x) *±<*<f>(x)f'(x)' therefore lim x = a d,{x) = /'(<0 +'(«) (2) This is exactly the same result as was obtained for. the form ; hence the procedure for evaluating the indeterminate forms -, g, is the same in both cases. ^ 166 DIFFERENTIAL CALCULUS When the true value of ■->--' is or go , equation (1) is satis- fied, independent of the value of ,~ ; but (2) still gives the correct value. For, suppose ™ ^ w = 0. Consider the function which has the form °°- when a; = a, and has the determinate GO 7 value c, which is not zero. Hence by (2) lim fix) + c<f> (x) = /' (a) + c<f>' (a) = /' (a) c * = a <f>(x) 4>'(a) <j>' (a) Therefore, by subtracting c, lim /(a?) = /(a) , • £ = «<£(&) </>'(a) If "F-fiS. =00, then "™ ^ = 0, which can be treated x ~ a <f>(x) x - a f(x) as the previous case. The forms • go and go-go can usually be evaluated by putting them in one or the other of the forms already dis- cussed. In the case of the others, in which the indetermi- nateness appears in the exponent, the logarithm of the function can be reduced to one of the preceding forms. EXERCISES Evaluate the following expressions: 1. lo g si » 2 * when x = 0. log sin x 2. !2££ ^ 4. tan x tan 5 x when x _ IT 5. sec 3 a: when X _ IT cot x sec 5 x 2 6. x s[nx when x = 0. 7. (cos ax) csc2cx when x = 0. ^n 6. x B[QX when x = 0. 3. — when x = go. IT) CHAPTER XII CONTACT AND CURVATURE 87. Order of contact. The points of intersection of the two are found by making the two equations simultaneous ; that is, by finding those values of x for which Suppose x = a is one v?lue that satisfies this equation. Then the point x = a, y = A(i) = if; (a) is common to the curves. If, moreover, the two curves have the same tangent at this point, they are said to touch each other, or to have contact with each other. The values of y and of — are thus the J dx same for both curves at the point in question, which requires that *(«)=</<(«), <j,\a)=4,\a). d 2 v If, in addition, the value of — " is the same for each curve dx~ at the point, then and the curves are said to have a contact of the second order with each other, provided cf>'"(a) =£ i//"(a). If <f>(a) =\f/(a), and all the derivatives up to the nth order inclusive are equal to each other, that is, <f>'(a) = if/' (a), <}>"(a) = i(,"(a), .-., cf> (n \a) =<A (n) (a), but <f> (n+1) (a) =£ f n+1) (a), the curves are said to have contact of the ?ith order. 167 168 DIFFERENTIAL CALCULUS 88. Number of conditions implied by contact. or j equation of the curve y = <f>(x) is given, and it is requir ,. to determine the equation of a second curve y = \\j (x) that shall have contact of any given order with y = <f>(x) at a specified point, then, from the definition given in the preceding article for contact of the nth order, n -f 1 conditions must be imposed upon the coefficients in \p(x). The required curve must therefore con- tain at least n -f- 1 arbitrary constants in >rder to fulfill the required conditions. A straight line has two arbitrary constants, which can be determined by two conditions ; accordingly a straight line can be drawn which touches a given curve at any specified point. For if the equation of a line is written y = mx -f b, then dy dr.; p. dx dx hence, through any arbitrary point x = a on a given curve y=cf>(x), a line can be drawn which has contact of the first order with the curve, but which has not in general contact of the second order; for the two conditions for first-order con- tact are ma + 6 = <£(«), m = <£'(«), which are just sufficient to determine m and b. In general no line can be drawn having contact of an order higher than the first with a given curve at a given point ; but there are certain special points at which this can be done. For example, the additional condition for second-order contact is = <f>"(a), which is satisfied when the point x = a is a point of inflexion on the given curve y=<j>(x). (Art. 49.) Thus the tangent at a point of inflexion on a curve has contact of the second order with the curve. CONTACT AND CURVATURE 169 The equation of a circle has three independent constants. It is therefore possible to determine a circle having contact of the second order with a given curve at any assigned point. The equation of a parabola has four constants, hence a parabola can be found which has contact of the third order with the given curve at a given point. The general equation of a central conic has five independent constants, hence a conic can be found which has contact of the fourth order with a given curve at any specified point. As in the case of the tangent line, special points may be found for which these curves have contact of higher order. 89. Contact of odd and of even order. Theorem. At a point where two curves have contact of an odd order they do not cross each other; but they do cross where they have contact of an even order. For, let the curves y = <f>(x), y=\f/(x) have contact of the ?ith order at the point whose abscissa is a ; and let y x , y 2 be the ordinates of these curves at the point whose abscissa is a + h. Then y 1 = cf>(a + h), y 2 =z ij,(a-\-h), and by Taylor's theorem Vl = *(a) + *'(a) • h + *^) • ft 2 + • • • + *^.». + *^« (a)+ .... n ! (?i -f 1) ! fc .^(o)+^(a).*+*^i.V + • n ! (n + 1) ! 170 DIFFERENTIAL CALCULUS Since by hypothesis the two curves have contact of the nth order at the point whose abscissa is a, hence 4(a) =$(a), <j>'(a)=ip' a), ..'., <y(a) = ^(a), and yi -y^-J^i r+1(a)+ ... _ r+ i (a) _ ... ]; but this expression, when h is sufficiently diminished, has the same sign as .,,, r „,.> , , , v _ Hence, if n is odd, y l — y 2 does not change sign when h is changed into — h, and thus the two curves do not cross each other at the common point. On the other hand, if n is even, ll\ — V* changes sign with 7i ; and therefore when the contact is of even order the curves cross each other at their common point. Geometrically, we may say that two curves having contact of the nth order pass through n -\- 1 common points which approach coincidence at the point of contact. For let y = <f>(x), y = xf/(x) touch each other at x = a. This means that they have two coincident points in common at (a, <f>(a)), and the condi- tions to be satisfied are <f>(a)=^(a), *'(a) = *'(a). If the curves also have a point in common for x = a -f- h, then <f>(a + h)=ip(a + h). Expanding by Taylor's series and making use of the preced- ing conditions, we may cancel the common factor Jr. If now this condition is still satisfied when h approaches zero, so that the third point of intersection approaches the position of the two coincident ones, then we must have the further condition CONTACT AND CURVATURE 171 <£"(a)= \J/"(a). Thus, three coincident points of- intersection imply contact of the second order. By repeating this argu- ment the above theorem results. For example, the tangent line usually lies entirely on one side of the curve, but at a point of inflexion the tangent crosses the curve. Again, the circle of second-order contact crosses the curve except at the special points noted later, in which the circle has contact of the third order. EXERCISES 1. Find the order of contact of the curves 4 y = x 2 and y = x — 1. 2. Find the order of contact of the curves x = y s and x — 2 y + 1 = 0. 3. Find the order of contact of the curves 4 y = x 2 — 4 and x 2 — 2 y — 3 — y 2 . 4. Determine the parabola having its axis parallel to the #-axis, which has the closest possible contact with the curve a 2 y — x 3 at the point («, a). (The equation of a parabola having its axis parallel to the y-axis is of the form y = Ax 2 + Bx + C.) 5. Determine a straight line which has contact of the second order with the curve y = x 8 - 3x 2 - 9 x + 9. 6. Find the order of contact of y = log(x — 1) and x 2 — 6x + 2y + 8 = at the point (2, 0). 7. What must be the value of a in order that the curves y=x+l-\-a(x— l) 2 and xy = 3 x — 1 may have contact of the second order? 172 DIFFERENTIAL CALCULUS 8. Determine the parabola which has its axis parallel to the y-axis and has contact of the second order with the hyperbola xy = 1 at the point (1, 1). 9. Determine the point and order of contact of the curves (a) y=x*, y = 6 x 2 - 9 x + 4 ; (6) y = x s , y = -Q x 2 -12x-8. 10. Determine the parabola which has its axis parallel to the y-axis, passes through the point (0, 3), and has contact of the first order with the curve y = 2 x 2 at the point (1, 2). Similarly for a parabola having its axis parallel to the x-axis. 11. Show that the curve y = sin x has contact of the sixth order with the curve x z , x 5 v = x * 6 120 at the origin. Show that y = sin x, y = sinh x, have contact of the second order at the origin. Draw these curves. 12. Find the order of contact of the curves y = cos x, y = cosh x at the point (0, 1). Sketch the curves. 13. Find the order of contact at the origin of the curves , ■ , , _ sinh x y = tan x, y = tanh x = . cosh x 90. Circle of curvature. The circle that has contact of the closest order with a given curve at a specified point is called the osculating circle or circle of curvature of the curve at the given point. The radius of this circle is called the radius of curvature, and its center is called the center of curvature at the assigned point. 91. Length of radius of curvature; coordinates of center of curvature. Let the equation of a circle be (X-ay + (Y-pf = R\ (1) in which R is the radius, and a, ft are the coordinates of the center, the current coordinates being denoted by X, Yto dis- CONTACT AND CURVATURE 173 tinguish them from the coordinates of a point on the given curve. It is required to determine R, a, /?, so that this circle may have contact of the second order with the given curve at the point (x, y). From (1), by successive differentiation, we deduce (X-o) + (F-/8)g=0, \dX K H 'dX 2 (2) If the circle (1) has contact of the second order at the point (x, y) with the given curve, then when X = x it is necessary that dY = c]y d 2 Y = d 2 y dX dx dX 2 dx 2 ' (3) Substituting these expressions in (2), we obtain dy (x- a) +(3,-^)^ = 0, whence y-P- H% dx 2 dy dx x~ a= — L + (D" <Py dx* (4) (5) and finally, by substitution in (1), R .MIT) dx 2 (6) 174 DIFFERENTIAL CALCULUS Ex. 1. For the curve y = sin x, show that a = x-\- cot x(l + cos 2 x), s j3 = — 2 cos a: esc x, it = — (1 + cos' 2 x) 2 esc x. Find the numerical values of a and B when a: = 0, — , — , — , and locate the corresponding r 6 3 2 points («, /3) on a drawing. Sketch roughly the path of this point as x varies. Write the equation of the osculating circle for the point x = - , and for x = -. Draw these circles. 3 2 Ex. 2. For the curve y = x s , find a, /?, R in terms of x. Compute their numerical values at x — 1, .7, .5, .3, 0. Show that i? is a mini- mum when x — — — = .39 •••, and that the value of R is .57 •••• 92. Limiting intersection of normals. Let P= (# 1? y-^) and P' = (a? 2 , y 2 ) be Wo points on a given curve f(x, y) = 0. The equations of the normals at these points are ( a ._ a . 2 ) + ( 2/ _ 2/2 )^? = 0< d# 2 If («', /?') is the point of intersection of these two lines, then dx x (^_ iK2 ) + ( /3 '_ 2/2 )^ == 0. Now consider the function if/(x) of a: denned by the equations f(x) = (x- a') + (y- /?') £, f(x, y) = 0. Since i/a(^ 1 ) = and ^(# 2 ) = 0, hence by Rolle's theorem (Art. 79) it follows that t'(x) = 0, CONTACT AND CURVATURE 175 in which x is defined by the inequalities U/j <^, X < C X 2 > Hence a', ft' may be determined by the simultaneous equations ^(^) = 0, f(5) = 0. When P' = (x 2 , y 2 ) approaches coincidence with the point P=(x x , y x ), then if/' (x) = i// (a;,), and therefore from (4), the point («', /8') becomes the center of curvature, hence : The center of curvature at a point Pon a curve is the limiting position of the point of intersection of the normal at P with the normal at the point P', when P' approaches the position of P. 93. Direction of radius of curvature. Since, at any point P on the given curve, the value of -* is the same for the curve dx and the osculating circle for that point, it follows that they have the same tangent and normal at P, and hence that the radius of curvature coincides with the normal. Again, since the value of — is the same for both curves at P, it follows dx 2 from Art. 50, that they have the same direction of bending at that point, and hence that the center of curvature lies on the concave side of the given curve (Fig. 43). It follows from this fact and Art. 87 that the osculating circle is the limiting position of a circle passing through three points on the curve when these points move into coincidence. The radius of curvature is usually regarded as positive or negative according as the bending of the curve is positive or d 2 v negative (Art. 49), that is, according as the value of — -~ is ctx~ positive or negative ; hence, in the expression for R, the radi- cal in the numerator is always to be given the positive sign. 176 DIFFERENTIAL CALCULUS The sign of R changes as the point P passes through a point of inflexion on the given curve (Fig. 44). It is evident from the figure that in this case R passes through an infinite value ; Fig. 43 Fig. 44 for the circle through the points N, P, Q approaches coinci- dence with the inflexional tangent when N and Q approach coincidence with P, and the center of this circle at the same time passes to infinity. 94. Total curvature of a given arc ; average curvature. The total curvature of an arc PQ (Fig. 45) in which the bending is in one direction, is the angle through which the tangent swings as the point of contact moves from the initial point P to the terminal point Q ; or, in other words, it is the angle between the tan- gents at P and Q, measured from the former to the latter. Thus the total curvature of a given arc is positive or negative according as the bending is in the positive or negative direction of rotation. The total curvature of an arc divided by the length of the arc is called the average curvature of the arc. If the length of Fig. 45 CONTACT AND CURVATURE 177 the arc PQ is As centimeters, and if its total curvature is A<£ radians, then its average curvature is — — radians per centimeter. As 95. Measure of curvature at a given point. The measure of the curvature of a given curve at a given point P is the limit which the average curvature of the arc PQ approaches when the point Q approaches coincidence with P. Since the average curvature of the arc PQ is — — , the As measure of the curvature at the point P is lim A<£ _ dcf> As=0 As d and may be regarded as the rate of deflection of the arc from the tangent estimated per unit of length ; or, as the ratio of the angular velocity of the tangent to the linear velocity of the point of contact. To express k in terms of x, y, and the derivatives of y, we observe that whence and tan <£ = dx ' * = dx d<f>_ ds ds \ dxj dx\ dxj ds <l 2 y dx 2 1 \dxj dx cPy defy dx 2 K ~ds = Mtw therefore k = ^ = "^ 7 9 • [Art. 41. EL. CALC. — 12 178 DIFFERENTIAL CALCULUS, 96. Curvature of an arc of a circle. In the case of a circular arc the normals are radii ; * hence As = r.A0, ^ = -, (1) As r and therefore k = -. r It follows that the average curvature of all arcs of the same circle is constant and equal to - radians per unit of length. r For example, in a circle of 2 feet radius the total curvature of an arc of 3 feet is § = 1.5 radians, and the average curva- ture is .5 radian per foot. It also follows from (1) that in different circles, arcs of the same length have a total curvature inversely proportional to their radii. Thus on a circumference of 1 meter radius, an arc of 5 decimeters has a total curvature of .5 radian, and an average curvature of .1 radian per decimeter ; but on a circumference of half a meter radius, the same length of arc has a total curvature of 1 radian and an average curvature of .2 radian per decimeter. 97. Curvature of osculating circle. A curve and its osculat- ing circle at P have the same measure of curvature at that point. For, let k, k' be their respective measures of curvature at the point of contact (x, y). Then from Art. 95, &y dx 2 \>+m\ CONTACT AND CURVATURE 179 But this is the reciprocal of the expression for the radius of curvature (Eq. (6), p. 173) ; hence 1 K = • R That is : the measure of curvature k at a point P is the recipro- cal of the radius of curvature R for that point. Since a curve and its osculating circle have the same radius of curvature (Art. 90) at their point of contact, it follows from this result that the measure of curvature is also the same for both; *=*'. It is on account of this property that the osculating circle is called the circle of curvature. This is sometimes used as the denning property of the circle of curvature. The radius of curvature at P would then be denned as the radius of the circle whose measure of curvature is the same as that of the given curve at the point P. Its value, as found from Art. 95 and Art. 96, accords with that given in Art. 91. EXERCISES 1. Find the radius of curvature of the curve y 2 = 4 ax at the origin. 2. Find the radius of curvature of the curve y s + x z + a(x 2 + y 2 ) = cfiy at the origin. 3. Find the radius of curvature of the curve ahj — bx z + cx 2 y at the origin. Find the center and the radius of curvature for each of the following curves at the point (x, y) and their numerical values at the special point indicated. Find where the curvature is greatest and least on each curve. 4. Rectangular hyperbola x$ = m 2 at (m, m). 5. Hyperbola — - -^ = 1 at (a, 0). a 2 b 2 180 DIFFERENTIAL CALCULUS 6. General parabola a n ~ x y = x n at (a, a). 7. Parabola Vx + Vy = Va at (a, 0). 1 2 2 8. Hypocycloid x 3 -f y s = of 3 " at the point at which x = y. 9. Cissoid y" a,t x = a. 10. Catenary y—-( e a _j_ e <*) at £ = 0. 11. In what points of the curve y = x 3 is the curvature greatest ? 98. Direct derivation of the expressions for k and R in polar coordinates. Using the notation of Art. 58, we have hence But _d$_dB_ ~ ds~ ds dO d O ds dO ('+2) ■'% tan^ = p — , ^=tan _1 dp P_ dp dO (1) [Art. 44. therefore, by differentiating as to and reducing, we obtain 'dp^ 2 dij/ _ \d6 d0~ P 2 + (IT ' CONTACT AND CURVATURE 181 which, substituted in (1), gives HUT Since k = — , it follows that H R = Mm f j *d0 2 + 2 (U This result should be compared with that of Art. 72. When u = - is taken as dependent variable, the expres- P sion for k assumes the simpler form K = »r Since at a point of inflexion k vanishes and changes sign, hence the condition for a point of inflexion, expressed in polar 72 coordinates, is that u -\ shall vanish and change sign. cW EXERCISES Find the radius of curvature for each of the following curves : 1. p = a . 3. p = 2 a cos $ - a. 5. p 2 cos 2 = a 2 . 2. p 2 = a 2 cos 2 6. 4. p cos 2 J $ = a. 6. p = 2 a(l - cos 0) . 7. P = a. 182 DIFFERENTIAL CALCULUS EVOLUTES AND INVOLUTES 99. Definition of an evolute. When the point P moves along the given curve, the center of curvature C describes another curve which is called the evolute of the first. Let f(x, y) = be the equation of the given curve. Then the equation of the locus described by the point C is found by eliminating x and y from the three equations x— a dy dx i+(& \dx dx 2 *-p- — it ' dx 2 and thus obtaining a relation between a, /3, the coordinates of the center of curvature. No general process of elimination can be given ; the method to be adopted depends upon the form of the given equation /(*,y) = o. Even when the elimination cannot be performed, the evolute can be traced from point to point by computing successive values of (a, /3) corresponding to successive values of (x f y). Ex. 1. Find the evolute of the parabola ?/ 2 = 4 px. Since «=2pW, ^ = »H <?t> = _ M*-! , dx dx 2 2 hence x — a = — p^x~^(l + px~ *)2 p~?x% — — 2 (x + p), and y - p = (1 + px~ x )2 p"~W = 2(p~%x% + pW) ; therefore a = 2p + 3 a:, ft = — 2 p*x*. CONTACT AND CURVATURE 183 Fig. 46 The result of eliminating x between the last two equations is i.e., 4(«- 2 P y = 27 p/3 2 , which is tho equation of the evolute of the parabola, a, {3 being the current coordinates. Use the expressions for a and /3 to compute their values, and to locate the points («, /3) when x — 0, P Ex. 2. Find the evolute of the ellipse -2 ?<2 Here dx 2 ir- x ir a 2 b 2 (1) y_ .dy = o 'hi- - b 2 X b 2 dx dx dy dx -ft 2 / , b 2 x 2 \ -b 2 , 2 „ , , 2 „. -ft* r a # \ a 2 y / a 4 #° a 2 // 3 184 DIFFERENTIAL CALCULUS whence y H aW \ b* a>Y J \ ¥ by- Therefore - (3 = ^ ~ b ' V - (2) b* Similarly, a = a —^x^. (3) On eliminating x, y between (1), (2), (3), the equation of the locus described by (a, (3) is (aa)f + (bp)i = (a 2 - 6 2 )l (Fig. 51) Use (2), (3) to locate various values of («, /3), and trace the evolute. Take a = 2b; a =—. 3 100. Properties of the evolute. The evolute has two im- portant properties that will now be established. I. The normal to the curve is tangent to the evolute. The relations connecting the coordinates (a, ft) of the center of curvature with the coordinates (x, y) of the corresponding point on the curve are, by Art. 91, x- a + (y-/J)^ = 0, (1) 1+ (!)' + (*-«g=°- < 2 > By differentiating (1) as to x, considering a, (3, y as functions of*,weobtain + ^_^_<L«_ M = . (3) \dxj K HJ dx 2 dx dx dx w Subtracting (3) from (2), we obtain da + dgdy = ^ ^ dx dx dx , dB dx whence — = — — • da dy CONTACT AND CURVATURE 185 But c -l=- is the slope of the tangent to the evolute at (a, /?), da and is the slope of the normal to the given curve at dy (x, y). Hence these lines have the same slope ; but they pass through the same point (a, f3), therefore they are coincident. II. TJie difference between two radii ' of curvature of the given curve, which touch the evolute at the points C lt C 2 Fig. 47 (Fig. 47), is equal to the arc C X C 2 of the evolute. Since R is the distance between the points (x, y), (a, (3), hence (aj - a) 2 + (y - /3) 2 = Br. (5) When the point (x, y) moves along the given curve, the point («, /?) moves along the evolute, and thus a, /?, R, y, are all functions of x. Differentiation of (5) as to x gives (-^-SVMs-i?)-*^ (6) hence, subtracting (6) from (1), we obtain ( X - a) p +(y -f }) f = -l t ™. (7) dx dx dx Again, from (1) and (4), da d(3 dx dx x — a y — (3 (8) 186 DIFFERENTIAL CALCULUS Hence, each of these fractions is equal to 4 da\ 2 ,fdg\ 2 do- dx) \dxj dx V(a-a) 2 + (^=^ _± ^' in which <r is the arc of the evolute. (Compare Art. 41.) Next, multiplying numerator and denominator of the first member of (8) by x — a, and those of the second member by y — /?, and combining new numerators and denominators, we find that each of the fractions in (8) is equal to (x-af + iy-py ■ which equals — R 0B dx R 2 by (7) and (5). By combining with (9), we obtain d<r_ dR dx dx' that is, ^(cr±R)=0. dx Therefore <r ± R = constant, (10) wherein o- is measured from a fixed point A on the evolute. Now, let (?!, C 2 be the centers of curvature for the points P ly P 2 on the given curve ; let P X C X = 7^, P 2 C 2 = R 2 ; and let the arcs AC X , AC 2 be denoted by p- lt <r 2 . Then <t 1 ±R 1 = <t 2 ± R 2 , by (10) ; that is a- 1 — (T 2 =± (R 2 — R } ) ; hence, arc C X C 2 — R 2 — R v (11) CONTACT AND CURVATURE 187 Fig. 48 Thus, in Fig. 48, P 2 C 2 +CA = P S C 3 , etc. Hence, if a thread is wrapped around the eve-lute, and then is un- wound, the free end of it can be ^ N made to trace out the original curve. From this property the locus of the centers of curvature of a given curve is called the evolute of that curve, and the latter is called the involute of the former. When the string is unwound, each point of it describes a different involute; hence, any curve has an infinite number of involutes, but only one. evolute. Any two of these involutes intercept a constant distance on their common normal, and are called parallel curves on account of this property. Ex. Find the length of that part of the evolute of the parab- ola which lies inside the curve. From Fig. 46 the required length is twice the difference between the tangents C 3 P 3 and P Q C , both of which are normals to the parabola. To find the coordinates of the point P 3 , write the equation of the tangent to the evolute at C 3 , and find the other point at which it intersects the parabola. The coordinates of C 3 , the point of intersection of the two curves, are (8p, 4joV2), and the equation of the tangent at C 3 is 2x -V2y - Sp = 0. This tangent intersects the parabola at the point (2p, — 2 V2 />), which is Po. 188 DIFFERENTIAL CALCULUS The value of the radius of curvature is ~{ x ~*~P) f hence P C =2p, Vp P3C3 = GV3p, hence the arc C C S is 2p(3V3 - 1), and the required length of the evolute is therefore 4jt>(3V3 — 1). EXERCISES Find the coordinates of the center of curvature for each of the following curves : 1. x 2 + y 2 = a 2 . 3. y 3 ^ a 2 x. 2. x = a log a + ^ a ' 2 ~ y2 - Va 2 - y 2 . 4. y=-^( e a + e «). Find the equations of the evolutes of the following curves : 5. xy = a 2 . 6. a 2 y 2 - b 2 x 2 = - a 2 b 2 . 7. x% + y* = ah 8. Show that the curvature of an ellipse is a minimum at the end of the minor axis, and that the osculating circle at this point has con- tact of the third order with the curve. Fia. 49 This circle of curvature must be entirely outside the ellipse (Fig. 49). For, consider two points Pi, P 2 , one on each side of B, the end of the minor axis. At these points the curvature is greater CONTACT AND CURVATURE 189 than at B, hence these points must be farther from the tangent at B than the circle of curvature, which has everywhere the same curva- ture as at B. 9. Similarly, show that the curvature at A, the end of the major axis, is a maximum, and that the circle of curvature at A lies entirely within the ellipse (Fig. 49). 10. Show how to sketch the circle of curvature for points between A and B. The circle of curvature for points between A and B has three coincident points in common with the ellipse (Art. 93), hence the circle crosses the curve (Art. 89). Let K, P, L be three points on the arc, such that K is nearest A and L nearest B. The center Fig. 50 of curvature for P lies on the normal to P, and on the concave side of the curve. The circle crosses at P, lying outside of the ellipse at K (on the side towards .4), and inside the ellipse at L; for the bend- ing of the ellipse increases from B to P and from P to K, while the bending (curvature) of the osculating circle remains constant (Fig. 50). 11. Two centers of curvature lie on every normal. Prove geo- metrically that the normals to the curve are tangent to the e volute. 190 DIFFERENTIAL CALCULUS 12. Show that the entire length of the e volute of the ellipse is 4 (- — — ). [From equation (11) above, take B v R 2 as the radii of \b a J curvature at the extremities of the major and minor axes.] 13. If E is the center of curvature at the vertex A (Fig. 51), prove that CE = ae 1 , in which e is the eccentricity of the ellipse; and hence that CD, CA, CF, CE form a geometric series whose common ratio is e. Show also that DA, AF, FE form a similar series. 14. If H is the center of curvature for B, show that the point 77 is with- out or within the ellipse, according as a > or < bV2, or according as e 2 > or < J. Sketch the evolute when b = -— -• o 15. Show by inspection of the figure that four real normals can be drawn to the ellipse from any point within the evolute. 16. Find the parametric equations of the evolute of the cycloid x = a(d — sin $), y = a (I — cos 6). CHAPTER XIII SINGULAR POINTS 101. Definition of a singular point. If the equation f(x, y) = is represented by a curve, the derivative -^, when it has a cix determinate value, measures the slope of the tangent at the point (x, y). There may be certain points on the curve, how- ever, at which the expression for the derivative assumes an illusory or indeterminate form ; and, in consequence, the slope of the tangent at such a point cannot be directly determined by the method of Art. 5. Such values of x, y are called sin- gular values, and the corresponding points on the curve are called singular points. 102. Determination of singular points of algebraic curves. When the equation of the curve is rationalized and cleared of fractions, let it take the form /(a?, y = 0. This gives, by differentiation with regard to x, as in Art. 65, df + dfdy^Q dx dy dx 3f du dr whence Tx = ~W' (1) By In order that — may become illusory, it is therefore neces- dx sarytohave # = 0, ^ = 0. (2) dx dy 191 192 DIFFERENTIAL CALCULUS Thus, to determine whether a given curve f(x, y) = has singular points, put J- and f- each equal to zero and solve dx dy these equations for x and y. If any pair of values of x and y, so found, satisfy the equa- tion f(x, y) = 0, the point determined by them is a singular point on the curve. To determine the appearance of the curve in the vicinity of a singular point (x lf yx), evaluate the indeterminate form dy _ dx dx = ~aj = ^ by finding the limit approached continuously by the slope of the tangent when x = x x , y = y v "** dy dx\dx y Hence dx d fdf dx dx 2 dx dydx d 2 f , d*fdy dx dy dy 2 dx [Arts. 64, 85. at the point (x ly y{). This equation cleared of fractions gives, to determine the slope at (ajj, ?/i)> the quadratic dy 2 [dxj dxdy\dx) dx 2 V } This quadratic equation has in general two roots. The only exceptions occur when simultaneously, at the point in SINGULAR POINTS 193 in which case -^ is still indeterminate in form, and must be dx evaluated as before. The result of the next evaluation is a cubic in — , which gives three values to the slope, unless all dx the third partial derivatives vanish simultaneously at the singular point. The geometric interpretation of the two roots of equation (3) will now be given, and similar principles will apply when the quadratic is replaced by an equation of higher degree. The two roots of (3) are real and distinct, real and coinci- dent, or imaginary, according as H f a 2 /y d 2 f d*f \dx dyj dx 2 By 2 is positive, zero, or negative. These three cases will be con- sidered separately. 103. Multiple points. First let 77 be positive. Then at the point (x, y) for which -J- = 0, -J- = 0, there are two values of dx dy the slope, and hence two distinct singular tangents. It fol- lows from this that the curve goes through the point in two directions, or, in other words, two branches of the curve cross at this point. Such a point is called a real double point of the curve, or simply a node. The conditions, then, to be satis- fied at a node (x lf y{) are ox v oy L and H(x u y x ) > 0. Ex. Examine for singular points the curve 3 x 2 _ xy _ o f + x s -8y* = 0. EL. CALC. — 13 194 DIFFERENTIAL CALCULUS Here $f=6x-y+3x*, & = - x - 4 y - 21 v*. dx dy 9 9 The values x = 0, y = will satisfy these three equations, hence (0, 0) is a singular point. Since |^=6+ Qx = 6 at (0,0), Fig. 52 hence the equation determining the slope is, from (3), -*a)"-«e)+-* of which the roots are 1 and — f . It follows that (0, 0) is a double point at which the tangents have the slopes 1, — f. Find the equation of the real asymptote, and the coordinates of the finite point in which it meets the curve. 104. Cusps. Next let H=0. The two tangents are then coincident, and there are two cases to consider. If the curve recedes from the tangent in both directions from the point of tangency, the singular point is called a tacnode. Two distinct branches of the curve touch each other at this point. (See Fig. 53.) SINGULAR POINTS 195 If both branches of the curve recede from the tangent in only one direction from the point of tangency, the point is called a cusp. Here again there are two cases to be distinguished. If the branches recede from the point on opposite sides of the double tangent, the cusp is said to be of the first kind ; if they recede ( on the same side, it is called a cusp of the second kind. The method of investigation will be illustrated by a few examples. . Ex. 1. f(x, y) = aY - <* 2 x A + * 6 = 0. ^ = -4a2z« + 6a? s ; & = 2 a*y. dx By The point (0, 0) will satisfy /(x, y) = 0, $£■ = 0, |^= 0; hence it dx dy is a singular point. Proceeding to the second derivatives, we obtain §y = - 12 a 2 * 2 + 30 x* = at (0, 0), dx 2 BY = dxd/ a 2 / = o, The two values of -* are there- fore coincident, and each equal to Fig. 53 zero. From the form of the equation, the curve is evidently sym- metrical with regard to both axes ; hence the point (0, 0) is a tacnode. No part of the curve can be at a greater distance from the y-axis than ± a, at which points -^ is infinite. The maximum value of y corresponds tox = ± aVf. Between x = 0, x = aVj there is a point of inflexion (Fig. 53). 196 DIFFERENTIAL CALCULUS Sketch the curves obtained by giving larger and larger values to the parameter a. Ex.2. f(x,y) = y*-x* = 0; dx~ 6X ' d y- 2y ' Hence the point (0, 0) is a singular point. Further, ^ = - 6a: = at (0, 0) ; d 2 f =0 . dY =2 dxdy ' dy 2 Therefore the two roots of the quadratic equation defining -^ are both equal to zero. So far, this case is exactly like the last one, but here no part of the curve lies to the left of the axis y. On the right side, the curve is symmetric with regard to the x-axis. As x increases, y increases ; there are no maxima nor minima, and no inflexions (Fig. 54). . p]x. 3. f(x, y) - x 4 - 2 ax' 2 y - axy 2 -f ahf- = 0. The point (0, 0) is a singular point, and the roots of the quadratic defining -j- are both equal to zero, hence the origin is a cusp, and the cuspidal tangent is the a; -axis. To show the form of the curve near the cusp, solve the equation for ?/. Then ._ a — x\ y a I First suppose that a is positive. When x is negative, y is imaginary ; when x = 0, y = ; when x is positive, but less than a, y has two positive values, therefore two branches are above the a>axis. When x = a, one branch becomes in- finite, having the asymptote x = a ; the other branch has the ordinate \ a. The origin is therefore a cusp of the second kind (Fig. 55). Next suppose that a is negative. When x is positive, y is imagi- rary; when x is negative, y is real. The same reasoning as before SINGULAR POINTS 197 shows that there is a cusp of the second kind in the second quarter, with the #-axis as a cuspidal tangent. Examine the transition case in which a = 0. Fig. 54 Fig. 55 105. Conjugate points. Lastly, let H be negative. In this case there are no real tangents ; hence no points in the im- mediate vicinity of the given point satisfy the equation of the curve. Such an isolated point is called a conjugate point. Ex. 1. /(*, y) = ay 2 - x* + bx 2 = 0. Here (0, 0) is a singular point of the locus, and at this point we find dy = dx both roots being imaginary if a and b have the same sign. To show the form of the curve, solve the given equation for y. Then y=±x\ x — b Fig. 5G and hence, if a and b are positive, there are no real points on the curve between x = and x - b. Thus O is an isolated point (Fig. 56). Examine the cases in which a or b is negative. 198 DIFFERENTIAL CALCULUS These are the only singular ities that algebraic curves can have, although complicated combinations of them may appear. In each of the foregoing examples, the singular point was (0, 0) : but for any other point, the same reasoning will apply. Ex. 2. f(x, y) = x 2 + 3 y* - 13 y 2 - 4 x + 17 y - 3 = 0, 3£=2.r-4, 3/"= 9 y 2 - 26 y + 17. dx dy At the point (2, 1), /(2, 1) = 0, &=0, ^=0; hence (2, 1) is , . 4. dx dy a singular point. Also^ = 2; JE£-=0; |2f= 18y - 26, = - 8 at (2, 1). dx 2 da; cty d// 2 Hence — = ± - ; and thus the equations of the two tangents at the dx 2 node (2, 1) are y - 1 = \(x - 2), y - 1 = - | (a: - 2). When JT is negative, the singular point is necessarily a con- jugate point, but the converse is not always true. A singular point may be a conjugate point when H = 0. [Compare Ex. 4 below.] EXERCISES ON CHAPTER XIII Examine each of the following curves for multiple points and find the equations of the tangents at each such point ; also find the asymptotes and sketch the curve : 1. a 2 x 2 = b 2 y 2 + *V- 2 a — x 3. x3 + y\ = «f ; or, in rational form, (x 2 +y 2 — a 2 ) 3 + 27 a 2 x 2 y 2 = 0. 4. y 2 (x 2 - a 2 ) = x 4 . 5. y— a+ x + bx 2 + cx% ; or, in rational form, ( y - a -x- bx 2 ) 2 - c 2 s« = ( ) . SINGULAR POINTS 199 When a curve has two parallel asymptotes it is said to have a node at infinity in the direction of the parallel asymptotes. Apply to Ex. 6. 6. (x*-y*y—4y* + y = 0. 7. x*-2 ay* - 3 a 2 y 2 - 2 d 2 x 2 + a 4 = 0. J. y 2 _ x ( x + a )2. a > o . a < o. 9. x s — 3 axy + y z = 0. Find the asymptote and sketch the curve. 10. y' 2 = x* + x 5 . 11. Show that the curve y = x log a: has a terminating point at the origin. Find the minimum value of y and sketch the curve. 12. y — x 2 log x. CHAPTER XIV ENVELOPES 106. Family of curves. The equation of a curve, /(»,y) = o, usually involves, besides the variables x and ?/, certain coeffi- cients that serve to fix the size, shape, and position of the curve. The coefficients are called constants with reference to the variables x and y, but it has been seen in previous chapters that they may take different values in different problems, while the form of the equation is preserved. Let a be one of these "constants." Then if a is given a series of numerical values, and if the locus of the equation, corre- sponding to each special value of a is traced, a series of curves is obtained, all having the same general character, but differ- ing somewhat from each other in size, shape, or position. A system of curves so obtained is called a family of curves. For example, if h, k are fixed, and }> is arbitrary, the equa- tion (y—k) 2 = 2p(x — h) represents a family of parabolas, each curve of which has the same vertex (h, k), and the same axis y = k, but a different latus rectum. Again, if k is the arbitrary constant, this equation represents a family of parab- olas having parallel axes, the same latus rectum, and having their vertices on the same line x = h. The presence of an arbitrary constant a in the equation of a curve is indicated in functional notation by writing the 200 ENVELOPES 201 equation in the form ,/(#, y, a) = 0. The quantity a, which is constant for the same curve but different for different curves, is called the parameter of the family. The equations of two neighboring curves are then written f(x, y, a) = 0, />, y, a + h) = 0, in which h is a small increment of a. These curves can be brought as near to coincidence as desired by diminishing h. 107. Envelope of a family of curves. A point of intersection of two neighboring curves of the family tends toward a limit- ing position as the curves approach coincidence. The locus of such limiting points of intersection is called the envelope of the family. Let f(x, y, a) = 0, f(x, y, a + h) = (1) be two curves of the family. By the theorem of mean value '(Art. 39) f(x, y, a + h) = f(x, y, a) + h d /- (as, y, a + 6k), (2) da which, on account of equation (1), reduces to d £{x,y,« + 6h)=(). Hence, it follows that in the limit, when h = 0, J-(x, y, «)=0 da is the equation of a curve passing through the limiting points of intersection of the curve /(a?, y } a) = with its consecutive curve. This determines for any assigned value of a definite limiting points of intersection on the corresponding member of 202 DIFFERENTIAL CALCULUS the family. The locus of all such points is then to be obtained by eliminating the parameter a from the equations f(x, y, a) = 0, -f (x, y, a) = 0. da The resulting equation in x and y represents the fixed enve- lope of the family. 108. The envelope touches every curve of the family. I. Geometrical proof. Let A, B, C (Fig. 57) be three consec- utive curves of the family ; let A, B intersect in P, and B, C inter- sect in Q. When P, Q approach coincidence, PQ will be the direction of the tangent to the envelope at P; but since P, Q are two points on B that approach coincidence, hence PQ is also the direction of the tangent to B at P, and accordingly B and the envelope have a common tangent at P. Similarly for every curve of the family. II. More rigorous analytical proof. Let — f(x, ?/, a) = da be solved for a, in the form a = <£(#, y). Then the equation of the envelope is f(x, y, <f>(x, y)) = 0. Equating the total avderivative to zero, we obtain dx dy dx d<f>\dx By dxj ENVELOPES 203 but -J- = -J- = 0, hence the slope of the tangent to the enve- d<f> da lope at the point (x, y) is given by df + dfdy = dx dy dx But this equation defines the direction of the tangent to the curve /(a?, y, a) — at the same point, and therefore a limit- ing point of intersection on any member of the family is a point of contact of this curve with the envelope. Ex. Find the envelope of the family of lines Differentiate (1) as to m, y = mx+£-, (1) • m obtained by varying m. =*-■£. (2) To eliminate m multiply (2) by m and square ; square (1) and sub- tract the first from the second. The envelope is found to be the parabola « A r y 2 = Ipx. Draw the lines (1) corresponding to m = 1, 2, 3, 4, oo ; m - - 1, - 2, - 3, - 4. 109. Envelope of normals of a given curve. The evolute (Art. 99) was defined as the locus of the centers of curvature. The center of curvature was shown to be the point of intersec- tion of consecutive normals (Art. 92), whence by Art. 107 the envelope of the normals is the evolute. Ex. Find the envelope of the normals to the parabola y 2 = ±px. The equation of the normal at (x v y\) is 204 DIFFERENTIAL CALCULUS or, eliminating x\ by means of the equation yf = ipx v we obtain J Ul 8^ 2 p K } The envelope of this line, when y x takes all values, is required. Differentiate as to ?/,, n „ _ i - ^JL\ _ JL 8p* 2p' On substituting this value for y l in (1), the result, 27 pif = 4(> - 2 jo) 3 , is the equation of the required evolute. Show that this semi-cubical parabola has a cusp at (2p, 0). Trace the curve. 110. Two parameters, one equation of condition. In many cases a family of curves may have two parameters which are connected by an equation. For instance, the equation of the normal to a given curve contains two parameters x Yi y x which are connected by the equation of the curve. In such cases one parameter may be eliminated by means of the given rela- tion, and the other treated as before. When the elimination is difficult to perform, both equations may be differentiated as to' one of the parameters, a, regard- ing the other parameter /? as a function of a. This gives four equations from which a, 8, and — may be eliminated, the da resulting equation being that of the desired envelope. Ex. 1. Find the envelope of the line a b the sum of its intercepts remaining constant. ENVELOPES ■ 205 The two equations are - -\- -J- — 1, a b a + b = c. Differentiate both equations as to a ; — x _y_db _ „ • a 2 H 2 da~ ' 1+* da Eliminate Then — = *-. which reduces to a 2 b 2 db i da x y x + y a b a h - — - ; whence a = Vex, b = Vey. a b a + b c Therefore Vx + Vy = Vc is the equation of the desired envelope. [Compare Ex. p. 87.] This equation when rationalized is (x- yy-2c(x + y) + c 2 = 0. By turning the coordinate axes through 45°, show that this repre- sents a parabola whose axis bisects the angle between the original axes. Show that the curve touches both these axes. Draw different lines of the family, corresponding to a = 4, b = 4 ; a = 5, b = 3 ; a = 6, Z> = 2 ; a - 7, b = 1 ; a = 8, 6 = 0; etc. Ex. 2. Find the envelope of the family of coaxial ellipses having a constant area. Here x - + f- = \- a* l b 2 a& = £ 2 . For symmetry, regard a and 6 as functions of a single parameter /. 206 DIFFERENTIAL CALCULUS Then by differentiation as to t, &<]±j r t_db _ a s dt b* dt.~ ' hence -.da . db A b h a — = dt dt & = T i 2 6 2 2 a = ±xV'2, b =±yV2, and the envelope is the pair of rectangular hyperbolas xy = ± \ k' 2 . Y Fig. 58 Note. A family of curves may have no envelope; i.e., consecutive curves may not intersect; e.g., the family of concentric circles x 2 +y 2 = r 2 , obtained by giving r all possible values. ENVELOPES 207 If every curve of a family has a node, and the node has different positions for different curves of the family, the enve- lope will be composed of two (or more) curves, one of which is the locus of the node. Ex. Find the envelope of the system in which X is a varying parameter. Here 3L = — 2(y — A.) = 0; by combining with /= to eliminate dX X, we obtain x 2 = 0, x — 1 = 0, x + 1 = 0. From Art. 103 it is seen that the point x = 0, y = X is a node on /; moreover, the various curves of the family are ob- tained by moving any one of them parallel to the y-axis. The lines x — 1=0, £ + 1 = form the proper envelope, and x = is the locus of the node. EXERCISES ON CHAPTER XIV Find the envelope of each of the following families of curves ; draw to scale various members of the family, and verify that the en- velope has been correctly found. 1. The family of straight lines x cos a + y sin a = p, when « is a parameter. 2. A straight line of fixed length a moving with its extremities in two rectangular axes. 3. Ellipses described with common centers and axes, and having the sum of the semi-axes equal to c. 4. The straight lines having the product of their intercepts on the coordinate axes equal to k 2 . 5. The lines y - ft = m (x -«)+ rvT+m*, m being a variable parameter. 208 DIFFERENTIAL CALCULUS 6. A circle moving with its center on a parabola whose equation is y 2 = 4 ax, and passing through the vertex of the parabola. 7. A perpendicular to any normal to the parabola y 2 = 4 ax, drawn through the intersection of the normal with the x-axis. 8. The family of circles whose diameters are double ordinates of the ellipse b 2 x' 2 + a 2 y 2 = a 2 // 2 . 9. The circles which pass through the origin and have their centers on the hyperbola x 2 — y 2 = c 2 . 10. The family of straight lines y = 2 mx + m A , m being the vari- able parameter. 11. The ellipses whose axes coincide, and such that the distance between the extremities of the major and minor axes is constant and equal to k. 12. From a fixed point on the circumference of a circle chords are drawn, and on these as diameters circles are described. 13. With the point (xi, //i) on a given ellipse as center, an ellipse is described having its axes equal and parallel to those of the given ellipse. Let (xi, y{) describe the given ellipse. 14. Show that if the corner of a rectangular piece of paper is folded down so that the sum of the edges left unfolded is constant, the crease will envelop a parabola. 15. In the " nodal family " (y - 2 a) 2 =(x - «) 2 + 8 x s - y 3 , show that the usual process gives for envelope a composite locus, made up of the "node-locus" (a line) and the envelope proper (an ellipse). 16. The family of curves (y — x 2 ) + a (x — y 2 ) = 0. INTEGRAL CALCULUS :>>*< CHAPTER I GENERAL PRINCIPLES OF INTEGRATION 111. The fundamental problem. The fundamental problem of the Differential Calculus, as explained in the preceding pages, is this : Given a function f(x) of an independent variable x, to deter- mine its derivative f'(x). It is now proposed to consider the inverse problem, viz. : Given any function f'(x), to determine the function f(x) hav- ing f'(x) for its derivative. The solution of this inverse problem is one of the objects of the Integral Calculus. The given function f(x) is called the integrand, the func- tion f(x) which is to be found is called the integral, and the process gone through in order to obtain the unknown function f{x) is called integration. The operation and result of differentiation are symbolized by the formula c j Tx f{x) = m ' (1) or, written in the notation of differentials, df(x)=f\x)dx. (2) el. c/xc. — 14 209 210 INTEGRAL CALCULUS The operation of integration is indicated by prefixing the symbol j to the function, or differential, whose integral it is required to find. It is called the integral sign, or the sign of integration. Accordingly, the formula of integration is written thus : /(*)=//'(*) dx. Following long established usage, the differential, rather than the derivative, of the unknown function /(a?) is written under the sign of integration. One of the advantages of so doing is that the variable, with respect to which the integration is performed, is explicitly mentioned. This is, of course, not necessary when only one variable is involved, but it is essential when several variables enter into the integrand, or when a change of variable is made during the process of integration. 112. Integration by inspection. The most obvious aid to integration is a knowledge of the rules and results of differen- tiation. It frequently happens that the required function /(a?) can be determined at once by recollecting the result of some previous differentiation. For example, suppose it is required to find / cos x dx. It will be recalled that cos x dx is the differential of sin x, and thus the proposed integration is immediately effected ; that is, / cos xdx = sin x. Again, suppose it is required to integrate x n dx, /• GENERAL PRINCIPLES OF INTEGRATION 211 in which n is any constant (except — 1). This problem sug- gests the formula for differentiating a variable affected by a constant exponent [(6), p. 44]. The formula referred to may be written . n+1 d(- )= x n dx, \n + y /» zy.n + 1 and hence we conclude, s, f x n dx ?i + l An exception to this result occurs when n has the value — 1. For in that case we deduce from (8), p. 44, the formula of integration I x L dx = I — = log x. The method used in the above illustrations may be designated as integration by inspection. This is, in fact, the only practical method available. The object of the various devices suggested in the subsequent pages is to transform the given integrand or to separate it into simpler elements in such a way that the method of inspection can Ipe applied. 113. The fundamental formulas of integration. When the formulas of differentiation, pp. 44-45, are borne in mind, the method of inspection referred to in the preceding article leads at once to the following fundamental integrals. Upon these, sooner or later, every integration must be made to depend. clu = n + l n+ 1 I. f u tl n.Jf = ,. g „. III. Ca^du=~ lU - J log a 212 INTEGRAL CALCULUS IV. Ce u du = e u . V. I cos u du = sin u, VI. I sin u du = — cos u. VII. I sec 2 udu = tan w. VIII. I esc 2 «* rfi* = — cot u, IX. J sec u tan ^ d«* = sec u. X. I esc u cot ?^ f/«j = — esc u. XI. f rfM =sin-iM. or-cos-t^. XII. f ^ M , = tan-i w, or - cot-i t*. J 1 + w 2 114. Certain general principles. In applying the above for- mulas of integration certain principles which follow from the rules of differentiation should be made use of. (a) Tlie integral of the sum of a finite number of functions is equal to the sum of the integrals of the functions taken separately. This follows from Art. 10. For example, I dx = I x dx — I — ■ = - — lo J x *s J x "J sx. GENERAL PRINCIPLES OF INTEGRATION 213 (b) A constant factor may be removed from one side of the sign of integration to the other. For, since d(cu) = cdu, it follows that I cdu—cu-c I du. To illustrate, let it be required to integrate x 2 dx. /< The numerical factor 5 is first placed outside the sign of integration, after which formula I is applied. Accordingly, n ^3 I 5 x 2 dx = 5 I x 2 dx = -—. Again, suppose the integral /, x dx is to be found. We notice that if the numerator had an addi- tional factor 2, it would be the exact differential of the denominator, and formula II would be applicable. All that is required, then, in order to reduce the given integral to a known form, is to multiply inside the sign of integration by 2 and outside by \. This gives Cxdx 1 C2xdx lAI(x 2 + l) li /o liN In this connection it must not be forgotten that: An expression containing the variable of integration cannot be moved from one side of the sign of integration to the other. (c) An arbitrary constant may be added to the result of integration. 214 INTEGRAL CALCULUS For, the derivative of a constant is zero and hence du = d(u + c), from which follows d(u + c) = u + c. f au =f' This constant is called the constant of integration. From the preceding remark it follows that the result of integration is not unique, but that any number of functions (differing from each other, however, only by an additive con- stant) can be found, each of which has the same given expres- sion as its derivative. [Compare Art. 10, Cor.] Thus, any one of the functions x 2 — l, x* + l, tf + a 2 , (x—a)(x-\-a) may serve as a solution of the problem of inte- grating f 2 x dx. It often happens that different methods of integration lead to different results. All such differences, however, can occur only in the constant terms. For example, f 3 (x + 1) 2 dx = 3 |f(aj + l) 2 d{x + 1) = (x + l) 3 = aj 8 + 3a 2 -f3a;+l. Integration of the terms separately gives Csx 2 dx-\- C$xdx+ ^3^ = ^ + 3^4-3^, a result that agrees with the preceding except in the constant term. Again, from formula XII, dx s tan -1 x, or — cot -1 x. ar' + l GENERAL PRINCIPLES OF INTEGRATION 215 It does not follow from this that tan _1 & is equal to — cot -1 #. But they can differ at most by an additive constant. In fact, it is known from trigonometry that — cot -1 x = tan -1 x -f for + j , _ in which k is any integer. In a similar manner the different results in formula XI can be explained. EXERCISES Integrate the following : 1. (Vxdx. 11. f CSc2j: ^. J J cot X [Hint. For the purpose of in- tegration this may be written 12. f - sin (x% dx.~] J ~ dx fw*. 13. f * L/ * J J x log x\_ t/ log x + cos a; r 2. \ x a dx. 3 log x L «' log a: f ,/.* 14 C 5x 2 dx ' J 3-' ' J X * + l ' V x 4 Cmt*JJz m 15. Jtana:^[ = -| -^ s ^ ]. 5. jV -**)•**. 16 ^otxdx. 6 C5x*-^x + l dx 17 C eaxdx , 7. (x(x 2 + a 2 )' 2 dx. H8. (e**xdx. 8. J(aar+6)»rfar. * 19. f («-!-&) »»+n*a*a:. 20. I cos 2 xdx. 9 ' J^i- 20- J' , n C(n -x)dx r . u " J Ti 7 * 21. i sin nz rt.r. ^ 2 ax - x 1 J 216 INTEGRAL CALCULUS , 22. fco**xdx[= P+ c ™ 2x dx~]. 23- \sln 2 xdx. i24. i sin(m -f n)x dx. 25. Kxsmx^dx. 26. ( co$?xdx\ = i (1 — sin 2 a:)cos xdx . 27. ( sin 3 a;G?a;. 28. rtan2xr/xT= f(sec 2 a; - l)r/af|. 29. f tan 2 x sec 2 ztf* 30. \ esc 2 (ax -\- b)dx. 31. ( Vcot a: • esc 2 x dx. 32 f ^ r f — r sec 2 .rr/.r ~j •/ sin x cos a; L J tan a: J 33. \ sec 3 a: tan a; dx. 34. r tnTirr/r . 35. r J sec a- ./ r/s Va 2 - x 2 [Hint. Divide numerator and denominator by a and then write in the form J ^w 36. J dx Vl - 4 x 2 37. J a 2 + m 5 38. f-^- J a 2 a- 2 + fta 39. f ** r = f ^- 2 > i. J x-2 _ 4 3 + 5 L J (a: - 2) 2 + 1 J 115. Integration by parts. If u and v are functions of x, the rule for differentiating a product gives d(uv) = v du -J- u dv, whence, by integrating and transposing terms, we have I u dv — uv — | v du. GENERAL PRINCIPLES OF INTEGRATION 217 This formula affords a most valuable method of integration, known as integration by parts. By its use a given integral is made to depend on another integral, which in many cases is of a simpler form and more readily integrable than the original one. Ex. 1. J \ogxdx. Assume u = log x, dv = dx. Then du = — , v = x. x By substituting in the formula for integration by parts, we obtain, i log xdx — x log x — \ dx = x log x — x = xQog x — 1 ) = x(\og x — log e) = x log - . e Ex. 2. (xe x dx. Assume u = x, dv = e* dx. Then du = dx, v = e x , and ( xe x dx = xe x — ( e x dx = e?(x — 1). Suppose that a different choice had been made for u and dv in the present problem, say u = ^ rfy = ^ From this would follow r 2 du — e x dx. v = — , 2 and j xe x dx = | x 2 e x — I —e x dx. C x 2 It will be observed that the new integral j —e*dx is less simple in form than the original one ; hence the present choice of u and dv is not a fortunate one. No general rule can be laid down for the selection of u and dv. Several trials may be necessary before a suitable one can be found. 218 INTEGRAL CALCULUS It is to be remarked, however, that dv should be so chosen that its integral may be as simple as possible, while u should be so chosen that in differentiating it a material simplification is brought about. Thus in Ex. 1, by taking u — log x, the transcendental function is made to disappear by differentiation. In Ex. 2, the presence of either x or e x prevents direct integration. The first factor x can be removed by differentiation, and thus the choice u = x is naturally suggested. Ex. 3. ( x 2 a x dx. From the preceding remark it is evident that the only choice which will simplify the integral is u = x 2 , dv = a x dx. a Hence du = 2 xdx, v = - , log: a and (x 2 a*dx = -^- — (xa x dx. J lotr a log: a J Apply the same method to the new integral, assuming u = x, di' = a x dx, whence du = dx, v — , log a and f xa x dx = ^- - — !— f a x dx ■J log a log a J x(t ar log a (log a)' 2 By substituting in the preceding formula, we have J log a L log a (logo) 2 J GENERAL PRINCIPLES OF INTEGRATION 219 EXERCISES 1. i sin~ i xdx. 7. i x cot -1 x dx. 2. f «* tan" 1 (€*)</*. 8. \ x sin 3 x dx. 3. 1 a: 2 cos xdx. 9. \ e* cos x dx. 4. \x n \ogxdx. 10. \ e x sin xdx. 5. 1 a* 2 tan -1 xdx. 11. \ cos a: cos 2 a: /7a*. 6. f sec a: tan x log cos a: dx. 12. I x sec 2 a- </a\ 116. Integration by substitution. It is often necessary to simplify a given differential f'(x)clx by the introduction of a new variable before integration can be effected. Except for certain special classes of differentials (see, for example, Arts. 127-129) no general rule can be laid down for the guidance of the student in the use of this method, but some aid may be derived from the hints contained in the problems which follow. Ex. 1. | xdx . Va 2 - x 2 Introduce a new variable z by means of the substitution a-— x 2 = z. Differentiate and divide by - 2, whence xdx— — — . Accordingly, The details required in carrying out this substitution are so simple that they can be omitted and the solution of the problem will then take the following form : f XdX = (V - **)"* xdx = - I ( (« 2 - x*y*( - 2 xdx) J Vf/ 2 - a- 2 J • - J 220 INTEGRAL CALCULUS In this series of steps the last integral is obtained by multiplying inside the sign of integration by — 2 and outside by — I, the object being to make the second factor the differential of a 2 — x 2 . Think- ing of the latter as a new variable, the integrand contains this variable affected by an exponent (- i) and multiplied by the differ- ential of the variable, in which case formula I can be applied. Ex. 2. r !°£* dx . J x Assume log x = z.. Then -- = dz, x and fl^?^ = f^ = l 2 = -Q^. J x J 2 2 Here again it is not necessary to write out the details of the sub- stitution, as it is easy to think of log a; as a new independent variable and to perform the integration with respect to it. It is then readily seen that the expression to be integrated consists of the variable dx log # multiplied by its differential — : , and that the integration is x accordingly reduced to an immediate application of the first formula of integration. Thus k2 ij\ogx.d(\ogx) = Off*)' Ex. 3. j Ex. 3. \ e ian * dx gives 1 + x 2 Think of tan -1 x as a new variable and apply formula IV. This r e tan-'*_^ _ fgtan-'x, /(tan-la-) = ^tan^x. J 1 + x 2 J Ex. 4. C*™- 1 ***. dx j- Ilegard sin -1 a: as a new variable and - as the differential Vl — x 2 of that variable. Apply formula I. GENERAL PRINCIPLES OF INTEGRATION 221 Ex. 5. j*0 2 + 2 x + 3)0 + l)dx. Multiply and divide by 2. The integral then takes the form If (a? + 2 x + 3) • (2 x + 2)rfar. Observing that (2 x -\- 2)dx is the differential of x 2 + 2 x + 3, and using the latter expression as a new variable, we see that formula I is directly applicable, leading to the result Ex. 6. flog cos (x 2 + 1) sin (a; 2 + 1) • xdx. Make the substitution x" 2 + 1 = z. The given integral takes the form - I log cos z sin z dz. Make a second change of variable, cos z — y. Then sin zdz = — dy. The transformed integral is -lijlogydy, to which the result of Ex. 1, p. 217, can be at once applied. It will be observed that two substitutions which naturally suggest themselves from the form of the integrand are made in succession. The two together are obviously equivalent to the one transformation, cos(x 2 + 1) = y. Ex.7, f du ■ Ex.8, f du . J y/cfi _ u * J u- + a 2 -[Hint. Substitute u = as.] 222 INTEGRAL CALCULUS Ex. 9. f dx /„.9 — • Ex. 10. ^ dx V2 ax — x* Hint. Substitute x = - • [Hint. Substitute x = z + a.] Ex. 11. | esc m du. Multiply and divide the integrand by esc u — cot u. It will then be seen that the integral has the form i — J z Another method would be to use the trigonometric formula n • u u sin u = 2 sin -cos -, 2 2 \ 2 V2/ Cdt =1—1 tan^ J l 2 Ex. 12. \ sec u du. Put u = z and use Ex. 11. 2 Solve the problem also by means of substitutions similar to those used in the preceding example. sin 3 x Ex.13. (x*y/a*-x*dx. Ex.15. f£°l£ J J sin- Ex.14, f * a ' 7ar • Ex.16, f J(a;-1) 3 J dx cos 2 a; -f 2 sin 2 x Put tan x — z. Ex. 17. Prove that I — — can be integrated by a substitution, J (a + bx) n when m is a positive integer. 117. Additional standard forms. The integrals in Exs. 7, 8, 11, 12 of the preceding article, and in Exs. 15, 16 of Art. 114, are of such frequent occurrence that it is desirable to collect GENERAL PRINCIPLES OF INTEGRATION 223 the results of integration into an additional list of standard forms. Two other very useful formulas are also included, the derivation of which we now give. du Integration of f du Make the substitution u + Vw 2 4- « = z. From this equation, we obtain, by differentiation, (l + U } du = dz ; \ Vw 2 + a J du that is, ( VV + a + u) — — = = dz, Vw* + a whence, du dz dz Vm 2 + a Vw 2 + a-\-u z This gives, on integrating, >_* ,r*_ log , J Vm 2 +« ^ z = \og(u -f Vw 2 + a). Integration of f f u o • J w 2 — a 2 The fraction — may be written as the sum of two u 2 — a" simpler fractions, _l_ = j_[^ l_i u- — a 2 2a\_u —a u + a J 224 INTEGRAL CALCULUS whose denominators are the factors of u 2 — a 2 . Hence, r du _ 1 rr_du_ du 1 J u 2 — a 2 2 a J [_u — a u -f a J li / x .1 1 , u— a J a it 4- ct = 2^ log (w - a)- log (w 4- a) XIII. du =S in-^ Va 2 /; XIV. f **" - log (u 4 VttHo). a u — a u + a XVII. I tan udu = - log cos u - log sec u. XVIII. J cot udu = log sin u. XIX. J sec w du = log (sec ™ + tan u) = log tan (tt + jY XX. J esc m r Jw = log (esc u — cot w) = log tan „ • 118. Integrals of the forms / •(Ax + B)dx and /» (^g; + B)dx ax 2 + bx + c J ■ y /ax*+ bx + c Such integrals occur so frequently that they deserve special mention. The integration is facilitated by the substitution of a new variable t which reduces the affected quadratic ax 2 4- bx 4- c to a pure quadratic of the form mt 2 + n. The mode of procedure will be readily understood from the follow- ing illustrative problems. . GENERAL PRINCIPLES OF INTEGRATION 225 x , , C xdx I,x. 1. \ — -. J 2 x 2 + 2 x + 3 The first step is to complete the square of the x terms in the denominator. After the factor 2 has been placed outside the integral sign, the quadratic expression may be written (x* + x + 1) + a - \) = (x + \y + 1 . Now substitute a new variable t in place of x+\. Since x = t — \ and dx = dt, we obtain for the new form of the given integral 1 r(t-\)dt = \ r'ltdt If dt **+* 1L = ilog(* 2 +-^ — tan 2V5 V5 Ex.2, f (2^-1)^ J VI +2^-3^' Divide out V3 from the denominator; since the coefficient of x 2 is negative, put the x terms in parentheses preceded by the negative sign and complete the square. The integral then becomes \)dx. V'SJ v'| - (x - i)* Now make the substitution x — \= t. Since dx = <//, the integral reduces to V ( -^=-^(M 4 (-- -^- V.jJ V| - ** ViiJ \9 / 3V3J V-f - t' 2 2 '3A9 ; 3 V3 V 2 / 2 /r~2 I 1 • J3x-1\ -v/- -f - :r — a; 2 sin -1 - Vo - ^ 3 3 V3 V 2 ; - -a/1 + 2 a: - 3 x 2 - — siu-*f 3 * ~ * V 3^ 3V3 V 2 J EL. CALC — 15 226 INTEGRAL CALCULUS It is seen from the two preceding examples that the met) here used contains two essential steps : (1) Completing the square of the x terms in ax 2 -\- bx + c ; (2) Substituting a new variable for the part in parentheses. If the numerator of the new integral contains two terms, separate into two integrals and integrate each one separately. EXERCISES 1 C dx 8 C (' 2x ~ 3 ) rix J3x 2 -2x + 5' 9. )\ }~ X dx. dx J 8 + 4 x - 4 x 2 4 r dx ^ V30 x - 9 x i - 24 5. C x dx * Vx* + 2 x + 2 6. J VI + 2 x - x 2 7 C (4x + 5)dx * V8 - 4 x - 4 x 2 119. Integrals of the forms + [Rationalize the numerator.] 1Q r (3x + 2)rfz , 11. ]•# 12 f (2 x _+_l),/a . J V-2^- 3z- 1 13 r (x-Z)dx ' J V-3s 2 -2a; + l f- ** and f dx (Ax + ByWax* + bx + c Integrals of these types can be reduced to forms given in the preceding article by means of the reciprocal substitution GENERAL PRINCIPLES OF INTEGRATION 227 I x EXERCISES C dx 7 f_. dx * xVx* + a 2 ^ (x + 2) V- a: 2 - 10x-7 8. f dx . J ar 2 Va 2 — x 2 ^ 6 xVx 2 + a 2 rfx xVa 2 — x 2 dx x Vo x- — 4 x + 1 rte (x + l)Vx 2 + 2 X + 3 tfs (x + l)Vx 2 + x + 1 tfx ^ x 2 v x 2 - a 2 10. f_ rfj; 5. f- dx ' JxWx* + a* r ^/x ii r - f i x * (\ - x^ \f'2 x 2 - 4 r. + 1 ^ f2 x - (1 - x) v/2 x 1 - 4 x + 1 J (2 x- - 1) V4 x' 2 - 3 EXERCISES ON CHAPTER I 1. ( e** e x dx. 7. \x(a' 2 - x 2 )* dx. 8. f *•« ■ 9. f •" ■ J y/ X + 1 + Vx - 1 10. i cos 8 x dx. 11. ( sec 3 x dx. 12- 1 f r sin e* rfx. 2. f 5 x z dx J 4 + x 8 3 f (2 + 3x 2 ) f /x J 6 x 3 + 12 x + 5 4. f 1 + *rf*. J Vx 5. f * . J ^8 - 2 « 6 r //x 228 INTEGRAL CALCULUS dx 13 C sin x dx 23 f_ J a cos x + b * 6xVl - log a; C dx ' 1 , «/. C e x dx J Vl - e 2 * 24. \ -• 14 . [Put e— = «.] 15 r s<fo . 25. j* e x + e- cos dO y/l _ ^4 vl + cos' 2 - sin ^ 16. S , dX ■ 26. f- dx V'4 a: 4 + 8 a; 2 J a: (log a:) 2 + x 17 l e T^- 27. f( seC * Vd,. e e J \a - b tan x/ 18. \% 4 tan -1 a; da;. 28. J. (a: — a) dx C x~ dx J Va*-a 2 (x — a) 2 —(x-a) 4 J a x 2,j_ ™ f ^0 r fl-sin^./n 20 - )m^L = 3^s^ rfe ]- "«-v8^ +2 «+i ._ f tan (9^ r Ja + fttan'0" 31. J sin a: log tan x dx. 32 r ^ r— f s * n x ^ x J 1 + cot a: L J sin a: 4- cos x — 1 f ( sm x + cos x) — (cos x — sin a:) , 2 J sin a; + cos x = 1 rA.cosar-sinarX^l 2 J V sin a; -f cos a;/ J [Another method would be to multiply numerator and denominator by sin a:(cos x — sin x) and express in terms of the double angle.] CHAPTER II REDUCTION FORMULAS 120. In Arts. 118, 119 the integration of certain simple ex- pressions containing an irrationality of the form ^/atf+bx+c was explained. As was shown in Art. 118, the radical can be reduced to the form V ± x 2 ± a? by a change of variable. It remains to show how the integration can be performed in in such cases as, for example, x n dx j z w V ± x 2 ± a 2 dx, J - V ± a? ± ci- ri being any integer. For this purpose it is convenient to consider a more general type of integral of which the preceding are special cases, viz., Car (a + bx n ) p dx, (1) in which m, n, p are any numbers whatever, integral or frac- tional, positive or-negative. It is t$> be remarked in the first place that n can, without loss of generality, be regarded as positive. For, if n were negative, say n = — n', the integrand could be written x mf a + IV = x m f axn 'j' 5 V = x m -* n '(b + ax*'y This expression, which is of the same type as x m (a + bx n ) p , is such that the exponent of x inside the parentheses is positive. 229 230 INTEGRAL CALCULUS It will now be proved that an integral of the type (1) can in general be reduced to one of the four integrals (a) A j x m ~ n (a + bx n ) p dx, (b Ai x m+n (a + bx n ) p dx, (c) A I x m (a + bx n ) p -*dx, (d) A ( x m (a + bx n ]*+ l dx f plus an algebraic term of the form Bx K (a + bx n y. Here A, B, \, fx are certain constants which will be deter- mined presently. Observe that in each of the four cases the integral to which (1) is reduced is of the same type as (1), but that certain changes have taken place in the exponents, viz., the exponent m of the monomial factor is increased or dimin- ished by n, or, the exponent p of the binomial is increased or dimin- ished by unity. The values of A. and fi are determined by the following rule : Compare the exponents of the monomial factors in the given integral and in the integral to which it is to be reduced. Select the less of the two members and increase it by unity. The result is the value of\. In like manner, compare the exponents of the * binomial factors in the two integrals, select the less, and increase it by unity. This gives /x. Thus, if it is desired to reduce the given integral to A j x m ~ n (a 4- bx n ) p dx, first write down the formula j x m (a + bx n ) p dx — A \ x m ~ n (a 4- bx n ) p dx + Bx x (a 4- &»■)*. REDUCTION FORMULAS 231 The exponents of the monomial factors in the two integrals are m and m — n respectively, of which m — n is the less. This, increased by unity, gives the value of X; that is, X = m — n + 1. Again, the exponent of the binomial factor in each integral is the same, namely p, so that there is no choice as to which of the two is the less. Increase this number p by unity to obtain the value of /x. Hence /x = p -f 1. The above formula may now be written J x m (a -(- bx n ) p dx = A Cx m -' l (a + bx n ) p dx + Bx m - n+1 (a + bx n ) p+ \ (2) In order to determine the values of the unknown constants A and B, simplify the equation by differentiating both mem- bers. After being divided by x m ~ n (a + bx n ) p the resulting equation is reduced to x n = A + Ba(m — n + 1) + Bb(m + np + l)a,- n . By equating coefficients of like powers of x in both members, we find the values of A and B to be A _ a(m - m + 1) B= 1 b(m + np -h 1) ' b(m + np + l)' When these values are substituted in formula (2), it becomes J x m (a +bx n ) p dx ci{ m — n + 1) + np+l)J v 7 6(m+np + l) LJ Notice that the existence of formula (2) has been proved by showing that values can be found for A and B which make the two members of this equation identical. 232 INTEGRAL CALCULUS There is one case, however, in which this reduction is impossible, viz., when m + np + 1 = 0, for in that case A and B become infinite. [See Ex. 4, p. 235.] In a similar manner the three following formulas may be derived : J x m (a + bx n ) p dx a(m + l) J k -r j a (m + l) I £ m (a + bx n ) p dx : l I x m (a + bx n ) p x dx H *■—-*- f- • TC] m + np + 1 •/ m + ?ip + 1 I x m (a-\-bx n ) p dx : — ! — i — — I x m (a + bx n ) p +hlx i — ■ - TDl an(j> + l) J awQj + 1) The cases in which the above reductions are impossible are, For formulas [A] and [C], when m + np + 1 = 0; for formula [B] , when m + 1 = ; for formula [D] , when _p + 1 = 0. .. (x s Va Ex.1. \ x 8 Va* - x* <lx. If the monomial factor were x instead of x 3 , the integration could easily be effected by using formula I. Since in the present case m — 3, n = 2, formula [A], which diminishes in by n, will reduce the above integral to one that can be directly integrated. REDUCTION FORMULAS 233 Instead of substituting in [A], as might readily be done, it is best to apply to particular problems the same mode of procedure that was used in deriving the general formula. There are two advantages in this. First, it makes the student independent of the formulas, and second, when several reductions have to be made in the same problem, the work is generally shorter. [See Ex. 4.] Accordingly assume f x\d 2 - xrf dx = A ( x(a 2 - x 2 y dx + Bx 2 (a 2 - x 2 )\ the values of X and ft having been determined by the previously given rule. Differentiate, and divide the resulting equation by x(a 2 — x 2 y. This gives . 6 x 2 = A + B(2a 2 -5x 2 ), from which, on equating coefficients of like powers of x, o 5 hence, (x 3 ^d 2 -x 2 dx = —( (a 2 ~ x 2 ) * xdx-\ x\a 2 - z 2 )* = - ^(2 a 2 + 3x 2 ) (a*-xrf. :. 2. (Vx 2 Ex. 2. \ Vx 2 -2x -6 dx. By following the suggestions of Art. 118, this integral can be re- duced to the form C Vz 2 - 4 dz, in which z = x — 1. Assume jV - 4)Klz = A§(z 2 - 4)~^ dz + Bz(z 2 - 4)i In determining A notice that m = in both integrals, so that X = + 1 = 1. Also, fji = - £ + ! = £. 234 INTEGRAL CALCULUS Ex.3. ( V2 ax - x l dx. The mode of procedure of Ex. 2 may be followed. Another method can also be used, as follows. On writing in the form fa* (2 a -xy dx, and observing that the integration of jV* (2 <z -aO"~*tf*= J V2 ax - x 2 can be performed (see Ex. 10, p. 222), it will be seen that integration may be effected in the present case by reducing each of the exponents m and p by unity. This is possible since n = 1 and m can accordingly be diminished by 1. Hence assume (V (2 a- xy dx = A ' jV * (2 a- x)* dx + B'x? (2 a- x)% . The exponent of the binomial in the new integral may be reduced in turn by assuming f aT* (2 a- a?)* dx = A" ( x~? (2 a- x)~^ dx + B" x?(2a-x)K When this expression is substituted for the integral in the second member of the preceding equation, the result takes the form ( V2 ax - x 2 dx = A ( dx + Bx*(2 a - a:)* + Cx?(2 a - xf, J . a/2 ax - x 1 in which A, B, C are written for brevity in the place of A' A", A'B", B' respectively. The values of A, B, C are calculated in the usual manner by differentiating, simplifying, and equating coefficients of like powers of x. The method just given requires two reductions, and hence is less suitable than that employed in Ex. 2, which requires but one reduction. The rule for determining the values of A. and /x may now be advantageously abbreviated. Let m,p be the exponents of the REDUCTION FORMULAS 235 two factors in the given integral, and m', p' the corresponding exponents in the new integral. Of these two pairs, m, p and m', p', one of the numbers in the one pair is less than the cor- responding number in the other pair. This fact will be ex- pressed briefly by saying that the one pair is less than the other pair. With this understanding the preceding rule may be expressed as follows ; Select the less of the two pairs of exponents m, p and m', p'. Increase each number in the pair selected by unity. Tliis gives the pair of exponents A., /jl. Ex.4, f * dx . (x 2 + a 2 )* Assume successively ( x\x 2 + a 2 )~~ 2 dx = A' (x\x 2 + a 2 )~2 dx + B'x\x 2 + a 2 )"^, (x 4 (x 2 + a 2 )~* dx = A" (x 2 (x 2 + a 2 )~* dx + B"x*(x* + a 2 )*, (x 2 (x 2 + a 2 )"* dx = A'"1j(x 2 + a 2 ) - ^ tte + i?'".r(.r 2 + a 2 )i These equations may be combined into the single formula ( x\x 2 + a*)~*dx = A fj (x 2 + a 2 )~^dx + Bx(x 2 + a 2 y + Cx*(x 2 + a 2 )* + Dx*(x 2 + a 2 )""i The values of the coefficients are found to be A=-$a*, B=-, C = -—. D = —- 2 a 2 a 2 Hence ( x\x 2 + a*)~* dx = ** +%<** * _ 3 a -2 log ( X + VX 2 + a 2). J 2Vx 2 •+- a' 2 236 INTEGRAL CALCULUS In this example three reductions were necessary; first, a reduction of type [X)], second, and third, a reduction of type [-4]. Can these reductions be taken in any order? The different possible arrangements of the order in which these three reductions might succeed each other are (1) UL Ul [2>] ; (2) 01], [Z>], [A] ; (3) [Z>], Ul [A], of which number (3) was chosen in the solution of the problem. Of the other two arrangements, (2) can be used, but (1) cannot. For, after first applying \_A.~] (which would be done in either case), the new integral is „ _ 3 \ x 2 (a 2 + x 2 ) J dx. If \_A~\ were now applied it would be necessary to assume f x\a 2 + x 2 )~i dx = A ( (a 2 + x 2 )~i + Bx(a 2 + x 2 )~\. This equation, when differentiated and simplified, becomes x 2 = A + Ba 2 , a relation which it is clearly impossible to reduce to an identity by equating coefficients of like powers of x, since there is no x 2 term in the right member to correspond with the one in the left member. It will be observed that this is the exceptional case mentioned on page 232, in which m + np + 1 = 0. EXERCISES 1. f (a 2 - x 2 y 2 dx. 5. ( Va 2 - 2. f— ^ 6. f_ J(> 2 + 4)2 J x * x 2 dx. dx 3. f ** • 7 C—** (x 2 +a) . C x 2 dx /» 3 REDUCTION FORMULAS 237 9. (Vx^T^dx. 12. ( dx • 10. f a;v2ox - x*dx. 13 f ^f J ' J (y+ + i x + .3)3 r y/'^nx # 2 , /• 11. J -^— dx. i4. j VI - 2 x - x 2 dx. 15. Show that f dx _ 1 f a? (9 _o\ C dx _~| J 2 + c)» ~ 2 c(> - 1) L (a; 2 + c)»-i x K " U ' } J (x 2 + c)«-U ' 16. f_^ 19. f 17 "^ sin0<70 (1 + e sin 2 0)t [Substitute cos = z.] C xdx J (* 2 +7) 2 ' 20. jWa ia J : — Hh^" 21. f (« 2 - x 2 )'2 2 - s 2 rfa:. dx. CHAPTER III INTEGRATION OF RATIONAL FRACTIONS 121. Decomposition of rational fractions. The object of the present chapter is to show how to integrate fractions of the form <f>(x) wherein <f>(x) and if/(x) are polynomials in x. The desired result is accomplished by the method of sepa- rating the given fraction into a sum of terms of a simpler kind, and integrating term by term. If the degree of the numerator is equal to or greater than the degree of the denominator, the indicated division can be carried out until a remainder is obtained which is of lower degree than the denominator. Hence the fraction can be re- duced to the form iM =aX n + bx n-l + ... +./M, xP{x) x];(x) in which the degree of f(x) is less than that of if/(x). As to the remainder fraction ^ > ' . it is to be remarked in the first place that the methods of the preceding articles are sufficient to effect the integration of such simple fractions as A A' _. Mx + n M'x+N ' . P^±_Q _ (1) x—a(x — a)' 2 ' ' x ,2 ±a 2 ' (# 2 ±a 2 ) 2 ' ' x 2 +mx+ri Now the sum of several such fractions is a fraction of the kind under consideration, viz., one whose numerator is of 238 INTEGRATION OF RATIONAL FRACTIONS 239 lower degree than its denominator. The question naturally arises as to whether the converse is possible, that is : Can f(x) every fraction ^-^ be separated into a sum of fractions of as simple types as those given in (1)? The answer is, yes. Since the sum of several fractions has for its denominator the least common multiple of the several denominators, it fol- lows that if -7^- can be separated into a sum of simpler frac- tions, the denominators of these fractions must be divisors of $(x). Now it is known from Algebra that every polynomial xf/(x) having real coefficients (and only those having real coeffi- cients are to be considered in what follows) is the product of factors of either the first or the second degree, the coefficients of each factor being real. This fact naturally leads to the discussion of four different cases. I. When if/(x) can be separated into real factors of the first degree, no two alike. E.g., xf,(x) = (x-a) (x - b) (x - c). II. When the real factors are all of the first degree, some of which are repeated. E.g., if; (x) = (x-a)(x- b) 2 (x - cf. III. When some of the factors are necessarily of the sec- ond degree, but no two such are alike. E.g., + (x) = (x 2 + a 2 ) (x 2 + x + 1) (x - 6) (x - c) 2 . IV. When second degree factors occur, some of which are repeated. E.g., t(x) = (x 2 + a 2 ) 2 (x 2 -x + l)(x-b). ■ 240 INTEGRAL CALCULUS 122. Case I. Factors of the first degree, none repeated. When ij/(x) is of the form ijz (xj = {x — a) (x — b)(x — c) • • • (x — n), f(x) A . B G . N assume J -\J- = f- H h • • • H > j/a(.t) a; — a x — b x — c £ — ft in which .4, 5, C, •••, N are constants whose values are to be determined by the condition that the sum of the terms in the right-hand member shall be identical with the left-hand member. Ex. C*-** + *dz. J x\ - 3 x + 2 Dividing numerator by denominator, we obtain x 3 — 3 x" 2 -f a: a: .r- a; + 2 x 2 - 3 ar + 2 Assume — — ■ = — - — - -f (x-l)(a:-2) x-1 x - 2 By clearing of fractions, we have (1) x = A(x -2) +B(x- 1). In order that the two members of this equation may be identical it is necessary that the coefficients of like powers of x be the same in each. Hence 1 = ^+5, = - 2 A - B, from which A = - 1, B = 2. Accordingly the given integral becomes $( x+ ^--^dz = ^+]og(x-l)-21og(x-2) Alog *-* . 2 & (a; - 2) 2 INTEGRATION OF RATIONAL FRACTIONS 241 A shorter method of calculating the coefficients can be used. Since equation (1) is an identity, it is true for all values of x. By giving x the value 1 the equation reduces to 1 = A(— 1), or A = — 1. Again, assume x = 2. Whence 2 = B. EXERCISES 1 r dx . r (x 2 - ab) dx J x 2 - a 2 ' " J (x - a) (x - b) ' 2. f 1 - 3 * ^. 5. r__^f J x s - x J x 2 - 4 x + 1 3 r (x*-V2)dx 6 f_(x 2 - l)dx J x 2 + 4 x + 3 " ' J (x* ? f x 2 -2cx+flc- q& + & J (x — a)(x — &)(x — c) 8. fx 2 (x + a)- 1 ^*^) -1 ^- 9 r (3s + l)tfar 12 f__^£_ J2x 2 + 3x-2' " J x 2 + 7 a; + 12 1Q f (x 2 + qft)<7x 13 f tf* Jx(x -a)(x+ &)' ' J a 2 x 2 - i 2 ' 4)(4x 2 - 1) Jx. 11 f (' + *)*« . 14. f_! J 2 x - x 2 - x 3 J 1 sec 2 x e/x tan* 2 x [Put tan x = t.~\ 123. Case II. Factors of the first degree, some repeated. Fx r (ox 2 -3x+l)^x J x(x - l) 3 Assume (1) 5 x 2 - 3 x + 1 ^ ,1 [ B , C , D x(x-l) 3 x x-1 (x-1) 2 (x-1) 3 To justify this assumption, observe that: (a) In adding the fractions in the right-hand member, the least common multiple of the denominators will be x(x — l) 3 , which is identical with the denominator in the left-hand member. el. calc — 16 242 INTEGRAL CALCULUS (6) Further, the expressions x, x — 1, (x — l) 2 , (a; — l) 3 are the only ones which can be assumed as denominators of the partial fractions, since these are the only divisors of x (x — l) 3 consisting of powers of a prime factor. (c) When equation (1) is cleared of fractions, and the coefficients of like powers of x in both members are equated, four equations are obtained, exactly the right number from which to determine the four unknown constants A, B, C, D. Instead of the method just indicated in (e) for calculating the coefficients, a more rapid process would be as follows. By clearing of fractions, the identity (1) may be written 5 x 2 - 3 x + 1= A(x - l) 3 + Bx(x - l) 2 + Cx(x - 1) + Dx. Putting x = 1 gives at once 3 = D. Substitute for D the value just found, and transpose the corre- sponding term. This gives 5a;2_ 6z + 1 = A(x-1)* + Bx(x- 1)-+ Cx (x - 1). It can be seen by inspection that the right-hand member of the result is divisible by x — 1. As this relation is an identity, it follows that the left-hand member is also divisible by x — 1. When this factor is removed from both members, the equation reduces to 5 x - 1 = A (x - l) 2 + Bx (x - 1) f Cx. Now put x = 1. Then C = 4. Substitute the value found for C, transpose, and divide by x — 1. The result is 1 = A (x - 1) + Bx. By giving x the values and 1 in succession, we find that A =-1, 5=1. Accordingly, we have r (5s»-3ar+l)tf* = (7 1 + _L_ + 4 » ) dx J X(X-1Y J\ XX- (x-1)* (X- = log ; x(x-iy J\ x x-\ o-i) 2 (x-iy x - l Sx - 5 2(x - l) 2 ' INTEGRATION OF RATIONAL FRACTIONS 243 EXERCISES C dx 4 f (V2r+l)</g " y{x-m* + v ' J^ (x+ V2)2 Jx 3 (x-1) J 38(3 + 1)3 T xr/x 6 C 2(.r 3 + ft 2 *)*/* J (x 2 - a 2 ) 2 ' ' J a: 4 - 2 a 2 x 2 f a 4 ' 7 f V2^ f — J (2 + (2 + V2 - V2 x) 8 8 r ax* + n' 2 x 2 + (a + \)x+a dx J x 2 (a + x) 9 r(.r 3 - \)dx ±1 C (x 2 -Ux+2G)dx J x 3 +3x 2 '* 'J 10. f (ax 2 + fa 8 )- 1 rfx. [Substitute .r — 3 = 2.] 12. r x ' 2<lx , T .r 2 r/x J (x-aY (- [Substitute a; — a = 2.] 124. Case III. Occurrence of quadratic factors, none repeated. Fx ± C (4x 2 + ox + ±)dx ' J (x' 2 + l)(x 2 + 2x + 2)' Assume .jn 4 x 2 + 5 a: + 4 _ J x + i? Cx + /> O 2 + l)(x 2 + 2 x + 2) x 2 + 1 x 2 + 2 3 + 2 Then (2) 4x 2 + 5x+4=(^x + £)(z 2 + 2a; + 2) + (Ca; + Z))(x 2 + 1). By equating coefficients of like powers of x = A + C, 5 = 2 ,4 + 2 5 + C, 4 = 2/1+5 + A 4 = 2/3 + Z), from which 4 = 1, 5 = 2, C - - 1, Z> = 0. Hence the given integral becomes C(x + '2)r/x C X'lx „. , 1/ , ix, 11 * 2 +l 1 —5—1 J ~ — ; ; = 2 tan- 1 ^ + tan" 1 ^ + l)+£log- — -^ — -. J x 2 +\ Jx 2 +2x + 2 x 2 + 2.r + 2 244 INTEGRAL CALCULUS To make clear the reasons for the assumption which was made con- cerning the form of equation (1), observe that since the factors of the denominator in the left member are x 2 + 1 and x 2 + 2 x -f 2, these must necessarily be the denominators in the right member. Also, since the numerator of the given fraction is of lower degree than its denominator, the numerator of each partial fraction must be of lower degree than its denominator. As the latter is of the second degree in each case, the most general form for a numerator fulfilling this re- quirement (i.e., to be of lower degree than its denominator) is an ex- pression of the first degree such as Ax + B, or Cx + D. Notice, besides, that in equating the coefficients of like powers of x in opposite members of equation (2), four equations are. obtained which exactly suffice to determine the four unknown coefficients A,B, C,D. dx Ex.2. ( J (x 2 + 2 +l)0 2 +2) We can assume in this case - — («+!)(« + 2) (.r 2 + l)(a; 2 + 2) x 2 + 1 x 2 + ;ion x 2 = t, the , to which Case I is applicable. For if we make the substitution x 2 = t, the given fraction becomes 1 EXERCISES ± C 4dx 5 C (ix-6)dx J x 8 + 4 x J x 4 + 2 x' 2 „ C xdx 6 C x dx J (x + \){x 2 +1)' J x* + x 2 + 1* 3 f fix 7 C xdx Jx 3 + a 8 ' ' J (x - a)\x 2 + a 2 )' 4 C (a 2 -b 2 )dx 8 C (x* + 2x + 2)dx J (x 2 + a 2 )(x 2 + b 2 )' ' J (x - l)(x 2 + 2x + 2) 9. f 2d * J (x-lX^ + 1) INTEGRATION OF RATIONAL FRACTIONS 245 125. Case IV. Occurrence of quadratic factors, some repeated. This case bears the same relation to Case III that Case II bears to Case I, and an exactly analogous mode of procedure is to be followed. Ex. f 2 * 5 -* 4 * 8 * 8 + *,/*. J O 2 + 2) 3 Assume 2x 5 -x 4 + 8x* + i = Ax + B Cx + D Ex + F (x' 2 + 2)* " x 2 + 2 (x a + 2) a (^ + 2) 3 ' Whence, by clearing of fractions, 2x*-x* + 8xS+4=(Ax+B)(x 2 + 2) 2 +(Cx + D)(x 2 + 2)+Ex + F. (1) Instead of equating coefficients of like powers of x, as might be done, we may calculate the values of A, B y C, ••• by the following briefer method. Substitute for x 2 the value — 2, or, what is the same thing, let x = V— 2. This causes all the terms of the right member to drop out except the last two, and equation (1) reduces to • - 8V^2 = EV^2 + F. By equating real and imaginary terms in both members, we obtain - 8 = E, = F. Substitute the values found for E and Fin (1), and transpose the corresponding terms. Both members will then contain the factor x 2 -f 2. On striking this out the equation reduces to 2x* -x 2 + ±x + 2 = (Ax + B)(x 2 + 2) + Cx + D. Proceed as before by putting x 2 = — 2. Whence 4 = CV^2 + D, and therefore = C, 4 = D. 246 INTEGRAL CALCULUS Substitute these values, transpose, and divide by x 2 -f 2. This gives 2x -1 -Ax + B, whence A = 2, i? = - 1. The given integral accordingly reduces to J x 2 + 2 J (x- 2 + 2) 2 ^ (** + 2) s The first term becomes J x 2 + 2 ^ z 2 + 2 yo v 2 The second, integrated by the method of reduction (Chap. II), 8" ives x 1 . . x tan -1 & + 2 V2 V2 Finally, by using formula I the last term is integrated immediately. Hence f 2 x b - x A + 8 x s + 4 , , / o , \ , a: J (x l -4- 2 V ir- 2 -4- 2 (z 2 + 2) 3 ° v x 2 + 2 (x- 2 + 2)2 EXERCISES + 1/ •> a: 2 (ar 2 +l) s J V^ 2 + 1/ J x 2 (x 2 + l) 2 2. f^ + ^' + ^fe. 5. f^ + ^'^rft J (x 2 -ha 2 ) 2 J (x 2 + a 2 ) 2 3 C 2xdx 6 C x 6 dx J (1+ x)(l+x 2 ) 2 ' * J(l + x 2 ) 8 " [Ex. 6 can also be integrated, and more easily, by means of the substitution 1 + x 2 = £.] The principles used in the preceding cases in the assump- tion of the partial fractions may be summed up as follows : Each of the denominators of the partial fractions contains one and only one of the prime fac'ors of the given denominator. INTEGRATION OF RATIONAL FRACTIONS 247 When a prime factor occurs to the nth power in the denominator of the given fraction, all of its different powers from the Jirst to the nth must be used as denominators of the partial fractions. TJie numerator of each of the assumed fractions is of degree one lower than the degree of the prime factor ichose power occurs in the corresponding denominator. 126. General theorem. Since every rational fraction can be integrated by first separating it, if necessary, into simpler frac- tions in accordance with some one of the cases considered above, the important conclusion is at once deducible : Hie integral of every rational algebraic fraction is expressible in terms of algebraic, logarithmic, and inverse-trigonometric functions. CHAPTER IV INTEGRATION BY RATIONALIZATION At the end of the preceding chapter it was remarked that every rational algebraic function can be integrated. The question as to the possibility of integrating irrational func- tions has next to be considered. This has already been touched upon in Chapter II, where a certain type of irrational functions was treated by the method of reduction. In the present chapter it is proposed to consider the sim- plest cases of irrational functions, viz., those containing ■y/ax -f- b and yW 2 -f bx -4- c, and to show how, by a process of rationalization, every such function can be integrated. 127. Integration of functions containing the irrationality •y/anc + b. When the integrand contains -y/ax + b, that is, the nth root of an expression of the first degree in x, but no other irrationality, it can be reduced to a rational form by means of the substitution ■y/aoc -\-b = z. Ex. 1. Ki dx x + 3 - - 1 Assume V2 x + 13 = z, that is, 2 x + 3 = z 2 . Then dx = z <h, and f dx _ C zdz z+ log(z-l) V2 x + 3 - 1 sz-i = V2 .rT~3 + log (V2T+3 - 1). 248 INTEGRATION BY RATIONALIZATION 249 j-l + «* r **- 1 + X* — X 1 — VI x z + a: It would appear at first sight that this integrand contains several irrationalities, viz., Vx, Vx, Vx. It is readily seen, however, that they are all powers of Vx, and hence the substitution Vx = z will rationalize the expression to be integrated. EXERCISES 4. 5. f dx K Vx + 1 S; fix /x+ Vx f dx dx (x - l)Vx dx (x — a — b 2 ) Vx — a 5- 2 Vx - 1 -f x x 7 + x i l+^l dx . When two irrationalities of the form Vaa 4- b, -Vex -f d occur in the integrand, the first radical can be made to dis- appear by the substitution Vax -\-b = z. The second radical then reduces to V C a (z*-b) + d, and the method of the next article can be applied. 128. Integration of expressions containing Vaoc 2 +bx + c. Every expression containing Va.T 2 + bx + c, but no other irrationality, can be rationalized by a proper substitution. Two cases are distinguished. (a) When ax 2 -f bx + c has real factors. We may then write the quadratic expression in the factored form ax 2 -f bx + c = a (x — a) (x — /?), (1) 250 INTEGRAL CALCULUS in which a and ft are real. Introduce a new variable t by means of the formula VaX 2 + bx + c = t(x — a) . (A) Square both members of this equation and replace the left member by means of (1). This gives a(x-a)(x-(3) = t 2 (x-a) 2 . On canceling x — a and solving for x we obtain as the equa- tion of transformation .9 x= «tr-aP m (2) Hence x (and therefore dx) is rationally expressible in terms of t, while the radical reduces to [ at 2 -a(S 1 _ t 2 -a at(a — /?) f-a (3) which is also rational in t. The substitution of these expres- sions in the proposed integrand gives a rational fraction which may be treated by the methods of the preceding chapter. (b) When a, the coefficient of x 2 , is positive. Make the substitution Vase 2 + bx + c = Va- x + t. (B) By squaring both members and solving for x we obtain b - 2Vat while the radical is expressible in the form ■Vat 2 — bt-\-^fac /K x -p > \») 2Vat-b INTEGRATION BY RATIONALIZATION 251 and hence the integrand becomes rational when expressed in terms of t. The only case that is not included in (a) or (6) is that in which the factors of ax 2 + bx + c are imaginary and the coeffi- cient a is negative ; the radical is then imaginary for all values of x. Although the integral can be obtained (in an imaginary form) by either of the preceding substitutions, this case does not arise in practical applications of the calculus and will not be considered further. Ex. 1. f dx J x + vV-2 + 2 x - 1 Formula (B) gives Vx 2 + 2x-l = x + t, whence, by solving for x, we obtain ,_ ' 2 +l 2(1 - _ /2 i 9 j J. J and accordingly dx = — — — - dt, h y 2(1 - 2 ^ + Dx-U - fi + 2< + 1 2(1 - When these expressions are substituted in the above integral it reduces to r (- t * + 2t+l)dt 2(1 + 2 The work of integrating may be facilitated by means of the trans- formation 1 -f t = 2. The result, in terms of x, is h(x - VV 2 + 2x - 1) + 1 - x + Vx 2 + 2 x - 1 + 2 log (1 - x + Vx 2 + 2 x - 1 ). 252 INTEGRAL CALCULUS Ex.2, r Vl +*</*_. ^ (1 - *) VT - X By rationalizing either numerator or denominator we obtain Vl — x 2 as the radical part of the integrand. Formula (A) gives Vl — x' 2 = 7(1 — x), whence JI±* = *, (1) '1 — x or J- ± ^ = <», (2) 1 — x and hence, by differentiation, 2dx (\- x y = 2tdt. (3) Add 1 to both members of (2) and combine the two terms of the left member. The result is (4) o 1 -X = t 2 +l. 1) ividing (3) by (4), we have rfx _2tdt Now multiply (1) and (5) together and integrate. We obtain f /l + x dx r 2 t 2 dt J^l -x ' 1 -x~ J t' 2 + 1 = 2 jrr^_ 2tan -iJr±Z '1 — a: '1 — a: -J EXERCISES (1 -x)(l -Vl -x' 2 ) dx r dx J V2 x 2 - 3 x + 1 [V2x a -3x-+ 1 + V2(a; - 1)] INTEGRATION BY RATIONALIZATION 253 We can rationalize also by means of a trigonometric substi- tution. First reduce ax 2 -\-bx + c to the form ±t 2 ±k 2 , as in Art. 118, and then make one of the following transformations: In k 2 — t 2 put t = k sin 0, in t 2 — k 2 put t = k sec 0, in t 2 + k 2 put t = k tan 0. Since V — t 2 — 1& is imaginary, we shall exclude this case from consideration. The resulting trigonometric functions can then be integrated by methods to be explained in the next chapter. 129. There is one case in which a different transformation leads more rapidly to the desired result. If, after reducing the terms under the radical sign to one of the simple forms mentioned in the preceding paragraph, the integrand can be expressed as the product of t dt and a function containing only even powers of t, then we may substitute y/ ± t 2 ±k 2 = Z. For this gives t 2 = ± (z 2 ± k 2 ) and tdt = ±zdz, and hence the integral takes a rational form in z. EXERCISES ON CHAPTER IV 1 f (— ^ s + 4rx)dx [Notice that Art. 129 is applicable.] 2. f t(x-a)%-\-\dx 3 C y/J+ldx Jo(x-a)$-(x-a)i' V7+1 + 2 254 INTEGRAL CALCULUS 4. f dx 8. f ,lx J x + Vx-1 J * + Vx 2 - 1 5- j-^4- 9. fi+^tf* *^ (« + *)* J 1 + %x 6. f(2- 3 *-*)<**. 10 . f_ x 3 dx x — 3 x 6 + 5 x* v a' 2 — x 2 7 f dx -,, ( y/a*-x*dx [Use trigonometric substitutions in the following exercises.] </x 12 13. 14. *" x 2 "" (x 2 + a 2 )s f ^ . 17. f rf * . J (x' 2 +a 2 )^ J (a?-x*)% CHAPTER V INTEGRATION OF TRIGONOMETRIC FUNCTIONS 130. In regard to the integration of trigonometric functions, it is to be remarked in the first place that every rational trigo- nometric function can be rationally expressed in terms of sine and cosine. It is accordingly evident that such functions can be inte- grated by means of the substitution sin x — z. After the substitution has been effected, the integrand may involve the irrationality Vl — 2T [ = COS X ]. This can be removed by rationalization, as explained in the preceding chapter, or the metnod of reduction may be employed. The substitution cos x — z will serve equally well. It is usually easier, however, to integrate the trigonometric forms without any such previous transformation to algebraic functions. The following articles treat of the cases of most frequent occurrence. 131 . Csec 2n x dx, f csc 2w x dx. In this case n is supposed to be a positive integer. If sec 2n # dx is written in the form sec 2n ~-.r • setfxdx = (1 + tan 2 x) n ~ l d (tan a?), 255 256 . INTEGRAL CALCULUS the first integral becomes f (tan 2 cc + l) n - 1 d(tan x). If (tan 2 a? + 1)" -1 is expanded by the binomial formula and integrated term by term, the required result is readily obtained. In like manner, I csc 2n x dx = f csc 2 "~ 2 # «csc 2 a; dx = — f(cot 2 ^ + l) n ~V/(cot x). This last form can be integrated, as in the preceding case, by expanding the binomial in the integrand. The same method will evidently apply to integrals of the form I tan m .T sec 2n # dx, I cot TO # csc 2h x dx, in which m is any number. EXERCISES 1 f ( Jx 5 f (1 -cosxydx J cos 4 z * sin 4 .*: r 6 C tjx 2. J CSC 4 X dx. ' J sin 4 x . CQS 4 X ( CQS 4 X _ S j u V)4 3. \sec e xdx. 7. i - [ = 1 tan~ 8 .r sec 4 xdx'] . J J sin 3 x cos x J 4 C ^ x q r cos 2 ar dx J ain^x cos 8 * J siii°x INTEGRATION OF TRIGONOMETRIC FUNCTIONS 257 132. fsec w x tan 2n+1 sc dx, fcsc m ^ cot 2n+1 sc da. In these integrands n is a positive integer, or zero, so that 2)i + l is any positive odd integer, while m is unrestricted. The first integral may be written in the form I sec m-1 aj tan 2n a; • sec x tan x dx = I sec m_1 a;(sec 2 # — l)'V?(sec x), which can be integrated after expanding (sec 2 .c — l) n by the binomial formula. Similarly, I csc TO # cot 2n+1 # dx = I csc m_1 # cot 2n # • esc x cot x dx = — I csc m_1 x(csc 2 a; — l) n c?(csc x). EXERCISES 1. \ sec 2 * tan 3 x dx. 5. I tan 5 * dx. 2. f csc 8 * cot 5 * dx. 6. f sm3 - r dx [ = f sec*" 8 * tan 8 * tfx] . •J J cos"* J f sec a* , r Jcot 5 a* 7/ J tan a: dx. 4. l : in x cot 3 * (/*. 8 i cot x dx. EL. CALC. 17 258 INTEGRAL CALCULUS 133. CtsiVL n acdx, Ccot n ocdx. The first integral can be treated thus : I tsin n x dx = M tan n ~ 2 • tan 2 # dx = f tan n_2 ^(sec 2 ic — l)dx ^""^ 'Un-'xdx. When n is a positive integer, the work of integration may- be rapidly carried out by writing t for brevity in place of tan x and then putting t n dx in a different form by means of the following process. First, divide t n by £ 2 + 1; the quotient is a polynomial of the form t n ~ 2 — t n ~~ A + t n ~ G — •••, while the re- mainder R is either ± 1 or ± t according as n is even or odd. Then, since the dividend equals the product of divisor and quotient plus the remainder, we have «•=(«•-" - r- 4 + r~ 6 )(t 2 + 1) + R. But since (tan 2 # + l)dx = sec 2 x dx = d(tan x) = dt, we have Ctsm n x dx = C(t n ~ 2 - r~* + * n ~ 6 )^ + f ^ da; - For example, f tan 8 # dx = C{f -t* + t 2 -- l)dt + Cdx, and J tan 7 x dx = \ (f — f + t) dt — (tan x dx. INTEGRATION OF TRIGONOMETRIC FUNCTIONS 259 The integral J cot n # dx can be treated in a similar manner, in case n is a positive integer. For any value of n we have j cot n # dx = I cot n-2 # aot 2 xdx = I cot n_2 #(csc 2 # — l)dx = _ COt- 1 .T _ C cot n-2 xdXt n-1 J Since tan x and cot x are reciprocals of each other, the above method is sufficient to integrate any integral power of tan x or cot x. Another method of procedure would be to make the substi- tution tan x = z, whence j tan n xdx= j — - (h If the exponent n is a fraction, say n = -, the last integral can be rationalized by the substitution z = u q . It is evident from this that any rational power of tangent or cotangent can be integrated. EXERCISES 1. j cot*xdx. 3. ( (tan x - cotx) 3 dx. 2. \ tau s axdx. 4. ( (tan n x + ta,n n - 2 x)dx. 5. j tan 8 x dx. 260 INTEGRAL CALCULUS When n is a positive integer show that ^ d. 9n i tan 2n_1 x tan 2 " -3 a; , , , n „ 1A x 6. 1 tan 2 " x dx = h ••• 4- (— l) n-1 (tan x — x). J 2n- 1 2n-3 V J K ' r> Ct. 9„a.i j tan 2n x tan 2 " -2 a: 7. \ tan 2 " +1 xdx = + ••• + (— l) n_1 (J tan 2 a; + log cos a;). 134. fsiu m x cos** x dx. (a) Either m or n a positive odd integer. If one of the exponents, for example m, is a positive odd integer, the given integral may be written j sin" 1-1 x cos n x sin xdx = — I (1 — cos 2 x) 2 cos n x d (cos x). Since m is odd, m — 1 is even, and therefore — - — is a Z positive integer. Hence the binomial can be expanded into a finite number of terms, and thus the integration can be easily completed. E x. ( sin 5 xVcos xdx. According to the method just indicated this integral can be re- duced to — ( sin 4 xVcos x d(cos x) = — \ (1 — cos 2 a;) 2 (cos a;) 2 c?(cos x) = — | cos 2 X + \ COS * x — T 2 T cos *~ X. EXERCISES 1. (sin* xdx. 3. (- 2. j*sin3a:cos 4 a:rfx. 4> J cos 2 x j/^ _ C sin 8 x dx J VI - cos a; 'cos 5 X sin x dx. ' sin f ' x dx INTEGRATION OF TRIGONOMETRIC FUNCTIONS 261 (b) m + n an even negative integer. In this case the integral may be put in the form J^H^. cos TO+n x dx = f tan m a; sec- (m+n) x dx, COS m X J which can be integrated by Art. 131, since the exponent — (in -+- n) of sec x is an even positive integer. fcx. \ dx. COS^ X The integration is effected in the following steps : C y/s\\\xdx Ct- h 4 / J — ^= = \ tan 2 x sec 4 x dx Vcos x cos 4 x J = \ tan^ x(tan 2 x + 1) d (tan x) = 2 tan^2(! + \ tan 2 a;). EXERCISES dx 1. (^dx 4. f_ J sin 4 ar ./ sm 4 x cos- 6 x 2. r^_. 5. f J sin 6 a: * Vsin 3 x cos 5 x 3. f ^*. 6. p""^**,. J sin 8 x J cos n+2 x (c) Multiple angles. When m and ?i are both even positive integers, integration may be effected by the use of multiple angles. The trigono- metric formulas used for this purpose are 1 — cos 2 x sin- x = cos-a; sin x cos x 2 1 4- cos 2 x 2 sin 2 # 262 INTEGRAL CALCULUS Ex. ( sin 2 x cos 4 x dx. \ sin 2 x cos 4 x dx = I (sin a: cos x) 2 cos 2 x dx sin 2 2x1 + cos 2 x -J = - f sin 2 2 x rfx + — f sin 2 2 x cos 2 x rf(2 x) 111 - =§J- - dx+ 1 f 1 — cos 4 x , ,1 sin 3 2 x dx -\ 2 16 a = j & x - e¥ sin 4 x + 3 X 8 - sin 8 2 x. EXERCISES 1. ( cos 2 x sin 2 x </x. 3. j sin 4 x cos 4 x tfx. 2. ( sin 2 x cos 6 x g?x. 4. I (sin 4 x — cos 4 x) 4 </x. 5 C^lldx= f C 1 -cos 2 x) 2 ^ = f(sec 2 x-2 + cos 2 xyx. J cos 2 X J COS 2 x J (d) Reduction formulas. Integrate J sin m # cos n x dx by parts, taking u = cos* _1 a*, cfa = sin m # cos x dx, sin m+1 a* whence du = — (n — 1) cos n -a; sin x dx, v = , m+- 1 and therefore frill** cos'a! dx = sin " + ' a;cos "'" 1 - c + '^4 f sin"* 2 cos"" 2 * dft c/ m + 1 7/1 + 1./ In the last term replace sin 2 a- by 1 — cos 2 # and separate the integral into the two terms I sin m # cos n_2 x dx — I sin m a; cos n # dx. INTEGRATION OF TRIGONOMETRIC FUNCTIONS 263 Transpose the second integral and unite with the similar integral in the left member. After dividing the resulting equation by we obtain the formula of reduction m + 1 /„ j sin m+1 # cos M_1 x . n—1 C - m n-i i sm m # cos w # ax = 1 I sm m # cos n z x dx ra + ?i m + n J by means of which the exponent of the cosine factor may be diminished or increased by 2 according as the integral in the left member or that in the right member is taken as the given integral. In like manner a reduction formula may be deduced which decreases or increases the exponent of the sine factor by 2. The details are left to the student as an exercise. The result is /- m n i sin w_1 x cos" +1 x . m—1 C - m-2 n ^ sm m x cos n x ax = 1 I sm m l x cos n # ax. m 4- n m 4- nJ The two preceding formulas, when solved for the integrals in the right members, and m (or n) increased by 2, become /• m n i sin m+1 # cos n+1 . / c . m-\-n + 2 C . m n+2 j sm m aJcos n a;aa;= ! ! — I sm m xcos n +'xax, n + 1 n + l J Csin m x cos n xdx = smW+la? cos n+1 a; + m + n + 2 C sin m + 2 X G0S n xdx , J m + 1 m-j-1 J Whenever the values of m and n are such that one of the three preceding cases, (a), (6), (c), is applicable, the integration can generally be performed more quickly by one of those methods. 264 INTEGRAL CALCULUS EXERCISES 1. (sin*xdx. 2. (99&*dx. J J sin 2 x [In Ex. 2 after one reduction, diminishing the exponent of cos x by 2, Art. 133 may be applied.] 3 r sin 4 :r , . f dx _ C co^xdx J cos x J sin 3 2 a; J sin 4 a; 135. / eta r doc r a+bnosna J a + bsmnw J a + bcosncc' J a + bsinnw' J a + b sin noc+c cos nx These forms can be integrated by expressing them in terms of the half angle and then in terms of tan - -. Ex.l. J dx 5 + 4 cos x By making use of the trigonometric relations cos 2 - + sin 2 - = l, 2 2 cos x = cos 2 - — sin 2 - , 2 2 the denominator may be written in the form 5^cos 2 |+sin 2 |)+ 4(cos 2 |-sin 2 -V which becomes sin 2 - + 9 cos 2 - on collecting the terms ; whence 2 2 dx 1 sin 2 ! +9 ,,x cos 2 - 2 INTEGRATION OF TRIGONOMETRIC FUNCTIONS 265 Now divide numerator and denominator by cos 2 - and bear in mind 1 x that = sec 2 -. This gives „ t 2 cos 2 - i'(S) J - = ? tan- 1 (i tan- tan 2 - +9 3 Vd 2 Ex.2, J- t/x sin 3 a: + 1 Express the denominator in the form 4 sin — cos — + ( sin 2 — - + cos 2 — - J . 3 a: Then, after dividing both terms of the fraction by cos 2 — , the given integral becomes r^ 3 x sec 2 — dx 2 J tan 2 ^+4 tan— + 1 Now make the substitution tan— = t and apply Art. 118. It will be observed from these two problems that the aim is to put the denominator in the form of a homogeneous quadratic expression in sine and cosine functions. Then, when both terms of the fraction are divided by the square of the cosine, the denominator becomes quadratic in the tangent function while the numerator can be expressed as the differential of the tangent. EXERCISES 1. ( ^ . 5. f ^ . J 5 + 3 cos 2 x J (a sin x + b cos x) 2 2. (—«* . 6. f <" J 5 — 3 sin x J a 2 sin 2 x + b 2 cos 2 x dx „ C dx t. ( — dx . 7. f— J 1 - 2 sin 2 x J 1 + t. f * 8. f— J a sin x -f b cos x J 1 + cos 2 a: ax sin x + 2 cos a: 266 INTEGRAL CALCULUS 136. I e ax sin nx dx, I e ax cos nx dx. Integrate | e ax sin nx da; by parts, assuming u = sin nx, and cfa = e ax dx. This gives ^. * /e oz sin nx c?x = - e ax sin nx | e ox cos nx dx. (1) Integrate the same expression again, assuming this time u = e ax , dv = sin nx dx. Then /e ax sin nxdx = e ax cos nx + - ( e ax cos nx dx. (2) n nJ Multiply (1) by - and (2) by - and add. The integrals in the right members are eliminated, and the result is J„ . 7 e ax (a sin nx — n cos nx) e ax sin nx dx = — * ■ L • a 2 + n 2 By subtracting (1) from (2), the formula /, e ax (n sin nx 4- a cos nx) e ax cos nx dx = — * — L a~ + iv is obtained. EXERCISES ON CHAPTER V 1. Derive the reduction formula C, ec n x dx = tanssec*-** + n-2 r ecn _ 2jc d ^ J n — 1 n — \J [Integrate by parts, taking u = sec M ~ 2 :r, dv - sec 2 a: dx.~] 2. Derive f csc»x & = - cot * csc ;~ 2 * + ^? f ce-'* <** J W — 1 n — JV J sin x cos a; J cos 5 x INTEGRATION OF TRIGONOMETRIC FUNCTIONS 267 5. f **, 9. ( S ™^dx. J cos x sin 2 x J e x 10. \ e 2x sin 2 x dx. 6 - y^-^dx. 10 . f J cos 3 x J 7- I • 11. i <? x sin 2 a: sin a: efa. J (1 -x)Vl - a.- 2 J [Put a; = cos $] . [Hint. 2 sin 2 a: sin a: = cos x — cos 3 a\] f - a: 8. i e 2 cos~dx. 12. Show that f sin ax sin fta: tfa; = sin ( a " & ^ - sin + *>*. J 2(a - ft) 2(a + ft) Use the trigonometric formula sin a sin ft = i [cos(cc — j3)— cos(ct + /?)]. 13. Show that r • 7 7 cos (a — ft)ar cosfa + b)x \ sin aa: cos bx dx = * *- * — ^ ' • J 2(a - ft) 2 (a + ft) 14. Show that f cos a* cos bxdx = sin (g- ft) a; sin(q + ft)* J 2(a-ft) 2(a + ft) 15. i sin n x cos 3 a; da:. 19. I— — ■ - :~~; J J sin x cos 3 x — sin 3 x cos a: 16. j"^L_. 20. f^^-. ^ '.A 4? ^ sin 4 .r oos 4 sin^a: cos-* a: sin* a- cos* x 17. j*(tan a: + cot a-) 6 dx. 21. j* 4 ^ <fo. 18. ( «« f a sin a: + ft co s a: ■J (1 + cos a:) 3 zz ' J a Bin x + o cos ~ dx - ft COS X [Hint. Assume a sin x + 6 cos a: = .4 (a sin x + ft cos x) -f- i?(« cos x — ft sin a:) and determine J. and Z? by equating like terms. Treat Ex. 23 in like manner.] 23. ( aeX + he ~ X dx. 24. f sin (* + q > <fa. ^ «e* + fte~ x J sin (a; + ft) CHAPTER VI INTEGRATION AS A SUMMATION. AREAS 137. Areas. The problem of calculating the area bounded by given straight or curved lines can be solved by means of the Integral Calculus provided that the equations of the boun- dary curves are known and satisfy certain restrictions. Suppose it is required to determine the area limited by a continuous arc of a curve whose equation, in rectangular coor- dinates, is written in the form y =/(»), (i) by the two ordinates x — a and x = b, and by the it*-axis; that is, the area APQB (Fig. 59). We proceed as follows. It is as- sumed in the first place, for the sake of simplicity, that/(#) is always increasing (or always decreasing) between x = a and x = b, so that a vari- able point on the arc PQ is continually rising (or falling) as its abscissa x increases. Suppose, further, that every ordinate between x = a and x = b cuts the arc PQ in but one point. Let 2G8 Fig. 59 INTEGRATION AS A SUMMATION. AREAS 269 the interval A to B (Fig. 59) be divided into n equal intervals AA V A l A 2 '", A n _ 1 B, each of length Ax, so that interval AB= b — a = n • Ax. At each of the points of division A, A lf A 2 , • ••, B erect ordi- nate^ and suppose that these meet the curve in the points P, P x , P 2 , • ••, Q. Through the latter points draw lines PPi, P 1 R 2 ) PzR& ' ' ' Pn-i Rn parallel to the x-axis. A series of rectangles PA X , P X A>, • • • is thus formed, each of which lies entirely within the given area. These will be re- ferred to as the interior rectangles. By producing the lines already drawn, a series of rectangles SA 1} SiA 2J ••• is formed which will be called the exterior rectangles. It is clear that the given area will always be greater than the sum of the in- terior rectangles and always less than the sum of. the exterior, or, expressed in a formula, "^ PA X + iVl 2 + ••• +P n - 1 B< Area APQB < SA 1 + S 1 A i + ••• ■+S.-1-B. (2) The difference between the sum of the exterior and the sum of the interior rectangles is SP 1 + ^fi 2 +- + £„_! R n = rectangle S n _ x T = TQ • Ax. (3) As we suppose the curve to be continuous between P and Q, the line TQ is of finite length. If the number n of equal parts into which AB is divided is increased, the first sum in (2) increases in value and the second sum in (2) decreases. Moreover, as their difference TQ • Ax, given in (3), approaches the limit zero, it follows that the limit of the sum of the exterior rectangles is equal to the limit of the sum of the interior rectangles when n = cc, that is, when Ax = 0. 270 INTEGRAL CALCULUS Since the required area always has a value intermediate between the two sums, it follows that the area is equal to the limit of either sum. So that, for example, we have ^^ = A^O t PA > + P * A * + - + P n~l B l ( 4 ) The second member of this equation may be expressed in terms of the function f(x) which appears in the equation (1) of the given curve. For, area PA± = AP • Ax = f(a)Ax, since AP is the ordinate y when x = a. Similarly, area PiA 2 = A X P X • Ax = f(a + Ax) • Ax, area P 2 A 3 = A 2 P 2 • Ax = f(a + 2 Ax) • Ax, area P n - Y B = A n _iP n _i • Ax =f(a + n — 1 Ax) • Ax. If these expressions are substituted in (4), it takes the form area = A^o[/( a )+/( a + A ' T ) + /( a + 2Aa; )+ - + f(a + n-lAx)-]Ax. (5) As it now stands, the formula just derived is of little prac- tical value for computing areas. This is due to the fact that there is no general method for calculating the sum of the n terms given in brackets in the second member of (5). Fortunately, the value of the limit of this sum when n = cc and Ax = can be calculated by integration as we shall now proceed to show. 138. Expression of area as a definite integral. Denote the function arising from the integration of f(x) by F(x), that is, INTEGRATION AS A SUMMATION. AREAS 271 let F(x) = ff(x) dx, By definition of the derivative of F(x) we have lim F(x + Ax)-F(x) _ f( . A The quotient F ( x + Ax ) F ( x ) may be written in the form f(x) -J- <f>, in which <f> approaches zero at the same time as Ax, otherwise the limit of the quotient when Ax = could not be f(x). From this relation follows, on multiplying by Ax, F (x + Ax) - F (x) =f(x) • Ax + <£ • Ax. (6) Next, in equation (6) substitute for x the successive values a, a + Ax, a + 2 Ax, • • •, a + (n — 1) Ax. We thus deduce the following series of n equations, in which <£i> 02? ••* are used to denote the different values which <£ may take: F(a + Ax)-F(a)=f(a) -Ax+fa-Ax, F(a + 2 Ax) — F(a + Ax)=f(a + As) • Ax + <£ 2 • Ax, F(a + 3 Aaj)-2P(a + 2 Ax)=f(a + 2 Ax) • Ax -J- <£ 3 ■ Ax, F(a + n — l • Ax)-F(a + n-2 • Ax)=f(a + n- 2 ■ Ax)Ax + <£ n -i • Ax, F(a + nAx)—F(a + n — 1 • Ax) = /(a -fn- 1 • Ax)Ax 4- <£* • Ax. Let these n equations be added ; then all but two of the terms in the left member of the sum cancel each other and the 272 INTEGRAL CALCULUS result may be written F(b)-F(a) = [f(a)+f(a+Ax) + ... +/( a + n-l • Ax)]&x + W>i + 02 + -+0 B ]Aa>, in which b is written for a-\-n A#, since ?i A# = 6 — a. Now let Aa; approach zero. The expression (<fc + <fe-h ... +0 B )Aa> vanishes at the limit. For, let <£ denote the positive value of the numerically largest term of the set <f> u <f> 2 , • ••, cf> n ; then we have evidently |(0i + <f>2 H +0») Aa;|^ (^-|-<I> ...(w terms)) A# = n<£ • Aa; = wAa; • <£ = (6 — a) • 3>. Hence, from the fact that ^ m 3> = and that b — a is finite, it Ax=0 ' follows from Art. 3 that Hm(<k + 4> 2 + .••0 n )Aa J = O; and therefore 2^(6) - F(a) = J^J[/(a) +/(a + Aa>) + ... +f(a+n - 1 • Aa;)]Aa;. (7) Now the right member of this equation is exactly the ex- pression previously derived for the area APQB; hence, area APQB = F(b) - F(a). (8) To compute the value of the right member of (8), first obtain F(x) by integrating f(x) dx. Having determined F(x), substi- tute the values b and a which x takes at the extremities of the arc bounding the given area and then subtract the second from the first. This result may conveniently be represented by the symbol ff( x ) dx > INTEGRATION AS A SUMMATION. AREAS 273 which indicates both the integration to be performed and the substitution of the two limiting values a and b for x. It is called the definite integral of the function f(x) between the limits a and b. We thus obtain, as a final formula for area, area APQB = f / (») 'dx. (9) 139. Generalization of the area formula. Instead of taking the limit of the sum of the interior (or exterior) rectangles, a more Y Fig. 60 general procedure would be to take a series of intermediate rectangles. Let x 1 be any value of x between a and a + Ax, x 2 any value between a -{-Ax and a -+- 2 Ax, etc. Then / (x r ) Ax would be the area of a rectangle KLA X A (Fig. 60) intermediate between PA X and SA X ; that is, PA.Kf^AxKSA,. Likewise P X A 2 <f(x 2 )Ax < S^A 2 , etc. EL. CAI.C 18 274 INTEGRAL CALCULUS Hence, sum of interior rectangles < [/(a?i) +/(^) + •••]^ aJ < sum of exterior rectangles, and therefore (cf . Fig. 59), area APQB = ^o [/(&) + f(x 2 ) + ... + /«)] Ax. (10) This result combined with (9) gives for the definite integral the more general formula : fj(x)dx=£l [f{x 1 )+f(x 2 )+ -f(x„)-]Ax. (11) 140. Certain properties of definite integrals. From the defini- tion of the definite integral | f(x)dx as the limit of a par- ticular sum, certain important properties may be deduced. (a) Interchanging the limits a and b merely changes the sign of the definite integral. For, if x starts at the upper limit b and diminishes by the addition of successive negative increments (— Ax), a change of sign will occur in formula (7), giving F(a)-F{b)=£f{x)dx. Hence, f a f(x) dx = - Cf{x) dx. (b) If c is a number between a and b, then f b f(x) dx = Cf(x) dx + Cf(x) dx. INTEGRATION AS A SUMMATION. AREAS 275 (c) Tlie Mean Value Theorem. The area APQB (Fig. 61), which represents the numerical value of the definite integral may be expressed as follows. Let an ordinate MN be drawn in such a position that area PSN = area NRQ. If £ denotes the value of x cor- responding to the point N, then MN = f($), and area APQB = rectangle ASRB = MN-AB=f{Z){b-a). Hence, Cft*)** = f(gft> -a), (12) %J a in which £ is some value of x between a and b. This result is known as the Mean Value Theorem (compare Art. 39), dx is called the mean ordinate f/(*)< and the ordinate /(£) =^ a - b — a between x = a and x = b. This is also called the mean value of the function /(a?) between these limits. The theorem may be expressed in words as follows : The value of the definite integral £f{x)dx is equal to the product of the difference between the limits by the value of the function f(x) corresponding to a certain value x = f- between the limits of integration. (d) It is frequently desirable to make a change of variable in the definite integral in order to facilitate the work of inte- gration. It is obvious, from the nature of the definite integral, 276 INTEGRAL CALCULUS that the limits of integration must be changed so that in the new integral the limits shall be the values of the new variable corresponding to those of the old variable. Ex. Evaluate \ v a 2 — x 2 dx. j>~ 2 Make the change of variable x — a sin 0, whence dx = a cos dO, and therefore n f ° y/d'-x* dx = a 2 ( * cos 2 d6. Here the limits for the new integral are determined by inspection of the equation connecting x and 0, namely, sin 6 = -. It is seen that, a as x varies from to a, sin 6 varies from to 1. This corresponds to a variation of 6 between the limits and -. The indefinite integral is, by Art. 134 (c), The substitution of the limits gives the value — • 4 141. Maclaurin's formula. As an application of the mean value theorem (Art. 140 (c)), we derive Maclaurin's formula with the remainder term. Let s and t be independent variables. Suppose J[s — 1\ to- gether with its first n derivatives with respect to t, to be continuous within the interval to t v Then we have by inte- gration £f\s-t)dt = -f{s-t) =f(s)-f(s-t l ). On the other hand if we integrate by parts, taking u =f'( s — 0> dv = dt, we obtain £ l f'(s - t)dt =f'(s - t) • tT + jTV"(« -t)-tdt */o INTEGRATION AS A SUMMATION. AREAS 277 Integrate the last term by parts, taking u =/"{s — t), do = t dt. By successive applications of this process we deduce the formula /(*)_/(* - tl )=f'(s - (,)«. +/'(•-■ ttfl +f'"(s - trfl + - + ~^-r-. Cf\s-t)t^dt. (n — 1) \Jo (n-l) By the mean value theorem we have f V°(* - t n ~ l dt =f n) (s - et^idhf- 1 • t x , Jo in which 6 is a positive fraction and Qt x is the same as £ of (12). Inserting this in the preceding equation and substituting s = x, t-L — x — a (hence s — t x = a) we obtain as a final form f(x)=f(a) +f' { a)(x-a) +-^(x - af + ... + (iS)l /< " )(a; _ 6{X - a ^ X - °>"- If we replace by 1 — 0', the remainder term takes the form given on p. 153, with 6' written in the place of 0. 142. Remarks on the area formula, (a) It is noticed that the formula Cf(x)dx= *™ [/(a) + /(a + Ax) + - +/(o + ^=1 . Az)]A* indicates two steps, — a summation, and a process of passing to a limit. The differential /(x) dx which appears under the integral sign may be regarded as representing the general term/(x) Ax of the series to be summed, while the process of taking the limit of this sum is indicated by replacing Ax with the differential dx and prefixing the sign of integration. 278 INTEGRAL CALCULUS The general term f(x) Ax represents the area of an arbitrary rectangle (of the set of interior rectangles) whose altitude is the ordinate corresponding to an arbitrary x and whose width is Ax. This is called an element of area. The definite integral may then be thought of as indicating the limit of the sum of all contiguous elements of area between x — a and x = b. This notion of summation (followed by passing to the limit Ax = 0) is a very useful one in applying the calculus to prob- lems of geometry, mechanics, and physics. In each case an application of this notion consists in finding the general ex- pression for an element of the given magnitude (element of area, element of mass, element of moment of inertia, etc.) and then indicating the two steps of summation and taking the limit by changing Ax to dx and prefixing the symbol * of the definite integral. It must not be forgotten that in every case it is necessary to prove that the limit of the sum gives pre- cisely the desired result, f This we have already done in case of the area formula. (6) The element of area/(x) • Ax is positive when the cor- responding rectangle is above the x-axis, since in that case f(x) is positive, while Ax is positive if b>a. Accordingly, the formula \ f(x) dx gives a positive value for an area above the x-axis provided we take b>a. Similar considerations show that the same formula gives a negative value for an area below the x-axis. (c) If the curve y=f(x) crosses the x-axis between the two points A, B, then the area consists of a positive part APC, * This symbol originated historically from the initial of the word sum. f In some cases the limit of the sum is used as a definition of the magnitude in question, as, for example, in the definition of the length of arc. (Art. 151.) INTEGRATION AS A SUMMATION. AREAS 279 B Fig. (52 Q represented by the integral J f(x) dx, and a negative part CBQ represented by the integral I f(x) dx. The sum of these two integrals, which (by Art. 140 b) is Y equal to I f(x) dx, would accord- ingly give the algebraic sum of the O positive and the negative area, ((f) Some of the restrictions placed upon the function f(x) in Art. 137 can be removed. In the first place, suppose that f(x) is not always increasing (or decreasing) as x increases from a to b. Let ordinates be drawn at the maximum and minimum points of the given arc PQ (Fig. 63). These divide the required area into several parts A', A", A'" for each of which the ordinates satisfy the original condi- tion of Art. 137, hence we conclude that area = A' + A" + A"' =Cf{x) dx -f C*f(x) dx + Cf(x) dx = Cf(x) dx, by Art. 140 (6). A discussion of the methods to be employed in case f(x) be- comes discontinuous, or is not singly valued in the assigned interval, is postponed to Art. 143. (e) Since f(x)=y, formula (9) may be written more briefly area APQB= Cydx. (13) 280 INTEGRAL CALCULUS X B y=b r / / q {' V=a' L p X Fig. 04 (/) By exactly the same process used in deriving (9), or (13), it may be shown that the Me&A'PQB' (Fig. 64) bounded by the curve PQ, the ?/-axis, and the two lines y = a', y = b' is given by the formula area A'PQB' = | x dy, (g) If it is required to find the area bounded by several arcs such as PQ, QR, ES, etc. (Fig. 65), we may calculate by formula (9) the simple areas APQB, BQRC, etc., and by proper additions and sub- tractions obtain the desired area. Thus the area in Fig. 65 would be expressed by Xb /*c r*c S*d f Y {x) dx + J & f 2 (x) dx - | f 3 (x) dx -J f 4 (x) dx. 1. Find the area bounded by the curve y = log x, the ar-axis, and the ordinates — o <r — * The symbol la - Area APQB (Fig. 66) = /*8 -i8* I \ogxdx = x(logx — 1) = 3(log3-l)-2(log2-l) ( Fig. 66 =log-^-l. indicates that the values 3 and 2 are to be substituted for x in the expression which precedes the symbol and the second result sub- tracted from the first. INTEGRATION AS A SUMMATION. AREAS 281 2. Find the area bounded by the arc of the parabola y 2 = 4px measured from the vertex to the point whose abscissa is a, the x-axis and the ordinate x = a. From the result show that the area of the parabola cut off by a line perpendicular to the axis of the curve is two thirds the area of the rectangle circumscribing this segment. Does this result hold good for all parabolas? 3. Find the area between the x-axis and one semi-undulation of the curve y = sin x. 4. Find the area bounded by the semicubical parabola y 2 = 25 x 3 and the line x = 3. 5. Find the area bounded by the curve y 2 = 4(x-f 5) 3 and the^-axis. 6. Find the area bounded by the cubical parabola y = x 3 , the y-axis, and the line y = 1. 7. Find the area bounded by the curve x + y 3 = 2 and the coordi- nate axes. 8. Find the area bounded by the parabola y = 2 x 2 and the line y = 2x. 9. Find the area bounded by the parabola y = x 2 and the two lines y — x and y =2 x. 10. Find by integration the area of the circle x 2 + y 2 = r 2 . 11. Find the area between the curve y = x(x — l)(x — 3) and the x-axis. 12. Find the area bounded by the coordinate axes, the witch 8a 3 . y = — — - — -, and the ordinate x = x v By increasing x\ without limit, x 2 -f 4 a 2 ftnd the area between the curve and the x-axis. 13. Find the area of the ellipse — + y — = 1. a 2 b 2 (* 14. Find the area included between the hyperbola xy = 36 and the - line x -f y = 15. 282 INTEGRAL CALCULUS 15. Find the area bounded by the logarithmic curve y = a x , the x-axis, and the two ordinates x = x v x = x 2 . Show that the result is proportional to the difference between the ordinates. 16. Find the area between the curve y = (x 2 — l)(x 2 — 2) and the x-axis. 17. Find the area cut off from the parabola (x — l) 2 = y — 1 by the line y = x. 18. Find the area of the oval in the curve y 2 =(x — a)(x — b) 2 , given a < b. 19. Prove that the area of the curve a 2 y 2 = x 3 (2 a — x) is equal to that of a circle of radius a. Draw figures of the two curves (center of the circle at the point (a, 0)) and compare. 20. Find the area of the loop of the curve y 2 = x 4 4- x B . 21. Given the curve of damped vibrations y = e~ x sin x. Show that the areas contained between successive semi-undulations of the curve, and the positive x-axis form a geometrical series of alternately positive and negative terms. Find the sum of this infinite series and verify that the same result may be obtained by integrating between the limits and go. Find the total area included between the positive x-axis and the curve (changing the negative areas to positive). 22. Find the area bounded by the hyperbola xy = a 2 , the x-axis, and the two ordinates x = a, x — na. From the result obtained, prove that the area contained between an infinite branch of the curve and its asymptote is infinite. 23. Find the area contained between the curves y s = x and x 8 = y. 24. Take the segment of the equilateral hyperbola xy = k 2 , be- tween two points P and Q. Show that the area between this arc and the x-axis is the same as that between the same arc and the y-nxis. 25. Find the area bounded by the parabola Vx + Vy = Va and the coordinate axes. INTEGRATION AS A SUMMATION. AREAS 283 26. Find the area between the curve y 2 (y 2 — 2) = x — 1 and the coordinate axes. 27. Find the area common to the two ellipses a 2 b* ' b* a 2 28. Find the area enclosed by the curves y = sin x, y — cos x be- tween two consecutive intersections. 29. Find the mean ordinate of the curve y = tan x between the limits x = and x = - (see p. 275). 4 30. Find the mean value of the function sin x between the limits and -; also of the function e^sinar. 31. Find the area of the loop of the curve o a — x V 2 = x 2 . * a + x 143. Precautions to be observed in evaluating definite integrals. The method given above for determining plane areas in rec- tangular coordinates involves two essential steps : (1) To find the integral of the given function f(x) ; (2) To substitute for x the two limiting values a and b, and subtract the first result from the second. Erroneous conclusions may be reached, however, by an in- cautious application of this process. The case requiring par- ticular attention is that in which f(x) becomes infinite for some value of x between a and b, or at a or b. When that happens, a special investigation must be made. The method of procedure will be brought out in the following examples. Ex. 1. Find the area bounded by the curve y(x - l) 2 =c, the coordinate axes, and the ordinate x = 2. 284 INTEGRAL CALCULUS A direct application of the formula gives n cdx ^]: 2 c, ]6 is a sign of substitution, indicating that the values b, a are to be inserted for x in the expression immediately preceding the sign, and the second result subtracted from the first. This result is incorrect. A glance at the equation of the curve shows that f(x)\ = - becomes infinite for x = 1. It is L (x — 1)' 2 J Fig. 67 accordingly necessary to find the area OCPA (Fig. 67) bounded by an ordinate AP corresponding to a value x — x', which is less than 1. For this part of the area/(x) is finite and positive, and formula (9) can be immediately applied, with the result area OCPA = (* cdx , = c - T = c — -c. 0<x'<l. J Q (x _l } 2 O-l)J X'-I If now x' is made to increase and approach 1 as a limit, the value of the expression for the area will increase without limit. A like result is obtained for the area included between the ordi- nates x = 1 and x = 2. Hence the required area is infinite. Ex. 2. Find the area limited by the curve y 8 (x 2 — a 2 ) 2 = Sx 3 , the coordinate axes, and the ordinate x = 3 a. INTEGRATION AS A SUMMATION. AREAS 285 Since /(x) = — becomes infinite for x = a, it is necessary L O 2 - a 2 )t J in the first place to consider the area OPA (Fig. 68) and determine Fig. 68 what limit it approaches as AP approaches coincidence with the ordinate x = a. Accordingly area OPA = £* 2xdx ^ = 3(x 2 - a 2 )$T (x 2 — a 2 )f -* whence = 3(>' 2 -a 2 )* + 3a*, [area OP.4] = 3ai lim x 1 = a In the same manner, the area A'P'QB has the value 3a 2 xdx P 2 l 6 a^ — 3(x 2 — a 2 ) s, £ 0<x'<a a < x' < 3 a. (x 2 - a 2 )* As x' diminishes towards a, the area increases to the limiting value 2 6 aL Hence, by adding the two results, the required area is found tobe 8al + 6a*=9ai The same result is found by a direct application of (9), viz. : J-_2^ = 3(x2 _ a2)r |" = 9nS) •" (x 2 -a 2 )f Jo so that in this case an immediate use of the area formula gives the correct result. 286 INTEGRAL CALCULUS Some of the details in such problems as the two preceding may be omitted. It is unnecessary first to put x — x', a value ] ess than the critical one, and, after integration and substitu- tion of limits, to let x' approach the critical value as a limit. For this is clearly equivalent to taking the critical value at once as the upper limit for the portion of the area to the left of the infinite ordinate (or as the lower limit for the area to the right of this ordinate). Thus, in case of an infinite ordinate, the rule of procedure becomes : Calculate separately, by formula (9), the two portions of area on each side of the infinite ordinate and add the two results. If one of these portions is infinite, it is not necessary to calculate the other ; the required area is infinite. The formula (9) for area has been deduced under the as- sumption that the limits a and b are finite. It may happen, however, that the curve y =f(x) approaches the x-axis as an asymptote. It .might then be required to determine the strip of area extending to infinity between the curve and its asymp- tote. The method of procedure for such a case will be ex- plained in the following example. Ex. 3. Find the area bounded by the curve y(x 2 + 1) = 1 and the ar-axis. INTEGRATION AS A SUMMATION. AREAS 287 This curve being symmetrical with respect to the y-axis, it is sufficient to calculate the area in the first quadrant. As our formula of integration does not take account of the case b = co, we integrate from to x' and in the result cause x' to increase without limit. This limit will be defined to mean the area between the arc in the positive quadrant, its asymptote, and the y-axis. It is evident that these steps in the evaluation amount to a direct application of the area formula, using the limits and co . The half area is, accordingly, V = tan- ] a; = tan -1 co — tan- 1 0. Jo 1 + x 2 Jo We are here confronted with the difficulty that the anti-tangent is a many-valued function and there is a question as to which of its values should be chosen. It is necessary in such a case to go back and examine the limiting process just explained. The area OPQN is equal to tan -1 a;' — tan-*0. If x' approaches zero, this expression should approach zero; and as x' increases continuously the area also increases continuously. Accordingly, whatever value we choose for tan _1 0, the limit of tan- 1 a:' should be the value obtained by a continu- ous increase in this function as x' increases without limit. The sim- plest value for tan- 1 is 0. If tan- x' increases continuously from 0, it reaches the limit — when x' becomes infinite. Hence 2 . lim (tan-V-tan^O) = £• x' = co v 2 If we choose tan _1 = n7r, 7i any integer, then , . tan -1 a;' = nT+ — » J x' = x 2 and the difference gives * , as before. Ex. 4. Find the area bounded by the curve y(x 2 + a 2 ) 2 = x and the positive x-axis. Ex. 5. Find the area bounded by the curve y — tan -1 x, the coordi- nate axes, and the line x — 1. In this problem we have to deal with a many-valued function of x. In fact, to each value of x corresponds an infinite number of values of 288 INTEGRAL CALCULUS tan -1 a:. The problem, accordingly, has an indefiniteness, which must be removed by making some additional assumption. The curve y = tan -1 x consists of an infinite number of branches, corresponding ordinates of which differ by integer multiples of tt. Each branch is continuous for all finite values of x (see Fig. 70). It is evidently necessary to select one of these branches for the boundary of the proposed area, and discard all the others. Suppose, for example, the branch A B is selected. The ordinate to this branch has the value it when x is zero, and increases con- _ 5 _ tinuously to it -\ — = - — as x increases 4 4 continuously to 1. Hence the required area is Y y^" A ~B C X — ^ X -1 Fig. 70 | tan -1 x dx = \ x tan -1 x— \ log (a; 2 -f 1) 5* -1 log 2. 4 2 s EXERCISES 1. Find the area bounded by the curve y 2 (x — 1)= 1, the asymp- tote x = 1, and the line x = 2. 2. Find the area bounded by the curve y 3 (x — l) 4 = 1 and its asymptota, the x-axis. 3. Find the area bounded by the curve of Ex. 2, the x-axis, and the ordinate x = 2. 4. Find the area inclosed by the curve x 2 y 2 = a 2 (y 2 — x' 2 ) and its asymptote. 5. Find the area bounded by the curve a 2 x — y(x — a), the rr-axis, and the asymptote x = a. INTEGRATION AS A SUMMATION. AREAS 289 x 3 6. Find the area between the cissoid y 1 2a and its asymp- tote x = 2 a. 7. Find the area between the curve y 2 (l — x 2 ) = 1 and its asymptotes. 144. Calculation of area when x and y are expressible in terms of a third variable. When the rectangular coordinates of any point of the boundary arc of the required area are given as functions of a third variable 6, we may substitute in I ydx the expressions for y and dx in terms of 6 and integrate be- tween the corresponding new limits for in accordance with Art. 140(d). Area of the cycloid. This curve is traced by a point P in the circumference of a circle of radius r as the circle rolls on a straight line, without sliding. Y O M Fig. 71 Let the point Pbe in contact with the given line at when the circle begins to roll. Suppose that an arbitrary arc PQ has rolled over the segment OQ. Let (x, y) denote the rec- tangular coordinates of P, and let 6 represent, in radian meas- ure, the angle at the center C subtending PQ ; then, OQ = arc PQ = rO. EL. CALC. — 19 290 INTEGRAL CALCULUS Dropping a perpendicular PR on the line CQ, we have PR = r sin 6, RC = r cos 0. Accordingly, x=OM=OQ-MQ = rO-r sin 6 = r (0 - sin 0), y = MP= QC- RC= r - r cos = r(l - cos 0). These are called the two parametric equations of the cycloid, being a varying parameter. One complete arch of the cy- cloid is generated as 6 varies from to 2 7r, that is, as x varies from to 2 ttt. The maximum ordinate for this arc occurs at x — -n-r, and the arc is symmetrical with respect to this ordinate. The area inclosed by the arc OP A and the cc-axis is ydx= \ r(l - cos 0) • r(l - cos 0)d6 = 3 ttt 3 . The area is three times that of the rolling circle. EXERCISES 1. Find the area of the ellipse when x and y are expressed in terms of the eccentric angle, x = a cos <£, y = b sin <f>. What is the meaning of the negative sign in the result? 2. 2 2 2. Find the area of the hypocycloid x 3 + y* = a 3 by expressing x and y in the form x = a cos 3 0, y = a sin 3 6. 3. Find the area of the loop of the folium of Descartes x s + y s — 3 xy = 0. This area may be calculated either by expressing x and y in the 6*+ 1 3 + 1 obtained by putting y = Ox and solving for a: and ?/, or by transform- ing to polar coordinates and using the polar formula for area, Art. 145. INTEGRATION AS A SUMMATION. AREAS 291 4. Find the area within the curve y 2 = (1 — a; 2 ) 8 by assuming x = cos 0, y = sin 3 0. 5. Find the area of (aa:)*+ (by)* = (a 2 - b 2 )*, the evolute of the ellipse. (See Fig. 51, p. 190.) Express x and y in the form, ax = (a 2 - b 2 ) sin 8 0, by = (a 2 - b 2 ) cos 3 0. 145. Areas in polar coordinates. Let PQ be an arc of a curve whose equation is given in polar coordinates (p, 0). It is re- quired to find the area bounded by this curve and the two as- signed radii QP and OQ. Let A and B be any two points of the curve with coordi- nates (p, 0) and (p + Ap, + A<9) respectively. Through ^4 draw an arc AC of a circle with radius p and center 0. The element of area OAC is a sector of a circle of angle A0. The arc AC is, FlG> 72 therefore, p A0 and the sectorial area is | p 2 A0. The limit of the sum of all such elements contained between OP and OQ is !fV« (14) That this is the actual area sought remains to be proved by showing that the sum of the elements of area has the required area for its limit. This may be done by steps exactly analo- gous to those used in Art. 137, which would consist in proving that the sum of all interior sectors, such as OAC, has the same limit as the sum of all exterior sectors, such as ODB. The details are left to the student as an exercise. 292 INTEGRAL CALCULUS EXERCISES 1. Find the area of the three loops of the curve p = a sin 3 0. From the symmetry of the figure it is seen that one sixth of the total area is described as varies from to — . Hence the area is 6 j* Q 6 ± a 2 sin 2 S$ d$= % a 2 J * (1 - cos 6 0)dO r_a? This is one fourth the area of the circumscribing circle. 2. Find the area of the lemniscate p 2 = a 2 cos 2 0. 3. Find the area of the circle p = 2 r cos 0. 4. Find the area of the cardioid p = r(l — cos $). 5. Find the area of the circle p = 10 sin 0. 6. Find the area bounded by the hyperbolic spiral p$ = c . and radii drawn to two arbitrary points (p v #,) and (p 2 , 2 ). Show that the area is proportional to the difference between the radii. 7. Find the area of the four loops of the curve p = a sin 2 0. 8. Find the area of the loop in the spiral of Archimedes p = aO generated between the limits — - and + - for 6. 9. Find the area bounded by the lituus p 2 6 = k and two arbitrary radii, making angles 0\ and 6 2 with the polar axis. 10. Find the area of one loop of the curve p 2 = a 2 cos nO. 11. The radius vector of the logarithmic spiral p = e~ e starts at the angle = and rotates positively about the origin an infinite number of times. Determine the area swept over by the radius vector. 12. Find the area of the curve p 4 = sin 2 cos 0. 13. Find the area within the curve p = cos 2 0. 14. Find the area of the innermost loop of the double spiral p = 2 . 146. Approximate integration. The trapezoidal rule. As shown in Art. 138, the numerical value of the definite integral INTEGRATION AS A SUMMATION. AREAS 293 f y dx is the same as that of the area bounded by the curve y z=f(x), the z-axis, and the two ordinates x = a, x = b. When a, b, and the coefficients in f(x) are numerically given, the approximate value of this area, and therefore of the defi- nite integral, can be found by adding the n terms of the series [/(«) + f( a + Aa + ' * * + f( a + n — 1 • Ax)] Ax. The close- ness of the approximation improves with increasing values of n. A much more rapid method of approximation is now to be considered. Instead of forming rectangles, as in Fig. 59, p. 268, draw the chords PP X , P X P 2 , • • ., P n _!.Q, thus making trapezoidal elements of area, APP X A X , A x P x P 2 £ 2i etc. Denote the ordinates at A, A ly A 2 , 5—p, A A x A z Fia. 73 A^B' A n _ ly Bby y , y lf y 2 , >- ?/„_!, y n respectively. Also for brevity write Ax = h. the areas of the several trapezoids are APP X A X = 1^ + y x )h, A X P X P 2 A 2 = l( yi + y 2 )h, Then A>-iPn-iQB = i(y n _ 1 + y n )h. Hence, by adding, we obtain for the approximate value of the definite integral the expression *[*¥* + * + » + ••• + !,„-,]. This is known as the trapezoidal formula for the approximate value of J ydx and this method of computing its numerical value is called the trapezoidal rule. 294 INTEGRAL CALCULUS 147. Simpson's rule. With three ordinates. Instead of draw- ing the chords PP 1} P X P 2 pass a parabola, having its axis ver- tical, through the three points P, P x , P 2 and determine the area of the double strip bounded by the two ordinates y , y 2 , the #-axis, and the parabolic arc. The equation of the parabola is of the form y = k -+- Ix -f- mx 2 . For convenience take the origin at the foot of the middle ordinate y x . Then the abscissas of the three ordinates may be represented by — h, 0, + h, and the area under the para- bolic arc is given by the formula x \k + Ix + mx 2 )dx = | (6 k + 2 mh 2 ). This result can be expressed in a simple form in terms of the three ordinates 2/ , y l} y 2 . For, y =k — lh-\- mh 2 , Vi = k, y 2 = k + Ih + mh 2 ; therefore, y •+- y 2 = 2 k + 2 ra/i 2 , hence, 6 ft + 2 m/i 2 = y + 4 ?/j + ?/ 2 , and, accordingly, parabolic area APP X P 2 A 2 — - (y -f- 4 y x -f ?/ 2 ). (IS) o This is Simpson's parabolic formula for three ordinates. With n ordinates. In like manner the area bounded by the two ordinates y 2 , y 4 and a parabolic arc through P 2 , P 3 , P 4 is | (2/2 + 4 2/3 + 2/*), (16) INTEGRATION AS A SUMMATION. AREAS 295 and so on. If the number of ordinates y , y u •••, y n is odd, we obtain, by adding together the expressions (15), (16), etc. J[2/o + * Z/i + 2 2/2 + 4 2/3 +2 1/4 + • + 2 2/ n -2 + * V n -l + 2/J- o This is Simpson's formula for the approximate value of I ydx. 148. The limit of error in approximate integration. The ap- proximate value obtained for I f(x)dx by means of Simpson's s formula differs from the true value by an amount which does not exceed* _ { Q — a ) j l \£)rr j 180 in which f IV (£) is the value of the fourth derivative of f(x) when x is given a certain value £ between a and b. The limit of error for the trapezoidal rule is * (b-a)f"(£W 12 Since £ is not definitely known, in applying the above formulas to find the limit of error it is necessary to choose £ so that / IV (£) or /"(£) has its greatest value in the interval from a to b. The result so obtained may be considerably larger than would be given by the formula if £ were actually known. In some cases the result will be so large as to give no useful information in regard to the closeness of our approxi- mation. In other cases it will be small enough to indicate that the required degree of approximation has been attained. For example, suppose it is required to evaluate 'login a dx. * See Markoff, "Differenzenrechrmnij," § 14, pp. 57, 59. 296 INTEGRAL CALCULUS Since f(x) = x~ l \og l0 x, we obtain by successive differentiation /^(a;)=a;- 6 (241og 10 a;-50*iW"), M =log 10 e== 0.4343, very nearly. As we cannot readily determine by inspection the largest nu- merical value of f lv (x) in the interval 20 <^ x <J 30, we obtain the next derivative . , s . /rt _. ,_„«-.. /v(a?) = ar 6 (274 Jf- 120 log 10 x). The first factor x~ 6 is positive. The second factor takes a nega- tive value for x ^20 and hence / v (<r) is negative in the given interval. Therefore, / IV (x) is a decreasing function for all the values of x under consideration. But f lv (x) is positive for x = 30, and accordingly its greatest numerical value occurs for x = 20, which is / Iv (20) = 0.000003. The limit of error for Simpson's formula is, therefore, _ 10(0.000003) y = _ (a0OO0OO2)y . If we use 3 ordinates, then h=5 and the error does not exceed — 0.0001 + ; that is, the error is less than two units in the fourth place of decimals. EXERCISES In the following problems use Simpson's formula whenever an odd number of ordinates is given. Determine the limit of error and, when possible by direct integration, the exact error. Also evaluate by using the trapezoidal rule, and compare the degree of accuracy attained by the two different methods. 1. Evaluate ( x 2 dx by the trapezoidal rule, using 5 ordinates ; 9 ordinates. In the case of 9 ordinates, n = 8 and h = — ^ — = - , y = 0, 3/i = G) 2 >!/2 = l,y 3 = (§)V--, </8 = 42. 2. Prove that Simpson's rule gives the exact value of f x 2 dx y ( h x s dx, C\ ax* + /?.r 2 + yx + 8) dx. INTEGRATION AS A SUMMATION. AREAS 297 3. Evaluate ( cos xdx, using 3 ordinates ; 5 ordi nates ; 7 ordi- nates; 9 ordinates. Notice the variation of error with increasing values of n. 4. Evaluate ( Vxdx, using 5 ordinates. 5. Evaluate ( Vl + X s dx, using 4 ordinates ; 7 ordinates. 6. Evaluate f cos x dx, using 7 ordinates. Jo C 12 7. Evaluate \ log 10 x dx, using unit intervals. r to ,/ r 8. Evaluate \ '— , using 7 ordinates. Jio loo- r 9. Evaluate ( Vl — x i dx, using 6 ordinates. Jo ° 10. Evaluate ( e~ x idx, using 11 ordinates. This integral (with any upper limit) is called the Probability Inte- gral since it plays an important role in the theory of probabilities. IT 11. Evaluate f Vl — 3 sin' 2 xdx, using ? ordinates. r 10 12. Evaluate \ x 2 dx by the trapezoidal rule, using 11 ordinates. 13. Calculate the value of 7rfrom the formula — = ( ( x , using 1 J» 1 + x' 2 11 ordinates. Determine the error by comparison with the known value of 7r. r l 14. Evaluate i Vcos6d$, taking at intervals 15°, 10°, 0°. This, like Ex. 11, is an Elliptic Integral and cannot be integrated by any formula given in the present volume. It occurs in the prob- lem of calculating friction in journals. (See " Engineering Mathe- matics " by Prof. V. Kakapetoff, Part I, p. 16. Wiley, 1912.) 15. Evaluate ( °%w x dx, using 3 ordinates. J20 x & CHAPTER VII GEOMETRICAL APPLICATIONS 149. Volumes by single integration. The volumes of various solids may easily be calculated by a summation process exactly similar to that used in computing areas. The following prob- lems will make the mode of procedure clear. Ex. 1. A woodman fells a tree 2 ft. in diameter, cutting halfway through on each side. The lower face of each cut is horizontal and the upper face makes an angle of 60° with the lower. How much wood does he cut out? The portion cut out on one side forms a solid bounded by a cylindri- cal surface whose equation may be taken in the form x 2 + 1, and by two planes whose intersec- tion may be chosen for the y-axis. Imagine this wedge-shaped solid divided into thin plates by means of planes parallel to the a;2-plane and at equal distances Ay. The volume of an arbitrary plate PQRP'Q'R' is approximately equal to the area of the triangular face multiplied by the thickness Ay. Area PQR = \ RP ■ PQ= \xz =^x i GEOMETRICAL APPLICATIONS 299 since - = tan 60° = V3. The element of volume is therefore x V3 x 2 A Since the figure is symmetrical with respect to the arz-plane, it is suffi- cient to calculate the volume between the limits and 1 for y and double the result. The limit of th3 sum of all elements of volume in the first octant is ^j/^ = ^j-; ( i- y s )dy= -L. That this limit is the volume to be determined may be seen on observing that the element of volume falls short of the total amount contained in the plate PQRP'Q'R' by the prismatic piece PNP'QMQ'. The sum of all these neglected portions, in the first octant, is less than the volume of the maximum plate (having the a;z-plane for base), and hence approaches zero as Ay diminishes. Therefore the total volume of wood cut out is cu. ft. V3 Ex. 2. Calculate the volume in Ex. 1, by dividing the solid of Fig. 74 with equidistant planes parallel to the yz-plane. Ex. 3. Find the volume of the ellipsoid Imagine the solid divided into a number of thin plates by means of planes perpendicular to the x-axis and at equal distances Ax. Re- gard the volume of each plate as approximately that of an elliptic cylinder of altitude Ax, whose base is the section of the ellipsoid by one of the cutting planes. If the equation of this plane is x = A, the equation of the elliptic base of the plate is (in y, z coordinates) 300 INTEGRAL CALCULUS ** z* _ A 2 + -=1- r^ 2 c 2 a 2 X 2 Dividing by 1 — — , we obtain a 2 The semiaxes of the ellipse are = 1. Since the area of the ellipse is the product of the semiaxes multi- plied by 7r (Ex. 13, p. 281), it follows that the area of the elliptic base is irbcy 1 -J. On replacing A. by x, the element of volume may be written The sum of all such elements for values of x varying by equal increments Ax between and a differs from the volume of the half ellipsoid by a series of ring-shaped portions, the total sum of which is less than the volume of the maximum plate of the figure. It readily follows from this that the total volume of the ellipsoid is C a I x 2 \ 4 2 \ irbc 1 ) dx = - rrabc. Jo V a 2 / 3 Ex. 4. Solve Ex. 3 by taking the cutting planes parallel to the xz- plane and at equal distances At/. Ex. 5. Solve Ex. 3 by taking the cutting planes parallel to the x^-plane. Ex. 6. Find the volume of the portion of the elliptic paraboloid v 2 z 2 '— H — = x cut off by the plane x — 1. a 2 b 2 GEOMETRICAL APPLICATIONS 301 Ex. 7: Find the volume of the elliptic cone ^ -\ — = (x — l) 2 raeas- , a 2 b- ured from the ^2-plane as base to the vertex (1, 0, 0). Ex. 8. Find the volume of a pyramid of altitude h and of base area A. [Hint. Take the base on the ar^-plane, the altitude coinciding with the 2-axis. Cut the solid into thin plates by planes parallel to the base.] e=i. b 2 On the major axis a plane D Fig. 75 Ex. 9. Given an ellipse — a 2 rectangle A BCD is con- structed perpendicular to the plane of the ellipse. Through any point P of the line CD a plane is constructed perpendicu- lar to CD. The two points ii and 5 in which the latter plane meets the ellipse are joined to P by straight lines. The totality of all lines so determined forms a ruled surface called a conoid. Given A C — p, find the volume of the above conoid. Ex. 10. A rectangle moves from a fixed point P parallel to itself, one side varying as the distance from P, and the other as the square of this distance. At the distance of 2 ft., the rectangle becomes a square of 3 ft. on each side. What is the volume generated? Ex. 11. The center of a square moves along a diameter of a given circle of radius a, the plane of the square being perpendicular to that of the circle, and its magnitude varying in such a way that two oppo- site vertices move on the circumference of the circle. Find the vol- ume of the solid generated. 302 INTEGRAL CALCULUS Ex. 12. A right circular cone having an angle 2 at the vertex has its vertex on the surface of a sphere of radius a and its axis passing through the center of the sphere. Find the volume of the portion of the sphere which is exterior to the cone. Ex. 13. Find the volume of the paraboloid — + ^— = z cut off by the plane z = c. Ex. 14. A banister cap is bounded by two equal cylinders of revo- lution of radius r whose axes intersect at right angles in the plane of the base of the cap. Find the volume of the cap. 150. Volume of solid of revolution. Let the plane area, bounded by an arc PQ of a given curve (referred to rectangular axes) and the ordinates p n-\ ^-^L at the extremities P and Q, be revolved about the x-axis. It is required to find the volume of the solid so generated. Let the figure APQB be divided into n strips of width Ax by means of the ordinates A l P tJ A 2 P 2 ,.-, A-iPn-v In revolving about the x-axis, the rectangle APR X A X generates a cylinder of altitude Ax, the area of whose base is tt • AP . Hence 2 volume of cylinder = -k • AP • Ax. The volume of this cylinder is less than that generated by the strip APP X A X by the amount contained in the ring gen- erated by the triangular piece PR X P V Imagine this ring Fig. 76 GEOMETRICAL APPLICATIONS 303 pushed in the direction of the x-axis until it occupies the posi- tion of the ring generated by CDE. If every other neglected portion (such as is generated by P^P^) is treated in like manner, it is evident that the sum is less than the volume generated by the strip A^P^QB, and hence has zero for limit as Ax approaches zero. Therefore the sum of the n cylin- ders generated by the interior rectangles of the plane, viz. tt(.IF + A^ 2 + -+A n _ 1 P n J)Ax, has for limit the volume required. But the limit of this sum is the definite integral I wy 2 dx, and hence volume = 7r I y 2 dx. The volume generated by revolution about the y-axis is found by a like process to be expressed by the definite integral rj x?dy, in which a' and b' are the values of y at the extremities of the given arc. When the axis of revolution does not coincide with either of the coordinate axes, a similar procedure will usually give at once the element of volume, y Q y Examples 1-3 will illustrate. Ex. Find the volume of revo- lution of the segment of the parabola y 2 = x cut off by the line y = x, the axis of revolu- tion being the given line. Let OQ be the axis, and P any point of the parabolic arc. °\ Fig. 77 304 INTEGRAL CALCULUS If v denotes the perpendicular distance PR from P to OQ and u the length of the line OR, then the element of volume is 7T?> 2 Am. The formula of analytic geometry for the distance from a point to a line gives _ y — x _ Vx — x V ~~ V2~~ V2 ' in which (x, y) are the coordinates of P. The second form for v is obtained by substituting for y the expression given by the equation of the parabola. Since Am is measured on a line making an angle of 45° with the .r-axis, it follows that Am = V'2 ■ Ax. Hence the required volume is 1 [Vx- xV i 7T V2 dx V V2 / 30 V2 EXERCISES 1. A quadrant of a circle revolves about its chord. Find the volume of the spindle so generated. [Hint. Take the equation of the circle in the form x 2 + y 2 = r 2 and the equation of the chord x + y = r.] 2. Find the volume of revolution of the segment of the circle x 2 + y 2 — r 2 cut off by the line x = a, this line being the axis of revolution. 3. Find the volume of the truncated cone obtained by revolving about the ?/-axis the segment of the line 3 # + # = 5 between the points (2, - 1) and (1, 2). 4. Find the volume generated by the revolution of the cissoid x 3 = about the x-axis from the origin to the point (x v y^). 2a — x What is the limit of this volume as xi approaches 2 a? GEOMETRICAL APPLICATIONS 305 5. Find the volume obtained by revolving the entire cissoid about its asymptote, the line x = 2 a. [Hint. The element of volume is tt(2 a — x)' 2 Ay. For the pur- pose of integration express x and y in terms of a third variable t by means of the equations z = 2asin 2 *, 2/ = 2a— 1 ' * cost J 6. Find the volume of the oblate spheroid obtained by revolving 2 b* the ellipse — -f 2_ = 1 about its minor axis. 7. Find the volume of the sphere obtained by revolving the circle x' 2 + (y — k) 2 — r 2 about the y-axis. 8. The arc of the hyperbola xy = k 2 , extending from the vertex to infinity is revolved about its asymptote. Find the volume generated. What is the volume generated by revolving the same arc about the other asymptote? 9. Find the entire volume obtained by rotating the hypocycloid xi 4- y* = a* about either axis. 10. Find the volume obtained by the revolution of that part of the parabola Vx -f Vy = Va intercepted by the coordinate axes about one of those axes. 11. Find the volume generated by the revolution of the witch 8 a* ' 2 +4« 2 about the a: -axis. 12. Find the volume generated by the revolution of the witch about the #-axis, taking the portion of the curve from the vertex (x = 0) to the point (xi, y\). What is the limit of this volume as the point (x v y { ) moves toward infinity? EL. CALC. 20 306 INTEGRAL CALCULUS 13. Find the volume obtained by revolving a complete arch of the cycloid x = a(0 — sin 6), y = a(l — cos 6) about the a>axis. Volume = Tri ifdx = ira 3 l (1 - cos 0) 3 dO. 14. Find the volume obtained by revolving the cardioid p = a(l — cos 6) about the polar axis. Assume x = p cos 0, y = p sin B. Then dx = d(p cos 6) = d[a{\ — cos 0)cos &] = a sin 0( - 1 + 2 cos 6) dO. Hence volume = it f ifdx = - 7m 3 f "sin 3 0(1 - cos 0) 2 (1 - 2 cos 0) dd. 151. Lengths of curves. Rectangular coordinates. Let it be required to determine the length of a continuous arc PQ of a curve whose equation is written in rectangular coordinates It is first necessary to define what is meant by the length of a curve. For this pur- pose, suppose a series of points Pj, P 2 , '•', P„_i taken on the arc PQ (Fig. 78), and imagine the lengths of the chords PP X , P X P 2 , ---to have been determined. The limit of the sum of these chords as the length of each chord ap- proaches zero will be taken, in accordance with accepted usage, as the definition of the length of the arc PQ ; * that is, arc PQ = Lt (chord PP, + chord P,P 2 + •• • + chord P n _ x Q). (1) * That this limit is always the same no matter how the points Pi are chosen, as long as the curve lias a continuously turning tangent, and the distances Fig. 78 GEOMETRICAL APPLICATIONS 307 This definition is immediately convertible into a formula suitable for direct application. For, let the points P x , P.,, • •• be so chosen that PB 1 = P 1 R 2 = .. = Ax, the lines PBi, etc., being drawn parallel to the ic-axis. Denote by Ay the increment R x Px of y. Then the length of the chord PP X is V(A*)» + (Ayy = yjl+ ( '^Y A* = X /l + ( |^) V (2) \AxJ * \&yj Now — ^ is the slope of PP V It is, therefore, equal to the l\X slope of that tangent to the arc PP X which is parallel to the chord. If (x 1} i/i) denote the coordinates of the point of con- tact of this tangent line, then we have Ay _ dy l Ax dx x Hence the length of chord PP X may be expressed in the form /(x^Ax, in which 'w-V 1 *® 1 - (3 > Similarly P,P 2 =f(x 2 ) Ax, P 2 P S =f(x 3 ) Ax, • • • , in which x 2 is the abscissa of a certain point on the arc P X P 2 , and so for x s , •••. When these expressions are substituted in (1), it becomes arC PQ = Ax™ ^"W +/&) + • ' ' +/(*-)] **• Pi-lPi are all made to tend towards zero, admits of rigorous proof. The proof is, however, unsuitable for an elementary textbook. (See Rouche et Comberousse, " Traite de geome'trie," Part I, p. 189, Paris, 1891). 308 INTEGRAL CALCULUS But, by (11), p. 274, this limit is J f(x)dx. -Substituting for f(x) from (3), we obtain the formula -"e-rv+ffi)'** w day in which a and b are the abscissas of P and (^respectively. Taking for PP X the second form in (2), namely, vwiw we deduce in like manner -^-xv+sy* \d>j; in which a' and b f are the ordinates of P and Q. EXERCISES 1. Find the length of arc of the parabola y 2 = 4 px measured from the vertex to one extremity of the latus rectum. In this case ^=-W", dx y x hence length of arc = f P \il + £ dx = (" x+p dx. J0 " x > Vx*+px 2. Find the length of arc of the semicubical parabola ay 2 — x z from the origin to the point whose abscissa is -• 3. Find the length of arc of the curve y = log cos x, measured from the origin to the point whose abscissa is — • 2 2 2 4. Find the entire length of the hypocycloid x* + y^ = a 5 . GEOMETRICAL APPLICATIONS 309 6. Find the length of arc of the catenary y = - (e a + e a ) from the point (0, a) to the point whose abscissa is a. 6 ' 2x x 3 1 7. Find the length of arc of the curve y = — + — - between the limits x = 1 and x = 2. 8. Find the length of the logarithmic curve y = log x from a: = 1 to x = V3. 9. Find the length of arc of the evolute of the ellipse (ax)s + (byfs = (a*-b 2 )l 10. Find the length of arc of the curve y = a log (a 2 — x 2 ) from x = to x = - . 152. Lengths of curves. Polar coordinates. The polar formulas for length of arc may be derived from those of the previous article by transformation from rectangular to polar coordinates. Since x = p cos 0, y = p sin 6, we obtain by differentiating with respect to 6 dx = (dp cos 0-p sin dXie, dy = ( d ? sin 6 + p cos $) cl0, hence V i +© , *-va^v-v@y+7* Therefore the length of arc is arcPe = £^ + (|y<W, (5) the limits of integration being the values of 6 at P and Q. 310 INTEGRAL CALCULUS If p instead of is taken as the independent variable, we deduce in like manner arcPQ=jf;Vl + (^)V the limits being the values of p at P and Q. EXERCISES 1. Find the length of arc of the logarithmic spiral p = e a between the two points (p v 0j) and (p 2 , 2 ), and show that it is proportional to the difference of the two radii p\ and p 2 . 2. Find the length of arc of the circle p = 2 a sin 0. 3. Find the entire length of the cardioid p = a(\ — cos 6). Q 4. Find the length of the parabola p = a sec 2 - between the points (Pv #]) and (p 2 , (9 2 ). 5. Find the length of the spiral of Archimedes p = aO between two arbitrary points. 6. Find the length of arc of the spiral p = 0' 2 measured from = to 6 = 7t. 7. Find the entire length of the curve p = cos' 2 0. A 8. Find the entire length of the curve p = a sin 3 r - o 3. Find the length of arc of the cissoid p = 2 a tan sin 6 between the limits and — • 4 [Hint. For the purpose of integration, express the integrand in terms of sec as the independent variable.] 153. Measurement of arcs by the aid of parametric representa- tion. Suppose the rectangular coordinates of a point on a given curve are expressed in terms of a third variable t. Then, GEOMETRICAL APPLICATIONS 311 since in rectangular coordinates — =-%/( — J +( — ) (Art. 41), we have in which s = arc PQ, and t lt t 2 are the values of t corresponding to the points P and Q. In like manner, if the polar coordi- dates (p, 0) are expressed in terms of t, the formula for length of arc is -JWaW'Sr' dt since iwijH'f)' EXERCISES 1. Find the length of a complete arch of the cycloid x = a(t — sin t), y = a(l — cos /). 2. Find the length of the epicycloid x = a(m cos t — cos mt), y = a(m sin t — sin m/) from t = 0tot= ' m — 1 3. Find the length of arc of the hypocycloid x* -\- y^ = a? by ex- pressing x and y in the form x — a sin 3 t, y — a cos 3 t. 4. Find the length of the involute of the circle x = a(cos t + t sin <), y = a(sin t - t cos f) from < = to t = t v 5. Find the length of arc of the curve x* — yt = a% from (a, 0) to (x v yj by assuming x = a sec 3 t, y = a tan 3 /. 312 INTEGRAL CALCULUS 6. Find the length of arc of the curve x = e* sin t, y = e l cos t from t = to t = t\. 7. Find the length of arc of the curve x = a + t 2 , y = b 4 t 3 , meas- ured from the point t = to the point t = t v 154. Area of surface of revolution. Let AQ be a continuous arc of a curve whose equation is expressed in rectangular coordi- nates x and y. It is required to determine a formula for the area of the surface generated by revolv- ing the arc AQ about the a>axis. It has been shown in Art. 44, if p. 81, that if S denotes the area of the surface generated by the FlG - 79 rotation of AP (P being a variable point with coordinates (x, y)), then A# satisfies the conditions of inequality 2 7t y As < AS < 2 7r(y + Ay) As. (6) Let the arc AQ be divided into n equal parts of length A.s. For each segment of arc there will be a set of conditions such as (6), the values of y, Ay, AS being in general different for the different segments. Let the n sets of inequalities thus obtained be added. In what follows, the symbol ^, will be used as an abbreviation of the expression, "The sum of the n terms of the form." Since 2j A>S = S (in which S now denotes the entire surface generated by arc AQ), we have 2 7r ^yAs<S<2Tr^(y + Ay)As. (7) Now let As (and hence Ay) approach zero. The first mem- ber of (7) becomes 2-k | yds, which changes to GEOMETRICAL APPLICATIONS 313 on making x, or y, the independent variable. The limit of the last member of (7) may be written lim As. ™o X t y As + Ay As ~i = fy ds + lim S A ^ As - The last term is zero. For, let 8 represent the maximum value of Ay in any of the terms of 2 Ay As. Then follows ]T Ay As ^ 8 ]P As = 8 • arc AQ, and since 8 approaches zero, we conclude that lim Va^As = 0. Hence lim ^?y As = lim ^P (y -f Ay) As, and therefore In like manner the area of the surface obtained by revolving arc AQ about the y-axis is EXERCISES 1. Find the surface of the catenoid obtained by revolving the X _x catenary y = ~ (e a +e a ) about the y-axis, from x — to x = a. , XX Since -^- — \{e a -e a ), ax it follows that 1 + (dyy_(e*+e~°)\ + \dx) ' 4 314 INTEGRAL CALCULUS hence, by using the first formula of (8), the required surface has the area [e a + e a )dx. r jo *(< 2. Find the surface obtained by revolving about the y-axis the quarter of the circle x 2 -\ y 2 + 2x + 2y + l = contained between the points where it touches the coordinate axes. 3. Find the surface generated by revolving the parabola y 2 = 4px about the £-axis from the origin to the point (p, 2 p). 4. Find the surface generated by the revolution about the ?/-axis of the same arc as in Ex. 3. 5. Find the surface generated by the revolution of the ellipse a 2 + b 2 ~ ' (a) about its major axis (the prolate spheroid) ; (b) about its minor axis (the oblate spheroid). 6. Find the surface generated by the revolution of the cardioid p = a(l + cos 0) about the polar axis. Regarding the figure as referred in the first place to rectangular axes such that x — p cos 0, y = p sin we have surface = 2 tt jj y ds = 2 tt f "" p sin B^p 2 +(^Y d 0> ds =^p 2 + (& YdO by Art. 45. 7. Find the surface of the cone obtained by revolving that por- tion of the line - + £ = 1 which is intercepted by the coordinate axes, a b («) about the x-axis ; ((S) about the y-axis. 8. Find the surface of the sphere obtained by revolving the circle p = 2 a cos about the polar axis. [Cf. Ex. 6.] GEOMETRICAL APPLICATIONS 315 9. Find the surface generated by the revolution of a complete arch of the cycloid x — a(0 — sin 0), y = a(\ — cos 0) about the z-axis. 10. Find the surface of the ring generated by revolving the circle x 2 + (y — k) 2 = a 2 , k>a, about the x-axis. Also find the vol- ume of this ring. 11. Find the surface generated by the rotation of the involute of the circle x = a (cos t + t sin /), y = a (sin t — t cos t) about the x-axis from t — to t = t v 155. Various geometrical problems leading to integration. Ex. 1. A string AB of length a has a weight attached at B. The other extremity A moves along a straight line OX, drawing the weight in a rough horizontal plane XOY. The path traced by the point B is called the tractrix. What is its equation ? Let OF be the initial position of the string and AB any intermedi- ate position. Since at every instant the force is exerted on the weight 316 INTEGRAL CALCULUS B in the direction of the string BA, the motion of the point must be in the same direction ; that is, the direction of the tractrix at B is the same as that of the line BA and hence BA is tangent to the curve. The expression for the tangent length is (Art. 48, p. 86) dy dx =wir + l = a. dx Solving for • — , we obtain dy dx V a - Integrating with respect to y gives Vo2 -J- y dy = Va* - f - a log « + Va 2 - y* + c y The constant of integration is determined by the assumption that (0, a) is the starting point of the curve. Substituting these coordi- nates in the above equation, we find C = 0. Ex. 2. The equiangular spi- ral is a curve so constructed that the angle between the ra- dius vector to any point and the tangent at the same point is constant. Find its equation. Ex. 3. Determine the curve having the property that the line drawn from the foot of any ordinate of the curve per- pendicular to the correspond- ing tangent is of constant length a. If the angle which the Fig. 81 GEOMETRICAL APPLICATIONS 317 tangent makes with the a>axis is denoted by <f>, it is at once evident (Fig. 81) that a . 1 1 - = cos </> VI + tan** A /l+(^) ! From this follows x - i g ( ?y + Vf - a 2 ) + C. When the tangent is parallel to the x-axis, the ordinate itself is the perpendicular a. If this ordinate is chosen for the y-axis, the point (0, a) is a point of the curve, and hence C = - log a. The equation can accordingly be written v + ^v ~ a ' 2 = c «. (i) a From this follows, by taking the reciprocal of both members, y + Vyt- a 2 whence, on rationalizing the denominator, y - Vy2 _ q a _ ^ a o Adding (1) and (2) and dividing by -, we obtain (2) y = |(c« + « «), which is the equation of the catenary. Ex. 4. Find the equation of the curve for which the polar subnor- mal is proportional to (is a times) the sine of the vectorial angle. 318 INTEGRAL CALCULUS Ex. 5. Find the equation, in rectangular coordinates, of the curve having the property that the subnormal for any point of the curve is proportional to the abscissa. Ex. 6. Find the equation in polar coordinates of the curve for which the angle between the radius vector and the tangent is n times the vectorial angle. What is the curve when n = 1 ? When n = | ? Ex. 7. Find the rectangular equation of the curve for which the slope of the tangent varies as the ordinate of the point of contact. Ex. 8. Find the equation of the curve for which the polar sub- tangent is proportional to the length of the radius vector. Ex. 9. Find the volume generated by the revolution of the trac- trix (see Ex. 1) about the positive x-axis. Ex. 10. Find the area of the surface of the revolution described in Ex. 9. Ex. 11. Find the length of the tractrix from the cusp (the point (0, a)) to the point (an, yi). Ex. 12. Derive the following formulas for the length of arc s of a twisted curve, in space of three dimensions, limited by the points (#i, yi, zi), (#2, yi-, Z2), the coordinates being rectangular: -i:v 1+ (gr + (i) a -£v 1+ (i)v(i)> Ex. 13. Using the formula of Ex. 12, find the length of the helix x = a cos t, y = a sin t, z = bt, in which a and b are constants, and t is a variable parameter. Ex. 14. A plate of steel is \ inch thick and has the form of a right segment of a parabola. It weighs 490 lb. per cubic foot. Find the total weight of a plate 30 in. broad and 16 in. long. Take the equation of the parabola in the form y 2 = ±px. Since y= 15 when x — 16, we may find the value of p by substituting these GEOMETRICAL APPLICATIONS 319 coordinates in the assumed equation, namely, 4/> the parabolic plate is therefore *tf. The area of /*16 1 2 \ l £ x^ dx sq. in. The volume and hence the weight are now readily obtainable. Ex. 15. A plate of wrought iron of heavi- ness 480 lb. per cubic foot is \ in. thick and is bounded by three straight edges at right angles to each other, as shown in the figure, while the curved boundary is a hyperbola Fig. 82 with the equation (x + 5) y = 40, the base of the figure being on the ar-axis. Calculate the weight. Ex. 16. A metal plate, in the form of an equilateral triangle, is \ in. thick and has an altitude of 4 in. Any very narrow vertical strip, as AB, of length 2y and width Ax, is of nearly uniform density- The density varies from one strip to another in such a way that Fig. 83 the weight y per cubic inch is determined by the condition = 0.26 1 + 100 \ 9 + xV' Find the weight of the plate. [Hint. Calculate the weight of the strip AB, then take the limit of the sum of all such strips con- T tained in the figure.] r — Ex. 17. A trapezoidal plate ABCD is | in. thick. The weight y per cubic inch is constant along any vertical line, but varies with x according to the law y = 0.05 x 1 oz. per cubic inch. 320 INTEGRAL CALCULUS The first strip DA is 4 in. from the origin. What altitude h must be adopted for the trapezoid in order that the total weight of the plate may be just three ounces? Ex. 18. The frustum of a paraboloid of revolution has vertical par- allel bases five inches apart. The equation of the meridian curve, with the inch as the linear unit, is y = Vx. The heaviness y is constant over a vertical plane section, but varies with x according to the law y = 0.06 VlOO — x 2 lb. per cubic inch. Find the total weight from x = 4 to x = 9. CHAPTER VIII SUCCESSIVE INTEGRATION 156. Functions of a single variable. Thus far we have con- sidered the problem of finding the function y of x when -&■ dx only is given. It is now proposed to find y when its nth. d"v derivative — & is given. dx n The mode of procedure is evident. First find the function — '" which has — - for its derivative. Then, by integrating dx n ~ l dx n d n ~ 2 v the result, determine -„, and so on until after n successive dx n ~ 2 integrations the required result is found. As an arbitrary constant should be added after each integration in order to obtain the most general solution, the function y will contain n arbitrary constants. Ex.1. Given ^ = I,findy. dx 3 x 3 Integration of — with respect to x gives dx 1 2 & T l ' A second integration gives, dx 2 x and finally y = \ log x + \ Cix 2 + C 2 x + C 8 . EL. CALC— 21 321 322 INTEGRAL CALCULUS The triple integration required in this example will be symbolized by which will be called the triple integral of — with respect to x. x s Ex. 2. Determine the curves having the property that the radius of curvature at any point P is proportional to the cube of the secant of the angle which the tangent at P makes with a fixed line. If a system of rectangular axes is chosen with the given line for x-axis, it follows from equation (6), p. 173, and from Art. 42, that \dxl J i r 1 ,(dyyii ~y =«L + UJ J ' dx 2 in which a is an arbitrary constant. This equation reduces to £* = «, dx* from which follows y = jjf[a(dxy = «[f + Cix + e 2 ], Ci and Cz being constants of integration. Hence the required curves are the parabolas having axes parallel to the ?/-axis. The existence of the two arbitrary constants Ci, C 2 in the preceding- equation makes it possible to impose further conditions. Suppose, for example, it be required to determine the curve having the prop- erty already specified, and having besides a maximum (or a minimum) point at (1, 0). Since at such a point ( -2L = 0, it follows that dx = «(1+ Ci), whence C\ = — 1. SUCCESSIVE INTEGRATION 323 Also, by substituting (1, 0) in the equation of the curve, = a{\ - 1 + CV), Accordingly the required curve is y=l(x-iy. Ex. 3. Find the equation (in rectangular coordinates) of the curves having the property that the radius of curvature is equal to the cube of the tangent length. [Hint. Take y as the independent variable.] Ex. 4. A particle moves along a path in a plane such that the slope of the line tangent at the moving point changes at a rate pro- portional to the reciprocal of the abscissa of that point. Find the equation of the curve. Ex. 5. A particle starting at rest from a point P moves under the action of a force such that the acceleratiop (cf. Ex. 14, p. 77) at each instant of time is proportional to (is k times) the square root of the time. How far will the particle move in the time /? Ex. 6. In connection with a certain curve referred to rectangular axes, we know in advance that it passes through a point A on the y-axis at a distance 1.12 in. above the origin. It also passes through a point B of the first quadrant which is at a distance of 12 in. from the y-axis, and the slope of the tangent to the curve at this point is 0.09. At each point P of the curve the second derivative of y satis- fies the relation — { = 0.0012 x. It is required to find the general expression (in terms of x) of the ordinate and the slope of the tangent line for any point P of the curve. In particular, find the ordinate and slope when x = 20 in. 324 INTEGRAL CALCULUS Ex. 7. For a certain curve ADN situated in the first quadrant we have given 1000^ = 1.5 -0.276 x. dx 2 The point A has the coordinates (0, 0.04) and the abscissa of D is 10. At the point B of the curve, whose abscissa is 5, the slope of the tangent line is 0.002. A second curve DC is tangent to the first at the point D, and for each point of it we know that 1000 ^ = 0.2 x -0.115. dx 2 Find the equations of both curves. 157. Integration of functions of several variables. When functions of two or more variables are under consideration, the process of differentiation can in general be performed with respect to any one of the variables, while the others are treated as constant during the differentiation. A repeti- tion of this process gives rise to the notion of successive partial differentiation with respect to one or several of the variables involved in the given function. [Cf. Arts. 62, 67.] The reverse process readily suggests itself, and presents the problem : Given a partial (first, or higher) derivative of a function of several variables with respect to one or more of these variables, to find the original function. This problem is solved by means of the ordinary processes of integration, but the added constant of integration has a new meaning. This can be made clear by an example. Suppose u is an unknown function of x and y such that dx J SUCCESSIVE INTEGRATION 325 Integrate this with respect to x alone, treating y at the same time as though it were constant. This gives u = x 2 -f- 2 xy -f- <f>, in which <f> is an added constant of integration. But since y is regarded as constant during this integration, there is nothing to prevent <£ from depending on it. This depend- ence may be indicated by writing <f>(y) in the place of <f>. Hence the most general function having 2x + 2y for its partial derivative with respect to x is u = x 2 + 2xy+<f>(y), in which <f>(y) is an entirely arbitrary function of y. Again, suppose d 2 u 2 2 = x z y . dxdy Integrating first with respect to y, x being treated as though it were constant during this integration, we find where i{/(x) is an arbitrary function of x, and is to be regarded as an added constant for the integration with respect to y. Integrate the result with respect to x, treating y as constant- Then u = i «V + *0) + $>(?/). Here ®(y), the constant of integration with respect to x, is an arbitrary function of y, while *(x) = J ij/(x)dx. Since if/(x) is an arbitrary function of x, so also is <l'(x). 326 INTEGRAL CALCULUS 158. Integration of a total differential. The total differential of a function u depending on two variables has been defined (Art. 63) by the formula , du 7 . du - du = — dx-\ ay. dx dy The question now presents itself: Given a differential ex- pression of the form Pdx+Qdy, (1) wherein P and Q are functions of x and y, does there exist a function u of the same variables having (1) for its total differential f It is easy to see that in general such a function does riot exist. For, in order that (1) may be a total differential of a function u, it is evidently necessary that P and Q have the forms P=* Q = p. (2) dx ay What relation, then, must exist between P and Q in order that the conditions (2) may be satisfied? This is easily found as follows. Differentiate the first equation of 2 with respect to y, and the second with respect to x. This gives dP = d 2 u dQ = d 2 u dy dy dx' dx dxdy from which follows (Art. 68) 8P = 8Q (3 By dx This is the relation sought. The next step is to find the function u by integration. It is easier to make this process clear by an illustration. SUCCESSIVE INTEGRATION 327 Given (2 x + 2y + 2)dx +(2y + 2 x + 2)dy, find the function w having this as its total differential. Since P=2x + 2y + 2, Q=2y + 2x + 2, it is found by differentiation that ^=2 and 98-2, dy ox hence the necessary relation (3) is satisfied. From (2) it follows that p = 2x + 2y + 2. ox Integrating this with respect to x alone gives u = x 2 + 2xy + 2x + <£(?/). (4) It now remains to determine the function <j> (y) so that d f[=Q]=2y + 2x + 2. (5) Differentiating (4) with respect to y alone gives 1^ = 2*+^), dy where <t>'(y) denotes the derivative of <f>(y) with respect to y. The comparison of this result with (5) gives 2y + 2x+2 = 2x + <l>\y) 1 or <f>'(y) = 2y + 2, (6) whence, by integrating with respect to y, 4>(y)=y 2 + 2y + C, in which O is an arbitrary constant with respect to both x and y. Hence u = x 2 + 2 xy + 2 x + if + 2 y -f C. 328 INTEGRAL CALCULUS EXERCISES Determine in each of the following cases the function u having the given expression for its total differential : 1. y dx + x dy. 2. sin x cos y dx + cos x sin y dy. 3. y dx — x dy. 4 ydx-xdy xy 5. (3 x 2 - 3 ay)dx + (3 y 2 - 3 ax) dy. g y dx x dy x' 1 + y' 1 y' 2 + x' 2 7. (2 x 2 + 2xy + 5) dx + (x 2 -\- y 2 - y) dy. 8. O 4 + i/ 4 + x 2 - y 2 ) dx + (1 y s x - 2 xy + y - y 2 + 2) dy. <~ 159. Multiple integrals. The integration of — — was con- dx ay sidered in Art. 157. If F(x, y) is written for the given func- tion, the required integration will be represented by the symbol = || F(x, y).dx dy, and the function sought will be called the double integral of F(x, y) with respect to x and y. Likewise I F{x, y, z) dx dy dz will be called the triple integral of F(x, y, z). It represents d 3 u the function u whose third partial derivative — is the dx dy oz given function F(x, y, z). It will be understood in what fol- lows that the order of integration is from left to right, that is, SUCCESSIVE INTEGRATION 329 we integrate first with respect to the left-hand variable x, then with respect to y, and lastly with respect to z. Such integrals (double, triple, etc.) will be referred to in general as multiple integrals. 160. Definite multiple integrals. The idea of a multiple integral may be further extended so as to include the notion of a definite multiple integral in which limits of integration may be assigned to each variable. Thus the integral I I x 2 y s dy dx will mean that x 2 )/ is to be integrated first with respect to y between the limits and 2. This gives ►2 x-y* dy =4 x 2 . j: The result so obtained is to be integrated, with respect to x between the limits a and b, which leads to x h 4:X 2 dx = ±(b 3 -a 3 ) as the value of the given definite double integral. In general the expression x'j; Fix, y)dydx will be used as the symbol of a definite double integral. It will be understood that the integral signs with their attached limits are always to be read from right to left, so that in the above integral the limits for y are b and b', while those for x are a and a'. Since x is treated as constant in the integration with re- spect to y, the limits for y may be functions of x. Consider, 330 INTEGRAL CALCULUS for example, the integral J J xydydx. The first integra- tion (with respect to y) gives Jxy dy = x , x 2 x\ X s — X 2 By integrating this result with respect to x between limits and 1 the given integral is found to have the value — ^. EXERCISES Evaluate the following definite integrals : 1. f * ( 2 xcos (xy)dydx. 5 - j/jo ' p 2 sin dp dO. 2. j o j o x 2 ^ <Zx. 6. J o j y Vx*, - y^ dx dy. 3 r« f l °ev dxdy 7 f 2 f « C xs/z xdy dxdz ' Ji Jo y ' Ji J<> Jo a; 2 + # 2 * 4 'dzdydx . J j^sec 2 ^),/,^. 8. J ( J x J o - +y + 161. Plane areas by double integration. The area bounded by a plane curve (or by several curves) can be readily ex- pressed in the form of a definite double integral. An illus- trative example will explain the method. Ex. 1. Find by double integration the area of the circle ( x _ fl )2 + (y - b) 2 = r 2 . Imagine the given area divided into rectangles by a series of lines parallel to the y-axis at equal distances Ax, and a series of lines parallel to the x-axis at equal distances A?/. The area of one of these rectangles is Ay • Ax. This is called the element of area. The sum of all the rectangles interior to the circle SUCCESSIVE INTEGRATION 331 will be less than the area required by the amount contained in the small subdivisions which border the circumference of the circle. All these neglected portions are contained within a ring bounded by the given circle and a circle concentric with it, whose radius is less than r by the length of diagonal of an element of area, that is, of radius r - V(A;r) 2 + (Ay)K In other words, the amount neglected is less than the area of a circular ring whose width is V(Ax) 2 + (A*/) 2 and which therefore approaches zero simul- ! r " 5 taneously with Ax and Ay. Hence the area of the circle is the limit of the sum of all the elements of area included within it. To find the value of the limit of this sum it is convenient first to add together all the elements contained between two con- secutive parallels. Let P X P 2 be one of these parallels having the direction of the ar-axis. Then y remains constant while x varies from a — vV 2 - (y — b) 2 (the value of the abscissa at P\) to a + vV 2 —(y — b)' 1 (the value at P 2 ). The limit, as Ax approaches zero, of the sum of rectangles in the strip from PiP 2 is evidently A#[limit of sum (Ax + Ax +••■)] = Ay I a+vV-Cy-fi) 2 ;-\/r2-(y-6) 2 dx. (1) Now find the limit of the sum of all such strips contained within the circle. This requires the determination of the limit of the sum of terms such as (1) for the different values of y corresponding to the different strips. Since y begins at the lowest point A with the 332 INTEGRAL CALCULUS value b — r, and increases to b + r, the value reached at B, the final expression for the area is rb+r i- /*a+v'r 2 -(2/-&)2 ~1 fb+r r a +\Zr^-(y~b^ \ \ . . dx \dy = \ \ , dx dy. J b-r LJa-VrZ-(y-b)* J Jh-r ^a-Vr 2 -(y-b) 2 Integrating first with respect to x gives e-w*=s=ir i>^p^ = 2V r ,_ (y _ 6)2 , J a -vV2-(y-&)2 Ja-\/»*-(y-6)' w ' This result is then integrated with respect to y, giving f >+r 2Vr 2 - (y - b) 2 dy = (y-b)Vr 2 - (y - by + r 2 sm ^ —irr r J»- If the summation had begun by adding the rectangles in a strip parallel to the y-axis, and then adding all of these strips, the expres- sion for the area would take the form ) ) , dydx. It is seen from the last result that the order of integration in a double integral can be changed if the limits of integration are properly modified at the same time. Ex. 2. Find the area which is included between the two parabolas y 2 = 9 x and y 2 = 72 - 9 x. Ex. 3. Find the area between y 2 = 5a; and y = x. Ex. 4. Find by double integration the area of the segment of the circle x 2 + y 2 = 16 cut off by the line x + y = 4. Ex. 5. Find the area between the two curves y 8 = x and y = x z . Ex. 6. Find the area between the two curves y 2 = x s and y 2 = x. Ex. 7. Find by double integration the area _J2 of one loop of the polar curve p = a sin 2 0. Fi<;. W Imagine the area divided into small ele- SUCCESSIVE INTEGRATION 333 merits by means of concentric circles whose radii vary by equal increments Ap and by means of radii drawn from the origin, the angle between two consecutive radii being A0. (See Fig. 86.) The area of an arbitrary element may be expressed as the differ- ence of two circular sectors with a common angle A0 and with radii p + Ap and p respectively. That is, element of area = \(p + Ap) 2 A0 - \ p 2 A0 = pA0Ap + iA0(Ap) 2 . The sum of all the complete elements within the loop may then be represented by the formula ^pA0Ap + i^A0(Ap) 2 . Reasoning precisely as in Ex. 1, we find the limit of the first sum to be r\ /-a sin 2 The second sum may be written \ Ap2 A0 Ap, hence its limit is n I • lira Ap • lim ^A(9 Ap= i * ° * J ~ Jo d P dB = °' Following the analogy of Ex. 1, we can easily see that all the neglected incomplete elements of area lie within a narrow band along the boundary of the given area, the width of which band approaches 0. Their sum therefore approaches zero in passing to the limit. • It follows from the preceding discussion that the general formula for area in polar coordinates is §§pdpdO, the limits of integration being determined by the boundary of the given area. Ex. 8. Find by double integration the area of the cardioid p = a(l — cos 0). 334 INTEGRAL CALCULUS Ex. 9. Find the area of the lemniscate p 2 = a 2 cos 2 0. Ex. 10. Express by double integrals the three areas between the cardioid (Ex. 8) and the circle p = a. Ex. 11. Find by double integration the area of the triangle whose vertices have the rectangular coordinates (5, 2), (—3, 6), (7, 6). Ex. 12. Find the area common to the two circles x 1 - 8 x + y 2 - 8 y + 28 = 0, x 2 - 8a:+ y 1 - ±y + 16 =.0. 162. Volumes. The volume bounded by one or more surfaces can be expressed as a triple integral when the equations of the bounding surfaces are given. Let it be required to find the volume bounded by the surface ABC (Fig. 87) whose equation is z—f(x, y), and by the three coordinate planes. Imagine the figure divided into small equal rectangular parallelopipeds by means of three series of planes, the first series parallel to the ?/z-plane at equal distances Ax, the second parallel to the ccz-plane at equal distances Ay, and the third parallel to the a?2/-plane at equal distances Az. The volume of such a rectangular solid is Ax Ay Az ; it is called the element of volume. The limit of the sum of all such elements con- tained in OABC is the volume required, provided that the bounding surface ABC is continuous. For the sum of the neglected incomplete elements, which border the surface, is less than the volume of a shell whose outside boundary is the given surface and whose thickness is V(A#) 2 -f-(Ay) 2 -f-(Az) 2 , the diagonal of the element of volume. Hence the error ap- proaches zero as the three increments diminish. To effect this summation, add first all the elements in a SUCCESSIVE INTEGRATION 33; vertical column. This corresponds to integrating with respect to z (x and y remaining constant) from zero to f(x, y). , Then add all such vertical columns contained between two consecu- tive planes parallel to the ?/z-plaiie (x remaining constant), which corresponds to an integration with respect to y from y = to the value attained on the boundary of the curve AB. Fig 87. This value of y is found by solving the equation f(x, y) = 0. Finally, add all such plates for values of x varying from zero to its value at A. The result is expressed by the integral XX X dzd ' Jdx > 336 INTEGRAL CALCULUS in which <f>(x) is the result of solving the equation f(x, y) = for y, and a is the ^-coordinate of A. Ex. 1. Find the volume of the sphere of radius a. The equation of the sphere is x 2 + ?/ 2 + z % _ a2? or z = Va 2 — x 2 — y'K Since the coordinate planes divide the volume into eight equal portions, it is sufficient to find the volume in the first octant and multiply the result by 8. The volume being divided into equal rectangular solids as described above, the integration with respect to z is equivalent to finding the limit of the sum of all the elements contained in any vertical column. The limits of the integration with respect to z are the values of z corresponding to the bottom and the top of such a column, namely, z — 0, and z = Va 2 — x 2 — y 2 , since the point at the top is on the sur- face of the sphere. The limits of integration with respect to y are found to be y = (the value at the x-axis), and y — Va 2 — x 2 (the value of y at the cir- cumference of the circle a 2 — x 2 — y 2 = 0, in which the sphere is cut by the a^-plane). Finally, the limiting values for x are zero and a, the latter being the distance from the origin to the point in which the sphere inter- sects the x-axis. Hence V[= volume of sphere] = 8 j^ £ V " 2 -* 2 ^ V " 2 -**-** dz dy dx. Integration with respect to z gives V = 8 i \ Va 2 - x 2 - y 2 dy dx; then with respect to y and x, V = 8 j> [| V^^V, + ^^-^-J^ SUCCESSIVE INTEGRATION 337 Ex. 2. Find the volume of one of the wedges cut from the cylinder x 2 + y 2 = a 2 by the planes 2 = and z = mx. Ex. 3. Find the volume common to two right circular cylinders of the same radius a whose axes intersect at right angles. Ex. 4. Find the volume of the cylinder (x — 1) 2 + (y — l) 2 = 1 limited by the plane z = 0, and the hyperbolic paraboloid z = xy. Ex. 5. Find the volume of the ellipsoid a 2 b 2 c 2 Ex. 6. Find the volume of that portion of the elliptic paraboloid z=l-*-£ a 2 b 2 which is cut off by the plane z = 0. Ex. 7. Find by triple integration the volume of the tetrahedron formed by the three coordinate planes and the plane x + 2 y + 3z — 1. Ex. 8. Find the volume of the elliptic paraboloid 2 y 2 -f 3 z' 1 — 6 x cut off by the plane x = 2. el. calc. — 22 Jf- CHAPTER IX SOME APPLICATIONS OF INTEGRAL CALCULUS TO PROBLEMS OF MECHANICS 163. Liquid pressure on a plane vertical wall. The pressure exerted by the liquid upon any point of a plane vertical wall is proportional to the depth Surface B 7 of that point below the sur- f / face of the fluid. To calculate \ i y the pressure upon the entire wall we divide it into nar- row horizontal strips of equal areas Ai. Denote the breadth FlG - 88 ' of the Jcth strip PQ (Fig. 88), counting from the top, by h k . The pressure exerted on the ftth strip is equivalent to the weight of a column of fluid standing on a base of the same area AA and having an altitude intermediate between the least depth x and the greatest depth x + h k of points on the given strip. This altitude may be represented by x 4- k h k in which k has a value between and 1. If w denotes the weight of a cubic unit of the fluid, the pressure on PQ is w(x + k h k ) Ai. Summing the pressures for all the strips of the wall, we obtain for the total pressure ^w(x + 6 k h k ) AA. 338 SOME APPLICATIONS TO MECHANICS 339 In order to evaluate this sum we take its limit as &A approaches zero. This gives, by separating into two terms and observing that w is constant, Km y x &A + w lim V #A A A w A.4=b0 The second term reduces to zero. For, ]£0A A.4 = AA^e k h k < ^A • H (since 6 k < 1), in which H denotes the total altitude of the wall ; as A^l = the right member of this inequality approaches zero. Hence pressure = iv I x dA. In order to evaluate the integral, it is most convenient to make x the variable of integration. Denote by y the width of the wall at the depth x. Then &A = y k Ax in which y k is a certain value of y between y and 2/+A?/. (Compare Art. 40.) Dividing by A£ and passing to the limit we obtain, since lim y k = y, dA = dx dt ~ V dt ' or in the differential notation, dA — y dx. The substitution of this in the above integral gives pressure = w j xy dx, the limits of integration being the values of x at the top and the bottom of the given wall or surface. If the liquid is water and the unit of length is a foot, then w = 62^ lb. 340 INTEGRAL CALCULUS EXERCISES 1. Find the pressure on the end of a rectangular tank full of water that is 10 ft. long, 8 ft. wide, and 5 ft. deep. 2. A watermain 6 ft. in diameter is half full of water. Find the pressure on the gate that closes the main. 3. A vertical masonry dam in the form of a trapezoid is 200 ft. long at the surface of the water, 150 ft. long at the bottom, and 60 ft. high. What pressure must it withstand ? 4. A vertical cross section of a trough is a parabola with vertex downwards, the latus rectum lying in the surface and being 4 ft. long. Find the pressure on the end of the trough when it is full of water. 5. One end of an unfinished watermain, 4 ft. in diameter, is closed by a temporary bulkhead and the water is let in from the reservoir. Find the pressure on the bulkhead if its center is 40 ft. below the surface of the water in the reservoir. 164. Center of gravity. (1) For a system of n particles. Let P 1? P 2 be two particles of matter of masses (or weights) m x and p m 2 , respectively, and let x lf x 2 be their O " f *~~ distances from a chosen point O on the Fig. 89 straight line through them. There exists a point P such that the segments P X P and PP 2 are in- versely proportional to the masses of the two points, that is, PiP = %_ ^ PP 2 %' Let x represent the distance OP. Then formula (1), expressed in terms of the abscissas of the points, is x — x l _m 2 SOME APPLICATIONS TO MECHANICS 341 whence, by solving for x, m l x 1 -\-m 2 x 2 /ON m! + m 2 The point P is called the center of gravity, or, the center of mass, of the system formed by the two points P ly P 2 . If we imagine the line P X P 2 to consist of a rigid, weightless rod with the two given particles fastened at its extremities, and if we suppose this object to rest on the point P as a base, it will remain in equilibrium, without any tendency in either of the end points to move downward under the force of gravity. In other words, the system of two particles is equivalent, as far as the action of gravity is concerned, to a single particle, of mass m x -+- m 2 , placed at the point P. Let P 3 be a third point of mass m 3 situated on the same line with P x and P 2 . Then the abscissa x of the center of gravity of the system of three points may be found by calculating the center of gravity of the pair P 3 and P (the center of gravity for P lf P 2 ), the mass of P being taken as m 1 -f ra 2 , the sum of the masses of P x and P 2 . This gives Ox + m 2 ) 1 l ; — =2 + m 3 x 3 w&i + m 2 m^ 4- m 2 x 2 -f- m 3 X3 m (m x -+- WI2) + m s wh + m 2 + n h x = In like manner the center of gravity for any number n of particles situated on a straight line is given by the formula - _ m^ + m 2 x 2 -{ + m n x n ^ ,^ m 1 + m 2 + ••• +m n If the n particles are not on a straight line but are situated in the same plane at the points (a^, y^), (x 2 , y 2 ), • ••, (x n1 y n ), 342 INTEGRAL CALCULUS then the center of gravity of the system has its abscissa given by (3) and its ordinate y is y = WW + m 2 y 2 + h m^ ri m^mz-l \-m n If the n particles are not situated in one plane, there will be a third and similar formula for z. (2) For a continuous solid. Imagine the solid divided up into small elements, precisely as in determining its volume, by means of three series of planes parallel to the coordinate planes and at distances Ax, Ay, Az. If we regard any par- ticular element as being very nearly of uniform density, then the mass of an arbitrary element is approximately p Ax Ay Az, in which p is the weight of a cubic unit of homogeneous mat- ter having the same density as the given element. This num- ber p is usually called the density. For a finite number of elements the ^-coordinate of the center of gravity is determined approximately by (3) in the form (pigq + P1F2 + — + P A) Aa; Ay Az 0>i + p 2 + >~ + p n )AxAyAz in which x ly x 2 , •• are the abscissas for the different elements and p lt p 2 , ••• are their densities. The abscissa of the center of gravity of the given continuous solid is obtained by making Ax, Ay, Az approach zero as a limit.* This gives J I I pxdxdydz I I Ipdxdydz * A proof of this statement will be found in Art. 166. SOME APPLICATIONS TO MECHANICS 343 the limits of integration being determined just as in calculat- ing the volume of the solid. If the solid is homogeneous, p is constant and cancels out of numerator and denominator. Otherwise, it is a function of x, y, z. In precisely the same manner the values of y and z are obtained. The coordinates of the center of gravity are thus found to be x M m j j j 9 x dx dy dz ' y = mSSS p y dx dy dz ' '^SSS pzdxdydZi in which p is the density at the point (x, y, z) and M is the total mass of the given solid, that is, M= J J (pdxdydz. The coordinates of the center of gravity of a plane area are found in like manner to be x = — j Jpxdxdy, T/ = ~JJpydxdy, M=Jjpdxdy. EXERCISES In the following problems p is understood to be constant unless otherwise specified. The abbreviation C. G. will be used for " center of gravity." 1. Find the C. G. of the tetrahedron whose faces are the three coordinate planes and the plane x + 2y + 3z = 6. 2. Find the C. G. of the volume bounded by the coordinate planes the plane x -f y = 1, and the surface z = xy. 344 INTEGRAL CALCULUS 3. Find the C. G. of the volume bounded by the hyperboloid - — ^ — — = 1 and the plane x = k, k > a. a* b 2 c 2 r 4. Find the C. G. of the semiellipsoid on the positive side of the xy-plane, the equation of the ellipsoidal surface being — -f ^- -\ — = 1. a 2 b 2 c 2 5. Find the C. G. of a thin hemispherical shell of thickness h bounded by two concentric hemispheres of radii a and a + h. 6. A hemispherical iron bowl of uniform thickness a is filled with water. If the density of iron is seven times that of water, find the C. G., supposing the radius of the interior of the bowl to be r. [Hint. Find the C. G. of the iron bowl by means of Ex. 5. Find the C. G. of the hemisphere of water and combine the centers of grav- ity of the iron and the water by means of (2).] 7. Show that the C. G. of a triangular plate one inch thick is one half inch below the intersection of the medians of the upper face. 8. Find the C. G. of a T-iron one inch thick, the vertical bar being a inches wide and b inches high, and the horizontal bar a' inches wide and V inches long. 9. Find the C. G. of a sector of a circle of radius a and angle 0. 10. Find the C. G. of the segment of the circle x 2 + y* = r 2 cut off by the line x = a, < a < r. 11. Find the C. G. of the quadrant of an ellipse. 12. Find the C. G. of the segment of an ellipse cut off by the chord joining the extremities of the major and minor axes. 13. Find the C. G. of the area bounded by the parabola Vx + y/y = Va and the line x + y = a. 14. Prove that the volume of a solid of revolution is equal to the product of the generating area by the length of path described by its center of gravity. SOME APPLICATIONS TO MECHANICS 345 15. Find the C. G. of an octant of an ellipsoidal mass. 16. Find the C. G. of the preceding mass when the density varies directly as the distance from the plane x = 0. 17. Find the C. G. of an octant of a sphere. From this result find the C. G of an octant of a spherical shell of thickness h and inner radius a. 18. Find the C. G. of an octant of a sphere if the density varies directly as the distance from the center of the sphere. [Hint. Divide up into thin concentric shells of equal thickness h, the density of a particular shell being regarded as constant. Let A. denote the radius of an arbitrary shell, X the distance of its C. G. from the origin, and m its mass. Calculate X in terms of A by means of Ex. 17, measuring it on a line equally inclined to the x, y, z axes. Then use the different values of X in place of xi, X2, ••• , formula (3), and pass to the limit. 19. Find the C. G. of a right circular cone of altitude h and base- radius r. This problem can be solved by single integration if we suppose the solid divided up into thin plates of equal thickness by means of planes parallel to the base. Then find the approximate expression for the C. G. of any plate, apply (3), and pass to the limit. 20. Find the C. G. of the portion of the elliptical cone — a + f- = (z — l) 2 between the vertex (0, 0, 1) and the zv-plane. a 2 b 2 21. A cone of vertical angle 2 has its vertex on the surface of a sphere, its axis passing through the center of the sphere. (a) Find the C. G. of the mass outside the cone and inside the sphere. (b) Find the C. G. of the mass inside the sphere and inside the cone. 346 INTEGRAL CALCULUS 165. Moment of Inertia. The moment of inertia of a small particle of matter of mass m about an axis is defined as the product of the mass by the square of the distance of the particle from the axis. It measures the resistance of the par- ticle to rotation about the axis. To find the moment of inertia of a homogeneous solid body, imagine it divided up into small rectangular blocks (or ele- ments) of dimensions Asc, Ay, Az. Then the moment of inertia of a single element about the x'-axis is approximately p(y 2 + z~)& x ty Az, in which p is the density, that is, it is the weight of a cubic unit of the given solid. Summing up these elements over the whole body and taking the limit of the sum, we find the moment of inertia to be * ///< p(y 2 + z 2 )dxdydz f (4) the triple integral being extended over the entire solid, just as was done in finding its volume. If the solid is not homogeneous, then p is variable. Its value at a specified point P of the given body is equal to the weight of a homogeneous cubic unit of matter having the same density throughout as the particle of matter at the point P. It is a function of x, y, z which is to be determined by the conditions of the given problem. Similarly, the moment of inertia of a plane area about the a>axis is defined as the limit of the sum of terms formed by multiplying each element of area by the square of its distance from the axis. This gives the formula y 2 dx dy. //• * See the next article for a completion of the proof. SOME APPLICATIONS TO MECHANICS 347 EXERCISES In the following problems M.I. is used for brevity to denote '•• moment of inertia." Unless the contrary is stated, the body is homogeneous and of density p. 1. Find the M. I. of a rectangular parallelopiped of dimensions a, b, c about an edge a. Take three edges a, b, c meeting in a common point as the x, y, z axes, respectively. Then by formula (4) the M. I. is "!oTJ> + -~ 2) *"- yrf *- 2. Find the M. I. of a circular cylinder of radius a and altitude h about its axis. 3. Find the M.I. of the cylinder of Ex. 2 about a line perpendicu- lar to, and bisecting, the axis. 4. Find the M. I. of a circular cone of altitude a and base-radius r about its axis. Hint. If the axis of the cone is taken for the x-axis and its vertex at the origin, the equation of the conical surface is x 2 _ y 2 + z 2 "I a 2 r 2 J 5. Find the M. I. of an elliptical right cylinder about its longi- tudinal axis, the axes of the elliptical bases being 2 a, 2 b and the altitude h. 6. Find the M.I. of the preceding solid about the minor axis of an elliptical base. 7. Find the M.I. of the same body about a line bisecting the longitudinal axis and parallel to the major axes of the elliptical 348 INTEGRAL CALCULUS 8. Find the M. I. of a sphere about a diameter. Hence find the M. I. of a spherical shell of uniform thickness h about a diameter, as- suming that the M. I. of a solid consisting of two parts is the sum of the moments of the separate parts. 9. Find the M.I. of a spherical solid of radius r about a diameter if the density varies directly as the nth power of the distance from the center. [Hint. Imagine the sphere divided into concentric shells of equal thickness AA and denote by X the interior radius of any shell. Using the preceding problem, write down the element of M. L, that is, the M.I. of the shell of radius X and thickness AA.. Take the limit of the sum of all such elements as AA. = 0. The required M. I. is thus obtained by a single integration.] 10. Find the M. I. of a cube of edge a about its diagonal. [Hint. Take three faces of the cube as coordinate planes. Obtain an expression for the square of the distance from any point (#, y, z) to the diagonal of the cube that passes through the origin. This, multiplied by AxAyAz, will be the element of M.I. Then take the limit of the sum.] 11. Find the M. I. of a cylindrical shell, of length a, about its axis, the radius of the inner surface being r and that of the outer surface being R. 12. Find the M. I. of a rectangle of sides a, b about the side b. 13. Find the M. I. of a triangle of base b and altitude h about an axis through a vertex parallel to the opposite side. 14. Find the M. I. of a circle of radius a about a diameter. 166. Duhamel's Theorem. In order to complete the proof of the formulas for center of gravity and moment of inertia, we make use of the following theorem which is of very general use in applications of the Integral Calculus. SOME APPLICATIONS TO MECHANICS 349 Duhamel's Theorem. Let a ly ol>, •••, a n be positive variables, each of which approaches zero as n increases without limit, and suppose that the sum ai + c^-j- ••• + « n approaches a finite limit as n = oc . Let /3i, (3 2 , •••, ft be variables having the same prop- erty as the a's and such that ™ ^ = 1 for k=l, 2, • •-, n. Then "* ^(ft + ft* - +« = n 1 ^ oo (« 1 + « 2 + ...+« B ). Since lim & = 1, we may write & in the form l + e k in « = «>«* « fc which e k approaches zero as n = oc. Hence, ft = a k + e t %, and therefore 2 ft = S "* + S **<**• Let e denote the positive value of the numerically greatest term of the series e y , c 2 , •••, e„. Then we have the inequalities — €a 1 '^€ l a l < 4-««u — ca 2 5^ e 2 a 2 ^ 4- ca 2 , and by adding we obtain — «(«! + «o 4- • • • 4- «„) ^ 2 e *«* ^ + «(«i 4- « 2 4- h <*»)• Now let n increase without limit. Since by hypothesis e = and («!4-«2+ ••• 4- a n ) has a finite limit, it follows that the first and last members of the preceding inequalities vanish at 350 INTEGRAL CALCULUS the limit and therefore . Hence Jf^A- J?.S* # As an application of the above theorem, consider the sum occurring in the approximate formula for center of gravity, nam6ly ' <«% + *%+•••+ PA) AF in which AF= A# A?/ Az. Let p k , x k be the minimum, and p k ", x k " the maximum values of p, x in the ftth element of volume. For brevity write Pk 'x k '*V=a k , Pk "x k ''±V=(3 k . Then we have a k <^ p k x k AV^ /3 k hence, by taking the sum, 6 o "x " But £-* = ^* * which approaches 1 as n increases since p k ', x k approach equality with p k ", x k ". Hence lim X a * = lim ]£ A = lira ^ p a AF. In obtaining this result no restriction is placed on x k and p k * A variable which has zero as a limit is often called an infinitesi- mal. Hence ai, a2, •••, a n are infinitesimals. If we write Sj. = e^, then „ V^L — = lim e k = 0. When two infinitesimals, 5 and a, are so related that the ratio of 5 to a has the limit zero, then 5 is said to be infinitesimal with respect to a, or it is called an infinitesimal of a higher order than a. Since, by Duhamel's Theorem, lim ^(a*. -f 5*) = lim 2 a *i tn * 8 theorem is equivalent to saying that the limit of a sum of infinitesimals is not affected by dropping from each term an infinitesimal of a higher order. ANSWERS DIFFERENTIAL CALCULUS Page 23. Art. 7 2x-2; 2; 0; 1. 6. wx"- 1 . 2. 6x-4. 3. 1 4 a* 4. 4x 3 - 5. 3 X*' 7 x 2 + 2 a; (x + l) 2 ' « 1-x 2 (X* + 1)2 9. _L. 2Vx io. -fx-i Page 24. Art. 8 (6«-4)6*. 3. -i(10x-2). 4. (e—A) (rf-2) Page 32. Art. 13 1. 10X9. g 2 - 6 x - x 2 2. - 8 x-9. (a: 2 + 2) 2 3. 4. c 2Vx 1 Vx3 _ 5 4/ — 4 VX" 1 ^9. 10. 11. 3x-f 5 Vx + 2 Va(Vx,— y/a) 5. 2 y/x ( y/x -fa) ( Va + Vx) 3 6. n(x + a) nx* 1-1 . ,n-l. 12. 1 7. Vl-a; 2 (l-x) 8. a 2 3 13. 1 (a 2 - x 2 )s 2 x(l - x 2 ) + vT^x 2 355 356 ANSWERS 14. 15. 16. 17. 18. 19. 20. 21. 4 oft + 3 aft 4A/x^ a ! + a i 2 x 3 - 4 a (l-x 2 )^(l+z 2 )* — 2 nx"- 1 (x n -l) 2 ' _ m (ft + x)+n(g + x) (a + x)™+! • (b + x)«+! x 2 (x 3 + 1)T 56x 3 (x 2 + l)i 6 dx 12(m 2 - t* + l) — . ax I. 60 M 5(1 4-^2)2^?. ax 24. w + x^. ox du 25. (2m + 6xw) — + 3w 2 + 4x 3 . dx dx nu n (a+z) n (a + se) n + 1 27- 2 wx 3 w ^ + w 2 x 3 — + 3 uWw. dx dx 30. 6 2 x &x Va 2 32. (0,0), (i,--il \9o 27 a/ \9a' 27a/' 34 (21m 3 - 19m)10x (7 w 2 + 5)* 35. At right angles at (3, ± 6). y 2 + 2 xy 2xy + x 2 ' x + a a ax-}- b 8x- ■7 4x 2 -7 x + 2 2 1-X 2 ~; 4x Page 33. Art. 14 3 6(2x-3y) 2 +2(x-y)+l 9(2 x - 3 y)*+ 2(x - y) - 1 ' Page 37. Art. 18 7. ?ix n_1 log x + x M_1 . 8. nx"- 1 log x m + mx n_1 . x 1 -x* log x + 1 9. 10. x 2 -l 1 2 (\/x+l) 11. log e- 12. log 10 e 12xV2 + x — 1 2 V2+x (3 x 2 - V2+x) 2x + 7 *x 2 + 7x* xlogx 14. ae ax . 15. 4 e 4x + 5 . 16. i (i+x) 2 " 17. e x (1 + e*) 2 18. 2/ - 3 x 2 e* 19. 1 - y 2 . ANSWERS 25. 21ogx 26 1 xlogx 27. - (log x + 1) 28. a logxlPg«. X 357 e? + e~ 3 -(x-l)^(7x 2 +30x-97) 12(x-2)^(x-3)^ 2 + x - 5 x 2 e 1 2V1-X 21 1 + e * 31 1 + 3 x 2 - 2 x* ' X+e *' * (l-x 2 )f . 5 x*(a + 3 x) 2 (a - 2 x) (a 2 + 2 ax - 12 x 2 ). 22. wx"- 1 ^ + x n a x log a. 23. V« Vx(a — x) 24. 1 33. («-2«)vY x(logx) 2 ' • (x-a)* Page 41. Art. 22 1. 7 cos 7 x. 12. - 20 x (3 - 5 x 2 ) sec 2 (3- 5 x 2 2. — 5 sin 5 x. 13. 2 tan x sec 2 x — 2 tan x. 3. 2 x cos x 2 . 14. secx. 4. 2 cos 2 x cos x — sin 2 : t sin x. 15. cot Vx 5. 3 sin 2 x cos x. 2Vx 1 1 6. 10 x 09s 5 x 2 . 16. log a • a x • sec 2 (a x ). X 2 7. 14 sin 7 x cos 7 x. 17. w sin w_1 x sin (n + l)x. 8. sec 2 x (tan 2 x— 1). 18. cos2 W ^. 9. 3 sin 2 x cos 2 x — sin 4 x. ax 10. sec x (tan x + sec x). 19. mn sin m_1 nx • cos(w — n)x COS n+l mx 11. -16x(l-2x 2 )sin(l cos(l-2x 2 ) 2 . -2x 2 ) 2 20. 2 1 4- tan x 358 ANSWERS 21. cos (sin «) cos u^ 26 ' ~ 8 csc2 4 * cot 4 *■ dx J. 2 ae 8 * sin e 8 * • cos e 8 *. 26. 8 (4x - 3) sec (4x - 3)2 tan (4x - 3) 2 . 27. -2a?oatfrf+? eoV5tanVS - cose^logx+^i-. 2Vx 28. y cos xy xcosx 2 1-xcosxy VsTnx 2 29. - csc 2 (x + y). Page 43. Art. 23 4 * 6. 1 11. * \/l-4x 4 2Vsin- 1 xVl — x 2 Vl — x 2 1 7. 1 • 12. ^*- Vl -.x 2 e * + erx * Vx 2 - 1 3 3 8. - 1 13. -ZJL. V 6x-9x 2 xVl-(logx) 2 l+.x 2 o n SPC 2 X . 4. * . 9 /t t • 14. |Vl + cscx. ^1 _ x z VI - tan 2 x 5. _^_. 10. — J=- 15. h l + x 2 vT^T 2 16. sec 2 x.tan-ix + -^i£L. 17. sin-ix+- 1 + x 2 18. gtan 1 x 21, . _ 1 + x 2 " 2 Vx (x + 1) 2 22. 19. — - e x + e~ x VI - x* 23 r 25. L-x 2 2 sin x 26. 27. Vl — 4 cos 2 - 1 X 2(1 + x 2 ) -1 ,. — 2 cos -2 x + w 2 sin 2 x V 1 — x 2 ' " x 2 + 1 ' 24. 2. 28. 0. Page 45. Exercises on Chapter II 1. 6x+ 15 x 2 . 4 a 2 - 2 x 2 2 - 6 15 V^^x 2 x 3 x 4 5. log sin x + x cot x. 3 x - 1 « - a 8 3. 2Vx-3 *Va 2 -x 2 7, 8. 19. 1 J a + a x 'x — a ANSWERS 359 x V xj 1— iC 4 2 x 2 - 2 x 4- 1 2(x-x 2 ) 2 ' ""' x _ e^ 22. 1. 9> = ' cot «• «« „ 2 Vu 23. 2 tan « + e 8ec x • sec x tan x. JO. i-loga. 24. 2^ST. x 5 a* - y 2 11 -(3x4- a*) 25 2xy 2 + 3x 2 3x 2 + l (i + x 2 )i ' ,12. e*(cos x - sin x). 26. - 3 r + 1 13 - 1^3=^- 27. y+a^-i. 2 xy — x 2 ^Vx^n: 27. £ 14. J 9R 4 cos (21ocrx 2 -7) 5 + 3cosx 28 k — ^ l ' 15. tan-i J*. 29. x^ = 2y. y a dx y 16. 1 30. x = rnr. (- 1 + ^) 32. x, y are determined from 17. 4 tan* x. a 2 y _ ± b 2 x and equation 18 logx of curve. ' (1-x) 2 ' 33. x = kit ± -. 4 4 5 + 3 cos x 34. tan- 1 2 V2. Pages 49, 5Q Exercises on Chapter III 1. 72 x. 8. 8 tan x sec 2 x (3 sec 2 x - -1) 2. 0. 9. 2 cot x esc 2 x. 3. 3! X 4 ' 5! X6 ' 6 sec 4 x - 4 sec 2 x. 10. 11. 12. 16 sin x cos x. 24 4. 5. (1-x)* 48 X 6. 7. e* logx + Ai*—^!. x x 2 2 log x + 3. 13. 14. sinx. _ 8(e* - e-*) 360 ANSWERS 15. 16. 17. 18. 19. 21. 8 z 2 e2*. _4J x 2 ' a n e ax . (-l)"w! (x -!)"+!' \lmx + ra^Y (- !)"• (m+ n-1)! (w— 1) !(a + sc) w +" m ( - l)n-i . ( W _ 1) ; (a + x) n 3jp a 2/ 6 ' ft 4 24 - 2 a3 *y (y 2 -ax) 3 ' 25. -yr(*-i) 2 + Q/-n 2 i a:2( y _i)3 e 2 *. 34. 2"- 1 cos['2a;+ — V 3- -?/ (2- -2/) 3 (n- -1)! 2. and 1 3. r , max. ; . — , min. V3 V3 Page 53. Art. 28 Inc. from -co to i ; dec. from 1 to 1; inc. from 1 to + oo ; | Two. + 1 at x = \ ± V^ ; - 1 at x = J ± VJ. 4. ± tan-* ^. Page 60. Art 34 6. — 1, max. ; — £, min. 7. — 2, min. ; 1, max. 2, max. ; 3, min. 8. e, max. 2, min. ; f, max. 9. 2n7r, min. ; also tan-i ± V| (2 w+J)w, max. ; (2ra + £)7r, for angles in 2d and 3d min. for all integral values quarter. (2 n + 1) 71- , of w. tan -1 ± V|, 1st and 4th quarter, max. 10. 2, min. ; — 1, max. 11. x = 3m + 4 m 2 + 44w + 4=0. 2(u-l)' -, mm. 4 Pages 63-67. Exercises on Chapter IV Two thirds the length of the segment. The parts are equal. h 5. -*=• 7. 2r V3 6. 3 inches. V3 8. Area is ~ 2 ANSWERS 361 9. The side parallel to the wall is double each of the others. 10. The altitude is equal to the diameter of the base. 11. 8 inches. 12. One mile from stopping point. 13.— Most economical per hour at 15 knots. 14. fa. 15. The altitude of the rectangle is equal to the radius. 16. The altitude is equal to the radius of the base. 20 17. — - yards from the nearest point. > 4 2 - 12 2 19. 15 V2 feet. 20. The diameter of the sphere equals the edge of the cube. 21. ilfeet. V2 22. Circular arc is double the radius. 23. — , T> being the distance between the centers of the spheres. 24. Arc = 2 *r(l — Vf) . 25. Angle at center of variable circle denned by d — cot 6. 26. The line should be bisected at the given point. 27. The altitude is f the slant height of the cone. 28. (a* + &*>*. 31 - x = a 'P- 34. *=J2jD?feet 29. | a. 32. 20 ft. X » 35. tan 6 = sec 0— tan </>• 30. a + 6. 33. aV\. 36. 6 = 35° 20'. Pages 76, 77. Art. 39 3. About 3° 58' per second. 5. (3, -\ e ). 4. 120 feet per minute. 6. At 5a/2 miles per hour. 7. (3,6). 9. 2ab. 11. 5 w. 8. At 60°. 10. i 2. 12. 2. 362 ANSWERS 13. 1 and 5. 16. S ~64' '~32 17. a 16' Page 19. ± 16, =F 12 feet per second. 20. sin • d<t>. 36 21. — :rrL radians per second. V37* Page 83. Exercises on Chapter VI 1. -J^+JE, 2V^x, 4tt\/^+ ax, 4wax. 2. 2, 2. * x y x 3. secx. 5. ,-^(a 2 -a: 2 ). 7. V2 ap. a z 4. Tra 2 ^ 6 p Vl + (loga) 2 . p 9. 30 7t, 72 ir. 10. f sin d, 90°, 270° ; 2, - 2. Pages 87-89. Art. 48 I Xix . yiy _ 1 4. (a) x + 2 */ = 4 a, a 2 6 2 ' y-2x + 3a = 0. y yi Mb/ y = T2x±3. 2. y = x. (7) */ = x+p, x + y-3i>=0. 3. 2y = 9x -3, 9y + 2x = 29. 5. 3. 6. 4^17. 7. (a) Parallel at points of intersection with ax + hy = 0. Perpendicular at points of intersection with hx + &?/ = 0. (j8) Parallel at f—^-z, 8 *y 2 > ) ; perpendicular at X = 0. (7) Parallel at (—, fL^iiA . perpendicular at (0, 0); (2 a, 0). 8. _ = , i.e. they must be confocal. a b a' b' IT 12. 2 2 nx 13. 2c + a . 19. (2p, ±2pV2). o Page 95. Art. 51 1. An inflexion at x = y = 2. , /2o 3a\ /-2o 3o\ ANSWERS 363 8. Point of inflexion at (a, $a), tangent is x + y=-£. Bending o changes from negative to positive. 10. (- 1 ,l),(2 ± V3,^^g). Page 103. Art. 57 1. y = 0, x = a, x = — a. 2. x = 0, x = 2 a, 2/ = a, y = — a. 3. y = a, y=—a ; two imaginary. 4. y = « ; x = c twice. 5. y = — x + - ; two imaginary. o 6. x = 1 ; one parabolic branch. 8. a* = twice ; one parabolic branch. 9. x - 0, y - 0, x + y = 0. 10. y = # ; two imaginary. 11. x + y + a=0; two imaginary. 12. y -f sr, = ; two imaginary. 13. x = twice ; x = y, x = — y. 14. y=x, y=— x; two imaginary. 7.- x = — a, y = — 6, y=x + b-a. 15. a; + 2y-0,x+y=l,a--y = — 1. Page 107. Art. 60 1. yff = 6. 2. Polar subtangent = — , Polar normal = y/a?+p\ Polar subnormal = a. 3. yff = - + 2 0, Subtangent = - p cot 2 0, Tangent = q2p . 2 Va 4 -V Subnormal =- a ' 2sin2 * , Normal = £ P P 5. ^,2asin2^.tan^. 2 2 2 7. They have a common tangent at the pole ; elsewhere, -. 3. 1. Page 111. Art. 62 4. (ar + y) cos xy. 5. 1. Page 115. Art 63 5. \ square units. 6. 5Vlo! Differs by dxdy. 364 ANSWERS Page 117 Art. 65 3 _ ax + hy + g . ? 3 hx + by +f ' VlT^' 4. x ~. 8. *=1. y 3 &> + x 6 . 2l + V y+U. 9. yCco8(gy)-e^-2or1 > s s x [x + e 2 * — cos {xy)~\ Page 121. Art. 66 2. 8 a; + 8 y - z - 12 = 0. 5. — + JL + JL = at • £Ci* ^ z\^ 3. »=J s= t^=«=» 7 . 008 -i_i<L. 8 . JL. 1 - 4 3 V119 V17 9. 2* 1 * + „»-« 1 . = 0; ^fl = r=H= s t^l. 2 5Ci «/i — *i 10. 2x + 2V3y + 3s = 25, x + s = 5. Pages 128-130. Art. 72 2. **-2y^ = 0. 11. ^ + ^+y=0. dy' 2 dy dy 2 dy Q r h(f)T d 2 x dy 2 4. fd % x cPx\ dx_Q \dy s dy 2 ) dy 5. £* = cos's + 2(-T dx 2 \dxJ 6. dy* 8. ^L + y = 0. du 2 9. SUO. 12. 2z d 2 z , n f dz + 2(^\ dx 2 \dxj + (l-z 2 )2z — +z* = 0. dx 13. ff+«/=0. dt 2 14. ^ + ^ = 0. d£ 3 d« 2 < d« 16. dl> 17. -6. Page 137. Art. 74 6. Divergent. 7 Convergent. 8. Convergent in both cases. ANSWERS 365 Page 140. Art. 75 6. -1<x<1. 7. |s|>l. 8. -a<x<a. Page 145. Art. 77 2. ' /(s) = (x-l) 3 +(x-l)2 + 4(a:-l)-3. /(1.02) = - 2.919592, /(1.01) = - 2.959899. /(.99) = - 3.038901, /( .98) = - 3.079608. 3. 3(y-3) 2 +4(y-3)-8. 4. sin 31° = .51508. Page 149. Art. 78 1. x+ ??+2-x*+B. 3. .000002. 3 15 6. .017452. 7. 1 — — — — + ^. 2 8 y/S l/„ tt\ V3 2 2 9. ? + e?K + —h* + B. 10. 15 + 24 (a: - 2) + 13 (x - 2) 2 + 3 (a - 2) 8 . U.log« + *- ". + -»-».+*. 8 . a; 2 a; 2 3 X s 4 a: 4 12. - 4 (a; + 1) + 6 (x + l) 2 - 4 (x + 1)3 + (x + l) 4 . 14. 5.013. 16. 3.433987. 15. 11.0087. 17. .0127.-.. 18. 1 - (* - 1) + (x - l) 2 - (x - \f + B. to 2. Pages 159, 160. Art. 83 2. 2 a 2 a 2 + 6 2 3. 13 7 ' 4. JL. 2a Pages 163, 164. Art. 85 3. f 5. f 8. -4, 4. 4. 7. *■ 366 ANSWERS Page 164. Art. 85 1. 0. 6 9. 1. 2- 2. 6. I. 10. _?, n 3 3- 3. 7. 1. 11. I ]oga log b ' 4. !^?. * 1. Page 166. Art. 86 1- 1- 3. 0. 5. -$. 2. 0. 4. 5. 6. 1. «• e 2c2. Page 171. Art. 89 1. First. 6. Second. 2. They do not touch. 7. a = — 1. 3. Third. 8. y = 2 a; 2 - 5 x + 4. 4. 3i(x-a) = a(2/-a). 9 a. (— 2, - 8), First. 5. y + 12x = 10. 96. (- 2, -8), Second. 10. y=-x 2 + 2x + 3, $ x =-3y 2 + Uy-36. 2. First. 13. Second. 2 a. a . 2* Page 179. Art. 97 6 (* 2 + rcV)2 n(n — l)xy i. 2. 9 aVx(8a-3x)* 3(2 a - xy 3. 4. 00 . (z 2 + ?/ 2 )* 2 m 2 7 2(z + 2/)^ Va 10. «!. a 5. (e 2 x 2 - a 2 )^ a& 8. 3(aa*/)*- Page 181. Art. 98 3 a(5- 9- 11. ( * _j_y V3V2 54V2"/ 1. 2. pVl + (log a) 2 - - 4 cos 0)* - 6 cos 3, ANSWERS 367 2p* 6 4x/ ^ 3 Va 5. -if. a 2 1. a = 0, /3 = 0. 7 q(l + gg)* Page 188. Art. 100 2. a _alog-— - , p--. b (« + i8 )!-(«-/S)i = (4a)t. . a=x-^(e a -e a ),p = 2y. . (a + )3)f-(a-/3)§ = (4a; . , 1K . , . 6. (aa)*- (6/3)'- = ( a 2 + 6 2 )t. 3 ce = q + 15 y 8 = ay ~ ° y 6 a-V 2 a* 7. (a + B)% + (a - /3)t = 2 ai 16. a = a(0' - sin 6"), p = a(l - cos #'), 0' = - tt. Pages 198, 199. Exercises on Chapter XIII 1. (0, 0) ; ax±by = 0. 6. Two nodes at infinity ; the asymptotes arex = v-|-l, x+y=±\. 2. (O,0);cuspoffirstkind,y=0. m , n x , ' / _ V ' . *" 7. (0, -a); (+a, 0); (-«,0:) 3. Four cusps of first kind ; the tangents are, respectively, (0, ± a), (± a, 0) ; y = 0,a; = 0. V% + a) = ±V2a; 2(.z+a) = ±\/3y; 4. (0, 0) ; conjugate point with 2 (x -a) = ±VSy. real coincident tangents, y = 0. / Ax ° ' " 8. (— a, 0) ; conjugate points. 5. (0, a); y = a + *J cusp of 9 (0, 0) ; » = 0, y =0. second kind. 10. (0, 0) ; is a tacnode ; y = 0. 12. Terminating point at (0, 0) . Pages 207, 208. Exercises on Chapter XIV 1. x 2 + y l = p 2 . 7. y 2 = 4 a(2 a - x). 2. xf + i/f = ai »• & 2 x 2 +(a 2 fW=6 2 (a 2 + & 2 ). 3. x$ + yl = cf. 9 - (* 2 + ?/ 2 ) 2 = 4 c2 (^ 2 " y' 2 )' 4. 4xy = k*. 10 - 16y 3 + 27^ = 0. 5. (x - a) 2 + (y - py = r 2 . 11 - y±s±fc = 0. 6. y 2 (x + 2a) 4-a* = 0. 12. (x 2 +y 2 -ay) 2 =a 2 (x 2 +0/+a) 2 ). 13. 6 2 x 2 + a 2 y 2 = 4 a 2 6 2 . INTEGRAL CALCULUS Pages 215-216. Art. 114 1. „ 3 21. — cos nx. 2. x a+l n a+ 1 22. x . sin 2 x 2 4 3. 4. 2 to X «H am ~\L ,. 23. 24. x sin 2 x 2 4 cos (to 4 n) x 5. ax - \ aH* + | a *05* - £ x 2 . m+n 6. 5kJx-f -J* JL. e 2x 2 3x 8 25. 26. — | COS X 2 . sin x — ^ sin 3 x. 7. H^ 2 + « 2 ) 3 - 27. — cos x + I COS 3 X. 8. (atx + &) n+1 28. tan x — x. JoT+1) 29. | tan 3 x. 9. 10. log;(x 4 a). | log (2 ax - x 2 ). 30. — cot (ax 4 6). a 11. — log cot X. 31. - f (cot x)l 12. — log (1 + cos a). 32. log tan x. 13. log (logx). 33. \ sec 3 x. 14. flog(x 3 4l). 34. — cos X. 15. 16. — log COS X. log sin x. 35. sin-i^. a 17. - e ax . 36. i sin- 1 2a;. 18. a 37. a a 19. (a 4- b) m + nx w log (a 4- 6) 38. A tan-i^- ab b 20. | sin 2 x. 39. tan- 1 (x-2). Page 219. Art. 115 1. x sin^x + Vl — x 2 . 3. _X* Sill x -f- Z X COS X - 2. e x tan -1 e x — \ log (1 4- e *). 4. J2±i(log* * n + 1 V n + 1 ANSWERS 369 5. A[2z 3 tan-ix-z2 + log(l+*2)]. Q £e*(sin x + cos x). 6. sec x [log cos x + 1J. 10. | e* (sin x - cos x) . 7. J [(*» +1)001-1* + *;]. 11. icosxsin2x-icos2xsinx. 8. Ksin3*-3*cos3*]. 12. x tan x + log cos x. Pages 220-222. Art. 116 4. Ksin-i*)2. i, , ♦ « 11. log tan-. 6. Jcos(*2 + l)[l-logcos(*2 + i)]. 2 12. log tan g + |). 13. -§(a 3 -x*)i 7. sin-i « a 8. I tan-i « a a 9. ±cos-i«. a * 10. sin-i^-^ a 14. log (x - 1) - 15. 1 2 x-1 2(x-l) 2 sin 2 * 16. -^tan-^v^tan*). V2 Page 226. Art. 118 1. -i-logV^Cg+l)-!. 3 . _± lo|f *=*. 2V2 V2(x+1)+1 12 °x+l 2. -i_ tan-ii^l. 4. £sin-i(3* - 5). Vl4 V14 ' 5. Vx 2 + 2 x + 2 - log(x + 1 + Vx 2 + 2 * + 2). 6. - V-*2 + 2 x + 1 + sin-i^ni. V2 7. .-V / 8-4*-4*2 + 3 sin -i2*_+l - 2 3 8 ^V3x^ + x-'2-~ log(aj + | + Vx 2 + i x - f ). 9- ^Vl+*-2*2 + -JL_ sin -i 43-l t 2 4V2 3 10. ^ log (x- 5) -| log (x-1). 11. Vax - x 2 + ? sin-i 2-^J? . 2 a el. calc. — 24 370 ANSWERS 12. - V- 2 x 2 - 3 x - 1 - sin" 1 (4x + 3). 13. -lVi-2*-3^--^-sin-i 3a; + 1 3 2 V2 10 3V3 i log («_±^) Page 227. Art. 119 2 3. -lo 4. l-2x + V5^-4a;+ 1 / a+V« 2 -a 2 \ 1 , /V2 + Vx 2 + 2z + 3\ ~VS lQg ( "¥+1 J lo< / 1 — a; + 2 Vz 2 + x+J \ 6. sin- V2 (x - 1) 7 . _l ]o(f /l-x+V-x*-10x-r 9. a 2 z 10 5C + 2 Vx 2 + a' 2 — Va 2 — x 2 11. -JLrin-i 3 ~ 2 *_ V2 V3(2 x - 1) Pages 227-228. Exercises on Chapter I 3. i log (6 x^ + 12 a; + 5) 2. 5 tan-i-. 8 2 4. *£(! + **). 5. -f(3-2, )t 6. ^ 7. _| (d «_rf) 10. I sin 8 x. 12. — cos e*. V, 2 + 1 Ks + l) 1 - *(*-!)*. 11. 1 log tan ^^ + -1 3 s V 2 4J 13. log (a cos se + b). a 14. - log (e~ x + Ve~ 2x - 1) . 15. I sin-i x 2 . 2V2 L * J 2 & e* + l 18. i a* tan-i x - ^ * 4 + rV * 2 - tV log (* 2 + 1) ■ ANSWERS 3T1 19. -f2 + 2xloga + (x\ogaf}^ 2Q tiind _ secd . a* (log a) 3 .21. -cot?. 22. logCgcoB'g + ftBin'^ 2 2(6 -a) 23. -JVl-logx. 24. ilog(e'^ + l). 25. sin -i / 2 sin ^ + 1 V 26. tan- 1 (logic). 27 6(a — 6 tana;) 2' 28. irin-ir ^-^+^ 1 L V5 a 2 J V 3x 2 + 2x+l + log p + l+V3*» + 2s+r 30. - — log (a; + Vx 2 - a 2 ) + Vx 2 — a 2 2 x 2 2 a 2 x 31. — cos x log tan x + log tan - . 32. \x — \ log (sin x + cosx). Pages 236-237. Exercises on Chapter II 1. *(s a -2 _ 2 x 2 ) Va 2 - x 2 + — sin-i? . 2. ir_JL_ + ltan-i2l. 5. - Va 2 - x 2 + ^sin-i?- 8|_x 2 + 4 2 2j 2 2 a 3. 2a; - 1 -f-i-tan-i^l. 6. - ^ZEZ. 3(x 2 -x + l) 3V3 V3 4. _?v / a 2 -x 2 +— sin-i-. 7. + ' 2x _ 8. ~ (2 x 2 + 5 a 2 ) Vx 2 + «" + — log (x + Vx 2 + a 2 ) . 9. - Vx 2 + a + ^log(x -f Vx 2 + a). 10. -(2 x 2 - ax - 3 a 2 ) V2 ax - x 2 + ^ sin-i — - • 6 2 a n (2ax-x 2 )? 12> _Vl 3 ax 3 2 x*< 13 3fa + 2)«-5(g + 2) 3 t x + 1 8(x 2 + 4 x + 3) 2 16 ° x + 3 372 ANSWERS 14. l( x + l)Vl-2x-x 2 + sin-i «±i 2 V2 V^EI-lrin-ll. 18- 17. SILL • / - r- 2 x 2 2 x aVa + bx 2 -1 19. - cos * 2(x 2 + 7)' (1 + e) Vl + esin 2 20. -(2x 2 - a 2 )Va 2 - x 2 + ^siiT^- 8 8 a 21. — (33 a 4 - 26 a 2 x 2 + 8 x*) Va 2 - x 2 + ^- sin-i - • 48 16 a Page 241. Art. 122 1. JLlog^. 3. ^-4,+^logi^I 3 2c x + a 2 T 2 ° x + 1 2. w ( x + 1 ) 2 ■ 4- x + log(x-a)«(x-&)». x(x — 1) 5. 2 _±^§log(x-2-V3)-^^log(x-2+V3). 2V3 2V3 6 li 0(y (^-l)^-2) . 7 . i og («-«)(« - 6 > . 20 & (2x+l)(x + 2) x-c 8. x + ^— [a 2 log(x + a)-& 2 log(x + 6)]. & — a 9. log[(x + 2)V2x-l]. 12. ^-7x + 641og(x + 4) 10 . log (x- a )(, + 6 J, -271og(x + 3). 2 a& ax + & t1 1. x 6 14 I i og ! + ^ . u - 3 l0g (2-T^xr^^ 2 log i-. Page 243. Art. 123 L "2(x-l) + I l0g ^l* 3 " 2(a 2 - x 2 ) ' 2 - ^ + log ^ 4 ' v^Tvii* 5 . ^_2x + ^^ + logx(x4-l) 2 . 2 x 2 + x ANSWERS 373 6. log(x 2 -a 2 )--^ a2 x*-a* 4(V2 + l-a;)2 8. ax-- + log x x + a 9. x +-L-l[28\og(x + S)-\ogxl ox y 10. Aiog^±_^_±. ll. a;__A^_51og(x-3). a 2 x ax x — S Page 244. Art. 124 1. log *— . 2. llog aj, + 1 a + -tan-ig. °V^T4 4 *> + l) 2 ^2 3. — [log(x + a)--log(x*-ax+a 2 ) + V3 tan-* 2 * ~_ a ~| . 3a 2 L 2 aV3 J 4. - 1 tan-* - + - tan-i ? . 7. — tan-i* a aft b 2 a (x — a) 2 a 2 a 6. __Ltan-i-^5— . 9. log x ~ 1 -tan-* a. V3 2x 2 + l V^+T Page 24b. Art. 125 1. tan-i* + _^-. 2. Atan-i«U ?7 2< * x 2 + 1 2 a a 2 (x 2 + a 2 ) 3. llog * 2 + 1 4- z - 1 . 4 & (x + l) 2 2(z 2 +l) 4 -g + Slog^^- 3 + 2 * -Stan-ix. x x 2(.r 2 + l) 5. + log (x 2 + a 2 ) - J_ tan -i? # 2 (x 2 + a 2 ) 5 V J 2 a a 6. — - +ilog(x 2 + l). x*+l 4(x 2 + l) 2 2 &K ' 374 ANSWERS Page 249. Art. 127 1. log Vx + 1 ~i . 2. 2Vx-3?/x + Qfyx-6log(fyx + l). y/x + 1 + 1 3. 2 log ( Vx - 1 + 1) + Vx - 1 + 1 4. 2tan-iVx-2. 5. llog ^ a ^ - & & Vx — a -f 6 6. 14 (xT? - i -xt + J xt? - | x7 + i XT?). Page 252. Art. 128 L »l-xj x-l + Vl-a: 2 2 . _ 21og [-V2 + V^]. Pages 253-254. Exercises on Chapter IV 1. 2 V3 tan-i V^V^ ~ V ^ ~ 1# . 3 l 3 V2(x-a)*-l 2. o(x-a)^ -log *> '- . * 4V2 V2(x-a)^ + l 3. x-4Vx+ 1 + 81og(Vx + l +2). 2 2Vx^T + l 4. log(x+Vx-l) -tan" 1 V3 V3 5. M2x-3a)(a + x)i ? 1 1q „ V^IT^r + sVS 6. 6 log (** - 3xU 5). " 2 ^« 2 ° V * 2 - ^ " * V * 8. i[x 2 -xVx 2 - 1 +log(x+ Vx 2 -1)]. 9. f x^ - f x^ + f x* + 2 x^ - 3 x* - 6 x^ + 3 log (x* + 1) + 6 tan" 1 !*;*. 10. -$(2a* + s 2 )V^V. v gr^ ,, . a, „ Va 2 — x' 2 — a a 2 x 11. Va 2 -x 2 + o l0 S 2 Va 2 -x 2 4-a Va 2 - x' 2 • _, x 14. 12. - — ^_sin-i-. a 2 Vx 2 + a 2 ANSWERS 375 15 x(2x 2 + 3a 2 ) _ lg 1^ Va*+z*+x Va 2 + 3a 4 (x 2 + « 2 )^ 2 ° Va^+x^-x x 17. «Va 2 - x 2 Page 256. Art. 131 5. f esc 3 x — cot x — | cot 3 X. 6. - 64 [cot 4 x + -|- cot 3 4 a] . 3. tan x + | tan 3 a; + A tan 5 x. 1 3 7. — + log tan x. 4. — 128 [cot 2x + cot 3 2 a; 2 tan- a; + |cot 5 2x+ 1 cot 7 2x]. 8. - \ cot 3 x - i cot 5 x. Page 257. Art. 132 1. £ sec 4 £ — \ sec 2 x. 5. \ sec 4 x — sec 2 x + 2. — 4. esc 7 x + | esc 6 x — | esc 3 a:. 6. sec" -1 x sec n-3 x n — 1 w — 3 3. - [ -sec 5 ax — sec 3 ax + sec ax ) . a\5 3 / 7. log sec x. 4. — (sinx + cscx). 8. — log CSC X. Pages 259-260. Art. 133 1. _i C ot 3 x + cotx + x. tan-ix 4. 2. — tan 2 ax log sec ax. w 2a a ° 3. i (tan 2 x + cot 2 x) 5 - i tan? K - I tan5 « + i tan3 « + 4 log (sin x cos x) . — t ' an x + x - Page 260. Art. 134 (a) 1. — COS X + | COS 3 X. _4 2 8 _ . , 4. | cos 3 x + 3 cos 3 x — | cos' x. 2. — |cos 5 x + \ cos' x. 8 3. log sin x - sin 2 x + J sin 4 x. 5. |(1 - cosx) z - |(1 - cosx)*. Page 261. Art. 134 (6) 1. — $cot 3 x. 2. -cotx-fcot 3 x-icot 5 x. 5 - !Vtanx(tanx-3cotx). 3. _ cot 5 x (i + \ cot 2 x). 6. tann " ly + taD " +lx - w — 1 w + 1 4. — I cot 3 x — 2 cot x + tan x. 376 ANSWERS Page 262. Art. 134 (c) 1. | x — ■£% sin 4 x. 2- xis (^ # + t s i n3 2 £ — sin 4 x — \ sin 8 x) . 3. T |s(3x — sin 4 a; + fsin8x). 4. |(3 x + sin 4 a + | sin 8x). 5. tan x + £ sin 2 x — § x. Page 264. Art. 134 (d) 1. |(x — sin x cos x) . 2. ^ cot x(cos 2 x — 3)— §x. 3. — | sin 3 x — sin x + log tan (- + -]. 4. — \ cot 2 x esc 2 x + J log tan x. 5. | x + 2 cot x + \ sin x cos x — £ cot 3 x. Page 265. Art. 135 1. Jtan-i(^tanx). 2. - tan-i f~2 tan (*■ - -\~|. 1 , tanx-2-V3 3. 2 V3 tan x - 2 + V3 x b tan - - o + vV + b 2 4. — — log . v a 2 + 6 2 6 tan ^ - a - Va 2 + 5 2 -1 7. J-tan-if^Y V2 W2 / a (a tan x + &) 6. i-tan-if^i^y 8 ' 2 a& \ b J l i 2 log tan - - 3 2 Pages 266-267. Exercises on Chapter V 3. 2 Vtan x. . 4. J tan 4 x + \ tan 2 x. 5. - cscx + log tan I- • + - j. 6. - tan 2 x sin x + § fsin x - log tan (* + *\~] . 7. ^ + x • • ANSWERS 377 ?, x 11. I e x (sin x + cos a: — f sin3x 8. 62^sin- + cos-j. _; C os3x). 9. — $ e _I (sin 2 a; + 2 cos 2 x). 15 sin^x _ sin"+ 3 x 10. i e 2x( 2 - sin 2 x - cos 2 x). w + X n + 3 16. §tan*x — 2Vcotx. 17. - 32 cot 2 x(l + 1 cot/ 2 2 x + $ cot 4 2 x). 18. J- tan | x(l + | tan 2 \ x + \ tan 4 § x). 19. log tan 2 x. 20. - 8 [cot 2 x + ^ cot 3 2 x] . 21 1 / Va 2 -x 2 \« 22 ^^ + ^6^^^, + ^^ -• i(S + |)- + i(5-|) ta «^ + ^>- 24. x cos (« — &) + sin (a — b) log sin (x + 6). Pages 281-283. Art. 142 2. fp*A 3. 2. 4. 36 V3. 5. 40V5. 6. f. 7. fv^. 8. i. 9. £. 10. Trr 2 . 11. f|. 12. 4 a 2 tan- 1 ^i- ; 4 Tra 2 . 13. *•«&. 14. if* - 72 log 2. 2a' 2 15. *.-<* 16 24-8^ 17 f lg< ^ fl) | log a 5 20. 3 2 OJ * . ^ -|- I To3 ' ' 2' 2(6'- 1) " 22. a 2 logra. 23. 1. 25. -. & 6 26. A. 27. 4a&tan- 1 5. 28. T5 a 2V2. 29. I°£i. 30. * . IzlT" 7T 7T 7T 31. ^(4-tt). 2 V ; Pages 287-289. Art. 143 Ex. 4. 2-. 1. 4. 2. oo. 3. 3. 4. 4 a 2 . 2 5. oo. 6. 3tt« 2 . 7. 2tt. 378 ANSWERS Pages 290-291. Art. 144 1. vab. 2. lira 2 . 3. f. 4 lZ. 5 3y(g2-68)a | 4 8a6 Page 292. Art. 145 2. a\ 7. 2 n. h 3. 4. Ti-r 2 . 3 7rr 2 # 8. « 2 7r3 24 12. !• 5. 2 25 7T. 9. -log 2 * ft ;)• 13. 3tt 8 ' 6. c / 2 (P1 " -/>2). 10. n 14. 7T 5 5 ' Pages 296-297. Art. 148 In the following answers the values are given for Simpson's formula only, unless the trapezoidal formula is called for in the problem. 1. 22 ; 21.5. 3. 0.500014 ; 0.500002 ; 0.5000014 ; 0.5000011. 4. 5.2523. 5. 37.8555; 36.5261. 6. 0.9996. 7. 8.0047. 8. 39.6465. 9. 0.7593. 10. 0.7468. 11. 0.4443. 12. 335. 13. 3.006. 14. 1.1873; 1.1830; 11931. 15. 0.5633. 6. \irab. 7. \irab. 8. \AK. Pages 300-302. Art. 149 a irabp 9. _. 10. 4£ cu. ft. 12. | wa 8 cos 4 6. 13. 1 trabc*. 11. fa 3 . 14. fr 8 . Pages 304-306. Art. 150 1. jrL(io_3 7r). 2. 27rrl(2r 2 +a 2 )Vr 2 -^-ar 2 sin-i^2^1. 6V2 L3 « J 3. 7tt. 4. -7T 3 |«i« + ax{* + 4 eflxt + 8 a 3 ] og (~=^\ ] ANSWERS 379 5. 2 7r 2 a 3 . 6 . i*^. 7. firr 3 . 8. x/fc 3 ; oo. o 9. ^^- 10. Sf. 11. 4 «■*««. 105 15 12. 7rr8a 3 log^-4a 2 (2a-?/i)]; oo. 13. 5*- 2 a 3 . 14. ^~- 5. 7rr8a 3 log^-4a 2 (2a-yi)l; oo. 13. 57r 2 a 3 . 14. Pages 308-309. Art. 151 1. p[V2 4-log(l+ V'2)]. 6la 3 logVS '216' • 5 • 4. 6 a. 5. 2irr. 6. - (e - e" 1 ). 7. }f- 8. 2-V2 + log 1 i^. 9 . 4 (« 3 - &3 > . 10. alogS-^a. V3 aft & * Page 310. Art. 152 1. (p 2 -pi)Va 2 + 1. 2. 2ira. 3. 8 a. 4. a [tan* sec I + log tan (1+^~|* 2 . 5. ^[flV^ + l + log (5 + V^ + l)] \ c (7r 2 + 4)t 8 w A , 2 ,_ •• — f- -3 7. 4+ — log(V 3+2). 9. 2a[~V5-2-v / 31og V5+V5 .T L \/2(2 + V3)J 8 . 3ira a o-f,/5 o ,/5u„ V3 + V5 1 2 Page 311 Art. 153 1. 8a. 2. JJWL. 3. 6 a . 4. la*! 2 . ra — 1 7. ^[(4 + 9ti«)*-8]. Pages 313-315. Art. 154 1. Tra^l-^V 3. f^Cv/S-l). 2.^-2).'' 4 ' f[3V2-log(l + V2)]. 380 ANSWERS 5. (a) 2irb(b + - — cos- 1 - V Va 2 -& 2 a (6 ) 2 ^ + -Jgg- log r « + vtf'-y -|. Va 2 - b 2 La - Va 2 - 6 2 J «• ¥« 2 - 7. («) tiWoH^; 9 ¥»*• 8. 4*-d* (/S) 7ra\/« 2 + & 2 . 10. 4 7r 2 aA:; 2w 2 a 2 k. 11. 2 7ra 2 (3 sin f i — 3 ii cos <i — t\ 2 sin £i). Pages 316-320. Art. 155 2. p = e a . 4. p = — a cos + C. 5. y*= ax 2 + 6. 6. p n = c sin n0. Straight line. Cardioid. 7. y = e ax + c . 8. p: -h). _ ce a 9 ™1 10. 2 7ra 2 . 11. a log -2. 3 */i 13. Va 2 + b 2 (t 2 14. 22.7 lb. 15. 0.9627 lb. 16. 4.4312. 17. h = 5.28 in. 18. h = 43.17 lb. Pages 323-324. Art. 156 3. xy = ay 2 + by — \. 4. y = kx (log x - 1) + ax + b. 5. T 4 3 &«£. 6, y = 0.0002 x 3 + 0.0036z + 1.12, slope = 0.0006 a 2 + 0.0036 ; a =20, y = 2.792, slope = 0.2436. 7. 1000 y= -0.046 x 3 + 0.75 x 2 - 2 05 x + 40; 1000?/= ^ a 8 -0.0575 x 2 -9.7x+117.91§. Page 328. Art. 158 1. xy+C. 3. Impossible. 5. x 3 + y 3 — 3 axy + C. 2. - cos a cos y + C. 4. log-+C. 6. tan-^+C. y y 7. | a 3 + x 2 ?/ + 5 x + £ ?/ 3 - \y 2 + <7. 8. £ x 5 + xy 4 + $ x 3 - xy 2 + | y 2 - \ y* + 2 ?/ + (7. Page 330. Art. 160 1. — =■ 3. J. 5. | a". 7 - «• 2\/2 Z 3 2 2. 4. 4. 1. 6. 6& 3 . 8. J log 2. ANSWERS 381 Pages 332-334. Art. 161 6. &. c Srfl; ■Kd 2 2. 64. 4. 4 7T - 8. 3. 4£. 5. 1. '8 9. a 2 . /•y /» a /"a /*7r /*ir /*o(l — cos 0) 10. 2l"f pdpdtf;2j I _ Vi p .pfZ^p;2l | prfpdfl. JO Ja(l-co8 0) JO Jcos {}-*) JnJa a 2 1L 20. 12. §JH— 2V3. o Pages 336-337. Art. 162 3 3 * 2 ^ Page 340. Art. 163 1. 62251b. 2. 1120.51b. 3. 9337.5 tons. 4. 66.41b. 5. 15.645 tons. Pages 343-345. Art. 164 i. (t,M). 2 (fcfA). 4 . fo,o,^- c V 3. x = 8 (*+ a > g ,y = g = 0. V 8; 4(& + 2a)'^ - 3|"(a + ft) 4 a 4 "! from cen ter of sphere. ' 8|_(a + ft) 3 -a 3 J 3r7(r + «) 4 -6^-| from herical 8L7(r + a) 3 -6^J center. g af + 2 fl W + a^ above tne base . 2(ab + a'b') o 9. from the vertex. 4 a sin T ~3 2(r 2 — a' 2 ^ 10. from center of circle. 3 (V 2 cos- 1 --aVr 2 - a 2 ) » (& £)• 1S (¥• t)- 10 / 2fl 2 6 \ « /3a 36 3c\ 12 [S(^Y) ' 3(7325)- 15 ' U ' 8 ' 8 j 382 . ANSWERS 16 (*u l 6 -^ 16 c\ V15' 157r' 15 tt/ 17 (<L* 8a 3«\. - - = ^ = 3r (« + ft)*-g* -i ' \ 8 ' 8 ' 8 /' 8L(a + ^) 3 -a 3 J 18. x = y = z = — - 5 19. (0, 0, — j, the base of the cone being in xy-pl&ne. 20. (0, 0, \) 21. (a) a cos 2 ; (6) a ( 1 H — ] both measured from the vertex, \ 1 + cos 2 d ) Pages 347-348. Art. 165 1. «^(&2 + c 2). 6. !!^(3« 2 + 4^ 2 ). o 12 2. ^ ira^h. ^ Tra&fr /g &2 i #2) 3. 7 ^(3 a 2 + /l 2) < 12 8. *£; ^[(r + fc).-,*]. 4 ttoH 15 15 10 ' g 8 7IT"+ 5 5. \ Trabh(a 2 + tf) . ' 3(n + 5)' 10. ia 5 . 11. ^ira(2?*-r*). 12. ^a 3 6. 13. \bh*. 14. iira* INDEX (The numbers refer to pages) Absolute value, 134. Absolutely convergent, 134. Acceleration, 77. Actual velocity, 68. Approximate integra- tion, 292. Arc, length of, 306. Area, by double integra- tion, 330. derivative of, 79. formula for, 273, 280. in polar coordinates, 291. in rectangular coordi- nates, 273. Asymptotes, 96. Average curvature, 176. Bending, direction of, 90. Cardioid , area of, 292, 333. Catenary, 180, 316. length of arc, 309. volume of revolution, 312. Catenoid, 313. Cauchy's form of remain- der, 153. Center of curvature, 172. Center of gravity, 340. Change of variable, 124. Circle, area by double integration, 331. of curvature, 172. Cissoid, 180. area of, 289. Component velocity, 70. Concave, 89. toward axis, 94. Conditionally conver- gent, 134. Conditions for contact, 168. Conjugate point, 197. Conoid, 301. Constant, 15. factor, 25, 213. of integration, 214. Contact, 167. of odd and even order, 169. Continuity, 19, 109. Continuous function, 19. Convergence, 132. Convex, 94. to the axis, 94. Critical values, 55. Cubical parabola, 281. Cusp, 194. Cycloid, length of, 311. surface of revolution, 315. Decreasing function, 51. Definite integral, 270. geometric meaning of, 273. multiple integral, 329. Dependent variable, 15. Derivative, 21, 22. of arc, 79. of area, 78. of surface, 81. of volume, 81. Determinate value, 158. Development, 131, 160. Differentials, 74, 210. integration of, 326. total, 112. Differentiation, 23. of elementary forms, 44, 45. Direction of curvature, 175. Discontinuous function, 20. Divergent series, 132. Duhamel's theorem, 348. 383 Ellipse, area of, 281. evolute of, 190, 309. parametric form, 290. Ellipsoid, volume, 299. Envelope, 200. Epicycloid, length of, 311. Equiangular spiral, 316. Evaluation, 160, 165. Evolute, 182. of ellipse, 190, 183, 291. of parabola, 183. Expansion of functions, 131. Exterior rectangles, 269. Family of curves, 200. Formula for integration by parts, 216. Formulas of differentia- tion, 44, 45. of integration, 211, 224. of reduction, 229, 262. Function, 15. Helix, 318. Hyperbolic branches, 95. spiral, area of, 292. Hypocycloid, area of, 290. length of arc of, 308, 311. volume of revolution of, 305. Implicit function, 33. Impossibility of reduc- tion, 232. Increasing function, 51. Increment, 21. Independent variable, 15. Indeterminate form, 157. Infinite, 20. Infinite limits of integra- tion, 287. ordinates, 97. Infinitesimal, 350. 384 INDEX Integral, 209. definite, 270. double, 330. multiple, 329. of sum, 212. triple, 328, 334. Integration, 209. by inspection, 211. by parts, 216. by rationalization, 248. by substitution, 219. formulas of, 211, 224. of rational fractions, 238. of total differential, 326. successive, 321. summation, 268. Interior rectangles, 269. Interval of convergence, 138. Involute, 187. of circle, 311. Lagrange's form of re- mainder, 152. Lemniscate, area of, 334. Length of arc, 306. of e volute, 185. of space curve, 318. polar coordinates, 309. rectangular coordi- nates, 306. Limit, 15. change of, in definite integral, 276. Limits, infinite, for defi- nite integral, 287. Liquid pressure, 338. Logarithm, derivative of, 34. Logarithmic curve, 280. spiral, length of arc, 310. Maclaurin's series, 141, 276. Maximum, 53. Mean value theorem, 74, 275. Measure of curvature, 177. Minimum, 53. Moment of inertia, 346. Multiple points, 193. Natural logarithms, 36. Normal, 85. Notation for rates, 72. Oblique asymptotes, 99. Order of contact, 167. of differentiation, 122. of infinitesimal, 350. Osculating circle, 172. Osgood, 132. Parabola, 108, 89, 171. semi-cubical, 308. Parabolic branches, 95. Paraboloid, 314. Parallel curves, 187. Parameter, 201. Partial derivative, 110. Point of inflexion, 90. Polar coordinates, 104. subnormal, 106. subtangent, 106. Problem of differential calculus, 21. of integral calculus, 209. Radius of curvature, 172. Rates, 68. Rational fractions, inte- gration of, 238. Rationalization, 248, 249. Rectangles, exterior and interior, 269. Reduction, cases of im- possibility of, 236. formulas, 229, 262. Remainder, 150. Rolle's theorem, 150. Simpson's rule, 294. Singular point, 191. Slope, 16. Solid of revolution, 81. Sphere, volume' by triple integration, 336,. Spheroid, oblate, 305, 314. prolate, 314. Spiral, of Archimedes, 107. equiangular, 108 s 316, 310. hyperbolic, 292. logarithmic, 292. Standard forms, 211, 224. Stationary tangent, 90. Steps in differentiation, 22. Stirling, 141. Subnormal, 86. Subtangent, 86. Summation, 268. Surface of revolution, 81. area of, 312. Tacnode, 194. Tangent, 85. Tangent plane, 119. Taylor, 141. Taylor's series, 148. Tests for convergence, 133. Total curvature, 176. differential, 112. Tractrix, 315. length of, 318. surface of revolution of, 318. volume of revolution of, 318. Transcendental func- tions, 34. Trapezoidal rule, 292. Trigonometric functions, integration of, 255. Variable, 15. Volume of solid* of revo- lution, 302. Volumes by triple inte- gration, 334. Witch, area of, 286. volume of revolution of, 305. \ KETOKN TO DESK FROM i-l^&BOKKOWBO # ; m LOAN DEPT. ESS * MAR 27 '69-^ (J6057S101476-A-32 «&8BF- % * b UNIVERSITY OF CALIFORNIA LIBRARY * | f ^ s| * v * *