ALBERT R. MANN LIBRARY . New York State Colleges OF Agriculture and Home Economics AT Cornell University DATE Duf DEI J 4 la/z # m^ ly/b # ^-;2'7c Mil' All n Niiiliir \ )': t^ iD ^P^ ^ i^/X fl, ■^^'^s IL) f^cgre aL umHKMBmmin n^ipt Ut-v w^, ~'\ «ni^ «a4Hll ■*"1*V:.^ '^**'*'-#*air _^»rtM*fc,(jJ i CAYLORD 1 PNINTEDINU.S.A. Cornell University Library QA 303.G77 Elements of the differntial and integral 3 1924 003 686 221 Cornell University Library The original of tiiis bool< is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924003686221 Sir Isaac Newton ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS (REVISED EDITION) BY WILLIAM ANTHONY GEANVILLE, Ph.D., LL,D. PRESIDENT OF PKNNSTLVANIA COLLEGE WITH THE EDITORIAL COOPERATION OF PEECEY E. SMITH, Ph.D. PKOFESSOR OP MATHEMATICS IN THE SHErFIELD SCIENTIFIC SCHOOI TALK UNIVERSITY GINN AND COMPANY BOSTON ■ NEW YORK • CHICAGO • LONDON ATLANTA , • DALLAS • COLUMBUS • SAN FRANCISCO Col'VRIGIIT, 1004, 1911, BY TFiLLiAM Anthony Gkanvillb and Percey F. Smith ALL EIGHTS EESEBVKD 910.2 GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. PKEFACE That teachers and students of the Calculus have shown such a gen- erous appreciation of Granville's "Elements of the Differential and Integral Calculus " has been very gratifying to the author. In the last few years considerable progress has been made in the teaching of the elements of the Calculus, and in this revised edition of Granville's " Calculus" the latest and best methods are exhibited, — methods that have stood the test of actual classroom work. Those features of the first edition which contributed so much to its usefulness and popu- larity have been retained. The introductory matter has been cut down somewhat in order to get down to the real business of the Calculus sooner. As this is designed essentially for a drill book, the pedagogic principle that each result should be made intuitionally as well as analytically evident to the student has been kept constantly in mind. The object is not to teach the student to rely on his intuition, but, "in some cases, to use this faculty in advance of analytical investigation. Graphical illustration has been drawn on very liberally. This Calculus is based on the method of limits and is divided into two main parts, — Differential Calculus and Integral Calculus. As special features, attention may be called to the effort to make per- fectly clear the nature and extent of each new theorem, the large number of carefully graded exercises, and the summarizing into working rules of the methods of solving problems. In the Integral Calculus the notion of integration over a plane area has been much enlarged upon, and integration as the limit of a summation is con- stantly emphasized. The existence of the limit e has been assumed and its approximate value calculated from its graph. A large num- ber of new examples have been added, both with and without answers. At the end of almost every chapter will be found a col- lection of miscellaneous examples. Among the new topics added are approximate integration, trapezoidal rule, parabolic rule, orthogonal VI PEEFACE trajectories, centers of area and volume, pressure of liquids, work done, etc. Simple practical problems have been added throughout; problems that illustrate the theory and at the same time are of interest to the student. These problems do not presuppose an ex- tended knowledge in any particular branch of science, but are based on knowledge that all students of the Calculus are supposed to have in common. The author has tried to write a textbook that is thoroughly modern and teachable, and the capacity and needs of the student pursuing a first course in the Calculus have been kept constantly in mind. The book contains more material than is necessary for the usual course of one hundred lessons given in our colleges and engineering schools ; but this gives teachers an opportunity to choose such subjects as best suit the needs of their classes. It is believed that the volume con- tains all topics from which a selection naturally would be made in preparing students either for elementary work in apphed science or for more advanced work in pure mathematics. WILLIAM A. GRANVILLE Pennsylvania College Gettysburg, Pa. CONTENTS DIFFERENTIAL CALCULUS QHAPTER I COLLECTION OF J'ORMULAS SECTION PAGE 1. Formulas from Algebra, Trigonometry, and Analytic Geometry . 1 2. Greek alphabet ........... 3 3. Rules for signs in the four quadrants ....... 3 4. Natural values of the trigonometric functions . . . ... 4 5. Tables of logarithms . .5 CHAPTER II VARIABLES AND EUNCTIONS 6. Variables and constants ......... 6 7. Interval of a variable ......... 6 8. Continuous variation . . . . . . ■ . . . .6 9. Functions . 7 10. Independent and dependent variables . . . ... . .7 11. Notation of functions 8 12. Values of the independent variable for which a function is defined . 8 CHAPTER III THEORY OF LIMITS 13. Limit of a variable 14. Division by zero excluded 15. Infinitesimals 16. The concept of infinity (oo) 17. Limiting value of a function . 11 12 13 . 13 . ♦ 14 18. Continuous and discontinuous functions 14 19. Continuity and discontinuity of functions illustrated by their graphs . 16 20. Fundamental theorems on limits .18 21. Special limiting values .20 22. The limit of ?HL£ as a; = -21 X 23. The number e ' 22 24. Expressions assuming the form ^ 23 vn VIU CONTENTS CHAPTER IV DIFFERENTIATION SEOTIOK 25. Introduction 26. Increments ..... 27. Comparison of increments 28. Derivative of a function of one variable 29. Symbols for derivatives . 30. Differentiable functions . 31. General rule for differentiation 32. Applications of the derivative to Geometry . PAGE 25 25 26 27 28 29 29 31 CHAPTER V RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS 33. Importance of General Rule 34. Differentiation of a constant .... 35. Differentiation of a variable with respect to itself . 36. Differentiation of a sum ... . . 37. Differentiation of the product of a constant and a function . 38. Differentiation of the product of tv^o functions 39. Differentiation of the product of any finite number of functions 40. Differentiation of a function with a constant exponent . 41. Differentiation of a quotient . 42. Differentiation of a function of a function 43. Differentiation of inverse functions 44. Differentiation of a logarithm 45. Differentiation of the simple exponential function . 46. Differentiation of the general exponential function 47. Logarithmic differentiation 48. Differentiation of sin b . 49. Differentiation of cos u . 50. Differentiation of tan v . 51. Differentiation of cot t) 52. Differentiation of sec v . 53. Differentiation of esc v 54. Differentiation of vers v . 55. Differentiation of arc sinu 56. Differentiation of arc cosu 57. Differentiation of arc tan v 58. Differentiation of arc cot u 59. Differentiation of arc sec v 60. Differentiation of arc esc v 34 36 37 37 37 38 38 39 40 44 45 46 48 49 50 54 55 56 56 56 57 57 61 62 62 63 63 64 CONTENTS ix SECTION PA(JE 61. DifEerentiation of arc vers v 65 62. Implicit functions . . 69 63. DifEerentiation of implicit functions 69 CHAPTER VI SIMPLE APPLICATIONS OF THE DERIVATIVE 64. Direction of a curve .......... 73 65. Equations of tangent and normal, lengths of subtangent and subnormal. *■ Rectangular coordinates 76 66. Parametric equations of a curve . 79 67. Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point . . .83 68. Lengths of polar subtangent and polar subnormal . . .86 69. Solution of equations having multiple roots 88 70. Applications of the derivative in mechanics. Velocity ... 90 71. Component velocities ... ..... 91 72. Acceleration ... . . . . 92 73. Component accelerations ...... .93 CHAPTEE VII SUCCESSIVE DIFEERENTIATION 74. Definition of successive derivatives . . . . 97 75. Notation . . . 97 76. The «th derivative 98 77. Leibnitz's formula for the nth derivative of a product . . 98 78. Successive differentiation of implicit functions ... . 100 CHAPTEE VIII MAXIMA AND MINIMA. POINTS OP INFLECTION. CURVE TRACING 79. Introduction . . ... . . 103 80. Increasing and decreasing functions . .... 106 81. Tests for determining when a function is increasing and when de- creasing ...... .... 108 82. Maximum and minimum values of a function . . 109 83. First method for examining a function for maximum and minimum values ... . . .... Ill 84. Second method for examining a function for maximum and minimum values ......... 112 85. Definition of points of inflection and rule for finding points of in- flection . . .... . . 125 86. Curve tracing . . . . ... 128 CONTENTS CHAPTEE IX DIFFERENTIALS SECTION 87. Introduction 88. Deflnitions ....... 89. Infinitesimals ...... 90. Derivative of the arc in rectangular coordinates 91. Derivative of the arc in polar coordinates . 92. Formulas for finding the differentials of functions 93. Successive differentials PAGE 131 131 132 134 135 187 139 CHAPTEE X RATES 94. The derivative considered as the ratio of two rates 141 CHAPTEE XI CHANGE OF VARIABLE 95. Interchange of dependent and independent variables . . . 148 36. Change of the dependent variable . ... . 149 97. Change of the independent variable ....... 150 98. Simultaneous change of both independent and dependent variables . 152 CHAPTEE XII CURVATURE. RADIUS OF CURVATURE 99. Curvature . 100. Curvature of a circle . 101. Curvature at a point . 102. Formulas for curvature 103. Radius of curvature 104. Circle of curvature 155 155 156 156 159 161 CHAPTEE XIII THEOREM OF MEAN VALUE. INDETERMINATE FORMS 105. RoUe's Theorem .... 106. The Theorem of Mean Value . 107. The Extended Theorem of Mean Value 164 165 166 CONTENTS XI SECTION 108. Maxima and minima treated analytically .... 109. Indeterminate forms 110. Evaluation of a function taking on an indeterminate form 111. Evaluation of the indeterminate form - . . . . 112. Evaluation of the indeterminate form ^ . . . . 113. Evaluation of the indeterminate form ■ co 114. Evaluation of the indeterminate form oo — co . 115. Evaluation of the indeterminate forms 0°, 1", oo" PAGE 167 170 170 171 174 174 175 176 CHAPTER XIV CIECLE or CURVATURE. CENTER OF CURVATURE 116. Circle of curvature. Center of curvature 178 117. Second method for finding center of curvature . . . 180 118. Center of curvature the limiting position of the intersection of nor- mals at neighboring points .... ... 181 119. Evolutes . - ... 182 120. Properties of the evolute ......... 186 121. Involutes and their mechanical construction ..... 187 CHAPTEE XV PARTIAL DIFFERENTIATION 122. Continuous functions of two or more independent variables 123. Partial derivatives 124. Partial derivatives interpreted geometrically 125. Total derivatives 126. Total differentials 127. Differentiation of implicit functions 128. Successive partial derivatives 129. Order of differentiation immaterial 190 191 192 194 197 198 202 208 CHAPTER XVI ENVELOPES 130. Family of curves. Variable parameter 205 131. Envelope of a family of curves depending on one parameter . . 205 132. The evolute of a given curve considered as the envelope of its normals 208 133. Two parameters connected by one equation of condition . 209 xu CONTENTS CHAPTER XVII SERIES SECTION 134. Introduction ...... 135. Infinite series .... 136. Existence of a limit .... 137. Fundamental test for convergence 138. Comparison test for convergence 139. Cauchy's ratio test for convergence 140. Alternating series ..... 141. Absolute convergence 142. Power series PAGE 212 213 215 216 217 218 220 220 223 CHAPTER XVIII EXPANSION OF FUNCTIONS 143. Introduction . . . . 144. Taylor's Theorem and Taylor's .Series ... 145. Maclaurin's Theorem and Maclaurin's Series 146. Computation by series .... . . 147. Approximate formulas derived from series. Interpolation 148. Taylor's Theorem for functions of two or more variables 149. Maxims, and minima of functions of two independent variables 227 ■228 230 234 237 240 243 CHAPTER XIX ASYMPTOTES. SINGULAR POINTS 150. Rectilinear asymptotes ...... 151. Asymptotes found by method of limiting intercepts 152. Method of determining asymptotes to algebraic curves 153. Asymptotes in polar coordinates 154. Singular points ... .... 155. Determination of the tangent to an algebraic curve at a given point by inspection . ... 156. Nodes ... . . 157. Cusps ... . . 158. Conjugate or isolated points . . . . 159. Transcendental singularities 249 249 250 254 255 255 258 259 260 260 CHAPTER XX APPLICATIONS TO GEOMETRY OF SPACE 160. Tangent line and normal plane to a skew curve whose equations are given in parametric form ........ 262 161. Tangent plane to a surface ........ 264 CONTENTS xm SECTION PAGE 162. Normal line to a surface ... 266 163. Another form of the equations of the tangent line to a skew curve . 268 164. Another form of the equation of the normal plane to a skew curve . 269 CHAPTER XXI CURVES FOE REFERENCE INTEGRAL CALCULUS CHAPTEE XXII INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 165. Integration ..... . 279 166. Constant of integration. Indefinite integral .... 281 167. Rules for integrating standard elementary forms .... 282 168. Trigonometric differentials . . . . . ' . . . 298 169. Integration of expressions containing V a^ _ x'- or V a;^ ± a^ by a trigo- nometric substitution ...... . . 304 . CHAPTER XXIII CONSTANT OF INTEGRATION 170. Determination of the constant of integration by means of initial con- ditions .......... 171. Geometrical signification of the constant of integration 172. Physical signification of the constant of integration .... 307 307 309 CHAPTEE XXIV THE DEFINITE INTEGRAL 178. Differential of an area .... 174. The definite integral 175. Calculation of a definite integral 176. Calculation of areas 177. Geometrical representation of an integral . 178. Mean value of (^ (a;) 179. Interchange of limits .... 180. Decomposition of the interval . 181. The definite integral a function of its limits 182. Infinite limits 183. When y = (x) is discontinuous 314 314 316 318 319 320 320 320 321 321 322 XIV CONTENTS CHAPTER XXV INTEGRATION OF RATIONAL FRACTIONS SECTION 184. Introduction 185. Case I 186. Case II 187; Case III . 188. Case IV PAGE 325 325 327 329 331 189. 190. 191. CHAPTER XXVI INTEGRATION BY SUBSTITUTION OF A NEW VARIABLE. RATIONALIZATION Introduction . . .... Differentials containing fractional powers of x only . Differentials containing fractional powers of a + 6a; only . 192. Change in Umits corresponding to change in variable 193. Differentials containing no radical except va + bx + x^ 194. Differentials containing no radical except Va + bx — x'^ 195. Binomial differentials .... 196. Conditions of integrability of binomial differentials 197. Transformation of trigonometric differentials 198. Miscellaneous substitutions 335 335 336 336 338 338 340 341 343 845 CHAPTER XXVII INTEGRATION BY PARTS. REDUCTION FORMULAS 199. Formula for integration by parts . . . 347 200. Reduction formulas for binomial differentials .... 350 201. Reduction formulas for trigonometric differentials . . . 356 202. To find fe"^ sin nxdx and C e°^ cos nxdx . . . . 359 CHAPTER XXVIII INTEGRATION A PROCESS OF SUMMATION 203. Introduction ....... 204. The fundamental theorem of Integral Calculus . 205. Analytical proof of the Fundamental Theorem . 206. Areas of plane curves. Rectangular coordinates 207. Area when curve is given in parametric form 208. Areas of plane curves. Polar coordinates 209. Length of a curve . . . . 210. Lengths of plane curves. Rectangular coordinates 211. Lengths of plane curves. Polar coordinates 361 361 364 365 368 370 872 373 375 CONTENTS XV SBOTJON 212. Volximes of solids of revolution 213. Areas of surfaces of revolution 214. Miscellaneous applications . PAGE 377 381 385 CHAPTEE XXIX SUCCESSIVE AND PAKTIAL INTEGRATION 215. Successive integration ........ 216. Partial integration 217. Definite double integral. Geometric interpretation 218. Value of a definite double integral over a region 219. Plane area as a definite double integral. Rectangular coordinates 220. Plane area as a definite double integral. Polar coordinates 221. Moment of area . ... 222. Center of area . . . . 223. Moment of inertia. Plane areas 224. Polar moment of inertia. Rectangular coordinates 225. Polar moment of inertia. Polar coordinates '226. General method for finding the areas of surfaces 227. Volumes found by triple integration .... 383 395 396 400 402 406 408 408 410 410 411 413 417 CHAPTER XXX OEDINAEY DIFFERENTIAL EQUATIONS 228. Differential equations. Order and degree .... 229. Solutions of differential equations .... 230. Verifications of solutions . . ... 231. Differential equations of the first order and of the first degree 232. Differential equations of the nth order and of the first degree 421 422 423 424 432 CHAPTER XXXI INTEGRAPH. APPROXIMATE INTEGRATION. TABLE OF INTEGRALS 233. Mechanical integration 234. Integral curves 235. Theintegraph . 236. Polar plaiumeter 237. Area swept over by a line . 238. Approximate integration . 239. Trapezoidal rule 240. Simpson's rule (parabolic rule) 241. Integrals for reference INDEX ....'. 443 443 445 446 446 448 448 449 451 461 GOTTFEIED WlLHELM LeIBNITZ DIFFERENTIAL CALCULUS CHAPTER I COLLECTION OF FORMULAS 1. Formulas* for reference. For the convenience of the student we give the following list of elementary formulas from Algebra, Geome- try, Trigonometry, and Analytic Geometry. 1. Binomial Theorem (n being a positive integer) : (a + 6)" = a» + na'-i6 -^Vll^tZ^a—^V' ^. '^(^ -!)("- 2) ^^.^^s + . . . [2 [3 n(„-l)(,»-2)...(rt-r + 2 )^„_^,,,,^_, \r-l 2; ji! = [)i = l-2-3.4-.-(n-l)n. - . . 3. In the quadratie .equation ax^ + bx + c = 0, when V — 4.ac > 0, the roots are real and unequal ; when 6^ — 4 ac = 0, the roots are real and equal ; when 6^ — 4 ac < 0, the roots are imaginary. 4. When a quadratic equation is reduced to the form x^ + px + q = 0, p = sum of roots with sign changed, and q = product of roots. ' 6. In an arithmetical series, I = a + (n -l)d ; s = 1(a + T) = ^[2a + (n -l)d]. 6. In a geometrical series, rl— a a (r» — 1) '- Z = ar«-i: s = = — ^ '-■ , r — l r — 1 7. log a& = log a + log 6. 10. logVa = -loga. 13. log- =- log a. 8. log - = log a — log 6. 11. Iogl:t0. 14. Circumference of circle =2 irr. 9. log a» = n log o. 12. loga a = 1. 15. Area of circle = irr^. * In formnlas 14-25, r denotes radius, a altitude, B area of base,' and s slant height. 1 DIFFEEENTlAL CALCULUS 16. Volume of prism = Ba. 17. Volume of pyramid = J Ba. 18. Volume of right circular cylinder = irr^a. 19. Lateral surface of right circular cylinder = 2 irra. 20. Total surface of right circular cylinder = 2irr(r + a). 21. Volume of right circular cone = \irr^a. 22. Lateral surface of right circular cone = irrs. 23. Total surface of right circular cone = 7rr(r + s). 24. Volume of sphere = ^trfi- 25. Surface of sphere = iirr^. 26. sin X = : cos x = ; tan x = — -— . cscx seox cotx sinx ^ cosx 27. tanx = — — ; cotx = -: cos X sm X 1 28. sin^x + oos^x = 1 ; 1 + tan^x = sec^x ; 1 +" cot^x = csc'x. /ir \ 31. sin (x + 2/) = sin X cos 2/ + cosx sin y. 29. smx = cos( xh 32. sin (x — 2/) = sin X cos 2/ — cosx sin y. cos X = sin ( X I : . . \2 / 33. cos(x ±y) = cosx cos?/ ^ sinx sm^. tanx = cot(--x). „. , . tanx + tanj/ \2 / 34. tan(x + 2/) = —■ 1 — tan X tan y 30. sin (ir — x) = sin x ; cos (tt — x) =— cosx: „,- , , ^ tan x — tan « ^ ' ' 35. tan (x — y) = ■ — tan (ir — x) = — tan X. 1 + tanxtany 3. sin2x = 2sinxcosz; cos2x = cos^x — sin^x; tan2x: 1 — tan^x „_. n.a; X „x .,x^ 2tanlx 37. sm X = 2 sin - cos- ; cos x = cos-' sin'' - : tan x = = 2 2' 2 2 l-tan2Jx 38. cos^x = J + J cos 2 X ; sin^x = J — | cos 2 x. 39. 1 + cos X = 2 cos^ - ; 1 — cos x = 2 sin^ - . 2' 2 .„ . X /l — cos X X /l + cos X X /l — ( 40. sii)- = ±-\ : cos-=±-\ ; tan- = + -v/ 2 \ 2 ' 2 \ 2 ' 2 \H-, 41. sinx + sin2/ = 2sin^(x + 2/) cos J (x — y). 42. sinx — sin 2/ = 2cosJ{x + y)sin^(x — y). 43. cos X + cos 2/ = 2 cos J (x + y) cos ^(x — y). 44. cos X — cos 2/ = — 2 sin i(x + y) sin J (x — 2/). ^,. <"' & c ^ „ „. 45. -: — - — - — - = -^ — - : Law of Sines, sin .4 sinB sinC 46. a2 = 62 + c2 - 2 6c coSyl ; Law of Cosines. 47. d = V(Xi - x^y + {y^ - y^Y ; distance between points (Xj, y^ and (x,^ Ax 4- Bv A- C 48. d = — ' , ^ ^ - ; distance from line 4x + JSi^ + C = to (x,, « •> ±VA2 + B2 COLLECTION OF FOEMULAS 5. Logarithms of numbers and trigonometric functions. Table of Mantissas of the Common Logarithms of Numbers No. 1 2 ' 3 4 5 6 7 8 9 1 0000 0414 0792 1139 1461 1761 2041 2304 2553 2788 2 3010 3222 3424 3617 3802 3979 4150 4314 4472 4624 3 4771 4914 5051 5185 5315 5441 5563 5682 5798 5911 4 6021 6128 6232 6335 6435 6532 6628 6721 6812 6902 5 6990 7076 7160 7243 7324 ' 7404 7482 7559 7634 7709 6 7782 7853 7924 7993 8062 8129 8195 8261 8325 8388 7' 8451 8513 8573 8633 8692 8751 8808 8865 8921 8976 8 9031 9085 9138 9191 9243 9294 9345 9395 9445 9494 9 10 11 9542 9590 9638 9685 9731 9777 9823 9868 9912 9956 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041' 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 Table of Logarithms of the Trigonombtkig Functions Angle in Radians Angle in Degrees log sin log COS log tan log cot .0000 0° 0.000 90° 1.5708 .0175 1° 8.2419 9.9999 8.2419 1.7581 89° 1.5533 .0349 2° 8.5428 9.9997 8.5431 1.4569 88° 1.5359 .0524 3° 8.7188 9.9994 8.7194 1.2806 ■ 87° 1.5184 .0698 4° 8.8436 9.9989 8.8446 ■ 1.1554 86° 1.5010 .0873 5° 8.9403 9.9983 8.9420 1.0580 85° 1.4835 .1745 10° 9.2397 9.9934. 9.2463 0.7537 80° 1.3963 .2618 15° , 9.4130 9.9849 9.4281 0.5719 75° 1.3090 .3491 20° 9.5341 9.9730 9.5611 0.4389 70° 1.2217 .4363 25° 9.6259 9.9573 9.6687 0.3313 65° 1.1345 .5236 30°, 9.6990 9.9375 9.7614 0.2386 60° 1.0472 .6109 35° 9.7586 9.9134 9.8452 0.1548 55° 0.9599 .6981 40° 9.8081 9.8843 9.9238 0.0762 50° • 0.8727 .7854 45° 9.8495 9.8495 0.0000 0.0000 45° 0.7854 Angle in Angle in log COS log sin log cot log tan Degrees Radians CHAPTER n VARIABLES AND FUNCTIONS 6. Variables and constants. A variable is a quantity to which an unlimited number of values can be assigned. Variables are denoted by the later letters of the alphabet. Thus, in ,the equation of a straight line, 5 + 1 = 1, a b X and y may be considered as the variable coordinates of a point moving along the line. A quantity whose value remains unchanged is called a constant. Numerical or ahsolute constants retain the same values in all prob- lems, as 2, 5, Vy, tt, etc. Arbitrary constants, or parameters, are constants to which any one of an unlimited set of numerical values may be assigned, and they are supposed to have these assigned values throughout the inves- tigation. They are usually denoted by the earlier letters of the alphabet. Thus, for every pair of values arbitrarily assigned to a and 6, the equation - + ! = ! a b represents some particular straight line. 7. Interval of a variable. Very often we confine ourselves to a portion only of the nuraber system. For example, we may restrict our variable so that it shall take on only such values as lie between a and b, where a and h may be included, or either or both excluded. We shall employ the symbol [a, 5], a being less than Sj to represent the numbers a, h, and all the numbers between them, unless otherwise stated. This symbol [a, 6] is read the interval from a to b. 8. Continuous variation. A variable x is said to vary continuously through an interval [a, 6], when a; starts with the value a and increases until it takes on the value b in such a manner as to assume the value 6 VARIABLES AKI) FUNCTIONS 7 of every number between a and 6 in the order of tbeir magnitudes. This may be illustrated geometrically as follows : The origin being at 0, lay off on the straight line the points A and B corresponding to the numbers a and b. Also let the point F corre- spond to a particular value of the variable x. Evidently the interval [a, 6] is represented by the segment AB. Now as a; varies continuously from a to 5, inclusive, i.e. through the interval [«, 6], the point P gen- erates the segment AB. 9. Functions. When two variables are so related that the value of the first variable depends on the value of the second variable, then the first variable is said to be a function of the second variable. Nearly all scientific problems deal with quantities and relations of this sort, and in the experiences of everyday life we are con- tinually meeting conditions illustrating the dependence of one quan- tity on another. For instance, the weight a man is able to lift depends on his strength, other things being equal. Similarly, the distance a boy can run may be considered as depending on the time. Or, we may say that the area of a square is a function of the length of a side, and the volume of a sphere is a function of its diameter. 10. Independent and dependent variables. The second variable, to which values may be assigned at pleasure within limits depending on the particular problem, is called the independent variable, of argument ; and the first variable, whose value is determined as soon as the value of the independent variable is fixed, is called the dependent variable, or function. Frequently, when we are considering two related variables, it is in our power to fix upon whichever we please as the independent variable; but having once made the choice, no change of independent variable is allowed without certain precautions and transformations. One quantity (the dependent variable) may be a function of two or more other quantities (the independent variables, or arguments). For example, the cost of cloth is a function of both the quality and quantity ; the area of a triangle is a function of the base and altitude; the volume of a rectangular parallelepiped is a function of its three 8 DIFFERjiiJNTiAL, CALCULUS 11. Notation of functions. The symbol /(x) is used to denote a function of a-, and is read f of x. In order to distinguish between different functions, the prefixed letter is changed, as F(x), ^(^)» f'(x), etc. , During any investigation the same functional symbol always indi- cates the same law of dependence of the function upon the variable. In the simpler cases this law takes the form of a series of analytical operations upon that variable. Hence, in such a case, the same func- tional symbol will indicate the same operations or series of operations, even though applied to different quantities. Thus, if f(x) = x'-'dx + 14., then fiy-) = f-9y + U. Also /(«) = «'- 9 a +14, /(6+l) = (6 + iy-9(i+l) + 14 = J^-7S+6, /(0) = 0^-9 0+14=14, /(-l) = (-iy-9(-l) + 14 = 24, /(3) = 3^-9-3+14=-4, /(7) = 7'-9-7 + 14 = 0, etc. Similarly, ^ (.r, ^) denotes a function of x and y, and is read ^ of X and y. If 4> {x, y') = sin (x + y), then (^ (a, J) = sin (a + 5), and ^(-' OJ=sin^=l. Again, if F(x, y, z)=2x+ 3 y —12z, then FQn, — m, m}= 2 m — Bni —12 m= — lSm, and i?'(3, 2, 1)= 2-3 + 3- 2-12-1=0. Evidently this system of notation may bo extended indefinitely. 12. Values of the independent variable for which a function is defined. Consider the functions a:^—2x+ 5, sin a;, arc tan a: of the independent variable x. Denoting the dependent variable in each case by y, we may write y = x^—2x+5, y = sinx, ?/ = arctana;. VAEIABLES AND FUNCTIONS 9 In each case y (the value of the function) is known, or, as we say, defined, for all values of x. This is not by any means true of all functions, as the following examples illustrating the more common exceptions will show. ' (1) y = : " ■ h Here the value of y (i.e. the function) is defined for all values of x except x = b. When x.= h the divisor becomes zero and the value of y cannot be computed from (1).* ' Any value might be assigned to the function for this value of the argument. (2) y=^. In this case the function is defined only for positive values of x. Negative values of x give imaginary values for y, and these must be excluded here, where we are confining ourselves to real numbers only. (3) y = \og^x. a>0 Here y is defined only for positive values nf x. For negative values of X this function does not exist (see § 19). (4) y = arc sin a;, y = arc cos x. Since sines and cosines cannot become greater than + 1 nor less than — 1, it follows that the above functions are defined for all values of x ranguig from — 1 to + 1 mclusive, but for no other values. EXAMPLES 1. Given f(x) = x^ - lOx^ + six - 30 ; show that /(0)=-30, f(y) = y'-10y' + 31y-30, /(2) = 0, f{a) = aS - 10 a2 + 31 a - 30, /(8) = /(5), f(yz) = 2/%' - 10 2/%2 + 31 yz- 30, /(I) >/(_ 3), f(x - 2) = x^ - 16x2 + 83x - 140, /(-I) =-6/(6). 2. If f(x) = X' - 3x + 2, fliid/(0), /(I), /(- 1), /(- i), /(IJ). 3. If fix) = x3 — lOx^ + Six — 30, and (x) = x* — 55x2 _ 2i0x — 216, show that /(2) = .^(-2), /(3) = 0(-3), /(5) = 0(-4), /(0) + 0(0) + 246 = 0. 4. If ^(x) = 2^ find F(0), F(- 8), F(i), F(- 1). 5. Given F(x) = x(x- 1) (x + 6) (x - ^) (x + f ) ; show that ^(0) = F{1) = F(- 6) = F(i) = F{- i) = 0. * See § 14, p. 12. 10 DIFFEEENTIAL CALCULUS 6. If f(m,) = ^~ ■ , show that m^ + l 1 +f(m,i)f(m^) 1 + OTiWij 7. If 1^ (x) = a»^, show that (y)- (z) =.^ (y + z). J X 8. Given 0(x) = log ; show that 0(x) + 0(3/) = 0(fJ^). 9. If f() = cos 0, show that /(*) =/(- 0) = -/(t - 0) = -/(T + *) . 10. If F(S) = tan 6, show that 11. Given ^ (x) = x^" + x^"* + 1 ; show that 12. If /(x) = ^^^ , find /(V2). ^ns. - .0204. X + 7 13 ■ever CHAPTER III THEORY OF LIMITS 13. Limit of a variable. If a variable v takes on successively a series of values that approach nearer and nearer to a constant value I in such a manner that |v — Z|* becomes and remains less than any assigned arbi- trarily small positive quantity, then v is said to approach the limit I, or to converge to the limit I. Symbolically this is written limit V = 1, or, v = I. .The following familiar examples illustrate what is meant: (1) As the number of sides of a regular inscribed polygon is indefi- nitely increased, the limit of the area of the polygon is the area of the circle. In this case the variable is always less than its limit. (2) Similarly, the limit of the area of the circumscribed polygon is also the area of the circle, but now the variable is always greater, than its limit. (3) Consider the series The sum of any even number (2 w) of the first terms of this series is ^"2 4 8 2^"-^ 2^"-^ J--1 9?^ 2 1 (5) ^^„=__^ = --^-^5^. By6, p. 1 Similarly, the sum of any odd number (2 w+1) of the first terms of the series is 111 11 '^2''+i~ 2'*"4~8'' 2^"^'*' 2^' -J— 1 921+1 o -1 (C^ S =— = = - + — — By6, p. 1 * To be read the numerical value of the difference between v and I. 11 10 DIFFERENTIAL CALCULUS Writing (5) -and (C) in the foriiiis we have and limit f^_ a „ = 00 U ^ limit n = cc S.. I^a variable, it is seen that both a limit as the number of Hence, by definition of the S„, and ^2n+i ^^6 variables ap icreases without limit. ling up the first two, ti^ggj^our, e^^ terms of (^), the sums by (jB) and (C) tOD^»a4^CTfea4gly^s and greater than |, illustratiijg the case when the vanable^'^i^^is case the sum of the terms )dtemateli/ less and greater than wi^mit. In the emmples shown WSS^^Me^jiever Caches its limit. This is not by any ^eans always the case5^^^®«ithe definition of the limit ^of a variable it is cmar that the essenN^^Obe*definition is simply that ; numerical vg/luaolthe difference beW^en the variable and its limit dmately'WcpmeSiiid remaJH"less thaj^Miy positive number we may choosS?*ba^ever small. (4) As an "^""^p*i=^?'|'>*"»t.^.of^jj^^ <^t.^ fact that the variable may reach its limit, consider the folros«£g!**T9B€«aBBBftBies of regu^lar polygons be inscribed in a aijjgl&the numtJstof sides increasing indefinitely. Choosing any oJW^of tnS8j*|SftnsteiiCT**the circumscribed polygon whose sides touch \Btei^circle atrfeeyettiseg/oS^e inscribed polygon. Let p^ and i^ be th\peSmeters of n»e inscnbed and circumscribed polygons of n sides, and C the circumf ertoce of the circle, and sup- pose the values of a variable x to,be as foi^ws _ ^? ^to +1' ^n + 2' ^1^' '" -i-o, etc. ^„. Then, evidently, and the limit is reachea^y the variable, every thirS value of the variable being C. 14. Division by zero excluded. - is in^i^erminaie. For the quotient of two numbers is that number which mulclWiedr by the divisor will give the dividend. But any number whatever maltiplied by zero gives THEORY OF LIMITS 13 zero, and the quotient is indeterminate ; that is, any number whatever may be considered as the quotient, a result which is of no value. - has no meaning, a being different from zero, for there exists no number such that if it be multiplied by zero, the product will equal a. Therefore division hy zero is not an admissible operation. Care should be takeu not to divide by zero inadvertently. The following fallacy is an illustration. Assume that a = 6. Then evidently ab = a'. Subtracting b^, ab-b^ = a^- b^. Factoring, 6 (a — 6) = (a + 6) (a — 6). Dividing by a — 6, 6 = a + 6. But o = 6, therefore 6 = 26, or, 1 = 2. The result is absurd, and is caused by the fact that we divided by a — 6 = 0. 15. Infinitesimals. A variable v whose limit is zero is called an infinitesimal.* This is written limit V = 0, or, i; = 0, and means that the successive numerical values of v ultimately become and remain less than any positive number however small. Such a variable is said to leeome indefinitely small or to ultimately vanish. If limit v = l, then limit (y — l')=Q; that is, the difference between a variable and its limit is an infinitesimal. Conversely, if the difference between a variable and a constant is an infinitesimal, then the variable approaches the constant as a limit. 16. The concept of infinity (oo). If a variable v ultimately becomes and remains greater than any assigned positive number however large, we say v increases without limit, and write limit w = + oo, or, v = + co. If a variable v ultimately becomes and remains algebraically less than any assigned negative number, we say v decreases without limit, and write limit t) = -oo, or, v = -x. * Hence a constant, no matter how small it may be, is not an infinitesimal. 14 DIFFERENTIAL CALCULUS If a variable v ultimately becomes and remains in numerical value greater than any assigned positive number however large, we say v, .in numerical value, increases without limit, or v becomes infinitely great,* and write U^^i^. ^ ^ ^ ^ ^^^ ^ ^ ^ _ Infinity (oo) is not a number; it simply serves to characterize a particular mode of variation of a variable by virtue of which it increases or decreases without limit. 17. Limiting value of a function. Given a function /(a;). If the independent variable x takes on any series of values such that limit x = a, and at the same time the dependent variable /(x) takes on a series of corresponding values such that limit /(a;)=^, then as a single statement this is written and is read the limit off(x), as x approaches the limit a in any manner, is A. 18. Continuous and discontinuous functions. A function /(a;) is said to be continuous for x = a ii the limiting value of the function when x approaches the limit a in any manner is the value assigned to the function for x= a. f[n symbols, if then f(x) is continuous for x = a. The function is said to be discontinuous for a; = a if this condition is not satisfied. For example, if the function is discontinuous for a; = a. The attention of the student is now called to the following cases which occur frequently. *Oii account of the notation used and for the sake of uniformity, the expression I) = +00 is sometimes read v approaches the limit phis infinity. Similarly, i> = - oo is read V approaches the limit minus infinity, and w =s= oo is read w, in numerical value, approaches the limit infinity. While the ahove notation is convenient to use in this connection, the student must not forget that infinity is not a limit in the sense in which we defined a limit on p. 11, for infinity is not a number at all. THEORY OF LIMITS 15 Case I. As ^ example illustrating a simple case of a function con- tinuous for a particular value of the variable, consider the function /(^) = x-2 For a; = 1, 'f(x) =f(l) = 3. Moreover, if x approaches the limit 1 in any manner, the function f(x) approaches 3 as a limit. Hence the function is contiuuous for x = \. Case II. The definition of a continuous function assumes that the function is already defined for x = a. If this is not the case, how- ever, it is sometimes possible to assign such a value to the function for x = a that the condition of continuity shall be satisfied. The following theorem covers these cases. Theorem. If f(x) is not defined for x = a, and if then fQt) will be continuous for x = a, if B is assumed as the value of f^x) for x=a. Thus the function x-2 is not defined for a; = 2 (since then there would be division by zero). But for every other value of x, X— 2 and . li-i*(.+ 2) = 4; therefore ^^l^^4. x=2 x-2 Although the funttion is not defined for a; = 2, if we arbitrarily assign it the value 4 for a; = 2, it then becomes continuous for this value. A function f(x) is said to he continuous in an interval when it is continuous for all values of x in this interval.* * In this book we shall deal only with functions which are in general continuous, that is, continuous for all values of x, with the possible exception of certain isolated values, our results in general being understood as valid only for such values of x for which the function in question is actually continuous. Unless special attention is called thereto, we shall as a rule pay no attention to the possibilities of such exceptional values of x for which the function is discontinuous; The definition of a continuous function /(i) is sometimes roughly (but imperfectly) summed up in the statement that a small change in x shall produce a small change inf(x) . We shall not consider functions having an infinite number of oscillations in a limited region. J^ DIFFERENTIAL CALCULUS 19. Continuity and discontinuity of functions illustrated by their graphs. (1) Consider the function 3?, and let If we assume values for x and calculate the corresponding values of y, we can plot a series of points. Drawing a smooth line free-hand tlirough these points, a good representation of the gen- eral behavior of the function may be obtained. This picture or image of the function is called its graph. It is evidently the locus of all points satisfying equation (^). Such a series or assemblage of points is also called a curve. Evidently we may assume values of x so near together as to bring the values of y (and therefore the points of the curve) as near together as we please. In other words, there are no breaks in the curve, and the function 3? is continuous for all values of x. (2) The graph of the continuous function sin a; is plotted by draw- ing the locus of y = sin X. It is seen that no break in the curve occurs anywhere. . (3) The continuous function e" is of very frequent occurrence in the Calculus. If we plot its graph from y = e', (e = 2.718 • • ■) we get a smooth curve as shown. From this it is clearly seen that, (a) when x = 0, ^™'t y^^^-^^^. (b) when x>Q, y(= e") is positive and increases as we pass towards the right from the origin ; (c) when x<0, y (= e"^) is still positive and decreases as we pass towards the left from the origin. y (4) The function log^a; is closely related to the last one discussed. In fact, if we plot its graph "o' f^«°^ 2/ = log.:., it will be seen that its graph has tha same rela- tion to OX and OY as the graph of e'' has to F and OX. THEOEY OF LIMITS 17 Here we see the following facts pictured : (a) For x = l, log^a; = log^ 1=0. (b) For x>l, log^x is positive and increases as x increases. (c) For 1 > 2: > 0, log^a; is negative and increases in numerical value as X diminishes, that is, ^'^'' log a; = — 00 . (d) For a; s 0, log^a; is not deiined ; hence the entire graph lies to the right of OY. (5) Consider the function -> and set X 1 If the graph of this function be plotted, it will be seen that as x approaches the value zero from the left (negatively), the points of the curve ultimately drop down an infinitely great distance, and as x approaches ' the value zero from the right, the curve extends upward infinitely far. The curve then does not form a continuous branch from one side to the other of the axis of Y, showing graphically that the function is discontinuous for x=0, but continuous for all other values of x. , (6) From the graph of y = 1-a? it is seen that the function : ± 1, but continuous for all r l-x" is discontinuous for the two values x other values of x. (7) The graph of y = td.nx shows that the function tana; is dis- continuous for infinitely many values of the independent variable x, namely, x=^ — , where n denotes any odd positive or negative integer. (8) The function arc tan * 18 DIFFERENTIAL CALCULUS has infinitely many values for a given value of x, the graph of equation y = arc tan x consisting of infinitely many branches. If, however, we confine our- selves to any single branch, the function is continuous. For instance, if we say that y shall be the arc of smallest numeri- cal value whose tangent is x, that is, y shall take on only values between — ^ and ^) then we are limited to the branch passing through the origia,; and the condition for contiauity is satisfied. 4-7r X (9) Similarly, arc tan - , X is found to be a many-valued function. Confining ourselves to one branch of the graph of i ° ^ V = arc tan-, we see that as x approaches zero from the left, y approaches the limit - 1, and as x approaches zero from the right, y approaches the Y limit + 1 -S 2 ^ Hence the function is discon- tinuous when x=0. Its value for x= can be assigned at pleasure. " Functions exist which are discontinuous for every value of the independent vari- able within a certain range. In the ordinary applications of the Cal- culus, however, we deal with functions which are discontinuous (if at all) only for certain isolated values of the independent variable; such functions are therefore in general continuous, and are the only ones considered in this book. 20. Fundamental theorems on limits. In problems involving limits the use of one or more of the following theorems is usually implied. It is assumed that the limit of each variable exists and is finite. Theorem:. The limit of the algebraic mm of a finite number of vari- ables is equal to the like algebraic sum of the limits of the sever4 variables. . Theorem II. The limit of the product of a finite number of variablei is equal to the product of the limits of the several vanables. ] Theorem III. The limit of the quotient of two variables is equal to the quotient of the limits of the separate variables, provided the limit of the denominator is not zero. THEORY OF LIMITS 19 Before proving these theorems it is necessary to establish the fol lowing properties of infinitesimals. (1) The sum of a finite numher of infinitesimals is an infinitesimal. To prove this we must show that the numerical value of this sum can be made less than any small positive quantity (as e) that may be assigned (§ 15). That this is possible is evident, for, the limit of each infinitesimal being zero, each one can be made numerically less than - (n being the number of infinitesimals), and therefore their sum can be made numerically less than e. (2) The product of a constant c and an infinitesimal is an infinitesimal. For the numerical value of the product can always be made less than any small positive quantity (as e) by making the numerical value of the infinitesimal less than -. c (3) The product of any finite numher of infinitesimals is an infinitesimal] For the numerical value of the product may be made less than any small positive quantity that can be assigned. If the given product contains n factors, then since each infinitesimal may be assumed less than the mth root of e, the product can be made less than e itseK. (4) Ifv is a variable which approaches a limit I different from zero, then the quotient of an infinitesimal hy v is also an infinitesimal. For if limit v = l, and k is any number numerically less than I, then, by defini- tion of a limit, v will ultimately become and remain numerically greater than &, Hence the quotient -, where e is an infinitesimal, will ulti- V mately become and remain numerically less than -, and is therefore by (2) an infinitesimal. Proof of Theorem I. Let v^, v^, v^, ■■■ be the variables, and l^, l^, l^, ••■ their respective limits. We may then write "l- ■h- «1' \- ■h = «2' ^3- ■h = «3' where e , e , e , ■•■ are infinitesimals (i.e. variables having zero for a limit). Adding 20 DIFFEEENTIAL CALCULUS Since the right-hand member is an infinitesimal by (1), p. 19, we have, from the converse tlieorem on p. 18, hmit (y^+v,+ v^+ • ■ ■) = ^1+ l,+ l,+ ---, or, limit (y^+ v^+ v^-\ ) = limit v^ + limit v^ + limit v^A , which was to be proved. Proof of Theorem II. Let v^ and v^ be the variables, \ and l^ their respective limits, and e^ and e^ infinitesimals; then v,= \+e^ and ^2=^2+V -Multiplying, V2=(^:+0a+0 -Kh+k%+h^r+',^: or. (5) V2-^A=^l«2+ ^2^1+^2- Since the right-hand member is an infinitesimal by (1) and (2), p. 19, we have, as before, limit (y^^) = l^^= limit v^ ■ limit v^, which was to be proved. Proof of Theorem III. Using the same notation as before, or. Here again the right-hand member is an infinitesimal by (4), p. 19, ^ > j.^. /^^_Z,_limi^_ if ?« =^ ; hence 1imif, -- \v^/ l^ limit V, which was to be proved. It is evident that if any of the variables be replaced by constants, our reasoning still holds, and the above theorems are true. 21. Special limiting values. The following examples are of special importance in the study of the Caleulhs. In the following examples a > and c^a. THEORY OF LIMITS , 21 Written in the form of limits. Abbreviated form often used. (1) (2) limit « ^. e ■ oo = oo. (3) 00 — = CO. c (4) limit ^ _o- ^ = 0. 00 (5) limit , when a < 1 a-"=+oo. (6) liniit ^a^o, X = + oo ' when a<\ a+" = 0. (7) X^i"-oo«^=0' when a > 1 a-- = 0. (8) x'ifi'.«^=+-' when fli > 1 «+"=+oo. (9) ^'rGl0g»^=+-' when a < 1 log„0 = + 00. (10) X = + »l0g«.^=--' when rt < 1 l0g„ (+ OO) = - CO. (11) Si°g»-=--' when a > 1 logaO=-<»- (12) x=t»l«g«-=+-' when a > 1 l0ga(+ QO) = + 00. The expressions in the second column are not to be considered as expressing numerical equalities (oo not being a number); they are merely symbolical equations implying the relations indicated in the first column, and should be so understood. no oi. ^1. i limit sin x ' ^ 22. Show that „ =1.* x = X Let be the center of a circle whose radius is unity. Let arc AM= arc AM' =x, and let MT and M'T be tangents drawn to the circle at M and M'. From Geometry, MPM' < MAM' < MTM'; or 2 sin a; < 2 a; < 2 tan x. Dividing through by 2 sin x, we get l<-^<— ■ sm X cos X * If we refer to the table on p. 4, it will be seen that for all angles less than 10° the angle in radians and the sine of the angle are equal to three decimal places. If larger tables are consulted, five-place, say, it will be seen that for all angles less than 2.2° the sine of the angle and the angle itself are equal to four decimal places. From this we may well suspect that limit sin a; _ ^ 10. DIFFERENTIAL CALCULUS If now X approaches the limit zero, limit X a; = sin a; limit must lie between the constant 1 and limit a; Therefore ^""''' -^ = 1, or, „ a; = U sm x x=v x=0 cos X limit sin x X 5 which is also 1. 1. Th. Ill, p. 18 It is interesting to note the behavior of this function from its graph, the locus of equation • gjj^ ^ Although the function is not defined for a; = 0, yet it is not discon- tinuous when a; = if we define sin ~0~ = L Case II, p. 15 23. The number e. One of the most important limits in the Cal- culus is limit ,^ _^ .i _ 2.71828 ••• = «. a;= 0^ ^ To prove rigorously that such a limit e exists, is beyond the scope of this book. For the present we shall content ourselves by plotting the locus of the equation i y = (\+xy and show graphically that, as a; = 0, the function (1 + a;)»=(= y) "/ y X V l(i 1.0096 5 1.4310 2 1.7320 1 2.0000 .5 2.2500 -.5 4.0000 .1 2.5937 -.1 2.8680 .01 2.7048 -.01 2.7320 .001 2.7169 -.001 2.7195 takes on values in the near neighborhood of 2.718 ■■• , and therefore e =',2.718 • • • approximately. THEORY OF LIMITS As a; == frbih the left, y decreases and approaches « as a limit. As a: 4 from the right, y increases and also approaches e as a hmit. As a; =b 00, 2/ approaches the limit 1 ; and as a; == — 1 from the right, y injereases without limit. In Chap. XVIII, Ex. 15, p. 233, we will show how to calculate the value of e to wa.j number of decimal places. Natural logarithms are those which have the number e for base. These logaritluns play a very important rSle in mathematics. When the base is not indicated explicitly, the base e is always understood in what follows in this book. Thus log^v is written simply log v. Natural logarithms possess the following characteristic property: If a; — in any way whatever, ,. ., logCl + ar) 1 lunit _EJi_ L = limit log (1 + x'yi = log e = 1. 24. Expressions assuming the form ^ • As oo is not a number, the .expression go -f- oo is indeterminate. To evaluate a fraction assuming this form, the niimerator and denominator being algebraic functions, we shall find useful the following Rule. Divide both numerator and denominator by the highest power of the variable occurring in either. Then substitute the value of the variable. 1 17 1 t Hmit 2a:8- 3x^ + 4 Illustrative Example 1. Evaluate ^ _ 7 i ir-^ • Solution. Substituting directly, we get "™'' —— = ^ , which is indeter- x = oo5x — x^ — 7x2 oo minate. Hence, following the above rule, we divide both numerator and denominator by x'. Then 2-5 + 4 limit 2x8-3x^ + 4 ^ limit z x^ _ i ^^ x = QO 5x — x2 — 7x3 ^ = <»5_1_ 7 t2 r. EXAMPLES Prove the following : ^ limit / x + l \ _ J ■ x = co\ X / _ . limit /a; + 1\ limit /, , 1\ p«»*- x=«,(-^j=x=«(i+5; _ limit ,j- limit /1\ Th. I, p. 18 ■ X = 00 '• ' X = 00 \X, =1+0=1 22 DIFFEEENTIAL CALCULUS 2 limit / z'' + 2a: \ 1 ,1 + -' limit / x^ + 2x \ ^ limit | 1 \a;'' [Dividing both numerator and denominator by x^.} limit / 2\ j; = 00 \ ^ xj imit /^ _ „\ limit X limit i-t\ , limit /2' X = 00 ^ ' "'" X = QC ^x, limit X imit /^\ _ limit ,„, = 00 \j;2/ X = 00 ^ ' Th. Ill, p. 18 Th. I, p. 18 _ l+0 _ 1 ~0-3~ 3' 3 limit x^-2x+5 ^1 limit a | - ^. „ 4 limit 3xS + 6xg _ _ 2 j^ limit 2x3 + Sa" ■ x = 02a;4-l5x2^ 5' ■x-=0 a;S *• P limit x' + l _ jg limit 5xg-2x ^ 2;=-2a; + 3~ •x = co x 6. 1'°"'' (3 0x2 - 2 Ax + 5 ^2) = 3 ^^.s. jg. "'"'* ^^ = 1. 7. i^™'' (ax2 + 6x + c) = 00. 17 limit njn + V, ^ ^ " ' n = <»(n + 2){n + 3) ■ „ limit (x-A:)2-2fcx3 ^ ,. . „s 1 8. i._o-^^ , ,■ ,, = 1. -iQ limit s'-l „ «-" x(x + i;) 18. ^^j— -^ = 3. limit X^ + 1 1 1- -4. /^ _L l\n ^n Q = -. in limit (X + /l)» — x" »• x = co3j;2 + 2x-1 3 ^^-^ = 0^ 1 = 'M"-i- ^Q limit 3 + 2x — = 0. on limit r la , ,>sin7i"| ^ 5x 20. ^^q|cos(^ + /i)-^I = oos(9. ,, limit cos(a — a) ^ limit 4 ir^ _ -r 4 11. a_ir ^^ ^=— tana. 2I = 2 cos (2 a: — a) 'x = oo4 3J.2 3' j^2 limit ox' + 6x + c _ a limit 1 - cos g _ 1 ■ ^ = «'dx2 + ex+/ d' -^ = g2 -2' „„ limit 1 .... . . , , X = a x — a = ~ "") " ^ IS mcrgasing as it approaches the value a. „. limit 1 , ■« ■ J . . , , **• X = a X — a = + ■">" ^ 's decreasing as it approaches the value a. CHAPTER IV DIFFERENTIATION 25. Introduction. We shall now proceed to investigate the man- ner in which a function changes in value as the independent variable changes. The fundamental problem of the DifPerential Calculus is to establish a measure of this change in the function with mathematical precision. It was while investigating problems of this sort, dealing with continuously varying quantities, that Newton* was led to the discovery of the fundamental principles of the Calculus, the most scientific and powerful tool of the modern mathematician. 26. Increments. The increment of a variable in changing from one numerical value to another is the difference found by subtracting the first value from the second. An increment of x is denoted by the symbol A.x, read delta x. The student is warned against reading this symbol delta times x, it having no such meaning. Evidently this increment may be either positive or negative ^ according as the variable in changing is increas- ing or decreasing in value. Similarly, Ay denotes an increment of «/, A(/> denotes an increment of (j), A/(a;) denotes an increment of /(a;), etc. If in 1/ —f(x) the independent variable x takes on an increment Aa;, then Ay is always understood to denote the corresponding increment of the function /(a;) (or dependent variable «/). The increment Ay is always assumed to be reckoned from a definite initial value of y corresponding to the arbitrarily fixed initial value of x from which the increment Ax is reckoned. For iaistance, consider the function y = jcl * Sir Isaac Newton (1642-1727), an Englishman, was a man of the most extraordinary- genius. He developed the science of the Calculus under the name of Fluxions. Although Newton had discovered and made use of the new science as early as 1670, his first published work in which it occurs. is dated 1687, having the title Philosophiae Naturalis Principia Mathematica. This was Newton's principal work. Laplace said of it, '" It will always remain preeminent above all other productions of the human mind." See frontispiece. t Some writers call a negative increment a decrement. 25 26 DIFFERENTIAL CALCULUS Assuming a; = 10 for the initial value of x fixes y — 100 as the initial , value of y. Suppose X increases to a; = 12, that is, Aa;=2; then y increases to y = 144, and Ay = 44. Suppose X decreases to x = 9, that is. As; = — 1 ; then y decreases to z/ = 81, and Ay = — 19. It may happen that as x increases, y decreases, or the reverse ; in either case Ax and Ay will have opposite signs. It is also clear (as illustrated in the above example) that if y =f(x) is a continuous function and Ax is decreasing in numerical value, then Ay also decreases in numerical value. 27. Comparison of increments. Consider the function Assuming a fixed initial value for x, let x take on an increment Ax. Then y will take on a corresponding increment Ay, and we have y + Ay = (x + Axf, or, y + Ay = x^+2x-Ax + (^Axy. Subtracting (^), y =0? (B) ^ Ay= 2 a; ■ Az + (Axy we get the increment'Ay in terms of x and Ax. To find the ratio of the increments, divide (5) by Ax, giving ^=22; + Aa:. Ax If the initial value of x is 4, it is evident that limit Ay _ g Aa; = Aa: Let us carefully note the behavior of the ratio of the increments of X and y as the increment of x diminishes. Initial New Increment Initial New Increment; Ay value of X value of X Aa: value of y value of y Ay AX 4 5.0 1.0 16 25. 9. 9. 4 4.8 0.8 16 23.04 7.04 8.8 4 4.6 0.6 16 21. 1« 5.16 8.6 4 4.4 0.4 16 19.36 3.36 8.4 4 4.2 0.2 16 17.64 1.64 8.2 4 4.1 0.1 16 16.81 0.81 8.1 4 4.01 0.01 16 16.0801 0.089i 8.01 DIFFERENTIATION 27 It is apparent that as Ax decreases, Ay also diminishes, but their ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01 ; Ay illustrating the fact that — ^ can be brought as near to 8 in value as Ax we please by making Ax small enough. Therefore limit ^ ^ g * Aa; = Aa; * 28. Derivative of a functioii of one variable. The fundamental definition of the Differential Calculus is : The derivative ^ of a function is the limit of the ratio of the increment of the function to the increment of the independent variable, when the lat- ter increment varies and approaches the limit zero. When the limit of this ratio exists, the function is said to be differ- entiable, or to possess'a derivative. The above definition may be given in a more compact form symbol- ically as follows : Given the function (^) y=fi^\ and consider x to have a fixed value. Let x take on an increment Ax; then the function y takes on an increment Ay, the new value of the function being (E) y+Ay=f(x + Ax~). To find the increment of the function, subtract (^) from (JB), giving (0 Ay =f(x + Ax-) -fix). Dividing by the increment of the variable. Ax,- we get Ay _ f{x + A^)-f(x) ^ ^ Aa; ' Aa; The limit of this ratio when Aa; approaches the limit zero is, from our definition, the derivative and is denoted hy the symbol -^- Therefore , „ dy _ limit f(x-\-Ax)-f{x) defines the derivative of y[orf(x)'\ with respect to x. *The student should guard against the common error of concluding that because the numerator and denominator of a fraction are each approaching zero as a hmit, the limit of the value of the fraction (or ratio) is zero. The limit of the ratio may take on any numerical value. In the above example the limit is 8. t Also called the differential coefficient or the derived function. 28 DIFFEEENTIAL CALCULUS From CD) we also get ^ * dy^ limit Ay dx Ajr = OAj:' The process of finding the derivative of a function is called differ- entiation. It should be carefully noted that the derivative is the limit of the ratio, not the ratio of the limits. The latter raj:io would assume the form — , which is indeterminate (§ 14, p. 12). 29. Symbols for derivatives. Since Ay and Ax are always finite and have definite values, the expression Ay Ax is really a fraction. The symbol dy dv' however, is to be regarded not as a fraction but as the limiting value of a fraction. In many cases it will be seen that this symbol does possess fractional properties, and later on we shall show how meanings may be attached to di/ and dx, but for the present the symbol -— is to be considered as a whole. Since the derivative of a function of x is in general also a function of X, the symbol /'(a;) is also used to denote the derivative of /(a;). Hence, if ^ =/(a:), we may write — =f'(x'), which is read the derivative of y with respect to x equals f prime of x The symbol ^ dx when considered by itself is called the differentiating operator, and indicates that any function written after it is to be differentiated with respect to x. Thus ^ or — y indicates the derivative of y with respect to x ; ■^/(«) mdicates the derivative oif(x) with respect to x; — (2x'+ 5) indicates the derivative of 2a;^+ 5 with respect to x. dx dy y' is an abbreviated form of — . dx DIFFERENTIATION 29 The symbol D^ is used by some writers instead of — If then ax y =/(^)' we, may write the identities 30. Differentiable functions. From the Theory of Limits it is clear that if the derivative of a function exists for a certain value of the independent variable, the "function itself must be continuous for that value of the variable. The converse, however, is not always true, functions having been discovered that are continuous and yet possess no derivative. But such functions do not occur often in applied mathematics, and in this hook only differentiable- functions are considered, that is, functions that .possess a derivative for all values of the independent variable save at most for isolated values. 31. General rule for differentiation. From the definition of a deriv- ative it is seen that the process of differentiating a function y =f(x) consists m taking the following distinct steps : General Rxilb for Differentiation* First Step. In the function replace x by x + Ax, giving a new value of the function, y + i^y. Second Step. Subtract the given value of the function from the new value in order to find A?/ (the increment of the function^. Third Step. Divide the remainder At/ (the increment of the function) by Aa; (the increment of the independent variable). Fourth Step. Find the limit of this quotient, when Ax (the increment of the independent variable) varies and approaches the limit zero. This is the derivative required. The student should become thoroughly familiar with this rule by applying the process to a large number of examples. Three such examples will now be worked out in detail. Illustrative Example 1. Differentiate Sx^ + 5. Solution. Applying the successive steps in the General Rule, we get, after placing y = Sx^ + 5, First step. y + Ay = 3(x +Axy + 5 = 3x2 + 6a;.Aa; + 3(Ax)2 + 5. • Also called the Four-step Rule. 30 DIFFEEENTIAL CALCULUS Second step. y + A?/ = Sa;^ + 6a; • Ax + 3(Aa;)2 + 5 y =3x"- -l-_5 Ay = 6x-Ax + 3(Ax)^. Third stgj. :^ = 6 a; + 3 • Ax. Ax dy Fourth step. — = 6x. Ajis. dx We may also write this — (3x2 + 5) = 6x. iLLnsTRATivE ExAMFLE 2. Differentiate x' — 2x + 7. Solution. Place y = x^ — 2x + 7. First step. y + Ay = (x + Ax)" - 2 (x + Ax) + 7 = x' + 3 x2 . Ax + 8 X • (Ax)2 + (Ax)3 -2x-2.Ax + 7. Second step. y + Ay = x^ + Sx'' . Ax + Sx ■ (AxY + (Ax)" — 2x-2-Aa; + 7 y =x" — 2x +7 Ay= 3x2Tax+"3xT(Ax)2"4~(Ax)8 — 2 • Ax. Third stej). — = 3.^2 + 3 x ■ Ax J- (Ax)^ - 2. Fourth step. — = 3 x^ - 2. Ans. dx Or, — (x'-2x+'7) = 3x2-2. dx Illustrative Example 3. Difierentiate — x" Solution. Place y = — x^ Krsi step. 7j + Ay =: Second step. y + Ay = y = (X + Ax)2 c (X + Ax)2 c Aj/: c^ _ — c ■ Ax (2 X + Ax) (X + Ax)2 x2 x2 (X + Ax)2 Third step. ^=-c 2"= + ^^ Fourth step. Or, Ax x2 (X + Ax)2 #__ 2x (fe~ ' x2 (x)2 = r- • ^TIS. x" dx\xV xT' DIFFERENTIATION 31 EXAMPLES Use the General Rule, p. 29, in differentiating the following functions : 1. v = Sx^. . 2. y = x^ + 2. 3. y = 5- 4. s = 2i2. 4x. Ans. ^: dx dy dx dy dx :.6x. = 2x. = -4. 5. y = -■ X e.y = X + 2 dt dy dx dy dx ■ = it. 2^ ' x^' 7. y = x' 8. y = 2x^-3. 9. 2/ = 1-2x3. 10. p=a9^. 11. 1/ 12. 2/ Ans. ^-. dx 3x2 2^ x2 x2_i ^2/ , -i = 4x. dx ^=-6x^ dx dy ___4 cfcc x^ dy dx 6x 13. y = 7 x2 + X. 14. s = aP-2bi. 15. r = 8< + 3i2. 16.. = i. 17. s = -- 18. y = bx^ — ex. 19. p = 3^3 -2^2. 20. 2/ = fx^ — Jx. . 21. y = X 22. p: (X2 - 1)2 23. y = ix2^2x. 24. z = 4x-3x2. 25. p = Z9 + e^. ax + h 26. 2/ = : 1+^ 27. 2 = x'' x3 + 2 Ans. y' = 2x — 3. s' = 4i + 5. p' = 156l2-2. y' = 2 ax + 6. 2i + 3 28. 2/ = x2-3x + 6. 29. s = 2i2 + 54-8. 30. p = 56^-20+6. 31. 2/ = as'' + 6a; + c. 32. Applications of the derivative to Geometry. We shall now consider a theorem which is fundamental in all applications of the Differential Calculus to Geometry. Let 3, (A) 2/=/(^) be the equation of a curve AB. Now differentiate (^) by the G-eneral Rule and interpret each step geometrically. First Step. Second Step. Third Step. y + ^y =fQc + Aa;) y + £^y=f(x-ir^x) y =f(P) A^ =f(x + Ace) -fix) = BQ. A.y _ fix + ^x•) -f(x) ^RQ _BQ Ax Aa; MN PB = tdiXL BP Q = tdjx (\> = slope of secant line PQ. 32 DIFFERENTIAL CALCULUS Fourth Step. (-B) limit A^^ limit /(a: + Aa;) — /(:t / Aa; = Aa; Aa; = Aa; = -^ = value of the derivative at P. dx But when we let Aa; = 0, the point Q will move along the curve and approach nearer and nearer to P, the secant will turn about P and approach the tangent as a limiting position, and we have also "°^t ^=,^°^* tan V -^v. V By the General Rule, u + Am First Step. «/ + A«/ = Second Step. A?/ = u + Am u _v- Am — u-^v V +^v V v(v + Aw) Third Step. Am Av V u — Ay _ Ax Ax Ax v(y + Av') du dv Fourth Step. c^y "^ dx "" dx dx v^ [Applying Theorems II and III, p. 18.] du dv V u — d /u\ dx dx VII ••• — - = 1 dx\v/ v" The derivative of a fraction is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. When the denominator is constant, set w = c in VII, giving du d /u\ dx Vila _ _ =_. dx \c/ c rsinoe*=* = 0.1 L dx dx 1 We may also get VII a from IV as follows : du d /m\ _ldu _dx dx\c I c dx The derivative of the quotient of a function hy a constant is equal to the derivative of the function divided hy the constant. All explicit algebraic functions of one independent variable may be differentiated by following the rules we have deduced so far. RULES FOE DIFFEEENTIATING 41 EXAMPLES* Differentiate the following : 1. y = a;'. Solution, f = £W = 3X^ Ans. By VI a [» = 3.] 2. y = ax^ — bx^. = 4aa;S — 26s. 4ns. By Via 3. y = xi + 5. ''■'"*^"- l = |("'> + |(') i^yin = |si Ans. By VI a and I = ^xt + |x H 2^x-^. Ans. By IV and VI a 5. 2/ = (i2 - 3)5 dx dx Solution. — = 6(a;2-3)* — (x2-3) bv VI [w = a;2 - 3 and n = 5.] = 5(x2-3)*.2s = 10x(x2_3)4. Ans. We might have expanded this function by the Binomial Theorem and then applied III, etc., but the above process is to be preferred. 6. 2/=Va2_x2. Solution. — = — (a2-z2)* = i(o2-x2)-* — (a2-i2) by VI [j; = a' - x^ and n = }.] = l(a2_j.2)-J(_2x) = ^ Am. 2 Va2 - x2 7. 2^ = (3x2 + 2)Vl + 5l2. Solution. — = (3 x2 + 2) — (1 + 5 x2)* + (1 + 5 x^)*-^ (3 x^ + 2) by V dx dx dx [u = 3x^ + 2, and » = (1 + 5 a;^)^] = (3x2 + 2)1(1 + 5x2)-i — (1 + 5xn + (i + 5x2)*6x by VI, etc. 2 dx = (3x2 + 2)(l + 5x2)"*5x + 6x(l+ Sx^)* 5x(3x2 + 2) „ n r-J 45x3 + 16x = — ; ' + 6xVl + 5xg = ~ Ans. Vl + 5x2 Vl + 5x2 *'When learning to differentiate, the student should have oral drill in differentiating simple functions. 4-2 DIFFEEENTIAL CALCULUS a' + ifl ■y/a" - x2 (0= - a;2)* i. (a2 + a;S) _ (a« + a^) — (a^ - s^)* (2x cia; 8. y: Solution, -i = by VII dx a' — x^ _ 2x{a^ - x^) + x{a^ + x^) [Multiplying both numerator and denominator by (a? — a:*)*.] 3 aH - x' (a2 _ a;2)i 4jis. 9. 2/ = 5x* + 3a;2 _ 6. -^ = 20a;3 + 6a;. cZx 10. y = 3cx^-8dx+ 6e. — = 6cx—8d. dx 11. 2/ = x'+i: ^ = (a + 6)s'' + i'-i. (ix 12. y = x" + nx + n. — = 7ia;»-i + n. ax 13. /(z) = |x' - I x" + 5. /'(x) = 2x2 - 3j._ 14. /(x) = (a + 6)x2 + ex + d. f{x) = 2 (a + 6)x + c. 15. —(a + bx + ex') = 6 + 2cx. 21. — (2x' + 5) = 6x2. dx dx 16. ■^{5r'-3^ + 6) = 5m2/"— 1-3. 22. — (Si^ - 2«2) = 15<< - 4<. ay dt 17. -^(2x-2+3x-3) =-4x-3-9x-*. 23. ^(ad^ + 6i9) = 4a(?8 + 6. 18. |-(3.s-'-s)=-12s-''-l. 24. — (5-2a2)=_3ar2. 19. -^(4x* + x2) = 2x"* + 2x. 25. — (9tt + i-i) = 154! _ ds 'da ,-_ ,_, , .„. — --vJ«^ + «-i) = 15if-<-^. dx di 20. ^(y-^-iy-^ = -2y-'' + 2y-l. 26. -^(2xi2 - x') = 24x" - 9x8 d2/ dx ' 27. r = 06*8 + d^2 ^. g|9_ ^ V'=3ce2 + 2de + e. 28. 2/ = 6x^ + 4xJ + 2xi j/'=21xi + 10xi +3xi 29. 2/=V3x + v^ + i. ^'=__!_. + . ^ 1 X 2V3x 3v^ a;'* a + 6s + cx2 a X " X2 (x — V\^ 31. 2/ = i — -i-. y'=|xJ-5xt + 2i-s- + J x"J. X - X - x^ + a , 2x^+x + 2x^-3c 32. y = "'-----^" . ,'=: ^^ ■ 2xi 33. 2/ = (2 xS + x2 - 5)3. ,/= 6 X (3 z + 1) (2 x^ + x2 _ 5)2, EULES FOR DIFFERENTIATING 43 34. f(x) = (a + 6x2)i f{x) = ^ (« + to^)! . 35. f(x) = (1 + 4x3) (1 + 2x2). f\x) = 4a;(l + 3a; + 10s'). 36. /(i) = (a + a;) Va — x. 37. /(s) = (a + x)'»(6 + x)». 38. y = -- a?" 39. 2/ = X (a2 + x2)Va2.-x2. 40. Differentiate the following functions (a) — (2xS-4x + 6). (b) I(at7 + 6J5_9). at (c) A(3(9i-26'H6e). etc/ (d) |-(2x' + x)i ax A-> 2x* 41. y 42. y = 43. s'= f'ix\ ""^"^ 2Va — X fix) = (a + x)'»(b + dy n dx x"+i X)" + ; la + x b + x. dy a* + a2x2 _ 4 :X4 dx .Va2 — x2 (e) |(6 + «f')^- (i)|(xt-at). (f)£(^=-«''P- (i)|(5 + 2t)i. (g)^(4-0^. (k) AVa + fiVi. da (1) ^(2x^2x1) dx (h)|Vl + 9t2. dy 862x8-4x5 62 - i2 dx (62 _ x2)2 a—x dy 2 a a + X dx (a + x)2 (x+Ax)~(x) Av Av Ax Ax EULES FOE, DIFFERENTIATING 4 The left-hand members show one form of the ratio of the incremen ,- of each function to the increment of the corresponding variable, and the right-hand members exhibit the same ratios in another form. Before passing to the limit let us form a product of these two ratios, choos- ing the left-hand forms for this purpose. This gives — 5 which equals — • Av ^x Ax Write this ^ = ^-~ Ax Av Ax Fourth Step. Passing to the limit. This may also be written ^^^y.dv Th.II,p.l8 ^ ^ dx dv dx (B) ^=f'(V)-i>'(x). dx If y =f(v') and V = 4> (of), the derivative of y with respect to x equals the product of the derivative of y with respect to v and the derivative of v with respect to x. 43. Differentiation of inverse functions. Let y be given as a function of x by means of the relation y —f(x'). It is usually possible in the case of functions considered in this book to solve this equation for x, giving x = (y'); that is, to consider y as the independent and x as the dependent variable. In that case f(x) and ^ («/) are said to be inverse functions. When we wish to distinguish between' the two it is customary to call the first one given the direct function and the second one the inverse function. Thus, in the examples which follow, if the second members in the first column are taken as the direct functions, then the corresponding members in the second column will be respectively their inverse functions. y = x^ + l, a;=±Vy-l. «/ = a% x = log„y. y = sm.x, x = arc sin y. Let us now differentiate the inverse functions y=f(x) and x = ^(y') simultaneously by the G-eneral Rule. 4 46 DIFFERENTIAL CALCULUS First Step. y+Ay=f(x+Ax) a: + Aa; = (^ («/ + At/) Second Step. y+Ay=f(x+Ax) a; + Aa; = («/ + A«/) .y =/(«:) ^ =4>iy') Ay=f(x+Ax)-f(x)' Ax = (i/+Ai/-)-{y) thxkbStep. %^ /C^+a;)-/(-) , A.^ K,y+Ay)-Ky) Aa; Aa; Ay Ay Taking the product of the left-hand forms of these ratios, we get Ay Aa; _ ^ Aa; Ay or, ^=J_. Aa; Ax ^y FouitTH Step. Passing to the limit, dy .1 dy or. The derivative of the inverse function is equal to the reciprocal of the derivative of the direct function. 44. Differentiation of a logarithm. Let y = log„v.* Differentiating by the General Rule, p. 29, considering v as the independent variable, we have First Step. y + Ay = log„ (y + A«). Second Step. Ay = log„(v + Av') — log„v t •j -H.("-±^)..o,.(x+^). [By 8, p. 1.] * The student must not forget that this function is defined only for positive values of the base o and the variable v. ^ t If we take the third and fourth steps without transforming the right-hand member, there results: Third step. ^ = log<.(» + A«)-log„» At) A« fourth step. -^=7.' ^'^'"'^ '^ indeterminate. Hence the limiting value of the right-hand member in the third step cannot be found by direct substitution, and the above transfor- mation is necessary. RULES rOR DIFFERENTIATING 47 V = -log fl + — .• V X V / rDmding the logarithm by v and at the same time multiplying the exponent of the] . [parenthesis by v changes the form of the expression but not its value (see 9, p. !)■ J Fourth Step. — = — loff.e, dv V .,. V [ When Ao, = 0, — = 0. Therefore J'"" n (^ '*' ~~) ^^ = «. *roni P- 22, placing a; = — • 1 Hence d d / \ 1 Since v is a function of x and it is required to differentiate log„a with respect to x, we must use formula (A), § 42, for differentiating a function of a function, namely, dy _dy . dv dx dv dx Substituting value of -^ from (J), we get dy , 1 dv -2. = log„e — • dx V dx A; d ^, s , dx vm .-. — (log„i') = log<,e dx V When a = e, log„e = log^e =1,' and Till becomes dv^ d ,-, . dx Villa — G°Si') = — dx V The derivative of the logarithm of a function is equal to the product of the modulus * of the system of logarithms and the derivative of the function, divided by the function. * The logarithm of e to any base a (= logae) is called the modulus of the system whose base is a. In Algebra it is shown that we may find the logarithm of a nmnber N to any base a by means of the formula loga N= loga e • loge iV= ^^ • The modulus of the common or Briggs system with base 10 is log,oe=. 434294 ■••. 48 DIFFERENTIAL CALCULUS 45. Diiferentiation of the simple exponential function. Let «/ = *"• a > Taking the logarithm of both sides to the base e, we get \ogy = v\oga, log V or, V = —2-^ log a Differentiate with respect to y by formula Villa, *! = _!_ i- dy log a y and from (C), § 43, relating to inverse functions, we get dy or. ^^ = loga.y, Since w is a function of x and it is required to differentiate a" with respect to x, we must use formula (.4.), § 42, for differentiating a function of a function, namely, dy _ dy dv dx dv dx Substituting the value of -^ from (^), we get dv , , dv -~ = log«- a" dx dx IX .-. — (a") = log a a"- — . dx^ -" dx When a = e, log a = log e = 1, and IX becomes II a —(€"') = e«—. ■ dx^ ^ dx The derivative of a constant with a variable exponent is equal to the product of the natural logarithm of the constant, the constant with, the variable exponent, and the derivative of the exponent. RULES FOR DIFFEREKTIATING 49 46. Differentiation of the general exponential function. Let -" —- »'" * Taking the logarithm of both sides to the base e, log,i^= V log.M, or, y = e';'"^". DifEerentiating by formula IX a, = m" - -— + log M-— • \u ax ax I I .-. — (a'') = iw-^ t-logu-u" — dx dx dx The derivative of a function with a variable exponent is equal to the sum of the two results obtained by first differentiating by VI, regarding the exponent as constant; and again differentiating by II, regarding the function as constant. Let v = n, any constant ; then X reduces to d , ^^ „ ,du —-(u") = nu''-^-^- dx dx But this is the form differentiated in § 40; therefore VI holds true for any value of n. Illustrative Example 1. Difierentiate y = log(x2 + a). — {x' + a) Solution. . ^^dx by VIII a dx x' + a [v = afl + a.] 2x x^ + a Ans. Illusteative Example 2. DifEerentiate y = logVl — x^^. - (1 - x2)^ Afl-x=l* Solution. $- = — by VIII a dx (1 _ 3.2)J i(l-xif)-*(-2x) (1-X2) X _X2)* Atis. by VI X2-1 • u can here assume only positive values. 60 DIFFEJtEXTIAL CALGLTLUS Illlstkative Example 3. Differentiate y = a'^. ' Solution. — = logaa'"^— (Sa^) by IX dx 4 ). a^ = i°g°«- 2^'^a.3 - 7. v = xloga. V'=logx + l. 8. /(x) = logx'. ■^'^^^ = x' 8 loe^ X 9. /(x) = log« X. f(x) = — I — Hint. logS z = (log x)s. Use first VI, « = log x, re = 3 ; and then VIII a. 10. /(x) = log^!^- /'(a:) = dv 1 dx X + a dy a dx ax + b dv ix dx 1-x* y' = 2x + l x^ + x tt' 3x2- -2 x'-2x + 5 71' rz : loff„ e • - 2 + 3x2 a — X a-' — x'' 1 11. /(x) = log (X + Vl + X2). /'(x) = Vl + X^ 12. — e<« = ae»^. 17. — e !■'+»='= 2 xe»'+=''. dx dx 13. Ae4:>: + 6 = 4e4l+6. dx d 14. —081 = 3 aS'' logos, dx 15. ^log(3-2i2) = ^ ** 18. — oi°K« = -ai°B«loga. 19, -6»==2slog6.6='^ ds 20. d» 2V» di ' ' 2 £2- 3 16. ^ log lil^ = -A^ . 21. — a'' = logo ■ a^ • fF. dy ^1-y 1-y^ dx ^ 22. 2/ = 7^' + 2x. y' = 21og7.(x + 1)7^ + 21. 23. v = c"-"^. V'=-2xlogo-c'''-^. 24. v = iog-^^ ^y- ^ 1 + e=^ dx 1 + e=^ 25. — [e=t(l-x2)] ='e^(l-2x-x2\. dx 28. a - -- 39 (f— e-» gL^e-x 30. y = x»a^. 31. y = x='. 82. 1 y = x?'. 33. y = slog":. 34. f{y) = \osy-ey. 35. ns) = y. J \ ■^1 — '"5 \^ + 1 + X Hint. First rationalize the denominator. 1 41. y = xi°s='. 42. y = e^. 43. y 44. y ^r 45. w = rfi\ 46. e =(!)■• 47. y = a^. 48. y 1 EULES FOE DIFFEEENTIATING 53 d^ _ 4 dX ~ (gl + e-a:)2 ' y'= a=^»-i{n+ xloga). y' = x^(loga; + l). 1 x^O-Jogx) 2/' = logx^ • xios":-!. /'(!/) = e» /log 2/ + -V 1 — s log s 36. /(x) = log (logs). /'(a;) = 1 xlogx 37. F(x) = log^aogx). F'{z) = iMil££^. xlogx 38. 0(x) = log(log4x). ,^'(x) xlogx 39. ^ (2/) = log ^~- ■ ^ '(y) = ^ 1-2/2 Vl + X2 49. y = a' — /yVaa — x2 ^ = 0. dx — = e»^(l + logx)x»'. dx V T^ = ' dx \x/ \ X / s="(r(--;)- d» \ V / f.gyaoga-log.-l). ^ = x*"+"-i(™logx + l^. dx J^ = x'^x': (log X + log2 X +- ) dx \ a;/ dj/ _ xy log a <^ (a2-x2)5 54 DIFFERENTIAL CALCULUS 60. Difierentiate the following functions : (a) ^aj'logx. (f) ^^e'loga;. (k) i- log (a» + 6*) ax ax ax (b) 4-(^'- !)*■ fe) T "^ ^- W r ^°Sio (** + 5 '') ox ox ox (c) —log ^— . (h) ; (m) ^ ' dx ^x + 3 ^'dxxlogx ^ ' dx e""^ (d) i.logA^£.. (i) i-logx«Vl + x«. (n) |-(x2 + a2)e-' + «-. <^ Vl + x "* '"^ = (^ + 1)' ^ =: _ (z + l)(Sx' + 14x + 5) ■ ^'^(x + 2)»(x + 3)«' dx (x + 2)*(x + 3)« Hint. Take logarithm of both sides before differentiating in this and the following examples. , , _ (X - 1)8 dy _ (x-l)a(7x'+30x-97) (x-2)i{x-3)* ^ 12(x-2)J(x-3/i CO A /i , X ''2' 2 + x — 5x' 63. y = xvl — x(l + x). — = — . <^ 2VI-X x(l + x2) d2^ H-3x2-2x* 64. y = - . — = Vl-x2 <^ (l-x2)t 55. y = x5(a + 3i)»(a-2x)2. ^ = 5x*(a + 3x)2(a-2x)(a2 + 2ai-12x2). 48. Difierentiation of sin v. Let y = sin ti. By General Rule, p. 29, considering v as the independent variable, we have First Step. y + Ay = sin (w + Av). Second Step. At/ = sin (v + Av) — sin v* o / , Av\ . Av + = 2 cos / 1) +— \ • sm— .T *If we take the third and fourth steps without transforming the right-hand member, there results : Third step. ^ = «'°(» + A")-sin« . , Ac Av Fourth step. ^^7,> winch is Indeterminate (see footnote, p. 46). tLet A = v + Av A = v + Av and B^v B=v Adding, A + B^iv + Av Subtracting, A-B = Av Therefore -{A + B)=v + —- -(A-B-i^—. 2' ' 2 2^ ' 2 Substituting these values oiA,B,i{A + S),HA-B)hi terms of » and A» in the formula from Trigonometry (42, p. 2) , sin 4 - sin B =• 2 008 1 (-4 + iB) sin i (^ - B), weget sin(j; + Aii)-8in«=.2cos (u + — I sin — • \ 2 / 2 RULES FOR DIFFERENTIATING 65 (sin— ^ Av 2 Fourth Step. -^ = cos v. av s™«« l'"Jo(-Sr-) = ''''y522.p.21. and lSoo8(« + f )-co8 .. Since t» is a function of x and it is required to differentiate sinw with respect to x, we must use formula (A"), § 42, for differentiating SL function of a function, namely, di/ _ dy dv dx ' dv dx Substituting value -j- from Fourth Step, we get dy dv -^ = cost)-— • dx dx d dv XI .*. — Csin w) = cos » — dx^ ^ dx The statement- of the corresponding rules will now be left to the student. 49. Differentiation of cos i;. Let y = cos v. By 29, p. 2, this may be written y = sin/|-A Differentiating by formula XI, dy Iv \ d lir \ = eos(|-.)(-|) dv t — smv-;— dx rsJnoe coa^j-o J-slno, by 29, p. a I d do m .*. — Ccosw) = — sm»— • dx^ ^ dx 56 DIFFERENTIAL CALCULUS 50. Differentiation of tan v. Let y = tan v. By 27, p. 2, this may be written sint; w = . cosw Differentiating by formula VII, c? , . . . d . , cos V — - (sm v) — 8m.v-—- (cos v) dy _ dx^ ^ dx^ ^ dx cos^w „ dv , . „ dv cos V -- + sm « -— dx dx cos^w dv dx dv - = sec t) — . COS^l ; dx .] = sec2 Vl — a; • i (1 — x)"*(— 1) sec" Vl — a; 2 Vl - a" S. y = cos'a;. This may also be written y = (oosz)'. — = 3 (cos x)^—- (cos X) by VI ax ax [» = cos a; and » = 3.] = 3 cos^a; (— sin x) by XII = — Ssinxcos^x. 4. y = sin nx sin'x. dy . d . . d — - = sin nx — (sm x)» + sin»x — (sm nx) " by V ox ax dx [«= sinna; and v= sin"^.] = sin nx ■ n (sin x)» -i — (sin x) + sin»x cos nx — (nx) by VI and XI dx dx = n sin nx • sin" -1 X cos X + n sin''x cos nx = n sin« -1 X (sin nx cos x + coS nx sin x) = nsin»-ixsin(n + l)x. e dy o. y = sec ax. Ans. — ■ = a sec ox tan ax. ' (ix d?/ 6. y = tan (ax + 6). — = a sec^ (ox + 6). dx ds 7. s = cos 3 ox. — =— 3a sin 3 ax. dx dx 8. s = cot(2i2 + 3). —--4tcsc^{2t^ + 3). 9. /(y) = sin2y cos2^. /'(j^) = 2 cos22/ cosy — sin2y siny 10. J'(x) = cot25x. f"(x)=-10cot5xcsc2 5x. 11. F(e) = tB.n0-d. J"(fl) = tan'tf. 12. /(0) = sin + cos 0. /' (0) = cos 0. 13. /(O = sln»«cos«. /'(«) = Bin=i(3cos2<-sina<). dr 14. r = aoos2^. — -=— 2asin2S. dff 60 DIFFERENTIAL CALCULUS i-fi., .„ „„ d a a . a 15. — sin^x = sin2a;. 23. — cos- = — sin-- dx dt t V' t 16. — cos=a;2 __ ea;(;og2x2sinx2. , 24. —sin — = — — cos — . dx dff e^ e^ e^ d t^ f- i^ d 17. — CSC - = — i cso — cot — . 25. — e '""^^ = e ™»' cos x. dt 2 2 2 dx io <^ / TT asm2s no ^ ■ n s cos(loga;) 18. — avcos2s = • 26. — sm(loga) = • ds Vcos2s ^"^ * 19. Aa(l-cos(9) = asin(9. 27. — tan(logx) = ^^°'''°^^\ d9 dx ^ ' X 20. — (log cos x) =— tan x. 28. — a sin^ - = a sin^ - cos - dx^ ^ ' (to 3 3 3 21. — (log tan x) = -^ 39. — sin (cos or) = — sin a: cos (cos a) dx sin2x da — (log sin^x) = 2 cotx. 30. — - dx dx sec x no "^ /I ■ ■> \ n i n/v -b<:o!.t_ dx dx dt (b) - cos (o - 6s). (g) A sin3 2 x. (1) ^ sin - cos^ - • ^•^^ £'""?■ (h)|cosMlogx). (-)|<=°t^- (d)— cotV^. (i) |-tan2Vl-x2. (n) Avi + cosV- "X ox dd> dx"- ■ ^■" ^-sv-"-;. w-1 (e) seoes-. (j) |-log(sin2ax). (o) ^logVl- 2sin2s. 38. — (x'e""^) = x"-iesmi(„ ^. j. cosx). dx dx ~ 39. ^ (e<^ cos mx) = e<" (a cos mx — to sin mx) . 40./«?) = l±^. ^.(^)^ 2sin^ 1-cosS -^ ^ ' (l-cos(9)2 ^1 j-/.\ e»* (a sin — cos 0) •^(*) = a^ + i -■ f'(t>) = ^ sin 0. 42. /(s) = (s cot s)2. /'(s) = 2 s cot s (cot s - s csc^ s). 43. r-ita,n^g-ta.nd + EULES FOE DIFFEEENTIATING dr 61 dO ■ tan* e. 44. y = x""^. 45. y = (sin a;)":. 46. y = (sin k) '«■"■. -« -r. f^ . dv 47. Prove — cosu = - sin b — , using the General Rule. ux dx 48. Prove — cot d = - osc^ « -^ by replacing cot v by °°^-" . ax ax sin » 55. Differentiation of arc sin v. Let «/ = arc sin v -, then w = sin y. Differentiating with respect to y by XI, dy .;„,/sina; , , \ y' = (sin o;)^ [log sin a; + x cot x] . y' = (sin x)>^^ »(1 + sec^ x log sin x). .* therefore dv -— = cos y : dy ^■ dy_^ 1 c?v cos ?/ By (C), p. 46 But since v is a function of x, this may be substituted in dy _ dy dv dx dv dx {A), p. 45 giving dy _ \ dv dx cos «/ dx 1 (?« Vl— t)^ dx [cosy="\/l — sin2^ = v'l — v^ the positive sign of the radical being taken,"] since cos 2/ is positive for all values of y between — ^ and ^ inclusive. I A; IVIII . • . — (arc sin v) = dir Vl- z/" * It should be remembered that this function is defined only for values of v between — 1 and + 1 inclusive and that y (the function) is many-valued, there being infinitely many arcs whose sines all equal v. Thus, in the figure (the locus of ^= arc sin u), when v= OM, y = MPi, MP^, MP^, ■■■, MQi, MQ2, In the above discussion, in order to make the function single- valued, only values of y between - - and - inclusive (points on arc QOP) axe considered ; that is, the arc of smallest numerical value whose sine is v. 62 DIFFERENTIAL CALCULUS 56. Differentiation of arc cos v. Let y = arc cos V ; then v = cos y. Differentiating with respect to y by III, dv dy ^^ 1 dv --—smy; therefore By (C), p. 46 But since formula givmg sm^ w is a function of x, this may be substituted in the dy _ dy dv dx dy dx dv dx 1 (A), p. 45 sm.y 1 dv dx dv ^/l — v^dx XIX [sin y =^1- cos* y =Vl- v\ the plus sign of the radical heing taken, since sin y is positive for all values of y betveen and n- inclusive. dv d ^ , dx .". — Care cos »)= • dx^ Vl-ir" Differentiation of arc tan v. y = arc tan v ; ^ V = tan y. Differentiating with respect to y by IIV, dv „ — = sec'y; dy :] therefore ^ = dy^ 1 c?v sec^t/ By (C), p. 46 * This function is defined only for values of « between -1 and +1 inclu- sive, and is many-valued. In the figure (the locus oiy = arc cos a), when V = OM, y = MPx, MP2, ■■■, MQp MQ2, ■■ ■ ■ In order to make the function single-valued, only values of y between and T inclusive are considered; that is, the smallest positive arc whose cosine is v. Hence we confine ourselves to arc QP of the graph, t This function is defined for all values of v, and is many- valued, as is clearly shown by its graph. In order to make it single-valued, only values of y between - — and yr are con- sidered ; that is, the arc of smallest numerical value whose tangent is v (branch AOB). Y B 2 ^^ 5 —7^ 0_ F liULEiS FOE, DIFFERENTIATING . (i;j But since w is a function of x, this may be substituted in the formula giving ^=J_.*: dx sec'i/ dx 1 dv XI 1 + v^dx [sees J, = 1 + tau2 y = 1 + j;!.] A; .. — (aictanv)= dx^ ^ l + ir" 58. Differentiation of arc cot v.* Following the method of the last section, we get dv XXI '^ , * N dx — (arc cot V) = dx^ ^ l+v" 59. Differentiation of arc sec v. Let y = arc sec v ; ^ then V — sec y. *■ This function is defined for all values of a, and is many-valued, as is seen from its graph (Fig. a). In order to make it single-valued, only values of y between and tt are considered ; that is, the smallest positive arc whose cotangent is v. Hence we confine our- selves to branch AB. =-ii;:|j v,__ r~, <; ^ B "^ V > — -2ir Fig. a Fig. & t This function is defined for all values of v except those lying between -1 and +1, and is seen to be many-valued. To make the function single-valued, y is taken as the arc of smallest numerical value whose secant is v. This means that if v is positive, we confine ourselves to points on arc AB (Fig. V),y taking on values between and ;?■ (0 may be Inclnded) ; and if » is negative, we confine ourselves to points on arc DC,y taking on values betrreen - ir and - -^ (- IT may be included) . 64 DIFFEEENTIAL CALCULUS Differentiating with respect to y by XV, dv -z- = sec y tan y ; dy therefore dy _ By (C), p. 46 dv sec y tan y But since « is a function of x, this may be substituted in the formula (^), p. 45 givmg nil dy _ dy dv dx dv dx dy _ 1 ^ dx sec y tan y dx 1 dv v^/v^ — l dx sec y = », and tan )/ = Vsec^ j/ - 1 = Vj)2 - 1, the plus sign of the radical heing taken, since tan y is positive for all values of y between and — and between — ir and > including and — tt. do d . , Ix .-. — (arcseci;) = — • dx vy/v^-l 60. Differentiation of arc esc v.* Let then y = arc esc v ; V = CSC y. Differentiating with respect to y by IVI and following the method of the last section, we get dv mil — (arc CSC v) = , • dx v\'v^ - 1 * This function is defined for all values of v except those lying between -1 and +1, and is seen to be many-valued. To make the function single-valued, y is taken as the are of small- est numerical value whose cosecant is v. This means that if v is positive, we confine ourselves to points on the arc AB (Fig. a), y 2[l may be included | ; and -HIT Fis. a taking on values between and if V is negative, we confine ourselves to points on the arc CD, y taking on values between - ir and - t (- — may be included). EULES FOR DIFFEKEXTIATIXG 61. Differentiation of arc vers v. Let y = arc vers v ; * then V = vers «/. Differentiating with respect to y by IVII, dv - = sm2,; ^ = J_. cZ?> sin «/ 65 therefore By (C), p. 46 ;he formula (^), p. 45 But since v is a function of x, this may be substituted in the formula dy _ dy dv dx dv dx dy 1 dv givmg -^ = — dx sin y dx _ 1 dv ["sin !/- Vi _ cos2 y = Vi-(i- vers j/)2 = V2 «-»!!, the plus sign of the radicall Lheing taken, since sin y is positive for all values of y between and ir inclusive.! — dx IXTF — Care vers v) = , V EXAMPLES Differentiate the following : 1. y = arc tan ax^ Solution. Solution. dy £<"■' dx 1 + (ax^Y ')• 2ax -4x 1 + aH*' dy_ d dx (3 a;- 4a;S) dx vr -(3x- -4a:8)2 [»= = 3a:-4a;s.] 3- 12 l2 Vl- 9x2 + 24a;* -16x6 Vl-x^ * Defined only for values of v between and 2 inclusive, and is many- valued. To make the function continuous, y is taken as the smallest positive arc whose versed sine is v ; that is, y lies between and ir inclusive. Hence we confine ourselves to arc OP of the graph (Fig. a) . by XX by XVIII ZT Tv 60 DIFFERENTIAL CALCULUS , x^ + 1 3. V = arc sec d /x2 + 1' \) dy dx V — 1/ !,■„ TVTT Solutjon. -^ = , = Dy XXII " I^V{l^)"-' r a;2 + l 1 (a2-l)2a;-(x2 + l)2a; (X2 - 1)2 x2 + l 2x S2_l x2_l ^ d . X = 1 9. '^ dx, a Va2 - x2 (tc x2 + l 9. — arctanVl— x = — 2 VI -X (2 -I) = "^ w 2 =. -2x ,. d 3 2 5. — arc cot (j? — 6) = 10. — arc cosec — = - dx l + (x2— 5)2 dx 2i VQ — 4x2 „ d , 2x 2 „ d . 2x2 2 6. — arctan = 11. — arcyers = dx l-x2 l + x2 dx l + x2 l + x2 „ d 1 2 ,„dxo 7. --arccosec — - — — ■ 12. — arctaii- = dx 2 x2 — 1 yx_ -53 dx a a2 + x' 8. — arc vers 2 x' — — 13. — are sin- — *= Vl-x^ *; V2 Vl-2x- 14. /(x) = X Va2 - x2 + a2 arc sin - . /' (x) = 2 Va^ — x^. 15. /(x) = Vo2 - x2 + a arc sin - . /' (x) = /^^-II^^* . a ^ ' \a + x/ 16. X = r arc vers- — V2n/— i/2. x^ 17. tf=arcsin(3j- — 1). r + a — c 19. s = arc sec _ Vl-t2 di VT^ 20. — (x arc sin x) = arc sin x + - ^ 18. = arc tan - 1 — ar dx 2/ dj/ V2ry- -2/2 de_ 3 dj- V6r- 9r2 d0 1 dr l + r2 d«_ 1 *= vr •x= 21. ^ (tan e arc tan 6) = sec2 « arc tan 9 + *^"^ . dS 1 + ^ 22. - [log (arc cos «)] = ^ . ™ arccosiVl — f2 23. /(2/) = arooos(Iogy). /'(y) =- . 2/Vl-(logy)S! 24. /{#) = arc sin vWe. /' (g) = ^ Vl + csctf. RULES FOE DIFFERENTIATING 67 25. /(0) = arc tan^ /l-cos» f, /w,\ = 1 . \l + cos0' ^ ' 2 26. p = e>"=t«'7. dy^ e-^f-? -, . e" — e- " du 2 27. u=arctan- 2 dv e" + e-' e'— e-' ds 2 28. s = arc cos e«+e-* dt e'+e-« 29. y = a;a«Bma:. y, ^ 3..„ri„^ /aTcsinx ^ logz \ V a; Vl - a;2/ 30. j/ = e^arctana. 2/' = e^ ^ + a;"^ arc tan x(l+ logs) 1 31. y = arc sin (sin a;). y' = \. on ^ 4 sins 4 32. y = arc tan 3 + 5cosa; 5 + 3cose 33. y = arc cot- + log, / ^-" . y' = _if^ . a; \x + a a^ - a* 34. y = log(- 1 arc tan a;. \1 — a;/ 2 a' S' = l-aji 35. 2/ = Vl -x" arc sin a; - a;. ^, ^ _ a: arc sing Vl-a;2 36. Differentiate the following functions : (a) -— arcsin2a;2. (f ) — i^ arc sin - . (k) — arcsinVl — y==. ax at. Z dy (b) — arctana^s. (g) —e^" «'■'»'. (1) — arc tan (log 3 az). da; dt dz (c) — arc sec - ■ (h) — tan d>^ ■ arc tan (hi. (m) — (a^ + s^) arc sec - . da; a ^ ' dif ^ ds^ ' 2 ,-,, d •,., d .8 , ^ (^ i2a (Q) — a; arc cos a;. (i) — arc sin a . (n) — arc cot ^ ' da; ^' d9 ^ ' da 8 (e) — a;2 arc cot eta;. (j) — -arctan Vl + 8'^. (o) — Vl — i^ arc sin t. da; d9 dt Formulas (A), p. 45, for differentiating a function of a function, and (C), p. 46, for differentiating inverse functions, have been added to the list of formulas at the beginning of this chapter as XXV and XXVI respectively. dy dv In the next eight examples, first find — and — by differentiation and then dv dx substitute the results in ^ = ^.^ by XXV dx dv dx to find ^' da; * As was pointed out on p. 44, it might be possible to eliminate v between the two given expressions so as to find y directly as a function of x, but in most cases the above method is to be preferred. 68 DIFFERENTIAL CALCULUS In general our results should he expressed explicitly in terms of the independent variable : that is, — in terms of x, ■— in terms of y, — in terms of 0, etc. dx dy dff 37. 2/ = 2»2 — 4, M = 3a;2 + 1. — = iv; — = 6x; substituting in XXV, dv dx ^ = 4v-6x = 2ix(3x' + l). dx 38. J/ = tan 2b, D = arctan(2x — 1). ^ = 2 sec2 2 1! ; — = ; substituting in XXV, dv dx 2x^-2x + l dy _ 2 sec2 2 » _ tan^ 2 d + 1 _ 2x^-2x4-1 da~2a;2_2a; + l~ 2x^-2x + l~ 2{x-xY [2a: — 1 1 Since 11= arc tan (21 — 1), tanti = 2a;-l, tan2» = ^ — —•> -i ' 39. 3/ = 3 1)2 - 4 B + 5, « = 2 a;3 - 5. -^ = 72 a;^ _ 204 a;2. 40. y = dx 2 1) a; dy 3»-2 2x-l (Ja;(x-2)2 41. y = log (a'' — i)^), B = a sin X. — = — 2 tan x. (2x dy e^ 42. y = arc tan (a + v),v = e^. dx 1 + (a + e^)2 dr 43. r = e2« +'e», s = log (t - «2). _ = 4 fs _ g ja + i. di dx . In the following examples first find — by diiferentiation and then substitute in dy dy^l_ dx dx to find ^. dy dx by XXVI .. /TT— dy 2Vl + y ) 2x 44. X = 2/vl + !/. — = -+ i— ^= dx 1+3^ 22/ + 32/2 . c fr~, dy 2 Vl + cos y 2 45. X = vl + cosj/. -i = ^ " = (ix sin y ^2-x^ 46. x = dy _{l + \ogyy 1 + logy dx logy 47. x = alog^±^^«^HZ. dy^_y_V^^ y dx a' 48. X = r arc vers V2 ry — y^. — r dx V2r- 49. Show that the geometrical significance of XXVI is that the tangent makes complementary angles with the two coordinate axes. RULES FOR DIFFERENTIATING 69 62. Implicit functions. When a relation between x and y is given by means of an equation not solved for y, then y is called an implidt function of x. For example, the equation x'-Ay=0 . defines y as an implicit function of x. Evidently x is also defined by means of this equation as an implicit function of y. Similarly, x' + f + g'-a' = defines any one of the three variables as an implicit function of the other two. It is sometimes possible to solve the equation defining an implicit function for one of the variables and thus change it into an explicit function. For instance, the above two implicit functions may be solved for y, giving ^ ^=4 and y = ± '^a^ — ^^ — 2^ ; the first showing y as an explicit function of x, and the second as an explicit function of x and 2. In a given case, however, such a solution may be either impossible or too complicated for convenient use. The two implicit functions used in this article for illustration may be respectively denoted by yz-g.^ v) = and F(x, y, z) = 0. 63. Differentiation of implicit functions. When y is defined as an implicit function of x by means of an equation in the form it was explained in the last section how it might be inconvenient to solve for y in terms of x ; that is, to find y as an explicit function of x so that the formulas we have deduced in this chapter may be applied directly. Such, for instance, would be the case for the equation (^) aa;'+ 2a;V — /» — 10 = 0. We then follow the rule : Differentiate, regarding y as a function of x, and put the result equal to zero.* That is, / (C) / |/(-^^) = o. » This process will be justified In § 7. Only corresponding values of x and y which satisfy the given equation may be substituted in the derivative. 70 DIFFERENTIAL CALCULUS Let us apply this rule in finding -j- from (5). -^(ffla:'+2a;=«/-2^'a;-10) = 0; by (C) ax 6 aa^+1o^^ + 6 a;V -y' -7 xf^ = 0; ax ax (2 3?-7xy'') ^ = y- 6 ax'- 6 a:''?/; dy _y'' — 6 ax'' — 6 x'y . dx" 2a?—7xy^ The student should observe that in general thg result will contain both X and y. EXAMPLES Differentiate the following by the above rule : 1. y^ = ipz. 2. x2 + y« = r2. 3. 62x2 + a?yi. = a2ft2 4. 3/S-32/ + 2aa; = 0. 5. K* + J/* = a*. 6. s^ + !/^ = a^. '■©'HI)'-'- 8. 2/2-2a;3/ + 62 = 0. 9. x» + y8 - 3 asv = 0. 10. xf = y. 11. p2 = a2cos2S. 12. /o2cosff=a2sin3e. 13. cos (uo) = a>. 14. *=cos(S + 0). _ d 1 + sin (5 + ) dy_ ^. dx V ' dy_ X dx y dy_ i^x dx ci?y' dy _ 2a dx 3(1-2/2) dy _ - /^ dx yJx dy _ dx -. (m) xcoty + y=0. (d) 2/2 = x« + a. (i) xV + 4 y = 0. (n) y" = log a. (e) s2 - y2 = 16. (j) 2/2 = sin 2a;. (o) 6==" + 2j/» = 0. 16. A race track has the form of the circle x^ + y^ = 2500. The directions OX and OY are east and north respectively, and the unit is 1 rod. If a runner starts east at the extreme north point, in what direction will he be going (a) when 25 V2 rods east of Oy ? Ans. Southeast or southwest. (b) when 25 V2 rods north of OX ? Southeast or northeast. (o) when 30 rods west of Or ? E. 36° 52' 12" N. or W. 36° 52' 12" N. (d) when 40 rods south of OX? (e) when 10 rods east of OY? 17. An automobile course is elliptic in form, the major axis being 6 miles long and running east and west, while the minor axis is 2 miles long. If a car starts north at the extreme east point of the course, in what direction will the car be going (a) when 2 miles west of the starting point ? (b) when J mile north of the starting point ? MISCELLANEOUS EXAMPLES Differentiate the following functions : 1. arc sin Vl — 4x2. Ans. -2 Vl-4a;2 v2. xe^. e^(2a;2+l). o , ■ " 1 t" 3. logsin-- -cot-. ^2 2 2 a a 4. arc cos-. y V2/2 _ a^ X a2 Va2 - x2 (a2 - x2)t log a: 6 ^ 1 + logX (l + l0gX)2 7. log sec (1-2 X). -2tan(l-2x). g_ a;2e2-3^. xe2-3a:(2 — 3x). 9. log vs — COS i . CSC t. cost 10. arcsinVj(l-cosx). i- 2s 11. arc tan 12. (2x (l-5s2)Vs2-l 7 + 4x »/ 2 3(1 + x) Vl + i x ' arc sin x (x2 + 2 ) VI — x" , x2arcsinx. Xo. r I 72 DIFFERENTIAL CALCULUS „0 , ,6 31. (log tan 3- xV- 14. tan'^ + logsec^-. i i' 3 3 ^^ 2-3i^+iti + t\ 15. arotanJ(e2=: + e-2=:). ' t /Sy^ 33 (l + a)(l-2a)(2 + a;) 16- y • ■ (3 + a;)(2-3x) 17 ajtana:. 34. arc tan (log 3 x) . (x + 2) J (x2 - 1)^ 35. -5/(6 -OX"-)". 18. ■ * 21. oosz 26. 27. x* 36. logV(a2-te2)n 19. gsced-Sx). 37. log / y' + i 20. arc tan Vl — x^. "Vj/i Z^ 38. gaTCBec29_ 39 ;(2-3x)8 22. e'">< \ l + 4x " 23. logsin^i^. -y^iTT^ 24. e<" log sin ox. ' cosx 25. sin=0cos(^. 41. e^^ log sin x. J5 42. arc sin - X 2V(6-cx")'» Vl + x^ 771 + X gm arc tana: 4o. arC tan tt^. 1 + ™ Vl + x2 44. a'™'''^. 28. tan^x-logsec^x. 45. cotS(logax). 29 31og(2cosx + 3sinx) + 2x 46. (1 - 3 x^) gS. 30. arc cot - + log J- — - ■ 47. log X ° _\x + a VT+X3 CHAPTER VI SIMPLE APPLICATIONS OF THE DERIVATIVE 64. Direction of a curve. It was shown in § 32, p. 31, that if is the equation of a curve (see figure), then dx ::tanr = slope of line tangent to the curve at any point P. The direction of a curve at any point is defined to he the same as the direction of the line tangent to the curve at that point. From this it follows at once that — = tanr = slope of the curve at any point P. dx At a particular point whose coordinates are known we write [— = slope of the curve (or tangent) at point (x^, y^. dx\^=,,^ At points such as Z>, F, H, where the curve (or tangent) is parallel to the axis, of X, j r = 0° ; therefore — = 0. dx At points such as A, B, G, where the curve (or tangent) is per- pendicular to the axis of X, du T = 90° ; therefore — = oo. dx 73 74 DIFFERENTIAL CALCULUS At points such as E, where the curve is rising,* r = an acute angle ; therefore — = a positive number. dx The curve (or tangent) has a positive slope to the left of B, between D and F, and to the right of G. At points such as C, where the curve is falling* r = an obtuse angle ; therefore — = a negative number. dx The curve (or tangent) has a negative slope between B and D, and between F and G. Illustrative Example 1. Given the curve y = — — x^ + 2 (see figure). „ X' (a) Find t when a; = 1. O H^ ^ (b) pind T when a; = 3. (o) Find the points where the curve is parallel to OX. (d) Find the points where r = 45°. (e) Find the points where the curve is parallel to the line 2a;-3y = 6 (line 4 B). Solution. Differentiating, — = x^ — 2 x = slope at any point, dx (a) tan t = (b) tan T : = 1 - 2 = - 1 ; therefore t = 135°. A-M. ar=l = 9 — 6 = 3: therefore t = arc tan 3. Ans. (c) T = 0°, tan T = — = ; therefore x^ — 2 x = 0. Solving this equation, we'ftnd dx ' that X = or 2, giving points C and D where the curve (or tangent) is parallel to OX (d) T = 45°, tan T = — = 1 ; therefore x^ — 2 x = 1. Solving, we get x = 1 ± V2, dx giving two points where the slope of the curve (or tangent) is unity. (e) Slope of line = f ; therefore x^ — 2 x = |. Solving, we get x = 1 ± V|, giving points E and F where curve (or tangent) is parallel to line AB. Since a curve at any point has the same direction as its tangent at that point, the angle between two curves at a common point will be the angle between their tangents at that point. Illustrative Example 2. Find the angle of intersection of the circles (A) x2 + j^s-4» = l, (B) x» + y2-2i/ = &. * When moving from left to right on curve. SIMPLE APPLICATIONS OF THE DERIVATIVE 75 Solution. Solving simulteneously, we find the points of intersection to be (8, 2) and dy 2 — X • from (4). By§63, p. 69 ^ dx y dy X dx \ — y from (B). By § 63, p. 69 !/=2 = — i = slope of tangent to (^1) at (3, 2). : — 3 = slope of tangent to (B) at (3, 2). ]X= 8 j,= 2 The formula for finding the angle between two lines whose slopes ?Lre m, and m„ is tan e = —J ?- • 55, p. 3 1 + m,m„ Substituting, tan 6 = — 5-!— = 1 ; therefore 9 = 45°. Ans. - - 1 + 1 This is also the angle of intersection at the point (1, — 2). I EXAMPLES The corresponding figure should be drawn in each of the following examples : Find the slope ofy — at the origin. Ans. 1 = tan t. 1 + x^ 2. What angle does the tangent to the curve x^y^ = a^{x + y) at the origin make with the axis of X ? Ans. t = 13-5°. 3. What is the direction in which the point geti6rating the graph of y = 3x^ — x tends to move at the instant when x = l? Ans. Parallel to a line whose slope is 6. dy 4. Show that -^ (or slope) is constant for a straight line. dx 6. Find the points where the curve y = x^ — Sx^ — 9x+5is parallel to the axis of JT. Ans. X = 8, x = — 1. V 6. At what point on y^ = 2 x^ is the slope equal to 3 ? Ans. (2, 4). 7. At what points on the circle ifi + y^ = r^ is the slope of the tangent line equal to -I? , / 3r 4r\ * Ans. (±^ -> ' / 3r 4r\ 8. Where will a point moving on the parabola y = x^ — 7x + She moving paral- lel to the line 2/ = 5 1+ 2 ? .4ns. (6,-3). 9. Find the points where a particle moving on the circle x^ + y^ = 169 moves per- pendicular to the line 52 -|- 122/ = 60. ^"«- (±12, +5). 10. Show that all the curves of the system y = log fa have the same slope ; i.e. the slope is independent of k. 11. The path of the projectile from a mortar cannon lies on the parabola y = 2 a; — X? ; the itnit is 1 mile, OX being horizontal and OY vertical, and the origin being the point of projection. Find the> direction of motion of the projectile (a) at instant of projection ; (b) when it strikes a vertical cliff IJ miles distant. (c) Where will the path make an inclination of 45°'with the horizontal ? ■ (d) Where will the projectile travel horizontally ? Ans. (a) arc tan 2; (b) 13.5°; (c) (J, J); (d) (1, 1). 76 DIFFERENTIAL CALCULUS 12. If the cannon in the preceding example was situated on a hillside of inclination:^ 45°, at what angle would a shot fired up strike the hillside ? -4™s- 45°. 13. At what angles does a road following the line 3y — 2x — 8 = intersect a rail- way track following the parabola y^ = 8x. Ans. arc tan J, and arc tan J. 14. Find the angle of intersection between the parabola y^ = Sx and the circle x^ + y^ = 16. ^'^- ^^^ tan f Vs. x^ 2/2 15. Show that the hyperbola x^ — y^ = 5 and the ellipse r^ + -5- = 1 intersect at lo O right angles. 16. Show that the circle x^ + y^ = 8ax and the cissoid y^ 2a — X (a) are perpendicvilar at the origin ; (b) intersect at an angle of 45° at two other points. 17. Find the angle of intersection of the parabola x^ = iay and the witch y = ^" Ans. arc tan 3 = 71° 33'.9, " a;2 + 4 a2 18. Show that the tangents to the folium of Descartes x^ + y^ = 3 axy at the points where it meets the parabola y'^ = ax are parallel to the axis of Y. 19. At how many points will a particle moving on the curve y = x^ — ix^ -{■ x — i be moving parallel to the axis of -Z" ? What are the points ? Ans. Two ; at (1, - 4) and (J, - Y?*)- 20. Find the angle at which the parabolas y = Zx'^ — I and y = 2x'^ + Z intersect. Ana. arc tan ^. 21. Find the relation between the coefficients of the conies a^K^ + h^y^ = 1 and a^^ + 622/" = 1 when they intersect at right angles. , 1 1 _ 1 1 Oj 6^ flj \ 65. Equations of tangent and normal, lengths of subtangent and subnormal. Rectangular coordinates. The equation of a straight line passing through the point (a;^, y^) and having the slope m is y — y^=m(x — x^). 54, (c), p. 3 j^ If this line is tangent to the curve AB at the point P(x^-, «/j), then from § 64, p. 73, JII N X :tanT=r^l =^*. Hence at point of contact I^Cx^, y^ the equation of the, tangent line TJl is * By this notation is meant that we should first find — , then in the result substitute x-i dx , for X and y^ for y. The student is warned against interpreting the symbol -^ to mean tlm derivative of y-^ with respect to x-^, for that has no meaning whatever, since ij and y^ are both constants. SIMPLE APPLICATIONS OF THE DERIVATIVE 77 The normal being perpendicular to tangent, its slope is -^ = -^- By 55, p. 3 And since it also passes through the point of contact ^(z^, y ), we have for the equation of the normal ^iV" (2) y-y, = -p:(^x-x,y That portion of the tangent which is intercepted between the point of contact and OX is called the length of the tangent (= TiJ), and its projection on the axis of Xis called the length of the mhtangent (=TM). Similarly, we have the length of the normal (= ^iV") and the length of the subnormal (= MIT). In the triangle TI^M, tan t = — 3- ;■ therefore TM MP dx (3) ^iW* = ^ = ^1 ^ = length of subtangent. In the triangle ilfi^iV, tan t = ; therefore (4) MN^ = MI^ tan T = ^1 -^ = length of subnormal. dXj^ The length of tangent (= T^) and the length of normal (= ^iV) may then be found directly from the figure, each being the hypotenuse of a right triangle having the two legs known. Thus TF^ = VTM' + MI^' = ^(y^^J+Cy;)^ (5) = Ui ^|(^) + 1 = length of tangent. Ji^=y/Mll' + M2^' = ^(y^y + L^J (6) =y^^i + 1 J^ J = length of normal. The student is advised to get the lengths of the tangent and of the normal directly from the figure rather than by using (5) and (6). Whai the length of subtangent or subnormal at a point on a curve is determined, the tangent and normal may be easily constructed. * If subtangent extends to the right of T, we consider it positive ; if to the left, negative, t If subnormal extends to the right of M, we consider it positive ; if to the left, negative. 78 DIEFEEENTIAL CALCULUS EXAMPLES 1. Find the equations of tangent and normal, lengths of subtangent, subnormal tangent, and normal at the point (a, a) on the cissoid y^ = ■ Solution. 2a — X dy _ 3 ax^ — x^ dx~ y{2a — xY' ^ = r^] = _Scfi-^ ^ 2 = slope of tangent. dx-f L*i;Jx=a a(2a— a)2 Substituting in (1) gives y = 2x — a, equation of tangent. Substituting in (2) gives 2y + x = Sa, equation of normal. Substituting in (3) gives TM = - = length of subtangent. Substituting in (4) gives MN = 2 a = length of subnormal. Also FT = V(TM)^ + (MP)^ =^^^ ^ a^- ^ ^^^^^ ^^ ^^^^^^^^ and FN = V{MN)^ + {MP)^ = VIoMT^ = a Vs = length of normal. 2. Find equations of tangent and normal to the ellipse x" + 2 j^'^ — 2 X2/ — x = at the points where x = 1. Ans. At (1, 0), 2 y = x — 1, ^ + 2 x = 2. At (1,1), 2y = x + l,y + 2x = B. 3. Find equations of tangent and normal, lengths of subtangent and subnormal at the point (x^, y^) on the circle x^ + y^ = r^.* 2 Ans. XjX + y^y = r^, x^y — y^x = 0, -, —x^. Xj 4. Show that the subtangent to the parabola y^ = 4 px is bisected at the vertex, and that the subnormal is constant and equal to 2p. 5. Find the equation of the tangent at (Xj, j/j) to the ellipse 1- — = 1. y 6. Find equations of tangent and normal to the witch y = at the point where x = 2 o. 4 a^ + x= Ans. x + 2y = 4:a, y = 2x — Sa. X _x 7. Prove that at any point on the catenary y = -(fe«+ e °) the lengths of sub- 2x _2£ 2 2 normal and normal are-(e'' — e ») and — respectively. 8. Find equations of tangent and normal, lengths of subtangent and subnormal, to each of the following curves at the points indicated : Aa) 2/ = x^ at (I, \). (e) 2/ = 9 - x^ at (- 3, 0). i^b) 2/2 = 4x at (9, - 6). (f) x^ = &y where x =- 6. (c) x2 + 5^2_i4^1iere2/ = l. (g) x^-xj/ + 2x- 9= 0, (3, 2). (d) x2 + 2/2 = 25 at (- 3, - 4). (h) 2x2 - yi ^ 14 ^t (3, _ 2). • In Exs. 3 and 6 the student should notice that if we drop the subscripts in equations oJ tangents, they reduce to the equations of the curves themselves. SIMPLE APPLICATIONS 0¥ THE DEKI\'ATIVE 79 9. Prove that the length of subtangent toy = a^ is constant and equal to . ■ log a 10. Get the equation of tangent to the parabola y^ = 20x which makes an angle of 45° with the axis of X. Arts, y = a + 5. Hint. First find point of contact by method of Illustrative Example 1, (d), p. 74. 11. Find equations of tangents to the circle x' + y^= 52 which are parallel to the line 2a; + 32/ = 6. Arts. 2a; + 32/ ± 26 = 0. 12. Find equations of tangents to the hyperbola ix^—9y^ + 36 = which are perpendicular to the line 2y + 5x = 10. • Ans. 2a;— 5^±8 = 0. 13. Show that in the equilateral hyperbola 2xy = a^ the area of the triangle formed by a tangent and the coordinate axes is constant and equal to a?. 14. Find equations of tangents and normals to the curve y^ = tix^ — x' at the points where X = 1. .4ns. At (1, 1), 2?/ = x + 1, ^ + 2a; = 3. At (1, — 1), 2 2/ = — a; - 1, 2/ - 2x = — 3. 15. Show that the sum of the intercepts of the tangent to the parabola xi + yi = ai on the coordinate axes is constant and equal to a. 16. Find the equation of tangent to the curve x^ (x + y) = a^ (x — y) at the origin. Ans. y = x. 17. Show that for the hypocycloid xs + j/s = a^ that portion of the tangent included between the coordinate axes is constant and equal to a. X 18. Show that the ciirve y = a^ has a constant subtangent. 66. Parametric equations of a curve. Let the equation of a curve be If a; is given as a function of a third variable, t say, called 2i param- eter., then by virtue of (^) y is also a function of t, and the same func- tional relation (A) between x and y may generally be expressed by means of equations in the form ryi each value of t giving a value of x and a value of y. Equations {E) are called 'parametric equations of the curve. If we eliminate t between equations (5), it is evident that the relation {A) must result. For example, take equation of circle s^ + f = r», ory = V7^-a^. Let x = rcost; then y = r sin <, and we have x = r cos t. ^ ^ I -< = r sm t. as parametric equations of the circle in the figure, t being the parameter. so DIFFERENTIAL CALCULUS If we eliminate t between equations (C) by squaring and add- ing the results, we have x' + y^^ r" (cos" t + sin'' f) = r", the rectangular equation of the circle. It is evident .that if t varies from to 2 TT, the point P (x, y) will describe a complete circumference. In § 71 we shall discuss the motion of a point P, which motion is defined by equations such as We call these the parametric equations of the path, the time t being the parameter. Thus in Ex. 2, p. 93, we see that x==v^ cos a ■ t, are really the parametric equations of the trajectory of a projectile, the time t being the parameter. The elimination of t gives the rectan- gular equation of the trajectory 9^ y = x tan a ■ 2 v^ cos" a Since from (5) y is given as a function of t, and i as a function of X, we have j j j^ ay _ ay at dx dt dx dy 1 dt that is, dy _dt _'(J) (P~) dx dx fi(V) dt Hence, if the parametric equations of a curve are given, we can find equations of tangent and normal, lengths of subtangent and subnor-, mal at a given point on the curve, by first finding the value of ^ at dx that point from (D) and then substituting in formulas (1), (2), (3), (4) of the last section. SIMPLE APPLICATIONS OF THE DEKIVATI\E 81 Illustrative Example 1. Find equations of tangent and normal, length? of subtangent and subnormal to the ellipse (E) at the point where (p = r X = a cos ip, \y = bsin^,* dx Solution. The parameter being 0, — = — a sin 0, dy , ^ = 6 cos < dy dip 6cos0 Substituting in (D), — = dx a sin = slope at any point. Substituting = - in the given equations (E), we get/ -^, — ^) as the point of ^ VV2 V2/ contact. Hence Substituting in (1), p. 76, y — or. Substituting in (2), p. 77, y — dy^ __ b da;, a V2 "V V2/' bx + ay = \/2 ab, equation of tangent. 6 V2' or, V2 {ax — by) = a' — 6^, equation of normal. Substituting in (3) and (4), p. 77, — = ( ) = = length of subnormal. V2\ o,/ .aV2 V^ ) = = length of subtangent. 6/ V2 *As in the figure draw the major and minor auxiliary circles of the ellipse. Through two points S and C on the same radius draw lines parallel to the axes of coordinates. These lines will intersect in a point P (x, y) on the ellipse, because a; = OA = OB cos 1^ = a cos and y = AP= OD=OCs,\nsia.= 2 iro is called the base of one arch of the cycloid, and the point 7 is caUed the vertex. Eliminating 0, we get the rectangular equation 3;= a arc cos (^)- ^^ ay-y^ SIMPLE APPLICATIONS OF THE DEEIVATIVE 83 6. X = l-t,y = t^; t = 3. 8. s = t«, v = «; « = 2. 9. 1 = t», J/ = t2; «=_1. 10. s = 2-i, 2/ = 3t2; t = l. 11. a; = cos «, y = sin 2 « ; < = - . 3 12. £ = 3e^', 2/ = 2e«; t = 0. 13. a; = sini, y = 2cost; « = -. 4 14. x = 4cos<, j/ = 3sin«; t = -. 15. x = log(« + 2), y = i; t = 2. In the following curves find lengths of (a) subtangent, (b) subnormal, (c) tangent, (d) normal, at any point : Cx = a (cos t'+ t sin t), ja 16. The curve (sint — ioosi). ns. (a) a; = 4 a cos' t, Ans. (a) ycott, (b) y tani, (c) -^, (d)-^- sin t cos t . ^ 3/ ^^ 4 CK cos' t 17. The hypocycloid (astroid) i ' Ly = 4asin^i. Ans. (a) -ycoti, (b) -j/tani, (c) -^, (d) -i^. sin t cos t 18. The circle 19. The cardioid 20. The folium x = r cos t, r sin i. rx = a(2 cost — cos2i), \y = a(2sin t — sin2i). St ' 1 + to' ' 1 + «8 ' 21. The hyperbolic spiral a! = - cos t, t y = -sm t. 67. Angle between the radius vector drawn to a point on a curve and the tangent to the curve at that point. Let the equation of the curve in polar coordinates be p =f(Q'). Let P be any fixed point (/?, ^) on the curve. i^^. If 0, which we assume as the independent vari- able, takes on an increment A^, then p will take on a corresponding increment A^o. Denote by Q the point (jp + A/d, Q + A^). Draw PR perpendicular to OQ. Then OQ^p + t^p, PR = p sin A^, and OB = p cos A^. Also, '-^«^ = l| = o/'' /J sin A^ 0^ p + Ap — p cos A^ 84 DIFFERENTIAL CALCULUS Denote by yjr the angle between the radius vector OP and the tangent FT. If we now let A0 approach the limit zero, then (a) the point Q will approach indefinitely near P ; (b) the seeant PQ will approach the tangent PT as a limiting pod- tion ; and (c) the angle PQR will approach yjr as a limit. Hence , , ^ limit p sin. Ad ^^'^ A0 = O p + Ap — pcosAd 5^/ _ limit psinA^ 2psin^— +A/3 I Since from 39, p. 2, p — pcosA0 = p (1 — cos A9) = 2psin2 — • I p sin A^ ^ limit Ae Ad '^ Ad tDiTiding both numerator and denominator by Aff.] sin A0 limit ^' Ad Ae = Q a5 sm— — . A^ 2 Ap ''^"^T-^^ + A^ c- limit /Ap\ dp , limit/- A6\ . , limit /sin A^' S"^^^ A^=o(Aij=i "'^^ A^=o(^"^TJ=^' ^^^° A^=o(^r . A6I sm — and ^^^^——=1 by § 22, p. 21, we havg- T de From the triangle OPT we get =1 SIMPLE APPLICATIONS OF THE DEEIVATIVE 85 Having found t, we may then find tan t, the slope of the tangent to the curve at P. Or since, from (5), , i ^a . ,-s tan 6 + tan -^ir tan T = tan (9 + -ylr) = — i_ , ^ ^^ 1- tan ^ tan i|r we may calculate tan i/r from (A) and substitute in the formula (C) slope of tangent = tan r = r-^ £- . ^ ^ ^ 1 - tan fl tan ^ Illustrative Example 1. Find ^ and t in the cardioid p = a(l — costf). Also find the slope at ^ = — . 6 dp Solution. -C = o sin 9. Substituting in (A) gives Q 2 a sin^ - p a(l - cos S) 2 ^ „ „„ . tan i/- = X. = _i . ^ = = tan - . By 39, p. 2, and 37, p. 2 -!- 2asm-cos- d9 2 2 9 9 ti '\ R Since tan ^ = tan -, f = -. Ans. Substituting in (B), t = 9 + - = Ans. tanT = tan- = l. Ans. 4 To find the angle of intersection of two curves C and C" whose equations are given in polar coordinates, we may proceed as follows : angle TPT' = angle OPl"— angle OFT, or, ^ = yjr' — -^Jr. Hence (D) tan m = 1- i— , 1 + tan^'tan^ where tan i^' and tan -), i i j. tan = ' = |. .•. = arc tan |. Ans. 86 DIFFERENTIAL CALCULUS 68, Lengths of polar subtangent and polar subnormal. Draw a line NT through the origin perpendicular to the radius vector of the point P on the curve. If PT is the tangent and' PiV the normal to the curve at P, then 0T= length of polar subtangent, and 0-ZV= length of polar subnormal of the curve at P. OT In the triangle OPT, tan ■^jr = Therefore OT = D tan -«|r = /)" — = length of polar subtangent.* dp In the triangle OPN, Un-f = --^- Therefore (8) 0N= — ^ — = — = length of polar subnormal. tani/r d6 The length of the polar tangent (=PT} and the length of the polar normal (=P-ZV") may be found from the figure, each being the hypot- enuse of a right triangle. Illtjstkative Example 3. Find lengths of polar subtangent and subnormal to the lemniscate p^ = a^ cos2d. Solution. Differentiating the equation of the curve as an implicit function witli respect to 0, 2p^ = -2a.'sm2e, '^ de dp a' sin 26 09= -p Substituting in (7) and (8), we get length of polar subtangent : length of polar subnormal = — a? sin 2 6 a^sinSfl If we wish to express the results in terms of 8, find p i n term s of 6 froni the given equation and substitute. Thus, in the above, p = ±aVcos2^; therefore length of polar subtangent = ± a cot 2S Vcos2S. de * When 6 increases with p, — is positive and ^ is an acute angle, as in the above flgnie. Then the subtangent T is imsitive and is measured to the right of an observer placed at and dB looking along OP. When — Is negative, the subtangent is negative and is measured to the left of the observer. SIMPLE APPLICATIONS OF THE DERIVATIVE 87 EXAMPLES 1. In tlie circle p = r sinS, find f and t in terms of 9. Ans. \l/ = 6, t=26. a 2. In the parabola p = a sec' - > show that t + ^ = ir. 3. In the curve p^ = a? cos 2^, show that 2^' = w + 45. 4. Show that \j/ is constant in the logarithmic spiral p = ef^. Since the tangent makes a constant angle with the radius vector, this curve is also called the equi- angular spiral. 5. Given the curve p= a sin' - , prove that r = 4 ^. o 6. Show that tan ^ = 5 in the spiral of Archimedes p = aO. Find values of f when 9=2-ir and 4 it. Ans. f = 80° 57' and 85° 27' 7. Find the angle between the straight line p cosS=2a and the circle p = 6 a sin S. , Ans. arc tan j. 6 8. Show that the parabolas p= a seo^ - and p = b csc^ - intersect at right angles. 9. Find the angle of intersection of p= a sin d and p = a sin 2 0. Ans. At origin 0° ; at two other points arc tan3 VS. 10. Find the slopes of the following curves at the points designated : (a) p= a(l — oosd). 9 = —- Ans. —1. (b) p=as&c^e. p = 2a. 3. (c) p=:asin45. origin. 0, 1, oo, — 1. (d) p^ = a^ sin 4 9. origin. 0, 1, oo, — 1. (e) p = a sin 3 9. origin. 0, Vs, — Vs. (f ) p = acosS9. origin . (g) p=acos20. origin. (h) p = asm29. ^ = 7- a (i) p=asm30. ^^r' (i) p = <^ff- « = \- (^)p9=a. S = |- (I) p = ffi. 9=0. 11. Prove that the spiral of Archimedes p = a9, and the reciprocal spiral P = i ' intersect at right angles. 12. Find the angle between the parabola p = a sec^ - and the straight line p sine = 2a. ^ns. 45°. 13. Show that the two cardioids p = o(l + cos^ and p = o(l — cos 5) cut each' other perpendicularly. 14. Find lengths- of subtangent, subnormal, tangent, and nonnal of the spiral of Archimedes p = oB. ^^ ^^^t. = f! , tan. = ? V^T?, a a, subn. = a, nor. = VaT^^. The student should note the fact that the subnormal is constant. 88 DIFFEEENTIAL CALCULUS 15. Get lengths of subtangent, subnormal, tangent, and normal in the logarithmic spiral p = a«. Arts. subt. = — ^ I tan. = p -t /l + logo V log a \ log2 o subn. = p log a, nor. = p Vl + log^ a. When a = e, we notice that subt. = subn., and tan. = nor. 16. Find the angles between the curves p = a(l + cos 5), p = h{l — co&6). Ans. Oand-. Oj 2 17. Show that the reciprocal spiral /o = - has a constant subtangent. 9 18. Show that the equilateral hyperbolas p'' sin 2^ = a^, p^ cos2^ = 6'' intersect at right angles. 69. Solution of equations having multiple roots. Any root which occurs more than once in an equation is called a multiple root. Thus 3, 3, 3,-2 are the roots of (A) a;"- 7 a;* +9 a;' +27 a; -54 = 0; hence 3 is a multiple root occurring three times. Evidently QA) may also be written in the form ' (a;-3)'(a:+2)=0. Let /(a;) denote an integral rational function of x having a multiple root a, and suppose it occurs m times. Then we may write (-B) f(x) = (x~ayix), where ^(a;) is the product of the factors corresponding to all the roots of /(a;) differing from a. Differentiating (5), f(x) = (x — a)"'^'(a;) + ^{x')m(x — a;)'"-\ ' (<7) f(x) = (x- ay-^ l(x - a) 4>'(x') + (x) w]. Therefore /'(a;) contains the factor (x — a) repeated m — 1 times and no more ; that is, the highest common factor (H.C.F.) of f(x) and /'(a;) has m — 1 roots equal to a. In case /(a;) has a second multiple root /3 occurring r times, it is evident that the H.C.F. would also contain the factor (a; — /3)''~^, and so on for any number of different multiple roots, each occurring once more in /(a;) than in the H.C.F. We may then state a rule for finding the multiple roots of an equation fQc) = as follows : First Step. Findf'(x'). Second Step. Find the KQ.F. of f(x) andf{x). Third Step. Find the roots of the K O.F. Fach different root of the H.Q.F. will occur once more inf(x) than it does in the M.O.F. SIMPLE APPLICATIONS OF THE DERIVATIVE 89 If it turns out that the H.C.F. does not involve x, then f(x) has no multiple roots and the above process is of no assistance in the solution of the equation, but it may be of interest to know that the equation has no equal, i.e. multiple, roots. / ' 'i Illustkative Example 1. Solve the equation a;^ — Sx^ + 13 a; — 6 = 0. Solution. Place f[x) = x^ — %x'^-\-\Zx-Q. First step. f'{x) = 3 x^ - 16 a; + 13. Second step. H.C.r. = x — 1. Third step. x — 1 = 0. .■. x = 1. Since 1 occurs once as a root in the H.C.F. , it will occur twice in the given equa- tion ; that is, (X — 1)2 will occur there as a factor. Dividing x^ — Sx^ + 13x — 6 by (X — 1)2 gives the only remaining factor (x — 6), yielding the root 6. The roots of our equation are then 1, 1, 6. Drawing the graph of the function, we see that at the double root x = 1 the graph touches OX but does not cross it.* EXAMPLES Solve the first ten equations by the method of this section : Ans. 2, 2, 3. -1,-1,-1,3. 3, 3, 3, - 2. 3, 3, 3, - 4. 1, 1, - 4, - 4. 3, 3, - 1, 4. 2, 2, 1±V3. -1,-1,-1, 2, 2} 2, 2, 2, - 3, - 3. -1,-1, -1, 3±V=1. 1. x» - 7x2 + 16x- 12 = 0. 2. x*-6x2-8x-3 = 0. 3. X* - 7x3 +'9x2 + 27x- 54 = 0. ;4. X* - 5x5 - 9x2 + 81 x - 108 = o. 5. x^ + 6x3 + a;2 _ 24x + 16 = 0. 6. X* - 9x3 + 23x2 - 3x - 36 = o. 7. X*- 6x3 + 10x2-8 = 0. 8. x5 - X* — 5x3 + x2 + 8x + 4 = 0. 9. x5 - 15x3 + 10x2 + 60x- 72 = 0. 10. x^ — Sx^— 5x3 + 13x2 + 24x + 10 = 0. Show that the following four equations have no multiple (equal) roots : 11. x3 + 9x2 + 2 X - 48 = 0. - 12. x« - 15x2 _ lox + 24 = 0. 13. x«-3x3-6x2 + 14x + 12 = 0. " 14. x» — a» = 0. 15. Show that the condition that the equation x' + 3 gx + )• = shall have a dotible root is 4 g3 + r? = 0. 16. Show that the condition that the equation x3 + 3px2 + r = shall have a double root is r (4p3 + y) = 0. * Since the first derivative vanishes for every multiple root, it follows that the axis of -X"is tangent to the graph at all points corre- sponding to multiple roots. If a multiple root occurs an even number of times, the graph will not cross the axis of X at such a point (see figure) ; if it occurs an odd number of times, the graph will cross. 90 DIFFERENTIAL CALCULUS 70. Applications of the derivative in mechanics. Velocity. Recti- linear motion. Consider the motion of a point P on the straight line AB. Let 8 be the distance meas- u — 2 -^p 5 ured from some fixed point as A to any position of P, and let t be the corresponding elapsed time. To each value of t corresponds a position of P and therefore a distance (or space) s. Hence « wUl be a function of t, and we may write Now let t take on an increment A< ; then » takes on an increment As,* and As rA') — = the average velocity of P during the time interval At If P moves with uniform motion, the above ratio will have the same value for every interval of time and is the velocity at any instant. For the general case of any kind of motion, uniform or not, we define the velocity (time rate of change of «) at any instant as the As limit of the ratio — as A< approaches the limit zero; that is, At _ limit As "' At=^QAt or, (9) i; = — . dt The velocity is the derivative of the distance (= space') with respect to the time. To show that this agrees with the conception we already have of velocity, let us find the velocity of a falling body at the end of two seconds. By experiment it has been found that a body falling freely from rest in a vacuum near the earth's surface follows approximately the law (5) « = 16.1f, where « = space fallen in feet, t — time in seconds. Apply the Q-en- eral Ride, p. 29, to (£). » As being the space or distance passed over in the time A*. SIMPLE APPLICATIONS OJ^' THE DERIVATIVE 91 First Step. s + ^8 = 16.1(t + Aty=16.ie + S2.2t-At + lQ.l(Aty. Second Step. A8= 3-2.2 «-A< + 16.1 (A0°- As Third Step. — = S2,2t+lQ.l At = average velocity throughout the time interval At. Placing < = 2, As (»^_ at • M^ feet. Find the velocity, acceleration, component velocities, -q^^— and component accelerations (a) at any instant ; (b) at the end of the first second, having given v-^ = 100 ft. per sec, = 30° ; (c) find direction of motion at the end of the first second. 94 DIFFEEENTIAL CALCULUS Solution. From (10) and (11), (a) Vx = »j cos ; v^ = v^Bm

/3 (2 = - TT^a 6 18 = 2, D = = 6, a = 18. . , . 1 . . , Do: 86.6 of motion with the horizontal. 3. Given the following equations of rectilinear motion. Find the distance, velocity, and acceleration at t.^"- instant indicated : (a) s = iS + 2 i2 ; t = 2. Ans. s = 16, « = 20, a = 16. (b) s = «2 + 2 i ; f = 3. s = 15, » = 8, a = 2. (c) s = 3-4i; t = 4. s =- 13, « =- 4, q:= 0. (d) x = 2t-t2; t = i. a; = l,» = 0, a = -2. (e) J/ = 2 J - ts ; t = 0. y = 0,v = 2,a=0. (f) A = 20t + 16«2 ; « = 10. h = 1800, v = 340, a= 32. (g) s=2sinJ; f = ^- s = V2, » = V?, a = - VS. (h) y = a cos — , t = 1. o (i) s = 2e»'; t = 0. (j) s = 2t2_3«; t = 2. (k) s = 4 + J8 ; t = 3. (1) y = 5cos2t; * = -- (m) s = 6 sin — ; i = 2. (n) a; = ae-2'; i = 1. (o) s = I + 6J2 ; i = i„. (p)s = 101og^;i = l. 4. If a projectile be given an initial velocity of 200 ft. per sec. in a direction inclined 45° with the horizontal, find (a) the velocity and direction of motion at the end of the third and sixth seconds ; (b) the component velocities at the same instants. Conditions are the same as for Ex. 2. Ans. (a) "When i = 3, « = 148.3 ft. per sec, t = 17° 35', when i = 6, v = 150.5 ft. per sec, t = 159° 53' ; (b) when t = 3, Da; = 141.4 ft. per sec, Vy = 44.8 ft. per sec. when t = 6,Vx = 141.4 ft. per sec, »j, = — 51.8 ft. per sec. 6. The height (= s) in feet reached in t seconds by a body projected vertically upwards with a velocity of v^ ft. per sec. is given by the formula ,s = y-i6.i<2. SIMPLE APPLICATIONS OP THE DERIVATIVE 96 Find (a) velocity and acceleration at any instant; and, if b,= 300 ft. per sec, find velocity and aoceleration (b) at end of 2 seconds; (c) at end of 16 seconds. Eesist^ KQoe of air is neglected. Ans. (a) » = Oj — 32.2 1, a = — 32.2 ; (b) » = 236.6 ft. per sec. upwards, or = 82.2 ft. per (sec.)* dovimwards; (c) = 183 ft. per sec. dovmvyards, a = 32.2 ft. per (sec.)* downwards. 6. A cannon ball is fired vertically upwards with a muzzle velocity of 644 ft. per sec. Find (a,) its velocity at the end of 10 seconds ; (b) for how long it will continue to. rise. Conditions same as for Ex. 5. Ans. (a) 322 ft. per sec. upwards; (b) 20 seconds. 7. A train left a station and in t hours was at a distance (space) of s = t8 + 2«2 + 3t miles from the starting point. Find its acceleration (a) at the end of t hours ; (b) at the end of 2 hours. Ans. (a) a = 6 1 + 4 ; (b) a = 16 miles per (hour)*. 8. In t hours a train had reached a point at the distance of J t* — 4 i' + 16 i* miles from the starting point, (a) Find its velocity and acceleration, (b) When will the train stop to change the direction of its motion ? (o) Describe the motion during the firstlOhours. Ans. (a) b = i' - 12i* + 324, a = 3t*~ 24S + 32; (b) at end of fourth and eighth hours ; (c) forward first 4 hours, backward the next 4 hours, forward again after 8 hours. 9. The space in feet described in t seconds by a point is expressed by the formula s = 48t-16<*. Find the velocity and acceleration at the end of 1^ seconds. Ans. ■!) = 0, a = — 82 ft. per (sec.)*. 10. Find the acceleration, having given (a) B = <* + 2i; i = 3. Ans.a=8.- (b) 1) = 8 J - iS ; < = 2. a = -9. (c)» = 48in-;<=: — ■ a = Vs. ^ ' 2 3 (d) t) = a cos 3 < ; i = - • a = — 3 o. \ I '6 (e) n= 5e2'; i = 1. a:=10e*. 11. At the end of t seconds a body has a velocity of 3 4* + 2t ft. per sec; find its acceleration (a) in general ; (b) at the end of 4 seconds. Ans. (a) a=6t + 2tt. per (sec.)* ; (b) a = 26 ft. per (sec.)* 12. The vertical component of velocity of a point at the end of t seconds is Dy = 3t* - 2< + 6 ft. per sec. Find the vertical component of acceleration (a) at any instant ; (b) at the end of 2 seconds. Ans. (a) a,= 6J — 2; (b) 10 ft. per (sec.)* 13. If a point moves in a fixed path so that s=VJ, show that the acceleration is negative and proportional to the cube of the velocity. 96 DIFFERENTIAL CALCULUS 14. If the space described is given by s = ae' + be-', show that the acceleration is always equal in magnitude to the space passed over. 15. If a point referred to rectangular coordinates moves so that X = a cos t + b, and y = a sin t + c, show that its velocity has a constant magnitude. 16. If the path of a moving point is the sine curve fa; = at, \y = 6sinai, show (a) that the s-component of the velocity is constant ; (b) that the acceleration of the point at any instant is proportional to its distance from the axis of X. 17. Given the following equations of curvilinear motion, find at the given instant Bx, "», " ; <^i, cTj), o: ; position of point (coordinates) ; direction of motion. Also find the equation of the path in rectangular coordinates. {a) x = t^,y = t; t = 2. (g) X = 2 smt, y = S cost ; t = IT. (b) x = t,y = t^; t = l. (^. x = smt,y = cos2t; « = ^. (c) x = t',y = t^;t = 3. ^ ' 4 (d) x = 2t,y = t^ + 3;t = 0. W ^ = 24, y = 3e'; t = 0. (e) x = l-t^,y = 2t; t = 2. Q) x = 3t, y = logt; t = l. (f) X = asini, y = aoost; t = (k) x = t,y = 12i-i; t = 3. CHAPTER VII SUCCESSIVE DIFFERENTIATION 74. Definition of successive derivatives. We have seen that the derivative of a function of x is in general also a function of x. This new function may also be differentiable, in which case the derivative of the first derivative is called the second derivative of the original function. Similarly, the derivative of the second derivative is called the third derivative; and so on to the wth derivative. Thus, if dy_ dx = 12i U^4\ = mx\ ax \dxj dx d Idy dx\dx = 72 X, etc. 75. Notation. The symbols for the successive derivatives are usually abbreviated as follows: dx\dx] dx^ L dx ' d Idy dx \dx ,±(d^\^^^ dx\dx''/ dx"' liy d / d''-^y \_d''y _ ' dx\dx"-^j d3? —f(x), the successive derivatives are also denoted by f'Qo-), /"(^), /"'(^). /X^). • • •' /'"'(^); y', y", y"\ r. ,/»). or, #/(-)' S/(-)' l^/(-)' |i/(^)' ■••' ;£^(-) dx dx- dx"'' ^"^' dx*' 97 98 DIFFERENTIAL CALCULUS 76. The nth derivative. For certain functions a general expression involving n may be found for the nth derivative. The usual plan is to find a number of the first successive derivatives, as many as may be necessary to discover their law of formation, and then by induction write down the wth derivative. Spy Illustrative Example 1. Given y = e^, find Solutun. dy_ dx ■ aef^, cPy _ da;2~ ■■ a^ef^,_ d"y _ dx" : we^. Ans. Illustrative Example 2. Given y = logx, find—. dx," Solution. dy _ di~ 1 ■ — ) X d^_ dx2 1 X2' d'y_ dx^~ 1-2 X3 ' d*y _ 1.2.3 X* ' doy _ dx" |n-l Illustrative EXAMPLi: 3. Given y = sinx, find"^"^. dx" Solution. dy -2- = cos X = dx sin(x + |), Ati&. d^y d . / Tr\ I n-X . / 27r\ ^ = s^'H" + 2) = '=°n" ": 2) = "" (^ + t)' d^y d . I 27r\ / 2ir\ . / ZtA di^ = S"H^ + T) = ''°H^"'T) = ^"(" + T) d"y . I nv\ .-. -— = sin(xH ). Ans. dx" \ 2 / 77. Leibnitz's Formula for the nth derivative of a product. This formula expresses the wth derivative of the product of two variables in terms of the variables themselves and their successive derivatives. SUCCESSIVE DIFFEKENTIATION 99 If u and V are functions of x, we have, from V, d . . du dv -(uv-) = — v + u-- Differentiating again with respect to x, d^ . - _ ^u dudv^ dudv d^v _ d^u ^dudv d% do? d7? dx dx dx dx dx^ dx'' dx dx da? Similarly, c?' . _ d\ d^u dv „ d^u dv ^du d^v du d^v dh di? d^ di? dx da? dx dx dx' dx da? da? _ d^u o d^u dv „du cPv d% da? da? dx dx da? dx^ However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond to the expo- nents of the Binomial Theorem.* Reasoning then by mathematical induction from the >nth to the (to + l)th derivative of the product, we can prove Leibnitz s Formula d" , d"a (p-'^udv n(n-i) d''-^u - 4 sec^^. 13. r = log sin 0. ■>•"' = 2 cot csc^ 0. 14. /(<) = e- 1 cos i. f" («) = - 4 e- ' cos i = - 4/(«). 15. /(e) =V^i^2^. /"(e) = 3 [/(«)]= -f{0). 16. p = (g2 + a2)arotani. T^= , 2 , 2^2 " " ' a dg^ (o^ + ?2)^ 17. y = a^. -i = (log a)»a»^. dx" d"!/ |k — 1 18.^ = log(l + x). _. = (_l)»-i__^. d"2/ / , Kir\ 19. 2/ = cosax. di^~""''°^V'^ T/' any \n-l 20. 2, = x»-Mogx. ^-^T' [TC= a positive integer.] 2'^^ = TT^- ^»-'( 'Ni + x)n+r 2 HiKT. Reduce fraction to form - 1 + :; before differentiating. 1 + x d^v dv 22. lfy = e^sinx, prove that _|_2— + 2y = 0. " ' ^ dx^ dx d^v dy 23. If 2/ = a cos (log x) + b sin (log x), prove that ^^^ + ^^ + 2' = ^- 102 DIFFERENTIAL CALCULUS Use Leibnitz's Formula in the next four examples : 24. y = x^a^. — = oi»(loga)»-2[(a;loga + n)^ - n] dx" dx" 26. /(x) = e' sill X. /'"'(») = (v^"e'sin (x + ^V 27. f(0) = cos aS cos 6(?. /<»)(«) = ^±±-^ cos r(a + b) 6* + ^1 (a - 6)» + ^^ '- cos 2 ' 28. Show that the formulas for acceleration, (14), (15), p. 92, may be written d^s dH d^y a = — . aa; = — , a„ = — - • 29. y^^iax. 30. 62^2 ^ ^22,2 = a262. 31. X2 + 2/2 _ ^2_ 32. 2/2 + ^ = 1=. 33. 0x2 + 2 te2^ + fjy2 - 1 34. j/2-2xj/ = a2. _ , _ dx2 (2/-x)s' dx8 (y-xf „, ^ ^ (Pe tan20 — tan^A 35. sec0cos» = c. = -. d 37. Find the second derivative in the following : (a) log(u + ») = «-«. (e) 2/S + xS-3ax2/ = 0. (b) e« + u = e" + i). (f) j,2 _ 2mx2^ + a;2 _ „ ^ 0. (c)s = l + te'. (g) y = sin(x + 2/). (d) e» + St - e = 0. (h) e' + y = xy. iPy__4^ dx^ y^ d2^___6«_ _ d^y _ 36°x (ix2 yS d^y _ 24 X dxs ~ (1 + 23^)=' d2y_ A2-a6 (1x2 ^ (Ax + 6j/)3 ' d^y _ a2 dSj/ Sa^x CHAPTER Vin MAXIMA AND MINIMA. POINTS OF INFLECTION. CURVE TRACING 79. Introduction. A great many practical problems occur where we have to deal with functions of such a nature that they have a greatest (maximum) value or a least (minimum) value,* and it is very important to know what particular value of the variable gives such a value of the function. For instance, suppose that it is required to find the dimensions of the rectangle of greatest area that can be inscribed in a circle of radius 5 inches. Consider the circle in the following figure : Inscribe any rectangle, as BD. Let CD = X ; then BE = VlOO — a;^, and the area of the rectangle is evidently (1) ' A = x^\^(i-x\ That a rectangle of maximum^ area must exist may be seen as follows : Let the base CD (= x) increase to 10 inches (the diameter) ; then the altitude i>^ = VlOO— a;^ will decrease to zero and the area will, become zero. Now let the base decrease to zero ; then the altitude will increase to 10 inches and the area will again become zero. It is therefore intuitionally evident that there exists a greatest rectangle. By a careful study of the figure we might sus- pect that when the rectangle becomes a square its area would be the greatest, but this would at best be mere guess- work. A better way would evidently be to plot the graph of the function (1) and note its behavior. To aid us in drawing the graph of (1), we observe that (a) from the na.ture of the problem it is evident that x and A must both be positive ; and (b) the values, of x range from zero to 10 iaclusive. * There may be more than one of each, as illustrated on p. 109. 103 104 DIFFERENTIAL CALCULUS Now construct a table of values and draw the graph. What do we learn from the graph? X A 1 9.9 2 19.6 3 28.6 4 ■ 36.6 5 43.0 • 6 48.0 7 49.7 8 48.0 9 39.6 10 0.0 (a) If carefully drawn, we may find quite accurately the area of the rectangle corresponding to any value of x by measuring the length of the corresponding ordinate. Thus, when X = 0M= 3 inches, then A = MP = 28.6 square inches ; and when x = 0N= 4l\ inches, then A =NQ = about 39.8 sq. in. (found by measurement). (b) There is one horizontal tangent (i?(S). The ordinate TIT from its point of contact T is greater than any other ordinate. Hence this discovery : One of the inscribed rectangles has evidently a greater area than any of the others. In other words, we may infer from this that the function defined by (1) has a maximum value. "We cannot find this value (= ST) exactly by measurement, but it is very easy to find, using Calculus methods. We observed that at T the tangent was horizontal; hence the slope will be zero at that point (Illustrative Example 1, p. 74). To find the abscissa ofT we then find the first derivative of (1), place it equal to zero, and solve for x. Thus (1) A = xVlOO - x\ dA _ 100~2a? dx ~ VlOO-a;' 100 - 2 a;^ Vioo^ 0. Solving, MAXIMA AND MINIMA a;=5V2. 105 Substituting back, we get DE = VlOO — x"" = 5 V2. Heiice the rectangle of maximum area inscribed in the circle is a square of area A = CD X DE = 5V2 X 5V2 = 50 square inches. The length of HT is therefore 50. Take another example. A wooden box is to be built to contain 108 cu. ft. It is to have an open top and a square base. What must be its dimensions in order that the amount of material required shall be a minimum ; that is, what dimensions will make the cost the least ? Let a; = length of side of square base in feet, and y = height of box. Since the volume of the box is given, how- ever, y may be found in terms of x. Thus volume = x\ = 108 ; .'.«/ = — ^ • ar We may now express the number (= Jf) of square feet of lumber required as a function of a; as follows : area pf base = a^ sq. ft.. y= 108 and area of four sides = 4 a;?/ = 432 (2) .V M 1 433 2 220 3 153 4 124 5 111 6 108 7 111 8 118 9 129 10 143 Xl(^ is a formula giving the number of square feet required in any such box having a capacity of 108 cu. ft. Draw a graph of (2). 106 DIFFERENTIAL CALCULUS What do we learn from the graph f (a) If carefully drawn, we may measure the ordinate correspond- ing to any length (= x) of the side of the square base and so deter- mine the number of square feet of lumber required. (b) There is one horizontal tangent (^BS). The ordinate from its point of contact T is less than any other ordinate. Hence this dis- covery: One of the boxes evidently takes less lumber than any of the others. In other words, we may infer that the function defined by (2) has a minimum value. Let us find this point on the graph ex- actly, using our Calculus. Differentiating (2) to get the slope at any point, we have ^j^ 4.32 dx x^ At the lowest point T the slope will be zero. Hence that is, when a; = 6 the least amount of lumber will be needed. Substituting in (2), we see that this is il/ = 108 sq.ft. The fact that a least value of M exists is also shown by the follow- ing reasoning. Let the base increase from a very sm.all square to a very large one. In the former case the height must be very great and therefore the amount of lumber required will be large. In the latter case, while the height is small, the base will take a great deal of lumber. Hence M varies from a large value, grows less, then increases again to another large value. It follows, then, that the graph must have a " lowest " point corresponding to the dimensions which require the least amount of lumber, and therefore would involve the least cost. We will now proceed to the treatment in detail of the subject of maxima and minima. 80. Increasing and decreasing functions.* A function is said to be increasing when it increases as the variable increases and decreases as the variable decreases. A function is said to be deereadng when it decreases as the variable increases and increases as the variable decreases. *The proofs given here depend chiefly on geometric intuition. The subject ol Maxima and Minima will be treated analytically in § 108, p. 1R7. MAXIMA AND MINIMA 107 The graph of a function indicates plainly whether it is increasing or decreasing. For instance, consider the function ar° whose graph (Fig. a) is the locus of the equation y = a''. a > 1 As we move along the curve from left to right the curve is rising ; that is, as x increases the function (= y) always increases. Therefore (f is an increasing function for all values of x. \ Fig. a Pig. 5 On the other hand, consider the function (a — a;)' whose graph (Fig. J) is the locus of the equation y=(o- xy. Now as we move along the curve from left to right the curve is faMng ; that is, as x increases, the function (= y") always decreases. Hence (a — xf is a decreasing function for all values of x. That a function may be sometimes increas- ing and sometimes decreasing is shown by the graph (Fig. c) of As we move along the curve from left to right ■ pio. c the curve rises until we reach the point A, then it falls from A to B, and to the right of B it is always rising. Hence (a) from x = —^~^) = + . V(+)(+) Whenx>5V2, /'(x) = -^^^Mi]= = -. V(+)(+) Since the sign of the first derivative changes from + to — at x = 5 V2, the function has a maximum value /(5V2) = 5V2.5V2 = 50. Ans. 84. Second method for examining a function for maximum and mini- mum values. From (19), p. 110, it is clear that in the vicinity of a maximum value of /(a), in passing along the graph from left to right, /'(a;) changes from + to to — . ^ Hence /'(a;) is a decreasing function, and by § 81 we know that its derivative, i.e. the second deriv- ^F ative [=/"(2;)] of the function itself, is negative or zero. Sunilarly, we have, from (20), p. 110, that in the vicinity of a minimum value of /(a;) /'(a;) changes from ~ to to +. Hence /'(a;) is an increasing function and by § 81 it follows that /"■(a;) is positive or zero. MAXIMA AND MINIMA 113 The student should observe that/" (2;) is positive not only at mini- mum points (as at A) but also at points such as P. For, as a point passes through P in moving from left to right, slope = tan T = -f- =f'Qc) is an increasing function. At such a point the curve is said to be concave upwards. Similarly, /"(«) is negative not only at maximum points (as at JS) but also at points such as Q. For, as a point passes through Q, dtt slope = tan t = -r^ =/'(«) is a decreasing function. At such a point the curve is said to be concave downwards.* We may then state the(^fficieirr]bonditions for maximum and mini- mum values oifCx) for certain values of the variable as foUovs^s : (21) f(x') is a maximum ii/'(jr) = and/"(jr) = a negative number. (22) f(x) is a minimum if /'(x) = and /"(jr) = a positive number. Following is the corresponding working rule. First Step. Find the first derivative of the function. Second Step. Set the first derivative equal to zero and solve the result- ing equation f or (^e^^oots in order to find the critical values of the variable. Third Step. Find the second derivative. Fourth Step. Substitute each critical value for the variable in the second derivative. If the result is negative, then the function is a maximum for that critical value; if the result is positive, the function is a minimum. When /"(a;) = 0, or does not exist, the above process fails, although there may even then be a maximum or a minimum ; in that case the first method given in the last section still holds, being fundamental. Usually this second method does apply, and when the process of find- ing the second derivative is not too long or tedious, it is generally the shortest method. Let us now apply the above rule to test analytically the function X found in the example worked out on p. 105. ' * At a point where the curve is concave upwards we sometimes say that the curve has a positive bending, and where it is concave downwards a negative bending. 114 DIM^EilENTIAL CALCULUS Solution. ., 432 /(X) = .,;=+—■ X First step. /'(x) = --tf. Second step. 432 „ 2x - = 0, a = 6, critical value. Third step. /..«.»-. Fourth step. /"(6) = + . Hence /(6) = 108, minimum value. The work of finding maximum and minimum values may f requentty be simplified by the aid of the following principles, which follow at once from our discussion of the subject. (a) The maximum and minimum values of a continuous function must occur altematelif. (b) When c is a positive constant, c ■fQsi') is a maximwn or a minimum for swch values of x, and such only, as ma]cef(x') a maximum or a minimum. Hence, in determining the critical values of x and testing for max- ima and minima, any constant factor may be omitted. When c is negative, o-f(^x') is a maximum whenf(x) is a minimum, mid conversely. (c) If c is a constant, /.^ -, j , x^ n ^ -^ -^ ' f(x) and c +f(x) have maximum and minimum values for the same values of x. Hence a constant term may be omitted when finding critical values of X and testing. In general we must first construct, from the conditions given in the problem, the function whose maximum and minimum values are required, as was done in the two examples worked out on pp. 103- 106. This is sometimes a problem of considerable difficulty. No rule applicable in all cases can be given for constructing the function, but in a large number of problems we may be guided by the following General directions. (a) Express the function whose maximum or minimum, is involved in the problem. (b) If the resulting expression contains more than one variable, the conditions of the problem will furnish enough relations between the variar hies so that all may be expressed in terms of a single one. MAXIMA AND MINIMA 115 (c) To the resulting function of a single variable apply one of our two rules for finding maximum and minimum values. (d) In practical problems it is usually easy to tell which critical value will give a maximum and which a minimum value, so it is not always necessary to apply the fourth step of our rules. (e) Draw the graph of the function (p. 104) in order to check the work PROBLEMS 1. It is desired to make an open-top box of greatest possible volume from a sqviare piece of tin whose side is a, by cutting equal squares out of the corners and then fold- ing up the tin to form the sides. What should be the length of a side of the squares cut out ? Solution. Let x = side of small square = depth of box ; then a — 2x = side of square forming bottom of box, and volume is V= {a— 2xYx; ■ which is the function to be made a maximum by varying »; Applying rule, dV First step. — = (a -2x)2 — 4x (a - 2x) = a^ _ 8 ax + 12x2. dx ■ Second step. Solvins,' a- — Sax -\- 12x^ = gives critical values x = - and — . ° 2 6 It is evident from the figure that z = - must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, X = - is found to give a maximum volume Hence the side of the square to be cut out is one sixth of the side of the given square. The drawing of the graph of the function in this and the following problems is left to the student. 2. Assuming that the strength of a beam with rectangular cross section varies directly as the breadth and as the square of the depth, what are the dimensions of the strongest beam that can be sawed out of a round log whose diameter is d ? Solution. If X = breadth and y = depth, then the beam will have maximum strength when the function xy^ is a maximum. From the figure, y^ = d^ — x^ ; hence we should test the function f(x) = x{d^-x^). First step. f'{x)=-2x'' + d^-x' = d^-3x^. Second step, d^ — 3x^ = 0. .: x = = critical value which gives a maximum. Therefore, if the beam is cut so that depth = Vf of diameter of log, and breadth ^V^^ of diameter of log, the beam will have maximum strength. 116 DIFFERENTIAL CALCULUS 3. What is the width of the rectangle of maximum area that can be inscribed in a given segment OAA' of a parabola ? Hint. If OC=h, JBC=h-x and FP'= 2 y ; therefore the area of rectangle FDD'P' is But since P lies on the parabola y''=2px, the function to he tested is 2(ft-a;)V2pi. Ana. Width = I A. 4. Find the altitude of the cone of maximum volume that can be inscribed in a sphere of radius r. Hint. Volume of cone = i irx^y. Batx^= BC x CD=y (2r- y) ; there- tore the function to be tested is f(.y) = ~yH2r-y). Ans. Altitude of cone = ^r. 5. Find the altitude of the cylinder of maximum volume that can be inscribed in a given right cone. Hint. Let ^IC- ;• and BC = /i,. Volume of cylinder = Tra:^?/. But from similar triangles ABC and BBG I r:x::h:h-y. Hence the function to be tested is r (h - ?/) h "^ Ans. Altitude = J ft. 6. Divide a into two parts such that their product is a maximum. Ans. Each part = - ■ 7. Divide 10 into two such parts that the sum of the double of one and square of the other may be a minimum. Ans. 9 and 1. 8. Find the number that exceeds its square by the greatest possible quantity. Ans. \. 9. What number added to its reciprocal gives the least possible sum ? Ans. 1. 10. Assuming that the stiffness of a beam of rectangular cross section varies directly as the breadth and the cube of the depth, what must be the breadth of the stiffest beam that can be cut from a log 16 inches in diameter ? Ans. Breadth = 8 inches. 11. A water tank is to be constructed with a square base and open top, and is to hold 64 cubic yards. If the cost of the sides is SI a square yard, and of the bottom $2 a square yard, what are the dimensions when the cost is a minimum 1 What is the minimum cost ? Ans. Side of base = 4 yd., height = 4 yd., cost $96. ^ 12. A rectangular tract of land is to be bought for the purpose of laying out a quarter-mile track with straightaway sides and semicircular ends. In addition a strip 35 yards wide along each straightaway is to be bought for grand stands, training quarters, etc. If the land costs •'?200 an acre, what will be the maximum cost of the land required ? ^l„s. $856. MAXIMA AND MINIMA 117 13. A torpedo boat is anchored 9 miles from the nearest point of a beach, and it is desired to send a messenger in the shortest possible time to a military camp situated 15 miles from that point along the shore. If he can walk 5 miles an hour but row only 4 miles an hour, required the place he must land. Ans. 3 miles from the camp. 14. A gas holder is a cylindrical vessel closed at the top and open at the bottom, where it sinks into the water. "What should be its proportions for a given volume to require the least material (this would also give least weight) ? Ans. Diameter = double the height. 15. What should be the dimensions and weight of a gas holder of 8,000,000 cubic feet capacity, built in the most economical manner out of sheet iron -j^j of an inch thick and weighing 2J lb. per sq. ft. ? Ans. Height = 137 ft., diameter = 273 ft., weight = 220 tons. ^ 16." A sheet of paper is to contain 18 sq. in. of printed matter. The margins at the top and bottom are to be 2 inches each and at the sides 1 inch each. Determine the dimen- sions of the sheet which will .require the least amount of paper. Ans. 5 in. by 10 in. *17. A paper-box manufacturer has in stock a quantity of strawboard 30 inches by 14 inches. Out of this material he wishes to make open-top boxes by cutting equal squares out of each corner and then folding up to form the sides. Find the side of the square that should be cut out in order to give the boxes maximum volume. Ans. 3 inches. 18. A roofer wishes to make an open gutter of maximum capacity whose bottom and sides are each 4 inches wide and whose sides have the same slope. What should be the width across the top ? Ans. 8 inches. 19. Assuming that the energy expended in driving a steamboat through the water varies as the cube of her velocity, find her most economical rate per hour when steam- ing against a current running c miles per hour. Hint. Let v = most economical speed ; then av^ = energy expended each hour, a being a constant depending upon the partic- ular conditions, and v-c= actual distance advanced per hour. Hence -^^ is the energy expended per mile of distance advanced, and it is therefore the function whose minimum is wanted. . _ s „ 20. Prove that a conical tent of a given capacity will require the least amount of canvas when the height is V2 times the radius of the base. Show that when the caayas is laid out flat it will be a circle with a sector of 152° 9' cut out. A bell tent 10 ft. high should then have a base of diameter 14 ft. and would require 272 sq. ft. of canvas. 21. A cylindrical steam boiler is to be constructed having a capacity of 1000 cu. ft. The material for the side costs S2 a square foot, and for the ends $3 a square foot. Find radius when the cost is the least. , 10 ^^ 22. In the corner of a field bounded by two perpendicular roads a spring is situated 6 rods from one road and 8 rods from the other. How should a straight road be run by this spring and across the corner so as to cut off as little of the field as possible ? Ans. 12 and 16 rods from comer. What would be the length of the shortest road that could be run across ? .4ns. (6^-1- 8^)'"roda 118 Dli^FEEENTIAL CALCULUS 23. Show that a square is the rectangle of maximum perimeter that can be inscribed in a given circle. 24. ' Two poles of height a and 6 feet are standing upriglit and are c feet apart. Find the point on the line joining their bases such that the sum of the squares of the distances from this point to the tops of the poles is a minimum. Ans. Midway between the poles. AYhen will the sum of these distances be a minimum ? 25. A conical tank with open top is to be built to contain V cubic feet. Determine the shape if the material used is a minimum. 26. An isosceles triangle has a base 12 in. long and altitude 10 in. Find the rec- tangle of maximum area that can be inscribed in it, one side of the rectangle coincid- ing with the base of the triangle. 27. Divide the number 4 into two such parts that the sum of the cube of one part and three times the square of the other shall have a maximum value. 28. Divide the number a into two parts such that the product of one part by the fourth power of the other part shall be a maximum. 29. A can buoy in the form of a double cone is to be made from two equal circular iron plates of radius r. Find the radius of the base of the cone when the buoy has the greatest displacement (maximum volume). Ans. rVf. 30. Into a full conical wineglass of depth a and generating angle a there is care- fully dropped a sphere of such size as to cause the greatest overflow. Show that the radius of the sphere is d gjn a sin a + cos g a 31. A wall 27 ft. high is 8 ft. from a house. Find the length of the shortest ladder that vrill reach the house if one end rests on the ground outside of the wall. Atis. 18 Vl3. 32. A vessel is anchored 3 miles offshore, and opposite a point 5 miles further along the shore another vessel is anchored 9 miles from the shore. A boat from the first vessel is to land a passenger on the shore and then proceed to the other vessel. "What is the shortest course of the boat ? Ans. 13 miles. 33. A steel girder 25 ft. long is moved on rollers along a passageway 12.8 ft. wide and into a corridor at right angles to the passageway. Neglecting the width of the girder, how wide must the corridor be ? Ans. 5.4 ft. 34. A miner wishes to dig a tunnel from a point A to a, point B 300 feet below and 500 feet to the east of A. Below the level of A it is bed rock and above A is soft earth. If the cost of tunneling through earth is SI and through rock S3 per linear foot, find the minimum cost of a tunnel. Ans. S1348.53. 35. A carpenter has 108 sq. ft. of lumber with which to build a box with a square base and open top. Find the dimensions of the largest possible box he can make. Ans. 6x6x3. 36. Find the right triangle of maximum area that can be constructed on a line of length h as hypotenuse. ft Ans. — = = length of both legs. V2 37. What is the isosceles triangle of maximum area that can be inscribed in a given circle ? Ans. An equilateral triangle. 38. Find the altitude of the maximum rectangle that can be inscribed in a right triangle with base b and altitude h. " ^ Ans. Altitude = - • MAXIMA AND MINIMA 119 39. Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse Vhfl + a^y^ = a^l^. Ans. a V2 and 6 V2 ; area = 2 ab. 40. Find the altitude of the right cylinder of maximum volume that can be inscribed in a sphere of radius r. 2 r Ans. Altitude of cylinder = Vi 41. Find the altitude of the right cylinder of maximum convex (curved) surface that can be inscribed in a given sphere. Ans. Altitude of cylinder =,r V2. 42. What are the dimensions of the right hexagonal prism of minimum surface whose volume is 36 cubic feet ? A-ns. Altitude = 2 V3 ; side of hexagon = 2. • 43. Find the altitude of the right cone of minimum volume circumscribed about a given sphere. Ans. Altitude = 4r, and volume = 2 x vol. of sphere. 44. A right cone of maximum volume is inscribed in a given right cone, the vertex of the inside cone being at the center of the base of the given cone. Show that the altitude of the inside cone is one third the altitude of the given cone. 45. Given a point on the axis of the parabola v^ = 2px at a distance a from the vertex ; find the abscissa of the point of the curve nearest to it. Ans. x = a — p. 46. What is the length, of the shortest line that can be drawn tangent to the ellipse Vx^ + aV = a,^^ and meeting the coordinate axes ? Ans. a + b. 47. A Norman windovT consists of a rectangle surmounted by a semicircle. Given the perimeter, required the height and breadth of the window when the quantity of light admitted is a maximum. Ans. Radius of circle = height of rectangle. 48. A tapestry 7 feet in height is hung on a wall so that its lower edge is 9 feet above an observer's eye. At what distance .from the wall should he stand in order to obtain the most favorable view ? Ans. 12 feet. Hint. The vertical angle subtended by the tapestry in the eye of the observer must be at a maximum. 49. What are the most economical proportions of a tin can which sliall have a given capacity, making allowance for waste ? 2 Vs Ans. Height = x diameter of base. IT Hint. There is no waste In cutting out tin for the side of the can, but for top and bottom a hexagon of tin circumscribing the circular pieces required is used up. Note 1. If no allowance is made for waste, then height = diameter. Note 2. We know that the shape of a bee cell is hexagonal, giving a certain capacity for honey with the greatest possible economy of wax. 50. An open cylindrical trough is constructed by bending a given sheet of tin of breadth 2 a. Find the radius of the cylinder of which the trough forms a, part when the capacity of the trough is a maximum. 2 a Ans. Rad. = — ; i.e. it must be bent in the form of a semicircle. IT 51. A weight Wis to be raised by means of a lever with the force F at one end and the point of support at the other. If the weight is suspended from a point at a distance a from the point of support, and the weight of the beam is w pounds ^ ^ J. per linear foot, what should be the length of the lever in order that w/M/MJM/MMm'///M tiie force required to lift it shall be a minimum ? r — 777 Ji Ans. X = -*/ feet. 120 DIFFEEEE"TIAL CALCULUS 52. An electric arc light is to be placed directly over the center of a circular plot of grass 100 feet in diameter. Assuming that the intensity of light varies directly as the sine of the angle under which it strikes an illuminated surface, and inversely as the square of its distance from the surface, how high should the light be hung in order that the best possible light shall fall on a walk along the circumference of the plot? ^ ^'^ * + U3.715. — leet. V2 53. The lower corner of a leaf, whose width is a, is folded over so as just to reach the inner edge of the page, (a) Find the width of the part folded over when the length of the crease is a minimum, (b) Find the width when the area folded over is a minimum. Ans. (a)fa;(b)Ja. 54. A rectangular stockade is to be built which must have a, certain area. If a stone wall already constructed is available for one of the sides, find the dimensions which would make the cost of construction the least. Ans. Side parallel to wall = twice the length of each end. 55. A cow is tethered by a perfectly smooth rope, a slip noose in the rope being thrown over a large square post. If the cow pulls the rope taut in the direction shown in the figure, at what angle will the rope leave the post ? Ans. 30°. 56. When the resistance of air is taken into account, the inclination of a pendulum to the vertical may be given by the formula d = ae- *' cos (nt + e) . Show that the greatest elongations occur at equal intervals — of time. n 57. It is required to measure a certain unknown magnitude x with precision. Suppose that n equally careful observations of the magnitude are made, giving the results „ „ „ . . . „ The errors of these observations are evidently X— ttj, X— flji * — "^si •■•. x—a„, some of which are positive and some negative. It has been agreed that the most probable value of x is such that it renders the sum of the squares of the errors, namely (X - ai)2 + (X - 02)2 + (X - ag)^ + . . . + (x _ a„)2, a minimum. Show that this gives the arithmetical mean of the observations as the most probable value of x. 58. The bending moment at B of a beam of length I, uniformly loaded, is given by the formula M = i wlx — i wx^, where w = load per unit length. Show that the maximum bending moment is at the center of the beam. 59. If the total waste per mile in an electric conductor is r where c = current in amperes, r = resistance in ohms per mile, and t = a constant depending on the interest on the investment and the depreciation of the plant, what is the relation between c, r, and t when the waste is a minimum ? Ans. cr = t. MAXIMA AND MIli^IMA 121 60. A submarine telegraph cable consists of a core of copper wires with a covering made of nonconducting material. If x denote the ratio of the radius of the core to the 'thickness of the covering, it is known that the speed of signaling varies"as x^ log - • X 1 Show that the greatest speed is attained when x = ~p ■ ve 61. Assuming that the power given out by a voltaic cell is given by the formula (r + Rf where E = constant electromotive force, r = constant internal resistance, B = exter- nal resistance, prove that P is a maximum when r = B. 62. The force exerted by a circular electric current of radius a qn a small magnet whose axis coincides with the axis of the circle varies as X {a? + x2)t where x = distance of magnet from plane of circle. Prove that the force is a maxi- mum when a; = - • 2 63. We have two sources of heat at A and B with intensities a and 6 respectively. The total intensity of heat at a distance of x from A is given by the formula Show that the temperature at P will be the lowest when %-_-.«--lJ- a «' (Z — X _ v'fe that is, the distances BP and AP have the same ratio as the cube roots of the corre- sponding heat intensities. The distance of P from A is 64. The range OX of a projectile in a vacuum is given by the formula v^ sin 2 . P R ^'i \ where u^ = initial velocity, g = acceleration due to grav- -^ ity, = angle of projection with the horizontal. Find the angle of projection which gives the greatest range for a given initial velocity. Ans. = 48°. 65. The total time of flight of the projectile in the last problem is given by the formula ^ _ 2 ti, sm ~ g At what angle should it be projected in order to make the time of flight a maximum ? Ans. = 90°. 122 DIFFERENTIAL CALCULUS 66. The time it takes a ball to roll down an inclined plane AB is given by the foniiula — \gs\n%(t> Neglecting friction, etc., what must be the value of to make the quickest descent ? Ans. = 45°. 67. Examine the function (x — l)2(x + l)^ for maximum and minimum values. Use the first method, p. 111. Solution. f(x) = (X - 1)2 (x + 1)'. First step. f'(x) = 2(x - l)(x + 1)3 + 3(x - l)Hx + 1)" = (x - l)(x + l)2{5x - 1). Second step, (x - 1) (x + 1)^ (5 x - 1) = 0, X = 1, — 1, J, which are critical values. Third step. f'(x) = 5 (x - 1) (x + 1)^ (x - i) . Fourth step. Examine first for critical value x = 1 (C in ^S^^^)- whenxl,/'(x) = 5(+)(+)2(+)= +. Therefore, when x = 1 the function has a minimum value/(l) = (= ordinate of C). Examine now for the critical value x = ^ (B in figure). WhenxJ,/'(x) = 5(-)(+)2(+)=-. Therefore, when x = J the function has a maximum value /(J) = 1.11 (= ordinate of B). Examine lastly for the critical value x =— 1 (A in figure). Whenx<-l,/'(x) = 5(-)(-)2(-)= +. Whenx>-l,/'(x) = 5(-)( + )2(-)=+. Therefore, when x = — 1 the function has neither a maximum nor a minimum value. 68. Examine the function a — b(x — c)' for maxima and minima. Solution. /(x) = (X — 6 (x — c)*. f{x)= -■ 3 (x - c)* Since x = c is a critical value for which /'(x) = oo, but for which /(x) is not infinite, let us test the function for maximum and minimum values when x = c. When X < c, /'(x) = + . When X > c, /'(x) = — . Hence, when x = c = OM the function has a maximum value /(c) : Examine tlie following functions for maximum and minimum values : 69. (x — 3)2(x — 2). Ans. x = J, gives max. = ^7 ; X = 3, gives min. = 0. 70 (x - 1)S (x - 2)2. X = f , gives max. = .03456 ; X = 2, gives min. = ; X = 1, gives neither. :MP. MAXIMA AND MINIMA 123 71. (x- 4)5(2 + 2)*. Ans. x =- 2, givos max.; ^ = h gives it|in.; S = 4, gives neither. 72. (x-2)^{2x + l)K a; =- J, gives max.; X = }J, gives min.; a; = 2, gives neither. 73. (X + l)*(x - 5)2. x=i, gives max.; s = — 1 and 5, give min. 74. (2x — a)*(x — a)f. x = — , gives max.; o X = a, gives min.; X = - , gives neither. 75. x(x — 1)2 (x + 1)'. a; = J, gives max.; X = 1 and — 1^, give min.; X = — 1, gives neither. 76. X (a + x)'' (a — x)'. x=— a and -, give max.; o a X = , gives min. ; X = a, gives neither. 77. 6 + c (x — a)^. X = a, gives min. = 6. 78. a — 6(x — c)^. • No max. or min. x2 _ 7 X + 6 • . . 79. ■ X = 4, gives max.; x-10 (a — x)^ x'^ X = 16, gives min. a 80. i '—■ x = -, gives mm. a— 2x 4 1 T. 4- x^ 81 t — ^ . X = J, gives min. 1 + X — s^ gg xg-3x + 2_ x=V2, givesmin. = 12V2-17; x^ + Sx + 2 r- r X = — V2, givesmax. = — 12 V2 — 17; X = — 1, — 2, give neitliev. (x — a) (6 — x) 2 a6 . (a - 6)^ 83_ ^ •' ' ZL. x = . gives max. = -^^ '—• + 6 4ab 84.— + -- a; = — --, gives mm. ; X a — X 0, f"(z)— — . .-. curve is concave upwards to the left and concave downwards to the right of x (A in figure) . -wiien x < |, f"(x) = -; and when x > f , f"{x) = + . .-. curve is concave downwards to the left and concave upwards to the right of X = I (JB in figure) . The curve is evidently concave upwards everywhere to the left of A, concave down- wards between A (0, 1) and B (f, J^), and concave upwards everywhere to the right of B. 2. (2/-2)3 = (x-4). y = 2 + {x — 4)^. y Solution. First step. dy 1, -^ = - (x ■ dx 3^ ■dx2 9^^' ■4)- ■4)- Second step. When x = 4, both first and second derivatives are infinite. Third step. d^v d^v When X <4, — ^ — + : but when x > 4, — 4 = dx' dx^ We may therefore conclude that the tangent at (4, 2) is perpendicular to the axis of X, that to the left of (4, 2) the curve is concave upwards, and to the right of (4, 2) it is concave downwards. Therefore (4, 2) must be considered a point of inflection. Z. y = x^. Ans. Concave upwards everywhere. 4- y = 5- B. y = X? ■2x — x2. . y = x3-3x2-9x + 9. y = a + {x — by. 8. ah/ = - ■ax^ + 2 a'. 9. y = x*. 10. y = x*-12x« + 48x2 11. y = sinx. 50. Concave downwards everywhere. Concave downwards to the left and concave up- wards to the right of (0, 0). Concave downwards to the left and concave up- wards to the right of (1, — 2). Concave downwards to the left and concave up- wards to the right of (6, a). Concave downwards to the left and concave up- wards to the right of (o^) -bi- concave upwards everywhere. Concave upwards to the left of x = 2, concave downwards between x = 2 and x = 4, concave upwards to the right of x = 4. Points of inflection are x = nir, n being any integer. 128 DIFFERENTIAL CALCULUS 12. y = tanx. Arts. Points of inflection are x = rvn;n being any integer. 13. Show that no conic section can have a point of inflection. 14. Show that the graphs of e= and log x have no points of inflection. 86. Cuxve tracing. The elementary method of tracing (or plotting) a curve whose equation is given in rectangular coordinates, and oise with which the student is already familiar, is to solve its equation for y (or x), assume arbitrary values of x (or y"), calculate the correspond- ing values of y (or x), plot the respective points, and draw a smooth curve through them, the result being an approximation to the required curve. This process is laborious at best, and in case the equation of the curve is of a degree higher than the second, the solved form of such an equation may be unsuitable for the purpose of computation, or else it may fail altogether, since it is not always possible to solve the equation for y or x. The general form of a curve is usually all that is desired, and the Calculus furnishes us with powerful methods for determining the shape of a curve with very little computation. The first derivative gives us the slope of the curve at any point ; the second derivative determiaes the intervals within which the curve is concave upward or concave downward, and the points of inflection separate these intervals ; the maximum points are the high points and the minimum points are the low points on the curve. As a guide in his work the student may follow the Rule for tracing curves. Rectangular coordinates. First Step. Find the first derivative ; •place it equal to zero ; solving gives the abscissas of maximum and minimmn points. Second Step. Find the secmtd derivative; place it equal to zero; solv- ing gives the abscissas of the points of inflection^ Third Step. Calculate the corresponding ordinates of the points whose abscissas were found in the first two steps. Calculate as many more points as may be necessary to give a good idea of the shape of the curve. Fill out a table such as is shown in the example worked out. Fourth Step. Plot the points determined and sketch in the curve to correspond with the results shown in the table. If the calculated values of the ordinates are large, it is best to reduce the scale on the T-axis so that the general behavior of the curve will be shown within the limits of the paper used. Coordinate plotting paper should be employed. MAXIMA AND MINIMA 129 EXAMPLES Trace the following curves, making use of the above rule. Also find the equations of the tangent and normal at each point of inflection. \.y = rS-Oa 2 + 24 z-1. Solation. Use the above rule. First step.. ?/' = 3x2-16 x + 24. 3x2- 18a; + 24 = 0, a;.= 2, 4. Second step. 2/"=6x-18 6a;- 18 = 0, x = 3. Third step. .7! y y y" Kemarks Direction of Curve 2 -7 13 + - max. > concave ^own 3 11 — pt. of infl. i- concave up 4 9 + min. 6 29 + + J Fourth step. Plotting the points and sketching in the curve, we get the figure shown. To find the equations of the tangent and normal to the curve at the point of inflec- tion Pj(3, 11), use formulas (1), (2), pp. 76, 77. This gives 3 x + 2/ = 20 f or the tangent and 3 2/ — X = 30 for the normal. 2. 2/ = x3-6x2-36x + 5. I^+ Ans. Max. {- 2, 45) ; min. (6, - 211) ; pt. of infl. (2, - 83) ; tan. y + 48 x - 13 = ; nor. 48 y - X + 3986 = 0. 3. 2/ = x*- 2x2 + 10. A-ns. Max. (0, 10); min. (±1, 9); pt. of infl. / 1 85\ 4. ?/ = |x*-3x2 + 2. Ans. Max. (0, .2) ; min. (± VI, — |); pt. of infl. (± 1, - \). 5. y = 6x l + x2 Ans. Max. (1, 3) ; min. (- 1, - 3) ; pt. of infl. (0, 0), 6. s^ = 12x — x8. -Ans. Max. (2, 16); min. (-2, —16); pt. of infl. (0, 0). 7. 42/ + X* - 3x2 + 4 = 0. Max. (2, 0); min. (0, - 1). 130 DIFFEKENTIAL CALCULUS 8. i/ = a;S_3a;2_9a; + 9. 9. 23/ + x3- 9x + 6 = 0. 10. y = »» - 6z2 _ I6i + 2. 11. y(l + x'') = x. 12. 2/ = 13. V = er'*. 15. y = {x + l)*(a; - 5)2. a; + 2 16. y = — V- x' 17. y = x'-Sx^-2ix. 18. v = 18 + 36i-3x2_2x8. 19. y = x — 2 cos X. 20. y = Sx-x'. 21. ?/ = a;3_9x2 + 15s -3 22. s2y = 4 + x. 23. 4y = !*- 6x=+ 5. 24. y = ? x^ + Sa' 25. y = sinz H 27. y = 5x-2z2- Jx3 28.. = 1±^. 2x 29. 2/ = X — 2 sin X. 30. y = log cos X. 31. y = log(l + x2). CHAPTER IX DIFFERENTIALS 87. Introduction. Thus far we have represented the derivative of .y=f(x) by the notation ^^ dx ^ =/'(-)• We have taken special pains to impress on the student that the symbol ^ dx was to be considered not as an ordinary fraction with dy as numerator and dx as denominator, but as a single symbol denoting the Umit of the quotient ^y Air as Aa; approaches the limit zero. Problems do occur, however, where it is very convenient to be able to give a meaning to dx and-cZ?/ separately, and it is especially useful in applications of the Integral Calculus. How this may be done is explained in what follows. 88. Definitions. If f'(x) is the derivative of f(x) for a particular value of X, and Aa; is an arbitrarily chosen increment of x, then the differ- ential off(x), denoted by the symbol d/Qc}, is defined by the equation (A) • df(x-)=f'Cx-)AT. If now /(a-) = X, then /'(a;) = 1, and (^) reduces to dx = Ax, showing that when x is the independent variable, the differential of x(= dx) is identical with Aa;. Hence, if y =/(a;), (A) may in general be written in the form (S) dy=r(x)dx* . y * On account ol the position which the derivative f'(z) here oconpies, it is sometimes called the differential coefficient. The student should observe the important fact that, since dx may be given any arbi- trary value whatever, dx is independent of x. Hence, dy is a function of two independeiit variables x and dx. 131 132 DIFFEEENTIAL CALCULUS The differential of a function equals its derivative multiplied hy the differential of the independent variable. Let us illustrate what this means geometri- P'\ cally. p/\dy Let /'(a;) be the derivative of y=f{oS) at P. --^f^'!^ Take ia; = J'Q, then / M M'X' dy =f'(x) dx = tanT-PQ==^-PQ = QT. Therefore dy, or df(x), is the increment (= QT^ of the ordinate of the tangent corresponding to dx* This gives the following interpretation of the derivative as, a fraction. If an arbitrarily chosen increment of the independent variable x for a point P (x, y') on the curve y =f(x) be denoted by dx, then in the derivative -^=f'(x) = tanT, dx dy denotes the corresponding increment of the ordinate drawn to the tangent. 89. Infinitesimals. In the Differential Calculus we are usually con- cerned with the derivative, that is, with the ratio of the differentials dy and dx. In some applications it is also useful to consider dx as an infinitesimal (see § 15, p. 13), that is, as a variable whose values remain numerically small, and which, at some stage of the investiga- tion, approaches the limit zero. Then by (5), p. 131, and (2), p. 19, dy is also an infinitesimal. In problems where several infinitesimals enter we often make use of the following Theorem. In problems involving the limit of the ratio of two infinites- imals, either infinitesimal may be replaced by an infinitesimal so related to it that the limit of their ratio is unity. Proof. Let a, /3, a', /3' be infinitesimals so related that ( C) limit - = 1 and limit ^ = 1. ^ ^ a /3 * The student should note especially that the differential (= dy) and the increment (= &y) of the function corresponding to the same value of dx (= Ax) are not in general equal. For, in the figure, dy = QT, but Ay=QP'. DIFFERENTIALS 133 We have — = —. — .— identically, ■ limit -, ■ limit ^ Th. II, p. 18 a' p . 1 . 1. By (C) aad limit — = limit — -, /3 ^' = limit — /3' ' cc cc (D) ■ ■ . limit — = limit --■ Q. E. D. p /3' Now let us apply this theorem to the two following important limits. For the independent variable x, we know from the previous section that Aa; and dx are identical. \x Hence their ratio is unity, and also limit — = 1. That is, by the clx above theorem, (^) In the limit of the ratio of Aa; and a second infinitesimal, Ak mdy- he replaced hy dx. On the contrary it was shown that, for the dependent variable y, Az/ and dy are in general unequal. But we shall now show, however, that in this case also Aw limit =^=1. dy Since . ^™^ ^ —^=fCx'), we way write Aa; = Aa; ^ If =/'(.) + . where e is an infinitesimal which approaches zero when Aa; = 0. Clearing of fractions, remembering that Aa; = dx, . A?/ =f{x) dx + e- Aa;, or Ay = dy + €- Ax. (B), p. 131 Dividing both sides by Ay, ^ dy , Ax Ay Ay ■ . limit '^y ^1 ' ' Ax = Ay and hence }^^^r. ^ =1. That is, by the above theorem, Aa; = dy (F') In the limit of the ratio of Ay and a second infinitesimal. Ay may he replaced hy dy. 134 DIFFERENTIAL CALCULUS 90. Derivative of the arc in rectangular coordinates. Let s be the length* of the arc AP measured from a fixed point A on the curve.^ Q. Denote the increment of « (= arc PQ) by A^. 0^£^y' The definition of the length of arc depends on yihx', the assumption that, as Q approaches P, 1 I limit /5l^2£dZe\i. If we now apply the theorem on p. 132 to this, we get ((?) In the limit of the ratio of chord PQ and a second infinitesimal, chord PQ may he replaced hy arc P§(=As). From the above figure {H') (chord PQf^ (Aa:)' + (Az/)^ Dividing through by (Aa;)^, we get (:'^)-Ht} Now let Q approach P as a limiting position ; then Aa; = and we \dx) \dxj r«i„„» limit /• chord PQ\ limit /As\ ds l Similarly, if we divide (-H") by (^Ay^ and pass to the limit, we get (25) ^=aI(^^Vi. &_ WdxV dy~^\dy) Also, from the above figure, a Ax . a Ay cos a = — , sm p = f chord P^ chord PQ Now as Q approaches P as a limiting position ==r, and we get dx . du (26) cosr = — , sinr = — . ds ds [since from («) limit —-^£-—= limit — = ^, and limit ^ = limit ^' = ^ .1 L ahoriPQ As ds chord PQ As dt 1 * Defined in § 209. DIFFERENTIALS 135 Usmg the notation of differentials, formulas (25) and (26) may be written (27) (28) ds dx. dy. Substituting the value of ds from (27) in (26), (29) cos r = . h(l)' sin r ==: . dx '+^f] K"'! An easy way to remember the relations (24)-(26) between the differentials dx, dy, ds is to note that they are correctly represented by a right triangle whose hypotenuse is ds, whose sides are dx and dy, and whose angle at the base is t. Then Y / X y dx ' O X ds = -V(dxy+(dyy, and, dividing by dx or dy, gives (24) or (25) respectively. Also, from the figure, cos T = ■ dx ds' dy sm r = — ; ds the same relations given by (26). 91. Derivative of the arc in polar coordinates. In the derivation which follows we shall employ the same figure and the same notation used on pp. 83, 84. From the right triangle PBQ (chord P^)' = <:FEy+cBQy = (p sin Aey +(p + Ap-p cos A^)". -^ Dividing throughout by (A^)^ we get /chord P(g Y_ ,/ sin Ad V /Ap l-cosAg V. V Ae AO \Ae Ad 136 DIFFERENTIAL CALCULUS Passing to the limit as A^ diminishes towards zero, we get ' ds dd) ^'^W' (30) de M'^ \do) In the notation of differentials this becomes (31) ds = p'+{'4 -,1 2 These relations between p and the differentials ds, dp, and d6 are correctly represented by a right triangle Ijjj whose hypotenuse is ds and whose sides are dp and pd6. Then "P -^ ds = V(^pddy+(^dpy, -^ and dividing by d8 gives (30). Denoting by -yjr the angle between dp and cZs, we get at once de tan "^ = p which is the same as (-4), p. 84. dp iLLnsTRATi VE ExAiMPLE 1. Find the differential of tlie arc of tlie circle x^ + y^ = /■*. Solution. Differentiating, -2. = dx y To find ds in terms of x we substitute in (27), giving L yH L y^ J W'\ -vA^iTT^ To find ds in terms of y we substitute in (28), giving Illustrative Example 2. Eind the differential of the arc of the cardioicl p = a (1 — cos 9) in terms of 6. Solution. Differentiating, — = a sin 9. Substituting in (31), gives ds = [a2(l - cos^)2 + a? sin^^J^di? = a [2 - 2 cos^J^dS = a r4sin2 -l*d^ = 2 a sin - dj9. , limit chord PQ _ Uniit As ^ ds limit sinAS^ By (ff), p. 134 By §22, p. 21 2 31^2 Aff ^^^ limit Ll^osAS limit ?._ limit „;_, A9 2^ n i n R^™ o .j«oo oi A9=o ^9 A9=o 49 -A9=oS™-^ Afl~° By 39, p. 2, and §22, p. 21 DIFFEKENTIALS 137 KXAMPLES Find the differential of arc in eacli of the following curves : 2. y = ax'. ds = Vl + 4 0^X2 (ij. 3. y = x^. d!8=Vl + 9x*dx. 4. y' = x2. (Zs = jv'4 + 9ydy. 5. E* + V* = a*. d. = ^2d2/. 6. 6^x2 + aV - a%^. HINT, e^ = «'^-*^ a2 \ a2 - x2 7. e^cosx = 1. ds = sec X dx. 8. /D = a cos ^. ds= add. 9. p' = a?oos2e. ds = aVseG20d0. 10. p = aeS""'". ds = p CSC a dff. 11. p = a9. ds = ao ■Vl + log' ade. 1 12. p = aS. ds = - Va' + p' dp. 13. (a) x2 - 2/2 = a2. (h) x^ + y^ = a*. (b) x2 = 4 ay. (i) y' = ax^. (c) 2/ = e^+ e-^. (J) 2/ = logx. (d) xy = a. W 4 X = 2/^. (e) 2/ = log sec X. (f) p = 2atanSsine. (1) p p = a sec^ - . '^ 2 (g) p = aseo3g. (m) (0 = l + sinS. (n) p(9 = a. 92. Formulas for finding the differentials of functions. Since the differential of a function is its derivative multiplied by the differen- tial of the independent variable, it follows at once that the formulas for finding differentials are the same as those for finding derivatives given in § 33, pp. 34-36, if we multiply each one by dx. This gives us I (^(e)=0. ^ II dQr) = dx. III d(u + v — w')=du + dv — dw. IV d(ov) = cdv. C i^ V d(uv) = udv + v du. VI d(v''')=nv''-'^dv. 138 DIFFERENTIAL CALCULUS Via d(3f) = nx''-''dx. ,-.-, v/mX vdu — udv VII d{-\= ; Vila dC^^^^- VIII d(\og„v') = log„e ^ • IX d( + sec 0. 11. r = JtanS^ + tan^. 12. /(x) = (loga;)8. 13. 0(i) = - dy = 2na;2»-i , (l + a;2)» + i dy = Zx^dx 2 (x3 _ 1) dy = ;2(e2^— e-2»:)(te. dy = : e^^hogx + -\dx. ds = t'-^ydt. dp = C0S2 (t> dr = -. seo* Sde. f'{x)dx^ SQogxYdx X '{t)dt^ Zt^dt (1-<2)S (1-*'')^ rxjogx 1 logxdx_ L 1 - X ^ 'J {l-xf 15. d Faro tan log w] = ^"^ y[l + (log2/)2] 16. d \r arc vers - - V2n/-y2] = ^"^ L *■ J ^2ry-y^ 17. d r^2£^ _ hog tan^l = _ Jl- . L2sin2 2 2 J Bin*^ CHAPTER X RATES 94. The derivative considered as the ratio of two rates. Let be the equation of a curve generated by a moving point P. Its coordi- nates X and y may then be considered as functions of the time, as explained iu § 71, p. 91. Differentiating y^ with respect to i, by IIV, we have (32) ^ = /'(;c)-- dt ^ '' dt At any instant the time rate of change ^ of y (or the function) equals its derivative multiplied hy the time rate of change of the independent variable. Or, write (32) in the form dy (33) dx ^ -^ dx dt The derivative measures the ratio of the time rate of change of y to that of X. ds — being the time rate of change of length of arc, we have from (12), p. 92, (34) ds dt -A^HW which is the relation indicated by the above figure. As a guide ia solving rate problems use the following rule : FiEST Step. Draw a figure illustrating the problem. Denote by x, y, z, etc., the quantities which vary with the time. Second Step. Obtain a relation between the variables involved which will hold true at any instant. 141 142 DIFFERENTIAL CALCULUS Third Step. Differentiate with respect to the time. Fourth Step. Make a list of the given and required quantities. Fifth Step. Substitute the hnown quantities in the result found by differentiating (third step), and solve for the unknown. EXAMPLES 1. A man is walking at the rate of 5 miles per hour towards the foot of a tower 60 ft. high. At what rate is he approaching the top when he is 80 ft. from the foot of the tower ? Solntian. Apply the above rule. First step. Draw the figure. Let x = distance of the man from the foot and y = his distance from the top of the tower at any instant. Second step. Since we have a right triangle, 2/Z = a" + 3600. Third step. Differentiating, we get „ dy „ dx 2y-^ = 2s — 1 or, dt dt ' ., dy xdx (A) — = 1 meaning that at any instant whatever dt y dt {Eate of change ofy) = (-) {rale of change of'x). ■ Fourth step. a; = 80, — = 5 miles an hour, dt = 6 X 5280 ft. an hour y = -Vx^ + 3600 dy _ ^ = 100. dJ ~ ' Fifth step. Substituting back in {A), dy 80 — = — X 5 X 5280 ft. per hour dt 100 = 4 miles per hour. Ans. 2. A point moves on the parabola dy = x^ in such a way that when a; = 6, the abscissa is increasing at the rate of 2 ft. per second. At what rates are the ordinate and length of arc increasing at the same instant ? Solution. First step. Plot the parabola. Second step. 6y = x^. Third step. „dy dx 6-i = 2x — , or, dt dt ' dy _x dx 'dt~3 ' Tl' This means that at any point on the parabola (Rate of change of ordinate) -. - 1 (rote of change of abscissa). RATKS 14;- Fourth step. — =2 ft. per second. dt x = 6. ^ = ? 1 dt y = — =&. — = ? 6 (it Fifth step. Substituting back in (JB), dy 6 -- = - X 2 = 4 ft. per second. Ana. dt Z Substituting in (34), p. 141, ds — = V(2)2 + (4)2 = 2 VS ft. per second. Ans. at From the first result we note that at the point P (6, 6) the ordinate changes twice as rapidly as the abscissa. If we consider th? point P' (— 6, 6) instead, the result is — = — 4 ft. per second, the dt minus sign indicating that the ordinate is decreasing as the abscissa increases. 3. A circular plate of metal expands by heat so that its radius increases miiformly at the rate of .01 inch per second. At what rate is the surface increasing when the radius is two inches ? • Solution. Let x = radius and y = area of plate. Then y = ir'3?. dv dx (C) =^ = 27rx — ■ ^ ' dt dt Thai is, at any instant the area of the plate is increasing in square inches 2irx times as fast as the radius is increasinj? in linear inches. dx dy X = 2, _ = .01, -^ = 1 ' dt dt Substituting in (C), — = 27r X 2 X .01 = .04 7rsq. in. per sec. Ans. 4. An arc light is hung 12 ft. directly above a straight horizontal walk on which a boy 5 ft. in height is walking. How fast is the boy's shadow lengthening when he is walking away from the light at the rate of 168 ft. per minute ? Solution. Let x = distance of boy from a point directly under light X, and y = length of boy's shadow. From the figure, y.y + x:: 6:12, or y = ^x. T»-i» ^. i. dy 6dx Diflerentiatine, — = ; .» ._ ^' dt 7 dt «r M i.e. the shadow is lengthening f as fast as the boy is walking, or 120 ft. per minute. 5. In a parabola y^ — 12 x, if x increases uniformly at the rate of 2 in. per second, at what rate is y increasing when x — 3 in. ? Ans. 2 in. per sec. ds_ dt" -.VI. dt~ -l-M ds_ = Vi. 144 DIFFERENTIAL CALCULUS 6. At what point on the parabola of the last example do the abscissa and ordinate increase at the same rate ? Ans. (3,6). 7. In the function y = ix^ + Q, what is the value of x at the point where y increases 24 times as fast as a; ? Ana. x=±2. 8. The ordinate of a point describing the curve x^ + y' = 25 is decreasing at the rate of IJ in. per second. How rapidly is the abscissa changing when the ordinate is 4 inches ? a '^ n ■ Avs. — = 2 in. per sec. dt 9. Find the values of x at the points where the rate of change of a;' — 12a;2 + 45X-13 is zero. Ans. x = 3 and 5. 10. At what point on the ellipse 16 x^ + 9 2/^ = 400 does y decrease at the same rate that X increases ? Ans. (3,'/-). H. Where in the first quadrant does the arc increase twice as fast as the ordinate ? 4ms. At 60°. A point generates each of the following curves. Find the rate at which the arc is increasing in each case : 12. y^ = 2x; — = 2, x = 2. Ans. ' dt ' 13. xy = 6;^ = 2,y = 3. dt , ' dx 14. x2 + 42/2 = 20 ; — =- 1, 2/ = 1. dt az 15. 2/ = x8 ; — = 3, X = - 3. dt 16. y^ = x'; ^ = 4,2/ = 8. dt 17. The side of an equilateral triangle is 24 inches long, and is increasing at the rate of 3 inches per hour. How fast is the area increasing ? Ans. 36V3 sq. in. per houi. 18. Find the rate of change of the area of a square when the side 6 is increasing at the rate of a units per second. Ans. 2 ai sq. units per sec. 19. (a) The volume of a spherical soap bubble increases how many times as fast as the radius ? (b) When its radius is 4 in. and increasing at the rate of J in. per second, how fast is the volume increasing ? Ans. (a) 4 ttt" times as fast ; (b) 32 TT cu. in. per sec. How fast is the surface increasing in the last case ? 20. One end of a ladder 50 ft. long is leaning against a perpendicular wall stand- ing on a horizontal plane. Supposing the foot of the ladder to be pulled away from the wall at the rate of 3 ft. per minute ; (a) how fast is the top of the ladder descending when the foot is 14 ft. from the wall ? (b) when will the top and bottom of the ladder move at the same rate ? (c) when is the top of the ladder descending at the rate of 4 ft. per minute ? Ans. (a) | ft. p'er min. ; (b) when 25 V2 ft. from wall ; (c) when 40 ft. from wall. 21. A barge whose deck is 12 ft. below the level of a dock is drawn up to it by means of a cable attached to a ring in the floor of the dock, the cable being hauled in by a windlass on deck at the rate of 8 ft. per minute. How fast is the barge moving towards the dock when 16 ft. away ? Ans. 10 ft. per minute. RATES 145 22. An elevated car is 40 ft. immediately above a surface car, their tracks inter- secting at right angles, li the speed of the elevated car is 16 miles per hour and of the surface car 8 miles per hour, at what rate are the cars separating 5 minutes after they meet ? Ans. 17.9 miles per hour. 23. One ship v^as sailing south at the rate of 6 miles per hour ; another east at the rate of 8 miles per hour. At 4 p.m. the second crossed the track of the first where the first was two hours before ; (a) how was the distance between the ships changing at 3 P.M. ? (b) how at 5 p.m. ? (c) when was the distance between them not changing ? .4ns. (a) Diminishing 2.8 miles per hour ; (b) increasing 8.73 miles per hour ; (c) 3 : 17 P.M. 24. Assuming the volume of the wood in a tree to be' proportional to the cube of its diameter, and that the latter increases uniformly year by year when growing, show that the rate of growth when the diameter is 3 ft. is 36 times as great as when the diameter is 6 inches. 25. A railroad train is running 15 miles an hour past a station 800 ft. long, the track having the form of the parabola y^ = 600 X, and situated as shown in the figure. If the sun is just rising in the east, find how fast the shadow S of the locomotive L is moving along the wall of the station at the instant it reaches the end of the wall. Solution. y '2 -600 a;. dx kNorth 2j^^ = 600 dt dt dz _ y dy ' dt ~ 300 dt ' Substituting this value of — in dt ds Tt' W Now /dsy- ^ /y_ dyV , (^Y \dt) \300 dt) \dtj we get da di' : 15 miles per hour = 22 ft. per sec. : 400 and ^ = ? dt Substituting back in (D), we get dy dt = 13J ft. per second. Ans. 26. An express train and a balloon start from the same point at the same instant. The former travels 50 miles an hour and the latter rises at the rate of 10 miles an hour. How fast are they separating ? ^ns. 61 miles an hour. 146 DIFFERENTIAL CALCULUS 27. A man 6 ft. tall walks away from a lamp-post 10 ft. high at the rate of 4 miles an hour. How fast does the shadow of his head move ? Arts. 10 miles an hour. 28. The rays of the sun make an angle of 30" with the horizon. A ball is thrown vertically upward to a height of 64 ft. How fast is the shadow of the ball moving along the ground just before it strikes the ground ? Ans. 110.8 ft. per sea 29. A ship is anchored in 18 ft. of water. The cable passes over a sheave on the bow 6 ft. above the surface of the water. If the cable is taken in at the rate of 1 ft. a second, how fast is the ship moving when there are 30 ft. of cable out ? Ans. If ft. per sec. 30. A man is hoisting a chest to a window 50 ft. up by means of a block and tackle. If he pulls in the rope at the rate of 10 ft. a minute while walking away from the building at the rate of 5 ft. a minute, how fast is the chest rising at the end of the second minute ? Ans. 10.98 ft. per min. 31. Water flows from a faucet into a hemispherical basin of diameter 14 inches at the rate of 2 ou. in. per second. How fast is the, water rising (a) when the water is halfway to the top ? (b) just as it runs over ? (The volume of a spherical segment = ^ Trr^ A 4- ^ ir ^5, where h = altitude of segment.) 32. Sand is being poured on the ground from the orifice of an elevated pipe, and forms a pile which has always the shape of a right circular cone whose height is equal to the radius of the base. If sand is falling at the rate of 6 cu. ft. per sec, how fast is the height of the pile increasing when the height is 5 ft. ? 33. An aeroplane is 528 ft. directly above an automobile and starts east at the rate of 20 miles an hour at the same instant the automobile starts east at the rate of 40 miles an hour. How fast are they separating ? 34. A revolving light sending out a bundle of parallel rays is at a distance of ^ a mile from the shore and makes 1 revolution a minute. Find how fast the light is traveling along the straight beach when at a distance of 1 mile from the nearest point of the shore. Ans. 15.7 miles per min. 35. A kite is 150 ft. high and 200 ft. of string are out. If the kite starts drifting away horizontally at the rate of 4 miles an hour, how fast is the string being paid out at the start ? Ans. 2.64 miles an hour. 36. A solution is poured into a conical filter of base radius 6 cm. and height 24 cm. at the rate of 2 cu. cm. a second, and filters out at the rate of 1 cu. cm. a second. How fast is the level of the solution rising when (a) one third of the way up ? (b) at the top? Ans. (a) .079 cm. per sec; (b) .009 cm. per sec. 37. A horse runs 10 miles per hour on a circular track in the center of which is an arc light. How fast will his shadow move along a straight board fence (tangent to the track at the starting point) when he has completed one eighth of the circuit ? Ans. 20 miles per hour. 38. The edges of a cube are 24 inches and are increasing at the rate of .02 in. per minute. At what rate is (a) the volume increasing ? (b) the area increasing ? 39. The edges of a regular tetrahedron are 10 inches and are increasing at the rate of .3 in. per hour. At what rate is (a) the volume increasing ? (b) the area increasing? 40. An electric light hangs 40 ft. from a stone wall. A man is walking 12 ft. per second on a straight path 10 ft. from the light and perpendicular to the wall. How fast is the man's shadow moving when he is 30 ft. from the wall ? Ans. 48 ft. per sec. RATES 147 41. The approach to a drawbridge has a gate whose two arms rotate about the same axis as shown in the figure. The arm over the driveway is 4 yards long and the arm over the footwalk is 3 yards long. Both arms ro- tate at the rate of 5 radians per minute. At what rate is the distance between the ex- tremities of the arms chang- ing when they make an angle of 45° with the horizontal ? Ans. 24 yd. per min. 42. A conical funnel of radius 3 inches and of the same depth is filled with a solu- tion which filters at the rate of 1 cu. in. per minute. How fast is the surface falling when it is 1 inch from the top of the funnel ? , 1 Ans. i-ir 111. per min. 43. An angle is increasing at a constant rate. Show that the tangent and sine are increasing at the same rate when the angle is zero, and that the tangent increases eight times as fast as the sine when the angle is 60°. CHAPTER XI CHANGE OF VARIABLE 95. Interchange of dependent and independent variables. It is some- times desirable to transform an expression involving derivatives of y with respect to x into an equivalent expression involving instead deriv- atives of X with respect to y. Our examples will show that in many cases such a change transforms the given expression into a much simpler one. Or perhaps x is given as an explicit function of y in a problem, and it is found more convenient to use a formula involving dSj ui J (m11 Cm U -T-' ^r-;» etc., than one involvinsr -^i -r4) etc. We shall now proceed dy dy^ dx dx^ ^ to find the formulas necessary for makuag such transformations. Given y =/(.-r), then from IXTI we have <^^'i^ dy 1 dx (35) _ = _ , 3- =^ ^ dx dx ay Ty dii dx giving -^ in terms of —-■ Also, by XXV, dx dy ' dly^^d^/dy\^d^/dy\dy^ dx' dx \dxj dy \dx) dx or ^^±n\dy^ da? dy\dx\ dx \dyj- d^x But ^ l-^\ = - --^; and ^ = i- from (35). dxV dx dx dy) dy Substituting these in (^), we get d^x d^y dy^ (36) dx^ /dry [dyj 148 CHANGE OF VARIABLE 149 . . d^y . , . dx , d^x c.. ., -, giving -7^ in terms of — and ■— ■ Similarly, d^y_ dy' dy W) and so on for higher derivatives. This transformation is called changing the independent variable from x to y. Illustrative Example 1. Change the independent variable from x to y in the equation 3 /^Y_ ^ ^ _ ^ /'^Y= \dxV dx dx' dx^ \dx) Solution. Substituting from (35), (36), (37), dy^ I / 1 \ I ^y^ ^y ~ V— V I (—11 /'— V \dyj I \dy/ ^ W Eeducing, we get d^x dH _ dy^ dy^ a much simpler equation. 96. Change of the dependent variable. Let and suppose at the same time y is a function of z, say TT7- ii dy d^y . ■ . ^ dz d^z , We may then express -j-, -y|» etc., m terms 01 — » --^> etc., as follows. In general, s is a function of y by (B), p. 45 ; and since y is a func- tion of X by (-4), it is evident that g is a function of x. Hence by XXV we have 150 DIFFEEENTIAL CALCULUS Similarly for higher derivatives. This transformation is called changing the dependent variable from y to 2, the independent variable remaining x throughout. We will now illustrate this process by means of an example. Illustrative Example 1. Having given the equation change the dependent variable from y to z by means of the relation {F) y — tan z. Solution. From {F), dy „ dz d^y , dH ^ . . ^ /dzV -^ = sec^z — > — ^ = sec^'z h 2sec2ztanz( — 1 • dx dx dx^ dx^ \dx/ Substituting in (E), „ d^z , „ . ^ IdzV , , 2(l + tanz)/ „ dzV sec^z f- Ssec^ztanzl — ) = 1 H — ^ — ■ ^(seo^z — | , dx^ \dx/ l + tan^z \ dxJ dPz /dz\^ and reducing, we get — - — 2 I — ) = cos%. Ans. CuC \uX/ 97. Change of the independent variable. Let 3/ be a function of -x, and at the same time let x (and hence also ^) be a function of a new variable t. It is required to express dy d^y dx dx^ in terms of new derivatives having t as the independent variable. ^^y^^ dy^dydx^^^ dt dx dt du dt * d (dy\ Also d'y_d/dy\ d /dy\dt dt\dx) do? dx\dxj dt\dx) dx dx But differentiating (Ji) with respect to t. l(dy\_d ldy\ dx d^y dy d^x dt dt df dt df dt\dx) dt dx /dxV \dtl \dt Therefore CHAJNGE OF VARIABLE , 151 dx d^y dy d^x d^y_dt'^~'Jt'dF and so on for higher derivatives. This transformation is called changing the independent variable from x to t. It is usually better to work out examples by the methods illustrated above rather than by using the formulas deduced. Illustrative Example 1. Change the independent variable from i to f in the equation. (C) dx' dx by means of the relation W x = e'. Solution. dx — = e'; therefore dt (B) dt di^'" Also ^ = ^*; therefore dx dt dx (F) dy_^_,dy dx dt ■> ^'^° dx2 dx\dt/ dt dx dt\dljdx dt dx Substituting in the last result from (E), ^^' dx' dt' dt Substituting (Z>), (F), (G) in (C), \ dt' dt I \ dt/ d'y and reducing, we get -z-j + y = 0. Atis. Since the formulas deduced in the Differential Calculus generally involve derivatives of y with respect to x, such formulas as (.4) and (B) are especially useful when the parametric equations of a curve are given. Such examples were given on pp. 82, 83, and many others will be employed in what follows. 152 DIFFERENTIAL CALCULUS 98. Simultaneous change of both independent and dependent variables. It is often desirable to change both variables simultaneously. An im- portant case is that arising in the transformation from rectangular to polar coordinates. Since x=p cos 6 and y = psm6, the equation f(x,y')=(i becomes by substitution an equation between p and d, defining /j as a function of 6. Hence p, x, y are all functions of Q. Illustrative Example 1. Transform the formula for the radius of curvature (A) R: into polar coordinates. Solution. Since In (A) and (B), pp. 150, 151, t is any variable on which x and y depend, we may in this case let t = 0, giving dy (B) {C) dy d9 . — = — , and dx 6x d9 dx d^y dy d^x d'y _ ded£^~d0de^ d^ ~ /di\3 W/- Substituting (B) and (C) in (A), we get \21? dx d^y dy dH m R. R. d9 dff^ de ( /dxV dx d^y dy d'^x MWde m But since x= p cos9 and y = p sin 5, we have ^; = -psin^+cos^|;g = pcos^+sin^|; dx dff g=-pcos.-.sin.g,cos.g;g = -,sin..2cos.g Substituting these in (D) and reducing. ,d'P p^ + 2 \d9/ 2 d^p P — - Ans. CHANGE OF VAEIABLE 153 EXAMPLES Change the independent variable from xtoy in the four following equations : d^ dH_ dH dx „ _2y— = 0. dy^ dy dx^ \dx/ dx dy^ \dyl \ dx I \dxy \ dx J dxdx" \dy^/ \dy / dy^ Change the dependent Variable from y to z in the following equations : . , , ^.dH dz d^z , „ , „ Ans. (z + 1) — : = + 2^ + 22. ^ ' dx^ dxdx2 da;2 1 + 2/2 \dxj ' (ii2 \dxj 7.,.£!|_(3v^ + 2x,2)J| + |2(^y+2x,? + 3xv|? + x3,3 = 0,2/ = e^ (Ja;3 \ dx I dx^ y_ \dx/ dx j dx . d'z ^ d'z , „ „dz , , . Ans. 2x \-3x^ l-x' = 0. dx' dx^ dx Change the independent variable in the following eight equations : 8.^ ^^ + _J^ = 0, x = cosi. ^„s. ^ + 2/ = 0. (to2 l-x^cfe l-x^ ' di2 " 9. (i_x^)^_x^ = 0, x = cos.. ^ = 0. ^ 'dx2 dx dz- 10. (1 _ y2) ^ _ y ^ + a^M = 0, y = sinx. ^' + a^u = 0. dy^ dy dx- U.x^^ + 2x^ + ^, = 0, x = l. §^r + «^^ = 0. dx2 dx x2 z dz2 12.,3^ + 3x2^ + xf^+« = 0, x = .« S + « = 0- dx3 dx2 dx (i«^ 13.^ + ^^^ + _^_ = 0, x = tan^. S + '-^ = «• dx2 1 + x2 dx (1 + x2)2 d52 14. ^ + sit— + sec^s = 0, s = arc tan i. ds^ ds (j2 „ dii Ans. (1 + i2) ^- + (2 1 + M arc tan <) — + 1 = 0. d2j/ 1 , d'^y 2 dy . 15.x^g + a2, = 0, . = -. An. -+-- + a2. = 0. 154 DIFFERENTIAL CALCULUS 111 the following seven examples the equations are given in parametric foim. Find — and — - in each case : dx dx'' 16. x = 7 + P, y = S + i^-Si*. Ans. ^ = 1 - 6*2, ^ =- 6. dx ax' 17. X = cott, y = sinH. Ans. — = - SsinHcost, —^=SBin^t(4— 6mnH). dx da? dy d?y 1 18. I = a(cosJ + isini), 2/ = a(smt — icosi). Ans. -^ = tant, — = ^ " ^ ' dx dx^ atcosH 19. X = , y = 1+t 1+t 20. x = 2t, y = 2- P. 21. x = l-t', y = t". 22. X = o cos t, y = b sin t. dy 23. Tran.sf orm — - by assuming x = p cos^, y = p sin 6. aRI Atis. 24. Let f{x, y) = be the equation of a curve. Find an expression for its slope I — ) in terms of polar coordinates. pcosO + sitid — \dx/ ^ dy '^ dff Ans. -^ — psinff + costf — CHAPTER XII CURVATURE. RADIUS OF CURVATURE 99. Curvature. The shape of a curve depends very largely upon the rate at -which the direction of the tangent changes as the point of contact describes the curve. This rate of change of direction is called curvatute and is denoted by K. "We now proceed to find its analytical expression, first for the simple case of the circle, and then for curves in general. 100. Curvature of a circle. Consider a circle of radius It. Let T = angle that the tangent at P makes with OX, and T + At = angle made by the tangent at a neighboring point P'. Then we say At = total curvature of arc PP'- If the point P with its tangent be supposed to move along the curve to P', the total curvature (= At) would measure the total change in direction, or rotation, of the tangent; or, what is the same thing, the total change in direction of the arc itself. Denoting by s the length of the arc of the curve measured from some fixed point (as ^)to P, and by As the length of the arc PP', then the ratio At As measures the average change in direction per unit length of arc* Since, from the figure, \g=R At At 1 or — = — ' As B * Thus, if At = - radians (= 30°) , and As = 3 centimeters, then -^ = ^ radians per centi- meter = 10° per centimeter = average rate of oliange of direction. 155 156 DIFFERENTIAL CALCULUS it is evident that this ratio is constant everywhere on the circle. This ratio is, by definition, the curvature of the circle, and we have (38) K=-. The curvature of a circle equals the reciprocal of its radius. 101. Curvature at a point. Consider any curve. As in the last At = total curvature of the arc PP', section, and At — = average curvature of the arc PP'. More important, however, than the notion of the average curvature of an arc is that of curvature at a point. This is obtained as follows. Imagine P' to approach P along the curve ; then the limiting value of the average curvature ( = — I as P' ap- proaches P along the curve is defined as the curvature at P, that is. Curvature at a point = . „ ( — 1 = — . ^ As = \^As/ ds „ dr (39) .■.K=— = curvature. ds Since the angle At is measured in radians and the length of arc A« in units of length, it follows that the unit of curvature at a point is one radian per unit of length. 102. Formulas for curvature. It is evident that if, in the last sec- tion, instead of measuring the angles which the tangents made with OX, we had denoted by t and t 4- At the angles made by the tangents with any arbitrarily fixed line, the different steps would in no wise have been changed, and consequently the results are entirely independent of the system of coordinates used. However, since the equations of the curves we shall consider are all given in either rectangular or polar coordinates, it is necessary to deduce formulas for K in terms of both. We have or tan t = — 5 dx T = arc tan 32, p. 31 dx CURVATURE Differentiating with respect to x, using XX (^) -T- = TT^o- Also 157 ^^(S (^) From (24), p. 134 Dividing (^) by (£) gives dr dx But (40) s h(i)T t^2;_ dr ds ds dx K=. -- K. Hence 1 + If the equation of the curve be given in polar coordinates, K may be found as follows : From (5), p. 84, r = 6 + ■^. Differentiating, (C) dr i .^"^ de~ dd But tan i/r = -|- • From (A}, p. 84 dp Te p .-. ii- = arctanT;-- ^ dp dd Differentiating by XX with respect to 6 and reducing, d^lr [ddj ^ de^ dd ' it \^y = a(m sm i — sin ml), i = J, 17. The curve . "' _ 4.ma . /m + 1\ ^ Ans. R = sm ' — t„. m-1 \ 2 / ° 21. rind the radius of curvature for each of the following curves at the point indicated ; draw the cuiTe and the corresponding circle of curvature : (a) x = f,2y = t; J = 1. (e) x-= t,y = 6«-i; t = 2. {h) x = f,V = fi; t = l. {i) x = 2e>,y = e-'; t = 0. (c) X = sint, y = cos2t; t = - ■ (g) x = smt,y = 2cost; t = —■ {d) x = l-t,y = 't^;t = 3. ^ (h) x = t^,y = t^ + 2t; t = l.^ 22. An automobile race track has the form of the ellipse x^ + 16y^ = 16, the unit being one mile. At what rate is a car on this track changing its direction (a) when passing through one end of the major axis ? (b) when passing through one end of the minor axis ? (c) when two miles from the minor axis ? (d) when equidistant from the minor and major axes ? Ans. (a) 4 radians per mile ; (b) ^ radian per mile. 23. On leaving her dock a steamship moves on an arc of the semicubical parabola iy^ = x^. If the shore line coincides with the axis of y, and the unit of length is one mile, how fast is the ship changing its direction when one mile from the shore ? Ans. -f^-g radians per mile. 24. A battleship 400 ft. long has changed its direction 30° while moving through a distance equal to its own length. What is the radius of the circle in which it is moving ? Ans. 764 ft. 25. At what rate is a bicycle rider on a circular track of half a mile diameter changing his direction ? Ans. 4 rad. per mile = 43' per rod. 26. The origin being directly above the starting point, an aeroplane follows approximately the spiral p = 6, the unit of length being one mile. How rapidly is the i aeroplane turning at the instant it has circled the starting point once ? 27. A railway track has curves of approximately the form of arcs from the follow- ing curves. At what rate will an engine change its direction when passing through the points indicated (1 mi. = unit of length) : (a) y = x", (2, 8) ? (d) j/ = e'-,x = 0? ^ (b) y = x=, (.3, 0) ? (e) 2/ = cosx, x = -? (c) x^- 2/2 =8, (.3,1)? (t) pe = i,e = i CHAPTER XIII THEOREM OF MEAN VALUE. INDETERMINATE FORMS (.b.o) 105. RoUe's Theorem. Let y =f(£) be a continuous single-valued function of x, vanishing for x= a and x = h, and suppose that f'(x) changes continuously when X varies from a to 6. The function will then be rep- X resented graphically by a continuous curve as in the figure. Geometric intuition shows us at once that for at least one value of x be- tween a and h the tangent is parallel to the axis of X (as at P); that is, the slope is .zero. This illustrates Rolle's Theorem : If f(£) vanishes when x = a and x = b, and f(x) and f'(x) are con- tinuous for all values of x from x= a to x = b, then f'{x) will be zero for at least one value of x between a and b. This theorem is obviously true, because as x increases from a to J, /(x) cannot always increase or always decrease as x increases, since f(a) = and /(&) = 0. Hence for at least one value of x between a and 6, f(x) must cease to increase and begin to decrease, or else cease to decrease and begin to increase ; and for that particular value of x the first derivative must be zero (§ 81, p. 108). That Rolle's Theorem does not apply when f(x) or f'(x) are discontinuous is illus- trated as follows : Fig. a shows the graph of a function which is discontinuous (= co) for a; = c, a value lying be- tween a and h. Pig. 6 shows a continuous func- tion whose first deriyative is discontinuous (= oo) for such an intermediate Fig. a Fig. 6 value X = c. In either case it is seen that at no point on the graph hetween a; = a and x = h does the tangent (or curve) become parallel to OX 164 THEOREM OF MEAN VALUE 165 106. The Theorem of Mean Value.* Consider the quantity Q defined bv the equation (^) /(5)-/(a)-(6-a)a=0. Let F(^x') be a function formed by replacing 6 by a; in the left-hand member of (5) ; that is, ((7) F(a:-) =/(x) -/(a) - (^x - «) Q. From (5), F(b-) = 0, and from (C), F(a) = ; therefore, by Rolle's Theorem (p. 164) F'Qc) must be zero for at least one value of x between a and h, say for x^. But by differentiatuig (C) we get Therefore, since and F'(x)=f(x)-Q- F'(x^-) = 0, then also /'(a:) - § = 0, Substituting this value of Q in (^), we get the Theorem of Mean Value, (44) /(&)-/(a) ^f,^^^-^^ aIFFEEENTIAL CALCULUS Equating (D) and (-£■), we get b — a ^ which is the Theorem of Mean Value. The student should draw curves (as the one on p. 164) to show that there may be more than one such point in the interval; and curves to illustrate, on the other hand, that the theorem may not be true if /(«) becomes discontinuous for any value of x between a and h (Fig. a, p. 164), or \if'(x') becomes discontinuous (Fig. h, p. 164). Clearing (44) of fractions, we may also write the theorem in the form (45) f(b-)=f- «)/'(«) - 4 (^' - «)V"(") - ^ (^ - ^ys= 0, we can derive the equation (I)) f(h) =fia) + (h- a-) /'(a) + 4 (* - «)V"(«) + ri(6-«y/"'(a;3), a) will not change sign, and the difference f(x)-fia') will have the same sign for all values of x in the interval [a — h, a + A], and, moreover, this sign will be the same as the sign off"(a'). It therefore follows from our definitions (^) and (5) that (^) f(a) is a maximum iff'(ci) = andf"(a') = a negative number; (i^) f{a) is a minimum iff'(a~) = andf"(a') = a positive number. These conditions are the same as (21) and (22), p. 113. III. Letf'(a)=f"(a)=^, and f'^a) ^ Q. From (X>), p. 167, replacing b hj x and transposing /(a), (G) fix-) -f(a-) = ,1 (x - ayf'\x^). - a 1 lin^xl _0 cosx_L:=o 1 /'(O) x_ x=0 F'(0) — cscx cot X. /"(O) 2 sin X cos x "1 ~ — ' — - = 0. Ans. . indeterminate. xji F"(0) cos X — X sin x]x = o I 113. Evaluation of the indeterminate form Ooo. If a function f{x) ■ ^(x) takes on the indeterminate form • oo for a; = a, we write the given function ■ /(a;)-^(;i) = -^ or = i:i_Z 1 r 1 so as to cause it to take on one of the forms 77 or ^, thus bruaging it under § 111 or § 112. ^ INDETEEMINATE FOEiVhS 175 Illustrative Example 1. Evaluate sec 3 a; cos 5 x for a; =- . • . 2 Solution. .sec3«cos5x"| „ = oo • 0. .■.indeterminate. Substituting for ,seo 3 x, the function becomes = ^-^ . cos3x cosSx F{x) f(-) \2/ cos 5x1 „ . , , mdeteruunate. _/7r\ cos3x \2/ ft — sinSx 1 =«. ...h- i=^ •3l=5~ 3' sin 5 X • 5~1 5 Ans. 114. Evaluation of the indeterminate form » — oo. It is possible in general to transform the expression into a fraction which will assume either the form j. ov ^■ Illustrative Example 1. Evaluate secx — tanx for x = - . 2 Solution, sec X — tanx] „ = oo — oo. .-.indeterminate. _ _. ^ , 1 sin X I— sin X fix) By Trigonometry, sec x — tan x — = = —^-^ . cosx eosx cosx F{x) f(-) \2 l-sinx1 1-10 . ^ ^ = = = - • .-. indeterminate. /7r\ cosx X„!: %) f'(-) W_-cosx-| ^_0_^o. ^n.. i^ ■ sin X EXAMPLES Evaluate the following expressions by differentiation : * 1 limit ^1+^. j^^ a 6. limit'ogsin2x ^^^^ x_oocx'' + (J c X — V log sin x 2 limit -9 limit cot X 7 limit -°..^^ 3 ^■^ = 0i^- -"• ^•^ = ^tan3^ 3 limit logx ^ ■ X = CO x" " ■ 8. .^ ^-Z tan (^ d limit ^^ n , *• x = o=>i;- ^- 9 limit logx ■ x = Ocotx S. "!!!'lr^- »• 10. "'""xlogsinx. 0. X _ CO log X . X = ° * In solving the remaining examples in this chapter it may be of assistance to tlie student to refer to §24, pp. 23, 24, where many special forms not indeterminate are evaluated. 176 DIFFERENTIAL CALCULUS 1 ta limit r ^ 1 An') —- 0. 19 limit r_l ^1. _i. a;=iLlogx logxj 11. ^™'^xcot™. j^o limit y ' y = cDgay limit ,. . 13. , _ E (t - 2 z) tan X. 2. 20. "™'i. fsec 6 - tan ^] . 0. 14. "™*xsin^. X = 00 X , _ limit .1 r •*• T A ^ = Lsin^ * 1 — cos ^ J 2 15. _f,X"log'X. [re positive.] " i_ v- t-j 21 limit r_J 1 1 ' ^ = Lsin^ 1 — cos J limit 22. limit r_^ ?-1. 1 16. ^ ^ £ (1 - tan ^) sec 2 e. 1. 2/ = 1 Lz/ - 1 log2/J 2 17. li™'(a^-0^)tan!^. ^. 23. "'"'!; [f- ^ , ^ ,1 \ 115. Evaluation of the indeterminate forms 0° , 1", oo°. Given a func- tion of the form /('a;)*^. In order that the function shall take on one of the above three forms, we must have for a certain value of x /(x) = 0, («) log/(a;). In any of the above cases the logarithm of y (the function) will take on the indetermiaate form 0.00. Evaluating this by the process illustrated in § 113 gives the limit of the logarithm of the function. This being equal to the logarithm of the limit of the function, the limit of the function is known.* Illustkative Example 1. Evaluate x^ when x = 0. Solution. This function assumes the indeterminate form 0° for s = 0. Let y = x^; then logj/ = xlogx = 0- — 00, when x = 0. By§n3, p. 174, log2/ = i2ii = ZLi5, when x = 0. 1 00 X * Thus, if limit logey=a, then ?/ = e». INDETERMmATE FOEMS 177 By § 112, p. 174, logy = -£- = - » = 0, when a; = 0. Since 2/ = x"', this gives log,x» = ; i.e., a;^ = 1. Ans. 1 Illustrative Example 2. Evaluate (1 + s)^ when x = 0. Solution. This function assumes the indeterminate form 1" for x = 0. 1 Let 2/ = (1 + x)i ; tlien log2/ = -log(l + x)= oo-O, when x = 0. By § 113, p. 174, logy = ^^0- + ^) ^ ^ ^^^^ ^^^ X 1 By § 111, p. 171, logy = —tI.= = x, when x = 0. 1 1 + X 1 11^ Since y = (l + x)^, this gives loge(l + x)=" = 1 ; i.e. (1 + x)^ = e. Ans. Illustrative Example 3. Evaluate (cot x)=™«' for x = 0. Solution. This function assumes the indeterminate form oo" for x = 0. Let y = (cotx)»™^; then logy = sin X log cotx = • oo, when x = 0. By § 113, p. 174, log y = i^^^^ = ^ , when x = 0. CSC X CO — cso'^x „ „ , , „ , » , n cot X sin X „ , By § 112, p. 174, logy = = — — = 0, when x = 0. — CSC X cotx cos^x Since y = (cotx)»'"=', this gives loge(cotx)'™^ = 0; i.e. (cotx)'™== = 1. Ans. EXAMPLES Evaluate the following expressions by difierentiation : 1- ^Z\ ^ ^- Ans. l- 7. ^™'* (e- + xf. Ans. eK „ limit /lY""" 1 8. "™^* (ootx)i5f5. I. a; = \x/ ■ .1 ^ limit ^- i™ (1 + "^)''- ^• 3. ^™';(sine)fne. 1. ^-" „*• ^ = ¥ 10. J™* (tan ^y'"^. 1. ^•"=^(1%-)"- - 11.^^-^* (cos J)^. ■ I-^«-. ^•x = o(l + *'"^)°°'" ^- 12. ^™^ (cotx)-. 1. 6_ limit /ay ^, limit / xy.n|L: ^|_ CHAPTER XIV CIRCLE OF CURVATURE. CENTER OF CURVATURE 116. Circle of curvature.* Center of curvature. If a circle be drawn through three points P^, ij, -^ on a plane curve, and if JJ and ij be made to approach i^ along the curve as a limiting position, then the circle will in general approach in magni- tude and position a limiting circle called the circle of curvature of the curve at the point P^. The center of this circle is called the center of curvature. Let the equation of the curve be P„(.!IC2,Vs) (1) y =fQo') ; PiC»v'yi"> and let x^, x^, x^ be the abscissas of the points ^, Py, P^ respectively, (a', yS') the coordinates of the center, and R' the radius of the circle passing through the three points. Then the equation of the circle is and since the coordinates of the points ij, ij, P^ must satisfy this equa- tion, we have (2) '(^o-«T+(2/o-/3'y-^"=0, Xx-ay+(jj-^y-R<-'=<). Now consider the function of x defined by Fix) = (x-a'y+Qy- ^y - R'\ in which y has been replaced hj f(x) from (1). Then from equations (2) we get * Sometimes called the osculating circle. The circle ol curvature was defined from another point of view on p. 161. 178 CIECLE AND CENTER. OP CURVATUEE 179 Hence, by RoUe's Theorem (p. 164), F'^x) must vanish for afc least two values of x, one lying between x^ and x^, say x', and the othfet lying between x^ and x^, say x" ; that is, F'(x') = 0, F'(x":) = 0. Again, for the same reason, F"(x) must vanish for some value of x between x' and x", say x^ ; hence F"(x^)^0. Therefore the elements a', /8', R' of the circle passing through the points ij, -^, JJ must satisfy the three equations F(x^) = 0, F'C^') = 0, F"(x^=0. Now let the points J^ and i^ approach -^ as a limiting position ; then x^, x^, a/, x", x^ will all approach aj^i as a limit, and the elements a, /3, B of the osculating circle are therefore determined by the three equations FCx^) = 0, n^„) = 0, F"Cx;)=0; or, dropping the subscripts, which is the same thing, (5) (a: - a) + (2/ - /3) ^ = 0, differentiating (A). Solving (5) and (C) for x— a and y — ^, we get ( -j4 =5^= ^[l + Z^M cZa; L \dxj (^) a;— a; = ' dx"" y-^=- ^-C hence the coordinates of the center of curvature are (^) a = x — ih(i; 1+ dx" ; p = y + dx^ ^0 180 DIFFEEENTIAL CALCULUS Substituting the values oi x — a and y — ^ from (D) in (^), and solving for B, we get Mm i? = ±: which is identical with (42), p. 159. Hence Theorem. The radius of the circle of curvature equals the radius of curvature. 117. Second method for finding center of curvature. Here we shall make use of the definition of circle of curvature given on p. 161. Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature (a, /8) corresponding to the point P(^, «/) on the curve. Then a = 0A = 01>-AD = 0D-BP = x-BP, ^ = AC=AB+BC=DP+BC = y+BC. But BP = Rsm.T, BC = R cos r. Hence {A) a = x—BsmT, /3 = y+iJ cost. From (29), p. 135, and (42), p. 159, dy SU1T= ^, COST y> — "--. \ \ \ c B y ^ X. y 2 1 1 ) X h(l)T . " Ml) Substitutiug these back in (^), we get R = Mm d^ dx^ (50) o = jr- P = y + £y From (23), p. 126, we know that at a point of inflection (as Q in the next figure) ^ = 0. dx'' CIECLE AND CENTER OF CUEVATUEE 181 Therefore, by (40), p. 157, the curvature K= ; and from (42), p. 159, and (50), p. 180, we see that in general a, /3, R increase without limit as the second derivative approaches zero. That is, if we suppose P with its tangent to move along the curve to P\ at the point of inflection Q the curvature is zero, the rotation of the tangent is momentarily arrested, and as the direction of rotation changes, the center of cur- vature moves out indefinitely and the radius of curvature becomes infinite. Illustrative Example 1. Find the coordinates of the center of curvature of the parabola ■y' = ipx corresponding (a) to any point on the curve ; (b) to the vertex. Solution. dy _ 2j V y^ (a) Substituting in (E), p. 179, a = x + y^ + 4p2 y2 2p -f y 4^2 = 3x + 2p. /S = y- 2^ y2 4p2 yZ Therefore ( 3x + 2b, 1 is the center of curvature ' corresponding to any point on the curve, (b) (2j3, 0) is the center of curvature corresponding to the vertex (0, 0). 118. Center of curvature the limiting position of the intersection of normals at neighboring points. Let the equation of a curve be The equations of the normals to the curve at two neighboring points ^ and .^ are ■ dy^ Pp(a,j3) r)^ = o. (x-X-) + iy^- P^ljBo'Vo) If the normals intersect at C'(a', /S'), the coordinates of this point must satisfy both equations, giving (^) (^„-«') + (yo-^')^=0. dx^ (p-cc':,:^iy-^'~)^=^ dx. ' From (2), p. 77, Xand T being the variable coordinates. 182 DIFFEEENTIAL CALCULUS Now consider the function of x defined by f in which y has been replaced by /(a;) from QA). Then equations (5) show that 0(a;„)=O, (x,~)=0, (t>Xx,)=0; and C'(^a', /8') will approach as a limiting position the center of cur- vature C(a, /3) corresponding to ^ on the curve. For if we drop the subscripts and write the last two equations in the form ^-(l)'-(-«3-' it is evident that solving for a' and /3' will give the same results as solving (5) and (C), p. 179, for a and /S. Hence Theorem. The center of curvature C corresponding to a point P on a curve is the limiting position of the intersection of the normal to the curve at P with a neighboring normal. 119. Evolutes. The locus of the centers of curvature of a given curve is called the evolute of that curve. Consider the circle of curvature corre- sponding to a point P on a curve. If P moves along the given curve, we may suppose the corresponding circle of curva- ture to roll along the curve with it, its ( ^'^)^-'''\^^~)'/Pi ° ^^^^^ varying so as to be always equal to \v \ \'-^yP^ t^s radius of curvature of the curve at the ^ — J<>Pi ■ pomt P. The curve CC, described by the center of the circle is the evolute of PH;. It is instructive to make an approximate construction of the evolute of a curve by estimating (from the shape of the curve) the lengths CIRCLE AND CENTER OJ CUEVATUEE 183 of the radii of curvature at different points on the curve and then drawing them in and drawing the locus of the centers of curvature. Formula (£), p. 179, gives the coordinates of any point (a, /8) on the evolute expressed in terms of the coordinates of the corresponding point (x, y) of the given curve. But «/ is a function of x ; therefore a = x- v-m dy_ dx dx"- ^=y^ dx"" give us at once the. parametric equations of the evolute in terms of the- parameter x. To find the ordinary rectangular equation of the evolute we elimi- nate X between the two expressions. No general process of elimination can be given that will apply in all cases, the method to be adopted depending on the form of the given equation. In a large number of cases, however, the student can find the rectangular equation of the evolute by taking the following steps : General directions for finding the equation of the evolute in rectangular coordinates. First Step. Find a and ^ from (50), p. 180. Second S^ep. Solve the two resulting equations for x and y in terms of a and /8. Third Step. Substitute these values of x and y in the given equation. This gives a relation between the variables a and /3 which is the equation of the evolute. Illusteatite Example 1. Find the equation of the evolute of the parabola y'-' = ipx. Solution. First step. Second step. Third step dy dx 2p d^y y da? (4p2/S)S = 4p(- a-_^y PP^ = -{oc-2py Eemembering that a denotes the abscissa and p the ordinate of a rectangular system of eoSrdinates, we see that the evolute of the parabola AOB is the semicubical parabola DC'E; the centers of curvature for 0, P, Pj, Pj being at C, C, Cjj G^ respectively. 184 DIFFERENTIAL CALCULUS Illustkative Example 2. Find the equation of the evolute of the ellipse Solution. dy _ l^x d^__ dx~ a^y' dx^ 6* First step. (a2-62)a;5 Second step. / a^^a \J "'-W-W' y- i "'^ t Third step, (aa)* + (6/3)' = (a^ - }fi)^, the equa- tion of the evolute EHJE'H' of the ellipse ABA'B'. E, E\ S\ H are the centers of curvature corresponding to the points A, A', B, B', on the curve, and C, C, C" corre- spond to the points P, P', P" When the equations of the curve are given in parametric form, we proceed to find -^ and — -^j as on p. 160, from (^) d^/ dx d'^y dy d^x dy _~dt d^'y _~dt~d^~~di~df _ dx dx da? /dx\ dt \dti and then substitute the results in formulas (50), p. 180. This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations. Illustrative Example 3. The parametric equations of a curve are X- — ■ — , y = -. 4 Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where t = 1, and draw the corresponding cir- cle of curvature. &_ i d^x _1 dJ~2' d^~2' dy_f d^y _ Solution. dt 2 ■ df2 Substituting in above formulas (A) and then in (50), p. 180, gives CIECLE AND CENTER OF CURVATURE 185 the parametric equations of the evolute. Assuming values of the parameter t, we cal- culate x,y; a, p from (B) and (C) ; and tabulate the results as follows : Now plot the curve and its evolute. The point (J, 0) is common to the given curve and its evolute. The given curve (semicubical parabola) lies entirely to the right and the evo- lute entirely to the left of a; = J-. The circle of curvature at A (J, J), where < = 1, will have its center at A' (— J, |) on the evolute and radius = AA'. To verify our work find radius of curvature at A. From (42), p. 159, we get ^(1 + i -= V2, when t = 1. B = This should equal the distance t X y a ;3 3 f 1 2 i ^ -'f ¥ 1 if A -H 3 1 i i -\ i i i -1 1 2 -\ -i -J -f H -A -u -3 -2 i -* -¥ -¥ -3 ^ -1 4^' = V(| -I- 1-)^ -I- (i - f)2 = V2. Illustrative Example 4. Find the parametric equations of the evolute of the cycloid, rx = a{t-smt), ^' {y:=a(l-oost). Solution. As in Illustkative Example 2, p. 160, we get dy _ sint d^y 1 dx" 1 — Gost' dx''~ a(l— cos«)2 Substituting these results in formulas (50), p. 180, we get ra = a(t + sini), ^ ' \/3 =— a(l— cosi). Ans. 186 DIFPEEENTIAL CALCULUS Note. If we eliminate t between equations (D), there results the rectangular equa^ tion of the evolute OO'Q,'' referred to the axes Cor and CjS. The coordinates of with respect to these axes are (— ira, —2a). Let us transform equations (D) to the new set of axes OX and OT. Then a = x — ira, p = y — 2a, Substituting in (D) and reducing, the equations of the evolute become fx = a(if — sini'), i(l — cosi^. fx = a iy = a Since (E) and (C) are identical in form, we have : The eoolvie of a cycloid is itself a cycloid whose geTierating circle equals that of the given cycloid. 120. Properties of the evolute. From (A), p. 180, (u4) a = a; — ^sinT, ^ = y +B cos t. Let us choose as independent variable the lengths of the arc on the given curve ; then x, y, B, t, a, /3 are functions of s. Differentiating (^) with respect to s gives (C) da dx „ dr ^- = -; B cos T -— ■ as ds as ■SUIT tZ/3 dv T^ • dr , — - = -p- — -Ksinr— - + cos r as ds ds dB ds dB But — - = cosT, ^ = sinT, from (26), p. 134; and — =— , ds ds \ y r ds B (38) and (39), p. 156. Substituting in (5) and (C), we obtain from (D) (^) da 1 . dB -z—= COST— jffi COST smT : ds B ds sm.T dB ds d^ . „ . 1 , dB dB -— = smT— ^smT- — + C0ST — = cost ds B ds ds Dividing (-E) by (Z>) gives (i?) -^=— COtT = : da tan t 1_ dy dx CIRCLE AND CENTER OF CURVATURE 187 But — = tan t' = slope of tangent to the evolute at C, and -J- = tan T = slope of tangent to the given curve at the corre- sponding point F(x, y). Substituting the last two results in QF), we get 1 tanT'=- tanr Since the slope of one tangent is the negative reciprocal of the slope of the other, they are perpendicular. But a line perpendicular to the tangent at J" is a normal to the curve. .Hence A normal to the given curve is a tangent to its evolute. Again, squaring equations (i>) and (-E') and addiag, we get • m-m- /d£Y \ds)' But if «' = length of arc of the evolute, the left-hand member of ds' (G) is precisely the square of -— (from (34), p. 141, where t= s, (XS s =«', x=a, y = y8). Hence (D) asserts that (fj= \ds/' ds ds That is, the radium of curvature of the given curve increases or decreases as fast as the arc of the evolute increases. In our figure this means that JJCi-PC=arcCCj. The length of an arc of the evolute is equal to the difference between the radii of curvature of the given curve which are tangent to this arc at its extremities. Thus ia Illustrative Example 4, p. 186, we observe that if we fold Q'P''(= 4 a) over to the left on the evolute, P^ will reach to 0', and we have : The length of one arc of the cycloid (as OO'Q"} is eight times the length of the radius of the generating circle. 121. Involutes and their mechanical construction. Let a flexible ruler be bent in the form of the curve C^C^, the evolute of the curve ^i^, and suppose a string of length B^, with one end fastened at C,, to 188 DIFFERENTIAL CALCULUS be wrapped around the ruler (or curve). It is clear from the results of the last section that when the string is unwound and kept taut, the free end will describe the curve j^^. Hence the name evolute. The curve ^I^ is said to be an invo- lute of C^C^. Obviously any point on the string will describe an involute, so that a given curve has an infinite number of involutes but only one evolute. The involutes P^P^, P^'Pj, P^"P," are ca&%A. parallel curves since the distance between any two of them measured along their common normals is con- stant. The student should observe how the parabola and ellipse on pp. 183, 184 may be constructed in this way from their evolutes. EXAMPLES Find the coordinates of tlie center of curvature and the equation .of the evolute of each of the following curves. Draw the curve and its evolute, and draw at least one circle of curvature. 1. The hyperbola ?- _ ^ = l a? 6^ Ans. (a? + b^) x^ P-- (a2+ 62)3/8 a* 6* evolute (aa)i - (6;8)i- = (a" + b^)i. Ans. cz = X + 3xiy^, p = y + Zxiyi> evolute (a + /3)l- + (a - jS)f = 2 ai. 3. Find the coordinates of the center of curvature of the cubical parabola y^ = a^x. 2. The hypocycloid xt +yi = ai. Ans. a = a* + 15y* P-- a*2/ — 9^' 6 aV 2 a* 4. Show that in the parabola xi + yi-=: ai we have the relation a + p = 3{x + y). 5. Given the equation of the equilateral hyperbola 2xy = a^; show that a + ^-- . {y + xY , a — tl — (y - xY From this derive the equation of the evolute (or + |8)f — (a — /3)l- = 2 ai. Find the parametric equations of the evolutes of the following curves in terms of the parameter t. Draw the curve and its evolute, and draw at least one circle of curvature. 6. The hypocycloid 7. The curve (x = a oos^ t, \y = a sin' J. X = 3 i2, y = St- i3. . ( a = a cos' t + 3 a cos t sirfi, 1^ /3 = 3 a oos'^ i sin i + a sin' t. \ (3 = -4<3. •<*), CIRCLE AND CENTEE OF CUEVATURE 189 8. The curve 9. The curve 10. The curve 11. The curve 12. The curve 13. The curve 14. The curve 15. The curve 16. The curve X = a (cost + isint), y = a(smi — icosJ). Ans. ( a = a cos (, \ ^ = o sin t. :3i, ■■P-6. 'x = \v = = 6 - «2, -.21. 'x - = 2i, = i2 _ 2. fx = '.y = = 4(, = 3 + t2. -x = = 9-i^ = 2*. fx - = 2«, ' y = 3 X - 3 Vy-- = £2. rs=2i, 17. x=4-i2, y = 2t. 18. x = 2«, y = 16-i^ 19. X = t, y = sin t. 20. x = -, 2/ = 3t. 21. x = P, y = \t'. A , = 3.^-1 '. j3 = -2t3. '_^=3«2. ' a = - «', 'j3 = ll + 3«2. r Of = 7 - 3 «2, {p = -2tK r 4i-<5 „ 12+5t^ f _4iS + 12t 3 2 i2 + i* [''= 2 r 12i* + 9 "" 4*3 ' 27 + 4t^_ 6« 22. X = i, 2/ = «'. 23. X = sin t, y = S cos i. 24. X = 1 — cos i, 2/ = t — sin i. 25. X = cos*i, y = sin^i. 26. X = a sec t, y = b\ an£. CHAPTER XV PARTIAL DIFFERENTIATION 122. Continuous functions of two or more independent variables. A function f(x, «/) of two independent variables x and y is defined as continuous for the values (a, V) of (a-, y) when limit x = af(x, y^=f(a, J), y = h no matter in what way r and y approach their respective limits a and I. This definition is sometimes roughly summed up in the state- ment that a very small change, in one or both of the independent variahleg shall produce a very small change in the value of the function.* We may illustrate this geometrically by considering the surface represented by the equation ^ _ ^^^ n Consider a fixed point P on the surface where x = a and y = h. Denote by Ax and Ay the increments of the independent variables X and y, and by Az the corresponding increment of the dependent variable z, the coordinates of P' being (x + Ax, y + Ay, z + Az}. At P the value of the function is z =f(a, b) = MP. If the function is continuous at P, then however Ax and Ay may approach the limit zero, A^ will also approach the limit zero. That is, M'P' will approach coincidence with MP, the point P' approaching the point P on the surface from any direction whatever. A similar definition holds for a continuous function of more than two independent variables. In what follows, only values of the independent variables are considered for which a function is contmuous. * This will be better understood if the student again reads over § 18, p. 14, on eontinuou? functions of a single variable. 190 PARTIAL DIFFERENTIATION 191 123. Partial derivatives. Since x and y are independent in X may be supposed to vary while y remains constant, or the reverse. The derivative of z with- respect to x when x varies and y remains constant * is called the partial derivative of z with respect to x, and is dz denoted by the sjrmbol — We may then write dx /•i^ ?!.- li^it l f(.^ + '^^' y) -/(^' y) 1 <^^^ g2=~A2;=0L Ax I Similarly, when x remains constant* and y varies, the partial derivative of z with respect to y is ^""^ Sy Ay=0l Ay J" — is also written — f(x, y), or — • dx dx^ ^ ^■' dx Similarly, — is also written ■ — f(x, tf), or — • dy dy dy In order to avoid confusion the round d ^ has been generally adopted to indicate partial differentiation. Other notations, however, which are in use are (i)' \£)'' ■^^^''' ^^' ■^^^*'' ^^' ■^-^'^' ^)' •^^^^' ^^' •^-'^' ^''■^' ^-' ""■ Our notation may be extended to a function of any number of independent variables. Thus, if u = F(x, y, g), then we have the three partial derivatives du du du. dF dF dF — ) — 5 — 5 or, — J — ) — dx dy dz dx dy oz Illustkative Example 1. Find the partial derivatives of z = ox^ + 2 6xj/ + cy^. dz Solution. — = 2ax + 2by, treating y as a constant, dx — = 2bx + 2cy, treating a; as a constant. dy * The constant values are substituted in the function before differentiating. t Introduced by Jacobi (1804-1851). 192 UIFFEKENTIAL CALCULUS Illustkative Example 2. Find the partial derivatives of m = sin (ax + by 4- cz). Solution. — = a cos (ax + by + cz), treating y and z as constants, Bx — = 6 cos {ax + iy + cz), treating x and z as constants, sy — = c cos {ax + by + cz), treating y and x as constants. dz Again turning to the function 8z we have, by (A), p. 191, defined — as the limit of the ratio of the Bx increment of the function (y being constant) to the increment of x, as tlie increment of x approaches the hmit zero. Similarly, (.B), p. 191, dz has defined :— • It is evident, however, that if we look upon these partial derivatives from the point of view of § 94, p. 141, then 8z dx may be considered as the ratio of the time rates of change of z and X when y is constant, and gg dy as the ratio of the time rates of change of z and y wlien x is constant. 124. Partial derivatives interpreted geometrically. Let the equa- tion of the surface shown in the figure be 2 ^fQe, y)- Pass a plane EFGH through the point P (where x= a and y — V) on the surface parallel to the XO-Z^-plane. Since the equation of this plane is y = i, the equation of the section JPK cut out of the surface is 2 =f(x, J), if we consider EF as the axis of Z and EH as the axis of X. In this plane -- means the same as — -? and we have ox dx dz — = tan MTP ox = slope of section JK at P. PARTIAL DIFFERENTIATION 193 Similarly, if we pass the plane BCD through P parallel to the rO^-plane, its equation is x = a 8z dz and for the section DPI, — means the same as -^ • Hence dy dy 3z dz — = -r- = — tan MT'P = slope of section DI at P. dy dy ^ -^ j;2 y2 ^ Illustrative Example 1. Given the ellipsoid 1- — ^ — = 1 ; find the slope of the section cf the ellipsoid made (a) by the plane j/ = 1 at the point where x = 4 and z is positive ; (b) by the plane x = 2 at the point where y = Z and z is positive. Solntian. Considering y as constant, 2 X 2 z 3z „ dz X = 0, or — = 24 6 3x ax 4z 2 ?/ 2 z dz dz y When X is constant, — ^H = 0, or — = — —■ 12 6 ay Sy 2z (a) When y =1 and x = 4, z = -»/- • .-. — = — -\\- ■ Ans. \ ^ ox \ o (b) When X = 2 and 2/ = 3, z = ^. .-. — = V2. Avs. ^ ' V2 5y 2 EXAMPLES 1. u = x' + 3xV — ^- -^™s- — = 3x^ + 6x2/; 3x — = 3x2-3^2. Sy 2. u = Ax^ + Bxy + Cy'^ + Dx + Ey + F. — = 2Ax + By + D; 8x du Bx + 2Cy + E. sy Z. u = {ax2 + hy^ + cz^Y- X 4. tt = arc sin - - Su 2anxu ax Su ax2 2bnyu dy du _ ax2 + by'' + cz^ 1 dx du_ V^ 2-x2 X ^ y Vy^ — x^ 8m ■ 5. u = xi'. — = 2/x!'-i; ax — = xJ'logx. 6. u = axVz + 6a;y'z* + cy* + <^^- — = 3 axVz + ^Z/'z* + ^^ ; — = 2ax8yz + SteyV + 6cy5 ; — = axV + 4 6xj^z' + 3 dxz^. dz 194 DIKFEEENTIAL CALCULUS 7. M = vh/'- — 2xy* + SxV ; show that x [■ y — = 5m. dx dy 8. M = ; show that x \- y — = u. X + y dx dy 9. u= (y — z) (z — x) (x — y); show that 1 1 = 0. dx dy dz 10. u = log (e^ + e*) ; show that — -\ =1. dx dy 11. !t = ; show that = (z + « — 1) u. e'+en dx dy 12. u = xi'j/^'; show that a; 1- y — = (x + 2/ + \ogu)u. dx dy 13. u = log (x* + v' + z^ — 3x«z) ; show that 1 1 = dx dy dz x + y + z 14. u = e^'siny + efsinx ; show that /— ) + (— ) = e^^ + e^y + 2e=» + i'sin (x + y): 15. M = log (tanx + tan j/ + tanz) ; show that . „ du du cm „ sm 2 X f- sin 2y f- sin 2 z — = 2. 3x Sj/ 3z 16. Let y be the altitude of a right circular cone and x the radius of its base. Show (a) that If the base remains constant, the volume changes \ in? times as fast as the altitude ; (b) that if the altitude remains constant, the volume changes \ my times as fast as the radius of the base. X^ Ip" 17. A point moves on the elliptic paraboloid z = \-— and also in a plane par- 9 4 allel to the XOZ-plane. When x = 3 ft. and is increasing at the rate of 9 ft. per second, find (a) the time rate of change of z ; (b) the magnitude of the velocity of • the point ; (c) the direction of its motion. Am. (a) Mj = 6 ft. per sec. ; (b) u = 3 Vl3 ft. per sec. ; (c) T = arc tan |, the angle made with the XOY-plane. 18. If, on the surface of Ex. 17, the point moves in a plane parallel to the plane YOZ., find, when 2/ = 2 and increases at the rate of 5 ft. per sec, (a) .the time rate of change of z ; (b) the magnitude of the velocity of the point ; (c) the direction of its motion. ' Am. (a) 5 ft. per sec. ; (b) 5 V2 ft. per sec. ; IT (c) T = - , the angle made with the plane XOY. 125. Total derivatives. We have already considered the differ- entiation of a function of one function of a single independent variable. Thus, if y=^f(y) and t; = ^(a;), it was shown that dy _ dy dv dx dv dx PARTIAL DIFFERENTIATION 195 We shall next consider a function of two variables, both of which depend on a single independent variable. Consider the function where x and y are functions of a third variable t. Let t take on the increment Ai, and let Aa;, Aj/, Am be the corre- sponding increments of x, y, u respectively. Then the quantity Am =f(x + Ax, y + Ay) -/(«, y) is called the total increment of u. Adding and subtracting f(x, y + A«/) in the second member, (A) Au=[f(x+Ax, y+Ay-)-f(x, y+Ay)^ + lfCx, y+Ay-)-f(x, j^)]. Applying the Theorem of Mean Value (46), p. 166, to each of the two differences on the right-hand side of (^), we get, for the first difference, (5) fCx + Ax,y + Ay') -fix, y -I- Ay) ^fX^ + e^.Ax,y + Ay~)Ax. ra=x, Aa= Aa;, and since x varies while y + Ay remains'! [constant, -we get the partial derivative -with respect to a:. J For the second difference we get (C) fCx, y + Ay-) ~f(x, y-) =fjCx, y + d^- Ay) Ay. [a = yt Aa = Ay, and since y varies while x remains con-l stant, we get the partial derivative with respect to y.J Substituting (5) and (C) in (A) gives (D) Am =fj(x + e^. Ax, y + Ay) Ax +fj(ix, y + 6^. Ay) Ay, where 6^ and 6^ are positive proper fractions. Dividing (X>) by At, (^) ^^=f:(x + e^.Ax,y + Ay)^^+fXx,y + d,Ay)^^. Now let At approach zero as a limit, then [Since Ax and Ay converge to zero with At, we get /^(x, y) anAfy'{x, y) heing assumed continuous. Replacing /(a;, y) by m in {F), we get the total derivative du du dx du dy ^"^ H^dxH^'dyW 196 DIFFERENTIAL CALCULUS In the same way, if w =f(,x, y, s), and a;, y, z are all functions of t, we get du du dx du dy du dz <^^^ di^dxdi'^'dydi^didi' and so on for any number of variables.* In (51) we may suppose t = z; then «/ is a function of x, and u is really a function of the one variable rr/ giving du du du dy ^^^^ di^si-^-^di' In the same way, from (52) we have du du du dy du dz ^^*^ '^'"Vx^Vy'dx^Vzdx The student should observe that — and -- have quite different ox ax du meanings. The partial derivative — ■ is formed on the supposition that dx the particular, variable x alone varies, while du limit /Am\ dx Aa;: where Am is the total increment of u caused by changes in all the vari- ables, these increments being due to the change Ax in the independent variable. In contradistinction to partial derivatives, — i — are called dt dx total derivatives with respect to t and x respectively.^ * This is really only a special case ol a general theorem which may be stated as follows : If u is a function of the independent variables x, y, z, ■ ■ ., each of these in turn being a function of the independent variables r, s, t, ■ • ., then (with certain assumptions as to continuity) du^duSx_^SuSy_^Sudz_^^_ dr dx dr dy dr dz 3r , . ., • , , , , S« 3m and similar expressions hold for — i — , etc. t It should be observed that — has a perfectly definite value for any point (x,y), while —- depends not only on the point {x,y), but also on the particular direction chosen to reach that point. Hence ' g — is called a point function ; while du . — is not called a point function unless it is agreed to approach "^ the point from some particular direction. PAETIAL DIFFERENTIATION 197 Illustrative Example 1. Given u = sin-, x = e' y = f; find — . y dt ,,,.. 3ul X du X X dx . dy „ Solution. — = - cos - , — = cos - ; — = e*, — = 2 i. dx y y By y^ y dt dt Substituting in (51), _ = (t _ 2) - cos - • Ans. Illustrative Example 2. Given u = e"^ {y — z), y = a sinx, z = cosx ; find — . Solution. — = ae^(y — z), — = e<", ^ =_ e<^; J^ = acosi, — =— sins. 8x dy dz dx dx Substituting in (54), du — = ae^ (V — z) + ae^ cos k + e°»^ sin x = e^(a^ + 1) sin x. Ans. Note. In examples like the above, u could, by substitution, be found explicitly in terms of the independent variable and then differentiated directly, but generally this process would be longer and in many cases could not be used at all. Formulas (51) and (52) are very useful in all applications involv- iag time rates of change of functions of two or more variables. The process is practically the same as that outliaed in the rule given on p. 141, except that, instead of differentiating with respect to t (Third Step), we find the partial derivatives and substitute in (51) or (52). Let us illustrate by an example. Illustrative Example 3. The altitude of a circular cone is 100 inches, and decreasing at the rate of 10 inches per second ; and the radius of the base is 50 inches, and increasing at the rate of 5 inches per second. At what rate is the volume changing ? Solution. Let X = radius of base, y = altitude ; then u = -^ mc^y = /^ volume, — = -'TXy, — = -ttx^. Substitute in (51), du 2 dx 1 „dy — = - irxy 1- - TTX' dt 3 dt 3 dt But X = 50, y = 100, dz_ dt " 5 ^^- = - 10. du dt 2 • 5000 . 5 1 -3" ■ 2500 10 = ] 126. Total differentials. Multiplying (51) and (52) through by dt, we get (55) du = — dx + — dy, dx dy du , du , du , (56) du = — dx-{ dy-\ dz; ' dx dy dz and so on.* Equations (55) and (56) define the quantity du, which is Called a total differential of u ov a, complete differential, T du T du T du ^ and — dx, — dv, — dz dx dy dz * A geometric interpretation of this result will be given on p. 264. 198 DIFFERENTIAL CALCULUS are called partial differentials. These partial differentials are some- times denoted by dji, d^u, d^u, so that (56) is also written du = d^u + d^ + djiL. y Illustkative Example 1. Given u = arc tan - , find du. X du y Su X Solution. ~ ~ Sx x" + y^ dy x^ + y'^ Substituting in (55), , xdy — ydx . du = Ans. X2 + 2/2 iLLnSTRATivE EXAMPLE 2. The base and altitude of a rectangle are 5 and 4 inches respectively. At a certain instant they are increasing continuously at the rate of 2 inches and 1 inch per second respectively. At what rate is the area of the rectangle increasing at that instant ? Solution. Let x = base, y = altitude ; then u = xy = area, — = y, — = i. Sx Sy . Substituting in (51), ... du dx ^ dy (A) — = y \- x-^. ^ ' dt dt dt But 1 = 5 in., y = i in., — = 2 in. per sec, — = 1 in. per sec. dt dt .-. — = (8 + 5) sq. in. per sec. = 13 sq. in. per sec. Ans. dt Note. Considering du as an Infinitesimal increment of area due to the infinitesimal increments dx and dy, du is evideatly the sum of two thin strips added on to the two .sides. For, in du = ydx + xdy (multiplying (A) by dt), ydx = area of vertical strip, and dy i xdy = area of horizontal strip. But the total increment Au due to the increments dec and V dy is evidently Au = ydx + xdy + dxdy. Hence the smaU rectangle in the upper right-hand comer (= dxdy) is evidently the difference between Au and du. This figure illustrates the fact that the total increment and the total differential of a function of several variables are not in general equal. 127. Differentiation of implicit functions. The equation defines either a; or «/ as an implicit function of the other.* It repre- sents any equation containing x and y when all its terms have been transposed to the first member. Let ,, du du , du dy . \ ^r,„ * We assume that a small change in the value of x causes only a small change in the value of y- PAETIAL DIFFEEENTIATION 199 du dx dtt But from (^), f(x, y) = 0. .■. m = and — = ; that is, -' to dy dx Solving for -^* we get tu (57) ■ 1^ = -^, ^=^0 a formula for differeutiating implicit functions. This formula in the form (C) is equivalent to the process employed in § 62, p. 69, for differentiating implicit functions, and all the examples on p. 70 may be solved by using formula (57). Since (-D) /(^, y)=0 for all admissible vaMes of x and y, we may say that (57) gives the relative time rates of change of x and y which keep f(x, y~) from changing at all. Geometrically this "means that the point (x, «/) must move on the curve whose equation is (J>), and (57) determines the direction of its motion at any instant. Since •w =/(a^> «/)> we may write (57) in the form ^y (57a) ^=._^. ^^0 „ , dy Illustrative Example 1. Given x^?/* + sin y = 0, find— • Solution. 'Letf(x,y) = x^y* + smy. — = 2xy*, ^=4a;V + oosy. . . f rom (57 a), -^ = - ^ „ -^f Ans. dx ey » ^ " ^ " dx ixV + cosy Illustrative Example 2. If x increases at the rate of 2 inches per second as it passes through the value x = 3 inches, at what rate must y change when y = l inch, in order that the function 2xy^ — 3 x^y shall remain constant ? Solution. Let/(x, 2/) = 2X2/2 — Sx^y; then ^=2y^-exy, ^=4x2/ -3x2. dx dy Substituting in (57 a), ^y dy _ 2y^-&xy ^^'di _ 2y^-&xy By (33), p. 141 dx ixy-3x^ dx^ ixy-Bx^ di But X = 3, y = 1, — = 2. ••■ ^ = - 2,% ft- per second. Ans. dt at du 8u * It is assumed that t— and -r- exist. ox By 200 DIFrERENTIAL CALCULUS Let P be the point (x, y, z) on the surface given by the equation , (^) u = F(x, y, z) = 0, and let PC and AP be sections made by planes through P parallel to the YOZ- and XO^-planes respectively. Along the curve AP, y is constant; therefore, from (^), z is an im- plicit function of x alone, and we have, from (57 a), (58) dz dx' dF 'dx W dz / / s giving the slope at P of the curve AP, § 122, p. 190. "dz dz . — is used instead of -- in the first member, since z was origmally, dx dx from (jE'), an implicit function of x and y, but (58) is deduced on the hypothesis that y remains constant. Similarly, the slope at P of the curve PC is SF (59) '^=Jy.. dy dF 'dz EXAMPLES Find the total derivatives, using (51), (52), or (53), in the following six examples: 1. It = z^ + )/2 + zj/, 2 = sin s, 2/ = e^. Ans. — = Se^a: -|- e^(sinx + cosx) + sin2a;. dx 2. u = arc tan (xy), y = g^. 3. u = log(a^— p'), p = a sin 5. 4. u = v'^-\-iiy,v = log s,y = e-'. 5. It = arc sin (r — s), r = 3 i, s = 4t^. gaxly _ g\ e. u = — ;f -, y = asmx, z = cosx. a^ + 1 du e^(l + x) Ans. ~=—^ — ■ — '-■ dx 1 + x^e^'' du ~de dM _ 2 ?) + ds s du 8 2 tan 5. dt Vl - P' du — = e"^ sm X. dx Using (55) or (56), find the total differentials in the next eight examples: 7. M = hyH +cx^ + gy^ + ex. Ans. du = {hy^ + 2cx + e)dx + {2byx + 3gy^)dy. S. u = logx". du = -dx + log axly. X PARTIAL DIFPEEENTIATION 201 9. u = y^':. Ans. du = y''^^ log y cos xdx + -?^^^dy. yCOVCTSX 10. u = x^o,,. du = u(^-2lldx+ ^-^^dy\ . 11. « = i±i. du = ^Ml^^. s-t (s-i)^ 12. M = sin {pq) . ' du = cos (pq) {qdp + pd/j] . IZ. u = xy. du^xu"-^ (yzdx + zx, log xdy + X2^ log xdz). 14. M = tan^^tan^StanV- tiw = 4u ('-^ + ^^ + _*L_\ . \sm 2 sin 2 ^ sin 2 fj 15. Assuming the characteristic equation of a perfect gas to be vp — Rt, where v = volume, p = pressure, t = absolute temperature, and B a constant, what is the relation between the difierentials dv, dp, dt ? 4ns. vdp + pdv = Edt. 16. Using the result in the last example as applied to air, suppose that in a given case we have found by actual experiment that t = 300° C, p = 2000 lb. per sq. ft., v = 14.4 cubic feet. Pind the change inp, assuming it to be uniform, when t changes to 301° C, and v to 14.5 cubic feet. B = 96. Ans. — 7.22 lb. per sq. ft. 17. One side of a triangle is 8 ft. long, and increasing 4 inches per second ; another side is 5 ft., and decreasing 2 inches per second. The included angle is 60°, and increasing 2° per second. At what rate is the area of the triangle changing ? Aivs. Increasing 71.05 sq. in. per sec. 18. At what rate is the side opposite the given angle in the last example increasing ? Ains. 4.93 in. per sec. 19. One side of a rectangle is 10 in. and increasing 2 in.' per sec. The other side is 15 in. and decreasing 1 in. per sec. At what rate is the area changing at the end of two seconds ? Avs. Increasing 12 sq. in. per sec. 20. The three edges of a rectangular parallelepiped are 3, 4, 5 inches, and are each increasing at the rate of .02 in. per min. At what rate is the volume changing ? 21. A boy starts flying a kite. If it moves horizontally at the rate of 2 ft. a sec. and rises at the rate of 5 ft. a sec, how fast is the string being paid out ? Ans. 5.38 ft. a sec. 22. A man standing on a dock is drawing in the painter of a boat at the rate of 2 ft. a sec. His hands are 6 ft. above the bow of the boat. How fast is the boat moving when it is 8 ft. from the dock ? Ans. \ ft. a sec. 23. The volume and the radius of a cylindrical boiler are expanding at the rate of 1 cu. ft. and .001 ft. per min. respectively. How fast is the length of the boiler changing when the boiler contains 60 cu. ft. and has a radius of 2 ft. ? Am. .078 ft. a min. 24. Water is running out of an opening in the vertex of a conical filtering glass, 8 inches high and 6 inches across the top, at the rate of .005 cu. in. per hour. How fast is the surface of the water falling when the depth of the water is 4 inches ? 202 DIFFERENTIAL CALCULUS 25. A covered water tank is made of sheet iron in the form of an inverted cone of altitude 8 ft. surmounted by a cylinder of altitude 5 ft. The diameter is 6 ft. If the sun's heat is increasing the diameter at the rate of ;002 ft. per min., the altitude of the cylinder at the rate of .003 ft. per min., and the altitude of the cone at the rate of .0025 ft. per minute, at what rate is (a) the volume increasing ; (b) the total area increasing ? In the remaining examples find — , using formula (57 a) : 26. (a;» -\- y^Y - a^ (k^ - 2/^) = o. 4ns. ^ = - "^ '^^^ -+ ^'^ ~ "' 27. e* — e=» + X2/ = 0. da; 2^ 2 (a;2 + y^) + a* dy _ (?^—y dx e« + x dy y fcos (xy) — ef^ — 2 a;l 28. sin (xy) - e»T - x^j/ = 0. — = ^ 5__»Z — ^ Ifl. dx X [x + e^ — cos {xy)] 128. Successive partial derivatives. Consider the function then, in general, du , dv, — and — dx dy are functions of both x and y, and may be differentiated again -with respect to either independent variable, giving successive partial deriva- tives. Regarding x alone as varying, we denote the results by d'u 8'u dx"' 8'u dx'" d"u dsif or, when y alone varies, 8\ df d'u d'u the notation being similar to that employed for functions of a single variable. If we differentiate u with respect to «, regarding y as constant, and then this result with respect to y, regarding x as constant, we obtain :;^l ^T" I' which we denote by — — Sy\dx) ^ dydi dydx Similarly, if we differentiate twice with respect to x and then once with respect to y, the result is denoted by the symbol 8»M PARTIAL DIFFERENTIATION 203 129. Order of differentiation immaterial. Consider the function /(a;, y). Changing x into x-\-b.x and keeping y constant, we get from the Theorem of Mean Value, (46), p. 166, {£) f(x+Lx,y-)-f(^x,y-)=^x.fl(ix+d.^x,yy QS 207 Note. In ease the rectangular equation of the envelope is required we may either eliminate the parameter from the parametric equations of the envelope, or else eliminate the parameter from the given equation (IT) of the family and the partial derivative (H'). Illustrative Example 1. Find the envelope of the family of straight lines s cos a + y sin a = p, a being the variable parameter. Solution. (A) X cos a + ysma = p. First step. DifEerentiating {A) with respect to a, (B) —xsina + y cos a = 0. Second step. Multiplying {A) by cos a and (B) by sin a and subtracting, we get x — p cos a. Similarly, eliminating x between (A) and (B), we get y = psin a. The parametric equations of the envelope are therefore 'x=p cos a, ' = p sin a, a being the parameter. Squaring equations (C) and add- ing, we get a;2-l- 2/2=^)2, the rectangular equation of the envelope, which is a circle. Illustrative Example 2. Find the envelope of a line of constant length a, whose extremities move along two fixed rectangular axes. Solution. Let AB = a in length, and let (A) X cos a + ysina—p = be its equation. Now as AB moves always touching the two axes, both a and p will vary. But p may be found in terms of a. For AO = AB oosa = a cos a, and p = A sin or = a sin a cos a. Substituting in {A), ^ (B) X cos a + J/ sin a: — a sin or cos a = 0, •where a is the variable parameter. Differentiating (B) with respect to a, (C) —xsina + ycosa+'a sin^ a — a cos^ a = 0. Solving (B) and (C) for x and y in terms of a, we get (C) m (x = asin^or, \y = acos^a. the parametric equations of the envelope, a hypocycloid. The corresponding rectangular equation is found from equations (D) by eliminat- ing a as follows : • i. i. ■ o ^' xt = at sm^ a. yi = ai CDs' x. Adding, xi + yi = of, the rectangular equation of the hypocycloid. 208 DIFFERENTIAL CALCULUS Illhstrativb Example 1. Find the rectangular equation of the envelope of the V straight line y = mx + — , where the slope m is the variable parameter. Solution. First step. y = ma; + ; = x- '» = ±A-- P. Solving, Substitute in the given equation, y=±^l.X±-yjyp=±2V^, and squaring, y^ = ipx, a parabola, is the equation of the envelope. The family of straight lines formed by varying the slope m is shovyn in the figure, each line being n tangent to the envelope, for we know from Analytic Geometry that y = mx H — is the tangent to the parabola y^ = ipx expressed in terms of its own slope m. 132. The evolute of a given curve considered as the envelope of its normals. Since the normals to a curve are all tangent to the evolute, § 118, p. 181, it is evident that the evolute of a curve may also he defined as the envelope of its Twrmals ; that is, as the locus of the ultimate intersections of neighboring normals. It is also interesting to notice that if we find the para- metric equations of the envelope by the method of the previous section, we get the coordinates X and y of the center of curvature ; so that we have here a second -method for finding the coor- dinates of the center of curvature. If we then eliminate the variable parameter, we have a relation between x and y which is the rectan- gular equation of the evolute (envelope of the normals). Illdstkative Example 1. Find the evolute of the parabola y^ = ipx considered as the envelope of its normals. Solution. The equation of the normal at any point (x', y^ is from (2), p. 77. As we are considering the normals all along the curve, both i' and if will vary. Eliminating x' by means of y-^ = i^px', we get the equation of the normal to be '''en cur«^ .(^) V — y = — — . Sp^ 2p ENVELOPES 209 Considering y' as the variable parameter," we wish to find the envelope of this family of normals. Differentiating (A.) with respect to y\ l = it-^ and solving for x. X 2p' Sy'^ + 8p^ 4p Substituting this value of x in (A) and solving for y, (C) y= — 4p2 (B) and (C) are then the coSrdinates of the center of curvature of the parabola. Taken together, (B) and (C) are the parametric equations of the evolute in terms of the parameter y'. Eliminating y' between (B) and (C) gives 27^)2/2 = 4 (x-2p)3, the rectangular equation of the evolute of the parabola. This is the same result we obtained in Illustrative Example 1, p. 183, by the first method. 133. Two parameters connected by one equation of condition. Many problems occur where it is convenient to use two parameters con- nected by an equation of condition. For instance, the example given in the last section involves the two parameters x' and y' which are connected by the equation of the curve. In this case we ehminated a;', leaving only the one parameter y'. However, when the elimination is difficult to perform, both the given equation and the equation of condition between the two param- eters may be differentiated with respect to one of the parameters, regarding either parameter as a function of the other. By studying the solution of the following problem the process will be made clear. Illustrative Example 1. Find the envelope of the family of ellipses whose axes coincide and whose area is constant. Solution. {A) + ^- = 1 is the equation of the ellipse where a and b are the variable parameters connected by the equation (B) TTob = k, Tcah being the area of an ellipse whose semi- axes are a and 6. • Differentiating (A) and (B), regarding a and 6 as variables and x and y as constants, we have, using differentials, - + = 0, from (A), and bda + adb ■■ : 0, from {B). 210 DIFFEKliNTIAL CALCULUS Transposing one term in each to the second member and dividing, we get x2 ^ yi x^ 1 ,V^\ Therefore, from (4), — = - and — = - , giving a = ± » V2 and 6 = ± 2/ V2. Substituting these values in (B), we get the envelope A; X2^ = ± — . a pair of conjugate rectangular hyperbolas (see last figure). EXAMPLES 1. Find the envelope of tlie family of straight lines y = imx + m*, m being the variable parameter. Am. x =— 2m?, y =— 3m* ; ot Wy^ + 27x* = 0.* 2. Find the envelope of the family of parabolas y^ = a(x — a), a being the variable parameter. ^tis. x = 2a,y = ±a; ovy = ±^x. 3. Find the envelope of the family of circles x'' + (y — pf = r^, /3 being the variable parameter. Ans. x = ±r. 4. Find the equat ion of the curve having as tangents the family of straight lines y = mx ± ^a^nfi + V^, the slope m being the variable parameter. A-ns. The ellipse ft^x^ + aV = a^^- 5. Find the envelope of the family of circles whose diameters are double ordi- nates of the parabola y^ — ipx. Ans. The parabola y^ = 4p{p + x). 6. Find the envelope of the family of circles whose diameters are double ordi- nates of the ellipse 6^x2 + aV = a^*^- . rr,, „. x^ y^ Ans. The ellipse 1- _ = 1. 7. A circle moves with its center on the parabola y^ = 4ax, and its circumference passes through the vertex of the parabola. Find the equation of the envelope of the circles. Ans. The cissoid y^ (x + 2 a) + x' = 0. 8. Find the curve whose tangents are y = lx ± VoP + bl + e, the slope I being supposed to vary. Ans. i(ay^ + bxy + cx^) = iac — V. 9. Find the evolute of the ellipse 6^x2 + a^y^ = a^lfi, taking the equation of nor- mal in the form b^/ = ax tan - (a2 _ ft2) gjn ^^ the eccentric -angle 1, — + — — + — - — -_!_ 2p Sp 2p 2p~ 2p~ 2p-^' 1+1+1+1 <1+1+1+1^1^/J_V ip bP &p IP ip ip ip 4.P 4p \2p-^/ 11 1 1 1 1 1 1 1 18/1\3 Sp 15p 8p 8^ 8* 8p 8? 8^ "'" 8p "*" 8p ~ 8p ~ \2p^/ ' and so on. Construct the series '-> ^-^,H^:}'H^)'* When p>l, series (H) is a geometric series with the common ratio less than unity, and is therefore convergent. But the sum of (G) is less than the sum of (J?), as shown by the above inequalities ; therefore (G) is also convergent. When p = l, series (G) becomes the harmonic series which we saw was divergent, and neither of the above tests apply. When p 1 (or /a = oo). Following the same line of reasoning as in I, the series (A) may be shown to be divergent. III. When p = l, the series maybe either convergent or divergent; that is, there is no test. For, consider the p series, namely, l+2; + 3^ + 4; + '"'+^+(w+l)i'"'"""' The test ratio is ^^^ =/^L^Y=( 1- ^)" n+lj \ n + ^^ limit fu^\ limit L 1 y^(i^.^l^_ ). Hence /> = 1, no matter what value p may have. But on p. 218 we showed that ^^^^^ p>l, the series converges, and when ^ ^ 1, the series diverges. Thus it appears that p can equal unity both for convergent and for divergent series, and the ratio test for convergence fails. There are other tests to apply in cases like this, but the scope of our book does not admit of their consideration. * When examining a series lor convergence we are at liberty to disregard any finite number of terms; the rejection of such terms would affect the value but not the existence of the limit. 220 DIFFERENTIAL CALCULUS Our results may then be stated as follows : Given the series of positive terms M1+M2+M8+ ■■■+"«+""+!+■■■' find the limit limit /«^ w = co\ u I. When /> 1, the series is divergent. III. When p=l, there is no test. 140. Alternating series. This is the name given to a series whose terms are alternately positive and negative. Such series occur fre- ' quently in practice and are of considerable importance. If Mj-1(,,+ M3- «,+ ■•• is an alternating series whose terms never increase in numerical value, and if ^^"""^ M„ = 0, then the series is convergent. Proof. The sum of 2 n (an even number) terms may be written in the two forms {A) S^^ = (u^-u^) + (u^-u^) + (u^-u^)+---+(u^^_-,-u^,), or Since each difference is positive (if it is not zero, and the assump- tion ™ ■"„ = excludes equality of the terms of the series), series (A) shows that S,^^ is positive and increases with w, while series (5) shows that S^^ is always less than u^; therefore, by Theorem I, p. 215, 62„'must approach a limit less than u^ when n increases, and the series is convergent. Illustrative Example 4. Test the alternating series 1 1 !-■■■• Solution. Since each term is less in numerical value than the preceding one, and 71 = 00 ^^' »i = CO \^ri/ the series is convergent. 141. Absolute convergence. A series is said to be absolutely^ or unconditionally convergent when the series formed from it by making all its terms positive is convergent. Other convergent series are said * It is not enough that Un + i/Un becomes and remains less than unity for all values of n, but this test requires that the limit of Wn + i/un shall be less than unity. For instance, in the case of the harmonic series this ratio is always less than unity and yet the series diverges as we have seen. The limit, however, is not less than unity but equals unity. t The terms of the new series are the numerical (absolute) values of the terms of the given series. SEEIES 221 to be not absolutely convergent or conditionally convergent. To this latter class belong some convergent alternating series. For example, the series 1111 22 ^33 44 -t- 56 is absolutely convergent, since the series (C), p. 217, namely, is convergent. The series 1+22 + 38 + 44+55 + - 2^3 4^5 is conditionally convergent, since the harmonic series i+-2n+i4+- is divergent. A series with terms of different signs is convergent if the series deduced from it by making all the signs positive is convergent. The proof of this theorem is omitted. Assuming that the ratio test on p. 219 holds without placing any restriction on the signs of the terms of a series, we may summarize our results in the following General directions for testing the series «!+ "2+ M3+ M^H 1- 1«„+ w„+iH • When it is an alternating series whose terms never increase in numer- ical value, and if limit , . _ n n. = CO " ' then the series is convergent. In any series in which the above conditions are not satisfied, we deter- mine the form of u^ and m„^i and calculate the limit limit / m„ + i \ n = c\ u„ j I. When\p\<\, the series is absolutely convergent. II. When 1/3 1 > 1) the series is divergent. III. When |p| = l, there is no test, and we should compare the series with some series which we know to be convergent, as a + ar -\- ar^ -\- ar"^ -\ ; r < 1, {geometric series) l + ^ + ^ + i-, + '--; F>1' (i> series) 222 DirFEEENTIAL CALCULUS or compare the given series with some series which is known to be divergent, as 1,1. 1 + 1-1... .• (harmonic series) "^2 3 4 ' l4-i--|- — 4- — +•••; p\lOn-i-i JnJ n = a>\ 10 / and by II, p. 221, the series is divergent. Illustrative Example 3. Test the series 1,1,1, (^) 1:2 + 3:4 + 5:6+ ■■■• Solution. Here ^ = —-1^^, ,^ + , = __^^^^-^ . limit /y^ + i\ ^ limit [" (2m-l)2rt 1 ^ co '■" = «'\'tn/ " = «=L(2n + l)(2Ti + 2)J 00 This being an indeterminate form, we evaluate it, using the rule on p. 174. limit /Sn — 2\ oo Difierentiatxng, ^L-^t^ (__) = _ . Differentiating again, ^^™'^ Q = 1 ( = p) . This gives no test (III, p. 221). But if we compare series (C) with (G), p. 218, making p = 2, namely, W ^ + i5 + 3i + 5 + ---' we see that (C) must be convergent, since its terms are less than the corresponding terms of (D), which was proved convergent. SERIES o-v?. EXAMPLES .Show that the following ten series are convergent ; 12 22 32 ' 2 "*" 22 "*" 28 """ 2^ "^ " '•r2 + 3^ + 5^ + ' 1 1^3 1,3^ 3 3-6 3.6.9 ^■| + [| + [?+-- 6. 1 + -!= + ^ + -^ + 2V2 3V3 4Vi 7.i-Jt+JL-JL + JL_.... 8.1-1. 1 + 1. l_l.i + . 2 2 22 3 2' 42* 9. JL_J_ + J . log 2 log 3 log 4 10.1+1+1 + .... 22 32 42 Show that the following four series are divergent : 11.1 + 1 + 1+.... 2 4 6 i2.a + l±A + l±l + l±l + . 1 + ! 1 + 32 1 + 42 2 3 [i 13. ^ + i=i + -b. + . 10 102 103 ^ 14.1 + 1 + 1 + 1 + .. 3 5 7 142. Power series. A series of ascending integral powers of a vari- able, say X, of the form \ (.4) a^ + a^z + a^x'' + a^ -\ , where the coefficients a^, a^, a^, ■■■ are independent of x, is called a power series in x. Such series are of prime importance in the further study of the Calculus. In special cases a power series in x may converge for all values of x, but in general it will converge for some values of x and be divergent for other values of x. We shall examine (^) only for the case when the coefficients are such that limit /^+i w = 00 \ a, where X is a definite number. In (A) limit / m„ + i\ limit / a„ + i^ = i, = ^^^('^\.x=Lx. n=cc\ a Referring to tests I, II, III, on p. 221, we have in this case p = Lz, and hence the series (^) is I. Absolutely convergent when | ia; | < 1, or | a; | < — II. Divergent when | ia; | > 1, or | a; | > — III. No test when I ia; I = 1, or \x\ = 224 DIFFERENTIAL CALCULUS We may then write clown the following General directions for finding the interval of convergence of the power series, (^) a„ + "i'' + «2* + «s«= + ■ FiKST Step. Write down the series formed hy coefficients, namely, a„+ai+«2+«3+ ■•■ + «„+«„ + ! + ■•■• Second Step. Calculate the limit limit / «„ + ! w = CO I a. L. Third Step. Then the power series (A) is I. Absolutely convergent for all values of x lying between and + ■ II. Divergent for all values ofx less than — 1 III. No test when x= ± or greater than + ; but then we substitute these two values of X in the power series (^A) and apply to them the general directions on p. 221. Note. When L = 0, ± — =±qo and the power series is ahsolutely J-/ convergent for all values of x. Illustrative Example 1. Find the interval of convergence for the series (B) X'- a;= X' 22 '32 42 ^ Solution. Firssl step. The series formed by the coefficients is 1 (C) 11- 1 1 1-.. 22 32 42 ^ Second step. "'"'t (?:^ = l™it [ n=_ 1 ^ „ i = <»\a„/ n=coL (w + l)2j 00 Differentiating, 1™'* / ^" \ = fE. n = \ 2(ji + l)/ ■" Differentiating again, ^'°"* ( — _ 1 = — 1 f = il 7l = ao\ 2/ '■ Third step. \L\ 1-1 = 1. By I the series is absolutely convergent when x lies between — 1 and + 1. By II the series is divergent when x is less than — 1 or greater than + 1. By III there is no test when x = ± 1. SERIES 225 Substituting x = 1 in {B), we get l_iH.i_i + 22 ^32 42 ^ which is an alternating aeries that converges. Substituting a; = — 1 in (B), we get 22 32 42 ' which is conrergent by comparison with the p series (p>l). The series m "the above example is said to have [— 1, 1] as the interval of conver- gence. This may be written — 1 ^ s ^ 1, or indicated graphically as follows : J^ EXAMPLES Per what values of the variable are the following series Graphical representations of . n intervals of convergence * convergent ? . 15. 1 + x + x^ + x' -\ . Ans. -l0. I T~ Hint. Neither the sine nor cosine can exceed 1 numer- ically. 24. l + xlog» + -^ + -jg- + .... , ^ J ^ Ans. All values of x. 2B. 1 1 H . Ans. x > 1. 1 ® » l + xl + x^l + x' o-tl 2 3 2-4 5 2-4.6 7 -1 +i .4ns. — l=x^l. 27. 1 + X+ 2x2 + 3xs + ---. 28.x-^ + ^-^ + .... 3 5 7 29. lOx + 100x2 + lOOOxS + • ■ •. 30. 1 + X + [2x2 + |3x' + • ■ • . * End points that are not included in the interval of convergence have circles drawn about them. , CHAPTER XVIII EXPANSION OF FUNCTIONS 143. Introduction. The student is already familiar with some methods of expanding certain functions into series. Thus, by the Binomial Theorem, (A-) (a + xy=a'+4. a'x + 6 aV+ 4 ax'+x', giving a finite power series from which the exact value of (a + x")* for any value of x may be calculated. Also by actual division, (-B) -J—=l + x + x'+x'+-..+x''-' + (-l-)3^, i —X \l—x/ we get an equivalent series, all of whose coefficients except that of 3f are constants, n being a positive integer. Suppose we wish to calculate the value of this function when x= .5, not by substituting directly in 1 T^x' but by substituting a; = .5 in the equivalent series ((7) (l + x + x'+x'+.-. + x'^-'y + L-^x^. Assuming w = 8, ((7) gives for x = .5 (m -^ = 1.9921875 + .0078125. 1—x If we then assume the value, of the function to be the sum of the first eight terms of series (C), the error we make is .0078125. However, in case we need the value of the function correct to two decimal places only, the number 1.99 is as close an approximation to the true value as ifve care for, since the error is less than .01. It is evident that if a greater degree of accuracy is desired, all we need to do is to lise more terms of the power series (^) l + x + x^ + x'+--.. 227 228 DIFFEEENTIAL CALCULUS Since, however, we see at once that LI — a;Jx=.5 there is no necessity for the above discussion, except for purposes of illustration. As a matter of fact the process of computing the value of a function from an equivalent series into which it has been expanded is of the greatest practical importance, the values of the elementary transcendental functions such as the sine, cosiue, logarithm, etc., being computed most simply in this way. So far we have learned how to expand only a few special forms into series ; we shall now consider a method of expansion applicable to an extensive and important class of functions and called Taylor's Theorem. 144. Taylor's Theorem * and Taylor's Series. Replacing J by a; in (-2^), p. 167, the extended theorem of the mean takes on the form (61) /(x) =/(a) + ^^fXd) + l£^V"(a) + (£^V'"(a) + • • • Li Lf If where x^ lies between a and x. (61), which is one of the most far- reaching theorems in the Calculus, is called Taylor s Theorem. We see that it expresses /(j-) as the sum of a finite series in (a; — a). ^^ (jr\^ The last term in (61), namely -^-j — —/^"Xx^}, is sometimes called I!!: the remainder in Taylor's Theorem after n terms. If this remainder converges toward zero as the number of terms increases without limit, then the right-hand side of (61) becomes an infinite power series called Taylor's Scries.'^ In that case we may write (61) in the form (62) fCx)^f(a) + ^^/'(a) + ^^^/"(a) + (£^/"'(a) + . . . , and we say that the function has been expanded into a Taylor's Series. For all values of .;■ for which the remainder approaches zero as n increases without limit, this series converges and its sum g'ives the exact value of /(a;), because the difference (= the remainder) between the function and the sum of n terms of the series approaches the limit zero (§ 15, p. 13). * Also known as Taylor'a Formula. t Puljlished by Dr. Brook Taylor (1685-1731) in his Methodus Incrementorum. London, 1715. EXPANSION OF FUNCTIONS 229 If the series converges for values of x for which tlie remainder does not approach zero as n increases without hmit, then the Hmit of the sum of the series is not equal to the function /(x). The infinite series (62) represents the function for those values of x, and those only, for which the. remainder approaches zero as the num- ber of terms increases without limit. It is usually easier to determine the interval of convergence of the series than that for which the remainder approaches zero ; but in simple cases the two intervals are identical. When the values of a function and its suecessive derivatives are known for some value of the variable, as x = a, then (62) is used for iinding the value of the function for values of x near a, and (62) is also called the expansion off(x') in the vicinity of x = a. Illustrative Example 1. Expand log a; in powers of (a; — 1). Solution. /(z) = log X, /(I) = ; f'(x) = - ~) X /'(I) = = 1; rw = 1 /"(I) = =-1; /'"(x) = 2 /"'(I) = = 2. Substituting in (62), log x = x - 1 - J (a; - 1)^ + J (x - l)^ . Ans. This converges for values of x between and 2 and is the expansion of log x in the vicinity of x = l, the remainder converging to zero. When a function of the sum of two numbers a and x is given, say/(a + a;), it is frequently desirable to expand the function into a power series m one of them, say x. For this purpose we use another • form of Taylor's Series, got by replacing a; by a + a; in (62), namely, (63) /(c + xy^ /(a) + ^/'(a) + |/"(a) + |/"'(a) + • • ■ • Illtjsteative Example 1. Expand sin (a + x) in powers of x. Solution. Here /(a + x) = sin (a + x). Hence, placing x = 0, f(a) = sin a, f'(a) = oos.a, f"{a) = — sin a, f"'(a) =— coso, Substituting in (61), sin(a + x) = sina + -cosa — |— sina — rr-cosoH- ■ ■•. A.ns. 230 DIFFERENTIAL CALCULUS EXAMPLES * 1. Expand & in powers otx-2. Ans. eF = e^ + e^ (x - 2) + r^(x - 2f + ■ ■ ■ 2. Expand x' — 2x2 + 5a; — 7 in powers of a; — 1. Ans. - 3 + 4(s - 1) + (a; - 1)'' + (E - 1)3. 3. Expand 32/2 _Uy^j in powers of y - 3. Ans. -8 + 4(y - 3) + S{y - 3)2. 4. Expand 6z^ + 7z + 3 in powers of z - 2. Ans. 31+ 27(« - 2) + 6(z - 2)2. 5. Expand 4x= — 17x2 + n x + 2 in powers of x — 4. 6. Expand by* + 6y^ — 17 2/2 + ISy — 20 in powers of y + 4. 7. Expand e^ in powers of x + 1. 8. Expand sin x in powers of x — a. 9. Expand cosx in powers of x — a. 10. Expand cos (a + x) in powers of x. 2 8 Ans. COS (a + x) = cos o — x sin a — r— cos o + i— sin a + ■ • • . 11. Expand log (x + ft) in powers of X. 2 s Ans. log(x + ft) = logft + |-^ + ^ + .... 12. Expand tan (x + ft) in powers of ft. Ans. tan (x + ft) = tanx + ft sec2x + ft2 sec2x tanx + • • ■ . 13. Expand tlie following in powers of ft. (a) (X + ft)» = x» + 7ia;»-ift + ??^^?p^x«-2ft2 + '^('^-l)("-2) ^„_3^s + . . .. (b) e»+* = 6="/ 1 + ft + 1^ + 1^ + .. = .(l+. + | + |^...). 145. Maclaurin's Theorem and Maclaurin's Series. A particulax ease of Taylor's Theorem is found by placing a = in (61), p. 228, giving (64) /(a:) =/(0) + jf /'(O) + JLV"(0) + ^/"'(O) + • • • + r^r"-''(o) + r-f'"\xj, n-l F where a;^ lies between and x. (64) is called Maclaurin's Theorem. The right-hand member is evidently a series in x in the same sense that (61), p. 228, is a series in a; — a. Placing a = IQ (62), p. 228, we get Maclaurin's Series,^ (65) ^ /(^)=/(0) + j^/'(0) + J^/"(0) + ^/'"(0) + ..., * In these examples we asstune that the functions can he developed into a power series, t Named after Colin Maclaurin (1698-1746), heing first published in his Treatise of Fluxions, Edinburgh, 1742. The series is really due to Stirling (1692-1770). •expansion of functions 231 a special case of Taylor's Series that is very useful. The statements made concerning the remainder and the convergence of Taylor's Series apply with equal force to Maclaurin's Series, the latter being merely a. special case of the former. The student should not fail to note the importance of such an expansion as (65). In all practical computations results correct to a. certain number of decimal places are sought, and since the process in question replaces a function perhaps difficult to calculate by an ordinary polynomial with constant coefficients, it is very useful in sim- plifying such computations. Of course we must use terms enough to give the desired degree of accuracy. In the case of an alternating series (§ 139, p. 218) the error made by stopping at any term is numerically less than that term, since the sum of the series after that term is numerically less than that term. Illustrative Example 1. Expand cose into an infinite power series and determine for what values of x it converges. Solution. DifEerentiating first and then placing x = 0, we get /(a;) = coss, /(O) = 1, /(x)=-sina;, /'(O) = 0, /"(»)•=- cos X, /"(0)=-l, /'"(x) = sinx, /'"(0) = 0, /'''(x) = cosx, /i''(0) = l, p(x)=-smx, /^(0) = 0, /■ri(x) = - cos X, /^(O) = - 1, etc., etc. Substituting in (65), T 3! IE (4) eosx = l-^ + j^-^+.... Comparing with Ex. 20, p. 225, we see ihat the series converges for all values of x. In the same way for sinx. x" . X' ;5 y^ which converges for all values of x (Ex. 21, p. 225).* *Since here/<'"(a:) = sin(a; + — j and/C'Caii) =siiija;i + ^|' we have, by substituting in the last term of (64) , p. 23i, » / \ remainder =|— sin I a!i+ — V 0), ((?), (<7), (JT), (i), i^(0)=/(2;, ?/), i.e. F(t} is replaced by /(a;, ^), ^ ^ dx dy i?"(0) = A^^ + 2 M-^ + fJ^, ^ -' da? dxdy dy" F"'CO-) = h'^+ 3 A^;fc-^ + 3 AF-^ + f|^, and so on. Substituting these results in (^), we get (66) fix + ht,y + kt) = fix, y) + f (|ft g + ftgj \2\ dx' dxdy^ di/'J^ To get/(2; + h, y + k}, replace i by 1 in (66), giving Taylor's Theorem for a function of two independent variables, (67) f(ix + h,y + K) = fix,y~) + h^^+k^ f"' ajr" dxdy dy'l ' which is the required expansion in powers of h and k. Evidently (67) is also adapted to the expansion of f(x + k, y + k') in powers of x and y by simply interchanging x with h and y with k. Thus (67a) f{x + h,y+k) = fCh,K) + x^+y^^ + '(x^':£ + 2xy^ + y''^)+.... Similarly, for three variables we shall find (68) f(x + h, y + k, z+r) = f(x, y, z} + h^ + k^ + /^ dx dy dz [2\ dJ^^ djt Sz' dxdy ^f "■" dz and so on for any number of variables. dzdx dydzj EXPANSION OF FUNCTIONS 243 EXAMPLES 1. Given /(s, y) = Ax^ + Bxy + Cy'^, expand f{x + h,y-{-k) in powers of h and k. Solution. —=i2Ax-\-By, — = Bx + 2Cy; dx dy dx^ dxdy Sy^ The third and higher partial derivatives are all zero. Substituting in (67), f(x + h,y + k)~Ax^ + Bxy + Cy^ + (2Ax + By)h + (Bx + 2 Cy)k + AK' + Bhk + Ck^. Ans. 2. Given fix, y, z)b E Ax^ + By^ + Cz^ expand /(x + 1, y-^ ■ m, z + n I, m, n. Solution. 3/_ Sx -.2 Ax, ' Sy -2 By, 3/_ 3z :2Cz; dx^' = 2A, ^f _ --2B, d^f_ 8z2 :2C, dxdy dydz czdx = 0. The third and higher partial derivatives are all zero. Substituting in (68), f(x + l,y + m, z + n) = Ax^ + By''+ Cz'^ + 2 Axl + 2 Bym + 2 Czn + AP + Bm? + On". Ans. 3. Given /(x, y) = Vx tan y, expand /(x + K, y + k) in powers of h and k. 4. Given /(x, y, z) = Ax^ + By" + Cz" + Dxy + Eyz + Fzx, expand /(x + 7j, 2/ + A, 2 + ™ powers of h, k, I. 149. Maxima and minima of functions of two independent variables. The function f(x, y) is said to be a maximum at x=a, y — h when f(a, by is greater than f(x, y) for all values of x and y in the neigh- borhood of a and b. Similarly, /(a, J) is said to be a minimum at a; = a, y = b when /(a, J) is less than /(a;, «/) for all values of x and 1/ in the neighborhood of a and 5. These definitions may be stated in analytical form as follows : If, for all values of h and k numerically less than some small posi- tive quantity, (^A) f(a + h, b + Jc) —/(a, h~)= a negative number, then f(a, J) is a maximum value oif(x, «/). If (By f(a + h, b + ky —f(a, by= a positive number, then /(«, by is a minimum value oif(x, yy. These statements may be interpreted geometrically as follows: a point P on the surface ^ _ ^/-^ % 244 DIFFEEENTIAL CALCULUS is a maximum point when it is " higher " than all other points on the surface in its neighborhood, the coordinate plane XO Y being assumed horizontal. Similarly, P' is a minimum point on the surface when it is " lower " than all other points on the surface in its neighborhood. It is therefore evident that all vertical planes through P cut the surface in curves (as APB or DPU in the figure), each of which has a maximum ordinate z (= MP) at P. In the same manner all vertical planes through P' cut the surface in curves (as BP'C or FP'G}, each of which has a minimum ordinate z(=NP') at P'. Also, any contour (as HIJK) cut out of the surface by a horizontal plane in the immediate neighborhood of P must be a small closed curve. Similarly, we have the contour LSBT near the minimum point P'. It was shown in §§ 81, 82, pp. 108, 109, that a necessary condition that a function of one variable should have a maximum or a minimum for a given value of the variable was that its first derivative should be zero for the given value of the variable. Similarly, for a function /(a;, y) of two independent variables, a wecessary, condition that /(a, 6) should be a maximum or a minimum (i.e. a turning value) is that for x= a, V = h, dy Proof. Evidently (^) and (B) must hold when h=Q; that is, /(« + A, 6)-/(a, 6) is always negative or always positive for all values of h sufficiently small numerically. By §§ 81, 82, a necessary condition for this is that -— f(x, V) shall vanish for a; = a, or, what ambunts to the same ax thing, —f(x, y) shall vanish for a; = a, y = h. Similarly, (^) and (S) must hold when A = 0, giving as a second necessary condition that — f(x, y) shall vanish iov x = a, y = h. (^) ^ = 0, dx 0. EXPANSION OF FUNCTIONS 245 In order to determine sufficient conditions that /(a, 5) shall be a maximum or a minimum, it is necessary to proceed to higher deriva- tives. To derive sufficient conditions for all cases is beyond the scope of this book.* The following discussion, however, will suffice for all the problems given here. Expanding f(a + h, b + k) by Taylor's Theorem, (67), p. 242, re- placing xhj a and t/ by b, we get (D) /(« + h,h + Jc-) =f(a, l')+-h^I+k^ ox cy [2 \ c^ dxdy dy^j where the partial derivatives are evaluated for x=a, y = h, and li denotes the sum of all the terms not vn'itten down. All such terms are of a degree higher than the second in h and Ic. Since t^= and t^= 0, from (C), p. 244, we get, after transpos- ing /(«, J), W/(a + M + ^)-/(.,5) = l(A^g+2M^ + F|)-fi^. If f(a, 6) is a turning value, the expression on the left-hand side of (E^ must retain the same sign for all values of Ji and k sufficiently small in numerical value, — the negative sign for a maximum value (see (^), p. 243) and the positive sign for a minimum value (see (5), p. 243) ; i.e. f(a, V) will be a maximum or a minimum according as the right- hand side of (^) is negative or positive. Now i? is of a degree higher than the second in A and k. Hence as h and k diminish in numerical value, it seems plausible to conclude that the numerical value of R will eventually become and remain less than the numerical value of the sum of the three terms of the second degree written down on the right-hand dde of (E). t Then the sign of the right-hand side (and therefore also of the left-hand side) will be the same as the sign of the expression (J.) A^2^+2M^-fF|C- ^ ^ cx^ cxdy dy^ But from Algebra we know that the quadratic expression ¥A+2hkC + k^B always has the same sign as A (or B") when AB— C^> 0. * See Oours d' Analyse, Vol. I, by C. Jordan. t Peano has shown that this conclusion does not always hold. See the article on " Maxima and Mmima of Functions of Several Variables," by Professor James Pierpont in the Bulletin of the American Mathematical Society, Vol. IV. 246 DIFFERENTIAL CALCULUS o2j? o2j? q2^ Applying this to (F), A= -r^, B = -4' C^ = :^-^' and we see that (i^), and therefore also the left-hand member of (^), has the same sign as -^( or -fA when d7?\ df) 8'/ ay /gyy^o da? df \dxdyj Hence the following rule for finding maximum and minimum values of a f unction /(jr, y). FiBST Step. Solve the simultaneous equations dx dy Second Step. Calculate for these values of x and y the value of ay8y_/_eyY da? df [SxdyJ Third Step. The function will have a dj/ dj maadmum if A > Q and -=^( or -^ ) < ; ■^ da?\ df) • minimum if A > and da?\ df) neither a maximum nor a minimum if A<(). The question is undecided i/" A = 0.* The student should notice that this rule does not necessarily give all maximum and minimum values. For a pair of values of x and y determined by the First Step may cause A to vanish, and may lead to a maximum or a minimum or neither. Further investigation is therefore necessary for such values. The rule is, however, sufficient for solving many important examples. The question of maxima and minima of functions of three or more independent variables must be left to more advanced treatises. Illustrative Example 1. Examine the function Zaxy — x' — y^ for maximum and minimum values. Solution. /(i, y) = 3 axy — x' — 2^. FirstsUp. — = 3a^-3a;2-o, ^ = Zax-Zy^ = (i. dx dy Solving these two simultaneous equations, we get x = 0, x = a, y = 0; y = a. * The discussion of the text merely renders the given rule plausible. The student should observe that the case A = Is omitted in the discussion. EXPANSION or FUNCTIONS 247 Third step. When a; = and y = 0, A = - Oa", and there can be neither a maxi- mum nor a minimum at (0, 0). When,s = a and y = a, A = + 27 a^ ; and since — ^ = — 6 a, we have the conditions for a maximum value of the function fulfilled at (a, a). Substituting a; = o, j/ = a in the given function, we get its maximum value equal to a". Illustrative Example 2. Divide a into three parts such that their product shall be a maximum. Solution. Let x = first part, y = second part ; then a— {x-\-y) = a — x — y = third part, and the function to be examined is /(a;, y) = xy{a-x — y). First step. ~ = ay ~ 2xy — y"^ = 0, — — ax — 2xy — x^ = 0. Sx cy Solving simultaneously, we get as one pair of values x = - , y = -• o o Secondstep. -^ = —2y, — ^ = o-2x — 2w, -J.=-2x: dx^ "' dx&y "' dy" A = ixy — (a — 2x — 2y)2. d da? d^f 2 a Third step. When x = - and y = - , A = — ; and since —^= , it is seen that 3 3 3 Sx^ 3 our product is a maximum when x = -, y = -■ Therefore the third part is also - , and o o 3 a' the maximum value of the product is — EXAMPLES 1. Find the minimum value ot x^ + xy + y^ — ax — by. Ans. -J- {ai> — a^ — 6^). 2. Show that sin x + siny + cos (x + y) is a minimum when x = y = — , and a , IT 2 maximum when x = y = —■ 6 3. Show that xe!' + ^=™J' has neither a maximum nor a minimum. (ax + hu 4- c)^ 4. Show that the maximum value of -^ ^is a^ + If' + c^. x2 + y2 + 1 5. Find the greatest rectangular parallelepiped that can be inscribed in an ellipsoid. That is, find the maximum value of 8xyz(= volume) subject to the condition x^ «2 g2 8 oJc HiHT. Let u = xyz, and substitute the value of z from the equation of the ellipsoid. This where m is a function of only two variables. *x = 0, y = are not considered, .since from the nature of the problem we would then have a minimum. 248 DIFFERENTIAL CALCULUS 6. Show that the surface of a rectangular parallelepiped of given volume is least when the solid is a cube. 7. Examine x* + y^— x^ + xy — y^ for maximum and minimum values. Ans. (Maximum when x= 0, y = ; minimum when x = y = ± i, and when x =— y =± ^ Vs. 8. Show that when the radius of the base equals the depth, a steel cylindrical standpipe of a given capacity requires the least amount of material in its construction. 9. Show that the most economical dimensions for a rectangular ta,nk to hold a given volume are a square base and a depth equal to one half the side of the base. 10. The electric time constant of a cylindrical coil of wire is mxyz u = , ax + oy +CZ where x is the mean radius, y is the difference between the internal and external radii, z is the axial length, and m, a, b, c are known constants. The volume of the coil is nxyz = g. Find the values of x, y, z which make u a minimum if the volume of the coil is fixed. ifabcci Ans. ax = iy = cz = \j — —■ \ n CHAPTER XIX ASYMPTOTES. SINGULAR POINTS 150. Rectilinear asymptotes. An asymptote to a curve is the limit- ing position* of a tangent whose point of contact moves off to an iafinite distance from the origin, t . Thus, in the hyperbola, the asymptote AB is the limiting position of the tangent PT as the poiat of contact P moves off to the right to an infinite distance. In the case of algebraic curves the following definition is useful: an asymptote is the limiting position of a secant as two points of intersection of the secant with a branch of the curve move off in the same direction along that branch to an infinite distance. For example, the asymptote AB is the limiting posi- tion of the secant PQ a& P and Q move upwards to an iafinite distance. 151. Asymptotes found by method of limiting intercepts. The equa- tion of the tangent to a curve at (x^, y^ is, by (1), p. 76, First placing y = and solving for x, and then placing a; = and solving for y, and denoting the intercepts by x^ and y. respectively, we get dx. -y^dy^ ■■ intercept on OX ; y^ =zy^— x^ -j-^ = intercept on Y. Since an asymptote must pass within- a finite distance of the origin, one or both of these intercepts must approach finite values as limits when the point of contact (x^, y^) moves off to an infinite distance. If limit (a;;) = a and limit (z/,) = S, *A line that approaches a fixed straight line as a limiting position cannot be whoUy at infinity ; hence it follows that an asymptote must pass within a finite distance of the origin. It is evident that a curve which has no infinite branch can tave no real asymptote. t Or, less precisely, an asymptote to a curve is sometimes defined as a tangent whose point of contact is at an infinite distance. 249 250 DIFFERENTIAL CALCULUS then the equation of the asymptote is found by substituting the Hmit- ing values a and h in the equation - + ? = !• a If only one of these limits exists, but \\m\t i-^\ = m, \dxj then we have one intercept and the slope given, so that the equation of the asymptote is y y = m.v + 5, or x = — + a. m Illustrative Example 1. Find the asymptotes to the hyperbola — r = 1- „ , ,. dy b^x b 1 , limit /dy\ , 6 Solution. — = — — = ± ^=^ , and m= (-— 1 = ±-. dz a?y a I gi x _ co \^/ ^ Also x; = — and Vi= ; hence these intercepts are zero when x = y = co. X y Therefore the asymptotes pass through the origin (see figure on p. 249) and their equations are . 2/ — = ± -(x — Ox), or ay = ±bx. Ans. a This method is frequently too complicated to be of practical use. The most convenient method of determining the asymptotes to alge- braic curves is given in the next section. 152. Method of determining asymptotes to algebraic curves. Given the algebraic equation in two variables, If this equation when cleared of fractions and radicals is of degree n, then it may be arranged according to descending powers of one of the variables, say y, in the form (£) ay- + (i.c + c-)t/'-' + (^cW + ex +/) t/" -+... = ().* For a given value of i- this equation determines in general n values of y. * For use In this section the attention of the student is called to the following theorem from Algebra : Given an algebraic equation of degree n, Ay' + Btj^-i-^ Cy"-^ + Dy''-^ + --- = 0. When A approaches zero, one root (value of y) approaches x . When A and B approach zero, two roots approach oo . When A, B, and C approach zero, three roots approach oo , etc. ASYMPTOTES 251 Case I. To determine the asymptotes to the curve (B) which are parallel to the coordinate axes. Let us first investigate for asymptotes parallel to OY. The equation of any such asymptote is of the form (C) x = h, and it must have two points of intersection with (5) having infin ity, ordinates. First. Suppose a is not zero in (^), that is, the term in y" is present. Then for any finite value of x, (S) gives n values of y, all finite. Hence all such lines as (C) will intersect (5) in points having finite ordinates, and there are no asymptotes parallel to O Y. Second. Next suppose a = 0, but h and c are not zero. Then we know from Algebra that one root (=«/) of (5) is infinite for every iinite value of x; that is, any arbitrary line (C) intersects (5) at only one point having an infinite ordinate. If now, in addition, hx + c = Q, or (i)) a: = -^, then the first two terms in (5) will drop out, and hence two of its roots are infinite. That is, (i>) and (5) intersect in two points having infinite ordinates, and therefore (J>) is the equation of an asymptote to (E) which is parallel to Y. Third. If a = 6 = e = 0, there are two values of x that make y in (B') infinite, namely, those satisfying the equation (E~) dx^ + ex +/= 0. Solving (E^ for x, we get two asymptotes parallel to Y, and so on in general. In the same way, by arranging f(x, y") according to descending powers of x, we may find the asymptotes parallel to OX. Hence the following rule for finding the asymptotes parallel to the coSrdinate axis : First Step. Equate to zero the coefficient of the highest power of x in the equation. This gives all asymptotes parallel to OX. Second Step. EquMte to zero the coefficient of the highest power of y in the eqwatim,. This gives all asymptotes parallel to Y. Note. Of course if one or loth of these coefficients do not involve X (or y~), they cannot be zero, and there will he no corresponding asymptote. 252 DIFFEEENTIAL CALCULUS Illustrative Example 1. Find the asymptotes of the curve a^x = y (x — (£f. Solution. Arranging the terms according to powers of x, yx^ — (iay-^ a?)x + a^y = 0. Equating to zero the coefBcient of the high- est power of X, we get y = as the asymptote parallel to OX. In fact, the asymptote coin- cides with the axis of x. Arranging the terms according to the powers of y, (x — a)^y — a^x = 0. Placing the coefficient of y equal to zero, we get X = a twice, showing that AB is a double asymptote parallel to OT. If this curve is examined for asymptotes oblique to the axes by the method explained below, it will be seen that there are none. Hence 2/ = and x = a are the only asymptotes of the given curve. Case II. To determine asymptotes oblique to the coordinate axes. Given the algebraic equation Consider the straight line ((?) y = mx + k. It is required to determine m and k so that the line ((?) shall be an asymptote to the curve (-F). Since an asymptote is the limiting position of a secant as two points of intersection on the same branch of the curve move off to an infinite distance, if we eliminate y between (i^) and ((?), the resulting equa- tion in X, namely, (^) f(x, mx + k')= 0, must have two infinite roots. But this requires that the coefficients of the two highest powers of x shall vanish. Equating these coeffi- cients to zero, we get two equations from which the required values of m and k may be determined. Substituting these values in (G) gives the equation of an asymptote. Hence the following rule for finding asymptotes oblique to the coordinate axes : First Step. Replace y hy mx + k in the given equation and expand. Second Step. Arrange the terms according to descending powers ofx. Third Step. Equate to zero the coefficients of the two highest powers * of X, and solve for m and k. * If the term involving a;"-! is missing, or if the value of m obtained by placing the first coefficient equal to zero causes the second coefficient to vanish, then by placing the coeffi- cients of x» and x"-2 equal to zero we obtain two equations from which the values of m and Tc may be found. In this case we shall, in general, obtain two Ic's for each to, that is, - pairs of parallel oblique asymptotes. Similarly, if the term in a;»-2 is also missing, each value of m furnishes three parallel oblique asymptotes, and so on. ASYMPTOTES 253 Fourth Step. Substitute these values of m and k in y = mx + Te. This gives the required asymptotes. x^ for asymptotes. .A Y Illustkative Example 2. Examine y^ = % w? - Solution. Since none of the terms involve both X and y, it is evident that there are no asymptotes parallel to the coordinate axes. To find the oblique asymptotes, eliminate y between the given equation . and y = otk 4- &. This gives (ttm; + A:)' = 2 ox^ — x^ ; and arranging the terms in powers of x, (1 + m^)x3 + (Sm^fc— 2a)x2 + Sfe^ma; + P = 0. Placing the first two coefficients equal to zero, 14-m3 = and Sm^A; — 2a = 0. 2a Solving, we get m = — 1, i; = Substituting in y = mx, + /c, we have y -. the equation of asymptote AH. ■x+- EXAMPLES Examine the first eight curves for asymptotes by the method of § 150, and the remaining ones by the method of § 151 : 1. y — ff:. Atis. y = 0. 2. y = e-^^. Ans. y = 0. 3. y'=\ogx. i.y = (^ + lj . X = 0. y = e, X: 1. 5. y = tanx. 1 6. y = eP= — l. 7. 2/2 = 6x2 + x^ 8. Show that the parabola has no asymptotes. 9. y^ = a^ — x'. 10. The cissoid y^ = 2r — X 11. yM = y^x + x^. 12. j/2(x2 + l) = x2(x2-l). 13. 2/2 (a; _ 2 a) = x« - d^ 14. x^y^ = a^(x^ + y^). 15. 2/(x2-36x + 262) = x3-3ax2 + a8. 16. 2/ = c + - (X - by 17. The folium x^ + y^ — Saxy = 0. 18. The witch x^y = 4 a^ (2 a - 2/) • 19. X2/2 + x^y = a^. 20. x' + 2x^y - xy^ - 2y'' + iy^ + 2xy + y = 1. x + 2y =^0, x + y = 1, x-y --L n being any odd integer, x = X = 0, 2/ = 0. 2/ = X + 2. y + x = 0. X — 2r. x = a. y = ±x. x = 2a, y =±{x + a). x=± a, y =±a. x = b, x = 2 6, y + 3a = x + 3b. y = c, X = 6. y + x + a = 0. 2/ = 0. X = 0, 2/ = 0, X + 2/ = 0. 254 DIFFEEENTIAL CALCULUS 153. Asymptotes in polar coordinates. Lefc/(/>, ^) = be the equa- tion of the curve PQ having the asymptote CD. As the asymptote must pass within a finite distance (as OE") of the origin, and the point of contact is at an infinite distance, it is evident that the radius vector OF drawn to the point of contact is parallel to the asymptote, and the subtangent OE is perpendicular to it. Or, more precisely, the distance of the asymptote from the origin is the limiting value of the polar subtangent as the point of contact moves off an infinite distance. To determine the asymptotes to a jwlar curve, proceed as follows : First Step. Find from the equation of the curve the values of 6 which make p = co* These values of 6 give the directions of the asymptotes. Second Step. Find the limit of the polar subtangent PYp' by (7), p. 86 as 6 approaches each such value, remembering that p approaches oo at the same time. Third Step. If the limiting value of the polar subtangent is finite, there is a corresponding asymptote at that distance from the origin and parallel to the radius vector drawn to the point of contact. When this limit is pos- itive the asymptote is to the right of the origin, and when negative, to the left, looking in the direction of the infinite radius vector. EXAMPLES 1. Examine the hypertolic spiral p = - f or asymptotes. 6 dp a jB Solution. When 0=0, p ,de ■■ 00. Also - dS ; ; hence subtangent = p^ dp e" ■ 0^1 [p' f ] = - «. ^l"0h is finite. dp It happens in this case that the subtangent is the same for all values of 6. The curve has therefore an asymptote BC parallel to the Initial line OA and at a dis- tance a above it. * If the equation san be written as a polynomial in p, these values of 9 may be found by equatrag to zero the coefficient of the highest power of p. SINGULAR POINTS 255 i Examine the following curves for asymptotes : 2. pcos0= acos2 6. Ans. There is an asymptote perpendicular to the initial line at a distance a to the left of the origin. 3. p = a tan S. Ans. There are two asymptotes perpendicular to the initial line and at a dis- tance o from th^ origin, on either side of it. 4. The lituus pdi — a. Ans. The initial line. 5. p = a sec 2 9. Ans. There are four asymptotes at the same distance - from the origin, and inclined 45° to the Initial line. 6. (p — a) sin fl = 6. Ans. There is an asymptote parallel to the initial line at the distance 6 above it. 7. /} = a(sec2^ + tan2S). Ans. Two asymptotes parallel to ^ = — , at distance a on each side of origin. 4 8. Show that the initial line is an asymptote to two tranches of the curve p'^ime = a?cos2e. 9. Parabola p = Atis. There is no asymptote. 1— cosw 154. Singular points. Given a curve whose equation is Any poiat on the curve for which ^1=0 and ^=0 dx dy is called a singular point of the curve. All other points are called ordinary points of the curve. Since by (57 a), p. 199, we have dy__^ dx~ df it is evident that at a singular point the direction of the curve (or tangent) is indeterminate, for the slope takes the form - • In the next section it will be shown how tangents at such points may be found. 155. Determination of the tangent to an algebraic curve at a given point by inspection. If we transform the given equation to a new set of parallel coordinate axes having as origin the point in question on the curve, we know that the new equation will have no constant term. Hence it may be written in the form (A) f(x, y') = ax + ly + (ex' + dxy + ef) + (fx^ + gx'y + hxf + if) + ■■■ = (). 266 DIFFERENTIAL CALCULUS The equation of a tangent to the curve at the given point (now the origin) will be (B) ^={^y By (1), p. 76 Let y = mx be the equation of a line through the origin and a second point P on the locus of {A). If then P approa.ches along the curve, we have, from (5), (C~) limit OT = -^- ax Let O be an ordinary point. Then, by § 155, a and h do not both vanisli, since at (0, 0), from (^), p. 255, (if 3/ J, — = a, _ = J. dx dy Replace y in (^) by mx, divide out the factor x, and let x approach zero as a limit. Then (A) will become * a + bm = 0. Hence we have, from (i?) and (C), ax + hy = 0, the equation of the tangent. The left-hand member is seen to consist of the terms of the first degree in (J.). When is not an ordinary pomt we have a = b = Q. Assume that c, d, e do not all vanish. Then, proceeding as before (except that we divide out the factor a:^), we find, after letting x approach the limit zero, that (^) becomes c + dm + em' = 0, or, from (C), Substituting from (5), we see that (-£■) ex' + dxy + ef=^0 is the equation of the pair of tangents at the origin. The left-hand member is seen to consist -of the terms of the second degree in (^). Such a singular point of the curve is called a double point from the fact that there are two tangents to the curve at that point. * After dividing by x an algebraic equation in m remains wliose coefficients are functions of X. If now a; approaches zero as a limit, tbe theorem holds that one root of this equation in m will approach the limit — a-j- 6. SINGULAR POINTS 267 Since at (0, 0), from i(^), dj = 2, ay = d. dj = 2g, dx" dxdy dy" it is evident that (D) may be written in the form ^ ' dx^ dxdy\dx) dy^\dxj In the same manner, if there is a triple point at the origin, the equation of the three tangents feeing /2;3 _^ g^y ^ ^^2 ^ ^y ^ Q^ and so on in general. If we wish to investigate the appearance of a curve at a given point, it is of fundamental importance to solve the tangent problem for that point. The above results indicate that this can be done hy simple inspection after we have transformed the origin to that point. Hence we have the following rule for finding the tangents at a given point. EiBST Step. Transform the origin to the point in question. Second Step. Arrange the terms of the resulting equation according to ascending powers of x and y. Third Step. Set the group of terms of lowest degree equal to zero. This gives the equation of the tangents at the point (origin'). iLLnsTEATiTE EXAMPLE 1. Eind the equation of the tangent to the ellipse 5x2+ 52^2+ 2x?/-12x-12y = at the origin. . Solution. Placing the terms of lowest (first) degree equal to zero, we get — 12x-12y = 0, or X + 2/ = 0, which is then the equation of the tangent PT at the origin. Illustkatite Example 2. Examine the curve 3"a;2 — xy — 2 ^2 + x^ — 8 2/3 = for tangents at the origin. Solution. Placing the terms of lowest (second) degree equal to zero, 3x2.-x2/-2 2/2 = 0, or (x-y)(3x + 2y) = 0, I- 2/ = being the equation of the tangent AB, and 3x + 22/ = the equation of the tangent CD. The origin is, then, a double point of the curve. 258 DIFFEEENTIAL CALCULUS Since the roots of the quadratic equation (jF), p. 257, namely, ^/^Y-4-2-^/^V^=0 dy\dx) d,x»y\dx) dx'' may be real and unequal, real and equal, or imaginary, there are three cases of double points to be considered, according as \dxdy/ dx dy is positive, zero, or negative (see 3, p. 1). 156. Nodes. /^y_£^£^>0. In this case there are two real and unequal values of the slope -p ) found from (i^), so that we have two distinct real tangents dxl to the curve at the singular point in question. This means that the curve passes through the point in two diflFerent directions, or, in other words, two branches of the curve cross at this point. Such a singular point we call a real double point of the curve, or a node. Hence the conditions to be satisfied at a node are dx dy \Sxdy/ dx^ dy' I LLDSTKATivE EXAMPLE 1. Examine the lemniscate y^ = x- — x* for singular points. Solution. Here /(x, y) = y" — x'' + x* = 0. Also ^=_2x + 4x3 = 0, — = 2y = 0. Sx ' 8y The point (0, 0) is a singular point, since its coordinates satisfy the above three equations. We have at (0, 0) ^=-2 -?^ = ^ = 2 '^ ^ ax2 ' Bx&y • ' ey2 \dxdy) dx^Sy^ ' Af and the origin is a double point (node) through which two branches of the curve pass in different directions. By placing the terms of the lowest (second) degree equal to zero we get 2/2 — x^ = 0, or y = X and y =— x, the equations of the two tangents AB and CD at the singular point or node (0, 0). SUSTGULAE, POINTS 259 157. Cusps. [dxdyj dx^ dy^ In this case there are two real and equal values of the slope found from (F^; hence there are two coincident tangents. This means that the two branches of the curve which pass through the point are tangent. When the curve recedes from the tangent in both directions from the point of tangency, the singular point is called a pmnt of osculation ; if it recedes from the point of tangency in one direction only, it is called a cusp. There are two kinds of cusps. First kind. When the two branches lie on opposite sides of the common tangent. Second Mnd. When the two branches lie on the same side of the common tangent.* The following examples illustrate how we may determine the nature of singular points coming under this head. Illustrative Examplb 1. Examine a*y^ = a%* — x" for singular points. Solution. Here /(x, y) — a*y^ — a^x^ + x« = 0, dx = —iaH^+ 6x5 32/ and (0, 0) is a singular point, since it satisfies the above three equations. Also, at (0, 0) we have -y 8^ 3x2 = 0, 0, a2/ = 2 a*, Sxdy ' Sy^ and since the curve is symmetrical with respect to OY, the origin is a point of osculation. Placing the terms of lowest (second) degree equal to zero, we get y^ = 0, showing that the two common tangents coincide with OX. Illtistrative Example 2. Examine y^ = x' for singular points. Solution. Here /(x, y) = y^ — x^ = 0, 3x .=-3x2 = 0, sy showing that (0, 0) is a singular point. Also, at (0, 0) we have - = 0, «^/. = 0, ^ = 0, ^ = 2. Wsy/ 3x2 32/2 3x2 ■ gj-gy This is not a point of osculation, however, for if we solve the given equation for ;/ we get y = ± VxS, * Meaning in the neighborhood of the singular point. 260 DIFFERENTIAL CALCULUS which shows that the curve extends to the right only of 0¥, for negative values of x make y imaginary. The origin is therefore a cusp, and since the branches lie on oppo- site sides of the common tangent, it is a cusp of the first kind. Placing the terms of lowest (second) degree equal to zero, we get y^ = 0, showing that the two common tangents coincide with OX. Illustrative Example 3. Examine {y — x'')^=x^ for singular points. Solution. Proceeding as in the last example, we find a cusp at (0, 0), the common tangents to the two branches coinciding with OX. Solving for y, y = x^ ± xi. If we let X take on any value between and 1, y takes on two different positive values, showing that in the vicinity of the origin both branches lie above the common tangent. Hence the singular point (0, 0) is a cusp of the second kind. 158. Conjugate or isolated points. | — — ) 4 —^ < 0. In this case the values of the slope found are imaginary. Hence there are no real tangents ; the singular point is the real intersection of imaginary branches of the curve, and the coordinates of no other real point in the immediate vicinity satisfy the equation of the curve. Such an isolated point is called a conjugate point. Illdstkative Example 1. Examine the curve y^=x^—x^ for singular points. Solution. Here (0, 0) is found to be a singular point of the curve at which — = ± V— 1. Hence the origin is a conjugate point. Solving the equation tor y, y = ±x Vx^. This shows clearly that the origin is an isolated point of the curve, for no values of X between and 1 give real values of y. 159. Transcendental singularities. A curve whose equation involves transcendental functions is called a transcendental curve. Such a curve may have an end point at which it terminates abruptly, caused by a discontinuity in the function; or a salient point at which two branches of the curve terminate without having a common tan- gent, caused by a discontinuity in the derivative. Illustrative Example 1. Show that y = x\ogx has an end point at the origin. Solution. X cannot be negative, since negative numbers have no logarithms; hence the curve extends only to the right of OY. When x = 0, y = 0. There being only one value of y for each positive value of x, the curve consists of a single branch terminating at the origin, which is therefore an end point. SINGULAR POINTS 261 Illustrative Example 2. Show that y = 1 1 + e^ J has a salient point at the origin. Solution. Here — = 1- - dx 1 e' 1 + e=" X (1 + e^)2 If X is positive and approaches zero as a limit, we have ultimately 3/ = and -^ z= 0. dx If X is negative and approaches zero as a limit, we get ultimately y = and ^ = 1. dx Hence at the origin two branches meet, one having OX as its tangent and the other, AB, making an angle of 45° with OX. EXAMPLES 1. Show that ^2 — 2 x^ + x^ has a node at the origin, the slopes of the tangents being ±V2. 2. Show that the origin is a node of y^{a? + x^) = x^ (a^ — x^), and that the tan- gents bisect the angles between the axes. 3. Prove that (a, 0) is a node oty^ = x(x— a)^, and that the slopes of the tangents are ±Vo. 4. Prove that a^y^ — 2 ahx^y — x'' = has a point of osculation at the origin. 5. Show that the curve y^ = x^ + x* has a point of osculation at the origin. 6. Show that the oissoid y^ = - - has a cusp of the first kind at the origin. 2a — X 7. Show that j/^ = 2 ax^ — x' has a cusp of the first kind at the origin. 8. In the curve (y — x^Y = x" show that the origin is a cusp of the first or second kind according as n is < or > 4. 9. Prove that the curve x* — 2 aa?y — axy^ + a?y^ — has a cusp of the second Mnd at the origin. 10. Show that the origin is a conjugate point on the curve y^ (x^ — a?) = x^ 11. Show that the curve y^ = x (a + x)^ has a conjugate point at (— a, 0). 12. Show that the origin is a conjugate point on the curve ay^ — x^ + 6x^ = when a and 6 have the same sign, and a node when they have opposite signs. 13. Show that the curve x* + 2 ax^y — ay^ = has a triple point at the origin, and that the slopes of the tangents are 0, + V2, and — V2. 14. Show that the points of intersection of the curve ( -| + (-) = 1 with the axes are cusps of the first Mnd. ^"'^ ^^' 15. Show that no curve of the second or third degree in x and y can have a cusp of the second kind. 16. Show that y = e ^ has an end point at the origin. 17. Show that y = x arc tan - has a salient point at the origin, the slopes of the tangents being ± — . CHAPTER XX APPLICATIONS TO GEOMETRY OF SPACE 160. Tangent line and normal plane to a skew curve whose equations are given in parametric form. The student is already familiar with the parametric representation of a plane curve. In order to extend this notion to curves in space, let the coordinates of any point P (x, y, z) on a skew curve be given as functions of some fourth variable which we shall denote by t, thus, (^) ^ = .^(0, 2/=t(0. « = x(0- \ [ztAx.^-fAi/.z-l-AzI i^y The elimination of the parameter t between these equations two by two will give us the equations of the pro- jecting cylinders of the curve on the X/ coordinate planes. Let the point P(x, y, z) correspond to the value t of the param- eter, and the point F'(^x + Ax, y + Ay, z + Az) correspond to the value t+At; Ax, Ay, Az being the increments of x, y, z due to the incre- ment At as found from equations {A). From Analytic Geometry of three dimensions we know that the direction cosines of the secant (diagonal) FF' are proportional to Ax, Ay, Az; or, dividing through by At and denoting the direction angles of the secant by a', /3', 7', I 01 , Ax Ay Az cos a' : cos B' : cos 7' : : — : — ^ : ' At At At Now let P' approach P along the curve. Then At, and therefore also Ax, Ay, Az, will approach zero as a limit, the secant PF' will approach the tangent line to the curve at P as a limiting position, and we shall have dx dy dz dt dt ' dt 262 cos a : cos /8 : cos 7 : APPLICATIONS TO GEOMETEY OF SPACE 263 where a, /3, 7 are the direction angles of the tangent (or curve) at F- Hence the equations of the tangent line to the curve at the point (x, y, z) are given hy X-x Y-y Z-z (69) dt dz Tt and the equation of the normal plane, i.e. the plane passing through (x, y, z) perpendicular to the tangent, is X, Y, Z being the variable coordinates. Illustrative Example 1. Find the equations of the tangent and the equation of the normal plane to the helix * (d being the parameter) (x = acos6, ■iy = a Bind, Iz = bS, (a) at any point ; (b) when 5 = 2 ir. dx Solution. dy = — asinff =—y, ^ = acos6 d0 d0 dz , ■ X, — = 0. d9 Substituting in (69) and (70), we get at (x, y, z) X-x Y—y Z-z -y tangent line ; and -y{X-x) -^xij— y) + 6(Z — z) = 0, normal plane. When 9 = itr, the point of contact is (a, 0, 267r), J- ^"^"^ X-a _ r-0 _ Z-2hrr d " a ~ b or, X=a, bY = aZ — 2 (Onr, the equations of the tangent line ; and aY+bZ-2b^ir = 0, the equation of the normal plane. ♦ The helix may be defined as a curve traced on a right circular cylinder so as to cut all the elements at the same angle. Take OZ as the axis of the cylinder, and the point of starting in OX at Pq- Let o = radius pi base of cylinder and 6= angle of rotation. By definition, PIT PN ' k (const.) , or z=akd. SN arcPo-^ «* Let ak=h; then z=be. Also y = MK= a sin B, x= 0M= a cos 6. 264 DIFFERENTIAL CALCULUS EXAMPLES Find the equations of the tangent line and the equation of the normal plane to each of the following skew curves at the point indicated : x — 2_y — l z — 4. \. x = 2t,y = P,z = i.ti; t = l. Ans. 2 2 16 x + y + 8z-35 = 0. 2. a; = t2-l, 2/ = i + l, z = i8; J = 2. Ans. ?^ = ^ZL? = ?J1^ ; 4 1 12 4x + 2/ + 12z — 111 = 0. 3. x = t^-l,y = t + f^,z = it^-it + l; < = 1. Ana. - = l^ = ^^; ' ' ' 3 3 9 a; + jr + 3z-8 = 0. A , -4 44'^ A 4x — IT V2« — 1 V2z — 1 4. X = i, ^ = sm i, z = cos i ; t = - . Ans. = i = ; 4 4 1 _ 1 I6X+V2?/— V2z-47r = 0. 5. x = at,y — U^, z = ct^; t = 1. 6. x = t, y = l~t', z = 3i^ +'4i; <=— 2. 7. x = t,y = e',z = e-'; t = 0. 8. X = asint, y = b cos t, z = t; i = — ■ 9. Find the direction cosines of the tangent to the curve x = t", y = t^, z = i* at point X = 1. 161. Tangent plane to a surface. A straight line is said to be tan- gent to a surface at a point P if it is the limiting position of a secant through F and a neighboring point P' on the surface, when P' is made to approach P along the surface. We now proceed to establish a theorem of fundamental importance. Theorem. All tangent lines to a surface at a given point* lie in general in a plane called the tangent plane at that point. Proof. Let be the equation of the given surface, and let P (x, y, z) be the given point on the surface. If now P' be made to approach P along a curve C lying on the surface and passing through P and P', then evidently the secant PP' approaches the position of a tangent to the curve C at P. Now let the equations of the curve C be (5) a;= V.di^^dt+^Jt='- By (52), p. 196 This equation shows that the tangent line to C, whose direction cosines are proportional to dx dy dz dt dt dt is perpendicular * to a line whose direction cosines are determined by the ratios ^^^ ^^ g^ ox dy dz and since C is any curve on the surface through P, it follows at once, if we replace the point T(x, y, a) by Ii(x^, y^, z^), that all tangent hues to the surface at P^ lie in the plane ^ dF^ dF, dF. . (71) — i(x-jrO+r-i (y-^i) +77(^-^1) =0,t ox^ oy^ dz^ which is then the formula for finding the equation of a plane tangent at (x , y , z ) to a mirfaee whose equation is given in the form F(x, y, 2) = 0. In case the equation of the surface is given in the form z =f(x, «/), let (-D) F(x, y, Z-) =f(x, y)-z=0. dx dx dx dy dy dy dz * From Solid Analytic Geometry we know that il two lines having the direction cosines coscti, cos;3i, C0S7X ^■•"i '^°^ '^i' ""^^zt <=os 72 are perpendicular, then cos OTi cos a2 + cos ^1 cos P2 + cos 7i cos 72 = 0. dF-i BF-i SFi t The direction cosines of the normal to the plane (71) are proportional to g^ > g— > -gj- ■ Hence from Analytic Geometry we see that (C) is the condition that the tangents whose direction cosines are cos a, cos (3, cos 7 are perpendicular to the normal; i.e. the tangents must lie in the plane. } In agreement with our former practice, dF\ dF\ dFi Sz^ dz^ denote the values of the partial derivatives at the point (xi, 2/1, Zi). 266 DIFFERENTIAL CALCULUS If we evaluate these at (x^, t/^, z^ and substitute in (71), we get ('2) S(^-'i) + 5(y-yi)-(^--^i) = o, which is then the. formula for finding the equation of a plane tangent at (x^ ^j, Zj) to a surface whose equation is given in the form z =f(x, y). In § 126, p. 197, we found (55) the total differential of a function u (or z) of x and y, namely, (E) dx dy We have now a means of interpreting this result geometrically. For the tangent plane to the surface z =/(i, y) at (x, y, z) is, from (72), {F) Z-z = |(X-.)-,|(r-.), X, Y, Z denoting the variable coordinates at any point on the plane. If we substitute . X = x + dx, and Y=y + dy P(.X,Y,Z) in (F), there results (G) Z-z = ~dx + —dy. dx dy Comparing (E) and (G), we get (H) dz = Z — z. Hence Theorem. The total differential of a function f(x, y) corresponding to the increments dx and dy equals the corresponding increment of the z-coordinate of the tangent plane to the swrfoice z =/(x, y). Thus, in the figure, PP' is the plane tangent to surface PQ at ^Y P(x, y, z). Let AB = dx and CD = dy ; then (Jz = Z - z = DP' - DE = EP'. 162. Normal line to a surface. The normal line to a surface at a given point is the line passing through the point perpendicular to the tangent plane to the surface at that point. The direction cosines of any line perpendicular to the tangent plane (71) are proportional to dF, gjr ai-; dx. (73) dz. APPLICATIONS TO GEOMETRY OF SPACE 267 an the equationg of the normal line* to the surface F(x, y, 2)=0 at Similarly, from (72), (74) x-x^ ^ y-y^ ^ z-z^ azj azi - 1 are the equations of the normal line * to the gurfdce z =/(a;, y')at(x,y,z). EXAMPLES 1. Find the equation of the tangent plane and the equations -of the normal line to the sphere x^ + y^-\-z^ = 14 at the point (1, 2, 3). Solution. Let F(x, y, z) = x^ + y^ +z^ - 14 ; . dF dF dF then =2x,-- = 2y,— = 2z; x^^ 1, y^ = 2, z^ = 3. ox oy oz asi ^y■^ Sz, Substituting in (71), 2(x - 1) + 4(y - 2) + 6(z - 3) = 0, x + 22/ + 3z = 14, the tangent plane. substituting in (73), ?^ = ^^^ = ?^ , giving z = 3 a; and 2z = 3y, equations of the normal line. 2. !E1nd the equation of the tangent plane and the equations of the normal line to the ellipsoid 4a;2 + 9^" + 362" = 36 at point of contact where x = 2, y = 1, and z is positive. Ans. Tangent plane, 8{x-2) + 9(y -1) + 6 VTl (z - ^ Vll) = ; ,,. x-2 y-1 z-iVu normal line, = = ^^ni — 8 9 eViT 3. Knd the equation of the tangent plane to the elliptic parabola z = 2 x^ + 4 j/^ at the point (2, 1, 12). Ans. 8x + Sy — z = 12. 4. Eind the equations of the normal line to the hyperboloid of one sheet x2 _ 43/2 + 2z2 = 6 at (2, 2, 3). Am. y + 4x = 10, 3x-z = 3. 5. Find the equation of the tangent plane to the hyperboloid of two sheets i" «2 z" . x,x y,y z,z , 6. Find the equation of the tangent plane at the point (x„ y.^, Zj) on the surface 0x2 + 6j,2 ^ 0^2 ^ (J _ 0. Ans. oXjX + iyi_y + czjZ + d = 0. 7. Show that the equation of the plane tangent to the sphere a;2 + 2/2 + z2 + 2ix + 2Jlf?/ + 2JVz + -D = at the point (Xj, y.^, z^) is a^iX + ViV + z.^z + L{x + Xi) + M{y + y^) + N(z + z{) + D = 0. * See second footnote, p. 265. 268 DIFFEEENTIAL CALCULUS 8. Pind the equation of the tangent plane at any point of the surface x^ + y^ + z^ = o^, and show that the sum of the squares of the intercepts on the axes made by the tangent plane is constant. 9. Prove that the tetrahedron formed by the coordinate planes and any tangent plane to the surface xyz = a' is of constant volume. 10. Find the equation of the tangent plane and the equations of the normal line to the following surfaces at the points indicated : (a) 2x2 + 42/2_z = 0; (2, 1, 12). {d) 3x^ + y^ - 2z = 0; x = 1, y = 1. (b) x^ + iy'-z' = 16; (1,2, -1). (e) x^y^ + 2x + z' = 16; x - 2, y = 1. (c) x^ + y^ + z^ = 11; (3, 1, 1). (f) a;2 + 32/2 + 2z2 = 9; 2/ = l, z = l. 163. Another form of the equations of the tangent line to a skew curve. If the curve in question be the curve of iatersection AB of the two surfaces F(^x, y, 2) = and u^ G (x, y, z) = 0, the tangent line PT at -P(a;j, y^, z^) is the iatersection of the tangent planes CD and GE at that point, j^ for. it is also tangent to both surfaces and .(^ ' hence must lie in both tangent planes. The equations of the two tangent planes at P are, from (71), (75) dF, OF, dF, aG, SG, SG, Taken simultaneously, the equations (75) are the equations of the tangent line PT to the skew curve AB. Equations (75) ia more com- pact form are (76) or. (77) x—x^ y-Vx z — z. dF\dG^_dF\dG^ dF\dG^_dF\dG^ dF^dG^ dF^dG^ dy^ dz^ dz^ ey^ dz^ dx^ dx^'dz^ dx^'dy^~'dy^'dx^ X— X, y-Ui z-z. dF^ dF^ dy^ dz^ 5Gi 5Gi dy^ dz^ dF^ dF^ dz^ dx^ 5Gi aGi dz^ dx^ dF^ aFj dx^ dy^ eGi 5Gi dx^ dy^ using the notation of determinants. APPLICATIONS TO GEOMETRY OF SPACE 269 164. Another form of the equation of the normal plane to a skew curve. The normal plane to a skew curve at a given point has abeady been defined as the plane passing through that point perpendicular to the tangent line to the curve at that point. Thus, in the above figure, FHI is the normal plane to the curve AB at P. Since this plane is perpendicular to (77), we have at once dF^ dF^ dy^ dz^ dG^ 5Gi az/i dz^ (^-^i) + dF^ dF^ azi dx^ dG^ aGi dz^ dx^ (jj-yO + dF^ dF^ dx^ dy^ dG^ dGj^ dx^ dy^ (z — z^') = 0. (78) the equation of the normal plane to a skew curve. EXAMPLES 1. Eind the equations of the tangent line and the equation of the normal plane at (r, r, r V2) to the curve of intersection of the sphere and cylinder whose equations are respectively x^ + y^ + z^ = 4: f^, x" + y^ = 2rx. Solution. Let i?' = x2 + y^ + z^ — ir'' and G = x^ + y^-2rx. ^ = 0. m^2r, ^^ = 2 dx ' '- ■^1 dx. -0, Substituting in (77), x — r _y — r _z — r V2 . or, 2/ = r, a; + V2 z = 3 r, the equations of the tangent PT at P to the curve of intersection. Substituting in (78), we get the equation of the normal plane, - V2 (x - r) +0 (y - r) + (z - r V2) = 0, or, Vix— z = 0. 2. Find the equations of the tangent line to the circle a;2 + yi + z2 = 25, E + z = 5, at the point (2, 2 Vs, 3). A-ns. 2x + 2 VS?/ + 3z = 25, x + z = 5. 3. Knd the equation of the normal plane to the curve at (Xj, 2/j, Zi). x^ - rx + 2/2 = 0, Am. 2 y^z^x - (2 Xi - r) z^y - ry^z = 0. 270 DIFFEEENTIAL CALCULUS 4. Find the equations of the tangent line and the normal plane to the curve at (1, - 1, 2). 5. Find the direction of the curve xyz = 1, y^ = x at the point (1, 1, 1). 6. What is the direction of the tangent to the curve y = x^, z^ = l — y at (0, 0, 1) ? 7. The equations of a helix (spiral) are x2 + y2 = r'-i, y = X tan - . c Show that at the point (Xj, j/j, Zj) the equations of the tangent line are c(s-Xi) + 2/i(2-Zi) = 0, c(2/-2/i)-a;i(z-Zi) = 0; and the equation of the normal plane is yjX — Xjy — c{z — z^) = 0. fg2 y1 -2 8. A skew curve is formed by the intersection of the cone V- = and 02 62 c2 the sphere x^ + y^ + z2 = ^2. ghow that at the point (Xj, y^, Zj) the equations of the tangent line to the curve are c2 (a? - 62) xj (x - X,) = - a2 (ft^ + c^) z^ (z - z,), c2 (a2 - 62) y^ (y - i/j) = + 6^ (c^ + a^) z, (z - z^) ; and the equation of the normal plane is efi (62 + c2) j/jZiX - 62 (c2 + a2) z^x^y - c^ (a^ - 62) x^y^z = 0. CHAPTER XXI CURVES FOR REFERENCE For the convenience of the student a number of the more common curves employed in the text are collected here. Cubical Pakabola y Semicubical Pakabola ' = aa^. i/^=cu^. The Witch of Agnesi The Cissoid op Diocles Y T^ --' V 'J ^^^ o X se'y==4a\2a-i/'). f(2a-x) = : 271 272 DIFFERENTIAL CALCULUS The Lemniscatb of Beknoulli The Conchoid of Nicomedes Y p^ = a^ cos 2 6. p =a CSC 6 + b. Cycloid, Oedinaky Case Y 1 'a ^ a ^^^v ^ 1 J ■ a \/ X x = a arc vers — — y/zay- X-V2. X = a(d — sin 6'), y=a(l — cos ^). Cycloid, Vertex at Origin Y Ic x= a arc vers - + V2 ay — y^. x — a(6 + sin ^), r/=a(l — cos^). Parabola a;'^+z/*=a* CUKVES EOE EEFERENCE 27S Hypocycloid of Four Cusps Evolute of Ellipse - Y\ x=a cos' 6, y = a sin' Q. {axf+(lyf=(a^-l?-f Caedioid /3 = a (1 — cos ^). Folium of Descartes Y a?+i^~ Saxi/ = 0. Sine Curve Cosine Curve t/ = sin X. y = cos X. 274 DIFFEKENTIAL CALCULUS LiMAcosr Stbophoid ¥ p = b — a cos 0. r-^ ,a + x Spiral ( 3F Archimedes r X / \ f \ hs ^ ' r p = a6. Logarithmic ok Equiangular SflSAI. p = e"*, or log p = a6. Hyperbolic ok Eeciprocal Spiral Lituus pB = a. P^e = a\ OUEVES FOR EEFERENCE 276 Parabolic Spiral Logarithmic Curve (p — a)^ = 4 ac6. y = log X. Exponential Curve i/ = e'. Probability Curve Y Secant Curve |t 1 if . I¥ ^Uir « 1 ll 1 ^ y = sec X. Tangent Curve y = tan x. 276 DIFFERENTIAL CALCULUS Theee-Lbaved Eosb Three-Leaved Eosb 3 y p = a sin 3 d. p = a cos 3 6. Four-Leaved Eosb Foub-Lbaved Rose p= a sin 2 0. p = a cos 2 0. Two-Leaved Rose Lemniscate p''=a'sm2e. Eight-Leaved Rose p = a sin 4 6. CUEVE8 FOR EEPEEEJiTCE 27T Cttkve with End Point AT Origin Curve with Salient Point AT Origin JO y = x log X. i(l + ^ = x. CuKVB WITH Conjugate (Isolated) Curve with Cusp op Second Point at the Origin Kind at Origin (iy-xy=a^. p = a sec - . Equilateral Hyperbola • IT I X xy = a. IliTEaRAL CALCULUS CHAPTER XXII INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 165. Integration. The student is already familiar with the mutu- ally inverse operations of addition and subtraction, multiphcation and division, involution and evolution. In the examples which fol- low, the second members of one column are respectively the inverse of the second members of the other column : y = «% a; = log„?/; y — sin X, x= arc sin y. From the Differential Calculus we have learned how to calculate the derivative /'(a;) of a given function f(x), an operation indicated by or, if we are using differentials, by df(x)=fix)dx. The problems of the Integral Calculus depend on the inverse operation, namely : To find a function fQc) whose derivative (A) f'(x) = (x) is given. Or, since it is customary to use differentials in the Integral Calculus, we may write (£) ' 'df(x-) =f'(a>') d^ = 'l> (.<"} <^^^ and state the problem as follows : Saving given the differential of a function, to find the function itself. 279 280 INTEGRAL CALCULUS The function /(a;) thus found is called an integral* of the given differential expression, the process of finding it is called integration, and the operation is indicated by writing the integral dgn^ j in front of the given differential expression ; thus (C) jf(x)dxt^f(x). read an integral of f'(x) dx equals f(x). The differential dx indicates that X is the variable of integration. For example, (a) If f(x') = x\ then /'(«) dx=B 3?dx, and /^ %x^dx = x\ (b) If/(a;) = sina;, th.QTX f (x) dx = cos xdx, and / cos xdx = sui X. (c) If /(a;) = arc tan x, then /'(a;) dx = z ^ > and dec ; = arc tan x. Let us now emphasize what is apparent from the preceding expla- nations, namely, that Differentiation and integration are inverse operations. Differentiating (C) gives (1)) djf<(x)dx=f(x)dx. Substituting the value of /'(») dxl= df(x')'] from (S) in ( C), we get W Jdf(x-)==f(xy d r Therefore, considered as symbols of operation, -j- and I ■■■dx wcq inverse to each other ; or, if we are using differentials, d and / are inverse to each other. "^ * Called anti-differential by some writers. t Historically this sign is a distorted S, the initial letter of the -word sum. Instead of defining integration as the inverse of differesitiation, we may define it as a process of sum- mation, a very important notion which we will consider in Chapter XXVIII. X Some authors write this D~ f'(x) when they wish to emphasize the fact that it is an inverse operation. INTEGRATION 281 When d is followed by / they annul each other, as in (Z)), but when I is followed by c?, as in (^), that will not in general be the case unless we ignore the constant of integration. The reason for this will appear at once from the definition of the constant of integration given in the next section. 166. Constant of integration. Indefinite integral. From the pre- ceding section it follows that since d(a?') = 3 x^dx, we have / 3 a?dx — a?; since ci(a;'-|- 2) = 3 x^dx, we have / 3 x^dx = x'+2; since d(af— 7) = 3 x'dx, we have I 3 afdx = a;'— 7. In fact, smce ^^^a^ g,^ _ 3 ^^^^ where C is any arbitrary constant, we have / dx''dx = x''+C. A constant C arising in this way is called a constant of integration.* Since we can give C as many values as we please, it follows that if a given differential expression has one integral, it has infijiitely many differing only by constants. Hence P \f(x)dx=fix) + C; and since C is unknown and indefinite, the expression is called the indefinite integral of f'(x) dx. It is evident that if ^(x) is a function the derivative of which is /(a;), then ^ (x) + C, where C is any constant whatever, is likewise a function the derivative of which is f{x). Hence the Theorem. If two functions differ hy a constant, they have the same It is, however, not obvious that if ^ (x) is a function the derivative of which is f(x), then all functions having the same derivative fCx) are of the form <^(x) + C, where C is any constant. In other words, there remains to be proved the * Constant here means that it is independent of the variable of integration. 282 USTTEGEAL CALCULUS Converse theorem. If two functions have the same derivative, their difference is a constant. Proof. Let (x) and ■^•{x') be two functions having the common derivative /(a;)- Place F(z') = (x) = -^l d f ^^ + (7j = V'dv, VI, p. 34 ifdv = + C. This holds true for all values of n except w = — 1. For, when w = — 1, (4) gives / ^-1+1 1 which has no meaning. The case when n = — l comes under (5). Proof of (5). Since , d(logv+C) = —, Vma, p. 35 we get f ~~ ^°S v+C. The results we get from (5) may be put in more compact form if we denote the constant of integration by log c. Thus / — = log V + log c = log ev. Formula (5) states that if the expression under the integral sign is a fraction whose numerator is the differential of the denominator, then the vntegral is the natural logarithm of the denominator. 286 ESTTEGEAL CALCULUS EXAMPLES * For formulas (l)-{5). Verify the following integrations : 1. fx'^dx = — 1- C = — + C, by (4), where v = x. and n = 6 J 6 + 1 7 2. fv^di = fxi dx = y + O = |xi + C, where v = x and n = i. where v = x and n = — 3. 4. fax^dx = aCx^dx = —- + C. 5. rx2di = ^ + C. by (4) by (4) By (2) and (4) /2 , 3 XT XT dx = -— + C. J 3x2 3a; J ax^ 2 a 10. j'5ydy = ^ + C. dx — -X V2px + C. o 3xi .3. Cs-ids = 2Vs + C. .. f 5m^z^dz = ^^ + G. 1-n 1 r(nx) " (ix = (nx)»4-C. [. Cy-m-idy= + (7. 11. rV2px 19. r(2x'-5x2-3x + 4)dx= C%x^dx- C bx^dx- Cixdx + C i = 2 Tx^dx - 5 Tx^dx - 3 Cxdx + 4 fdx dx by(i) by (2) X* 5xs 3x2 , : — h 4 X + C. 2 3 2 Note. Although each separate integration requires an arbitrary constant, we write down only a single constant denoting their algebraic sum. 20. Ci^-^ + ZcVlAdx= f2ax-idx- l'bx-^dx+ Cscx^dx = 2a Cx-^dx — b Cx-^dx + 3e fx^dx :2a.^-6.^ + 3c.?! + C * - 1 i by(i) by (2) by (4) = 4aVxH 1--CXT + C. » 5 * Wlien learning to integrate, the student should have oral drill in integrating simple functions. INTEGRATION 287 5 7 22. r(-5/ii_ 1 +lW^i^_3xJ-JL + c. J \ -5^x5/ 5 2x* . r(ot - x^Ydx = aH + ~ahi --aix^ - — + C. «/ 7 5 3 /x= 23. " ' Hint. First expand. 24. /(a= - ,V V^d^ = 2yf (I - i^ + i^* - g)+ C. 25. /(VS - Vt)»d« = alt - 2 ait + ?^ - ^ + C. (x^ - 2)8x8da! = _ _ ^ + 2x6 - 2x« + O. 10 a 27. J(a2 + 6V)ixdx = ("' + ^°'«')* + c. Hint. This may be brought to form (4) . For let v=a^ + fi^ajZ and n=i; then d« = 2 b^dx. If we now Insert the constant factor 2 6^ before xdx, and its reciprocal —-rz before the integral sign (so as not to change the value of the expression) , the expression may be integrated, using (4), namely, /> yn +1 .1 n + 1 Thus, r(o2+63a;2)4a;(2a; = -L C {a^ -1-1,2x^)^2 b^dx^—^ r(a2 + ?>2a;2)4d(a2 + 62a;2) 1 (a2+ 62a;2)i ^ ^ (ag + 62a!2)i ^ 2 62 3 3 62 Note. The student is warned against transferring any function of the variable from one side of the integral sign to the other, since that would change the value of the integral. ■ 28. rVo2 - x^xdx =[(0? - x^ixdx=- i(a^ - x^)! + C. 29. f(3ax2 + 4&x')t(2ax + 46x2)*E=^(3ax2 + 45x')l + C. Hint. Use (4) , making v = Sax!^ + 4 bx^, cJw = (6 aa; + 12 6a;2) dx and n = i- 30. fb{6ax^ + 86xS)i(2 ax + 46x2)dx = — {6ax^ + 86xS)f + C. J 16 3 ._^^^2 ^^3^_^^ Hint. Write this f (a^ + x^)~^x^dx and apply (4). 32. f ^_ =-2VT^r^ + 0. ■^ Vl-x 33. /2,n/(^ + l^^dy = |^(?'' +i'¥ + C ■ 34. r(l + e!>;)i^>dx = 1(1 + e»:)i + C. sini'x cosxcfe = / (sin x)^ cos x A? = ^ ^ + C = — - — 1- C. Hint. Use (4) , making « = sin a, dv = cos xdx, and n=-2. 288 INTEGRAL CALCULUS „„/■«. , oos'x _ 36. I cos'x sinxdx= l-C J 6 37 . / sin' ax cos axdx = — sin* ax + C. J 4a cos*3x sin Sxdx = cos=3a; + C. 15 39. f /^ =-Vaii-x^ + C. Va2 - a;2 5 adi _ a 1)6 ~ (6 - «)s 40. r_i^^ = _^^_ + c. 41. fv'l + x^xtJx = J(l + x2)t + c. •^ Vl - s2 4 M»- Id-it _ (a +6u»)i t + 6M»)">~ 6w(l- 2 osds _ a ,2 _ c2s2)2 ~ c' Saxdx 3a 62 + e^x 3axdx „ /■ stix 43. r ^" -- ^ v-^-/- +c. J (a + 6m»)"> 6w (1 — m) 44 r 2'^'^« = ° I c 45. I = — - log(62 + e^x^) + C. „ , , . /> 3 oxdx „ /■ stix -, ,„. Solution. ( — — - = 3 a I — -— ■ By (2) This resembles (5). For let v = b^ + ^3? ; then d» = 2 e^xdx. If we introduce the factor 2 e^ after the integral sign, and — - bfef ore it, we have not changed the value of the expression, but the numerator is now seen to be the differential of the denom- inator. Therefore 3 a \ — — I = — / — ^^ — ■ = log (62 + e2x2) + G. By (5). J 62 + e2x2 2 e2 J 62 + e2i2 2 e2 J 62 + e2x2 2 e2 ^ ' ^ ' 50. ''^'*" . rx^Ox x2 x= , , ,. ^ '•i^TT=^-"¥+3-^°^(^ + '^ + ^- Hint. First divide the numerator by the denominator. 51- ^^k+l^ " "^ ~ "^o^K^x + 3)2 + C. 52. INTEGRATION 289 /2;n — 1 I 2 dx = -log (a?*.— nx) + C. x''— nz n 53 r (.--2)^^. ^2_6 j^^_ -/ a + 6f» n6 55. r(log a)3 — = 1 (log a)i + C. -^l~dr = — + r + 21og{r - 1) + C. 57. J — -^ = 21og(e>=+l) + C. ' sin xdx 1 , + baosx sec^ 6dd 1, 58. I = log (a + ft cos x) + C. ./ a + ocosx ^°- /S =•¥ + '' + ^"s^""- ^> + ^• 61. f?^^dr = \og(ef+lY-r + C. 62. Integrate the following and verify your results by differentiation : {a)fUx^-^dx. Solution. fUx^--\dx = 4 fx^dx - 2 f— = — - 2 log x + C. renficution,. d(^ — 21ogx + c\ = (^:3x^-2 ■-\dx = Ux^-?\dx. (b)/x^&. (h)/s™ + «ds. (n)J|£l^. (t)/sin3^cos^dx. (e)/5^x<^. (i)/..i... (0)/,-^. (u)/^. (f)/75fa^. (l)/6^cfe. (r)j(?^hil^. (x)/(J + l)^6^dx. <^)/^- <'"^/^\ ^^^/v5f3^^- (^)/<^°^*)'t- Proofs of (6) and (7). These follow at once from the corresponding formulas for differentiation, II and Ha, p. 35. 290 INTEGRAL CALCULUS- EXAMPLES For formulas (6) and (7). Verify the following integrations : ba?^dx = — — + C. 2 log a Solution, fftas^ffe = ft frt- "^dj-. By (2) This resembles (6). Let« = 2a;; then tli} = 2 ilx. If we then insert the factor 2 before (fe and the factor ^ before the integral sign, we have bCa^-dx = -fa^-2dx = lfa^-d{2x) = l-:^ + C. By (6) J 2J iJ ^ ' 2 loga 2. Cz&^ = %^-irC. 7. Je-»*i;=-e-^+C. . f^dx = n^+C. 8. J e^dx = — + C. 3 fax /g2cosx * /* 5^ ^ e-2cosa:sini(fe= ^ + ^- ^^' J ^ \^'^ " 21og3 •■' _l + loga /I / a'^ \ (esx + a6'»)dx = -{^'^ + j + C. 13. rg^ + 4»^ + 8(x + 2)(ix^ie^ + *^ + » + C. Ignx _ femx) cfe = — + C. nloga mlogo /X _3C X _X (e" + e ") *c = a (e" — e ") + C 16. r(e!'+ e-!')2dy = l(e2i'-e-2!') + 22/ + C. J a^f log o — log b iei^+a^x+3i,^2x)clx = —- + — — — - + C. 4 51oga ■ 2 log 6 /I fpSat . g— 8a("l (ea< + e- ■«)Sdt = ± 1_ + 3 ea! -1 3 e- <" -— + C. a\_ 3 3 J 20. Integrate the following and verify your results by differentiation : (a) Ce^'ds. (e) fe-^'^dx. (i) Cbe^dx. (m) fa^xdx. 2'^t. (i) J e " *i:. (n) j e '■ dS. (c) fc"''^. (g) fs^e^dx. (k) ri^- <^Uvii- ^^^-/ftf^- INTEGEATION 291 (q) ra2»™*cos0(i0. (s) Cef^'esingdO. (u) fe^^^'sec^tdt. (r) j{^+e"^Ydx. (t) fe^-ixdx. (y) Ca^ogx^. Proofs of (8)-(13). These follow at once from the corresponding formulas for differentiation, XI, etc., p. 35. Proof of (14). I tan vdv = ( sin vdv -/ -P cost) sin vdv cos V ' d (eos v") cosv = — log COS v + C by (5) = logsecv+C [Since — log cos v = — log = — log 1 + log sec v — log sec v. S6C 1} J Proof of (15). 1 cot vdv = l —. = | — ^ ^ - J J sm.v J smv = log sin V + C- By (5) „ , ,^ ^ „. secw+tant) Proof of (16). Since sec?; = secv ; ^ ^ sec V.+ tan v _ sec V tan v + sec^v sec V + tan v //"sec « tan v + sec v , sec v.dv = / av J sec V + tan » / d (sec t> + tan v') sec V + tan v = log (sec t) + tan v^ + C. By (5) „ ^ , ^ ^ _. CSC w— cot » Proof of (17). Since esc i; = esc v CSC V — cot V — CSC V cot V + csc^'v / cscv. (k) I Job J sin24^ , . r dx (e)/eos(6 + ax)dx. (1)/-^. •^ 1 — si (t) fseo^2axdx. (m) C (sec2d - osc-jdd. ,^s /■ 2gdi ^ 'J sin6i ^^^/^^fs^- (n)/(tan0 + sec0)^d0. (v)J^*'^^ Proof of (18). Since cos a cosSe d" c?(-arctaii-+C)= fL_=-^^, by HII, p. 35 . v^+a^ /dv 1 V g , ^ = - are tan - + C* + a a a Proof of (19). Since -^-i-^ = A. (A L_\ r dv ^ J^ r/_i 1 J v^~a^ 2aJ \v~ra v + dv + a/ 1 Za v + a 'Also (i/- arc cot- + c'| = — r^ and f— ^^=_larccot- + (7'. Hence \o a I v^ + a'' J v^ + a^ a a r dv 1 / =-a i/ 1)2 + a2 a - are tan — + C= — arc cot - + C". Since arc tan - + arc cot — = — , we see that one result may be easily transformed into the other. a a 2 V V The same kind of discussion may be given for (20) involving arc sin - and arc cos - , and for (23) 'involving arc sec- and arc esc -. o a t By breaking the fraction up into partial fractions (see Case I, p. 325) . 294 INTEGRAL CALCULUS Proof of (20). Since d"' d fare sin - + cU , ^"^ = a^ , ^ XVIII, p. 35 = arc sm - + C- a r dv Proof of (21). Assume v^atanS, where g is a new variable; differentiating, dv = a sec^ zdz. Hence, by substitution, /dv _ r a seo^ zdz _ P sec' zdz ^7+7^" J VaHan^z + a' J Vtan's+l = I sec zdz = log (sec z + tan s) + C by (16) = log (tan 2 + Vtan^2 + l) + c. By 28, p. 2 But tan 2 = - ; hence, a -. v+^v^+a' , = log h c a = log(v +Vv^+ a^)— log a + c. Placing C =— log a + c, we get f-^^ = log{v+V7T^^)+C. J Vv^+ar In the same manner, by assuming v = a sec z, dv = a sec s tan zdz we get f dv fa sec 2 tan 2^2 r I = I = I sec 2a2 = log (sec 2 + tan 2) + c by (16) = log (sec 2 + Vsec^ 2 — l) + c by 28, p. 2 = log('^ + >J^-lVc = log(t;+V7^^)+C. Proofs of (22) and (23). These follow at once from the corre- sponding formulas for differentiation, XXII and XXIV, p. 36. INTEGKATION 295 A large number of the fractional forms to be integrated have a single term in the numerator, while the denominator is a quadratic expression with or without a square root sign over it. Tte following outline will assist the student in choosing the right formula. Numerator of First Degree Numerator of Zero Degree No radical in denominator f^ = losv + C J V r dv 1 ^ « , „ -' 1)2 + a2 a a r do 1 . u — a . ^ Eadical in denominator / «"dK = + C J n + 1 r dv • " . ^ 1 = am sin - + (V, or, ■' Va^-'!)^ « / -— i^= = log(u + VD2±a2) + C ' ■■' Vv^±a^ Students should be drilled in integrating the simple forms orally and to tell by inspection what formulas may be apphed in inte- grating examples chosen at random. EXAiyiPLES For formulas (18)-(23). Verify, the following integrations : 1 r lix 1 ^ 2a; , - 1. I ^ = - arc tan f- G. J 4x2+ 9 6 3 Solution. This resembles (18) . For, let v^ = ix^ and a^ = 9; then v ^2x,dv = 2dx, and a = 3. Hence if we multiply the numerator by 2 and divide in front of the integral sign by 2, we get 2dx _ 1 f d(2x) /dx _ 1 r 4x2+ 9~2J (2s)2+ (3)2 2 J (2x)2+(3)2 = - arc tan 1- G . 3 „ /• dx 1 , 3x-2 ,, 2. I = — log h ( J 9x2-4 12 • 3x + 2 (2x dx 1 . 3x „ z= - arc sin 1- G . Vl6-9z2 3 4 V9 — x- dx Vx2-9 = = arc sin - + C. i2 3 By (18) 6. r — ^ — =:log(x+Vx2+9)+6'. •*' Vx2 + 9 /- 5dx " J X2 + ' 8 C '^ = ■ J a2x2 - c2 i :- arc tan- + C. 3 3 b , ax — c -log — +C. 2ac ax + c ■.log(x + Vx'-9) + C. 9 .jl^^^log'^^ + ^j + C. S-x" eVs x" -V5 296 INTEGRAL CALCULUS 10. f ^"^ =garcsmx^ + C. 13. f— ^= = arc vers ? + C J Vi-K* 2- ■'' Vex- x^ *> 11. "^ =-arcsec— + C. 14. j ^^ — ^^ = ^^loKrr +C ^- arf. t.a.ll - 4- (7. 15. | - X* + e* 2 f "^ =-arcsec^ + C. 14. I „ ,.2^ = ^7:;^ l°g » — ^ 12. f^?^ = JL arc tan ?? + C. 15- /^|^ = ^^^^ ^'^ «' + ^■ 6 16. f = — = arc sin a /^ s + C. •^ V3-5s2 VB ^'S 17. T— =^= = — log(Vau + VoB^^ + C. ■^ Vat!^ — 6 Va ,„ r cos ardor 1 ^ /sinaN , „ 18. ( = - arc tan ) + C. J a^ + sin^ a a \ a / 19. I — . = arc sin (log x) + G. •^ X Vl— log^x 20. f ^ = - log (ex + V62 + e%2) + c. •^ Vb^ + e^x- « 21. f , ^^ =\\og(bv + Vf^v^-a^) + G. 22. I — — = arc sm \-G. adz a ^ z — e = - arc tan ■ e)2 + 62 6 6 23. r '^ = :iarctan^^^^ + C. J (z_,-" -" - 24. I = - arc tan Y G. Jx^+2x+b 2 2 Hint. By completing the square in the denominator, this expression may be brought to a form similar to that ol Ex. 17. Thus, /dx r dx r dx l.a; + l,^ _ ,,„, = I = I =- arc tan + C. By (18) a;2 + 2s + 5 J (x2 + 2a: + l)+4 J (a: + l)2 + 4 2 2 Here v = x + l and a = 2. n- r dx .2x — 1 25. I — — = arc sm 1- G. •^ V2 + X — x2 3 Hint. Bring this to the form of Ex. 16 by completing the square. Thus, /dx r dx r dx r dx . 2^-1 , I , I , == I =arcsm— - + C, By (20) V2+i-a:2 J V2-(a:2-a:) J V2-(x^-x+l)+l J Vf-(a;-J)2 ^ Here v = x-\ and a = j. /dx 2 2x + 1 == — - = -— arctan ^ +C. 1 + X + x2 VS V3 „„ r dx 1 r dx 1 ^3x— 1._, 27. I = - I — = —= arc tan =^ + O. J3x2-2x + 4 3^x2-|x + f Vll Vll INTEGRATION 297 Vi-|X-x2 2^ V^-(x2+iX+^j)+^j 1 . 8x + 3 , -, ; - arc sin J- C 2 Vil 29. . f ^=: = arcsm(2g — 3) + C. "^ V3x-x2-2 30. I = - log h C. 31. f -^^ ^J_iog^^ + g-^+c: Jy2 + 32, + l V5 22/ + 3 + V5 32. C—=^=z=\og{t + \ + VW+t + i\^-G. J Vl + J + i2 V 2 / 33. r ^ = arc tan (2 z - 1) + 0. 34. r , ^ - = log(s+ a + V2g„s + .■jgj + f;. 35. ( - = — arc sec 1-0. 36. r_4^^^ = i arc vers 18 x»+C. •-' Vx^— 9x8 3 r(b + ex)dx 6 ^ x e, , „ ». „ 37. ^ — i— = -arctan- + -log a2 + a;2) + c. Hint. A fraction with more than one term in the numerator may be broken up into the sum of two or more fractions having the several terms of the original numerator as numer- tors, all the denominators being the same as the denominator of the original fraction. Thus, the last example may be written /{b + ex)dx_ r bdx r exdx _ r dx r_xdx_ a2 + s2 ~J a^ + x^ J a2 + a;2~ J a^ + x'' ^J a^ + x^' each term being integrated separately. 38. r(i^^:I)^ = ?log(x2+9)--arctan- + C. J x^+^ 2^^ '3 3 39. r^^dx = llog(3x2-2)--i-log ^^-^ +C. JZx^-% 3 ^^ 2V6 XV3 + V2 40. r_j j — g- ds=-3V9-s2-2arcsin- + C. •J VO _ s2 3 41 / ^ + 3 ■ {-^^p^=ax = -y/^fiTi + 31og(x + Vx2 + 4) + O. Vx2 + 4 2. rlil^llf = 5 Vst^ir^-^ log («V3+V3t^39) + a. 42. . , ._. . Vse^-g 3 V3 298 INTEGRAL CALCULUS 43. Integrate the following expressions and verify your results by difEerentiatioij : dx ,.> r 2dx , , f Sdx (a) r "^ ■ (i) f- •' v'4 — 25SC2 "^ V25a;2_, (c) f—^ (k) f^^= (d)/— ^=. (I)/- <^ aV9a2-4 -' sVGx" — 16 sin^dS , V r dz I (e) / . (m) / V9-4cos2tf -^ zV4 — (logz)"' /' (2a:-3)(fc /•(« + 2)d« ^ ' J x^ + i ' "• ' .1 it^-s (h)r ^ (P)/ , "" 168. Trigonometric differentials. We shall now consider some trigo- nometric differentials of frequent occurrence which may be reaAQy integrated by being transformed into standard forms by means of simple trigonometric reductions. y^i 1 J ■^ V5x2 + 1 (r) C ''"' . ^ 'J 12w2_3 ■ J cos' a . „ „ /-sin^ ardor „ 4. ( cos^a sin ada= h C. 8. I = sec ar + cos cr + C. •' Z J cos^or 9. / cos*x sin' xdx = — J cos^x + |cos'x + C. tn r ■ r. J .5^1! COS''X , 10. I sin^idx =— COSX + -cos°x \- ■I 3 5 3 5 c. ■.I r r. J • ^ • B , smox . _, 11. I cos^xox = sm I sin"x -\ [■ C. J 3 5 12. fsin^ ^ cos' 4>d4> = ^ sin'V ^ — ^ sinV + 0. * This was integrated by the power formula taking n = l, w = sin a;, dv = cos xdx. To illus- trate how an answer may take on different forms when more than one method of integration is possible, let us take n=l, «; = cosx, du=-sinxda;, and again integrate by the power formula. Then / sin X cos xdx = - I (cos x) (- sin x dx) = — + C", a result which differs from the first one in the arbitrary constant only. For, eosiix „, l-sin2x ^ ^,_ 1 ^ sin^x ^ p/^ 8"'''a; 1 ^ ^, 2 2 2 2 22' Hence, comparing the two answers, C=-i+ C" 300 INTEGRAL CALCULUS 13. fsini e cos= eae = § slnt e-^ sin¥ + ^sm'^6 + C. 14. r ^ '°°^ d;y = - 2 Voosy fl - ? cos^y + ^ cos*2/) + 0. •' Vcosy \ 5 9 / ,_ /^oos^idt 3.2./, 1 . o. , 1 . iA , /-. 15. I — = -smti/l sm2« + -sm*i) + C. 16. Integrate the following expressions and prove your results by differentiation : (a) fs,in^26dff. (f) f cos^ ax sin axdx. (k) jsin!'mt coe,^ mtdt. (b) Ccos^-de. ig) fsin^-^ cos -^dx. {\) fsin^ntdt. (c) I sin 2 X cos 2 xdi. (h) / cos^ 3 1 sin 3 xctc. (m) | sin*x coszda. (d) I sin^icos^tdi. (i) j sin^bsdosbsds. (n) | cos*j/sinydy. (e) fcos-sin-dx. (j) Tcos^lsin^^d^. (o) f co^ (a + bt)d!l J a, d «/ 2 2 J Example II. To find j tan"xdx, or I eofxdx. These forms can be readily integrated, when n is an integer, on somewhat the same plan as the previous examples. Illdstrative Example 1. Find j tan%dx. Solution. rtan*xdx = rtan2x(sec2x — l)dx by 28, p. 2 = I tan^xsec^xtJx— | tan^xdx = I (tan x)^d (tan x)— j (sec'' x — l)dx tan'x „ = — tan X + X + C. Example III. To find j secTxdx, or j csd"xdx. These can be easily integrated when w is a positive even integer, as fgllows: Illustrative Example 2. Find fsec'xdx. Solution. jseo^xdx=j(ta,n^x + l)^sec^xdx by 28, p. 2 = J (tanx)*sec2xdx + 2 r(tanx)2sec2xdx + Csec^xdx tan^x , „tan'x = — z [-2— -— + tanx + (7. 5 o When n is an odd positive integer greater than unity, the best plan is to reduce to sine or cosine and then use reduction formulas on p. 303. nSTTEGRATION 301 Example IV. To find \ tari'xsed^xdx, or j eot"'xc8(fxdx. When w is a positive even integer we proceed as in Example III. Illdstkative Example 3. Find j tan'xsec*xdx. Solution. I ta.n^xsec'^xdx = j tan^a;(tan2a; + l)sec^xdx by 28, p. 2 = I (tanx)*sec2a^ + I tsm.'^xsec^xdx 9 Here v = tans, dv = sec^xdx, etc. tan»x_^ta^^p_ By (4) When m is odd we may proceed as in the following example. Illustrative Example 4. Find j tan^xsec'xttc. Solution. I tan* x sec' xdx = T tan* x sec^ x sec x tan xdx = j (seo^x — 1)^ sec^x sec x tan x»2xdx. 31 •/ 2x^dx {af-x^)i -{a + xf 32. y" '"' dx. 33 ■/ Vx log^xdx 34. I e-<^x2dx. 35, ax — b x2 + 4m2 dx. 36. ririi^dx. 37. I cos'ox sin axdx. 38. f cot* 3 aydy. fsin^Ba 39 3 xdx. 306 LKTTEGEAL CALCULUS 40. The following firnctions have been obtained by differentiating certain func- tions. Find the functions and verify your results by differentiation. (a) 5x^ + sin 2s. Solation. In this example (5 a;' + sin 2 a!) (is is the differential expression to be iate- grated. Thus f(5x^ + sia2x)dx = -^ cos2z+C. Aiis. , J 4 2 d /6x* 1 Vacation. — ( cos2z + C) = 5x' + sin2x. dx\ 4 2 (b)5a;8-6a;. mx + n (l-2y)s (c)2x2-3z-4. -^g^-Khfl Vy (d) cos^ax-l-sin-. (k) ' ^-"^ . (s) (g) « ' V3T4^' ^'x^-\-ix-\ (e) VST^. 6t+c (t) sec*^. *^ + » , , o-6s (1) , =^- X . ("'^ ^34^- V4-X2+2X 6+2x" (n) ^° . (v)(J-e-% (M 3 + 2x ,.^~^^ (w)a;'(l + a;2)i \"' ^2 I 1 (o) Sin mx cos mx. ^ ' «i-Sn7- (q)tan»-. (y)x^VrT^2. CHAPTER XXin CONSTANT OF INTEGRATION 170. Determination of the constant of integration by means of initial conditions. As was pointed" out on p. 281, the constant of integration may be found in any given case when we know, the value of the integral for some value of the variable. In fact, it is necessary, in. order to be able to determiae the constant of integration, to have some data given in addition to the differential expression to be integrated. Let us illustrate this by means of an example. Illustrative Example 1. Find a function whose first derivative is 3 x^ — 2 x + 5, and whicli shall have the value 12 when x = 1. Solution. (3x2 — 2x + 5)dx is the differential expression to be integrated. Thus r(3x2-2x+ 5)(Jx = x3-x2 + 5x+ C, where C is the constant of integration. From the conditions of our problem this result must equal 12 when x = 1 ; that is, 12 = 1 - 1 + 5 + O, or O = 7. Hence x*— x^ + 5x + 7 is the required function. 171. Geometrical signification of the constant of integration. We shall illustrate this by means of examples. Illustkative Example 1. Determine the equation of the curve at every point of which the tangent has the slope 2x. Solution. Since the slope of the tangent to a curve at dv any point is — , we have, by hypothesis, dx dy =:2x, Integrating, dx dy = 2 xdx. y — 2 I xdx, or, (A) y = x^+C, where G is the constant of integration. Now if we give to a series of values, say 6, 0, — 3, {A) yields the equations y = x^+6, y=x\ y = x^-3, whose loci are parabolas with axes coinciding with the axis of y and having 6, 0, — 3 respectively as intercepts on the axis of T. 307 308 INTEGRAL CALCULUS + C, All of the parabolas (A) (there are an infinite number of them) have the" same value of — ; that is, they have the same direction (or slope) for the same value of x. ax ' It will also be noticed that the difference in the lengths of their ordinates remains the same for all values of x. Hence all the parabolas can be obtained by moving any one of them vertically up or down, the value of C in this case not affecting the slope of the curve. If in the above example we impose the additional condition that the curve shall pass through the point (1, 4), then the coordinates of this point must satisfy (A), giving 4 = 1 + C, or G = 3. Hence the particular curve required is the parabola y = x^ + 3. Illustrative Example 2. Determine the equation of a curve such that the slope of the tangent to the curve at any point is the negative ratio of the abscissa to the ordinate. y- Solution. The condition of the problem is expressed by the equation ^j^ ^ dx y or, separating the variables, ydy =— xdx. Integrating, — = or, a;2 + j/2 = 2 C. This we see represents a series of concentric circles with their centers at the origin. If, in addition, we impose the condition that the curve must pass through the point (3,4), then 9 + 16 = 20. Hence the particular curve required is the circle x^ + y^ — 25. The orthogonal trajectories of a system of curves are another sys- tem of curves each of which cuts all the curves of the first system at right angles. Hence the slope of the tangent to a curve of the new system at a point will be the negative reciprocal of the slope of the tangent to that curve of the given system which passes through that point. Let us illustrate by an example. Illustrative Example 3. Find the equation of the orthogonal trajectories of the system of circles in Illustrative Example 2. yik Solution. For the orthogonal system we will then have dy_y_ dx X or, separating the variables, Ay _dx y X Integrating, log 2/ = log x + log c = log ex, or, y = ex. Hence the orthogonal trajectories of the system of circles x^^y'^ = is the system of straight lines which pass through the origin, as shown in the figure. CONSTANT OF INTEGEATION 309 172. Physical signification of the constant of integration. The fol- lowing examples will illustrate what is meant. Illtjstrative Example 1. Find the laws governing the motion of a point which moves in a straight line with constant acceleration. Solution. Since the acceleration = — from (14), p. 92 I is constant, say /, we have L dt J or, dv=fdt. Integrating, (A) v=ft+C. To determine C, suppose that the initial velocity be »,, ; that is, let 1! = »o when t = 0. These values substituted in (A) give ?)„ = + C, or, C = »„. Hence (A) becomes (B) v=fl + Vg. ds Since d = — [(9), p. 90], we get from {B) or, ds = ftdt + Vgdt. Integrating, (C) s = ifP + v,t + C. To determine C, suppose that the initial space (= distance) be «„ ; that is, let 8 = 85 when i = 0. These values substituted in (C) give So = + + 0, or, C = Sa. Hence (0) becomes (D) s = ^/!2+V + V By substituting the values /= gr, Uq = 0, «„ = 0, s = A in (B) and (D), we get the laws of motion of a body falling from rest in a vacuum, namely, {Ba) V = gt, and (Da) h = \gt^. Eliminating t between {B a) and (D a) gives « = V2 gh. Illustkative Example 2. Discuss the motion of a projectile having an initial velocity »„ inclined at an angle a with the horizontal, the resistance of the air being neglected. y Solution. Assume the Xy-plane as the plane of mo- tion, OX as horizontal, and OF as vertical, and let the projectile be thrown from the origin. Suppose the projectile to be acted upon by gravity alone. Then the acceleration in the horizontal direc- ^ "ocosa tion will be zero and in the vertical direction — g. Hence from (15), p. 93, dfx „ J dvy —5 = 0, and -^ =— g. dt ' dt " 310 INTEGRAL CALCULUS Integrating, Da: = Cj, and Vj,=— gt+ G^. But Bq cos a = initial Telocity in the horizontal direction, and Vq sin a = initial velocity in the vertical direction. Hence C-^ = »(, cos a, and Oj = Vg sin a, giving (E) Vx = ■Oq cos a, and By = — ffJ + Oo ^'° ^- But from (10) and (11), p. 92, Dj, = — , and i>y = -£ ; therefore {E) gives dx ,dy ... — = D„ cos q:, and -^ = — grt + b„ sin a, or, dx = Wj cos adt, and dy =— gtdt + Wq sin adt. Integrating, we get (F) x = VgCOsa-t + Oj, and y =— i gt^ + VgSiji a ■ t + C^. To determine Cj and C^, we observe that when i = 0, x = and y = 0. Substituting these values in (F) gives Cg = 0, and C^ = 0. Hence (G) !" = «(, cos a ■ t, and (fl") y=-igt' + VoSma-t. Eliminating t between (G) and (B"), we obtain fin'' (I) y = xtajia ^- — -— , ^ ' 20o2cos2q: which is the equation of the trajectory, and shows that the projectile will move in a parabola. EXAMPLES 1. The following expressions have been obtained by differentiating certain functions Find the function in each case for the given values of the variable and the function : Derivative of function Value of variable Corresponding value of function Answers (a)x-3. 2. 9. ^-3x + 13. 304 + 3x + ^^ T (b) 3 + x-5x2. 6. -20. (c) yi - b^. 2. 0. (d) sin a + cos a. TT 2' 1. 2. sin or — cos a + 1. (^>7 2-t- 0. log (2 i- 42). (f) sec2S + tanS. 0. 5. tan S + log sec ff + 6. ^'K^ + a^- a. 2a,' arc tan- + — . a 4a (h) &x» + ax + 4. b. 10. (i)V^ + ^. 4. 0. (i) COt^— CSC^(l>. TT 2' S. (k) Se"\ 0. 7 4* CONSTANT OP INTEGEATION 311 2. Find the equation of the system of curves such that the slope of the tangent at any point is : (a) X. Ans. Parabolas, y — — }■ C. Parabolas, y = x^—2x + G. 1/2 (b) 2x-2. (c) 1 y' (d) V (e) X (f) 3a;2. (g) a;2+5i. (t) 1 (i\ s V) ^' (i) z (k) 62x (1) Ifly' (m) xy. (n) y- (0) m. (P) 1 + x ^ y Parabolas, — = x + 0. ' 2 Semicubical parabolas, — = — (- C. ^ o Semicubical parabolas, — — — ■\- C. Cubical parabolas, y = x' + 0. Cubical parabolas, y = — + -x2+ 0. o ^ Cubical parabolas, ^ = x + C o Equilateral hyperbolas, y^—x'' = C. Equilateral hyperbolas, xy = C. Hyperbolas, aV - ft^x^ = C. Ellipses, 62j/2 + aH^ = C. x2 ^ logy — — (- C, or y = ce^. logy = X + C, or j^ = ce^. Straight lines, y = mx + G. Circles, x^ + y^ + 2x - 2y + G = 0. 3. Find the equations of those curves of the systems found in Ex. 2 (a), (c), (d), (i), (i), (m), V7hich pass through the point (2, — 1). x'-i Ans: (a)x2-2y-6 = 0; (m) y=-e 2 ; etc. 4. Find the equations of those curves of the systems found in Ex. 2 (b), (e), (g), (h), (o), (p), vrhich pass through the origin. Ans. Qi) y = x^- 2x; (o) y = mx; etc. 5. Find the equations of the orthogonal trajectories of the following systems of curves found in Ex. 2 : (a) 2/ = — + C, Ex. 2 (a). ^ms. j^ = - log x + C. (b) ?^ = X + C, Ex. 2 (c). logy=-x + C. (o)|- = |-+0,Ex.2(d). (d) y2-x2=0, Ex. 2(i). (e) xy = 0, Ex. 2 (j). (f) y = cef^, Ex. 2 (n). (g) 1/ = mx + C, Ex. 2 (o); (h) x^ + y^+2x-2y + C = 0, Ex. 2 (p). log 2/ = X xy = G. y^-x^ = G. 2 r + C. my + x = C. v-\ = c(x + l). 312 INTEGRAL CALCULUS 6. Find the equation of the curve whose subnormal is constant and equal to 2 a. dy Ans. 2/2 = 4 as + C, a parabola. Hint. From (4), p. 77, subnormal = J/ -7- • 7. Find the curve vrhose subtangent is constant and equal to a (see (3), p. 77). Ans. alogy = x + G. 8. Find the curve whose subnormal equals the abscissa of the point of contact. Ans. y^ — x^ = 2C, an equilateral hyperbola. 9. Find the curve whose normal is constant (= B), assuming that y = B when X = 0. Ans. x^ + y^ = B^, a circle. Hint. From (6), p. 77, length of normal = 2/-4/l+(^j , or dx=^(,B^-y^~^ydy. 10. Find the curve whose subtangent equals three times the abscissa of the point of contact. Ans. x = q/'. 11. Show that the curve whose polar subtangent (see (7), p. 86) is constant is the reciprocal spiral. 12. Show that the curve whose polar subnormal (see (8), p. 86) is constant is the spiral of Archimedes. 13. Find the curve in which the polar subnormal is proportional to the length of the radius vector. Ans. p = ce^. 14. Find the curve in which the polar subnormal is proportional to the sine of the vectorial angle. Ans. p = c — a cos6. 15. Find the curve in which the polar subtangent is proportional to the length of the radius vector. Ans. p = C€°*. 16. Determine the curve in which the polar subtangent and the polar subnormal are in a constant ratio. Ans. p = ce°«. 17. Find the equation of the curve in which the angle between the radius vector and the tangent is one half the vectorial angle. Ans. p = c(l— cos8). 18. Determine the curves in which the subtangent is n times the subnormal ; and find the particular curve which passes through (2, 3) . Ans. Vny = x + C ; Vn{y — 3) = a; — 2. 19. Determine the curves in which the length of the subnormal is proportional to the square of the ordinate. Ans. y = c^. 20. Find the curves in which the angle between the radius vector and the tangent at any point is n times the vectorial angle. Ans. p" = c sin nS. Assuming that v = Vg when i = 0, find the relation between v and t, knowing that the acceleration is : 21. Zero. Ans. v = v„. 22. Constant = k. v = v„ + kt. 23. a + U. K = u^ ^. at + _ . Assuming that s = when t = 0, find the relation between s and t, knowing that the velocity is : 24. Constant (= e„). Ans. s = u„«. 25. m + ni. s = mt + —- 2 26. 3 + 2f-SJ2 s = 3t + t^-f CONSTANT OF INTEGKATION 313 27. The velocity of a body starting from rest is 5 i^ feet per second after t seconds, (a) How far will it be from the point of starting in 3 seconds ? (b) In what time will it pass over a distance of 360 feet measured from the starting point ? Ans. (a) 45 ft. ; (b) 6 seconds. 28. Assuming that s = 2 when t = 1, find the relation between s and 4, knowing that the velocity is : (a) 3. Ans. s = 3 « - 1. (b)2«-3. s = j2_3j + 4_ (c)i2+2J-l. s = - + i2_t + l ■^ 3 3 (d) -• s = logt + 2. (e) 4<3-4. s = i4-4«+5. (f)^- s=-^ + 4 + 2. <2 ■ t 29. Assuming that » = 3 when i = 2, find the relation between » and t, knowing that the acceleration is : (a) 2. Ans. v = it — l. (b)3<2 + l. „^i3 + i_7_ (c) t5-2i. » = ^_i2 + 3. 4 (d)- + f. K = log- + _ + i. 30. A train starting from a station has, after t hours, a speed .of i' — 21 i^ -|- 80 < miles per hour. Find (a) its distance from the station ; (b) during what interval the train was moving backwards ; (c) when the train repassed the station ; (d) the dis- tance the train had traveled when it passed the station the last time. Ans. (a) if- W + 40 1^ miles ; (b) from 5th to 16th hour ; (c) in 8 and 20 hours ; (d) 4658^ miles. 31. A body starts from the origin and in t seconds its velocity in the X direction is 12 i and in the Fdirection 4*^ — 9. Find (a) the distances traversed parallel to each axis; (b) the equation of the path. /o \ T Am. (a)x = 6i2, y = -fi-Qt; (b) j, = (-x- 9J^5. 32. The equation giving the strength of the current i for the time t after the source of E.M.F. is removed is (K and L being constants) dt _m Find i, assuming that I = current when i = 0. Ans. i = le ^■ 33. Find the current of discharge i from a condenser of capacity C in a circuit of resistance R, assuming the initial current to be'/j, having given the relation (C and E being constants) di dt ' — - = -— ■ Ans. i = 7oe''-". 34. If a particle moves so that its velocities parallel to the axes of X and Y are h/ and fee respectively, prove that its path is an equilateral hyperbola. I 35. A body starts from the origin of coordinates, and in t seconds its velocity parallel to the axis of X is 6 1, and its velocity parallel to the axis of r is 3 i^ - 3. Find (a) the distance traversed parallel to each axis in t seconds ; (b) the equation of the path. Ans. (a) X = 3t% y = fi-3t; (b) 27 y^ = x{x - 9)^. CHAPTER XXIV THE DEFINITE INTEGRAL 173. Differential of an area. Consider the continuous function (x), and let ^ ^ ^(^^-^ be the equation of the curve AB. Let CD be a fixed and MP a variable ordinate, and let u be the measure of the area CMPD.* When X takes on a sufficiently small increment Ax, u takes on an increment Am (= area MNQP^. Completing the rectangles MNBP and MNQS, we see that area JfTViJP < areailfiV^P < axe&MNQS, or, MP • Aa; < Am < NQ ■ Ax ; and, dividing by Ax, MP<^< NQJ Ax Now let Ax approach zero as a limit ; then since MP remains fixed and NQ approaches MP as a limit (since ^ is a continuous function of r.), we get ^^ du = 1/dx. Theorem. The differential of the area bounded by any curve, the axis of X, and two ordinate^ is equal to the product of the ordinate ter- minating the area and the differential of the corresponding abscissa. 174. The definite integral. It follows from the theorem in the last section that if AB is the locus of y = 4>(x), then du = ydx, or (^) du = ^ (a;) dx, * We may suppose this area to be generated by a variable ordinate starting out from CD and moving to the right; hence u -will be a function of z which vanishes when x=a. t In this figure MP is less than NQ; if MP happens to be greater than NQ, simply reverse the inequality signs. 314 or, using differentials, THE DEFINITE INTEGRAL 315 where du is the differential of the area between the curve, the axis of X, and any two ordinates. Integrating (^A), we get u= j (x)dx exists (it is here repre- sented geometrically as an area), denote ithjf(x-) + C. (B) .•.M=/(^),+ (7. We may determine C, as in Chapter XXIII, if we know the value of u for some value of x. If we agree to reckon the area from the axis of y, i.e. when (C) x = a, u = area OCDG, and when x=b, u = area OEFG, etc., it follows that if (i)) a; = 0, then m = 0. Substituting (D) in (B'), we get ^=/(0)+C, or, (7=-/(0). Hence from (5) we obtain (E) M=/(x)-/(0), giviag the area from the axis of y to any ordinate (as MP'). To find the area between the ordinates CD and EF, substitute the values (C) in (^), giving (-F) area CDG =fCa) -/(O), ( (?) area OEFG =/ (6) -/(O). Subtracting (J") from ((?), (F) area CEFD =f(V)-f{d)* Theorem. The difference of the values of j ydx for x = a and x = h gives the area bounded by the curve whose ordinate is y, the axis of X, and the ordinates corresponding to x= a and x=b. This difference is represented by the symbol ^ (O J^b rib ydx, or, j <^{x)dx, a %J a *The student should observe that under the present hypothesis /(a:) will he a single- valued function which changes contimioiisly from/(a) to/(6) as x changes from a to 6. t This notation is due to Joseph Fourier (1768-1830). 316 INTEGRAL CALCULUS aad is read "the integral from a to J of ydx." The operation is called integration between limits, a being the lower and b the upper limit.* Since (I) always has a definite value, it is called a definite integral. For, if cf>(x)dx=f<:x}+C, then £cl,(x^dx = ^f(:x)+C^^ or (\(:x')dx=fib-)-fia-), the constant of integration having disappeared. We may accordingly define the symbol J(j} (.(•) dx or I ydx a ^ o. as the numerical measure of the area, bounded by the curve y = 4> (a;),t the axis of X, and the ordinates <^ the curve at x = a, x=b. This definition presupposes that these lines hound an area, i.e. the curve does not rise or fall to infinity, and both a and b are finite. We have shown that the numerical value of the definite integral is always fib')-f(a), but we shall see in Illustrative Example 2, p. 324, that /(6)— /(a) may be a number when the definite iutegral has no meaning. 175. Calculation of a definite integral. The process may be sum- marized as follows : First Step. Mnd the indefinite integral of the given differential ex- Second Step. Substitute in this indefinite integral first the upper limit and then the lower limit for the variable, and subtract the last result from the first. It is not necessary to bring in the constant of integration, siuee it always disappears m subtracting. " The word limit in this connection means merely the value of the variable at one end of its range (end value), and should not be confused with the meaning of the word in the Theory of Limits. t (x) <)> is continuous and single-valued throughout the interval [a, 6]. THE DEFINITE INTEGRAL 317 Illustrative Example 1. Find I x^dx. J"* fx'T* 64 1 x^dx = — = — — - = 21. Ans. 1 L 3 J 1 *^ *^ iLLtrsTKATivB EXAMPLE 2. Find r sinxdx. Solution. ( sin xdx = — cos a; = — (— 1) — — 1 = 2. Atis. Illustrative Example 3. Find | — -• Jo a^ + x^ r" dx ri i a;"l<» 1 ^ , 1 ^ . Solution. I = - arc tan - = - arc tan 1 arc tan Jo o^ + x^ La ajo a a = = Ans. ia ia EXAMPLES y.j\x^dx = SS. 13. r^sec*ed^ = |. Jo 2. f (aH-x')dx = —- ,^ /■2'-V2r Jo 4 14. I — —■ Jo a/^ dx = ir. Vx 1 -l~"' 15. r''(|Vt-/;t2)dt = 2V5-5. Jo x^ 'dx rdx Trr ■^1 ^ Jo Vr2 - x2 2 5. £' (x2 - 2x + 2) (X - 1) dx =- i. /.2r2 V27# _ Jo V2r-y /»1 ox _ /q 1 J-m* 315 0* •^o 31 + 1 3 19.2a/ (2'+2cos^)id;^ = 8a. J_ n r,/i dx TT ff «■ /o^ V2— 3x2 4V3 20. r^sin3Q:cos^a:da = j'ij. 9. f'_^^^=^yi25. - •'a 2 ■J'x2-4 21. J*, tan a(x)dx=f(h')—f(a'), and j\ (X) dx =f(a) -f(b) = - [/(J) -/(a)], (j:) dx——i (x) dx. a Jb Theorem. Interchanging the limits is equivalent to changing the sign of the definite integral. 180. Decomposition df the interval of integration of the definite integral. Smce / 4> (x) dx =f(x^) -f(a), and I 4i(x)dx=f(h')--f(x^), we get, by addition, r '4>(x)dx+ f (x)dx=f(h}-f{ay But r (ix~)dx. J a J a Jx^ THE DEFINITE INTEGRAL 321 Interpreting this theorem geometrically, as in § 174, p. 315, we see that the integral on the left-hand side represents the whole area CEFB, the first integral on the right- hand side the area CMPD, and the second integral on the right-hand side the area MEFP. The truth of the theorem is there- fore obvious. Even if x^ does not lie in the interval between a and S, the truth of the theorem is apparent when the sign as well as the magnitude of the areas is taken into account. Evidently the definite integral may be decom- posed into any number of separate definite integrals in this way. 181. The definite integral a functioA of its limits. From f\(x)dx=f(h-)-f(ia) U a we see that the definite integral is a function of its limits. Thus ^(z)dz has precisely the same value as / (j)(x)dx. Theorem. A definite integral is a function of its limits. 182. Infinite limits. So far the limits of the integral have been assumed as finite. Even in elementary work, however, it is some- times desirable to remove this restriction and to consider integrals with infinite limits. ' This is possible in certain cases by making use of the following definitions. When the upper limit is infinite, / ^(x)dx = ^ ^™j*„ / (a;) dx, and when the lower limit is infinite, {\{x)dx = ^'^■'2j\{x)dx, provided the limits exist. J"»-f- CO ^JTT 1 X2 Solution. r^= limt /-'^^ Umit f 1^ ^ limit r_l + i1 1. An,. 6 = +<»L \ 322 Illustrative Example 2. Find INTEGRAL CALCULUS /. a;2+ 4a2 o,^ /" + " ^a''^ limit /■' Sa^cfe limit rA„2,rnt=,r, ^ 1' Solution. I = 1, , I = 1 , 4 a'' arc tan -— ^ limit r4a2arctan — I = 4a2-- = 27ra2. Arw, 6 = +a>L 2aJ 2 Let us interpret this result geometrically. The graph of our function is the witch, the locus of ^"~x« + 4a2' = 4 a^ arc tan x^ + ia? 2 a ,0PQ6=J' Now as the ordinate Q6 moves indefinitely to the right, b_ '2a & is always finite, and limit 4 a^ arc tan ; a? arc tan - 'f r4a^arctanAl = + " L 2 a J = 2ira2, which is also finite. In such oases we call the result the area bounded by the curve, the ordinate OP, and OX, although strictly speaking this area is not completely bounded. — 1 X Solution. /•+"^_ limit r''dx_ 111 Ji X ~b = + "Ji X ~b = limit + CC (log 6). The limit of log 6 as 6 increases without limit does not exist; hence the integral has in this case no meaning. 183. When y = (jt) is discontinuous. Let us now consider cases when the function to be integrated is discontinuous for isolated values of the variable lying within the limits of integration. Consider first the case where the function to be integrated is con tinuous for all values of x between the limits a and 6 except x=a. , li a<.b and e is positive, we use the definition CA-) fcf, (X) dx = "-^^'l f (r.) dx, J a J a + 6 and when <^ (x) is continuous except at a; = 6, we use the definition (^) fV C^) dx = f^i* f" V (^) dx, provided the limits are definite quantities. Illustrative Example 1. Find 1 THE DEFINITE INTEGRAL dx 323 Solution. Here ind r ^ . becomes infinite for x = a. Therefore, by (B), Va2 - x2 " dx limit /•""' ^ limit r" dx _ limit/-"-' (fo _ limit [" . xl"-' =Sh-.(i-3]= arc sin 1 = - . Ans. Illtjstkative Example 2. Find / — Jo a;2 Solution. Here — becomes infinite for a; = 0. Therefore, by (A)^ ■ '■', M /■i^_ limit /"I (fa; _ limit /I .\ ' "J^ Jo a;''~f = OJe ^~' = 0(,e"~ /■ In this case there is no limit and therefore the integral does not exist. If c lies between a and 6, and ^ (x) is continuous except at a; = e, then, e and e' being positive numbers, the integral between a and h is defined hy (C} f (a;) dx = I'^'o* f" V (^) ^^ + e' ='o r <^ (^) '^^^ provided each separate limit is a definite quantity. X3o 2 xdx ' )(t+g) ^ t 11. \- x^ • = =log = + -log ;r + C. Jl x2 + 4x 4 ^< ,„ /■« dx , 11 13. I = log — Jo l + 3x + 2x2 ° 6 14. / ^^ ' = log J 3 x" — 4 X 186. Case II. When the factors of the denominator are all of the first degree and some repeated. To every n-f old linear factor, such as (a; — a)", there corresponds the n partial fractions ^ . S _ , L -1 "I" ' (x — ay {x — ay-^ x—a The last one is integrated as in Case I. The rest are all integrated by means of the power formula. Thus -i+^- 328 INTEGRAL CALCULUS Illustkative Example 1. Find ( —, ttt:^- gg + l x(x-l)5' /x' + x(x — '. Solution. Since x — 1 occurs three times as a factor, we assume x' + l A, B , G ^ D : = — + 77 TTi + 7: — T^ + r — 7- X(X-I)' X (X-1)8 (X-1)2 Clearing of fractions, x8 + 1 = ^ (x - 1)8 + Bx + Cx (X - 1) + Dx(x - 1)2. x' + l=:(A + D)x3 + (-3A + C-2D)x^ + (SA + B-G + D)x— A. Equating the coefl&cients of like powers of x, we get the simultaneous equations A + D = l, -3A + C-2D = 0, 3A + B-C + D:=0, -A = l. Solving, A=-l, B = 2, C = l, D = 2, and x3 + l ^ 1 ^ 2 ^ 1 ^2 x(x-l)3 X (x-l)s {x-l)2 ■ r f "'"i\3 '^=-'°g'^- / \,„ -^-r + 21og(x-l) + C ■/ X (x — If (x — 1)2 X — 1 X , (X-1)2 ^ (X-l)2 EXAMPLES ■ J (X - i)2(x - 2) ~ i - 1 "*■ °^^^ri "*" " ^/ (x + ;?x + l) =^ + "'^^- + ^) + ^- x2 + 1 ^ 1 2 . r x'-' + 1 , 1 2 , , r(x5-x3 + l)dx x2 1 ,1,, x-1 „ ■J x(x + l)8 2(x + l)2"^ ='(x + l)2"'" /• x'dx _ _ 5x + 12 /x + 4\2 J (x + 2)2(x + 4)2- x^ + ex + s"*" ^^V^l^/"^*^' INTEGEATION OF EATIONAL REACTIONS 329 r, r ^i « , 1 , «+V2 , _, (i2_2)2 4(i2_2) 8v^ t-V2 - as^ds , , . ^ . 2a2 a' = a log (s + a) + {s+af ^^ ' s+a 2(s + a)2 11. f (-^ , "^ ,J dz = log(z + m)'»(z + n)-" - -!?L- + 0. J \z + m (z + n)'^/ z + v 12. r° , '^ =l-log2, Ji x2(l + a;) ^ x2(l + a;) ,„ f° dt , . 1 13. I = log2 Ji ((l + t)^ ^ 2 14. r'ii±M^=iog?+i Ji a + 2x2 + x3 ^5 5 187. Case III. When the denominator contains factors of the second degree but none repeated. To every nonrepeated quadratic factor, such as x^+px + q, there corresponds a partial fraction of the form Ax + B a?+px+ q This may be integrated as follows : r(Ax+^-^+Adx r {Ax+E)dx _ I \ 2 2/ J x^+px + q J a^+px+q Adding and subtracting — in the numerator. ■^+B)dx xr+px + q «/ x+px + q _A r (2x+p')dx ^ (2B-Ap\^ ^ <& a?+px+q V 2 / (^^P\\(^ P ^ + 2J+l^ 4 [CJompleting the square in the denominator of the second integral.] = 9 log (x^+px + q-)+ arc tan ^^ + ^• Since a;^+^a; + ^ = has imaginary roots, we know from 3, p. 1, that4?-/>0. 330 INTEGEAL CALCULUS 4dx Illustrative Example 1. Find /4ax x' + 4a! 4 A Bx+C Solution. Assume ——: — — — F— = :-• x{x'^ + i) X , x^ + i Clearing of fractions, 4: = A(x^ + 4:) + x{Bx + C) = {A + B)x^ + Cx + 4:A. Equating the coefBcients of like powers of x, we get A+B = 0, G = 0, 4:A = i. 4 1 X This gives ^ = 1, ;3 = -l, G = 0, so that -— —- -• /■ 4dx _ rdx r xdx x(x2 + 4) ~J X Jx^ + i = log X log (x^ + 4) + log c = log — ^=: • ^ns. 2 Vx2 + 4 EXAIUPLES rdr , X « — — — = log - + a. oT »<^ li x2 + 4,2 ,x,_ 2. f = — log [- - arc tan - + C J (X + 1) (x" + 4) 10 ^ (X + 1)2 ^ 5 2 „ /■ (2x2-3x-3)dx , (x2-2x + 5)i 1 ^ x-1 ^ 3. / — ^ = log -^ '— + - arc tan \- C. •/ (x-l)(x2-2x+ 5) ^ x-1 2 2 ^ /• x^dx 1 , 1 + X 1 . , _ 4. I = - log arc tan x + G. Jl-x< 4^1-x 2 ^ r dx 1 , X* 1 5- I -^ :: = - log — arc tan x + ( . ■/ (x2 + 1) (x2 + x) 4 °(x + l)2(x2 + l) 2 ^ (x3 - 6) dx , x2 + 4 3 ^ X 3 x 6- I -^ -^ — t: = log , + - arc tan arc tan 1- C. J X* + 0x^ + 8 ^V^T2 2 ■ 2 V2 V2 _ r (5x2-l)dx , x2-2x + 5 , 5 ^ j,-l 2 x „ '■■/ (x2 + 3)(x''-2x+5) "^°°^-^^Ti~- + 2"'^''^"^-^-Vl""'^"^Vl^^- Jx^ + 1 6 *x2-x + l^V3 VS . r z^dz 1, /z-l\ , V2 ^ 2 9. ( = - log ( 1 + arc tan + C. J z*-\-z^-2 6 ^ \z + 1/ 3 V2 Jo (l + x2){3 + x2) ^Xlg- 13. r 2X2 + X + 3 ^1 ,^^. Jo (x + l)(x2 + l) '=4 INTEGEATION OF RATIONAL FRACTIONS 331 188. Case IV. When the denominator contains factors of the second degree some of which are repeated. To every n-fold quadratic factor, such as (x'^+px + qy, there cor- respond the n partial fractions (A) ^ + B Cx+B Lx+M. {x^ + px + qy (x'^+px + qy-)^'"^ x'+px + q To derive a formula for integrating the first one we proceed as follows : r /-+^ dx= / V i_j L J Qr+px + qy J (a;'+j92; + ^)" Adding and subtracting S- in the nnmevator. J (3?+px + qy J (x^+px + qy dx (x'^+px + qy The first one of these may be integrated by (4) p. 284; hence Ax + B J A •dx = (x^+px + qy 2(1 -n')(x^+px + qy-'^ Qt'+px+qy hi Let us now differentiate the function Thus (.^+p- + 9y-^ £\ -» or dx\(y+px + qy-^l (x' + px + qy-' (x' + px + qy + '-4g) + (2-g-^j>)(2jc + j>) (X' +px+ qy 2 (72 - 1) (4 g - /)") (j:^ + px + qy-' dx It is seen that our integral has been made to depend on the inte- gration of a rational fraction of the same type in which, however, the quadratic factor occurs only n— 1 times. By applying the formula (£) n — 1 times successively it is evident that our integral may be made ultimately to depend on dx /; x' + px + q and this may be integrated by completing the square, as shown on p. 296. In the same manner all but the last fraction of (^) may be inte- grated. But this last fraction, namely, Lx + M a?+px + q may be integrated by the method already given under the previous case (p. 329). * 4 5 -p2>0, since a:2 +pa; + g = has imaginary roots. INTEGEATIOJST OF RATIONAL FEACTIONS 333 Illustkative Example 1. Find C- — — — — ' J {3? + 2)2 Solution. Since x^ + 2 occurs twice as a factor, we assume x^ + x^ + 2 _ Ax + B Cx + D {x^ + 2)^ ~ (a;2 + 2)2 a;^ + 2 ' Clearing of fractions, we get x^ + x^ + 2 = Ax + B + {Cx + D) (x2 + 2). x^ + x^ + 2 = Cx' + Bx^ + {A + 2 C)x + B + 2 D. Equating the coefficients of like powers of x, = 1, D = l, ^ + 20 = 0, B + 2D = 2. This gives A =-2, B = 0, C = 1, D = 1. „ x« + x^ + 2 2x x + 1 Hence = ' — , and (x2 + 2)2 (x2+ 2)2^x2 +2 (x' + x2 4- 2) (Jx. _ r 2 xdx r xdx r dx /(x» + x'' 4- 2) ax. _ r 2 xdx r xdx r (x2 + 2)2 ~~'J (z" + 2)2 "*" J x2 + 2"*"J I x2 + 2 11 X 1 ; + — =arotan— -:-l--log(x2 + 2)+ C. a;' + 2 V2 V2 2 Illustrative Example 2. Pind / :^dx. nnd f "^ +^ J (x2 + : .1)2 Solution. Since x2 + 1 occurs twice as a factor, we assume 2x^ + x+3 _ ^x + B Cx + D (x2 + 1)2 ~ (x2 + 1)2 x2 + 1 ■ Clearing of fractions, 2x3 + X + 3 = ^x + B + (Cx + Z)) (x^ + 1). Equating the coefficients of like powers of x and solving, we get A=-l, B=3, G = 2, D = 0. Hence / ■ — dx = | — dx + ( J (X2 + 1)2 J(X2+1)2 J X^ + 1 = log(x2 + 1) + r ~ ^ "*" ^ dx. ^^ 'J (X2 + 1)2 Now apply formula (E), p. 332, to the remaining integral. Here A=-l, B = 3, p = b, q = l, n = 2. Substituting, we get x + 3 , l + 3s Sr dx l + 3x 3 r^:x + 3_^^J^+Sx_ S r dx -' (X2 + 1)2 2(X2 + 1) 2^x2 + 1 + - arc tan x. (x2 + l)2 2(x2 + l) 2^x2 + 1 2(x2 + l) 2 Therefore 334 INTEGRAL CALCULUS EXAMPLES 1. f — = +iarctans + C. t+1^^ = ^-^ + log(x^ + 2)i L= arc tan ^ + C. ' " ■ 2)2 4(x2 + 2) ^' 4V2 V2 2 <- x» + x J (X2 + J (1 2xdx 1, x2 + l , x-1 „ + x)(l + xY 4 °(X + 1)2 2(x2 + l) 2a2x 4. / / :^^ — 1 dx = X H 2 a arc tan - + C. J Vx^ + aV x^ + a^ a g H4X + S ■ J (4x2 + (4x + 3)dx _ 4x» + 5x -2 1 ^,„,,,2x^^ -' ° ■ 3)8 8(4x2 + 3)2 leVs V3 „ r 9x'(Jx 3x 1, (x + l)2 , /5 ^ 2x-l , ^ I _!:;: _ L _ log _i — : — l \. V3 arc tan — h C. J (x3 + 1)2 xs + 1 2 * i2 - X + 1 V3 ^x^+xi+x^ + ^ ^^^_ 1^ 19 -/ (x2 + 2)2 (x2 + 3)2 2(x2 + 2) x2 + 3 2 °^ ' b\ / „ r (4x2-8x)dx 3x2 -X (x-l)2 , ^ , „ 8. / — i '- = \- log ^^ — + arc tan x + C J (x - 1)2 (x2 + 1)2 (I _ 1) (a;2 + 1) ^ x2 + 1 „ r (3x + 2)(to . 13X-24 26 , 2x-3 , ^ 9. I — ^ ^ — '- = arc tan — h C. J(x2-3x + 3)2 3(x2-3x + 3)^3V3 Vs Since a rational function may always be reduced to the quotient of two integral rational functions, i.e. to a rational fraction, it follows from the preceding sections in this chapter that any rational function whose denominator can be broken up into real quadratic and hnear factors may be expressed as the algebraic sum of integral rational functions and partial fractions. The terms of this sum have forms all of which we have shown how to integrate. Hence the Theorem. The integral of every rational function whose denominator can he broken up into real quadratic and linear factors may he found, and is expressible in terms of algebraic, logarithmic, and inverse-trigono- metric functions ; that is, in terms of the elementary functions. CHAPTER XXVI INTEGRATION BY SUBSTITUTION OF A NEW VARIABLE. RATIONALIZATION 189. Introduction. In the last chapter it was shown that all rational functions whose denominators can be broken up into real quadratic and linear factors may be integrated. Of algebraic functions which are not rational, that is, such as contain radicals, only a small number, relatively speaking, can be integrated in terms of elementary functions. By substituting a new variable, however, these functions can in some cases be transformed into equivalent functions that are either in the list of standard forms (pp. 284, 285) or else are rational. The method of integrating a function that is not rational by substituting for the old variable such a function of a new variable that the result is a rational function is sometimes called integration hy rationalization. This is a very important artifice in integration and we will now take up some of the more important cases coming under this head. 190. Differentials containing fractional powers of x only. Such an eaypression can be transformed into a rational form hy means of the substitution x = z^ where n is the least common denominator of the fractional exponents of x. For X, dx, and each radical can then be expressed rationally in terms of z. dx. ^ Solution. Since 12 is the L.C.M. of the denominators of the fractional exponents, we assume x = z^^. Here dx = 12zii0, it may he written Vs -»/- + - a: + 3;2, « » \9 ? and therefore comes under the ahove head, where a--, 5=- • q q I t If the radical is of the form vn+px-qx^, g >0, it may he written Vq\j—^Ex-x^, and therefore comes under the above head, where a = - , 6 = — • g i t If the factors ota + bx-x^ are Imaginary, Va + bx- x^ is imaginary for all values of a;. For if one of the factors isx-m + in, the other must be - (a; - m - in) , and therefore 6 + ax - a;2 = - (a; - m + ire) (a; - m - in) = - [(a; - m)2 + ra2], which is negative for aU values of a;. We shall consider only those cases where the factors are real. DSTTEGEATION BY EATIONALIZATION 339 For if ^a-\-hx-x^= -^ (x - a)(/3 - a;) = (a; - a)z, by squaring, caacelling out (x — a), and solving for a;, we get . = ^; then ^.^^(^ ^).^. and -s/a + hx-x\=(x-a)z\=^^^^^-^. sr + l Hence a;, dx, and va + 6a; — a;*" are rational when expressed in terms of 2. Illustkativb Example 1. Find ind f- (Zx V2 + X - X2 Solution. Since . 2 + x — x^ = (a; 4. i) (2 — x), e assume -v/(x + 1) (2 — x) = (x + 1) z. 2 2^ Squaring and solving for x, x = — Hence dx = -— — — - , and -^2 + x — x^ [= (x + 1) zl : /dx r dz — = — 2 I — = — 2 arc tan z + C V2 + X - x2 -^ z^ + 1 1^^ = — 2 arc tan -t f 1- G, \x +1 when we substitute back the value of z In terms of x. EXAMPLES _ 1 ,„„Vx2 — x + 2 + X-V2 t + 2 dx Ti dx 1 , ■\/2 + 2x — V2 — X 1. I , . = — =log— ^ -+C. •f xVx^ — x + 2 V2 Vx2 — X + 2 + X+V2 2. r — - = 2 arc tan (x + Vx^ + 2 x - l) + C. ■' xVx2 + 2x — 1 3. f ^ °- =^log •J X V 2 4- X — x2 V2 X V2 + X — x2 V2 V2 + 2X + V2-X + C. ^ C dx . x+Vx2 + 4x— 4 , _ 4. I — — = arc tan ■ 1- C. ■' X Vx2 4-4x — 4 2 5. fX^^-±-^dx = ,^ + log(x + 2 +Vx2 + 4x) + C. ^ x^ x+ vV + 4x 6. f ^ = -|±|^=+C. (2 ax - x2)l a2 V2 ox - x^ 8. r(^^±^!)i^=iog(x+i+V2^i:^) 4=+c. ■^ a;^ , x+V2x + x2 a r ax , X — i + vx' + x + j. 9. / , = log- , +C- •^ X Vx^ + X + 1 X + 1 + Vx2 + X + 1 1A r *: ^ * |2(3-x) , „ 10. / — = — -./- arc tan -* /— ^ + C. "'^.V'firr_fi_^2 V3 \3(x-2) 340 INTEGEAL CALCULUS The general integral treated in the last two sections has then the ^^^^ E{x,-Va + bx + cx')dx, where B denotes a rational function. Combining the results of this chapter with the theorem on p. 334, we can then state the following Theorem. Every rational function of x and the. square root of a poly- nomial of degree not higher than the second can he integrated and the result expressed in terms of the elementary functions* 195. Binomial differentials. A differential of the form x'^ia + hx'^ydx, where a and h are any constants and the exponents^, n,p are rational numbers, is called a binomial differential. Let x = z°; then dx = az" ~ ^ dz, and x^Qx + hx'ydx = a^^''-^(a + h^ydz. If an integer a be chosen such that ma and no. are also integers,^ we see that the given differential is equivalent to another of the same form where m and n have been replaced by integers. Also «'"(« + bx'Ydx = x'"->-'"'(ax-"+bydx transforms the given differential into another of the same form where the exponent w of a; has been replaced by — n. Therefore, no matter what the algebraic sign of n may be, in one of the two differentials the exponent of x inside the parentheses will surely be positive. When p is an integer the binomial may be expanded and the dif- ferential integrated termwise. In what follows p. is regarded as a fraction ; hence we replace it by - , where r and s are integers. * s We may then make the following statement : Every binomial differential may be reduced to the form x'^ia + hapydx, where m, n, r, s are integers and n is positive. *As before, however, it is assumed that in each case the denominator of the rational function can be broken up into real quadratic and linear factors. t It is always possible to choose a so that ma and na are integers, for we can take a as the L.C.M. of the denominators of m and n. t The case where p is an integer is not excluded, but appears as a special case, namely, r-p, s = l. INTEGRATION BY EATIONALIZATION 341 196. Conditions of integrability of the binomial differential (A) x'"(^a + bx''ydx. Case I. Assume a + bx"=^. 1 r Then (a + 6af)' = 2, and (a + Ja;")' = z' ; 1 also x = ( — - — 1, and x'" hence dx= —^~''^l — :; — ) dz. on Substituting in (^), we get n + I x'"(a + barydx = -^z'-+''-''(^—^\ " dz. The second member of this expression is rational when m +1 n is an integer or zero. Case II. Assume a + bx"= sfaf: Then af= -, and a + bx" = z!'af = ■ ^ ^-b a-b r r T Hence (a + bx^'J = a' {f-b^'z^; 11mm also X = ct (f— 6)""'', 7^= a" (f— 6)" "; s i -1-1 and ti2;==--aV-V2'-6) " &. w ' Substituting in (^), we get 971 -f- 1 T The second member of this expression is rational when h - is an integer or zero. Mence the binomial differential r oT'^a + bx^ydx can he integrated by rationalization in. the following oases : * •Assuming as before that the denominator of the resulting rational function can he broken up into real quadratic and linear factors. 342 INTEGRAL CALCULUS Case I. When ^ = an integer or zero, hy assuming n Case II. When -\--=an integer or zero, by assuming n s a + haf=z:'x\ EXAMPLES C x^dx _ c , 12a + 6x2 ■^ (a + W)^ -^ 6' Va + 6x2. Solution, m = 3, 71 = 2, r = - 3, s = 2 ; and here — = 2, an integer. Hence '71 this conies under Case I and we assume a + 6x2 = z2 ; whence x = i^—^ , dx = -— ? -, and (a + bx2)t = z^. \ 6 / bi(z2-a)4 /• x»dx _ /■ ('^jzi^^ zi^z 2^ "•' {a + 6x2)f ~-^ V ^ / bi{z^-a)i ^' = ij(l - az-^)dz = i (z + az-i) + O 1 2a + 6x2 _ + O. 2. r^^ •^ x4vT+ dx _{2x^-l)(l + x')i 3x8 + C. 7* 1 7/1 -4" X )" Solution. m=— 4, n = 2, -= ; and here — — 1- - = — 2, an integer. Hence s 2 n s this comes under Case II and we assume l + x^ = zV, ,z=ii±^; X whence x^ = — , 1 + x^ = — , Vl + x^ = -; z2-l z2-l (z2_l)i 1 1 zdz also X = , a;* = -— ; and dx= — (z2_])i (^^-1)^ (z2_l)l /_zdz_ dz 3xs 3. /x3(l + x^M = (3x--2)(l + x2)f ^ ^ 4.r^^=_^+c. •^ (1 + X2)t Vl + X2 INTEGRATION BY EATIONALIZATION 343 5. r^^=(i+x^)i^^+c. 8. r •^ Vl + s" 3 ./o Vo^-T^ 15 '• / "^ 4 =- »(i + =^'>~* (^^ + S + ^- 10. rv v^^^dx = !:^. •'s2(l + x2)5 \ '^f Jo 16 11. rV(a2-x2)i^dx = — . Jo 32 12. r ^^— = ilog^^^Il^ + C. y(a2 + »2)* ^" Va2 + 3/2 + a 13. ft^l + 2P)icU, = (1 + 2P)i ^^^~^ + G. 14. rii(l + u)^(iu = y^(l + u)4(5M-2)+ C. 16. r_^!^^= ^! . + C. ."' (a + fia^)^ 3a(ol+6a:?)^ 16. fe^(i+ 0^)^d9 = #j(l + 5==)^- 1(1 + «^)* + A(l + ^)*+ C dx 3x' + 2a "■/; ■+C. e2 (a + x8)7 2 a^x (a + x?)^ 197. Transfonnation of trigonometric differentials From Trigonometry n T. — • ^. sin , ,v,., 2 2 (J[) sin a; = 2 sin -cos-, 87, p. 2 (^) cos But . X 1 sm-_ esc id X 1 cos- = _ sec cos X = cos" - — sm' -. 37, p. 2 tan- I 4°*1+^ a(i^1 f >|l + tan^i If we now assume tan ^ = Sj or, x=2 arc tan z, we get ' sin| = -^, cos| = -^==. 344 INTEGRAL CALCULUS Substituting in (A) and (5), 2z 1-2= sin a; =5 ;i cosa; = - r- 1 + 2^ 1 + 2' O J- Also by differentiating x—2 arc tan s we have cia; = :j -■ Since sin x, cos a;, and dx are here expressed rationally in terms of 2, it follows that A trigonometric differential involving sin x and cos x rationally only can be transformed by means of the substitution tan 75 = 2, or, what is the sam,e thing, by the substitutions 22 1-2= , -Idz sm X = ; J cos X = , dx = ■ 1 + 2^ 1 + 2= 1+2= into another differential expression which is rational- in z. It is evident that if a trigonometric differential involves tan x, cot x, sec X, esc X rationally only, it will be included ia the above theorem, since these four functions caii be expressed rationally in terms of sin X, or cos x, or both. It follows, therefore, that any rational trigono- metric differential can be integrated.* 1. r-ii±i J sina;(l EXAMPLES "^^^ '- = - tan^ - + tan- + - log tan- + C. + cosx) 4 2 22^ 2 Solution. Since this differential is rational in sinx and cosx, we make the above substitutions at once, giving \ 1 + Wi- r (l + sinx)dx _ I \ 1 + zV 1+1 J sinx{l + cosx) I 2z /■, , 1 — z= 'J 1 + Z2\ l + z2 _ r (1+^ + 2 z)dz ) r l + z2 + 2z)dz 1 r, 1 /z' \ = -(- + 2z + logz) + C = ^tan2| + tan| + llog(tan|) + C. *See footnote, p. 34:1. INTEGEATION BY RATIONALIZATION 345 2. pi_^ = l. 8.r^= ^ +0. Jol + sinx J 1 — sina; - ^ » 1 — tan- fir 2 3. / V-^^— = 1. Jo 5 + dx SCOSK dy '■/ir dx 5sinx 3 rir ay _ TT rf 1 / \ „ /•« dor TT Jo 2 + cosa a-v/s H- I ;^ : = - arc tan (3 tan x) + C •/ 5 — 4 cos 2x3 ,2 dx J -2 ox _ 1 _ ^ 2„l+cosx ■ 12. r = — -arctan/ Vstan-l+C. y -/2-cos« Vi \ 2/ ,„ /• dx 1 ^ /5tanx4-4\ „ 13. I = -arctan( H^j+o. J 5 + 4sin2x 3 \ 3 / , . c cosxdx „ ^ (^ ^\ ^ ^ , r, . X, „ 14. I = 2 arc tan I tan- I— tan- + C = x — tan- + C. -/ 1 + cosx \ 2/ 2 2 15. Derive by the method of this article formulas (16) and (17), p. 284 16 /sinxdx 2 .r.^/^x\_ 2 „ = h 2 arc tan ( x tan- 1 + C = \-x + G. ^ + ''^^ 1 + tan? V 2; i + tan? 2 2 198. Miscellaneous substitutions. So far the substitutions considered have rationalized the given differential expression. In a great number of cases, however, integrations may be effected by means of substitu- tions which do not rationalize the given differential, but no general rule can be given, and the experience gained in working out a large number of problems must be our guide. A very useful substitution is 1 , dz z z called the redprocal substitution. Let us use this substitution ia the next example. /-^^2 2j2 dx. X* 1 dz Solution. Making the substitution x = - , dx = -, we get 346 INTEGRAL CALCULUS EXAMPLES 1. • r_^_=J_ log ^i-+c. J (X - 2)8 ^ ^ ' (X - 2)2 „ /- x8(ix 18x2 + 27x + 11 , , , , ,, , „ 3. I = ■ -r hlog(x + l)+C. J (x + 1)* 6(X + 1)' SV -r ;-r 6. f- -. , •^ sV 1 + x + x2 2 + X + 2 Vl + X + x2 x(7 (a^ + x2)t a^Va^ + x^ dx 1, = log " a + ex xVa^ + x^ Va2 + x2 cfc = = log — ex Assume s' < = z. AsBtune x -2 = :Z. Assume x + 1 = z. Assume x _1 z Assume x a ~ z' Assume X : _1 " z' -!. f^^l+J^ax = l(l + logx)i + C. Assume 1 + log x = z. 8. / = — (3 e^ - 4) (e" + 1)1 + C. Assume e'' + 1 = z. ■'' (e:.: + i)¥ ^1 nT*" 1X1,, „, „ ^•/t;: — S-; = S-;-7 + 7l°S'(e^-2) + C'- Assumee^ = z. Je2x_2e^ 2e^ 4 4 ,- /• xtfe 1, 1 1 ^ 1 r 2x ,1 „ 10. ( T = 5log 1 -=arctaii— = ^+1+0. •^(iH-x^)! 2 (xs + l)*-x V8, V3L(xS4.l)* J z' Assume x' = 11. r (X - x«)idx _ ^ 1 Ji 7 — "• Assume x = - • y X* z /•4(sin^+ cos^)d5 log 3 Jo 3 + sm2i9 = T' ' ^^"""^^ ™'^- °°^'' = ^ ^^•X ex + e-:. = ^''°*^°^-^- Assumee^ = z. . »« dx Vox — X^ Assume x = a sin^z. 15 f'°^W^:ri Jo ^ + s dx = 4-w. Assume e^-l = z2. 16. f^^V2t + im = Vi - J log(2 + Vi). Assume 4 + 1 = z. ^ Ve^-ldx = -^— . Assume e- - 1 = z. ^2 + Vi (s2 + i)dx , „ ■ ' 1 18. i=^=4= = log3. AssnmB^_i r !Vx*4- Tx2 + 1 Assume x = z. X CHAPTER XXVII INTEGRATION BY PARTS. REDUCTION FORMULAS 199. Formula for integration by parts. If u and v are functions of a single independent variable, we have, from the formula for the dif- ferentiation of a product (V, p. 34), d (uv) = udv + vdu, or, transposing, udv = d(uv') — vdu. Integrating this, we get the inverse formula, (Ay j udv = uv— I vdu, called the formula for integration by parts. This formula makes the inte- gration of udv, which we may not be able to integrate directly, depend on the integration of dv and vdu, which may be in such form as to be readily integrable. This method of integration hy parts is one of the most useful in the Integral Calculus. To apply this formula in any given case the given differential must be separated into two factors, namely, u and dv. No general directions can be given for choosing these factors, except that (a) dx is always a part of dv; (b) it must be possible to integrate dv; and (c) when the expression to be integrated is the product of two func- tions, it is usually best to choose the most complicated looking one that it is possible to integrate as part of dv. The following examples will show in detail how the formula is apphed : Illust^rative Example 1. Find j xooBxdx. Solution. Let u = x and dv = cos xdx ; then du = dx and «= | cosxdx = sinx. Substituting in (A), , ) , I X cos xdx = X sin X — / sin x dx = X sinx + cosx + C. 847 348 INTEGRAL CALCULUS Illustrative Example 2. Find I xlogxdx. Solution. Let U - zlog X am i dB = xdx t then du = X and » = 1 xdx = a2 " 2 Substituting in w. V f' logxda; = zlog x2 -/!■ dx, X = _logx--+G Illustrative Example 3. Eind / x^^dx. Solution. Let u = e^ and dv = xdx ; xdx = — 2 Substituting in (A), /x' rx' xef^dx = e<" I — ef^adx 2 J 2 r n ■ / x^e"™(ix. 3?0>^ a '~2 2. But x^ef^dx is not as simple to integrate as xe^^dx, which fact indicates that we did not choose our factors suitably. Instead, let u = x and dv = e^dx ; en Substituting in (A), ef^dx = a /(fix n gO^E xef^dx = X / — dx a J a a a?' a \ a/ It may be necessary to apply the formula for integration by parts more than once, as in the following example : Illustrative Example 4. Find fx^e^dx. Solution. Let u = x^ and dv = e<"dx ; then du = 2xdx and v= fe^dx — Substituting in (.4), fx^0^dx = x2 • — - r — . 2 J a J a gax ixdx x^e^ 2 <^ =T-.7^ xe^dx. INTEGRATION BY PAETS 349 The Integral in the last term may he found hy applying formula (A) again, which -/ a \ a/ Suhstituting this result in {B), we get I x^e""^ = s-0» - - + C = — (x2 + ) + 0. J a a^ \ a/ a \ a a^l Among the most important applications of the method of iategration by parts is the integration of (a) differentials involving products, (b) differentials involving logarithms, (c) differentials involving inverse circular functions. EXAMPLES 1. rx21ogxda; = — (logx — -1 + C 2. I asinada =— aoosar + sina + C. 3. fare sin xdx = x arc sin x + Vl — x^ + C. Hint. Let u = arc sin a: and dv = cbx, etc. 4. I logxdx = X (logx — 1) + C. 5. Tare tanxdx = x arc tan x — log (1 + x^)' + G. /xt + i / 1 \ x» log xdx = ( logx r I + C. n + l\ n + 1/ /x*^ 4- 1 X • X arc tan xdx = arctanx — - + C. 2 2 8. fare cot ydy = y arc cot j/ + -J log (1 + v'') + C. 9. fxoFdx = cF\-^ — ^ +C. J Llog a log oJ 10. CPam = a< r Ji- - ^ + -^1 + C. •/ Llog a log^o log^aj 11. Tcos 6 log sin Odd = sin 6 (log sin 5 — 1) + C. 12. fx^efdx = &:(x^- 2x + 2) + C. , 13. rxsinxcosxdx = isin2x — ixcos2x + C. 14. rx2e-=^ = e-=^(2-2x-x2) + C. 15. rarctan-v^dx = xarctan-v^ — Vx + arotan Vx + C. 350 BSTTEGEAL CALCULUS 16. /--^ = if-i-i3^]+^- 2«- ^^log^ = |(log.-l) + 0. 17. £'cBlog«to = - i. 21- X'^"^" sinztfe = I - 1. . 18. f\ogydy=-l. 22./^'arctanedS = |-logV2. t/O 19. . JVsinada = TT- 2. 23. jVlogsds =- i- J a \ a a' a'/ 25. fip^ sin .^d^ = 2 cos + 2 (^ sin ^ — .^2 cos ^ + C. 26. f (log x)2dx = X [log2 X - 2 log X + 2] + 0. /Q.2 atan2 0.^0- = o-tan or — — + log cos a + C. 28. f '°g^ = ^logx - log(x + 1) + C. J (x + l)2 x + 1 Hint. Let «=loga; and dw=- -—.etc. (3;+ 1)2 /J.S x^ + 2 / r -. x^ arc sin xdx = — arcsinxH — vl — x^ + G. 30. Tsec^ ff log tan Sdff = tan 6 (log tan fl — 1) + C. 31. Tlog (log x) — = log X ■ log (log x) - log X + C. 32. f]£S^^+J)^ =' 2 ^^TI [log(x + 1) - 2] + C. ■^ Vx + 1 fx' (a - x2)i dx =- Jx2 (a - x2)l - T% (a - x^)^ + C. 33. Hint. Let u = x^ and dv = {a- x^)^ xdx, etc. 34. r Va2 - x^dx = - Vo2 - x2 + — arc sin - +, C. J 2 2 a 35. r_^^=_l(x2 + 2)(l-x2)i + C. 36. fVa^ + x^dx = - Va2 + x^ + — log(x + Va^ + x^) + C. J 2 2 I „„ r a^^dx X /-;; j . a^ . x . ^ 37. / , = Wa? — x^ + — arc sin - + C. J Va2 - x2 2 2 a /•(logx)2dx 2 r, „ 4, , 81 , - 200. Reduction formulas for binomial differentials. It was shown in § 195, p. 340, that any binomial differential may be reduced to the form x''(a + bx''ydx, where ^ is a rational number, m and n are integers, and n is positive. Also in § 196, p. 341, we learned how to integrate such a differential expression in certain cases. REDUCTION FORMULAS 351 In general we can integrate such an expression by parts, using (il), p. 347, if it can be integrated at all. To apply the method of integra- tion hy parts to every example, however, is rather a long and tedious process. When the binomial differential cannot be integrated readily by any of the methods shown so far, it is customary to employ reduc- tion formulas deduced by the method of integration by parts. By means of these reduction formulas the given differential is expressed as the sum of two terms, one of which is not affected by the sign of integration, and the other is an integral of the same form as the origi- nal expression, but one which is easier to integrate. The following are the four principal induction formulas : (A) \ x^(a-\-bx''ydx = —-—^ — , ..\ ^ ■' J ^ ^ (n/> + 7n + l)6 anp r i,xny-'. Ox. (m-l-l)a J ^ ^ r x'"+'-(a-\-bx"y+^ ^ ^ J ^ ^ n(/» + l)a While it is not desirable for the student to. memorize these formulas, he should know what each one will do and when each one fails. Thus : Formula (A) diminishes m hy n. (A) fails when np-^-m -{-1=0. Formula (J?) diminishes p by 1. (J?) fails when np-i-m -{-1=0. Formula (C) increases m by n. (C) fails when m-\-l=0. Formula (D) increases p by 1. (P) fails when ^+1=0. 352 INTEGRAL CALCULUS I. To derive formula (A~). The formula for iiitegration by parts is (A) ' j udv = uv— j vdu. (4), p. 347 We may apply this formula in the integration of \ x'^(a + ha^ydx by placing m = 2;'"-"+^* and dv = (a + hx''ya^-'^dx; (a + lx"Y*^ then du = (m — n-\-V)x'^-"dx and v= ^ _/ - Substituting in (^A), But far- " (a + bafy +^dx= jar - " (a + hx")" (a + haf^dx = a I ar-''(^a + bx"ydx + b Cx''(a + bx"ydx. Substituting this in (5), we get — ^^— _^ I a;"" - " ( a + bx"ydx nb(p+l} J Transposing the last term to the first member, combining, and solv ing for I 2;'"(a + bx^ydx, we obtain ^ ^ 6(n/> + m + l) -:ry ^!— 4- I x'"-"Ca + bx"ydx. b(np + m + l}J y ^ J * In order to integrate dv by (4) it is necessary that x outside tlie parenthesis shall have the exponent n - 1. Subtracting n-1 from m leaves m - n + 1 for the exponent of a: in m. EEDUCTION FOEMULAS 353 It is seen by formula (A) that the integration of a;"" (a + bx^ydx is made to depend upon the integration of another . differential of the same form in which m is replaced hjm — n. By repeated applications of formula (4), m may be diminished by any multiple of n. "When np + m +1=0, formula (^1) evidently fails (the denominator vanishing). But in that case m+l [-p = 0; n hence we can apply the method of § 196, p. 841, and the formula is not needed. II. To derive formula (B'). Separating the factors, we may write (0) jx'"Qa + bx"ydx= jx''(a + hx"y-\a + bx')dx = a j x''(a + bx''y-^dx + b jx'^+''(a + bx"y-'dx. Now let us apply formula (.4) to the last term of (C) by substi- tuting in the formula m + n for m, and p — 1 for p. This gives I fx-^X-+bxr-'dx= ^'^'"^^+^''"y - <^+'^] Cx-ia + bx'^y-^dx. J np + m+1 np+m+\j Substituting this in (C), and combining like terms, we get (J5) f:^ (a + bx^ydx = - np + m + 1 np + m + ij ^ ^ Each application of formula (5) diminishes p by unity. Formula (5) fails for the same case as (.4). III. To derive formula (C). Solving formula (4) for f' x'"-"(a + bx''ydx, and substituting m + n for m, we get x'"+\a + b3^y+^ a(m + l) + n + 1(772 + 1) (C) fx'"(^a + bx"ydx = - HnP + n + m + i} C ^r.^n^a+bx-ydx. a(777 + l) J ^ ^ ^ 354 INTEGEAL CALCULUS Therefore each time we apply (C), m is replaced hj m + n. When m + 1 = 0, formula (C) fails, but then the differential expression can be integrated by the method of § 196, p. 341, and the formula is not needed. IV. To derive formula (D). Solving formula (5) for / x'"(a + hx''y-'^dx. and substituting ^ + 1 f or p, we get + — —- I x"'(a + bx"y+'-dx. Each application of (i)) increases p by unity. Evidently (7)) fails when ^+1=0, but then p = — l and the expression is rational. EXAMPLES 1. f-^^=- l(x2 + 2) (1 - x^)i + G. Solution. Here m = 3, n = 2, p = — J, a = 1, 6 = — 1. We apply reduction formula (A) in this case because the integration of the differen- tial would then depend on the integration of / a;(l — x^)~idx, which comes under (4), p. 284. Hence, substituting in (A), we obtain fx^l_,.yi^^-^I^110^^r}ll 1(3-2 + 1) r,a-.(i_,.)-i (J5) 1 sui"'2; cos"a;aa: = + m + 1 ^^ I sin"'+ In the same way, if we w — 1 W T I sin"'+''a;eos°~''a;^a;. let u = sin™ ^ar, and dv = cos" a; siu a;c?a;, we obtain siQ"'~^a;cos""''^a; (0) / sm^a; cos"a;c?a; = — • M + 1 H ;-Y I sin'" ^a;cos"+^a:cZa;. But I sin"" + ^ a; cos" ~ ^ xdx = | sin"" a; (1 — cos^ a;) cos" ~ ^ xdx = I sia"'a;cos''~^a;(?a;— | s sia"'a;cos''~^a;(?a;— | siu" a; cos" a:c?a;. Substituting this in (5), combiuiug like terms, and solving for I sin" a; cos''a;cZa;, we get m / sin^jccos" jrdr = sin^+^jccos""*;!: m + n n-1 -J' sin'^jrcos""" Jfdjr. m + Making a similar substitution ia (C), we get sin"'~^Jccos"+^jr cn Jsi sin'"j:cos"jrd[r = m + n -I I sin'"~^jf cos"jr A J sinxcos'x 1 /• . , (A) I sm^x cos* xdx = — ■ 1- - / cos* xdx. [Here m = 2, n= 4.] Applying formula (E) to tlie integral in the second member of (A), we get , /B\ r d J sinxcos'x 3 /• . , (P) j cos* xdx = 1- - / cos'' xdx. [Herem=0, »=4.] Applying formula {E) to the second member of (B) gives //-,> r ■> J sinxcosx X (C) ( cos^xdx = 1 J 2 2 Now substitute the result (0) in (B), and then this result in (A). This gives the answer as above. 2 r„:-,i» „„„!!„j oosx/sin^x sin'x sin rsin*x cos^xdx = 5^ (?^ - ?12!^ _ !l^\ + - + C . I = tanx — 2cotx cot^x + C. ■/ sin* X cos'' X 3 SEDUCTION FORMULAS 359 4. I — V-; — = -— (3— oos^k) i-C. J sm^'x 2 2 /I 1 seo'xdx = -seoajtani + -log(seca; + tana;) + C. A 2 /I 1 csc'xda; =— - esc x cot a; + - log (esc a; — oota;) + C. A 2 _ r cos*adar cos a 3, a 7. / —^-, =-;r-r-; eosar--logtan- + C. J sm'a 2sin2Q: 2 2 o r • « J cosa/sin^a , 5 . . 5 . \ ha 8. / smeada =_ __ (^__ + _ sm»<. + - sma) + i| + C. 9. rcscas e^ 2e^ , /■ cos 2 xdx 1 ,„ . „ „ „ , „ 5. I = (2sm2x- 3cos2x) + G. „ C ■ 1 J, ^ /-i 2 sin 2 X + cos 2 x\ „ 6. I e^sm2x(ix = — (1 1+0. tv r ^ 9 J e^ /, , 2 sin 2 or + cos 2 q:\ ^ 7. I e*cos2a;da: = — IIH )+G. /- X -/ X x\ e^cos-tix = e2(sin- + cos-| + C. 2 \ 2 2/ 9. I e"" (sm aa + cos aa) da = — ■ 1- C. e'^(sin2x— cos 2 x) c?x =• — (sin 2 x — 5cos2x) + C 11. I e-''smx(Zx = -. | e-sa= Jo 2 Jo 1 r" (x^), Cx^), ■■■, ^QcJ. (3) The areas of the successive rec- tangles are obviously ^ 3mb x" axi X. (ix^)Ax^, (,x^)Ax^, (x^)Ax^, •••, <^(a;„)Aa;„. (4) The area under the curve is therefore equal to n = »['^(^:)^^i + '^(^.) A2;, + (x) dx. Therefore our discussion gives (-5) jT <^ (^) '^^ = n = « [<^ (^J ^y + (^,) Ax^+---+ (xj A J . * This eonstrafetion Includes the previous one as a special case, namely, when the point i chosen at one extremity of a subinterval. INTEGEATION A PEOCESS OF SUMMATION 363 This equation has been derived by making use of the notion of area. Intuition has aided us in estabhshing the result. Let us now regard (5) dmply as a theorem in analysis, which may then be stated as follows : Fundamental Theorem of the Integral Calculus Let (a;) be continuous for the interval x = atox=b. Let this interval be divided into n svhintervals whose lengths are Ax , Ax , • • ■ , Ax^, and points be chosen, one in each subinterval, their abscissas being x,x , • • • , x^ respectively. Consider the sum (C) cl>(x;)Ax^ + 4>(x^)Ax^+ ■ ■ ■ +(ixjAx„=^cl>(x,)Ax,. t=i Then the limiting value of this sum when n increases without limit, and each subinterval approaches zero as a limit, equals the value of the definite integral /•<> I ^ (x) dx. Equation (5) may be abbreviated as follows : (D) , f (^x-)dx = ^'^l^(x,)Ax, The importance of this theorem results from the fact that we are able to calculate by integration a magnitude which is the limit of a sum of the form (C). It may be remarked that each term in the sum (C) is a differen- tial expression, since the lengths Ax^, Ax^, • ■ • , Ax^ approach zero as a limit. . Each term is also called an element of the magnitude to be calculated. The following rule will be of service in applying this theorem to practical problems. Fundamental Theorem. Eule FiEST Step. Divide the required magnitude into similar parts such that it is clear that the desired result will be found by taJdng the limit of a sum of such parts. Second Step. Find expressions for the magnitudes of these parts such that their sum will be of the form (C). Thied Step. Having chosen the proper limits x = a and x=b, we apply the Fundamental Theorem and integrate. 364 INTEGEAL CALCULUS As in the last section, divide 205. Analytical proof of the Fundamental Theorem the interval from x = atox = b into any number n of subMitervals, not necessarily equal, and de- note the abscissas of these points of division by 61, 62, • • •, 6n-i, and the lengths of the suhinter- vals by Ax^, Ax^, • • • , Ax„. Now, however, we let xi, X2, • ■ • , < denote abscissas, one in each inter- val, determined by the Theorem of Mean Value (44), p. 165, erect ordinates at these points, and through their extremities draw horizontal lines to form rectangles, as in the figure. Note that here ^(a;) takes the place of (p'(x). Applying (44) to the first interval {a = a, 6 = b^, and xi lies between a and b^), we have ™.' nr' t' fiW)- :^ = 0(xf), or, since Also fej— a 6j^ — a = Ax^, /(6i)-/(a) = 0(xl)AXi. /(b^) _/(6^) = ,^(a^ Axj, for the second Interval, /(6g) — /(ftg) = 0(^3) Aaig, for the third interval, etc., f(p) — /(6„_i) = 0(xQ Ax„, for the rith interval. Adding these, we get (E) f(b) -f{a) = 0(xl)AXi + .^(xaO AXg + ■ • . + 0(xi)Ax„. But {x'l) ■ AXj = area of the first rectangle, 0(X2) • AXj = area of the second rectangle, etc. Hence the sum on the right-hand side of (-B) equals the sum. of the areas of the rectangles. But from (4), p. 361, the left-hand side of (-B) equals the area between the curve y = {x), the axis of X, and the ordinates at x = a and x = 6. Then the sum (F) ^0(xOAxi equals this area. And while the corresponding sum n (G) ^0(x.)AXi i = l [Where Xi is any aliscissa of the subintenral A(.] (formed as in last section) does not also give the area, nevertheless we may show that the two sums (F) and (G) approach equality when n increases without limit and each subinterval approaches zero as a limit. For the difference (xQ — (x;) does not ex- ceed in numerical value the difference of the greatest and smallest ordinates in Axj. And furthermore it is always possible * to make all these diflerenoes less in numerical value than any assignable positive number c, however small, by continuing the process of subdivision far enough, i.e. by choosing n sufficiently large. Hence for such a choice of n the difference of the sums (F) and (G) is less in numerical value than e(6 — a), * That such is the case is shown in advanced works on the Calculus. INTEGEATION A PROCESS OF SUMMATION 365 i.e. less than any assignable positive quantity, however small. Accordingly as n in- creases without limit, the sums (F) and (G) approach equality, and since (F) is always equal to the area, the fundamental result follows that in which the interval [a, 6] is subdivided in any manner whatever, and Xi is any abscissa in the corresponding subinterval. 206. Areas of plane curves. Rectangular coordinates. As already explained, the area between a curve, the axis of X, and the ordinates x = a and a; = 5 is given by the formula (^) area -£ ydx, the value of ^ in terms of x being substi- tuted from the equation of the curve. Equation (A~) is readily memorized by observing that ydx represents the area of a rectangle (as CE) of base dx and altitude y. It is convenient to think of the required area ABQP as the limit of the sum of all such rectangles (strips) between the ordinates AP and BQ. Let us now apply the Fundamental Theorem, p. 363, to the calcu- lation of the area bounded by the curve x=j> Qy'), QAB in figure), the axis of Y, and the horizontal hues y = c and y = d. FiKST Step. Construct the n rectangles as in the figure. The required area is clearly the limit of the sum of the areas of these rectangles as their number increases with- out limit and the altitude of each one ap- proaches zero as a limit. Second Step. Denote the altitudes by Ay^, Ay^, etc. Take the point in each inter- val at the upper extremity and denote their ordinates by y^, y^, etc. Then the bases are ^(^j), ^(iy^)-! etc., and the sum of the areas of the rectangles is iVi) ^^1 +iy^^y,+--- + 'i> iyn) ^yn=X (yd ^yr Third Step. Applying the Fundamental Theorem gives limit n ■■ 366 INTEGRAL CALCULUS Hence the area between a curve, the axis of F, and the horizontal hnes y—e and y = d\& given by the formula ^ (■B) area =x xdy. the value of a; in terms of y being sub- stituted from the equation of the curve. Formula (5) is remembered as indicating the limit of the sum of all horizontal strips (rectangles) within the required area, x and dy being the base and altitude of any strip. Illustrative BxamplB 1. Find the area included between the semicubical parab- ola y^ = 3? and the line x = i. Solution. Let us first find the area OMP, half of the required area OPP' For the upper branch of the curve y = V^, and summing up all the strips between the limits 1 = and a; = 4, we get, by substituting in (A), area OMP = f ydx = f x^dx. = ^ = 12f . J fj Jo Hence area OFF' = 2 ■ -^ = 25f . If the unit of length is one inch, the area of OPP' is square inches. Note. For the lower branch y =— xi; hence area OifP'= f (-a;^)(ir =- 12f Jo — X This area lies below the axis of x and has a negative sign because the ordinates are negative. In finding the area OMP above, the result was positive because the ordinates were positive, th£ area lying above the axis of x. The above result, 25|, was the total area regardless of sign. As we shall illus- trate in the next example, it is important to note the sign of the area when the curve crosses the axis of X within the limits of integration. Illustrative Example 2. Find the area of one y arch of the sine curve y = sina;. Solution. Placing y = and solving for x, we find x = 0, TT, 2 IT, etc. "0" Substituting in {A), p. 365, area OAB = f ydx= f smxdx = 2. Ja Jo area BCD = f ydx = f sinxdx = — 2, Ja J-ir a,Tea, OABCB = f ydx= f " sinxdx = 0. J a Jo This last result takes into account the signs of Ihe two separate areas composing the whole. The total area regardless of these signs equals 4, Also and ESTTEGRATION A PKOCESS OF SUMMATION 3G7 Illustrative Example 3. Find the area included between the parabola x^ = iay and the witch _ Sgg ^~ x^ + ia'^' Solution. To determine the limits of integra- tion we solve the equations simultaneously to find where the curves intersect. The coordinates of A are found to be (— 2 a, a), and of C (2 a, a). It is seen from the figure that area AOCB = areaX>^CB^ — aveaDECOA. /» 2 a g a^dx But area DECBA = 2 x area OBCB = 2 | 2 ira?, Jo a;2 + 4a2 and area DECOA = 2 x area OEC = 2 I — Jo 4a 2ax2 4a2 3 Hence area 40CB =27ra2 ^ = 2a2(ir-|). A-aa. o Another method is to consider the strip PS as an element of the area. If i/ is the ordinate corresponding to the witch, and y" to the parabola, the differential expression for the area of the strip PS equals (j/' — y")dx. Substituting the values of y' and y" in terms of x from the given equations, we get area AOCB = 2 x area OCB '>2a = 2f {y'-y")dx Jo Jo \x^ + 4:a^ 4a/ = 2o2(ff-f). X^ «2 Illustrative Example 4. Find the area of the ellipse 1- — = 1 Solution. To find the area of the quadrant OAB, the limits are x = 0, x = a ; and 2, = - Va2 - x2. Hence, substituting in (A), p. 365, areaO^JB = - fia^ — x^idx aJo = — (a^ — x^)^ + — arc sin - L2a^ '2 oJo TTOb Therefore the entire area of the ellipse equals iraJ. Y / ^ -^ ^ I ' /4 / N^ ^ — a J 368 INTEGEAL CALCULUS 207. Area when equation of the curve is given in parametric form. Let the equation of the curve be given in the parametric form We then have y = ^ (f), and dx =f'(t') dt, which substituted* in (^), p. 365, gives (^) area = r>l>(ty'(t^dt, where t = t^ when x = a, and t = t^ when x=b. We may employ this formula (A} when finding the area under a curve given in parametric form. Or we may find y and dx from the parametric equations of the curve in terms of t and dt and then substitute the results dkectly in (.A), p. 365. Thus in finding the area of tlie ellipse in Illustrative Example 4, p. 367, it would have been simpler to use the parametric equations of the ellipse a; = o cos 0, y = b sin 0, where the eccentric angle is the parameter (§ 66, p. 81). Here 2/ = 6 sin u area OAB = I ydx = — I a6 sin^ 0(j0 = Hence the entire area equals iraft. Ans. EXAMPLES 1. }?ind the area bounded by the line y = hx, the axis of X, and the ordinate g. _ 2. Ans. 10. 2. Find the area bounded by the parabola y^ = ix, the axis of F, and the lines 2/ = 4 and 2/ = 6. ^ws- 12f. 3. Find the area of the circle x^ + j/^ = r^, Ans. ttt^. 4. Find the area bounded hj y^ = Qx and y = 3x. . Ans: i. 5. Find the area bounded by the coordinate axis and the curve y = eF. Ans. 1. 6. Find the area bounded by the curve y = log x, the axis of y, and the lines 2/ = and 2/ = 2. Ans. e^-l. 7. Find the entire area of the curve x^ + y^ = a». Ans. %Trefi. 8. Find the area between' the catenary ?/ = - [e* + e "], the axis of T, the axis of X, and the line x = o. Ayia ^ \e^ - 11 * For a rigorous proof of this substitution the student is referred to more advanced treatises on the Calculus. INTEGEATION A PEOCESS 0¥ SUMMATION 369 9. Find the area between the curve y = logx, the axis of X, and the ordinates j; = lands = a. ^Tis. a(loga-l) + l. 10. Find the entire area of the curve 11. Find the entire area of the curve a^y^ = x^{2a — x). Ans. ira?. 12. Find the area bounded by the curves x{y — eF) = sin x, and 2 x?/ = 2 sin a; + x', the axis of Y, and the ordinate x = 1. Ans. f (e»= — ^ x^) cix = e — | = 1 .55 + . . .. J 13. Find the area betv?een the Vfitch y = — and the axis of X, its asymptote. x^ + 4 a^ . . n Ans. iira^. 14. Find the area between the cissoid y' = and its asymptote, the line x = 2a. 2a- X ^^_ g^^2_ 15. Find the area bounded by ^ = x"^ ^^ = g^ and the axis of T. Ans. 12. 16. Find the area included between the two parabolas y^ = 2px and x^ = 2py. . 4p2 3 17 . Find the area included between the parabola y' = 2x and the circle y^ = 4x — x^, and lying outside of the parabola. Ans. 0.475. 18. Find the area bounded hj y = x^, y = x, y = 2x. Ans. I. 19. Find an expression for the area bounded by the equilateral hyperbola x' — y^ = a^, the axis of X, and a line drawn from the origin to any point (x, y). Ans. ^log^±I. 2 a 20. Find by integration the area of the triangle bounded by the axis of T and the lines 2x + y + S = and y =— i. Ans. 4. 21. Find the area of the circle (x = r cos d, \y = rsin^, 9 bemg the parameter. Ans. irfi 22. Find the area of the ellipse X = a cos 0, 2/ = 6 sin 0, where the eccentric angle ^ is the parameter. Ans. irab. 23. Find the area of the cardioid X = a (2 cos t — cos 2 1), y = a(2 sin t— sin 2 1). 24. Find the area of one arch of the cycloid rx = a(8 — sin^), \y = a{l— cosO), 9 being the parameter. Hint. Since x varies from to 2 ira, 6 varies from to 2 tt. Ans. 3 ira?' ; that is, three times the area of the generating circle. 370 INTEGKAL CALCULUS 25. The locus of A in the figure, p. 82, is called the "companion to the cycloid." Its- equations are x = a£ y = a{l-cosd). Y Pind the area of one arch. Ans. 2ira^. 26. Find the area of the hypocycloid fx = acos'O, y = asw?6, $ being the parameter. Ans. ; that is, three eighths of the area of the ° circumscribing circle. 27. Mnd the area of the loop of the folium of Descartes a;' + 2/* = 3 axy. -■ ix ; then x = , Hint. Let w = . and da;= Zadt. l+«3 (1+ x^i^x' [Di-riding inside the radical ty (Aa:')^ and multiplying outside ty Aa:'.] But from the Theorem of Mean Value, (44), p. 165 (if Ay is denoted by/(6) -f(cC) and Aa;' by 6 - a), we get ^-/'W' X being the abscissa of a point I^ on the curve between P' and P" at which the tangent is parallel to the chord. Substituting, P'P" = [1 +f(xyfA7^ = length of first chord. Similarly, P"P"' = [1 +f'(x^'''\^Ax" = length of second chord, jKniQ = I"! +/'(a;„)^]^Aa;^"'= length of wth chord. The length of the inscribed broken line joining P' and Q (sum of the chords) is then the sum of these expressions, namely, [1 +/'(^,)n^A:^ + [1 +f'(ix;)'f/^" +•■• + [! +/'(^„)n*^*'"' 874 INTEGEAL CALCULUS . Thied Step. Applying the Fundamental Theorem, Hence, denoting the length of arc F'Q by s, we have the formula for the length of the arc ^b s=j ll+fXxy}idx, ov (^) dp -XT-(I dx, where — must be found in terms of x from the equation of the ax given curve. Sometimes it is more convenient to use y as the independent variable. To derive a formula to cover this case, we know from (35), p. 148, that dy 1 ^ -, dx ^ -— = -—; hence dx = —- ay. ax ax ay dy Substituting this value of dx in QA.'), and noting that the corre- sponding y limits are c and d, we get* the formula for the length of the arc. (5) -m +1 dy, dx . where — in terms of y must be found from the equation of the given curve. Illustrative Example 1. Mnd the length of the circle x^ + y^ = r'^. Solution. Differentiating, — = Substituting in (^), *^ ^ arcB.4 dx r Substituting ij^=r'-x'' from the equation of the] [eirole in order to get everything in terms of x.\ ... arcB4 = rr— ^=f;arcsi„?T=^. Jo Vr2-a;2 L r\ 2 Hence the total length equals 2 irr. Ans. •"X'[»(i)'J**-x'[(S)Mi)"(i)>=i:'[(S)""T*- INTEGRATION A PEOCESS OF SUMMATIO.N 375 EXAMPLES 1. Find the length of the arc of the semicubical parabola aj/^ = ^a fjom y^g origin to the ordinate x = 5 a. 336 a Ans. 27 2. Find the entire length of the hypocycloid x^ + y^ = a^. Ans. 6 a. n - —- 3. Rectify the catenary y = - (e" + e~"°) from x = to the point (x, y). Ans. -(e^—e "). 4. Find the length of one complete arch of the cycloid ^ Hint. Use (-B). Here x = r arc vers - — V2 ry — y^. Ans. 8r. r ^ dx y '^y V2 1-y - 2/2 5. Find the length of the arc of the parabola y"^ = 2x0, from the vertex to one extremity of the latus rectum. n Vs n , /-, Ans. ^ + |log(l+V2). 6. Rectify the curve 9 ay^ = x (x — 3 a)^ from x = to x = 3 a. ^ns. 2 a Vs. 7. Pind the lenffth in one quadrant of the curve (-) +(r =1- «2 i «/. i 7.2 W \W ^ns. " +«'' + '' . a + 6 • e^ + 1 8. Find the length between x = a and x = 6 of the curve 0i = • ^ e' — l e26 _ I Ans. log \-a—b. 9. The equations of the involute of a circle are e^"— 1 Cx = a{cos0 + 6 sinff), \^y = a (sin S — ^ cos 0). Find the length of the arc from ^ = to ^ = 5^. Ans. iaOf. f X i^ c^ sin TT 10. Find the length of arc of curved „ „ from ^ = to S = - ■ ^ \y = eicosd 2 i; Ans. .V2(e2-1). 11. Find the lengths of arcs in the following curves : gx \ (a) y = log ; X = 1, X = 2. (d) y = logx ; x = 1, x = 4. (b) 2/ = logJ-x2); x = 0,x = ^. (e) 2/ = log sec x; x = 0,x = ^. X^ 1 TT TT (c) 2/ = — --logx; x = l, x = 2. (f) 2/ = logcscx; 3; = -, 3; = g' 211. Lengths of plane curves. Polar coordinates. Formulas (.A) and (5) of the last section for finding the lengths of curves whose equa- tions are given in rectangular coordinates involved the differential expressions U+mp. and U^J+1 In each case, if we introduce the differential of the independent vari- able inside the radical, they reduce to the form dy. 376 INTEGEAL CALCULUS Let us now transform this expression into polar coordinates by means of the substitutions x = pcosO, i/ = psin6. Then dx = — psia Odd + cos 6dp, dy = p cos OdQ + sin 6dp, and we have [&'+ df'f= [(^ p sin ede + cos ddpf^ (p cos 0d0 + sm ddpyf = [p'de'+dp^f. If the equation of the curve is then dp=f'ce}de = ^d0. Substituting this in the above differential expression, we get If then a and ^ are the limits of the independent variable 6 corre- sponding to the limits in (^) and (5), p. 374, we get the formula for the length of the arc, (^) -fXt dd, where p and — ^ in terms of 6 must be substi- da tuted from the equation of the given curve. "^ ^ In case it is more convenient to use p as the independent variable, and the equation is in the form ^ = '^^^^' de then dd = <^'(/3) dp = — dp. dp Substitutmg this in ■ [p^d0^+dp^f WV dp) Hence if p^ and p^ are the corresponding limits of the independent variable p, we get the formula for the length of the arc, gives /''(3^)+l dp. ''^ "X."KI)"^']'*' where — in terms of p must be substituted from the equation of the given curve. INTEGEATION A PROCESS OF SUMMATION 377 Illustrative Example 1. find the perimeter of the cardioid p = « (1 + c(is^). dp Solution. Here - d0 ■ a sin 0. I If we let 6 vary from to ir, the point F will generate one half of the curve. Substituting in {A), p. 376, - = /""[a^ll + cos(9)2 + a^m&iB^ae 2 «/ : 8 a. Ans. EXAMPLES 1. Find the length of the spiral of Archimedes, p = ad, from the origin to the end of the first revolution. ^^ TO.VrT4^ + -log(2,r + vT+l^). 2. Rectify the spiral p = e^ from the origin to the point (p, 6). Ans. — Vo^ + T. Hint. Use (B). a 3. Find the length of the curve p = a sec" - from ^ = 2 Ans. /V2 + logtan — \a. 4. Find the circumference of the circle p = 2r sin ^. Atis. 2 ■nr. 5. Find the length of the hyperbolic spiral p9 = a from (P i, gi) to (p^, 9^). Ans. Va2 + p2 — V^M^ + a loi og Show that 04, 6. Show that the entire length of the curve p = a sin= - is AB, BC are in arithmetical progression. 7. Find the length of arc of the cissoid p = 2 a tan ff sin 8 from ^ = to ^ = - • 212. Volumes of solids of revolu- tion. Let V denote the volume of the solid generated by revolviQg the plane surface ABCD about the axis of X, the equation of the plane curve DC beiag FiEST Step. Construct rectangles within the plane area ABCD as in the figure. When this area is re- volved about the axis of X, each rectangle generates a cylinder of revolution. The required volume is clearly equal to the limit of the sum of the volumes of these cylinders. 378 INTEGRAL CALCULUS Second Step. Denote the bases of the rectangles by Ax^, Ax^, etc., and the corresponding altitudes by r/^, t/^, etc. Then the volume of' the cylinder generated by the rectangle AEFB will be irylAx^ and the sum of the volumes of all such cylinders is TrylAx^ + -^yl^^i + • • ■ + 7r«/„A2:„ = ^iry^AXi. Third Step. Applying the Fundamental Theorem (using limits OA = a and OB = J), ^^^^^ » />;> 1=1 -ya Hence the volume generated by revolving, about the axis of X, the area bounded by the curve, the axis of X, and the ordinates x= a and a; = S is given by the formula QAy F,= ttJ fdx, where the value of «/ in terms of x must be substituted from the equation of the given curve. This formula is easily remembered if we consider a slice or disk of the solid between two planes perpendicular to the axis of revolution as an element of the volume, and regard it as a cylinder of infinitesimal altitude dx and with a base of area iry'^, and hence of volume iry^dx. Similarly, when OY is the axis of revolution we use the formula (B~) V^=7r J'x'dy, where the value of x in terms of y must be substituted from the equation of the given curve. iLLnSTKATivE EXAMPLE 1. Fiud the volume generated by revolving the 'ellipse 1-^ = 1 ahout the axis of X. 62 b\ Solution. Since y^ =— (a^ — x^), and the re- quired volume is twice the volume generated by OAB, we get, substituting in (.4), y /.a . /»a?>2 2 Jo Jo a^^ ' Fa: = 3 i-n-ab^ To verify this result, let b = a. Then V^ iTra^ the volume of a sphere, which is only a special case of the ellipsoid. "When the ellipse is revolved about its major axis, the solid generated is called a prolate spheroid ; when about its minor axis, an oblate spheroid. INTEGRATION A PROCESS OF SUMMATION 379 EXAMPLES 1. Find the volume of the sphere generated by revolving the circle a;^ ^. j,2 _ ^a about a diameter. Aiis. iirfi. 2. Find by integration the volume of the right cone generated by revolving the triangle whose vertices are (0, 0), (a, 0), (a, 6) about OX. Also find the volume gen- erated by revolving this triangle about OY. "Verify your results geometrically. 3. Find the volume of the torus (ring) generated by revolving the circle a;2 + (2/ - 6)2 = a? about OX. Ans. 2 Tr^a^b. 4. Find by integration the volume of the right cylinder generated by revolving the area bounded hj x = 0, y = 0, x = 6, y = i (a) about OX; (b) about OY. Verify your results geometrically. 5. Find by integration the volume of the truncated cone generated by revolving the area bounded hjy = 6 — x, 2^ = 0, x=0, x = 4 about OX. Verify geometrically. 6. Find the volume of the paraboloid of revolution generated by revolving the arc, of the parabola y^ = 4ax between the origin and the point (x-y, y^ about its axis. 2 Tn/lx, Ans. AiraXj^ = : ; i.e. one half of the volume of the circumscribing cylinder. 2i 7. Find the volume generated by revolving the arc in Ex. 6 about the axis of Y. Alls. ^ = -irx^y.; i.e. one fifth of the cylinder of altitude y, and radius of base x,. 80 a^ 5 8. Find by integration the volume of the cone generated by revolving about OX that part of the line 4a; — 52/ + 3 = which is intercepted between the coordinate axes. Sir Ans. 9 Find the volume generated by revolving about OX the curve 100 (a; — 4 a) 2/2 = ox (x — 3 a) between the limits x = and x = 3 a. Ans. (15 — 16 log 2). 10. Find the volume generated by revolving about OX the areas bounded by the following loci : 2 2 2 32 TTOi (a) The hypocycloid xt + y'a = a^, Ans. — -— - . 105 (b) The parabola x^ + yi = ai, x = 0, y = 0. ^ ■ 15 (c) One arch ot y = sin x. 2 ' (d) The parabola 2/^ = 4 x, x = 4. 32 ir. — fe2 — 11 (e) y = xe^, x = l, y = 0. 4 "• '■ 2 (f) y^ = 9x,y = 3x. (g) The witch y — — -— ' y = 0. iir'^a?. (h) y2(4 + j.2) = i^ j, = o, x,= 0, x=a>. (1) 2/(1 + x2) = X, y = 0, X = 0, X = 8. (i) 2/ = x^ 2/ = 0, X = 1. (m) y{x- if = 1, ?/ = 0, x = 3, x = 4. (j) 2/2(6 - X) = x2, 2/ = 0, X = 0, X = 4. (n) 2/2 = (x + 2)^, i/ = 0, x =- 1, x = 0. (k) 42/2 = x3, X = 4. (o) (X - 1)2/ = 2, 2/ = 0, X = 2, X = 5. 380 INTEGEAL CALCULUS 11. Find the volume generated by revolving the areas bounded by the f oUovsring loci : About OX About OY (a) 2/ = e», X = 0, 2/ = 0. (b) y = x^,x = 2, y = 0. (c) ay^ = x', 2/ = 0, X = a. (d)^ + ^=l. ^ ' 16 9 <•'©'-©* Ar^. -. 27r. 128 TT 64 TT 7 5 iffO^ fTO^ 48 TT. 6iw: 35 Ka^b 5 (I) 2/2 = 9x, ?/ = 0, x = 9. - (j) x2 = 16 - y, 2/ = 0. (g) 2^2 = 4 - X, X = 0. (k) x2 + 92/= = 36. (h) y^ = x-\-Q,x = Q. (1) 2/ = 2 X, 2/ = 0, X = 3. (i) x2 = 1 + 2/, 2/ = 0. (m) 2/ = X + 2, 2/ = 0, X = 0, X = 3. 12. Pind the volume generated by revolving one arch of the cycloid X = r arc vers V 2 ry — y- about OX, its base. ' Hint. Substitute dx = ' > and limits ;/ = 0, ?/ = 2 r, in (.4) , p. 374. ^„g_ g ^2^ V 2 j-t/ ■ j/2 13. Find the volume generated by revolving the catenary 2/ = - (e" + e ") about the axis of X from x = to x = b. ■" ^^ ^j 8 ^ '2 14. Find the volume of the solid generated by revolving the oissoid y^ = about its asymptote x = 2a. ..no, 15. Given the slope of tangent to the tractrix — =- . find the solid generated by revolving it about OX. v a- — y Ans. fjra^. 16. Show that the volume of a conical cap of height a cut from the solid generated by revolving the rectangular hyperbola x'^ — y^ = a^ about OX, equals the volume of a sphere of radius a. 17. Using the parametric equations of the hypocycloid ( X = a cos' 6, ly = a sin' 5, find the volume of the solid generated by revolving it about OX. Ans. ^ • " 105 18. Find the volume generated by revolving one arch of the cycloid fx = 0(6 — sin^), _ [ ^ = a(l— cos 5), about its base OX. ^ Ans. 5irV. Show that if the arch be revolved about OY, the volume generated is 67:^0'. 19. Show that the volume of the egg generated by revolving the curve x'y" + (x - a) (X - ft) = 0, (a< 6) about OX is ir-l (a + ft) log- - 2{& - a)i. I, a J 20. Find the volume generated by revolving the curve x* — a^x^ + a?y'^ = about OX. ^ 4 to' INTEGKATIOK A PKOCESS OF SUMMATION 381 213. Areas of surfaces of revolution. A surface of revolution is generated by revolving the arc CD of the curve y =f(.^') about the axis of X It is desired to measure this sur- face by making use of the Funda- mental Theorem. First Step. As before, divide the interval AB into subintervals AKj, Ax^, etc., and erect ordinates at the points of division. Draw the chords CS, EF, etc., of the curve. When the curve is revolved, each chord generates the lateral surface of a frustum of a cone of revolu- tion. The required surface of revolution is defined as the limit of the sum of the lateral surfaces of these frustums. Second Step. For the sake of clearness let us draw the first frus- tum on a larger scale. Let M be the middle point of the chord CE. Then (^) lateral area = 2 irNM- CE* In order to apply the Fundamental Theorem it is necessary to express this product as a function of the abscissa of -some point in the interval Aa;^. As in § 210, p. 373, we get, using the Theorem of Mean Value, the length of chord (J?) CE^ll+f(x;)'fAx^, where x^ is the abscissa of the point Ii(x^, y^) on the arc CE, where the tangent is parallel to the chord CE. Let the horizontal line through M intersect QI[ at Bj and denote BP^ by e^.^ Then (C) NM=y^-e^. Substituting (5) and (C) in (^), we get 2 7r(t/^— e^) [1 +f'(xy''YAx^= lateral area of first frustum. * The lateral area of the frustum of a cone of revolution is equal to the circumference of the middle section multiplied by the slant height. t The student will observe that as Aajj approaches zero as a limit, e^ also approaches the limit zero. 382 INTEGRAL CALCULUS Similarly, 2 7r(«/ — e ) [1 +f'(^x^'yAx^= lateral area of second frustum, 2 7r(y„— e„)[l4-/'(2'„)''] Aa;„= lateral area of last frustum. Hence V 2 TT (t/j— e,.) [1 +/'(2;i)^]*Aa;,.= sum of lateral areas of frustums. i = l This may be written 1=1 !=1 Third Step. Applying the Fundamental Theorem to the first sum (using the limits OA = a and OB = 6), we get limit n 'I X 2 -^i/, [1 +f'C^^^^^^^ ,^10 work =:62ir j (100 - x^) xdx = 155,000 tt ft. lb. |^ 2. A trough 2 ft. deep and 2 ft. broad at the top has semielliptioal ends. If it is full of water, find the pressure on one end. Ans. 165J lb. ' 3. A floodgate 8 ft. square has its top just even with the surface of the water. Find the pressure on each of the two portions into which the square is divided by one of its diagonals. Ans. 5290f lb., 10,581^ lb. 4. Find the pressure on one face of a submerged vertical equilateral triangle of side 4 ft., one side lying in the surface of the water. Ans. 496 lb. 5. A horizontal cylindrical oil tank is half full of oil. The diameter of each end is 4 ft. Find the pressure on one end if the oil weighs 50 lb. per cubic foot. Ans. 266flb. 390 INTEGKAL CALCULUS 6. Find the work done in pumping out a semielliptical reservoir filled with water. The top is a circle of diameter 6 ft. and the depth is 5 ft. Ans. 34871 tt ft. lb. 7. rind the pressure on the surface of the reservoir in Example 1. 8. Find the pressure on the surface of the reservoir in Example 6. 9. A conical reservoir 12 ft. deep is filled with a liquid weighing 80 lb. per cubic foot. The top of the reservoir is a circle 8 ft. in diameter. Find the energy expended in pumping it out. Ans. 15,360 tt ft", lb. 10. The cross section of a trough is a parabola with vertex downward, the latus •rectum lying in the surface and being 4 feet long. Find the pressure on one end of the trough when it is full of a liquid weighing 62J lb. per cubic foot. Ans. 66 lb. 11. Find the pressure on a sphere 6 feet in diameter which is immersed in water, its center being 10 feet below the surface of the water. Hint. Pressure = 2 td" / y(\(i + x)ds,a,r\6.dx = -ih;. -» ■' Ans. 22320 TT lb. 12. A board in the form of a parabolic segment by a chord perpendicular to the axis is immersed in water. The vertex is at the surface and the axis is vertical. It is 20 feet deep and 12 feet broad. Find the pressure in tons. Ans. 59.52. 13. How far must the board in Example 12 be sunk to double the pressure ? Ans. 12 feet. 14. A water tank is in the form of a hemisphere 24 feet in diameter, surmounted by a cylinder of the same diameter and 10 feet high. Find the work done in pumping it out when filled within 2 feet of the top. 15. The center of a square moves along a diameter of a given circle of radius a, the plane of the square being perpendicular to that of the circle, and its magnitude varying in such a way that two opposite vertices move on the circumference of the circle, tind the volume of the solid generated. Ans. fa'. 16. A circle of radius a moves with its center on the circumference of an equal circle, and keeps parallel to a given plane which is perpendicular to the plane of the given circle. Find the volume of the solid it will generate. 2 a' ° ^ Ans. _!^(3'n- + 8). o 17. A variable equilateral triangle moves with its plane perpendicular to the x-axis and the ends of its base on the points on the curves 2/^ = 16aa; and y^ = iax respec- tively above the x-axis. Find the volume generated by the triangle as it moves from the origin to the points whose abscissa is a. .Jg Ans. a'. 2 18. A rectangle moves from a fixed point, one side being always equal to the dis- tance from this point, and the other equal to the square of this distance. What is the volume generated while the rectangle moves a distance of 2 ft.? Ans. 4 cu. ft. X^ 7/2 19. On the double ordinates of the ellipse — -f- 2- = 1, isosceles triangles of verti- cal angle 90° are described in planes perpendicular to that of the ellipse. Find the volume of the solid generated by supposing such a variable triangle moving from one extremity to the other of the major axis of the ellipse. iab^ Ans. 3 INTEGEATION A PROCESS OF SUMMATION 391 20. Determine the amount of attraction exerted by a thin, straight, homogeneous rod of uniform thickness, 6i length (, and of mass M, upon a material point P of mass m situated at a distance of a from one end of the rod in its line of direction. Solution.* Suppose the rod to be divided into equal infinitesimal portions (ele- ments) of length dx. — = mass of a unit length of rod ; hence — da; = mass of any element. , Newton's Law for measuring the attraction between any two masses is product' of masses force of attraction = ; (distance between them)^ therefore the force of attraction between the particle at P and an element of the rod is — rrub: m w—x—J^"] -l- (x + a) which is then an element of the force of attraction required. The total attraction between the particle at P and the rod being the limit of the sum of all such elements between a; = and x = I, we have — mdx I _ Mm r' dx _ (X + a)2 ~ IT Jo (X + a)2 ~ i , „ ^.*..„ , ™, Mm, , force of attraction = | = | = H Ans. aifl-\- I) 21. Determine the amount of attraction in the last example if P lies in the per- pendicular bisector of the" rod at the distance a from it. . 2 mJf ^ I '^ Ans. arc tan al 2 a 22. A vessel in the form of a right circular cone is filled with water. If ft is its height and r the radius of base, what time will it require to empty itself through an orifice of area a at the vertex ? Solution. Neglecting all hurtful resistances, it is known that the velocity of dis- charge through an orifice is that acquired by a body falling freely from a height equal to the depth of the water. If then x denote depth of water, u = V2 gx. L^ ^,.,^__,_£ __^^i Denote by dQ the volume of water discharged in time dt, \^ ~A_' and by dx the corresponding fall of surface. The volume of Vp^^:::^, — .„__----^| water discharged through the orifice in a unit of time is .4W _ -^ \ a-Jigx, V V| h being measured as a right cylinder of area of base a and alti- \ / x \ tude «(=V25rx). Therefore in time di, X/ ] | {A) dQ = aVigxdt. Denoting by S the area of surface of water when the depth is x, we have, from Geometry, S _x^ vr^ * The two following examples indicate commonly employed " short methods," the detailed exposition followed in the preceding sections being omitted. The student should however supply this. 392 INTEGEAL CALCULUS But the volume of water discharged in time dt may also be considered as the vol- ume of cylinder AB of area of base S and altitude dx; hence {B) dQ = Sdx = —^ Equating (A) and (B) and solving for dt, Trr^z^dx dt = - Therefore * = I "' '" . = " " ' . A-ns. Jo ah?V2gx - " irrVdx _ 2 in''' Vh ah^ 'V2gx 5 a 'v2g 23. A perfect gas in a cylinder expands against a piston head from the volume Dj to the volume Uj, the temperature remaining constant. Pind the work done. Solution. Let c = area of cross section of cylinder. If dv = increment of volume, dv then — = distance piston head moves while volume takes on the increment dv. By Boyle's Law, pv = k {= const.) . k .-. p = - z= pressure on piston head. Hence element of work done = (= pressure x dist.). , . , , , Cikdv k /•■"•dv k, v, . . total work done = / = - | — = - log -1 . CHAPTER XXIX SUCCESSIVE AND PARTIAL INTEGRATION 215. Successive integration. Corresponding to successive differenti- ation in the Differential Calculus we have the inverse process of successive integration in the Integral Calculus. We shall illustrate by means of examples the details of this process, and show how problems arise where it is necessary to apply it. ' Illustrative Example 1. Given — ^ = 6x, to find y, dx Solution. We may write this \dxV dx -""' or, d(^)=6xdx. Integrating, s=/--- or, s=--^- This may also be written 4r) ^^' 3x^ + c dx ^''^'''' or, d{^ = {Zx^ + c^)dx. Integrating again, ^=f(3x^ + c,)dx,or. w ^ = x^ + c^a; + c,. Again % = (a;3 + CjX + Cj) dx, and integrating, {B) -w4 /. 2*2 y = — + -y- + CgX + Cg. Ans. The result (^) is also written in the form = I I 6 xdxdx (or = l I 6 xdi?'). dy dx and is called a dovhle integral, while (^) is written in the form y = 1 I { Q xdxdxdx (or = 1 I I 6 xdx^'), 393 394 INTEGRAL CALCULUS and is called a triple integral. In general, a multiple integral requires two or more successive integrations. As before, if there are no limits assigned, as in the above example, the integral is indefinite ; if there are limits assigned for each successive integration, the integral is definite. Illustrative Example 2. Find the equation of a curve for every point of whicli the second derivative of the ordinate with respect to the abscissa equals 4. Solution. Here — ^ = 4. Integrating as in Illustrative Example 1, (C) | = ''- + «^- (D) y = 2x2 + Cja; + Cj. Ans. This is the equation of a parabola with its axis parallel to OT and extending upward. By giving the arbitrary constants of integration Cj and Cj all possible values, we obtain all such parabolas. In order to determine Cj and c^, two mpre conditions are necessary. Suppose we say (a) that at the point where x = 2 the slope of the tangent to the parabola is -zero ; and (b) that the parabola passes through the point (2, — 1). (a) Substituting x = 2 and — = in (C) dx gives = 8 + Cj. Hence c^ = — 8, and (D) becomes y = 2x^ — Bx + Cj. (b) The coordinates of (2, — 1) must satisfy this equation ; therefore -1 = 8-16+ c,, or, C2 = + 7 Therefore the equation of the particular parabola which satisfies all three con- ditions is 2/ = 2x2-8x + 7. EXAMPLES 1. Given ^ = 0x2, find y. Ans. y = ^ + ^ + c„x + c,. (Jx^ 60 2 ' 2- Gi^en g = 0, find y. ^ = ¥ + "^'^ + '«• 3. Given d^y = — ^ , find j/. 2/ = logi + £l^ + c^x + c,. 5. Giveng. 3.2-1, find. a = ^1 - llog* + £^ + c,i + c, 6. Given d^p = sin cos^^d^^, find p. p = 5l^ - 1 sin + c^^ + c^. 7. Determine the equations of all curves having zero curvature. Hint. —^ = 0, from (40), p. 157, since ^=0. Ans. y = c^x + Cj, a doubly infinite system of straight lines. SUCCESSIVE AND PARTIAL llNTEGEATION 895 8. The acceleration of a moving point is constant and equal to /; find the distance (space) traversed. Hint. -— =/. Ans. s = :!— -\- cA + c„. dfi 2 9. Show in Ex. 8 that Cj stands for the initial velocity and Cj for the initial distance. 10. Find the equation of the curve at each point of which the second derivative of the ordinate with respect to the abscissa is four times the abscissa, and which passes through the origin and the point (2, 4) . Ans. Sy = 2x(x^ — 1). 11. Given — _ = a; cos a;, find y. Ans. w = a; cosx — 4sinx + -J ^— \- c„x + c,. 12. Given — - = sin»x, find y. Ans. y — y — y c„x + c„. 216. Partial integration. Corresponding to partial differentiation in the Differential Calculus we have the inverse process of partial integration in the Integral Calculus. As may be inferred from the connection, partial integration means that, having given a differ- ential expression involving two or more independent variables, we integrate it, considering first a single one only as varying and all the rest constant. Then we integrate the result, considering another one as varying and the others constant, and so on. Such integrals are called double, triple, etc., according to the number of variables, and are called multiple integrals.* Thus the expression ^^ « = ( //(a:, y'ydydx indicates that we wish to find a function m of 2; and y such that dxdy =/(-P' yy In the solution of this problem the only new feature is that the constant of integration has a new form. We shall illustrate this by means of examples. Thus suppose we wish to find u, having given Integrating this with respect to x, considering y as constant, we '^^ u = T'-hxy + Sx + cj), * The integrals of the same name in the last section are special cases of these, namely, when we integrate with respect to the same variable throughout. 396 INTEGRAL CALCULUS where

will in general be a function of y. We shall then indicate this dependence of <^ on z/ by replacing by the symbol <^ (?/). Hence the mo^t general form of u is where ^(y) denotes an arbitrary/ function of y. As another problem let us find iA) u=JJ(x^+f}dydx. This means that we wish to find u, having given d\ „ „ dxdy ^ Integrating first with respect to y, regardmg x as constant, we get - = ^y^\ + ^ix), where -^(x) is an arbitrary function of x and is to be regarded as the constant of integration. Now integrating this result with respect to a;, regarding y as con- stant, we have a 3 " = ^ + ^ + ^(^)+(j/), where ^ («/) is the constant of integration, and ■^(ix')=(-<^(x)dx. 217. Definite double integral. Geometric interpretation. Let/(a;, y) be a continuous and single-valued function of x and y. Geometrically, is the equation of a surface, as KL. Take some area S m the XY- plane and construct upon >S as a base the right cylinder whose elements are accordingly parallel to OZ. Let this cylmder intersect KL in the area ;S', and now let us find the volume V of the solid bounded by S, S', and the cylindrical surface. We proceed as follows : At equal distances apart (= Ax} m the area S draw a set of hues parallel to OY, and then a second set parallel to OX at equal distances apart (= Ay'). Through these lines pass planes parallel to YOZ and SUCCESSIVE AND PAETIAL INTEGE^TION 397 F^ XOZ respectively. Then within the areas 8 and S' we have a net- work of lines, as in the figure, that in S being composed of rectangles, each of area Lx-tl^y. This construction divides the cylmder into a number of vertical columns, such as MNPQ, whose upper and lower bases are corresponding portions of the networks in S' and S respectively. As the upper bases of these columns are curvi- liaear, we of course can- not calculate the volume of the colunms directly. Let us replace these col- umns by prisms whose upper bases are found thus: each column is cut through by a plane parallel to Xr passed through that vertex of the upper base for which x and y have the least numerical values. Thus the column MNPQ is replaced by the right prism MNPB, the upper base being in a plane through P parallel to the XOZ-plane. If the coordinates of P are (x, y, z), then MP = z =f(x, y}, and therefore (B) volume of MHPR =f(x, y') Ay • Ax. Calculating the volume of each of the other prisms formed in the same way by replacing x and y in (5) by corresponding values, and adding the results, we obtain a volume V approximately equal to V; that is, ((7) r'=^'^fCx,y')Ay-Ax; where the double summation sign ^V indicates that there are two variables in the quantity to be summed up. If now in the figure we increase the number of divisions of the network in S indefinitely by letting Ax and Ay diminish indefinitely, and calculate in each case the double sum (C), then obviously F' will approach F as a limit, and hence we have the fundamental result limit __■__. (Z>) F = Ay = y y /(a;, y-) Ay • Ax. 398 INTEGRAL CALCULUS The required volume may also be found as follows : Consider any one of tlie successive slices into which the solid is divided by the planes parallel to YZ; for example, the slice whose faces are FIGH and TLJK. The thickness of this slice is Aa;. Now the values of z along the curve HI are found by writing x = OB in the equation z=f(x, y); that is, along HI z=fiOB,y-). fiPD, y)dy. DF The volume of the shoe under discussion is approximately equal to that of a prism whose base is FIGH and altitude Ax; that is, equal to r^(f Ax ■ area FIGH = Ax j /( OD, y) dy. J DF The required volume of the whole sohd is evidently the limit of the sum of all prisms constructed in like manner, as x (= OD) varies from OA to OB ; that is, ^^ ^^ (^) r= f dxf f{x,y-)dy. J OA JdF Similarly, it may be shown that Jr^ov pEU dy] f(x,y')dx. oc Jew The integrals (^) and (#) are also written m the more compact form (^OB pBG rtOV pEU / / f(p,y^dydx and | I f(x,y')dxdy. JoA J DF Joe Jew In {E') the limits DF and DG are functions of a;, since they are found by solving the equation of the boundary curve of the base of the solid for y. Similarly, in (i^) the limits FW and FT! are functions of y. Now comparing (X)), {E'), and (J^) gives the result limit ,^,^ /•"i r^ (G) F= A2/ = ^y^%S(x, y)Ay.Ax= / fQx, y) dydx A^ — " Ja^ Ju^ I /(»' y) dxdy, &, Jv„ where v^ and v^ are, in general, functions of y, and u^ and u^ functions of X, the second integral sign applying to the first differential and being calculated first. SUCCESSIVE AND PARTIAL INTEGIIAT1(3N 399 Our result may be stated in the following form : The definite double integral may he interpreted as that portion of the volume of a truncated right cylinder which is included between the plane XOY and the mirface the base of the cylinder being the area bounded by the curves y = Uj, y==u^, x = a^, x=a^. Similarly for the second integral. It is instructive to look upon the above process of finding the vol- ume of the solid as follows: Consider a column of infinitesimal base dydx and altitude z as an element of the volume. Summing up all such elements from y = DF to y= DG, X in the meanwhile being consjiant (say = 0-D), gives the volume of an indefinitely thin slice having FGHI as one face. The volume of the whole solid is then found by summing up all such slices from x = OA to 2; = OB. In partial integration involving two variables the order of integra- tion denotes that the limits on the inside integral sign correspond to the variable whose differential is written inside, the differentials of the variables and their corresponding limits on the integral signs being written in the reverse order. Wa'-=? Illustrative Example 1. Find the value of the definite double integral Jo Jo Jo Jo (x + y) dydx. Solution. / I (a; + y) dydx, L Vo^-a^ (x + y)dy\dx 2/2 "1 -v „ _ XJ/ + 2- dx ^ Jo 2 a' . Aw. Interpreting this result geometrically, it means that we have found the volume of the solid of cylindrical shape standing on OAB as base and bounded at the top by the ■ surface (plane) z = x + y. 400 INTEGRAL CALCULUS The attention of the student is now particularly called to the manner in which the limits do hound the base OAB, which corresponds to the area S in the figure, p. 397. Our solid here stands on a base in the XF-plane bounded by y = (line OB) y =Va^ — x^ (quadrant of circle AB) x = aineOA)'] . ,. ., ,,. -r,™ MromalinutB. ► from y limits; x = a (line BE) _ 218. Value of a definite double integral over a region S. In the last section we represented the definite double integral as a volume. This does not necessarily mean that every definite double integral is a vol- ume, for the physical interpretation of the result depends on the nature of the quantities represented by .r, y, z. Thus, i£x, i/, z are simply con- sidered as the coordinates of a point in space, and nothing more, then the result is indeed a volume. In order to give the definite double iiitegral in question an interpretation not necessarily involving the geometrical concept of volume, we observe at once that the variable z does not occur explicitly in the integral, and therefore we may confine ourselves to the JTY-plane. In fact, let us consider simply a region S in the XY-plane, and a given function ¥ f(x, y). Then, drawing a network as be- fore, calculate the value of f(x, y')^.y^x for eacih point* of the network, and sum up, findmg in this way and finally pass to the limit as Aa; and A«/ approach zero. This opera- tion we call integrating the function f (x, y^ over the region S, and it is denoted by the symbol rr S If S is bounded by the curves x = a^,x = a^,y = u^, y - u^, then, by (G}, jjfi^^ y') ^y^^ =f J f(.^, y) dydx. s "^ "' * More generally, divide the interval on OX into subintervals Aa:i, Axg, • ■ ■ , Aas», and on Orinto AiJi, Aj/2, ■ ■ • , Aj^m. Draw the network, and in each rectangle AxtAi/t (not necessa- rily a comer) choose a point Xi, yt. Then it Is clear intuitionally that jjf (a:, y) dxdy = ™ ^ ^ ^ X"^ ^^'' ^'' ^'^<^V''- SUCCESSIVE AND PAKTIAL INTEGRATION 401. We may state our result as follows : Theorem. To integrate a given function f (x, y) over a given region S in the XOY-plane means to calculate the value of limit ^ ^ Aj/ = ^ ■^ a% explained above, and the remit is equal to the definite double integral f f f(pc,y')dydx, or, I / f(x,y')dxdy, the limits being chosen so that the entire region S is covered. This process is indicated briefly by pp jjf(.^' y) ^y^'^- s In what follows we shall show how the area of the region itself and its moment of inertia may be calculated in this way. Before attempting to apply partial integration to practical problems it is best that the student should acquire by practice some facility in eyaluating definite multiple integrals. I {a — y) x'dyclx — Jr»26 /.a /*^*r y^~\^ r^^a^ I (a — y) xMydx = ( \ay — — \ x^dx — \ — x^da; 6 •'0 Jh \_ 2 Jo Jb % /» a /. Va^ — o? 2 d^ Illustrative Example 2. Verify ( / xdydx = Ja'^ — x" [•a c ■\la^ - a? /■of ~\-\/a?-ay' Solution. I ( xdydx = | \xy \ dx -'o J_^y^r^ Jo I J_v5r3 • = p2xVa2-x2dx =["- ?/a2 - »tT= -a^- In partial integration involving three variables the order of inte- gration is denoted in the same way as for two variables ; that is, the order of the limits on the integral signs, reading from the inside to the left, is the same as the order of the corresponding variables whose differentials are read from the inside to the right. I / xy^dzdydx = — 2 Ji Ji 2 Xa y->a rtij nZ n^r /»G ~| rtS /»2r "15 / / xyHzdydx = I | / xyHz dydx = ( I xy'z dydx Jl Ji •/2 •'1 \_J2 J J2 J\ \_ J2 = 3 r r zyHydx = sf If xyHy\dx 35 2 ' 402 Verify the following : 1. I I xy(x-y)dydx = -—-(a-0). Jo Jo o 2. ( ''rdedr = L^. Jb_ Ja 24 2 na ply 11a* 3. I / zydadv = ---- Jo Jy-a 24 & t/o •^a 6 INTEGRAL CALCULUS. EXAMPLES pa pV^ 2 S 10. I / (i2/da; = -o2. Jo Jo o r" r^^ 2 ,/ai 4\ ll.££x.^d.^ = -a=(---) pa px py a' 12. I I I x^yHdzdydx — — Jo Jo Jo 9( Jo Jo 09 90 ■ 13 -a /.V5^ (!z(2x "20 p^2ax—3? x'W Vox — Z^ f7.S = 4a2 3ira3 5. I I I " dzdydx=^ Jo Jo Jo 4 /.n- /ia(l + C08fl) „ 4a^ 6. f f r^sva.9drde = ^- Jo Jo 3 /»6 /»10( ^ 7. I I Vst-t2dsctt = 6 6^ 143 «« 3 16 a* 8. ("" r«(u) + 2B) pl prP^ <£ 9. I ( e^dwdv = Jo Jo dwdv: 1 2' 30 ^(X'^y)dxdy = -^ -a«/0 5 nir /»acoB0 a^ 15. I I psmmpde=i-~ Jo c/0 3 16. r " f''(x^-\-y^)dydx-. Jo t/o JoJ^x2 + 2,2 2 a -/•■/.:./'*»«=('-ii)s- Jn a /^ a: (j^ — ^3 / r^ sin S'JSdr = — - — (oos p — cos a), b Ja 3 20 Jb Jp il px pX + I j ef^+v + 'dzdydx^ Jo Jo e*-3 3e2 h e. 8 4 I I (x2 + 2/2 + z2)d2dyda; = -^(a2 + 62 + c2). «/o t/o 3 /» 6 /• 10 ?/ , 22. ( I -Vxy - y^dxdy = 61)^. Jo J II ■ Ji Jo Jo x2 + J/2 ~ 2 ■ 219. Plane area as a definite double integral. Rectangular coordinates. As a simple application of the theorem of the last section (p. 401'j, we shall now determine the area of the region S itself in the XOF-plane by double integration.* * Some of the examples that -will be given in this and the following articles may be solved by means of a single integration by methods already explained. The only reason in such cases for using successive integration is to familiarize the student with a new method for solution which is sometimes the only one possible. SUCCESSIVE AND PAETIAL INTEGEATION 403 As before, draw lines parallel to OX and OY at distances Aa; and Ay respectively. Now take any one of the rectangles formed in this way, then element of area = area of rectangle PQ = t^y- Aa;, the coordinates of P being (x, y). Denoting by A the entire area of region S, we have, using the motion of a double summation, (^) limit Ay iXX'^y^''- We calculate this by the theftrem on p. 401, setting /(a;, «/) = !, and get (^) OA J CD dydx. Y T, F , ^ ' \ " j ! Q u / A f / ' ,/ \M y D^^ ^ ' .. A U B ^ where CD and CU are, in general, functions of x, and OA and OB are constants giving the extreme values of x, all four of these quantities being determined from the equations of the curve or curves which bound the region S. It is instructive to interpret this double integral geometrically by referring to our figure. When we integrate first with respect to y, keeping x (= OC) constant, we are summing up all the elements in a vertical strip (as BPy. Then integrating the result with respect to x means that we are summing up all such vertical strips included in the region, and this obviously gives the entire area of the region iS. Or, if we change the order of inte- gration, we have (C) / OK J HO dxdy. i 7- __^ -^ r- _\ _) Q Jt -^ M M ^ ^ m m M hi : / vl / K - — ' — ■ U- V- C where HG and HI are, in general, func- tions of y, and OK and OL are constants giving the extreme values of y, all four of these quantities being determined from the equations of the curve or curves which bound the region S. Geometrically, this means that we now commence by summing up all the elements in a horizontal strip (as GJ~), and then find the entire area by summing up all such strips within the region. 404 INTEGRAL CALCULUS Corresponding to the two orders of summation (integration), the following notation and figures are sometimes used: (^) A = j I dyclx, ^ = / I dxdy. S S Referring to the result stated on p. 401, we may say : The area of any region is the value of the double integral of the function f(x, y) = 1 taken over that region. Or, also, from § 217, p. 396, The area equals numerically the volume of a right cylinder of unit height erected on the base S. Illustrative Example 1. Calculate the area of the circle x^ + y'' = r^ by double integration. Solution. Summing up first the elements in a vertical strip, we have from (B), p. 403, ^OA nMR A= \ \ dydx. Job J MS From the equation of the boundary curve (circle) we get Hence JlfJS=Vr2-x2, MS OB =—r, OA = r. -V, -Vr- -x% A= I I dydx XVr^ — x'^'dx = Trr'. Ans. ■r When the region whose area we wish to find is symmetrical with respect to one or both of the coordinate axes, it sometimes saves us labor to calculate the area of only a part at first. In the above example we may choose our limits so as to cover only one quadrant Qf the circle, and then multiply the result by 4. Thus A ^r' dyd.> '-[ ^? 2 7 " ' 4 '■ A= 7rr^. Ans. SUCCESSIVE AND PARTIAL INTEGRATION 405 Ilmsteative Example 2. Calculate that portion of the area which lies above OX bounded by the semlcubioal parabola y^ = x' and the straight line y = x. Solution. Summing up first the elements in a horizontal strip, we have from (C), J'-UJJ pAV I dxdy. J AB From the equation of the line, AB = y, and from the equation of the curve, ^C = y^, solving each one for x. To determine OD, solve the two equations simultaneously to find the point of intersection £. This gives the point (1, 1); hence 0D=1. Therefore A=f r'^y=C\yi-y)ay=\'4^yI\ Jo Jy Jo L 5 2jo EXAMPLES 1. Find by double integration the area between the straight line and a parabola with its axis along OX, each of which joins the origin and '^/^ i thepoint (a, 6). ^^s. Cf Jo Jbx dydx = — 6 2. Find by double integration the area between the two parabolas Sy^ = 25 x and 5a;2 = 9y. Ans. 5. 3. Required the area in the first quadrant which lies between the parabola y^ = ax and the circle y^ = 2ax — x^. , ira?' 2 a^ Ans. 4. Solve Problems 2 and 3 by first summing up all the elements in a horizontal strip, and then summing up all such strips. TTO? 2ar ~3~ Ans. Ex. 2, ( I V i^j^^ ^ g ex. 3, | / " dxdy = — - Jo Jsy^ Jo Ja_VS5Z^ 4 5. Find by double integration the areas bounded by the following loci : (a) xi + yi = aJ, x + y = a. (b) 2/2 = 9 + x, 2/2 = 9_3x. (o) y = sin x,y = cos x, x = 0. 8 a' Ans. (d)y = - 2^ = K, x = 0. x2 + 4a2 (e) x^ + y^ = a^,x-\- y = a. (f) 2/2 = x + 4, s^2 = 4-2x. (g) y^ = ia^ — x'^, y' = i.a^ — i ax. (h) x2 + 2/2 = 25, 27 2/2 = 16 x^. (i) iy^ =x', y = x. (k) a;2 - 2/2 = 14, x2 + 2/2 = 36. V^- -1. o2(7r -!)• a2 2 37ra2 32 406 INTEGRAL CALCULUS 220. Plane area as a definite double integral. Polar coordinates. Suppose the equations of the curve or curves which bound the region S are given in polar coordinates. Then the region may be divided into checks bounded by radial lines drawn from the origin, and concentric circles drawn with centers at the origin. Let PS=Ap and angle POS = A0. Then arc PE = pA0, and the area of the shaded check, considered as a rectangle, is pAO ■ Ap. The sum of the areas of all such checks in the region will be ^pApAe. Since the required area is evidently the limit of this sum, we have the formula (A) A=CCpdpde. S Here, again, the summation (integration) may be effected in two ways. When we integrate first with respect to 0, keeping p constant, it means that we sum up all the elements (checks) in a segment of a circular ring (as ABCU), and next integrating with respect to p, that we sum up all such rings within the entire regfon. Our limits then appear as follows: '■^ pOF f^ angle XOB (5) A= [ { pd0dp, Joe JimgleXOA the angles XOA and XOB beuig, in general, functions of p, and OH and OF constants giving the extreme values of p. Suppose we now reverse the order of integration. Integrating first with respect to p, keeping constant, means that we sum up all the elements (checks) in a wedge-shaped ,/n. l h strip (as GKLH^. Then integrating with respect to 0, we sum up all such strips within the region S. Here J jangle A'07 pOH ] pdpd0, OH and OG being, in general, functions of 0, and the angles XOJ"and XOI being constants giving the extreme values of 0. SUCCESSIVE AND PAETIAL INTEGEATION 407 Corresponding to the two orders of summation (integration), the following notation and figures may be conveniently employed : (■O) A= ffpdpde, A=ffpdddp. s s These are easily remembered if we think of the elements (checks) as being rectangles with dimensions pd6 and dp, and hence of area pdOdp. Illustrative Example 1. Find tile area, of the circle p = 2r aosd by double integration. Solution. Summing up all the elements in a sector (as OB), the limits are and 2 r cos 6 ; and summing TT up all such sectors, the limits are and — for the q semicircle OXB. Substituting in (D), = f I pdpd9 = ^—, or, A = m-^. Ans. EXAMPLES 1. In the above example find the area by integrating first with respect to 9. 2. rind by double integration the entire areas in Examples 1-16, pp. 368, 369. /? 3. rind by double integration the area of that part of the parabola p = a sec^ - intercepted between the curve and its latus rectum. ir ,» „ , Ans. 2f^^£'^-^pdpdd = ^-f. 4. Find by double integration the area between the two circles p = a cos ^, p=b cos^, b>a: integrating first with respect to p. , ^^ nbcoae „ tt „ Ans. 2 ( -= ( pdpdd = —{b^ — a^). Jo Jacose 4 5. Solve the last problem by first integrating with respect to 0. 6. Find by double integration the area bounded by the following loci : (a) p = 6 sin 9, p = 12 sin 9. Ans. 27 tt. (b) p cos 9 — 4:, p = S. (c) p = a sec^ - , (0 = 2 a. (d) (0 = a (1 + COS ^), p = 2 a cos ^. (e) psax9 = h, p = 10. (f) /Q = 8 cos ^, p cos 9 — 2. (g) p = 2 cos ^, p = 8 cos ^. 64 TT 3 2aV- -16V3. 8a2 .3 ■KO? 2 408 INTEGRAL CALCULUS - 221. Moment of area. Consider an element of the area of the region S, as FQ, the coordinates of P being (x, y). Multiplying the area of this element (= AyAa;) by the distance of P from the T-axis (=»), we get the product which is called the moment of the element PQ with respect to the Y-axis. Form a similar product for every element within the region and add all such products by a double summation. Then the limit of this sum, namely, limit __v_-\ rr (5) A2/ = X Xa^AyAa; = I I xdydx, defines the moment of area of the region S with respect to the Y-axis. Denotiag this moment by M^, we get Y y1 p ■>, s / r— p^ r y r \ a; Q J I 1 I / 7 / P"^ ■C+A^ r 2/^ ^ -1 U' i (C) M^=jj xdydx, the limits of iutegration beiag determined in the same way as for finding the area. In the same manner, if we denote the moment of area with respect to the X-axis by M^ we get (2?) M^=jjydydx* the limits being the same as for (C). 222. Center of area. This is defined as the point (i, y^ given by the formulas (-E) (F) x = - 3l, area y = > or, area fC xdydx ffi ydydx From (£), Ifdydx ffdydx area ■x = M^ and area • y = M^. Hence, if we suppose the area of a region to he concentrated at (x, y^, the moments of area with respect to the coordinate axes remain unchanged. * From the result on p. 401 we may say that Mx is the value ol the double integral of the function/(3;, y) = rj taken over the region. Similarly, My is the value when/(a;, y) = x. SUCCESSIVE AND PAETIAL INTEGRATION 409 The center of area of a thin homogeneous plate or lamina is the same as its center of mass (or center of gravity^.* If a coordinate axis is an axis of symmetry of the area, it is evident that the corresponding coordinate of the center of area will be zero. In polar coordinates x = p cos 6, y = p sin 6, and element of area = pA.p/\d replaces Ay^x. Hence formulas (^F) become (G) . //' yo" COS ddpd6 x = - If' p" sin Sdpde , y = ffpdpdO ffpdpdO the limits being the same throughout and determined (as before) in the same way as for finding the area. Illustrative Example 1. Find the center of the area bounded by j/^ = 4 x, x = 4, ?/ = 0, and lying above OX. 1 •.4 /.2a:' Solution. By'(C),p.408, Jlfj,= f f "^ zdy6x = ^^. By (D), p. 408, M^=f f "^ ydydx = 16. Area = | I dydx = */. Substituting in (E), p. 408, x = i|i-^^^ = Y, and 1/ = 16-Y = I- ^^■ EXAMPLES 1. Find the centers of the areas bounded by the following loci : (a) The quadrant of a circle. (b) The quadrant of an ellipse. .(c) y = sinx, y = 0, from x = to x = tt. (d) A quadrant of x^ + y^ = a^. (e) y^ = iax, X = h. (f) y = 2x,y = 0,x = S. (g) y^ = Sx,y = 0,y + x = 6. (h) (2a-x)y^ = x^,x = 2a. (i) y^(a^-x^)-a\x = 0. (j) xi + yi = a^,x = 0,y = 0. (k) Cycloid x = a (i? - sin i9), 2/ = a (1 - cos (9). » If the plate is supported loosely on a horizontal axis through its center of gravity, there ■will be no tendency to rotate, whatever the position of the plate may be. Ans. X ir ~3t = y X _4a ~ Sir V 46 "Stt X IT ~2' y = TT ' X 256 315 a IT V- X -¥• y = :0. X = 2 = V- X = 2.48, y = 1.4 X 5a v^ = 0. X 2a y-- = 0. X X a = air. y- y = ha ■410 INTEGRAL CALCULUS 2. Find flie centers of the areas bounded by the following curves : (a) One loop of p^ = a^ cos 2 5. (b) One loop of p = a sin2 tf. Avs. X = , y = 0. 128 a 105 IT 5a (f) p = 8cosS, /ocos5 = 2. (g) p = 2 cos 9, p = 9icos0. = y- y = 0. (c) Cardioid p = a(l + cosS). (d) p = 6 sin e, p = 12 sin d. (e) p cos ff = 4s, p = S. 223. Moment of inertia of plane areas. Consider an element of the area of the region S, as FQ, the coordinates of P being (x, y). Mul- tiplying the area of this element (= At/Aa;) by the square of the distance (= x) of P from the F-axis, we get the product (J) o^AyAx, which is called the moment of inertia* of the element PQ with respect to the F-axis. Form a similar product for every element within the region and add all such products by a double summation. Then the hmit of this sum, namely limit (^) Ax : Ay ■- 'XX^ ■^AyAx =11 x^dydx. defines the moment of inertia of the area of S with respect to- the Y-axis. Denoting this moment by I^, we get (C) Iy=jjx'dydx, the limits of mtegration bemg determined in the same way as for finding the area. In the same manner, if we denote the moment of inertia of the area with respect to the X-axis hy I^, we get (-0) ,=jjfdydx, the limits being the same as for (C). 224. Polar moment of inertia. Rectangular coordinates. Consider an element of the area of region S, as P Q. I f the coordinates of P are (x, 2/), the distance of P from is V?+ y^. Multiplying the area of * Because the element of area is multiplied by the square of Its distance from the I'-axis it is sometimes called the second moment, to conform with the definition of moment of area (§221, p. 408). SUCCESSIVE AND PARTIAL INTEGRATION 411 element (= Aj/Aa;) by tlie square of the distance of P from- the origin,' we have the product q^^ + y^ ^y^^ which is called the polar moment of inertia of the element PQ with respect to the origin. The value of the double sum (^) £= lZXi^+ /)A2/A2; =jj(x'+f) dydx defines the polar moment of inertia of the area within the region S with respect to the origin. Denoting this moment of inertia by /„, we get the limits of integration being determmed in the same way as for finding the area. From (F), /„ = / / (a;^ + y^) dydx = I ix^dydx + j j y^dydx. By comparison with (C) and (U) we get (S" lying on the surface. Denote by S the region on the XOF-plane, which is the- orthogonal projection of S' on that plane. Now pass planes parallel to YOZ and XOZ at com- mon distances Aa; and ^y respectively. As in § 217, these planes form truncated prisms (as P5) bounded at the top by a portion (as PQ) of the given surface whose projection on the XOF-plane 414 INTEGRAL CALCULUS is a rectangle of area Aa;Az/ (as ASy, which rectangle also forms the lower base of the prism, the coordinates of F being (ps, y, z). Now consider the plane tangent to the surface KL at P. Evidently the same rectangle AB is the projection on the -XOZ-plane of that portion of the tangent plane (-P-B) which is intercepted by the prism FB. Assuming 7 as the angle the tangent plane makes with the XOF-plane, we have area AB = area FB ■ cos 7, ["The projection of a plane area upon a second plane is equal to the area of the] [portion projected multiplied by the cosine of the angle between the planes. J or, But AyAx = area FB ■ cos 7. 1 cos 7 = ■ Hmm ["Cosine of angle between tangent plane, (72), p. 266, and A'O Y-l [plane found by method given In Solid Analytic GeometryJ hence or, A«/Aa; = ■ area FB ■V / dx) '^\8y^ area FB =[-(£HI)' AyAa;, which we take as the element of area of the region S'. We then define the area of the region S' as limit , , r Ay=oyy\i+ ZnAA Aa; = (£J- dz Ai/Ax, the summation extending over the region ;i5, as in § 217. Denoting by A the area of the region (S^', we have -//.h(S)'-(S)T*-. the limits of integration depending on the projection on the XOY-plane of the region whose area we wish to calculate. Thus for (fi) we choose our limits from the boundary curve or curves of the region S in the XOr-plane precisely as we have been doing in the previous four sections. SUCCESSIVE AND PARTIAL INTEGRATION 415 If it is more convenient to project the required area on the XOZ- plane, use the formula s where the limits are found from the boundary of the region 8, which is now the projection of the required area on the JTO^-plane. Similarly, we may use ' -//[-(l)'HS)t dzdy, the limits being found by projecting the required area on the YOZ- plane. In some problems it is required to find the area of a portion of one surface intercepted by a second surface. In such cases tlie partial derivatives required for substitution in the formula should be found from the equation of the surface whose partial area is wanted. Since the limits are found by projecting the required area on one of the coordinate planes, it should be remembered that To find the projection of the area required on the XO Y-plane, elimi- nate z between the equations of the surf aces whose intersections form tJie loundary of the area. Similarly, we eliminate y to find the projection on the XOZ-plane, and X to find it on the YOZ-plane. This area of- a surface gives a further illustration of integration of a function over a given area. Thus in (E), p. 414, we integrate the function ^ ,2 /a„\2lt' HMD] ■ over the projection on the XO y-plane of the required curvilinear surface. Illusikative Example 1. Find the area of the surface of sphere x^ + y^ + z^ = r^ by double integration. Solution. Let ABC in the figure be one eighth of the surface of the sphere. Here dz _ X 8z __y 8x~ z' Sy~ z' j;2 yi x^ + y^ + e'^ _ r- Wl \dyj z" z^ ; r' — x^ — 'i 416 INTEGRAL CALCULUS The projection of the area required on the XOT-plane is AOB, a region bounded by a; = 0, {OB) ; y = 0, (OA) ■ x^ + y^ = r\ {BA). Integrating first with respect to y, we sum up all the elements along a strip (as DEFG) which is projected on the XOZ-plane in a strip also (&s MNFG ); that is, our y-limits are zero and MF ( = Vr" — x^) . Then integrating with respect to X sums up all such strips composing the surface ABC; that is, our x-limits are zero and OA{=r). ^ Substituting in (B), we get or. A _ r'- /-v^ 8 ~ Jo Jo A = iTTT^. Ans. rdydx Vr^- ^X /B FG Illustrative Example 2. The center of a sphere of radius r is on the surface of a right cylinder, the radius of whose base is -• Pind the surface of the cylinder intercepted by the sphere. Solution. Taking the origin at the center of the sphere, an element of the cylinder for the z-axis, and a diameter of a right section of the cylinder for the x-axis, the equation of the sphere is x^ ■\-y^ + z^ = r^, and of the cylinder x^ ■\-y^ = rx. OBAPB is evidently one fourth of the cylindrical surface required. Since this area projects into the semicircular arc ODA on the XOy-plane, there is no region S from which to determine our limits in this plane ; hence we will project our area on, say, the XOZ-plane. Then the region S over which we integrate is OACB, which is bounded by z = 0, (OA) ; x = 0, (OB) ; z2 + rx = r^, {ACB) ; the last equation being found by eliminating y between the equations of the two surfaces. Integrating first with respect to z means that we sum up all the elements in a verti cal strip y^ (as PD), the z-limits being zero and Vr^ — rx. Then on integrating with respect to x we sum up all such strips, the x-limits being zero and r. Since the required surface lies on the cylinder, the partial derivatives required for formula (C), p. 415, must be found from the equation of the cylinder. Hence dy _r —2x dx 2j/ — = 0, dz Substituting in (C), p. 415 A T Substituting the value of y in terms of x from the equation of the cylinder, A=2r I =2r '■^ d.r. = 2r ^-dx = 4:r-^. Jo Jo Vrx — x^ ■'0 Vrx — x^ -^o ^^ ^ i < SUCCESSIVE AND PAETIAL INTEGRATION 417 Ans. 8W / •■ -' dV^ ^srK EXAMPLES 1. In the preceding example find the surface of the sphere intercepted by the cylinder. r /.v';jZ^ dydx J Jq 'vr^ x^ y^ 2. The axes of two equal right circular cylinders, r heing the radius of their bases, intersect at right angles. Pind the surface of one intercepted by the other. Hint. Take x'^ + z^=r' and a!^ + j/2 = 7-2 as equations of cylinders. rTf ., Jo f/o y'j.a /p2 3. Find by integration the area of that portion of the surface of the sphere x^ + y^ + z^ = 100 which lies between the parallel planes x = — 8 and x = 6. 4. Find the surface of the cylinder x^ + y^ = r^ included between the plane z = mx and the XOF-plane. Ans. ir^m. 5. Find the surface of the cylinder z^ + (x cos a + y sin a)^ = r' which is situated in the positive compartment of coordinates. Hint. The axis of this cylinder Is the line z = 0, x cos cc + y sin or = ; and the radius of base is r. . r^ Ans. sin a cos a 6. Find the area of that part of the plane - + ^ + - = 1 which is intercepted by the coordinate planes. Ans. i^b^c^ + cV + a^t^. 7. Find the area of the surface of the paraboloid y^ -\- z"^ = iax intercepted by the , parabolic cylinder 2/2 =. CKC and the plane X = 3 a. Ans. ^-^-ira^. 8. In the preceding example find the area of the surface of the cylinder inter- cepted by the paraboloid and plane. . /- „ /To . Y a^ V3 9. Find the area of that portion of the surface of the cylinder j/* + z* = a* bounded by a curve whose projection on the XF-plane is x*' + ^^ = a^. Ans. ^ a?. 10. Find the area of that portion of the sphere x^ + y^ ^ ^^ = 2 a^/ cut out by one nappe of the cone x^ + «^ = 2/^- ^™«- 2 ttb^. 227. Volumes found by triple integration. In many cases the vol- ume of a solid bounded by surfaces whose equations are given may be calculated by means of three successive integrations, the process being merely an extension of the methods employed in the preceding sections of this chapter. Suppose the solid in question be divided by planes parallel to the I coordinate planes into rectangular parallelepipeds having the dimen- sions As, At/, Aa;. The volume of one of these parallelepipeds is A2 • Ay • Aa;, and we choose it as the element of volume. Now sum up all such elements within the region B bounded by the given surfaces by first summing up all the elements in a column 418 INTEGRAL CALCULUS parallel to one of the coordinate axes ; then sum up all such columns in a slice parallel to one of the coordinate planes containing that axis, and finally sum up all such slices within the region in question. The volume V of the solid will then be the limit of this triple sum as Az, At/, Aa; each approaches zero as a limit. That is, limit the summations being extended over the entire region E bounded by the given surfaces. Or, what amounts to the same thing. =/// dzdydx, R the limits of integration depending on the equations of the bounding surfaces. Thus, by extension of the principle of § 218, p. 401, we speak of volume as the result of integrating the function f(x, y,z) = l through- out a given region. More generally, many problems require the integra- tion of a variable function of x, y, and z throughout a given region, this being expressed by the notation ///■ f(x, y, z) dzdydx. R which is, of course, the limit of a triple sum analogous to the double sums we have already discussed. The method of evaluating this triple integral is precisely analogous to that already explained for double integrals in § 218, p. 401. Illustrative Example 1. Find the volume of that portion of the ellipsoid a2 &2 + ^2 - which lies in the first octant. Solution. Let — ABC be that portion of the ellipsoid whose volume is required, the equations of the bounding surfaces being (1) (2) (3) (4) ^ + | + ^ = M^i*C), z = 0, (OAB), y = 0, {OAG), x = 0, (OBC). Y/b~ PQ is an element, being one of the rectangular parallelepipeds with dimensions Az, Ay, Ax into which the planes parallel to the coordinate planes have divided the region. G E SUCCESSIVE AND PARTIAL INTEGRATION 439 Integrating first with respect to z, we sum up all suc li elements in a column (as MS), the z-limits being zero [from (2)] and TR = c -t /l — — - 1^ [from (1) by BoMngforz]. ^ "^ ^^• Integrating next with respect to y, we sum up all such column s in a slice (as 1 [from equation X- u- "'^ of the curve AGB, namely h — = 1, by solving for yl. ' a' 6^ Lastly, integrating with respect to x, we sum up all such slices within the entire region — ABC, the x-limits being zero [from (4)] and OA = a. Hence ^ °= "'-dzdydx Jo Jo Jo Therefore the volume of the entire ellipsoid is 4 7ra6c Illustrative Example 2. Eind the volume of the solid contained between the paraboloid of revolution the cylinder and the plane x^ + j/2 = az, x^ + y^ = 2 ax, z = 0. 3.2 _(. „2 Solution. The z-limits are zero and iVP(= , found by solving equation of paraboloid for z). The y-limits are zero and Jf2V( = V2ax — x^, found by solving equation of cylinder for y). The x-limits are zero and OA (= 2 a). ' The above limits are for the solid ONAB, one half of the solid whose volume is required. ^.^,^. y /.-la n^-lax-xi /■ ti , , , SttO? Hence — = | dzdydx = ^^ — 2 Jo Jo Jo 4 Zira? 2 Therefore Y — - EXAJHPLES 1. Knd the volume of the sphere x'' + y^ + z^ = r^ by triple integration. jLns. 3 2. Find the volume of one of the wedges cut from the cylinder x^ + y^ = r' by the planes z = and z = mx. /.r [.^r^-sn prKc 2r^m Ans. 2 I ( ( dzdydx = — -— ■ J J Jo o 3. Find the volume of a right elliptic cylinder whose axis coincides with the »-axis and whose altitude = 2 a, the equation of the base being cV + b'^z'^ = 6%^. Am. 8 I dzdydx = 2 irahc. Jo Jo Jo 420 INTEGEAL CALCULUS 4. Find the entire volume bounded by the surface (-1 + (r) + (") ~ ^' ^""^ ^^^ coSrdinate planes. ^"^ ^''' ^''^ Ans. —. 5. Find the entire volume bounded by the surface x^ + y' + z^ = a». 4 ^qS Ans. 35 6. Find the volume cut from a sphere of radius o by a right circular cylinder with 6 as radius of base, and whose axis passes through the center of the sphere. Ans. — [a^ - (a^ - b^)i]. o 7. Find by triple integration the volume of the solid bounded by the planes x = a,y = b, z = mx and the coordinate planes XOY and XOZ. Ans. imba'. 8. The center of a sphere of radius r is on the surface of a right circular cylinder T the radius of whose basis is - • Find the volume of the portion of the cylinder inter- cepted by the sphere. Ans. f (tt — J)r2. 9. Find the volume bounded by the hyperbolic paraboloid cz = xy, the XOY- plane, and the planes x = a^, x = a^, y = b^, y = b^. ("^ — a?) (pi — b?) Ans. 4c 10. Find the volume common to the two cylinders x^ + y^ = r^ and x^ + z^ =>^. 16r3 Ans. 3 11. Find the volume of the tetrahedron bounded by the coordinate planes and the plane - + - + - = 1. Ans. - abc. a b c 6 12. Find the volume bounded by the paraboloid x^ + y" — z = 1 and the XF-plane. Ans. - ■ • 2 13. Find the volume common to the paraboloid y^ + z^ = 4:ax and the cylinder x^ + y^ = 2ax. Ans. 2Tra^ + -J/ a'. 14. Find the volume included between the paraboloid y^ + z'^ = iax, the parabolic cylinder y^ = ax, and the plane x — 3a. Ans. (6 tt + 9 Vs) a'. 15. Find the entire volume within the surface x^ + yi + z'^ = a^. 16. Compute the volume of a cylindrical column standing on the area common to the two parabolas x = y^,y = x^ as base and cut off by the surface z = 12 + y — x^. 17. Find the volume bounded by the surfaces y^ = x + 1, y'' =— x + 1, z =—2, z = x + i. 18. Find the volume bounded by z = x^ + 2y'', x + y = 1, and the coordinate planes. 19. Given a right circular cylinder of altitude a and radius of base r. Through a diameter of the upper base pass two planes which touch the lower base on opposite sides. Find the volume of the cylinder included between the two planes. Ans. (tt — f) wfi. CHAPTER XXX ORDINARY DIFFERENTIAL EQUATIONS* 228. Differential equations. Order and degree. A differential equa- tion is an equation involving derivatives or differentials. Differential equations have been frequently employed in this book, the following being examples : ^ '' \ dx /\dxy \ dx^ jdx (3) tan' ^ dd P (4) d-^y dx"- = 12(2a; -!)• (5) dy = = 2 dx. (6) dp = a^sin P ^'dO. (7) d'y- = (20«'- -12x')dx- (8) 8u dx , du bu. {A-), p. 84 Ex. 1, p. 101 Ex. 2, p. 138 Ex. 3, p. 138 Ex. 1, p. 139 Ex. 7, p. 194 ^'>"a^ + ^3^='a^- Ex. 8, p. 204 (10) -^ = n+2.xyz + a; VV) u. Ex. 7, p. 204 oxdydz In fact, all of Chapter XI in the Differential Calculus and all of Chapter XXIII in the Integral Calculus treats of differential equations. An ordinary differential equation involves only one independent variable. The iirst seven of the above examples are ordinary differ- ential equations. * A few types only of differential equations are treated in this chapter, namely, such as the student is litely to encounter in elementary work in Mechanics and Physics. 421 422 IXTKURAL CAL(!ULUS A partial differential equation involves more than one independent variable, as (8), (9), (10). In this chapter we shall deal with ordinary differential equations only. The order of a differential equation is that of the highest derivative (or differential) iu it. Thus (3), (5), (6), (8) are of the first order; (1), (4), (7) are of the second order; and (2), (10) are of the third order. The degree of a differential equation which is algebraic in the derivatives (or differentials) is the power of the highest derivative (or differential)- in it when the equation is free from radicals and fractions. Thus all the above are examples of differential equations of the first degree except (2), which is of the second degree. 229. Solutions of differential equations. Constants of integration. A solution or integral of a differential equation is a relation between the variables involved by which the equation is identically satisfied. Thus (^) , y = '^i sin X is a solution of the differential equation (^) dx' .+y = 0. For, differentiating (^), d^y d.r' : — Cj sin X. Now, if we substitute (^A) and (C) in (£), we get — Cj sin X + Cj sin x= Q, showing that {A} satisfies (5) identically. Here c^ is an arbitrary constant. In the same manner (-0) y = c^eosx may be shown to be a solution of {£} for any value of c.,. The relation C-E) g = e^siax + c^ cos x is a still more general solution of (5). In fact, by giving particular values to c^ and c^ it is seen that the solution (^) includes the solu- tions (4) and (Z)). The arbitrary constants -c^ and c^ appearing in these solutions are called constants of integration. A solution such as (^), which con- tains a number of arbitrary essential constants equal to the order of OEDINARY DIFFEEENTIAL EQUATIONS 423 the equation (in this case two), is called the general solution or the complete integral.* Solutions obtained therefrom by giving particular values to the constants are called particular solutions or particular integrals. •' The solution of a differential equation is considered as having been effected when it has been reduced to an expression involving integrals, whether the actual integrations can be effected or not. 230. Verification of the solutions of differential equations. Before taking up the problem of solving differential equations it is best to further familiarize the student with what is meant by the solution of a differential equation by verifying a number of given solutions. Illustrative Example 1. Show that (1) y = c-^x CQS log X + CgX sin log x + x log x is a solution of the differential equation (2) x^g-x| + 2, = xlogx. Solution. Differentiating (1), we get (3) -^ = (Cj — Ci)sinlogx + (Cg + Cj)ooslogx + logx + 1. dx Substituting (1), (3), (4) in (2), we find that the equation is identically satisfied. EXAMPLES Verify the following solutions of the corresponding dii|erential equations : Differential equations Solutions ^ m'-^-x^ + y = 0. y = cx + c-c'. \dx/ dx dx 2. J^y+2x^-3/ = 0. y^ = 2cx + cK \dx/ ax dx l + c e-' 4. ^ + ^^"^=0. j/ = c,x + ^ + C3. dx^ X dx^ X dx2 dx ("^ — ^) * It is shown in works on Differential Equations that the general solution has n arbitrary constants when the differential equation is of the ntb order. 424 INTEGEAL CALCULUS Differential equations Solutions °- T7- 4?^ + 6^- *T^ + 2/ = <'• y = {c^ + v + e,x^ + c,xO)e'. dx* dx^ dx^ dx 7. x2^- 5x^ + 52/ = -. Ay = ^ + c^x' + c^. dx^ dx X 3x 8. x — — y + X Vx^ — y^ = 0. arc sin- = c — x. dx X „ dy sin2x . 1 , . , 9.-^ + 2/ cosx = y = sinx — 1 + ce-^^^. dx 2 10. (1 — x^) — — X— — a% = 0. y = Cie<"»"=»™^ + c^e-'"-^''^^. dx^ dx ii.^ + !J = o. , = ^ + c,. (Jx^ X dx X 231. Differential equations of the first order and of the first degree. Such an equation may be brought into the form Mdx + Ndt/ = 0, in which M and N are functions of x and y. Differential equations coming under this head may be divided into the following tjrpes : Type I. Variables separable. When the terms of a differential equation can be so arranged that it takes on the form CA) fQr:)dx+i\y-)dy=Q, where /(a;) is a function of x alone and i^(y) is a function of y alone, the process is called separation af the variables, and the solu- tion is obtained by direct integration. Thus integrating (^), we get the general solution (5) jfQx) dx +jF(y^ dy = c, where e is an arbitrary constant. Equations which are not given in the simple form (^) may often be brought into that form by means of the following rule for separating the variables. TiKST Step. Clear of fractions, and if the equation involves deriva- tives, multiply through by the differential of the independent variable. Second Step. Collect all the terms containing the same differential into a single term. If, then, the equation takes on the form XYdx+X'Y'dy = 0, where X, X' are functions of x alone, and Y, Y' are functions of y alone, it may be brought to the form (^) by dividing through by X'Y. Thikd Step. Integrate each part separately, as in (5). ORDINARY DIITERENTIAL EQUATIONS 425 Illustkative Example 1. Solve the equation dy 1 + 2/2 dx (1 + X?) xy Solution. First step. (1 + x^) xydy = (l + y^)dx. Second step. (1 + y^)dx — x(l + x'^) ydy = 0. To separate the variables we bow divide by x (1 + x^) (1 + y"), giving (fe ydy _ Q a; (1 + x2) 1 + y' Third step. r^ fJ^ = C, •/ X (1 + x2) J 1 + y^ rd^_rxd._r_yd^ p. 329 J X J 1 + x^ ■) 1 + y^ logs - hog(l + x2) - ^log(l + y^) = G, log (1 + x2) (1 + 2/2) = 2 logx - 2 C. This result may be written in more compact form if we replace — 2 C by logc, i.e. we simply give a new form to the arbitrary constant. Our solution then becomes log (1 + x2) (1 + 2/2) = logx2 + log c, log(l + x2) (l + 2/2) = logcx2, (1 + x2) (1 + 2/2) = cx2. Arvs. I dy ^ \ c alx-— + 2y] = xy- \ dx I c Illustkative Example 2. Solve the equation dy dx Solution. First step. axdy + 2 aydx = xydy. Second step. 2 aydx + x(a~ y)dy — 0. To separate the variables we divide by xy, 2adx {a — y)dy _ X y ~ ' Third step. 2aj'^ + af^-fdy = G, 2a\ogx + a\osy — y = G, alogx^y = C + y, log.x22/ = -| + |. By passing from logarithms to exponentials this result may be written in the form c y_ c Denoting the constant e« by c, we get our solution in the form « x^y — ce". Ans. 426 INTEGRAL CALCULUS EXAfflPLES Differential eqiwiions Solutions 1. ydx-xdy = 0. y = <^- 2. (1 + y)dx - (1 —x)dy = 0. (1 + V) (1 - x) = c. 3. (\ + x)y6x-+{}.-y)xdy = (i. logxy + x-y = c. 4. (x2 - a'')dy -yclx = 0. y^" = «|^- 5.(x^-yx^)^ + y^ + xy^ = 0. ^ + log| = c. dx xy X 1 6. uHv + (v-a)du = 0. v-a = c0'. du _ l + u' u-l±A. ' dv~ 1 + vi^ 1 — CT 5. {1+ s^)dt — t^ds = 0. 2t*— arctans = c. 9. dp + p ta,n8de = 0. p = c cos^. 10. sin 6 cos (pdO — cos sin 0d0 = 0. cos ^ = c cos 6. 11 . sec^ 6 tan 0dS + sec^ (^ tan M(^ = 0. tan tan = c. 12. sec^ tan 0(l0 + sec" tan SdS = 0. sin" + sin" = c. 13. xydx - (a + x){b + y)dy = 0. x-y = c + log(a + x)<^. 14. (1 + x") dy — Vl — 2/"(i!; = 0. arc sin 2/ — arc tan x = c. 15. Vl - x'^dy + Vl - y'^dx = 0. j/ Vl - x^ + xVl- 2^" = c. 16. Septan 2/cJx + (1 - e^) sec" 2/dj/ = 0. tanv = c(l- e^^)'. 17. 2 x"2/d2/ = (1 + x") dx. 2/" = — - + X + c. ' x 18. (x - i/"x)dx + (!/ - x"y)di/ = 0. x" + y" = x"2/" + c. 19. (x"2/ + x)d3/ + (X2/" — y)dx = 0. xj/ + log- = c. Type II. Homogeneous equations. The differential equation Mdx+Ndy = Q is said to be homogeneous when M and N are homogeneous functions of X and y of the same degree.* Sucli differential equations may be solved by making the substitution y = vx. This will give a differential equation in v and x in which the vari- ables are separable, and hence we may follow the rule on p. 424. * A function of x and y Is said to be homogeneous in the variables if the result of replacing X and y by Xx and 'Ky (X being arbitrary) reduces to the original function multiplied by some power of X. This power of X is called the degree of the original function. ORDINARY DIFFERENTIAL EQUATIONS 427 Illustrative Bxamplk 1. Solve the equation Solution. j,2(fc + (s2 _ ^y) ay = 0. Since this is a homogeneous differential equation, we transform it hy means of the substitution ^^^_ Hence dy = .dx, u and logeW = — / Pdx +C, giving Equation (D) then becomes dz dx To find z from the last equation, substitute in it the value of u from (^) and integrate. This gives (i^) CjZ = fQef'''"'dx + C. The solution of (A) is then found by substituting the values of u and 2 from (^) and (i?") in (5). This gives («) y = e-I'-'-lj Qe!''''-dx + cY OEDIKAEY DIPFEEENTIAL EQUATIONS 429 The proof of the correctness of (G) is immediately established by- substitution in (^). In solving examples coming under this head the student is advised to find the solution by following the method illustrated above, rather than by using (G) as a formula. Illustrative Example 1. Solve the equation ^ ' ■ dx x + 1 ^ ' Solution. This Is evidently in the linear form (A), where P = ?— and Q = (a; + l)i x + 1 Let y = vz; then — = a — + z — • Substituting in the given equation (1), we get dx dx dx dz du 2uz , , ,,i dz /du 2u \ , , ,.5 Now to determine u we place the coefficient of z equal to zero. This gives du 2 M _ » dx 1 + X ' du _ 2dx u 1 + X l0geM = 2l0g(l + x), (3) u = ei»E (1 + ^)2 = (1 + x)2.* Equation (2) now becomes, since the term in z drops out, u — = (x + l)S- dx Replacing u by its value from (3), J = (x + l)i ox d2 = (x + 1)^ *>;, 2(x + l)t (4) z = -^-^ + 0. Substituting (4) and (3) in y = uz, we get the solution y^^J^±^ + G{x + lY. Ans. » Since lege « = loge elosd + ^'^ = log (1 + a:)2 • lege e = log (1 + a;)2, it follows that «= (1 + a;)2. For the sake of simplicity we have assmned the particular value zero for the constant of integration. 430 INTEGEAL CALCULUS EXAMPLES mfferential equations SoMions 1. ^ _ _E^ = (a + 1)3. 2y = (x + l)* + c(z + lf. dx X + 1 dy^^^xj^ y^cx<' + X 1 dx X X 1 — " ^ 3. x(l - x^)dy + (2x2 - i^^jjx = ax'^dx. ' y = ax + ex Vl - x^. i.dy-^^^ = ^^. y = ax + c(l + x^h " 1 + x^- 1 + x^ 5. — cosi + ssint = l. s = sini + ccost. dt 6. — + scos< = isin2«. s = sint- 1 + ce-»"'. dt 7. ^_5:j/ = e^n. 2/ = x''(g^+c). dx X 8. ^ + -3/ = — . x«3/ = ax + c. (ix X x" 9. ^ + 2/ = -- ' e^ = x + c. dx e' , 10. ^ + ^— #^y = 1. 2/ = x2(l + c^). dx x^ Type IV. Equations reducible to the linear form. Some equations that are not linear can be reduced to the linear form by means of a suitable transformation. One type of such equations is where P, Q are functions of x alone, or constants. Equation (^) may be reduced to the linear form (A), Type III, by means of the substitution 2 = y~° + '- Such a reduction, however, is not necessary if we employ the same method for finding the solution as that given under Type III, p. 427. Let us illustrate this by means of an example. Illustrative Example 1. Solve the equation (1) ^ + y- = a\ogx.yK dx X Solution. This is evidently in the form (A), where P = -, Q = alogx, 71 = 2. X T i iv dy dz , du Let y = uz; then -S- = u \- z — dx dx dx Substituting in (1), we get dz , du uz u— + z—--\ = alogj;- «^2^ ox dx X dz /du u\ (2) u-+(- + -y = alosx.uH^ ORDINARY DIFFERENTIAL EQUATIONS 431 Now to determine u we place the coefficient of z equal to zero. This gives ^ + "^ = 0, dx X du _ dx u X logu =— logx = log-, X (3) u = l. X Since the term in z drops out, equation (2) now becomes u — = alogX'M^'', dx dz _ dx~ : alOgX-MZ^. Replac ing u by its value from (3), dz _ dx" 2^ : a logx-—, X dz z2 dx -. alogx — , X 1_ z a{logx)2 1 ^ 2 (4) z - 2 a(logx)2 + 20 Substituting (4) and (3) in 2/ = uz, we get the solution y = 1 2 _ _ /I ™.\'2 1 O /^ X a(logx)2 + 2C xj/ [o(logx)2 + 2 C] + 2 = 0. Ans. EXAMPLES Differential equatiows Solutions l.^ + xy = xV- y-^ = x^ + l + ce^. dx. 2.n-x^)^-xy = axy^' y = (c Vl^^x^- a)-i. dx 3.Sy^^-ay^ = x + l. 2,a = e6--^-i- 4. ^(xV + xy) =1. ^[(2 - 2'')«' + <=] = *'■ 5. (2/ logx - l)2/cix = xdy. y = {cx + logx + 1) dv tanx + secx 6. y-cosx^ = 2/2cosx(l-sinx). y= ^^^^ ^ (. rr ^2/ , !!1,,„o. «-! = lOgX + 1 + CX. dx -1 432 ESTTEGEAL CALCULUS 232. Differential equations of the nth order and of the first degree. Under this head we will consider four types which are of importance in elementary work. They are special cases of linear differential equar tions, which we defined on p. 427. Type I. The linear differential equation d''v d''-'^y d"-'y in which the coefficients p^, p^, ■■■, p„ are constants. The substitution of e™ for y in the first member gives This expression vanishes for all values of r which satisfy the equation (5) r^+p/''-^+p/'-^+---+p„=0; and therefore for each of these values of r, e""' is a solution of (A). Equation (5) is called the auxiliary equation of (.A). We observe that the coefficients are the same in both, the exponents in (5) cor- responding to the order of, the derivatives in (^), and y in (.4) being replaced by 1. Let the roots of the auxiliary equation (5) be r^, r^, ••■,»•„; then (C) e-'i^ ev, ..., e'"" are solutions of (^). Moreover, if each one of the solutions (C) be multiplied by an arbitrary constant, the products (D} c^ev, c^ev, ..., oy^ are also found to be solutions.* And the sum of the solutions (D), namely, (^) 2/ = c/'" + o^ev + . . . + c„e'-"", may, by substitution, be shown to be a solution of (.4). Solution (^) contains n arbitrary constants and is the general solution (if the roots are all different), while (C) are particular solutions. Case I. When the auxiliary equation has imaginary roots. Since imaginary roots occur in pairs, let one pair of such roots be r^=a + bi, r^=a — bi. i = V^ * Substituting Cie'i'lor y in (A), the left-hand member becomes (ri" +Piri"--' + pa?-!""^ + ■ • ■ + Pn) c^e'^'. But this vanishes since r^ is a root of (B); hence Oie''" is a solution of (A). Similarly for the other roots. OEDINAKY DIKFEREJSTIAL ECiUATiO:NS 433 The corresponding solution is = e'^\ ej(cos hx + i sin hx) + e^Ccos bx — i sin 6a;) P = e"^! (Cj+ ejj)cos 5a; + i(ej— C2)sin 6a;|, or, y = ■e'^(^ cos Ja; + 5 sin 62;), where A and S are arbitrary constants. Case II. When the auxiliary equation has multiple roots. Consider the linear differential equation of the third order where p^, p^, p^ are constants. The corresponding auxiliary equation is (G) r'+p/+p^r+p=0. If j-j is a root of (G), we have shown that e'"''" is a solution of (i^). We will now show that if r^ is a double root of ( G), then xe''^" is also a solution of (i^). Replacing y in the left-hand member of (i^) by xeT'", we get (if) xe-^Xr^+p^r^ +p/^ +P3) + e'-xX3 r^ + ^f^^+f^)- But since r^ is a double root of ((?), and 3 rl^ ^P^r^+P^= 0- By § 69, p. 88 * Replacing x by ibx in Example 1, p. 232, gives ^ , ., ^2a;2 iftSjjS 64a;4 jftSajS e<^=l + ,to_-^_-^ + _ + -^-.... or, (1) '^'"=i-i2-+ir--^Y"-"i3:+^--j' and replacing^ by -i6. gives ^^^^ .^^_^ ^^^ .^^^ (2) But, replacing a by b« in (^), (5), p- 231, we get (3) cos6a; = l--jy + -T| ■ (4) sin 5a! = ox - -ij- + -rg • Hence (1) and (2) become e**= cos Sa; + i sin bx, e-""- cos bx-i sin bx. 434 INTEGRAL CALCULUS Hence (IT) vanishes, and xe^^'^ is a solution of (F'). Corresponding to the double root r, we then havie the two solutions More generally, if r^ is a multiple' root of the auxiliary equation (B), p. 432, occurring s times, then we may at once write down s distinct solutions of the differential equation (^), p. 432, namely, eje'"i% o^xe^'r', c^e'^", ■■■, c^x'~^e''i\ In case a + hi and a — hi are each multiple roots of the auxiliary equation, occurring s times, it follows that we may write down 2 s distinct solutions of the differential equation, namely, c^e"^ cos hx, c^e"^ cos hx, c^e"" cos hx, • • • , cpf~^e'" cos hx ; cje"" sin hx, c^xe"^ sin hx, c'^x^e"^ sin hx, • • • , clx^~'^e'^ sin hx. Our results may now be summed up in the following rule for solving differential equations of the type d"y d"-^y d"-^y where p^, p^, • ■ • , jo„ are constants. EiKST Step. Write down the corresponding auxiliary equation r^+p^r''-'-+p/'-^+ ■ ■ . +p^= 0. Second Step. Solve completely the auxiliary equation. Third Step. From the roots of the auxiliary equation write down the corresponding particular solutions of the differential equation as follows : Auxiliary Equation Differential Equation (a) Uach distinct real 1 V gives a particular solution e'''^'. (b) Uach distinct pair 1 . j two paHicular solutions e"^ cos hx, (c) A multiple root occur- ring s times r ■s' of imaginary roots a±hi J ^"^^^ | e<^ sin hx. s particular solutions ohtained hy multiplying the particular solutions (a) or (h'),hyl,x,x% .--..x'-K- FouETH Step. Multiply each of the n* independent solutions hy an arbitrary constant and add the results. This pives the complete solution. * A check on the accuracy of the work is found in the fact that the first three stepS must give n independent solutions. . . ,, OEDINAKY DIFFERENTIAL EQUATIONS 435 Illustrative Example 1. Solve — ^-3—^ + 42/ = 0. Solution, follow above rule. First step, r* — 3 r^ + 4 = 0, auxiliary equation. Second step. Solving, the roots are — 1, 2, 2. Third step, (a) The root — 1 gives the solution er". (b) The double root 2 gives the two solutions e^"", xe^'^. Fourth step. General solution is y = Cjg-^ + Cje^"^ + CjXe^^. Ans. Illustrative Example 2. Solve — -4^ + 10—^-12^ + 5y = 0. /2 + C2e-="^+C3sin2x + CjCOs2x. dx* dx2 dt' dt2 dt 12 3 9. ^_6— + 13a = 0. i4 = (CiSin2u + C2COs2u)eS". du^ dv 10 ^ + 2n2^ + 71*2/ = 0. y = (Ci + C2x)cosKX + (C3 + c,x)sinM. dx* dx^ ,0 --/ iVi «V3\ 11 ^ = s s = Cie* + e s/casin-— + C3COS-— |. ■ di" ■ \ / ^ / 43U INTEGllAL CAL(!ULU,S Differential equations General solutions 12. ^ - 7 - + 6s = O'. s = c,e^* + e„e' + c.e-s'. dt^ dt 12 8 14. ^ + 3^-10?/ = 0. 2/ = c,e2^+c,e-s»'. dx^ dx 15. —+ 2-^ + 10?/ = 0. ' 4/ = e-«(c, cos3a; + c„sin3a;). 16. 2— ^-3— + 2— + 2^ = 0. y = e,eri=' + e!'{c„cosx + e,smx) dx'' dx^ dx Type II. The linear differential equation where X is a function of x alone, or constant, and p , p , ■■■,p„ are constants. When X= 0, (/) reduces to (A), Type I, p. 432, The complete solution of (J") is called the complementary function of (/). Let u be the complete solution of (t7), i.e. the complementajy function of (/), and v any particular solution of (/). Then Adding, we get -^^(u + v-)+p^^^^^(u + v-)+p^^^^(u + v^+...+p^(u + v-) = X, showing that m + v is a solution* of (/). To iind a particular solution w is a problem of considerable diffi- culty except in special cases. For the problems given in this book we may use the following rule for solving differential equations of Type II. First Step. Replace the right-hand member of the given equation (/) hy zero and solve hy the rule on p. 434. This gives as a solution the complementary function of (J), namely, y = u. * In works on differential equations it is shown that u^■vis the complete solution. ORDINAEY DIFFEEENTIAL EQUATIOHS 437 Second Step. Differentiate successively the given equation (J) and obtain, either directly or by elimination, a differential equation of a higher order of Type I. Third Step. Solving this new equation by the rule on p. 434, we get its complete solution ■ „ _ ,, _i_ y — u -\- V, where the part u is the complementary function of (/) already found in the first step,* and v is the sum of the additional terms found. Fourth Step. To find the values 'of the constants of integration in ■ the particular solution v, substitute y = v and its derivatives in the given equation (/). In the resulting identity equate the coefficients of like terms, solve for the constants of integration, substitute their values back in ^ _ . ^ _l ^ giving the complete solution of (/). This method will now be illustrated by means of examples. Note. The solution of the auxiliary equation of the new derived differential equa- tion is facilitated by observing that Ihe left-hand member of that equation is exactly divisible by the left-hand member of the auxiliary equation used in finding the com- plementary function. Illustrative Example 1. Solve dx? dx Solution. First step. Replacing the right-hand member by zero, ^ ' dx^ dx Applying the rule on p. 434, we get as the complete solution of {L) (M) y = c^g' + c^e-^'' = u. Second step. Differentiating (K) gives ^ ' djp dx? dx . ' Multiplying (K) by 2 and adding the result to (N), we get «°) S"S— »■ a differential equation of Type I. TAird step. Solving by the rule on p. 434, we get the complete solution of (0) to be y_= Cjg"' -f- CgE-^^ -f- CgXe- 23--, or, from (M), y = u + c^xe-^^ = u + v. * From the method of derivation it is obvious that every solution of the original equation must also be a solution of the derived equation. 438 INTEGEAL CALCULUS Fourth step. "We now determine c, so that CgXe-"^ shall be a particular solution v of (JT). , M Substituting y = c^xe-^^, — = c^e-^='(l-2x), -^ = c^e-^^{4:X- i) in (-ST), we get .-. —3c^ = a, or, Cj^— 4a. Hence a particular solution of (K) is » =— -J-axe-^^, and the complete solution is Illustrative Example 2. Solve (P\ — - + re^2/ = cos ax. ^ ' dx" Solution. First step. Solving (Q) 5^ + "'^ = ''' we get the complementary function (B) y = 0^ sin nx + Cj cos rue = m. Second step. Differentiating (P) twice, we get (S) — ^ + »2 — ^ = — a^ cos ax. Multiplying (P) by a^ and adding the result to (S) gives (T) ?? + («' + «') ?! + «'™'2' = 0- Third step. The complete solution of (T) is y = Cj sin nx + c^ cos nx + Cg sin ax + c^ cos ax, or, 2/ = M + Cg sin ax + c^ cos ax = m + d. Fourth step. Let us now determine Cg and t;^ so that Cg sin ax + c^ cos ax shall be a particular solution d of (P). Substituting dy ■ iPy o ■ o w = c, sin ax + c. cos ax, -^ = Cga cos ax — c.a sm ax, — ^ = — CgU^ sm ax — c.a^ cos ax in (P), we get (n^c^ — a^c^) cos ax + (n%g — a^Cg) sin ax = cos ax. Equating the coefficients of like terms in this identity, we get n\ — a\ = 1 and n\ — a\ = 0, Hence a particular solution of (P) is cos ax V = and the complete solution is cos ax y = u + V = e^smnx + c^ cos nx'+ — OEDINAEY DIFFERENTIAL EQUATIONS 439 EXAMPLES SWerential equations Complete solutions 3- ^ - a's = < + 1. s = de-' + c^e-"' - i±i. . d'p od^p , dp ^ I g2\ "• ^ - '='2' = a; • y = c,e«^ + c^e-'^ + Cg sinaa; + c^ cosax - — , 6- T-^ + a^s = cosaa;. « = Cj sin ox + Cj cos ox + xsmax *;^ ' • " • 2a '''■ ^ ~ ^" di "^ "'* " ^'" s = (Ci + Cat) E»« + e' (a -1)2 d^v dy 1 9-^-y = 5x + 2. 2/ = Cie»'+ c^e-^- 5x-2. dx^ dx ' " "1" ' "2" ^„2_5„+6 dx^ dx « 1 2 ,i2_3,i + 2 (ji2_3n + 2)2 13, d^s „ ds d«2 dt 9^ + 20s = y = T ; 1- c,x''-i + h c„_ix + c„. |?n.-+ n ox = log(2/+V2/2 + Ci) + Cj, or, 3 i = 2 ai (s* - 2 c^) (s^" + Cj)^ + Cj. (c^t + Cg)^ + a = Ci2/2. 5. df' ^' 6. ^ = x». dx" 7. dx^ 8. d^s 1 9. d'^y _ a dt^^y^' 10. = 0^. dt^ 11. — ^ = x" sill X. dx'^ , — ^c?e^ + 1-1 i •V2 71 = Cj log h f"- Vcf e"^ + 1 + 1 ■j/ = c^ + c^x + (6 - x^) sin x - 4x cosx. 12 -^ = _ ^ . Find «, having given that — = and s = a, when t = 0. ■ d«2 g2 dt Ans. t = J—\-(s.rcvers— - -A-Vas- s^l 442 INTEGRAL CALCULUS MISCELLAIfEOUS EXAMPLES Solve the following difierential equations : dx* dx^ dx^ dx ^ — H 1 ! It IS seen that — indicates the number of linear units in the ordinate M^^Pi, and also the number of units of area in the shaded area OM^P^. Also, since from (E) dy x^ ^ . xf — = — . or, tan t = — , dx 3 3 and from (JF) X,' M,P, = -i ^ ^ 8 f=m it is seen that the same number — indicates the length of ordinate M^P■^ and the slope of the tangent at Pf. Evidently the origin is a point of inflection of the integral curve and .i, point with minimum ordinate on the original curve. 235. The integraph-. The theory of this instrument is' exceedingly simple and depends on the relation between the given curve and a corresponding inte- gral curve. The instrument is con- ^structed as follows: Arec- ^ tangular carriage G moves on rollers over the plane in a direction parallel to the axis of X of the curve y =/(»)■ Two sides of the carriage are parallel to the axis of X; the other two, of course, are perpendicular to it. Along one of these perpendicular sides moves a small carriage Cj bear- ing the tracing point T, and along the other a small carriage G^ bearing a frame F which can revolve about an axis perpendicular to the surface, and which , carries the sharp-edged disk D, to the plane of which it is perpendicular. A stud S-^ is fixed 446 LNTEGEAL CALCULUS in the cai'riage Cj so as to be at the same distance from the axis of X as is the trac- ing point T. A second stud S^ is set in a crossbar of the main carriage C so as to be on the axis of X. A split ruler B joins these two studs and slides upon them. A crosshead JET slides upon this ruler and is joined to the frame J" by a parallelogram. The essential part of the instrument consists of the sharp-edged disk D, which moves under pressure over a smooth plane surface (paper). This disk will not slide, and hence as it rolls must always move along a path the tangent to which at every point is the trace of the plane of the disk. If now this disk is caused to move, it is evident from the figure that the construction of the machine insures that the plane of the disk D shall be parallel to the ruler E. But if a is the distance between the ordi- nates through the studs Sj, S^, and t is the angle made by B (and therefore also plane of disk) with the axis of X, we have (A) tan T = ?^ ; and if y' = F(x') is the curve traced by the point of contact of the disk, we have {B) tanT = ^'.* dx Comparing (A) and {B), — = -, or, dx a (C) ]/ = - fydx = i Jf{x) dx = F{x')A That is (dropping the primes),the curve y = F(x) is an integral curve of the curve The factor - evidently fixes merely the scale to which the integral curve is drawn, and does not affect its form. A pencil or pen is attached to the carriage Cj in order to draw the curve y = F{x). Displacing the disk D before tracing the original curve is equivalent to changing the constant of integration. 236. Polar planimeter. This is an instrument for measuring areas mechanically. Before describing the machine we shall take up the theory on which it is based. 237. Calculation of the area swept over by a moving line of con- stant length. Consider the area. ABQB'A'PA swept over by the line AB of constant length I. Let PQ and P'Q^ be consecutive positions of the line, de = angle POP' = change in * Since x = x' + a, where d = width of machme, and therefore ^ = ^ . ^ ^ ^' _ dx" dx dx' dx ' t It IS assumed that the instrument is so constructed that the abscissas of any two corre- spondmg points of the two curves differ only by a constant; hence a is a function of x'. INTEGEAPH 447 direction of PQ, and ds = circular arc described about by the middle point R of the line. Using difierbntiajs, we have ^' area of 0Q(^= i OQ^dO* area of OPP'= i OP^dd. .-. areaof PQQ^P'^i OQ^dS-i OP'^dd A = i{OQ+OP){OQ-OP)dg = OB ■ pqae = I ■ 0Bd6 = Ms. O Summing up all such elements, (A) a^rea, ABQB'A'PA = fids = lCdsz= Is, where s = displacement of the center of the line in a direction always perpendicular to the line.t To find s, let the line be replaced by a rod having a small wheel at the center B, the rod being the axis of the wheel. Now as the rod is moved horizontally over the surface (paper), the wheel will, in general, both slide and rotate. Evidently s = distance it rolls = circumference of wheel x number of revolutions. (B) .-.8 = 2 mrn, where r = radius of wheel, and n = number of revolutions. Substituting (B) in (A), we get (C) area swept over = 2 irrln. So far we have tacitly assumed that the areas were swept over always in the same direction. It is easy to see, however, that the results hold true without any such : restriction, provided areas are taken as posi- tive oi negative according as they are swept over towards the side of the line on which di is taken positive, or the reverse. Choose signs as indicated in the figure. If the line AB .i/.^Zi//, returns finally to its original position, A and '\' //^ B having described closed curves, it is evi- , ^ ^ -^ ^ dent that" the formula above will give (taking f account of signs) the excess of the area in- ^ closed by the path of A over that inclosed by the path of B. For ABQB'A'PA = ABBB'A'PA + closed curve jBQB'Z)-B, B'A'GABBB' = ABDB'A'PA + closed curve ^P^'C^. positive area negative area Finding the difEerence, we have net area = closed curve BQB'-D-B — closed curve ^P^'C^. * Area of circular sector = f radius x arc = i Q • Q dS = i Q dB. t It should be observed that s wiU not be the length of the path described by the center R xoUbssAA' and BB' are the ares.of circles with the center at 0. 448 INTEGRAL CALCULUS tracing point Now if the area of one of these closed curves (as APA'CA) is zero, that is, A keeps to the same path both going and returning, the area swept over by the line will equal the area of the closed curve BQB'DB. A simple and widely used type of polar planimfeter was invented by Amsler, of Schafihausen, in 1854. This consists essentially of two bars OA and AB, freely jointed at A, OA rotating about a fixed point O and AB being the axis of a wheel situated at its center E, and having a tracing point at B. Now if the tracing point completely describes the closed curve, A will oscillate to and fro along an arc of a circle (as CD), describing a contour of zero area. Hence the area swept over by the bar AB exactly equals the area of the closed curve, and is given by the formula (D) area of closed curve = 2 irr/n, where I = length of bar AB, r = radius of wheel, n = number of revolutions indicated on the wheel after the tracing point has made one complete circuit of the curve. 238. Approximate integration. Since the value of a definite integral is a measure of the area under a curve, it follows that the accurate measurement of such an area will give the exact value of a definite integral, and an approximate measurement of this area will give an approximate value of the integral. "We will now explain two approx- imate rules for measuring areas. 239. Trapezoidal rule.' Instead of inscribing rectangles within the area, as was done in § 204, p. 361, it is evident that we shall get a much closer approximation to the _.i area by inscribing trapezoids. Thus divide the interval from x = a to x=h into n equal parts and de- note each part by Ax. Then, the area of a trapezoid being one half the product of the sum of the parallel sides multiplied by the o] a ' ' b S altitude, we get i (^0 + ^i) ^^ = area of first trapezoid, i (2^1 +2/2)^^= area of second trapezoid, i (2/»-i — 2/«) ^^ = area of nth trapezoid.. Adding, we get K2/0+ 2 t/j-l- 2 ?/jj-|- . ■ . -I- 2 t/„_j-|-yj Aa; = area of trapezoids. APPROXIMATE INTEGEATION 449. Hence trapezoidal rule is (4) area = 0i/, + y, + y, + . . ■ +y„_^ + \y„-)Lx. It is clear that the greater the number of intervals (i.e. the smaller Aa; is) the closer wiU the sum of the areas of the trapezoids approach the area under the curve. Illustkatite Example 1. Calculate C x^clx by the trapezoidal rule, dividing X = 1 to a; = 12 into eleven intervals. ^ Solution. Here = — — — = 1 = Aj. The area in question is under the curve Ti 11 y = x2. Substituting the abscissas x = 1, 2, 3, • ■ • , 12 in this equation, we get the ordi- nates y = 1, 4,9, ■ ■ ■, 144. Hence, from {A), area = (i + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + J. 144) • 1 = 577^. By integration J x^dx = — = 575f . Hence, in this example, the trapezoidal rule is in error by less than one third of 1%. 240. Simpson's rule (parabolic rule). Instead of drawing straight lines (chords) between the points of a curve and forming trapezoids, we can get a still closer approximation to the area by connecting the points with arcs of parabolas and summing up the areas under these arcs. A parabola with a vertical axis may be passed through any ' three points on a curve, and a series of such arcs will fit the curve more closely than the broken line of chords. We now divide the inter- val from x=a=OM„ to x=b=OM^ into an even number (= n) of parts, each equal to Ax. Through each successive set of three points ij, ij, P^; P^, P^, P^; etc., are drawn arcs of parabolas with vertical axes. From the figure area of parabolic strip M^P^P^P^M^ = area of trapezoid M^^P^P^M^ + area of parabolic segment P^P^Pi- But the area of the trapezoidilf„^^i!f^ =\(iya + y^'^ ^^ = (2^0+^2)^^' and the area of the parabolic segment J^iji^ = two thirds of the circumscribing parallelogram P^PjP^P^ 450 INTEGEAL CALCULUS Hence area of first parabolic strip M^ll^I^M,^ Similarly, second strip = — (y^ + 4 «/g + ?/ J, Apt third strip = -^ (y, + 4 2/5 + y^}. wth strip = — («/„_2+ 4 i/„_-, + y„). Adding, we get A-y -3- (2/0 + 4^1+ 22/^+42/^+ 2y^+ ..- + 2y„_,+ 42/„_i + yJ as the sum of these areas. Hence Simpson's rule is (n being even) Ax (E) area = _ (i/„ + 4 y^ + 2 y, + 4 1/3 + 2 y, + • ■ • + y„) . As in the case of the trapezoidal rule, the greater the number of parts into which M^M^ is divided, the closer will the result be to the area under the curve. Illustrative Example 1. Calculate | x'ds by Simpson's rule, taking ten intervals. Ja Solution. Here = = 1 = Ax. The area in question is under the curve n 10 y = a;'. Substituting the abscissas a; = 0, 1, 2, • • • , 10 in ?/ = i', we get the ordinates 2/ = 0, 1, 8, 27, • ■ • , 1000. Hence, from (5), area = J (0 + 4 + 16 + 108 + 128 + 500 + 432 + 1372 + 1024 + 2916 + 1000) = 2500. ^10 ra;*"]!" By integration, I x^dx = -r = 2500, so that in this example Simpson's rule happens to give an exact result. EXAMPLES 1. Calculate the integral in Illustrative Example 1 (above) by the trapezoidal rule, taking ten intervals. Ans. 2525. 2. Calculate / — by both rules when n = 12. •^' ^ ^ns. Trap. 1.6182; Simp. 1.6098. 3. Evaluate I x^dx by both rules when n = 10. ^ Ans. Trap. 3690 ; Simp. 3660. ■ log^„a:da; by both rules when n = 10. Ana. Trap. 6.0656 ; Simp. 6.0896. 5. Evaluate / -„ by both rules when n = 6. " "'"'^ ^ns. Trap. 1.0885; Simp. 1.0906. TABLE OF INTEGRALS 451 sin xdx by both rules for ten-degree intervals. • ■ 7. Evaluate ( x^dx by both rules for n = 12. 8. Find the error in the evaluation of I x^dx by Simpson's rule when n = 10. 9. Evaluate | e^dx by Simpson's rule when n = 10. 241. Integrals for reference. Following is a table of integrals for reference. In going over the subject of Integral Calculus for the first time, the student is advised to use this table sparingly, if at all. As soon as the derivation of these integrals is thoroughly understood, the table may be properly used for saving time and labor in the solution of practical problems. SOME ELEMENTARY FORMS 1. C(du ±dxi±dw ±---)— Cdu ± jdv ± jdw ± • ■ • . 2. fadv = a Cdv. 4. ix'^dx = + 0,n^—lL. 3. fdf{x) = ff{x) dx=f(x)+G. 6. J ^ = log X + C. Forms containing Integral Powers of a + bx „ r c^ 1 1 / , i_\ , /^ M - <5 *■ ^ ^ ■'£ ' "'V t' 6. ( — = -log(a + 6x) + C. y -~ A^ J a + ox b 7. r(. + ;^)ndx= (° + ^>;"V c.n^-l. J 6 (m + 1) 8. Cf{x, a + bx)dx. Try one of the substitutions, z = a + bx, xz = a + bx. 9. f-!^^ = -la + bx-alog(a+bx)-\ + C. J a + bx b^ ,0. C-^^ = -U{a + bxf- 2a(a + bx) + anog(a + bx)] + C. J a + bx V , r dx 1. a + bx 1. I = log f- C. J x{a+bx) a x 2. r_^_=_l + liog^+^ + c. J x^(a + bx) ax a^ x J (a-^bxf 6^ ' a + 6xj 4 r_?!^_ = ira + 6x-2alog(a+te)--^l+C. r_^_ :3 _J__ _ 1 log ^:!^ + c. Jx{a + 6x)^ a(a + ta) o^ x r xdx _i.r ,J__ + 1 1+c. J (a + bxf~b'^l a + bx 2(a+bx)''^ OM 452 INTEGEAL CALCULUS Forms containing a^ + x^, (fi — x% a + hx", a + hx^ 17. r^ = Itan-i? + C ; f-^ = Un-i'x + C. J a'-x^ 2a ^a-x Jx^-a? 2a "x + a 19. r ^ =J-Un-^xJ^+C. J a + bx^ Va6 ^« 20. r '^ =±.iog^±^+c. J a^-b^x^ 2ab ^a-bx 21. fx^(a + bxf')Pdx b(np + m + 1) b{np + m + 1)^ 22. f^(a + bxn).dx = '^^"^'' + ^^' + ""^ J x'^(a + bxn).-iax. J np + m + 1 np + m + IJ 23. f ^ _ 1 {m— n + np — l)b r dx ~ (m — l)ax'»-i(a+ 6a;»)P-i (m— l)a J a™ - » (a + fee")*' 24. r — ^ — J a"' (a + 6X")* _ 1 m— n + np — lr dx ~ an{p — l)i"'-i(a + hx^y-'*^ an{p — 1) J x'^(a + bx'')p-^ /■(a + 6x»)Pdx _ {a + 6a;»)i' + i 6 (m — » — np — 1) /^ (a + 6x»)^ da; ./ a;"> a(m— l)!"-! a(?n — 1) J z™-n r{a + bxf')Pdx _ (a + 6x")p anp /^(a + 6x")?-i(ic J x"* (rip — m + l)x'»-i rtp — m + lJ x™ /" x'^'dx _ a!"'-'> + i a{m — ji + 1) /- x"'-'^dx ■ J (o + ftx")* ~ 6 (m — np + 1) (a + 6x»)»' - 1 b(m, — np + l)J {a + bx'')p' C ^'"'^ _ x^ + i m+n — np + l/- x""*!); ' •/ (a + 6x»)P~ a7i(p — l)(a + 6x")p-i (in{p — 1) J(a + 6x»)p-i' ■ -/ (a2 + x2)» ~ 2(n- l)a2L(a2 + x^)"-! "^ ^ "~ 'J (a^ + x^)"-!]' ' J (a + 6x2)" ^ 2{n - l)a\_(a + te^)"-! ^ "~ 'J (a + 6x2)»-ij' „, /^ xdx 1 f dz , , 31. I ^— = - I , where z = x^. J (a + 6x^2)" 2 J (a + fiz)" 32 r_^^?_ ^ r^ , 1 r dx 'J (a + bx^)" 2b(n-l)(a + bx^)n-i^ 2b(n-l)J (a + bx^)»-l' 33. r ^ =±log_^L_ + c. J X (a + ftx") an a + &x» 34 r '^ -^ r '^ '' r ■ J x^(a + 6x2)" a J x^ (a + ftx^)"- 1 aJ (a + bx^)" 35. r^^ = JLiog(x2 + «Uc. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46 47 48. 49. 50. TABLE OF INTBGEALS x^dx X a r dx r X'ax _x a r ■I a + bx^~b~bJ a + bx^ f-J^ =J_log^L_-+c. J x{a + bx^) 2a ^a+bx^ r ax _ \ b r a. ■> x^ (a + 6x2) ~ oa; aJ a-\- /dx, (ST 6x2 (a + 6x2)2 2a(a + 6x2) 2 1 r dx 2aJ a + 6x2' Forms containing Va + bx rxV^T^^^^^+^+c. J 1562 I x2 Va + 6x dx : 2(8a2- 12 a6x + 1562x2) V(a + bxf 1056= + 0. •^ Va + 6x 362 /- x2^ 2(8a2-4a6x + 362x2) , — -— - ^ I -~^^^ = — i ^V a + 6x + C ■^ Va + 6x 156' r dx 1 , Va + bx — Va , „ , ^ „ I ^ = — ;= log- , — ;= + C, for a > 0. /: X Va + 6x Va Va + 6x + Va (Zx 2 , , /a + 6x , „ , ^n tan- i-v / — ■ 1- C, for o < 0. \ — a dx a X Va + 6x "> dx — Va + 6x 6 r aa •^ .T-Va + bx ax or d! 2a'^xVa + 6a rVa + bxdx „ / , , , r . I ■ = 2Va + 6x+a/ J x J . dx X Va + 6x Forms containing Vx2 + a^ C(x^ + a?')^dx = -Vx2+a2 + ^log(x + Vx' + o2) + C. 453 3 a* 51 52 58 54, f{x2 + a^i^dx = -(2x2 + 5a2) Vx2 + a? + ^log(x + Vx^ + a2) + O. J 8 8 r(,. + „.)fcfe = ^(?!±^ + -ij-/(x2 + a2)t ' J ^ n + 1 n + 1*/ n . rx(x2+a2)2dx = dx. (x2 + a2) K + 2 -+C. rx2 (x2 + a2)idx = I (2 x2 + a2) v^M^ - ^ log (x + VSm^) + C. "^ (x2 + a2)i •/ :log(x + Vx2+a2) + 0. =+(7, (x2 + a2)t a2V'a!2 + a2 454 mTEGllAL CALCULUS 55. f — ?^?— = VxM^+C'. 56. f ^'^ = ? Vi^T^ - ^ log(x + V^^T^) + C. 57. r ^'^ =_ ■^_ + iog(^ + ^rr^) + c. 58. / r = -log , +g- ''x(x2+a2)4 " a + vx2 + a2 59. r_^=-^!±^+c. . 'K^x- + a^)i ^a^x'' ^2a3 ^ /'(x2+a2)idx / „ , „ , a + Va^ + x^ . I i '- = V a^ + x^ — o log 1- C. J X X r(x^+a^)idx Vx" + a^ , , / , / , ," o \ , ^ . / ^ — - — '- = ■ 1- log (x + Vx^ ^- a') + C. J x^ X Forms containing V a;^ — a^ 63. f{x^ - a2)idx = -Vx2-o2- ^log (x + Vx^-a^) + C. 64. C{x^ - a2)t dx = -(2 x2 - 5 o2) Vx^-a^ + ^ log (x + Vx^-a^) + C. «/ 8 8 n 65. r(g'- ag)'dx = ^(^'- "'')' _ J^ r(x'i + a?Y~''dx. J ^ 71 + 1 re + 1-' x(x2 - a'^fdx = y^ ^ + C. n + 2 67. Cx^(3? - a^)idx = - (2x2 - ^2) Vx^-o^ - —log (x + Vx^-az) + C. «/ 8 8 68. 69. r — = log (x + Vx2 - a2) + C. (x2-a2)4 (2x 70. I" — 5^? — = Vx2-(i2+C. ■^ (x2-a2)i 71. r_^^ = |V^^3^ + ^log(x + V^^^^=) + Cf. •^ /™2 _ ^2\i .^ 2 (x2 - a")^ 72. x'^dx ;. r '""^ 3 — - y^ + log(x + V^^T^) + C. •^(x2-a2)t Vx2 - a2 r dx 1 ,x,_,/-dx , „ I = -sec-i- + C; I ___-=sec-ix + C. ■^x(x2-ani " " *' xVx2-l 73, . x(x2-a2)t TABLE OF INTEGRALS 455 74. f ^ = ^5E»!+c. X' :2(s2_a2)i "'^ 75. /; dx -Vx" - a^ 1 X + T-^sec-i-+ G. 76. C {x^-a^)idx r-. , a J ~ = va;2 — d' — a cos- 1 - + 77 -+C. X ^- J ^ = - -^^ + log (x + Vx^-a^) + C. FOKMS CONTAINING Va^ — X^ 78. f(a^ - x^)idx = Va2 - x2 + ^ sin- 1 - + C. "^ 2 2 a 79. r(a2 - a;2)^dx = 5 (5 a2 _ 2 x^ Va^-x^ + ^ sin- 1 ? + C. ■' 8 8 o 80. /(a^- x^)t x^ X a 106. rV2ax-x^^^_(2ax-x2). •^ x8 3ax8 107. r — ^? — = ^-" + c. • V2 OX - x° __(2ax-x2)l x' ~ 3ax8 dx _ X— a (2 ox - x2)l a^ V2ax-x2 108./^^ = — ^=H-C. (2ax-x'')t aV2ax-x2 109. Jf{x, V2ax-x2) dx = Jf(2 + a, Va^- 32) dz, where z = x-a. 110. f , = = log(x + a + V2 ox + x2) + C. ■^ V2 ax + x2 111. JF{x, V2ax + iE2)dx =/^(z - a, Vz--=-a2)dz, wher? ? = x + a. TABLE or INTEGEALS 457 Forms containing a + bx ± cx^ 2 ^ , 2ca; + 6 tan-i - ■ + ox + cx2 V4ac-62 y/iaa-^ 112. I — - tan-i- — 4- 0, when 6^ < 4 gc. J a + Ox + CX^ -v/4. /»/• _ W a/4 n/! _ W 113. I = - log ; — ;z=^ + C, when 6^ > 4 ac. J a + bx + cx' v'62_4ac 2cx + 6 + Vft^- 4ac 114. r ^5 = 1 ,^gV g+4ac + 2ex-6 _|^ ^ ■^a + 6x-cx2 V62 + 4ac V62 + 4ac - 2cx + 6 115. f ■ ^ = ^lng(2fix4-h4-2v^Va + 6x4-ex2>4-0. •^ Va + 6x + ca;2 Vc 116. fVa + 6x +cx2dx = ^"'^ "*" '' Va + Ox + ex'' - ^^~ '^'^ log(2cx + 6 + 2 Ve Va+ to + cx'^) + O. ,,_ r dx 1-1 2cx— 6 , „ 117. I =— -an-i— — +(7. "^ Va + 6x — cx^ vc vP+4ac ,,„ /^ / ; ;, 2cx — 6 / 7 3 b^+iac . , 2cx — 5 118. / Va + bx—cx^dx = Va + 6x— cx^ H sm-i — — + V. 119. I — — = ■ log(2cx + 6 + 2 VcVa + bx + cx^)+ V. '' -y/a + bx + cx^ '^ lA 120. r ^ _ ^-^'^ + ^-I^^Asin-i ^^-^ ^O. '1-^ •^ Va + 6x - cx2 « 2ct V62 + 4 ac Other Algebraic Forms 121. f^h±ldx. = V(a + X) (6 + x) + (a - 6) log (Va + x + Vb + x) + G. J \ b + x 122. r /^L:r^^ = V(a - X) (6 + X) + (a + 5) sin- i-i/^-i^ + <"• ^\6+x \a+o 123. r^/^+-5 (Zx = - V(a + x)(6-x) - (a + 6) sin- '•x/^^ + C. J\6 — X f \a + o 124. rJl±^dx = -Vl-x2+sin-ix,+ C. ' •^ V(x-aW-x) >'/3-« Exponential and Trigonometric Forms ' 126 Ca^dx = -^—+ C. • 129. / sinxdx =— cosx + C. J log a -^ 127. fe^^iZx = e^+ C: 130. fcosxiix = sinx + C. 128. re^dx = —+ C. 131. Ttanxcfe = logsecx =— logcosx + C. 132. fcot xdx = log sin x+ C. secxox = j = log(.secx + tanx) = log tan \^^ + 2 j + '^- 458 INTEGRAL CALCULUS coseo xdx = I = log (cosec x — cot x) = log tan - + G. ^ sina; 2 136. 137 135. fsec^xdx = ta,nx + C. 138. J coseo soot s(i)5=- cosec a; + C. .. rcosec2 xdx,=-cotx+C. 139. Jsin^ ado; = | - - sin 2 a; + C. . Csecxtanxdx = seca; + G. 140. Jcos^xxtx = | + -sin2z + C. , ., r . , slnx-ixcosa; n— 1 /" . „ , 141. I sin»X(Jx = 1 I sin''-2a^. J n ji ■/ ,.„ /• , cos«-ixsina; n — 1 /■ _ „ , 142. I cos»a;da; = — -i | cos»-2xdx. ■/ n n >/ ' -/ sin^x n — lsin"-ix n—1-/ sin /cte _ 1 sin X 71 — 2 /" cos"x n— lcos»-ix n — l-zc 144 sm»-2x dx 1 sinx , n — 2 f dx cos"-^x cos^-iisino + ix m — 1 ,,- r . , cos^-J^isinn + ix m — 1 r _„ ., . , 145. / cos"'xsin»xax — 1 ( cos™-^xsin''xax. J m+ n m + nJ ■•An r __ • J sin»-ixcos™+ix n — 1 r ■ , « j 146. I cos™xsm»iax = 1 ( cos'»xsm»j-2xax. ■' m + n m + nJ /dx _ 1 1 m + n — 2/" dx sin^xcosnx n—1 sin" - 1 x cos" - ^ x n — 1 -/ ; 147 148 sin^xcos^-^x dx 1 1 m + n—2 r dx /dx _ 1 1 m + n—2r sin^xcos"! m— 1 sin^-ixcosn-^x m— 1 -/si sin'"-2xcos"x ..„ /'oos"xdx_ cos'n + ix m — n + 2 /"oos^xdx J sin»x (n — l)sin»-ix n — 1 -•' sin'-i^x /cos^xdx _ cos^-^x m — 1 fcos^-'-'xdx sin»x (m — m)sin»-ix m—nJ sin»x 150 ^ sm»x (m — 151. I sinx cos»xdx = f- G. J n+ 1 /sin" i" ^ X sin»x coBxdx = 1- G. n + 1 sin" + 1 x n + tan" - 1 X 154 tan»xdx = | tan»-2a;dx + C. n — 1 J /cot** — ^ X /" oot»xdx= ( cot»-2xdx + C. n — 1 •) ICC C- -J sin(m + n)x sin(m— n^x „ 155. I sin Tnx sin nxdx = i — '- — i 'sL+a. J 2(m + n) 2(m-n) icn C J sin(m + n)x sin(m — n)x „ 156. I cosmxcosnaxix = — i — - — '- — \- ■ — i l!iz + c •■' 2(m + n) 2(m-n) .-„ C . - cos (m-\- rC)x cos (m — nl x 157. I sm mx cos jixdx = ^ — —i ^ ZlsL+G J 2(m + n) 2(m-n)- 158. / — -4 = r^ tan- 1( -1/^^ tan ^ ) + C, wheno>6. J a + bcosx ^gp. _ j2 \\a + 6 2/ TABLE OF INTEGEALS 459 V6 — a tan - + V6 + a 159. f — — =z log + C, when a