DIFFERENTIAL AND INTEGRAL CALCULUS NICHOLSON 'cs I 23 1626e A 57804 2 UNIA PUB.CO ARTES LIBRARY 1837 SCIENTIA VERITAS OF THE UNIVERSITY OF MICHIGAN REMURIBUS UNUM TUEBOR SI QUÆRIS-PENINSULAM·AMINAM. CIRCUMSPICE FROM THE LIBRARY OF PROFESSOR W.W.BEMAN AB,1870: AM, 1873 TEACHER OF MATHEMATICS 1871-1922 7 ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS With Examples and Practical Applications mes iliam BY Ja W. NICHOLSON, A.M., LL.D. = . President and Professor of Mathematics Louisiana State University and Agricultural and Mechanical College NEW YORK AND NEW ORLEANS UNIVERSITY PUBLISHING COMPANY 1896 + Copyright, 1896, BY UNIVERSITY PUBLISHING CO. All Rights Reserved * * 1673 [ Giad.3 Prq. W. W. Beman 10-13-1923 PREFACE. IN many respects this work is quite different from any other on the same subject, though in preparing it there has been no attempt at originality beyond presenting the principles in a more tangible form than usual, and thus securing a better text-book for the ordinary student of mathematics. The aim has been to prepare a work for beginners, and at the same time to make it sufficiently comprehensive for the requirements of the usual undergraduate course. The chief distinction of the treatise is that it is based on the conception of Proportional Variations. This method has been employed as the most elementary and practical, and none the less rigorous or general, form of presenting the first principles of the subject (see the following Note). Differentiation and Integration are carried on together, and the early introduction of practical applications both of the dif- ferential and integral calculus, which this mode of presenting the subject permits, is intended to serve an important purpose in illustrating the utility and potentiality of the science, and arous- ing the interest of the student. The formulas for differentiating and integrating are, as a rule, expressed in terms of u and v instead of x, u and v being functions of x. The advantages thus secured are obvious. Among the additional features of special interest may be mentioned the following: (1) The treatment of da as a variable independent of x (Art. 68, and Appendix, A,); (2) a rigorous deduction of a simple test of absolute convergency, without recourse to the remainder in Taylor's formula (Arts. 115 to 119); (3) an extension of the ordinary rules for finding maxima and minima (Arts. 140 to 143); (4) a chapter on Independent Integration (Chap. IX); (5) integration by indeterminate co- iii + iv PREFACE. efficients (Arts. 211 to 216); (6) the introduction of turns in curve-tracing (Arts. 175 to 179); and (7) a new proof of Taylor's formula, which is believed to be as rigorous as, and less artificial than, those in general use (Appendix, A). In preparing the book the best available authors have been consulted, and many of the examples have been taken from the works of Todhunter, Williamson, Courtenay, Byerly, Rice and Johnson, Taylor, Osborne, Loomis, and Bowser. I improve this opportunity to tender my thanks to Prof. William Hoover of the Ohio University, Prof. Alfred Hume of the University of Mississippi, and Prof. O. D. Smith of the Polytechnic Institute of Alabama for valuable assistance in the reading of proofs. Their corrections and suggestions have relieved the treatise from various imperfections it would other- wise have contained. Further acknowledgments of indebtedness are also due to my colleague, Prof. C. Alphonso Smith, of the Department of English, who has aided me with his scholarly criticisms. BATON ROUGE, LA., 1896. JAMES W. NICHOLSON. NOTE.-The method of Proportional Variations, which is the suggestion and outgrowth of work in the class-room, is believed to possess the follow- ing merits: (a) The conception is one with which the student is already familiar, for the principle of proportional changes is among the first that he encoun- ters, even in the lower mathematics. (b) It affords finite differentials, and, without introducing infinitesimals, or infinitely small quantities, or the foreign element of time," has all the advantages of the differential notation. (c) In many cases the proportional variations (or differential) can be detected by inspection (see Arts. 31, 32, 35), and in all cases they may be deduced by the theory of limits. Hence the method has all the lucidity of finite differences and all the rigor of the doctrine of limits. (d) It is a method to which the doctrines of Infinitesimals and Rate of Change are easy corollaries. (e) In general, the form and properties of the increments of all quan- tities are due to proportion and acceleration, or to proportional and dispro- portional changes; hence, a system of Calculus based on such changes adapts itself naturally to questions in Geometry, Mechanics, and Physics. CONTENTS. CHAPTER I. FUNDAMENTAL PRINCIPLES. [Pages 1 to 20.] Quantity, Variables and constants, Art. 1.* Dependent and indepen- dent variables, 2. Functions, 3. Increasing and decreasing functions, 4. Explicit and implicit functions, 5. Algebraic and Transcendental func- tions, 6. Continuous functions, 7. Notation of functions, 8. Examples. Increments. Increments of independent variable and function, nota. tion, and illustrations, 9. General formula, 10. Examples. Variation. Proportional variation, 11. Principles, 12 to 15. Dispro. portional variation, 16. Principles, 17 to 19. Composition of increments, illustrations, 20. Theory of Limits. Definition and illustrations of limits, 21. Prin- ciples, 22, 23. Proof of the formula, Дy = m₁h + m,h², and geometrical illustration, 24. Differentials and Accelerations. Definition and notation of differ- ential and acceleration, 25. Corollaries. Derivatives, 26. How the pro- portional variations or differential of y = f(x) may be found, 27 to 29. Differentials of Geometric Functions. Differentiation, 30. Plane areas in rectangular co-ordinates, 31. Solids of revolution, 32. Arcs of curves in rectangular co-ordinates, 33. Surfaces of revolution, 34. Plane areas in polar co-ordinates, 35. CHAPTER II. ELEMENTARY DIFFERENTIATION AND -INTEGRATION. [Pages 21 to 42.] Differentiation. Rules for differentiating, 36. Differential of c or C, 37. Differential of cv, 38. Differential of vy, 39. Differential of vyz, 40. * Similarly, the following numbers refer to articles and not to rages. V vi CONTENTS. V Differential of 41; also of Y Differential of v+y-2, 45. с Y 42. Differential of v", 43; also of √o, 44. Examples. Slope of Curves. Direction and slope of a line, 46. Direction of a curve at any point, 47. Slope of a curve, 48. tan ø, sin ò̟, cos ☀, where is the angle of direction of a curve, 49. Examples. Integration. Definition and sign, 50. Integral of 0, 52. Integral of cdv, 53. Sordo == Sv= n dv= C Dependent integration, 51. Integral of v"dv, 54, 55. vrcdv, 56. Integral of (a + bx²)²x²-1dx, 57. Integral of du + dv - dz, 58. Examples. Problems. The problem of integration, 59. Definite value of the con- stant C, 60. Examples. Applications to geometry, 61. Areas of curves, 62. Definite integrals, 63. Length of curves, 64. Areas of surfaces of revolution, 65. Volumes of sòlids of revolution, 66. CHAPTER III, SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. [Pages 43 to 54.] Successive Differentials. Definition and Notation, 67. Why do may be treated as a constant in successive differentiation, 68. Examples. Liebnitz's theorem, 69. Rate of Change. Uniform change, 70. Variable change, 71. Ex- amples. Applications to geometry, 72. Given the rate of change of a curve to find the rates of change of its co ordinates, 73. Application to Mechanics. Velocity, 74. Examples. Positive and negative velocity, 75. Examples. Uniformly accelerated motion, 76. Formulas for the free fall of bodies in vacuo, 77. Problems in Mechanics. CHAPTER IV. GENERAL DIFFERENTIATION. [Pages 55 to 83.] Logarithms. LEMMA, The limit of (1+ (1 + 1 ) Differential of loga v, 79. Examples. Exponential Functions. Differential of y', 81. Examples. Trigonometric Functions. ferential of sin v and cos v, 83. of tan v, 85. d(cot v), 86. Examples. 2 2 > as zapproaches ∞, 78. The differential of av, 80. Examples. Circular measure of angles, 82. Dif- Sin dv = dv, cos dv Sin dv = dv, cos dv = 1, 84. Differential d(sec v), 87. d(cosec v), 88. d(vers v), 89. . CONTENTS. vii Inverse Trigonometric Functions. d(sin-1 y), 90. d(cos-1 y), 91. d(tan-1 y), 92. d(cot-¹ y), 93. d(sec-1 y), 94. d(cosec-1 y), 95. d(vers-1y), 96. Examples. Differential of an arc in polar co-ordinates, 97. tan 4, sin, cos, where is the angle formed by the tangent and radius vector, 98. Functions of two or more independent variables. How such functions may vary, 99. A partial differential, 100. Total differential, 101, 102. A partial derivative, 103. The total derivative, 104. Examples. Function of Functions, 105. Examples. Successive partial differentials d2u d2u and derivatives, 106. 107. Examples. To find the suc- dx dy dy dx' cessive differentials of a function of two independent variables, 108. plicit functions, 109. Examples. Successive derivatives of an implicit function, 110. Examples. Change of the independent variable, 111. To dy dx Im- find the successive derivatives of when neither x nor y is independent, 112. Examples. Miscellaneous Examples. CHAPTER V. SERIES, DEVELOPMENT OF FUNCTIONS, AND INDETERMINATE FORMS. [Pages 84 to 105.] Series. Definition, 113. To develop a function, 114. To develop a function, 114. An absolutely convergent series, 115. Test of absolutely convergent series, 116. Coral- laries, 117 to 119. Examples. Development of Functions. Two formulas, 120. Taylor's formula, 121, 122. Binomial theorem, 123. Developments of sin (y + x), sin x, cos v, cos (y + x), 124. Development of log (y + x), 125. Maclaurin's formula, 126. Developments of ax, ex, e, 127. Development of tan-1 x, 128. Examples. To find the value of 7, 129. To compute natural loga- rithms, 130. To compute common logarithms, 131. Q Indeterminate Forms. How the form arises, 132. To evaluate 0 0 functions of the form 133. Examples. Of the form 134. Examples. 0' Of the forms 0 X and ∞ > and 1 136. Examples. ∞, 135. Examples. Of the forms 0°, °, Implicit functions, 137. Examples. CHAPTER VI. MAXIMA AND MINIMA. [Pages 106 to 121.] Definitions and Principles, Definitions and illustrations, 138. If f(a') is a max. or min. then f'(a') = 0 or ∞, 139. f(a') is neither a max. + viii CONTENTS. nor a min., if an even number of the roots of ƒ'(x) = 0 and f(x) = ∞ are equal to a', 140. f(a') is a max. or min. if an odd number of the roots of f(x) = 0 and f(x) = ∞ are equal to a', 141. Max. and min. occur alternately, 142. Rules for Finding Maxima and Minima. When all the roots of f(x) = 0 and ∞ are known or can be conveniently found, 143. I. By sub- stituting a' h and a'+h for x, 144. II. By Taylor's formula, 145. Self- evident principles which serve to facilitate the solution of problems, 146. Examples and problems. - Functions of two Independent Variables. Definition, 147. Con- ditions for maxima and minima, 148. Examples. CHAPTER VII. APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO PLANE CURVES. [Pages 122 to 159.] Tangents, Normals, and Asymptotes. Equations of the tangent and normal, 149. Lengths of tangent, normal, subtangent, and subnormal, 150. Examples. Lengths of tangent, normal, subtangent, and subnormal in polar co-ordinates, 151. Examples. Asymptote, 152. General equa- tion of an asymptote, 153. Relation of y to x when they are infinite, 154. Examples. Asymptotes determined by inspection, 155. Examples. Curvature. Total curvature, 156. Uniform curvature, 157. Variable curvature, 158. Radius of curvature, 159. Examples. Radius of curva- ture in polar co-ordinates, 160. Examples. Contact of Different Orders. Definitions, 161. When two curves cross or do not cross at their point of contact, 162. Examples. Osculating curves, 163. Osculating straight line, 164. Osculating circle, 165. 167. Involutes and Evolutes. Definitions, 166. Elementary Principles, To find the equation of the evolute of a given curve, 168. Examples. Envelopes. Definition, 169. The envelope is tangent to every curve of the series, 170. To find the equation of the envelope of a given series of curves, 171. Examples. Tracing Curves. The general form of a curve, etc., 172. Direction of curvature, convex and concave arcs, 173. Point of inflection and prin- ciples, 174. Examples. Singular Points. Definition (etc.). x-turns and y-turns, and multiple points, 175. To determine the positions of the singular points of curves, 176. To determine the character of the multiple points of curves, 177. Examples. Tracing polar curves, 178. Examples. The character of multiple points often more easily determined by changing to polar co ordi- nates, 179. Examples. CONTENTS. ix J CHAPTER VIII. GENERAL DEPENDENT INTEGRATION. [Pages 160 to 190.] Fundamental Formulas. Twenty-two formulas, 180. Reduction and Integration of Differentials. Reduction of Differ- entials, definition, and how effected, 181. By constant multipliers, 182. Examples. Reduction of Differentials by Decomposition. How effected, 183. Elementary differentials, 184. Examples. Trigonometric differentials, 185. Examples. How trigonometric differentials may often be more conveniently integrated, 186. Rational fractions, 187. When the simple factors of the denominator are real and unequal, 188. Examples. When some of the simple factors of the denominator are real and equal, 189. Examples. When some of the factors of the denominator are imaginary and unequal, 190. Examples. When some of the simple factors of the denominator are imaginary and equal, 191. Example. Reduction and Integration by Substitution. Irrational differen- tials, 192. Examples. When a + bx is the only part having a fractional aº is the exponent, 193. Examples. When a + bx + x² or √ a + bx only surd involved, 194. Examples. Binomial differentials, 195. Con- ditions of integrability, 196. Examples. Integration by Parts. Reduction Formulas. Fundamental formula, 197. Examples. Definition, 198. Reduction formula fc: xº log² xdx, 199. Examples. Reduction formula for a¤xdx, 200. Examples. Reduction formulas for x cos ax dx and xn sin ax dx, 201. Reduction formula for X sin-1 xdx, 202. Examples. dx Examples. dx dx 203. Integral of and 204. sin x COS X Integral of a + b cos x Approximate Integration. Last resort in separating a differential into its integrable parts, 205. Examples. Development of functions by exact and approximate integration, 206. Examples. CHAPTER IX. INTEGRATION CONTINUED. [Pages 191 to 210.] Independent Integration. Increments deduced from differentials, 207. Examples. Increments as definite integrals, 208. Examples. A more convenient series, 209. Examples. Bernouilli's series, 210. Integration by Indeterminate Coefficients. Explanation and formu- la, 211. When k = 0 or the integration is independent, 212. Illustra- tions and examples. Application to fsinm x cos x dx, 213. When ✯ is not X CONTENTS. • ÷ ? = 0 or when the integration is partly dependent, 214. Illustrations and examples. Reduction formulas for binomial differentials, 215. Approx- imate integration and the elliptic differential, 216. CHAPTER X. INTEGRATION AS A SUMMATION OF ELEMENTS. [Pages 211 to 246.] Elements of Functions. Differentials may be as small as we please, 217. Elements, 218. Signification of a definite integral as a sum, 219. Illustrative examples. When do is not a constant, 220. Signification of a definite integral as the limit of a sum, 221. Illustrative example. Inte- gration equivalent to two distinct operations, 222. Application to Geometry. Length of curves, rectangular co-ordi- nates, 223. Examples. Polar co-ordinates, 224. Examples. To find the equation of a curve when its length is given, 225. Areas of curves, rect- angular co-ordinates, 226. Examples. Generatrix of area, 227. Polar co-ordinates, 228. Examples. Areas of surfaces of revolution, 229. Ex- amples. Volumes of solids of revolution, 230. Successive Integration. A double integral, 231. Definite double integrals, 232. A triple integral, 233. Examples. Areas of Surfaces. Plane surfaces, rectangular co-ordinates, 234. Polar co-ordinates, 235. Examples. Surfaces in general, 236. Examples. Volumes of Solids Determined by Triple Integration. Formula, 237. Examples. Application to Mechanics. WORK, how computed, 238. Example. Centre of Gravity. Centre of gravity, moments, etc., 239. Centre of gravity of a plane area, 240. Centre of gravity of a plane curve, 241. Centres of gravity of solids and surfaces of revolution, 242. Examples APPENDIX. [Pages 247 to 256.] Differentiable functions, A,. Another illustration of the formula ▲y = m¸h+m,h², A₂. The differential of an independent variable is, in general, a variable, A.. Another method of finding the differentials of av and loga v, A4. A rigorous proof of Taylor's formula, A5. Completion of Maclaurin's formula. A.. The values of m, and m₂ in the formula Дy = m₁h + m₂l², A7. LIMITED COURSES. (a) The first three chapters. This course is complete as far as it goes, since differentiation and integration are carried on together. It embraces the notation, fundamental principles, and some of the most important applications of the Calculus. The student who understands elementary algebra and geometry, and the construction of elementary loci, should find but little dif- ficulty in mastering it. (b) The first six chapters. This adds to the former course the transcendental functions, development of functions, evalua- tion of the indeterminate forms, and maxima and minima. Suggestions. (1) It is recommended to omit the more difficult examples and problems in passing over the book the first time. (2) A, of the Appendix may be substituted for Arts. 116 to 122, at the discretion of the teacher. xi DIFFERENTIAL AND INTEGRAL CALCULUS. . : CHAPTER I. FUNDAMENTAL PRINCIPLES. QUANTITY. 1. THERE are two kinds of quantities employed in Calculus, variables and constants. Variables are quantities whose values are to be considered as changing or changeable. They are usually represented by the final letters of the alphabet. Constants are quantities whose values are not to be consid- ered as changing or changeable. They are usually represented by the first letters of the alphabet. Particular values of vari- ables are constants. 2. Dependent and Independent Variables. A Depend- ent Variable is one that depends upon another variable for its value, and an Independent Variable is one that does not depend on another variable, but one to which any arbitrary value or change of value may be assigned. In the elementary differential and integral calculus, the independent variable is usually restricted to real values. Thus, in u= x² - 7x + 5, v = (1 − x²), y = log (1 + x), u, v and y are dependent variables, since they depend on the 1 i ; 2 DIFFERENTIAL AND INTEGRAL CALCULUS. variable x for their values; but x is an independent variable, since, as we may suppose, any value may be assigned to it without reference to any other variable. 3. Functions. Dependent variables are usually called func- tions of the variables on which they depend. Hence, one vari- able is a function of another when the first depends upon the second for its value, or when the two are so related that changes in the value of the latter produce changes in the value of the former. Thus, the area of a varying square is a function of its side; the cost of cloth is a function of the quality and quantity; the space described by a falling body is a function of the time; every mathematical expression depending on x for its value, as x², (3x-7)³, 5x²-6x+11, etc., is a function of x. 2 4. Increasing and Decreasing Functions. An Increasing Function is one that increases when the variable increases, as (x+1)², 3x³, log (5 + x); and a Decreasing Function is one 5 that decreases when the variable increases, as 10-22, etc. x² A function of x may be increasing for certain values of x, and decreasing for other values. Thus, y a 4x+5 is a decreasing function for all values = of x 2, but increasing for all values of x > 2. X 5. Explicit and Implicit Functions. An Explicit Function is one whose value is directly expressed in terms of the variable and constants. Thus, in the equations y = (ax)³, y = x² + 3x + 5, y is an explicit function of x. An Implicit Function is one whose value is implied in an equation, but not expressed directly in terms of the variable and constants. Thus, in the equation a² + 2xy +5y= 10, y is an implicit function of x, and x is an implicit function of y. By solving the equation for x or y, the function becomes explicit. 1 FUNDAMENTAL PRINCIPLES. 3 6. Algebraic and Transcendental Functions. One variable is called an Algebraic Function of another when the two are connected by an algebraic equation; that is, an equation which contains a finite number of terms involving only constant integral powers of the variables, or an equation which admits of being reduced to this form. + - Thus, in y = x² - 5x, or x²y² — xy³ — 8xy — 50, or y Vax² + y = 7, y is an algebraic function of x, and vice — versa. If two variables are connected by an equation which is not algebraic, each is called a Transcendental Function of the other. Thus, if y sin x, y is a transcendental function of x, and x of y. The following are the elementary transcendental functions: A Logarithmic Function is one that involves the logarithm of a variable; as, log x, log (a + y). An Exponential Function is one in which the variable enters as an exponent; as, a*, y*. A Trigonometric Function is the sine, cosine, tangent, etc., of a variable angle; as, sin x, cos y. -1 An Inverse-Trigonometric Function is an angle whose sine, cosine, tangent, etc., is a variable; as, sin-¹x, cos˜¹ y, tan-1 t, etc., which are read, an angle whose sine is x,' angle whose cosine is y," etc. 99 66 an 7. Continuous Functions. A function of a variable is con- tinuous between certain values of the variable (1) when it has a finite value for every value of the variable, and (2) when the changes in its value corresponding to indefinitely small changes in the value of the variable are themselves indefinitely small. Thus, in y = ax + b, or y = sin x, or y = e*, y is continu- ous for all finite real values of x; so also in y = √ a² — x², x², but as a real quantity only for real values of x> a and BCDP and < BCP'Q; that is, Au y▲x<▲u< (y +▲y)4x; :. y< V Since u = we have uy = v. У Differentiating, udy + ydu = dv. (Art. 39) บ dv dy dv udy У Whence, du = y Y ydv - vdy y² 2 .. RULE.—Multiply the denominator by the differential of the numerator and from the product subtract the numerator multiplied by the differential of the denominator, and divide the result by the square of the denominator. 42. COR. I. The differential of a fraction whose numerator is a constant is minus the numerator into the differential of the denominator divided by the square of the denominator. For, Art. 41, 41, α(2) = C yoo – cây y² cdy y² since dc 0. 2 iy 43. To differentiate a variable having a constant ex- ponent. Let u = v², where v is any function of x, and n is any con- stant integer or fraction. I. When the exponent is a positive integer. Since vn = v.v. v.v.v... to n factors, d(v") = v²¬¹dv+v-¹dv. . . to n terms (Art. 40) = nvn-¹dv. 24 DIFFERENTIAL AND INTEGRAL CALCULUS. II. When the exponent is a positive fraction. α Let u v², then Differentiating, > Substituting for u, Reducing, гис = гесь. cuc-¹du — ava-¹dv. CV (21) 0-1 Cv α a du — ava-¹dv. du = ava-1dv. . Dividing by co α a а α -1 du= dv. C III. When the exponent is negative. Let u = v-m, then 26 === 1 ут Differentiating, Art. 42, du = Mvm-1 v2m dv mv-m-¹dv. Hence, for all the cases, we have this RULE.-Take the product of the exponent, the variable with its exponent diminished by 1, and the differential of the variable. -8 Thus: d(x) = 5x+dx, d(u') = — 7u¯³du, d(v³) = {v*dv, d(z¯³) — — fz¯³dz, d(y³) = {y˜³dy dy 3 3 Vy² The same rule holds when the exponent n is irrational or imaginary. 44. COR. I. The differential of the square root of a variable is the differential of the variable divided by twice the radical. * ELEMENTARY DIFFERENTIATION AND INTEGRATION. 25 For, d( √/v) = d(v*) = {v¯‡dv dv = 2 No 45. To differentiate the algebraic sum of several variables. Let u = v + y − %, where v, y, and z are functions of x. (1) Let / represent any variable increment of x estimated from any particular value of x, and du, dv, dy, and dz the correspond- ing proportional increments of u, v, y, and z; then (2)(1), u + du = v + dv + y + dy = du dvdy - dz. (z+dz). .. RULE.-Take the algebraic sum of their differentials. EXAMPLES. Differentiate the following: 1. y = x² + 5x² - 3x + 7. dy = d(x³) + d(5x) — d(3x)+d(7) = d(x³) +5d(x²) — 3d(x) + d(7) = 3x²dx + 10xdx 3dx +0 = (3x²+10x3)dr, Ans. 2 y = x + 5x +3. 3. Y = x* 4x³ 3 3x². 3x5 x²-9. 4. Y 5. y=x+2x-7x+10. 6. y = x² + x−². 7. y = x−³ + x³. 8. y = Y બન 2:3 ax-¹ + b. 9. y = 6x* — 4x + 2x*. 10. y = 3x 11. y = axm 6x+5. bx-n. 12. y = √3x + 4√%. (2) (Art. 45) (Art. 38) (Arts. 37, 43) dy = (2x+5)dx. dy (4x³- 12x²-6x)dx. = 4 dy =(15x etc.)dx. = dy (6x+10.x + etc.)dx. dy = (2x (2x-2x-3)dx. 4 dy − (− 3x−¹ + 3x¯¹)dx. = dy = (§x³ + 4ax−5)dx. -3 dy = (21x¹³ — etc.)dx. dy = (− x ¯¹ + 4x¯*)dx. dy = (maxm-1+bnx-n-1)dx. 3dx 2dz dy = + 2√3x Nz 26 DIFFERENTIAL AND INTEGRAL CALCULUS. : $ * 1 13. y = √x +2. (= x²+x¯¹.) dy = (fx¯³ — x¯*)dx. X a b 14. y = x + x²² 18 ياه (= ax¯'+bx².) dy= dy=( a 2a 15. y²= 4ax. dy y dx. IC (- 24 - 20 ) dx. 26 X 16. y²+x²= R². dy dx. Y b²x 17. a³y² + b²x² = a²b³. dy= dx. a²y Find the following: 18. d(x+1)(x²+2x). = (x² + 1)d(x² + 2x) + (x² + 2x)d(x² + 1) (Art. 39) (x² + 1)(2x + 2)dx + (x² + 2x)2xdx (2x³ + 2x² + 2x + 2)dx + (2x² + 4x²)dx = (4x³ + 6x² + 2x2)dx, Ans. 19. d(x1)(x² + x + 1). 20. d(ax¹y³). 21. d(x)(y²+1). 23. d(6.x¹y¹). 22. d(x² - 1)(x* + 1). 24. d[x-*(1+x-³)]. 25. d(1+x)(x + x²)x. 3x²dx. 4ax³y'dx + 3axʻy❜dy. (y² + 1)dx + 2xydy. (6x-4x+2x)dx. 3x¯*y³dx + 2x*y¯³dy. (-2x-³ — 5x )dx. 3 · 26. d(1+ 2x²)(1 + 4x³). 3 27. d(x + 1)(x³ — x² + x − 1). 4x³dx. 28. d(x³ + a)(3x² + b). 29. d(12x³y¹+15abc). Differentiate the following: 1 + x (15x+3bx²+6ax)dx. 18x*y*dx + 16x*y*dy. 30. u = 1 X ELEMENTARY DIFFERENTIATION AND INTEGRATION. 27 (1 − x)d(1 + x) − (1 + x)d(1. — x) du = 2dx (1 − x)² › α X || Ans. (1 - x)" (Art. 41) adx 32. u = 31. u = 2 1 + x 1 + x²° 3 2. du x² du = (12xx³)dx (1 + x*) * 3x²dx 33. u = du = (1 + x) ³ · 3 (1 + x) * * 4 хо x² 2xdx 34. u = du x² - 1 x — 1 (* _1}* XC³ 35. u = a² X²® 2x² 3 8x²+6x+12 36. u = du = dx. 4x + x²° 3 (4x + x²)² 37. u = (a + bx)³° x³ 38. u = 3x² + x³ du (1 + x)²° (1 + x)³dx. 39. u = √x² 3x + 5. d(x² - 3x+5) (2x-3)dx du = (Art. 44) 2√x² - 3x + 5 2√x2 3x + 5 3xdx 40. u = √3x² - 5. du √3x² 5 2 + 9x 41. u = √4x + √9x³. du = dx. 2 √x 2+3x 42. u = x√1+x• du dx. 2√1+x X dx 43. u = √1 — x² 1 + x 44. u = 1 − x du = du = (1 − x²) (1 − x)√1 − x² dx 28 DIFFERENTIAL AND INTEGRAL CALCULUS. 45. u = (x³-3x² + 4x — 5)°. 2 du=5(x³-3x² + 4x-5)*d(x³-3x² + 4x-5) (Art. 43) = 5(x³ — 3x² + 4x — 5)*(3x² — 6x + 4)dx. 46. u = (x³- 7x+5)³. 47. u = (x² + 5x − 9)*. 48. u = +5 – (2√x+3)-³. du=3(x³—7x+5)² (3x²—7)dx. du=}(x²+5x—9)³ (2x+5)dx. 3dx √x(2√x+3)** du = m X mxm-1dx 49. u = du = X 50. u = (a + x)√ a − x. du = : 3x³ + 2 dy 51. y = x(x² + 1)* dx (1 − x)' (a - 3x)dx 2√a α X 2 x² (x³ +1)*° m +1 √(x + a)³ dy (x − 2a)√x + a 52. y = √x dx ɑ (x − a)* SLOPE OF CURVES. 46. Direction. The direction of a straight line is deter- mined by its angle of direction, which is the angle formed by the axis of x and the line. The Slope of a line is the tangent of its angle of direction. Thus, in Fig. 4, the angle of direction of the line TP is Dt dy XTP, and the slope of the line is tan XTP = PD dx 47. The direction of a curve at any point is the same as that of a tangent to the curve at that point. For, at the point P (Fig. 4), the deviation of the secant SPP' from the curve PP', arising from the former's cutting the latter, diminishes indefinitely as P' approaches P; and since the tangent TPt, which touches the curve at P, is the limiting position of the secant, the tangent has the same direc- tion as the curve at the point P or (x, y). ELEMENTARY DIFFERENTIATION AND INTEGRATION. 29 48. COR. I. The slope of a curve at any point is the slope of its tangent at that point. Therefore the slope of a curve at dy the point (x, y) is The differential of the arc of a curve at dx any point is a straight line laid off on the tangent to the curve at that point, Art. 33. 49. The angle of direction of a curve at the point (x, y) is usually denoted by ; that is, = XTP, Fig. 4. dy dy dx ... tan sin and cos dx ds' ds' where ds = the differential of the arc of the curve, Art. 33. EXAMPLES. x³ 1. The equation of a curve is y = 2 - 2x; find the slope of the curve at the point (x,y). Differentiating the equation and dividing by dx, we have dy dx = 3x² 3x² — 2, Ans. 2. In the same curve find the slope at the point where x = 1. dy = 1. dx 3. Find the slope of the parabola y² = 9x at the point where x = 4. dy dx 4 3 4. In the same parabola find the point where the curve makes an angle of 45° with the axis of x. Since tan 45° 1, we make dx dy 9 2y 1; hence 2y = 9, or y = 41, and x = y² 9 9 4 24. 5. In the circle y²+x= R', find the point where the slope of the curve is Where y #R, and x = R. 30 DIFFERENTIAL AND INTEGRAL CALCULUS. г 6. In the same circle find the point where the curve is par- allel with the line whose equation is 5y + 12x = 60. Where x = R. 1 7. At what angles does the parabola y = 6x cut the circle. y² + x² = 16? Find their slopes at their points of intersection; then find the angles between the lines having these slopes. Thus: solving the two equations, we find (for one of the points of intersection of the two curves) x 2 and y = 23. The slope of the parabola at this point is (= √3), and that of the circle x = 3 y is — ~ (= −√3). Therefore the tangent of the required Y angle is √ √ 3 + √ √ 3 Lako 1 − 1 √ 3 × √3 1 √3 11 100 LO 5 1/3 = 2.88 +. 3 8. At what angles does the line 3y 2x 80 cut the parabola y² = 8x ? — Tan-1 .2 and tan-1 .125. 9. The equation of a curve is y = x³ 9x² + 24x - 11. (1) Find the slope of the curve at the point where x = 3. (2) Find the values of x at the points where the slope of the curve is 45. (3) Find the values of x at the points where the curve is parallel with the axis of x. (1) — 3; (2) x = 7 and — 1; (3) ≈ = 2 and 4. 10. Find the point where the curve y=x7x+3 is par- allel to the line y = 5x + 2. Where x = 6. 11. Find the point where the parabola y = 4ax is parallel 2 Where x R – 2a. = with the circle y² = 2Rx x². 12. Show that the ellipse x² y² + 18 8 = 1 intersects the hyper- bola 2² = y²+ 5 at right angles. 13. At what angles does the circle x² + y² = 8ax intersect 2:3 the cissoid y' ? 2a X At the origin, 90°; at the other two points, 45°. ELEMENTARY DIFFERENTIATION AND INTEGRATION. 31 . INTEGRATION. 50. Integration is the inverse of differentiation. Thus, while differentiation is the process of deriving the differential of a function from the function, integration is the process of de- riving the function from its differential. The function is called the integral of the differential. Thus, the differential of a being 3x dx, x is the integral of 3x dx. The Sign of integration is f; thus 3.x'dz indicates the operation of integrating 3x'dæ, therefore √3a²dx = x", which is read “ the integral of 3æªdæ is x." The two signs d and Sannul each other;* Sd(x²) = 2ª and dƒ (3x°dx) = 3x°dx. 51. Dependent Integration. When the process of inte- grating depends on reversing the corresponding process of dif ferentiating, as is usually the case, it is called dependent inte- gration. This process will now be employed in establishing rules for integrating elementary differentials. 52. To integrate 0. Since dC = 0, Art. 37, we have fo= C. Therefore, as 0 may be added to any differential without changing its value, the general form of its integral will not be complete without an indeterminate constant term. This con- stant term, as will be seen, may be eliminated, or determined from the data of any particular problem. 53. To integrate the product of a differential and a constant. Since d(cv) = cdv, Art. 38, we have fcdv = c f dv = cv + C. * To be exact, dƒƒ(x)dx = f(x)dx, but ſåƒ(x) = f(x) + C. 32 DIFFERENTIAL AND INTEGRAL CALCULUS. • .. RULE.-Integrate the differential, and multiply the result by the constant. 54. To integrate v"dv, where n has any positive or nega- tive, integral or fractional value except Since d vn+1 \n + 1 = vªdv, Art. 43, we have S vrdv = vn+1 n + 1 + C. 1. .. RULE.-Remove dv, increase the exponent by 1, and divide the result by the new exponent. Thus, ƒ x'dx = fx' + C'; Sv'dv = − ±v³+C; Sy³dy = ży³ + C; C. Sзu¬ìdu = 6u+ + 0. 55. Cor. I. The above rule applies also to (v)"dv, where v is any function of a variable. Thus, ƒ (2ª — 3x + 5)³[2x — 3]dx = ‡(x² — 3x + 5)* + C. = In this example v x2 3x + 5, and dv = (2x-3)dx; that is, the rule applies whenever the factor without the (), viz., [2x - 3]dx, is the differential of the quantity within the ( ). 1 56. Cor. II. fv"dv = f (v)"cdv; introducing a constant C thus is sometimes necessary to render the quantity without the ()the differential of the one within. Thus, S'(4.x + 2x²)³(1 + x)dx = 4 ƒ (4x + 2x²)³(4 + 4x)dx 3 = 2³ (4x + 2x²)³ + C. 57. COR. III. Any differential of the form (a + bx")³ xdx, where n m + 1, can be integrated by the above rule; thus, = xn 1 1 S (a + bx")³ x™• ¹ dx = — S (a + bx")³nbx"¹dx nb ELEMENTARY DIFFERENTIATION AND INTEGRATION. 33 1 nb S (a + bx")ºa(a + bx”) (a + bx²)p+1 = + C. nb (p + 1) = Io(3+5) + C. (5 +7x*)* =S (5 + 7x°)¯*x*dx = o³¹³(5 +7x³)−³ + C. Thus, S x √3+5x* dx = ƒ (3 + 5x²)*xdx S x²dx The rule, Art. 54, does not hold for n = 1, for the reason that d(vº) is not v¹dv, but 0. The formula for this case will be derived subsequently, Art. 180, formula 2. 58. To integrate the algebraic sum of two or more dif- ferentials. we Since d(u + v― z) = du + dv - dz, have S (au S (du + dv − dz) = S du + Sav - Saz = u + v −z + C. .. RULE.—Take the algebraic sum of their integrals. (Art. 45) Thus, S (3x² − 2x+5)dx = ƒ 3x²dx − ƒ 2xdx +f5d Integrate the following: 1. dy (6x-4x+5)dx. = y S 2xdx+S x³ = x + 5 +C EXAMPLES. = S 6x'dx − S 4xdx + Södx = 6 ƒ x²dx − 4 f xdx + 5 fax (Art. 58) dx (Art. 53) (Art. 54) y = x² — x³ + C. X³ x² = = 6· 3 4. +5.x + C 2 =2x³- 2x²+5x+ C, Ans. 2. dy = (7x® 5x)dx. 3. dy = (5x³ — 3x¯¹)dx. 3 y = { x² + x² + C. 34 DIFFERENTIAL AND INTEGRAL CALCULUS. ! 4. dy = (3x* + 2x¯¾³)dx. y = 2x² x² + C. y = {x³ + fx² + C. 5. dy = (x³ + x¹)dx. 6. dy 7. dy dx = or x¯³dx. Vx dx 8. dy = (ax² a or xdx. + 1 =)dx. 2 √x ) dx. b X 9. dy = (→ 1/2 + 2)dx. 2+)dx. X 10. dy = (1 + x¹)³x³dx. 3 y = &x* + C. y = − z x³ + C. y = §ax³ + √x+ C. a х y = + etc. 3x³ y = ƒ 4(1 + x*)³4x³dx = ‡ƒ (1+x')°d(1+x*) (Art. 56) 11. dy = (1 + x)'dx. = √(1 + x*)* + C, Ans. (Art. 55) y = (1 + x) * + C. y = f(1 + x) * + C. -1 12. dy = (1 + x)³dx. 13. dy = (1 − x)˜³dx. y = (1 − x)¯¹+ C. 14. dy = (a + xyxda. y = (a + x²)² + C. 15. dy = (a — x³)¯*x²dx. y = − }(a − x³)* + 0. C. 16. dy = (b+x˜³)*x¯³dx. y = − - 3 (b + x˜¯²³) ³ + C. Find the following: 17. S (x* — 3x¯¹)dx. =√(1~ 18. S 19. 20. 21. xdx √/1 − x² Slat + x² √1 + x³ 3cx St S (va + sex ) 2x ·√(√3 = 3 Sve )dx. dz dx. √2x³-6x+5 3)dx. 5 -3 ‡x³ + x³ + C. 2012 √1 − x² + C. = 2² ² + (1 + x²) ³ + 0. √ ax + c √x³ + C. √2x²-6x+5+ C. ELEMENTARY DIFFERENTIATION AND INTEGRATION. 35 √1 − x² 22. S (x² + 3x² − 6)*(x² + 2x)dx. f(x² + 3x² — 6)³ + C. 23. S xdx 1 + C. (1 − x²) ³ 24. S 4aydy (b — 2y³)** S. 25. 26. S 6x2dx √4x* 5 (x − 1)dx √3x² 6x+7 27. S (1 − x²)³dx. S(1- 28. (a+x) dx. √(a X x² 2 29. S(a + bx + cx²)dx. Nx α b = 2y+ + C. 3 1/4x³-5+ C. 1/3x²-6x+7+ C. x = x² + 3x² - 4x² + C. X³ a² +2ax + + 0. X 3 2 √/x(a +‡bx+‡cx²)+C. - 30. S √x(a³ — x²)³dx. 2x³ (ª²º 3a²x² 3α²x² ry 11 + 2025) + α. X 15 The following differentials may be reduced to the form of (a + bxn) xn-¹dx, and then integrated by Art. 57. 31. S x-2dx √ a² + x* =S. x-³dx √a²x²+1 √a² + x² +C. a*x Multiply the binomial under the radical sign by x-2, and the numerator of the fraction by x-¹. 2 dx 32. f√x² = a' de 33. 34. 35. S X S√2ax — x² dx S S X 3 dx (a² + x²) * xdx (2ax − x²)* - Зах 3 (x² — a²)³ + C. Зах з (2αx — x²)³ + c. + C. a √ zax + C. x² X a² √ a² + x² X 36 DIFFERENTIAL AND INTEGRAL CALCULUS. Ba PROBLEMS. 59. The Problem of Integration is the inverse of that of differentiation; if the latter is "given a curve to find its slope," the former is " given the slope of a curve to find the curve"; if the latter is "given a function to find its rate of change, the former is "given the rate of change of a function to find the function." Since the general problem of differentiation is "given a function and its variable to find their proportional changes," that of integration is "given the proportional changes of a function and its variable to find the function." 60. Definite Value of the Constant C. To complete each integral as determined by the preceding rules, we have added a constant quantity C. While the value of this constant is un- known, it is said to be indefinite; but it becomes definite when its value is assigned or determined by the conditions of the problem under consideration. The signification of C and the manner of determining its definite value are illustrated in the following examples. EXAMPLES. 1. Required the equation of the curve whose slope at the the point (x, y) is 4x³ — 2x + 3. By Art. 48, the slope of a curve at (x, y) is dy dx dy = 4.x³ 2x + 3, dx or Integrating, dy = (4x³ (4x³-2x+3)dx. y = x* — x² + 3x + C, which is the indefinite integral or equation required. 0 To determine the value of C we may make x = 0, which gives yo = C, where y。 indicates what y becomes when x = 0, and is therefore the distance from the origin to where the curve cuts the axis of y. ELEMENTARY DIFFERENTIATION AND INTEGRATION. 37 Therefore the equation may be written Y = x² x² + 3x + Y., which becomes definite when y, is known. When we know any corresponding values of an integral and its variable, C can be determined. Thus, if it is given that the last curve passes through the point x' = 2, y' = 10, then when x = 2, y = 10, and we have x′ 102' 2'+32 +0, or C, C - 8. 2. What is the equation of a curve which passes through the origin, and whose slope at the point (x, y) is (a + x)²? Here dy = (a + 2)², or dy = (a + x)'dx. dx Integrating, y = {(a + x)³ + C. Since the curve passes through the origin, y₁ = 0; hence, making x = 0, we have 0 = {a³ + C; :. C= .. 0 fa³, and the required equation is y = f(a + x)³ — §a³. Y = x³ 3. Required the equation of a curve which passes through the point (23, y' = 11), and whose slope at (x, y) is 3x² - 10x+1. - 5 +x+26. 4. The differential of an integral is (3 + x*)*xdx, and the integral is 0 when x = 0; required the value of C. C = √3. 5. The differential of a function is (1 + 2x)dx, and the function is 0 when x = 0; find the function. 8 27 *(1+ £x)* — *. 6. A function and its variable vanish simultaneously, and their proportional changes are as x²(a² + x³)−¹ to 1; find the — (a² + x³) — fa. Take da for the increment of x, then x²(a² + x) * dx will be the proportional increment of the function. function. 38 DIFFERENTIAL AND İNTEGRAL CALCULUS. 7. Find the area of a plane curve whose differential is √4ax dx, and whose value = 0 when x = 0. zx √4ax. 8. When x is increased by dx the proportional increment of the arc of a certain curve is (9+x*)*x'dx; find the length of the arc, supposing it and x to start from the same point. ¿(9 + x¹)* − 4†. 9. The area of a surface of revolution is increased propor- tionally by (1 + x − x²)¯*(1 − 2x)dx when x is increased by da; find the area of the surface, if it = 0 when x = 0. [(1 + x − x²)* − 1]2π. 10. The differential of the volume of a solid of revolution is π(2Rx - x²)dx; find the volume supposing it and x to vanish simultaneously. π[Rx² — }x³]. 61. Applications to Geometry. The last four problems indicate the possibility of finding the length and area of plane curves and the area of surfaces and volume of solids of revolu- tion when their differentials are known. Now Arts. 31, 32, 33, 34 enable us to find the differentials when the equation of the curve is given. A 62. Areas of Curves. 11. Find the area of OBPA, the equation of the curve APE being y = x² 8x + 15, P C E G O а въ F FIG. 8. Hence, where x = OB and BP. y = By Art. 31, the differ- ential of OBPA (= u) is x ydx, and in the present ex- ample y = x² - 8x + 15. du ydx (x²- 8x+15)dx; :. u = f (x² =S = = f (x² − 8x + 15)dx 8x+15)dx = x³ – 4x² + 15x + C. Evidently the area u = 0 when x = 0, hence C' = 0. ELEMENTARY DIFFERENTIATION AND INTEGRATION. 39 63. Definite Integrals. If, in any indefinite integral, two different values of the variable be substituted, and the one result subtracted from the other, C will be eliminated, and the result is called a definite integral. Thus, Fig. 8, if Oa 1 and Ob 2, to obtain the area of the 23 4x+15x+C and 3 section abcd we substitute 2 for x in get 163 + C (= area of ObcA), and then substitute 1 for x and obtain 11+ C (= area of Öad A); subtract the latter from the former and we have 53 (= area of abcd). The following is the notation by which these operations are indicated: 02 √(x² - 8x+15) dx = [ 23 2 a ab 3 4x² + 152 + C ]* = 163–113 = 53. +15x In general, cates (1) that the differential following it is to be integrated; (2) that b and a are to be substituted successively for the variable in the indefinite integral; and (3) that the second of these re- sults is to be subtracted from the first. The operation is called integrating between the limits a and b. is the symbol of a definite integral, and indi- 12. In the last example find the area of EFG. Since the roots of x² 8x+15=0 are 3 and 5, OE = 3 and OG = 5. Hence we have 25 Lydx ydx = [ {} 3 — 4x² + 15x + C +0] 5 - 11 = area of EFG. The result is negative because the area lies below the axis of x. 13. In the same example find the area of OEA. 13 du 18. La 14. Find the general area of the curve y 3x² 2x + 3 estimated from the origin. 15. Find the areas of the enclosed by the curve y = x³ - S" ydx = x* — x² + 3x. positive and negative surfaces x and the axis of x. Lydx = 4; Sydx = −1. 1 3 * 40 DIFFERENTIAL AND INTEGRAL CALCULUS. • : = 16. Find the entire area of y² = (1 + x²)x between the origin and the point whose abscissa is x. }(1 + x³)¹ — }. 17. Find the positive area of the parabola y = 4ax. zxy. 18. Find the area of y' x' (a' - x) between the limits y³ = xº (a³ — x³) x = 0 and x = α. 64. Lengths of Curves. P na £* (a° — x*)*x*dx = ‡a'. 19. Find the length of the arc OP, the equation of the curve OP being y² = ax³. By Art. 33, the differential of OP (=s) is 11+ (dy) dx. (dy)`'da. To apply this to the present curve, we differentiate y² = ax³, and obtain 2ydy = 3ax❜dx. 2 .2 3αx² [dy 9α²x² 9ах = dx 4y² X B FIG. 9. dy Hence, dx 2y ds = √1+ (dy 2 dx .. s = √ (4 + 9ax) * dx dx = √1+ dx = 2 4 9ах 114+9ax dx. 4 (4+9ax) — 8 * - 27a 3 20. The differential of the equation of a certain curve is dy = √x³- 1dx; find the length of the curve, beginning at the origin. Here dy dx √x 3 - 1; .. \dx (dy)* +1 = x²; s = f.x+. 21. Find the length of the arc of a curve whose equation is y = f(x-1), measured from the point where x = 1. 3 x* 22. Find the length of a curve the differential of whose equation is dy = √x² + 2x dx, beginning at the origin. 23. Find the length of the curve y = limits (1), x = 2, x = 3; (2), x = a, x = b. x² + x. X 1 + between the > 12 Xx b³ — a³ b a (1) }; (2) + 12 ab ELEMENTARY DIFFERENTIATION AND INTEGRATION. 41 " 24. Find the general length of the curve y = (1) √x, estimated from the origin, (1 + 3 1 ) √ √ == Nx. X. • 65. Areas of Surfaces of Revolution. 25. AP is the arc of the circle y² = R²— x²; find the area of the zone generated by re- volving AP about the axis OX. By Art. 34, dS = 2ny 1+ (d) dx; in dx A P this example, since y' R' - x², = R² x², dx dy X Y 0 B E FIG. 10. X Hence, dS= 2πyV 1+ xx . . S = 2xR S Sa x² y² dx = 2π Rdx. dx = 2π Rx. 26. In ex. 25 find the area of the zone generated by the arc PP', supposing OB = a and OE = b. [S]] = 2xR(b − a). α 27. In ex. 25, find the surface of the generated sphere. [S]+R = 4Rπ². -R 28. Find the surface of the cone which is generated by revolving the line y = mx about the axis OX. [S]*"= пy√x² + y². 29. Find the surface of the paraboloid, the generating curve being the parabola y² = 4ax. [S]] = {x√ã [(a + x)ª — (a)*]. 0 π a 30. Find the surface generated by the revolution of the curve y = ax³ about the x axis. [8]* = π [(1 + 9a²x*)* − 1] 27a 42 DIFFERENTIAL AND INTEGRAL CALCULUS. ! : A 66. Volumes of Solids of Revolution. 31. Find the volume generated by revolving OBPA about B E FIG. 11. -X F X ... :: v = π. S (3x² − the axis OX, the equation of the curve AP being y² = 3x² 36x + 105. By Art. 32, dvпy'de; in = π this example y³-3x²-36x+105, hence dv = π(3x² -36x+105)dx. 36x + 105)dx = π(׳ - 18x2105x). 32. In the same example find the volume of the solid generated by the revolution (1) of OEA; (2) of EGF. 5 Απ. (1) [v] = 200π; (2) [v]'= — 4%. 33. Find the volume of the cone which is generated by revolving about x the triangle whose base is x and whose altitude is y (= mx). v = π S" (m²x²)dx = π m²x² 2 3 0 3 X πY'( 13 ). 34. Find the volume of the solid which is generated by revolving y = x²-4 about OX. x f (x* — 8x² + 16)dx = V = π π[‡x® π[‡x' — §×³ + 16x] + C. 35. In ex. 34, find the volume generated by the negative area of the given curve. +2 [v]** = 34fzπ. 36. Find the volume of the spherical segments generated (1) by OBPA and (2) by BEP'P, Fig. 10; also find (3) the volume of the sphere (see Ex. 26). a b (1) [v]¶= x(R²a — fa³); (2) [v] = x[R²(b − a) — §(b³—a³)]; +R (3) [v]** = 4πR³. -R a — 37. Find the volume of a prolate spheroid, the generatrix being the ellipse a'y' a'b' - b²x². α = [v]*ª = $rab². -α CHAPTER III. SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. SUCCESSIVE DIFFERENTIALS. 67. The differential of any function, as u = f(x), denoted by du, is the first differential. The differential of the first differ- ential, viz., d(du), denoted by d'u, is the second differential. The differential of the second differential, viz., d(d'u), denoted by d³u, is the third differential; and in general the differential of d¹-¹u, denoted by dru, is the nth differential. du, d³u, d³u, etc., are the Successive Differentials of u. 68. If x is the independent variable, its differential (dx) is altogether independent of x,-the increment given x at any in- stant being entirely independent of the value which x may have at that instant. It at once follows, therefore, that the differential of da with respect to x, like the differential of any other variable which is independent of x, is 0. Hence (dx)= d'x = 0. For example, take the function u = x³; then (1) du = 3x³dx; (2) d'u = d(3x²)dx + 3x²d(x) = 6xdx²; (3) d'u = d(6x)dx² + 6xd(dx') = 6dx³; (4) d*u = d(6)dx³ +6d(dx³) = 0. 2 Therefore, in finding the successive differentials of a func- tion, we treat the differential of the independent variable as a constant. See Appendix, A¸. The student should observe the difference between expres- sions like d'y, dy', and d(y'); d'y is the second differential of 1, dy' is the square of the differential of y, and d(y') is the differ- ential of y', or 2ydy. 43 44 DIFFERENTIAL AND INTEGRAL CALCULUS. If u = f(x), the successive derivatives or differential coef- du d'u d³u dx' dx² dx³, ficients of u with respect to x are etc. The successive derivatives of f(x) are also denoted by f'(x), ƒ''(x), ƒ'''(x), . . . ƒn(x). Therefore, when u = f(x), we have du d'u dx = f'(x), dx² =ƒ''(x), EXAMPLES. dru dxn = ƒn(x). 1. Find the successive differentials of y = x*. dy = 4x³dx. d'y = d(4x'dx) = 4dxd (x³) = 12x²dx². d³y = d(12x³dx³) = 12dx²d(x²) == 24xdx³. d'y = d(24xdx³) = 24dx³d(x) = 24dx*. d³y = d(24dx*) =0. 2. Find the successive derivatives of x³- 4x²+3x-5. Let f(x) = x²-4x²+3x-5; then f'(x) d = dx :(x² 4x² + 3x-5)= 3x² — 8x+3; d f'(x) dx (3x² (3x² - 8x + 3) = 6x — 8; d dx 0. ƒ'''(x) = (6x (6x — 8) = 6; fiv (x) = 3. Find the successive differentials of u = 2x³ — Yx³. = du (6x14x)dx; d'u = (12x 14)dx'; d³u = 12dx*. 4. u = x² 3x³ + 5x. 5. u = x-³. = = - du 3x dx; d'u 12x-'dx'; d'u= 60x dx'; etc. 6. u =x-2 du 1 x−¹ + 5. (2x³-x-2)dx; d'u= (6x-* — 2x-³)dx; etc. 7. u = (x + 1)³. du = (x+1)¯³dx; d³u — — f(x + 1)¯³dx²; etc. 1 8. If f(x) x + 2' show that f(x) = − 6 (x + 2) *° SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 45 4* 3* 1 9. If f(x) show that fiv(1) 3x + 4' 1 X 1 10. If f(x) show that ƒ''(1) 1 + x² اچه 3x + 2 11. If f(x) = show that f'''(0) x² 4' 영​ㅇ ​69. Leibnitz's Theorem. To find the successive differentials of the product of two variables. Let u and v be functions of x; then d(uv) = udv+vdu. (1) Differentiating (1), regarding u, v, du, dv, as functions of x, we have or d² (uv) =ud'v + dudv+ dvdu + vdu, d² d' (uv) = ud'v + 2dudv+ vdu. Differentiating (2), we get d³ (uv) = ud³v + 3dud'v + 3d³udv + vď³u, (2) from which we see that the law of the coefficients is evidently the same as in the binomial formula. RATE OF CHANGE. 70. Uniform Change. When a variable changes uniformly it experiences equal changes in equal intervals of time, whatever the magnitude of these intervals. Thus, we may suppose the line ak (= x) to be generated by a point moving over each of the equal distances ab, bc, cd, etc., a b с d FIG. 12. in a second of time. If ab = bc = cd ... = h in., the rate of change of x is h in. per second. Denoting the number of seconds In general, * |4, called factorial 4, means 1 × 2 × 3 × 4. n = 1 × 2 × 3... x n. - 46 DIFFERENTIAL AND INTEGRAL CALCULUS. : + . by t, we have x = ht. Differentiating this equation, regarding h as constant, we have dx dt dx = hdt, and =h. Therefore, I. In uniform change the increment of the vari- able varies as the increment of the time. dx dt II. The derivative expresses the rate of change of x per second, when t represents seconds. 71. Variable Change. When a variable does not change uniformly, its rate of change is ever changing, and is measured at any instant by what the change or increment would be during the next unit of time if it (the rate of change) were to remain unchanged during that time. Thus, when a train leaves a station it usually goes, for some distance, faster and faster; that is, its rate of change or velocity is constantly accelerated. If at any instant during this time we say "It is going at the rate of 20 miles per hour," we mean that it would travel that distance the next hour if its present rate should remain unchanged, or, as in uniform change, if the in- crement of the distance should vary as the increment of the time, that is, uniformly. The same is true of any other variable; that is, u being a function of time (t), the rate of change of u is measured by what Дu would be in the next interval of time if it varied as At or dt; but du is what Au would be if it so varied, Art. 25, hence the differential of u measures its rate of change. COR. I. Since du ∞ dt, du = mdt, or du dt =m, where m du) is the rate of change of u per second, and, be- ing a function of t, is in general variable. See Appendix, A„. SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 47 EXAMPLES. 1. Required the rate of change of the area of a square. the side and u the area; then Let du dx u = x², and du = 2xdx. ... =2x dt dt 'du That is, the rate of change of the area dt is equal to the rate of change of the side (dx dt (a) multiplied by twice the side (2x). We may omit dt and regard du and dx as the measures of the rates of change of u and x. Thus, du = 2xdx signifies that the area is increasing in square inches 22 times as fast as the side is increasing in linear inches. 2. In the function u = x² 4x+5, (1) at what rate is u increasing when x is 5 in. and increasing at the rate of 3 in. per second? (2) At what rate is x increasing when it is 10 in. and u is increasing at the rate of 40 in. per second? (3) What is the value of x at the point where u is increasing 10 times as fast as x ? Differentiating the given function, we have du = (2x-4)dx. (1) In this equation we have three quantities, viz.: du, the rate of increase of u; dx, the rate of increase of x; and x; hence, when either two of these are given the third may be found. (1) x = 5 and dx = 3; substituting in (1) and we have du (104)3 = 18, Ans. = (2) du = 40, x = 10; substituting in (1) and we have (3) du = 40 (204)dx, whence dx = 24, Ans. = 10dx; substituting in (1), we have 10dx = (2x 4)dx, or 10 = 2x 4, whence x = 7, Ans. - 3. If x increases uniformly at the rate of 5 inches per second, at what rate is ux-4x increasing when x 10 = inches? 1480 in. per second. = 4 + I 48 DIFFERENTIAL AND INTEGRAL CALCULUS. . ► : 4. In the function y = 2x³ +6, what is the value of x at the point where y increases 24 times as fast as x? x = ± 2. 5. If the side of a square increases uniformly at the rate of 3 inches per second, what is the length of the side at the time the area is increasing at the rate of 20 sq. inches per second? Let the side, u = the area; then u = x². x 6. In the last example, supposing the area to increase uni- formly at the rate of 10 sq. inches per second, at what rate will the side be increasing when the area is 22 sq. inches? Take x = Vu. 5 √/22 in. per second. 7. A circular plate of metal expands by heat so that its dia- meter increases uniformly at the rate of 2 inches per second; at what rate is the surface increasing when the diameter is 5 inches? π Let x = the diameter, u = the area; then u — x². 4 5π sq. in. in. per second. 8. In the last problem, if the surface increases uniformly at the rate of 50 sq. inches per second, at what rate will the diam- eter be increasing when it becomes 5 inches? 20 in. per second. π 9. The volume of a spherical soap-bubble increases how many times as fast as the radius? When its radius is 4 in., and increasing at the rate of in. per second, how fast is the volume increasing ? Let the radius, u volume; then u‡πx³. x = = = (1) 47x² times as fast. (2) 32π cu. in. per second. 10. A ladder 50 ft. long is leaning against a perpendicular wall, the foot of the ladder being on a horizontal plane x ft. from the base of the wall. Suppose the foot of the ladder to be pulled away from the wall at the rate of 3 ft. per minute. (1) How fast is the top of the ladder descending when x = 14 ft.? (2) How fast is it descending when x 30 ft? (3) What is the value of x when the top of the ladder is descending at the rate = SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 49 of 4 ft. per minute? (4) And what at the time the bottom and top of the ladder are moving at the same rate ? Let y = the distance from the base of the wall to the top of the ladder; then y = 1/2500 — x³• (1) ft. per minute. (4) x = 25 √2 ft. (2) 24 ft. per minute. (3) x = 40 ft. 11. What is the value of x at the point where x-5x+17x 3 X and x³ 3x change at the same rate? 12. Find the values of 3 change of x 12x² + 45x x = 2. at the points where the rate of 13 is zero. 2 x = 3,x=5. 13. In a parabola whose equation is y = 12x, if x increases uniformly at the rate of 2 in. per second, at what rate is y in- creasing when x = 3 inches? 2 in. per second. 14. In the same parabola, at what point do y and x vary at the same rate? When y = 6. 15. In the ellipse whose equation is 444y' + 25x² = 11111, at what point of the curve does y decrease at the same rate that x increases ? When y = 3 and x = 5§. 16. Find the points where the rate of change of the ordinate y = x³- 6x² + 3x + 5 is equal to the rate of change of the slope хо of the curve. Where x = 1 and x = = 5. 17. Two straight roads intersect at right angles; a bicyclist travelling the one at the rate of 10 miles per hour passes the intersection 2 hours in advance of another travelling the other road at the rate of 8 miles per hour. At what rate were they separating (1) at the end of 14 hours after the first man passed the intersection? (2) At the end of 2 hours? (3) Required the distance (y) between them when it is not changing. (1) dy dy dt = 5; (2) = 10; (3) y = 100 1/41 mi. 41 72. Applications to Geometry. The rates of change of the areas and lengths of curves and of the areas and volumes of surfaces and solids of revolution are given by Arts. 31, 32, 33, 34. The applications are made as in Arts. 62, 64, 65, 66. 50 DIFFERENTIAL AND INTEGRAL CALCULUS. = རྩྭ y² Rate of Change of Curves. Formula, ds = 1+ (dy) ax. 18. Find the rate of change of the arc of the parabola. = 4ax. α ds = √1+ 2 dx. X 19. In the previous example, if a = 9, and x increases at the rate of 12 inches per second, at what rate will the arc be increas- ing when x = 16 inches ? ds= 15 in. per second. 20. Show that the rate of change of the arc of the circle x² + y² = R² is ds = Rdx VR2-22 21. In a circle whose radius is 20 in., the abscissa changes at the rate of ʼn in. per sec.; at what rate is the arc increasing when x = 12 in. ? in in. per second. 73. Given the rate of change of the arc of a curve (ds), to find the rates of change of its co-ordinates x and y. Take the parabola y = 4ax, then ydy = 2adx; between this equation and ds" = dx² + dy eliminate (1) dx and (2) dy, and we have 2a X dy= ds √4a² + y² and dx = 1 ds. a + x 22. In the parabola y 4x, if the arc increases uniformly at the rate of 5 inches per second, at what rates are y and x in- creasing when x = 9 inches? peop 1/10 and 10 inches per second. 23. If the arc of the circle x² + y² = 100 increases at the rate of 5 inches per second, at what rates are y and x changing when x 6 inches? - 3 and 4 in. per second. APPLICATION TO MECHANICS. 74. Velocity is the rate of change of the distance described by a moving body. Hence, if s = the distance, v = the velocity and t = the time, we have I. v = ds ; dt =S'.' Ivat, våt, and t=ƒª ย SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 51 Again, denoting the rate of change of the velocity by a', we have II. dv d's a' dt dt²; ..v= = Sa'at, and t =Sav dv a'' EXAMPLES. 1. If s 3ť, what is the velocity and its rate of change? = ds Since s = 31, =6tv, the velocity; dt dv and since v = 6t, = 6 = a', the rate of change of v. dt Thus, if the unit of s is one foot, and the unit of t one second, v = ds dt second. dv dt [= 6t] ft. per second, and a′ = [= 6] ft. per 2. A body passes over a distance of ct in t seconds; find v and a', (1) in general, and (2) at the end of 9 seconds. с C C с (1) and ; (2) and 2 Nt 4 √ t³ 6 108' 3. A body after moving t seconds has a velocity of 3t+2t ft. per second; find its distance from the point of starting. S våt Svdt = t' + t². 4. The velocity of a body after moving t seconds is 5 ft. per second; (1) how far will it be from the point of starting in 3 seconds? (2) In what time will it pass over a distance of 360 feet? (1) 45 ft.; (2) 6 sec. 5. A body moves from A, and in t seconds its velocity is 14t ft. per second; (1) how far is the body from A? (2) In how many seconds will the body have gone 847 feet? (1) 7 ft.; (2) 11 seconds. 75. The velocity is positive or negative according as s is in- creasing or decreasing, and a' sustains the same relation to v; 52 DIFFERENTIAL AND INTEGRAL CALCULUS. › therefore, if s increases as the moving body goes forward and decreases as it goes backward, the body is moving forward or backward according as v is positive or negative. 6. A train left a station and in t hours was at a distance of at — 4ť³ + 16ť miles from the starting-point; required the velocity and its rate of change, also when the train was moving backward, when the velocity or rate per hour was decreasing, and the entire distance travelled in 10 hours. S= ds dt = t³ dv = 3ť² dt - - - 4ť" + 16ť² = the distance from station. 12t² + 32t = v = the velocity. The roots of t³ 24t + 32 = a′ = the rate of change of v. = 12ť² + 32t = 0 are 0, 4 and 8; therefore v is negative, and the train was moving backward from the 4th to the 8th hour. Again, the roots of 3ť — 24t + 32 = 0 are 1.7 — and 6.3 +; hence a' is negative, and therefore v was decreasing, from the 1.7th to the 6.3th hour. The roots of i̟ť 4t³ + 16ť² = 0 are 0, 0, 8 and 8; that is, s = 0 when t = 8; hence the train was at the starting-point at the end of 8 hours, having gone backward as far as it had forward. Since the train moved forward the first four hours, then backward the next four hours, and then forward, the entire dis- tance passed over in 10 hours was 4 8 10 [8]*— [8] + [8]¸®: = 64 + 64 + 100 = 228 (miles). 0 # 4 7. A train left a station and in t hours was moving at the rate of t³ 21ť + 80t miles per hour; required (1) the distance ts from the starting-point; (2) when the train was moving back- ward; (3) when its rate per hour was decreasing; (4) when the train repassed the station; and (5) how far it had travelled when it passed the starting-point the last time. SUCCESSIVE DIFFERENTIALS AND RATE OF CHANGE. 53 S (1) 8 = ƒ vat = ƒ (tª — 21ť² + 80t)dt = ‡ť − 7ť + 40ť. 4t¹ — (2) From the 5th to the 16th hour; (3) from the 2.27th to the 11.72th hour; (4) in 8 and 20 hours; (5) 5 16 20 [8] - [s] + [s] = 46584 miles. 0 5 16 8. A traveller left a point A at 12 M., and in t hours after his rate per hour was 5 - t miles; (1) how far forward did he go? (2) At what times was he 8 miles from A? (3) What were his rates per hour when at a distance of 10 miles from A? (1) 12 miles; (2) 2 P.M. and 8 P.M.; (3) + 2 mi. and 2 mi. per hour. 76. Uniformly Accelerated Motion is that in which the rate of change of the velocity (a') is constant. That is, v changes Δυ dv uniformly or v o t; hence, Art. 70, = a'the rate of ΔΙ dt change of the velocity. Formulas. = faʼdt = a't + C = a't + v。 (1) (2) and v =S vdt = S (a't+v.)dt = ža′t +v¸t +8, S= in which v, and s, represent the initial velocity and distance; that is, the values of v and s when t=0. If v。 = 0 and s, = 0 when t = 0, then (1) and (2) become 28 a' v=a't, s=fa't²; :. t = 1 and v = √2a's. (3) 77. The increment of v, or acceleration, produced by gravity is about 32.17 ft. per second, and is usually represented by g. Hence, by substituting g for a' in (3), we obtain the four for- mulas for the free fall of bodies in vacuo near the earth's sur- face. When the bodies are not in vacuo the formulas generally are slightly inaccurate, on account of the resistance of the at- mosphere. 54 DIFFERENTIAL AND INTEGRAL CALCULUS. | PROBLEMS. 1. A rifle-ball is projected from O in the direction of Ywith a velocity of b ft. per second; required its path, knowing that its velocity in t seconds along the action-line of gravity (OX) will be gt ft. per second. Let OX and OY be the axis of x and y, X FIG. 13. respectively; then dy dx = b, and gt; dt y = bt, dt bt, .. (1) and X = žgť. . (2) Eliminating t between (1) and (2), we have 262 y² X, g that is, the path of the ball is an arc of a parabola. Im 2. A body starts from O, and in t seconds its velocity in the direction of OX is 2abt, and in the direction of OY is a³ť² — b²; find its velocity along its path Onm, the distances in the direction of each axis and along the line of its path, and the equation of its path, the axes being rectangular. Y -X 12 FIG. 14. dx = 2abt; dt dy V v dt V s 8 ds dt — a³t² — b²; = 8 Let v, v₁ and v, represent respectively V y the velocities in the directions of the axes x and y and the path s. .. x = = S2abtdt = abt'; Then :: y = f (a°t² = – – ・S (a²t — bº)dt = za \dt (1) — ·b²)dt — }a²t³ — b³t; (2) = √a^tª + 2a²b²t² + b* = a²t² + b². dx dt 2 (dy S = =S (a³t² + b²)dt = fa²t³ +b²t. . (3) Now to find the path of the body, we eliminate t between (1) and (2) and obtain y= ах 3b X b2 abⓇ CHAPTER IV. GENERAL DIFFERENTIATION, LOGARITHMS. 78. LEMMA. The limit of (1+ is the sum of the infinite series 2 + 2 , as z approaches infinity, 1 1 + etc. +++ 13 14 Assuming the binomial theorem for positive integral values of %, we have Z ( 1 + 1 ) = 1 + 2 + (² = 1) + z(z − — 2) 1 1 z(z — 1 2 22 - 1 =1+1+ + 1 (1 − 1 ) + which evidently approaches 2 + 2 + (-1) (= - 1 13 13 +- etc. ( 1 − 1 ) (1-2)+ etc., + etc., as z approaches infinity. The sum of this series is represented by e. It is the base of the natural system of logarithms, which is equal to 2.718281, approximately. limit (1+ (1 + 212) 3 = e. ช 79. To differentiate the logarithm of a variable. Let u = logav, where v is a function of x. 55 56 DIFFERENTIAL AND INTEGRAL CALCULUS. F ! When x is increased by h we have X Au = loga(v + 4v) — logɑ v = Δυ loga ("+4v) Δυ Δυ 213 1 loga(1+ Δυν V Av. V Au ข Δυ 4 = log (1+r)² Δυ 1 V V 이 ​Passing to the limit, remembering that as h approaches 0, V approaches 0, and approaches ∞, we have, as in the pre- ceding lemma, Δύ du 1 dv loga e, or du = (loga e)** dv * v บ RULE.—Divide the differential of the variable by the variable itself, and multiply the quotient by the constant loga e. The factor loga e, usually represented by m, is called the modulus of the system whose base is a. When a = e the mod- ulus is unity and we have du = simply. Herein lies the dv V advantage of the natural system of logarithms, whose base is e, in all discussions of a theoretical nature. Hereafter when the base of a given logarithm is not indicated, it is to be understood that the base is e. EXAMPLES. 1. Differentiate u = log(x³-2x+5). du = d(x³-2x+5) x³-2x+5 (3x² — 2)dx x³-2x+5 * For another method, see Appendix, A., Cor. III. } GENERAL DIFFERENTIATION. 57 2. u = log(12). au == df} = α (1 + x) 1 + x 2dx 1- X 2dx ÷ X 1 - x (1 − x)² 1+x 1-x² · 3x²dx 3. u = log √x — a³. Vx³ du = 2(x³ — a³)* 4. uloga (52² — x³)*. du = 4m (10 — 3x)dx 5x dx x² 5. u = log(x + √1 + x²). du √1 + x² 6. u = log [(a−x) va +x]· du 7. y = log³ x. 8. y,= log¹ (log x). dy 9. y = log (log*x). dy = (a + 3x)dx 2(a² - x²) dy3logx dx 4 log³ (log x)dx x log x 4dx x log x X dx 10. y = log(x+a+ √2ax + x³). dy √2ax + x² EXPONENTIAL FUNCTIONS. 80. To differentiate u = αº. Passing to logarithms, we have log u = v log a. du = dv log a. Differentiating, Multiplying by u, Substituting a" for u, И du = u log a dv, dua" log a dv.* * For another method, see Appendix, A4. 58 DIFFERENTIAL AND INTEGRAL CALCULUS. Hence, the differential of an exponential function with a constant base is equal to the function itself into the natural logarithm of the base into the differential of the exponent. COR. I. d(eº) = edv, since log e = 1. EXAMPLES. 1. Differentiate y = m1+x². dy √1+x² log (m)d( √1+ 20²) = 2. y = elog x. = m m √1+x² log (m)x dx √1+x² elosa đọ dy X 3. y = ex log x exc 1 + x ex(1 − x³). ex 4. Y 5. y = 6. y = log ex e 7. y 1+ exco ex +e-x· 81. To differentiate u = yº. Passing to logarithms, we have x dy=ex log (log x + 1)dx. dy = exx dx (1+x)²° dy=e"(1-3x² - x³)dx. dx dy = 1 + ex :)" dy = (~120--) dz. ex te-x Differentiating, Multiplying by u, Substituting y" for u, log u = v log y. du U = dv log y + v dy Y dy Y du = u log ydv + uv duy log ydv + vyv-¹dy. Hence, the differential of an exponential function with a variable base is the sum of the results obtained by first differen- tiating as though the base were constant, and then as though the exponent were constant. GENERAL DIFFERENTIATION. 59 ↑ The method of differentiating a function by first passing to logarithms, as in the two preceding demonstrations, is called logarithmic differentiation. It may be used to great advantage in differentiating many exponential functions and those involv- ing products and quotients. 1. Differentiate EXAMPLES. y = (x² + 1)x+¹ dy = (x² + 1)x+¹ log (x² + 1)d(x + 1) + (x + 1)(x² + 1)ºd(x² + 1) = (x² + 1)*[(x² + 1) log (x² + 1) + 2(x² + x)]dx. 2. y = x². 3. y = 1 Vx x or xx. ex 1 4. y ex + 1 dy = x*(log x + 1)dx. x (1 — log x)dx x² Vx dy 2ex dy = (ex + 1)² jzdx. dy 5. y = p²² (1+log 2). *dư Make ux, differentiate, and replace u and du by their values. Xx 6. y = x²². dy = +log x + log² x\xˆdx. 20 x) xªdx. 7. y = يع انح 8 dy = (#)"(log-1)dx. TRIGONOMETRIC FUNCTIONS. 82. Circular Measure of Angles. If v = the length of the circular arc BP, and r = the length of the radius OB in terms of the same unit, the ratio is the circular measure of the angle V グ ​BOP. When r = 1, the measure of the an- gle is simply v, which is the length of the This method of measuring an angle is arc. P α B FIG. 15. called the circular or analytic system, as distinguished from the degree or gradual method. 60 DIFFERENTIAL AND INTEGRAL CALCULUS. 83. To differentiate sin v and cos v. With the radius OB (= 1) describe the circle whose centre is 0, and let the angle BOP or its measuring arc BP be any value of v, which is a function of x, then PE = sin v and OE cos v. G T F FIG. 16. P Let us suppose BP or v to receive an in- crement. Take PT, a part of the tangent line at P, for the differential (dv) of the arc BP (Art. 48), then the proportional incre- ments of sin v and cos v will be GT and – GP, EB respectively. Therefore GT = d(sin v), and - GP d(cos v). = The angle GTP = BOP = v; hence in the triangle GTP we have (1) (2) - = GT cos v dv, .. d(sin v) = cos v dv. .. d(cos v) = sin v dv. GP sin v dv, Hence, the differential of the sine of an angle is the cosine of the angle into the differential of the angle. The differential of the cosine of an angle is minus the sine of the angle into the differential of the angle. 84. COR. I. TF = sin (v + dv) and OF= cos (v + dv), which approach, respectively, PT and OB as v diminishes, and when v = 0 we have sin dv = dv and cos dv = 1. That is, the sine of the differential of an arc is the differential of the arc itself, and the cosine of the differential of an arc is 1. 85. The differential of the tangent of an angle is equal to the square of the secant of the angle into the differential of the angle. For, tan v = d tan v = sin v cos v cos vd sin v sin vd cos v cos² v (Art. 41) (cos' v + sin² v)dv cos² v = sec² v dv. GENERAL DIFFERENTIATION. 61 86. The differential of the cotangent of an angle is equal to minus the square of the cosecant of the angle into the differ- ential of the angle. 1 For, cot v = tan v d cot v = d tan v tan² v sec² vd v cosec² v dv. tan² v 87. The differential of the secant of an angle is equal to the secant of the angle into the tangent of the angle into the differ- ential of the angle. 1 For, sec v = COS V d cos v .. d sec v = Cos² v sin v dv cos² v sec v tan v dv. 2 88. The differential of the cosecant of an angle is equal to minus the cosecant of the angle into the cotangent of the angle into the differential of the angle. 1 For, cosec v = sin v' d sin v cos v dv .. •• d cosec v = = cosec v cot v dv. sin² v sin' v = sin v dv. EXAMPLES. 89. d vers v = d(1 d(1 — cos v) Differentiate the following: 1. y sin (x² - x). dy = = cos (x² — x)d(x² — x) = cos (x² — x)(2x − 1 )dx. 2. y = tan¹ (x³). dy = 4 tan³ (x)d (tan x³) = 4 tan³ (x³) sec² (x³)d(x³) = 12 tan³ (x³) sec² (x³)x²dx. 62 DIFFERENTIAL AND INTEGRAL CALCULUS. 4 -: ར༔ 3. y = cos (ax). 4. y = cos³ x. 3 5. Y = sin x cos x. 6. y = tan² 5x. dy=a sin (ax)dx. dy 3 cos² x sin x dx. dy = (cos* x – sin* x)dx. 10 tan (5x) sec² (5x)dx. 3 sec³ x tan x dx. (sin x tan x sec x)dx. dy = 2x cot x(cot x dy = x cos x dx. dy= 2 sin² x dx. dy 7. y = sec³ x. dy 8. y = sin x tan x. dy 9. y = (x cot x)². 10. y = x sin x + cos x. 11. y=x - sin x cos x. 12. y = tan x X. 13. y sin (cos x). sin x-sin³ x. 14. y = sin x 15. y = † cos³x COS X. = tan² x dx. x cosec² x)dx. dysin x cos (cos x)dx. dy dy = cos³ x dx. dysin' x dx. 16. y = tan³ x-tan x+x. dy = tan¹ x dx. • 17. y=xsin x. 18. y = (sin x)*. dy sin x = X sin x X dy = +log x cos x d Jdx (sin x)[log sin x + x cot x]dx. Prove the following by differentiating both members (see Art. 50): 19. Scot x dx = log (sin x) + C. tan x dx = log (cos x) + C. 20. S- S- tan 21. S sec 22. sec x cosec x dx — log (tan x) + C. S- sec x cosec x dx = log (cot x) + C. tan x dx - log (sec x) + C. 23. Stan 24. S. S- 25. 26. S - cot x dx = log (cosec x) + C. sin x dx 1 - COS X Ssin log (vers x)+ C. sin x cos³ x dx = † sin² x — ‡ sin* x + C. 27. S sin³ x cos³ x dx sin' x cos x dxsin' x sin x + C. = GENERAL DIFFERENTIATION. 63 30. If f(x) 31. If f(x) 3 'sin' x 28. Ssi 2 2dx cos² x dx sec x + cos x + C. Prove the following: 29. If f(x) = (x²-6x+12)e*, ƒ'''(x) = x²e*. cos x, fiv(x) = e e-* cos 4e-x cos x. = tan x, ƒ'''(x) = 6 sec¹ x 4 sec² x. 7 cos x cos' x d'y 32. If y = = sin³ x. 9 3 27, dx³ 33. If f(x) = x³ log x, fiv(1) = 6. 34. If f(x) = log sin x, ƒ""' fm() = 4. − 35. If f(x)= sin x, f'(0) = 1; f""'(0) = — 1. 36. If f(x) = log (1 + x), ƒ"(0) = −— 1; ƒ'(0) = |4. 37. If f(x) = aº, ƒ"(0) = log" a. 4 38. If f(x)= log x, fiv (c) = €º log x, fiv(c) = e°º log c+ + с 68 c² 6 INVERSE TRIGONOMETRIC FUNCTIONS. 90. To differentiate sin-¹ y. Let dy v = sin-1 sin-¹ y, y, then y = sin v. = cos v dv = √1 — y³dv. — dv dy 'I — y² dy* or d(sin-¹ y) = √1 — y³ 91. To differentiate cos-¹y. Let v = cos-¹y, then y = cos v. sin v dv = — √1 — y³dv. dy dy dy dv= or d(cos-¹y) = √ī — y³³ √1 — y² which always has the sign of — sin v. * To avoid the double sign ±, we shall suppose 0 < v < other quadrant the sign will be that of cos v. π 201 ; for any 64 DIFFERENTIAL AND INTEGRAL CALCULUS. f. 92. To differentiate tan-¹ y. Let v = tan-¹y, then y = tan v. dy = sec² v dv = (1+ y²)dv. dv= dy 1+ y²⁹ or In a similar manner we find (tan-¹y)=1+ y² dy · dy 93. d(cot-¹y) 1+ y² dy 94. d(sec-¹y) y vy² — 1 dy 95. d(cosec-¹ y) y vy-1 dy 96. d(vers-¹y) √ Zy - Y y² EXAMPLES. Differentiate the following: 1. y = sin¹√x. X. dx dy = d Nx √1 − ( √x )² 2 V x dx X √1 − x 2 V x ха - x\ tan-1(1 2. y = tan- dy \1 · +x a G 3. y = sec-¹ nx. X 1+ + x G 2dx 2 (1 + x)² dx 1+ - X 2 (1 + x)² + (1 - x)' 1 + x² \1+x (1 + x)² dx 2x 4. y = ver-¹ 9 dy= dy x √ n²x² - 1 dx √9x-x² GENERAL DIFFERENTIATION. 65 3dx = = 5. y sin¹ (3x — 4x³). dy √1 − x² dx 6. y -1 sin ¹ (2x — 1). dy= √x ха 7. y = sin-¹ (sin x). dy = dx. 8. y = sin-¹ (sin x). dy = ¿(√1 + cosecx)dx. 2x 9. y = tan-¹ dy= 1 − x²° 1 COS X 10. y = tan-¹ 1+ cos x 11. y = (x + 1) tan-¹ Vx-√x. 12. y log = 1 + x + tan-¹ x. 1 − x dy = 1 − x* 4 2dx 1 + x²· dy = {dx. 1 dy = tan-¹ √x dx. dx 1 X tan-1 a a 1 x V x² a² a 13. 14. Prove the following by differentiating both sides: S S α dx +2² a² + dx + C. X -1 -sec + C. α 15. S dx X sin-1 + C. Va² X2 α S dx X S 16. 17. S 18. Na² - x² dx x²dx = vers-1 + C. √2αx x² a - x Na² a² = + sin-12+ C. 2 2 √ 2αx x² x + 3a 2 α 2ax-x+a vers-1 12 + C. a 19. j madx = m sin- √c² — b² — Qabx a²x² 20. S madx 21. S macdx = m tan √ c² + b²+2abx + α²x² a²x² + 2abx + b² ÷ c² C = m log(ax+b+√c²+(b+ax)²+C. 며 ​-1 (ax + b) + C. 2-1 (ax + b -²) + C. C 66 DIFFERENTIAL AND INTEGRAL CALCULUS. - 22. S (ax + b) (cx + d) m(ad — bc)dx = m log(ax+3)+ + C. cx d The last four equations may be conveniently employed in integrating a certain important class of differentials, of which the following are illustrations: 23. Required the integral of Here x² 3x 28 = (x 5dx x²-3x-28" 7)(x+4); hence we may inte- grate by formula 22, thus: Make ax + b = x-7, cx + d = x + 4, and m(ad-bc) = 5; we then have a 1, b = −7, c = 1, d=4 and m; hence, substituting in 22, we have S 5dx √ x³-3x² - 28 X 3х 24. Required the integral of 5 11 log 3dx = (x+4)+0. C. = 4x² + 3x + 1° Integrating by Ex. 21, we have a 4, 2ab3, b' + c² 1 and mac = 3; whence a = 3; whence a = 2, b = &, c = 4 √/7 and m = 6 :: √ 4x² + 3x + 1 3dx 6 tan-1 2x + 1 ) + 0. NJ 5dx 25. Find the integral of 16 12x 4x² — = Integrating by Ex. 19, we have a² 4, 2ab12, c² - b² = 16 and ma = 5; hence a = 2, b = 3, c = 5, m = §. 5dx 5 -1 √16 — 12x 4x² =sin(2x+3)+ C. 5 26. √x² + z+1 Sa 27. S₂ dx 2 2x + 1 tan-1 + C. √3 √3 dx 1 2+ 5x2* x √5 tan-1 +a 1/10 √2 GENERAL DIFFERENTIATION. 67 dx S 2x² - 4x - 7 √2 2x – 2 – 3 √2 log 12 2x 2312 28. 29. S 30. S 31. S dx √1 – 3x dx 3+2x sin-1 + C. 22 X 13 Vm + nx + rx² dx + C. 17 log (x Vr + 2 + Vm + nx + rx³) + C. NT V m + nx rx² 32. Sm+nx + rx². n V r 1 2rx n sin-1 + C. Vr √4mr + n² 2 tan-1 √ 4mr n² 2rx + n √4mr n² + C. 97. To find the differential of an arc in polar co-ordinates. Let AP (s) be the arc of a curve, O the pole, OP (= r) the radius vector, and PT a tangent to the curve at P. Let 0=XOP and = OPT. Increase arc PP' by POP' (= 40), then 4s and OP'r + Ar. Draw PD perpendicular to OP', then PD = r sin 40 and DP' = r + Ar rcos 40. The chord PP' = √PD' + DP" chord PP' 40 = O A P D T S FIG. 17. DP'\2 40 √ (PD)² + (DP)²* ΔΗ X Passing to the limit, remembering that as 40 approaches 0, and sin 40 ΔΗ the limits of chord PP' arc PP' (Art. 33), also of cos 40, is each unity, we have ds= r²+(dry)')ão. 68 DIFFERENTIAL AND INTEGRAL CALCULUS. T- r 98. COR. I. As P' approaches P, the angle OP'P approaches the angle OPT (=); therefore limit PD tan = = - 400 P'D rdo dr rdo sin = ds و dr cos & ds FUNCTIONS OF TWO OR MORE INDEPENDENT VARIABLES. 99. A function of two variables, as u = x³y + y³ and u = sin (x + y), is represented by f(x, y); and similarly f(x, y, z) represents a function of the three variables x, y and z. F Since x and y are independent of each other, the function u = f(x, y) may change in three ways. Thus, let u = xy = area of OBPA, where OB = x and OA = y. (1) x may change and y not, which would give du = BCDP = ydx; (2) y may change and x not, which would give du = APFG = xdy; (3) both x and y may change, which would give (understanding by du the portion of the increment G dy A P y dx x B FIG. 18. D which is of the first degree in dx and dy) duydx+xdy. Hence du may have three different values, and it is desirable to employ a notation by which they may be represented. 100. A Partial Differential of a function of two or more variables is the differential obtained on the hypothesis that only one of its variables changes, as ydx and xdy in the previous du example, and these are denoted respectively by dx and or du and du. dx du dy, dy 101. Total Differential. In a function of two or more in- dependent variables, if each variable receives an increment, that portion of the corresponding increment of the function which is of the first degree with respect to the increments of the variables is the total differential of the function. GENERAL DIFFERENTIATION. 69 102. PROP. The total differential of a function of two or more independent variables is the sum of its partial differentials. Let u represent any function of x and y. When x becomes x+7, the part of the corresponding incie- ment of u which involves the first power of h or dx is du dx dx. Hence, omitting the terms involving the higher powers of dx, Art. 29, the new value of u is du u + dx. dx In this new value of u when y is increased by k (= dy) the du parts of the corresponding increments of u and dx which in- dx du d /du volve only the first power of dy are dy and dy dy dx dx)dy; hence, omitting the terms involving the higher powers of dy, the second new valne of u is du du u + -dx + dx dy dy + d (du dx) dy, dy dx which result is the same as if x and y had changed simulta- neously, for the result of increasing x by dx and y by dy is evi- dently the same whether the changes be made separately or simultaneously. Hence, since the last term involves the product of dx and dy, we have du = du dx du dx + dy. dy In a similar manner it may be shown that the theorem is true of functions having any number of variables. COR. I. The total differential of a function is the sum of those parts of its increment which vary as the increments of the variables, respectively. COR. II. The theorem is also true of functions whose variables are not independent of each other. 70 DIFFERENTIAL AND INTEGRAL CALCULUS. - COR. III. The total differential of a function of two or more variables, as x, y and z, is the sum of the differentials obtained by first differentiating as though x only were variable, and then as though y only were variable, and then as though z only were variable. 103. A Partial Derivative of a function of two or more variables is the ratio of the partial differential of the function to the differential of the variable supposed to change. 104. The Total Derivative of a function of two or more variables, only one of which is independent, is the ratio of the total differential of the function to the differential of the inde- pendent variable. To prevent confusion we shall distinguish the total differen- tial or derivative of two or more variables by enclosing it in brackets. Thus, if u = f(x, y), [du] = dx du du dx + -dy, dy where du -dx and dx du dy are the partial differentials with respect dy to x and y, respectively. EXAMPLES. Find the total differential of— 1. u = x² - 3xy + 2y³. du dx dx = (2x — 3y)dx; du dy dy (3x — 4y)dy. .. [du] = (2x-3y)dx — (3x — 4y)dy. x + y 2. u = [du] X y X 3. u = sin-¹ [du] y' -1 4. u tan- y [du] X: = ydx- 2 y Vy² — x² ydx xây — yo x² + y² 2(xdy – ydx) (x − y)² xdy GENERAL DIFFERENTIATION. 71 5. u = log y*. 6. u = ysin x [du] = 2dy + log ydx. Y [du] = ysin x log y cos x dx+ 105. Function of Functions. If u = F(y) and y sin x dy. ycovers x = = f(x), u is indirectly a function of x through y. In such cases the value du of may be obtained by finding the value of u in terms of x, dx and differentiating the result; but it is often more easily found by the formula du dx du dy dy dx (1) That is, the derivative of u with respect to x is equal to the derivative of u with respect to y multiplied by the derivative of y with respect to x. Thus, if u = tan-¹y, and y = log x, then du dy 1 dy 1 1+ y²' dx X du 1 and .. • dx x(1+ y³) ' If u = F(v, y), v = f(x) and y = f(x), to obtain du dx total derivative of u with respect to x, we may proceed thus: the du du Since [du] = dv + dv dy dy? du du dv du dy dividing by dx, + dx dv dx dy dx' du which gives dx • (2) in terms of derivatives which can be reckoned out from the given equations. Thus, if u = v² + vy, v = log x and y = ex, then du du dv 1 dy dv = 2v+y, v, and ex; dy dx X dx 72 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 substituting in (2), we have du 2v+y + vex. dx X If u = F(x, v, z), v = f(x) and z = f(x), we have du du du [du] = dx + ·dv + dz. dx dv dz du du du dv du dz = + + dx dx dv dx dz dx > • O (3) du du du du where and are partial derivatives, and dx' dv dz dx [date] the total derivative of u. EXAMPLES. Find [de] in the following: 1. u = e*(y — z), y = sin x, and z = cos x. du dx ] = 2e sin x. du 2. u tan tan−12, and y = √'r² — 2º. dx 2 1 202 du -1 3. u = tan-¹ (xy), and y = ex. dx ex (1 + x) 1 + x²e²x² x* 4. If y = uz and ue, z = x² - 4x³ + 12x² - 24x + 24, find the slope of the curve of which y is the ordinate and x the abscissa. dy = exx¹. dx 106. Successive Partial Differentials and Derivatives. We have seen that the differential of u (= f(x, y)) (1) with re- d spect to x is denoted by (u)dx, and (2) with respect to y by dx d (u)dy. dy GENERAL DIFFERENTIATION. 73 du Similarly, the differential of dx dx d du (1) with respect to x is denoted by dx dx dxdx, Jax, d / du dx dy, dy\ dx by al (2) with respect to y is denoted by Hence d³u dy dx dy dx d²u dx² dx² d²u dx dy dx dy is a symbol for the result obtained by dif- ferentiating u two times in succession: once, and first, with respect to y, and then once with respect to x. Similarly, d³ u dx dy² dx dy indicates the result of three successive differentials of u: first, once with respect to x, and then twice with respect to y• In finding these successive partial differentials, we treat dy and do as constants, since y and x are regarded as independent variables, see Art. 68. The symbols for the partial derivatives are ď²u d²u d² u d³ u 27 dx² dx dxdy' dy²' dx 39 d³u dx dy etc. 107. PRINCIPLE. If u = f(x, y), For, Art. 102, changing x and then y to obtain d² u d² u dx dy dy dx ď² u or dx dy' changing y and then x to obtain d²u dy dx' is equivalent to changing : x and y simultaneously; and therefore the results are equal. COR. I. If u be differentiated m times with respect to x, and n times with respect to y, the result is the same whatever be the order of the differentiations. EXAMPLES. 1. Given u= x²y³; find ď²u dx dy d²u and dy dx 6xy³. 74 DIFFERENTIAL AND INTEGRAL CALCULUS. ! ..... d³ u d³u 2. Given u = x²y + xy'; verify dx dy² 2 d'u dy³ dx d'u · 3. If u = y log (1 + xy), show that dy dx dx dy 4. Given u = x² - y² x² + y² d'u d² u 2 ; verify dx dy dy dx* 108. To find the successive differentials of a function of two independent variables. Let u = f(x, y); then du [du] = dx + dx dy du ·dy. du (1) du Differentiating (1) and observing that and are usually dx dy functions of x and y, and that x and y are independent, Art. 68, we get d²u d'u [du] dx² + dx dy + dx2 dx dy d²u dy dx d'u dy dx + dy dy², or, Art. 107, ď² u d² u d²u [d'u] = -dx²+2· dx dy + dx² dx dy dyzdy? (2) Differentiating (2), remembering that each term is a function of x and y, we have d³ u [d³u] -dx³ +3 2 dx³ d³u dx dy ď³u d³u 3 dx² dy + ³dx dydx dy² + dy³ dy dy³ . . . ; and similarly may [d'u], [d'u], etc., be found. By observing the analogy between the values of [d'u] and [d³u], and the de- velopments of (a + x)² and (a + x), the formula for the value of [dru] may be easily written out. 109. Implicit Functions. f(x, y) = 0, the formula [du] In functions of the form du du = dx + dy is often useful in dx dy dy dx finding the value of the derivative, or slope, GENERAL DIFFERENTIATION. 75 Thus, take f(x, y) = ax³ +x sin y = 0. 3 Making u ax + x sin y, we have du [du] = ax dx+ du dx++ dy = (3ax² + sin y)dx + x cos y dy. dx dy But since u = 0, [du] = 0. du dy dx 3ax² + sin y dx du x cos y dy EXAMPLES. dy Find the derivative, of the following: dx' dy 3x² 2ax 1. (y — b)² — x³ + ax² = 0. dx 2(y- b) - dy 2. x* + 2ax³y — ay³ = 0. 3. x³ +3axy + y³ = 0, dx $$ 4x³ + 4axy dy 3ay² — 2ax² x² + ay dx y² + ax® 110. Successive Derivatives of an Implicit Function. The following examples will serve to illustrate how the succes- sive derivatives of implicit functions in general may be deter- mined. EXAMPLES. dy dx² 1. Find when y² — 4ax = 0. Here dy 2a (1) dx Y Differentiating (1), we have d'y 2ady (2) dx .2 76 DIFFERENTIAL AND INTEGRAL CALCULUS. **- Eliminating dy in (1) and (2), we have d'y dx² 4a² 2 y" 2. Given a'y' + b²x² — a²b² = 0, show that dy b₁ dx² a²y" d'y 2a³xy y³ 3. Given y'+x-3axy = 0, show that dx² (y² — ax) ³° dy d'y d³y dx' dx²' dx³' 111. Change of the Independent Variable. After ob- taining the derivatives etc., on the hypothesis that x was the independent variable and y the function, it is some- times desirable to change these expressions into their equivalents with y for the independent variable and x the function, or with x and y for the functions and some other variable, as t, for the independent variable, and so on. 112. To find the successive derivatives of 213 dy when dx dy dx neither x nor y is independent. Under this hypothesis is to be differentiated as a fraction having both terms variable. ď³y d (dy dx d'y - dy d³x = (1) dx² dx dx dx³ Similarly, 3 d³y (d² dx = d (dv) (dx d³y — dy d³x)dx dx dx² dx³ 3(dx d³y — dy d³x)d²x (2) In like manner we obtain the other successive derivatives. COR. I. If y is independent, then d'y d'y = 0, and we have = dzy dx² dy d'x dx³ (3) d'y 3 (d'x)'dy 2 = dx³ d³x dy dx dx (4) GENERAL DIFFERENTIATION. 77 Formulas (1) and (2) give us the values to be substituted for ď²y d³y and when neither x nor y is independent; and formulas 3 dx² dx³ (3) and (4) give us the values of the same derivatives when y is independent. If a new variable t, of which x = f(t), is to be the inde- pendent variable, in Art. 111, we replace x, dx, d²x, etc., by their values as determined from x = = f(t). EXAMPLES. 2 1. Given yd¹y+dy' + dx² = 0, where x is independent, to find (1) the transformed equation in which neither x nor y is independent; also (2) the one in which y is independent. đ²y dx² (1) Dividing by dx2, substituting for from (1), and multi- plying both members by dx³, we have y(d³y dx d²x dy) + dy³ dx + dx³ = 0. (2) Making d'y= 0 in this last equation, and dividing by - dy, we have d²x dx³ dx Y = 0. dy² dy³ dy 2. Change the independent variable from x tot in d'y 1 dy + x dx dx² +y=0, when x = 21t. dzy Substituting for from (1), multiplying by x dx³, and mak- ing x = 2√t, dx = dx² t¯*dt, and dx = — t¹dť², we obtain d'y dy t + + y = 0. dt² dt Change the independent variable from a to y in the two fol- lowing equations: d²x + = 0. dy dy 3 [day dy d³y dy/dy 2 d³x 3. 3 \dx² = 0. dx dx³ dx² dx dy dy + = 0. dx dx d'y 4. x √x² X d2x 2 dx² + dy' T dy² -1= 0. 78 DIFFERENTIAL AND INTEGRAL CALCULUS. ୮ : Change the independent variable from x to t in the two fol- lowing equations: d'y 5. + dx² 2x dy y + - 0, where x = tan t. 1 + x² dx (1 + x²)² d³y dt² +y=0. d²y dy dzy 6. (1 − x²) X = 0, where x = cos t. = 0. dx² dx dt 7. Find the value of R = dx* 29 where x is in- dependent, supposing neither x nor y to be independent. R = (dx² + dy³)* dx d³y — dy d³x (1+dy) dy 2 MISCELLANEOUS EXAMPLES. Differentiate the following: 1. y = 3(x² + 1)*(4x² — 3). dy 56x³ (x² + 1)¹dx. 1 dy 2. y = x + √ 1 + x² dx Na a + x + √ a X dy = √ a + x 3. y 4. y = x + log cos = 5. y log Na + √ x Na x log x 1 - X 2 X x + 1 Na dx X π dy 4 dx 6. y 7. y = log V 2 x² + x + 1 Nx dy +log (1 − x). dx 8. y = sin (x + a) cos (x − a). - 9. y = log tan ( X πT. 2 10. sin 2x = 2 sin x cos x. 2l8 28 28 28 28 28 2şla X 1. √1 + x² a²+a√ a² — x² x²Va X 2 dy dx = 1 + tan x Nã (α − x) √x log x (1 − x)²° x² 2 - 1 x² + x² + 1° dy = cos 2xdx. dy = sec x. dx cos 2x = cos² x — sin² x. 2 GENERAL DIFFERENTIATION. 79 I 11. sin 2x= 1 - tan² x cos 2x = 1+tan' x 2 tan x 1+ tan² x 12. sin 3x = 3 sin x - 4 sin³ x. 13. y = tan-¹ e*. = = 14. y cos -1 X 6 6 12° + 1/ 1 cos 3x4 cos³ x-3 cos x. dy dx dy dx 33 33 33 3ls dy dx 1 ex + e-x +e-x° 6x2 x² + 1 2 15. y = sec-1 2x² 1' x² 1 X 16. y = tan-¹ x + tan-¹ -1 dy = 0. 1+x dx Prove the following by differentiation: ndx 2 cos² x + n³ sin* x 2ax²dx 17. S 18. Sax α = tan-1 19. log (1x) = x — + tan-1 (n tan x) + C. X X a) log a (x + α) + C. x² + 3 4 + etc. хо x² 20. tan-¹ x = x + + etc. 3 5 ry Find the slopes of the following curves: 21. The quadratrix, y = (ax) tan πα 2a dy π πα πX (α — x) sec² tan dx 2a 2α 2a 1 y - r √/2ry — y². dy dx = 2r - y y 22. The cycloid, x = r vers 80 DIFFERENTIAL AND INTEGRAL CALCULUS. 23. The catenary, y = 201 (co + e¯³). dy dx 1/2 = 1 (o² - 6-7). e 24. The tractrix, x = a log a + √ a² — y² y √a² y². dy Y dx Na² y² 25. Find the following: 12 S (13-5)dx. 26. S 4xdx 3(x²+2)** 27. S ($24 28. Sva². x² dx. 29. S 30. S dx dx 6 5 x² 3x³ + + C.. (x² + 2)² + C. - 5x³ + 2x¯¹ + ‡x¯³)dx. (x 1)³ + α Vx (a² — x²)³ 3a²x³ + C. X + C. x² + C. ах X *| ૭ 6 x² + 2 6 log x + 0. x √2ax x² - 31. f(x² - 2) dx 32. 33. 34. 35. 36. S X •2(x + 1)dx x² + 2x + 3 ار S S S (2x ( 1) 2x + 3 9x² dx dx 4 √1 + 4x² dx √x² 37. S a²√ a² 2 - √2αx-x² log (x²+2x+3)+ C. dx. X 4x + 13 2dx 3x²+10x + 3' 2 log (2x+3) + C. 3x 2 1½ log 3x + 2. + C. † log (2x + √1 + 4x²) + 0. log (x−2+√x²-4x+13)+0. 3x C. + log (32 + 1) + 0. x 3 GENERAL DIFFERENTIATION. 81 38. S dx 2x 3 sin-1 + C. √1 + 3x x² 1/13 39. S dx 1 X 3 tan- -1 + 0. √2 S 2. 3xdx 41. V125. 2 - 4 40. x² 1 1. S 42. S 0 6x+11 (x² −2x+2)(x−1)dx. x³dx x+1 2 43. / 3 sin a cos ada. 23 - log 3. xdx. sin³ x + C. (a − b cos² x)*+C. 44. ƒ 3b(a — b cos² x)* sin x cos xdx. 45. S 'sec² x + 1 Ssec² x tan x + x dx. 46. S (tan x + cot x)³dx. log (tan x + x)+C. tan x cot x + C. ď²y dx² (y - 2x) (y x) 3 dzy c(a² — 1) 48. Given y' - 2axy + x²= c, to find y² dx² - (y ax) ³° 3 47. Find when y² — 2xy + c = 0. 49. In yd'y + dy' + dx = 0, change the independent vari- able from x to y. 3 d²x dx³ dx Y = 0. dy² dy³ dy to z in 50. Change the independent variable from x (2x − 1)³d³y dx³ 3 dy dx + (2x-1) x = 2y, where 2x = 1 + e*. 4. dz³ dy - 1244 +9 dy =y. dz² dz 51. If y² = sec 2x, prove that = 3y-y. d²y = dx² i. . 82 DIFFERENTIAL AND INTEGRAL CALCULUS. 52. Given st³ — 5ť 6t; find the velocity (v) and its = 5ť²+6t; rate of change (a') when t = 10. v = 206; a′ = 50. 53. In t seconds after a body leaves a certain point the rate of change of its velocity is 6t - 12 feet per second; required its velocity and distance from the point of starting. v = 3t2 12t; s=t³ 6t2. 54. In the last example how far will the body travel in 10 seconds ? − [8]*+ [s]] = 32 + 432 = 464 (ft.). 10 4 55. How many times faster is x increasing than log x, when x = n ? n times. 56. Required the value of x at the point where the slope of the curve y = tan x is 2. π 4' 57. A man is walking on a straight path at the rate of 5 ft. per second; how fast is he approaching a point 120 ft. from the path in a perpendicular, when he is 50 ft. from the foot of the perpendicular? 11 ft. per sec. 58. A vertical wheel whose circumference is 20 ft. makes 5 revolutions a second about a fixed axis. How fast is a point in its circumference moving horizontally, when it is 30° from either extremity of the horizontal diameter ? 50 ft. per sec. 59. A buggy wheel whose radius is rolls along a horizontal path with a velocity v'; required the velocity (ds) dt of any point (x, y) in its circumference; also the velocity of the point hori- zontally (de) (da) and vertically dt (dy dt The curve described by the point in the circumference of the wheel is a cycloid whose equation is xr vers-1 differentiating this and dividing by dt, we have dx dt = y √2ry — y² dy dt Y - y √2ry—y²; T (1) : GENERAL DIFFERENTIATION. 83 ま ​2' Again, the abscissa of the center of the wheel is 7 vers-¹ 2; differentiating this and dividing by dt, and we have v' = p dy √2ry — y³dt ' Again, since ds² = dx² + dy², we have ds dt 2 dx (dy + dt/ dt From (1), (2), and (3) we readily obtain 2 dy W2ry y v'. ? dt I' dx dt (2), and ds dt de = (√²). 2Y dx dy ds (2) (3) 60. In the last example find the values of dt' dt and at dt the point (1) where y = 0; (2) where y = r; (3) where y = 2r. 61. Water is poured at a uniform rate into a conical glass 3 inches in height, filling the glass in 8 seconds. At what rate is the surface rising (1) at the end of 1 second? (2) At what rate when the surface reaches the brim? (1) in. per sec. (2) in. per sec. 1 1 .. CHAPTER V. SERIES, DEVELOPMENT OF FUNCTIONS, AND INDETERMINATE FORMS. SERIES. 113. A Series is a succession of terms following one another according to some fixed law. If the sum of the first n terms of an infinite series approaches a definite limit as n increases indefinitely, the series is Conver- gent; if not, it is Divergent. The sum of a finite series is the sum of all its terms; and the sum of an infinite convergent series is the limit which the sum of the first n terms approaches as n increases. An infinite divergent series has no definite sum. 114. To Develop a function is to find a series, the sum of which shall be equal to the function. Hence the development of a function is either a finite or an infinite convergent series. For example, (x + 1)³ = x³ + 3x² + 3x + 1. This finite series is the development of the function (x + 1)³ for any value of x. Again, by division we obtain 1 1-x = - 1 + x + x² + x³ +... x²-1. . (1) 1 Now this series is the development of only for values 1-x : 84 SERIES. 85 of a numerically less than 1, for the omitted remainder is xn 1 − x ; hence, denoting the series by s, we have 1 =8+ 1 — x 1 - x 1 xn Therefore s can be the value of only when 0, 1 - x 1 - x and s the development of cause xn 1 X 1 1 - X only for such values of x as will to approach 0, as n increases indefinitely, and this can be the case only when x < 1. Thus: (1) For x = 2 we have −1=1+2+4+8+... 2"-12", which would be absurd were the remainder 2" omitted. (2) For x = we have 1 1 2 2=1+ +=+ + 1 2n 1 1 + 2n-1 + Qn' in which the remainder decreases as n increases, indefinitely. 115. A series is said to be absolutely convergent when it remains convergent on making the signs of all its terms positive; but only conditionally convergent when it becomes divergent on such a change of signs. The series 11+}-{+... is an example of a condi- tionally convergent series. 2 116. PROP. The infinite series u, + u₂+... un-1 + Un + ... will be absolutely convergent if the terms are all finite, and the limit of the ratio , as n is indefinitely increased, is numerically less than unity. Ип Un-1 86 DIFFERENTIAL AND INTEGRAL CALCULUS. F 1 : 1 Since the limit of, as n is indefinitely increased, is less than Un Un-1 1, there must exist some finite integer, c, such that for all values of n which are greater than c, Ип Un-1 is less than 1. u。 is a definite finite The sum of the terms u, +u₂+ 2 quantity, and it only remains to show that the sum of the remain- ing terms, uc+1 + Wo+2 + ., is also. Let r be a number less than 1 but greater than any of the ratios Wo+1 Wo+2 Uc+3 ; then Uc Uc+1 Wc+2 Uc+1 k the series is divergent, and when xk the series in some cases is convergent, and in others di- vergent. 119. COR. III. The series ƒ'(x) = a,+2α¸x +3α¸x² + ... (n − 1)an-12n−2+na-1, obtained by differentiating f(x) = a. + a‚x+α„x² +··· An-1x²-¹ + A₂, is convergent for the same values of x as the last-mentioned series. For in the former k = the limit of same as the limit of an-1 An (n − 1)an-1, which is the nan > It is also evident that the limits of convergence of the series obtained by integrating the individual terms of f(x)dx are the same as those of the series f(x) itself. EXAMPLES. Find the values of x which will render the following con- vergent: x² хо xn-1 1. 1+x+ 2 + + Xn + +... 3 N 1 n 1. 1 and an = n-1 N Here an-1 which 1 when n =∞ Hence (117), - 1 < x < 1; that is, x lies between 1 and +1. M An-1 an n 1 1+ n 1 n — l' x² X³ x⭑ X"-1 Xn 2.1+x+ + Q2 1 13 + 14 n = I + +. Here An-1 n an n - 1 =n, which = ∞ when n = ∞; hence ∞ < x <∞; that is, the series is convergent for all finite values of x. 3x 5x2 2 r23 3. + + +...+ 5 10 2(n-1)+1 (n−1)²+1 -20%-1+ Xn-1 2n+1 -Xn n² + 1×” +... 88 DIFFERENTIAL AND INTEGRAL CALCULUS. : 2n+1' An-1 Here an (n − 1)² + 1 2(n-1)+1 n² + 1 X which = 1 when ʼn =∞. .. − 1 < x < 1. 4. x + 2²x² + 3³½³ + ... (n • - х x² 5. + 1+ √Ĩ 1+ √2 +. − 1)²x²-¹ + n²x" +... 1)²xn−1 +n²x² xr-1 Xh − 1 < x < 1. 1 + √ ñ − 1 + 1 + √ ñ DEVELOPMENT OF FUNCTIONS. +... - 1 < x < 1. 120. There are two common and useful formulas for de- veloping functions: Taylor's and Maclaurin's. We shall deduce Taylor's formula first, as Maclaurin's may be derived from it. 121. Taylor's Formula is a formula for developing f(y + x) in a series, where f(y + x) represents any function of the sum of two variables, such as (y + x)", log (y + x), sin (y + x), av+*. The derivation of Taylor's formula may be regarded simply as the process of finding the acceleration of a function. Thus, let u = f(y), and suppose y to be increased by x, we then have (Art. 24), or ▲u = f(y + x) − f(y) = Ax + m„x”, f(y + x) = f(y) + Ax + m„x², 1 (1) (2) where A (= m₁ = ƒ'(y)) is a constant with respect to x, and m, is the acceleration of u (Art. 25), and this is what we wish now to determine. Since m, is of such a character that ma vanishes with x, we will assume (see footnote, p. 10) ՊՈՆՉ = 3 B + Cx + Dx² +... Lx²-³, (3) where B, C, D,... L are independent of x, and x has such a value as to render the series (if infinite) convergent. Substitut- ing in (2), we have f(y + x) = f(y) + Ax + Bx² + Cx³ +... Lx²-¹. . (4) DEVELOPMENT OF FUNCTIONS. 89 Differentiating (4) successively with respect to x, we have Lxn−2; A+ 2Bx+3Cx²+... (n-1) Lx-2; (5) f'(y + x) = ƒ"(y + x) = 2B + 6Cx + . . . (n − 1) (n − 2) Lx-3; (6) ƒ'''(y+x) = 60 + . . . (n − 1) (n = 6C + . . . (n − 1) (n − 2) (n − 3) Lx-±; (7) ; ƒn-¹(y + x) = \n - 1 L. (8) These equations, (5), (6), (7), etc., are true for any value of x which renders equation (4) convergent (Art. 119); therefore they are true when x = 0, which gives f(y) = A, .. A = ƒ'(y); 1 ƒ"(y) = 2B, :. :: B = f(y); f(y) = 6C, .. C f(y); 13 1 f-(y). n 1 f"−1(y) = \n — 1L, .. L= Substituting these values for A, B, C, . . . L in (4), we have x² f(y+x) = f(y)+ƒ'(y)x+ƒ'' (y)/2+ƒ'''(y) X 3 XR-1 (A) +...fm-(y) n 1 +. This is the formula required, which was first published in 1715 by Dr. Brook Taylor, from whom it takes its name. The preceding is not a rigorous demonstration of Taylor's formula, inasmuch as the possibility of development in the proposed form is assumed. A rigorous proof, including the form of the remainder, has been inserted in the Appendix, A., to be used or not, according as the teacher or student may desire. 90 DIFFERENTIAL AND INTEGRAL CALCULUS. i 122. COR. I. To determine for what values of x the series is convergent. The nth and (n+1)th terms of the series are evidently fr-(y) x²-1 n - 1 Xn and ƒ*(y)\n' Therefore (Art. 117) the series is convergent for any value of a numerically less than fn-1 (y). f^(y) (F)), when ₂ =∞. f-(y) or n — 1 n1 n f(y) Hence, if fn-(y) f" (y) ∞, the series is con- vergent for all finite values of x; and if n is not zero when n =∞o, f(y) fr (y) = 0, when n =∞, the series is divergent for all values of x except 0. In deducing Taylor's formula we have supposed all the func- tions that occur to be continuous. Hence the formula is inap- plicable, or "fails," if the function, or any of its differential coefficients, be infinite for values of the variable lying between the limits for which the development holds. 123. To develop (y + x). Here Make x = 0, f(y + x) = (y + x)™. f(y) = ym. Differentiate, etc., f'(y) = mym-1, ƒ"(y) = m(m − 1)ym-2, ƒ'''(y) = m(m − 1)(m — 2)ym-3, etc. etc. Substituting these values in (A), we have m(m − 1) (y + x)m = ym +mx ym−1+ — 1) x²ym-2 m(m − 1)(m − 2) + 1) (m − 2) µ³ym-³, etc., 13 which is the Binomial Theorem. (B) DEVELOPMENT OF FUNCTIONS. 91 COR. I. Let us determine for what values of x the equation is true, supposing m negative or fractional. ƒn-¹(y) = m(m − 1)... (m − n + 2)ym−n+¹, ƒn(y) = m(m − 1) . . . (m (m − n + 1)ym-n; .. (Art. 122), fn-1(y) ny N- f(y) m n + 1 which y when n∞. Therefore formula (B) is true when x is numerically less than y. COR. II. Making y in (B) equal to 1, we have (1+x)m=1+mx+ m(m − 1) m(m—1) (m −2)µ³ + etc., (C) | ২৩ x²+ 3 - in which 1 < x < +1. 124. To develop sin (y + x). Here f(y+x) = sin (y + x). Making x = 0, and differentiating, we have f(y) = sin y; = f'(y) cos y; ƒ"(y) = - sin y; f""(y) = cos y; fiv(y) = sin y; etc. Substituting these values in (A), we have sin (y + x)=sin y (1- x² X⭑ + + etc. 2 6 (D) X5 х3 + cos y (x-+-+ etc.) 15 12 COR. I. In (D) by making y = 0, remembering that sin 00 and cos 0 = 1, we have sin x = x хо + 13 |5 x² + etc. 7 (E) 92 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 . COR. II. Differentiating (E), we have x² X* COS X = =1– + + etc. 4 16 (F) COR. III. For the quantities within the parentheses in (D), substituting their values from (E) and (F), we have sin (y + x) = sin y cos x + cos y sin x. (G) COR. IV. Differentiating (G), regarding x as constant and y as variable, we have cos (y + x) = cos y cos x sin y sin x. (H) 125. To develop log (y + x). f(y + x) = log (y + x); making x = 0, and differentiating, we have 1 f" f(y) = log (y); ƒ'(y) ==;; ƒ”(y)=- 2 f'"' (y) = ya; fiv (y) = = 13 y, etc. 1 y², Substituting in (A), we have log (y + x) = log (y) + which is the logarithmic series. X x² X 2 Y + Qy² 3y³ etc., (I) COR. I. The nth and (n + 1)th terms of (I) are, omitting x²-1 хт the signs, (n 1)y"-1 ny" and ; hence, Art. 117, an-1 an ny n — I' which = y when n∞. Therefore formula (I) is true when x is numerically less.than y. COR. II. In (I), by making y = 1, we have x² X³ X⭑ log (1 + x) = x + + etc., 2 3 which is true for all values of a numerically less than 1. X f (K) DEVELOPMENT OF FUNCTIONS. 93 126. Maclaurin's Formula is a formula for developing a function of a single variable, as y = a", y = log (1 + x), y = (a + x)". It may be derived from (A) by making y = 0, which gives x² X ƒ(x) = f(0) +ƒ'(0)x+ƒ''(0) +ƒ'''(0)- X³ · +... fn-¹(0). xn-1 n - 1 +..., (L) in which ƒ(0), ƒ'(0), ƒ'''(0), etc., represent the values which f(x) and its successive derivatives assume when x = = 0. COR. I. Formula (L) is true for all values of x numerically n f¹−1(0) less than when n∞. fr (0) 127. To develop a". Here f(x) = a*, f'(x) = a* loga, f'(x) = a log³ a, :: ƒ(0) = a° = 1; :: f'(0) = log a; . f'(0) = log" a; f'"'(x) = a log³ a, etc. · . f'"'(0) = log³ a; etc. Substituting these values in (L), we have a = 1 + log ax + log² a x² + log³ a 2 13 +log* a (M) 4 which is called the Exponential Series. COR. I. This series is convergent for all finite values of x, since f"-1(0)÷f*(0) is obviously finite and different from zero for all values of n. COR. II. Making a e, remembering that log e = 1; we have x² e²=1+x+ 12 + хо X⭑ + + etc.. 3 |4 (N) 94 DIFFERENTIAL AND INTEGRAL CALCULUS. COR. III. Making x = 1, we have 1 1 1 e=1+1+ 3 12 + 1 + 1 + 1/1/ + etc. (P) Hence e 2.718281 +. 128. Find the development of tan-¹x. In the applications of Maclaurin's formula the labor of find- ing the successive derivatives can often be lessened by taking the development of the first derivative, as follows: f(x) = tan-¹x, f'(x) - 1 1 + x² 1 ..ƒ(0) = tan-¹0 = 0; ..ƒ'(0) =1; − 1 − x² + ¿¹ −x + etc., 3 ƒ''(x) = -2x+4x³ — 6x + etc., ƒ'''(x) = − 2 +3.4. x²-5.6x+ etc., fiv(x) = 2.3.4x-4.5. 6x + etc., ƒ'(x) = |4 −3.4.5.6x² + etc., ..ƒ''(0) = 0; ..ƒ'''(0) = — 2; .. f¹v (0) = 0; ..ƒ˜(0) = 4; fvi(2) 2.3.4.5.6x + etc., .. fri(0) = 0; frii (x) 16+ etc., .. ƒvii (0) -16; etc., etc. Substituting in (L), we have 3 tan-1 x = x Xx хо x² + 3 5 + etc. (Q) EXAMPLES. Develop the following: 1. √1+x². x² 1+ 2 X® + 5x8 8 16 128 + etc. Put x²=y, and develop; then replace y by its value. 2. (α-+ x)-³. 3. (1+x)³. а-3 X 1+ 3 3a‍¹x+6α¯³x² 10a-6x+ etc. x² 5x³ 3 + etc. 81 DEVELOPMENT OF FUNCTIONS. 95 x² ха x5 X® 4. esin x. 1+x+ + + etc. 2 2.4 3.5 2.4.5.6 5. ecos x e(1 - 2352 + 4x¹ 31x 4 16 + etc.). 6. tan x. 3 7. sec x. + etc. 8. log (1 + sin x). X 2 + etc. 9. log (1e). log 2 + ૨૭/૪ x² + etc. 8 10. ex sin x 1 + x² + +etc. 3 2x³ 11. ex sec x. 1 + x + x² + + etc. 3 x² 2x³ 12. log (1 x + x²). x + 2 + 3 + etc. 4 2x5 x + + + etc. x² 1+ 2 + + 15 5x¹ 24 X3 6 12 Putxxy, and develop; then replace y by its value. In the two following examples put y for x², develop, and re- place y by its value. x² 13. √1 — x². 2 5x8 - 1. etc. 2 8 16 128 1 14. 1- + √1+x² 3x¹ 2.4 3.5x + etc. 2.4.6 129. To find the value of π. We find by development that sin-¹x = x + 1.3x 1.3.5x7 + 2.3 + 2.4.5 2.4.6.7 + etc., where x lies between 1 and + 1. 96 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 or Making, remembering that sin-1 = π =1(1+ 6 2 π we have 6 1 3 5 24 + + 640 7168 +), π = 3.141592 + 130. To compute Natural logarithms. We have found log (1 + x) 4 = x Xx 205 + 2 3 4 + 5 etc.,. (1) where x is numerically less than 1. We now proceed to modify this series so that it shall be true and convergent for larger values of x. x for x in (1), we have Substituting - x² X³ X⭑ log (1 - x) = X 2 3 4 에 ​etc. . (2) Subtracting (2) from (1), we have log (1 + 2) — log (1 − x) = 2(x + X3 X 3 + + ry 272+ etc.).. (3) 1 In (3) make x = 2z+1 و where z may have any positive 1 + x % +1 1 − x Z value; then and we have log (1 + x) — log (1 log (1-x)= log (z+1) - log (2), log (z+1) = log (z) 1 1 1 + 2 [ 22 +=+= + 3(2% +1)³ † 5(2% +1)° 2z+1 + etc.] (4) By this series we can compute the Natural logarithm of any number (z+1) when we know the logarithm of the number (z) less by unity. z Making = 1, remembering that log (1) = 0, we have log (2) = = 2 =[+ 1 3.33 1 + + + etc. 5.3° 7.37 1 : DEVELOPMENT OF FUNCTIONS. 97 3 Taking six terms of this series, we have log 2 = .693147 + Putting z = 2 in (4), we have log (3) = log 2 + 2(3 3 + = 1.098612 + 1 1 + + + etc.) 1 3.53 5.55 log (4) = 2 log 2 = 1.386294 +. Putting z = 4, we have 7.5' log (5) = log 4 + 2 4+2 + G+ 1 1 1 + 3.9³ 5.9° + + etc. 77.97 etc.) = 1.609437 +. log 10 = log 2+ log 5 = 2.302585 +. In this way we can compute the natural logarithms of all numbers. It is not necessary to use the formula in finding the logarithms of composite members, for they can be found by simply adding the logarithms of their factors. Thus log 15 = log 3 + log 5. 131. To compute common logarithms. The modulus of the common system is m = log₁e (Art. 79). Hence 10me, .. log (10m) = log e, or m log 10 = 1. .*. m = Let then 1 log 10 1 2.302585 = .434294 +. log₁, v = n and log v = n'; 10" v and en = v; 10"=e", n log 10 = n'. log (10") = log e"', or .. n = (n') .434294 +. Hence, to find the common logarithm of any number we multiply the natural logarithm of that number by the modulus of the common system. 98 DIFFERENTIAL AND INTEGRAL CALCULUS. INDETERMINATE FORMS. 132. In algebra is called a symbol of indetermination, since any number whatever may assume this form. Thus, n × 0 =0: divide both sides by 0 and we have n = 010 There are many fractions which assume the form of O - in 0 consequence of one and the same supposition, which makes both numerator and denominator = 0. Such fractions are called Vanishing Fractions, and their values, which appear under the 0 form of can generally be determined by the calculus. 0 Thus the fraction X X x² α a² becomes 0- when x = α. This form arises from the existence of a factor (x a) com- mon to both numerator and denominator, which factor becomes O under the particular supposition. Dividing both terms by this factor, we have x³ — a³ x² — a² x² + ax + a² which = x + a За 2 when x = α. ŷ 133. To evaluate a fraction that takes the form of = 0, Let u and v be functions of x such that when x = a, u = and v = 0. Let u and v be estimated from the point where their values are 0, that is, from where 2 = a; then when x (= a) is increased by h we shall have, identically, ՂՆ Au = V Δύ INDETERMINATE FORMS 99 Ash approaches 0, or x approaches a, the limit of du equal to dv (Art. 27); therefore น บ 1. du = dv α U du which is read, when x = a, is equal to V dv Applying this to the preceding example, we have X³ 3 a d(x³ — a³) 3x² За = x² a² d(x² a²) 2x 2. Δη is Δυ An easy deduction of the rule for reckoning forms like is obtained by the use of Taylor's formula, as follows: Let (x) and f(x) be two functions of x such that f(x) = 0 and 4(x) = 0, when x = a, then we shall have (a) 0 Evidently f(a) $(a) f(a) limit (a + h) 6(a h=0Lf(a + h) limit [¤(a)+0′(a)h+30″(a)h²+.. [? p² h = 0\ƒ(a)+ƒ'(a)h +¥ƒ''(a)h² + = p'(a) φ(α) f'(a) f(x). a COR. I. If ø′(a) = 0 and ƒ’(a) = 0, we obviously have Ф(х) = p" (a) ƒ"(a) f(x). ; a and if ø′′(a) = 0, and ƒ"(a) = 0, we have p'''(a) (x)7 ƒ'''(a)¯¯ ƒ(x). α ; and so on. Hence, RULE. Substitute for the numerator and denom- inator, respectively, their first derivatives, or their second derivatives, and so on, till a fraction is obtained whose terms do Um 100 DIFFERENTIAL AND INTEGRAL CALCULUS. not both become 0 when x = a; the values thus found will be the true value of the vanishing fraction. Compute the following: EXAMPLES. X$ 17 LO 5 → 1. x² — 1 x² 16 2. x² + x 5x2 — 20 8x +37 3. 7x² 9x+2 2 5 1 3x + 2 3 4. X x² x + 1 2 1 COS X 5. Ŏ. sin x a 6. X2 sin x 7. - 2011 2aⓇ X X sin x 8. 3 9. X x² sin 3x sin 2x 1 6 Hico ao laz 3 ex e 10. 11. (π sin x log sin x 2x) 2 嗣​。 2 sec² x — 2 tan x 1 až H100 mlaz 12. 1 + cos 4x 2' 4 sin x X COS X 13. 2. 14. X sin x (x-2)e* + x + 2- ex+x+2 (ex — 1)³ 21. 1 6 INDETERMINATE FORMS. 101 There are other indeterminate forms besides such as 88 ∞ × 0, ∞ -∞, 0°, ∞ °, 1°, which will be considered in succes- sion. 134. To evaluate a fraction that takes the form of Let u and v be functions of x such that u∞ and v∞o 1 1 when x = a; then for the same value of x, = 0 and = 0. : И ข Hence .. Art. 133, ala 212 1 V when x = a. U ( u³dv when xα. . v²du' (1) Dividing (1) by W d U we obtain , V udv И du 1= or = vdu' ว a dv α И (2) И is = 0 or ∞, it Now (2) is derived from (1) by dividing by ~; hence, if 10 ย finite, (2) is true for all finite values of; and if may be shown that (2) is true in these cases also. Suppose O when x = a, and k a finite quantity, И บ V then И +h + k = u + kv = k. ข V To this last fraction (2) evidently applies, hence u + kv du + kdv or V dv И V du + k = +k; dv И du that is, V dv' when x = α. 102 DIFFERENTIAL AND INTEGRAL CALCULUS. ! V If = ∞, then ย U Therefore the form =0, and we have the preceding case. 8|8 is to be evaluated in the same way as the form EXAMPLES. Find the values of the following: X 1. log x = -#-1-11--1 = d (log x) ax² + b 2. cx² + d ax 17 3. X log (x − 7 (z 4. tan x log x 5. cot x log cot x 6. cosec x log tan 2x 7. log tan x 8. 9. २७/६ sio C ∞. 0. N 2 0. 0. tan [77(x+1) πα tan 2 log cos (πα) log (1-x) . 1. - 2. 1. INDETERMINATE FORMS. 103 135. To evaluate a function that takes the form of 0x∞o or∞∞. Functions of this kind can be transformed so as to assume 0 the form of o or and then be evaluated by the previous ∞ methods. Find the following: 1. sec 3x cos 5x] ર X EXAMPLES. This takes the form of ∞o X 0; but sec 3x cos 5x which takes the form of o 2. sec x tan x રાત્રે This takes the form of ∞ ∞; but sec x 5 3 ن انت cos 5x cos 3x' 0. 1 tan x = COS X sin x 1 - sin x which takes the form of cos x cos x 1 0° 1 1 3. log x X - 2* 5. 4. cosec² x € ex - e X 6. (1 — tan x) sec 2x]„ 1 1 X 1 1 1 1. 4 X a² πα 4 77. tan a² 2a π a 8. x sin X 9. xm log" x]. (m and n being +.) 10. (1 - x) tan (7)]. ¿ Rik 0. 104 DIFFERENTIAL AND INTEGRAL CALCULUS. " -- 136. To evaluate a function that takes the form of 0°, ∞ °, or 1°. Take the logarithm of the given function, which will as- sume the form of 0 × ∞, and can be evaluated by Art. 135. From this the value of the function can be found. EXAMPLES. Find the following: 1. (1 + 2 ) * ] : X This takes the form of 1". Making y = log y = x log (1 + 2). ea =(1+ (1 + 2), we have a The value of x log (1 + 2)] is found by Art. 135 to be a. Hence, when x = ∞, log y = a; .. y = eª. 2. Vx], or Xx 1. 1 1 X This takes the form of °; the log of x is log x, the value 1 of which, when x = ∞, is 0; hence x-] = 1. x 3. (sin x)tan ∞]½· 4. (cot x)sin x]. 5. (e* + 1)²]. 6. (tan +7)=7] πχ tan 4 1 7. (cot x) log]. 8. Vlog (e + x)].. 9. Ve*+x]。. x2 10. Vcos 2.1. 1. 1. e. 1 1 • HIO HIO HIO OHIO e 1 ee. e². 1 INDETERMINATE FORMS. 105 11. log x 1. X n 12. (cos mx)x² Inm² e 137. In implicit functions, as f(x, y) = 0, the derivative dy can be evaluated by the previous methods when it assumes dx an indeterminate form for particular values of x and y. EXAMPLES. 1. Find the slope of x¹ — a³xy + b²y² at the point (0, 0). dy Here 4x³ a²y 0 when x = y = 0. dx a²x - 2b³y dy dy 12x² a² a? dy dx dx dy a² 262 2b² dy a² dy' when x = y = 0; 262 dx dx ady that is, dx dy dx a² Hedy or 26°(dy a² (dy dy = a² dy' \dx \dx dx; dx dy a² .. =0 0 or dx b2. 2. Find the slopes of x³ — 3axy+y=0 at (0, 0). 3. Find the slopes of x+axy — ay³ = 0 at (0, 0). 4. Find the slopes of y² = x(x+a)² at (— a, 0). 5. Find the slopes of x+2ax'y - ay³ = 0 at (0, 0). — O and ∞. O and ± 1. ± √ − a. 0 and ± 1/2. 1 CHAPTER VI. MAXIMA AND MINIMA. DEFINITIONS AND PRINCIPLES. 138. A Maximum Value of a function is a value that is greater than its immediately preceding and succeeding values, and a Minimum Value is one that is less than its immediately preceding and succeeding values. Thus, while x increases continuously, if f(x) increases up to a certain value, say f(a), and then decreases, f(a) is a maximum value of f(x); and if, while x increases, f(x) decreases to a certain value, say f(b), and then increases, f(b) is a minimum value of f(x). For example, sin x increases as x increases till the latter reaches 90°, after which sin a decreases as x goes on increasing; that is, sin 90° is a maximum value of sin x. Again, if x increases continuously from 0 to 5, f(x) = x² — 6x +10 will decrease until x becomes 3 and then it will increase; hence ƒ(3) = 1 is a minimum value of f(x), or x² 6x + 10. Let the student substitute 1, 2, 3,. . . 10, successively, for x in f(x) = x³ 18x² + 96x 20, and thus show that f(4) is a maximum and f(8) is a minimum. 139. Any value of x that renders f(x) a maximum or a minimum is a root of the equation f'(x) = 0 or ∞, if f(x) and f'(x) vary continuously with x. For, if we conceive x as always increasing, f(x) changes from an increasing to a decreasing function as it passes through a 106 MAXIMA AND MINIMA. 107 maximum value, say f(a), and from a decreasing to an increas- ing function as it passes through a minimum value, say f(b). Consequently f'(x) must change sign as a passes through a or b, Art. 25. But f'(x) can change sign only by passing through 0 or ∞. Therefore ƒ'(a) or ƒ'(b) = 0 or ∞ ; that is, a and b are roots* of f'(x) = 0 or ∞. A O a B b с d 1 de h m FIG. 19. To illustrate the preceding principles and definitions graphically, let y = f(x) be the equation of the curve Am; then f'(x) = the slope of the curve at the point P or (x, y), Art. 48. As x (OB) increases, the point P will move from A along the curve to the right, and y or f(x) will in- crease till it becomes aa', and then decrease till it becomes bb', and then increase, etc. Therefore aa',, cc,', ee' are maximum, and bb′, dd' are minimum, values of f(x). The slope of the curve, f'(x), is evidently positive before, and negative after, each maximum value of f(x); and negative before, and positive after, each minimum value of f(x). Moreover, at the points where f(x) is a maximum or a minimum, the curve is either parallel or per- pendicular to the axis of x, and therefore ƒ'(x) = 0 or ∞ . The converse of this theorem is not always true; that is, any root of ƒ′(x) = 0 or ∞ does not necessarily render ƒ(x) a max- imum or a minimum. It is our purpose now to determine which of the roots will render f(x) a maximum and which a minimum. 140. If the sign of f'(x) undergoes no change as ƒ'(x) passes through 0 or ∞, the corresponding value of f(x) will be neither a maximum nor a minimum. For, so long as the sign of f'(x) undergoes no change, f(x). * Here, and in what follows, the word root includes the real values of x which satisfy the equations ƒ'(x) = 0 or ∞, whether ƒ'(x) be an algebraic or a transcendental function. 108 DIFFERENTIAL AND INTEGRAL CALCULUS. " does not change from an increasing to a decreasing function, nor vice versa. COR. I. If an even number of the roots of ƒ′(x) = 0 or ∞o are equal to a, then x = a will not render f(x) a maximum of a minimum. 141. If the sign of f'(x) undergoes a change as f'(x) passes through O or∞, the corresponding value of f(x) is a maximum 0 or a minimuит. For, if f'(x) undergoes a change of sign, f(x) necessarily passes from an increasing to a decreasing function, or vice versa. COR. I. If an odd number of the roots of f'(x) =0 or ∞ are cqual to a, then x = a will render f(a) a maximum or a minimum. COR. II. Therefore, omitting the equal roots of which the number is even, every real root of f'(x) = 0 or ∞ will render f(x) either a maximum or a minimum. Now, of these let us find which will render f(x) a maximum and which a minimum. 142. Maxima and minima of a function occur alternately. For, suppose that ƒ(a) and ƒ(b) are maxima of f(x), where a < b. Just after passing through ƒ(a), f(x) is decreasing, and increasing just before it reaches f(b); but in passing from a decreasing to an increasing state it must pass through a mini- mum; hence there is one minimum between every two consecu- tive maxima. a 3 COR. I. Denote the roots of f'(x) = 0 and ƒ'(x) = ∞ which will render f(x) a maximum or a minimum by ɑ¸, ɑ,, ɑ¸, ɑ, etc., in ascending order of algebraic magnitude. Then, if f(x) is an increasing function for all values of x less than a,, that is, if f'(a,h) is positive, h being ever so small, f(a,), f(a.), f(a.), etc., are maxima, and f(a,), f(a.), etc., are minima; and if f(x) is a decreasing function for the same values of x, that is, if f'(a, — h) is negative, the maxima and minima will be inter- changed. 1. - MAXIMA AND MINIMA. 109 1 RULES FOR FINDING MAXIMUM AND MINIMUM VALUES OF FUNCTIONS. 143. The preceding principles indicate the following rule for finding the values of x which will render any function as f(x) a maximum or minimum: = Differentiate the function f(x); make f'(x) = 0 and f'(x) = ∞o; find the real roots of both equations, and arrange all of them in order of algebraic magnitude, as a,, a, a,, etc., omitting the equal roots when there are an even number of them; substitute - —∞ or a, -h, h being very small, for x in f'(x), and (1) if the result is +, a,, a,, etc., will each render ƒ(ɔ) a maximum, and a,, α, etc., will each render f(x) a min- imum; (2) if the result is -, f(a,), f(a,), etc., will be minima, and ƒ(a,), ƒ(α¸), etc., will be maxima. 00 144. The preceding rule requires that all the real roots shall be found; it is sometimes desirable to know independently whether any particular root as a' will render f(x) a maximum or minimum. This may be done thus: I. Substitute a'h and a'h for x in f'(x), h being a small quantity, and (1) if ƒ’(a' − h) is +, and ƒ'(a' + h) is f(a') will be a maximum; (2) if f'(a' h) is and ƒ'(a' + h) — is +, f(a') will be a minimum, and if f'(a' — h) and f'(a' + h) have the same sign, ƒ(a') will be neither a maximum nor a min- imum. 145. II. Developing f(xh) and f(x + h) by Taylor's formula, substituting a' for x, transposing ƒ(a'), and remember- ing that f'(a') = 0, we have — f(a′ — h) − f(a') = ƒ''('a')"="" h³ ht ƒ''' (a'). +fiv (a') (1) 13 and h² h³ h+ ƒ'(a' + h) − f(x') = ƒ''('a')" " " + f'"'"'('a')"},{" +ƒ'''' (a') == + fix (a') +. (2) 12 13 110 DIFFERENTIAL AND INTEGRAL CALCULUS. If h be taken very small, the sign of the second member of either (1) or (2) will be the same as the sign of its first term. Hence, if f''(a') is negative, ƒ(a') is greater than both ƒ(a′ – h) and ƒ(a' -+ h), and therefore a maximum; while if f''(a') is positive, f(a') is less than both f(a' – h) and ƒ(a' + h), and therefore a minimum. If f"(a') = 0, and f'"' (a') is not 0, f(a') is neither greater than both ƒ(a' — h) and f(a' + h) nor less than both, and is therefore neither a maximum nor a minimum. If ƒ'''(a') as well as f''(a') is 0, then, as before, f(a') will be a maximum or a minimum according as fiv(a') is negative or po- sitive; and so on. Hence if a' is a root of f'(x) = 0 or ∞, substitute it for x in the successive derivatives of f(x). If the first derivative that does not reduce to 0 is of an odd order, f(a') is neither a maxi- mum nor a minimum; but if the first derivative that does not reduce to 0 is of an even order, f(a') is a maximum or a mini- mum, according as this derivative is negative or positive. NOTE.-In many instances this rule is impracticable on ac- count of the great labor involved in finding the successive de- rivatives. 146. The following principles are self-evident, and often serve to facilitate the solution of problems in maxima and minima: (1) If c is positive, f(x) and c × f(x) are maxima or minima for the same value of x; hence a constant positive factor or divisor may be rejected in finding this value of x. (2) log f(x) and f(x) are maxima or minima for the same value of x; hence log may be rejected. (3) c+f(x) and f(x) are maxima or minima for the same value of x; hence the constant c may be rejected. (4) If n is a positive whole number, [f(x)]" and f(x) are n p maxima or minima for the same value of x; hence in [f(x)]² the denominator q may be rejected, or in f(x) the radical may be removed. 1 MAXIMA AND MINIMA. 111 3x² 24x+85 a EXAMPLES. 1. Find what values of x will render x³ maximum or a minimum. = 3x² Here f(x) = x³-3x² - 24x + 85, f'(x) 6x 24, ƒ''(x) = 6x — 6. Making f'(x) = 0, we have 3x²-6x-240, the roots of which are x = + 4, x = 2. Now to determine whether these values of x give maxima or minima values of f(x), we substitute them for x in ƒ''(x). Thus: X f''(4) = 6 x 4 − 6 = + 18, ƒ''(− 2) = 6 × −26 - - 18. Hence, Art. 145, when x = 4, f(x) is a minimum, and when x=2, f(x) is a maximum. 3 Let the student construct the locus of y = x³- 3x² and thus exhibit these results graphically. 2. Examine f(x) = x³ minima. — 2 242 +8%, 3x² + 3x + 7 for maxima and = Here f'(x)=3x² - 6x +3, ƒ"(x) = 6x — 6, ƒ""'(x) = 6. The roots of 36x30 are x 1, x 1. Substituting these = values of x in f'(x) and ƒ""(x), we have f"(1) = 0, ƒ""(1) = 6. Therefore the function f(x) has neither a maximum nor minimum value, which we also infer from the fact that the two roots of f'(x) = 0 are equal, Art. 140, Cor. I. 3. Find the maxima and minima of x5.— 5x¹ + 5x³ = 1. 20x³ 3 Here f'(a) 5x 20x + 15x'; .. f'(x) = 0 is 5x +1520, or (x²-4x+3)2= 0; the four roots of which are +15x² = 0, 0, 1, and 3. Rejecting the two equal roots, Art. 140, we have a,1,a,3, and since f'(- ∞) = 5(-∞) is +, the given function is a maximum when x = 1, and a minimum when x = 3. and Therefore f(1) = 0 is a maximum, ƒ(3) = — 28 is a minimum. 2 112 DIFFERENTIAL AND INTEGRAL CALCULUS ፡ 4. Examine (x-1)*(x+2)' for maxima and minima. Differentiating and reducing, we have ƒ'(x) = (x − 1)'(x+ 2)²(7x+5) The roots of f'(x) = 0 are those of (x 1) = 0, (x+2)= 0 and 7x+5= 0; hence there are three roots each equal to 1, two each equal to 2, and one equal to . - = Rejecting the two equal roots, we have a, and a, 1, and since f'(-∞) = ( — ∞)³ ( — ∞)²(— 7∞) is +, f(x) is a max imum when x = , and a minimum when x = 1. 5. Determine when b+c (x - a) is a maximum or mini- mum. By (3) of Art. 146 we may remove b, by (1) c, and by (4) 3; hence we have f(x) = (x − a)²; ·· ƒ (x) = 2(x—a), and x-a=0, or a₁ = a; and since f'(- ∞) is -, the given function is a min- imum when x = c. 1 6. Find the maximum and minimum values of f(x) = (x+3)³ (x+2) ³° Here f'(x) = x(x+3)² (x + 2)** I. f'(x) = 0 gives x(x+3)² = 0, of which one root is 0 and the other two are 3 and 3. II. f'(x) = ∞ gives (x+2) = 0, the three roots of which is each 2. Rejecting the two equal roots, we have a, 2, 2,0; and since f'(-∞) is +, f(− 2) = ∞ is a maximum value of f(x), and f(0) is a minimum. 7. If the derivative of f(x) is f'(x) = x² - 10x + 21, what values of x will render f(x) a maximum or minimum. The roots of x² - 10x+210 are 3 and 7; substituting these for x in f'(x) = 2x 10, we have f"(3) 610 = = 4, and ƒ"(7) 14 104; therefore f(3) is a maximum and f(7) is a minimum. Find the values of x which will give maximum and mini- mum values of the following functions: • MAXIMA AND MINIMA. 113 8. u = x² 8x+12. x = 4, min. 9. u = x³ 3x² 24 +85. 10. ux³-3x²+6x+7 x= 2, max. x = 4, min. Neither a max nor min. 3 11. u = 2x³ — 21x² + 36x - 20. 12. u = (x-9)³(x — 8)*. x² - 7x+6 X 10 x = = 1, max.. x = 6, min. X = 8, max., x = 84, min. x = 4, max.; x = 16, min. x = 3, max x = 13, min. 13. u = (x+2)³ 14. u = ( — 3)* x² + 3 15. u = x= x — 1° 1 x + x• 16. u = 1 + x (α — x)³ X 17 u = α 2x 4 18. u = + X 3 - x 9 19 u = sin x + cos x. 20. u = sin x (1 + cos x). 21. u = 22. u = sin x 1 + tan x X 1+x tan x , − 1, max.; x = 3, min. X = = ½ min. a X = min. 4' x= 9, max.; x = 13, min. π X = max. π X max. 3 > π X = max. 4 x= cos x, max. 23. u = (1 + x³) (7 — x)². x=1, max.; x=0 and x=7, min. 24. If x + y = n, what is the greatest possible value of xy? Make u = xy = x(n = x(n − x ) = n - x²· \n². 25. If y = mx + c, find the least possible value of x²+y C √m² + 1 Make u = √ x² + y² = √ (1 + m²)x² + 2mcx + c². 114 DIFFERENTIAL AND INTEGRAL CALCULUS. 26. A merchant bought a bolt of linen, paying as many cents for each yard as there were yards in the bolt, and sold it at 20 cts. per yard; required the greatest possible profit. $1.00. 27. A club of x members has x³ 12x² + 45x + 15 dollars in its treasury; how much is that apiece if the amount is (1) a minimum? (2) A maximum? (1) $13.00; (2) $23.00. 28. Find the value of when sin cos is a maximum. 4.= cos¹ — 2√2 = 135°. 29. Find the fraction that exceeds its square by the greatest possible quantity. 1. 30. Find the fraction that exceeds its nth power by the greatest possible quantity. 1 (~14) n-1 31. Find a number x such that its xth root shall be a max- imum. x = e = 2.7182 +. 32. Find the altitude of the maximum rectangle inscribed in a given triangle. Let ABC be the triangle and HKFE the required maxi- mum rectangle; let b=A B, h = CD, x=DG, and u = area of HKFE; then C E F G A B H D K FIG. 20. b (h is required. u = EF × DG = (EF)x. To express EF in terms of x, we have or • CD CG.: AB: EF, hh-x::b: EF; .. EF = — (1 − x), and u = Dropping h whence x = 201 b > b ñ (hx — x²), whose maximum value we have f(x) = hx - x², . f'(x) = h 2x = 0, ; that is, the altitude of the maximum rectangle is one half of the altitude of the triangle. MAXIMA AND MINIMA. 115 .. 33. Find the altitude of the maximum rectangle inscribed in a given parabola. Let BAC be the parabola, and IH the rectangle; let h= AD, x = AE, y = EH, and u = area of IH. Then and y² = 4ax, u = 2(ED)(EH) = 2(h - x)y =2(h― x)√4a x* = 4√a (hx* — x³). A G E H B D K FIG. 21. Hence f(x) = hx - x, and f'(x) = \hx¯¹ — fx* = 0, h or Nx = 3√x; whence x = 3h; . DE = {h. 34. Find the altitude of a maximum cylinder with respect to its volume that can be inscribed in a given right cone. F A B D E FIG. 22. Let BD = b, and v = vol- Let ED be the altitude of the cylinder inscribed in the cone DAC. AD = h, ED = x, EF = y, ume of the cylinder; then v = π(EF)'ED = X. To express y in terms of x, we have AD: BD:: AE: EF, or h:b::h-x:y; whence b y = h (h-x), and v = πzz (h−x)³x. = .. f(x) = (h—x)x h²x 2hx² + x³, which is to be a maximum. h² f'(x) = h² - 4hx+3x= 0; whence 3x0, or xh. That is, the altitude of the cylinder is of the altitude of the cone. 35. Find the altitude of the cylinder in Fig. 22, if the cyl- inder is a maximum with respect to its lateral surface. "י 116 DIFFERENTIAL AND INTEGRAL CALCULUS. Denoting the lateral surface of the cylinder by S, we have b S = 2x(EF)ED = 2xyx = 27(h-x)x, which is to be a max- b imum. Dropping the constant factor 27, we have † (x) = hx − x²; x²; : f'(x) = h 2x = 0, or x = h 201 36. Find the dimensions of a cylindrical open-top vessel which has the least surface with a given capacity. Let x = the radius of the base, y = the altitude, s = the surface, and c = the capacity. Then c = πx³у • (1) (1) and s = πχ + 2πχ. . (2) C 2c From (1), y = πλ .. 8= 7 + which is to be a min- X imum. 2c ··ƒ(x) = πx²+ , ƒ'(x) = = 2πα X 212 2c = 0; 3 C whence x = and y = , which is obtained by substitut- π π C ing for x in y = πχ 37. A rectangle is inscribed in a circle whose radius is R; find the sides of the rectangle when it is a maximum (1) with respect to its area, (2) with respect to its perimeter. Each side = R √2 in both cases. 38. The hypothenuse of a right triangle is h; find the ratio of the other sides when the triangle is a maximum (1) with re- spect to its area, (2) with respect to its perimeter. Ratio = 1 in both cases. 39. A cylinder is inscribed in a sphere whose radius is R; find the radius of the cylinder when it is a maximum (1) with respect to its volume, (2) with respect to its convex surface. (1) ± √бR; (2) † VTR. MAXIMA AND MINIMA. 117 40. A cone is inscribed in. a sphere whose radius is R; find the altitude of the cone when it is a maximum (1) with respect to its volume, (2) with respect to its convex surface. (1) K; (2) R. 41. Find the maximum isosceles triangle with respect to its area that can be inscribed in a given circle. An equilateral triangle. 42. Find the dimensions of a cone which has the greatest volume with a given amount of surface. The slant height is three times the radius of the base. 43 Find the shortest distance from the point (x′ = 1, y′ 2) to the line 3y == 4x + 12. Ans. 2. y' 44. Find the shortest distance from the point (x′ = 2, y′ = 1) to the parabola y² = 4x. 2 √2 45. A square sheet of tin has a square cut out at each corner, find the side of the square cut out when the remainder of the sheet will form an open-top box of maximum capacity. A side = the side of the sheet of tin. 46. A man is at one corner of a square field whose sides are each 780 yards and wishes to go to the opposite corner in the least possible time; (1) how far along the side must he go before turning across the field if he can travel along the side and through the field at the rates, respectively, of 65 and 25 yards per minute? (2) In what time will he reach the opposite (1) 455 yards; (2) 40 min. 48 sec. 47. Find the altitude of the least isosceles triangle circum- scribed about an ellipse whose semi-axes are a and b, the base of the triangle being parallel to the major axis. corner? 3b. 48. A steamer whose speed is 8 knots per hour and course due north sights another steamer directly ahead, whose speed is 10 knots and whose course is due west. What must be the course of the first steamer to cross the track of the second at the least possible distance from her? N. (cos-¹ 3) W. 49 If the statue of Washington on the cupola of the Capitol is a feet in height and b feet above the level of an observer's eyes, at what horizontal distance from the centre of the cupola Y 118 DIFFERENTIAL AND INTEGRAL CALCULUS. should the observer stand to obtain the most favorable view of the statue? vb(a+b) feet. MAXIMA AND MINIMA OF FUNCTIONS OF TWO INDEPENDENT VARIABLES. 147 Definition.-A function, u = f(x, y), of two independ- ent variables has a maximum or minimum value according as f(x+ h, y + k) < f(x, y), or f(x + h, y + k) > f(x, y), for all small values of h and k, positive or negative. 148. Conditions for maxima and minima.—In the function u = f(x, y) if we suppose x and y to vary simultaneously, it is obvious from Art. 139, that the maximum or minimum values of u will occur at the points where the total differential of u, [du], is equal to zero. That is, when du du [du] = dx dy dx+ dy = 0.. (1) - As dx (= h) and dy (= k) are independent of each other, each term of (1) must be equal to zero. Hence du du = 0, and = 0... dx dy (2) These equations express the first conditions essential to the existence of either a maximum or a minimum. Again, as u passes through a maximum or minimum value, [du] changes from + to, or to +, respectively; therefore, in the former case [du] is decreasing, hence [d'u] is, and in the latter [du] is increasing, hence [a³u] is +. But the signs of d²u dx² d'u [d'u] = a* dx² + 2d²u ·dx dy + dy dy². dyzdy? dx dy (3) must evidently be independent of the signs of dr and dy, how- ever large or small these differentials may be supposed to be. 'This can be the case only when d'u I d²u dx² -)(du > dx dy (4) MAXIMA AND MINIMA. 119 For, making A = ďu dx³ d'u ď³u B = Ꮯ 29 dx dy' dy² 2) we have 2 Ah² + 2 Bhk + Ck² = (5) A (Ah + Bk) +(AC − B*)l In order that (5) may preserve the same sign for all values of h and k, it is necessary that AC - B' should be positive; for if negative, the numerator of (5) will be positive when k = 0, and negative when Ah + Bk = 0. Hence we have as an additional condition for a maximum or a minimum, AC > B', or (4). With this condition, the sign of (5) depends on that of the Hence for a maximum we must have denominator A. d'u A or <0,. dx2 and for a minimum d'u dx² A or > 0.. (6) (1) 1 It should be noticed that AC> B2 requires that A and C should have the same sign. = The exceptional cases, where B' AC, or where A = 0, B = 0, C = 0, require further investigation, but we shall not consider them in this book. The conditions for a maximum or minimum value of u = f(x, y) are then, viz.: For either a maximum or a minimum, du du 0, and 0;. P (8) dx dy d³u d²u d'u also • (9) 2 dx² dy' dx dy ď³u d²u For a maximum, <0, < 0. (10) dx² dy² d²u d'u For a minimum, > 0, > 0. (11) dx2 dy² に ​120 DIFFERENTIAL AND INTEGRAL CALCULUS. -- EXAMPLES. 1. Find the minimum value of u = x³ + y³ — 3axy. = du du Here dx = 3x² 3ay; 3y² - 3ax; dy a²u d'u d'u also 6.x, = 6y, За. dx² dy' dx dy whence x Applying (8), we have x²-ay0, and y' x = 0, y = 0; and The values = 0, y = 0, give = = = y³ — ax = 0; x = a, y = a.. ď² u dx² d'u d'u = 0, = 0, За, dy2 dx dy which do not satisfy (9). Hence they do not give a maximum or a minimum. The values = a, y = a, give d²u 3a, ď² u ď²u dx² 6α, 6a, dx dy dy² which satisfy both (9) and (11). Hence they give a minimum value of u, which is — a³. 2. Find the minimum value of α. x² + xy + y² ax by. §(ab — a² — b³). 3. Find the maximum value of (a — x)(a − y)(x + y − a). аз 27* 4. Required the triangle of maximum area that can be inscribed in a given circle. The triangle is equilateral. 5. Divide a into three parts, x, Y, α X y, such that their continued product, xy(a — x − y), may be the greatest possible. a x = y = a - x — y y = 3 MAXIMA AND MINIMA 121 6. Divide 45 into three parts, x, y, 45 - x - y, such that x³y³ (45 — x − y)* may be a maximum. x = 10, y = 15, 45 — x y = 20. 7. Find the least possible surface of a rectangular parallelo- piped whose volume is a³. 6a², 8. Find the dimensions of the greatest rectangular parallelo- piped that can be inscribed in the ellipsoid X2 y2 z² + + = 1. a² b2 c² ža √3, §b √3, &c √3. 1 CHAPTER VII. APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO PLANE CURVES. TANGENTS, NORMALS, AND ASYMPTOTES. 149. Equations of the Tangent and Normal. In Fig. 23 let P(x,, y,) be the point of tangency of the tangent TP; then the equation of TP is yy, = m(x − x,), where m is the tangent of the angle BTP. But, Art. 26, tan BTP therefore the equation of the tangent PT is = dy₁. dx, 1 " dy, (A) y = y₁ = 1 dx (x − x₁), 1 where 1 dy is the value of dy with respect to the curve AP at dx dx₁ the point (x,, y₁). Since the normal PN is perpendicular to the tangent or curve at P, its equation may be obtained from (A) by substitut- dx dy, which gives for ing dy, dx, 1 y - Y₁ dx₁ (x − x₁). dy, • • (B) EXAMPLES. 1. Find the equations of the tangent and normal to the pa- rabola y¹ = 4ax. 122 DIFFERENTIAL CALCULUS AND PLANE CURVES. 123 : Here dy 2a ; dx y dy dx, .. 1 dy, dx, 2a Y₁ Substituting this value of in (A) and (B), we have 2a 1 Za (x- y-y₁ = 1 (x − x₁), tangent; 1 y₁ y — y₁ = — Y₂ (x − x¸), normal. - 2a (1) (2) 2. Find the equations of the tangent and normal to the pa- rabola y² = 18x at the point x, Here 4a 18 and y,² 1 2. 18x,; .. 2a9 and y₁ = 6. Substituting in (1) and (2), and reducing, we have 2y=3x+6, tangent, and 3y=2x+22, normal. Find the equations of the tangent and normal to the follow- ing curves: 3. The circle, y³ + x² = R³. (1) yy,+xx, = R²; (2) yx, — xy₁ = 0. 4. The ellipse, a³y² + b²x² = a²b³. (1) a³yy,+ b²xx₁ = a²b²; (2) y — y₁ = 1 5. The cissoid, y²(2a — x) = x³. (1) tangent, y-y₁ = ± 1 1 a² y 1 (x (x − x¸). b²x₁ ·(x − x¸). (3α — x,) √x, (x (2α — x¸)* 6. Find the equation of the normal to y" 6x5 at y₁ = 5. y = {x + 40. 1 124 DIFFERENTIAL AND INTEGRAL CALCULUS. 7. Find the equation of the tangent at y, = 4 to the cycloid x= 10 vers 1 y 10 - √/20y — y³. y = 2x + 20(1 — vers -¹). 150. Length of Tangent, Normal, Subtangent, and Sub- normal. Let PT represent the tangent at the point P(x,, y₁), P Y PN the normal; then y, PB, BT is the = subtangent, and BN is the subnormal. Let the angle BTP, then BTBP cot = Y₁dy, dx TOA B N (1) = & ; FIG. 23. dx, that is, subtangent= (C) 1 ¹dy, 1 (2) that is, BN = BP tan BPN = BP tan ; dy, (D) subnormal = Y ₁ λ x 、 f ds, = (3) TP (BP÷ sin p) 4) = BP (Art. 49) dy, tangent= (4) that is, that is, PN=(BP÷cos ) = BP 2 In formulas (E) and (F), ds, = √dx," + dy,, Art. 33. If the subtangent be estimated from the point T, and the subnormal from B, each will be positive or negative according as it extends to the right or left. ds, (E) Y₁dy, 1 ds, ; dx ds, (F) normal = Y₁ dx, 1 DIFFERENTIAL CALCULUS AND PLANE CURVES. 125 EXAMPLES. 1. Find the values of the subtangent and subnormal of the ellipse a²y² + b²x² = a²b². dy, b² x 1 Here dx, 1 a²y, -; substituting in (C) and (D) we have Subt. a²y, b* . X a³ b³x, subn. = a² X 1 2. Find the values of the subtangent and subnormal of the ellipse 9y² + 4x² = 36, at x, = 1. = 1 Here a 3, b = 2, x,= 1, which substituted in the preced- ing answers give subt. 8, subn.. = Find the values of the following: subtangents and subnormals of the 3. y³ 3 = ax. 4. Parabola, y² = 4ax. 5. y = a. απ. a Subt. 3x,; = subn.= 3y, Subt. 2x,; = subn. = 2a. 1 Subt. = log a' subn. = a**, log a. 6. Find the length of the tangent of the tractrix, x = a log ( a + √a² Y ― y" 2 (a² — y³)*. Tang. = a. 7. Find the lengths of the normal and subnormal of the cycloid x = r vers-12 — √2ry — y². p = Norm. (2ry); subn. (2ry — y³). = 126 DIFFERENTIAL AND INTEGRAL CALCULUS. N Y B 151. Lengths of Tangent, Normal, Subtangent, and Sub- normal in Polar Co-ordinates. Let AP (s) be a curve, O the pole, OP (r) the radius vector, PT a tangent at P. Let O 8= XOP. A & Draw OT perpendicular to OP and prolong it to meet the normal X NP at N; then PT is the polar tan- gent, PN the polar normal, OT the polar subtangent, and ON the polar subnormal. Evidently, T FIG. 24. ONP OPT 4. (Art. 97) (1) OT = OP tan OPT = r(rd); (Art. 98) r²do that is, the polar subtangent = (G) dr (2) ON = OP cot ONP = r(rão); dr =r dr that is, the polar subnormal = do • . (H) dr (3) TP = (OP ÷ cos OPT) = r ÷ ; ds rds that is, the polar tangent= (I) dr rdo (4) PN=(OP÷sin ONP) = r ÷ ; ds ds that is, the polar normal = (J) d Ꮎ " In formulas (I) and (J) ds = √dr² + r²d0², Art. 97. +9 DIFFERENTIAL CALCULUS AND PLANE CURVES. 127 3 EXAMPLES. Find the tangent, normal, subtangent, and subnormal of the following polar curves: 1. The spiral of Archimedes, r = a0. ав 1 dr a s From (G), subt. α from (H), subn. = a; from (1), tang. == √a² + p²; a from (J), norm. = √aª +r². 2 The logarithmic spiral, r = aº. dr do = aº log a. Substituting in (G), (H), (I), (J), we find p subt. = log a =mr; subn. = m tang. = r√1+m²; norm. r√1+ log² a. = Find the subtangent and subnormal of the following: 3. The hyperbolic spiral, r✪ = a. Subt. a; subn. a 4. The Lemniscate of Bernouilli, r² = a² cos 20. Subt. = p3 a² sin 20; subn. = 20³ ༤ sin 20. go 128 DIFFERENTIAL AND INTEGRAL CALCULUS. ! 152. An Asymptote to a curve is a tangent which passes within a finite distance of the origin and touches the curve at an infinite distance. A curve which has no infinite branch can have no real asymptote. In Fig. 23, let x, and y, represent the intercepts OT and OD, respectively; then, in (A), Art. 149, by making (1) y = 0 and (2) x = 0, we find = dx, (1) x = x, — Y₁ x。 = OT; ¹dy, 1 dy, (2) Y₁ = Y₁ = x ₁dx, = OD.. 0 (K) (L) Now, if the curve AP is of such a character that x, or y。, or both, remain finite when x, or y,, or both, become infinite (see Art. 154), the tangent TP will be an asymptote to the curve. EXAMPLES. 1. Examine y³=6x+x³ for asymptotes. x; Since dy dr 4x + x² dy, y² 4x, +2,, which substituted in • Ax₁ 1 1 yi (K) and (L) give (1) x。 = x₂ y, 1 3 2 1 4x, + x,² 4 +1 X which - 2 when x, ∞. 4x² + x,³ 0 (2) y₁ = y₁- 1 1 yi 1 which = 2 when x₁ = ∞. 11 2 6 X +1)*³ ་ DIFFERENTIAL CALCULUS AND PLANE CURVES. 129 Therefore the straight line whose x and y intercepts are 2 and +2, respectively, is an asymptote to the curve. Since the asymptote passes through the points (— 2, 0) and (0, 2), its equation is y = x+2. 153. General Equation of the Asymptote. Since the asymptote passes through the points (x, 0) and (0, y), its equa- tion is 1 dy y = Ax (x − x.), . . . (M), or y y = dy ¹ x + y o⋅ x + y。· · · (N) ax, This equation enables us to determine whether or not any given curve has an asymptote, and, if it has, to find its equation. dy₁ and y, assume when x Let us denote the values which dx, by m, and b,, respectively; then we have 1 = ∞ y = mx + b, m¸x+b₁. (P) 154. When the terms of the equation f(x, y) = 0 are of different degrees, to find the relation of y to x when they are infinite, we may omit all the terms except the group which are of the highest degree with respect to x and y. Thus, when x is infinite, the equation ay² - bx² + cy + dx = e +cy 16 b gives ay² - bx²= 0, or y = V ± -X. 2 α a²b², A curve like a²y²+b²x² = a'b', or y = x²(a²x²), etc., which has no infinite branch or branches, has no real asymptote; this is indicated by the fact that when x is infinite, y, as deter- mined above, will be imaginary. 2 2. Find the asymptote of the hyperbola a'y² - b²x² — — a²b³. b When x = ∞, y = ± o, b ± −x, or y₁ = ±x a a = ་་ 130 DIFFERENTIAL AND INTEGRAL CALCULUS. . 1 dy, b²x, b²x² b2 1 ; dx, 0 y₁ = y₂- a²y, a²y, y₁ 1 b :: m₁ = ± α and b, 0, which substituted in (P) gives = b y = ±-x, Ans. ņ 3. Find the asymptote of the parabola y² = 4ax. ∞, y = ± 2 √ax, or When ∞∞, or y₁ = ± 2√ax¸. dy, dx, 2α ; Yo = Y₁ Y₁ 2ax, Y₁ 1 √ax,, which = ∞ when x, = ∞. 1 Therefore the parabola has no asymptote. 4. Find the asymptote of y³ When x = ∞, we have y³ = ax² + x³. = x³; y = x or y₁ = X1· • dy, dx, 2αx, +3x,² ax 1 1 3y, and yo 1 3y, 2 ; a 1 hence m₁ = 1 and b, = 3' α and the asymptote is y = x + 33 155. Asymptotes Determined by Inspection. When an asymptote is perpendicular to the axis of x or y, it can often be determined by inspection. In the first case m₁ = ∞, or which, substituted in (M), gives x-x, 0 1 dx, = 0, dy, = 0, since, in this case, = = 1 x=x,; that is, if y, is infinite when x, is finite, xx, 0 is the equation of the asymptote. Thus, in the cissoid, y² = 2α x² y = ∞ when x = 2a; hence the line x 2a = 0, which is paral- DIFFERENTIAL CALCULUS AND PLANE CURVES. 131 lel to the axis of y and at a distance 2a from it, is an asymptote to the curve. α Again, in xy = a or y X x' when x = 0, y = ∞; therefore x = 0, or the axis of y, is an asymptote to the curve. Similarly, in y a*, when x = ∞, y = 0; hence y = 0, or the axis of x, is an asymptote to the logarithmic curve. 5. Find the asymptotes of xy-ay- bx = 0. (1) x α= 0; (2) y - b = 0. a 6. Find the asymptote of y' = аха X³ a y = −x + 33 · 7. Find the asymptotes of y=c+ as (x — b)²* y = c and x = b. 8. Find the asymptotes of y'(x² + 1) = x²(x² — 1). 3 9. Find the asymptotes of y'(x — α) = x² + ax³. CURVATURE. y = ±x. x = a and y = ± (x + a). 156. A point moving along an arc of a curve changes its direction continuously, and the total change of direction is called the Total Curvature of the arc. m P T Y As Δρ P S A О A E ዕ FIG. 25. The angle TP', Fig. 25, through which the tangent PT rotates as the point of tangency P moves from P to P', being 1 132 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 the total change of direction of the point P, is the total curva- ture of the arc PP'. 157. Uniform Curvature. The curvature is uniform when, as the point of tangency moves over equal arcs, the tangent turns through equal angles; that is, when the distance described by the point varies as its direction. Let APm be the curve, AP = s, PP' — 4s, XEP = Φ, Art. 49; then TtP' = 4p. Let PC and P'C' be normals meet- ing at C. Supposing 48 x 44, we have (Art. 12) 1 ΔΦ 48m4p, or = m As (1) Let us consider the meaning of 4. If the distance As As gives a total curvature of 40, since As ¤ 40, a distance of 1 will give a curvature of ΔΦ As That is, ΔΦ As is the curvature per distance of unity, or the rate of change of the direction of a curve with respect to that of its length, for which reason it is called the curvature of the curve. (2) Let us determine the value of m. The circle is the only curve of uniform curvature. Hence, supposing 4s ∞ 40, PP' is the arc of a circle whose radius (say R) is CP. The angle PCP' TtP' = 40; but arc PP' CP X angle PCP'; that = is, As R40; hence m = R, and we have = × ΔΦ 1 As RⓇ COR. I. The curvature of any circle is equal to the reciprocal of its radius; and the curvatures of any two circles are inversely proportional to their radii. COR. II. If R = 1, ΔΦ 48 1; that is, the unit of curvature is the curvature of a circle whose radius is unity. -- 靠 ​DIFFERENTIAL CALCULUS AND PLANE CURVES. 133 158. Variable Curvature. When the curvature is variable, we define the curvature at any point P of the curve as the value which would have were the curvature there to become uni- ΔΦ As аф ds form. Hence the curvature at P is the value of at that point. 159. Radius of Curvature. A circle tangent to a curve at any point, and having the same curvature as that of the curve at that point, is called the circle of curvature; its radius, the radius of curvature; and its centre, the centre of curvature. The curvature of this circle being that of the given curve, is do equal to ; therefore the radius of curvature of APm at P, Fig. 25, is ds R = = аф COR. I. To express R in terms of the differentials of x and y. hence we have tan = dy dx •: &= tan-1 dy ; dx аф = d'y dx dx² + dy²; also, ds = (dx² + dy³)*. ds (dx² + dy²)* (1 + (dy )")" ล :. R* = or аф dx d'y dzy dx² (1) * R will be positive or negative according as the curve is convex or concave (Art. 173), but its sign is often neglected. v ! A 134 DIFFERENTIAL AND INTEGRAL CALCULUS. EXAMPLES. 1. Find the radius of curvature of the parabola y³ = Here dy 2a dx y ď³y 4a² and dx² Substituting in (1), we have (y² + 4a³)* R = 4a² 2 = 4ax. At the vertex, where y = 0, we have R = 2a, which is evi- dently the minimum radius of curvature. 2. Find the radius of curvature of the ellipse a³y² + b²x² = a²b³. R = &&& dy b2x dx d'y a¹y' dx² b₁ a²y a³y³ (a*y³ +_b*x³) *__ (a*y* + b*x²)* a'y¹ a*b* b At the vertex x = a, y = 0, R = and at the vertex x = 0, α a² y = b, R = Ъ which are respectively the minimum and maxi- mum radii of curvature. 3. Find the radius of curvature of the cycloid xr vers- √2ry — y². - Y — r dx Here dy y √2ry — y² dy³ 2r ; .. 1+ dx² + DIFFERENTIAL CALCULUS AND PLANE CURVES. 135 ď³y dx² ጥ •. R = 2√2ry, y² which equals twice the normal. 4. Find the radius of curvature of the logarithmic curve y = ax. R = (m² + y²); my 5. Find the point on the parabola y = 8x at which the radius of curvature is 713. 6. Find the radius of curvature of y = x* the origin. y= 3, x 14. 4x³ 18x2 at 1 R = 36' 7. Find the curvature of the equilateral hyperbola xy = 12 at the point where x = 3. 8. Find the radius of curvature of the catenary 1 24 R 125' a y = +e y² R = a 9. Find the radius of curvature of the hypocycloid x³ + y²³ = a a³ R = 3(axy)*. 160. The radius of curvature in polar co-ordinates can be found by transforming the value of R given in the answer to Ex. 7, Art. 112, to polar co-ordinates. We thus obtain dr² + ᏧᎳ . R = dr² d'r (2) p² + 2 dtj d f² 136 DIFFERENTIAL AND INTEGRAL CALCULUS. EXAMPLES. 1. Find the radius of curvature of the spiral of Archimedes, r = al. Here dr å³r do =α, do = 0; substituting in (2), we have (2¹² + a²)³ _ a(1 + (²)³ R = 7.² +2α² 2+ y² Find the radius of curvature of the following: 2. The logarithmic spiral r = aº. R=r√1+ (log a)❜ 2 3. The cardioid 7 = a(1 — cos 0). R = ৩/ ২৩ √2ar. a³ 4. The lemniscate r² = a² cos 20. R = 3r' CONTACT OF DIFFERENT ORDERS. = 161. Let y = f(x) and Y (x) be any two curves referred to the same axes. Let the curves intersect at the point P, whose abscissa is a, then f(a) = $(a). If f(a) = $(a), and ƒ'(a) = p′(a), the curves are tangent at P, and are said to have a contact of the first order. If f(a) = p(a), ƒ'(a) = '(a), and and ƒ''(a) = p''(a), the curves have the same curvature at P, their contact is of the second order. If, in addition, f""(a) = '''(a), their contact is of the third order; and so on. Thus, contact of the nth order imposes n + 1 conditions. 162. Two curves cross or do not cross at their point of con- tact, according as their order of contact is even or odd. Let x = a be the abscissa of the point of contact of the DIFFERENTIAL CALCULUS AND PLANE CURVES. 137 = curves y = f(x) and y = (x), then f(a) (a). Let h be a small increment of x. By Taylor's formula, we have f(a + h) = f(a) +f'(a)h +f''(a) 4(a+h) = 4(a) + p′(a)h+ p''(a) Subtracting (2) from (1), we obtain ૦૨ | h³ 3 h³ +f'" (a)- +; (1) h³ + p'" (a) h² h³ +. (2) f(a + h)−4(a + h) = h[ƒ' (a)−¢′(a)] + [。 [ƒ'"' (a) − ¢″(a)] ht — + [ƒ''''(a) — ¢'''(a)]+[ƒª(a) — ع(a)]+. (3) 13 4 (a) If f(x) — Þ(x) changes sign as x increases from a − h to a+h, the two curves evidently cross at a; if not, the curves touch each other, but do not cross. (b) If the contact is of an odd order, the first term of the second member of (3), which does not vanish, contains an even power of h; hence the sign of the second member, and therefore the first, undergoes no change as a passes from a h to a +h, and the curves do not cross. — (c) If the contact is of an even order, the first term of the second member of (3), which does not vanish, contains an odd power of h; hence, in this case, f(x) — 4(x) changes sign as x passes from a h to a + h, and therefore the curves cross. COR. I. At a point of maximum or minimum curvature, the circle of curvature has contact of the third order with the curve, for it does not cut the curve at such a point. Cor. II. If two curves are tangent to, and cross each other at, a certain point, they have contact of at least the second order. { 138 DIFFERENTIAL AND INTEGRAL CALCULUS. ; 1 1 EXAMPLES. 1. Find the order of contact of the two curves y = x³ — 3x²+7 and y+3x= 8. By combining the two equations we find that (x = 1, y = 5) is a point of contact. Making f(x) = x³ — 3x² + 7 and p(x) = 8 — 3x, we have ƒ'(x) = 3x² — 6x, p′(x) = −3; .'. ƒ′(1) = p′(1) = — 3; ƒ''(x)=6x-6, =6x-6, p''(x) =0; .. ƒ''(1) = $′′(1) = 0; ƒ'''(x) = 6, ¤'''(x) = 0; :. ƒ''''(1) > p'''(1). Hence the contact is of the second order. 2. Find the order of contact of the parabola y' 4x and the line 3y=x+9. 3. Find the order of contact of the curves = First order. y = 3x x² and xy xy = 3x — 1. Second order. 4. Find the order of contact of y = log(x-1) and x²-6x+2y+8=0, at the point (2, 0). Second order. 5. Find the order of contact of the parabola y=4x+4 and the circle y²+x²=2x+3. Third order. 163. Osculating Curves. The curve of a given species that has the highest order of contact possible with a given curve at any point is called the osculating curve of that species. A curve may be made to fulfil as many independent condi- tions as there are arbitrary constants in its equation, and no more. Therefore, in order that y = f(x) may have contact of 7 DIFFERENTIAL CALCULUS AND PLANE CURVES. 139 A the nth order with a given curve at a given point, the equation must involve n + 1 arbitrary constants. Hence, as y = ax + b has two constants, the osculating straight line has contact of the first order. = As (x - a)²+(y b) has three constants, the osculat- ing circle has, in general, contact of the second order. 164. To find the osculating straight line at any point (x', y') of a given curve y = f(x). The equation of a line is y = ax + b. Since the line and curve pass through (x', y'), we have y' = ax′+b=ƒ(x'). dy' Also, = α =ƒ'(x'), dx' since f'(x)=′(x′). Solving (2) and (3) for a and b, we have α = dy' dx' and b = y' dy_x dx' which, substituted in (1), gives dy' y — y' = dx (x -(x − x'). Therefore the osculating straight line is a tangent to the curve, as would be inferred. 165. To find the radius of the osculating circle at any point of a given curve, y = f(x). The general equation of a circle whose radius is r is (x − a)² + (y — b)² = r². . - (1) (1) (2) (3) 140 DIFFERENTIAL AND INTEGRAL CALCULUS. .. 1. Differentiating twice successively, we have • dy x − a + (y − b) = 0, dx 1+ dx² dy²+(y-b) ď²y = 0. dx² dx² + dy³ From (3), y − b = - day From (2), X α = (dx² + dy³)dy d'y dx Substituting (4) and (5) in (1), we have r = (dx² + dy²)³ dx d³y • (2) (3) (4) (5) By comparing this result with formula 1, Art. 159, it will he seen that the osculating circle is the same as the circle of cur- vature. INVOLUTES AND EVOLUTES. 166. An Involute may be regarded as a curve traced by a point in a thread as it is unwound from another curve, called the Evolute. Thus, imagine a thread stretched around the curve A¸P¸m, with one end fastened at m,; if the thread is unwound by carry- ing the point at A above and around to the right, that point of the thread will trace the involute APm of which A,P₁m, is the evolute. 1 An evolute may have an unlimited number of involutes, for A may be any point on the curve A,m,. DIFFERENTIAL CALCULUS AND PLANE CURVES. 141 In what follows the chief object is to deduce certain prop- erties of the evolute from its involute, or vice versa, and for uniformity the co-ordinates of P (the involute) will be repre- sented by x, y, and those of P, (the corresponding point of the Mi P m A A1 FIG. 26. evolute) by x,, y,; the arc AP by s; the arc AP, by s,; and the angles of direction of AP and AP,, at P and P,, by o and o,, respectively. 167. Elementary Principles. I. PP, the arc AP, s,. = = 05 m ليا E X B₁ A T B F FIG. 27. II. PP, is tangent to AP,m, at P,, for it has the same direction as the curve at that point. III. The line PP, is a normal to the curve APm at P. 142 DIFFERENTIAL AND INTEGRAL CALCULUS. " : For, draw TP tangent to the curve AP at P. (P¸E)² + (EP)² = (P₁P)³, or (y,− y)² + (x − x¸)² = s‚². (1) .. (y, — y)(dy, — dy) + (x − x,)(dx — dx¸) = 8¸ds,. (2) Again, P,E = P‚P sin EPP,, or y₁- y = ε ₁ds,' dy, (3) 1 1 also, EP = P¸P cos EPP₁, or x X = x₁ = S dx, ds₁ · (4) 1 bering that da," + dy,' = ds,', we have Substituting in (2) from (3) and (4), and reducing, remem- 2 1 dy, dx, dx that is, dy' 1 π π cot ; :: ₁ tan ₁ hence P,P is perpendicular to the tangent PT. +, or PFX = + PTX; 2 dy dx (5) COR. I. Since sin ₁ = cos P, , 1 ds, ds dx, dy (6) also, since cos ₁ sin 0, ds, ds COR. II. The point P, is the centre of curvature of the curve APm at P. For, if circles be described from P, and P, as centres with PP' and P,P as radii, respectively, the arc PP' will lie within the one circle and without the other, since the straight line P''P' is equal to the partly curved line PP,P. Hence the circumference of the circle whose centre is P, crosses and touches the curve APm at P (Art. 162, Cor. II). COR. III. Since P,Ps, R, we have (Art. 159) S Ꭶ, 1 ds аф = (dx² + dy²) dxd y (7) DIFFERENTIAL CALCULUS AND PLANE CURVES. 143 COR. IV. From (4) and (6), x, x — § 1αs and from (3) and (5), dy = S ds' dx 1 y₁ = y + s₁ πs (8) (9) Substituting for s, in (8) and (9), from (7), we have (1 + dx² ãx dy² \ dy dxdx dy2 1+ dx² x₁ = x and y₁ = y + (10) 1 d³y dx² d'y dx² ** These values of x, and y, are the values of the co-ordinates of the centre of curvature at the point P. 168. To find the equation of the evolute of any given curve. By differentiating the equation of the given curve, and sub- stituting the results in (10), x, and y, may be expressed in terms of x and y. If, between the equations thus obtained and that of the given curve, x and y be eliminated, the resulting equation involving x, and y, will be the equation of the evolute. EXAMPLES. 1. Find the equation of the evolute of the parabola y² = 4ax. Here dy dx Y 2α dzy 4a² = dx² y³' Substituting in (10), we have x₁ = x + y²+4a² 2a y³ y² y 4a² =3x+2a; x = x' 3 2a 144 DIFFERENTIAL AND INTEGRAL CALCULUS. , Y₁ = y y²+4a² y³ y³ 4a² y³ 4a² y = — (2a)³y, ¹. These values of x and y substituted in y² = 4ax give 4 y₁² = 1 27a (x, — 2a)³, which is the equation required; hence the evolute is the semi- cubical parabola. COR. I. The length of the arc of the evolute s, may be found by formula (7), Art. 167. 2. Find the evolute of the ellipse a²y²+ b²x² = a³b³. (ax,)*+ (by,)* = (aª — b²)³. 3. Find the co-ordinates of the centre of curvature of the cubical parabola y³ = a*x. 3 = a² + 15y* 6a²y = , У 1 a'y - 9y' 2a¹ 4. Find the co-ordinates of the centre of curvature of the X catenary y = α 2 a e + e X₁ = X — Y √ y² — a², y₁ = 2y. a 5. Find the co-ordinates of the centre of curvature, and the equation of the evolute, of the hypocycloid x² + y² = a 3 2 x₁ = x + 3 √xy³, y₁ = y + 3 √x²y; (x, +y,)² + (x, − y,)ª 1 2a³. y, 1 6. Find the evolute of the equilateral hyperbola xy=m³. NOTE. First prove that (x, + y₁)³ — (x, —- y,)³ — (4m)³. = m /m 1 x₁+ y₁ = 1 2 X + X m and x₁y₁ = m / m X 2 X m/ and thence derive the equation of the evolute. " : 11 DIFFERENTIAL CALCULUS AND PLANE CURVES. 145 169. Let ENVELOPES. f(x, y, a) = 0 . (1) be the equation of a curve, a being some constant quantity. If we assign different values to a, we will obtain a series of distinct curves, but all belonging to the same system or family of curves. One of the curves of this family can be obtained by increasing a by h, thus converting (1) into f(x, y, a + h) = 0. (2) If h be supposed indefinitely small, the curves (1) and (2) are said to be consecutive. The points of intersection of the curves (1) and (2) approach definite limiting positions as h approaches 0, and the locus of these limiting positions, as different values are assigned a, is called the Envelope of the system f(x, y, a) = 0. The quantity a which remains constant for any one curve of the series, but varies as we pass from one curve to another, is called the variable parameter of the series. 170. The envelope of a series of curves is tangent to every curve of the series. Let A, B, C be any three curves of the series, A and B inter- secting at P, and B and C at P'. B A P P' FIG. 28. As these curves approach coincidence, the limiting positions of P and P' will be two consecutive points of the envelope and of the curve B. Hence the envelope touches B. As an illustration see example 1 under the next article. A 146 DIFFERENTIAL AND INTEGRAL CALCULUS. 7 171. To find the equation of the envelope of a given series of curves. The point of intersection of (1) and (2) will be found by combining the equations. Now, subtracting (1) from (2), we have f(x, y, a + h) − f(x, y, a) = 0.. h (3) When the curves approach coincidence, h approaches 0, and (3) becomes d daf(x, y, a) = 0. . (4) Thus, equations (1) and (4) determine the intersection of any two consecutive curves. Hence, by eliminating a between (1) and (4), we shall obtain the equation of the locus of these intersections, which is the equation of the envelope. EXAMPLES. 1. Find the envelope of y = ax + parameter. m a m " a being the variable a y = ax + is the equation of a line, as MN, Fig. 29. When a receives an increment h, the line takes a new position, say M'N', which intersects the former line at C. As h approaches 0, c approaches p, a point on the locus (APm) of all similar intersections. Differentiating with respect to a, x and y being constants, we have m 0 = x whence M whence a = ± XC y = ±[√ mx + √mx], or y² = 4mx. which is the equation of a parabola, - DIFFERENTIAL CALCULUS AND PLANE CURVES. 147 t Let it be observed that the problem is the same as that of finding the curve of which y = ax + m a is the tangent. M A p с FIG. 29. N' N m 2. Find the curve whose tangent is y = mx + a √m² + 1, m being the variable parameter. x² + y² = a², a circle. 3. If a right triangle varies in such a manner that its area is constantly equal to c, find the envelope of the hypothenuse, or the curve to which the hypothenuse is the tangent. Let OA = a, OB = b; then the equation of AB is 2c But ab = 2c, or b = a 818 220 α + + y = 1. b ay = 1, 2c where a is the variable parameter. (1) (2) 2cx Differentiating (2), we get a = which, substituted in y (2), gives xy = √, an equilateral hyperbola. 2 148 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 Solve the preceding problem on the hypothesis that the hy- pothenuse is constantly equal to c. x²+ y² = c³, the hypocycloid. Since a normal to a curve is tangent to the evolute of the curve, the latter is the envelope of the successive normals, or the locus of their intersections. 4. Find the evolute of the parabola y = 4ax, taking the equation to the normal in the form y = m(x — 2a) — am³, m y= being the variable parameter. 27ay 4(x-2a)³. = 5. Find the evolute of the ellipse a'y' + b²x²= a²b², taking the equation of the normal in the form by = ax tan a' — (a² — b²) sin a′, where the variable parameter a' is the eccentric angle. (ax)* + (by)³ = (a - b). See Ex. 2, Art. 168, 6. Find the curve whose tangent is y = mx + (a³m² + b²)+, m being the variable parameter. a). a²y² + b²x² = a²b³. 7. Find the envelope of the family of parabolas whose equa- tion is y² = a(x y = ± ±x. 8. Find the locus of the intersections of x cos a + y sin a = p with itself as a increases continuously. x² + y² = p². 9. Find the envelope of all ellipses having a common area (πc²), the axes being coincident. xy = ± √c². 10. Find the evolute of the curve x³ + y³ = a³, the equation of whose normal is y cos a' x sin a' = a cos 2a', where a' is the angle which the normal makes with the axis of x. (x + y)³ + (x − y) = 2a³. See Ex. 5, Art. 168. 11. Find the equation of the curve, the equation of its tan- gent being y = 2mx+m', where m is the variable parameter. (y) + (x) = 0. TRACING CURVES. 172. The Rudimentary Method of tracing a curve is to reduce its equation to the form of y = f(x); that is, solve the DIFFERENTIAL CALCULUS AND PLANE CURVES. 149 equation f(x, y) = 0 for y, assign values to x, find the correspond- ing values of y, draw a curve through the points thus determined, and it will be approximately the curve required. This process is laborious, and often impossible on account of our inability to solve f(x, y) = 0 for y. The General Form of a curve is usually all that is desired, and this can generally be found by determining its singular or characteristic points and properties, and these are embraced chiefly in the position of certain turning-points of the curve, the direction of curvature between these points, and where and how the branches intersect or meet each other. In addition, we may find, by previous methods, where the curve cuts the axes, whether or not it has infinite branches, asymptotes, etc. 173. Direction of Curvature. The terms Convex and Concave have their ordinary meaning when applied to the arcs of curves. P E Y A B₁ O FIG. 30. B X Thus, as seen from some point below, the arcs AB, and CD are concave, and B, C and DE convex. 174. A Point of Inflection is the point at which the curve changes from concave to convex, or from convex to concave; as the points B,, C, D. PRINCIPLES.-The slope (dy) dx of the curve evidently de- creases as the point P(x, y) moves from A along the curve to the right until P reaches B,, and then increases until P reaches dy dx C, etc. Therefore (1) when the arc is concave, decreases as d²y dx² x increases, hence (Art. 25) its derivative is ; (2) when dy dx the arc is convex, increases as x increases, hence dx² is +. Therefore, I. At any point of the curve y = f(x), the curve d²y is concave or convex according as is negative or positive. dx² 150 DIFFERENTIAL AND INTEGRAL CALCULUS. day dx² dy II. The roots of = 0 or ∞ which will render a maxi- dx mum or minimum are the abscissas of the points of inflections. EXAMPLES. Examine the following curves for concave and convex arcs, and for points of inflection. day 1. y = x²-6x+7. = 2. dx² day dx² Since is +, the curve is convex at every point. 2. y = x²-6x+17x-6. The root of x d'y = 6(x − 2). dx² 20 is 2, the point of inflection; the curve is concave when x < 2, convex when x > 2. 3. y = x² 12x³ + 48x² - 50. X Points of inflection, x = 2, x = 4; curve convex when x < 2 and > 4, and concave when 2 < x < 4. X 2 4. y = X 3 dy 2 dx² (x — 3) ³° Point of inflection at x=3; convex when x> 3, concave when x < 3. d'y 1 5. y = log(x-1). dx² (x − 1)³° 1 = ∞ gives (x — 1)² = 0, which has two equal roots; (x — 1)* hence, Art. 140, there is no point of inflection; eurve concave. DIFFERENTIAL CALCULUS AND PLANE CURVES. 151 6. Prove that the curve y = X³ has points of inflection a² + x² at (0, 0), (a √/3, §a √3), (− a √3, — ‡a √3). 7. Prove that the witch of Agnesi, x²y = 4a²(2a — y), has points of inflection at (±a √3, a), and is concave between these points and convex outside of them. 8. Find the points of inflection of y = sin 2x + cos 2x. SINGULAR POINTS. 175. The Singular Points of a curve are the turns and multiple points. A Turn in rectangular co-ordinates is a point at which a curve ceases to go (1) up or down, or (2) to the right or left, Y B H F D E X A FIG. 31. and begins to go in the opposite direction. The former, as B, E, F, G, are called y-turns, and the latter, as C, D, H, x-turns. The x-turns and y-turns evidently occur at the maximum or minimum values of x and y, respectively. ! Fr : 152 DIFFERENTIAL AND INTEGRAL CALCULUS. A Multiple Point* is one through which two or more branches of a curve pass, or at which they meet. A multiple point is double when there are only two branches; triple when only three, and so on. A Multiple Point of Intersection is a multiple point at which the branches intersect (Fig. 32, a). b d α с FIG. 32. e An Osculating Point is a multiple point through which two branches pass, and at which they are tangent (Fig. 32, b, c). A Cusp is a multiple point at which two branches terminate and are tangent (Fig. 32, d, e). A cusp or osculating point is of the first or second species according as the two branches are on opposite sides (Fig. 32, b, d) or the same side (Fig. 32, c, e) of their common tangent. A Conjugate Point is one that is entirely isolated from the curve, and yet one whose co-ordinates satisfy the equation of the curve. For example, in the equation y = (a + x) √x, if x is negative y is imaginary, yet the co-ordinates of the point (x = — a, y = 0) satisfy the equation. Hence ( a, 0) is a conjugate point. A conjugate point is, generally, the intersection or point of meeting of two imaginary branches of the curve, and may, in exceptional cases, also lie on a real branch of the curve. There are other singular points, such as Stop Points, at which a single branch of a curve stops suddenly, and Shooting Points, at which two or more branches stop without being tan- gent to each other. But as these rarely occur, they are omitted in this book. * See Taylor's Calculus. DIFFERENTIAL CALCULUS AND PLANE CURVES. 153 176. To determine the positions of the singular points of a curve. Let u = f(x, y) = 0 be the equation of the curve, free from radicals. Then (Art. 109) du dy dx dx du dy dy du (a) For the x-turns, ∞ ; = 0. dx dy dy du (b) For the y-turns, 0; . = 0. .. dx dx dy dx' (c) For multiple points, by definition, has two or more values; hence, since u contains no radicals, dy must be of the dx 0 form Therefore du dx du : 0 and 0. dy du Hence, to find the x-turns we have u = 0 and 0; to dy du find the y-turns, we have u = 0 and = 0; and the values of dx y and x which satisfy all these equations are the co-ordinates of the multiple points. 177. To determine the character of the multiple points of a curve. From the definitions of the multiple points it follows that: dy dx I. At a multiple point of intersection has two or more unequal real values. dy II. At an osculating point or a cusp has two equal values. dx dy III. At a conjugate point at least two of the values of imaginary. are dx 154 DIFFERENTIAL AND INTEGRAL CALCULUS. 1 1 * 1 EXAMPLES. Find the singular points of the following curves. 1. U = x² - x² — xy + y² xy + y² — 3 = 0.. (1) du dy du x + 2y = 0; 0;. (2) dx 2xy=0. 0. . (3) From (1) and (2) we find (2, 1), (— 2, − 1), the co-ordinates. of the x-turns A and A'. B *w B A D FIG. 33. X From (1) and (3) we find (1, 2), (1, 2), the co-ordinates of the y-turns B and B'. Since neither pair of these values satisfies (1), (2), (3), the curve has no multiple points. 2. u = u = 4y² — (25 — x²) (x² + 7) = 0. FIG. 34. du 8y = 0; dy du =2x(x²+7) - 2x (25-x²)=0. dx -X From these equations we find (a) the x-turns, (5, 0), (— 5, 0), (± √ — 7, 0); (b) the y-turns, (3, ± 8), (0, ± § √7), (− 3, ± 8). DIFFERENTIAL CALCULUS AND PLANE CURVES. 155 The figure (34) is only an approximate representation of the curve. 3. u = y* — 2(2 + 2*)y + (§ 2º) = = 0. x-turns, (± 4, 18), and (±17 4 √17, 18). 17/ 4. y-turns, (0, 4), (0, 0), and u = x* + 2ax²y — ay³ = 0. (± 16 17 √17, 1917). (1) Y du dy | = a(2x² 3y') = 0. (2) du 4x(x² + ay) = 0. . (3) dx FIG. 35. Sa -X y-turns, (0, 0), (a, − a), (— a, − a); x-turns, (0, 0), (1ª √6, — 8a), (-4a √6, 9 9 9 Now there appears to be an x-turn and a y-turn at the point (0, 0), and in a certain sense this is evidently true; but we should regard the result as signifying that (0, 0) is a multiple point of some kind, since x = 0, y = 0 satisfy equations (1), (2), and (3). Let us now determine the character of the point. (3) by (2), we have Dividing dy 4x³ + 4axy dx 3ay³ — 2xx² Our object now is to find the value of the slope dy dx at the multiple point (0, 0). For these values of x and y, the form of, hence the value required may be obtained by dy assumes dx Art. 137. 156 DIFFERENTIAL AND INTEGRAL CALCULUS. # dy dx We see from Ex. 5, Art. 137, that = 0 and ± √2 at the point (0,0). Hence the origin (0, 0) is a triple point, the three branches which pass through the point being inclined to the x-axis at the angles 0, tan-¹ √ and tan-¹ (— √2), respectively, as in the figure. See Art. 179. 5. y² = a²x² - x*. x-turns, (0, 0), (a, 0), (— a, 0); a a y-turns, (0, 0), (½ √2, ± 1 a²), ( − & √Ã, ± 1 a²). 2 The point (0, 0) is a double point of intersection, since at that point dy = ± a. dx 6. Examine y²(a² — x³) — x* = 0 for multiple points. dy dx · At the point (0, 0), = 0; that is, it has two equal values; hence (0, 0) is an osculating point or a cusp; and since the curve is symmetrical with respect to both axes the point is evidently an osculating point of the first species. 7. Determine the general form of the curve y = a²x³. When x = ∞, y = ±∞o; hence the curve has two infinite branches, one in the first and one in the fourth quadrant. ¡Y FIG. 36. -X When x is negative, y is imaginary; hence the curve does not extend to the left of the y-axis. When x = 0, y = 0; hence both branches start from the origin. dy At the point (0, 0), = ±0; hence, dx since the curve is symmetrical with respect to the x-axis, the origin is a cusp of the first species. DIFFERENTIAL CALCULUS AND PLANE CURVES, 157 Again, since d³y dx² За = ± > the upper branch is convex and 2 Nx x². the lower concave. 8. Examine the curve (y — x²)² = x³, or y x³, or y = x²± x Has two infinite branches, one in the first and one in the fourth quadrant, both starting from the origin. For every positive value of x, y has two real values, both of which are posi- tive as long as x < 1, but at the point where x = 1 the lower branch crosses the x-axis. The origin is a cusp of the second species. Let 178. Tracing Polar Curves. f(r, 0) 0 be the polar equation of the curve. FIG. 37. -X (a) By solving the equation f(«, 0) = 0 for 0, we find the direction of the curve at the point r = a. If a = 0, the values of will be the angles at which the curve cuts the polar axis at the pole. (b) By solving the equation dr do =0 for we find the values of 0 for which r is a maximum or minimum, or the r-turns, at which the curve is perpendicular to the radius vector. 9. Trace the curve r = a sin 30, Fig. 38. (a) Making r = 0, we have sin 30 = 0; hence 0 = 0, §π, §π, which are the angles at which the curve cuts the polar axis at Y the pole. dr do (b) = 3a cos 30 = 0; hence the values -X of 0 at the r-turns are π, π 2' π, at which FIG. 38. a, while points r = a, — a, a, respectively. dr do Since = 3a cos 30, r increases from 0 to increases from 0 to π; r decreases from a to 3 a, while while increases from 7 to increases from 7 to π; r increases from a to + a, 7; and r decreases from a to 0, 158 DIFFERENTIAL AND INTEGRAL CALCULUS. .: while increases from 7 to 7. Further revolution of the radius vector would retrace the loops already found. FIG. 39. -X 10. Trace the curve r = a sin 20, Fig. 39. From this and the previous example, we infer that the locus of r = a sin no consists of n loops when n is odd, and 2n loops when n is even. 11. Trace the curve ra cos e cos 20, or r = a(2 cos³ 0 cos 6), Fig. 40. 12. Trace the lemniscate r² = a' cos 20, Fig. 41. Y X Y -X FIG. 41. FIG. 40. 179. The character of multiple points in rectangular co- ordinates may often be more easily determined by changing to polar co-ordinates, and applying (a) of Art. 178. Thus, in Ex. 4, Art. 177, make y = r sin 0 and x = r cos 0, divide by, and we have r cos¹ 0 + 2a cos² ( sin 0 — a sin³ 0 — 0. Now making r = 0, and we have sin 00, and tan² 0 = 2; that is, the angles at which the curve cuts the x-axis at the origin are sin¹0, tan-1 1/2, tan-¹ — √2. Trace the following curves: 13. The Cissoid, 14. The Conchoid, - y² (2a — x) = x³. x²y² = (b² — y²)(a+y)². • DIFFERENTIAL CALCULUS AND PLANE CURVES. 159 15. The Witch, 16. The Lituus, 17. The Parabola, 18. The Curve, 19. The Cardioid, (x² + 4a³)y = 8a³. r√o = a. r = a sec² Ꮎ 201 Ꮎ r = a sin³ 3· 20. The Hypocycloid, r = a(1 cos 0). x³ + y³ = a a³. 21. Examine ay² = x³ — bx² for multiple points. y² (0, 0) is a conjugate point. x)² 22. Prove that y=x and (y-x)=x have cusps of the first species at the origin. 23. Prove that x¹ 2ax²y the second species at the origin. axy² + a³y² = 0 has a cusp of 24. Prove that x* ax³y — axy² + a³y² = 0 has a conjugate point at the origin. 25. Prove that the multiple point of ay = (x − a)" (x — b) at (a, 0) is (1) a conjugate point if a b, and (3) a cusp if a = b. ד' CHAPTER VIII. GENERAL DEPENDENT INTEGRATION. FUNDAMENTAL FORMULAS 180. The differentials in the following twenty-two formulas are the fundamental integrable forms, to one of which we endeavor to reduce every differential that is to be integrated by the dependent process (Art. 51): 1. S v"dv = 2. vn+1 n + 1 + C. fdv = log (v)+C, or log v + log c = log cv. 3. Sæváv = аг log a 4. Sedv=c+0. + C. cos v dv = sin v + C. 5. Scos 6. S sin v dv dv= cos v + C. sec² v dv = tan v + C. 7. S se 8. 9. S cosec" v dv= cot v + C. S se sec v tan v dv = sec v + C. 160 GENERAL DEPENDENT INTEGRATION. 161 : cosec v cot v dv = ose cosec v + C. 10. Sc 11. St 12. 13. 14. 15. 16. tan v dv = log sec v + C. Scot v dv = log cot v dv = log sin v + C. S se sec v dv log (sec v + tan v) + C. = cosec v dv = log (cosec v Sco Sv S 17. S 18. 19. 20. VI dv dv 1 =sin¹v+C. 22 cos¹v + C. √1 — v² dv 1 + v² SI+ √ = SF S dv 1 + v² dv v V v² — 1 dv v¥ v² – 1 S 21. S S 22. √2v dv dv √ v² ± m tan-¹v + C. 1 وح cot-¹ v + C. — sec-¹ v + C. ― cot v) + C. = cosec¹ v + C. 1 = vers v + C. = 2 log (v + √v²± m) +.C. In these formulas v may be the independent variable, or some function of this variable, and the process of integration consists largely in reducing or transforming any given differen- tial into one of the above forms. 162 DIFFERENTIAL AND INTEGRAL CALCULUS. REDUCTION AND INTEGRATION OF DIFFERENTIALS. 181. Reduction of Differentials is the process of reducing them to integrable forms, and is effected chiefly (1) by constant multipliers, (2) by decomposing or separating them into their integrable parts, and (3) by substitution. áv 182. Reduction of Differentials by Constant Multipliers. PRINCIPLE. The value of any differential of the form cav remains unchanged if du be multiplied and c be divided by the same constant. Find the following: EXAMPLES. 1. y = Sato ndv n V Ans. tan-1 + C. Na Na We reduce this to the form of 17, Art. 180; thus v N dv n v č d y = α v2 α a = Answer. 1+ 1+ a ( v a 2. y= y = S ndv Na (See 15, Art. 180.) Ans. n sin-¹ V + C. Na y = n a S dv n va / 1 - v2 a Na S ( a = Answer. 2 V 1- a Let the student compare these results with formulas (13) and (15), page 65, and in a similar manner deduce formulas (14) and (16) on that page. GENERAL DEPENDENT INTEGRATION. 163 3. y = S (3x+3)dx x² + 2x + 5° Ans. log (x²+2x+5)+C. 3 Y 2 ៖ ((2x+2)dx 3 'd(x² + 2x+5) = Ans. x² +2x+5 x²+2x+5 5x2dx Sra +4 dx 4. 5. S √2 — 9x² S 6. 7. S 8. S √6x 3dx 3dx 2+7x²° xdx 1+25° 4 9. S dz 1 5a² • 11. 1+52 dx 2+322° 922 (Formula 2, Art. 180.) ΣΤ log (7x+4)+ C. 3x sin-1 1 √2 + C. vers-¹ 3x + C. 3 √14 tan-1 x -1 tan-¹x² + C. 2 + C: tan-¹x/5+ C. tan-¹ x + C. sec-¹xv+0. 1 -1 1/5 10. S 1 16 S dx 1 -1 x√2x² — 3 √3 S 3dx √9x 4x² 13. S dx 1 -1 cos-¹ x√ + C. √3 2x² 1/2 14. S dx + tan-12 + C. x² + 16° 4 12. -1 vers 8x + C. 9 164 DIFFERENTIAL AND INTEGRAL CALCULUS. 부 ​15. S 5dx x√ 3x² — 5 16. S dx √5x - 3x² √5 sec¹ x√3 + C. X 3√3 sec¯¹x√§ + C. Some of the preceding examples may be conveniently solved by formulas (19) and (21), page 65. REDUCTION OF DIFFERENTIALS BY DECOMPOSITION. 183. The process of reducing differentials to integrable forms consists largely in separating them into their integrable parts. 184. Elementary Differentials. In these the necessary reductions are effected by the simple operations of algebra. EXAMPLES. Find the following: 1. S (320 (3x + 5) dx. 4x+1 S (30€ •(3x + 5) dx = · x = S 3xdx 2. f (x − 3. 4. S 3) √1 — x² S (2 S √4x −5x) - 4x² + 1 dx. log (4x² + 1) + § tan-¹ (2x) + C. √ 4203 +1 5dx + √ 1 + 4200 S 4x² − (1 − x²)* — 3 sin−¹ x + C. 5 3 - dx. 2x² (2x − x²)* — vers-¹x + C. √g √2 √x - 1 x² — 1 X x V x² 1 √ √ x² = 1 dx. X (x² − 1)* + cosec¯¹x + C GENERAL DEPENDENT INTEGRATION. 165 5. St dx. X √ a + x Va X Na + x Na a + x Va² 2 x² sin X -1 x² + C. α 6. f(x + 1) dx. X 3 1 x + 3 log x X 2x² + C. 7. S (2² = 1)dx. X³ ха 3 + + x + C. 2 8. S₂ 41dz. X x + 4 3 + 201 x² x+ log (1 + x) +C. 185. Trigonometric Differentials.-In reducing these we use the elementary formulas of Trigonometry, such as sin' x + cos x = 1, sin 2x = 2 sin x cos x, cos 2x = cos² x sin² x, etc. EXAMPLES. Find: 1. /sin' z dr. .. cos x + cos³ x + C. sin³ x = sin x (1 sin x (1 — cos² x) cos x) = sinx - (cos x) sin x. S(sin²x)dx = ƒ sin a)da S sin x dx + S (cos x)'d(cos x). 2. S'ain' z dr. [= √(1-00s' a)' sin z dz.] 7 5 x 3. Ssin' xdx. 4. S cos³x dx. cos² cos x + cos³ x — } cos³ x + C. cos x + cos³ x − & cos x + cos' x + C. sinx - sin x + C. cos³ x = cos x(1 — sin³ x). 166 DIFFERENTIAL AND INTEGRAL CALCULUS. 5. Sco cos³ x dx. sinx - sin³ x + ‡ sin³ x + C. x In like manner S cos" x da and Ssin" a da can be found where m is any odd positive integer. 6. S sir sin* x cos³ x dx. sin'xsin' x + C. 2 sin* x cos³ x = sin' x(1 - sin x) cos x. X In a similar manner Ssin” S sin" a cos" a dx may be found when x x either m or n is any odd positive integer. 7. S sin'x cos' x dx. - f cos³ x + cos' x + C. 8. S sin' x cos x dx. 3 9. S sin' x cos x dx. S 10. S cos² x dx. 11. S sin' x dx. 12. S cos² x = or sin*x + sin'x - sin x + sin' x – } sin³ x + C. - (cos x)+(cos x) + C. 8/02 + sin 2x + C. + cos 2x. 2018 2018 sin 2x+C, sin x cos x + C. dx sin x cos x dx 13. S sin² x cos² x 2 X 'sec² x [=/dr.] tan x tan x cot x + C. 1 sin' x + cos² x sin² x cos² x sin² x cos² x log tan x + C. 1 GENERAL DEPENDENT INTEGRATION. 167 14. Ssin' x 15. 2 dx cos² x x Ssin² a dx. 6 cos x sin² x 6 secx + cos x + C. tan x+tan³ x + C. tan³ x sec¹ x = tan³ x(1 + tan² x) sec² x. - cos x In like manner sinm x cos x dx or cosm x sin" x da may be integrated when- 16. S ever m n is even and negative. dx cost x tan x+tan³ x + C. 17. Sco 2 cos² x dx sin* x - cot³ x + C. 186. Trigonometric differentials can often be more conven- iently integrated as indicated by the following solutions. 4 18. fcos* x sin³ x dx. - † cos³ x + ‡ cos² x + C. dy Make cos x = y; then sin x = √ 1 — y² and dəc .. Sco cos* x sin³ x dx = S — y‘(1 — y³)dy 19. Stan tans x dx. Make tan x = y; then dx dx =S 5 = — fy° + fy' + C. tan*x - tan³ x + log (sec x) + C. dy y² + 1° 2 S (y² - Y .. Stanz de = Sydy = √ (y' −y + y² ² + 1 ) dy 1 = ‡y*— ży² + † log (y² + 1) + C. This method combined with that of Arts. 212 to 215 affords a complete solution of rational trigonometric differentials. J . 168 DIFFERENTIAL AND INTEGRAL CALCULUS. 187. Rational Fractions. A fraction whose terms involve only a finite number of positive and integral powers of the variable is called a Rational Fraction; as dy= X4 2x³ 13x² + 17 - dx. x² + 3x + 2 To separate this fraction into its integrable parts, we first divide the numerator by the denominator and obtain * dy = (x² (x² - 5x)dx + 10x + 17 x² + 3x + 2 dx. Again, by separating the fractional part of this quotient into two parts (its "partial" fractions), we obtain 10x+17 x² + 3x + 2 7dx 3dx dx = + x+1 x + 2 7dx 3dx x + 1 5x) + 7 dx + 7 S x + 1 dy= (x² - 5x)dx + = f (x² — 5x)dx y = + x + 2° dx + dx + 3 √ x + 2 ³f= X +7 log (x + 1) + 3 log (x + 2) + C. X 5x2 or Y 3 2 2x³ 15x2 y = 6 That is, + log (≈ + 1)(x + 2)³ + C. The first step in the above and similar operations is very simple, and it is our present purpose to show how the second step, the separation and integration of fractions whose denomi- nators contain a higher power of x than the numerator, may be effected; and to render the process as simple as possible we shall apply it to particular examples in each of the four cases that may occur. . GENERAL DEPENDENT INTEGRATION. 169 188. CASE I. When the simple factors of the denominator are real and unequal. EXAMPLES. 1. Integrate dy (x + 1)dx = 3 x³+6x²+8x The roots of x³ + 6x² + 8x = 0, are 0, − 2, and — 4; hence the factors of x³ + 6x² + 8x are x, x + 2, and x + 4. Assume 3 x+1 x²+6x²+8x A B C + + (1) X x + 2 x + 4 Clearing (1) of fractions, we have x+1=A(x+2)(x+4) + B(x) (x + 4) + C(x)(x+2), (2) or x+1=(A + B + C)x² + (6A + 4B + 20)x+84. Equating the coefficients of the like powers of x, we have A+B+C=0, 6A + 4B + 2C1, 8A = 1. Solving these equations, we find A, B, and C-. Substituting these values in (1), we have :. dy x+1 x³ + 6x² + 8x 8x dx = dx + 8x 4(x+2) •· · y = & fax + + .. dx pd x² + + √ x + 1 3 + 4(x+2) 8(x+4)* 3dx 8(x+4)* 2 $ √ dx x + 4 log x + log (x + 2) — § log (x+4)+1, log c. Vox (x cx(x + 2)² y = log v y= 3 (x+4)** 170 DIFFERENTIAL AND INTEGRAL CALCULUS. 2. The values of A, B, and C may be obtained from (2), thus: 0, we have 1= 8A; .. A= Making x = Making x Making x = = - 2, we have 1=-4B; .. B = 4, we have 3= 80; 1. 4. :.0= §. Principle. In this case, to every factor of the denominator, x-a, there corresponds a partial fraction of the form Find the following: X- - A α as 2. Sa 3dx 4 3. Sz 4. 5. adx ха a²° α S (5x + 1)dx x²+x-2° (22 f(x² + x − 1)dx x²+x²-6x log x + 2. (2 + 2)² + 0. C. X α log + C. x + a log [(x-1)³(x + 2)³c]. log √x(x − 2)³(x + 3)* + C. x² + x² 8 X 6. dx. 3 4x 100 + х3 2 +4x+ log x² (x — 2)³ (x+2)³ + C. 189. CASE II. When some of the simple factors of the de- nominator are real and equal. EXAMPLES. 1. Integrate dy = (x² + x)dx (x − 2)²(x — 1)' x² + x A B Assume + + (x − 2)²(x − 1) (x —. 2)² X ર 2 C x — 1° 1)+ Clearing of fractions, we have x² + x = A(x − 1) + B(x − 2) (x − 1) + C(x − 2)². 2)(x GENERAL DEPENDENT INTEGRATION. 171 ་ Making x = 2, we have 6 = A; .. A = 6. Making x = 1, we have 2 = C'; .. C = 2. Making x = 0, we have 0 = dy: = x² + x (x − 2)²(x − 1) 6dx (x − 2)² 1 − A + 2B +40; .. B = − 1. 6 A+ 2 + (x − 2)² X 2 x — 1° dx 2dx + X 2 y=6 65 dx (x 2)³ 6 X - 2 x — 1° dx Sa²² 2 + 2 S = X dx x — 1 log (x − 2) + 2 log (x − 1) + C (x-1)2 1)² 6 = log + α. (x − 2) X 2 - Principle. In this case, to every factor of the form (x − a)” there corresponds a series of n partial fractions of the form A B K -19 X a (x − a)n' (x x — α)n−1› Find the following: 2. S (32 (3x-1)dx 8 3 log (x − 3) (x — 3)³ · X + C. 3 3. S (929 (9x+9x128)dx (x − 3)² (x + 1) x²dx X LO 5 3 + 17 log (x − 3) — 8 log (x + 1) + C. 4 + √2+60 14-8x+1 2+2 + log (x + 1) + C. 4. S 5. 6. S S 3 x³ 5x² 4° dx (x + 2)²(x+3)²* 3x + 2 x(x+1)³dx. 2x+5 (x+2)(x+3) log (2+3) 4x + 3 2(x + 1)² + 2 log 10g (271) + 0. + + C. 172 DIFFERENTIAL AND INTEGRAL CALCULUS. 190. CASE III. When some of the simple factors of the de- nominator are imaginary and unequal. 1. Integrate EXAMPLES. xdx dy = (x + 1)(x² + 4)° - Here the two simple factors of x² + 4 are x + 2 1 and x-21; we may take these factors and proceed as in Case I, but the integrals obtained would involve the logarithms of imaginaries; to obviate this, we assume X (x + 1)(x² + 4) x + 1 Clearing of fractions, we have A. Bx + C + x² + 4 (1) (2) x = A(x²+4)+(Bx + C)(x+1). Differentiating (1), we have 1 = 2Ax + B(x + 1) + Bx + C. In (1) making x1, we have A. In (1) making x = In (2) making x = dx .. dy = - b ( d x 1) + 12 + 1/ 3 x dx 120 + + 0, we have C = −4A = +1. 0, we have B = 1− C = f. 4) dx (x+4) 12² + 4/ + } ada) ;)+ \20² dx +1 (706 2 \x² + 4/ +4/ † log (x + 1) + † log (x² + 4) + & tan-¹ .. y 10 = log +4 + & tan-¹ 2+ C. 2 X 212 + c : (x + 1)² Principle. In this case, to every factor of the denominator of the form (x − a)² + b² there corresponds a partial fraction of the form Ax + B (x − a)² + b²° GENERAL DEPENDENT INTEGRATION. 173 2. Száz 1 − x¹ (x² + 1)(x² +4)* x2dx x² + x² - 2° 3. S. 4. S 5. Sa 6. S 4dx x+1 X³ 1 x³ + 3x 1 + x log tan-¹x + C. - 1 − x dx X tan-¹x tan-1 201 + C. + log(2+1) + √2 X tan-1 + C. 3 √2 x² +3 dx. х x+log √3 tan-1 x² +C. 1/3 1 x² + x √2 + 1 log + √2 tan-¹ x V Z + C. √2 x² x √2+1 1 x² 191. CASE IV. When some of the simple factors of the denominator are imaginary and equal. 1. Integrate dy= 2 EXAMPLE. dx (x² + 3)²(x − 1)° 1 Ax + B Cx + D E Assume (x² + 3)²(x − 1) ¯ (x² +3)³ 2 + + x² + 3 XC 1° 1 x + 1 dy = 1 1 ..y = 8(x²+3) log (x²+3) Clearing of fractions, we have 1 = (Ax+B)(x−1) + (Cx + D)(x²+3)(x−1)+E(x²+3)². (1) Whence A = 1, B = −1, α=−16, −1, 16. D= −√, E = · 16, 1 (x + 1) 3)şdx 4 (x² + 3) dx — 16 x²+3 - 16 x - 1 tan-1 X 1 dx + 1 32 16 13 1/3 1 1 + log (x — 1) 16 √ dx (x² + 3) ³° dx The integration of differentials of the form be more conveniently obtained by Art. 211 or 215. may (ax² + b)m 174 DIFFERENTIAL AND INTEGRAL CALCULUS. . REDUCTION BY SUBSTITUTION. 192. Irrational Differentials. To integrate an irrational differential which is not of one of the known integrable forms, we first rationalize it, and then proceed according to the previ- ous methods. To show, in a simple manner, how rationalization is to be effected, we shall apply the process to a few particular examples. EXAMPLES. Find the following: 1. y = S_ ( 2 √ x + 3 ) dx 2 √x (x + 3 √x+5)* Make x = x²; .. dx = 2z dz. y = √ (2x+3)2z dz =S. 2. 2x(x²+3x+5) ((2 √x + 1)dx 41/x² + x √x dx 3. S 262² + 2) S 4. √ 5. √ S 2(x++ *° dx = log (z² + 3% + 5) = log (x + 3 √x + 5). √2 [Make z = 2.] 3 V x V 1 dx 1 − x³ √₂(1-2). √ x (1 − x) dx 6. √242 [Mako z = 2.] S x = √x + √x + C. tan-¹ √x + C. X sin-¹ Vx+C. log 8 (1 + √²) + + C. 2 √x − 3 V x + 6 Vx - 6 log (1 + Vx) + C. 7. S (x − 1)dx 2(x − 4) √x √x - 2 √x + & log + C. √x + 2 ! GENERAL DEPENDENT INTEGRATION. 175 ; 193. When a + bx is the only part having a fractional exponent. Assume a + bx = 2", where n is the least common multiple of the denominators of all the fractional exponents; then the values of x, dx, and each of the surds, will be rational in terms of z. Find EXAMPLES. 1. Y = x² dx. Assume 1+ x = x²; then 1+x √1 = %. Also X xz² z² — 1, x² 2. 3. (1) ၆ x =(z−1), . (2) dx = 2z dz. Multiplying (1), (2) and (3) together, we have Jav1+xd = ( −1) Sx S x dx √/1 + x dx SπNI+ x √ 1 + x 4. Sx(a+x)*dx. Z 8 to ¾ z¹ — — 2° + 3 z³ +0 (3) = q(1 + x)³ — §(1 + x)* + f(1 + x)³ + C. 2(x − 2)√1 + x + C. 3 √1+x-1 log + C. √1+x+ X z³ (a + x) * (4x — 3a) + C. 194. When a + bx+x² or Va+bx-x is the only surd involved. A differential containing no surd except a + bx+x³ can be rationalized by assuming a + bx+x² = z x; and one containing no surd except Va+bx-x can be rationalized by # 4 176 DIFFERENTIAL AND INTEGRAL CALCULUS. α assuming √a + bx − x² = (x − r)z, where r is one of the roots of a + bx − x² = 0. The process is illustrated in integrating the following im- portant differentials (see Ex. 30, 31, page 67). 1. Find y S Assume dx √ a + bx + x² √ a + bx + x³ = z − x; then 22 α a + bx = z³ — 2zx, x= 2z + b' dx 2(x² + bz+a)dz (2%+b)² (1) × (2), S 1 √ a + bx + x² dec √ a + bx + x² dx √ a + bx+x² (1) 2z+b (2) z² + bz + a S 2dz 2z+b тъ log (2z + b) + C. = log(2x+b+2 √ a + bx+x²)+C. Na + x² = log (2x + 2 √ α + x²) + C. S When b = 0, S dx 2. Find Y =S⋅ dx √ a + bx-x² = = Represent the factors of a + bx x² 0 by xr and x, and assume √ a + bx − x² = √(x − 1') (r' — x) = (x − r)z, = 1°)(r′ GENERAL DEPENDENT INTEGRATION. 177 rz² + pl then p' — x = (x — 1°)z², x 2² + 1 ³ 2(r-r')zdz 1 dx = (≈* + 1)³ 2 (1) √ a + bx x² z² + 1 (y' — 1') z° • (2) (1) × (2), S dx √a + bx - x² 2 fi dz 1 + z² -1 · 2 tan-¹z + C. S dx X 2 tan-1 + C. √ a + bx B x² XC ጥ S 2 tan-1 2 dx log √2 When b = 0, r Na dx - X² +va, and rl = 3. 12 12 12 108 (12+12+12- x + x Assume X 2 √2 + x − x² = √ (2 − x)(1 + x) = (2 − x)z; then X = 2z² 1 2² + 1' dx = 6z dz (≈² + 1)” and √2 + x - x² 2 S x² S 2dz 1 z √2 – 1 log 2z² — 1 - √2 z √2 + 1 2 3% ≈² + 1° Na, we have √ a + x √a-x + C. √/2 2x √2 — x + C. √2 2x √2 X 4. 5. S S dx √ z = x √2 + x- dx - x √ x² + 2x − 1 2 tan-¹(x + √x² + 2x − 1) + C. x² dx √3+2x-x² X 3 sin- -1 2 ર 1 _ (x + 3) √ 3 + 2x (x+3) 1/3 — x² 2 178 DIFFERENTIAL AND INTEGRAL CALCULUS. 195. Binomial Differentials. Differentials of the form xm(a + bxn) dx, where m, n, r, and s represent any positive or negative integers, are called binomial differentials. 196. To determine the conditions under which a binomial differential is integrable. S p I. When is a positive integer the binomial factor can be S developed in a finite number of terms, and the differential exactly integrated; and when is a negative integer the differential is r S a rational fraction whose integral can be obtained by the method of Art. 187, 212, 214, or 215. II. Assume a + bx" = 2³; :. (a + bxn) = x², 8 (1) 28 x = and b S a 1 > xm n dx = = ( - ) - bn -28–1 b 1 ( b dz.. Multiplying (1), (2) and (3) together, we have m an (2) (3) S xm (a + bx") dx = bn m + 1 α n b 1 dz.. (4) The second member of (4), and therefore the first, is in- tegrable when m + 1 n is a positive or negative integer, by Case I. III. Assume a + bx² = 2ºx"; ... an = a(zº — b)-¹, 1 x = añ(2º — b). a + bxn az8 G m 1 —=—, n xm = añ (z³ — b)¯ m n (a + bx")" = a"(z• — b)˜¯"zdz, (1) (2) GENERAL DEPENDENT INTEGRATION. 179 and dx = Sanys-1(28 — b) n dz. (3) N Multiplying (1), (2) and (3) together, we have S m+1 x¹(a + bx¹)sdx + - -a N n m + 1 n + +1) yr+s-¹dz. (4) By Case I, the second member of (4) is integrable when m + 1 N + is a positive or negative integer. S Hence, x™ (a + bx”) dx can be integrated by rationalization: I. When m + 1 is an integer or O, by assuming a + bxn = z³. N m + 1 II. When + is an integer or 0, by assuming n S a + bxn = z³xn. When the differential reduces to a rational fraction, which is m + 1 m+1 r the case when +1 is a negative, or + +1 a posi- n n S tive, integer, it is less laborious to integrate by a method to be subsequently given. EXAMPLES. 1. Find ƒx*(1+ 2ª²)*dæ. m + 1 5+1 Here ૭ =3, an integer, and s = 2; hence we n 2 assume 1 + x² = 2"; z²; .'. (1 + x²)* = %, • (1) x² = z² 1, x = (2² + 1)³.... (z+1).. (2) Differentiating (2), 6x³dx=6(x² + 1)'z dz.. 2 (3) 180 DIFFERENTIAL AND INTEGRAL CALCULUS. C 1 Multiplying (1) and (3) and dividing by 6, we have S'x'(1 + x°)*dx = ƒ (x² + 1)ºz°dz Find: 2. Sx*(1+x²)*dx. = S (2° + 2x* + z³)dz C = 4x² + fz² + 12² + α = 4(1+x²)³ + f(1 + x²)² + §(1 + x²)ª+ 0. (1 + 2¹³)¹³ (3 2 1 5 2) + 0. x²) 15 ‚³½ (1 + xº)³ — §(1 + x²)® + √³(1 + x²) § + C. 3. Sx*(1+x°)³ïx. 4. Sx²(1 + x²)¯*dx. (1 + 2º)* (2ª¸ 2 x²) + C. 2x²+1 x√1 + x² 5. Sx²(1+x²)¯*dx. Here m + 1 n 2+1 2 11Q an integer; hence we assume m + 1 r and + п S 11 — 1+x² = x²x²; .'. (1 + x²)¯* — (2² − 1)*, .3 1 2 ২৩। oo laz 3 2 || + C. - 2, (1) 2 x² = z² - 1; . (2) zdz dx = (3) - (≈² − 1)³ ° F GENERAL DEPENDENT INTEGRATION. 181 Multiplying (1), (2), and (3) together, we have ƒ x (1 + x²) ~ "dx = ƒ − (1 − )ds Sx¯²(1+x²)˜³dx A 1 where z = √1 + x³. х 6. /(1+x²)¯³dx. [m= 0.] 7. S x™‘(1 — 2x²)¯*dx. 1 dz = − ( x + 1 ) + α X 2 √1 + x² + C. (1 + 4x²)(1 − 2x²)+ 3x³ + C. 8. S x¯²(a+xº)¯³dx. - 3x³ + 2α 2a²x (a + x³)* + C. INTEGRATION BY PARTS. 197. Integrating both members of d(uv) = udv+vdu and transposing, we have S udv = uv - Si vdu, (A) which is the formula for integration by parts. It reduces the integration of udv to that of vdu, and by its application many differentials can be reduced to one of the elementary forms. 1. Find Sæ² log xdx. EXAMPLES. Assume u = log x; then dx dv = x²dx, du = X and v = Sxdx = x² } 182 DIFFERENTIAL AND INTEGRAL CALCULUS. Substituting in (A), we have Sæ² log 22:3 2 xdx Xx log x -fxdx = 3 x³ log x 3 + C. .9 -1 = 3 2. Find Ssin-¹x dx. Assume usin-¹x; then dv = dx, v = x, and du = Substituting in (A), we have dx √1-x² -1 S sin-¹x dx = x sin-¹x — 1 S x dx √1−x² = x sin-¹x + (1 − x²)*+C. Find the following: 3. Stan 4. S% - -1 X dx. x cos x dx. x tan-¹x — log (1 + x²)* + C. x sin x + cos x + C. eaal X 0. eaz (27 - 21/2) + α. 5. Sxe xeaxdx. Make dv = eaxdx; eax .. v= u = x. α 6. Sz'erdz. ax Sometimes two or more applications of the formula are re- quired, as in the next example. [22-23 +3] + 2x 2 eax + C. a α α eax Make dv = eаxdx; •• v = u = x², du = 2x dx. α eax x²eax dx = eаxx dx Szerdz=2-2 ferdz Searn Now apply the formula to the last term, as in Ex. 5, and we obtain the entire integral. a α + GENERAL DEPENDENT INTEGRATION. 183 2 — x²√ a² — x² — }(a² — x²)²+C. 7. S x³dx X x dx .'. v = Va² - x², and u = x². Make dv= Na² S x³ dx = Na² x² − x²√a² — x² + S 2x(a² — x²)*dx. In a similar manner we may integrate any binomial differen- tial, or by continued application of the formula reduce it to a simpler form, but by the method of Art. 211 the result may in general be obtained with less labor. 8. Sx² x³ cos x dx. Make dv = cos x dx; then v = sin x, u = x³, du = 3x²dx. Sx³ x³ cos x dx = x³ sin x S 3x² sin 3x² sin x dx. Again, make dv=- sin x dx; then v=cos x, u=3x², du=6xdx. 3x² sin x dx = 3x² cos x f3x² sin x dx J S 6x 6x cos x dx. Again, make dv——cos x dx, then v=-sin x, u=6x, du=6dx. .. - f 6x cos x dx = − 6x sin x + 6 sin x dx(= − 6 cos x). ·Sx² x³ cos x dx = x³ sin x+3x² cos x-6x sin x-6 cos x+C. 9. flog zdz. 10. fr' log a dr. x 11. Sæ² lóg² x dx. x(log x − 1) + C. 4 · X log x 16 + C. 3 x²(log² x — log x + ) + C. 184 DIFFERENTIAL AND INTEGRAL CALCULUS. 12. Serár. 3 ex (x*- 4x³+12x²-24x+24)+C. exx⭑dx. 13. Sxadr. ах X log a 14. Sx²e-*dx. 15. fr'e dr. eax 16. Sæ² sin-¹ x dx. S 17. S log x dx (x + 1)²° 18. Sx³ (log x)'dx. 20 1 1- a)+ log a) + C. - e-* (x²+2x+2)+ C. 6 (x² - 3x² + 6x - 8 ) + 0. 3 -1 α 2 a² α x² a 3 a C. sin" 2+2 + 2 √1-x+0. X x + 1 9 x² C. log x log (x + 1) + C. 2-[(log x)² — † log ≈ + §] + C. 19. S (a² (a² —- x³)¯*x³dx. − XC 2 − a² 2/ (a² − x²)*+ 2 o sin & + X C. a a² 2 20. S (a*+x*)¯*x*dx. 3 (aª+x²)* − a log (x+√x²+aª)+0. REDUCTION FORMULAS. 198. Reduction formulas are formulas by which the integral of a differential may be made to depend on the integral of a similar, but simpler, differential. Sxº (log x)* dx, 199. To find the reduction formula for S where n is a positive integer. Assume then dv = x² dx and W = (log x)"; XP+1 v= P+1 and du= n(log x)"- dx -1 X Substituting in (A), Art. 197, we have Sx³(log x)"dx = x²+¹(log x)" P+1 n p 7 1S x" (log x)"-¹dx, (1) + le GENERAL DEPENDENT INTEGRATION. 185 in which the proposed integral depends upon another of the same form, but having the exponent of log x less by one. By successive applications of this formula the exponent of log x is reduced to zero, and the proposed integral is made to depend upon the known form faºdx. XP SX (log x)"da, where X Sx COR. I. If the given integral were is any algebraic function of x, we should have S X (log x)"dx = X¸(log x)" where X₁ = SXdx. X, 1. Sx' log* x dx. = — n (log x)^-1 dx, XC EXAMPLES. Here p 3, p + 1 = 4, n = 3 and n 12. Substituting in (1), we have S'x' log³ x dx = ‡x* log³ x − ‡ƒ x° (log x)'dx. x³ 0 By applying the formula to the last term, etc., we obtain 3.2 ƒ'x' logʻ z = "=-[log" a — 23 log′ x +3+3 log x 3.2.1] +0. S 2 3 2. 3. 4 Sx' (log x)³ dx. log x dx Slog (1 + x) *° x 4 6 Х 43 C. (log x)² — § log x + rs + C. - 18 log x log (1 + x)+ C. X = 1 1 + x 29 (1 + x) ³ ³ .. X₁ X, 1 1 + x 186 DIFFERENTIAL AND INTEGRAL CALCULUS. : 200. To find the reduction formula for Sa"a"da, where n is a positive integer. a*xˆdx, Let then dv = a*dx and u = x"; v = ax log a .. Art. 197, fa*x"dx = 1. fa'r'dz. and dunx²-¹dx. axxn log a ገ iS ala log a axxn-1dx. (1) 6 logo a] + c. EXAMPLES. ax 3x² 6x XC³ + log a log a log³ a a∞x³ 3 log a log a af a²x²dx. Here n = 3, . Sa²x²dx By further applications of (1) we obtain the desired result. 2. Sax axx* dx. ах 4x³ 12x² 24x 24 XC4 + + log a log a log² a log³ a log¹ a + C. +0 eax 3x² 6x 6 3. eaxx³ dx. X³ + α a a² a³ 1 + C. 201. To find the reduction formula for Sx" cos Sx" sin ax dx, where n is a positive integer. xn x" cos ax dx and dv = cos ax dx; and Make u = xn sin ax then du = nx"-1dx and v v = a Sx cos az de = 12" sin az "fin az da. x² ax dx a GENERAL DEPENDENT INTEGRATION. 187 Similarly, we find Xn sin ax dx = - 1 a a +2 S x²-1 cos ax dx. x² cos ax+ Hence, in either case, the integral can be made to depend on the known form Scos ax dx or f sin ax dx. EXAMPLES. x cos x dx. x³ sin x + 3x³ cos x 1. S 2² 2. S'2' sin x dz. Sx x* 6x sin ? 6 cos x + C. 4 x¹ cos x + 4x³ sin x + 12x² cos x 24x sin x 24 cos x + C. x dx, 202. To find the reduction formula for Xsin-1 Xtan-¹x dx, etc., where X is an algebraic function of x. -1 Make u = sin-¹ x -1 and dv = Xdx; 7x then Zu and v = x² = Xda = X₁ (say). .'. S X sin¹x dx = X, sin-¹ x — S X,dx √1 — x² EXAMPLES. x² tan-¹x dx. x³ tan-¹ x 3 -1 x² log (1 + x²) + + C. 6 6 Sætan´¹2 1. J Here x² Sx² tan-¹x dx 1 + x² 2. X = x² and X₁ = 23 X, 3 x tan-¹x — (tan−¹ x)² — ½ log (1 + x²) + C. ƒ sec¨¹x—†(x². x² sec−¹x dx. ‡x³ sec-¹ x—† (x² — 1)*x— ‡ log (x+√x²−1+C. 3. Reduction formulas for binomial differentials are deduced in Art. 215. 188 DIFFERENTIAL AND INTEGRAL CALCULUS. 203. To integrate dx a + b cos x dx 11 a(cos alcos² 2 dx a + b cos x 8122 dx 2 + sin 2) + 6(cos² 2 — sin³ 2) (a + b) cos² X 20 | X dx + (a - b) sin² sec² 5 dx X n² 228 (a + b) + (a - b) tan² :: = 2 S a+b cos x ર ગ d(tan 2/2) 8/03 2 (a + b) + (a - b) tan² X 102 S S dv or joz + 203 S; dv , according as a> or : CHAPTER IX. INTEGRATION—(Continued). INDEPENDENT INTEGRATION. 207. Increments Deduced from Differentials. We have seen that the increment of a function is the sum of the differ- ential and the acceleration; hence, when the former is known, we can find the differential by simply removing the acceleration. Taylor's formula enables us to reverse this operation in many cases, and find the increment when the differential is known. Let u = f(x). = Increasing x by h, we have, by Taylor's formula, u+4u = f(x+h) hn = f(x) +ƒ'(x)h +ƒ'(x)" · · · ƒ"(x) * * ……. ▲u = f(x + h) − f(x) h² 2 = ƒ'(x)h +ƒ''(x)', + · · · ƒ¹(x) in which du f'(x)h. = = N h n n (A) X Therefore, when the differential of a function of x is known, the increment may be found by taking the successive derivatives. of the differential coefficient, and substituting them in (A). When f(x) = 0, and each of the subsequent derivatives of f(x) = 0, the series will be finite and express the exact value of Au; otherwise the series will be infinite, and, if convergent, will give the approximate value of Дu. 191 二 ​192 DIFFERENTIAL AND INTEGRAL CALCULUS. | EXAMPLES. 1. If du = (x² - 5x+6)dx, what is the value of 4u? Here f'(x) = x²-5x+6. Differentiating this, we obtain f'(x)=2x-5, ƒ""'(x) = 2, fiv(x) = 0. Substituting these values in (A), we have Δυ Au= (x²-5x+6)h + (2x h² hs 5x + 6)h + (2x − 5) ½ + ½ 2. If du= (x² - 3x-10)h, what is the value of 4u? Au = (x² - 3x-10)h + (2x-3) 3) ½ + 3. If du = (x³ — 7x² + 12x)dx, what is the value of au? h² 2 h² h⁹ 3 3 h₁ 3 4 4. Find au when du = sin x dx. au = cos x h² 2 h⭑ sin x COS X + etc. 6 24 au = (3x² - 14x+12)+(3x 5. If du = (√1+x)dx, by how much will be increased : X when x is increased by h? Au = (√1+x)h + (1 + x)' h³ ព (1 + x)− 4 24 + etc. h³ h³ h₁ h³ au = + 2x 6x2 12x³ 20.x* 3 + etc. 6. Find au when du = log x dx. 7. A function is increasing at the rate of 4x³dx; find its suc- ceeding increment. 2 ▲u = 4x³h + 6x³h² + 4xh³ +h*. 8. At the end of t seconds the velocity of a body is ds dt = (3ť² - 2t) ft. per second; − find the distance it will travel the following second, dt being the unit of time. 48 (3t2t)dt + (3t 1)dť² + dť³. 9. The rate of acceleration of the velocity of a body is = dv = (3t+4) ft. per second; dt INTEGRATION. 193 find the increment (1) of the velocity (v), and (2) of the distance (s) for the following second. Av = (3t+4) dt + gdť³. V = S (3t + 4) at = $t² + 4t + C. (3t+4) As = (3ť² + 4t + C) dt + (3t + 2)dt² + dt. 208. Increments as Definite Integrals. In Fig. 5, where u = area of OBPA, 4u = the area of BCP'P, which is evi- dently the integral of du between the limits x and x + h. In general S х x+h ƒ'(x)dx = f(x + h) − f(x). For, since_df(x) = ƒ'(x)dx, ƒƒ'(x)dx = f(x)+C. x+h دیر x ƒ'(x)dx = [f(x) + C Therefore (A) may be written x+h h2 x+h ત "= = f(x + h) − f(x). [ƒ'(x)dx ƒ'(x)h+ƒ'"(x) ƒ'(x)dx = ƒ'(x)h +ƒ''(x) +ƒ'''(x)· 12 h³ hn 13 · ƒn (x) = (B) n By this formula we can obtain exactly, or in the form of an infinite series, the definite integral of any function of a single variable, and the operation does not involve the reversing of any of the formulas for differentiating. But, in general, this method is much inferior to that of dependent integration, since by the latter many differentials can be integrated in finite terms which by the former could be expressed only in the form of an infinite series. However, it forms an important part of the theory of differentials and integrals, and is often useful as a method of approximation. More convenient formulas for practical purposes will be derived from (A), but before doing so let us apply (B) to the following illustrative examples. 194 DIFFERENTIAL AND INTEGRAL CALCULUS. . I # 1. Find the area BCP'P, the equation of APP' being A P P -X B C E F n FIG. 43. y = x² - 7x+12, where x = OB, y = BP, and BC h or dx. Let u= area of OBPA, then du = ydx. = — y =ƒ'(x) = x² - 7x+12, f'(x) = 2x — 7, f'(x) = 2, fiv(x) = 0. Substituting in (B), we have x+h √ — h² x (y)dx = (x² —— '7x+12)h+(2x−7)½+ ཚེ་ = area of BCP'P. 3 COR. I. To find the area of OEA we make x = 0 and h = OE = 3, and obtain 13. To find the area of EnF, we make x = OE = 3 and h = EF = 1, and get — ‡. Let the student solve the following in a similar manner. 2. The equation of a curve is y = x³ 6x² + 11x — 6; find the areas of the two sections enclosed by the curve and the axis of x. 209. A More Convenient Series. x = 0, then making h=æ, and writing we have [*ƒ′(x)dx = f(0) +ƒ'(0)x +ƒ”(0)2 + 1; -1. 一 ​In (A), by making х S f'(x)dx for f(x), ·ƒ"(0)- (C) n Xn This formula may be obtained by developing ƒ'(x) by Maclaurin's formula, multiplying by dx, and integrating each INTEGRATION. 195 term separately, but as we are now exemplifying the method of independent integration, we will apply (C) directly to one or two examples. xx 1. Find √ (3x² – 14x+5)dx. Here f'(x) = 3x² - 14x+5, .. f'(0) = 5; f'(x)=6x-14, ƒ'''(x) = 6, .. ƒ"(0) = 14; ..ƒ""(0) = 6. - - Substituting in (C), we have x 2. (2 3. (3x+ √ (32² — 14x + 5)dx = x* — 7xª + 5x. 6x²+7)dx. - 2x³ + x²)dx. 210. Bernouilli's Series. hx, æ, observing that f'(x)dx x² 4x² - 2x³ +7x. ¾x³ − {x¹ + ‡x³. In formula (B), by making S* f'(x)dx = ƒ′(x)x − ƒ"(x) * + ... 2 f'(x)dx, we have Xn + ... − (− 1)¹ƒn(x){ - 1)"ƒ" (x) · · · (D) This formula, called Bernouilli's Series, like formulas (B) and (C), shows the possibility of expressing the integral of every function of a single variable, in terms of that variable, since the successive derivatives f'(x), f'(x), etc., can always be deduced from f'(x). Hence, in all cases where the series are finite or infinite and convergent the integral may be ob- tained exactly or approximately. Sƒ' In finding f'(x)dx by (B), (C), or (D) the limits of the difference between the approximate value found and true value may be determined as in A, of the Appendix. 196 DIFFERENTIAL AND INTEGRAL CALCULUS. -1 INTEGRATION BY INDETERMINATE COEFFICIENTS. 211. The process of integrating binomial and trigonometric differentials by successive reductions is generally very tedious, and it is our purpose now to present a method which is gener- ally less laborious, and which is also applicable to many other classes of differentials. Let urvdx be the differential to be integrated, where u and v are functions of x, and let us assume ƒ w*vdx = u*+f(x) + kƒ°w°v¸dæ, · J (E) in which it is required to find the function f(x) and the con- stant k. In examples where Suvdx can be expressed under the form of ur+¹f(x), we shall find = 0; and when this is not the k case, uvda will be determined in terms of fuvda, which Surv we can generally make of a more elementary character by as- signing suitable values to s and v¸. Another advantage of expressing the required integral under the form of (E) arises from the fact that ur+f(x) often vanishes for the desired limits of integration, in which case the definite integral depends on the last term only. Differentiating (E) and dividing by uˆdx, we have v = (r + 1)du f(x) + uf'(x) + kv¸u³-*. dx (F) The simplest and easiest method of solving this equation for f(x) and k is by indeterminate coefficients, as illustrated in the following examples. 212. CASE I. When k=0;-Independent Integration. 5 1. Find √(1+x²)¯*x*dx. EXAMPLES. 4 ! INTEGRATION. 197 du dx Comparing this with (E) we have u=1+x³, r = 2x, and v = x³. Substituting in (F), we have - x³ = xƒ(x) + (1 + x²)ƒ′(x) + kv¸(1 + x²)³+‡. . . (1) The first member of this equation being of the fifth degree the second must be also; hence, f(x) must be of the fourth de- gree, and since u involves only the second power of x, we may assume have or f(x) = Ax' + Bx² + C, :. f'(x) = 4Аx³ +2Bx. Substituting in (1), and arranging in reference to x, we x* = 5Ax² + (3B+4A)x³ + (C + 2B)x + kv¸(1 + x²) 8++. 1=5A, 3B=-4A, C=— 2B, k=0, A = }, B = − 1, C = 1% • 18%. These values determine f(x), which substituted in (E) gives S(1 + x')¯*x*dx = (1 + x²)*[fæ* − ´ffaª + ffx] + C. 2. Find Sx-*(1+x*)*dx. 1 − (1 + 2')' ( 1 − 1832) + 0. 2²) 5x5 15x³ du Here u=1+x³, r = 2x, and v=x-; .. (F) gives dx x−6 = 3xƒ(x) + (1 + x²)ƒ′(x) + kv,(1 + x²)³-‡. . . (1) In order that the two members may be of the same degree we assume 5 -4 f(x) = Ax˜³+ Bx-³+ Cx-¹; .. f'(x) = - 5 Ax-°-3Bx˜^— Cx-². : Substituting in (1), and arranging in reference to x, we find A = − }, B = †, C= 0, k = 0; this determines f(x); which substituted in (E) gives the desired result. 15, 198 DIFFERENTIAL AND INTEGRAL CALCULUS. T ? , A careful inspection of the previous examples suggests the following rule for determining the form of f(x) for binomial differentials of the form (a + bx") xdx, v being xm: I. When m is positive the highest exponent of x in f(x) will n + 1. be m II. When m is negative the algebraically lowest exponent of x in f(x) will be m + 1. III. The remaining exponents decrease or increase alge- braically by n. The rule is also applicable when u is a polynomial in which n is the highest exponent of x, provided that the exponents of x in f(x) increase or decrease by the least difference between the exponents of x in u. When is a fraction and u a polynomial of a higher degree than the second, the differential cannot ordinarily be inte- grated; or, more accurately, its integral cannot ordinarily be finitely expressed in terms of the functions with which we are familiar. The exceptional or integrable cases are, in general, where u, v, and r are such that it is possible for f(x) to have as many coefficients as there will be independent equations between the coefficients in equation (F), and where k v¸u'dx is 0 or one of the integrable forms. In a differential of any given form the conditions of integrability may often be determined by the present method. 3. S₂- dx x* √x² + 6x + 15 √x² + 6x + 15 45 1 1 3 X 2x² 6x + 1) + C. Here u = x² + 6x + 15, r = lowest exponent of x in f(x) will be others will increase by 1, giving fx , and v=x-*; hence the - 4+1= 3, and the -3 2 + 1 Ax¬³ + Вx-² + Сx-¹ + D. The process can also be applied to many differentials in which v is a polynomial, as in the next example. 4. S 3x²+5x+5 dx. √x² + 2x+3 √ x * + 2x + 3(3x² + 1 ) + C. 2 INTEGRATION. 199 Here ux²+ 2x + 3, r = −, v = 3x² + 5x + 5, and f(x) is of the form Ax + B. dx 5. S (1 + x²)** XC (1 + 2) [18+ + + + 1 ] +0 x²) # 152: = C. Here u=1+ x², r = − }, v = 1; and making v, 1, s = -, (F) gives 1= 5xf(2)+(1+)f’(c)+k(1+x), where f (x) is evidently of the form Ax³ + Bx³ + Cx. m 213. We have seen (Art. 185) that sin x cos" x dx can be easily reduced to an integrable form when either m or n, or both, are positive odd integers, or when m+n is an even integer and negative. In such cases, m and n being integers, the inte- gration may be effected by the independent method, as in the two following examples; but this method of integrating such differentials is introduced and recommended chiefly for its bear- ing on the cases in which the above conditions do not exist, and which are usually solved by successive reductions. 6. f sin³ x cos¹ x dx. f'sin' 5 cos x 9 sin x 4 sin² x + 8 + + C. 63 315 du We may make u = cos x, r = 4; then sin x and dx 5 sin x. Substituting in (F), we have 4 sinº x = - 5 sin xf(x) + cos xf'(x) + kv, coss-¹ X, where f(x) is evidently of the form A sin¹ x + B sin² x + C, and hence f'(x) = (4A sin³ x + 2B sin x) cos x. ... sin³ x = 5A sin x 5 5B sin³ x 5 C sin x + (4A sin³ x + 2B sin x) cos² + etc. X Now, substituting 1 - sin' x for cosa, reducing, and arrang- ing with respect to sin x, we have 1 200 DIFFERENTIAL AND INTEGRAL CALCULUS. · ¿ - (1 + 9A) sin³ x + (7B − 4A) sin³ x + (5C − 2B) sin x + kv, cos³-4 x = 0. .. A, B, C, k = 0. dx 7. Sez cos x sin x 8 5 cos x 15 3 sin'z-sin'+1]+ sin❜ C. du Make u cos x, cos x, 1' = — 6, sin x, v = 1, s=0, dx 1 v₁ = 1, and we have from (F) Assume 1 = 5 sin x f(x) + cos xf'(x) + k cos x. f(x)= A sin x + B sin' x + Csin x. 4 2 .. ƒ'(x) = 5A sin x cos x + 3B sin x cos x + С cos x. Substitute, reduce, etc., and we find 8 4 A = B = C = 1, k=0. 15' 3' This method of integration can often be applied to other classes of differentials, as in the next two examples. 8. fx log' z dz dx. x 2 5 (log-logx+5)+0. C. Make u = x, 1' = 4, v = log³ x, and assume f(x) = A log² x + Blog x + C. 9. eаxx³ dx. 3 (203 3x² 6x eαx a a² + a³ α a¹ 6) + α. C. du = αeаx; dx Make u = eª×, r = 0, s = 1; then v = ex³, and substituting in (F) and dividing by eax, we have x³ = aƒ(x) +f'(x) + kv,, where we evidently have f(x) = Ax³ + Bx² + Cx + D. INTEGRATION. 201 Find: 10. S 3 x³ dx — √1 x² V1 — 2 ( x² 2 + √1 - x² 3 1 ) + 0. S x³ dx − - 1 − x² ( x + 3.5 1.2.4\ 1. 3x² + 1.3.1) + 0. 11. 12. S √1 − x³ x* dx √ 1 − x² √1 — x² 13. Sx-²(1+x²)¯³dx. Assume 1.6 1.4.6 + 1.2.4.6\ 5.7 3.5.72² + 1.3.5.7 + C. 1 2x - (+240) + 0. = A х + Cx. f(x) 14. S xdx Na + bx² 3x¹ b.x³ (3.xª 4αx² 8a2 1 + + C. √ a + bx² b b2/15b 15. S X 12bx² 16. dx (a + bx²) S (5x² + 3x² + 2)dx √1 + x³ 17. S x°(2 + 3x²)*dx. 18. Sx'(1-22)¯*dr. ار 19. 20. S (x* - 2x + 1)dx √x² - 4x+1 (x³ + x²)dx (a + bx²)² (3a² + 1) + C. (2 + 3x²)³/9 27 √5 3 + 4x + x² (2x-5x-6x+7)dx (* _ 5 + 2)* 21. S(32° - 22. S z 5 sin" x cos x da. Make X u = sinx, α 3 2√/1 + x³(x + 1) + C. 24 32 x² + + C. 35 105 4x² - (1 − 2x²)¹ (1 + 1¹² ) + 0. (- + C. 3x³ (x* — 4x + 1)³(~~) + C. (58 +4x+x²)*(´??² 7x 49 +19+0. 3 6 15 (2 – 5x + 2)(2 − 5) + 0. sin*x(cos² x + 3) + C. r = 5, v = cos³ x. 202 DIFFERENTIAL AND INTEGRAL CALCULUS. 23. f'sin° 2 dx. x 1 3 cos x [sin' x + 4 sin' x − 8] + C. 1 24. Sx® x fr log' z dr.log'-log+log-]+0 6 25. Seªxx'dx. caz[xt сах α 26. log da x dx. 1 2 4x³ 4.3x² + 36 4.3.2x 4.3.2.1 + 2.1] C. a³ 3 a |+c. 1 2x² log² x + log x + 1] + C. 2 ας 214. CASE II. When k is not = 0;-Dependent Integra- tion. EXAMPLES. X³ X 1 6 24 16 + sin-¹x+C. 1 16 1. ƒ'x'(1 − x²)*da. ¸ (1—2°)* [ ~* 4 The differential x¹(1 − x²)*dx has the same binomial factor -1 as (1 - x²)-*dx, whose integral is sin-¹x; hence, by expressing the former in terms of the latter, the required integral may be obtained in terms of (1 − x°)¯*dæ. Thus, x*(1 − x²)* = x*(1 − x²)(1 − x²) −✦ = (x*—x°)(1−x²)¯*. · du .. u=1 - x², r = − 1), dx = 2x, v = x¹ - x°, which sub- stituted in (F) gives (making s = r and v₁ = 1) 1 x¹ — x® = − xƒ(x) + (1 − x³)ƒ′(x) + k, · where f(x) = Ax³ + B׳ + Сx; and proceeding as before, we find A, B, C = - 2. S dx (1 + x²) ³° = Here v = 1+x³, r ki 16. fe, and k 1 (1 + x²)²[§x* + §x] + § tan‍¹ x + C. du 3, =2x, v = 1, and in order to dx INTEGRATION. 203 express / (1 + 2)² dx in terms of i√ 1 + x²² = tan-¹ x] we make 1+23 v₁ = 1, s = − 1, 8 - r = 2, and (F) becomes 1= s 4xf(x) + (1 + x²)ƒ′(x) + k(1 + 2x² + x³), where f(x) = Ax* + Bx. 1 3. S=√1-3 dx x* -√1-x² +32) + log (1 & (1- √1-x + C. X 4x¹ In order to express the integral in terms of dx x√1 − x² we make ur x√1 — x² = √x² — x², [= log(1 − 1 = 2*)] 4 = X S= – †, then u = x² — x¹, r = − 1, du dx 1 = 2x — 4x³, v = 1, and (F) becomes 2 X-4 — = ( − 2 )f() + ( * _ *)f? (2) +k, · -5 -3 X where f(x) = Ax¬³ + Bx¬³ + Сx-1, since x is a factor of u. x² + 8x + 21 4x+9) dx. 4. S (x² Here du u = x² - 4x + 9, r = − 2, &, =2x-4, and v=x+8x + 21; dx hence we have from (F), making s = 1, x³ +8x+21= (4 − 2x)f(x) + (x² - 4x+9)ƒ′(x) + kv,(x² — 4x + 9). - (1) The second member of this equation must be of the third degree; but if we make v, 1, the solution will be impossible, let us therefore assume that k = 1 and v₁ = Cx + D; we may then make f(x) = Ax+ B and f'(x) = A. Substituting in (1), 1 數 ​I 204 DIFFERENTIAL AND INTEGRAL CALCULUS. • * we find A, B = −, C= 1, and D = 4. Hence the re- quired integral is 3(x-7) 2(x² - 4x+9) + f ( x + 3 2¹²)dx X2 4x+9 1dx 3(x-7) + 2(x² - 4x+9) S X2 √x²x² - 4x + 9 (x − 2)dx + S - √ 2² = 4 x +9 3(x-7) 315 2 tan-¹ + C. 2 1/5 2(x² - 4x+9) + † log (x²-4x+9) + m The solutions of the three following examples illustrate the manner of integrating sin x cos x dx when the conditions stated in Art. 213 do not exist. [cos' x cos x]+ log tan x+C. 5. dx sin X 1 § 5 sin* x du = Here we make u — sin x, r = — 5; then v=1 and = cos x. X. dx Making v₁ = 1 and s-1, (F) gives 1 1=-4 cos xf(x) + sin xf'(x) + k sin* x. Making sin x = (1 cos² x)², ƒ(x) = A cos³ x + B cos x, and proceeding as in examples 6 and 7, Art. 213, we find A = §, B = — §, and k .. S dx I sin x = ૐ. 1 (§ cos³ x — § cos x) + § sin* x S dx (Art. 204) sin x This example may also be solved like the following one. dx 6. Sade z 7 cos x du Make u = sin x, r = 0, then = cos x, and v = cos˜” x. dx .. cos¬7 x = cos xf(x) + sin xƒ'(x) + kv, sin³ x, • (1) INTEGRATION. 205 where and -6 ƒ(x) = A cos¯ˆ x + B cos*x + С cos¬² f'(x)=(6A cos-x+4B cos-5x+2C cos-3x) sin x. Substituting in (1), making sin² x = 1− cos² x, reducing, etc., we have 6A 1 4B – 5A 20-3B C + 5 + cos' x cos x cos³ x 1 + kv, sin³ x = 0. COS X A A .. 4 = †‚, B = ‚†‚ C = 1§, and (making s=0, v, = 2, 49 5 8 1 0. cos x) k=C. .. S dx sin x 5.1 sin x + cos x 7 6 cos* x 6.4 cos¹ x 5.3.1 sin x 5.3.1 + + 6.4.2 cos² x 6.4.2 S dx To integrate the last term see Art. 204, page 188. 7. S sin' x cos x dx. cos x sin³ x[- + X 8 3 COS X cos³ x 16 3 cos x + 64 1-128 3 (sin x cos x-x)+C. Make usin x, r = 2, then v = sin² x cos¹ x = 2, then v = sin² x cos¹ x = cos¹ x—cos® x; also make s = 2 and v₁ = 1, then v, cos x cos x = 3 cos xf(x) + sin xf'(x)+k.. (1) Make f(x) = A cos x + B cos³ x+C cos x, find f'(x), sub- stitute in (1), for sin³ x write 1 cos² x, etc., and we find 3 A=}, B, C, and k = = S sin x cos x dx = sin³ x 3 = 64° 3 cos³ x 8 cos³ x + + 16 3 cos x 64 3 + 4S sin²x dx. To integrate the last term, see ex. 11, Art. 185. 64 • • 206 DIFFERENTIAL AND INTEGRAL CALCULUS. να 8. x²dx Sad [--+ sin-2] =↓ 2 α Va² х3 dx 9. x³ √ a³ x² · a Απα. [- 2 Na² - 20² 2α²x² + X +a 1 X 3 log 2a a + √ a² x² -α a² = log(-1) 2a³ 10. Svæ²+zdz. √x²+x* + log (x + √ a² + xº) + 0. 11. f(x² + x²)¹dz. S X 2 2 — * (2x² + 5a") √z" + a + a log (x + √x² + a¹) + C. X 8 12. S (a² — x²)¹³dx. } (5a³ — 2x°)√aª — xª + За 8 sin-1 -12 +0. a za x²dx 13. √2ax ха 14. 15. 16. 17. Love 02a 2α S S [-1/2ax - 2+ (2 + 3a) + 30° √2ax x²√2ax x² dx. x³√2ax-x dx. x⭑dx √1 − x² 1 xdx - 18. √ √3 + 2x + x² dx. x² 2 За vers -1 ४। 2 α 0 2a = πα. Επα. πα. 1.3/π 2.4\2. 1.3.5/π 2.4.6 2. √3 + 2x + x² (x² + ¹) + log (2x+2+2√3 + 2x + x²) + C. 19. 10+ 3x - xdx. 2x X 2x √10 + 3x (223)+2 sin-1 22-3+0. - 49 8 C. 7 INTEGRATION. 207 3х2 2x + 5 20. dx. 8177. √2 X -2 21. S dx (α² + x²)²° X 1 X 2a² (a² + x²) + tan-1 2a³ + C. a 22. Sxo • x − x² + 21 -dx. (x² + 3)* 6x+12x²+22x+27 1 4(x² + 3)² 2 +log (2²+3) 1 X + tan-1 + C. 2√3 √3 23. S (3x + 2)dx (x² − 3x + 3)²º 24. S sin' x dx. 4 13x 3(x²-3x+3) 24 26 + tan-1 3√3 2x - 3 + C. √3 sin' x( cos' x x 25. Ssin' & dx. sec² x cos x) - 5 cos x /sin' x 2 3 sin x cos x+x+C. sin³ x 12 sin* x sec-² x = sin¹ x cos² x = sin¹ x sin x X in ²) + 12/65 + c. 8 sin . Hence we may make u = cos x, r = 0, and v = sin* x sinⓇ x. 26. S sin? sin² x dx cos x sin x 4 cos¹ x sin x 8 cos² x log (secx+tan x) +C. sin' x cos³ x 1 cos² x 1 1 = 5 5 cos³ x cos x cos³ x 3 Hence we may make u = sin x, r = 0, and v = cos 27. S cos 28. S sin'x secʻx dx. 29. Ssecad dx. cos* x cosec³ x dx. x sin' x + cos-5 x -8 COS X. COS X X COS X & log tan + C. sin³ x 3 cos³ x 1 2 sin² x 5 sin³ x 2 3 +[x-sin x cos x]+C. 3 cos x 1 5 cos² x\3 sin³ x 3 sin x + log (sec x + tan x) + C. 20 102 5 sin x) 208 DIFFERENTIAL AND INTEGRAL CALCULUS. . 7. ++ 1 ! 215. Reduction Formulas for Binomial Differentials. These may be easily obtained by the method of indeterminate coefficients. I. Required S (a + bx")³x”dæ in terms of S (a + bxn)ºxm-ndx. du Make u = a + bx², r = p, nbxn−1, v = xm, s=r, dx and v₁ = xm-n. Substituting in (F), we have xm= (p + 1)nbx¹-¹ƒ(x) + (a + bx¹)ƒ'(x) + kxm-n¸ (1) where f(x) = Axm-n+¹, and f'(x) = (m − n + 1)Axm-n. Substituting these values in (1), arranging with reference to x, we find A = S 1 b(np + m + 1)' (a + bx)³xmdx = a(m − n + 1) and k = b(np + m + 1)' xm-n+1 (a + bxn) p+1 b(np+m+1) a(mn+1) b(np + m + 1) √ ( a + bx") pxm-ndx. (A) By a repetition of this formula m may be diminished by any integral multiple of n. m II. Required (a + bx”)³x”da in terms of Make .! S (a + bxn)р-¹xmdx. u = a + bx², r=p-1, v = (a + bxn)xm, and — s = p − 1, v₁ = xm. 1 1 INTEGRATION. 209 Substituting in (F) and proceeding as before, we get S (a + bxn\xmdx = flat xm+1 (a + bxn) p np + m + 1 + anp p S np + m + 1 √ (a + bx")»-¹x»dx. (B) Each application of this formula diminishes the exponent of a+ban by unity. When m or p is negative, we need formulas for increasing instead of diminishing them; hence the following: III. Required S(a + bx”)"x"dæ in terms of S (a + bx¹)ºxm+ndx. Solving (A) for ƒ (a + bx")"æm-nda, and substituting m +n for m, we get S (a + bx")³xmdx = xm+¹(a + bxn)p+1 a(m + 1) b(np + n + m + 1) S (a + bx”)»xm+"dx. a(m + 1) IV. Required S(a+ban)³½"dæ in terms of Sla (a + bxn)x+¹xmdx. Solving (B) for S (a for p, we find S (a + bx”)³x™dx = − + (C) + bx")-¹"da, and substituting p +1 bx")³¬¹™da, xm +1 (a + bxn) p+1 an(p + 1) np + n + m + 1 f (a + ba”)p+¹xmdx. S an(p+1) (D) 210 DIFFERENTIAL AND INTEGRAL CALCULUS. • 216. The approximate integral of many differentials may be conveniently obtained by the method of indeterminate coeffi- cients. The following important example will serve to illustrate the process. Integrate the Elliptic Differential dx ds = (a² — e²x²)* Va² x² Comparing this with (E), we may make u = a² — x³, whence 2x, and v = (a² — e²x²), which, when devel- du r = - dx oped by the Binomial Theorem, gives v = α e² 2a e¹ -X² 8a 3X $ 16952.6 Substituting in (F), Art. 211, making s=-, v, 1, we = 1 have a e²x² 2a 8a³ e*x eεx® 5 16a ... = − x f(x) + (a² — x³)ƒ'(x)+k, where f(x) is evidently of the form fx... Ax+ Bx³ + Сx. Proceeding as in Art. 212, we find A = 96a59 5 B=-Aa² + 4 e¹ 32a9 C = = Ba² + k = a - Ca². 4a x = eε /x 5e® e + + 96\a 384 32/ \a ( +65 5e® 3e¹ 256 + 64 4 ++) ) (~)] + a(1 - e² 3e¹ 5e 4 64 256 ..) sin-12 1 CHAPTER X. INTEGRATION AS A SUMMATION OF ELEMENTS. ELEMENTS OF FUNCTIONS. 217. Hitherto nothing has been said about the magnitude of differentials. Whether they are large or small does not affect the principles which have been deduced; hence we may regard them as small as we please. They are variables whose limits are zero. 1 2 218. In the present chapter increments are called and treated as Elements.* Thus 4y or 4f(x) (= m¸h + m¸h³, Art. 24) is an element of the function y = f(x). For convenience the element 4f(x) will often be represented by E, and the differential dy or f'(x)dx by Da, which may be called, respectively, the xth element and the ath differential of the function f(x). Since D¸ varies as da and approaches E, indefinitely as dx approaches 0, D, is called the differential value of E, with respect to dx. ༠ X1 The expression ments like Ex, or the sum of the successive values of Ex, be- tween the x-limits x, and x,. That is, supposing the increment of x to be always h, [E] represents the sum of all the ele- Xz Σ" [Ex] = Ex₂+ Ex₁+h+ Ex+2h+... Ex₂- h (or x² + (n − 1)h). X1 - 1 The prac- 219. A Definite Integral Regarded as a Sum. tical importance of integration consists chiefly in regarding it * Because sum and element are correlative terms. 211 1 ✔ 4 • :: + 212 DIFFERENTIAL AND INTEGRAL CALCULUS. as the summation of a certain series. For example, in seeking the area of a curve, we conceive it divided into an indefinite number of suitable elementary areas, of which we seek to deter- mine the sum by a process of integration. The solution of this fundamental problem is effected by the following formula and its corollaries. Suppose that in any function of x, as f(x), we change x from. x₁ to x, by giving to x successive increments. The whole change in the value of f(x), viz., f(x) -f(x,), must be the sum of the partial changes produced by the increments given to x. 1 is, or (Art. 208) ƒ(x,) − f(x,) = Σ**[4ƒ(x)], X2 X1 Sone D x Dx = Σ*'[Ex]• • That (A) X1 X1 Formula (A) is not only an expression of the simple fact that the whole is equal to the sum of its parts or elements, but it signifies that the integral of the differential value of an element, between certain limits, is the sum of the successive values of that element, between the same limits. p. m P D d3 d2 F d₁ A E а1 a α 2 A s B C G FIG. 43. us consider the area of EGQF, Suppose y = f'(x) to be the As an illustration of (A) let where OEx, and OG = x,. equation of AF'Q, where x = OB and y = BP. INTEGRATION AS A SUMMATION OF ELEMENTS. 213 Let u the area of OBPA, and let BC (= h) be an element of x, then BCDP = du = ƒ′(x)dx Dx, and RCP'P =Dx+ au = Ex. (a) Evidently, xq EGQF= Dx, Art. 208. X1 (1) (b) Divide EG (= x, x,) into n parts, each equal to h, and draw the ordinates a,d,, a,d,, ad,, etc., then EGQF = Ed,+ a‚¿‚ + α‚d¸+t, 1 (2) as in which Ed,, a‚d,, a‚d, etc., are the successive values of Ex, x increases by h from x, to x,. That is, Ed,=Ex,, a₁d₁ = Ex₁+h, a,d₁ = Ex₁+2h, etc. Hence (2) may be written. 1 X1 EGQF=Σ*[Ex]. Σ**[Ex].. (3) X1 Now, equating (1) and (3), and we obtain formula (A). In further illustration of formula (A), let us show that the signification which it expresses is true of (a) 3ªdx. •X2 £** 3x³dx = [x² + C']²² = x,' — x‚'. 3 1 (4) X1 X1 (b) Since f'(x) = 3x², Ex= 3x²h + 3xh² + h³, Art. 207, 2 Ex = 3x¸³h + 3x¸h² + h³; Ex₁+h=3x, h + 9x,h² + 7h³; Ex₁+2h = 3x,'h+15x,h² + 19h"; Ex₁+(n−1)h= 3x¸³h + 3x¸(2n − 1)h² + (3n²-· 3n + 1)h³. 1 1 Taking the sum, remembering that x,+ nhx,, and, by Algebra, 3+9+15+... 3(2n-1) = 3n', and 1+7+19+ (3n² - 3n+1)= n³, we have 2 >*[E]=3nh, +3(h) +(nh) 1 = (x,+ nh)' — x‚' = x‚³ — x‚”. (5) 214 DIFFERENTIAL AND INTEGRAL CALCULUS Comparing (4) and (5), we see that the results of the opera- tions indicated in (A), when applied to 3xdx, X1 3x dx, are the same. 220. Formula (A) is also true when x, x, (= EG) is not divided into equal parts. Let us suppose x, x, to be divided into the following equal or unequal positive parts: a 2 a, c—b,...l—k, x₂-l, b - α, the sum of which is evidently x, x,; then we have identically +. Sy dx = Lydx + y dx+ydz+ydz, (1) So a. k in which a x₁, b — a, c — b, etc., may be considered the suc- cessive values of x and y da, fyda, etc., the correspond- X1 ing successive values of E. Hence (1) is the general significa- tion of (A), which the student may easily illustrate with a figure. 221. A Definite Integral Regarded as the Limit of a Sum. In Fig. 43 let us suppose n to increase and h to decrease, nh being always equal to EG. Since the limit of E, D, as h approaches 0, is unity, the sum of all the rectangles like Dr approaches indefinitely the constant sum of all the elements like E. Therefore Ex limit h÷0 X Σ X1 [D] =Σ[E] Substituting in (A), Art. 219, we have X1 √ X2 Dx = limit2 Ex (B) Σ Dr]. h÷0x1 X2 √ Dx That is, the definite integral D, is equal to the limit of the sum of all the successive values of Da, as x increases by h # from x, to x₂ 2 י . INTEGRATION AS A SUMMATION OF ELEMENTS. 215 3x²dx. For example, let us find the value of ***da. Since Dx = 3x³h, we have Dx₁ = 3x¸³h, Dx₁+h = 3x¸³h + 6x¸h² + 3h³, 1 Dx₁+2h = 3x¸³h + 12x,h² + 12h³, 1 X1 Dx₁+(n − 1) = 3x¸³h + 6(n − 1)x¸h² +3(n − 1)³h³. - Taking the sum, remembering that 0 + 6 + 12 + ... 6(n − 1) = 3(n² — n), and 0+3+12+27+... 3(n − 1)² = : we have (n − 1)(n) (2n − 1)¸ (n − 1)(n) (2n-1) 2 1 X [Dx] 3nhx¸² + 3(n² — n)x¸h² + - 1) h³ 2 'nh h². 2 X2 2 = 3(nh) * + 3(nh)*x, − 3(nh) h+ (nh)® – }(nh)*h + Now, making nh = x, x,, and then passing to the limit by making h= 0, we have limit Xy 2 3 limit − − Σ [Dx] = 3(x, − x‚)x‚² + 3(x, − x‚)³x¸ + (x, − x¸)³ 4x÷0 X1 2 = x,³ — x‚³‚ X2 which is evidently equal to £ 3x dx. X1 222. It is important to observe that, whether an integral be regarded as a sum, or the limit of a sum, integrating is equiv- alent to two distinct operations: (a) If a sum, as in Art. 219, integrating f'(x)dx is equivalent to (1) increasing the differential f'(x)dx by the acceleration au to obtain the element E, and (2) finding the sum of the succes- sive values of Ex * 216 DIFFERENTIAL AND INTEGRAL CALCULUS. " 1 (b) If the limit of a sum, as in Art. 221, integrating f'(x)dx is equivalent to (1) finding the sum of the successive values of f'(x)h, and (2) taking the limit of the sum, as h approaches 0. In case (a) all the quantities involved are finite; but in case (3) the limit of each part is 0, and the limit of the number of parts is ∞. Both methods have their advantages, and hence both will be employed, more or less, in the applications which follow. Just here the student may profitably read Art. 238, which offers a simple illustration of the significations and practical im- portance of formulas (A) and (B). APPLICATIONS TO GEOMETRY.* 223. Lengths of Curves.-I. Rectangular Co-ordinates. To find the length (s) of the arc APQ between the limits OB = x, and OG=x,. (Fig. 44.) } d5 d4 d3 d2 di P t A B C FIG. 44. Here Ex = Pd, and D₂ = Pt=√1+ Dx -X G dy dx h, Art. 33, and * The previous applications of Calculus to Geometry, Arts. 62, 64, 65, 66, were limited to the most elementary rules for integration; in this chapter it is our purpose to extend these applications by the more advanced methods of integration with which the student is now familiar, and in doing so to impress upon him the important principle of integration as a summa- tion. INTEGRATION AS A SUMMATION OF ELEMENTS. 217 ! X2 Pd,+ d¸d₂+d„d¸ + etc. = Σ¸[Ex] or 2 X1 limit X2 Σ₁₂ [Dx]= √™ Dx. h÷0-21 X2 X 1 S x2 dy dx -)° dx or X1 Sv dx 2 1+ dy Jay. (C) X1 √1+ s = EXAMPLES. 1. Find the length of the arc of the curve y = tween the limits x, = 1 and x, = 2. 6 - Jdx dx = dx² dx + 2 x²dx dy 1 Here dx 2x² (1+ (1 + 1 ) + dy² 1 + x¹ ds 2x² dx 1+ +x S= 2 1 =[- 2x² S 1 3 2x2 2 1 X + = 11. 2x 6 1 1 2. Rectify the parabola y² = 4ax, using the formula ds dy dx 2 + 1. dy 1 +1= Vy² + 4a². θα dx Y ; dy 2α 1 ds = √ y² dy 4a² 2 [•]. Za S (4a² + y³)* dy + 1 be- 2x 0 2 y √ 4a² + y² 4a + a log ( y +√4a² + y²). Art. 214, Ex. 10. 2a 8 = √x² - 1+ C. 3. Rectify the curve y = log (x + √x² − 1). 4. Rectify the ellipse y² = (1 − e²)(a² — dy - (1 —— e²) 2 dx Y VI S x²). x √1 - e² 2 2 Va² - x² : ... か ​218 DIFFERENTIAL AND INTEGRAL CALCULUS. To find the length of a quadrant, we must integrate between the limits O and a; hence 8 = √ √ 1 + x²(1 − e³) a dx dx = √ a³ — e²x² 2 a² x² √ a² - x² π =α 2 (1- 1 1.3 1.32.5 e¹ 4 22.42 22.42.62 etc.). Art. 216. Y 5. Rectify the cycloid xrvers-¹ 3 - √2ry — y³. go Here dx y dy √2ry - Y y❜dy' 2ry — y² 2r dy V 2r Y G ds = V dy²+ S = · S (2r)* (2r − y)¯* dy B E = −2 √/2r(2r — y) + C. If we estimate the arc from the point B where y = 2r, we shall have, when s = 0, y = 2r; hence, A C FIG. 45. H 0 = 0 + C, .. C' = 0, and s = − 2 √2r(2r — y). Since BC= 2r and BE = 2ry, BG = √2r(2r — y). BD=s=2BG; or the arc of a cycloid, estimated from the vertex of the axis, is equal to twice the corresponding chord of the generating circle; hence the entire arc BDA is equal to twice the diameter BC, and the entire curve ADBH is equal to four times the diameter of the generating circle. хо 1 1 6. Rectify y = + 16 2x² S = + C. 16 2x² INTEGRATION AS A SUMMATION OF ELEMENTS. 219 NOTE. The value of C depends on the point from which s is measured. Thus, if s is estimated from x = 1, then s = 0 when x = 1, and we have 0 = − + C; that is, C = †• 7. Rectify y = xn+1 1 4(n + 1) † (n − 1)2~-1° 2 s = y (n хт x²-m 8. Rectify y = 2m 2(2 — m) * s = y + X2-m 2 ทะ 1)xn−1 +0. + C. 9. Rectify y = log (x² + 3x + 2). 10. Rectify y = ‡x² — log x. x 8 = x + + log (~2 + 1 ) + C. s x + 2 8 = x²+log x + C. S= A curve is said to be rectifiable when its length can be ex- pressed in finite terms by aid of the algebraic and elementary transcendental functions. 224. II. Polar Co-ordinates. To find the length (s) of the arc APQ between the limits 0₁ = AOP or r OP, and 6₂ = AOQ or r, 0Q. (Fig. 46.) Ꮎ = 1 N Here Ee = Pd, and De Pt = √ r² + hence or 002 Pd₁+d¸d₂+d„d¸ + etc. = limit 40÷0 2 Өг 1 (dr) de, Art. 97; Σ.“[Ee] Σ"[Do] = $*[Do]. 2 L'"*" (~" + (dr.')")'de, or £,'" (1 + ("de")')'dr. dr. (D) do 1 Here ; .. $ = ȧr α S" (1+ a² 11. Rectify the spiral of Archimedes, r = ab. r(a² + p¹²)* 2a α + 21 log p² 1 } dr = 1 ½ £˜ˆ (a² + r²)¹dr a r + va² + y²² a Art. 214, Ex. 10. S 220 DIFFERENTIAL AND INTEGRAL CALCULUS. Let this result be compared with the length of the parabola, Ex. 2. 12. Find the length of the logarithmic spiral loga r = 0. mdr do m Here Ꮎ do = Р dr jo and s = £*(1 + m²)³dr = r(1 + m²)³. ძვ da dit D A FIG. 46. 13. Find the length, measured from the origin, of the curve — У = a log a² x² a² a + x a log X. a X 14. Find the entire length of the arc of the hypocycloid x² + y³ = a³. 6a. 15. Find the entire length of the cardioid r = a(1 — cos 0). 8a. Ꮎ 3πα 16. Find the entire length of the curve r = a sin³ 3' 2 INTEGRATION AS A SUMMATION OF ELEMENTS. 221 1 17. Find the length, between x = a and x = b, of the curve его - ex + 1 ex — 1° e2b 1 log e2a + a − b. 1 18. Find the length of the tractrix, measured from (0, a), its differential equation being ds a dy Y α a log 19. Find the length of the arc, measured from the vertex, of the catenary a C y = +e 20. Find the length of a quadrant of the curve (22) * · + (~~) * = 1. 22 Ꮳ (---)) e a² + ab + b² a + b 225. To find the equation of a curve when its length is given. 1. Find the equation of a curve whose length is x² s' = log x + 4' ds 1 + x² ; √ ds\ 2 1 — x² dy -1= = dx 2x dx 2x dx 1 − x² Hence, dy = 1 dx, and y = lòg x − x² + C. 2x EXAMPLES. Find the equations of the curves whose lengths are: 1 2. s = + 4x² 8' y = s 27+C. See Note, p. 219. 4 222 DIFFERENTIAL AND INTEGRAL CALCULUS. 3. 8 == x + log (2 + 1) 4. slog (tan x) y = log (x² — 1) +C. y = log (sin 2x)+C. 5. s = 1 Xn+1 + 2(n-1)x"-1 2(n + 1)° 2n+1 y = s + C. n + 1 226. Areas of Curves.-I. Rectangular Co-ordinates. To find the area of the surface between a given curve, the axis of x and two ordinates whose abscissas are x, and x,, we have, Art. 219, W = Syda = [E] Σε . (E) For a definite area between the curve and axis of y, we have 32 yz u = xdy = Σ[Ey]. . Yı (F) EXAMPLES. Find the area of the curve y = 1 x₁ = 1 and x = 2. 1 + between the limits 16 2x2 A 2 Sydz 1 ydx = 16 2x² +212) dx :)dx = [ хо 1 51 80 ZX 80 2. Find the area of the circle y² — a² — x². P α x y B C FIG. 47. __ (OB)(BP) = 2 X х Area of OBPA = ydx = ƒ˜” (a² — x²)*dx = x (a² √ - x (a² — x²)+ a² X ૭ 2 + sin-1 2 a (Ex. 17, Art. 96) (OP)(arc AP) + = area OBP + area OPA. ૨૭ INTEGRATION AS A SUMMATION OF ELEMENTS. 223 ..1 To find the area of the quadrant OCA we have S ƒª (a² — x²)*dx = ±ña². The value of π is given in Art. 129. 3. Find the area of an ellipse, a¹y¹ — a¹b¹ — b²x². b b Y √ a² — x²; .. u = a√ √ a² — x² dx. a — x² b + (the area) == √ √a² = a* dx = ‡nab; a να entire area — яаВ. = 2 4. Find the area of the hyperbola, a'y' b'x' — a³b³. b y = √x a²; И α α a√ √x² - a" dx, or u = 2a 2 bx (x² - a²)* ab To find C, we know that when x = a, u = 0; hence log (x + √x² — a³) + C. 0 = ab 2 log a + C; .. C = ab 2 log a. ay b Substituting this value of C, and making √x — a² = az, we have Area of hyperbola xy ab 2 2 log (2/2+3). b 5. Find the area of the surface between the arc of the pa- rabola y² = 4ax and the axis of y. zxy. 224 DIFFERENTIAL AND INTEGRAL CALCULUS. line 227. It is often convenient and suggestive to regard a defi- P B C FIG. 48. D E nite integral like √**y(dx) = EGQF (Fig. 43) as signifying that "the ordinate y (or generatrix PB), moving perpen- dicularly to the axis of x from x=x, toxx,, generates EGQF. Thus, let it be required to find the area of the surface between the parabola y² -4ax and the straight Y =x. We at once have OPDC= X₂ = OE x)dx (PC) da = √(√4ax - x) de - $a³. This method can be employed, with equal facility, in finding the volumes of many solids, in which case the generatrix is a surface. 6. Find the area of one branch of the cycloid. dx = y dy .. U = S y❜dy √2ry — y² √2ry — y² For the area of one branch we have 2• u = 2 y❜dy √2ry — y² = 3πr², (Ex. 13, Art. 214) which is three times that of the generating circle. 7. Find the area of the curve xy = a. b [u]° = = a log // x+1=0 8. Find the area of the curve x³y − x + 1 = 0 between the x-limits 1 and 2. 9. Find the area of both loops of a¹y² = a²b²x² — b³xª. 11/ tab. INTEGRATION AS A SUMMATION OF ELEMENTS. 225 10. Find the area of both loops of the curve aºy² = a²x² 12 x¹. $. 11. Prove that the area of the curve a¹y³ = (a³ — x³)xº, be- tween the x-limits 0 and a, is the same for all values of a. 228. II. Polar Co-ordinates. In Fig. 46 the area of POD (= r²d0, Art. 35) is the differential value of the element POd¸; therefore źr²d0 = POd¸+d¸Ød¸ + ã‚Ød¸+... dn-1OQ, where 0 = AOP and 0, 2 = AOQ. u = ‡ƒª³r³d0 = area POQ. 12. Find the area of the spiral of Archimedes, r = a0. (G) dr 1 do = ; .. 21 = a Za S 2.3 r²dr = σα 1 COR. I. If a = If r = 1, or 0 = as is usual, u = }πr³. 2π, u = ‡π, which is the area described by one revolution of the radius vector. If r = 2, or 0 = 4π, U = 7, which is the area described by two revolutions of the radius vector, which includes the first spire twice; hence the area of the entire spire is π-ππ. 13. Find the area of the hyperbolic spiral, r = a Ө a² a² ȧu a²do ; .. 202 [u]*' = [ a² 20 +α = 20 20, 14. Find the area of the logarithmic spiral logar. mdr do = ... when m = 1, rdr du = and_u = r²; 2 226 DIFFERENTIAL AND INTEGRAL CALCULUS. that is, the area of the natural logarithmic spiral is equal to one fourth the square described on the radius vector. 15. Find the entire area within the hypocycloid 16. Find the area 2² = 4ay and the witch 2·³ + y* = a³. 2:3 πα of the surface between the parabola y 8a³ x² + 4a². (2π — §)a² 17. Find the entire area of the cardioid r = a(1 a(1 — cos (). 18. Find the area of a loop of the curve x* +y* = a³xy. επα. Απα 19. Find the area of the loop of the curve a³y³ = x¹(b + x). 326³ 105a 20. Find the area included between the axes and the curve (~2) * + (~~) ** = 1. ab 20° 21. Find the area between the curve x'y' + a²y² = a²x² and one of its asymptotes. 22. Find the area of the loop of the curve y³ + ax² axy = 0. 2a³. o'oa². 23. Find the area of the three loops of the curve } r = a sin 30. (See Fig. 38.) Απα. 24. Show that the whole area of r = a(sin 20+ cos 20) is equal to that of a circle whose radius is a. 229. Areas of Surfaces of Revolution. By Art. 34 the differential value of the xth element of surface is 27у(1+dy)*dx; ✔ INTEGRATION AS A SUMMATION OF ELEMENTS. 227 A therefore ~x2 S = 2π y 1 + dy² \ + dx² 2 ແງ dx = Σ**[E]. . (H) X1 EXAMPLES. 1. Find the area of the surface generated by revolving the X 1 arc of the curve y = + about the axis of x, between 16 2x² x₁ = 2 and x. = 4. 2 dy ds 1 + dx dx X¹ S = 2π 16 2x² + 1/( 1 + )dx. 4 X8 3x² 1 π + +α 256 16 4x¹ 2577024 271 π. 2 2. Find the area of the surface of a prolate spheroid, the generating curve being the ellipse a'y' b'(a' — y³). b y = a ... Area 2 2 √a² — x²³, and ds = S 0 a 2луds=4π- 2 = a² — e²x² dx. a² X2 b a 47 % S (a² e*x²)* dx α a² ex = 4π a + sin-1 2 2e a 0 = 2πb² + Σπαν e sin-1 e. bx (α² x(a³ — e*x²)* 3. Find the area of the surface generated by the revolution of the cycloid about its base. 2r Area = 2 2ñyds = 4π √/?r S y dy 0 √2r — Y 228 DIFFERENTIAL AND INTEGRAL CALCULUS. = } 1 Qr = 4π √/2r √27 [− 3(4r -}(4r + y) (2r y)+ 64 = ·πr². 3 4. Find the surface of the paraboloid between the limits x = 0 and x = a, the generating curve being y² = 4ax. §(√8 — 1)8πα³. 5. Find the surface generated when the cycloid revolves about the tangent at its vertex. 3,2 πρ 6. Find the surface generated by the revolution about the axis of x of the portion of the curve y = e*, which is on the left of the axis of y. π(√2 + log (1 + √½)). 7. Find the entire surface of the oblate spheroid produced by the revolution of the ellipse a'y + b²x²=a2b2 about its minor axis. ba 1+ e 1 – e επα + π- log e - 8. Find the surface generated when the cycloid revolves about its axis. 8πιπ 8π¹³ (π — §). 9. Find the surface generated by revolving the arc of the cardioid r = a(1 cos ) about the initial axis. 8 = 27 Syds = 27 Sr sin 0 √dr² + r²d0³. πα. 10. Find the surface generated by the revolution of a loop of the lemniscate 7² = a² sin 20 about the polar axis. 2πας. 2 230. Volumes of Solids of Revolution. In Art. 32 the volume of the cylinder (ny'dx) generated by the revolution of the rectangle BCDP is the differential value of the xth element of volume; therefore v = *²[Ex] or limit π X2 h = 0 < ; [xy'h] = x * y'dx. X1 X1 (J) INTEGRATION AS A SUMMATION OF ELEMENTS. 229 E. EXAMPLES. 1. Find the volume of a paraboloid, the generating curve be- ing the parabola y² = 4ax. π v = π xx So y³dx 4ña f*x dx = 2πax². = 4πα When the curve is revolved about the axis of y, we evidently have v = π Sx'dy. 2. Find the volume generated by revolving the surface be- tween the parabola y = +√4ax and the axis of y about that axis. y₁ [v]" = x fxdy = xf {{ dy = TW гобо 16a² 80a2 = πx'у. That is, the entire volume is one fifth of the volume of the circumscribing cylinder; therefore the volume generated by the surface of the parabola in revolving it about the axis of y is four fifths of that of the circumscribing cylinder. 3. Find the volume of the solid generated by the revolution of the cycloid about its base. dx = y dy ; dv dv = пу³dx y² пу'dу 9 √2ry - y' 2 √2ry To obtain half of the volume, we must integrate between the limits y = 0 and y = 2r. v = 2π S 21 0 y'dy √2ry — y² -572³ πρ That is, vof the circumscribed cylinder. 4. Find the volume generated by revolving the ellipse AA' about the tangent X'X as an axis. (Fig. 49.) Let OA = a, OY=b, O'B = x, BP = y, and BP' = y'; 2 b b then y = (a + √u² — x”), and y′ = (a — √ a² — x²). a y' a - i 230 ! DIFFERENTIAL AND INTEGRAL CALCULUS. Supposing P'n dx, the volume generated by P'm, viz., = b2 y')dx or 4π— √α² π (у'y')dx or a √ a² - x²dx, Α' O Ο' FIG. 49. a m d A C P n -X B is the differential value of the volume of the element generated by P'cdP, with respect to dx. of a -α b2 2 4× √ α² — x³dx = 2ñ²ab², (Ex. 17, p. 56) a -a which is the entire volume, being the sum of the volumes gen- erated by revolving all the elements like P'cdP between x = and x = a, or the limit of the sum of the volumes generated by all the rectangles like P'nmP. 3 5. Find the volume of the closed portion of the solid gener- ated by the revolution of the curve (y² — b²)² - a³x around the axis of y. 256 προ 315 a 6 6. Find the volume generated by revolving the curve (x-4α)y² = ax(x - 3a) about the axis of x, between the x-limits 0 and 3a. πα 2 (15-16 log 2). 7. Find the volume generated by revolving the cycloid round πληθ. the tangent at the vertex. 8. Find the volume and surface of the torus generated by revolving the circle x² + (y — b)² = a² about the axis of x. 272аb and 4π²аb. . ↓ INTEGRATION AS A SUMMATION OF ELEMENTS. 231 9. Find the entire volume and surface generated by revolv- ing the hypocycloid x3 + y = a³ about the axis of X. x³ y³ 32πα 12πα and 5 105 10. Find the volume generated by the curve xy² = 4a³ (2α-x) revolving about its asymptote. Απ'α. X 11. One branch of the sinusoid y = b sin the axis of x; find the volume generated. is revolved about a SUCCESSIVE INTEGRATION. 231. A Double Integral is the indicated result of reversing the operations represented by d²u dx dy' Thus, if dau dy dx = xy, then SS2 xy³dy dx, which indicates two successive integrations, the first with refer- ence to x, regarding y as constant, and the second with reference to y, regarding x as constant. Thus, S Szy'dy dz=S(+0)y'dy dx x² y³ 3 6 Cy³ + + C₁, 3 where C and C, are the constants of integration. 232. Definite Double Integrals. Here both the integra- tions are between given limits. For example, C Soc Soc x 3 "x³y'dx dy. This notation indicates that the integrations are to be taken in the following order: : 232 DIFFERENTIAL AND INTEGRAL CALCULUS. : " 2 C SS wydz dy = S. Say'dy) dz 3 Jax as dx = (c* — b*). 12 3 That is, as dy is written last, the y-integration is taken first. The limits of the first integration are often functions of the second variable. For example, a √x dx (√x)dx S.' S. " da dy = S" (W)dz = ŝa 0 As another example, S.S rdr do= (**) ar dr = b². ¡ 233. A Triple Integral is the indicated operations of three successive integrations, for which the notation is similar to that of double integrals. Thus, SSS" Sª £*£*£" x°yzdx dy dz = £“ [S*(S*x*yzdz)dy]dx. α a = £“ [S" (±x*y*)dy ] dx = £" [tox®]dx = v%a". J EXAMPLES. 1. Find the following: Ꮫ a £$£° (x² + y²)dx dy. 0 2. S. S. SS a2 x2 (x² + y²)dx dy. fab(a² + b²). Απα. • INTEGRATION AS A SUMMATION OF ELEMENTS. 233 3. 2a S&S 0 a 0 √ zax a2 ax dx dz Vax - x² x2 ах 4. S S - S 0 0 x² + y² α dx dy dz. 0 4a². πας. APEAS OF SURFACES DETERMINED BY DOUBLE INTEGRATION. 234. Plane Surfaces-(a) Rectangular Co-ordinates. In the formula u = Syda, Art. 226, we may make y = Say, which gives u = = Sfax dy. dx (J) 235. (b) Polar Co-ordinates. In the formula u = P Sao, P D q p m n K B C A FIG. 50. = Sr dr, and write = S Srdoar. u = Art. 228, we may make 2 1 (K) 234 DIFFERENTIAL AND INTEGRAL CALCULUS. . 3 As illustrative examples let us employ (J) and (K) in finding the area of a circle whose radius is a. Let (x, y) be the co-ordinates of the point m, and (x + dx, y+dy) of the point p, then mnpq = dxdy. Regarding x as constant, we have y=BP BCDP = Σ [mnpq] y=0 = dx S vody √2αx-x² = √2ax — x³dx. Again, since V2ax-xdx is the differential value of BP' with respect to dx, we have 2α Σ [BP']: 0 = 2a √2αx x²dx = ½ña² = area of OKP. (b) Let (r, 0) be the co-ordinates of m, and (r + dr, 0 + d0) of p', then mn = dr, mqrde, and rdedr (= area of mnpq) is = 19 20- P m X FIG. 51. the differential value of the element mnp'q' with respect to dr. Therefore, regarding as constant, we have ОР Ꮎ La cos 0 >°" [mnp'q'] = √ rd0dr = 2a² cos² #d0 = OPP¦. INTEGRATION AS A SUMMATION OF ELEMENTS. 235 Again, since the area of OPP,' is the differential value of the element OPP', with respect to do, we have Σ" [OPP'] = √** 2a² cos² Od0 = †πα³, which is one half the area of the circle. EXAMPLES. 1. Find the area (1) of a rectangle by double integration; (2) of a triangle. 2. Find the area between the parabola y² =ax and the circle y² = 2αx x². 2 (702 πα 2α² 4 3 r = a(1 3. Find by double integration the entire area of the cardioid a(1 — cos ()). 3πα 2 4. In a similar manner find the entire area of the Lemniscate r² = a² cos 20. a². 5. Find the whole area between the curve xy² = 4a² (2a — x) and its asymptote. Απα. 236. Surfaces in General.-To find the area (= S) of a surface whose equation is f(x, y, z) = 0. Let (x, y, z) be the co-ordinates of any point P of the surface, and (x + dx, y + dy, z + dz) the co-ordinates of a second point Q very near the first (Fig. 52). Draw planes through P and Q parallel to the planes XZ and YZ. These planes will intercept a curved quadrilateral PQ on the surface; its projection pq, a rectangle, on the plane of XY; and a parallelogram pʼq', not shown in the figure, on the tangent plane at P, of which pq is the projection. The area of p'q' dS, since = ds, since it is the dif- ferential value of PQ (= 4S) with respect to dx and dy. The projection of p'q' on XY is dxdy; similarly the projec tions of p'q' on XZ and YZ are dxdz and dydz; hence, denoting .. او י * Let 236 DIFFERENTIAL AND INTEGRAL CALCULUS. · the angles between the plane of p'q' and XY, XZ and YZ by a, ẞ and y, respectively, we have ZI cos adS = dxdy, (1) cos BdS= dxdz, . (2) cos ydS = dydz. . (3) 18 P B' K q -X FIG. 52. Squaring (1), (2), (3), and adding, remembering that we have hence, cos² a + cos' B+ cos y = 1, (dS)² = dx'dy' + dx³dz' + dy³dz'; ds = (1 + (dz')'+ (dz )')' dady, 8= dy dx dz 2 dxdy. (L) = SS (1 + (dy')' + (dz.)"')"'dady. dx INTEGRATION AS A SUMMATION OF ELEMENTS. 237 EXAMPLES. 1. Find the area of one eighth of the surface of the sphere x² + y² + z² = a³. Here 2 ༄།ི་ dz х dz dx dz² dz² 1+ + 1+ dx² dy' y אן א + y² 11 2 Substituting in (L), we have y S = = S Sadady = aSS; dxdy √ π² — x² - y² - Integrating first with reference to y between the limits - = 0 O and y = Va² — x", we get the differential value of the element B'C'KL; and then integrating with reference to x between the limits x = 0 and x = a, we get the sum of all the elements like B'C'KL between these limits, which sum is the area required. =af Sadzdy Hence S = a πα dx 0 0 Na² 2 x² — y² 2 2. The two cylinders x²+za² and x² + y² = a² intersect at right angles; find the surface of the one intercepted by the other. 8a². Here z = √α² −x, and for one eighth of the required sur- face the y-limits are 0 and Vax, and the x-limits 0 and a. 3. A sphere whose radius is a is cut by a right circular а cylinder, the radius of whose base is and one of whose edges 2' passes through the centre of the sphere; find the area of the surface of the sphere intercepted by the cylinder. 2α² (π-2). Take x² + y²+22=a2 for the sphere, and x² + y² = ax for the cylinder, then z = √a² — y² — x², and for one fourth of the I 7 238 DIFFERENTIAL AND INTEGRAL CALCULUS. -: 1 required surface the limits of y and x are 0, Vax - x², and 0, a, respectively. 4. In the preceding example find the surface of the cylinder intercepted by the sphere. (See Ex. 3, Art. 233.) 4a². 5. Find the area of the portion of the surface of the sphere x² + y² + z² = 2ay lying within the paraboloid y = mx² + nz². Σπα √ mn 237. To find the volume of a solid bounded by a surface whose equation is ƒ(x, y, z) = 0. Z h S PO ƒ' Ο a α a' A B m n FIG. 53. VOLUMES OF SOLIDS DETERMINED BY TRIPLE INTEGRATION. Letv the indefinite volume expressed by the product of x, y, z; then v = xyz, which may be written v=SSS az dy dz, Sax dx which becomes definite when the integrations are taken between certain limits, and we will now give the geometrical interpreta- tion of the formula, step by step. INTEGRATION AS A SUMMATION OF ELEMENTS. 239 Let (x, y, z) be the co-ordinates of the point P, and (x + dx, y + dy, z + dz) be the co-ordinates of the point Q; then PQ = dx dy dz. (a) Regarding x and y as constant and integrating between the z-limits 0 and id, we have ƒ“ id Σid[PQ] = ƒ dx dy dz = (id)dx dy = bh. (b) The volume of bh is the differential value of the element is' with respect to dy; hence Σam [is'] = √(id)dx dy = (afm)dx, which is the volume of the cylindrical segment afm-a'. (c) The volume of afm-a' is the differential value of the element ar with respect to da; hence OA Σ04 [ar] = √°^ (afm)dx 04 So Lam Lia £°* £ £* dx dy dz 0 0 = volume of OBC-A. dx dy dz = volume of OBC-A. . (M) COR. I. The limits of y and z are found thus: id is the posi- tive result of solving the equation f(x, y, z) = 0 for z, am is the positive result of solving f(x, y, 0) = 0 for y, and OA is the positive result of solving f(x, 0, 0) = 0 for x. =0 EXAMPLES. 1. Find the volume of one eighth of the ellipsoid a² The limits of z in this case are 0 and id = x² y¹ + + = 1. C² = c(1 – x² y^ \ ¥ a² b2 ; J 240 DIFFERENTIAL AND INTEGRAL CALCULUS. , the limits of y are 0 and am = =0(1– are 0 and a. Therefore the required volume is 6(1 — 237) * ; and the limits of x al 0 x2 Ca 1 x2 y¹ 1 12 b2 πανε dx dy dz= 0 6 2. Find the volume of the solid contained between (a) the paraboloid of revolution, x² + y² = az, (b) the cylinder, (c) and the plane, x² + y² = Qax, z = 0. 3πα 2 : The z-limits are 0 and 2+y, the y-limits are 0 and M a √2ax x², and the x-limits are 0 and 2a, for one half of the required volume. 3. Find the volume cut from a sphere whose radius is a by a right circular cylinder whose radius is b, and whose axis passes through the centre of the sphere. Απ - 3 ´(a' — (a² — b²)®). 4. Find the entire volume bounded by the surface whose equation is x³ + y³ + z³ = a³. Απα 35 5. Find the volume of the conoid bounded by the surface z²x² + a³y² = c²x², and the planes x = 0 and x = a. APPLICATION TO MECHANICS. παρ. 238. Work is said to be done when a body moves through space in opposition to resistance. A horse in drawing a cart or a plough does a certain amount of work, which depends on the resistance and the distance traversed. The force which the horse exerts, and the distance through which he moves, may be regarded as the two elements of the work done. If r lbs. is the constant resistance or force, and x feet the effective distance INTEGRATION AS A SUMMATION OF ELEMENTS. 241 through which the body moves, ra units of work will be done. By effective distance is meant the distance measured in the same direction as that in which the force is acting. Thus, when the resistance is constant, the amount of work may be represented by the area of a rectangle whose base is the distance (x) and whose altitude is the resistance (r). If the resistance or force is a variable dependent on the dis- tance x, it may be represented by f'(x), in which case the amount of work may be found by taking the sum of its elements, thus: In Fig. 43, if the force f'(x) (= BP) act through the small effective distance h (= BC), the work done will be in excess of f'(x)h (= BCDP) only by the acceleration of work (PDP') during that interval. Hence, f'(x)dx is the differential value of the xth element of work. Therefore the quantity of work between the limits x = x, and x = x, is, viz. : 1 limit Σx¹[ƒ'(x)h] = £**ƒ'(x)dx, **[E] or h÷0₁ in which the effective distance is x 2 X1° (N) COR. I. Effective distance, resistance, and work, and effective distance, force, and energy, bear the same relation to one another as the abscissa, ordinate, and area of a plane curve referred to rectangular co-ordinates, respectively. EXAMPLE. Let it be required to compute the quantity of work necessary to compress the spiral spring of the common spring-balance to any given degree, say from AB to DB.* Let the resistance (= f'(x)) vary directly as the degree of compression, and denote the distance AD' by x; then will f'(x) = mx, W FIG. 54. B where m is the resistance of the spring when the balance * Bartlett's Analytical Mechanics, page 39. A - + 4 ? * 242 DIFFERENTIAL AND INTEGRAL CALCULUS. is compressed through the distance unity. Substituting in (N), making x, 0 and x, AD, we have = = 1 the work = 0 St xz mx³ mx dx = + o 2 c]" = +mx.". 2 0 COR. I. If m = 10 pounds and î₂ 3 ft., then will the work = 45 units of work; that is, the quantity of work will be equal to that required to raise 45 pounds through a vertical height of one foot. CENTRE OF GRAVITY. 239. The bodies here considered are supposed to be of uni- form density; that is, equal quantities of a body have equal weights. The Centre of Gravity of a body is a point so situated that the force of gravity produces no tendency in the body to rotate about any axis passing through this point. The Moment of any element or particle of a body with reference to any horizontal axis is the product of the magnitude or weight of the element by the horizontal distance of its centre of gravity from the axis, and measures the tendency of the element, under the influence of gravity, to produce rotation about the axis. The moment of the body itself is the sum of the moments of its elements. If the axis of reference passes through the centre of gravity of the body, the moment of the body must be zero, otherwise the moments of the elements would not neutralize one another, and the body would rotate. 240. To find the centre of gravity of a plane area. In Fig. 43, suppose the plane curve placed in a horizontal position, and let A = the area of EGQF, x = OB, y = BP, x₁ = OE, x, OG. Also let (x', y') be the centre of gravity of A, and (x+a, y + B) be the centre of gravity of the rect- angle BCDP (= yh). Evidently the limit of a, as BC or h approaches 0, is 0. INTEGRATION AS A SUMMATION OF ELEMENTS. 243 The moment of BCDP with respect to an axis passing through (x', y') and parallel to the axis of y is (x + x − x')yh, which is the measure of the tendency of this rectangle (BCDP) to produce rotation about the given axis, and therefore the tendency of all the similar rectangles to produce rotation is Σ** [x+a- x'lyh. X1 The smaller the rectangles the nearer their sum comes to the whole area of the curve, and therefore the tendency of A to rotate about the given axis is 2x2 limit (x + x − x')yh (x − x')ydx; X1 h or a 10x₂ but as the axis of reference passes through the centre of gravity of A, this must equal zero. X1 xz (x − x′)ydx = Whence x' = 21 x2 CX2 x1 xydx ÷ In like manner we find - xydx S x2 XI — x'S ydx = 0. ydx = y' = √” y'dx ÷ A. X1 . X1 x2 xydx÷A. (P) 2 (Q) 241. To find the centre of gravity of a plane curve. In Fig. 44, suppose the plane curve PQ (= s) placed in a horizontal position; let x = OB, y = BP, x₂ = BP, x₂ = OG, and h = BC; also let (x', y') be the centre of gravity of s, and (x + α, y + ẞ) the centre of gravity of the tangent dy dx 2 h Pt(= √1+ (dv)'). The limit of a, as h approaches 0, is evidently 0. 1 • 244 DIFFERENTIAL AND INTEGRAL CALCULUS. I { The moment of Pt with respect to an axis passing through 2 (x', y′) and parallel to the axis of y is (x+a-x')√/1+ [dy h, dx and the sum of the moments of all the tangents like Pt is dy²+ Σ"*" [x + a − x'] (1 + d¹³ - du² ) h. The limit of this sum, as h and a approach 0, is equal to (x-x')ds, which is the moment of Xx limit X2 s with respect to the given axis, since s = Σ [Pt]. x2 九二​○ X1 [*³ (x − x′) ds = [" xds — x′ S ds — x' f ds = 0. Xx Whence x' Lhxds xds → s. In like manner we find y' = Jz. yds ÷ 8. (R) (S) 242. Let the student prove in a similar manner that the formula for finding the centre of gravity of a solid of revolution whose axis is the axis of x is x' =π xy'dx ÷ the volume; X1 (T) or x' = ["zy'dz + "'y'dx. X1 NOTE.-As the centre of gravity must evidently be on the axis of revolution, the formula given above entirely determines it. The same is true of the following formula. Prove that the formula for finding the centre of gravity of any surface of revolution whose axis is the axis of x is x' = Sexyds + S yds. હું (U) 1 INTEGRATION AS A SUMMATION OF ELEMENTS. 245 EXAMPLES. 1. Determine the centre of gravity (G) of an isosceles triangle. Let OD, the altitude, a, DC, half of the base, = b, OB = x, BP = y; then y mula (P), x' = OG - Soxy ax a bx y = and, by for- P B G D FIG. 55. α bx² dx 0 a = = area ADC Lab Za that is, the distance of the centre of gravity from the vertex of the triangle is equal to two thirds of the altitude of the triangle. 2. Determine the centre of gravity of the area bounded by the parabola y = 4ax and the double ordinate (2y) perpendicular to the axis of x. x' = {x, y' = 0. 3. Find the centre of gravity of the area of the curve xy² = b²(a — x). x' = a, y' = 0, 4. Find the centre of gravity of the area of the cycloid. x' = πr, y' = žr. 5. Find the centre of gravity of the area of 2³ + y lying in the first quadrant. a³ 256 a x' = y' = 315 π 6. Find the centre of gravity of the area of a'y² + b²x² = a²b³, lying in the first quadrant. 4a x′ = y' 3π' 4b 3π 7. Find the centre of gravity of the arc of the circle x² + y² = r² lying in the first quadrant. (See formulas (R) and (S).) x' = y' = 2r π 8. Find the centre of gravity of the arc of the curve x³+y = a³ lying in the first quadrant. x' = y' = za. 246 DIFFERENTIAL AND INTEGRAL CALCULUS. 9. Find the centre of gravity of the arc of the cycloid xr vers -1 Y p √2ry — y², x' = fr, y' = + fr. lying between (0, 0) and (πr, 2r). 10. Find the centre of gravity of the paraboloid generated by revolving y² = 4mx about the axis of x. (See formula (T).) x' = fx. 11. Find the centre of gravity of the segment of a sphere. generated by revolving y² = 2rx-x about the axis of x. When xr, x' = ğr. x' = x(8r - 3x) 4(3r — x) 12. A semi-ellipsoid is formed by the revolution of a semi- ellipse about its major axis; find the distance of the centre of gravity of the solid from the centre of the ellipse. x' = fα. za. 13. Find the centre of gravity of the convex surface of the cone generated by revolving the line y = mx about the axis of x. (See formula (U).) x' = zx. x′ = {x. 14. Find the centre of gravity of the surface of a spherical segment whose altitude is x. 15. Find the centre of gravity of the surface of the parabo- loid generated by revolving y² = 4mx about the axis of x. x′ = 1 (3x — 2m) (x + m)ª + 2m³ 5 (x + m)³ m² APPENDIX. A,. Differentiable Functions. A function, y = f(x), is said to be differentiable when 4 approaches a definite limit as 4x Va Thus, y = √ is differentiable, since 4x approaches zero. (Art. 10, ex. 5) Ду Ax 1 √x² + h + √x' approaches the definite limit, 1 as h approaches zero, x' 2 √x being any particular or definite value of x from which his estimated. All ordinary continuous functions are differentiable, but this does not follow from the mere fact that the functions are con- tinuous, for there are functions which are continuous and yet have no differential coefficients.* Functions of this kind, how- ever, are of such rare occurrence that the distinction between continuity and differentiability is seldom made in works on the Differential Calculus. That is, every function is regarded as continuous and differentiable between certain limits. Ду Ax The limit (m,) of in any particular case can often be conveniently determined by assuming that 4y = m,h+m,h², * See Harkness and Morley's Theory of Functions. 247 248 APPENDIX. which is true of all differentiable functions of a single variable, and then finding the value of m₁, as in A. The general values of m, and m,, and the exact conditions under which ▲y = m₂h+m,h' holds, are given in A,. 2 A. Another Illustration of the Formula 4y = m¸h+m„h². Suppose that a moving body has traversed a distance (s) in the time t, and that the value of s in terms of t is s = f(t). (1) Suppose we wish to find the actual velocity (v,) at the end of the time t₁. Let 4t, an increment of t estimated from t,, be any arbitrary period immediately succeeding the end of the time t₁, then the distance traversed by the body in that period will be the corresponding increment of s, viz., — 4s = f(t + 4t) − f(t) = m¸At + m„(4t)³. (2) The mean velocity (v) of a moving body, during any period of time, is the quotient obtained by dividing the distance traversed by the body by the length of the period. Therefore the mean velocity during the period 4t is As ΔΙ 1 = m₁ + m₂4t = v.. (3) Now this mean velocity evidently approaches the actual velocity v, as At approaches 0, indefinitely. Hence taking the limit of (3), we have As limit of = m₁ = v₁• At That is, the limit of As At or m,, is the actual velocity of the body at the end of the time t,, and hence m, 4t is what 4s would have been had it varied as 4t or had the actual velocity v, re- mained constant, and m,h2 is the acceleration of s during the period 4t. APPENDIX. 249 A,. The Differentials of Independent Variables are, in general, Variables. In differentiating y = f(x) successively dx is usually regarded as a constant; that is, as having the same value for all values of x. "This hypothesis," say Rice and John- · son,* “greatly simplifies the expressions for the second and higher differentials of functions of x, inasmuch as it is evidently equivalent to making all differentials of x higher than the first vanish." Again, "A differential of the second order or of a higher order,” says Byerly,† "has been defined by the aid of a derivative, which always implies the distinction between func- tion and variable, and on the hypothesis of an important dif- ference in the natures of the increments of function and vari- able; namely, that the increment of the independent variable is a constant magnitude, and that, consequently, its derivative and differential are zero. وو The impressions which these and similar statements in other excellent works are likely to make on the mind of the student are (a) that all independent variables vary uniformly, and (b) that they must vary in this manner in order that the differen- tials of their differentials shall be zero. That an independent variable may vary uniformly, as in Rate of Change, is granted; but differentials in general are variables whose limits are zero. Indeed one of the most important and essential properties of the differential of an independent vari- able is its independent variability. The imposition of any con- dition on a group of variables by which they may be expressed in terms of one another at once destroys the independence of the variables, and this is the case of variables under the hypothesis. of uniform variation, or rate of change. Thus, let u= f(x, y), and let us suppose x and y to vary simultaneously and uniformly; then dx = mdt and dy = m'dt; whence x = mit + C and y = m't+C'. Eliminating t and solving for y, we have y = p(x). Hence the supposition of uniform change renders the hypothesis of more than one in- * Dif. Calculus, Art. 79. + Dif. Calculus, Art. 204. L 250 APPENDIX. .. dependent variable impossible. Therefore, if independent variables (which are supposed to vary simultaneously) must vary uniformly in order that their higher differentials shall be zero, the successive differentials of u = f(y, x, z) can be obtained only by destroying the independence of all the variables x, y, z except one. Hence, in general, if the differential of dx, with respect to x, is zero, it is due to the fact that dx is a variable which is independent of x. A. To differentiate a" and loga v independently of Art. 78. Let y a", where v is a function of x. Increasing x by h, etc., and assuming that a" is differentiable (4,), we have Ду Δυ av(aso 1 Δυ 1 = m₂+m,Av. 2 Taking the limits, remembering that m,4v vanishes with and that a❞ is constant with respect to h, we have dy = a² dv ( limit α Δυ Av = 0 Δυ m. 19 where m, and a" are definite quantities. Therefore the limit of а Av 4 v 1 must be a definite quantity (m' say), not zero, and de- pendent solely on the base a, since it is evidently independent of x and h. dy = a'm', or dy or dy = a'm'dv. dv (1) COR. I. Since m' depends on the base and the base is arbi- trary, we may suppose the base to have such a value (say e) that m' = 1. Then de" — eºdv. - . APPENDIX. 251 = av aºm'dv = e“du. COR. II. To find the value of m′ in (1), let Differentiating (2), = er. (2) (3) du From (2) and (3), m' = (4) dv Taking loge of (2), v loge a = w. (5) Differentiating (5), dv log, a = du. (6) du Whence loge a = (7) dvⓇ Equating (4) and (7), m' = loge a. a.. (8) av d(aº): = aº log, a dv. COR. III. To differentiate y = log. v, we write it under the (9) form of ay = v. (10) Differentiating (10), av log, ady = dv; whence dy 1 dv loge a v dv or loga e (11) V 5 A.. A Rigorous Proof of Taylor's Formula. In what fol- lows, the function fly) and its n successive derivatives are sup- posed to be differentiable, and finite and continuous between the limits y and y+x. LEMMA. If F(z) is continuous betwen z = a and z = y, and if F(a) = F(y) = 0, then F'(z), if continuous, must equal zero for some value of z between ɑ and y. For, as z changes from a to y, F(2) passes from 0 to 0, that is, F(z) increases and then decreases, or vice versa; hence F'(z) must change from to or from to +, and therefore, since it is continuous, pass through 0. + 252 APPENDIX. In what follows will represent a positive proper fraction; that is, 0 <<1. Hence, 0 < x < x, and, by the lemma, F"[y+0(a− y)] = 0. or Under the given hypotheses, we have (A¸) fn-(y + x) = fr-1(y) tmt m ƒn−1(y + x) = ƒn−¹(y) + xs,. m₂x³, (1) where s (=m,+m,x) is continuous between y and y +x. Multiply (1) by dx, regard y constant, and integrate, and we have ƒn-2(y + x) = ƒn=2(y) + xfn-¹(y) + Sxs dx, . (2) the constant C being f-2(y), since ƒn-²(y + x) = ƒn−2(y) when x = 0. Multiply (2) by da, and integrate, denoting SS by ƒ“, and we have x² ƒn−³(y + x) = ƒn-³(y) + xfn-2(y) + fn-¹(y) + S*xs dx², (3) C being fn-³ (y). Continuing thus to n 1 integrations, we have f(y + x) = f(y) + xf'(y) + 置がし ​f(y) xn-1 +. n — 1· (y)+ƒ™¯xs dan-¹. (4) dx²-1. xs We wish now to find the value of the last term, "s da"-1, which is the remainder after n terms. In (4) put a y for x, and s' for the corresponding value of s, transpose, and we have У f(a) − f(y) – — a T -f'(y) (a — y)² + 12 1 (a — y)n−1 n-1 n_21ƒª-¹ (y) — ƒ™´¯`(a−y)s (—dy)=-1—0. (5) APPENDIX. 253 Let F(z) be a function such that F(2) = f(a) − f(x) — ª ¯ ½ƒ′(©) – − (a — z)n−1 ·1 (a — 2)² f¹¹ % on - 1 | 20 1 — 2 1 ƒn-¹ (2) — ƒ” (a−z)s′(−dz)”-¹….. (6) n - - Since s', having the same value as in (5), is independent of z, the last term of (6) is equal to (a — z)" s'. n Evidently F(2) = 0, first when z = y by (5), second when z = a; and since ƒ(y), ƒ′(y), ƒ”(y), etc., are all continuous from y to y + x (= a), ƒ(z), ƒ' (z), ƒ"(z), etc. (and therefore F(z) and F" (z)), are continuous between the same limits. Therefore, by the lemma, F(y + O(a− y)) = 0. Differentiating (6) to obtain F" (z), we have (α - z %) α % F' (z) = − ƒ' (2) +ƒ′(z) ƒ'' (%) + ·ƒ'' (%) 1 1 2 (a−2)² for (2) + (a—z)² 2 — ƒ''' (2) • • • (α — z)n-1 (a (α — z)n−1 + N - 1 fn(2) + That is, F" (z) = (α — z)n−1 n-1 n - 1 n−1 [ƒ"(z) — s′]. . -s'. (*) Now substituting y + O(a− y) for z, observing that for this value of z, F'(z) = 0, and dividing by (a (α- z)n-1 we have n 1 s' = ƒ“[y + A(à − y) ]. . In (6) substitute this value of s', and then put x for z (8) M 254 APPENDIX. ! - (whence F(x)=0), y+x for a (whence ay = x), transpose, and we have x² "(4) f(y + x) = f(y) + xƒ′(y) + 12" (y) xn-1 +.. n - Tfn··¹(y) + The last term Xn xn -ƒn (y + Ox). (9) -ƒ“(y + Ox) = R„, say, is called the remainder in Taylor's formula. It is obtained by differentiating f(y) n times and substituting y+0x for y in the final or nth derivative. When the function f(y + x) is such that this remainder ap- proaches 0 as n approaches ∞, the series will be convergent, otherwise it will be divergent. Hence the remainder enables us to ascertain the conditions under which any given function of the sum of two variables is developable by Taylor's formula, and also to find the limits of the error we make in stopping at any term of the series. EXAMPLE. 1. f(y + x)= log (y + x). See Art. 125. Since f(y) = log y, f"(y) = − (− 1)"- = -(−1)" ՊՆ 1 y" R₂ = − (− 1)” 1 X n\y + Ox! X Y + O x Now, since 0 < 0 < 1, is finite and a proper fraction if xory. Therefore R, approaches 0 as n approaches ∞, and log (y + x) is developable if x = or < y. . · • 4 · APPENDIX. 255 Again, since 0 < 0 < 1, the true numerical value of R, lies 1/x N between and n \y 1 X n\y + x n Hence if — (− 1)" 1 (~)* be 1/2 ԴՆ \ substituted for R, in (1), Art. 125, the series will be the value of log (y + x) to within less than 1 X n ny + x. A.. A Similar Completion of Maclaurin's Formula obtained by making y = 0 in (9), A.. Thus, may be # x² ƒ(x) = ƒ(0) + xƒʼ(0) + -ƒ''(0) +. xn-1 n-1· Xn N ƒn−1(0) + ƒn (Ox). (1) EXAMPLE. f(x)= e. X Since ƒ"(x) = e², R₂ сох eox. n For any finite value of x, (1) the fraction Xh evidently ap- N proaches 0 as n approaches, and (2), since 0 < 0 < 1, e☺x is finite. Hence the limit of R, as n approaches ∞, is 0, and ex is developable, for all finite values of x. Again, as 0 << 1, the true value of R, lies between хт Xn n and -ex Therefore the sum of all the terms after the nth in n series (N), Art. 127, is less than ex In ev A,. To determine the values of m, and m, in the formula Ay=m,h+m₂h (Art. 24). Since y = f(x), Ay = f(x + h) − f(x). • . ど ​256 APPENDIX. ' By Taylor's formula, h² 2 f(x + h) = f(x) + hf'(x) + ' f'(x + Oh). 4y = f'(x)h + ½ƒ''(x + Oli)h². Comparing this with the given formula, we have m, = ƒ′(x) and m₂ = {ƒ"(x+ Oh), where 0 < 0 < 1. 2 2 Therefore the formula 4y=m,h+m,h² is true in reference to y = f(x), (1) if f(x) and f'(x) are differentiable, and (2) if ƒ(x), ƒ'(x), and f'(x) are finite and continuous between the limits x and x + h. } : } E UNIVERSITY OF MICHIGAN וי 3 9015 06440 9314