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AN 
 
 ELEMENTARY TREATISE 
 
 DIFFERENTIAL AND INTEGRAL 
 
 CALCULUS 
 
 WITH NUMEROUS EXAMPLES. 
 
 EDWARD A. BOWSER, LL.D., 
 
 if 
 
 PROFESSOR OF MATHEMATICS AND ENGINEERING IN RUTGERS COLLEGE 
 
 NEW YORK 
 AN NOSTRAND COMPANY 
 
 23 Murray and 27 Warren Streets 
 1894 
 

 
 COPTRIGHT, 1880, by E. A. BOWSER. 
 
PREFACE 
 
 rpHE present work on the Differential and Integral Calculus is 
 designed as a text-book for colleges and scientific schools. The 
 aim has been to exhibit the subject in as concise and simple a manner 
 as was consistent with rigor of demonstration, to make it as attractive 
 to the beginner as the nature of the Calculus would permit, and to 
 arrange the successive portions of the subject in the order best suited 
 for the student. 
 
 I have adopted the method of infinitesimals, having learned from 
 experience that the fundamental principles of the subject are made 
 more intelligible to beginners by the method of infinitesimals than by 
 that of limits, while iu the practical applications of the Calculus the 
 investigations are carried on entirely by ihe method of infinitesimals^ 
 At the same time, a thorough knowledge of the subject requires tj 
 the student should become acquainted with both methods ; and $ 
 this reason, Chapter III is devoted exclusively to the method ol 
 limits. In this chapter, all the fundamental rules for differentiating 
 algebraic and transcendental functions are obtained by the method of 
 limits, so that the student may compare the two methods. Thl^chap- 
 ter may be omitted without interfering with the contu^fty of the 
 work, but the omission of at least the first part of the Siapter is not 
 recommended. 
 
 To familiarize the student with the principles of the subject, and 
 to fix the principles in his mind, a large number of examples is given 
 at the ends of the chapters. These examples have been carefully 
 selected with the view of illustrating the most important points of 
 the subject. The greater part of them will present no serious diffi- 
 culty to the student, while a few may require some analytical skill. 
 
 797960 
 
IT PREFACE. 
 
 In preparing this book, I have availed myself pretty freely of the 
 writings of the best American and English and French authors. 
 Many vol nines have been consulted whose titles are not mint ion. d, 
 as credit could not be given in every case, and probably 1 am Indebted 
 to those volumes for more tban I am aware of. The chief soured 
 upon which I have drawn are indicated by the references in the bod| 
 of the work and need not be here repeated. For examples, I have 
 drawn upon the treatises of Gregory, Price, Todhunter, Williamson, 
 Young, Hall, Rice and Johnson, Ray, and Olney, while quite a num- 
 ber has been taken from the works of De ISiorgan, Lacroix, Serret, 
 Courtenay, Loomis, Church, Byerly, Docharty, Strong, Smyth, and 
 the Mathematical Visitor; and I would hereby acknowledge my 
 indebtedness to all the above-named works, both American and 
 foreign, for many valuable hints, as well as for examples. A few 
 examples have been prepared specially for this work. 
 
 I have again to express my thanks to Mr. R. VV. Prentiss, Fellow 
 in Mathematics at the Johns Hopkins University, for reading the MS 
 and for valuable suggestions. 
 
 E. A. B. 
 Rutgers College, 
 New Brunswick, N. J., June, 1880. 
 
 ,.! 
 
TABLE OF CONTENTS 
 
 PART I. 
 DIFFERENTIAL CALCULUS 
 
 CHAPTER I. 
 
 FIRST PRINCIPLES. 
 
 <LRT. PAGE 
 
 1 . Constants and Variables 1 
 
 2. Independent and Dependent Variables 1 
 
 3. Functions. Geometric Representation 2 
 
 4. Algebraic and Transcendental Functions 4 
 
 5. Increasing and Decreasing Functions 5 
 
 6. Explicit and Implicit Functions 6 
 
 7. Continuous Functions ■, 6 
 
 8. Infinites and Infinitesimals 7 
 
 9. Orders of Infinites and Infinitesimals 8 
 
 10. Geometric Illustration of Infinitesimals 10 
 
 11 . Axioms. . . c , 12 
 
 Examples 13 
 
 CHAPTER II. 
 
 DIFFERENTIATION OF ALGEBRAIC AND TRANSCENDENTAL 
 FUNCTIONS. 
 
 1 2. Increments and Differentials 15 
 
 13. Consecutive Values 16 
 
 14. Differentiation of Sum of a Number of Functions 16 
 
 15. To Differentiate y = ax±b 17 
 
VI CONTENTS. 
 
 ART. PAGS 
 
 10 Differentiation of a Product of Two Functions 18 
 
 1 7. Differentiation of a Product 20 
 
 18. Differentiation of a Fraction 20 
 
 19 Differentiation of any Power 21 
 
 Examples 23 
 
 Illustrative Examples 26 
 
 20. Logarithmic and Exponential Functions 29 
 
 21 . Differentiation of an Exponential 31 
 
 22. Differentiation of an Exponential with Variable Base 32 
 
 Examples , S3 
 
 23 Logarithmic Differentiation. Examples 34 
 
 Illustrative Examples 35 
 
 TRIGONOMETRIC FUNCTIONS. 
 
 24. To Differentiate y = sin x 37 
 
 25. To Differentiate y = cos x 38 
 
 26. To Differentiate y = tan x 38 
 
 27. To Differentiate y = cot x 39 
 
 28. To Differentiate y = sec x , 39 
 
 29. To Differentiate y = eosec x 40 
 
 30. To Differentiate y = versin x. 40 
 
 31. To Differentiate y — covers x 40 
 
 32. Geometric Demonstration 41 
 
 Examples 43 
 
 Illustrative Examples. 44 
 
 CIRCULAR FUNCTIONS. 
 
 33. To Differentiate y = sin-' x 46 
 
 34. To Differentiate y = cos- 1 x 46 
 
 35. To Differentiate y — tan- 1 x 4? 
 
 36. To Differentiate y = cot" 1 x 4? 
 
 37. To Differentiate y = sec -1 x 47 
 
 3S. To Differentiate y = cosec-' x 47 
 
 39. To Differentiate y = vers -1 x 48 
 
 40. To Differentiate y = covers -1 x , 48 
 
 Examples , , 48 
 
 Miscellaneous Examples 49 
 
CONTENTS. VI) 
 
 CHAPTER III. 
 
 LIMITS — DER1V ED FUNCTIONS. 
 
 ./RT PAGB 
 
 41. Limiting Values 59 
 
 42. Algebraic Illustration 59 
 
 43. Trigonometric Illustration 60 
 
 44. Derivatives 62 
 
 45. Differential and Differential Coefficient 63 
 
 46. Algebraic Sum of a Number of Functions 63 
 
 47. Product of Two Functions 64 
 
 48. Product of any Number of Functions 65 
 
 49. Differentiation of a Fraction 66 
 
 50. Any Power of a Single Variable 67 
 
 51. Differentiation of log x 68 
 
 52. Differentiation of a x 68 
 
 53. Differentiation of sin x 68 
 
 54. Differentiation of cos x 69 
 
 CHAPTER IV. 
 
 SUCCESSIVE DIFFERENTIALS AND DERIVATIVES. 
 
 55. Successive Differentials. Examples 71 
 
 56. Successive Derivatives 74 
 
 56a. Geometric Representation of First Derivative. Examples . 76 
 
 CHAPTER V. 
 
 DEVELOPMENT OF FUNCTIONS. 
 
 57. Definition of Development of a Function 81 
 
 58. Maclaurin's Theorem 81 
 
 59 Tbe Binomial Theorem 85 
 
 60. To Develop y = sin x and y = cos x . 85 
 
 61. The Logarithmic Series. 86 
 
 62. The Exponential Series 90 
 
 63. To Develop y = tan- 1 x 91 
 
 64. Failure of Maclaurin's Theorem . 92 
 
 65. Taylor's Theorem. Lemma 93 
 
 66. To find Taylor's Theorem 95 
 
 67. The Binomial Theorem 97 
 
** T TAG! 
 
 68. To Develop u' — sin (x + y) 98 
 
 09. The Logarithmic Beiiei <js 
 
 70. The Exponential Seriea <is 
 
 71 Failure of Taylor's Theorem 99 
 
 Examples 100 
 
 CHAPTER VI. 
 
 EVALUATION OF INDETERMINATE FORMS. 
 
 72. Indeterminate Forms 103 
 
 73. Common Factors. Examples 104 
 
 74 Method of the Differential Calculus 105 
 
 oo 
 
 75. To evaluate Functions of the form — 108 
 
 oo 
 
 76. To evaluate Functions of the form Oxoo Ill 
 
 77. To evaluate Functions of the form oo — oo 112 
 
 78. To evaluate Functions of the forms 0°, oo°, and 1 ±Q0 113 
 
 79. Compound Indeterminate Forms 116 
 
 Examples 116 
 
 CHAPTER VII. 
 
 FUNCTIONS OF TWO OR MORE VARIABLES. — CHANGE OF 
 THE INDEPENDENT VARIABLE. 
 
 80. Partial Differentiation 120 
 
 81. Differentiation of a Function of Two Variables 122 
 
 82. To find the Total Derivative of u with respect to # 125 
 
 83. Successive Partial Differentiation 130 
 
 84. Proof that Order of Differentiation is indifferent 131 
 
 85- Successive Differentials of a Function of Two Independent 
 
 Variables 133 
 
 86. Implicit Functions 135 
 
 87. Differentiation of an Implicit Function 136 
 
 88. Second Derivative of an Implicit Function 138 
 
 89. Change of the Independent Variable 140 
 
 90. General Values of % , p{ t f^, etc 141 
 
 ax ax' ax 6 
 
 91. Transformation for Two Independent Variables 145 
 
 Examples 147 
 
CONTENTS. IX 
 
 CHAPTER VIII. 
 
 MAXIMA AND MINIMA OF FUNCTIONS OF A SINGLE 
 VARIABLE. 
 
 AKT. PAGE 
 
 92. Definition of a Maximum and a Minimum 151 
 
 93. Condition for a Maximum or Minimum 151 
 
 94. Geometric Illustration 152 
 
 95. Discrimination between Maxima and Minima 154 
 
 96. Condition given by Taylor's Theorem 154 
 
 97. Method of finding Maxima and Minima Values 155 
 
 98. Alternation of Maxima and Minima Values 156 
 
 99. Application of Axiomatic Principles 157 
 
 Examples 159 
 
 Geometric Problems 164 
 
 CHAPTER IX. 
 
 TANGENTS, NORMALS, AND ASYMPTOTES. 
 
 100. Equations of the Tangent and Normal 172 
 
 101. Length of Tangent, Normal, Subtangent, etc 175 
 
 102. Polar Curves. Tangents, Normals, Subtangents, etc 178 
 
 103. Eectilinear Asymptotes 181 
 
 104. Asymptotes determined by Expansion 184 
 
 105. Asymptotes in Polar Co-ordinates. Examples 186 
 
 CHAPTER X. 
 
 DIRECTION OF CURVATURE — SINGULAR POINTS — TRACING 
 OF CURVES. 
 
 106. Concavity and Convexity 191 
 
 107. Polar Co-ordinates 192 
 
 108. Singular Points 194 
 
 109. Points of Inflexion 194 
 
 110. Multiple Points 196 
 
 111. Cusps 199 
 
 112. Conjugate Points 201 
 
 113. Shooting Points. Stop Points 203 
 
 114. Tracing Curves 205 
 
 Examples 206 
 
 115. Tracing Polar Curves. Examples 210 
 
CONTENTS. 
 
 OHAPTEB XI. 
 
 RADIUS OF CURVATURE, EVOLUTES AND INVOLUTES — 
 ENVELOPES. 
 
 ART. PAGB 
 
 1 10. Curvature 216 
 
 117. Order of Contact of Curves 217 
 
 118. Dependence of Order of Contact on Arbitrary Constants 218 
 
 119. Radius of Curvature. Centre of Curvature 219 
 
 120. Second Method 220 
 
 121. Radius of Curvature in Polar Co-ordinates 222 
 
 122. Radius of Curvature at a Maximum or Minimum 222 
 
 123. Contact of Different Orders 223 
 
 Examples 224 
 
 124. Evolutes and Involutes 226 
 
 125. Equation of the Evolute 228 
 
 126 Normal to an Involute is tangent to Evolute 230 
 
 127. Envelopes of Curves 231 
 
 128. Equation of the Envelope of a Series of Curves 232 
 
 Examples 233 
 
 PART II. 
 INTEGRAL CALCULUS. 
 
 CHAPTEK I. 
 
 ELEMENTARY FORMS OF INTEGRATION. 
 
 129. Definitions 238 
 
 130. Elementary Rules for Integration 2:)9 
 
 131. Fundamental Forma 242 
 
 132. Integration by Transformation into Fundamental Forms 243 
 
 133. Integrating Factor. Examples 247 
 
 184, Transposition of Variable Factors. Examples 249 
 
 135. Trigonometric Reduction. Examples 254 
 
CONTENTS. xl 
 
 CHAPTER II. 
 
 INTEGRATION OF RATIONAL FRACTIONS. 
 
 A.RT. PAGE 
 
 136. Rational Fractions. 256 
 
 137. Case 1. Decomposition of a Rational Fraction . . 256 
 
 138. Case 2. " " " 259 
 
 139. Case 3. " " " 262 
 
 Examples. 263 
 
 CHAPTER III. 
 
 INTEGRATION OF IRRATIONAL FUNCTIONS BY 
 RATIONALIZATION. 
 
 140. Rationalization 269 
 
 141. Monomial Surds 269 
 
 142. Binomial Surds of the First Degree 270 
 
 rpltl -f- I W/w 
 
 143. Functions of the Form ; 272 
 
 (a + WY* 
 
 144. Functions containing only Trinomial Surds 272 
 
 145. Binomial Differentials 276 
 
 146. Conditions for Rationalization of x m {a + bx n )i dx 277 
 
 Examples 280 
 
 CHAPTER IV. 
 
 INTEGRATION BY SUCCESSIVE REDUCTIONS. 
 
 147. Formula of Reduction 285 
 
 148. Formula for Diminishing Exponent of a?, etc 285 
 
 149. Formula for Increasing Exponent of a 1 , etc 287 
 
 150. Formula for Diminishing Exponent of Parenthesis 288 
 
 151. Formula for Increasing Exponent of Parenthesis 289 
 
 Examples. Applications of Formulae 289 
 
 Logarithmic Functions 295 
 
 152. Reduction of the Form / X(log x) n dx 295 
 
 /x"dx 
 , — -- 297 
 log" x 
 
1 (t z dx 
 155. Reduction of the Form / — .- 800 
 
 Xli CONTESTS. 
 
 ART. PAGE 
 
 Exponential Fonns 299 
 
 154, Reduction of the Form / a^x^dx d9t 
 
 /<i J <l.v 
 
 156. Trigonometric Functions , , . 8 ( )1 
 
 157. Formulae of Reduction for / sin"' cos" Odd 303 
 
 158. Integration of sin'» cos" dd !...... 305 
 
 159. Reduction of the Form / sc" ccs ax dx 307 
 
 160. Reduction of the Form /V» cos" x dx 308 
 
 161. Integration of fix) sin- 1 x dx, f(x) tan- 1 x dx, etc 809 
 
 162. Integration of dy = z - a - 310 
 
 ° ° a + b cos 
 
 Examples 312 
 
 CHAPTER V. 
 
 INTEGRATION" BY SERIES — SUCCESSIVE INTEGRATION — IN- 
 TEGRATION OF FUNCTIONS OF TWO VARIABLES. 
 DEFINITE INTEGRALS. 
 
 163. Integration by Series 319 
 
 164. Successive Integration 321 
 
 /n 
 
 X dx n into a Series 323 
 
 166. Integrations of Functions of Two or More Variables 326 
 
 167. Integration of —^ =/(«, fft 326 
 
 168. Integration of Total Differentials of the First Order 329 
 
 169. Definite Integrals. Examples 331 
 
 170. Change of Limits 334 
 
 Examples 337 
 
 Formulas of Integration 340 
 
 CHAPTER VI. 
 
 LENGTHS OF CURVES. 
 
 171. Length of Plane Curves referred to Rectangular Axes 347 
 
 172. Rectification of Parabola 348 
 
CONTENTS. XI 11 
 
 ART. PAGE 
 
 173. Serai-cubical Parabola ' 349 
 
 174. The Circle 349 
 
 175. The Ellipse 350 
 
 176. The Cycloid 351 
 
 177. The Catenary 352 
 
 178. The Involute of a Circle 352 
 
 179. Rectification in Polar Co-ordinates 353 
 
 180. The Spiral of Archimedes 353 
 
 181. The Cardioide 353 
 
 182. Length o f Curves in Space 354 
 
 183. Intersection of Two Cylinders. Examples 355 
 
 CHAPTER VII. 
 
 AREAS OF PLANE CURVES. 
 
 184. Areas of Curves 360 
 
 185. Area between Two Curves 361 
 
 186. Area of the Circle 361 
 
 187. The Parabola 362 
 
 188. The Cycloid 362 
 
 189. The Ellipse ; 363 
 
 190. Area bat ween Parabola and Circle 363 
 
 191. Area in Polar Co-ordinates.' 364 
 
 192. The Spiral of Archimedes: .' 364 
 
 Examples 365 
 
 CHAPTER VIII. 
 
 AREAS OF CURVED SURFACES. 
 
 193. Surfaces of Revolution 369 
 
 194. Quadrature of the Sphere 370 
 
 195. The Paraboloid of Revolution 371 
 
 196. The Prolate Spheroid 372 
 
 197. The Catenary '. 372 
 
 98. The Surface of Revolution generated by Cycloid 373 
 
 ■id. Surface of Revolution in Polar Co-ordinates .* 374 
 
 200. The Cardioide 374 
 
 201. Any Curved Surfaces. Double Integration 375 
 
 202. Surface of the Octant of a Sphere 376 
 
 Examples 37? 
 
Xt? COAT 
 
 CHAPTER IX. 
 
 • VOLUMES OF SOLID3. 
 
 »3T. PAQH 
 
 203. Solids of Revolution 381 
 
 204 The Sphere 381 
 
 20"). Solid of Revolution of Cycloid 382 
 
 206. Solid of Revolution generated by Cissoid 383 
 
 207. Volumes of Solids bounded by any Curved Surface 383 
 
 208. Mixed System of Co-ordinates 388 
 
 209. Cubature in Polar Co-ordinates 289 
 
 Examples 390 
 
PART I. 
 
 DIFFERENTIAL CALCULUS, 
 
 CHAPTER I. 
 
 FIRST PRINCIPLES. 
 
 1. Constants and Variables. — In the Calculus, as in 
 Analytic Geometry, there are two kinds of quantities used, 
 constants and variables. 
 
 A constant quantity, or simply a constant, is one whose 
 value does not change in the same discussion, and is repre- 
 sented by one of the leading letters of the alphabet. 
 
 A variable quantity, or simply a variable, is one which 
 admits of an infinite number of values within certain limits 
 that are determined by the nature of the problem, and is 
 represented by one of the final letters of the alphabet. 
 
 For example, in the equation of the parabola, 
 
 f — %pv> 
 
 x and y are variables, as they represent the co-ordinates of 
 any point of the parabola, and so may have an indefinite 
 number of different values. 2p is a constant, as it represents 
 the latus rectum of the parabola, and so has but one fixed 
 value. Any given number is constant. 
 
 2. Independent and Dependent Variables. — An 
 
 independent variable is one to which any arbitrary value may 
 
4 \E OR MORE VARIABLES. 
 
 be assigned ;it \ leasare. A dependent variable is one whose 
 
 value varies in eoiiMMjuenee of the variation of the inde- 
 pendent variable or variables with which it is connected. 
 Thus, in the equation of the circle 
 
 s* + f = r>, 
 
 if we assign to X any arbitrary value, and find the correspond- 
 ing value of y, we make x the independent variable, and // 
 the dependent variable. If we were to assign to y any arbi- 
 trate \alue. and find the corresponding value of £, we would 
 make // the independent variable and x the dependent 
 variable. 
 
 Frequently, when we are considering two or more varia- 
 it is in our power to make whichever we please the 
 independent variable. But, having once chosen the inde- 
 pendent variable, we are not at liberty to change it through- 
 out our operations, unless we make the corresponding trans- 
 formations which such a change would require. 
 
 3. Functions.— One quantity is called a function of 
 another, when it is so connected with it that no change can 
 take place in the latter without producing a corresponding 
 change in the former. 
 
 For example, the sine, cosine, tangent, etc., of an angle 
 are said to be functions of the angle, as they depend upon 
 the angle for their value. Also, the area of a square is a 
 function of its side; the volume of a sphere is a function of 
 its radius. In like manner, any algebraic expression in x, as 
 
 X s — %hx* + bx + c, 
 
 is a function of x. Also, we may have a function of two or 
 more variables: a rectangle is a function of its two sides; 
 a parallelopiped is a function of its three edges ; the expres* 
 Ron tan (ax -f by) is a function of two variables, x and y; 
 *" + !/ 2 + z * i s a function of three variables, x, y, and z ; etc. 
 When we wish to write that one quantity is a function of 
 
NOTATION— GEOMETRIC REPRESENTATION, 3 
 
 one or more others, and wish, at the same time, to indicate 
 several forms of functions in the same discussion, we use 
 such symbols as the following : 
 
 y-=f(z); y = F(x); y = 0(a); y=f(x)- 
 
 y=f(x,z); <p(z,y) = 0; f{x } y,z) = 0; 
 
 which are read: "y equals the /function of x; y equals the 
 large F function of x ; y equals the <p function of x ; y equals 
 the /prime function of x; y equals the /function of x and 
 z; the function of x and y equals zero; the / function of 
 x, y, and z equals zero;" or sometimes " y •=. f of x, 
 y == i^of %," etc. If we do not care to state precisely the 
 form of the function, we may read the above, "y = a func- 
 tion of x ; y — a function of x and z ; a function of x and y 
 = ; a function of x, y, and z = 0." 
 For example, in the equation < 
 
 y = ax 2 + ~bx + c, 
 
 y is a function of x, and may be expressed, y =/(#). 
 Also, the equation 
 
 ax 2 -f fa^ -\- cy l = 
 
 may be expressed, f(x, y) = 0. 
 In like manner, the equations 
 
 y — ax? -f- bxh + C2 3 , 
 
 and ^ = ax 2 + te -f- dz\ 
 
 may be expressed, ?/ = f(x, z) and ?/ = <p (x, z). 
 
 Every function of a single variable may be represented geometri- 
 cally by the ordinate of a curve of which the variable is the cor 
 responding abscissa. For if y be any function of x, and we assign 
 any value to x and find the corresponding value of y, these two values 
 may be regarded as the co-ordinates of a point which may be con- 
 structed. In the same way, any number of values may be assigned to 
 X, and the corresponding values of y found, and a series of points con 
 
ALGEBRAIC AND TRANSCENDENTAL FUNCTIONS. 
 
 utructed. These points make up a curve of which the variable ordi. 
 nate is y and the corresponding abscissa is x. 
 
 In like nuinnt r it may be shown tltnt a function of two variables 
 may !>.' rep resen ted geometrically by theordinate of a surfaceoi which 
 i lie variables are the corresponding abscissas. 
 
 4. Algebraic and Transcendental Functions.— An 
 
 algebraic function is one in which the only operations indi- 
 cate! are addition, subtraction, multiplication, division, 
 involution, and evolution; as, 
 
 (a + bx*)»; (*-&£)*; y 
 
 x*-a* 
 
 etc. 
 
 (*» + £»*)*> 
 
 Transcendental functions are those which involve other 
 operations, and are subdivided into trigonometric, circular, 
 logarithmic and exponential. 
 
 * A trigonometric function is one which involves sines, tan- 
 osinee, etc., as variables. For example, 
 
 y = sin x; y — tan 2 x; y = cos £ sec z; etc. 
 
 A circular function is one in which the concept is a 
 variable arc, as sin -1 a:,* cos -1 *, sec -1 y, cot -1 .r, etc., read, 
 •• the arc whose sine is x, the arc whose cosine is x," etc. It 
 is the inverse of the trigonometric function ; thus, from the 
 trigonometric function, ?/ = sina.', we obtain the circular 
 function, z = sin -1 y. In the first function we think of the 
 right line, the sine, the arc being given to tell us which sine ; 
 in the second we think of the arc, the sine being given to 
 t 11 us which arc. The circular functions are often called 
 rse trigonometric functions. 
 
 * This Dotation was suggested by the use of the negative exponents in algebra. 
 If we have y = ax, we also have x = a l y, where y is a function of x, and x is the 
 corresponding Inverse function of y. It may be worth while to caution the begin- 
 ner against the error of supposing that sin ■ y is equivalent to --. — ; while it is 
 
 sin y 
 
 true that a ' is equivalent to - • 
 
INCREASING AND DECREASING FUNCTIONS. 
 
 A logarithmic function is one which involves logarithms 
 of the variables ; as, 
 
 y = log x; y = log y/a — x; 
 
 / a 2 _ x 2 
 
 y' = 31 °s\/^T^' ctc - 
 
 An exponential function is one in which the variable 
 enters as an exponent ; as, 
 
 y = a x ; y = x z ; u = x^; etc. 
 
 5. Increasing and Decreasing Functions. —An in- 
 
 creasing function is one that increases when its variable 
 increases, and decreases when its variable decreases. 
 For example, in the equations 
 
 y = ax 3 , y = log x, y = Va 2 + x 2 , y = a x , 
 
 y is an increasing function of x. 
 
 A decreasing function is one that decreases when its 
 variable increases, and increases when its variable decreases. 
 Thus, in the equations 
 
 y = x' ? = (* — £)*, y = hg-, x 2 + y 2 = r% 
 
 y is a decreasing function of x. In the expression, 
 
 y = (a- xf, 
 
 y is a decreasing function for all values of x < a, but in- 
 creasing for all values > a. In the expression 
 
 y == sin x, 
 
 y is an increasing function for all values of x between 0° 
 and 90°, decreasing for all values of x between 90° and 270°, 
 and increasing for all values of x between 270° and 360°. 
 
6 1721 I 0U8 i ' v.\ OTTO 
 
 6. Explicit and Implicit Functions.— An explicit 
 function is one whose value is directly expressed in terms of 
 the variable and constants. 
 
 For example, in the equations 
 
 y = (a — x) 2 , y = Va 2 — x 2 , y = tax* — 3^, 
 
 y is an explicit function of x. 
 
 An implicit function is one whose value is not dm 
 expressed in terms of the variables and constants. 
 
 For example, in the equations 
 
 y* — 3axy + x* = 4, x 2 — 3xy + 2y = 16, 
 
 y is an implicit function of x, or a; is an implicit fund ion 
 of y. If we solve either equation with respect to y, we shall 
 have y as an explicit function of x ; also, if we solve for x 3 
 we shall have x as an explicit function of y. 
 
 7. Continuous Functions. — A function of x is said to 
 be a continuous function of x, between the limits a and b, 
 when, for every value of x between these limits, the cor- 
 responding value of the function is finite, and when an 
 infinitely small change in the value of x produces only an 
 infinitely small change in the value of the function. If 
 these conditions are not fulfilled, the function is discon- 
 tinuous. 
 
 For example, both conditions are fulfilled in the equations 
 
 y = ax + b, y = sin x, 
 
 in which, as x changes, the value of the function also 
 changes, but changes gradually as x changes gradually, and 
 there is no abrupt passage from one value to another; if x 
 receives a very small change, the corresponding change in 
 the function of x is also very small. 
 
 The expression Vr 2 — x 2 is a continuous function of x 
 for all values of x between + r and — r, while Vx 2 — fl 
 is discontinuous between the same limits. 
 
INFINITES AND INFINITESIMALS. 1 
 
 £. Infinites and Infinitesimals. — An infinite quantity, 
 or an infinite, is a quantity which is greater than any assign- 
 able quantity. 
 
 An infinitesimal is a quantity which is less than any 
 assignable quantity. 
 
 An infinite is not the largest possible quantity, nor is an infinitesi- 
 mal the smallest ; there would, in this case, be but one infinite or 
 infinitesimal. Infinites may differ from each other and from a quan- 
 tity which transcends every assignable quantity, that is, from absolute 
 infinity. So may infinitesimals differ from each other and from abso- 
 lute zero. 
 
 The terms infinite and infinitesimal are not applicable to quantities 
 in themselves considered, but only in their relation to each other, or to 
 a common standard. A magnitude which is infinitely great in com- 
 parison with a finite magnitude is said to be infinitely great. Also, a 
 magnitude which is infinitely small in comparison with a finite mag- 
 nitude is said to be infinitely small. Thus, the diameter of the earth 
 is very great in comparison with the length of one inch, but very small 
 in comparison with the distance of the earth from the pole star ; and 
 it would accordingly be represented by a very large or a very small 
 number, according to which of these distances is assumed as the unit 
 of comparison. 
 
 The symbols qo and are used to represent an infinite 
 and an infinitesimal respectively, the relation of which is 
 
 co = jr and = 
 
 The cipher is an abbreviation to denote an indefinitely small 
 quantity, or an infinitesimal — that is, a quantity which is less than 
 any assignable quantity— and does not mean absolute zero ; neither 
 does oo express absolute infinity. 
 
 If a represents a finite quantity, and x an infinite, then 
 a 
 x 
 a 
 x 
 tesimal, and the reciprocal of an infinitesimal is infinite. 
 
 A number is infinitely great in comparison with another, 
 
 is an infinitesimal. If x is an infinitesimal and a is finite, 
 is infinite ; that is, the reciprocal of an infinite is infini- 
 
8 ORDERS OF I.XFIMThS AM) INFINITESIMALS. 
 
 win n no number can be found sufficiently large to express the 
 
 ratio between them. Thus, x is infinitely great in relation 
 
 to a, when no number can be found large enough to ex | 
 
 x 
 the quotient -• Also, a is infinitely small in relation to x 
 
 when no number can be found small enough to express the 
 
 quotient -; x and - represent an infinite and an iniini- 
 x x 
 
 tesiraal. 
 
 One million in comparison with one millionth is a very large nuni- 
 1>< t, tat not infinitely large, since the ratio of the first to the second 
 can be expressed in figures : it is one trillion : though a very large 
 number, it is finite. So, also, one millionth in comparison with one 
 million is a very small number, but not infinitely small, since a num- 
 ber can be found small enough to express the ratio of the first to the 
 second : it is one trillionth, and therefore finite. 
 
 9. Orders of Infinites and Infinitesimals.— But even 
 
 x . * 
 
 though - is greater than any quantity to which we can 
 a 
 
 assign a value, we may suppose another quantity as large in 
 relation to x as x is in relation to a ; for, whatever the mag- 
 nitude of x, we may have the proportion 
 
 a* 
 
 a i x : : x : — - 
 a 
 
 x 2 . 
 m which — is as large in relation to z as # is in relation to 
 a 
 
 x 2 
 a, for - will contain x as many times as x will contain a ; 
 
 x 2 
 hence, - may be regarded as an infinite of the second order, 
 
 x 
 
 - being an infinite of the first order. 
 
 Also, even though - is less than any quantity to which 
 x 
 
 in assign a value, we may suppose another quantity as 
 
 small in relation to a as a is in relation to x ; for we may 
 
 have the proportion, 
 
ORDERS OF INFINITES AND INFINITESIMALS. 
 
 a* 
 x : a :: a : — , 
 x 
 
 a 2 
 in which - is as small in relation to a as a is in relation to 
 x 
 
 a 2 
 a?, for - is contained as many times in a as a is contained 
 
 a 2 
 in x ; hence, - may be regarded as an infinitesimal of the 
 
 x 
 
 second order, - being an infinitesimal of the first order. 
 
 x 
 
 We may, again, suppose quantities infinitely greater and 
 infinitely less than these just named ; and so on indefinitely. 
 Thus, in the series 
 
 ax*, ax 2 , ax, a, -, ^, ^, etc., 
 
 if we suppose a finite and x infinite, it is clear that any 
 term is infinitely small with respect to the one that imme- 
 diately precedes it, and infinitely large with respect to the 
 one that immediately follows it; that is, ax*, ax 2 , ax are 
 infinites of the third, second, and first orders, respectively ; 
 
 -, -£, — 3 are infinitesimals of the first, second, and third 
 
 orders, respectively, while a is finite. 
 
 If two quantities, as x and y, are infinitesimals of the first 
 order, their product is an infinitesimal of the second order ; 
 for we have the proportion, 
 
 1 : x :: y : xy. 
 
 Hence, if x is infinitely small in relation to 1, xy is infinitely 
 small in relation to y; that is, it is an infinitesimal of the 
 second order when x and y are infinitesimals of the first 
 order. 
 
 Likewise, the product of two infinites of the first order is 
 an infinite of the second order. 
 
 The product of an infinite and an infinitesimal of the 
 game order is a finite quantity. The product of an infinite 
 
in RATIOa OF INFINITESIMALS. 
 
 ami an infinitesimal of different orders is an infinite or an 
 infinitesimal, according aa the order of the infinite is higher 
 or lower than that of the infinitesimal, and the order of the 
 prodncl is the sum of the orders of the factors. 
 For example, in the expressions 
 
 " o « o a a 2 
 
 the first product is finite; the second is an infinite of the 
 first order; the third is an infinitesimal of the second order. 
 
 Though two quantities are each infinitely small, they may have any 
 mtio whatever. 
 
 Tints, if a and 6 are finite and x is infinite, the two quantities 
 
 - and - are Infinitesimals j but their ratio is r , which is finite. In- 
 xx b 
 
 deed, two very small quantities may have a much larger ratio than 
 two very large quantities, for the value of a ratio depends on the rela- 
 and not on the absolute magnitude of the terms of the ratio. The 
 ratio of the fraction one-mUlionth to one-ten-millionth is ten, while the 
 ratio of <>/<■ million to ten million is one-tenth. The latter numbers are 
 respectively a million times a million, and ten million times ten mil- 
 lion, times as great as the first, and yet the ratio of the last two is 
 only one-hundredth as great as the ratio of the first two. 
 Assume the series 
 
 lo*' W)' viov' \fo7' (joy* \iov' etc 
 
 in which the first fraction is one-millionth, the second one-millionth 
 of the first, and so on. Now suppose the first fraction is one-millionth 
 of an inch in length, which may be regarded as a very small quantity 
 uf the first order ; the second, being one-millionth of the first, must 
 be regarded as a small quantity of the second order, and so on. Now, 
 if we continue this series indefinitely, it is dear that we can make the 
 terms besoms cm small as we please without ever reaching absolute zero. 
 It is also clear that, however small the terms of this series become, the 
 ratio of any term to the one that immediately follows it is one million. 
 
 10. Geometric Illustration of Infinitesimals. — The 
 following geometric results will help to illustrate the theory 
 of infinitesimals. 
 
GEOMETRIC ILLUSTRATION OF INFINITESIMALS. 11 
 
 Let A and B be two points on the 
 circumference of a circle ; draw the 
 diameter AE, and draw EB produced 
 to meet the tangent AD at D. Then, 
 as the triangles EAB and ADB are 
 similar, we have, 
 
 (i) 
 
 and 
 
 BE 
 
 AB 
 
 AE " 
 
 " AD' 
 
 AB 
 
 BD 
 
 AE " 
 
 " AD 
 
 (?) 
 
 Now suppose the point B to approach the point A till it 
 becomes infinitely near to it, then BE becomes ultimately 
 equal to AE; but, from (1), when 
 
 BE = AE, 
 we have AB = AD. 
 
 AB 
 
 Also, -£- becomes infinitely small, that is, AB becomes 
 
 an infinitely small quantity in comparison with AE. Hence, 
 from (2), BD becomes infinitely small in comparison with 
 AD or AB ; that is, when AB is an infinitesimal of the first 
 order, BD is an infinitesimal of the second order. 
 
 Since DE — AE < BD, it follows that, when one side of 
 a right-angled triangle is regarded as an infinitely small 
 quantity of the first order, the difference uetween the hypotli- 
 enuse and the remaining side is an infinitely small quantity 
 of the second order. 
 
 Draw BN perpendicular to KD ; then, since AB > AN, 
 we have, 
 
 AD - AB < AD - AN < DN ; 
 
 „ # AD - AB DN AD 
 
 therefore, — w _ < m < m 
 
 But AD is infinitely small in comparison with DE, there- 
 fore AD — AB is infinitely small in comparison with BD ; 
 
12 AXIOMS. 
 
 hut BD ifl mi infinitesimal of lln- second order (see above), 
 beuoe AD — AM is on infinitesimal of the third order* 
 
 In like manner it may be shown that BD — BN is an 
 i 11 ti ii i irs in ml of the fourth order, and soon. [The student 
 who wishes further illustration is referred to Williamson's 
 Dif. Cal., p. 35, from which this was taken.] 
 
 11. Axioms. — From the nature of an infinite quantity, 
 a finite quantify can have no value when added to it, and 
 must therefore be dropped. 
 
 An infinitesimal can have no value when added to a finite 
 quantity, and must therefore be dropped. 
 
 If an infinite or an infinitesimal be multiplied or divided 
 by a finite quantity, its order is not changed. 
 
 If an expression involves the sum or difference of infinites 
 of different orders, its value is equal to the infinite of the 
 highest order, and all the others can have no value when 
 added to it, and must be dropped. 
 
 If an expression involves the sum or difference of infini- 
 tesimals of different orders, its value is equal to tiie 
 infinitesimal of the lowest order, and all the others can have 
 no value when added to it, and must be dropped. 
 
 These axioms are self-evident, and, therefore, axioms in the strict 
 sense. For example, suppose we were to compare the mass of the sun 
 with that of the earth : the latter weighs about six sextillion tons, the 
 former weighs about 355000 times as much. If a weight of one grain 
 were added to or subtracted from either, it would not affect the ratio 
 appreciably ; and yet the grain, compared with either, is finite — it can 
 be expressed in figures, though on the verge of an infinitesimal. If 
 we divide this grain into a great many equal parts — a sextillion, for 
 instance — and add one of these parts to the sun or the earth, the error 
 of the ratio will be still less ; hence, when the subdivision is continued 
 indefinitely, it is evident that we may obtain a fraction less than any 
 assignable quantity, Twwever small, which, when added to the sun or 
 the earth, will affect the above ratio by a quantity less than any to 
 which we can assign a value. 
 
 By reason of the terms that may be omitted, in virtue of the prin- 
 ciples contained in these axioms, the equations formed in the solution 
 
EXAMPLES. 13 
 
 of a problem will be greatly simplified. It may be remarked that in 
 the method of limits* when exclusively adopted, it is usual to retain 
 infinitely small quantities of higher orders until the end of the calcu- 
 lation, and then to neglect them on proceeding to the limit ; while, in 
 the infinitesimal method, such quantities are neglected from the be- 
 ginning, from the knowledge that they cannot affect the final result, 
 as they necessarily disappear in the limit. The advantage derived 
 from neglecting these quantities will be evident when it is remem- 
 bered how much the difficulty in the solution of a problem is increased 
 when it is necessary to introduce into its equations the second, third, 
 and in general the higher powers of the quantities to be considered. 
 
 EXAMPLES. 
 
 3# 4- n 
 
 1. Find the value of the fraction - ■= , if x is infinite, 
 
 and a and b finite. 
 
 Since a and b are finite, they have no value in comparison 
 
 3x 
 with x, and must therefore be dropped, giving us — - = | 
 
 as the required value of the fraction. 
 
 2. Find the value of the fraction ~ 7 , if # is infini- 
 tesimal, and a and b finite. 
 
 Since x is an infinitesimal, it has no value in comparison 
 with a and b, and must therefore be dropped, giving us — -= 
 for the required value of the fraction. 
 
 $x* _i_ 2x 
 
 3. Find the value of -^- 5 , when x is infinite; als/ 
 
 when x is infinitesimal. 
 
 Ans. When x is infinite, 4; when infinitesimal, 2. 
 
 4. Find the value of — = ^ , when x is 
 
 mx 3 -f- nx 2 -f- px + q 
 
 infinite; and when infinitesimal. 
 
 n 6 
 
 Ans. When x is infinite, -; when infinitesimal, -• 
 m q 
 
 * For a discussion of limits, see Chapter IIL 
 
14 EXAMPLES. 
 
 5. Find the value of '"' , — r -7— r » when a; is infinites 
 
 ^' — ±x + 1 
 
 and when infinitesimal 
 
 Aus. \\ hen a* is infinite, oo; when infinitesimal, 2 
 
 n « la., i * 4** + 3a* + Hx — 1 
 
 C. Jbmu the value of — rt _ , , _ — H — , when x le 
 2./-' + -itc 2 + &E 
 
 infinite; and when infinitesimal. 
 
 Ans, When a; is infinite, 0; when infinitesimal, oo. 
 
 7. Find the value of -7-= , when a; is infinite; and 
 
 . . „ ., . 4a? — mx 
 
 when infinitesimal 
 
 Ans. When a; is infinite, 0; when infinitesimal, 7///. 
 
 X s 
 
 8. Find the value of - — — ^, when a; is infinite; and 
 when infinitesimal. 
 
 Ans. When a; is infinite, 00; when infinitesimal, 0. 
 
 y x 2y 
 
 9. Find the value of j «-> when a: and y are infini»- 
 
 tesimals. * 
 
 . I us. We do not know, since the relation between x and 3 
 is unknown. 
 
CHAPTER II 
 
 DIFFERENTIATION OF ALGEBRAIC AND TRANSCEN- 
 DENTAL FUNCTIONS. 
 
 12. Increments and Differentials— If any variable, 
 as x, be supposed to receive any change, such change is 
 called an increment ; this increment of x is usually denoted 
 by the notation Ax, read "difference x," or "delta z 9 " where 
 A is taken as an abbreviation of the word difference. If the 
 variable is increasing, the increment is + ; but if it is 
 decreasing, the increment is — . 
 
 When the increment, or difference, is supposed infinitely 
 small, or an infinitesimal, it is called a differential, and is 
 represented by dx, read " differential x," where d is taken as 
 an abbreviation of the word differential, or infinitely small 
 difference. The symbols A and d, when prefixed to a varia- 
 ble or function, have not the effect of multiplication ; that 
 is, dx is not d times x, and Ax is not A times x, but their 
 power is that of an operation performed on the quantity to 
 which they are prefixed. 
 
 If u be a function of x, and x becomes x + Ax, the cor- 
 responding value of u is represented by w + Aw ; that is, the 
 increment of u corresponding to a finite increment of x is 
 denoted by An, read " difference u." 
 
 If x becomes x-\-dx, the corresponding value of u is rep- 
 resented by u-\-du; that is, the infinitely small increment 
 of u caused by an infinitely small increment in x, on which 
 u depends, is denoted by du, read " differential uP Hence, 
 dx is the infinitesimal increment of x, or the infinitesimal 
 quantity by which x is increased ; and du is the correspond- 
 ing infinitesimal increment of u. 
 
1 6 COXSECUTIVE P O/.XTS— />/ / 7 7. // /■: \TTA TTON. 
 
 The differential du or dx is + or — according as the 
 variable is increasing or decreasing, i.e., the first \alue in 
 always to be taken from the second* 
 
 13. Consecutive Values. — Consecutive values of I 
 function or variable are values which differ from each other 
 1 1\ leas Hutu any assignable quantity. 
 
 Consecutive points are points nearer to each other than 
 any assignable distance. 
 
 Thus, if two points were one-millionth of an inch apart, they might 
 bo considered practically as consecutive points ; and yet we might have 
 it million points between them, the distance between any two of which 
 would be a millionth of a millionth of an inch ; and so we might have 
 ■ million points between any two of these last points, and so on ; that 
 is, however close two points might be to each other, we could still 
 suppose any number of points between them. 
 
 A differential has been defined as an infinitely small in- 
 crement, or an infinitesimal; it may also be defined as the 
 difference hetiveen hvo consecutive values of a variable or 
 function. The difference is always found by taking the first 
 value from the second. 
 
 In the Differential Calculus, we investigate the relations 
 between the infinitesimal increments of variables from given 
 relations between finite values of those variables. 
 
 The operation of finding the differential of a function or 
 a variable is called differentiation. 
 
 14. Differentiation of the Algebraic Sum of a 
 Number of Functions. 
 
 Let u = v -f y — z, (1) 
 
 in which u. v, y } z, are functions of x.* 
 
 * We might also, in a similar manner, find the differential of a function of sev- 
 eral variables ; butwe prefer to reserve the inquiry into the differentials of functions 
 ot several variable* for a later chapter, and confine ourselves at present to function! 
 of a single variable. 
 
DIFFERENTIATION OF A PRODUCT. 1? 
 
 Give to x the infinitesimal increment dx, and let du, di\ 
 iy, dz, be the corresponding infinitesimal increments of u, 
 v, y, z, due to the increment which x takes. Then (1) 
 
 becomes 
 
 u + du = v + dv -f y + dy — (z + dz). (2) 
 
 Subtracting (1) from (2), we have 
 
 du = dv + dy — dz, (3) 
 
 which is the differential required. 
 
 Therefore, the differential of the algebraic sum of 
 any number of functions is found' by taking the alge- 
 braic sum of their differentials. 
 
 15. To Differentiate 
 
 y = ax±b. (1) 
 
 Give to x the infinitesimal increment dx, and let dy be 
 the corresponding infinitesimal increment of y due to the 
 cerement which x takes. Then (1) becomes 
 
 y + dy = a(x + dx) ± b. (2) 
 
 Subtracting (1) from (2), we get 
 
 dy = adx, (3) 
 
 which is the required differential 
 
 Hence, the differential of the product of a constant 
 I) if a variable is equal to the constant multiplied by 
 the differential of the variable ; also, if a constant be 
 connected with a variable by the sign + or — , it dis~ 
 appears in differentiation. 
 
 Tin's may also be proved geometrically as follows: 
 
 Let AB (Fig. 2) be the line whose equation is y — ax+d, 
 and let (z, y) be any point P on this line. Give OM (= x) 
 
18 
 
 (iKitMKTItIC I I.I.I N77.M770 V. 
 
 dx), then the cor* 
 
 the infinitesimal increment MM' (= 
 responding increment of MP (=y) 
 will be OP' (= dy). Now in the tri- 
 angle OPP' We have 
 
 CP' = CPtanCPP';* 
 
 or letting a = tan CPP', and substi- 
 tuting for CF and CP their values dy 
 and dx, we have, 
 
 dy = adx. 
 
 It is evident that the constant b will disappear in differentiation, 
 from the very nature of constants, which do not admit of increase, and 
 therefore can take no increment. 
 
 16. Differentiation of the Product of two Func- 
 tions. 
 
 Fig. 2 
 
 Let 
 
 u = yz, 
 
 (1) 
 
 where y and z are both functions of x. Give x the infini 
 tesimal increment dx, and let du, dy, dz be the correspond- 
 ing increments of u, y, and z, due to the increment which 
 x takes. Then (1) becomes 
 
 u + du = (y + dy) (z + dz) 
 
 = yz -f zdy + ydz + dz dy. (2) 
 
 Subtracting (1) from (2), and omitting dzdy, since it is 
 an infinitesimal of the second order, and added to others of 
 the first order (Art. 11), we have 
 
 du = zdy + ydz, (3) 
 
 which is the required differential. 
 
 Hence, the differential of the -product of two func- 
 tions is equal to the first into the differential of the 
 second, plus the second into the differential of the 
 first. 
 
 * In the Calculus as in the Analytic Geometry, the radius is always regarded as 
 1, imfeM otherwise m ent i o ned. 
 
b c 4 
 
 D Oj/i 
 
 C 
 
 A l 
 
 "T" Fig. 3. BU 
 
 GEOMETRIC ILLUSTRATION, t 19 
 
 This may also be proved geometrically as follows : 
 
 Let z and y represent the lines AB 
 and BC respectively ; then will u rep- 
 resent the area of the rectangle ABCD. 
 Give AB and BC the infinitesimal in- 
 crements B# (= dz) and Cc (= dy) 
 respectively. Then the rectangle ABCD 
 will be increased by the rectangles BaCh, 
 DCbc, and Chcd, the values of which 
 are ydz, zdy, and dzdy respectively ; therefore 
 
 du = ydz + zdy + dz dy. 
 
 But dzdy being an infinitesimal of the second order and 
 connected with others of the first order, must be dropped 
 (Art. 11) ; if this were not done, infinitesimals would not 
 be what they are (Art. 8) ; the very fact of dropping the 
 term dz dy implies that its value, as compared with that of 
 ydz -\- zdy is infinitely small. 
 
 The statement that ydz + zdy + dz dy is rigorously equal to ydz + zdy 
 is not true, and yet by taking dz and dy sufficiently small, the error 
 may be made as small as we please. 
 
 Or, we may introduce the idea of motion, and consider 
 that dz and dy represent the rate at which AB and BC are 
 increasing at the instant they are equal to z and y respec- 
 tively. The rate at which the rectangle ABCD is enlarging 
 at this instant depends upon the length of BC and the rate 
 at which it is moving to the right + the length of DC and 
 the rate at which it is moving upward. If we let dz repre- 
 sent the rate at which BC is moving to the right, and dy the 
 rate at which DC is moving upward at the instant that 
 AB = z and BC = y, we shall have du = zdy -{-ydz as the 
 rate at which the rectangle ABCD is enlarging at this in- 
 stant. (See Price's Calculus, vol. i, p. 41.) 
 
hint i:i:\ii.\rn>.\ OF i FRACTION, 
 
 17. Differentiation of the Product of any Num- 
 ber of Functions. 
 
 Lei u = vyz, (1) 
 
 Then giving to x the infinitesimal increment dx, and letting 
 </h. th\ dy, dz be the corresponding increments of u, v, y, z, 
 (1) becomes 
 
 u + du = (v + dv) (y -f dy) (z + dz). (2) 
 
 Subtracting (1) from (2), and omitting infinitesimals* of 
 higher orders than the first, we have 
 
 du = yz dv + vz dy + vy dz, (3) 
 
 and so on for any number of functions. 
 
 Hence, the differential of the product of any num- 
 ber of functions is equal to the sum of the products of 
 the differential of each into the product of all the 
 others. 
 
 Cor. — Dividing (3) by (1), we have 
 du dv , dy dz 
 
 That is, if the differential of each function be di- 
 vided by the function itself, the sum of the quotients 
 will be equal to the differential of the product of the 
 junctions divided by the product. 
 
 18 Differentiation of a Fraction. 
 
 T 4. x 
 
 Let u = -, 
 
 y 
 
 then uy = x; (1) 
 
 therefore, by Art. 16, we have 
 
 udy f ydu = dx. 
 Substituting for it its value, we have 
 
DIFFERENTIATION OF A POWER. 21 
 
 or 
 
 - dy -f ydu = dx. 
 
 Solving for du, we get 
 
 . _ ydx — xdy 
 cu - p* > 
 
 which is the required differential. 
 
 Hence, the differential of a fraction is equal to the 
 denominator into the differential of the numerator, 
 minus the numerator into the differential of the de- 
 nominator, divided by the square of the denominator. 
 
 Cor. 1. — If the numerator be constant, the first term in 
 the differential vanishes, and we have 
 
 xdy 
 
 du = — 
 
 f 
 
 Hence, the differential of a fraction with a constant 
 num. erator is equal to minus the numerator into the 
 differential of the denominator divided by the square 
 of the denominator. 
 
 Cor. 2. — If the denominator be constant, the second term 
 vanishes, and we have 
 
 , dx 
 
 du = — , 
 
 y 
 
 which is the same result we would get by applying the rule 
 of Art. 15. 
 
 19. Differentiation of any Power of a Single Va- 
 riable. 
 
 Let y = x n . 
 
 1st. WTien n is a positive integer. 
 
 Regarding x n as the product x, x, x, etc., of n equal fac-' 
 tors, each equal to x, and applying the rule for differentiating 
 a product (Art. 17), we get 
 
\77 At or a I'owhi: 09 A VARIABLE 
 
 (I ij = x*~ l dx -f tf'- l dx + x n ~ l dx + etc., to n terms. 
 
 .-. dy = nx"~ x dx. (1) 
 
 2d. When n is a positive fraction. 
 
 m 
 
 Let y = #• ; 
 
 then y n = of 1 . 
 
 Differentiating this as just shown, we have, 
 n\f~ x dy = mx" 1 ' 1 dx. 
 
 Therefore, dy = — ; dx 
 
 _ rn x m ~ 1 y , 
 " n y n 
 
 m 
 
 m x m ~ l & -. g . n . 
 
 = — dx (since y n = x m ). 
 
 . m 2-i, 
 
 A dy = — x* dx. 
 
 3d. JPTtera n is a negative exponent, integral or 
 fractional. 
 
 Let y = ar n ; 
 
 then w = --• 
 
 Differentiating by Art. 18, Cor. 1, we have 
 
 7 nx n ~ l dx _- , 
 
 rf y = — ^sr— = - wa; <&• (3) 
 
 Combining the results in (1), (2), and (3), we have the 
 following rule : TJie differential of any constant power 
 of a variable is equal to the product of the exponent, 
 the variable with its exponent diminished by unity, 
 and the differential of the variable. 
 
EXAMPLES. 23 
 
 Ooit. — If n=i, we have from (1), 
 
 dx 
 
 dy = \x^~ x dx = \x~*dx = 
 
 2Vx 
 
 Henee, the differential of the square root of a varia- 
 nts is equal to the differential o£ the variable divided 
 by twise the square root of the variable. 
 
 EXAMPLES. 
 
 1. Differentiate y~9 + 2x-{-x B + x 2 y* — x 7 -f 2gxy. 
 
 By Art. 14, we differentiate each term separately, and 
 take the algebraic sum. By Art. 15, the constant 9 disap- 
 pears in the differentiation ; and the differential of 2x is the 
 constant 2, multiplied by the differential of the variable x, 
 giving 2dx. By Art. 19, the differential ofx 2 is 2xdx. The 
 term x 2 y z is the product of two functions ; therefore, Art. 16, 
 its differential is x 2 d{if) -\-y s d(x 2 ), which, Art. 19, gives 
 3x 2 y 2 dy + 2y z x dx. In like manner proceed with the 
 other terms, giving the proper rule in each case. The 
 answer is 
 
 dy = 2m + 3x 2 dx + 3x 2 y 2 dy + 2xy B dx — Wdx 
 -f 2ax dy + 2ay dx. 
 
 2. u = ax*y 2 . du — 3ax 2 y 2 dx + 2ax*y dy. 
 
 3. u = 2ax - 3x 2 — abx A — 7. 
 
 du = (2a — 6x — 4:abx*) dx. 
 
 4. u = x 2 y%. du = p 2 y% dy + 2xifi dx. 
 
 5. u = 2bz~ 2 + Saxhk 
 
 3ax%dz 4bdz 
 
 2^/ z 
 
 6. u = aW. du 
 
 du = hax*z* dx + -5-7=- 
 
 xdy + ydx 
 
 2xh£ 
 
!\ tAMPLBS. 
 
 7. u = ax* + -^ — &A 
 
 8. y 2 = 2jwa*, to find the value of e/#. 
 
 
 
 
 
 
 dy : 
 
 = P -dx. 
 
 y 
 
 9. 
 
 iff + Wx 2 = 
 
 a 2 b?, to find the value of 
 
 " dy. 
 
 
 
 
 
 
 dy - 
 
 
 Vx, 
 
 a 2 y 
 
 10. 
 
 3? + y 2 
 
 = ?<*, 
 
 to find the value of dy. 
 
 
 
 
 
 
 
 dy 
 
 = 
 
 x , 
 
 y 
 
 11. 
 
 
 a 
 
 
 du — 
 
 
 tydy 
 
 U ~ b- 
 
 -2y 2 
 
 -% 2 ) 2 
 
 12. 
 
 u = (a 
 
 4- bx + ex 2 ) 5 . 
 
 
 
 
 Regarding the quantity within the parenthesis as a varia- 
 ble, we have, by Art. 19, 
 
 du = 5 (a 4- bx + cx 2 yd(a + bx -{- ex 2 ) 
 = 5 (a + bx 4- a*?) 4 (J 4- 2cz) efo?. 
 
 2*2-3 
 
 4# 4- 3* 
 
 By Art. 18, 
 
 _ (4s -I- 3?) <Z (2s 8 - 3 ) - (23? - 3) d (Ax 4- 3?) 
 rfl * _ " (4a + 3?) 2 
 
 _ (43? 4- x 2 ) 4xdx — (23? — 3) (4 + 23:) dx 
 
 (43; + x 2 ) 2 
 _ (83? + Gx 4- 1 2) <fa 
 . — ~~ (4a; + a?) 2 
 
 ^ A 2x* , Sah? — Ax 5 • 
 
 a 2 — re 2 (^ 2 — 3?) 2 
 
y 
 
 EXAMPLES. 25 
 
 l+X 7 (1 — %X — X 2 ) dx 
 
 15 - » = !??• *" (i + V ' 
 
 1 W 
 
 1 7. y = {ax 2 — a; 3 ) 4 . 
 
 dy = 4 (rt£ 2 — a: 3 ) 3 (2aa; — dx 2 ) dx. 
 
 18. ^ == (a + to 2 )l dy =:i£{a + to*}* bx dx. 
 
 20. # = Vx* — a 3 (Art. 19, Cor.). 
 
 d(x s — a*) 3x 2 dx 
 
 dy 
 
 2V~x* — a* 2Vx* — aS 
 (a — x) dx 
 
 21. y = <\/2ax — x 2 . dy = 
 
 \/%ax — x 2 
 
 1 xdx 
 
 22. y = — • dy = r 
 
 23. 2/ = Val~c + V^ ¥ . ^ = a **** & 
 
 %yx 
 
 (2ax + b) dx 
 
 24. y = \/ax 2 + bx + c. dy : 
 
 2vtf# 2 '+ to + c 
 
 25. y = (a 3 + «)(3z 2 + 6) (Art. 16). 
 
 dy = (x* + a)d (3a: 2 + &) + (3a: 2 + b) d (a; 3 + a) 
 = (15a 4 + 3to 2 + 6aa;) fife. 
 
 26. y = (1 + 2a 2 ) (1 -f 4a?). 
 
 dy = ±x (1 + 3« + 10a- 3 ) da:. 
 (« + 3a;) dx 
 ZVaT+x 
 
 (a—3x)dx 
 
 28. y = (a + a) V« - jb. dy = J^Tri' 
 
 27, y = (a — x) y/a + x. dy — — 
 
ILL I STliA TIVE EXAMPLES. 
 
 ILLUSTRATIVE EXAMPLES. 
 
 1. In the parabola y 2 — 4./-, which is increasing the most 
 rapidly at x = 3, the abscissa or the ordinate ? How docs 
 the nhitive rate of change vary as we recede from the ver- 
 bal P 
 
 2 
 Differentiating y 2 = 4z, we get dy = - dx, which shows 
 
 that if we give to x the infinitely small increment dx, the 
 
 2 
 corresponding increment of y is - times as great ; that is, 
 
 if 
 
 2 
 
 the ordinate changes - times as fast as the abscissa. At 
 
 y_ 
 
 x = 3, we have y = \/l2. Hence, at this point, 
 
 dy = — ~=zz dx = — — dx : 
 
 that is, the ordinate is increasing a little over one-half as 
 fast as the abscissa at x = 3. 
 
 At *s=l 9 y = 2, and dy = dx; that is, x and y are 
 increasing equally ; in general, at the focus the abscissa and 
 ordinate of a parabola are increasing equally. At x = 4, 
 y = 4, and dy = \dx ; that is, y is increasing \ as fast as 
 x. At z = 9, y = 6, and dfy = -^ia ; that is, y is increas- 
 ing I as fast as x. At a; = 36, # = 12, and dy = \dx; 
 that is, y is increasing £ as fast as x, and so on. We see 
 
 2 
 from the equation dy = -dx, as well as from the figure of 
 
 (he parabola, that the larger x becomes, and therefore y, the 
 less rapidly y increases, while x continues to increase uni- 
 formly. 
 
 2. If the side of an equilateral triangle is increasing uni- 
 formly at the rate of \ an inch per second, at what rate is 
 its altitude increasing ? Is the relative rate of increase ol 
 
 lie and altitude constant or variable? 
 
ILLUSTRATIVE EXAMPLES. 2? 
 
 Let x = a side of the triangle and y = its altitude. Then 
 if = \x\ and dy — — dx, which shows that when x takes 
 the infinitely small increment dx, the corresponding incre- 
 ment of y is -5- times as great; that is, the altitude y 
 
 always changes -— times as fast as the side x. When x is 
 increasing at the rate of \ an inch per second, y is increas- 
 ing -— times \, or -— inches per second. 
 
 3. A boy is running on a horizontal plane directly towards 
 the foot of a tower 60 feet in height. How much faster is 
 he nearing the foot than the top of the tower ? How far is 
 he from the foot of the tower when he is approaching it 
 twice as fast as he is approaching the top ? When he is 
 100 feet from the foot of the tower, how much faster is he 
 approaching it than the top ? 
 
 Let x = the boy's distance from the foot of the tower, 
 and y = his distance from the top. Then we have 
 
 f- = x 2 + 6O 2 . 
 
 .*. dx = -dy ; 
 
 x 
 
 that is, the boy is nearing the foot - times as fast as he is 
 the top. 
 
 2d. When he is approaching the foot of the tower twice 
 as fast as he is the top, we have dx = 2dy, which in 
 
 dx = -dy 
 x J 
 
 gives us y = 2x, and this in y 2 = x z + GO gives us 
 
 3x* = 60 2 , or :r =5 --£■ = 34.64 
 
 V3 
 
\ ILLUSTRATIVE EXAMPLES. 
 
 Sd. When be is LOO fa t from the foot, 
 
 y = |/ 100 2 + GO 2 = 110.02, 
 
 . y 116.G2 .... , y , 
 and •- = -rrrx- , which in dx = J dy gives 
 j- 100 a; 
 
 dx = 1.16G2 dy; 
 
 that is, he is approaching the foot of the tower 1.1602 times 
 as fast as he is the top. 
 
 4. In the parabola y 2 = 12#, find the point at which the 
 ordinate and abscissa are increasing equally; also the point 
 at which the ordinate is increasing half as fast as the 
 abscissa. Ans. The point (3, 6); and the point (12, 12). 
 
 5. If the side of an equilateral triangle is increasing uni- 
 formly at the rate of 2 inches per second, at what rate is the 
 altitude increasing. j^ <\/3 inches per second. 
 
 6. If the side of an equilateral triangle is increasing uni- 
 formly at the rate of 5 inches per second, at what rate is the 
 area increasing when the side is 10 feet ? 
 
 Ans. ff\/3 sq. ft. per second. 
 
 7. A vessel is sailing northwest at the uniform rate of 
 10 miles per hour ; at what rate is she making north lati- 
 tude? Ans. 7.07+ miles per hour. 
 
 8. A boy is running on a horizontal plane directly toward 
 the foot of a tower, at the rate of 5 miles per hour ; at what 
 rate is he approaching the top of the tower when he is 00 
 feet from the foot, the tower being 80 feet high? 
 
 Ans. 3 miles per hour. 
 
LOGARITHMIC AND EXPONENTIAL FUNCTIONS. 'Z9 
 
 LOGARITHMIC AND EXPONENTIAL FUNC- 
 TIONS. 
 
 2Q To Differentiate y = log x.— We have 
 y + dy = log (x + dx) = log x(l + -~j 
 
 = lo g « + log (l + ^)- 
 Subtracting, we have, 
 
 dy = log (l + |) = m (f -<g- + etc.) 
 
 (from Algebra, where m is the modulus of the system). 
 
 .-. dy = d (log x) = m — (Art. 11). 
 
 Tliis result may also be obtained as follows: 
 Let y = ax. (1) 
 
 .-. log y = log a + log a. (2) 
 
 By Art 15, dy = a dx, (3) 
 
 and d (log y) = d (log #). (4) 
 
 Dividing (4) by (3), we get, 
 
 d (log y) = ^ (log a?) 
 
 d (log a) . * /<% 
 = v 6 -— , from (1) ; 
 
 *<fe 
 
 x 
 
 dy 
 
 ^ d (log y) _ jr^ 
 
 d (log a;) dz ' 
 a; 
 
30 ini'Fi:t: i:\ti a i. of a logarithm. 
 
 Multiply both terms of the second fraction by the arbi- 
 trary factor m, and we have 
 
 m dy 
 
 d (log y) _ y t (5) 
 
 d (log x) " m dx ' * ' 
 
 x 
 
 We may sii])pose m to have such a value as to make 
 
 d(\ogy) = m-2; 
 
 (<>) 
 
 dx 
 therefore, d (log x) = m — 
 
 X 
 
 01 
 
 Similarly, let y = bz. 
 
 
 .'. log y = log b + log z. 
 
 (8) 
 
 Differentiating, dy = bdz, 
 
 
 and d (log 7/) = d (log z). 
 
 
 Dividing and substituting, 
 
 dy 
 
 d 0°g y) _ . # t 
 
 rf (log z) " dz 
 z 
 
 
 But d(\ogy) = m^f- 
 
 
 £2 
 
 .-. fZ(log*) =m-. (9) 
 
 z 
 
 In the same way we may show that the differential of the 
 logarithm of any other quantity is equal to m times the 
 differential of the quantity divided by the quantity, and 
 hence .^e factor m is a constant, provided that the loga- 
 rithms be taken in each case in the same system ; of course, 
 if the logarithms in (8) be taken in a different system from 
 those in (2), the numerical values of log y in the two equa- 
 
DIFFERENTIATION OF AN EXPONENTIAL FUNCTION 31 
 
 tions are different, and therefore the m in (7) is different 
 from the m in (9). Since m is a constant in the same sys- 
 tem and different for different systems, it varies with the 
 base of the system, as the only other quantities involved in 
 logarithms are the number and its logarithm. That is, m is 
 a function of the base ; its value will be computed hereafter. 
 (See Rice and Johnson's Calculus, p. 39 ; also, Olney's Cal- 
 culus, p. 25.) 
 
 Hence, the differential of the logarithm of a quan- 
 tity is equal to the modulus of the system into the 
 differential of the quantity divided by the quantity. 
 
 7 
 
 J 
 
 Cor. — If the logarithm be taken in the Naperian * system, 
 the modulus is unity, and we have 
 
 7 dx 
 
 Hence, the differential of the logarithm of a quan- 
 tity in the Naperian system is equal to the different v^, 
 tial of the quantity divided by the quantity. 
 
 21. To Differentiate y = a x . 
 
 Passing to logarithms, we have, 
 
 log y = x log a. 
 Differentiating, we have 
 
 or dy = 
 
 m — = dx log a ; 
 
 if 
 
 y dx log a 
 
 m 
 
 a x 
 dy = d (a x ) = —\oga dx. 
 
 * So called from the name of the inventor of logarithms ; also sometimes called 
 natural logarithms, from being those which occur first in the investigation of a 
 method of calculating logarithms. They are sometimes called hyperbolic logarithms, 
 from having been originally derived from the hyperbola. 
 
82 DIFFERENTIATION 09 AH EXPONENTIAL FUNCTION* 
 
 lit ihv, the differential of an exponential function 
 
 icith a constant tmst is equal to the function into the 
 logarithm of the base into tin- differential of the ex- 
 ponent, divided by the modulus. 
 
 Cor. 1. — If we take Naperian logarithms, we have 
 dy = d (a x ) = a x log a dx (since m = 1). 
 
 Cor. 2. — If a = e, the Naperian base, then 
 
 log a = log e = 1, 
 
 and therefore dy = d(e x ) = e*dx. 
 
 Sch. — In analytical investigations, the logarithms used 
 are almost exclusively Naperian, the base of which system 
 is represented by the letter e. Since the form of the differ- 
 ential is the simplest in the Naperian system, we shall in 
 all cases understand our logarithms to be Naperian, unless 
 otherwise stated. 
 
 22. To Differentiate U = y x . 
 
 Passing to logarithms, we have, 
 
 log u = x log y. 
 
 Differentiating, we have, 
 
 du . , , dy 
 — =\ogydx + xj; 
 
 or, du = u log y dx -f ux~; 
 
 if 
 
 /. du = yx log y dx -f xy*- 1 dy. 
 
 Hence, to differentiate an exponential function 
 with a variable base, differentiate first as though the 
 base were constant and the exponent variable, and 
 second as though the base were variable and the expo- 
 nent constant, and take the sum of the results. 
 
EXAMPLES. 33 
 
 EXAM PLES, 
 
 1. y = x log x. dy = (log x + 1) dk 
 
 A ^logiz* dy = - X . 
 
 X 
 
 4. y = ««*. (ft/ = a «V log a <fo. 
 
 "*$f y = &?. dy = a?"\ \ogz(l + logz) + - L«<fe. 
 
 6. y = log Vt^l?. % = - ~p- 
 
 dx 
 
 7. */ = log(iB WlT~^). ^ = vTT^ ' 
 
 8. y = \og(^-±-^j = log (a + ib) - log (a - ib). 
 
 , 2adx 
 
 dy = 
 
 cP — x* 
 
 9. ^ = logy / ^ = ilog(l+ a ;)~ilo g (l-4 
 
 , dx 
 
 v^fO. y = log (log x). dy = 
 
 da: 
 
 a; log a; 
 dx 
 
 ^11. # = log 2 x. dy = 2 log a; 
 
 ^H^. ?/ — x x . dy = a* (log x + l)dx. 
 
 13. y = \og I 
 
 V IB 2 + 1 + IB 
 
 Multiplying both terms by the numerator to rationalize 
 the denominator, we get 
 
34 LOGARITHMIC DlFtVRBNTIATIOM* 
 
 y = log [a/^~+~1 — s] 8 ' 
 %dx 
 
 -yU. y = c* (a; — 1). <?y = *« *r 
 
 15. y = e* (a? - 2z + 2). dy = ^^c/a?. 
 
 e» _ 1 _ 2c* 7 
 
 16 - »;-JTfT * = (*+!?* 
 
 17. y = e*loga\ • tfy = **(loga? + -]<7ar. 
 
 ^18. v = c ,og 1/J37r *. 
 
 Then log y = log V# 2 + &, 
 
 y = 
 
 dy = 
 
 •. y = Va 2 + a 2 - 
 
 V« 2 + a? 
 c* 7 zePdx 
 
 19 - * = r+i- * = (T+i)'' 
 
 , a; t dx dx 
 
 20. y = log dy = —~ 
 
 Vtf + l + z x Vx* + l 
 
 21. y = log(a- + a+V2aa;+a^). rfy = ^==- 
 
 22. y = a**^ 1 . dy = av^^l afi^^dx. 
 
 23. Logarithmic Differentiation.— When the function 
 to be differentiated consists of products and quotients, the 
 differentiation is often performed with greater facility by 
 first passing to logarithms. This process is called logarith- 
 mic differentiation. 
 
EXAMPLES. 35 
 
 EXAMPLES. 
 
 1. u = x (« 2 + x 2 ) Va 2 — x\ 
 Passing to logarithms, we have 
 
 log u = log x + log (a 2 + x 2 ) + \ log (a 2 — a*). 
 <?z£ dx 2x dx xdx 
 
 u x a 2 + x 2 a 2 — x 2 
 
 r , , x 2 (a 2 + x 2 )-} 
 
 :. du = (a 2 + z 2 ) V« 2 -a; 3 +2a;Va 2 - a? — ~ y 
 
 L V « 2 — - a 2 _ 
 
 : a 4 + « 2 £ 2 — 4a; 4 . 
 or, du = — rta-. 
 
 Va 2 — x 2 
 
 1 + a 2 
 
 2. u = -j« Passing to logarithms, we have 
 
 j. — ~ x 
 
 log w = log (1 + a 2 ) — log (1 — x 2 ). 
 
 du 2xdx 2xdx kcdx 
 
 da 
 
 u~l+x 2 ^l-x 2 (1 + x 2 ) (1 - a; 2 ) 
 , 4:xdx 
 
 /8. w = (a* + l) 2 . du = 2a x (a* + 1) log a da\ 
 
 , a* — 1 7 2a x losadx 
 
 5. « = -, efa = 
 
 Vl — x (1 — «) Vl - a* 
 
 ILLUSTRATIVE EXAMPLES. 
 
 1. Which increases the more rapidly, a number or its 
 logarithm? How much more rapidly is the number 4238 
 increasing than its common logarithm, supposing the two 
 to be increasing uniformly? While the number increases 
 by 1, how much will its logarithm increase, supposing the 
 
//. /. ! ST/:. I 77 I /.' l.X. 1 MI'LES. 
 
 latter to Increase uniformly (which it does not) while the 
 number increases uniformly. 
 
 Let x = the number, and y its logarithm ; then we have 
 
 y as log a:; 
 
 .•. dy = - *fo, 
 
 which shows that if we give to the number (x) the infinitely 
 small increment (dx), the corresponding increment of y is 
 
 — times as great; that is, the logarithm (y) is increasing 
 
 - times as fast as the number. Hence, the increase in the 
 x 
 
 common logarithm of a number is >, =, < the increase 
 of the number, according as the number (x) <, =, > the 
 modulus (m). 
 
 When x = 4238, we have 
 
 m . .43429448, 
 
 4238 
 hence, dx = ^34^448 dy = ab ° ut 9758 dyl 
 
 .43429448 
 that is, the increment of the logarithm is ! — r^r— part of 
 
 the increment of the number, and the number is increasing 
 about 9758 times as fast as its logarithm. 
 
 While the number increases by ] , its logarithm will in- 
 crease (supposing it to increase uniformly with the number) 
 4.QJ.OQ448 
 
 Tooq — timcs 1 = .00010247; that is, the logarithm of 
 4/wOo 
 
 4239 would be .00010247 larger than the logarithm of 4238, 
 
 if it were increasing uniformly, while the number increased 
 
 from 4238 to 4239. 
 
 B km ark. — While a number ia increasing uniformly, its logarithm 
 is increasing more and more slowly; this is evident from the equation 
 
 $,y — — dx, which shows that if the number receives a very small in- 
 
TRIGONOMETRIC FUNCTIONS, 37 
 
 Crement, its logarithm receives a very small increment ; but on giving 
 to the number a second very small increment equal to the first, the 
 corresponding increment of the logarithm is a little less than the first, 
 and so on ; and yet the supposition that the relative rate of change of 
 a number and its logarithm is constant for comparatively small changes 
 in the number is sufficiently accurate for practical purposes, and is the 
 assumption made in using the tabular difference in the tables of loga- 
 rithms. 
 
 2. The common logarithm of 327 is 2.514548. What is 
 the logarithm of 327.12, supposing the relative rate of 
 change of the number and its logarithm to continue uni- 
 formly the same from 327 to 327.12 that it is at 327 ? 
 
 Ans, 2.514707. 
 
 3. Find what should be the tabular difference in the table 
 of logarithms for numbers between 4825 and 4826 ; in other 
 words, find the increment of the logarithm while the num- 
 ber increases from 4825 to 4826. Ans. .0000900. 
 
 4. Find what should be the tabular difference in the table 
 of logarithms for numbers between 9651 and 9652. 
 
 Ans. .0000450. 
 
 5. Find what should be the tabular difference in the table 
 of logarithms for numbers between 7235 and 7236. 
 
 Ans. .0000601. 
 
 TRIGONOMETRIC FUNCTIONS. 
 24. To Differentiate y = sin x. (1) 
 
 Give to x the infinitely small increment dx, and let dy 
 represent the corresponding increment of y; then we have 
 
 y + dy = sin (x + dx) 
 
 = sin x cos dx + cos x sin dx, (2) 
 
 Because the arc dx is infinitely small, its sine is equal to 
 the arc itself and its cosine equals 1 ; therefore (2) may be 
 written 
 
 y -j- dy = sin x 4- cos x dx, (3) 
 
TOMMTBiO FUNCTIONS. 
 
 Subtracting (i | (Vom (8), we have 
 
 ily =s cos g c/a;. (4) 
 
 Henoe, the differential of the sine of an arc is e(/(/n/ 
 to the cosine of the are into the differential of the arc 
 
 25. To Differentiate // = cob x. 
 
 Give to x the infinitely small increment dx, and we have 
 y -f dy = cos (x + dx) 
 
 = cos x cos dx — sin x sin dx < 
 = cos x — sin # dx (Art. 24). 
 .\ rfy = — sin x dx. 
 
 Otherwise thus : 
 
 We have y = cos 2 = sin (90° — x). 
 
 Differentiating by Art. 24, we have 
 
 dy = cos (90° - x) d (90° - x) 
 — sin x d (90° — x). 
 ,\ dy — — sin a; <£& 
 
 Hence, £7k? differential of the cosine of an arc is 
 negative and equal to the sine of the arc into the dif- 
 ferent 'nil of the arc. (The negative sign shows that the 
 cosine decreases as the arc increases.) 
 
 26. To Differentiate y = tan x, 
 
 __ sin a 
 
 We nave y = tan x = 
 
 a cos x 
 
 Differentiating by Arts. 18, 24, and 25, we have 
 
 cos x d sin x — sin a: dcosx 
 
 dy 
 
 COS 2 X 
 
 cos 2 a* + sin 2 x 7 dk 
 
 cos 2 a cos 2 # 
 
 = gee 2 # tfz. \ dy = sec 2 £ $c. 
 
TRIGONOMETRIC FUNCTIONS, 
 Otherwise thus: 
 
 Give to x the infinitesimal increment dx, and we have 
 y -f- dy =/ tan (x + dx) 
 ,\ dy = tan (x + afo) — tan x 
 tan # + tan dx 
 
 1 — tan x tan ate 
 
 tan # 
 
 tan x -\- dx , . , 
 
 = r.wij - tan a (smce tan * = & > 
 
 $£ 4- tan 2 x dx m , 
 
 = ^ 1 r~ = sec 2 a; dx 
 
 1 — tan re dx 
 
 (since tan # efa;, being an infinitesimal, may be dropped from 
 the denominator). 
 
 ,\ dy = sec 2 x dx. 
 
 Hence, the differential of the tangent of an arc is 
 equal to the square of the secant of the arc into the 
 differential of the arc. 
 
 27. To Differentiate y == cot x* 
 
 We have y = cot x = tan (90° — x). 
 
 .-. dy = sec 2 (90° - x) d (90° — z> 
 
 ,\ dy = — cosec 2 x dx. 
 
 The minus sign shows that the cotangent decrewwa *s the arc 
 increases. 
 
 Hence, the differential of the cotangent of an arc is 
 negative, and equal to the square of the cosecant of 
 the arc into the differential of the arc. 
 
 28. To Differentiate y = sec x, 
 
 "We have y = sec x = • 
 
 J cos x 
 
40 TKIGOMiME'lRIC Fl'M 'T/a.XS. 
 
 , d cos x sin x dx , 
 
 .-. (hi = = — = 7t — = sec a? tana; dx. 
 
 J C08 2 X COS 2 X 
 
 II. i ice, the differential of the secant of an are is 
 n in nl to the secant of the same arc into the tangi nl 
 of the arc, into the differential of the arc 
 
 29. To Differentiate y = cosec x. 
 
 We have y = cosec x = sec (90° — x). 
 
 .-. dy = d sec (90° — x) 
 
 = sec (90° - x) tan (90° - x) d (90° - x) 
 = — cosec x cot x dx. 
 
 Hence, the differential of the cosecant of an arc is 
 negative, and equal to the cosecant of the arc, into the 
 cotangent of the arc, into the differential of the arc. 
 
 30. To Differentiate y = vers x. 
 
 We have y = vers x = 1 — cos x, 
 
 .: dy = d (1 — cos x) = sin x dx. 
 
 Hence, the differential of the versed-sine of an arc 
 is equal to the sine of the arc into the differential of 
 the arc. 
 
 31. To Differentiate y = covers x. 
 
 We have y = covers x = vers (90° — x). 
 
 .\ dy = d vers (90° — x) = sin (90° — x) d (90° — x) 
 = — cos x dx. 
 
 Hence, the differential of the cover sed-sine of an 
 arc is negative, and equal to the cosine of the arc into 
 the differential of the arc. 
 
GEOMETRIC DEMONSTRATION. 
 
 II 
 
 32. Geometric Demonstration 
 
 at in the preceding Articles 
 admit also of easy demonstra- 
 tion by geometric construction. 
 Let P and Q be two consec- 
 utive points* in the arc of a 
 circle described with radius = 1. 
 Let x = arc AP ; then 
 
 dx = arc PQ. 
 
 Prom the figure we have, 
 
 PM = sin x; 
 
 -The results arrived 
 
 Fig. 4. 
 
 dx) ; 
 
 NQ = sin (x 
 .\ QR = d sin x. 
 
 OM = cos x ; ON = cos (x -f- dx) ; 
 
 % NM = — d cos x (minus because decreasing). 
 
 AT = tan x ; AT' = tan (x + dx) ; 
 
 /. TT' = d tan x. 
 
 OT = sec x; OT' = sec (x -f dx) ; 
 
 .-. DT' = d sec x. 
 
 Now, since EP and QP are perpendicular respectively to 
 MP and OP, and since DT and TT' are also perpendicular 
 to OT and OK respectively, the two infinitely small triangles 
 PQR and DTT' are similar to MOP. Hence we have the 
 following equations : 
 
 d sin x = RQ = QP cos PQR 
 
 z= cos x dx. 
 .'. d sin x = cos x dx. 
 
 * All that is meant here if? that P arid Q are to be reasoned upon as though they 
 were consecutive points ; of course, strictly speaking, consecutive points can never 
 be represented geometrically, since their distance apart is less than any assignable 
 distance. When we say that P and Q arc consecutive points, we may regard the 
 di-tance PQ in the figure as representing the infinitesimal distance between two 
 consecutive points, highly magnified. 
 
J-J bBOMXTRIO DEMONSTRATION, 
 
 d cos x = - PB = - PQ m PQll 
 
 = — sill X (Is. 
 
 A d cos x = — sin a; cfc. 
 
 rf tan x = TT' = DT sec DTT ss PT sec a; 
 = OTQP sec as (since DT - OT QP) 
 = sec 2 x dx. 
 
 A <? tan a; = sec 2 a: f/a;. 
 
 J sec 2; = DT' ss DT tan DTT' 
 
 = OT.QP tan a; = sec a; tans *fo. 
 
 A t? sec a; = sec a: tan a: <7a;. 
 
 Also, cb sb — £ (cot #), 
 
 and cd = — d (cosec #). 
 
 But the triangle cZ»c? is similar to the triangle OPM, since 
 cb and rf# are respectively perpendicular to MP and OP. 
 Hence we have 
 
 d cot x = —cb= —db cosec <lrb 
 
 = -- bO • QP cosec a; = — cosec 2 x dx, 
 
 A ^ cot a; = — cosec 2 a- t£& 
 
 d cosec x = —cd= — db cot Jc5 
 = -O&.QPcota; 
 = — cosec a: cot a- da\ 
 
 A «? cosec x = — cosec a: cot a? da\ 
 
 From the figure we see that the differential of the versed- 
 Bine is the same numerically as that of the cosine, but with 
 a contrary sign, t. e., as the versed-sine increases the cosine 
 decreases ; also the differential of the coversed-sine has the 
 same value numerically as that of the sine. 
 
EXAMPLES. 43 
 
 EXAMPLES. 
 
 1. y = sin mx. By Art. 24 we have, 
 
 dy = cos mx d (rnxf = wi cos »?a: „?#. 
 
 2. y -.- sin (a; 2 ). 
 
 c??/ = cos (a; 2 ) ^(« 2 ) = 2x cos (a; 2 ) a'a*. 
 
 3. y _= sin m a;. 
 
 aty = m sin m_1 a tf (sin x) = m sin" 1-1 x cos a. 6&. 
 
 4. y = cos 3 #. 
 
 dy = 3 cos 2 # (7 (cos a;) = — 3 cos 2 # sin x dx 
 = 3 (sin 3 x — sin #) dx. 
 
 5. y = sin 2a; cos x. 
 
 dy = sin 2x d cos # + cos a; r? sin 2a: 
 
 = — sin 2x sin x dx -f 2 cos 2a: cos a; ok. 
 
 6. y = cot 2 (a?), dy = — 6x 2 cot a? cosec 2 ^ Gfc. 
 
 7. ^ = sin 3 a; cos a:, aty = sin 2 a; (3 — 4 sin 2 a;) dx. 
 
 8. y = 3 sin 4 a\ «fy = 12 sin 3 a; cos a: dx. 
 
 9. y = sec 2 5a;. dy = 10 sec 2 5a; tan 5x dx. 
 
 10. i/ = log sin x. 
 
 7 d (sin a;) , A , _ AX cos a; 7 , , 
 
 dy = -A (Art. 20) = - — dx — cot x dx. 
 
 a sin x x ' sm a; 
 
 11. y = log (sin 2 x) = 2 log sin #. dy = 2 cot a; <fo. 
 
 12. «/ = log cos x. dy = — tan x dx. 
 
 n d tan x 2dx 
 IS. y = log tan*. ^ = _____ - ___ 
 
 14. ?/ = log cot x. 
 
 2dx 
 sin 2# 
 
 15. y = log sec x. dy = tan x dx. 
 
 16. ?/ = log cosec x. dy = — cot a; 6&. 
 
44 ILLUSTRATIVE EXAMPLES* 
 
 *m i A + COS 3 
 
 17. y = log\/r— — - 
 
 * to V 1 — COS X 
 
 = £ log (1 -f cos x) — J log (1 — cos x). 
 
 dx 
 * sin x 
 
 18. y = ef* cos x, dy = e* d cos a; -f cos a; fife* 
 
 = — e* sin x dx + e* cos x dx 
 = c* (cos x — sin a?) dx. 
 
 19. y = x sin a; -f cos a;. dy = x cos a; fifo. 
 
 20. y = xe eoi '. dy = e e " x (l — x sin a:) dx. 
 
 21. y as c 00 "* sin x. dy — e C0BI (cos x — sin 2 #) dx. 
 
 22. y sa log \/sin x + log Vcos x 
 
 = J log sin x + £ log cos x. 
 
 ,\ dy — \ (cot # — tan ») <fc as cot 2a: da*. 
 
 23. y = log (cos a; + V— 1 sin a:), dy = V— 1 dx. 
 
 dx 
 
 i /l + sin x . 
 
 24 - y = lo sVr^nrr rf * = 
 
 25. y as log tan (45° + \x). dy = 
 
 cos X 
 
 dx 
 cos x' 
 
 26. y = sin (log x). dy = - cos (log x) dx. 
 
 x 
 
 ILLUSTRATIVE EXAMPLES. 
 
 1. Which increases faster, the arc or its tangent? When 
 is this difference least, and when greatest? What is the 
 value of the arc when the tangent is increasing twice as 
 fast as the arc. and when increasing four times as fast as the 
 arc? 
 
 From y = tan x, we get dy = 8ec?xdx, which shows 
 that if we give to the aiv (x) the infinitesimal increment dx, 
 
ILLUSTRATIVE EXAMPLES. 45 
 
 the corresponding increment of the tangent (y) is sec 2 # 
 times as great ; that is, the tangent (y) is increasing secant 
 square times as fast as the arc, and hence is generally in- 
 creasing more rapidly than the arc. When x = 0, sec x = 1 ; 
 therefore, at this point, the tangent and the arc are increas- 
 ing at the same rate. When x = 90°, the secant is infinite ; 
 therefore, at this point, the tangent is increasing infinitely 
 faster than the arc. 
 
 * When the tangent is increasing twice as fast as the arc, 
 we have dy = 2dx, or sec 2 x = 2, which gives x = 45° ; 
 hence at 45° the tangent is increasing twice as fast as the 
 arc. « 
 
 When the tangent is increasing four times as fast as the 
 arc, we have dy = 4-dx, or sec 2 x = 4, which gives x = 60° ; 
 hence at 60° the tangent is increasing four times as fast as 
 the arc. 
 
 2. Assuming that the relative rate of increase of the sine, 
 as compared with the arc, remains constantly the same as 
 at 60°, how much does the sine increase when the arc in- 
 creases from 60° to 60° 20'. 
 
 Let x = the arc and y its sine ; then we have y = sin x, 
 ,\ dy = cosxdx, which shows that the increment of the 
 sine is cosine times the increment of the arc. Now the arc 
 
 3 14159 
 of 20 ' = ion = .0058177 = dx; therefore, 
 
 dy = cos 60° dx = ix .0058176 = .0029088, 
 
 which is the increase of the sine on the above supposition, 
 and is a little greater than the increase as found from a table 
 of natural sines, as it should be, since we have supposed the 
 sine to increase uniformly while the arc was increasing 
 uniformly from 60° to 60° 20', whereas the sine is increasing 
 'more and more slowly while the arc is increasing uniformly. 
 This is evident from the equation dy = cosxdx, and also 
 from geometric considerations. 
 
If, CIRCULAR Fr.\<ri(,\s. 
 
 3. The natural cosine of 5° 81' ifl .995368. Assuming 
 that the relative rate of change of the cosine and the arc 
 iviiiuins the same as at this point, while the arc increases to 
 
 ' . what is the cosine of 5° 32' ? Arts. .995340. 
 
 4. The logarithmic sine of 13° 49' is 9.3780633. Assum- 
 ing iliat tin- relative rate of change of the logarithmic sine 
 and the arc remains the same as at this point, while the arc 
 increases to 13° 49' 10 ", what is the logarithmic sine of 
 13° 49' 10"? Am. 9.3781489. 
 
 5. The log cot 58° 21' =9.789863. On the same sup- 
 position as above, what is the decrease of this logarithm for 
 1 second increase of arc. An*. .00000471. 
 
 & A wheel is revolving in a vertical plane about a fixed 
 cen t re. At what rate, as compared with its angular velocity, 
 is a point in its circumference ascending, wjien it is 60° 
 above the horizontal plane through the centre df motion. 
 
 Ans. Half as fast . 
 
 CIRCULAR FUNCTIONS. 
 33. To Differentiate y = sin -1 as.* 
 We have x = sin y ; 
 
 therefore, dx = cosy dy = v(l — sin 2 y)dy 
 
 = \/l~^~x*dy. 
 dx 
 
 •'• *y = vfzr^ = d ( sin_1 *)• 
 
 34. To Differentiate y = cos -1 x. 
 
 We have, x = cos y ; 
 
 therefore, dx = — s\ny dy = — Vl — cos'fy dy 
 
 — _ *J\~—~tfdy. 
 .-. dy = — = d (cos -1 x). 
 
 * This notation, as already explained, means y - the arc whose sine ia x. 
 
CIRCULAR FUNCTIONS. 4? 
 
 35. To Differentiate y = tan 1 x. . 
 
 We have x = tan y ; 
 
 therefore, «fc = sec 2 y <%>=? (1 + tan 2 ?/)eZy 
 
 = (1 + a?) <fy. 
 
 ••• ty = t+~x* ~ d ( tan_1 x ! 
 
 36. To Differentiate y = cot -1 a?. 
 
 We have x = cot ^ ; 
 
 therefore, <fo =^— cosec 2 ydy = — (1 + cot 2 y) tfy 
 
 ./ dy = - — ~ 2 = 6? (cot -1 aj). 
 
 37. To Differentiate y = sec -1 ac. 
 We have x = sec y ; 
 
 therefore, dx = sec y tan y dy = sec y y sec 2 y — 1 dy 
 
 = x^/x 2 — idy 
 
 ,\ eft/ = — == == d (sec -1 a;). 
 
 38. To Differentiate # = cosec -1 x. 
 
 We have x = cosec # ; 
 
 therefore, dx = — cosec?/ coty dy 
 
 = _ cosec y V cosec 2 y — 1 dy 
 = — #\/# 2 — 1 %• 
 
 dv = J!L=z=. = d (cosec -1 x\ 
 
is EXAM I' LIS. 
 
 39. To Differentiate // = vers -1 X* 
 
 \\V have x = vers y ; 
 
 therefore, dx = sin y dy = v 1 — cos 2 ?/ r/// 
 
 = Vl — (1 — vers y) 2 dy 
 
 = \/% vers ^ — vers 2 y dy 
 
 = v^z — x 2 dy. 
 
 .«. % = -~== = ^ (vers" 1 »} 
 
 V%x — x 2 
 
 40. To Differentiate y = covers -1 #. 
 
 We have a; = covers */ ; 
 
 therefore, dx — —cosy dy = — V 1 — sin 2 ?/ <fy 
 
 = — Vl — (1 — covers «/) 2 d# 
 ss — -y/2 covers y — covers 2 y dy 
 
 = — V%x — x 2 dy. 
 
 dx 
 
 dy = = d (covers -1 x), 
 
 VZx — x 2 
 
 EXAM PLES, 
 X 
 
 1. Differentiate «/ = sin -1 -• 
 9 a 
 
 We have, by Art. 33, 
 
 jX dx 
 
 a a dx 
 
 dy = 
 
 v-s V"-S 
 
 ^ ^/tf — x 2 
 
 x 
 2. Differentiate y' = a sin -1 — 
 
EXAMPLES. 
 
 dx 
 a — 
 
 We have, Art 33, dy' = 
 
 49 
 
 adx 
 
 \A-S 
 
 Va 2 — x 2 
 
 'Geometric illustration of Examples 1 and 2 ': 
 
 Let OA == 1, OA' = a, y = arc AB, y' = arc A'B', 
 x = M'B'. 
 
 WW _ x T > 
 
 OB' ~ a' 
 
 Now 
 
 BM 
 
 arc AB = sin -1 BM 
 
 . . B'M' 
 = sm~ 
 
 OB' 
 
 x 
 
 a 
 
 = sin 
 
 = y (see Ex. 1). 
 
 A'B' = A'O • arc AB = A'O • sin- 1 
 
 x 
 = a sin -1 - = «/' (see Ex. 2). 
 
 Fig. 5 
 
 B'M' 
 OB' 
 
 Also, A'B' =r sin -1 B'M' = sin -1 x (to radius a) 
 
 x 
 .'. a sin -1 - (to radius 1) = sin -1 x (to radius a). 
 
 Hence, in Example 1, y is the arc AB (to radius 1), and 
 
 x 
 is given in terms of the sine - (to radius 1) ; while in 
 
 a v 
 
 Example 2, y is the arc A'B' (to radius a), and is given in 
 
 x 
 terms of the sin - (to radius 1). 
 
 If we give B'M' (which is x in both examples) an incre- 
 ment (= dx), the corresponding increment in A'B' will be 
 a times as great as that on AB; that is, dy' in Ex. 2 is a 
 times dy in Ex. 1. 
 3 
 
50 EXAMPLES. 
 
 3. y = cos -1 — dy = — 
 
 a * t/a? — z* 
 
 ,x , adx 
 
 4. ^ = tan-i-. ^ = ^+^' 
 
 adx 
 
 a' ~ 9 ~ a' + z 3 ' 
 
 .x , ««« 
 
 6. y = sec -1 — ay 
 
 5. ?/ = cot -1 — dy = 
 
 a x^x* — # 2 
 
 x 7 adx 
 
 7. y = cosec -1 -• ay = 
 
 a * x^/tf _ a 2 
 
 .2 , ' dx 
 
 8. # = vers -1 — dy — 
 
 « V2«z — z 2 
 
 9. y = covers -1 — ay = 
 
 10. y = a cos -1 
 
 « A/2aa — x* 
 
 x 
 
 dy = 
 
 adx 
 a adx 
 
 V'-l 
 
 a? V« 2 — ^ 
 
 "2 
 
 az 
 
 11. y = «teni-. rf y = = __. 
 
 + « 2 
 
 12. 2/ = «cot-i-. % = _ = ______ 
 
 + « 2 
 
 13. y = a sec -1 -• «"?/ = 
 
 a; /x 2 
 
 a\l tf~ 
 
 a X /Z* * X^/ x 2_ a Z 
 
14. 
 
 MISCELLANEOUS EXAMPLES. 
 1 x 
 
 y = a cosec -1 -• 
 a 
 
 dx 
 a — 
 
 a 2 dx 
 
 
 uy — ■ ■■ 
 
 X 
 
 a 
 
 X 
 
 y = a vers -1 — dy = 
 
 1 x 
 y = a covers * - • 
 * a 
 
 dii — 
 
 v « 2 
 
 dx 
 
 a — 
 
 a 
 
 
 15 
 
 x \/x 2 — a 2 
 . adx 
 
 16. 
 
 /« x & 
 
 dx 
 
 a — 
 
 a 
 
 V2ax — x 2 
 adx 
 
 
 V 
 
 Lx X* 
 
 V%a% — x 2 
 
 51 
 
 MISCELLANEOUS EXAMPLES. 
 
 ., a 4- x 7 da — x 7 
 
 ~/l. y = -• dy = r^dx. 
 
 Wa — x 2 (a — x)? 
 
 S%. y = yx — Va 2 — x\ 
 
 (x 4- Va 2 — x 2 ) dx 
 
 UU Wa 
 
 2_z2(z_ Va 2 -^ 
 
 x 
 
 dip 
 
 o, y — — • dy — 
 
 * x + y i _ X 2 * 
 
 a»(i-^+Vrn? 
 
 1 A * 
 
 A a 4. «/ = . 
 
 3./ 2 ^ 
 
 * = (!-*•)* 
 
 y V(i-*y 
 
 K « 4 7 
 
 _ efi(tf~%x*)dx 
 
 • 7 " v«v-^ ' 
 
 Ix 2 (a 2 -.'*)■: 
 
M MISCELLANEOUS EXAMPLES. 
 
 6 v = ^1 + ! = -• 
 
 \ -' + i + x 
 
 In tractions of this form, the student will find it an ad 
 vantage to rationalize the denominator, by multiplying both 
 terms of the fraction by the complementary surd form ; that 
 is. in the present case, by Vx 2 + 1 — x. Thus, 
 
 <s/xi 4. i _ x Vx 2 + l—x 
 y = — — ■ — x — - 
 
 y/tf + 1 _|_ X Vtf +1—X 
 
 _ (yg~+T— xf 
 l 
 
 dx 
 
 b. » = 
 
 9. « 
 
 
 
 a# 
 
 — ' 
 
 2(1 + Vx)v 
 
 • 
 
 'x — X 2 
 
 vt 
 
 + « + 
 
 Vi- 
 
 X 
 
 
 
 Vi 
 
 + s- 
 
 vi- 
 
 X 
 
 
 
 
 
 
 dy 
 
 1 + Vi- 
 
 -a* 7 
 — — dx. 
 
 ■X 2 
 
 V^a; 
 
 v^ 
 
 p+i. 
 
 
 
 
 dy 
 
 = 
 
 iVx' + l 
 
 = dx. 
 
 12#a*>\/Vx + l 
 
MISCELLANEOUS EXAMPLES. 
 
 53 
 
 10. y = y 
 
 Vx j 
 
 Sb 
 
 ix 
 
 dy = _ _ggv^4f':zi3_- & 
 
 \l a 
 
 Vx 
 
 +. vv~ 
 
 11. y = \J 2x—\—\/ 2x—\ — V^x—\ — etc., arf inf. 
 Squaring, we have, 
 
 if = 2x— 1— Y 2a;— 1— \/2x— 1 — \/2z— 1— etc., ad inf. 
 y 2 = 2 x-l-y; 
 
 Hence, 
 and 
 
 y = -\± i V82; - 3. 
 2dx 
 
 A tfy = ± 
 
 12. y = 
 
 VSx — 3 
 
 yi + ^ + yr^ "^ 2 
 
 % 
 
 (-vrb) 
 
 dx. 
 
 13 - ^ ,og 7rT^ 
 
 dy ~ xjl + x*)' 
 
 1A 1 Vl +X +V1— X , 
 
 14, ?/ = log — £= — • ay — 
 
 Vl + x— Vl—x 
 
 dx 
 
 x'^/YZ-~ r 2 
 
 15. # = log 
 
 V^TT 2 -^' llJ x Vx* + a 2 
 
54 M/S( EL LA NJSO I S E.W 1 Ml'LES. 
 
 16. y = log [vT+^ + vT^z*]. 
 
 
 
 , dx dx 
 
 
 ~* " x x^/i _ tf 
 
 17. 
 
 y = log (z — a) - 
 
 
 18. 
 
 # = a*\ 
 
 dy = 2a x * log a xdx. 
 
 19. 
 
 y = e*(l — x*). 
 
 dy = c* (1 — 3z 2 — a: 3 ) dx. 
 
 20. 
 
 y = ef + e~* 
 
 7 4:dx 
 
 ay - (a + *-*)»' 
 
 21. 
 
 y = log (e* + fl _aj ). 
 
 ^ _ e - x 
 
 1 * = 5hhF»* 
 
 22. 
 
 y = *". 
 
 % _^(l-]og,)^ 
 
 23. 
 
 y = 2e ^(sl - 3a; + &S* - 6). % = ze ^<fo 
 
 24. 
 
 0* - 1) 1 
 
 V = s — « — — 
 
 (x - 2)* (x - 
 
 - T . (See Art 23.) 
 3)* 
 
 18(a;-2)t(a:-3)V 
 
 rfy= **(s + 3)*<fe 
 
 (a? + 2) 5 (a; + 1)*" 
 
 * + Va» - 1 * + Va? - 1 
 
 27. y = sin a; — J sin 3 a*. dy = cos 3 a; ffo. 
 
 28. y == i tun 3 a; — tan x + a;. dy = tan 4 ar <f& 
 
 29. y = i tan 3 a; f tan x. dy = sec 4 a; rfz. 
 
MISCELLANEOUS EXAMPLES. 
 
 55 
 
 30. y = 
 
 31. p = 
 
 32. y = 
 
 33. y = 
 
 34. y = 
 
 sin e*. 
 
 tan 2 sb + log (cos 2 #). 
 
 log (tan a; + sec x). 
 
 sin a; 
 
 dy = e* cos c® dx. 
 dy = 2 tan 3 a; d#. 
 rfy = sec a; dx. 
 
 , (cos 3 a; — sin 3 x) . 
 1 + tan a: * (sin x + cos a;) 2 
 
 log 
 
 V? 
 
 cos a; — b sin a: 
 
 cos a; -f J sin a; 
 
 — db dx 
 
 a 2 cos 2 x — W sin 2 a; 
 
 X 
 
 tan e*. 
 
 ^ = 
 
 JL 1 
 
 e* sec 2 e* da? 
 
 a^ 
 
 (sec 2 A/l —x)dx 
 %V~l—x 
 
 x i{a ". efy = # 9in * (cos a; • log a; -\ J da;. 
 
 2 3 cos x , _ , a: 
 
 -!-• ^-5 h 3 log tan -• 
 
 sin 2 x cos a; sin 2 x 2 
 
 2da; 
 ^ ~~ sin 3 a; cos 2 a; 
 
 35. y = 
 
 36. y = tan \/l — x. dy = — 
 
 37. 2/ = 
 
 38. y = 
 
 39. y = 
 We haye, 
 
 40. $f = 
 
 We have, 
 
 sin" 
 
 VI + z 2 
 
 a; 2 
 
 da; 
 
 da; 
 
 (1 + x*)§ ' (1 + xrf ' 1 + & 
 cos -1 a; Vl 
 
 ; 2 = COS -1 vs 
 
 a*. 
 
 _ tf Va; 2 - z 4 -T- Vl - (z 2 - «*) 
 (1 — 2a?)da; 
 
 Vl —a; 2 
 
 _^ Vl — a; 2 + x*. 
 
56 MISCELLANEOUS EXAMPLES. 
 
 (1 - 2z*) dx 
 
 .\ dy = — 
 
 V(l — a») (1 - x 2 + z 4 ) 
 
 , . x 2dx 
 
 41. y = Bm- l {2x VT^tf). dy = ^f^p' 
 
 42. y = sin" 1 (3* - 4Z 3 ). <fy = - ~ 3 ^ • 
 
 43. y = tan -1 5 . dy = :: =• 
 
 * 1 — x 2 y 1 +x* 
 
 44. y = s^ r ^. %= ____. 
 
 45. x s vers -1 1/ — \/2r# — y 2 (where vers -1 # is taken 
 to radius r). 
 
 We may write this (see Art. 40, Exs. 1 and 2), 
 x = r vers -1 ^ — V%ry — y 2 . 
 
 , _ r dy rdy — ydy 
 
 V%ry — y 2 V%ry — y 2 
 
 _ ydy 
 
 V%ry — y 2 
 
 46. y = xVa 2 — x 2 +a 2 sin -1 -• d> = 2Va 2 —x 2 dx. 
 
 47. y = log yj-^ + * tan_1 * ^ = j— y 
 
 ,2a; , dx 
 
 48. # = vers -1 -£■• ay = 
 
 49. y = e—'. gg a j*" —f 
 
/ MISCELLANEOUS EXAMPLES 
 
 50. y = ^ sla " 1 \ 
 
 dy = x™ 
 
 57 
 
 x log a; + (1 — x 2 )? sin _1 rc~| 
 x (1 _ &)\ 
 
 51. # = sec -1 nx. 
 
 52. y == sin -1 
 
 dy = 
 
 dy = 
 
 dx 
 
 <c^/n*x* — 1 
 adx 
 
 V«2 + X 2 * fl2 + X 2 
 
 53. # = sin" 1 Vsin x. dy = -J (Vl -f cosec a;) tfz. 
 
 k* i. 1 A — cos X 
 
 54. # = tan" 1 \ — dy = Ux. 
 
 V 1 + COS X u 2 
 
 55. y 
 
 x sin -1 x 
 
 + log Vl — a: 2 . % =s 
 
 sin -1 x 
 
 Vl — x 2 (l-z 2 )t 
 
 56. y =. (x + a) tan -1 \ / Vox. 
 
 % = tan- 1 ,* ri 
 xV& 
 
 dx. 
 
 dx. 
 
 57. y = sec -1 
 
 58. y = tan -1 
 
 59. y = sin -1 
 
 2 Va; 2 + x — 1 
 
 3 A — a 3 
 
 « 3 — 3«# 2 
 
 x\/a — b 
 
 Va(l +x 2 ) 
 
 dy 
 
 dy = 
 
 dx 
 
 dy 
 
 x<S/&+x-l 
 dadx 
 
 a 2 + x* 
 
 va 
 
 dx. 
 
 (1 + «*) Va + ^ 
 
 60. If two bodies start together from the extremity of the 
 diameter of a circle, the one moving uniformly along the 
 diameter at the rate of 10 feet per second, and the other in 
 the circumference with a variable velocity so as to keep it 
 always directly above the former, what is its velocity in the 
 
,>S MISCELLANEOUS EXAMPLES* 
 
 riivuinlVrence when passing the sixtieth degree from the 
 
 starting-point? . 20 „ . , 
 
 ft r Ans. — - feet per second. 
 
 Gl. If two bodies start together from the extremity of the 
 diameter of a circle, the one moving uniformly along the 
 tangent at the rate of 10 feet per second, and the other in 
 the circumference with a variable velocity, so as to be always 
 in the right line joining the first body with the centre of 
 the circle, what is its velocity when passing the forty-fifth 
 degree from the starting-point. Ans. 5 ft. per second. 
 
CHAPTER III 
 
 LIMITS AND DERIVED FUNCTIONS. 
 
 41. Limiting Values. — The rules for differentiation 
 have been deduced, in Chapter II, in accordance with the 
 method of infinitesimals explained in Chapter I. We shall 
 now deduce these rules by the method of limits. 
 
 The limit, or limiting value of a function, is that value 
 toward which the function continually approaches, till it 
 differs from it by less than any assignable quantity, while 
 the independent variable approaches some assigned value. 
 If the assigned value of the independent variable be zero, 
 the limit is called the inferior limit ; and if the value be 
 infinity, it is called the superior limit. 
 
 42. Algebraic Illustration. — Take the example, 
 
 1 
 
 y -1+"^ 
 and consider the series of values which y assumes when x 
 has assigned to it different positive values. When x = 0, 
 y = 1, and when x has any positive value, y is a positive 
 proper fraction; as x increases, y decreases, and can be made 
 smaller than any assignable fraction, however small, by giv- 
 ing to x a value sufficiently great. Thus, if we wish y to be 
 less than iqooooo ? we make x = 1000000, and get 
 
 1 
 
 y ~ 1 + 1000000 
 which is less than totooT¥- If we wish y to be less than 
 one-trillionth, we make x = 1000000000000, and the re- 
 quired result is obtained. We see that, however great x 
 may be taken, y can never become zero, though it may be 
 
I TRIGONOMMTRJC ILLUSTRATION. 
 
 made to differ from it by as small a quantity as we please. 
 
 Hence, the limit of the function t-t~- is zero when x is 
 infinite. 
 
 We are accustomed to speak of such expressions thus : 
 " When x is infinite, y equals zero." But both parts of this 
 sentence are abbreviations: "When x is infinite" means. 
 " When x is continually increased indefinitely" and not, 
 "When x is absolute infinity ;" and " y equals zero" means 
 strictly, "y can be made to differ from zero by as small a 
 quantity as we please." Under these circumstances, we say, 
 "the limit of y, when x increases indefinitely, is zero." 
 
 43. Trigonometric Illustration. — An excellent exam- 
 ple of a limit is found in Trigonometry. To find the values 
 
 of 7 — n and — ^— , when diminishes indefinitely. Here 
 
 we have 
 
 - — - = cos ; and when = 0, cos = 1. 
 tan0 
 
 Hence, if be diminished indefinitely, the fraction - — ^ 
 
 will approach as near as we please to unity. In other 
 
 words, the limit of 7 — ^, as continually diminishes, is 
 
 unity. We usually express this by saying, " The limit of 
 
 - — j: , when = 0, is unity ;" or, — ^ = 1, when = 0;" 
 tan J ' tan 
 
 that is, we use the words " when 6 = " as an abbreviation 
 
 for " when is continually diminished toward zero." 
 
 Since 7 — 2 = 1> when = 0, 
 
 tan0 
 
 we have also -. — „ = 1, when = 0. 
 sin 
 
tRIGONOM&TRiC ILLUSTRATION. 61 
 
 It is evident, from geometric considerations, that if be 
 the circular measure of an angle, we have 
 
 tan > > sin ; 
 
 tan ^ . ' 
 
 but in the limit, t, e., when = 0, we have 
 
 tan _ 
 sin (9 ~ 1 > 
 
 and therefore we have, at the same time, 
 
 ., , sin 
 
 -a — ^ ±= 1, and .*. — ^— = 1, 
 
 sm 6 9 
 
 which shows that, in a circle, the limit of the ratio of an arc 
 to its chord is unity. 
 
 "sin 
 In the expression, — — = 1, when 6 = 0," it is evident 
 u 
 
 that — — is never equal to 1 so long as 6 has a value dif- 
 ferent from zero ; and if we actually make 6 = 0, we render 
 the expression — — meaningless.* That is, while —^~ 
 
 approaches as nearly as we please to the limit unity, it never 
 actually attains that limit. 
 
 If a variable quantity be supposed to diminish gradually, till it be 
 less than anything finite which can be assigned, it is said in that state 
 to be indefinitely small, or an infinitesimal ; the cipher is often used 
 as an abbreviation to denote such a quantity, and does not mean abso- 
 lute zero ; neither does oo express absolute infinity. 
 
 Rem. — The student may here read Art. 12, which is applicable to 
 this method as well as to that of infinitesimals, which it is not neces 
 sary for us to insert again. 
 
 * See Todhuntcr's Dif. Cal., p. 6. 
 
DEh'l\ATlVES. 
 
 44. Derivatives.- Tli e ratio ol tin- increment of u to 
 
 Aw 
 
 that of x, when the increments are finite, is denoted by — -; 
 
 the ratio of the increment of u to that of x in the limit, 
 
 when both are infinitely small, is denoted by -j~, and 
 Ls called the derivatipe* of u with respect to x. 
 
 Thus, let ?/ =f(x) ; and let x take the increment h 
 (= Aa*), becoming x + //, while u takes the corresponding 
 increment A// ; then we have, 
 
 u -f Aw = /(# + A); 
 therefore, by subtraction, we have 
 
 tei=f(x + Ji)-f(x); 
 
 and dividing by h (= A a;), we get 
 
 *u _ f(x + h)-f (x) 
 
 Ax~ h l ; 
 
 It may seem superfluous to use both h and Ax to denote the same 
 thing, but in finding the limit of the second member, it will sometimes 
 be necessary to perforin several transformations, and therefore a sin- 
 gle letter is more convenient. In the first member, we use Ax on 
 account of symmetry. 
 
 The limiting value of the expression in (1), when h 
 is infinitely small, is called the derivative of u or 
 f{x) with respect to x, and is denoted by f (x). 
 
 Therefore, passing to the limit, by making h diminish 
 
 indefinitely, the second member of (1) becomes /' (x), and 
 
 dii 
 the first member becomes, at the same time, -=-; hence we 
 
 , (XX 
 
 have 
 
 * Called also the derived function and the differential coefficient. 
 
DIFFERENTIAL COEFFICIENT, 63 
 
 45. Differential and Differential Coefficient. 
 
 Let u = f(x) ; then, as we have (Art. 44), 
 
 we have du = df(x) = f (x)dx, 
 
 where dx and du are regarded as being infinitely small, and 
 are called respectively (Art. 12) the differential of x and the 
 corresponding differential of u. 
 
 f (x), which represents the ratio of the differential of the 
 function to that of the variable, and called the derivative of 
 f(x) (Art. 44), is also called the differential coefficient of 
 f(x), because it is the coefficient of dx in the differential 
 off(x). 
 
 Some writers * consider the symbol — only as a whole, and do not 
 assign a separate meaning to du and dx ; others.f who also consider 
 the symbol — only as a whole, regard it simply as a convenient nota 
 tion to represent ^ , and claim that du and dx are each absolutely zero. 
 
 46. Differentiation of the Algebraic Sum of a 
 Number of Functions. 
 
 Let y = au + bv + cw + z -f etc., 
 
 in which y, u, v, w, and z are functions of x. Suppose that 
 when x takes the increment h (= Ax), y, u, v, w, and z 
 take the increments Ay, Au, A?;, Aw, Az. Then we have, 
 
 y + Ay = a (u + Au) + b (v -f- Av) + c (w + Aio) + (z + Az) + etc. 
 / Ay = a Au + b Av + c Aiv + Az -f- etc. 
 Dividing by h or Ax, we have 
 
 * See Todhunter's Dif. Cal., p. 17 ; also De Morgan's Calculus, p. 14, etc. 
 t See Young's Dif. Cal., p. 4. 
 
A4 D1FFEREXTIAT10N OF A PRODUCT. 
 
 Ay lu , 7 Ay , Aw . Az , 
 
 -f- = fl — + & — - + c - -f - -f etc., 
 
 As Aa; Arc A./; A.e 
 
 which becomes in the limit, when h is infinitely small 
 (Art. 12), 
 
 dy du , ,dv , dw , dz , , . . - . N 
 
 1 = a Tx + h Tx + c dx~ + dx + etc - ( see Art - 14 >- 
 
 47. Differentiation of the Product of two Func- 
 tion* 
 
 Let y ss tnr, where w and y are both functions of a-, and 
 suppose Av, Aw, Ay to be the increments of y, w. v corre- 
 sponding to the increment Aa; in x. Then we have 
 
 y -f Ay = (w + Aw) (v + A?') 
 
 = wy -f w Ay + v Aw + Aw Aw. 
 
 .•. Ay = w Ay + v Aw + Aw Av ; 
 
 Aw Ay Aw Aw . 
 
 or, -* = u — + v - 1-7-^. 
 
 Aa; Aa; Aa; Aa; 
 
 Now suppose Aa: to be infinitely small, and 
 Aw Ay Aw 
 
 Aa;' Aa:' Aa;' 
 become in the limit, 
 
 dy dv du 
 
 dx } dx 9 dx 
 
 Also, since Ay vanishes at the same time, the limit of the 
 last term is zero, and hence in the limit we have 
 dy dv du /Q A . _,_. 
 
 Tx = u Tx + v Tx < See Art 16 "> 
 
 It can easily be seen that, although the last term vanishes, the 
 remaining terms may have any finite value whatever, since they con- 
 tain only the ratios of vanishing quantities (see Art. 9). For example, 
 
 — = - when x = ; but by canceling x we get — = a. But the 
 X u x 
 
 expression — x x, which equals - x when x = 0, becomes - x = 
 
 when x = 0. 
 
DIFFERENTIATION OF A PRODUCT. 05 
 
 Otherwise thus: 
 
 Let f(x) <p (x) denote the two functions of x, and let 
 
 u ==/($) (f)(x). 
 
 Change x into x + h, and let u + Au denote the new 
 product; then 
 
 u + Au = f(x + h) <f> (x + 7i) 
 
 Au _f ( x + 7Q <A (a; + ft) — / (a?) («) 
 
 Subtract and add f(x) <f* (x + h), which will not change 
 the value, and we have 
 
 A, = fix + K)-fiM ^ {x + h) +f(x) j^K^-m, 
 Now in the limit, when h is diminished indefinitely, 
 
 • *(* + h) -* & = #{*) (Art44) . 
 
 and (# + li) = <f> (x) ; 
 
 therefore, ~ = f (x) (x) +f(x) f (x), 
 
 which agrees with the preceding result. 
 
 48. Differentiation of the Product of any Number 
 of Functions. 
 
 Let y = uvw, 
 
 u, v, w being all functions of x. 
 
 Assume z = vw ; 
 
 then y = uz 9 
 
t'.'i />//// 7/ /:.Y'/7.r/7o.Y OP A FRACTION. 
 
 and by Art. 47 we have 
 
 dy _ udz zdu 
 dx ~~ dx dx 
 
 Also, by the same Article, 
 
 dz _ vdw ivdv % 
 dx ~' dx dx ' 
 
 hence, by substitution, we have 
 
 dy dw dv , du /a k .+„ x 
 
 I = «»£• + « B 5 + •" W (See Art 17 > 
 
 The same process can be extended to any number of 
 functions. 
 
 49. Differentiation of a Fraction. 
 
 Let 
 
 
 y = 
 
 u 
 
 V 
 
 Then 
 
 we shall have 
 
 
 
 
 y + Ay = 
 
 u + Au m 
 v + Av y 
 
 
 
 *y = 
 
 u -f Au u 
 
 v -f- lv v 
 
 
 
 
 v Au — u Av 
 
 
 
 
 v 2 + vAv 
 
 Dividing by Ax and 
 
 passing to the limit, 
 
 
 
 dy _ 
 
 du dv 
 dx dx 
 
 
 
 dx 
 
 V* 
 
 '(since 
 
 vdv vanishes). 
 
 (See 
 
 Art. 18.) 
 
 Cor.- 
 
 —If u is a constant, 
 
 we have 
 
 
 
 
 udv 
 
 
 
 dy 
 
 dx 
 
 dx 
 
DIFFERENTIATION OF ANY POWER. G7 
 
 50. Differentiation of any Power of a Single Va- 
 riable. 
 
 1st. When n is a positive integer. 
 
 Let y = x n ; 
 
 then we have y + by = (x + h) n ; 
 
 therefore, A?/ = nx n ~ x h -\ y — — — ' x n ~ 2 h 2 + etc. . . , h n . 
 
 1 • ii 
 
 Dividing by li or Ax, we get 
 
 Aw . w (» — 1) „ , _ . 
 
 AiB 2 
 
 Passing to the limit, we have 
 
 ^ = nx n ~\ (See Art. 19, 1st.) (1) 
 
 2d. When n is a positive fraction. 
 
 m 
 
 Let y = w* 
 
 where tf is a function of x ; then 
 y" = «*, 
 and d (?/ n ) =5 tf (ft-™) ; 
 
 hence, by (1), ny n ~ l -f- = mu m ~ x -^-* 
 
 dy _ w w m_1 c?w 
 dx ~ n y n ~ l dx 
 
 3d. When n is a negative exponent, integral or 
 fractional. 
 
 Let «/ = w~"; 
 
 then ii = — , 
 
68 DIFFi:i;i:.\ TTATIOA <>f LOGARITHMS, 
 
 and by Art. 49, Cor., we have 
 
 dy nu n ~ l du ,'/"/* L -.a .... /.,, 
 
 51. Differentiation of log x. 
 
 Let y = log x ; 
 
 therefore, y -f A// = log (a: -f- h), 
 
 and A// = log (x -\- h) — log x 
 
 = lo g (^) = ,o g (, + *) 
 
 therefore, — = m I zr^ + etc. ) ; 
 
 Ax \z 2x 2 r 
 
 therefore, passing to the limit, we get 
 
 dy m 1 
 
 i = — or 
 dx x x 
 
 (according as the logarithms are not or are taken in the 
 Naperian system. See Art. 20). 
 
 52. Differentiation of a x . 
 
 Let y = a x . 
 
 Proceeding exactly as in Art. 21, we get 
 
 dy a? 
 
 -f ■ = — log a or a x log a (Art. 21). 
 
 (XX ?ti 
 
 53. Differentiation of sin x. 
 
 Let y = sin a?; 
 
 therefore y + Ay =z sin (# 4- 7i) ; 
 
 hence, Ay = sin (# -f A) — sin x. 
 
DIFFERENTIATION OF A COSINE. 69 
 
 But from Trigonometry, 
 
 . A .-«'« A + B . A-B 
 sm A — sin B = 2 cos — - — sm — - — 
 
 </ 2 
 
 .". Ay = sin (x + h) — sin x 
 = 2 cos ^ + ^ j sm - ; 
 
 . h 
 
 Ay / h\ Sm 2 
 
 hence, _| = cos |* + -) __. 
 
 2 
 By Art. 43, when h is diminished indefinitely, the limit of 
 
 * 
 
 1 ; also, the limit of cos \x + -) = cos x. 
 
 sin^ 
 
 2 
 
 Therefore, -^- = cos x. (See Art. 24.) 
 
 dx v 
 
 54. Differentiation of cos x. 
 
 Let y = cos # ; 
 
 therefore, y -\- Ay = cos (# + A) ; 
 
 hence, A# = cos (x + h) — cos a; 
 
 = — 2 sin \x + ^ sm -, 
 
 « . /A + B\ . A-"B 
 
 because cos A — cos B = — 2 sin ^ — ^ — j sm — ^ 
 
 7 \ SH1 2 
 
 Therefore, g = -sin (x + J-g- 
 
70 COMPARISON OF Till-: TWO USTH0D8, 
 
 II. me. in I he limit, 
 
 -^ = — sin x. (See Art. 25.) 
 
 Of course this differentiation may be obtainod direct lv 
 from Art. 53, in the same manner as was done in the 2d 
 method of Art 25. 
 
 Since tan x, cot x, sec a, and cosec x are all fractional 
 forms, we may find the derivative of each of these functions 
 by Arts. 18 or 49, from those of sin x and cos a;, as was done 
 in Arts. 26, 27, 28, and 29 ; also, the derivatives of vers x 
 and covers x, as well as those of the circular functions, may 
 be found as in Arts. 30, 31, 33 to 40. 
 
 From the brief discussion that we have given, the student 
 will be able to compare the method of limits with the method 
 of infinitesimals ; he will see that the results obtained by 
 the two methods are identically the same. In discussing by 
 the former method, we restricted ourselves to the use of 
 limiting ratios, which are the proper auxiliaries in this 
 method. It will be observed that, in the former method, 
 very small quantities of higher orders are retained iill the 
 end of the calculation, and then neglected in passing to the 
 limit; while in the infinitesimal method such quantities 
 are neglected from the start, from the knowledge that they 
 necessarily disappear in the limit, and therefore cannot 
 affect the final result. As a logical basis of the Calculus, 
 the method of limits may have some advantages. In other 
 respects, the superiority is immeasurably on the side of the 
 method of infinitesimals. 
 
CHAPTER IV 
 
 SUCCESSIVE DIFFERENTIALS AND DERIVATIVES. 
 
 55. Successive Differentials. — The differential ob- 
 tained immediately from the function is the first differential. 
 Tne differential of the first differential is the second differ- 
 ential, represented by d*y, d 2 u, etc., and read, "second 
 differential of y," etc. The differential of the second dif- 
 ferential is the third differential, represented by d s y, d?u, 
 etc., and read, " third differential of y" etc. In like man- 
 ner, we have the fourth, fifth, etc., differentials. Differen- 
 tials thus obtained are called successive differentials. 
 
 Thus, let AB be a right line 
 whose equation is y — ax + b; 
 .-. dy = adx. Now regard dx as 
 constant, i. c., let x be equicres- 
 cent;* and let MM', M'M", and 
 M"M'" represent the successive 
 equal increments of x, or the data, 
 and R'P', R"P", R'"F" the corre- 
 sponding increments of y, or the 
 
 drj's. We see from the figure that R'P' == R'P'' = R'"P'" ; 
 therefore the di/s are all equal, and hence the difference 
 between any two consecutive %'s being 0, the differential 
 of dy, i. e., d 2 y = 0. Also, from the equation dy = adx we 
 have d?y = 0, since a and dx are both constants. 
 
 Take the case of the parabola y 2 = 2px (Fig. 7), from 
 pdx 
 
 M M M ' NT 
 
 Fig. 6. 
 
 which we get dy 
 
 Regarding dx as a constant, we 
 
 * When the variable increases by equal increments, i. e., when the differential is 
 constant, the variable is called an equicrescent variable. 
 

 KXAMl'LES. 
 
 have MM', M'M", M"M'" us the successive equal increments 
 of x y or the dx'a ; while we see from 
 Fig. 7 that R'P, R"P", R"P", or 
 the dy\ are no longer equal, but 
 diminish as we move towards the 
 right, and hence the difference be- 
 tween any two consecutive dtfs is a 
 negative quantity (remembering that 
 the difference is always found by 
 taking the first value from the second. 
 See Art. 12). Also, from the equa- 
 tion dy = - dx we see that dy varies inversely as y. 
 
 The student must be careful not to confound dhj wit h 
 dy 2 or d(y 2 ): the first is "second differential of ?y;" the 
 second is "the square of dy;" the third is the differential 
 of y 2 , which equals 2ydy. 
 
 EXAMPLES. 
 
 1. Find the successive differentials of y = x*. 
 Differentiating, we have dy = 5x i dx. Differentiating 
 
 this, remembering that d of dy is d?y and that dx is con- 
 stant, we have d i y = 20x 3 dx 2 . In the same way, differen- 
 tiating again, we have d 3 y— 6Qz~ dx 3 . Again, d 4 y = 120 r dx 4 . 
 Once more, d 5 y = 120dxP. If we differentiate again, we 
 have d G y = 0, since dx is constant. 
 
 2. Find the successive differentials of y = kx 3 — 3x 2 + 2x. 
 
 ( dy = (12x 2 — Gx -f 2) dx ; 
 Ans. < d*y = (24x — 6) dx 2 ; 
 ( d 3 y = 2UX 3 . 
 
 3. Find the first six successive differentials of y = sin x. 
 
 ( dy = cos x dx ; d?y = — sin x dx 2 ; 
 Ans. < (Py = — cos x d;i? ; d*y = sin x da 4 ; 
 
 ( d 5 y 
 
 cos x dx 5 ; tfy = — sin x dx G . 
 
EXAMPLES. 
 
 73 
 
 4. Find the first six successive differentials of y = cos x. 
 d\j = — sin x dx ; d 2 y = — cos x dx 2 ; 
 
 Ans. -J #?/ 
 dhj 
 
 sin # rfz 3 ; <-/ 4 // 
 sin a; rfz 5 ; d & y 
 
 cos a: c/.^ 4 ; 
 cos x dx*. 
 
 5. Find the fourth differential of y = «*« 
 
 i« t <i. tf 4 ?/ = w (w — 1) (w — 2) (n — 3) x n -*dx*. 
 
 6. Find the first three successive differentials of y = a x . 
 
 ( dy — a x log a dx ; 
 
 Ans. i d 2 y = a x log 2 « dx 2 ; 
 
 ( d s y = a* log 3 a ^r 8 . 
 
 7. Find the first four successive differentials of y = logx. 
 
 Ans. 
 
 rf y =--=-5 
 
 d»y 
 
 z 3 ' 
 
 <ty = - 
 
 to 4 
 
 X* ' 
 
 8. Find the first four successive differentials of y = 2« a/#« 
 
 ^4ws. 
 
 <fy = 
 
 atfo 
 
 flffo 8 
 
 y^s 
 
 » y — 
 
 ***' 
 
 3«6?o; 3 
 
 ** = 
 
 8z* 
 
 9. Find the first four successive differentials of 
 y = log (1 + x) in the common system. 
 
 mdx - wrfa; 2 
 
 ^fftS. 
 
 dy 
 y d*y- 
 
 1 + .?' 
 %mdx* 
 
 d 2 y = - 
 dhj = - 
 
 6m dx* 
 
 10. Find the fourth differential of y = e x . 
 
 Ans. d 4 y = ^dx 4 . 
 
H SSiV /■: I>ER1 VA Tl VES. 
 
 56. Successive Derivatives. — A first derivative* is 
 the ratio of the differential of a function to the different in I 
 of its variable. For example, let 
 
 y = z* 
 
 represent a function of x. Differentiating and dividing by 
 dx f we get 
 
 | = 6 *, (!) 
 
 The fraction ~ is called the first derivative of y with 
 
 respect to x, and represents the ratio of the differential of 
 the function to the differential of the variable, the value of 
 which is represented by the second member of the equation. 
 
 Clearing (1) of fractions, we have 
 
 dy = Qx^dx ; 
 
 du 
 hence, — or Qx 5 is also called the first differential coefficient 
 
 of y with respect to z, because it is the coefficient of dx. 
 
 A second derivative is the ratio of the second differential 
 of a function to the square of the differential of the variable. 
 Thus, differentiating (1) and dividing by dx, we get (since 
 dx is constant, Art. 55), 
 
 g = 30^, (2) 
 
 either member of which is called the second derivative of y 
 with respect to x. 
 
 A third derivative is the ratio of the third differential of 
 a function to the cube of the differential of the variable. 
 Tli us, differentiating (2) and dividing by dx, we get 
 
 * See Arts. 44 and 45. 
 
DIFFERENTIAL COEFFICIENTS. 75 
 
 g = 180* ' (3) 
 
 either member of which is called the third derivative of y 
 with respect to x. 
 
 In the same way, either member of 
 
 g = 360^ (4) 
 
 is called the fourth derivative of y with respect to x, and 
 so on. 
 
 Also, ~- , ~, -r~, ~, etc., are called respectively//^ 
 ' dx dx?- dx 3 dx 4 r J 
 
 first, second, third, fourth, etc., differential coefficients of 
 
 y with respect to x, because they are the coefficients of dx, 
 
 dx 2 , dx 5 , dx 4 , etc., if (1), (2), (3), (4), and so on, be cleared 
 
 of fractions. 
 
 In general, if y =f(x), we have 
 
 i - dJ *r = f{x) ( Art - 45 ); •'• d y = /to* 
 % = *£& =/-(,); xV-V-w** 
 
 3 = dJJ d!r - f" W ; ••• * = r {x) M 
 
 etc. == etc. ="etc. .*. etc. = etc. 
 
 % = ^^ - / ( "> to ; ••• ^ = / ( "> to ** 
 
 That is, the first, second, third, fourth, etc., derivatives 
 are also represented by /' (x), f" (ar), /'" (x), f" (r), etc. 
 
76 GEOMETRIC REPRESENTATION, 
 
 Strictly speaking, :' or J" (.>) are symbols representing 
 
 the ratio of an infinitesimal iucrement of the function to the 
 corresponding infinitesimal increment of the variable, while 
 the second member expresses its value. For example, in 
 the equation y = ax 4 , we obtain 
 
 §*/'(*) = *«* 
 
 dy 
 
 ~ or /' (x) is an arbitrary symbol, representing the value 
 
 of the ratio of the infinitesimal increment of the function 
 (ax 4 ) to the corresponding infinitesimal increment of the 
 variable (x), while kax* is the value itself. It is usual, 
 however, to call either the derivative. 
 
 56<7. Geometric Representation of the First De- 
 rivative. — Let AB be any plane curve 
 whose equation is y=f(x). Let P 
 and P' be consecutive points, and PM 
 and P'M' consecutive ordinates. The 
 part of the curve PP', called an ele- 
 ment* of the curve, does not differ 
 from a right line. The line PP' pro- 
 longed is tangent to the curve at the 
 point P (Anal. Geom., Art. 42). Draw PR parallel to XX', 
 and we have 
 
 MM' = PR = dx, and RP' = dy. 
 
 Denote the angle CTX by a, and since CTX = P'PR, we 
 have 
 
 tan a = -f- 
 ax 
 
 And since the tangent has the same direction as the curve 
 
 * In this work, the word "element" will be used for brevity to denote an "in- 
 finitesimal element." 
 
EXAMPLES. 77 
 
 at the tangent point P, « will also denote the inclination of 
 the curve to the axis of x. 
 
 Hence, the first derivative of the ordinate of a curve, 
 at any point, is represented by the trigonometric tan- 
 gent of the angle which the curve at that point, or its 
 tangent, makes with the axis of oc* 
 
 In expressing the above differentials and derivatives, we 
 have assumed the independent variable x to be equicrescent 
 (Art. 55), which we are always at liberty to do. This 
 hypothesis greatly simplifies the expressions for the second 
 and higher derivatives and differentials of functions of x, 
 inasmuch as it is equivalent to making all differentials of x 
 above the first vanish. Were we to find the second deriva- 
 tive of y with respect to x, regarding dx as variable, we 
 would have 
 
 cPy _ d (dy\ _ dx d 2 y — dydt x 
 dx 2 ~ dx \dx) ~ dx* ' 
 
 which is much less simple than the expression -j», obtained 
 by supposing dx to be constant. 
 
 EXAMPLES. 
 
 1. Given y = x n , to find the first four successive deriv- 
 atives. 
 
 d £=nx»->; 
 
 2 2 = n(n-l)x»-*; 
 
 dx* 
 
 cPy 
 dx* 
 
 = n(n — 1) (n — 2) x n ~ 3 ; 
 
 = n{n-l)(n-2)(n-3)x»- 
 
78 EXAMPLES. 
 
 If n be a positive integer, we have 
 
 g = «(»-l)(»-2).... 3-2.1. 
 
 and all the higher derivatives vanish. 
 
 If n be a negative integer or a fraction, none of the suc- 
 cessive derivatives can vanish. 
 
 2. Given y = X s log x ; find -~ 
 
 CtX' 
 
 j£ = Qx log x + 3x + 2x = 6x log x + bx ; 
 
 J = 61ogz + 6+5. g = -. 
 
 It can be easily seen that in this case all the terms in the 
 successive derivatives which do not contain log x will dis- 
 appear in the final result ; thus, the third derivative of x 2 is 
 zero, and therefore that term might have been neglected ; 
 and the same is true of bx, its second derivative being zero. 
 
 1+a; , , , dfy 240 
 
 4. y r= e aa? ; prove that -??-= flV*. 
 
 5. ?/ = tan x ; find the first four successive derivatives. 
 
 -t^ = 2 sec 8 a; tana;; 
 
 -t4 = 6 sec 4 x — 4 sec 2 a; ; 
 
 d 4 «/ 
 
 -r^ = 8 tan x sec 2 a; (3 sec 2 x — 1). 
 
EXAMPLES. 79 
 
 cl z v 
 6. y = log sin x ; prove that -7-^ = 2 cot x cosec 2 a;. 
 
 7. # 2 = 2pa> j find -^ 
 
 dy_ _P. 
 dx y y 
 
 'J ■**- 2? — 
 
 d*# _ d (p\ __ dx _ c y 
 
 dx* 
 
 d (p\ * dx v / > <%V P\ 
 
 -=- K) = 5- = f (since -/ =^l 
 
 dx \yf y* y i \ dx yl 
 
 2 ^ Q 2 P 
 
 <Py _ _ d^ /_ f\ _ U_~dx _ l_y _ 3£ 
 dx 3 ~ dx \ yv y 6 y* y h 
 
 8. y = x* ; prove that 
 d*y 
 
 dx* 
 
 x* (1 + log #) 2 + x*~K 
 
 9. ay + #a? =s a 2 #> ; prove that ^ = — J^. 
 9 } ' r tfV a 2 ?/ 3 
 
 11. «/ 2 = sec 2# ; prove that 2/ + -t4 = 3# 5 . 
 
 12. y = e~ x cos x ; prove that 4«/ + -— = 0. 
 
 ^ 5 w 48 
 13. y = x* log (x 2 ) ; prove that ~g 5 = — 
 
80 / XAMPLES. 
 
 14. y = ic 8 ; prove that 
 
 iPy = 6 (dxf + lSxdx&x + 33%, 
 when a is not equicrescent. 
 
 15. y = f(x); prove that 
 
 d*y = /'" (x) (dxf + 3/" (a?) dx&x + /' (a?) fc 
 when x is not equicrescent. 
 
 16. y = e* ; prove that 
 
 <Fy = e* (dx)* + 3(? dx d 2 x + e*^. 
 
 
 T 
 

 % 
 
 tfjfi 
 
 '< / 
 
 CHAPTER V. 
 
 DEVELOPMENT OF FUNCTIONS. 
 
 57. A function is said to be developed, when it is 
 transformed into an equivalent series of terms following 
 some general law. 
 
 For example, 
 
 y = (a + x)\ 
 
 when developed by the binomial theorem, becomes 
 
 y = a 4 + 4A- + 6a 2 a? + to 3 + x*, 
 which is a, finite series. Also, 
 
 1 + x 
 
 may be developed by division into the infinite series, 
 
 y = 1 + 2x + 2x* + 2x* + etc., 
 
 in which the terms are arranged according to the ascending 
 powers of x, each coefficient after the first term being 2. 
 
 One of the most useful applications of the theory of suc- 
 cessive derivatives is the means it gives us of developing 
 functions into series by methods which we now proceed to 
 explain. 
 
 MACLAURIN'S THEOREM. 
 
 58. Maclaurin's Theorem, is a theorem for developing 
 a function of a single variable into a series arranged 
 according to the ascending powers of that variable, with 
 constant coefficients. 
 
83 MACLAVIII\S THEOREM. 
 
 Let *=/(») 
 
 be the function to be developed ; and assume the develop- 
 ment of the form 
 
 y = f(x) = A + Bx+ Cx> + Dx* + Ex* + etc., (1) 
 
 in which A, B, C, D, E, etc., are independent of x, and 
 depend upon the constants which enter into the given func- 
 tion, and upon the form of the function. It is now re- 
 quired to find such values for the constants A, B, C, etc., 
 as will cause the assumed development to be true for all 
 values of x. 
 
 Differentiating (1) and finding the successive derivatives, 
 We have, 
 
 ^ = B + %Ox + 3Zte» + 4:Ex* + etc., (2) 
 
 -^ = 20 + 2 • Wx + 3 • ±Ex* + etc., (3) 
 
 H| = 2 • 3D + 2 • 3 • ±Ex + etc., (4) 
 
 g = 2.3.4^+etc, (5) 
 
 Now, as A y B, 0, etc., are independent of x, if we can 
 find what they should be for any one value of x, we shall 
 have their values for all values of x. Hence, making x = 
 in (1), (2), (3), etc., and representing what y becomes on 
 
 this hypothesis by (y) ; what -^ becomes by \-^\ ; what 
 -t4 becomes by \-r-J ; and so on ; we have, 
 (y) = A; .: A = (y). 
 
 -*•> - * = ©• 
 
MACLAUBIN'S THEOREM. 83 
 
 »-•'•»« *»-<3Jnfcr 
 
 Substituting these values in (1), we have, 
 
 + (S) 17^374 + ^ W 
 
 which is the theorem required. 
 
 Hence, by Maclaurin's Theorem, we may develop a func- 
 tion of a single variable, asy = f(%)> m ^° a series of terms, 
 the first of which is the value of the function when x = ; 
 the second is the value of the first derivative of the function 
 when x = into x ; the third is the value of the second 
 
 derivative when x = into ^-, etc.; the (n + l) th term is 
 
 We may also use the following notation for the function 
 and its successive derivatives : f(x), f (x), f" (x), f" (x), 
 f iv (x), etc., as given in Art. 56, and write the above 
 theorem, 
 
 9 = /(*) = /(0) + /' (0) I + /" (0) £ 3 + /'" (0) j-^-3 
 
 in which /(0), /'(0), /"(0), /"' (0), etc., represent the 
 values which f(x) and its successive derivatives assume 
 
SI MACLAl'h'IA'S THEOREM. 
 
 when x = 0. We shall use this notation instead of 
 -g, 5| , etc., for the sake of brevity. 
 
 This theorem, which is usually called Maclaurin's Theorem, was 
 previously given by Stirling in 1717 ; but appearing first in a work on 
 Fluxions by Maclaurin in 1742, it has usually been attributed to him, 
 and has gone by his name. Maclaurin, however, laid no claim to it, 
 for after proving it in his book, he adds, " this theorem was given by 
 Dr. Taylor." See Maclaurin's Fluxions, Vol. 2, Art. 751. 
 
 To Develop y = (a + x)* 
 
 Here 
 
 f(x) = (a + x)*; 
 
 hence, 
 
 /(0) = a\ ■ 
 
 
 f(x) = 6(a + xY; 
 
 
 /'(0) = 6a». 
 
 
 f"(x) = 5.6(a + xY; 
 
 
 /"(0) =5.6a*. 
 
 
 f"'(x) = 4-5. 6(0 + z) 3 ; 
 
 
 /'" (0) = 4 • 5 . 6a 8 . 
 
 
 f w (x) = 3.4.5.6(« + z)2; 
 
 
 f u (0) = 3 • 4 • 5 • 6a 2 . 
 
 
 f(x) = 2.3.4.5.6(0 + 3); 
 
 
 f (0) = 2 • 3 • 4 • 5 • 6a ; 
 
 
 f*(x) = 1.2.3.4.5.6; 
 
 
 /*(o) = 1.2.3.4.5.6; 
 
 Substituting 
 
 in (7), we have, 
 
 y = (a + xf = a? + 6a 5 x + 5 • 6a 4 j^ + 4 • 5 • QaZ j-2~-3 
 
 a*x* 2.3.4.5.6^ 1.2.3.4.5.6^ 
 
 + 3,4 * 5 ' b 1.2.3.4 + 1.2- 3-4. 5 + 1.2.3.4.5.6 
 
 = a 6 + Ga 5 x + IZaW + 20« 8 z 3 + UaW + 6a^ +x*, 
 
THE BINOMIAL THEOREM, 85 
 
 which is the same result we would obtain by the binomial 
 theorem. 
 
 THE BINOMIAL THEOREM. 
 
 59. To Develop y = (a + x)\ 
 
 Here f(x) — (a + x) n ; 
 hence, /(0) = a\ 
 
 f'{x) = n(a + a) 71 " 1 ; 
 
 f'(0) = na-K 
 
 f"(x) = n (n - 1) {a + as)-*. 
 
 /" (0) = ft (» — 1) a"" 9 . 
 
 /"' (») = » ( n — 1) (w - 2) (a + x) n ~*; 
 
 f" (0) = n (n - 1) (n - 2) a"- 3 . 
 f™{x) = n (n — 1) (n — 2) (w — 3) (« + ^) n_4 5 
 / iv (0) = % (n — 1) (w — 2) (n — 3) a"" 4 , etc. 
 
 Substituting in (7), Art. 58, we have, 
 
 n(n — l) cT*& 
 
 y = (a + z) n = a n + na^x + -4 — j_jL_ — 
 
 rc (ffl — 1) (n — 2) fl"-^ 3 
 + , _ ^ g 
 
 n (w - 1) (» - 2) (w - 3) d rht 
 + 1.2-3.4 + GtC - 
 
 ' Thus the truth of the binomial theorem is established, 
 applicable to all values of the exponent, whether positive 
 or negative, integral or fractional, real or imaginary. 
 
 60. 1. To Develop y = sin x. 
 
 Here f(x) = sin x ; hence, /(0) s= 0. 
 
 /' (a) = cos x ; " /' (0) = 1. 
 
 « /" (a;) = - sin a; ; " /"(0) = 0. . 
 
86 THE LOGARITHM TV SERIES. 
 
 Sere f'"{x) = —cos a:; hence, /'"(0) = — 1. 
 " / iv (z) = sin :c ; " f (0) = 0. 
 
 M /• (a;) = cos x; " f (0) s= 1. 
 
 Etc., etc. Etc., etc. 
 
 Hence, y = sin x = x 
 
 ./•• 
 
 1-2-3 ' 1.2.3.4.5 
 
 :r 7 - 
 
 + etc. 
 
 1.23.4.5.6.7 
 2. To Develop 1/ = cos x. 
 
 z 6 a« 
 
 1.2.3.4.5.6 ' 1.2.3.4.5.6.7.8 
 
 — etc. 
 
 The student will observe that "by taking the first derivative of the 
 series in (1), we obtain the series in (2), which is clearly as it should 
 be, since the first derivative of sin x is equal to cos x. 
 
 Since sin (— x) = — sin x, from Trigonometry we might have 
 inferred at once that the development of sin x in terms of # could con- 
 tain only odd powers of x. Similarly, as cos (— x) — cos x, the 
 development of cos x can contain only even powers. 
 
 By means of the two formula? in this Article we may 
 compute the natural sine and cosine of any arc. For exam- 
 ple, to compute the natural sine of 20°, we have x = arc of 
 
 20° = ^ = .3490652, which substituted in the formulae, 
 
 gives sin 20° = .342020 and cos 20° = .939693. 
 
 THE LOGARITHMIC SERIES. 
 
 61. To Develop y = log (1 + x) in the system irr 
 which the modulus is m. 
 
 Here f(x) = log (1 + x); hence, /(0) = 0. 
 
THE LOGARITHMIC SERIES. 87 
 
 >rw = -(1^; 
 
 hence, /" (0) = — m. 
 
 1 . 2m . 
 
 / ^-(1+^)3' 
 
 « f" (0) = 1 • 2m. 
 
 Etc. 
 
 " /*(0) = ~ L .2.3m. 
 Etc. 
 
 Substituting in (7), Art. 58, we have, 
 y = log (l-h#) = m (x— %x 2 + ^x?— Ja^-f^z 5 — etc.), (1) 
 which is called the logarithmic series. 
 
 Since in the Naperian system m = 1 (see Art. 20, Cor.), 
 we have, 
 
 Q% rfi3 /p4 q§ 
 
 if = log (1 + ») = « - g- + -j - j + 5 - etc. (2) 
 which is called the Naperian logarithmic series. 
 
 This formula might be used to compute Naperian loga- 
 rithms, of very small fractions ; but in other cases it is 
 useless, as the series in the second number is divergent for 
 values of x > 1. We therefore proceed to find a formula 
 in which the series is convergent for all values of x ; i. e., in 
 which the terms will grow smaller as we extend the series. 
 
 Substituting — x f or x in (2), we have, 
 
 loga-^-z-J-f-'J-f-etc. (3) 
 Subtracting (3) from (2), we have, 
 
 9,r3 Sr 5 2# 7 
 
 log (1 + x) - log (1 - *) = %x + ^ + ?j +"=■ ■+ etc. 
 
 or , og (^) = a (, + f + f + |' + otc). (4) 
 
88 CALCULATION OF LOGARITHMS. 
 
 Let r - 1 • ■ ! + *-' + 1 . 
 Substituting in (4), we have, 
 
 , *+l _ gf 1 1_ 1_ 
 
 10g 2 ~ L2z + 1 + 3(2z + 1) 3+ 5(22 + l) 5 
 
 + 7{2FT1? 4 "H 5 
 
 or log( 2 + l) = log, + 2[^ f+ 3 T ^- I? 
 
 + 5(2^1? + n^Tl? + H- (5) 
 This series converges for all positive values of z, and more 
 rapidly as z increases. By means of it the Naperian loga- 
 rithm of any number may be computed when the logarithm 
 of the preceding number is known. It is only necessary to 
 compute the logarithms of prime numbers from the series, 
 since the logarithm of any other number may be obtained 
 by adding the logarithms of its factors. The logarithm of 1 
 is 0. Making z = 1, 2, 4, C, etc., successively in (5), we 
 obtain the following 
 
 Naperian or Hyperbolic Logarithms. 
 
 3 
 
 log 2 = log 1 + 2 (I + 3I3 + J^ + J3, + ^ + jjI 
 
 1 1 1 \ 
 + 13~3 13 + 15. 3» + 17-3" + )'' 
 or, since log 1 = 0, 
 
 .33333333 
 .01234568 
 .00082305 
 log 2 = 2 { .00006532 ) = 2 (0.34657359) = 0.69 1& 
 .00000565 
 .00000051 
 .00000005 
 
CALCULATION OF LOGARITHMS. 8& 
 
 log 3 = log 2 + 2 g + A- 3 + - 1 - + J^ 7 + A- 9 + etc.) 
 
 = 1.09861228. 
 log 4 = 2 log 2 = 1.3862943G. 
 
 log 5 = log4 + 2g + ^3 + ^ 5 + ^ + A- 9 + etc.) 
 
 = 1.60943790. 
 log 6 = log 3 + log 2 = 1.79175946. 
 
 log 7 = log6 + 2(l + ^- 3 + ^- + -^- 7 - l -etc.) 
 
 = 1.94590996. 
 log 8 = 3 log 2 = 2.07944154. 
 
 log 9 = 2 log 3' = 2.19722456. 
 
 log 10 = log 5 + log 2 = 2.30258509. 
 
 In this manner, the Naperian logarithms of all numbers 
 may be computed. Where the numbers are large, their 
 logarithms are computed more easily than in the case of 
 small numbers. Thus, in computing the logarithm of 101, 
 the first term of the series gives the result true to seven 
 places of decimals. 
 
 Cor. 1. — From (1) we see that, the logarithms of the 
 same number in different systems are to each other 
 as the moduli of those systems; and also, that the 
 logarithm of a number in any system is equal to the 
 JVaperian logarithm of the same number into the 
 modulus of the given system. 
 
 Cor. 2. — Dividing (1) by (2), we have 
 
 Common log (1 -f x) _ m ^ 
 
 Naperian log (1 + x) 
 
 Hence, the modulus of the common system is equal 
 to the common logarithm of any number divided by 
 the Naperian logarithm of the same number. 
 
90 EX l'<> SKSTIAL SERIES. 
 
 Substituting in (G) the Naperian logarithm of 10 com- 
 pi it id above, and the common logarithm of 10, which is 1, 
 we have 
 
 m = o qaoLkao = .4342944819032518276511289 . . . 
 
 <c.«3U*cOoOuJ 
 
 which is the modulus of the common system. (See Serrefc's 
 Calcul Differentiel et Integral, p. 169.) 
 
 Hence, the common logarithm of any number is 
 equal to the Naperian logarithm of the same number 
 into the modulus of the common system, .43429448- 
 
 Cor. 3. — Representing the Naperian base by e (Art. 21, 
 Cor. 2), we have, from Cor. 1 of the present Article, 
 
 com. log e : Nap. log e (= 1) :: .43429448 : 1 ; 
 
 therefore, com. log e = .43429448 ; 
 
 and hence, from the table of common logarithms, we have 
 
 e = 2.718281 +. 
 
 EXPONENTIAL SERIES. 
 62. To Develop y = a x t 
 
 Here f(x) = a* ; hence, /(0) == 1. 
 
 /' (x) = a* log a ; " /' (0) == log a. 
 
 « /" (x) = a* (log a)*; " /" (0) = (log af. 
 « f" (x) = a* (log af ; " /"' (0) = (log a)\ 
 and the development is 
 
 9 S & == 1 + logfl ? + log^fl — + log 3 fl j-^-g 
 
 + log 4 a r7 ^ 7 3 T 4 + etc. (1) 
 
EXPONENTIAL SERIES. 91 
 
 Cor. — If a = e, the l^aperian base, the development 
 becomes 
 
 -. , * , & & & 
 
 y = e* = l + -+-- + — - + 
 
 1 ' 1-2 ' 1-2-3 ' 1.2.3-4 
 x n 
 + ■■•■ i.%.Z...n + etc " (2 > 
 
 Putting X = 1, we obtain the following series, which en- 
 ables us to compute the value of the quantity e to any 
 required degree of accuracy : 
 
 ^ = e = 2 + ^ + ^3 + ^ + 2^5 
 
 + ----2^.^ + etC - 
 
 = 2.718281828 + . 
 
 63. To Develop y = tan -1 x. 
 
 In the applications of Maclaurin's Theorem, the labor in 
 finding the successive derivatives is often very great. This 
 labor may sometimes be avoided by developing the first 
 derivative by some of the algebraic processes, as follows : 
 
 Here f(x) = tan -1 #; 
 hence, /(0) == 0. 
 
 = (by division) 1 — x 2 + x* — a 6 + z 8 ; 
 /'(0) = 1. 
 
 /" (x) == — 2x + 4Z 3 — 6X 5 + 8z 7 — 10a; 9 -fete. ; 
 /"(0) = 0. 
 
 f'"(x) = —2 + 3.4^-5.6^+7.8^ — etc.; 
 f'"(0) = -2. 
 f(z) = 2.3.4a; — 4.5.6a^ + etc.; 
 
 /"(0)=:0. 
 
92 FAILURE <>F KAVLAVKIN'B THEORBO* 
 
 /'(re) = 2-3.4 — 3.4.5-G:r 2 + etc.; 
 
 / v (0) = 2.3.4. 
 
 f^(x) = —2.3.4.5.Ga: + etc.; 
 / vi (0) = 0. 
 f*"(x) = -2.3.4.5.6 + etc.; 
 
 /""(o) = -2.3.4.5.6. 
 
 Substituting in (7) of Art. 58, we get 
 
 xP x? x^ 
 y = tan" 1 x = x — 3+5 ~f + etc - 
 
 64. It sometimes happens in the application of Maclau- 
 rin's Theorem that the function or some of its derivatives 
 become infinite when x = 0. Such functions cannot be 
 developed by Maclaurin's Theorem, since, in such cases, 
 some of the terms of the series would be infinite, while the 
 function itself would be finite. 
 
 For example, take the function y = log x. Here we 
 have 
 
 f(x) = logx; hence, /(0) = — 00. 
 
 /<•).= J5 " /'(°) = «>. 
 
 etc. etc. 
 
 Substituting in Maclaurin's Theorem, we have 
 
 x x^ 
 
 y = log re = — 00 + 00- — 00^- -f- etc. 
 
 J. £ 
 
 Here we have the absurd result that log x = 00 for all 
 values of x. Hence, y = log x cannot be developed by 
 Maclaurin's Theorem. 
 
 Similarly, y = cot x gives, when substituted in Maclau- 
 rin's Theorem, 
 
Taylor's theorem. 93 
 
 x 
 y = cot x = go — oo - + etc. ; 
 
 that is, cot x = co for all values of x, which is an absurd 
 result. Hence, cot x cannot be developed by Maclaurin's 
 Theorem. 
 
 Also, y = x* becomes, by Maclaurin's Theorem, 
 
 y = x 2 — + OO2 + etc. ; 
 
 that is, x* = oo for all values of x, which is an absurd 
 result. 
 
 Whether the failure of Maclaurin's Theorem to develop correctly is 
 due to the fact that the particular function is incapable of any devel- 
 opment, or whether it is simply because it will not develop in the 
 particular form assumed in this formula, the limits of this book will 
 not allow us to enquire. 
 
 TAYLOR'S THEOREM. 
 
 65. Taylor's Theorem is a theorem for developing a 
 function of the sum of two variables into a series arranged 
 according to the ascending powers of one of the variables, 
 with coefficients that are functions of the other variable and 
 of the constants. 
 
 Lemma. — We have first to prove the following lemma: 
 If we have a function of the sum of two variables x and y, 
 the derivative will be the same, whether we suppose x to 
 vary and y to remain constant, or y to vary and x to remain 
 constant. For example, let 
 
 u = (x + y)\ (1) 
 
 Differentiating (1), supposing x to vary and y to remain 
 constant, we have 
 
 g = »(z + 2 ,)»-' (2) 
 
'I TATLOBfS TBEOBEM. 
 
 Differentiating (1), supposing y to vary und x to remain 
 constant, we have 
 
 f y = n{ X + y y->; (3) 
 
 from which we see that the derivative is the same in both 
 (2) and (3). 
 
 In general, suppose we have any function of x + y, as 
 
 u=f(x + y). (4) 
 
 Let z = x + y ; (5) 
 
 A u=f(z). (6) 
 
 Differentiating (5), supposing x variable and y constant, 
 and also supposing y variable and x constant, we get 
 
 dz 
 
 and 
 
 dz 
 dy ~ 
 
 1. 
 
 Differentiating (6), we h 
 
 ave 
 
 
 
 du 
 dz 
 
 _ df{z) 
 dz 
 
 = /'(«)• 
 
 (See 
 
 Art 
 
 
 .\ du : 
 
 =/'(») 
 
 dz. 
 
 
 du _ 
 dx 
 
 */ \ dz 
 
 =/'(*) 
 
 (since 
 
 
 And similarly, 
 
 
 
 
 
 du _ 
 dy "" 
 
 du 
 dx 
 
 du 
 
 (since 
 
 
 -> 
 
 That is, the derivative of u with respect to x, y being 
 constant, is equal to the derivative of u with respect 
 to y, x being constant. 
 
METHOD OF TAYLOR' S THEOREM. 95 
 
 66. To prove Taylor's Theorem. 
 
 Let u' =/{% + y) be the function to be developed, and 
 assume the development of the form 
 
 u' = f{% + y) 
 
 = A+By+Cif + Df + % 4 + etc., (1) 
 
 m which A, B, C, etc., are independent of y, but are func- 
 tions of x and of the constants. It is now required to find 
 such values for A, B, C, etc., as will make the assumed 
 development true for all values of x and y. 
 
 Finding the derivative of u\ regarding x as constant and 
 y variable, we have 
 
 ^ = B + 2Cy + Wf + 4%3 + e t c . ( 2 ) 
 
 Again, finding the derivative of u\ regarding x as varia- 
 ble and y constant, we have 
 
 dm' dA (IB dC 2 dD , , Q . 
 
 ■& = te + i x y+ ite y+ te^ + ^- (3) 
 
 dm ' dm 
 By Art. 65, we have -=— = -5— ; therefore, 
 J ' dy dx ' 
 
 + etc. (4) 
 
 Since (1) is true for every value of y, it is true when 
 y — 0. Making y = in (1), and representing what u' 
 becomes on this hypothesis by u, we have 
 
 u=f(x) = A. (5) 
 
 Sines (4) is true for every value of y, it follows from the 
 principle of indeterminate coefficients (Algebra) that the 
 coefficients of the like powers of y in the two members 
 must be equal. Therefore, 
 
TA YLOR'S THEOREM. 
 
 * = %> ■'■ »■-£.;*-«! 
 
 *r- dB . >_ * A . 
 
 4/7=^ • J»- 1 *? 
 
 cte ' •* ~ 1.2.3-4* rf.T*' 
 
 Substituting these values of J, Z?, C, D, etc., in (1), we 
 have 
 
 , - , x , duy d?u y 2 ' d s u y z 
 
 u = fix A- v) = u A -A - — I - — 
 
 J\* + y) "At ^i + ^i.^ + rfa* 1.2.3 
 
 + 1* T^ki + etc - ^ 
 
 Or, using the other notation (Art. 56), we have 
 u' =/<*+*) =/(*) + f(x)\ + f"(x) ij +f"(x)^ 
 
 +/" (^1^74 + etc., (7) 
 
 which is Taylor's TJteorem. It is so called from its discov- 
 erer, Dr. Brook Taylor, and was first published by him in 
 1715, in his Method of Increments. 
 
 Hence, by Taylor's Theorem, we may develop a function 
 of the sum of two variables, as u =f{x + y), into a series 
 of terms, the first of which is the value of the function 
 when y = ; the second is the value of the first derivative 
 of the function when y = 0, into y; the third is the value 
 
 y 2 
 
 of the second derivative when «/ = 0, into —-, etc. 
 
 9 1-2 
 
 The development of f{x — y) is obtained from (6) or (7), 
 by changing -f y into — y; thus, 
 
or 
 
 BINOMIAL THEOREM. 97 
 
 v du y iPu y* d s u y 8 
 
 tZ 4 ^ y 4 , 
 
 + ^1.2.3-4 ~ etC * 
 
 , /(* -y)= f(x) -/' (x) \ +/" (x) /- -/'" (a;) ^ 
 
 +/ iv ^rfir4- etc - 
 
 Coe. — If we make x = in (7), we have 
 
 »' = /<*) =/(0)+/'(0)f +/"(0)^*3+/'"(0) I ^g 
 
 which is Maclaurin's Theorem. See (7) of Art. 58. 
 
 THE BINOMIAL THEOREM. 
 
 67. To Develop it' = (x + y)". 
 
 Making y = 0, and taking the successive derivatives, we 
 
 have 
 
 f(x) = x n , 
 
 f (x) = nx n ~\ 
 
 f"(x) = n(n-l)x n ~ 2 , 
 
 /'" (x) = * (» - 1) (w - 2) x n ~ 3 , 
 
 / iv (a?) = w (» — 1) (» — 2) (» - 3) x*- 4 , 
 
 etc. etc. 
 
 Substituting these values in (7), Art. 66, we have 
 
 wa?'- 1 ^ , n(n — l)x n ~ 2 y* 
 u' = (x + y) n = af + -^ + -i j^ *• 
 
 rc(rc--l)(rc-2)a"-V 
 "*" L2-3 ^ ' 
 
 which is the Binomial Theorem (see Art. 59). 
 5 
 
98 APPLICATIONS OF TAYLOR* S THE OH KM. 
 
 68. To Develop n' = sin (x + y). 
 
 Here f(x) = sin x, /' (x) = cos x, 
 
 f" (.<:) = — sin x, f" (x) = — cos x, etc. 
 
 Hence, 
 u' = sin (x + y) 
 
 = BiDr ( 1 -£» + i^i- l.8.3^.5.6 + ° tC -) 
 
 + C ° S * (l - l£i + r^fcs - 1^-576-7 + Ct0 ) 
 = sin x cos y + cos a: sin y. (See Art. 60.) 
 
 THE LOGARITHMIC SERIES. 
 
 69. To Develop u' = log (a? + y). 
 
 Here /(*) == log x, /'" (*) = |, 
 
 /<(*) = i, /"(*) = -J, 
 
 /"(*>= -^. etc. 
 
 Hence, w' = log (# + y) 
 
 . x y ly* , ly 3 l?/ 4 
 
 Cor. — If 2=1, this series becomes 
 
 log(l + 2/) =f -| 2 + | 3 -| 4 + etc, 
 which is the same as Art. 61. 
 
 EXPONENTIAL SERIES. 
 
 70. To Develop u' = a x+ y. 
 
 Here f(x) = a x , /" (2) = a* log 3 a, 
 
 f (3) = a* log a, /'" (a) = a* log 3 a, etc; 
 
FAILURE OF TAYLOR'S THEOREM. 99 
 
 Hence, 
 
 u' = a x+ y 
 
 = a?(l + logfl.y + log^a^- + logBfl ^-^ + etc. I 
 Cor. — If $ = 0, this series becomes 
 av = 1 + log a-y + log 2 a~ + lo g 3 « f^3 + etc., 
 which is the same as Art. 62. 
 
 71. Though Taylor's Theorem in general gives the cor- 
 rect development of every function of the sum of two 
 variables, yet it sometimes happens that, for particular 
 values of one of the variables, the function or some of its 
 derivatives become infinite ; for these particular values, the 
 theorem fails to give a correct development. 
 
 For example, take the function u' = Va + x + y. 
 
 Here, f(x) = Va + x f 
 
 1 
 
 2V« + x 
 
 ■ r (*) = - ,, ] _ \., 
 
 4 (a -+- xp 
 f" (x) = 3 £ , etc. 
 
 Substituting in (7) of Art. 66, we have 
 v! = a/« + # + y 
 
 =s V« + a H ~ — s ~ j H — — t — etc. 
 
 2Va + x 8(a + x)? 16(a + x)* 
 
 Now when x has the particular value — a, this equation 
 becomes 
 
 u' = Vy = + oo — co + oc — etc. ; 
 
100 EXAMPLES. 
 
 thai is, when x= —a, Vy = a> • But y is independent of 
 .r. and may have any value whatever, irrespective of the 
 value of .r, and hence the conclusion that when x = — a, 
 VV = <», cannot be true. For every other value of x, 
 however, all the terms in the series will be finite, and the 
 development true. 
 
 Similarly, u' =a + ya — x + y gives, when substituted 
 in Taylor's Theorem, 
 
 u' = a + v a — x + y 
 
 = a + Va — x - ^ + etc., 
 
 2ya — x 
 
 which, when x = a, becomes 
 
 v! = a + Vy = a — oo + etc. ; 
 
 and hence the development fails for the particular value, 
 x = a. 
 
 It will be seen that when Taylor's Theorem fails to give 
 the true development of a function, the failure is only for 
 particular values of the variable, all other values of both 
 variables giving a true development; but when Maclaurin's 
 Theorem fails to develop a function for one value of the 
 variable, it fails for every other value. 
 
 Many other formulae, still more comprehensive than these, 
 have been derived, for the development of functions; but a 
 discussion of them would be out of place in this work. 
 
 EXAMPLES. 
 
 1. Develop y = Vl + x 2 . 
 
 Put x 2 — z, and develop ; then replace z by its value. 
 
 Ans. y = yl + # 2 
 
 x 2 & x 6 5x* 
 
 = 1+ a-8 + i6-m + etc - 
 
2. y = 
 
 EXAMPLES. 101 
 
 1 
 
 1—X 
 
 y = = l-{-£-*-# 2 + £ 3 4-£ 4 + etc. 
 
 9 1 — x 
 
 3. y = (a + x)~\ 
 
 y = (a + x)~* = ar z — 3a~*x + 6«~ 5 ^ 2 — 10«" 6 ^ 
 
 f etc. 
 4:. y = e 8inx . 
 
 . r '- , a? x* . X s x*. 
 
 + etc. 
 5. y = #0®. 
 
 y = a?e* == a; + x* + ^- + ^ 3 + etc. 
 
 6. y = V22 — 1. 
 
 ($ x^ \ 
 
 1 ~~ X ~2~2~ etC 7* 
 
 1 y = (a? 4- a*)i 
 
 y = (a? + a; 2 )* = aV 1 + fritf + JflT*^ — A«" ^ 
 
 + etc. 
 
 8. y = — - 1 
 
 . " r 1 X* 5X* 5.9^2 
 
 5. 9- 13a: 16 . 
 
 + 4.8.12.16a" ~ 6tC ' 
 
 9. y = (a 5 + (fix — x 5 )^. 
 
 Put a 4 a; — x 5 = z, as in Ex. 1. 
 
 _ x 4 a* 4-9 a 3 
 
 y _ a + g - g^ • ^ ■+ 53^2 ' 1 .2 .3 
 
 4.9-14 x* , . 
 
 --5^'r2^4 + etc - 
 
.02 EXAMPLES. 
 
 10. u = (x + y)i 
 
 ?* = jrf + Ja;-ty - 4ar ty2 + ^t^ - etc. 
 
 11. u = cos (a; + y). (See Art. 68.) 
 
 w = cos (z + y) = cos re cosy — sinz sin y. 
 
 12. ?/ = tan x. 
 
 X s 2X 5 
 y = tanx = x + ~+ — + etc. 
 
 13. y ss sec x. 
 y = sec x 
 
 14. # = log (1 + sin x). 
 
 y = log (1 + sin x) = a; - | + ^ — g + etc. 
 
 w = sec x = 1 H 1- etc. 
 
 9 ^ 2 ^ 24 ^ 720 ^ 
 
CHAPTER VI. 
 
 EVALUATION OF INDETERMINATE FORMS. 
 
 72. Indeterminate Forms. — When an algebraic ex- 
 pression is in the form of a fraction, each of whose terms is 
 variable, it sometimes happens that, for a particular value 
 of the independent variable, the expression becomes inde- 
 terminate ; thus, if a certain value a when substituted for 
 
 fix) 
 x makes both terms of the fraction 44-r vanish, then it 
 
 <p{x) 
 
 reduces to the form - , and its value is said to be indetermi- 
 nate. 
 
 Similarly, the fraction becomes indeterminate if its terms 
 both become infinite for a particular value of x; also the 
 forms oo x and go — oo , as well as certain others whose 
 logarithms assume the form go x 0, are indeterminate forms. 
 It is the object of this chapter to show how the true value 
 of such expressions is to be found. By its true value is 
 meant the limiting value ivhich the fraction assumes when 
 x differs by an infinitesimal from the particular value which 
 makes the expression indeterminate. It is evident (Arts. 9, 
 43) that though the terms of the fraction may be infinitesi- 
 mal, the ratio of the terms may have any value whatever. 
 
 In many cases, the true values of indeterminate forms 
 
 can be best found by ordinary algebraic and trigonometric 
 
 processes. 
 
 X s — 1 
 For example, suppose we have to evaluate -g-— r when 
 
 x = 1. This fraction assumes the form - when x = 1 ; 
 but if we divide the numerator and denominator by x — 1 
 
104 EXAMPLES. 
 
 x 2 4_ x + 1 
 before making x = 1, the fraction becomes — — — j-~ 5 
 
 ami now if we make x — 1, the fraction becomes 
 
 1+1+1 3 
 
 1 + 1 ' ~ 2' 
 
 which is its true value when x = 1. 
 
 73. Hence the first step towards the evaluation of such 
 expressions is to detect, if possible, the factors common to 
 both terms of the fraction, and to divide them out; and 
 then to evaluate the resulting fraction by giving to the 
 variable the assigned value. 
 
 examples. . 
 
 %z i 
 
 1. Evaluate -^ — 5-5— — - — , when x = l. 
 
 a 3 — 2a; 2 + 2a; — 1 ' 
 
 This fraction may be written 
 
 (a; -1) (a? + a? + l) x 2 + x + 1 . 
 
 7 rfr-^ nr = -5 tt = ^> when a; = 1. 
 
 (a; — 1) (x 2 — x + 1) x 2 — x + 1 
 
 a; 
 
 2. The fraction — — = - = - , when x = 0. 
 
 V « + a; — v a — a u 
 
 To find its true value, multiply both terms of the fraction 
 
 by the complementary surd, Va + % + Va — x, and it 
 becomes 
 
 x (Va + x + V# — ^) V« + x + V« — a: 
 
 _ or - ; 
 
 and now making x = 0, the fraction becomes V#, which is 
 its true value when x = 0. 
 
 2a; - V5x 2 - a 2 . . 
 
 3. ■■ j when x = a. Ans. i. 
 
 a; _ V2a^ - a> 
 
METHOD OF EVALUATION. 
 
 a—Va? — x 2 
 x* 
 
 4. -5 , when x = 0. 
 
 
 105 
 
 Ans. 
 
 1 
 
 2a 
 
 Ans. 
 
 5. 
 
 Ans. 
 
 a 
 
 #5 I 
 
 5. -s , when 2 = 1. 
 
 a; — 1 
 
 6. a/z 2 + ## — x 9 when # = ao 
 
 There are many indeterminate forms in which it is either 
 impossible to detect the factor common to both terms, or 
 else the process is very laborious, and hence the necessity of 
 some general method for evaluating indeterminate forms. 
 Such a method is furnished us by the Differential Calculus, 
 which we now proceed to explain. 
 
 METHOD OF THE DIFFERENTIAL CAL- 
 CULUS. 
 
 74. To evaluate Functions of the form ^- 
 
 Let f(x) and <f> (x) be two functions of x such that 
 f{x) = and $ (x) == 0, when x = a. 
 
 Then we shall have Q4 = -x- 
 0(a) 
 
 Let x take an increment li, becoming x -+- h ; then the 
 fraction becomes 
 
 fix + h) 
 
 <p(x + h)' 
 
 Now develop f(x + h) and (j>(x + h) by Taylor's Theo- 
 rem ; substituting h for y in (7) of Art. 66, we have 
 
 f(x + A ) = /(*)+/>)? +/-(4 8+etc - . 
 
106 METHOD OF EVALUATION, 
 
 or when x = a, 
 
 /(fl + jfc) _ /w+/ > w*+rw? + ^ (i) 
 
 < fl + A) (a) + 0' (a) h + 0" (a) J + etc. 
 
 But by hypothesis / (a) = 0, and <f> (a) — 0. Hence, 
 dropping the first term in the numerator and denominator, 
 and dividing both by h, we have, 
 
 ♦<« + *> ♦'(«)+♦" «}+.**' 
 
 Now when h = 0, the numerator and denominator of 
 the second member become /' (a) and <p' (a) respectively ; 
 hence we have, 
 
 (a) 0' (a)> 
 
 f( x ) 
 
 as the true value of the fraction ■-— )-{ , when x = a. 
 
 (1.) If f (a) = and 0' (a) be not 0, the true value of 
 
 f(a) . 
 
 ~-r-i is zero. 
 <t>{a) 
 
 (2.) If/' (a) be not zero and 0' (a) = 0, the true value of 
 f(a) . 
 g>(a) 
 
 (3.) If /' (a) = 0, and 0' (a) = 0, the new fraction 
 
 .,) I is still of the indeterminate form -• Dropping in 
 
 this case the first two terms of the numerator and denomi- 
 
 h 2 
 nator of (1), dividing both by ^-, and making h = 0, 
 
 we have, 
 
EXAMPLES. 107 
 
 /(«) = /"M 
 
 as the true value of the fraction v4-? , when x = a. 
 
 (j>(x)' 
 
 If this fraction be also of the form - , we proceed to the 
 
 next derivative, and thus we proceed till a pair of deriva- 
 tives is found which do not both reduce to zero, when 
 x — a. The last result is the true value of the fraction. 
 
 EXAMPLES. 
 
 1. Evaluate g „ , when x = 1. 
 
 x — V 
 
 Here / (x) = log x, <j> (x) = x — 1 ; 
 
 •*• f'( x ) =-, and <f>'(x) = 1; 
 1 
 
 . £&>_£(*) -»_ 1 1 -i 
 
 0(a?) <?>» 1 xjf ' 
 
 That is, — sl— _ i when x = 1. 
 x — 1 ' 
 
 2. Evaluate s , when x = 0. 
 
 a; 2 
 
 /" (a:) sin a; , A 
 
 , when a; = 0; 
 
 IJo 
 
 <f>' (a;) 2a; 
 
 4>"(x) 
 
 m 
 
 <t>(x) 
 
 f" (x) _ cos a; "] _ i 
 ~ 2 Jo 
 
 * The subscript denotes the value of the independent variable for which the 
 function is evaluated. 
 
JUS EVALUATION. 
 
 x sin x — - 
 
 3. Evaluate ?, when a? = -. 
 
 cos a; ' 2 
 
 Here ^ '^ — x cos g + sin g "| 1 
 
 0' («) — sin a: J*— __ i — ~ 1 - 
 
 Hence ^ = - 1. 
 
 <p(x) 
 
 qX iLj; 
 
 4 - — — > w hen x = 0. ^s. i og « 
 
 f (i) = iw-sr^ °° or °> 
 
 according as 5 > or < 1. 
 
 a x — sin x , 
 
 6. -^ , when x = 0. ^^ ^ 
 
 7 ' -aT^Tn^-' when * = °- Aim. 2. 
 
 8 * ISqV' When * = °- ^ 2. 
 
 Q e* — 2 sin a? — er* _ 
 
 J * a 7 "-^^ » when * = 0. Take the third 
 
 derivative - Am. 4. 
 
 10 - 7 :r> when x = a. Cancel the factor (a — a?)i 
 
 ^rcs. (2a)i* 
 75. To evaluate Functions of the form g-. 
 
 oo 
 
 ^ *M=x' when* = «. 
 
EVALUATION. 109 
 
 Since the terms of this fraction are infinites when x = #, 
 their reciprocals are infinitesimals (Art. 8) ; that is, 
 
 0, and , ■. = 0, when x = a ; hence, 
 
 fix) ~ ' 4>{x) 
 
 1 
 
 m-±M-® w henz-« 
 
 (#) 
 
 and therefore the true values of ,j ' may be obtained by 
 
 W) 
 
 Art. 74 ; that is, by taking the derivatives of the terms, 
 thus, 
 
 /<*)_* (•) _ [»W]' _ »'(*)[/(*)]' when „ _ fl 
 
 /w _ »' («) l/»p 
 
 Dividing by —4-?, we get, 
 y \ a ) 
 
 ,_♦'(«) /(a) . 
 
 whence £® = £ft 
 
 oo . 
 Hence the true value of the indeterminate form — is 
 
 00 
 
 found in the same manner as that of the form -• 
 
 In the above demonstration, in dividing the equation by — -r, when 
 
 0(«) 
 would fail in either of these cases 
 
 x = a, we assumed that ^— is neither nor oo , so that the proof 
 
 0(a) 
 
110 EXAMPLES. 
 
 It may, however, be completed as follows : Suppose the true value 
 
 of £& to be ; then the value of 1&L±£±W iH h, where h may 
 
 9 («) 9 (a) 
 
 be any constant. But as this latter fraction has a value which is 
 
 neither nor x, its value by the above method is - , . - — ~ 
 
 9(a) 
 
 or , . . + h ; and since the value of this fraction is /a, the first term 
 9' («) < 
 
 ^ = ; i. e., where M _-, 0, £g is also 0. 
 
 9 («) 9 {a) 9' (a) 
 
 fix) 
 Similarly, if the true value of ^4-- be oo when x = a, then 
 
 9(x) 
 
 ■37- = 0; and therefore we have „ f/ x = 0, by what has just been 
 
 shown ; ,\ -T-^hr = a° . 
 9 (a) 
 
 Therefore, in every case the value of ^-~ determines the value of 
 
 9 («) 
 
 -— - for either of the indeterminate forms - or — . (See Williamson's 
 9(a) 00 v 
 
 Dif. Cal., p. 100.) 
 
 EXAMPLES. 
 
 1. Evaluate — — , when x = 00 . 
 
 x n 
 
 1 
 tt f(x) f'(x) x 1~| A 
 
 (a;) 0' (x) nx n ~ x narjae 
 
 lofif £ 
 
 2. Evaluate — ~ — , when x = 0. 
 
 cot # 
 
 1 
 
 Here -^ = - = - EE^ffl = ?. 
 
 0' (a:) — cosec 2 a; a; J 0' 
 
 /" (x) _ 2 sin x cos aTl 
 
 .-„ -^f-? = 0, when a; = 0. 
 cot a; 
 
EVALUATION. Ill 
 
 3. ~" ° — , when x = 0. .4 WS . o. 
 
 ex 
 
 4. , when x = 0. ^ws. — . 
 
 .to 8 
 
 cot y 
 
 log tan (2z) , 
 
 5. ° , v y , when a; = 0. Jws. 1. 
 log tan # 
 
 76. To evaluate Functions of the form x oo . 
 
 Let f(x) x <p(x) = x co 3 when x = a. 
 The function in this case is easily reducible to the form 
 - ; for if /(«) = 0, and <j> (a) = oo , the expression can be 
 
 written ^-M^ which = -; therefore /(#) x (p(x) = ^U 
 may be evaluated by the method of Art. 74. 
 
 EXAM PLES. 
 7TX 
 
 1. Evaluate (1 — x) tan — , when x = 1. 
 
 A 
 
 We may write this 
 
 
 , to 
 cot T 
 
 Here f ' (x) - 1 
 
 v ' — - cosec 2 -^r 
 
 2. Evaluate x n log a, when £ = 0. 
 
 1 
 
 i 1°£ x 
 x n log a; = 
 
 a; 
 
 o — — rcar"- 1 
 
 2 sin 2 
 
 (¥) 
 
 2 
 
 7r' 
 
112 EXAMPLES. 
 
 3. e*~* log x, when x = oo . Am. 0. 
 
 4. sec x Ix sin a; — T \ when x = -• -4hs. — 1. 
 
 77. To evaluate Functions of the form oo — oo . 
 
 Let f(x) and <p(x) be two functions of x which become 
 infinite when x = a. Then f(x) — <p (x) = oo — oo , when 
 *5 = a. 
 
 The function in this case can be easily reduced to the 
 
 form - , and may be evaluated as heretofore. 
 
 EXAMPLES. 
 
 2 1 
 
 1. Evaluate -^ — T ' T , when x = l. 
 
 x 2 — l x — l 
 
 This takes the form oo — oo , when x = \. 
 
 2 1 2— Z— 1 , 
 
 = p- , when x = 1. 
 
 a«_l z_i a 2 — 1 
 
 £(| = Tz- = -*' when* = l. 
 
 2 ' Evaluate xl^)-^^-' When * = ' 
 A'hich takes the form oo — oo , when x =. 0. 
 
 1 _ log (1 + a) _ z — (1 + s) log(l + x) 
 x(l + x) a* ^(l+z) 
 
 __ x — (l +x) log(l -I- a;) 
 
 "" x* 
 
 (remembering that 1 -|- x = 1, when a: vanishes). 
 
f(x) _l_log(l + x)-l 
 
 <t>' (x) %x 
 
 1 n 
 
 /"(*)_ 1 + x 
 
 *"(*)" 2 _ " * 
 
 113 
 
 = ^ , when a; = ; 
 
 3. Evaluate sec x — tan x, when # = -. 
 
 Z 
 
 1 — sin x . 7r 
 
 sec x — tan a; = ■ = ^ , when x = -• 
 
 cos # 2 
 
 /^) = - cos x ~\ = Q 
 0» -sin ay | 
 
 71 TT 
 
 Hence, sec 5 and tan - are either absolutely equal, or 
 Z Z 
 
 differ by a quantity which must be neglected in their alge- 
 braic sum.* 
 
 x 1 
 
 4. — = , when x = l. Ans. 4. 
 
 x — 1 log x * 
 
 1 x 
 
 5. , ; , when x = 1. -4ws. — 1. 
 
 log a log ar 
 
 78. To evaluate Functions of the forms 0°, 00 °, 
 and 1 ±0 °. 
 
 Let f(x) and (#) be two functions of x which, when 
 x = «, assume such values that [/(#)]* (a?) is one of the 
 above forms. 
 
 Let y=[f(x)]*W; 
 
 .-. logy = <l>(x)\ogf{x). 
 
 (1.) When f{x) = 00 or 0, and (x) = 0. 
 
 log ?/ = (a;) log /(a;) = (± 00 ), 
 which is the form of Art. 76. 
 
 * Price's Infinitesimal Calculus, Vol. I, p. 210. 
 
114 EXAMPLES. 
 
 Heuce, [/(#)]* (<P) becomes indeterminate when it is of 
 the form 0° or 00°. 
 
 (£.) When f(x) = 1, and <p (x) = ± oo . 
 
 log y = <)>(x) logf(x) = ± oo x 0. 
 Hence, [/(#)]* is indeterminate when of the forms 
 
 1 ± oo 
 
 Hence the indeterminate forms of this class are 
 
 0°,* oo o, 1 ±Q0 , 
 
 and may all be evaluated as in Art. 76, by first evaluating 
 their logarithms, which take the form x oo . 
 
 EXAMPLES 
 
 1. Evaluate a?, when x = 0. 
 
 We have log x* = x log x = —~- ; 
 1 
 
 fix)- - x ->- zJo-0, 
 •\ log x* = 0, when x = 0; 
 hence, x* = 1, when x = 0. 
 
 2. Evaluate ar", when x= cc. 
 
 i i 1 1 log a? 
 
 log ^ = a log x = x 5 
 
 <f>'(x)~ 1 "doo - " ' 
 
 * In general, the value of the indeterminate form 0° in 1. (See Note on Inde- 
 terminate Exponential Forms, hy P. Franklin, in Vol. I, No. 4, of American Journal 
 of Mathematics.) 
 
EXAMPLES. 115 
 
 \ log of = 0, when x = oo ; 
 
 hence, x x = 1, when x = go, 
 
 3. Evaluate (1 + - ) , when x = ao. 
 
 fr+ar. 
 
 Let a> = - , and denote the function by w. 
 
 Then u = (1 + «zj*] 
 
 (since when # = co , z = 0) ; and 
 
 log w — og V + a *v w i ien 2- = o. 
 
 z 
 
 Taking derivatives, we have 
 
 log w^ == rr-^- — ■ = a; 
 
 b 1 + az Jo 
 
 (a\ x 
 1 + - ) = a, when x = go ; 
 
 1 + - J = e a , when # = ao. 
 If a = 1, we have 
 
 that is, as a; increases indefinitely, the limiting value (Art, 
 41) of the function (l + -I is the Naperian base. 
 
 4. (-1 , when x = 0. Ans. 1. 
 
 5. af in *, when x = 0. -4»s. 1. 
 
 (v "■■" 
 
 2 ) a , when x = a. Ans. e*. 
 
116 WOUND i.xi.iti: i;uix ATE FOinis. 
 
 79. Compound Indeterminate Forms. — If an inde- 
 terminate form be the product of two or more expressions, 
 Bach of which becomes indeterminate for the same value of 
 ./. its true value can be found by evaluating each factor 
 separately; also, when the value of any indeterminate form 
 is known, that of any power of it can be determined. 
 
 EXAMPLES. 
 
 1. Evaluate — , when x = oo . 
 
 I x 
 This fraction may be written 
 
 x 
 We first evaluate — , when x = oo 
 
 i'n 
 
 <t> (x) 1 .. 
 
 = l = o. 
 
 00 
 
 Hence, - = 0" = 0. 
 
 2. Evaluate x m log" x, when x = 0, and m and n are 
 positive. 
 
 Here (J log *)• = (^Y- 
 
 We first evaluate -^f , when x = 0. 
 
 We have 
 
 x n 
 
 1 
 
 /'(*) = * 
 
 0'(z) m _*_i 
 
 = x- = 0. 
 
 m Jo 
 a™ log 71 x = 0" = 0. 
 
3. 
 
 x m — x n 
 
 EXAMPLES. 
 , when x = 1. 
 
 117 
 
 1 — a# 
 
 This function can be written in the form 
 x m 1 — x n 
 
 1 + zp 1 
 
 ^ 
 
 We have to evaluate only the latter function for x = 1, 
 since the former is determinate. 
 
 Here 
 
 f(x)_ __ — nx n ~ x 
 <j>' (x) ~~ —pxP~ l 
 
 P 
 
 x n-p 
 
 or 
 
 n 
 
 y 
 i 
 
 1 + xp ~ 2 ' 
 
 x n — x m+n _ n 
 = 2p' 
 
 4. 
 
 (x 2 — a 2 ) sin 
 
 1 — x-p 
 
 ttx 
 
 when x = l, 
 when x == 1. 
 when x = 1, 
 
 2« 
 
 # 2 cos 
 
 TTX 
 
 2a 
 
 when # = a. 
 
 <^-^ sin i 
 
 . 7T3? 
 
 x 2 -a 2 8m M 
 
 x 2 cos 
 
 2« 
 
 COS 
 
 TTX 
 
 2a 
 
 We have only to evaluate the first factor, 
 
 x 2 — a 2 ' 
 
 TTX 
 
 C0S 2a 
 
 2x 
 
 TT . TTX 
 
 -2~a Sm 2a 
 
 ±a 2 
 
 sin 
 
 and 
 
 TTX 
 
 2a 
 
 x 2 
 
U8 EXAMPLES. 
 
 a; 8 COS — 
 2a 
 
 EXAMPLES, 
 gfl; 0— x 
 
 1# log(l + s)' when a; = 0. Ans. 2. 
 
 * rf-rf-fo-s" ' When * = 3 - *• 
 
 Q /sin wa;\ m 
 
 6 - [—£—)> when a; = 0. n n . 
 
 (X q-x 2x 
 
 4 * (ex — 1)3 ( d * f * tnree ti me s)> when x = 0. 
 
 
 
 
 h 
 
 5. 
 
 1 — sin x -f cos # 7r 
 sin a; + cos a;- 1' When * = 2' 
 
 
 1. 
 
 6 
 
 tan a; — sin x , 
 
 
 
 
 sin'"* ' vvlieii x = °- 
 
 h 
 
 7. 
 
 log sin x f whenz = -. 
 
 
 a log a. 
 
 8. 
 
 z 2 + 2cosz — 2 ,„_ „ 
 
 ^ (air. four times), 
 
 when 
 
 x = 0. 
 
 
 
 
 *. 
 
 9. 
 
 a 1 * (pass to logarithms), when x 
 
 = 1. 
 
 1 
 
 10 « log (1+ a?) 
 U ' "ITr-coTF' when a: = 0. 2. 
 
 11. 2-e*, when x = 0. 
 
 oo. 
 
EXAMPLES. 119 
 
 12. (— — J (pass to logarithms), when x = oo. 
 
 Ans. 1. 
 
 13. 2x sin — , when x = oo . a. 
 
 4iX 
 
 14. e x sin x, when a; = 0. oo . 
 
 15. (cos ax) eosec2e * (pass to logarithms, and dif. twice), 
 
 tvhen x — 0. - f! 
 
 e *•. 
 
 16. z m (sin ic) tanx ( J - ^ ( see Art - ? 9 )> wnen » — s 
 
 v ' \2 sin 2a;/ v ' 2 
 
 7T W 
 
 17. (sin »)*" % when # = «• 
 
CHAPTER VII. 
 
 FUNCTIONS OF TWO OR MORE VARIABLES, AND 
 CHANGE OF THE INDEPENDENT VARIABLE. 
 
 80. Partial Differentiation. — In the preceding chap- 
 ben, we have considered only functions of one independent 
 variable; such functions are furnished us in Analytic 
 Geometry of Two Dimensions. In the present chapter, we 
 are to consider functions of two or more variables. A nalylic 
 Geometry of Three Dimensions introduces us to functions oi 
 the latter kind. For example, the equation 
 
 z = ax -f by 4- c (1) 
 
 represents a plane ; x and y are two independent variables, 
 of which z is a function. In' this equation, z may be 
 changed by changing either x or y, or by changing them 
 both, as they are entirely independent of each other, and 
 either of them may be considered to change without affect- 
 ing the other ; in this case z, the value of which depends 
 upon the values of x and y, is called a function of the inde- 
 pendent variables x and y. 
 
 A partial differential of a function of several variables is 
 a differential obtained on the hypothesis that only one of 
 the variables changes. 
 
 A total differential of a function of several variables is a 
 differential obtained on the hypothesis- that all the variables 
 change. 
 
 A partial derivative of a function of several variables is 
 the ratio of a partial differential of the function to the dif- 
 ferential of the variable supposed to change. 
 
PARTIAL DIFFERENTIATION. 121 
 
 A total derivative of a function of several variables is the 
 ratio of the total differential of the function to the differen- 
 tial of some one of its variables. (See Olney's Calculus, 
 p. 45.) 
 
 As all the variables except one are, for the time being, 
 treated as constants, it follows that the partial differentials 
 and derivatives of any expression can be obtained by the 
 same rules as the differentials and derivatives in the case of 
 a single variable. 
 
 If we differentiate (1), first with respect to x, regarding 
 y as constant, and then with respect to y, regarding x as 
 constant, we get 
 
 dz = adx, (2) 
 
 and dz = My. (3) 
 
 Dividing (2) and (3) by dx and dy respectively, we get, 
 
 and $ = b. (5) 
 
 dy 
 
 The expressions in (2) and (3) are called the partial 
 differentials of z with respect to x and y, respectively, while 
 
 ~ and — are called the partial derivatives of z with re- 
 dx dy 
 
 spect to the same variables. 
 
 Since a and b in (4) and (5) are the partial derivatives of 
 z with respect to x and y, respectively, we see from (2) 
 that the partial differential of z with respect to a;, is equal 
 to the partial derivative of z with respect to x multiplied 
 by dx, and similarly for the partial differential of y. 
 
 Hence, generally, if 
 
 / { x , V> z ) 
 
 denotes a function of three variables, x, y, z, its derivative 
 or differential when x alone is supposed to change, is called 
 6 
 
122 Partial differentiation. 
 
 tin' partial derivative or differential of the function with 
 respect to x, and similarly for the other variables, y and ... 
 If the function be represented by u, its partial derivatives 
 are denoted by 
 
 du du du 
 
 dx' dy' dz 9 
 
 and its partial differentials by 
 
 du j du 1 du , 
 
 Tx dx > Ty d v> Tz dz - 
 
 81. Differentiation of a Function of Two Varia- 
 bles. — Let u = f(x, y), and represent the partial differ- 
 
 du 
 ential of u with respect to x, by -=- dx, and with respect 
 
 to y, by -y- dy, while du represents the total differential. 
 
 Let x and y receive the infinitesimal increments dx and 
 dy, and let the corresponding increment of u be du. Then 
 we have, 
 
 du =f(x + dx, y + dy) —f(x, y). 
 
 Subtract and add f(x,y + dy), and we have 
 
 du =f(x+dx, y+dy)—f(x, y + dy) +f(x, y + dy) 
 
 -ffa y)- 
 
 du 
 Now f(x+dx, y + dy) —f(x, y-\-dy) = -r-dx, because 
 
 it is the difference between two consecutive states of the 
 function due to a change in x alone; that is, whatever dif- 
 ference there is between f(x + dx, y + dy) and/(#, y + dy) 
 is due solely to the change in x, as y + dy is the value of y 
 in both of them. For the same reason, 
 
 /fo y + dy) -f(x,y) = ^dy; 
 and therefore we have du = —dx + -j- dy, 
 
EXAMPLES. 123 
 
 du du 
 
 in which -r- dx and j- dy are the partial differentials of t*| 
 
 with respect to a; and y, respectively, while du is the total 
 differential of u when both & and y are supposed to vary. 
 In the same way, we may find the differential of any num- 
 ber of variables. 
 
 Hence, the total differential of a function of two or more 
 variables is equal to the sum of its partial differentials. 
 
 The student will carefully observe the different meanings 
 given to the infinitely small quantity du in this equation, 
 otherwise the equation will seem to be inconsistent with the 
 
 du 
 principles of algebra. Thus, in -y-cfc, du denotes the in- 
 finitely small change in u arising from the increment dx in 
 
 du 
 x, y being regarded as constant. Also, in -r- dy, du denotes 
 
 the infinitely small change in u arising from the increment 
 dy in y, x being regarded as constant, while du in the first 
 member denotes the total change in u caused by both x and 
 y changing. If the partial differentials of x and y be rep- 
 resented by d x u and d y u, respectively, the preceding equa- 
 tion may be written 
 
 du = d x u -f d y u. 
 
 EXAMPLES. 
 
 1. Let u = ay 2 + bxy + ex 2 + ey + gx + k, to find the 
 total differential of it. 
 Differentiating with respect to x, we have 
 
 d x ii = bydx + 2cxdx + gdx. 
 Differentiating with respect to y, we have 
 
 d y u = 2aydy + bxdy + edy. 
 Hence, du = (by -h 2cx + g) dx + (2ay +- bx + e) dy. 
 
124 EXAMPLES. 
 
 2. u = a* 
 
 Here d t u = yx v ~ ] dx, d ¥ u = x* \ogx dy* 
 
 Hence, du = yx»~ l dx + x* logx dy. 
 
 x 2 v 2 
 
 
 Here d x u = -^dx, 
 a 2 
 
 d y u = ^dy. 
 
 Hence, du = — 9 dx + -J c?w. 
 a 4 6* 
 
 4. w = tan -1 ". 
 
 Here d^w = 5 = 
 
 + x 2 
 
 4. 
 
 ydx 
 - x 2 + $■» 
 
 dy 
 
 xdy 
 
 d u — - 
 
 ~ x 2 + / 
 
 _ T sc^y — yds? 
 
 Hence, cm =■ %, , J . - 
 
 x* + y 2 
 
 5. m = sin -1 - + sin" 
 
 a o 
 
 Here d t u = — , rf,w = ,y • 
 
 Hence, d« = — - — : -\ — 
 
 sin x du 
 
 6. w = y 8,n ■. Jw = y 9i " * log y cos a <fe + ~~gy 
 
TOTAL DERIVATIVE, 125 
 
 m 1 x 7 ydx — xdy 
 
 7. u = vers -1 -• a u = • , — — 
 
 V V<s/%xy — x* 
 
 8. u = log Xf. du = - dx + log .r f/y. 
 
 82. To Find the Total Derivative of u with re- 
 spect to x, when u = f (y, z), and y = <P (a?), 
 
 8? = 01 (*»). 
 
 Since u=f(y, z), we have (Art. 81), 
 
 and since # = 0(#), we have d^ = -Jr-dx; 
 
 dz 
 since 2 = 0i (a;), we have dz = -r- dx. 
 
 (IX 
 
 Substituting these values for dy and dz in (1), we get 
 
 , dudy ., , du dz 7 ,_ N 
 
 du = Tutx dx + TzWx dx ' (2) 
 
 Dividing by dx, and denoting the M«£ derivative by ( ), 
 
 we have 
 
 ldu\ _ du dy du dz ,„* 
 
 Way — dy dx dz dx 
 
 Cor. 1. — If z = x, the proposition becomes u =f(x, y) 
 
 dz 
 and y = (#) ; and since y- = 1, (3) becomes 
 
 /du\ _ ^ 6?ft % 
 
 Way ~~ e?# dy dx 
 
 iy 
 
 Cor. 2.— If u —f{x, y, z), and y = (a:), and 2 = 0i («), 
 we have 
 
 , du 7 du , du . m 
 
 d " = dS* e + 3& rfJ ' + *' fc - (1) 
 
126 EXPLANATION OF TERMS. 
 
 dy — -f- dx, and dz = -r- dx. 
 * dx dx 
 
 Substituting the values of dy and dz in (1), and dividing 
 by dx, we get 
 
 idu\ _ du du dy du dz 
 \dxl dx dv dx dz dx 
 
 dy 
 
 Cor. 3.— If u =f(y, z, v), and y = <f> (x), and z = 
 0, (x), and v = fa (x), we have, 
 
 , du j du 3 dv , . 
 
 dw = T y d y + Tz' h + dv dv - w 
 
 dy ~ -j- dx\ dz =z ~ dx dv = j- dx. 
 
 CLX ClX (XX 
 
 Substituting the values of dy, dz, dv, in (1), and dividing 
 by dx, we get 
 
 ldu\ _ du dy du dz du dv 
 \dx) ~ dy dx dz dx dv dx 
 
 du 
 Cor. 4. — If u =f(y) and y = (x), to find -=-• 
 
 cix 
 
 du 
 Since u = / (?/), we have du = — dy. 
 
 Since y =. <f) (x), we have dy = ~ dx. 
 
 . , » , du dy . n du du du 
 
 therefore, du = -j--^dx, and .: -^ ss s-j u 
 
 dy dx dx dy dx 
 
 Sch. — The student must observe carefully the meanings 
 of the terms in this Art. Thus, in the Proposition, u is 
 indirectly a function of x through y and z. In Cor. 1, u is 
 directly a function of x and indirectly a function of x 
 through y. In Cor. 2, u is directly a function of x and 
 
EXAMPLES. 127 
 
 indirectly a function of x through y and z. In Cor. 3, u is 
 indirectly a function of x through y, z, and v. In Cor. 4, 
 w is indirectly a function of a; through «/. 
 
 The equations in this Article may seem to be inconsistent 
 with the principles of Algebra, and even absurd ; but a little 
 reflection will remove the difficulty. The du's must be 
 carefully distinguished from each other. In Cor. 1, for 
 
 example, the du in — is that part of the change in u 
 
 which results directly from a change in x, while y remains 
 
 constant ; and the du in -=- is that part of the change in u 
 
 which results indirectly from a change in x through y ; and 
 
 the du in l-r-J is the entire change in u which results 
 
 directly from a change in x, and indirectly from a change 
 in x through y. 
 
 
 EXAMPLES. 
 
 1. u 
 
 == tan" 1 - and y = (<r* - &% to find (^)- 
 
 Here 
 
 du _ y du x -i dy x 
 dx ~ r 2 ' dy ~ r 2 ' dx ~ y 
 
 Substituting iu (|) = * + J | (Art. 8., Co, 1), we 
 have ^ 
 
 (du\ _y_ ,(_x\/x\ 
 
 \dxl ~ r* "*" \ rV\ y) 
 
 _ y2 + X 2 ! 
 
 r 2 y Vr 2 -x* 
 
 and this value is of course the same that we would obtain 
 
 x 
 if we substituted in u = tan -1 - for y its value in terms of 
 
 x, and then differentiated with respect to x. 
 
L28 EXAMPLES. 
 
 2. u = tan -1 (£#) and y = e x , to find f-i-)* 
 
 Hero *"- y *?- * ^-e* 
 
 cfe-l + a^*' rfy""l+aY' <& "" # 
 
 .-. (Art. 82, Cor. 1), 
 
 (6?w\ _ y + e x x _ & (1 +#) ^ 
 Ixl " TT&f ~ 1 + a*^ ; 
 
 and this value is of course the same that we would obtain if 
 we differentiated tan -1 (xeP) with respect to x. 
 
 3. u = z 2 +y z + zy and z = sin x, y = &*, to find (-r-)- 
 
 it du 9 , du n t 
 
 Here ^ = 3y* + *, &= a » + * 
 
 ds . dy „ 
 
 — zh COS #, -jT = C*» 
 
 .-. (Art. 82), 
 
 (iD = (3y2 + *) ^ + ( 2 * + y) cos x 
 
 = (36 s * + sin x) e? + (2 sin a -f e a ) cos a? 
 
 = Se 3 * + e? (sin x + cos z) + sin 2x. 
 
 (See Todhunter's Dif. Oal., p. 150.) 
 
 Let the student confirm this result by substituting in w, 
 for y and z, their values in terms of x, thus obtaining 
 
 u = eP* -f e* sin # -f sin 2 z, 
 
 and then differentiate with respect to x. 
 
 4. w = sin -1 (y — z), y = 3x, z = 4a 8 . 
 
 du _ 1 
 
 dy ~ yl _ (y _ ^i' 
 
EXAMPLES. 129 
 
 du _ 1 
 
 da; da; 
 
 >«*»*© 
 
 Vi - (2/ - *) 2 
 
 3 - 12a; 2 3 
 
 VI — 9a? + 24a, 4 — 16^ Vl — x % 
 
 gCMC (y %\ 
 
 5. u = — )r — =-* and // = a sin #, 2 == cos x. . 
 
 a 2 + 1 J 
 
 du e™ 
 
 du e™ 
 
 dy ~ a 2 + 1 ' 
 
 lz~ ~ & + V 
 
 du 
 
 -r = a cos #, 
 aa; 
 
 dz 
 
 -=- = -sina 
 
 da; 
 
 r/w 
 
 a6? ax (^ _ ^ 
 
 da; ~ 
 
 « 2 + l 
 
 .-. (Art. 82, Cor. 2), 
 
 (du\ e ^ 
 — I = -= (a cos x + sin x + a 2 sin x — « cos a;) 
 
 = e ax sin #. (See Courtenay's Cal., p. 73.) 
 
 6. u = yz and y = e x , z = x* — Ax 3 + 12a: 2 — 24a; + 24. 
 
 © = <* 
 
 7. II u =f(z) and z = (a;, #), show that 
 
 _ dudz j du dz , 
 " ~ dz dx dz dy y ' 
 
 8. w = ^- - "/ + 32 and 2/ = lo S * 
 
130 PARTIAL DIFFERENTIATION, 
 
 83. Successive Partial Differentiation of Func- 
 tions of Two Independent Variables. 
 
 Let u=f(x,y) 
 
 be a function of the independent variables x andy; then 
 
 -j- and y- are, in general, functions of x and y, and hence 
 
 may be differentiated with respect to either x or y, thus 
 obtaining a class of second partial differentials. Since the 
 partial differentials of u with respect to x and y have been 
 
 du dijL 
 
 represented by -^- dx and -=- dy (Art. 81), we may repre- 
 sent the successive partial differentials as follows : 
 
 The partial differential of (-7- dx), with respect to x, 
 d {du 7 \ 7 
 
 = Tx\Tx dx ) dx > 
 
 which may be abbreviated into 
 
 The partial differential of \-j- dx), with respect to y, 
 
 d /du , \ , 
 = Ty\Tx dx ) d y' 
 
 which may be abbreviated into 
 
 d?u 
 
 dy dx 
 
 dydx. 
 
 cPu cPu 
 
 Again, both -=-^ dx 2 and -j — - dy dx will generally be 
 
 functions of both x and y, and may be differentiated with 
 respect to x or y, giving us third partial differentials, and 
 so on. Hence we use such symbols as 
 
PARTIAL DIFFERENTIATION. 131 
 
 ffiw dPu dftu 
 
 im iy9aS ^didx dxd ^ dx ' and ifTxW' 1 *' 
 
 the meaning of which is evident from the preceding re- 
 
 marks. For example, %- dx dy dx denotes that the 
 
 function u is first differentiated with respect to x, supposing 
 
 y constant; the resulting function is then differentiated 
 
 with respect to y, supposing x constant; this last result is 
 
 then differentiated with respect to x, supposing y constant; 
 
 and similarly in all other cases. 
 
 When u =f(x, y), the partial derivatives are denoted by 
 
 cPu d 2 u d?u dhi d?u d 3 u 
 
 _______ _tc 
 
 dx 2 ' dy 2 ' dxdy' dx?' dx 2 dy' dxdy 2 ' 
 
 84. If u be a Function of x and y, to prove that 
 
 cPu , , d 2 u 7 , 
 
 -= — 7- dx dy = _ — 7- dy dx, 
 dxdy * dy dx J 
 
 Take u = x 2 y% 
 
 — dx = 2xy*dx, 
 dx 
 
 ^dy = Sx 2 y 2 dy, 
 dy dx = 6xy 2 dydx 9 
 dx dy = 6xy 2 dx dy. 
 
 dy dx 
 
 d?u 
 dxdy 
 
 In this particular case, 
 
 d 2 u , , <Pu 7 , 
 
 -: — =- dx dy = -5 — =- dy dx ; 
 dx dy * dy dx a 
 
 that is, the values of the partial differentials are independ- 
 ent of the order in which the variables are supposed to 
 change. 
 
Vtt PARTIAL DIFFERENTIATION. 
 
 To sJww this generally: 
 
 Let u=f(x,y); 
 
 then ~^o dx - f( x + dx > V) ~f( x > V)' 
 
 This expression being regarded us a function of y, let y 
 become y + dy, £ remaining constant; then 
 
 !/X/r dx ) d y =f( x + dx > y+ d y) -/(*> y+fy) 
 
 y -[f(x+dx,y)-f(x, y y 
 
 = f(x+dx, y+dy) —f(x, y + dy) 
 
 -f(x + dx,y)+f(x,y). 
 In like manner, 
 
 jfcdy =f(x, y + dy) -f{x, y). 
 
 k-U;j d y) dx = f( x +dx, y+ d y) -f{x+dx, y ) 
 
 d J -U(*,y+dy)-f{x 7 y)} 
 
 = f(x + dx, y + dy) —f{x + dx, y) 
 
 -f{x,y + dy)+f(x i y). 
 These two results being identical, we have 
 d idu , \ 7 d (du 1 \ 7 
 
 Ty\Tx dx ) d y = dxXdy d y) dx > 
 
 that is, -z — j- dy dx = , , dx dy. 
 
 dydx * dxdy a 
 
 Dividing by dy dx, we get 
 
 cPu d*u 
 
 dy dx ~~ dx dy 
 
 In the same manner, it may be shown that 
 d? u j<> 7 d?u , 2 
 
 and so on to any extent. 
 
EXAMPLES. 133 
 
 EXAMPLES. 
 
 1. Given u = sin (x + y), to find the successive partial 
 derivatives with respect to x. 
 
 du , , \ d 2 u • / , \ 
 
 ^ = cos {x + y) y -^ = - sm (x + y) y 
 
 U 11 (lli 
 
 — = _ cos (x + y), -^ = sin (x + y), etc. 
 
 2. w = log (x + «/), to find the successive partial de- 
 rivatives with respect to x, and also with respect to y in 
 the common system. 
 
 du m d 2 u m 
 
 ~ ~¥Vy) 2 ' 
 
 m 
 
 (See Art. C5, Lemma.) 
 
 3. If u = x log «/, verify that 
 
 4. If w = tan -1 (-), verify that .i^ = 
 
 ^2 # + «/' 
 
 dx 2 
 
 dfo wi 
 
 d 2 u 
 
 % _ z + #' 
 
 dy 2 
 
 5. If w = sin («a; n + % ?l ), 
 
 d?u 
 
 2m 
 
 dx* ~ 
 
 ■ (X + y)»' 
 
 d 3 u 
 
 2m 
 
 dy 3 " 
 
 ' (a; + y) 3 
 
 d% 
 
 cPu 
 
 eft/rfiE " 
 
 ~ dxdy 
 
 cPu 
 dy 2 dx 
 
 d?u 
 ~ dxdy 2 
 
 d*u 
 
 d*u 
 
 .. cru cru 
 
 verif y that -&df = w^ 
 
 85. Successive Differentials of a Function of Two 
 Independent Variables. 
 
 Let u =f (x, y). 
 
 We have already found the first differential (Art. 81), 
 
 7 du . , du , n v 
 
 du = dx llx + Ty diJ - (1) 
 
134 SUCCESSIVE DIFFERENTIALS. 
 
 fill / 1 // 
 
 Differentiating this equation, and observing that -j-, — , 
 
 are, in general, functions of both x and y (Art. 83), aim 
 remembering that x and y are independent, and hence 
 thai dx and </// are constant, we have, 
 
 „ '*&», ^T^) , j *&**), 
 *" = —dz~- dx + ~3y— ^ + — &- & 
 
 '£*)' 
 
 dhi = ^ + ^ tfy«fe + ^drfy + ¥2 ^2 
 
 <fy- 
 
 ^ 7 O . rt ^ 7 7 ^ 7 O 
 
 -»*• + » 5***+'^* ( 2 > 
 
 (since -5— 5- dydx = y-j- dxdy y Art. 84). 
 
 Differentiating (2), remembering that each term is a 
 function of x and y, and hence that the total differential 
 of each term is equal to the sum of its partial differentials, 
 we get, 
 
 and so on. It will be observed that the coefficients and 
 exponents in the different terms of these differentials arc 
 the same as those in the corresponding powers of a bino- 
 mial; and hence any required differential may be written 
 out. 
 
 * The total differential of each of the terms (— dx) and (— <ly\ is equal to the 
 cum of its partial differentials. 
 
IMPLICIT FUNCTIONS 135 
 
 EXAM PLES, 
 
 
 i w=(^ + y 2 )k 
 
 
 
 du x 
 
 du 
 
 dy ~ 
 
 . y 
 
 dx ~ (38 + ffi ' 
 
 ' (* + y¥ 
 
 d 2 u y 2 
 dx* ~ (^T^ji ' 
 
 d 2 u 
 dxdy 
 
 -xy 
 (X 2 + jflt 
 
 d 2 u x* 
 dy 2 ~ (* ■+'#)* ' 
 
 dhi 
 dx* ~ 
 
 — 3^ 2 
 
 (* 2 + */ 2 )* 
 
 dhi (2x 2 — y 2 ) a 
 dxHy ~ y ( X i + tftyV 
 
 dhi 
 dxdy 2 
 
 3 (2y 2 — a?) 
 (x 2 + jf)t 
 
 d 3 u — dyx 2 
 
 
 
 dy* ~ (^ _ h y2 )f ' 
 
 
 
 .-. <Z%, = [— 3xy*do?-+ 3y (2x* - 
 
 - «/ 2 ) <fo% 
 
 4- &z (9,ifl — r? 
 
 \ fl/rrhfi _ a 
 
 !-j/'r2/7i/3n 
 
 (X 2 + ^)f 
 
 fc = [oW + 2ahdxdy + #% 2 ] e ox +&2 , 
 = 0$c + My] 2 e? xiby . 
 
 86. Implicit Functions (see Art. 6). — Thus far in this 
 Chapter, the methods which we have given, although often 
 convenient, are not absolutely necessary, as in every case by 
 making the proper substitutions we may obtain an explicit 
 function of x, and differentiate it by the rules in Chapter II. 
 But the case of implicit functions which we are now to 
 consider is one in which a new method is often indis- 
 pensable. 
 
 Let / {x, y) = be an implicit function of two varia- 
 bles, in which it is required to find -—• If this equation 
 
f;)6 IMPLICIT. FUNCTIONS. 
 
 can be solved with respect to y, giving for example 
 y — (x), then the derivative of y with respect to x can be 
 found by previous roles. But aa it is often difficult and 
 sometimes impossible to solve the given equation, it is 
 
 necessary to investigate a rule for finding -~ without 
 living the equation. 
 
 87. Differentiation of an Implicit Function. 
 
 Let f(x, y) = 0, 
 
 in which y is an implicit function of x, to find -^- 
 
 Let f(x, y) = u. 
 
 Then u =./(*, y) = 0. 
 
 Hence by (Art. 82, Cor. 1), we have, 
 
 _ du du dy 
 
 ~ dx dy dx 
 
 But u is always = 0, and therefore its total differential 
 = 0; hence l-j-j = 0, and therefore, 
 
 du du dy __ n 
 
 dx dy dx ~ ' 
 from which we get, 
 
 du 
 dy _ dx 
 dx ~ du' 
 dy 
 
 Sen. — It will be observed that while (~ 1 = 0, neither 
 -r- nor y- is in general ss 0. For example, 
 x 2 + y* — r 2 = 
 
 is of the form f(x, y) = 0. We see that if x changes while 
 y remains constant, the function changes, and hence is no 
 
EXAMPLES. 137 
 
 longer = 0. Also, if y changes while x remains constant, 
 the function does not remain = 0. But if when x changes 
 y takes a corresjjonding change by virtue of its dependence 
 on x f the function remains = 0. 
 
 EXAMPLES. 
 
 1. %f - 2xy + a* = 0, to find % 
 
 du a du a . 
 
 Tx = - % y> T y = *y-* x - 
 
 du 
 
 L , j. dy dx — 2y y 
 
 therefore, -/ = T - = — - v~ = — — • 
 
 dx du 2y — 2x y —x 
 
 dy 
 
 2. a*tf + bW - am = 0, to find ^- 
 
 (1) 
 
 Since y = - Vet? — x 2 , from the given equation, we may 
 solve this example directly by previous methods, and obtain 
 
 f= JL^ l (2) 
 
 dx a^/ a 2 _ x i 
 
 which agrees with (1) by substituting in it the value of y in 
 terms of x. 
 
 In this example we can verify our new rule by comparing 
 the result with that obtained by previous rules. In more 
 complex examples, such as the following one, we can find 
 
 ~ only by the new method. 
 
 
 du 
 
 dx " 
 
 Wx; 
 du 
 
 du 
 dy ~ 
 
 **y\ 
 
 therefore, 
 
 dy _ 
 dx " 
 
 dx 
 du 
 dy 
 
 2Wx 
 2a*y~ 
 
 
1:58 SECOND UK It IV A TIVK OF .1 .V 1 Ml' LICIT FUNCTIO.W 
 
 3. z 5 — ax*y + bx*y 2 — y 5 = 0, to find -X 
 
 g£ = 5a 4 — 302fy + 2% 2 ; 
 
 -7- = — flic 8 + 2fafy — 5?/ 4 ; 
 
 therefore, £ = ^*Zl + »*. 
 
 dx oy 4 — 2bx l y + ax 3 
 
 * * dx x*—3atf 
 
 5. 2/2 _ 2axy + a? - & = 0. ^ = 3^=*. 
 
 C. f - 3y + * = 0. 
 
 7. ^ + 3 flay + ^ = a d A = *±m, 
 
 * ■ u dx y 2 + ax 
 
 88. To Find the Second Derivative of an Im- 
 plicit Function. 
 
 Let u =f(x,y) = 0. 
 
 du 
 
 We have |=-|(ArtW), (1) 
 
 dy 
 
 it is required to find -^» 
 
 Differentiating (2), remembering that -=- » t~ ? are ^ uric * 
 tions of a: and y, we get ^ 
 
 d*tt dfy dy id?u dy d 2 u \ dy du d 2 y _ 
 i/# 2 dy dx dx \dy 2 dx dx dy) dx dy dx* 
 
' EXAMPLES. 139 
 
 <#% cT-u dy dhxdy 2 du d 2 y _ 
 
 dx 2 + dx~dydx + dfdx 2 + d^dx' 2 ~ {S) 
 
 du 
 Substituting the value of ~ from (1), and clearing of 
 
 fractions, we get 
 
 d 2 u du 2 d?u du du dhi du 2 du 3 d 2 y _ 
 
 dx 2 dy 2 dx dy dx dy dy 2 dx 2 dy z dx 2 ~ 
 
 Solving for ~ , we get 
 
 d?u /du\ 2 d 2 u dudu d?u/du\ 2 
 
 d 2 y _ dx 2 \dy) dx dy dx dy dy 2 \dx) . 
 
 dx 2 ~~ (du\* * ' 
 
 \dy) 
 Sch. — This equation is so complicated that in practice it 
 is generally more convenient to differentiate the value of the 
 first derivative immediately than to substitute in (4). The 
 third and higher derivatives may be obtained in a similar 
 manner, but their forms are very complicated. 
 
 Equation (2) is frequently called the first derived equation 
 or the differential equation of the first order ; and equation 
 (3) is called the second derived equation or the differential 
 equation of the second order. 
 
 EXAM PLES. 
 
 1. y - 2xy + a? = 0, to find g and g. 
 
 du du a n (Pu . 
 
 Tx = - % y' T y = % y- 2x ' d* = °> 
 
 &u 9 1 cPu _ 9 
 
 dxdy ~ ' dy 2 ~~ 
 
 Therefore, by (1), ' £ = ^ 
 
 and bv m ^ - ~ Uy (!l ~ x) -±- S - y ~ - SkUzM. 
 ana oy W , (fe2 _ - ^JZT^xf ~ (y-xf 
 
140 ( 1IANGE OF amSPSNDKNT VARIABLE. 
 
 2. f — 2axy + x 2 - & = 0, to find '';[ , ?|. (See Sch.) 
 
 &y _ 
 
 dz' dx 2 
 
 dy _ ay — x 
 dx ~ y — ax 
 
 <fc 2 (y — axy 
 
 _ (y — a.r) (a 2 // — y) — (ay — #) (rt 2 z — #) 
 
 (by substituting the value of -j) 
 
 - (« 2 — i) (y 2 — %<wy + ^) _ #* ( a 2 — 1) 
 (y — axf ' (y — axf' 
 
 3. z 3 + 3«a:y + # 3 = 0, to find ^ and ^. 
 
 ct ic- ax 
 
 Differentiating, we have 
 
 (x 2 + ay) dx + (if + ax) dy = ; 
 
 e?y _ x* + ay 
 dx ~ y 2 + «a; 
 
 . ?fy 
 
 ^c 2 ~ («/ 2 -f ax) 2 
 
 = 7 4^ XJ/ -Th' (See Price's Calculus, Vol. I, p. 142.) 
 (y 2 + a#) 3 v 
 
 89. Change of the Independent Variable. — Thus 
 
 far we have employed the derivatives -— , j¥, etc., upon 
 
 the hypothesis that x was the independent variable and y 
 the function. But in the discussion of expressions contain- 
 
VALUES OF DIFFERENTIAL COEFFICIENTS. 141 
 
 ing the successive differentials and derivatives of a function 
 with respect to x, it is frequently desirable to change the 
 expression into its equivalent when y is made the independ- 
 ent variable and x the function ; or to introduce some other 
 variable of which both y and x are functions, and make it 
 the independent variable. 
 
 90. To Find the Values of % g, % etc., 
 when neither x nor y is Equicrescent. (Art. 55.) 
 
 dn 
 The value of the first derivative, ~, will be the same 
 
 dx 
 
 whether x or y, or neither, is considered equicrescent. 
 
 d 2 v 
 The value of the second derivative, ~ , was obtained in 
 
 dx 2 
 
 dy 
 Art. 56 by differentiating ~y as a fraction with a constant 
 
 denominator and dividing by dx. 
 
 If we now consider that neither x nor y is equicrescent, 
 and hence that both dx and dy are variables, and differen- 
 tiate ~ 9 we have 
 dx 
 
 d ldy\ _ d 2 y dx — d?x dy ( , 
 
 ~ Tx\dx) ~~~ da? ' U 
 
 which is therefore the value of the second derivative when 
 neither variable is equicrescent. 
 
 Similarly, 
 
 tly = d_t(Py\ 
 dx 3 dx \dx 2 / 
 
 (d*y dx — d?x dy) dx — 3 (cPy dx — <Px dy) d?x /q . 
 ~ M ' ( } 
 
 which is the value of the third derivative when neither 
 variable is equicrescent, and so on for any other derivative. 
 
142 eXami'u:s. 
 
 Cor. — If & is equicrescent, these equations are identical 
 If y is equicrescent, <fiy = iPy = 0, and (1) becomes 
 
 
 
 tPy 
 dx* " 
 
 d 2 x dy 
 
 
 (2) 
 
 becomes 
 
 
 
 
 
 d*y 
 dx*' 
 
 _ 3(d 2 xfdy — (Pxdy 
 
 dx 
 
 
 
 dx* 
 
 
 (3) 
 
 (*) 
 
 which are the values of the second and third derivatives 
 when y is equicrescent. 
 
 Sch. 1. — Hence, if we wish to change an expression when 
 x is equicrescent into its equivalent where neither x nor y is 
 
 equicrescent, we must replace -~j -~ t etc., by their com- 
 plete values in (1), (2), etc.; but if we want an equivalent 
 expression in which y is equicrescent, we must replace 
 
 ~fe, 3' etc -> b y their values in ( 3 )> ( 4 )' etc - 
 
 Sch. 2. — If we wish to change an expression in which x 
 is equicrescent into its equivalent, and have the result in 
 terms of a new independent variable t, of which x is a 
 
 function, we must replace -~ , ~ , etc., by their complete 
 
 values in (1), (2), etc., and then substitute in the resulting 
 expression, in which neither x nor y is equicrescent, the 
 values of x, dx, d% etc., in terms of the new equicrescent 
 variable. 
 
 examples. 
 
 1. Transform x y| + U^J — JL = 0, in which x is 
 
 equicrescent, into its equivalents, (i) when neither x nor y 
 is equicrescent, (2) when y is equicrescent. 
 
EXAMPLES, 143 
 
 (fill 
 
 (_".) Replace t~ by its value in (1), and multiply by dx\ 
 and we have 
 
 x((Fy dx — d?x dy) + dy 3 — dy dx 2 = 0. 
 
 (#.) Put d?y = 0, divide by dy z to have the differential 
 of the independent variable in its proper position, the de- 
 nominator, and change signs, and we have 
 
 (Px , (dx\ 2 , n 
 
 dy 2 \dy 
 
 2, Transform -=4 — i s -^ + , ^ = °> m which re 
 
 dx 2 1 —x 2 dx 1 — x 2 
 
 is equicrescent, into its equivalent when is equicrescent, 
 having given x = cos 0. 
 
 Replacing -=-^ by its complete value in (1), the given 
 
 QiX 
 
 equation becomes 
 
 d 2 y dx — d 2 xdy x dy y _ ft 
 
 _____ _______ + ___ _ . 
 
 x = cos 0; 
 
 ,\ dx = — sin J0 and _$a? — — cos dd\ 
 l_ x 2 — s i n 2 0. 
 
 Substituting, we have 
 
 — d?y sin ft d0 + cos dO 2 d y cos0 <fy y_ _ Q 
 
 — sin 3 rf0 3 sin 2 (9 sin d0 + sin 2 ~ 
 
 ... ^ + y = o. (See Price's Calculus, Vol. I, p. 126.) 
 do 
 
 3. Transform -r|+-^ + V = 0, in which x is equi- 
 
 dx 2 xdx * 
 
 crescent, into its equivalent, (1) when neither y nor is 
 equicrescent; (#) when is equicrescent; (5) when y is 
 equicrescent, having given x 2 = 40. 
 
144 EXAMPLES. 
 
 Replacing in this equation the complete value of - •{, it 
 !h c.ines 
 
 cPy dx — d*x du 1 dy 
 
 dx 3 xdx u 
 
 X = 2(0)1; ... dx = 0-?dd. 
 
 d& d*0 
 (Px= = + — . 
 
 Substituting, we have 
 
 (1.) yd& + dydP + 0dtydO — 6dyeP0 = 0. 
 
 (#.) When is equicrescent, cPO — 0; therefore (1) be- 
 comes 
 
 y d0* + dyd& + OdHjdO = 0, 
 
 e^ + d 4 + y = o. 
 
 "m + te + y 
 
 (S.) When y is equicrescent, dpy — ; therefore (1) be- 
 comes 
 
 (Pd /dd\ 2 IdS 
 
 .(Co <wv /cmy . 
 
 6 df-\dy)-y\di)) = - 
 
 b*£ 
 
 4. Transform R = ^ into its equivalent, (1) in 
 
 dx* 
 the most general form; (2) when is equicrescent; (.9) 
 when r is equicrescent, having given x = r cos 0, and 
 y = r sin 0. 
 
 The complete value of R is 
 
 i? = _ (dx 2 + dy*)* 
 cPxdy — d?ydx 
 
 dx = dr cos — r sin d 0, 
 dy = dr sin + r cos g? 0. 
 
EXAMPLES. 145 
 
 d z x = cos (Pr — 2 sin drdd — r cos 6 d6 2 
 
 — r sin 0^0, 
 
 ^ = sin d 2 r + 2 cos ^n/0 — r sin d0 2 
 
 + r cos d 2 ; 
 
 .-. (dxf + (^)a = rfr 3 + rW, 
 
 ^a% — d 2 */^ = rd 2 rdi) — 2drW — rhld* — rdrdtd. 
 
 (1.) .'. R 
 
 [dr 2 + r 2 dd 2 ]l 
 
 
 tfr\ 3 
 
 (5.) j? 
 
 (See Serret's Calcnl Differential et Integral, p. 94.) 
 
 5. Transform 
 
 (dy 2 + dx 2 )§ + adxd 2 y = 0, 
 
 in which x is equicrescent, into its equivalents, (1) when 
 neither x nor y is equicrescent, (2) when y is equicrescent. 
 
 (i.) {dy 2 + efa 2 )* + « (d 2 ydx — d 2 xdy) = 0; 
 
 91. General Case of Transformation for Two Inde- 
 pendent Variables. — Let u be a function of the inde- 
 pendent variables, say u = f (x, y) ; and suppose x and y 
 functions of two new independent variables r and 0, 
 so that, 
 
 7 
 
And similarly 
 
 140 EXAMPLES. 
 
 x = cf>(r, 0) and y = t/> (r, 0) ; (1) 
 
 (.hen u may be regarded as a function of r and 0, through 
 x and y. It is required to find the values of y and -= in, 
 
 terms of derivatives of u, taken with respect to the new 
 variables r and 0. 
 
 Since « is a function of r through £ and y, we have 
 (Art. 82), 
 
 du _ du dx du dy . . 
 
 rfr ~~ dxdr dy dr ' 
 
 du _ du dx du dy # , . 
 
 dB ~ didS + dydd' 9 ^ 
 
 where the values of ^-, -/, -^, -^, can be found from (1). 
 dr dr dd dQ v ' 
 
 Whenever equations (1) can be solved for r and sepa- 
 rately, we can find by direct differentiation the values of 
 
 dr dr dd dd , . 
 
 -r,-r, -]-, -j-, and hence by substituting in 
 
 dx ciy cix cly 
 
 du dudr du dd 
 
 dx ~ dr dx dd dx' 
 
 , du du dr du dd ncx . 
 
 and Ty = drTy + T6Ty^-^' 
 
 we can obtain the values of -7- and - r -« 
 
 dx dy 
 
 When this process is not practicable, we can obtain their 
 values by solving (2) and (3) directly, as follows : 
 
 dis dx 
 
 Multiply (2) by -^ and (3) by -=- and subtract ; then 
 
 multiply (2) by -~ and (3) by ~ and subtract. We shall 
 then have two equations, from which we obtain, 
 
EXAMPLES. 147 
 
 (i) 
 
 du 
 
 du dy 
 drdd ' 
 
 dudy 
 ~dddr 
 
 dx " 
 
 ~ dxdy 
 
 dy dx 9 
 
 
 drdd 
 
 dr dd 
 
 
 du dx 
 
 du dx 
 
 du 
 
 dddr 
 
 drdd 
 
 dy " 
 
 dxdy 
 
 dy dx 
 
 
 drdd 
 
 dr dd 
 
 (5) 
 
 d?u d*u 
 The values of -z~ 2 , -r-j, etc., can be obtained from these, 
 
 but the general formulae are too complicated to be of much 
 practical use. (See Gregory's Examples, p. 35.) 
 
 Cor. — If x = r cos d and y = r sin d, (4) and (5) 
 become 
 
 du . du, tnn du du cos d du . n du 
 
 dx dr r dd dy r dd dr 
 
 £X AM PLES. 
 
 1. u = X -^l f to find du (Art 81). 
 
 x — y v ' 
 
 du _ Z(rdy-ydx) 
 {x - yf 
 
 2. u = sin ax + sin by + tan -1 -• 
 
 du = a cos axdx + b cos Jy^y H ^ — t- « 
 
 * * y 2 \- # 
 
 o • -1 X 
 
 3. m = sin ' - 
 
 y 
 
 , _ ^^c — xdy 
 
 y Vy 2 — % 2 
 
 4. u = sin (a; + y). 
 
 du = cos (x + y) (^t + dy). 
 
H8 EXAMPLES. 
 
 5. u = *' . 
 2^ — a 8 
 
 7 _ s ( g2 — ^ 2 ) (2y^» + ady) — 2xh/zdz 
 
 du -~ " (#-^y J — 
 
 j M = 2 y^=ffr. 
 
 y V* 2 — y 2 
 
 7. «< = cot xy to find /^) (Art. 82, Cor. 1). 
 
 (-=-) = — a? cosec 2 a? I- + log x -f\» 
 
 8. u = sin (y* - z), and y = log x, z=x* } to find (^') 
 (Art. 82). w 
 
 /<fa\ _ 2 (y — a?) co s (f — g) 
 Way " x 
 
 x 
 
 9. w = log tan -• 
 
 6 y 
 
 dy 
 
 /tfw\ _ y ~ dx 
 
 \Txl ~ 
 
 . X X 
 
 ir em- cos - 
 
 y y 
 
 10. w = log (x — « + Vz 2 — 2«#). 
 
 /tf«A 1_ 
 
 W " V^ — 2oaT 
 
 11. if w = a^z 4 + e'y 2 * 3 + ^y^ 2 , show that 
 
 d 4 ?t 
 
 »p-*»'.+* < Ari83 -> 
 
 12. If w = tan -1 , X y=. , show that 
 
 yl + a? + y 8 
 
 e£w 1 d A n lhxy 
 
 dxdy (i + X 2 + ^ijt' daft/y* ( X + ^ + y2 )| 
 
EXAMPLES. 143 
 
 13. u = a% 2 + ifx 2 to find d 2 u (Art. 85). 
 dhi = SxyWx 2 + (jx 2 ydxdy -f 2y 3 dx 2 + Gxy 2 dxdy 
 
 + 6x 2 ydxdy +• 2x 3 dy 2 + 6xy 2 dxdy + Gx 2 ydy 2 
 
 = (6xy 2 + 2f)dx 2 + 12 (afy + ?/^) <fe% + (0« 2 # + 2a; 3 )^ 2 
 
 14. a? -f #•— a = 0, to find ~|. (Art. 87.) 
 
 £?# _ ?/.^ _1 + y x log y 
 tfo ~~ xy x x + a* log # 
 
 u ax x* — xy log x 
 
 17. *-3^ + y . = 0. |=f5|- 
 
 18. we"* — «a?» = 0. -/ = ' "? — *• 
 
 a dx x (1 4- wy) 
 
 19. a* y + \/sec (ary) = 0. 
 
 dy _ y Vsec (a;?/) tan (xy) + 2a' y yx y ~ 1 log a? 
 
 ^ a; Vsec (a:?/) tan (xy) + 2a xy a; y log a log x 
 
 20. a; 4 + 2ax 2 y — ay 3 = 0, to find ^ and g. 
 
 ( Art 88) % = _^L±i^. 
 
 v J dx 2ax 2 — 3ay 2 
 
 g = - [(12a* + 4ay) (W - 3«jfl» 
 
 — 8«a: (4.x- 3 + 4oa;y) (2ax 2 —3ay 2 ) - (4a?+tei/) 2 6^ 
 -*■ (2«a- 2 — 'daiff = what ? 
 
 Show that —- — or ± a/2, when cc = and y = 0. 
 eta v 
 
150 IXAMPhES. 
 
 21. Change the independent variable from x to t in 
 
 (1 — x 2 ) — — x ~r = 0, when x = cos ,'. 
 v ' dx 2 dx 
 
 22. Change the independent variable from x to in 
 
 % + IT~Al + (T+^p = °' wllcn * = tan "• 
 
 d*y 
 
 Am - dip + y = °* 
 
 23. Change the independent variable from # to r, and 
 eliminate x, y, dx and </y, between 
 
 . xdy — ydx a . . a 
 
 l = j . ; , # = r cos 0, and y = r sin 0. 
 
 Arts, t = —7 — 
 dr 
 
 24. Change the independent variable from x to 2 in 
 
 p = z% % - 2x % + y % ,vhen *= f mi » = ^ a > 
 
 ^M. P = *[g + (* +l)^ + v(v + 2)]. 
 
CHAPTER VIII. 
 
 MAXIMA AND MINIMA OF ^UNCTIONS OF A 
 SINGLE VARIABLE. 
 
 92. Definition of a Maximum and a Minimum.— if, 
 while the independent variable increases continuously, a 
 function dependent on it increases up to a certain value, 
 and then decreases, the value at the end of the increase is 
 called a maximum value of the function. 
 
 ^f while the independent variable increases, the function 
 decreases to a certain value and then increases, the value at 
 the end of the decrease is called a minimum value of che 
 function. Hence, a maximum value of a function of 
 a single variable is a value which is greater than the 
 immediately pi^eceding and succeeding values, and a 
 minimum value is less than the immediately pre- 
 ceding and succeeding values. 
 
 For example, sin increases as increases till the latter 
 reaches 90°, after which sin decreases as increases ; 
 that is, sin is a maximum when <f> is 90°, since it is 
 greater than the immediately preceding and succeeding 
 values. Also, cosec decreases as increases till the latter 
 reaches 90°, after which cosec increases as increases ; 
 that is, cosec is a minimum when is 90°, since it is less 
 than the immediately preceding and succeeding values. 
 
 93. Condition for a Maximum or Minimum. — If y 
 
 be any function of x, and y be increasing as x increases, 
 the differential of the function is positive (Art. 12), and 
 
 hence the first derivative ~ will be positive. If the func- 
 tion be decreasing as x increases, the differential of the 
 
152 
 
 a EOMETRIC ILL USTRA TION. 
 
 function is negative, and hence the first derivative ~- will 
 , 6 dx 
 
 be negative. 
 
 Therefore, since at a maximum value the function 
 
 changes from increasing to decreasing, the first- derivative 
 
 must change its sign from plus to minus ; as the variable 
 
 increases. And since, at a minimum value, the functio" 
 
 changes from decreasing to increasing, the first derivative 
 
 nuist change its sign from minus to plus. But as a function 
 
 which is continuous* can change its sign only by passing 
 
 through or oo , it follows that the only values of the 
 
 variable corresponding to a maximum or a minim u in 
 
 value of the function, are those which make the first 
 
 derivative Ooroo. 
 
 94. Geometric Illustra- 
 tion. — This result is also 
 evident from geometric con- 
 siderations ; for, let y = f(x) 
 be the equation of the curve 
 AB. At the points P, P', P", 
 P lT , the tangents to the curve 
 are parallel to the axis of x, 
 and therefore at each of these points the first derivative 
 /' (x) = 0, by Art. 56°. 
 
 We see that as x is increasing and y approaching a 
 maximum value, as PM, the tangent to the curve makes 
 an acute angle with the axis of x ; hence, approaching F 
 
 At P the tangent becomes parallel 
 
 dit 
 
 to the axis of x ; hence, -~ = 0. Immediately after pass- 
 ing P the tangent makes an obtuse angle with the axk 
 
 Fig- 9. 
 
 from the left -,- is -f . 
 dx 
 
 of x ; hence, 
 
 dy 
 dx 
 
 is — . 
 
 * In this discussion the function is to be regarded as continuous. 
 
CRITERION OF MAXIMA AND MINIMA. 153 
 
 Also in approaching a minimum value, as P'M', from the 
 left, we see that the tangent makes an obtuse angle with 
 
 the axis of x, and hence -y- is — . At the point F, -f- = 0. 
 dx * ' fa 
 
 After passing P', the angle is acute and -~ is -f. 
 
 In passing P'", -M changes sign by passing through oo , 
 
 P'"M'" is a minimum ordinate. In approaching it from the 
 left the tangent makes an obtuse angle with the axis of x, 
 
 and hence -~ is — . At P"' the tangent is perpendicular 
 to the axis of x, and -~ = oo . After passing P"'M'", the 
 
 angle is acute and ~ is +. 
 
 dy 
 dx 
 
 While the first derivative can change its sign from + to 
 — or from — to + only by passing through or oo , it 
 does not follow that because it is or oo, it therefore 
 necessarily changes its sign. The first derivative as the 
 variable increases may be -+- , then 0, and then + , or it may 
 be — , then 0, and then — . This is evident from Fig. 9, 
 where, at the point D, the tangent is parallel to the axis of 
 
 x, and — is 0, although just before and just after it is — . 
 
 Hence the values of the variable which make -— = or oo , 
 
 dx 
 
 are simply critical* values, i. e., values to be examined. 
 
 As a maximum value is merely a value greater than that 
 
 which immediately precedes and follows it, a function may 
 
 have several maximum values, and for a like reason it may 
 
 have several minimum values. Also, a maximum value 
 
 may be equal to or even less than a minimum value of the 
 
 same function. For example, in Fig. 9, the minimum P'M' 
 
 is greater than the maximum P iv M iv . 
 
 * See Price's Cal., Vol. I, p. 237, 
 
151 CONDITION OIVEN BY TAYLOR'S THEOREM, 
 
 95. Method of Discriminating between Maxima 
 and Minima. — Since the first derivative at a maximum 
 
 state is 0, and at the immediately succeeding state is — , it 
 follows that the second derivative, which is the difference 
 between two consecutive first derivatives,* is — at a maxi- 
 mum. Also, since the first derivative at a minimum state 
 is 0, and at the immediately succeeding state is +, it fol- 
 lows that the second derivative is + at a minimum. There- 
 fore, for critical values of the variable, a function is at a 
 maximum or a minimum state according as its 
 second derivative at that state is — or +. 
 
 96. Condition for a Maximum or Minimum given 
 by Taylor's Theorem. — Let u =f(x) be any continuous 
 function of one variable ; and let a be a value of x corre- 
 sponding to a maximum or a minimum value of f(x). 
 Then if a takes a small increment and a small decrement 
 each equal to h, in the case of a maximum we must have, 
 for small values of h, 
 
 /(«) > f(a + h) and f(a) > f(a - h) j 
 and for a minimum, 
 
 /(«) < /(« + *) and /(«) < f(a - h). 
 Therefore, in either case, 
 
 f(a + h)-f(a) and f(a-h)-f(a) 
 have both the same sign. 
 
 By Taylor's Theorem, Art. 66, Eq. 7, and transposing, 
 we have 
 
 /(« + *)-/(«) = f («)*+/" (a) J + etc.; (1) 
 f(a - h) -f(a) = -/' (a) h +/" (a) J - etc. . (2) 
 
 * Remembering that the first value is always to be subtracted from the second.. 
 
FINDING MAXIMA AND MINIMA VALUES. 155 
 
 Now if li be taken infinitely small, the first term in the 
 second member of each of the equations (1) and (2) will be 
 greater than the sum of all the rest, and the sign of the 
 second member of each will be the same as that of its first 
 term, and hence f(a + //) —f(a) and f(a — h) —f{ci) 
 cannot have the same sign unless the first term of (1) and 
 (2) disappears, which, since h is not 0, requires that 
 f(a) = 0. 
 
 Hence, the values of x which maJxe f(x) a maxi- 
 mum or a minimum are in general roots of the equa- 
 tion, /' (x) = 0. 
 
 Also, when /' (a) = 0, the second members of (1) and 
 (2), for small values of h, have the same sign as /"(«); 
 that is, the first members of (1) and (2) are both positive 
 when /" (a) is positive, and negative when /" (a) is nega- 
 tive. Therefore, /(a) is a maximum or a minimum 
 according as f" (a) is negative or positive. 
 
 If, however, /" (a) vanish along with /' (a), the signs of 
 the second members of (1) and (2) will be the same as 
 /'" (a), and since f" (a) has opposite signs, it follows that 
 in this case f(a) is neither a maximum nor a mini- 
 mum unless f" (a) also vanish. But if /'" (a) = 0, 
 then f{a) is a maximum when / iv (a) is negative, and a 
 minimum when f iy (a) is positive, and so on. If the first 
 derivative which does not vanish is of an odd order, /(«) is 
 neither a maximum nor a minimum ; if of an even order, 
 f(a) is a maximum or a minimum, according as the sign 
 of the derivative which does not vanish is negative or posi- 
 tive. 
 
 97. Method of Finding Maxima and Minima 
 Values. — Hence, as the result of the preceding investiga- 
 tion we have the following rule for finding the maximum 
 or minimum values of a given function, f(x). 
 
 Find its first derivative, f (x) pat it equal to 0, 
 and solve the equation thus formed, /' (x) = 0. Sub- 
 
15G MAXIMA A.\l> Ml MM a VALUES ALTBRNATS, 
 
 stitnte the rallies of x (hits found for x in the second 
 derivative, f" (x). Each value of x which makes 
 the second derivative urn,, tire will, when substituted 
 in the function f (x) make it a maximum ; and each 
 value which makes the second derivative positive will 
 make t/ie function a minimum. If either value of 
 X reduces the second* derivative to 0, substitute in the 
 third, fourth, etc., until a derivative is found which 
 does not reduce to 0. // this be of an odd order, the 
 ral tie of x will not make the function a maximum 
 or minimum ; but if it be of an even order and nega- 
 tive, the function will be a maximum ; if positive, a 
 minimum. 
 
 Second Rule. — It is sometimes more convenient to 
 ascertain whether a root a of f" (a?) = corresponds to a 
 maximum or a minimum value of the function by substi- 
 tuting for x, in /" (x), a — li and a -f //, where h is infini- 
 tesimal. // the first result is -f and the second is — , 
 a coTTesponds to a maximum ; if the first result is 
 — and the second is +, it corresponds to a minimum. 
 If both results have the same sign, it corresponds to 
 neither a maximum nor a minimum. (See Arts. 93, 94.) 
 
 98. Maxima and Minima Values occur alternately. 
 
 — Suppose that f(x) is a maximum when x = a, and also 
 when x = b, where b>a; then, in passing from a to b, 
 when x = a + h (where h is very small), the function is 
 decreasing, and when x = b — h, it is increasing; but in 
 passing from a decreasing to an increasing state, it must 
 pass through a minimum value ; hence, between two maxi- 
 ma one minimum at least must exist. 
 
 In the same way, it may be shown that between two 
 minima one maximum must exist. 
 
 This is also evident from geometric considerations, for in 
 Fig. 9 we see that .there is a maximum value at P, a mini- 
 mum at P', a maximum at P", a minimum at P'", and so on. 
 
APPLICATIONS OF AXIOMATIC PRINCIPLES. 15? 
 
 99. The Investigation of Maxima and Minima is 
 often facilitated by the following Axiomatic Prin- 
 ciples : 
 
 1. If u be a maximum or minimum for any value of x, 
 and a be a positive constant, an is also a maximum or mini- 
 mum for the same value of x. Hence, before applying the 
 rule, a constant factor or divisor may be omitted. 
 
 2. If any value of x makes u a maximum or minimum, 
 it will make any positive power of u a maximum or mini- 
 mum, unless u be negative, in which case an even power of 
 a minimum is a maximum, and an even power of a maxi- 
 mum is a minimum. Hence, the function may be raised 
 to any power ; or, if under a radical, the radical may 
 be omitted. 
 
 3. Whenever u is a maximum or a minimum, hgu is a 
 maximum or minimum for the same value of x. Hence, 
 bo examine the logarithm of a function we have only 
 to examine the function itself. When the function con- 
 sists of products or quotients of roots and powers, its exam- 
 ination is often facilitated by passing to logarithms, as the 
 differentiation is made easier. 
 
 4. When a function is a maximum or a minimum, its 
 reciprocal is at the same time a minimum or a maximum; 
 this principle is of frequent use in maxima and minima. 
 
 5. If u is a maximum or minimum, u ± c is a maximum 
 or minimum. Hence, a constant connected by + or — 
 maybe omitted. 
 
 Other transformations are sometimes useful, but as they 
 depend upon particular forms which but rarely occur, they 
 may be left to the ingenuity of the student who wishes to 
 simplify the solution of the proposed problem. 
 
 It is not admissible to assume x = go in searching for 
 maxima and minima, for in that case x cannot have a suc- 
 ceeding value. 
 
158 EXAMPLES. 
 
 EXAMPLES. 
 
 / 1. Find the values of x which will make the function 
 f u = (jz + 'dx 2 — 4a? a maximum or minimum, and the cor 
 responding values of the function a. 
 
 Here ^ = 6 4- Gx - 12a?. 
 
 Now whatever values of # make u a maximum or mini- 
 
 dii 
 mum, will make -r- = (Art. 97) ; therefore, 
 iix 
 
 6 + 6a; — 12a: 2 = 0, or x 2 — \x = \\ 
 .*. « = i±i = +1 or -i 
 Hence, if ti have maximum or minimum values, they 
 must occur when x — 1 or — £. 
 
 To ascertain whetheFthese values are maxima or minima, 
 we form the second derivative of u\ thus, 
 
 ^ = 6-24*. 
 dx 2 
 
 cPu 
 When x = l, -j-^ = — 18, which corresponds to a maxi- 
 mum value of u. , 
 
 When a: — — J, -j~ 2 = + 18, which corresponds to a 
 
 minimum value of u. 
 \ Substituting these values of x in the given function, we 
 have 
 
 When x — I, u =6 + 3 — 4 = 5, a maximum. 
 
 When x— — \, n = — 3 -f f -f £ = — J? a minimum 
 
 2. Find the maxima and minima values of u in 
 
 u — x* — 8a? + 22a: 2 — 24a: + 12. 
 
 \ 
 
 ™ _ 4^3 _ 24.^2 + 44a; _ 24 = 0, 
 da; ' 
 
 J 
 
EXAMPLES. 159 
 
 or, a? — 6z 2 + 11a; — 6 = 0. 
 
 By trial, x — 1 is found to be a root of this equation ; 
 therefore, by dividing the first member of this equation by 
 x — 1, we find for the depressed equation, 
 
 x 2 — ox + 6 = ; .-. x = 2 or 3. 
 
 Hence the critical values are x = 1, x = 2, and x = 3. 
 
 ?JS = 12a* _ 48a; + 44 = + 8, when x = 1. 
 
 ax 1 i 
 
 = — 4, when a; = 2. 
 
 = + 8, when x = 3. 
 Therefore we have, 
 
 when a; = 1, u = 3, a minimum ; 
 
 when a; = 2, w = 4, a maximum ; 
 
 when x = 3, m = 3, a minimum. 
 
 3. Find the maxima and minima values of u in 
 w = (x - l) 4 (» + 2) 3 . 
 
 *!L = 4 (x - l) 3 (a; + 2) 3 + 3 (a; + 2f(x - If 
 
 ClX 
 
 - {x - 1)3 (x + 2)2 [4 (s + 2) + 3 (a; - 1)], 
 dx 
 
 or ^= (a;-l) 3 (a; + 2)2(^ + 5) = 0; (1) 
 
 .\ (x - I) = 0, (a; + 2) = 0, (7a; + 5) = 0. 
 
 ,\ *a? = 1, a; = — 2, and x = — 4j as the critical values 
 of & 
 
 In this case, it will be easier to test the critical values by 
 
 du 
 the second rule of Art. 97; that is, to see whether -^ 
 
 changes sign or not in passing through x = 1, — 2, and 
 — 4. in succession. 
 
160 EXAMPLES. 
 
 If we substitute in the second member of (1), (1 — h) 
 and (1 -\ //) for a:, where h is infinitesimal, we get 
 
 g = (i -h - 1)3(1 -u + a)»[7(i-A) + 5] 
 
 as -A3(3«*)»(ia-7/ 4 ) s _. 
 
 and ^ a (1 +*- 1) 3 (1 + * + ^[7(1 + A) + 5] 
 
 = A»(3-M)'(12 + 7*) = + 
 
 Therefore, as -,— changes sign from — to + at x = 1, 
 
 the function w at this point is a minimum. 
 
 du 
 When $ = — 2, -y- does not change sign ; /. u has no 
 
 clx \ * •"" 
 
 maximum or minimum at this point. 
 
 When x = — f , -5- changes sign from -f- to — ; /. u, 
 at this point, is a maximum. 
 Hence, when x = l, u = 0, a minimum. 
 
 124.93 
 when a; = — 4> w = — 77"" » a maximum. 
 
 It is usually easy to see from inspection whether -=- 
 
 changes sign in passing through a critical value of x, with- 
 out actually making the substitution. 
 
 4. Examine u = b -f (x — a) z for maxima and minima. 
 — = 3 (x — a) % = ; ,\ a? = a, and w = £. 
 
 Since a; = a makes -5-5 = 0, we must examine it by the 
 
 du 
 second rule of Art. 97, and see whether , changes sign at 
 
 x = a. 
 
EXAMPLES. 161 
 
 — = 3 (a — h — a) 2 = 3h 2 is the value of j- immediate^ 
 preceding x — a. 
 
 — = 3(a + h — a) 2 = 3h 2 is the value of -j- immediately 
 succeeding x = a. 
 
 Therefore, as y- does not change sign at x z= a, u = b 
 is neither a maximum nor a minimum. 
 
 5. Examine u = b + (x — a) 4 for maxima and minima. 
 
 du 
 
 -j- = 4 (x — of = ; •*• x = a and u — b. 
 
 (XX 
 
 It is easy to see that -r~ changes sign from — to f at 
 z-=.a\ .*. x = a gives u = b, a minimum. 
 
 (x 4- 2) 3 
 
 6. Examine u == ) rr= for maxima and minima. 
 
 (a; — dy 
 
 du (x + 2) 2 (x-13) n 
 
 — = - — ! — — -— = or oo : 
 
 dx (x-3f ' 
 
 •\ x = — 2, 13, or 3. 
 We see that when x = — 2, y does not change sign ; 
 
 .*. no maximum or minimum ; 
 
 du 
 when x = 13, -=- changes sign from — • to ■+• ; 
 
 dx 
 
 •\ a minimum ; 
 when x = 3, -y- changes sign from -f to — , 
 
 A a maximum; 
 hence when x = 13, w = *-p, a minimum value ; 
 and when a; = 3, u = oo , a maximum value. 
 
162 EXAMPLES. 
 
 7. Examine u = b + (x — a)^ for maxima and minima 
 
 — — §(z — a)$ = 0; /. x = a and w = #. 
 
 When x = a, -r- changes sign from — to -f . 
 
 x = a gives u = b, a minimum. 
 
 8. Examine M = b — (a — x)* for maxima and minima. 
 
 — = 4(«-rf = 0; /. x = a and u = 5. 
 da? ° v 
 
 (J,U 
 
 When x= a, -j- changes sign from + to — . 
 .\ a; = # gives u = b, a maximum. 
 
 9. Examine w = b + \/# 3 — 2a 2 x + aa; 2 for maxima and 
 minima. 
 
 If u is a maxima or minima, w — b will be so ; therefore 
 we omit the constant b and the radical by Art. 99, and get 
 
 u' = a? — 2«2a; + aa*; 
 -=— = — 2a 2 4- 2ax = 0; .: x = a and w = #. 
 
 When x = a, -T- changes sign from — to +. 
 .•. x = a gives u = b 9 a minimum. 
 
 a 2 x 
 
 10. Examine u = -, c- Q for maxima and minima. 
 
 (a — x) 2 
 
 Using the reciprocal, since it is more simple, and omitting 
 the constant a 2 (Art. 99), we have 
 
 (a — x) 2 a* , 
 
 u' = J '- = 2a + x; 
 
 x x y 
 
 du' a 2 , , , dV 2a 2 
 
 A <fc = -^ +1 = °' and ^ = ^5 
 
EXAMPLES. 163 
 
 d?u' , 2 
 .% a? = ± 0, and .*. -3-5 = ± — 
 
 - 1 - ' dx 2 a 
 
 Hence, x = + a makes u' a minimum, and x = — a 
 makes it a maximum; therefore, since maxima and minima 
 values of u' correspond respectively to the minima and 
 maxima values of u (Art. 99, 4), we have, 
 
 when x = a, u = oo , a maximum. 
 
 a . . 
 
 " x = — a, u = — T > a minimum. 
 4 
 
 i<Y?^ £7&e values of x which give maximum and 
 minimum values of the following functions: 
 
 1. u = X s — 3a; 2 — 24a; + 85. 
 
 Ans. x = — 2, max. ; a; = 4, min. 
 
 2. w = 2a? — 21a; 2 + 36a; — 20. 
 
 x = 1, max.; a? = 6, min. 
 
 3. u = a; 3 — 18a; 2 + 96a; — 20. 
 
 a; = 4, max. ; x = 8, min. 
 
 4. u = ^ ^r--- x = i«, min 
 
 a — 2a; 
 
 1 + 3a; . A 
 
 5. u = L a; = — 1^, max. 
 
 V4 + 5x 
 
 6. w = a? — 3a; 2 — 9a; -f- 5. 
 
 x = — 1, max. ; a; = 3, min. 
 
 7. u = x 9 — 3a; 2 + 6a; + 7. 
 
 Neither a max. nor a min. 
 
 8. u = (a; — 9) 5 (a; — 8) 4 . 
 
 a; = 8, max. ; x == 8$, min 
 
 9. m = ^ : a; = cos a;, max. 
 
 1 + x tan x 
 
 10. w = sin 3 a; cos a;, x = 60°, max. 
 
 ■a* Sin 3/ ,1 fO 
 
 11. u = — — t a; = 45 , max. 
 
 1 •}- tan a; 
 
164 
 
 QEOMETh'K ' PROBLEMS, 
 
 12. u = sin x -f cos x. 
 
 x =. 45°, max. ; x = 225 , min. 
 
 13. u = 
 
 log X 
 
 X n 
 
 x = e t max. 
 
 GEOMETRIC PROBLEMS. 
 
 The only difficulty in the solution of problems in maxima and 
 minima consists in obtaining a convenient algebraic expression for the 
 function whose maximum or minimum value is required. No gen- 
 eral rule can well be given by which this expression can be found. 
 Much will depend upon the ingenuity of the student. A careful ex- 
 amination of all the conditions of the problem, and tact in applying 
 his knowledge of principles previously learned in Algebra, Geometry, 
 and Trigonometry, with experience, will serve to guide him in form- 
 ing the expression for the function. After reducing the expression to 
 its simplest form by the axioms of Art. 99, he must proceed as in 
 Art, 97. 
 
 1. Find the maximum cylinder which 
 can be inscribed in a given right cone 1 ' 
 with a circular base. 
 
 Suppose a cylinder inscribed as in 
 the figure. Let AO = b, DO = a, 
 CO = x, CE = y. 
 
 Then, denoting the volume of the 
 cylinder by v, we have 
 
 v = ny 2 x. 
 
 E, 
 
 t 
 
 
 H 
 
 ■•'"" 
 
 c ^^/ 
 
 N \ 
 
 Fig. 10. 
 
 (i) 
 
 From the similar triangles DOA and DCE, we have 
 DO : AO :: DC : EC, 
 or a i h : : a ■ — x : y ; 
 
 which in (1) gives 
 
 v = - - ( a — x) 2 x. 
 
 (2) 
 
GEOMETRIC PROBLEMS. 
 
 Dropping constant factors (Art. 99), we have 
 u = (a — xf x = a 2 x — 2ax 2 + & ; 
 
 0, 
 
 Ja 2 ; .; x = a or \a. 
 
 la + Qx 
 
 
 or # 2 — f aa; 
 
 dx 2 
 
 = 2#, when x = a, 
 
 .'. minimum 
 
 2«, when x = \a, .'. maximum. 
 
 Hence the altitude of the maximum cylinder is one- third 
 of the cone. 
 
 The second value of x in (2) gives 
 
 tt& . . ^ a rs> 
 
 Volume of cone = \-nab 2 . 
 .«. Volume of cylinder = $ volume of cone. 
 
 y = - {a — J«) = |d = radius of base of cylinder. 
 
 2. What is the altitude of the 
 maximum rectangle that can be in- 
 scribed in a given parabola ? 
 
 Let AX = a, AH = x, DH = y, 
 and A = area of rectangle. Then 
 we have 
 
 A = 2y(a — x). - 
 
 But from the equation of the parabola, we have 
 
 y = V2px, 
 
 tfhich in (1) gives A = %*/%px {a — x). 
 
 u' = V% (a — x) = ax* — x%, 
 
 du' 
 
 • 
 
 dx 
 
 \ax~^ — \x? = 0. 
 
 \ x = £#. 
 
166 OBOMBTRlC r/:o/i/j:MS. 
 
 Since this value of x makes -=- change sign from -f- to 
 
 — , it makes the function A a maximum; therefore the 
 altitude of the maximum rectangle is fa. 
 
 3. What is the maximum cone that 
 can be inscribed in a given sphere? 
 
 Let ACB be the semicircle, and J 
 ACD the triangle which, revolved 
 about AB, generate the sphere and 
 cone respectively. Let AO = r, AD = x, and CD = y, 
 and v = volume of cone. 
 
 Then v = ^y 2 x. (1) 
 
 But y* = AD x DB = (2r — x).x, 
 
 which in (1) gives f == \n (2r — x) x 2 , (2) 
 
 or u = 2rx 2 — x* ; 
 
 A ^ = irx — 3x* = 0. 
 dx 
 
 ,\ x = and $ r. 
 
 du 
 The latter makes -7- change sign from + to — ; .*. n 
 
 makes v a maximum. 
 
 Hence the altitude of the maximum cone is § of the 
 
 diameter of the sphere. 
 
 The second value of x in. (2) gives 
 
 v = in (2r - $r) ftr)» = ff nr* = & x *7rr». 
 
 Volume of sphere = -f ^r 3 ; 
 
 .*. the cone = fc of the sphere. 
 
 4. Find the maximum parabola which can be cut from a 
 given right cone with a circular base, knowing that the area 
 of a parabola is f the product of its base and altitude. 
 
GEOMETRIC PROBLEMS. 
 
 16? 
 
 Let AB ^«,AC = b, and BH = x 
 then AH = a — x. 
 
 FE = 2EH = 2VAH"xTBH 
 
 = 2a/(« — #)#. 
 Also, BA:AO :: BH : HD, 
 
 or 
 
 a : b 
 
 x: RD = -x. 
 
 a 
 
 Calling the parabola A, we have 
 
 Fig. 13. 
 
 A = |FE x HD V = t-xV(a - x) x 9 
 
 or 
 
 du 
 dx 
 
 ax 5 — x*. 
 Sax 2 — ±x* 
 
 x = and x 
 du 
 
 o; 
 
 The second value makes ~r- change sign from + to — , 
 and .-. makes the function A a maximum. 
 
 A A = |.-.|«V(« - p) \a = lab\/3, 
 which is the area of the maximum parabola. 
 
 Rem. — In problems of maxima and minima, it is often more con- 
 venient to express the function u in terms of two variables, x and y, 
 which are connected by some equation, so that either may be regarded 
 as a function of the other. In this case, either variable of course may 
 be eliminated, and u expressed in terms of the other, and treated by , 
 the usual process, as in Examples 1, 2, and 3 It is often simpler, 
 however, to differentiate the function u, and the equation between x 
 and y, with respect to either of the variables, x, regarding the other, 
 
 y, as a function of it, and then eliminate the first derivative, -^. The 
 
 dx 
 
 second method of the following example will illustrate the process. 
 
168 
 
 GEOMETRIC PROBLEMS, 
 
 5. To find the maximum rectangle inscribed in a given 
 
 ellipse. 
 
 Let CM = x, PM = y, and 
 A = area of rectangle. Then we 
 have 
 
 A = ±xy, (1) 
 
 and ahf + Fx* = aW. (2) 
 
 F 
 
 
 
 
 A[ 
 
 Iff 
 
 c / M 
 
 u 
 
 
 
 
 J 
 
 
 
 
 
 1st Method. — From (2) we get 
 
 y 
 
 -Va* 
 a 
 
 x*, 
 
 Fig. 14. 
 
 which in (1) gives A = 4 - x "s/a 2 — x 2 , 
 
 or 
 
 u = a 2 x 2 — a 4 . 
 ~ = 2a 2 x — 42 s 
 
 0. 
 
 x= ± 
 
 V* 
 
 x = -\ — — makes -r- change sign from -J^ to — ; .*. it 
 makes A a maximum. 
 
 Hence, the sides of the maximum rectangle are a a/2 
 and o V%, and the area is 2ab. 
 
 2d Method. — Differentiate (1) and (2) with respect to x 
 after dropping the factor 4 from (1), and get 
 
 dA 
 dx 
 
 , + *g=0; 
 
 2a 2 y C -^ + 2b 2 x 
 
 dy _ 
 dx ~ 
 
 X 
 
 dy _ 
 
 Vx 
 
 dx 
 
 a 2 y 
 
 .\ -=- = -. or b 2 x 2 = a 2 y 2 ; 
 d l y x' u ' 
 
 which in (2) gives 
 
GEOMETRIC PROBLEMS. 169 
 
 2a 2 // 2 =3 aW\ .'. y = -~ and x = -~. 
 
 6. Find the cylinder of greatest convex surface that 
 
 can be inscribed in a right circular cone, whose altitude 
 
 is h and the radius of whose base is r. a , nhr 
 
 Surface = -jp- 
 </ 
 
 7. Determine the altitude of the maximum cylinder 
 which can be inscribed in a sphere whose radius is r. 
 
 Altitude = \r V3. 
 
 8. Find the maximum isosceles triangle that can be 
 inscribed in a circle. An equilateral triangle. 
 
 9. Find the area of the greatest rectangle that can be 
 inscribed in a circle whose radius is r. 
 
 The sides are each r V2. 
 
 10. Find the axis of the cone of maximum convex sur- 
 face that can be inscribed in a sphere of radius r. 
 
 The axis = $r. 
 
 11. Find the altitude of the maximum cone that can be 
 inscribed in a paraboloid of revolution, whose axis is a, the 
 vertex of the cone being at the middle point of the base of 
 the paraboloid. Altitude = %a. 
 
 12. Find the altitude of the cylinder of greatest convex 
 surface that can be inscribed in a sphere of radius r. 
 
 Altitude = r V%. 
 
 13. From a given surface s, a vessel with circular base 
 and open top is to be made, so as to contain the greatest 
 amount. Find its dimensions. (See Remark under Ex. 4.) 
 
 The altitude = radius of base = y — • 
 
 14. Find the maximum cone whose convex surface is 
 constant. The altitude = V% times the radius of base: 
 
 15. Find the maximum cylinder that can be inscribed in 
 
 an oblate spheroid whose semi-axes are a and b. 
 
 - 2 
 
 The radius of base = a Vf J the altitude = b —-• 
 
 V3 
 
170 GEOMETRIC PROBLEMS. 
 
 16. Find the maximum difference between the sine .and 
 cosine of any angle. When bhe angle = 135°. 
 
 17. Find the number of equal parts into which a must 
 be divided so that their continued product may be a 
 maximum. 
 
 Let x be the number of parts, and thus each part equals 
 
 -, and therefore u = (-) , from which we get x = - ; 
 
 therefore each part = e, and the product of all = (e)' e . 
 
 18. Find a number x such that the xth root shall be a 
 maximum. x = e =z 2.71828 -f-. 
 
 19. Find the fraction that exceeds its m* power by the 
 greatest possible quantity. / 1 \^i T 
 
 \m) 
 
 20. A person being in a boat 3 miles from the nearest 
 point of the beach, wishes to reach in the shortest time a 
 place 5 miles from that point along the shore ; supposing 
 he can walk 5 miles an hour, but row only at the rate of 
 4 miles an hour, required the place he must land. 
 
 One mile from the place to be reached. 
 
 21. A privateer wishes to get to sea unmolested, but has 
 to pass between two lights, A and B, on opposite head- 
 lands, the distance between which is c. The intensity of 
 the light A at a unit's distance is a, and the intensity of B 
 at the same distance is h ; at what point between the lights 
 must the privateer pass so as to be as little in the light as 
 possible, assuming the principle of optics that the intensity 
 of a light at any distance equals its intensity at the distance 
 one divided by the square of the distance from the light. 
 
 x 
 
 a? + rf 
 
 22. The flame of a candle is directly over the centre of a 
 circle whose radius is r ; what ought to be its height above 
 the plane of the circle so as to illuminate the circumfer- 
 ence as much as possible, supposing the intensity of the 
 
GEOMETRIC PR OBL EMS. 
 
 171 
 
 liglit to vary directly as the sine of the angle under which 
 it strikes the illuminated surface, and inversely as the 
 square of its distance from the same surface. 
 
 Height above the plane of the circle = rW\» 
 
 23. Find in the line joining the centres of two spheres, 
 the point from which the greatest 
 portion of spherical surface is 
 visible. 
 
 The function to be a maximum 
 is the sum of the two zones whose 
 altitudes are AD and ad; hence 
 we must find an expression for the areas of these zones. 
 
 Put CM = R and cm = r, Gc = a and CP = x. 
 
 The area of the zone on the sphere which has R for its 
 radius (from Geometry, or Art. 194) = 2rrRAD s= 2ttR 2 
 
 / R 3 \ 
 — 2rrRCD = 277 ( R 2 -J, and in the same way for the 
 
 other zone, from which we readily obtain the solution. 
 
 Fig. 15. 
 
 X = 
 
 Bl 
 
 3 
 
 + r* 
 
 24. Find the altitude of the cylinder inscribed in a sphere 
 of radius r, so that its whole surface shall be a maximum. 
 
 Altitude == 
 
 -n>-m 
 
CHAPTER IX 
 
 TANGENTS, NORMALS AND ASYMPTOTES. 
 
 100. Equations of the Tangent and Normal.— Let 
 
 P, (x\ y') be the point of tangency ; 
 the equation of the tangent line at 
 (x' f y') will be of the form (Anal. 
 Geom., Art. 25) 
 
 y — y' = a(x-x'), (1) 
 in which a is the tangent of the 2 
 
 M 
 Fig. 16. 
 
 angle which the tangent line makes )» 
 
 with the axis of x. It was shown in ' 
 
 Article 56« that the value of this tangent is equal to the 
 
 derivative of the ordinate of the point of tangency, with 
 
 r aspect to x, 
 
 dx r 
 
 or 
 
 He?jo-e 
 
 y-y 
 
 ■g(— '>> 
 
 m 
 
 is the equation of the tangent to the curve at the point 
 (x', y'), x and y being the current co-ordinates of the 
 tangent. 
 
 Since the normal is perpendicular to the tangent at the 
 point of tangency, its equation is, from (2), 
 
 y 
 
 dx' 
 'dy 
 
 -,(*-*')• 
 
 (3) 
 
 (Anal Geom., Art. 27, Cor. 2.) 
 
EXAMPLES. 173 
 
 Rem. — To apply (2) or (3) to any particular curve, we 
 
 d?/ dx' 
 substitute for —-, or -r-, , its value obtained from the equa- 
 tion of the curve and expressed in terms of the co-ordinates 
 of the point of tangency. 
 
 E X A M P LES. 
 
 1. Find the equations of the tangent and normal to the 
 
 ellipse 
 
 a*f + VW = aW. 
 
 Wefind dx=-tfy> " dx~'=~a^ 
 
 and this value in (2) gives, 
 
 Px' , „ 
 
 y-y = -^-*> ; 
 
 which by reduction becomes, 
 
 ahjy' + &xx' = a?b\ 
 which is the equation of the tangent; and 
 
 y-y =w {x ~ x) 
 
 is the equation of the normal 
 
 2. Find the equations of the tangent and normal to the 
 parabola tf = %px. 
 
 We find ~- = -, .*. -j-j = -,, 
 
 dx y ax y 
 
 and this value in (2) gives 
 
 y-y' = | ( x ~ x ')> 
 
 pr yy' -y' 2 = px—px\ 
 
 But y' % = 2pa! \ 
 
174 EXAMPLES, 
 
 ••• yy' = />(■* + x'\ 
 
 which is the equation of the tangent ; and 
 
 y-y' = -^(a-aO 
 is ///e equation of the normal. 
 
 3. Find the equations of the tangent and normal to an 
 hyperbola. 
 
 Tangent, a 2 yy' — b 2 xx' = — a 2 b 2 . 
 
 Normal, y — y' = - ^| (a; - *'). 
 
 4. Find the equation of the tangent to 3y 2 + x 2 — 5 = ? 
 at x = 1. 
 
 dv' x' 1 
 
 Here -^-, = — —-, = - — =F .29 about, 
 
 c?x' 3y ± 3.465 
 
 which in (2) gives 
 
 ^ =F 1.155 = =F.29(*-1), 
 
 or y = =f .29* ± 1.44. 
 
 Hence there are two tangents to this locus at x = 1, 
 their equations being 
 
 y = — .29* + 1.44 and y = -f .29* — 1.44. 
 
 5. Find the equation of the tangent to the parabola 
 y 1 = 9*, at * =: 4. 
 
 At (4, 6) the equation is y = f* + 3. 
 " (4, — 6) " " " y — — \x — 3. 
 
 6. Find the equation of the normal to y 2 = 2x 2 — X s , at 
 x = \. 
 
 At (1, +1) the equation is y = — 2x + 3. 
 « (i ? _ i) « « « y — ; 2x - 3. 
 
 7. Find the equation of the normal to y 2 = 6* — 5, at 
 v = 5, and the angle which this normal makes with the 
 axis of *. y = — f* + ^ ; angle = tan -1 (— |). 
 
LENGTH OF TANGENT, NORMAL, ETC. 
 
 175 
 
 101. Length of Tangent, Normal, Subtangent, 
 Subnormal, and Perpendicular on the Tangent from 
 the Origin. 
 
 Let PT represent the tangent at 
 the point P, PN the normal ; draw 
 the ordinate PM ; then 
 
 MT is called the subtangent, 
 
 MN " " " subnormal. 
 
 Let a = angle PTM; 
 
 then tan a = -p (Art. 56a). 
 
 7 1 
 
 % 
 
 Fig. 17. 
 
 1st. 
 
 2d. 
 
 3d. 
 
 TM = MP cot a = y 
 
 Subtangent = y 
 
 dy'' 
 
 dy'> 
 
 MN = MP tan MPN = y'tan «; 
 
 Subnormal = \j —-, 
 
 PT = VPM 2 + MT 2 
 
 Tangent = y'y 1 + \^) ' 
 
 4th, 
 
 PN = V PM 2 + MN 2 
 
 ■•■ Normal = y'^l+^ ■ 
 5th. The equation of the tangent at P {x',y') is (Art. 100) 5 
 
176 EXAMPLES, 
 
 *-# = %&-*)> 
 
 or xdy' — ydx — x'dy' + y'dx' = ; 
 
 which, written in the normal form, is 
 
 xdy'-ydx'-xdy ^fdx 
 V(dx')*+(dyy 
 
 i •^ y'dx' — x'dy' 
 
 hence, OD = ; —^ = = 0. 
 
 V(dx')* + (dy'f 
 
 .; Perpendicular on the tangent from the origin 
 
 y'dx' — x'dy' 
 
 = V(dxJ +W) V 
 
 Sch. — In these expressions for the subtangent and sub- 
 normal it is to be observed that the subtangent is measured 
 from M towards the left, and the subnormal from M towards 
 
 dy' 
 the right. If, in any curve, y' — is a negative quantity, it 
 
 denotes that N lies to the left of M, and as in that case 
 
 dx' 
 y' j-, is also negative, T lies to the right of M. 
 
 EXAM PLES, 
 
 1. Find the values of the subtangent, subnormal, and 
 perpendicular from the origin on the tangent, in the ellipse 
 a 2 y 2 + bW = aW. 
 
 tt dy* W 
 
 Here -/-, = 5 -,« 
 
 dx' a 2 y 
 
 Hence, the subtangent = y' -=-, = — -^- , 
 
 the subnormal = y' -?-, = „ x' ; 
 
 v dx a 2 ' 
 
EXAMPLES. 
 
 177 
 
 the perpendicular from origin on tangent 
 _ aW 
 
 " («Y 2 + Wx' 2 )^° 
 2. Find the subtangent and subnormal to the Cissoid 
 
 T = 
 
 2a — x 
 
 Here 
 
 (See Anal. Geom., Art. 149.) 
 
 dy' _ $b (3a — x) 
 M ' ~~ (2a - *)* " 
 
 Hence, the subtangent 
 
 the subnormal = 
 
 _ x (2a — x) 
 
 da — x 
 
 x 2 (3a — x) 
 (2a - xf ' 
 
 3. Find the value of the subtangent of y 2 = 3x 2 — 1% 
 at x = 4. Subtangent = d. 
 
 4. Find the length of the tangent to y 2 = 2x, at x — 8. 
 
 Tangent = 4VT?. 
 
 5. Find the values of the normal and subnormal to the 
 cycloid (Anal. Geom., Art. 156). 
 
 x =z r vers -1 - — \/2ry — y 2 ; V2^ 
 
 \ 
 
 ^ 
 
 dr is \/%ru iP- — L\ >^. 
 
 y \ 
 
 . «* # _ v *>' y y am c 
 
 > i 
 
 Fig. 18. 
 
 3 A' 
 
 rfy ^2ry-y 2 2r — y 
 
 dy _ 2r — y 
 d% \f2ru — u 2 
 
 
 .*. Subnormal = V2ry — f = MO. 
 
 Normal = \/%ry = PO. 
 
 It can be easily seen that PO is normal to the cycloid at 
 P ; for the motion of each point on the generating circle at 
 
L78 
 
 POLAR CURVES. 
 
 the instant is one of rotation about the poini of contact 0, 
 i.e., each point for an instant is describing an infinitely 
 small circular arc whose centre is at 0; and hence PO is 
 normal to the curve, i.e., the normal passes through the 
 foot of the vertical diameter of the generating circle. Also, 
 since OPH is a right angle, the tangent at P passes through 
 the upper extremity of the vertical diameter. 
 
 6. Find the length of the normal in the cycloid, the 
 radius of whose generatrix is 2, at y = 3. Normal = 2. 
 
 POLAR CURVES. 
 
 102. Tangents, Normals, Subtangents, Subnor- 
 mals, and Perpendicular on Tangents. 
 
 Let P be any point of the 
 curve APQ, the pole, OX the 
 initial line. Denote XOP by 
 0, and the radius-vector, OP, 
 by r. Give XOP the infinitesi- 
 mal increment POQ = dd, then 
 OQ = r + dr. From the pole 
 0, with the radius OP = r, de- 
 scribe the small arc PR, sub- 
 tending dd ; then, since dd = ab 
 is the arc at the unit's distance 
 from the pole 0, we have 
 
 PR = rdS and RQ = dr. 
 
 (1) 
 
 Let PQ, the element * of the arc of the curve, be repre- 
 sented by ds. 
 
 Ti7^2 
 
 or 
 
 PQ" = PR 2 + RQ 2 , 
 5? = r*~d& + 3?. (2) 
 
 Pass through the two points P and Q the right line QPT; 
 
 * See Art. 56a, foot-note. 
 
POLAR CURVES. 179 
 
 then, as P and Q are consecutive points, the line QP T is a 
 tangent to the curve at P (Art. 56a). Through P draw the 
 normal PC, and through draw COT perpendicular to OP, 
 and OD perpendicular to PT. The lengths PT and PC are 
 respectively called the polar tangent and the polar normal. 
 OC is called the polar subnormal; OT the polar subtangent ; 
 and OD, the perpendicular from the pole on the tangent, is 
 usually symbolized by p. The value of each of these lines is 
 required. 
 
 tanRQP = g = ^,from(l). (3) 
 
 Since OPT = OQT + dd, the two angles OPT and OQT 
 differ from each other by an infinitesimal, and therefore 
 OPT = OQT, and hence, 
 
 rdd 
 tan OPT = T -j-> from (3), (4) 
 
 sin OPT = sin OQP = ?? = ~, from (1). (5) 
 Hence, 
 OT = polar subtangent == OP tan OPT = -^ , 
 
 from (4). (6) 
 OC = polar subnormal = OP tan OPC == OP cot OPT 
 
 = |, from (4). (7) 
 
 PT = polar tangent = VOP 2 + OT 2 = r \/ 1 + r 2 ^, 
 
 from (6). (8) 
 
 PC = polar normal = VoP 2 + OC 5 * = \ r* + § 
 
 ao i 
 
 from (7). (9) 
 
180 
 
 KXAMPLES. 
 
 OD = p = OP sin OPD = ^ from (5) 
 
 ds 
 
 See Price's Calculus, Vol. I, p. 417. 
 
 Vr 2 d0*+dr*> 
 from (2). (10) 
 
 EXAMPLES. 
 
 1. The spiral of Archimedes, whose equation is r = «0. 
 (Anal. Geom., Art. 160.) 
 
 dO 1 „ , . r 2 
 
 Here 
 
 .\ Subt. = -, from (6). 
 dr a a x ' 
 
 Subn. =s a, from (7). 
 Tangent = r yl + - 2 , from (8), 
 Normal ss aA" 2 + a a , from (9). 
 
 p = -7======, from (10). 
 
 2. The logarithmic spiral r ss # 9 . 
 
 Art. 163.) 
 
 dv 
 Here -j- = a e log « = r log a ; 
 
 v 
 
 • Subt = :. = ???r, 
 
 log a 
 
 (where in is the modulus of the system 
 in which log a = 1). 
 
 (Anal. Geom., 
 
 Subn. 
 
 V 
 
 m 
 
 mr 
 
 Vi + h 
 
 Vm* + 1 
 
 tdr 2 ~~\* i 
 
 r 2 + -p = (r 2 + r 2 log 2 a)*. 
 
 Fig. 20 
 
RECTILINEAR ASYMPTOTES. 181 
 
 Tan. OPT == r di 
 
 dr log a ' 
 
 which is a constant ; and therefore the curve cuts every 
 radius-vector at the same angle, and hence it is called the 
 Equiangular Spiral. 
 
 If a = e, the Naperian base, we have, 
 
 tan OPT == .- 1 — = 1, and .\ OPT = 45°, 
 log e 
 
 and OT = OP = r. 
 
 3. Find the subtangent, subnormal, and perpendicular in 
 the Lemniscate of Bernouilli, r 2 = a 2 cos 26. (Anal. 
 Geom., Art. 154.) 
 
 Subtangent = 
 
 -r 3 
 
 a 2 sin W 
 
 Subnormal = — - sin 2d ; 
 r ' 
 
 Perpendicular = 
 
 Vr* + « 4 sin 2 20 a 
 
 4. Find the subtangent and subnormal in the hyper- 
 bolic* spiral rd = a. (Anal. Geom., Art. 161.) 
 
 Subt. = — a ; Subn. = • 
 
 a 
 
 RECTILINEAR ASYMPTOTES. 
 
 103. A Rectilinear Asymptote is a line which is 
 continually approaching a curve and becomes tangent to it 
 at an infinite distance from the origin, and yet passes 
 within a finite distance of the origin. 
 
 To find whether a proposed curve has an asymptote, we 
 must first ascertain if it has infinite branches, since if it 
 
 * This curve took its name from the analogy between its equation and that of 
 the hyperbola xy = a. (See Strong's Calculus, p. 145; also Young's Dif. Calculus, 
 p. 120.) 
 
182 RECTILINEAR ASYMPTOTES, 
 
 has not, there can be no asymptote. If it has an infinite 
 branch, we must then ascertain if the intercept on either 
 of the axes is finite. The equation of the tangent 
 (Art. 100) being, 
 
 if we make successively y = 0, x = 0, we shall find foi 
 the intercepts on the axes of x and y, the following : 
 
 dx 
 X °- X ~~ y dy' 
 
 (by putting x = x and y = y , and dropping accents), 
 
 dy 
 
 Now, if for x = oo both x and y are finite, they will 
 determine two points, one on each axis, through which an 
 asymptote passes. If for y = oo , x is finite and y infi- 
 nite, the asymptote is parallel to the axis of y. If for 
 x = oo , x Q is infinite and y finite, the asymptote is parallel 
 to the axis of x. If both x Q and y are infinite, the curve 
 has no asymptotes corresponding to x = oo. If both x 
 and y are 0, the asymptote passes through the origin, and 
 
 its direction is obtained by evaluating -y- for x = oo . 
 
 When there are asymptotes parallel to the axis, they may 
 usually be detected by inspection, as it is only necessary to 
 ascertain what values of x will make y = oo , and what 
 values of y will make x = oo . For example, in the equa- 
 tion xy — m, x = makes y = oo, and y = makes 
 x = co ; hence the two axes are asymptotes. Also in the 
 equation xy — ay — bx = 0, which may be put in. either 
 of the two forms, 
 
 bx ay 
 
 y = or x = — £-= ; 
 
 9 x — a y — b 
 
 y = go when x = a, and x = oo when y = b; 
 
&XAMPLUS. i83 
 
 hence the two lines x — a and y == I are asymptotes to 
 the curve. 
 In the logarithmic curve y = a* f 
 
 y = when x = — oo , 
 
 therefore the axis of x is an asymptote to the branch in the 
 second angle. 
 
 Also in the Cissoid y 2 = , 
 
 * 2a — x 
 
 y = oo when x = 2a; 
 hence # = 2a is an asymptote. 
 
 EXAM PLES 
 
 1. Examine the hyperbola 
 
 a 2 y 2 — W = — a 2 b 2 , for asymptotes. 
 
 Here 
 
 dy b 2 x a*y 2 a 2 _ . 
 
 -/■ = -=- ; .-. x = x — -rf- = - = for x = ± go . 
 
 ax a*y o 2 x x 
 
 b 2 x 2 b 2 
 
 y = y 5— = = f or y = ± oo . 
 
 v° v a 2 y y . 
 
 Hence the hyperbola has two asymptotes passing through 
 the origin. 
 
 ., dy b 2 x , b . 1 , b . 
 
 Also -f- = —- =z ± ■ = ± - f or x = oo . 
 
 dx a 2 y ~*~ a / 
 
 x l 
 
 Hence the asymptotes make with the axis of x an angle 
 whose tangent is ± - ; that is, they are the produced 
 diagonals of the rectangle of the axes. 
 
 2. Examine the parabola y 2 = 2px for asymptotes. 
 
1K4 ASYMPTOTES DETERMINED BY EXPANSION, 
 
 Here 
 
 -f- = - ; .-. x = — i - = — oo when a; or ?/ =s oo. 
 dx y' %p v 
 
 y = £ = go when y = oo or a: = oo . 
 
 Hence the parabola has no asymptotes. 
 The ellipse and circle have no real asymptotes, since 
 neither has an infinite branch. 
 
 3. Examine y s = ax 2 -f X s for asymptotes. 
 When z=±ao, y^^oo; .-. the curve has two 
 iii finite branches, one in the first and one in the third 
 angle. 
 
 dy _ %ax + Sx z m 
 dx ~ dy 2 ' 
 
 dy z ax 2 a 
 
 Xa = X 
 
 2ax + 3x 2 %az + 3z* 3' 
 
 when x = oo. 
 
 2ax 2 + 3a 3 '_ 3 (y 8 — g 8 ) — 2^ 
 
 y»-y- 3y2 - ■ - * 3*,* 
 
 , when # 
 
 3 (az 2 + a 3 }" 3 
 Hence the asymptote cuts the axis of # at a distance 
 
 — ^, and that of y at a distance ~ from the origin, and as 
 o o 
 
 it is therefore inclined at an angle of 45° to the axis of x, 
 
 its equation is 
 
 »-• + !' 
 
 (See Gregory's Examples, p. 153.) 
 
 104. Asymptotes Determined by Expansion. — A 
 
 very convenient method of examining for asymptotes con- 
 sists in expanding the equation into a series in descending 
 
EXAMPLES. 185 
 
 powers of x, by the binomial theorem, or by Maclaurin's 
 theorem, or by division or some other method. 
 
 EXAMPLES. 
 
 x^ -4- ctx 2 
 
 1. Examine if = for asymptotes. 
 
 x ~— a 
 
 Then 
 
 ±^(l + ^ + ^ + eto.) (1) 
 
 4 /x + a 
 
 = ±xV : 
 
 When x = oo (1) becomes 
 
 y = ± (a + a). (2) 
 
 We see that as # increases, the ordinate of (1) increases, 
 and when x becomes infinitely great, the difference between 
 the ordinate of (1) and that of (2) becomes infinitesimal; 
 that is, the curve (1) is approaching the line (2) and 
 becomes tangent to it when x = qo ; therefore, y = ±(#+«) 
 are the equations of two asymptotes to the curve (1) at 
 right angles to each other. 
 
 Another asymptote parallel to the axis of y is given hy 
 x = a. 
 
 2. Examine a? — xy 2 -f- ay 2 = for asymptotes. 
 Here y = ± y^ 
 
 Hence, y = ± \x + -- ) are the equations of the two 
 tisymptotes. 
 
 By inspection, we find that x = a is a third asymptote. 
 
 3. Examine y 2 = x 2 ~ z — - for asymptotes. 
 
18G EXAMPLES. 
 
 Here y = ± a?(l — ^ + etc.j 
 
 ••• # = ± Z are the two asymptotes. 
 
 105. Asymptotes in Polar Co-ordinates.- -When 
 the curve is referred to polar co-ordinates, there will be m 
 asymptote whenever the subtangent is finite for r = oc. 
 Its position also will be fixed, since it will be parallel to the 
 radius- vector. Hence, to examine for asymptotes, we find 
 what finite values of make r = oo ; if the corresponding 
 
 polar subtangent, ? ,2 -y-, which in this case becomes the 
 
 perpendicular on the tangent from the pole, is finite or zero, 
 there will be an asymptote parallel to the radius-vector. If 
 for r = oo the subtangent is oo , there is no corresponding 
 asymptote. 
 
 EXAMPLES. 
 
 1. Fiud the asymptotes of the hyperbola cPy* — Pa? = 
 *- aW by the polar method. 
 
 The polar equation is 
 
 a* sin 2 - b 2 cos 2 ft = — -j- (1) 
 
 ** 
 
 b 2 
 When r = oo , (1) becomes, tan 2 = — 2 
 
 Therefore the asymptotes are inclined to the initial line 
 ,ttan-(±^ 
 
 From (1) we get ^ = _ (5r __ 5 -_- g , 
 
 ana r atr _ ± (a 2 + i 2 ) sin cos ' w 
 
EXAMPLES. 187 
 
 which is equal to when 6 = tan~M ± -J ; hence both 
 asymptotes pass through the pole. 
 
 2. Find the asymptotes to the hyperbolic spiral rd = a. 
 (See Anal. Geom., Art. 161.) 
 
 Here r = -, .; r = oo, when 6 — 0. 
 
 u 
 
 dd a , 9 dd 
 
 -j- — 2 , and r 2 -y- = —a. 
 
 dr r l dr 
 
 There is an asymptote therefore which passes at a distance 
 a from the pole and is parallel to the initial line. 
 
 3. Find the asymptotes to the lituus rd? = a. (Anal. 
 Geom., Art. 162.) 
 
 Here r = ~, .*. r = oo , when = 0. 
 
 eh 
 
 M=-^, and **>= _2^ = 0, when = 0. 
 
 Therefore the initial line is an asymptote to the lituus. 
 
 4. Find the asymptotes of the Conchoid of Nicomedes, 
 r =p sec + m. (Anal. Geom., Art. 151.) 
 
 Here r = oo when = - ; and r 2 ^- = « when = -• 
 2 ' dr L 2 
 
 Therefore the asymptote cuts the initial line at right 
 angles, and at a distance ^? from the pole. 
 
 EXAMPLES. 
 
 1. Find the equation of the tangent to Si/ 2 — 2x>— 10 = 0. 
 at x = 4. ^4w». y = ± .7127a; ± .8909. 
 
 x 3 
 
 2. Find the equation of the tangent to y 2 = j , at 
 
 x = % 
 
 y = 2x — 2 and y = — 2x -f 2. 
 
188 EXAMPLES. 
 
 :;. Find the equation of the tangent to the Naperian 
 logarithmic curve. Ans. y = y' (x — x' -f 1). 
 
 At what point on y = X s — 3x 2 — 24a: + 85 is the 
 
 tangent parallel to the axis of a; ? 
 
 dv' 
 [Here we must put y -, = 0. See Art. 56a.] 
 
 At (4, 5) and (-2, 113). 
 5. At what point on y 2 = 2x 3 does the tangent make 
 with the axis of x an angle whose tangent is 3, and where is 
 it perpendicular? At (2, 4) ; at infinity. 
 
 G. At what angle does the line y = \x + 1 cut the carve 
 y 2 = \x ? [Find the point of intersection and the tangent 
 to the curve at this point; then find the angle between this 
 tangent and the given line.] 10° 14' and 33° 4'. 
 
 7. At what angle does y 2 = lOz cut x 2 + y 2 = 144 ? 
 
 71° 0' 58". 
 
 8. Show that the equation of a perpendicular from the 
 focus of the common parabola upon the tangent is 
 
 $r = - fa - 4rt- 
 
 9. Show that the length of the perpendicular from the 
 focus of an hyperbola to the asymptote is equal to the semi- 
 conjugate axis. 
 
 10. Find the abscissa of the point on the curve 
 
 y(x — l)(x — 2) = x — 3 
 at which a tangent is parallel to the axis of #. 
 
 x = 3 ± V%. 
 
 11. Find the abscissa of the point on the curve 
 
 f = (x — a) 2 (x — c) 
 
 at which a tangent is parallel to the axis of a-. 
 
 2c + a 
 
EXAMPLES. 189 
 
 X 
 
 12. Find the subtangent of the curve y 
 
 V2a — x 
 x (2a — x) 
 da — x 
 
 13. Find the subtangent of the curve y s —3axy + x* = 0. 
 
 2axy — X s 
 ay — x 2 
 
 14. Find the subtangent of the curve xy 2 = a 2 (a — x). 
 
 _ 2 (ax — x 2 ) 
 a 
 
 15. Find the subnormal of the curve y 2 = 2a 2 log x. 
 
 a 2 
 x 
 
 16. Find the subnormal of the curve day 2 + a 3 = 2a& 
 
 x 2 
 a 
 
 2 — 
 
 17. Find the subtangent of the curve y 
 
 a ~~~ x 
 
 2x (a — x) 
 
 3a — 2x~' 
 
 18. Find the subtangent of the curve 
 
 x 2 y 2 = (a + x) 2 (52 — x 2 ). 
 
 x (a + x) (b 2 — x 2 ) 
 x? + ab 2 
 
 19. Find the subnormal, subtangent, normal, and tangent 
 in the Catenary 
 
 y = l{ e ' +•"')■ 
 
 c l - ~-\ y 2 
 
 Subnormal = j[e c — e c I ; normal =- 
 
 Subtangent = — — : ; tangent 
 
 ifr 
 
 (y 2 - <*)* (y 2 - $Y 
 
190 EX AMP I 
 
 20. Find the perpendicular from the pole on the tangenl 
 in the lituus rd^ = a. 2a 2 r 
 
 P ~ (r4 + 4a 4)T 
 
 21. At what angle does y 2 = 2ax cut x 3 — 3axy+y s = ? 
 
 cot -1 ^4. 
 
 22. Examine y 2 = 2x -f 3a: 2 for asymptotes. 
 
 # as V3 a; + — - is an asymptote. 
 v3 
 
 23. Examine y 3 = Gx 2 + x 3 for asymptotes. 
 
 y = x + 2 is an asymptote. 
 
 24. Find the asymptotes of y 2 (x — 2a) = x 3 — a 3 . 
 
 x=2a; y = ± (x + a). 
 
 25 Find the asymptotes of y as — Z^^i 2 _±A 3 . 
 J * ^ ^ — 3^ + 2*2 
 
 x = b; x = 2b; y = x-3(a-b). 
 
CHAPTER X. 
 
 DIRECTION OF CURVATURE, SINGLE POINTS, 
 TRACING OF CURVES. 
 
 106. Concavity and Convexity. — The terms concav- 
 ity and convexity are used in mathematics in their ordinary 
 sense. A curve at a point is concave towards the axis of x 
 when in passing the point it lies between the tangent and 
 the axis. See Fig. 21. It is convex towards the axis of x 
 when its tangent lies between it and the axis. See 
 Fig. 22. 
 
 If a curve is concave down- 
 wards, as in Fig. 21, it is plain 
 that as x increases, a decreases, 
 and hence tan a decreases ; that 
 
 is, as x increases, ~ (Art. 56a) 
 
 decreases ; and therefore the de- 
 rivative of -— or ~ is negative. (Art. 12.) 
 
 In the same way if the curve is 
 convex downward, see Fig. 22, it is 
 plain that as x increases, a in- 
 creases, and therefore tan a in- 
 creases ; that is, as x increases, 
 
 - increases, and therefore the de- 
 
 Fig. 21. 
 
 d i or <?y 
 
 dx dx 2 
 
 dx 
 
 rivative of -,- or —^ is positive. 
 
 Hence the curve is concave or convex downward according 
 
 as ^ is - or +. 
 
192 
 
 POL AH CO-OR DIN A TES. 
 
 This is also evident from Fig. 23, where MM' = MM" 
 
 = dx; PF is common to the two curves and the common 
 
 tangent PR = PR' = dx\ and PR 
 
 = P,R'. But P "R > P,R' > P,R'. 
 
 Now P R and P"R' are consecutive 
 
 v.ihus of dy in the upper curve, 
 
 and Pit and PjR' are consecutive 
 
 values of dy in the lower curve, and 
 
 hence P "R' — PR = d (dy) = d 2 y is 
 
 +. , and P,R' — P'R = d 2 y is — ; that 
 
 is, d 2 y is — or -f-, according as the 
 
 curve is concave or convex downwards. 
 
 d?v 
 The sign of -~ 2 is of course the same as that of d 2 y, 
 
 since dx 2 is always positive. 
 
 We have supposed in the figures that the curve is above 
 the axis of x. If it be below the axis of x, the rule just 
 given still holds, as the student mav show by a course of 
 reasoning similar to the above. 
 
 d?y 
 
 If the curve is concave downwards, 
 
 dx 2 
 
 — ; if it be 
 
 above the axis of x, y is + ; therefore, y ~ is — ; if 
 
 the 
 
 d 2 y 
 
 curve be concave upwards, ~ is +; if it be below the 
 
 axis of x, y is — ; therefore, y - 
 
 is — when the curve is concave towards the axis of x. 
 
 d 2 y 
 the same way it may be shown that y ~ is +, when the 
 
 curve is convex towards the axis of x 
 
 ; that is, yg 
 In 
 
 dx 2 
 
 107. Polar Co-ordinates. — A curve referred to polar 
 co-ordinates is said to be concave or convex to the pole at 
 any point, according as the curve in the neighborhood of 
 that point does or does not lie on the same side of the tan 
 gent as the pole. 
 
EXAMPLES. 
 
 193 
 
 It is evident from Fig. 24, that when the curve is con- 
 cave toward the pole 0, as r increases'^ increases also, and 
 
 dv 
 therefore -j- is positive ; and if the curve is convex toward 
 
 the pole, as r increases p decreases, and 
 
 dv 
 therefore -=- is negative. If therefore 
 
 the equation of the curve is given in 
 terms of r and 6, to find whether the 
 curve is concave or convex towards the 
 pole, we must transform the equation 
 into its equivalent between r and p, by 
 
 means of (10) in Art. 102, and then find l ^- 
 
 dp 
 
 Fig. 24. 
 
 EXAMPLES. 
 
 1. Find the direction of curvature of 
 
 y 
 
 Here 
 
 ,_(*• 
 
 -l)(x- 3) 
 
 
 x — 2 
 
 cPy 
 
 dx> ~ 
 
 2 
 
 (x - 2) 3 ' 
 
 cPy 
 
 that is, -r^ is positive or negative, according as z< or >2; 
 
 and therefore the curve is convex downward for all values 
 of x < 2, and concave downwards for all values of x > 2. 
 
 2. Find the direction of curvature of 
 
 y = b -f- c (x + a) 2 and y = a 2 Vx 
 
 a. 
 
 Ans. The first is concave upward, the second is concave 
 
 towards the axis of x. 
 
 3. Find the direction of curvature of the lituus r = — . 
 9 
 
!!)! SINGULAR POINTS. 
 
 „ dr i\ )* 
 
 Here de = -w = -M> " 
 
 </,' r* ; 
 
 which in (10) of Art. 102 gives, 
 
 
 2a 2 r dr 
 
 (4^ + r*)* 
 
 Therefore the curve is concave toward the pole for values 
 of r < a V%, and convex for r > a a/2. 
 
 4. Find the direction of curvature of the logarithmic 
 spiral r = a e . 
 
 By Art 102, Ex. 2, 
 
 mr dr Vm 2 + 1 . 
 
 P """ vW+'l' " ^~ m " 
 
 which is always positive, and therefore the curve is always 
 concave toward the pole. 
 
 SINGULAR POINTS. 
 
 108. Singular Points of a curve are those points 
 which have some property peculiar to the curve itself, and 
 not depending on the position of the co-ordinate axes. 
 Such points are : 1st, Points of maxima and minima ordi- 
 nates ; 2d, Points of inflexion ; 3d, Multiple Points ; 4th, 
 Cusps ; 5th, Conjugate points ; 6th, Stop points ; 7th, 
 Shooting points. We shall not consider any examples of 
 the first kind of points, as they have already been illus- 
 trated in Chapter VIII, but will examine very briefly the 
 others. 
 
 109. Points of Inflexion. — A point of inflexion is a 
 
 point at which the curve is changing from convexity to 
 
 concavity, or the reverse ; or it may be defined as the 
 
 point at which the curve cuts the tangent at that point. 
 
 d 2 v 
 When the curve is convex downwards, y| is + (Art. 
 
EXAMPLES. 195 
 
 106), and when concave downwards, — | is ■— ; therefore, 
 
 ax 
 
 at a point of inflexion -~ is changing from + to — , or 
 
 from — to +, and hence it must be or oo . Hence to 
 
 d 2 y 
 find a point of inflexion, we must equate -~ to or oo , 
 
 and find the values of x ; then substitute for x a value a 
 little greater, and one a little less than the critical value ; 
 
 if -=-^ changes sign, this is a point of inflexion. 
 
 EXAM PLES. 
 
 1. Examine y = b + (x — a) 3 for points of inflexion. 
 Here g = 6(*-<») = 0; 
 
 .♦. x = a and hence y = b. 
 
 This is a critical point, i. e., one to be examined ; for if 
 there is a point of inflexion it is at x = a. For x > a, 
 
 -~ is +, and for x < a, ~ 2 is — . Hence there is a point 
 
 CIX (IX 
 
 of inflexion at (a, b). 
 
 2. Examine the witch of Agnesi, 
 
 x 2 y = 4« 2 (2a — y) } 
 for points of inflexion. 
 
 There are points of inflexion at ( ± — — , -— !• 
 
 ^ V3 z ' 
 
 3. Examine y = b -j- (x — «) 5 for points of inflexion. 
 
 There is a point of inflexion at (a, b). 
 
 4. Examine the lituus for points of inflexion. 
 
 By Art. 107, Ex. 3, -j- is changing sign from + to — 
 when r = a V2, indicating that the lituus changes at this 
 
196 METHOD OF FINDING MULTIPLE POINTS. 
 
 point from concavity to convexity, and hence there is a 
 point of inflexion at r = a y% 
 
 110. Multiple Points. — A multiple point is a point 
 through .which two or more branches of a curve pass. If 
 two branches meet at the same point, it is called a doubU 
 point; if three, a triple point ; and so on. There are two 
 kinds : 1st, a point where two or more branches intersect, 
 their several tangents at that point being inclined to each 
 other ; and 2d, a point where two or more branches are 
 tangent to each other. The latter are sometimes called 
 points of osculation. 
 
 As each branch of the curve has its tangent, there will 
 be at a multiple point as many tangents, and therefore as 
 
 many values of -~ as there are branches which meet in 
 
 this point. If these branches are all tangent, the values of 
 
 -,- will be equal. At a multiple point y will have but one 
 
 value, while at points near it, it will have two or more 
 
 values for each value of x. In functions of a simple form, 
 
 such a point can generally be determined by inspection. 
 
 After finding a value of x for which y has but one value, 
 
 and on both sides of which it has two or more values, form 
 
 civ 
 .-• If this has unequal values, the branches of the curve 
 
 intersect at this point, and the point is of the first kind. If 
 
 ~ has but one value, the branches are tangent to eaei 
 
 (IX 
 
 other at this point, and the point is of the second kind. 
 
 When the critical points are not readily found by inspec- 
 tion, we proceed as follows: 
 
 Let f(x,y) = (I) 
 
 1 e the equation of the locus freed from radicals. Then 
 
EXAMPLES. 
 
 197 
 
 dx 
 
 du 
 dx 
 
 dy 
 
 and as differentiation never introduces radicals when they 
 do not exist in the expression differentiated, the value of 
 
 -4- cannot contain radicals, and therefore cannot have sev- 
 ax 
 
 eral values, unless by taking the form 
 
 Hence we have 
 
 ' ° 
 
 dy du f . , du . . 
 
 -f-t=z- or -=- = 0, and -=- = 0, irom 
 dx dx dy 
 
 which to determine critical values of x and y. If these 
 
 values of x and y found from -=- = and — = are real 
 3 dx dy 
 
 and satisfy (1), they may belong to a multiple point. If y 
 
 has but one value for the corresponding value of x, and on 
 
 both sides of it y has two or more real values, this point is a 
 
 multiple point. We then evaluate 
 
 dy_0 
 dx~0 
 dy 
 
 , and if there 
 
 are several real and unequal values of -j- , there will be as 
 
 many intersecting branches of the curve passing through 
 the point examined. 
 
 EXAMPLES. 
 
 1. Determine whether the curve y = (x — a) yx + b has 
 a multiple point. 
 
 Here y has two values for every 
 positive value of x > or < a. When 
 x = or a, y has but one value, b; 
 hence there are two points to be ex- 
 amined. When x < 0, y is imagi- 
 nary ; hence the branches do not 
 pass through the point (0, b), and 
 
 Fig. 25. 
 
108 EXAMPLES. 
 
 therefore it is not a multiple point. When x > or < a, ) 
 has two real values, and therefore (a, b) is a double point 
 
 dy , 3x — a ,- , 
 
 -—- = ± = ± ya, when x = a. 
 
 ax ^yx 
 
 Therefore the point is of the first kind, and the tangents 
 to the curve at the point make with the axis of x angles 
 whose tangents are + Va and — Vet. 
 
 2. Examine x* + 2ax 2 y — ay 3 = for multiple points. 
 
 We proceed according to the second method, as all the 
 critical points in this example are not easily found by inspec- 
 tion. 
 
 du 
 
 ^ = lx(x> + ay) = 0; 
 
 (i) 
 
 g = «(^-3^) = 0; 
 
 (2) 
 
 dy 4.T 3 4- Aaxy 
 dx ~ 3ay 2 — 2ax* 
 
 (3) 
 
 Solving (1) and (2) for x and y, we find 
 
 (x = ()\ Ix = $aVQ\ (x s= — iaVl\ 
 
 Only the first pair will satisfy the equa- 
 tion of the curve, and therefore the ori- 
 gin is the only point to be examined. 
 
 Evaluating -f- in (3) for x = and 
 cix 
 
 y = 0, and representing ~ byjt?, and ~- ' 
 
 by p', for shortness, we have 
 
CUSPS. 
 
 199 
 
 dy _ _ &a? + 4=axy 
 
 dx= P = 
 
 
 0' 
 
 day 2 — 2ax 2 
 
 12x 2 + 4ay + 4:axp _ 
 
 when 
 
 Sayp - 4ax 
 
 _ 24# + Sap + laxp' 
 ~ 6ap 2 + 6ayp' — 4=a 
 
 - - , when 
 
 /s = 0\ 
 
 Vy = 0/ 
 
 P = °) 
 
 W = o/ 
 
 6ap 2 — ±a f 
 .: p (6ap 2 
 
 when 
 
 /* = 0\ 
 
 \t/ = 0/ 
 
 4#) = 8tf/> 
 
 i> = 
 
 rf# 
 
 0, + V%, or — V^. 
 
 Hence the origin is a triple point, the branches being in- 
 clined to the axis of x at the angles 0, tan -1 ^^), and 
 tan -1 ( — V%), respectively, as in the figure. (See Courte- 
 nay's Calculus, p. 191 ; or Young's Calculus, p. 151.) 
 
 3. Examine y 2 — x 2 (1 — x 2 ) = for multiple points. 
 Ans. There is a double point at the origin, the branches 
 
 being inclined to the axis of x at angles of 45° and 135° 
 respectively. 
 
 4. Show that ay % —x?y—ax % = has no multiple points. 
 
 111. Cusps. — A cusp is a point of a curve at which two 
 branches meet a common tangent, and 
 stop at that point. If the two branches 
 lie on opposite sides of the common tan- 
 gent, the cusp is said to be of the first 
 species ; if on the same side, the cusp is 
 said to be of the second species. 
 
 Since a cusp is really a multiple point 
 of the second kind, the only difference 
 being that the branches stop at the point, 
 instead of running through it, we exam- 
 ine for cusps as we do for multiple points; and to distin- 
 
 Fig. 27. 
 
200 cusps. 
 
 gnish a cusp from an ordinary multiple point, we trace the 
 curve in the vicinity of the point and sec if y is real on one 
 side and imaginary on the other. To ascertain the kind of 
 cusp, we compare the ordinates of the curve, near the point, 
 with the corresponding ordinate of the tangent; or ascertain 
 the direction of curvature by means of the second derivative. 
 In the particular case in which the common tangent to 
 the two branches is perpendicular to the axis of x, it is best 
 to consider y as the independent variable, and find the 
 
 values of -j- , etc. 
 dy' 
 
 EXAMPLES. 
 
 1. Examine y = x 2 ± x* for cusps. 
 
 We see that when x = 0, y has but one value, 0; when 
 x < 0, y is imaginary ; and when x > 0, y has two real 
 values; hence, (0, 0) is the point to be examined. 
 
 -^ = %x ± \x% = 0, when x = ; hence the axis of x is 
 
 a common tangent to both branches, and 
 there is a cusp at the origin. 
 
 ■~ = 2±^^ is positive when x = 0; 
 
 hence the cusp is of the second kind. 
 
 d?ii 
 The value of -^ shows that the upper 
 
 branch is always concave upward, while the lower branch 
 has a point of inflexion, when x = ■$£% ; from the origin to 
 the point of inflexion this branch is concave upward, after 
 which it is concave downward. 
 
 The value of -^ shows that the branch is horizontal 
 dx 
 
 when x = ££ . From y = x 2 — x% we find that the lower 
 
 branch cuts the axis of it' at x = 1. The shape of the curve 
 
 is given in Fig. 28. 
 
Fig.29 
 
 CONJUGATE POINTS. 201 
 
 2. Examine (y — h) 2 = (x — #) 3 for cusps. 
 
 Ans. The point («, &) is a cusp of the first kind. 
 
 3. Examine cy 2 = x z for cusps. 
 
 The origin is a cusp of the first kind. 
 
 112. Conjugate Points. — A conjugate point is an iso 
 lated point whose co-ordinates satisfy the equation of the 
 curve, while the point itself is entirely detached from every 
 other point of the curve. 
 
 For example, in the equation y = (a + x)\/x, if x is 
 negative, y is, in general, imaginary but for the particular 
 value x = — a, y = 0. Hence, P is a 
 point in the curve, and it is entirely 
 detached from all others. When x = 0, 
 y = 0, which shows that the curve p 
 
 passes through the origin. For positive 
 values of x, there will be two real values 
 of y, numerically equal, with opposite 
 signs. Hence, the curve has two infinite branches on the 
 right, which are symmetrical with respect to the axis of x. 
 
 If the first derivative becomes imaginary for any real 
 values of x and y, the corresponding point will be conjugate, 
 as the curve will then have no direction. It does not fol- 
 
 (lii 
 low, however, that at a conjugate point ~ ^*fl De imagi- 
 nary; for, if the curve y =f(x) have a conjugate point at 
 (x, y), from the definition of a conjugate point, we shall 
 have 
 
 f(%± h) = an imaginary quantity. But 
 
 *, . zx , dyh , $yW , d*yh* , , 
 
 therefore, if either one of the derivatives is imaginary, the 
 first member is imaginary; hence, at a conjugate point 
 some one or more of the derivatives is imaginary. 
 
 Since at a conjugate point some of the derivatives are 
 imaginary, let the n th derivative be the Jlrst that is imagi- 
 
EXAMPLES, 
 
 nary. Suppose the equation of the curve to be freed from 
 
 radicals, and denoted by u =f(x, y) = 0. Take the n lh 
 derived equation (Art. 88, Sch.) ; we have 
 
 du d n y d n u _ 
 
 dydx n '''' + d^~ ' 
 
 where the terms omitted contain derivatives of u with re- 
 spect to x and y, and derivatives of y with respect to x, of 
 
 du 
 lower orders than the n th . If, then, ~r- be not 0, the value 
 
 ay 
 
 d"v 
 of -r* obtained from the derived equation will be real, 
 ax 
 
 which is contrary to the hypothesis; hence, -=- = is a 
 
 necessary condition for the existence of a conjugate point. 
 But 
 
 du du dy _ 
 
 dx dy dx " 
 
 therefore, since -=- = 0, we must have -=- = 0. Hence, at 
 dy dx 
 
 a conjugate point we must have -j- = 0, and — = 0. 
 
 Rem.— Owing to the labor of finding the higher derivatives, it is 
 usually better, if the first derivative does not become imaginary, to 
 substitute successively a + h and a — h for x, in the equation of the 
 curve, where a is the value of x to be tested, and h is very small. If 
 both values of y prove imaginary, the point is a conjugate point. 
 
 EXAMPLES. 
 
 1. Examine ay 2 — X s + Aax 2 — 5a 2 x + 2a 3 = for con- 
 jugate points. 
 
 ^ = - %x 2 + Sax - 5a 2 = 0. (1) 
 
 I * %* = 0. (2) 
 
SHOOTING POINTS.— STOP POINTS. 203 
 
 Solving (1) and (2), we get 
 
 (x =z a\ , /x = £a\ 
 
 ( y =o) and {,j=iy 
 
 Only the first pair of values satisfies the equation of the 
 curve, and hence the point (a, 0) is to be examined. 
 
 d]L - - 3a a — Sax + 5a 2 _ Gx — 8 a 
 dx ~~ ^ ~ "lay "Zap 
 
 1 (x = a\ 
 
 = , when ( ); 
 
 p \y = 0/ 
 
 therefore, p 2 = — 1 ; ,\ p z= £ V — 1 = -^-. 
 
 This result being imaginary, the point (a, 0) is a conju- 
 gate point. 
 
 2. Show that a 4 — ax 2 y — axy 2 + a 2 y 2 = has a conju. 
 gate point at the origin. 
 
 3. Examine (c 2 y — x z ) 2 = (x — «) 5 (x — b) G for conjugate 
 points, in which ay b. 
 
 b* 
 
 The point Lb, —J is a conjugate point. 
 
 The first and second derivatives are real in this example ; hence the 
 better method of solving it will be to proceed according to the Remark 
 above given 
 
 113. Shooting Points are points at which two or more 
 branches of a curve terminate, without having a common 
 tangent. 
 
 Stop Points are points in which a single branch of a 
 curve suddenly stops. 
 
 These two classes of singular points but rarely occur, and 
 never in curves whose equations are of an algebraic form. 
 
204 
 
 EXAMPLES. 
 
 EXAMPLES. 
 
 1. Examine y = - L for shooting points. 
 
 Here 
 
 dy 
 dx 
 
 Am — ~ 1 + 
 
 1 + e* z(l+e*) 
 
 If x is + and small, y is + ; if 
 # is — and small, y is — . When 
 a is -f and approaches 0, y = 0, 
 
 and -^ = : when a is 
 ate 
 
 and ap- 
 
 Fig. 30. 
 
 proaches 0, y = 0, and -p 
 
 Hence, at the origin there is a shooting point, one branch 
 having the axis of x as its tangent, and the other inclined 
 to the axis of x at an angle of 45°. (See Serret's Calcul 
 Differentiel et Integral, p. 267. ) 
 
 2. Examine y = x log x. 
 
 When a; is -f, y has one real value ; when x = 0, y = ; 
 when x < 0, y is imaginary ; hence there is a stop point at 
 the origin. 
 
 3. Examine y 
 
 If 
 
 = x tan -1 — 
 x 
 
 *$L = tan-i 1 L, 
 
 ofc # iC 2 + 1 
 
 + or 
 
 0, # = 0; 
 
 dy _ n 
 dx ~~ 2 
 
 or 
 
 Hence the origin is a shooting point, the tangent being 
 inclined to the axis of x at angles tan -1 (1.5708) and 
 tan" 1 (- 1.5708). 
 
 4. Show that y = e x has a stop point at the origin. 
 
TRACING CURVES, 205 
 
 114. Tracing Curves.— We shall conclude this chap- 
 ter by a brief statement of the mode of tracing curves by 
 means of their equations. 
 
 The usual method of tracing curves consists in assigning 
 a series of different values to one of the variables, and cal- 
 culating the corresponding series of values of the other, thus 
 determining a definite number of points on the curve. By 
 drawing a curve or curves through these points, we are 
 enabled to form a tolerably accurate idea of the shape of the 
 curve. (See Anal. Geometry, Art. 21.) 
 
 In the present Article we shall indicate briefly the man- 
 ner of finding the general form of the curve, especially at 
 such points as present any peculiarity, so that the mind can 
 conceive the locus, or that it may be sketched without 
 going through the details of substituting a series of values, 
 as was referred to above. 
 
 To trace a curve from its equation, the following steps 
 will be found useful : 
 
 (i.) If it be possible, solve it with respect to one of its 
 variables, y for example, and observe whether the curve is 
 symmetrical with respect to either axis. 
 
 {2.) Find the points in which the curve cuts the axes, 
 also the limits and infinite branches. 
 
 (3. ) Find the positions of the asymptotes, if any, and at 
 which side of an asymptote the corresponding branches lie. 
 
 (4-.) Find the value of the first derivative, and thence 
 deduce the maximum and minimum points of the curve, the 
 angles at which the curve cuts the axes, and the multiple 
 points, if any. 
 
 (5.) Find the value of the second derivative, and thence 
 the direction of the curvature of the different branches, and 
 the points of inflexion, if any. 
 
 (6.) Determine the existence and nature of the singular 
 points by the usual rules. 
 
206 EXAMPLES. 
 
 EXAMPLES. 
 
 1. Trace the curve y = 5 « 
 
 * 1 + x* 
 
 When x = 0, y = ; .*. the curve passes through the 
 origin. 
 
 For all positive values of x, y is positive ; and when 
 x = qo , y = 0. For negative values of jc, ?/ is negative, and 
 when x = — oo , y = ; hence the curve has two infinite 
 branches, one in the first angle and one in the third, and the 
 axis of x is an asymptote to both branches. 
 
 ty - 1 ~ x2 . &y _ % x ( x * — 3) 
 
 dx~ (1+ a?)* 5 M ~~ (1 + Z 2 ) 3 " 
 
 dy 
 When a; = ± 1, -p = J .*. there is a maximum ordinate 
 
 at jc = + 1, and a minimum ordinate at a; = — 1, at 
 which points y = J and — £ respectively. 
 
 When x = 0, -p = 1 ; .*. the curve cuts the axis of x at 
 an angle of 45°. 
 
 Putting the second deriva- 
 tive equal to 0, we get x = 
 or ± a/3. Therefore, there 
 are points of inflexion at (0,0) 
 and at x = + a/3 and — a/3, 
 for which we have y = J a/3, 
 — £a/3. From x = — a/3 
 to a? = + a/3, the curve is concave towards the axis of x, 
 and beyond them it is convex. 
 
 .From this investigation the curve is readily constructed, 
 and ha3 the form given in the figure. 
 
 2. Trace the curve y % = %a& — x % . 
 
 y = x* (2a — x)* ; 
 
dy 
 dx 
 
 Aax 
 
 EXAMPLES, 
 
 3x\ d*y _ --8a 2 
 
 207 
 
 3tf 
 
 dx* 
 
 9# 3 (2a — x)* 
 
 When x = or 2#, «/ = 0; .*. the curve cuts the axis of 
 x at the origin and at x = 2rt. 
 To find the equation of the asymptote, we have 
 
 2a 
 ~ -) ~ Tx 
 
 -.(.-^=-4-!—-)' 
 
 therefore, y = — x + f « is the equation of the asymptote, 
 and as the next term of the expression is positive, the curve 
 lies above the asymptote. 
 
 Evaluating the first derivative for x = 0, y = 0, we have 
 dy _ Aax — 3a; 2 _ Aa — 6x < 
 
 % 
 
 ^ 
 o^ 
 
 /. (^ J = — = oo , when x = «/ = : 
 «fy _ /2a fc ft 
 
 " ^~ V% = ±00 ' y 
 
 Hence, at the origin there are 
 two branches of the curve tangent 
 to the axis of y ; and the value of 
 
 ^ shows that if y be negative as it 
 
 dx 
 
 dy 
 
 approaches 0, -~ will be imaginary ; 
 
 and hence the origin is a cusp of 
 
 the first species. 
 
 dy_ 
 
 dx 
 at x = fa. 
 
 When x = 2a,^=-^- = -<*>; » the curve cuts 
 the axis of x, at the point x = 2a, at right angles. 
 
 When x 
 
 %a, ^f- = ; .-. there is a maximum ordinate 
 
208 
 
 EXAMPLES, 
 
 Tatting the second derivative equal to oo, we get x = 2a. 
 When x < 2d, the second derivative is — , and when > 2a 
 it is -+- ; hence the left branch is everywhere concave down- 
 ward, and the right branch is concave downward from x = 
 to x = 2a. At this last point it cuts the axis of x at right 
 angles, and changes its curvature to concave upward; the 
 two branches touch the asymptote at x = + oo and — oo , 
 respectively, i. e., they have a common asymptote. 
 
 In the figure, OA = 2a, OB = \a, OC as fa. 
 
 3. Trace the curve y = x \ )• 
 
 J \x — af 
 
 Let x = ; .\ y = 0. 
 
 x < a; .*. y is positive. 
 
 x = a; # = <». 
 
 a? > a < 2a ; «/ is negative. 
 
 x = 2a; y = 0. 
 
 # > 2a ; y is positive. 
 
 & = go; y = go. 
 
 When # is — , ?/ is always negative. 
 
 To find the asymptote, we have 
 2a 
 
 Fig. 32.a 
 
 y 
 
 = # 
 
 i — 
 
 = 4-!)( 1 + ^etc.) 
 
 A a 2a 2 \ a 2 
 
 = # 1 1 — etc. ) = x — - a 
 
 V x x 2 I x 
 
 etc. 
 
 .♦. y = x — a is the equation of the asymptote. 
 
 Hence, take OB = a = OD, and the line BD produced 
 is the asymptote; also take OC = 2a. Then, since y = 0, 
 both when a; = and x = 2«, the curve cuts the axis of 3 
 
EXAMPLES. 209 
 
 at and C. Between and B, the curve is above the axis ; 
 at B the ordinate is infinite ; from B to 0, the curve is 
 below; from C to infinity, it is above OX. Also, if x is 
 negative, y is negative; therefore the branch on the left of 
 is entirely below the axis. 
 
 dy x 2 — 2ax + 2a 2 
 
 Also, 
 
 dx (x — a) 2 
 
 Let x = a ; .*. -~ = oo ; and the infinite ordinate at the 
 
 distance a to the right of the origin is an asymptote. 
 
 If x = 0, -*- = 2; if x = 2a, -f- = 2 ; i. e., the curve 
 
 cuts the axis of x at the origin and the distance 2a to the 
 right, at the same angle, tan -1 (2). 
 
 If x 2 — 2ax + 2a 2 or (x — a) 2 + a 2 = 0, x is impossible ; 
 hence there is no maximum or minimum ordinate. 
 
 Again, 
 
 &y _ 
 
 2{x 
 
 -a) 2 
 
 -2[(x- 
 
 a) 2 
 
 + 
 
 a 2 } 
 
 dx 2 
 
 (x- 
 
 2a 2 
 -af> 
 
 (x - af 
 
 
 
 
 (Pit 
 ,: y| is + if x < a, and is — if x > a. 
 
 But x < a, y is + ; and x > a < 2a, y is — ; and 
 x > 2a, y is -f- ; therefore, from to B, and B to C, the 
 curve is convex, and from C to infinity, it is concave to the 
 axis of x. 
 
 If x be — , ~ = t r^ is +, but y is — ; therefore 
 
 dx 2 (x + a) 3 J 
 
 the branch from the origin to the left is concave to the axis 
 
 of x. (See Hall's Calculus, pp. 182, 183.) 
 
 4. Trace the curve y 2 = a 2 x s . 
 
 The curve passes through the origin ; is symmetrical 
 with respect to the axis of x; lias a cusp of the first kind at 
 
210 TRACING POLAR CURVES. 
 
 the origin; both branches arc tangeni to the axis of x ; art 
 convex towards it ; are infinite in I he direction of positive 
 abscissas, and the curve has no asymptote or point of in- 
 flexion. 
 
 115. On Tracing Polar Curves.— Write the equation, 
 
 if possible, in the form r =f(6) ; give to 6 such values as 
 
 to make r easily found, as for example, 0, J^, "*, f n, etc. 
 
 (It 
 Putting -73 = 0, we find the values of 6 for which r is a 
 
 maximum or minimum, i. e., where the radius vector is 
 perpendicular to the curve. 
 
 Find the asymptotes and direction of curvature, and 
 points of inflexion. After this there will generally be but 
 little difficulty in finding the form of the curve. 
 
 EXAM PLES, 
 
 1. Trace the lituus r = -r- 
 
 When 6 = 0, r = oo ; when 6 = 1 (= 57°.3),* r = ± a ; 
 when (9 = 2 (= 114°.6), r = ± .Ha ; when 6 = 3, r = 
 ± .58rt, etc. ; when 6 = oo , r = 0. 
 
 civ t^ dv 
 
 ■jz = — s~5 » an( l when -j- = 0, r = ; hence, r and 
 
 are decreasing functions of each other throughout all their 
 values ; f and the curve starts from infinity, when 6 = 0, 
 and makes an infinite number of revolutions around the 
 pole, cutting every radius-vector at an oblique angle, and 
 reaching the pole only when 6 = oo . 
 
 The subtangent r 2 -j- = = 0, when r = oo ; hence 
 
 the initial line is an asymptote (Art. 105). 
 
 * The unit angle is that whose arc is equal to the radius, and is about 57°. 29578- 
 t If we consider alone the branch generated by the positive radius- vector. 
 
EXAMPLES. 211 
 
 di _ 2« 2 (4a 4 - _r*) , Ai± 1Q7 Ex 3 * hen(je there ig 
 
 dr (4^4 + r 4)f 
 a point of inflexion at r = a a/2 ; from r= to r = «_a/2 
 the curve is concave toward the pole, and from r = a\/2 to 
 r = oo it is convex. 
 
 2. Trace the curve r = a sin 30. 
 
 r = 0, when = 0, 60°, 120°, 180°, 240°, and 300°. 
 When = 2tt, or upwards, the same series of values recur. 
 
 If = 30°, 90°, 150°, 210°, 270°, and 330°, r = a, — a, 
 
 «, — a, a, and — a, successively. 
 
 dr 
 
 — = 3a cos 30, showing that r begins at when = 0, 
 
 (la 
 
 increases till it is a when = 30°, diminishes to as 
 
 passes from 30° to 60°, continues to diminish and becomes 
 
 — a when 6 becomes 90°, and so on. 
 
 -f- = , , which shows that 
 
 dr ( 9fl 2 _ gfaji 
 
 the curve is always concave towards the 
 pole. There is no asymptote, as r is 
 never oo . 
 
 Hence the curve consists of three 
 
 r ig. oo. 
 
 equal loops arranged symmetrically 
 around- the pole, each loop being traced twice in each revo- 
 lution of r. A little consideration will show that the form 
 of the curve is that given in the figure. (See Gregory's 
 Examples, p. 185 ; also Price's Calculus, Vol. I, p. 427.) 
 
 3. Trace the Chordel r = a cosec 
 
 UiJ' 
 
 If = 0, tin, 2mr, 3mt, 4=nn, bnn f etc., successively, 
 r = oo , a, oo , — a, — oo , a, etc. 
 
 dr a 0a , / 
 
 — = — — cosec — cot — = jr- cosec 2 5- 
 dO 2n In 2n 2n 2n 
 
 which Ms negative from = to d = mr, positive from 
 
212 
 
 KXAMl'LES. 
 
 = 7in to = Snn, negative from = 3«rr to = 5»tt, clc 
 Benoe pre Bee that r begins at oo when = 0; diminishes 
 till it becomes a when 6= nn; increases as passes from 
 nn to 2nn ; becomes oc when = 2wtt; when passes 2»7r, 
 r changes from + oo to — oo ; when increases from 2?m 
 to 3nn, r increases from — oo to — a; when increases 
 from 3wr to fam, r diminishes from — a to — oo; when 
 passes inn, r changes from — oo to + oo . When in- 
 creases beyond in, the same values of r recur, showing that 
 the curve is complete. 
 
 — x 
 
 Fig. 34. ""~""-^-— 
 
 ^coseo* ^ (-oob^-O gives = nn, Znn, 
 
 dr 
 
 d0~2n 
 
 hnn, etc. ; i. e., the radius- vector is a minimum at = nn, 
 dnn, hnn, etc. 
 
 rift 
 
 The subtangent = r 2 -^- = 
 
 and 
 
 2wa 
 
 6 
 
 C0S 2» 
 
 
 
 2/m when 
 
 0: 
 
 = 0; 
 
 2wa when 
 
 = 
 
 ± 2nn ; 
 
EXAMPLES. 213 
 
 therefore the curve has two asymptotes parallel to the initial 
 line, at the distances ± %na from the pole. 
 
 f z 2anr 
 
 *! + A* ( 4 «^ 2 - « 2 + r 2 )* 
 
 
 e?p _ 2a 3 n (4ra 2 — 1) 
 ^* ~ [a*(4n a ~l) +f*]*' 
 .*. the curve is always concave towards the pole. 
 
 Thus it appears that while 6 is increasing from to 2nn, 
 the positive end of the radius-vector traces the branch 
 drawn in Fig. 34; and while 6 increases from 2nn to 4w~, 
 the negative end of the radius-vector traces a second. branch 
 (not drawn), the two branches being symmetrical with 
 respect to the vertical line through the pole 0. 
 
 EXAMPLES. 
 
 1. Find the direction of curvature of the Witch of Agnesi 
 
 x % y = 4:0? (2a — y). 
 
 The curve is concave downward for all values of y between 
 2a and f «, and convex for all values of y between \a and 0. 
 
 2. Find the direction of curvature of y = h + (x — a) 3 . 
 
 Convex towards the axis of x from x > a to x = co ; and 
 from x — a — h^to x= — oo ; concave towards the axis 
 of x from x < a to x = a — b*. 
 
 3. Examine y = (a — x)* -f- ax for points of inflexion. 
 
 There is a point of inflexion at x = a. 
 
 4. Examine y = x + 36a; 2 — 2x 3 — x i for points of in- 
 flexion. Points of inflexion at x = 2, x = — 3. 
 
•Jl I I [MPLBS. 
 
 5. Find the co 
 
 -ordinates of the point of inflexion 
 
 i of the 
 
 curve 
 
 y = 
 
 x 2 (a 2 - a 2 ) 
 a 8 
 
 
 
 
 *=±^; ,= 
 
 A«- 
 
 G. Examine r -. 
 
 -02_] 
 
 : lor points of inflexion. 
 
 
 dr 2 
 Here ^ = 
 
 4r (r - 
 
 -«) 8 . 
 
 
 w 
 
 \ » 
 
 
 
 
 flr 2 
 
 /. etc. 
 
 ,\ p - 
 
 ' (4r 4 - 
 
 - 12a?- 3 + 13« 2 r 2 — 4«V)* ' 
 
 There are points of \ 
 
 inflexion at r = f a and r : 
 
 = $a. 
 
 7. Examine y 2 
 
 = (*- 
 
 Vfx for multiple points. 
 
 
 There is a multiple point at x = 1. 
 
 x 2 l a 2 x 2\ 
 
 8. Examine y 2 = — ^ a" ^ or mu ^iple points. 
 
 There is a multiple point at the origin, and the curve is 
 composed of two loops, one on the right and the other on 
 the left of the origin, the tangents bisecting the angles be- 
 tween the axes of co-ordinates. 
 
 9. Show that x* -f x 2 y 2 — Gatfy + a 2 y 2 = has a multiple 
 point of the second kind at the origin. 
 
 10. Show that y = a + x -f bx 2 ± ex? has a cusp of the 
 second kind at the point (0, a), and that the equation of the 
 tangent at the cusp is y = x + a. 
 
 11. Show that y z = ax 2 -f x? has a cusp of the first kind 
 at the origin. 
 
 12. Show that ay 2 — X s -f bx 2 = has a conjugate point 
 at the origin, and a point of inflexion at x = — • 
 
 o 
 
EXAMPLES. 
 
 215 
 
 13. Trace the curve y* = a 3 — x 3 . 
 
 The curve cuts the axes at (a, 0) and (0, a). 
 It has an asymptote which passes through the origin. 
 The points where the curve cuts the axes are points of 
 inflexion. 
 
 14. Trace the curve y = ax 2 ±. Vbx sin x. 
 
 For every positive value of x there are 
 two values of y, and therefore two 
 points, except when sin x = 0, in which 
 case the two points reduce to one. 
 These points form a series of loops like 
 the links of a chain, and have for a 
 diametral curve the parabola y = ax 2 , from which, when x 
 is positive, the loops recede and approach, meeting the 
 parabola whenever x = or rr, or any multiple of n. But 
 when x is negative, y is imaginary except when sin x = 0, 
 in which case y = ax 2 , so that on the negative side there is 
 an infinite number of conjugate points, each one on the 
 parabola opposite a double point of the curve. (See De 
 Morgan's Cal., p. 382 ; also, Price's Cal., Vol. I, p. 396.) 
 
 Fig. 35, 
 
CHAPTER XI 
 
 RADIUS OF CURVATURE, EVOLUTES AND INVO- 
 LUTES, ENVELOPES. 
 
 116. Curvature.— TJie curvature of a curve is its rate 
 of deviation from a tangent, and is measured by the external 
 angle between the tangents at the extremities of an indefi- 
 nitely small arc ; that is, by the angle between any infini- 
 tesimal element and the prolongation of the preceding 
 element. This angle is called the angle of contxngence of 
 the arc. Of two curves, that which departs most rapidly 
 from its tangent has the greatest curvature. In the same 
 or in equal circles, the curvature is the same at every 
 point ; but in unequal circles, the greater the radius the 
 less the curvature; that is, in diiferent circles the curvature 
 varies inversely as their radii. 
 
 Whatever be the curvature at 
 any point of a plane curve, it is 
 clear that a circle may be found 
 which has the same curvature as 
 the curve at the given point, and 
 this circle can be placed tangent 
 to the curve at that point, with 
 its radius coinciding in direction 
 with the normal to the curve at 
 
 the same point. This circle is called the osculating circle, 
 or the circle of curvature of that point of the curve. Tfie 
 radius of curvature is the radius of the osculating circle. 
 The centre of curvature is the centre of the osculating circle. 
 
 For example, let ABA'B' be an ellipse. If different 
 circles be passed through B with their centres on BB', it is 
 
 Fig. 36. 
 
ORDER OF CONTACT OF CURVES. 
 
 m 
 
 clear that they will coincide with the ellipse in very differ- 
 ent degrees, some falling within and others without. Now, 
 that one which coincides with the ellipse the most nearly 
 of all of them, as in this case MN, is the osculating circle 
 of the ellipse at B, and is entirely exterior to the ellipse. 
 The osculating circle at A or A', is entirely within the 
 ellipse ; while at any other point, as P, it cuts the ellipse, 
 as will be shown hereafter. 
 
 117. Order of Contact of 
 Curves.^-- Let y — f (z) and y 
 
 = </> (x) be the equations of the 
 two curves, AB and ab, referred to 
 the axes OX and OY. Giving to 
 x an infinitesimal increment //, and 
 expanding by Taylor's theorem, we 
 have, 
 
 Fig. 37 
 
 Jr 
 
 y x = f(x + h) = /(*) + f (x) h + /" (x) I 
 
 + /"'(s)^ + etc. 
 
 ¥■ 
 y, = (x + h) - <)> (x) + f (x) h + </>" (x) x 
 
 (1) 
 
 + t>' 
 
 ( x ) 37-3 + etc - 
 
 m 
 
 Now if, when x = a = OM, we have f (a) = $ (a), the 
 two curves intersect at P, i. e., have one point in common. 
 If in addition we have /' (a) 5= </>' («), the curves have a 
 common tangent at P, i. e., have two consecutive points in 
 common ; in this case they are said to have a contact of the 
 first order. If also we have, not only/ (a) = <p (a) and/' (a) 
 = 0' (a), but/" (a) = 0" (a) ; i. e., in passing along one of 
 
 the curves to the next consecutive point, ~ (♦. e., the curva- 
 ture), remains the same in both curves, and the new point 
 10 
 
218 CONTACT OF THE SECOND ORDER. 
 
 is ;tlso a point of the second curve ; /. o. t the curves have 
 three consecutive points in common : in this case the curves 
 are said to have a contact of the second order. If f {<() 
 = * (a), f (a) = </,- (a), f" (a) = ♦" (a), /'" (a) =f (a), 
 the contact is of the third order, and so on. It is plain 
 that the higher the order of contact, the more nearly do 
 the curves agree ; if every term in (1) is equal to the cor- 
 responding term in (2), then y x = y iy and the two curves 
 become coincident. 
 
 118. The Order of Contact depends on the num- 
 ber of Arbitrary Constants. — In order that a curve may 
 have contact of the n 01 order with a given curve, it follows 
 from Art. 117 that n + 1 equations must be satisfied. 
 Hence, if the equation to a species of curve contains n + 1 
 constants, we may by giving suitable values to those con- 
 stants, find the particular curve of the species that has 
 contact of the n th order with a given curve at a given point. 
 For example, the general equation of the right line has two 
 constants, and hence two conditions can be formed, f(x) 
 = (f> (x) and/' (x) = 0' (x), from which the values of the 
 constants may be determined so as to find the particular 
 right line which has contact of the first order with a given 
 curve at a given point, hi general, the right line cannot 
 have contact of a higher order than the first. 
 
 Contact of the second order requires three conditions, 
 f(x) = (a;), /' (x) = 0' (x), and /" (a?) = <j>" (x), and 
 hence in order that a curve may have contact of the second 
 order with a given curve, its equation must contain three 
 constants, and so on. The general equation of the circle 
 has three constants ; hence, at any point of a curve a circle 
 may be found which has contact of the second order with 
 the curve at that point ; this circle is called the osculating 
 circle or circle of cnrvatiire of that point ; in general, 
 the circle cannot have contact of a higher order than 
 the second. The parabola can have contact of the 
 
RADIUS OF CURVATURE. 219 
 
 third order, and the ellipse and hyperbola of the 
 fourth. 
 
 In this discussion we have assumed that the given curve is of such 
 nature as to allow of any order of contact. Of course the order of 
 contact is limited as much by one of tbe curves as by the other. For 
 example, if the given curve were a right line and the other a circle, 
 the contact could not in general be above the first order, although the 
 circle may have a contact of the second order with curves whose 
 equations have at least three constants. Also, we have used the 
 phrase in general, since exceptions occur at particular points, some of 
 which will be noticed hereafter. 
 
 119. To find the radius of curvature of a given 
 curve at a given -point, and the co-ordinates of the 
 centre of curvature. 
 
 Let the equation of the given curve be 
 
 rW.W « 
 
 and that of the required circle be 
 
 (x! — m) 2 + (y' — n) 2 = r 2 ; (2) 
 
 it is required to determine the values of m, n, and r. 
 
 Since (2) has three arbitrary constants, we may impose 
 three conditions, and determine the values of these con- 
 stants that fulfil them, and the contact will be of the 
 second order (Art. 118). 
 
 From (2), by differentiating twice, we have, 
 
 x '-m+(y'-n) l ^, = 0; (3) 
 
 If (2) is the circle of curvature at the point (x, y) of (1), 
 we must have, 
 
 x' == x, y' = y; 
 
 dy' _ dy d?y' _ d?y 
 dx drf dx' 2 dx A 
 
220 RADIUS OF CURVATURE. 
 
 Substituting these values in (2), (3), and (4), we have, 
 
 (S_ m )i+ (y-n)* = r*; (5) 
 
 x-m + (y-n) ( £ = (6) 
 
 Therefore, y — n= ^-— (8) 
 
 V^dxVdx /m 
 
 ".-?.. •= § (9) 
 
 By (5), (8), and (9), we have 
 
 <fo 2 
 
 From (9) and (8) we have 
 
 120. Second Method. — Let 6& denote an infinitely 
 small element of a curve at a point, and </> the angle which 
 the tangent at this point makes with the axis of x. Imagine 
 two normals to be drawn at the extremities of this elemen- 
 tary arc, i.e., at two consecutive points of the curve ; these 
 
RADIUS OF CURVATURE. 2?/l 
 
 normals will generally meet at a finite distance. Let r be 
 the distance from the curve to the point of intersection of 
 these consecutive normals. Then the angle included be- 
 tween these consecutive normals is equal to the correspond- 
 ing angle of contingence (Art. 116), /'. e., equal to d(p. Since 
 d4> is the arc between the two normals at the unit's distance 
 of the point of intersection, we have 
 
 ds = rd<p, or r = jt- (1) 
 
 Now this value of r evidently represents the radius of the 
 circle, which has the same curvature as that of the given 
 curve at the given point, and hence is the radius of curva- 
 ture for the given point, while the centre of curvature may 
 be defined as the point of intersection of two consecutive 
 normals. 
 
 To find the value of r, we have (Art. 56a), 
 
 tan (j) = -~- ; .:</> = tan" 1 -^ ; 
 
 dx ^ dx' 
 
 dy^ 
 
 and hence d<f> = -^ ; also, ds = Vdx* + dy\ 
 
 1 A — 
 
 + tit* 
 
 Substituting in (1), we have 
 
 dx* 
 which is the same as (10) of Art. 119. 
 
 As the expression (l + -A)* has always two values, the 
 
 one positive and the other negative, while the curve can 
 generally have only one definite circle of curvature at any 
 point, it will be necessary to agree upon which sign is to be 
 
222 h'ADTUS OF CURVATURE. 
 
 taken. We shall adopt the positive sign, and regard r as 
 positive' when the second derivative is positive, i.e., when 
 the curve is convex downwards. (Usage is not uniform on 
 this point. See Price's Calculus, Vol. I, p. 435. Todhun- 
 ter's Calculus, p. 339, etc.) 
 
 L21. To Find the Radius of Curvature in Terms 
 of Polar Co-ordinates. 
 
 We may obtain this by transforming (2) of Art. 120 to 
 polar co-ordinates, from which we find 
 
 \ r m) n* 
 
 K = — 
 
 where N is the normal. See Art. 102, Eq. 9. [See (2) of 
 Ex. 4, Art. 90.] 
 
 122. At a Point where the Radius of Curvature 
 is a Maximum or a Minimum, the Circle of Curva- 
 ture has Contact of the Third Order with the Curve. 
 
 Since r is to be a maximum or a minimum, we must 
 
 have T = 0. 
 dx 
 
 Differentiating (2) of Art. 120 with respect to x, we have 
 dr 2V" 1 " dx*J dx\dxV dx*\ "*" dx 2 ) 
 
 dx 
 
 W/ 
 
 = 
 
 d?y _ dx\dx 2 / 
 dx*~ dy 2 
 
 (1) 
 
DIFFERENT ORDERS OF CONTACT. 223 
 
 Differentiating (8) of Art. 119, we have 
 
 d 3 y _ dx\dx 2 ) . 
 
 d&-~ dtf' W 
 
 + dx> 
 Hence the third derivative at a point of maximum or 
 minimum curvature is the same as it is in the circle of 
 curvature, and therefore the contact at this point is of the 
 third order (Art. 117). 
 
 Cor. — The contact of the osculating circles at the 
 vertices of the conic sections is closer than at other 
 points. 
 
 123. Contact of Different Orders.— Let y = f(x) 
 and y = <f> (x) represent two curves, and let x v be the ab- 
 scissa of a point of their intersection ; then we have 
 
 /(*)■■-♦(*» 
 
 Substituting x x ± h for x x in both equations, and sup- 
 posing y x and «/ 2 the corresponding ordinates of the two 
 curves, we have 
 
 & =m±h) =/w+/'w(±a)+/"w^ 2 
 
 H-r'W^ + etc. (1) 
 
 y* = 0Oi ± *) = N + 0' (*) (± h) + <t>" {x x )^£ 
 
 + m (^^fr3- + etc. (2) 
 
 Subtracting (2) from (1), we get, for the difference of 
 their ordinates, corresponding to X\ ± h, 
 
 r. - y. = [/'(*.)-*'(*.)] (±h) + [/'•'(*.) - f (*)] {J j£ 
 
 + If'" (A) - f (*,)] ( -^f- + etc. (3) 
 
•.'•.' \ I I \Ml'LES. 
 
 Now if these curves have contact of the first order, the 
 lirst term of (3) reduces to zero (Art. 117). If they haw 
 contact of the second order, the first two terms reduce to 
 zero. If they have contact of the third order, the first three 
 terms reduce to zero, and so on. Hence, when the order of 
 contact is odd, the first term of (3) that does nut reduce to 
 zero must contain an even power of ± //, and y } — y. 2 doee 
 not change sign with h, and therefore the curves do not 
 intersect, the one lying entirely above the other ; but when 
 the order of contact is even, the first term of (3) that does 
 not reduce to zero must contain an odd power of ± h> and 
 V\ — y* changes sign with h, and therefore the curves inter- 
 sect, the one lying alternately above and below the other. 
 
 Cor. 1. — At a point of inflexion of a curve, the second 
 derivative equals ; also, the second derivative of any point 
 of a right line equals 0. Hence, at a point of inflexion,. 
 a rectilinear tangent to a curve has contact of the 
 second order, and therefore intersects the curve. 
 
 Cor. 2. — Since the circle of curvature has a contact of 
 the second order with a curve, it follows that the circle of 
 curvature, in general, cuts the curve as well as 
 touches it. 
 
 Cor. 3. — At the points of maximum and minimum curva- 
 ture, as for example at any of the four vertices of an ellipse; 
 the osculating circle does not cut the curve at its point o! 
 contact. 
 
 EXAMPLES. 
 
 1. Find the radius of curvature of an ellipse, 
 
 t . t - i 
 «2 "*" W ~ 
 
 „ dy _ Vx ., , dy 2 _ ay + frtf 
 
 Uere dz-~aY " + dx*~~ ay ' 
 
EXAMPLES. 225 
 
 ,. (Art. 120), r= X —g±- 
 dx* 
 = ^ T — " (neglecting the sign). 
 
 At the extremity of the major axis, 
 
 x = a, y = 0, .*. r = — • 
 
 At the extremity of the minor axis, 
 
 a 2 
 
 x = 0, y = h .'. r = £- • 
 
 2. Find the radius of curvature of the common parabola, 
 
 y z = 2px. 
 
 Here ^=^ &l^_t. 
 
 _ (y 2 + i? 2 )* _ (normal)* 
 •'• r ~ ^ - — -f 
 
 At the vertex, y = ; .*. r = p. 
 
 3. Find the radius of curvature of the cycloid 
 
 x = r vers -1 - — VSry — ?/ 2 . 
 Here - - — -3L__. , 1 + & = *! s 
 
 which equals twice the normal (Art. 101, Ex. 5). 
 
EV0LUTE8 AND INVOLUTES, 
 
 4. Find the radius of curvature of the parabola whose 
 latus-rectum is 9, at x = 3, and the co-ordinates of tin- 
 centre of curvature. r = 10.04 ; m = 13£, n = — 6.91. 
 
 5. Find the radius of curvature of the ellipse whose axes 
 are 8 and 4, at x = 2, and the co-ordinates of the centre of 
 curvature. r = 5.86 ; m = .38, n = — 3.9. 
 
 6. Find the radius of curvature of the logarithmic spiral 
 
 r = a . 
 d*r 
 
 dd* 
 
 R 
 
 dv 
 
 (r 2 -f r 2 log 2 a)* 
 2 + 2r 2 log 2 a — r 2 log 2 « 
 
 = a* log 2 a ; 
 
 (r 2 + r 2 log 2 a)l = N. 
 
 (See Ex. 2, Art. 102.) 
 
 7. Find the radius of curvature of the spiral of Archi- 
 
 medes, r = aB. B = ^r^r 
 
 8. Find the radius of curvature of the hyperbolic spiral, 
 
 _ r (« 2 + f 8 )* 
 rO = a. E = — * — -^ — £-. 
 
 124. Evolutes and Involutes.— The curve which is 
 the locus of the centres of all the osculating circles of a 
 given curve, is called the evolute of that curve ; the latter 
 curve is called the involute of the former. 
 
 Let P 1? P 2 , P 8 , etc., represent a series of w ■ 
 consecutive points on the curve MN, and 
 Ci, C 2 , C 3 , etc., the corresponding centres 
 of curvature ; then the curve 0„ C 2 , C 3 , 
 etc., is the evolute of MN, and MN is the 
 involute of 0„ C 2 , 3 , etc. Also, since the 
 lines C,P„ 2 P 2 , etc., are normals to the 
 involute at the consecutive points, the 
 points C„ C„ C 3 , etc., may be regarded as pi S' 38. 
 
EQUATION OF THE E VOLUTE. 227 
 
 consecutive points of the evolute ; and since each of the 
 normals PA, P 2 C 2 , etc., passes through two consecutive 
 points on the evolute, they are tangents to it. 
 
 Let r l} r. 2 , r 3 , etc., denote the lengths of the radii of 
 curvature at P„ P 2 , P 3 , etc., and we have, 
 
 n - Tx = P 2 C 2 - P,0, = P 2 C 2 - P*C, = C,0 a ; 
 
 also r z - r 2 = P 3 3 - P 2 C 2 = P 3 C 3 ~- PA = 2 3 ; 
 
 and r 4 — r 3 = C 3 4 , and so on to r n ; 
 
 hence by addition we have, 
 
 r n - n = c 1 o 2 + c 2 c 3 + a_,c, 
 
 This result holds when the number n is increased indefi- 
 nitely, and we infer that the length of any arc of the 
 evolute is equal, in general, to the difference between 
 the radii of curvature at its extremities. OJ^! b^j?> 
 
 It is evident that the involute may be < gener ated from its 
 evolute by winding a string round the evolute,~holding it 
 tight, and then unwinding it, each point in the string 
 will describe a different involute. It is from this property 
 that the names evolute and involute are given. While a 
 curve can only have one evolute, it can have an infinite 
 number of involutes. 
 
 The involutes described by two different points in the 
 moving string, are said to be parallel ; each curve being got 
 from the other by cutting off a constant length on it? 
 normal, measured from the involute. (Williamsons Dif< 
 ferential Calculus, p. 295.) 
 
 125. To find the Equation of the Evolute of a 
 Given Curve. — The co-ordinates of the centre of curva- 
 ture are the co-ordinates of the evolute (Art. 124). Hence, 
 if Ave combine (11) and (12) of Art. 119 with the equation 
 of the curve, and eliminate x and y, there will result an 
 equation expressing a relation between m and n, the co-or- 
 
228 
 
 EXAMPLES. 
 
 dinatefl of the required evolute, wbicli is therefore the 
 required equation ; the method can be best illustrated by 
 examples. 
 
 The eliminations are often quite difficult ; the following 
 are comparatively simple examples. 
 
 EXAMPLES. 
 
 1. Find the equation of the evolute of the parabola, 
 
 f = 2px. (1) 
 
 Here &=£. *2L = -t. 
 
 dx y' dx 2 y z 
 
 Substituting in (11) and (12) of Art 
 119, we have, 
 
 m = x 
 
 y*+tf y y* _ 
 
 + f y f~ dx + p > 
 
 n = y 
 
 _ m — p 
 
 y2 +p 2 yS 
 
 f 
 
 f 
 
 
 y = — p^rb 5 . 
 
 And these values of x and y in (1) give, 
 pin* = ip(m—p); 
 
 (3) 
 
 which is the equation required, and is called the semi-cubical 
 parabola. Tracing the curve, we find its form as given in 
 Fig. 39, where AO = p. 
 If we transfer the origin from O to A, (2) becomes 
 
 n 2 — =r- m 3 . 
 
 27p 
 
EXAMPLES. 
 
 229 
 
 2. Find the length of the evolute AQ', Fig. 39, in terms 
 of the co-ordinates of its extremities. 
 
 Let ON a^irq-=|fj ON' = m, N'Q' = n. 
 
 Then by Art. 123, Ex. 2, we have, 
 
 _ (f + y*)\ 
 f 
 Therefore, by Art. 124, we have, 
 
 Length of AQ' == Q'Q - AO = ^ t^ -p 
 = (n% + p%fi — p. (Since y 2 = ptffi, by Ex. 1.) 
 
 3. Find the equation of the evolute of the cycloid, 
 
 x = r vers -1 - — *J%ry — y\ (1) 
 
 Here ~ = 
 
 dx 
 
 dy _ v 2ry — ;/ 2 
 
 
 
 dx 2 
 
 f 
 
 m = x + 2 v%ry — y 2 and n — — y ; 
 
 or 
 
 # == m — 2 V— 2r« — w 2 and ?/ 
 which, in the equation of the cycloid, gives 
 
 m—r vers- 1 (—-,) + \/—%rn—n 2 , (2) 
 
 which is the equation of a cy- 
 cloid equal to the given cycloid; 
 the origin being at the highest 
 point. This will appear by re- 
 ferring the given cycloid 00 B 
 to the parallel axes O'X' and 
 O'Y'. 
 
 Y 
 
 Y 
 
 O' N M , 
 
 
 /£ F 
 
 
 
 fp 
 
 
 
 
 c.y 
 
 
 "ig.40. X 
 
 H ^ 
 
 A 
 
 
 B 
 
 X' 
 
= rvers 
 
 ■. in 
 
 880 PROPERTIES OF EVOLUTES. 
 
 Let (in, //) be any point P in the cycloid ; then we have 
 n= -MP= -ND, 
 m = O'M = PP = IIB - CB + DP 
 = arc PN + DP 
 
 = r vers-* (--2.) + ^ND x DC 
 
 \—") + V—n(2r+ n). 
 
 r vers - * ( — - ) + V — 2 r« — 5? ; (4) 
 
 which is the same as equation (2). Hence we see that the 
 equation of the evolute, OA (2), is the same as that of the 
 cycloid, OB (4). That is, the evolute of a cycloid is an 
 equal cycloid* 
 
 4. Find the equation of the evolute of the equilateral 
 hyperbola 2 xy = «A ^ (m + n) i_ (m _ n) i =u t 
 
 [Here find m + »= ^ + z )\ 
 
 (I 
 
 (u — x\^ 
 m — n = - ' ' ; .-.. etc.] 
 
 a 2 J 
 
 126. A Normal to an Involute is Tangent to the 
 Evolute. — This was shown geometrically in Art. 124. It 
 may also he shown as follows : 
 
 Let (x, y) be any point Q of the involute (Fig. 40), from 
 which the normal QQ' is drawn, and let {in, u) be the point 
 Q' on the evolute through which the normal passes. 
 
 The equation of QQ' is 
 
 y-n= - ( J*l{x-m)- (1) 
 
 or X - m + c £( ?/ - n ) = Q. (2) 
 
 !Now when we pass from a point Q to a consecutive 
 point on the involute, Q' also will change to a consecutive 
 
 * This property \\;is lirst discovered by Efaygeng. 
 

 &-»>§ 
 
 + 1 
 
 - + 
 
 dtf 
 
 dx* ~~ 
 
 which in 
 
 (3) gives 
 
 
 
 
 
 dm dy dn _ 
 dx dxdx 
 
 o, 
 
 or 
 
 dx 
 ~ dy 
 
 and this 
 
 in (1) gives 
 
 
 
 
 
 y — n = 
 
 dn 
 dm 
 
 (X- 
 
 -m); 
 
 ENVELOPES OF CURVES. 231 
 
 point of the e volute, therefore we differentiate (2) with 
 respect to x } regarding x, y, m, n, as variables, and get 
 
 dm d 2 y , > , dy 2 dy dn . , oh 
 
 But, since (m, n) is the centre of curvature corresponding 
 to Q, we have, by (8) of Art. 119, 
 
 0; 
 
 dm' 
 
 (i) 
 
 therefore (1) or (4), which is the equation of a normal to 
 the involute at Q (x, y), is also the equation of a tangent to 
 the e volute at Q' (m, n). 
 
 127. Envelopes of Curves. — Let us suppose that in 
 the equation of any plane curve of the form 
 
 f(x,y,a) = 0, (1) 
 
 we assign to the arbitrary constant a, a series of different 
 values, then for each value of a we get a distinct curve, 
 different from any of the others in form and position, and 
 (1) may be regarded as representing an indefinite number 
 of curves, each of which is determined when the correspond- 
 ing value of a is known, and varies as a varies. 
 
 The quantity a is called a variable parameter, the name 
 being applied to a quantity which is constant for any one 
 3urve of a series, but varies in changing from one curve to 
 another, and the equation / (x, y, a) = 0, is said to repre- 
 sent a family of curves. 
 
 If we suppose a to change continuously, i. e. t by 
 
232 EQUATION OF THE ENVELOPE. 
 
 infinitesimal increments, the curves of the series represented 
 by (1) will differ in position by infinitesimal amounts j and 
 
 in iv two adjacent curves of the series will, in general, inter- 
 sect; the intersections of these curves are points in the 
 lope. Hence an envelope may be defined as the locus 
 of the inter section of consecutive curves of a seritx. 
 
 It can be easily seen that the envelope is tangent to each 
 of the intersecting curves of the series; for, if we consider 
 four consecutive curves, and suppose P, to be the point of 
 intersection of the first and second, P 2 that of the second 
 and third, and P 3 that of the third and fourth, the line 
 PjP 2 joins two infinitely near points on the envelope and 
 on the second of the four curves, and hence is a tangent 
 both to the envelope and the second curve ; in the same 
 way it may be shown that the line P 2 P 3 is a tangent to the 
 envelope and the third consecutive curve, and so on. 
 
 128. To Find the Equation of the Envelope of a 
 given Series of Curves. 
 
 Let f(x, y, a) = 0, (1) 
 
 f{x,y, a + da) = 0, (2) 
 
 be the equations of two consecutive curves of the series; 
 then the co-ordinates of the points of intersection of (1) and 
 (2) will satisfy both (1) and (2), and therefore also will 
 satisfy the equation 
 
 /<*'»*)-£*>** + **) = o (Anal . Geom., Art. 30), 
 
 df(x, y, a) 
 da 
 
 and therefore the points of intersection of two infinitely 
 near curves of the series satisfy each of the equations (1) 
 and (3). Hence, to find the equation of the envelope, we 
 eliminate a between (1) and (3), i. e., we eliminate the varia- 
 ble parameter between the equation of the locus and its first 
 differential equation. 
 
 = 0, (3) 
 
EXAMPLES. 
 
 233 
 
 EXAMPLES. 
 
 711 
 
 1. Find the envelope of y = ax -\ , when a varies. 
 
 Differentiating with respect to a, x, and y, being constant, 
 jve have 
 
 ,\ y = ± [V^wsc + VW#J or y 2 = imx, 
 which is the equation of a parabola. 
 
 2. A right line of given length * 
 slides down between two rectangular 
 axes ; to find the envelope of the line 
 in all positions. 
 
 Let c be the length of the line, a 
 and b the intercepts OA and OB ; 
 then the equation of the line is 
 
 a + b ~ ** 
 
 (1) 
 
 ftgr. 41. 
 
 in which the variable parameters a and b are connected by 
 the equation 
 
 a 2 + b 2 = c 2 . (2) 
 
 Differentiating (1) and (2), regarding a and b as varia- 
 ble, we have 
 
 a 2 b 2 
 
 0, or 
 
 ada + bdb = 0, or 
 
 - rfa = f db. 
 i 2 IP 
 
 ada = bdb. 
 
 (3) 
 (4) 
 
 Dividing (3) by (4), we have 
 
 x 
 y a 
 
 V> or T* 
 
 y * + y 
 
 b_ _ a^ b 
 
 & ~ a 2 + W 
 
 1 
 
 C* 5 
 
 a — (%<?)* and b = (y(?) s , 
 
884 EXAMPLES. 
 
 which in (2) gives 
 
 (a*) 1 + (#)* = &\ 
 
 .\ x$ + y% = c$, 
 
 which is the equation required. 
 The form of the locus is given in 
 Fig. 42, and is called a hypo-cycloid, 
 which is a curve generated by a 
 point in the circumference of a 
 circle as it rolls on the concave 
 arc of a fixed circle. 
 
 Fig. 42. 
 
 3. Find the envelope of a series of ellipses whose axes are 
 coincident in direction, their product being constant. 
 
 •e ? + t-i 
 at + V*- 1 ' 
 
 
 (i) 
 
 a-b = c ; 
 
 
 (2) 
 
 .*. — Q da + f-o do = 0, or — 9 da = 
 a z b 3 a s 
 
 -p, 
 
 (3) 
 
 da db _ da db 
 
 h -T- = 0, or — = r-. 
 
 a b a b 
 
 Dividing (3) by (4), we have 
 
 X 2 1J 2 
 
 -j = "L = £, by substituting in (1). 
 
 ,\ a = ± x a/2, and b — ± yV2; 
 
 (4) 
 
 from (2), 
 
 xy = ± 
 
 r 
 
 which is the equation of an hyperbola referred to its asymp- 
 totes as axes. 
 
 This example may also be solved as follows : Eliminating 
 b from (1) and (2), we have 
 
EXAMPLES. 
 
 23i 
 
 
 1, 
 
 (5) 
 
 in which we have only the variable 
 
 parameter a. 
 
 
 z 2 .ay* _ 
 
 2 ^ 
 
 y 
 
 (6) 
 
 which in (5) gives 
 
 
 
 * + c ~ ' 
 
 .\ xy = \c. 
 
 
 4. Find the envelope of the right lines whose general 
 equation is 
 
 y = mx + (ahn 2 + b 2 )K (1) 
 
 where m is the variable parameter. 
 
 We find m = — , 
 
 a V« 2 - a 2 
 
 x 2 v 2 
 which in (1) gives — 2 + |- 2 = 1 for the required envelope. 
 
 Hence the envelope of (1 ) is an ellipse, as we might have 
 inferred, since (1) is a tangent to an ellipse. (See Anal. 
 Geom., Art. 74.) 
 
 EXAMPLES. 
 
 1. Find the radius of curvature of the logarithmic curve 
 x = log y. _ (ffl?-hyi)f 
 
 my 
 
 2. Find the radius of curvature of the cubical parabola 
 
 f = a2x - _ ( 9y* + « 4 ) f 
 
 T ~ 6a*y 
 
 3. Find the radius of curvature of the curve 
 
 y — x 3 — x 2 4- 1 
 
886 EXAMPLES. 
 
 where it cuts the axis of ?/, and also at the point of mini- 
 mum ordinate. 
 
 At the first point, f= —\\ at the second, r = J. 
 
 4. Find the radius of curvature of the curve 
 
 y* = fix 2 -f z 3 . 
 
 r - [y 4 + (4a? + x*)*]l 
 - Sx 2 y 
 
 5. Find the radius of curvature of the rectangular hyper- 
 bola xy = m 2 . r x z _|_ yi\i 
 
 r = ~2m 2 
 
 6. Find the radius of curvature of the Lemniscate of 
 Bernouilli r 2 = a 2 cos 20. p _ a 2 
 
 h -Zr 
 
 7. Find the equation of the evolute of the ellipse 
 
 a 2 y 2 + b 2 x 2 = aW. 
 
 (a?n)* + (fa)i = («2 _ ga)f. 
 
 8. Find the equation of the evolute of the hyperbola 
 
 a 2 y 2 — b 2 x 2 = — a 2 b 2 . 
 
 ( am )\ .- (£»)i = (a 2 + P)t. 
 
 9. Prove that, in Fig. 39, OM = 40A = 4p, and MP' 
 = 2pVt 
 
 10. Find the length of the evolute AP' in Fig. 39. 
 
 Am. (3* — l)p. 
 
 11. Find the length of the evolute of the ellipse. (See 
 
 Art. 123, Ex. 1, and Art. 124.) A a? — b 3 
 
 Ans. 4 z 
 
 ab 
 
 12. Find the length of the cycloidal arc OO'B, Fig. 40. 
 
 Ans. 8r. 
 
 13. Find the envelope of the series of parabolas whose 
 equation is y 2 = m (x — m), m being the variable parameter. 
 
 , x 
 
EXAMPLES. 237 
 
 14. Find „he envelope of the series of parabolas expressed 
 
 1 4- a 2 
 by the equation y = ax - — - x 2 , where a is the variable 
 
 parameter. 
 
 The result is a parabola whose equation is 
 
 This is the equation of the curve touched by the parabolas de- 
 scribed by projectiles discharged from a given point with a constant 
 velocity, but at different inclinations to the horizon. The problem 
 was the first of the kind proposed, and was solved by John Bernouilli, 
 but not by any general method. 
 
 15. Find the envelope of the hypothenuse of a right- 
 angled triangle of constant area c. c 
 
 Xy ~% 
 
 16. One angle of a triangle is fixed in position, find the 
 envelope of the opposite side when the area is constant = c. 
 
 c 
 
 17. Find the envelope of x cos a + y sin a = p, in 
 which a is the variable parameter. x 2 + y 2 = p 2 . 
 
 18. Find the envelope of the consecutive normals to the 
 parabola y 2 = 2px. 
 
 o 
 
 Ans. y 2 = nfr ( x — P) s > which is the same as was found 
 
 for the evolute in Ex. 1, Art. 125, as it clearly should be. 
 (See Art. 124.) 
 
 19. Find the envelope of the consecutive normals to the 
 ellipse a 2 y 2 + h 2 x 2 == a 2 b 2 . 
 
 A ns. (ax)% 4- (by)* = (a 2 — b 2 )i, which is the same as 
 was found in (7) for the evolute of the ellipse. 
 
PART II. 
 
 INTEGRAL CALCULUS, 
 
 CHAPTER I. 
 
 ELEMENTARY FORMS OF INTEGRATION. 
 
 129. Definitions. — The Integral Calculus is the inverse 
 of the Differential Calculus, its object being to find the 
 relations between finite values of variables from given 
 relations between the infinitesimal elements of those vari- 
 ables ; or, it may be defined as the process of finding the 
 function from which any given differential may have been 
 obtained. The function which being differentiated pro- 
 duces the given differential, is called the integral of the 
 differential. The process by which we obtain the integral 
 function from its differential is called integration. 
 
 The primary problem of the Integral Calculus is to effect 
 the summation of a certain infinite series of infinitesimals, 
 and hence the letter 8 was placed before the differential to 
 show that its sum was to be taken. This was elongated 
 
 into the symbol / (a long S), which is the sign of integra- 
 tion, and when placed before a differential, denotes that its 
 integral is to be taken. Thus, / ZxHx, which is read, " the 
 
 integral or* %x*clx" denotes that the integral of SMx is to 
 be taken. The signs of integration and differentiation are 
 
ELEMENTARY RULES FOR INTEGRATION. 239 
 
 inverse operations, and when placed before a quantity, 
 neutralize each other. Thus, 
 
 / d (ax) = ax, 
 and d I axdx = axdx, 
 
 130. Elementary Rules for Integration. — In the ele- 
 mentary forms of integration, the rules and methods are 
 obtained by reversing the corresponding rules for differ- 
 entiation. When a differential is given for integration, if 
 we cannot see by inspection what function, being differ- 
 entiated, produces it, or if it cannot be integrated by known 
 rules, we proceed to transform the differential into an 
 equivalent expression of known form, whose integral we 
 can see by inspection, or can obtain by known rules. In 
 every case, a sufficient reason that one function is the 
 integral of another is that the former, being differentiated, 
 gives the latter.* 
 
 (1.) Since 
 
 d(v + y — z) = dv + dy — dz ; (Art. 14.) 
 
 / (dv -f- dy — dz) = d (v + y — z) = v -\- y — z 
 
 = / dv + dy — / dz. 
 
 Hence, the integral of the algebraic sum of any 
 number of differentials is equal to the algebraic sum 
 of their integrals. 
 
 (2.) Since 
 
 d (ax ± b) = adx ; (Art. 15.) 
 
 * While there is no quantity whose differential cannot he found, there is a large 
 class of differentials whose integrals cannot be ohtained ; either because there is no 
 quantity which, being differentiated, will give them, or because the method for 
 their integration has not yet been found. 
 
i40 /:j,/:m/:.\ta/:)- RULES rOR TNTBGRATIOH. 
 
 I adx = I d (ax -f b) = ax ± b 
 = a I dx ± b. 
 
 Hence., a constant factor can be moved from one 
 ride of the integral sign to the other without affect- 
 in <J the value of the integral. Also, since constant 
 terms, connected by the sign ±, disappear in differentia- 
 tion, therefore in returning from the differential to the 
 integral, an arbitrary constant, as C, must be added, whose 
 value must be determined afterwards by the data of the 
 problem, as will be explained hereafter. 
 
 (3.) Since 
 
 d - [/(*)]" = « [/fa)]"" 1 df(*) ; ( Arts - 15 and 19 -) 
 
 ... fa [/<*)]-> <?/(*) = fd\ [/(*)]" = I [fix)-]" + c. 
 
 Hence, whenever a differential is the product of 
 three factors, viz, a constant factor, a variable factor 
 with any exponent except — 1, and a, differential 
 factor which is the differential of the variable factor 
 without its exponent, its integral is the product of 
 the constant factor by the variable factor with its 
 exponent increased by 1, divided by the new ex- 
 ponent.* 
 
 It will be seen that the rule fails when n = — 1, since 
 if we divide by 1 — 1 = 0, the result will be 00 . 
 
 (4.) Since d (a log x) = — - ; (Art. 20, Cor.) 
 •'• J -^ = f d (a log x) =a log x. 
 
 * The arbitrary constant is not mentioned since its addition is always under- 
 stood, and in the following integrals it will he omitted, as it can always be supplied 
 when necessary. 
 
EXAMPLES. 241 
 
 Hence, whenever a differential is a fraction whose 
 numerator is the product of a constant by the differ- 
 ential of the denominator, its integral is the product 
 of the constant by the J\ r aperian logarithm of the 
 denominator. 
 
 EXAMPLES. 
 
 1. To integrate dy = axHx. 
 
 y = I axhlx == / a • x 5 • dx — — • [by (3)]. 
 
 2. To integrate dy = (a + hx*yx 2 dx. 
 
 The differential of the quantity within the parenthesis 
 being 15x 2 dx, we write 
 
 * = /*(« + S* 3 ) 4 IS**** = ( * + £?- [by (J)} 
 
 This example might also be integrated by expanding the quantity 
 within the parenthesis, and integrating each term separately by (1), 
 but the process would be more lengthy than the one employed. 
 
 3. To integrate 
 
 dy — a (ax? + fa 3 )* 2xdx + 3fa 2 (ax? + fa 8 )* dx. 
 
 y = f[ a (ax 1 + fa 3 )* 2xdx + 3fa 2 (ax 2 + W)± dx] 
 s= J (ax 2 + fa 3 )* (2«» + 3fa 2 ) <fc = § («^ + fa 3 )* [by (5)]. 
 
 4. To integrate dy = — • 
 
 Since the numerator must be bdx tp be the differential of 
 the denominator, we must multiply it by b, taking care to 
 divide by b also ; hence, 
 
 /adx a P bdx a , , 7 x ri , ,.-. 
 
 T+y x = J aT y x = j lo g (« + fe )- ^ Wi- 
 ll 
 
242 DIFFERENTIAL FORMS. 
 
 131. Fundamental Forms. — On referring to the forms 
 of differentials established in Chap. 11. we may writedown 
 
 at once t lie following integrals from inspection, the arbitrary 
 constant being always understood. 
 
 1. 
 
 y = 1 ax 1l dx 
 
 az"+ l 
 
 " w+ 1* 
 
 2. 
 
 f*adx 
 
 a 
 
 
 (n — 1) x H ~ 1 ' 
 
 3. 
 
 fadx 
 
 = a log #.* 
 
 4. 
 
 y = J a x log a dx 
 
 = ax. 
 
 5. 
 
 y = / ePdx 
 
 = e*. 
 
 6. 
 
 y = J cos x dx 
 
 = sin x. 
 
 7. 
 
 y = 1 — sin x dx 
 
 = cos #. 
 
 8. 
 
 y = / sec 2 a; dx 
 
 = tan x 
 
 9. 
 
 y = 1 — cosec 2 x dx 
 
 = cot X. 
 
 10. 
 
 y = 1 sec x tan x dx 
 
 = sec #. 
 
 11. 
 
 y = / — cosec x cotan Z 
 
 cfo = cosec x. 
 
 12. 
 
 ?/ := / sin x dx 
 
 = vers x. 
 
 13. 
 
 y = 1 — cos x dx 
 
 = covers #. 
 
 * Since the constant c to be added is arbitrary, log c is arbitrary, and we may 
 write the integral in the form 
 
 r dx 
 
 log x + log c = log caj". 
 
INTEGRATION BY TRANSFORMATION. 243 
 
 dx 
 
 VI — x 2 
 dx 
 
 = sin -1 x. 
 
 = cos -1 x. 
 
 14. y = f 
 
 15. y = / — 
 * «/ Vl - a: 2 
 
 17. y = f-^- % = «*-«* 
 
 1Q /" ^ -1 
 
 18. y = — = sec 1 x. 
 
 J xVx*-l 
 
 19- , = / 
 
 cosec -1 #. 
 
 20. 'y = / — _ = vers -1 a:. 
 
 21. y = / 
 
 aVa* — 1 
 
 *s/%x — X 2 
 dx 
 
 V%% — x 2 
 
 = covers -1 x. 
 
 These integrals are called the fundamental or elementary 
 forms, to which all other forms, that admit of integration 
 in a finite number of terms, can be ultimately reduced. It 
 is in this algebraic reduction that the chief difficulty of the 
 Integral Calculus is found ; and the processes of the whole 
 subject are little else than a succession of transformations 
 and artifices by which this reduction may be effected. The 
 student must commit these fundamental forms to memory; 
 they are as essential in integration as the multiplication 
 table is in arithmetic. 
 
 132. Integration of other Circular and Trigono- 
 metric Functions by Transformation into the Fun- 
 damental Forms. 
 
 1. To integrate dy 
 
 Va 2 — Wx 2 
 We see that this has the general form of the differential 
 
244 INTEGRATION BY TRANSFORMATION. 
 
 of an arc in terms of its sine (see form 14 of Art. 131)-, 
 hence we transform onr expression into this form, as follows: 
 
 r dx r dx r S * 
 
 •\A-5 ! V-3 
 
 To make this quantity the differential of an arc in terms 
 of its sine, the numerator must be the differential of the 
 square root of the second term in the denominator, which 
 
 is - dx. Therefore we need to multiply the numerator by b, 
 
 which can be done by multiplying also by the reciprocal of 
 b, or putting the reciprocal of b outside the sign of integra- 
 tion. Hence, 
 
 = f dx _ f "^ = 1 f ^ 
 
 0-? 0"f 
 
 \ . x bx 
 
 = T sin -1 — • 
 b a 
 
 2. To integrate dy = 
 
 Va* - && 
 
 Here y = I — 
 
 dx 
 
 1 r a 
 
 Va*-VW 
 
 dx 
 
 cos 
 
 -i 
 
 bx 
 
 jjt# b a 
 
 J a- 
 
 1— ~- 
 
 dx 1 , , bx 
 
 tan -1 — • 
 
 2 + Pn* ab a 
 
 a ». C dx 1 <r-i hx 
 
PROCESS OF INTEGRATION, 245 
 
 dx 1 , bx 
 
 r dx i 
 
 5. y = I — = sec 
 
 f/ xVbW — a* a a 
 
 _ /» da: 1 .5a; 
 
 6, y — J — __=;_ cosec -1 — • 
 
 V^a 
 
 « « 
 
 f* dx \ .bx 
 
 dx 1 , fta; 
 
 covers 
 
 8 - V = /* — y- 
 
 ^ V%abx-bW & a 
 
 />, 7 /'sin .r rfa; /'d cos 35 
 
 9. y = / tan x ax = I = — / 
 
 u J J cos x i/ cos a: 
 
 = — log cos a; = log sec a*. 
 
 10. y = I cot x dx = I — : = log sin a;. 
 
 * «/ «/ sin a; & 
 
 /* da; /' </a: 
 
 11. y — I - — = / j-i — .- r- 
 
 17 «/ sin» «/ 2 sin £.z cos |a; 
 
 /i sec 2 (la:) da; , 
 2 . ^ — = log tan Ix, 
 tan £a; & 2 
 
 /* dx 1* dx 
 
 12. V = / = / : - 
 
 9 J COS X J . /7T \ 
 
 = log tan g + Is) [by (11)]. 
 
 ., rt f* dx /'sec 2 a; da; , 
 
 13. y = / = / —r = log tan x- 
 
 * t/ sin a; cos a; «/ tan re 
 
 P dx P (sin 2 a; + cos 2 a*) da; 
 
 y ~ J sin 2 a: cos 2 a; ~" «/ sin 2 a; cos 2 £ 
 
 (since sin 2 a; -f cos 2 a; = 1) = / (sec 2 a; + cosec 2 a;) da* 
 
 = tan x — cot x. 
 
 15. y ~ / tan 2 x dx = I (sec 2 a; — 1) da; = tan a — a*. 
 
246 EXAMPLES. 
 
 16. y = J cot 2 x dx = J (cosec 2 x —l)dx 
 
 = — cot X — X. 
 
 17. y = /cos 2 a; dx = J (I + b cos 2a;) dx (by Trig.) 
 
 = \x -f \ sin %x, 
 (See Price's Calculus, Vol. II, p. 68.) 
 
 18. y = J sin 2 a; t/a; = \x — \ sin 2a;. 
 
 Remark. — It will be observed that in every case we reduce the 
 function to a known form, and then pass to the integral by simple 
 inspection or by the elementary rules. Whenever there is any doubi 
 as to whether the integral found be correct or not, it is well to differen- 
 tiate it, and see if it gives the proposed differential. (See Art. 130.) 
 
 EXAMPLES. 
 
 1. ily = bx* dx. 
 
 Here y = / bx? dx = I b-x^-dx 
 
 = V>x* [by (3) of Art. 130]. 
 
 7 xdx 
 
 2. dy = 
 
 Va 2 + z 2 
 
 TT /» xdx n , 
 
 IIcre ! = fv^T* = A ^ + *>-***** 
 
 as (a 2 + a*)*. 
 
 3. dy = 2x* dx. y = $*#. 
 
 4. % = %x*dx. y = ■$£$. 
 
 5. dy = —x~*dx. y = 5x~K 
 
 6. ay = -— • # = 2\/#. 
 
INTEGRATING FACTOR. 24? 
 
 7. dy = — 5mx~* dx. y = — ±£mx%. 
 
 8. cly = (-§-«£* — f&tft) $e. «/ = ««§ — bxk 
 
 ). dy = [ 
 
 12 3 V _ 12 1 
 
 10 . **-fi=^* 
 
 Here y = f— (3ax* — x*)~*(2ax — x>) dx 
 
 = J — \ {^ ax% — x *)~^ ( Qax — % x2 ) dx 
 = - i (Boa* - »*)*. 
 
 11. dy = (2ax — x 2 )? (a — x)dx. y = \ (2ax — a*)&. 
 
 133. Integrating Factor. — It has been easy, in the 
 examples already given, to find the factor necessary to make 
 the differential factor the differential of the variable factor 
 [see (3) of Art. 130] ; but sometimes this factor is not easily 
 found, and often no such factor exists. There is a general 
 method of ascertaining whether there is such a factor or 
 not, and when there is, of finding it, which will be given 
 in the two following examples : 
 
 12. dy = 3 (Ux 2 — 2cx*)s (Ux — Sex 2 ) dx. 
 
 Suppose A to be the constant factor required ; then we 
 have 
 
 y = f-j (Ux 2 - 2«b»)* (12Abx - OAcafydx. 
 
 If A be the required factor, we must have 
 
 d (4bx 2 — 2cx*) = (\2Abx — 9Acx 2 ) dx, 
 
 or Sbx — 6cx 2 = \2Abx — 9Acx% 
 
 and since this result is to be true for every value of x, the 
 coefficients of the like powers of x must be equal to each 
 
348 EXAMPLES. 
 
 other, from the principle of indeterminate coefficients, giv- 
 ing ns two conditions, 
 
 12-44 = 8*, (1) 
 
 and — 9Ac = — 6c, (2) 
 
 or A = f. from (1), and ^4 = f, from (2) ; 
 
 hence § is the factor required, and we have 
 
 y = J\ (Ibx* - 2cx*ft (Sbx - 6ca?) dx 
 = -I (ibx* - 2cx*)k 
 
 13. dy = (2ax — a;2)i (5a — x) dx. 
 Let A be the required factor, then 
 
 y = / —r (2ax — x 2 )% (5Aa — Ax) dx, 
 
 •\ d (2ax — x 2 ) = (5 Act — Ax) dx ; 
 
 or 2a — 2x = 5Aa — Ax. 
 
 .\ 2a = hAa ; (1) 
 
 and — 2 = — A ; (2) 
 
 or ^4=f, from (1), and .4 = 2, from (2). 
 
 These values of A being incompatible with each other, 
 we infer that the differential cannot be integrated in this 
 form. 
 
 14. dy — (2b + Sax 2 — 5a 8 )"* (2ax — 5x 2 ) dx. 
 
 y — %(2b + 3ax* — ox*)\ 
 
 15. dy = (3ax* + 4fa»)* (2az + 4te*) <fo. 
 
 y — 4 (Sax 2 + 4fa*)i 
 
 16. dy = ax*dx + -^. y = a ^ + x \ m 
 
 2yx & 
 
EXAMPLES. 249 
 
 17. dy = 5 (5x 2 — 2x)* xdx — (5x 2 — 2x)% dx. 
 Here y = / (dx 2 — 2x)^ (5x — 1) dx. 
 
 y = 1 (538 - 2x% 
 
 134. A Differential may often be brought to the 
 
 form required in (3) of Art. 130, by transposing a variable 
 factor from the differential factor to the variable . factor, 
 or vice versa. 
 
 2adx 
 
 18. dy = 
 
 x v2ax — x 2 
 
 Here y = I (2ax — x 2 )~? 2ax~ i dx 
 
 — /'_ (%ax-\ _ 1)4 (_ 203-2) ax; 
 .. y = - 2 {2ax~ 1 — 1>* [by (3) of Art. 130] 
 
 __2 V2ax — x 2 
 
 x 
 
 — _ -, XUX X 
 
 19. dy = =• y = — — - ■ 
 
 (2ax — rz 2 ) 3 a y2ax — x 2 
 
 , axdx ax 
 
 20. efy = — — # = 
 
 (2bx + a 2 )* £ V2bx + a: 2 
 
 7 adz 2a \/Ux + 4c% 2 
 
 21. dy — r ' ■'• ■■ ' - ' V — — " ^ 
 
 a \/3fo + 4^ 2 3 ^ 
 
 23 - * = &Pfi§S" [See (4) of Art. 130.] 
 
 A? (a -\- bx + ex 2 ) . , - « 
 
 23 - * = T%H& y = log (0 + 2*«). 
 
250 EXAMPLES. 
 
 bx*dx 
 
 y = A log (3* + 7) = log (3x* + 7)*. 
 25. #«££j- » = log («• + *)*. 
 
 27. dy = (b — xfx^dx. 
 
 28 - ^ = Sfc- y = log («. +&f)* 
 
 <fa; 
 
 29. dy = 2 log 8 a: — # = £ log 4 #• 
 
 x 
 
 30. <fy = a 2 * log adx. [See (4) of Art. 131.] 
 
 Here y = J a* log adx = f%&* log ad (2x) = |aH 
 
 31. dy = 3a* 9 log axdx. y = \a x% . 
 
 32. dy = ba**dx. y = ~ 
 
 9 u 3 log a 
 
 33. dy = b&dx, y = 56*. 
 
 34. dy = cos 3a*fo. [See (6) of Art. 131.] 
 
 y = J sin 3#. 
 
 35. <fy = cos (a?) xdx. y •= \ sin (a£). 
 
 36. tf# = e 8in * cos xdx. y = e eln *. 
 
 37. dy — — e°' 08aj sin a*fc. y = ecos*. 
 
 38. dy = sin 2 (2a;) cos (2a;) dx. 
 
 Here # = f\ sin 2 (2a;) cos (2a;) d (2a;) = f sin 3 (2a;). 
 
EXAMPLES. 251 
 
 39. dy = cos 3 (2x) sin (2x) dx. y — — \ cos 4 (2a;). 
 
 40. % = sec2 ( x f x ^ x - y — i tan (^) 3 « 
 
 41. d?/ = sec (3a;) tan (3x) dx. y = % sec (3a;). 
 
 42. % = sin (ax) dx. y = - vers («#). 
 
 43. dy = tan ado;. «/ = log sec x. 
 
 44. dy = sin sec 2 Odd. y = sec 0. 
 
 45. dy = — J^rr=. [See (14) of Art. 131.1 
 
 y = sin -1 (2a;). 
 .» 7 axdx a . 1 , 9X 
 
 y = 'UTT3 y = 2 sm ^ 
 
 o;te 
 
 47. d# = 
 
 A/2 — 4o* 
 
 Here y = J 
 
 $dx 
 
 V2 . Vl - 2a>3 
 
 \/2 • fate? 
 
 = —-/'- 
 
 V2 • f ^ V2 • Vl - 2a? 
 
 = ^ a/F- 2aT 3 ' 
 .«. y = -J sin -1 a/2o?. 
 
 , , dx . 3a; 
 
 48. % = — = • y == 4 sm -1 — =• 
 
 A/2 -9a* * * a/2 
 
 — xdx 1 
 
 A/^5aT 4 ' ^ ~" 2^5 ~ Vyf 
 
 49. rfy = —_= y = — L — cos" 1 (^£ ^ 
 
 50. dy=- ;** .. . y = 2 cos"* ( x ft 
 Wax — x 2 V V a j 
 
EXAMPLES, 
 
 r 1 j 3«?:e g . 3a; 
 
 52. o> = Y+-&' y = 4 tan_1 a;8 - 
 
 53. ay = - — y = vers -1 3a;. 
 
 y V6a; -9a* * 
 
 ■ a j 2a# 
 
 64. ay = 
 
 x ^Sx* — 5 
 Here y . /" 2 -^== = A f_V§*L- 
 
 J x V5 vv - 1 V5* 7 vs vi^=i 
 
 2 
 •*• tf — 7/g sec -1 (a; v-f). 
 
 55. oty = ========== j y = V2 cosee -1 (a; Vf). 
 
 * VW — aa? 
 
 56. ay = "7 ' * # = - t= cosec -1 (a; V-M-). 
 
 * V14*— 9 V3 
 
 __ , 3aftfa? , , 9a* 
 
 58. ay = — • y = a covers -1 3a; 2 . 
 
 9 V6 - 9a* ^ 
 
 59. dy = Sar^^^ y _ 4 y^ vers" 1 (6**). 
 
 V 2a;* — 6a* 
 
 7 1 — sin a; 7 , 
 
 60# rfy = x+~qxkx V = g ^ + C ° S * )# 
 
 61. o> = r ~ ys:|tHr*jft 
 
EXAMPLES. 253 
 
 ** -r 3diB 3 ^ 
 
 63. ♦ -j-^- '"yg *■**■** 
 
 65. c?y = — • y — 2x V« 2 + a*. 
 
 * V« 2 + a* 
 
 66. t??/ = tan 2 x sec 2 #d#. # = i tan 3 x. 
 
 67. dy = ± -g* y = log 3 — a* + j- 
 
 ™ (# — 2) dx n /- , 4 
 
 68. rfy = -fi — y = 2 V% + ~t=- 
 
 69. dy = ^~-^ y = i^-a? + tan-»* 
 
 70. rfy = _%--*. 
 
 * 1 + a; + # 3 
 
 Here «/ = J 
 
 dx 
 
 i + (* + if 
 
 a j 
 
 «£ 
 
 
 17 
 
 „_ 7 (m + wz) da; 
 
 72. dn = * — I—— k 
 
 J a 2 + x 2 
 
 y = 7 tan_1 1 + I log ^ 2 + ^ 
 
254 TRIGONOMETRIC REDUCTION. 
 
 135. Trigonometric Reduction. — A very slight ac- 
 quaintance with Trigonometry will enable the student to 
 solve the following examples easily. After a simple 
 trigonometric reduction, the integrals are written out bv 
 
 inspection. 
 
 1. dy = tan 8 xdx. 
 
 Here y = / tan 3 xdx = I (sec 2 x — 1) tan xdx 
 
 = / [sec 2 x tan xdx — tan xdx] 
 
 = | tan 2 x — log sec #. [See (9) of Art. 132.] 
 
 2. dy = tan 4 a*fc. 
 
 y = i tan 8 a: — tan x + a\ [(15) of Art. 132.] 
 
 3. dy = tan 5 xdx. 
 
 y == J tan 4 a; — £ tan 2 x + log sec #. 
 
 4. dy = cot 3 a;d#. y = — J cot 2 a; — log sin #. 
 
 5. dy = cot 4 a*fc. . y = — J cot 3 a; + cot # -f- x. 
 
 6. rfy =s cot 5 xdx. 
 
 y •=. —\ cot 4 a; + \ cot 2 # + log sin x. 
 
 7. r/y = sin 3 xdx. 
 
 Here y = / \ (3 sin a; — sin 3a;) rfa; ; (from Trig.), 
 
 .*. y = -jij cos 3a; — | cos a; ; 
 
 or ^ = J cos 3 x — cos a;.* 
 
 8. dy = cos 3 a:rfa;. # = sin a: — \ sin 3 a\ 
 
 * By employing different methods, we often obtain integrals of the same 
 expression which are different in form, and which sometimes appear at first sight 
 not to agree. On examination, however, tley will always be found to differ only 
 by some constant ; otherwise they could not have the same differential. 
 
EXAMPLES. 255 
 
 9. dy — cos 4 xdx. 
 
 Here y = j (cos 2 xf dx = J \ [1 -f 2 cos 2# 
 
 + cos 3 &e] dx [See (17) of Art. 132.] 
 
 = i% + i sin 2x + -§ y cos 2 2a*2 (2a?) 
 
 = Ja? + I sin 2a; + J [a + J sin 4s] ; [by (17) of Art. 132.] 
 ,\ y = ^ sin 4a; + £ sin 2z + f a\ 
 
 10. % = sin 4 #cfa?. y = Jg sin 4a; — £ sin 2a; + fa;. 
 
 11. dy = sin 5 a;^. y = — cos a; + f cos 8 x — -J- cos 5 a;. 
 
 12. dy = cos 5 a;$z;. # = sin x — § sin 3 x + \ sin 5 a;. 
 
 13. dy = sin 3 a; cos 3 xdx. 
 
 y = y£ ? cos 6x — -fa cos 2a;. (From Ex. 7.) 
 
 dx 
 
 14. dy = - x— . V = I sec 2 a? + log tan a;. 
 
 * sin x cos 3 a; ^ ^ *> 
 
 15. dy = sin 3 a; cos 2 a;da;. «/ = i cos 5 a; — J cos 3 x. 
 
 16. f?y = cos 5 x sin 5 #$£. 
 
 y = _ ?2L? (J — i cos 2 a; + £ cos 4 a?). 
 
 17. dy = sin 6 x cos 3 a*?a: = (sin 6 x — sin 8 x) cos a;da;. 
 
 y = sin 7 x (| — | sin 2 a?). 
 
 . ., _ 7 sin 5 x 7 
 
 18. tf */ = — 5- da;. 
 
 ^ cos 2 x 
 
 TT /*(1 — 2 cos 2 x -f cos 4 a;) sin xdx 
 Here y = I - 4 
 
 * J cos 2 a; 
 
 == sec x -\- 2 cos x — \ cos 3 a;. 
 
 19. dy = cos 7 ayfe. 
 
 y = sin a; — sin 3 a; -f § sin 5 x — \ sin 7 a\ 
 
CHAPTER II. 
 
 INTEGRATION OF RATIONAL FRACTIONS. 
 
 136. Rational Fractions. — A fraction whose terms in 
 volve only positive and integral powers of the variable ia 
 called a -rational fraction. Its general form is 
 
 ax" 1 + bar- 1 -f cx" l ~ 2 + etc. . . . I . . 
 
 a'x n + b'x"- 1 + cV*- 2 + etc. . . . I" ^ ' 
 
 in which m and n are positive integers, and a, b, . . ., 
 a', b\ . . . are constants. 
 
 When m is > or = n, (1) may, by common division, be 
 reduced to the sum of an integral algebraic expression, and 
 a fraction whose denominator will be the same as that of (1) 
 and whose numerator will be at least one degree lower than 
 the denominator. For example, 
 
 X s , , 2^ + x 
 
 - x 2 + x 
 
 a« _ a? + a; + 1 a* — x 2 + x + 1 
 
 The former part can be integrated by the method of the 
 preceding chapter; the fractional part may be integrated by 
 decomposing it into a series of partial fractions, each of 
 which can be integrated separately. There are three cases, 
 which will be examined separately. 
 
 137. Case I. — When the denominator can be re- 
 solved into n real and unequal factors of the first 
 degree 
 
 f (oc\ (It 
 For brevity, let J \ ; . - denote the rational fraction 
 (p(x) 
 
 whose integral is required, and let (x— a) (x—b) ... (x—l) 
 
 be the n unequal factors of the denominator. Assume 
 

 DECOMPOSITION OF FRACTIONS. 25? 
 
 m = _A_ + J- + _£_ . JL m 
 
 <j>(x) x — a x — b x — c ' x — V K ' 
 
 where A, B, C, etc., are constants whose values are ;o be 
 determined. 
 
 Clearing (1) of fractions, by multiplying each numerator 
 by all the denominators except its own, we have 
 
 f(x) = A (x-b)(x- c). . . (x-l) + B (x—a)(x—c) . . . (x—l) 
 
 + etc. + L (x—a) {x—b) . . . (x—k), (2) 
 
 which is an identical equation of the (n—l) th degree. To 
 find A, B, C, etc., we may perform the operations indicated 
 jn (2), equate the coefficients of the like powers of x by the 
 principle of indeterminate coefficients in Algebra, and solve 
 the n resulting equations. The values of /J, B, 0, etc., thus 
 determined, being substituted in (1) and the factor dx intro- 
 duced, each term may be easily integrated by known 
 methods. 
 
 In practice, however, in this first case, there is a simpler 
 method of finding the values of A, B, etc., depending upon 
 the fact that (2) is true for every value of x. If in (2) we 
 make x = a, all the terms in the second member will re- 
 duce to 0, except the first, and we shall have 
 
 f(a) = A (a — b) (a — c) ... (a — I), 
 
 or A = 
 
 /<«) _ />) 
 
 (a — b) (a — c) ... (a — I) 0' (a) 
 
 In the same way, making x = b, all the terms of (2) 
 disappear except the second, giving us 
 
 f(b) = B(b-a)(b-c)...(b-l), 
 
 b = li i] - - £SL 
 
 \b — a)(b — c)...(b — l)~<p' (b) 
 
258 CASE T. 
 
 Or, in general, the value of L is determined in any one 
 
 T /It 
 
 of the terms, -, by substituting for# the corresponding 
 
 root / of <\> (x) in the expression ^rr\\ i. e., L = QL 
 
 EXAM PLES. 
 
 1. Integrate dy = gq ^+ ^ + , - 
 
 In this example, the roots * of the denominator are found 
 by Algebra to be — - 1, — 2, — 3. 
 
 A 2 s + 6Z 2 + lis + 6 = (x + 1) (a + 2) (x + 3). 
 Assume 
 
 2 s + 62* + llz + 6 2+1^2+2^2+3 ll 
 A 2* + l = .4(2 + 2)(2 + 3) +#(2+ l)(2 + 3) 
 
 + ff(s + l)(* + 2). 
 Making 2 =a — 1, we have 2 = 2^4, .•. ^4=1. 
 
 x = — 2, « " 5 as — #, a # as — 5, 
 " 2 = - 3, " " 10 = 2C, .-. G as 5. 
 
 Substituting these values of -4, 2?, C, in (1), and nralti' 
 plying by dx, we have 
 
 (a* + 1) dx 
 
 J 2 s + 
 
 6a; 8 +112 + 6 
 
 / (fo _ /* dx p dx 
 
 2 + 1 J 2 + 2 + V 2 + 3* 
 
 .'. ST = log (2 + 1) - 5 log (2 + 2) + 5 log (2 + 3) 
 
 _ (s + l)(2 + 3)» 
 
 - 10 & {x + 2)5 . 
 
 * If the factors of the denominator are not easily seen, put it eqnal to 0, and 
 solve the equation for x; the first root may be found by trial, x minus each of the 
 several roots in turn will be tbe factors. (Sec Algebra.) 
 
2. Integrate dy 
 
 DECOMPOSITION OF FRACTIONS. 259 
 
 adx 
 
 a" 
 
 a 
 
 y = i^hra = los \/^+ 
 
 3. Integrate A, = ^g^ 
 
 _ . x 2 — x— 2 
 
 4. Integrate * = gq^TT 
 
 y = kg (* - lj*(« + *)*. 
 
 5. Integrate dy = ftfe y = ± log g^g). 
 
 138. Case II.— When the denominator can be re- 
 solved into n real and equal factors of the first degree. 
 
 fix) 
 Let the denominator of the rational fraction \ ) { con- 
 
 (X) 
 
 tain n factors, each equal to x — a. ' 
 
 Assume 
 
 m _ _j__ . _?___ + ^ _ 
 
 <t> (x) (x — a)" (z — a)" -1 (x — a)" -2 
 
 Clearing (1) of fractions by multiplying each term by the 
 least common multiple of the denominators, we have 
 
 f(x) = A + B{x — a) + C(x-af 
 
 + . . . L (x-ay-\ (2) 
 
 which is an identical equation of the (n — l) ih degree. To 
 find the values of A, B, C, etc., we equate the coefficients 
 of the like powers of x, as in the preceding Article, and 
 solve the n resulting equations. The values of A, B, C, 
 
2G0 CASK //. 
 
 etc., thus determined, being substituted in (1), and the 
 factor dx introduced, each term may be easily integrated 
 bj known methods. 
 
 In this case we cannot find the values of A,B,C, etc, by the second 
 method used in Case I, but have to employ the first. When both 
 equal and unequal factors, however, occur in the denominator, both 
 methods may be combined to advantage. 
 
 1. Integrate dy = 
 
 EXAMPLES. 
 
 (2 — 3 i 2 ) dx 
 
 A 2-3^ A . B , C m 
 
 ASSUmG (^)-3 = (^)-3 + (^)^^' CI) 
 
 .•. 2-3^ = ^ + ^ + 2^+6^2 + 46^ + 4(7. 
 
 .\ A + %B + 46' = 2. (2) 
 
 B + 46 = 0. (3) 
 
 C=-3. (4) 
 
 Solving (2), (3), and (4), we get 
 
 A = — 10, B = 12, C = — 3. 
 
 Substituting these values of A, B, and C in (1), and 
 multiplying by dx, we have 
 
 (2 — 3a 2 ) dx _ Itidx 12dx ddx 
 
 (x + 2) 3 = (x + 2) 3 + (x + 2)2 x + % 
 
 (2 — 3z 2 ) dx 
 
 P (2-3x 2)d 
 * ~ J (x + 2) 3 
 
 {x + 2)2 a; + 2 
 2. Integrate <Z? = jJ*+^* 
 
 -3 1og(z + 2). 
 
 (**-»)•(* -I) 
 
DECOMPOSITION OF FRACTIONS. 261 
 
 Assume 
 
 (x - 2)2 {x-l)~ (x-2f' i ' (x-2)^ x- 1 W 
 
 .-. x* + x = A (x — l) + B(x-2) (x - 1) 
 
 + C(x-2y. (2) 
 
 Here we may use the second method of Case I, as follows : 
 
 Making x = 2, we find A = 6. 
 % = \> " « = 2. 
 
 Substituting in (2) for A and (7 their values, and making 
 r = 0, we find 
 
 = _ 6 + 2B + 8 ; .-. B = —1. 
 
 Substituting in (1), and multiplying by dx, we have 
 
 (x 2 + x) dx 
 
 y — j (x _ 2)2 (z 
 
 i) 
 
 /* 6dx P dx r 2dx 
 
 J ( x — 2f~ J x~—2 + J x~^l 
 
 — log (x — 2) + 2 log(# — 1). 
 
 a — 2 
 
 6 -, (z-1) 2 
 
 (3x — 1) ^ 
 
 3. Integrate dy = 
 
 (x-df 
 
 4. Integrate dy = -__-<&. 
 .V = log [jb (x — 3) 2 ]*. 
 
Mfl CASE m. 
 
 139. Case III. — Whsm sonic of the simple fme&n 
 of the denominator are imaginary. 
 
 The methods given in Arts. 137 and 138 apply to the 
 case of imaginary, as well as to real factors; but as the cor- 
 responding partial fractions appear in this case under an 
 imaginary form, it is desirable to give an investigation in 
 which the coefficients are all real. Since the (Jenorniiuitor 
 is real, if it contains imaginary factors, they must enter in 
 I >airs ; that is, for every factor of the form x ± a + bV — 1, 
 there must be another factor of the form x ± a — bV~- 1* 
 otherwise the product of the factors would not be real. 
 Every pair of conjugate imaginary factors of this form gives 
 a real quadratic factor of the form (x ± a) 2 + b'K 
 
 Let the denominator contain n real and equal quadratic 
 factors. Assume 
 
 f{x) __ Ax + B .Ox + D 
 
 (x) ~ [{x ± af + P] n + [(x ± af + b] n ~ l 
 
 Kx+L (1) 
 
 (x ± «) 2 4- & 
 
 If we clear (1) of fractions by multiplying each term by 
 the least common multiple of the denominators, we shall 
 have an identical equation of the (2n — \) th degree. 
 Equating the coefficients of the like powers of x, as in the 
 two preceding Articles, and solving the 2n resulting equa- 
 tions, we find the values of A, B, C, etc. Substituting 
 these values in (1), and introducing the factor dx, we have 
 a series of partial fractions, the general form of each being 
 
 (Ax + B) dx 
 
 l(x ±^7TPp 
 
 in which n is an integer. 
 To integrate this expression, put x±a = z; .: x = z^a, 
 
 * Called conjugate imaginary factors. 
 
EXAMPLES. 263 
 
 dx = dz, (x ± df = z\ Substituting these values, we 
 have, 
 
 r {Ax + B)dx _ j\A z qF Aa + B) dz 
 J [(x ± af + Pf ~ J (2 2 + Vf 
 
 r Azdz r (B q= Aa) dz 
 
 ~ J (? -t- Vf + J {f + ^2)n 
 
 ^ f A\A % 
 
 . m 2(n — 1){# + &) n -i + J (f + &)*' 
 
 (when .4' = .6 =F ^1«) ; 
 
 bo that the proposed integral is found to depend on the 
 integral of this last expression; and it will be shown in 
 Art. 151 that this integral may be made to depend finally 
 
 u P on ff^rp' s™ 1 ^ J tan_1 V ( Art 132, 3# ) 
 
 EXAMPLES. 
 
 1. Integrate dy = -g r-' 
 
 The factors of the denominator are 
 
 (x — 1) and (z 2 + x + 1). 
 therefore assume, 
 
 z A Bx + C 
 
 (1) 
 
 a*— l"~ x— 1 ^ a 2 + # + 1 
 z = Az* + Ax + A +Bx* + Cx — Bx - C. 
 A + B = ; yl ~ £ + C = 1 ; ^ - C = 0. 
 
 1 . 
 3 ? 
 
 * = -*; ^=f 
 
 r zdz r dx r ( x — i ) <%g 
 
 y — «/ ^~~ 1 — ./ *a> — 1 t/ *aj» + a + 1 
 
164 EXAMPI. 
 
 -W-i-iffcSfv (2) 
 
 (by changing the form of the denominator.) 
 
 Put x -f $ = z, then x — 1 = z — f , and dx = dz, and 
 the second term of (2) becomes 
 
 i A g ~ t) d z _ r zdz p dz 
 
 = — i log (* 2 4- 1) + 4= tan_1 ~F ( Art m > 3 ") 
 V3 V3 
 
 = _ i lo g ( 3 , + , + l) + -Ltan->^ ; ±i, 
 
 (by restoring the value of z). 
 Substituting in (2), we have, 
 
 y = i [log (* - 1) - i log & + x + i) 
 
 V3 J 
 2. Integrate <fy = — ^—y 
 
 To find the factors of the denominator, put it = and 
 Bolve with respect to x 2 ; thus, 
 
 at + & — 2 = 0, or a 4 + a 2 = 2. 
 
 .-. z 2 = - \ ± f = 1 or — 2. 
 
 ... & + s 2 — 2 = (a? - 1) (z 2 + 2). 
 
 A™ s* + *-2 = ^+i + ^ri + ?+a- ' (1) 
 
 Hence x* = A (x — 1) (z 2 + 2) + B (x + 1) (x 2 + 2) 
 
 + (G« -h 2?) (a - 1) (a + 1). (2) 
 
EXAMPLES. 265 
 
 We may equate the coefficients of the like powers of x, 
 to find the values of A, B, C, D, or proceed as follows : 
 
 Making x a= — 1, we find A = — £. 
 
 " a? = 1, u " B = f 
 
 Substituting these values of yl and 5 in (2) and equating 
 the coefficients of X s and x 2 , we have 
 
 6C=0 and 6Z> = 4; 
 
 .-. O = and 2> = £ 
 
 , r dx , r dx 9 r 
 
 h 1 b t/ 2 — 1 * e/ # 2 + 2 
 
 a; 
 
 
 EXAM PLES. 
 
 7 (a; — 1) ffe . (a; + 4)* 
 
 (z-l)t 
 
 3 - %-^-r-fcr^;- y = log 
 
 4 
 
 a 2 + 6a: + 8 
 
 (2x + 3) dx 
 
 X* + x* — 2x 
 
 (3a; 2 — 1) dx 
 
 x (x — 1) (x + 1) 
 
 «$£ 
 
 a; 2 + te 
 
 (2 + 3a; — 4a; 2 ) tfa; 
 
 ' 4:X — X S 
 
 y = 
 
 (5* - 2) <fc 
 
 a;* (a; + 2)* 
 ?/ = log [a;* (2 + a;)* (2 — a;)]. 
 
260 EXAMPLES, 
 
 7. *i& + »* 
 
 y the — 3* 
 
 t/ = alogx- ?-+J. i og ( a a _ ^ 
 
 y **- 6x + 8 ,-log (7 -^j. 
 
 9. <fr = , ^___. 
 
 y = - 2iF^D + lo s (*-«>* (■ + ■)*■ 
 
 10. * = ^±i±IL*. 
 
 v x(x — If (x + If 
 
 3 + x %i 
 
 11. dy = 
 
 12. dy = 
 
 (» + 2)»(c+..l) 
 
 3 
 
 x + 3 
 dx 
 
 ■ 13. <fy = -= — 
 
 * (x- 2)* (a? + 3)2 
 
 + t! 7 log (a + 3). 
 
 14. dW = (^- 4g + 3 l^. 
 
 * ^ — 6.t* + 9x 
 
 y = log \x (x - 3ff. 
 
15. dy 
 
 EXAMPLES. 867 
 
 xhlx 
 
 (x + 2) 2 (x + 4)2 
 
 5a; + 1 2 , , /H 4\* 
 
 a^fa; 
 
 16 - * as <. + i)<*+ij 
 
 i "A 1 1 (^ + 1)* 
 
 # = i tan -1 x + log ^ — J—-L. . 
 
 17 ' ^-^3_^ + 2 ^_2' 
 
 (».— 1)1 l a? 
 
 y = log -* — '— tan" 1 -— < 
 
 (z 2 + 2)* 3a/2 a/2 
 
 no J ( x% + #) ^ 
 
 y = § tan-* a; + log f- -i-r 
 
 (a? + 1)tV 
 
 19. dy = ^ — ■ — L — -• 
 
 * x s + x 2 + x + 1 
 
 (a; -J- 1)1- " .- . . 
 
 y = lo s f; — di - i tan * 
 (a* + iy 
 
 OA , 9z a + 9x — 128 7 
 
 20. dy = -- ~ dx. 
 
 * ic 3 — 5a; 2 + Sx + 9 
 
 . (x - 3)» 5 
 
 «- •• /iXClX 
 
 21. % 
 
 (a* + l) (^ + 3) 
 , ^ + l\i 
 
 * = log U+t) • 
 
268 EXAMPLES, 
 
 22. ft s (*-9/* 
 
 * a 8 — 4 
 
 y = I' + log [(z + 2)f (a: - 2)1], 
 
 23. dy = 
 
 (* + l)(3 + 2)(rf + l)' 
 
 ,,„[a±|to*j, A ^ 
 
CHAPTER III. 
 
 INTEGRATION OF IRRATIONAL FUNCTIONS B\ 
 RATIONALIZATION. 
 
 140. Rationalization. — When an irrational function, 
 which does not belong to one of the known elementary 
 forms, is to be integrated, we endeavor to rationalize it ; 
 that is, to transform it into an equivalent rational function 
 of another variable, by suitable substitutions, and integrate 
 the resulting functions by known methods. 
 
 141. Function containing only Monomial Surds.— 
 
 When the function contains only monomial surds, it can 
 be rationalized by substituting a new variable with an 
 exponent equal to the least common multiple of all the 
 denominators of the fractional exponents in the given 
 function. 
 
 For example, let the expression be of the form, 
 
 , ax 171 + hx™ , 
 a'x c + b'x e 
 Put x = z™ nce ; 
 
 .^ rj^rn — gm'nce : %n — ^n'mce j %c — gmnc'e : %e — z mnce' 
 
 dx = mncez m ™ e -Hz. 
 
 d Z m'nce _i famn'ce 
 
 Hence dy = -j-^—,^ mnce^^dz ; 
 
 which is evidently rational. 
 
270 FUNCTIONS CONTAINING BINOMIAL SURDS. 
 I x \ 
 
 1. Integrate dy = ? dx. (1) 
 
 1 — x* 
 
 Put x = z«; 
 
 then a;^ = 2 s , a£ = z 2 , and *fo = §zHz ; 
 
 (1 - 2?) 62?<fe 
 
 efy = 
 
 1 -z* 
 
 dz. 
 
 \ 1 -f- Z / 
 
 Integrating by known methods, and replacing z by its 
 value, we have 
 
 * = 6 Lt + 5 " 1 + 1 " a + ** ~ ^ (1 + ^J ' 
 
 o t . i. ^ 3a£<te 
 2. Integrate rf«/ == — j - 
 
 2a: 5 — a;* 
 y = - 18 [jL + | + ~ + ^ + 16^ + 32 log(2-s*)J . 
 
 142. Functions containing only Binomial Surds 
 
 of the First Degree.— When the function involves no 
 
 surd except one of the form (a + bx) n , it can be rationalized 
 as in the last Article, by treating a + bx as the variable. 
 And therefore can be integrated. 
 For example, let the expression be of the form, 
 
 , x n dx 
 dy = — , 
 
 va + bx 
 
 vhere n is a positive integer. 
 
 Put a + bx — z* ; 
 
 ., , 2zdz z 2 — a , n (z* — aY 
 
 then a# = —5— , x s= — =— , and af = — — =- — -• 
 £ b b n 
 
FUNCTIONS CONTAINING BINOMIAL SURDS. 271 
 
 x n dx 2(z* — a) n dz 
 
 V a + bx ° 
 
 This may be expanded by the Binomial Theorem, and 
 
 each term integrated separately. It is also evident that the 
 
 expression 
 
 x ll dx 
 
 (a + IxY 
 
 can be integrated by the same substitution. 
 dx 
 
 1. Integrate dy 
 dy 
 
 xVl + x 
 Put 1 + x = z 2 ; then dx — 2zdz and x = z 2 
 dx _ 2dz 
 
 . f — . I ' Vl +» — 1 
 
 or y = log = log — * 
 
 2. Integrate dy = — -.• 
 
 * (1 + 4ar)t 
 
 Put (1 + 4a) = z 2 ; 
 
 then <fe = ^, a* = ( *^* y , (1 + 4a)t = * 
 
 . , _ 1 (*»-!)'<** 
 - tfy -128 z* 
 
 i r 3 
 
 = 128 ! ** " 3 + z 2 
 
 a* 
 
 = rr 8 [*-- 3 -! + »]. 
 
j;-J FUNCTIONS CONTAINING TRINOMIAL SURDS. 
 or 
 
 143. Functions of the Form , , where n 
 
 is a Positive Integer. ( a + to2 )~ 
 
 Put a + &* s= '-'# 5 
 
 , , , zdz 9 z 2 — a tB (z 2 — «) n 
 
 then xdx = -r- , z 2 = — j— , a 2 " = - — F — — • 
 
 6 0" 
 
 a^ +1 <fc (z 2 — &J" ds 
 
 (« + to 2 )* 
 
 which may be expanded by the Binomial Theorem, and each 
 term integrated separately. It is also evident that the 
 expression 
 
 x 2nn dx 
 
 (a + hx 2 Y 
 
 can be integrated by the same substitution. 
 
 x^dx 
 1. Integrate dy = 
 
 A/1 -a 2 
 Put 1 — x* = z 2 \ then xdx = — zdz, x 2 = 1 — A 
 
 .\ dy = -0L= = - (1 - z 2 ) dz. 
 
 /. y = J*» - * = ${1 - z 2 )i - (1 - z 2 )i 
 
 2. Integrate ay = - • w = * —„ • 
 
 (« + cz 2 )* * 3c 2 (a + ex 2 )* 
 
 144. Functions containing only Trinomial Surds 
 of the Form Va + bx ± x 2 . 
 
 There are two cases, according as z 2 is + or — . 
 
FUNCTIONS CONTAINING TRINOMIAL SURDS. 273 
 
 Case I.— When x 2 is +. 
 
 Assume Va -f bx + x 2 = z — x ; 
 
 then a 4- bx = z 2 — 2zx : .\ x = 7 — — • 
 
 i + 20 
 
 2 (z 2 -f bz + «) tfz 
 
 efo; = 
 
 and v « + bx -f- a; 2 = 2 
 
 (b + 2*) 2 
 
 2 2 — a _ z 2 -\- bz -\- a 
 b~+ 2z ~ 2z + b 
 
 The values of x, dx, and Vci -\- bx -\- x 2 being expressed 
 in rational terms of z, the transformed function will be 
 rational, and may therefore be integrated and the z replaced 
 by its value Va + bx -f x 2 -f x. 
 
 Case II. — IVlien x 2 is — . 
 
 Let a and (3 be the two roots of the equation 
 x 2 — bx — a = ; 
 then we have # 2 — bx — a = {x — a) (x — (3); 
 .*. a -{- bx — x 2 = — (x — a) (x — j8) 
 = (x - a) 03 - a;). 
 
 Assume a/« + bx — x 2 = V(x — a) ((3 — x) 
 = (x — a) z; 
 .'. (x — a) ((3 — x) = (x — a) 2 z% 
 or (]3 — x) = (x — a) z 2 ; 
 
 az 2 + (3 
 
 whence, 
 
 and 
 
 - z 2 + 1 ' 
 
 - « - * ( Z 2 + 1)2 • 
 
 ^a + bx — x* = [ z2+1 «| 
 
 _ (£-«)* 
 
 " 2 2 + 1 
 
274 EXAMJ'/./s. 
 
 The values of fc, dx, va -f bx — x 2 , being expressed in 
 rational terms of z, the transformed function will be rational. 
 
 1. Integrate dy = 
 
 V(i + bx + x 2 
 
 Assume v a + bx + x 2 = z — x; 
 
 then, as in Case I, we have 
 
 a -f- bx = z 2 — 2zx ; .*. x = = — • 
 
 b + 2z 
 
 7 _ 2 ( z * + bz + a) dz 
 
 clx ~ {bTW 
 
 ,7 - 2 Qg 2 + ^ + «) dz x (2z + 5) 
 •'• "y ~ {}, + 2 Z )2 x (z 2 + bz + a) 
 
 2dz 
 
 b + 2z b 
 
 2 + Z 
 
 2 + * 
 = log |J + a; + V« + te + aM. 
 
 If 5 = 0, we have 
 
 y = f-y== = lo g(* + Va + a 2 )? 
 v« + a 3 
 
 and if a = 1, we have 
 
 da 
 
 ?/ 
 
 ^== = log (x + vrr* 2 ). 
 
EXAMPLES. 275 
 
 2dz 
 Had we integrated the expression ^ — - without dividing both 
 
 terms of it by 2, we would have found for the integral the following : 
 y = \ag(b + 2z) = log [b + 2x + 2<\/a + bx + x*], which differs from the 
 above integral only by the term, log 2, which is a constant. (See 
 Note to Art. 135 ) 
 
 2. Integrate dy = — 
 
 \a + bx — x 2 
 
 Let a and 13 be the roots of x 2 — bx — a = ; then, as in 
 Case II, we have 
 
 Vet + bx — x 2 = V(x — «) (j3 — x) = (x — a) z. 
 
 az 2 + j3 
 
 (j3-s) = (a -«)**; 
 
 s 2 + 1 
 
 (s 2 -f- l) 2 s 2 + 1 
 
 , 2 (« - j3) «f« (s 2 + 1) 2dz 
 
 (z 2 + l) 2 (0-«)s 1 +*' 
 
 y ~ J <s/a + dx-x* ~ ~~ J H 7 ? 
 
 = — 2 tan -1 s = — 2 tan -1 
 dx 
 
 \— x - 
 
 \ x — a 
 
 fir, 
 
 3. Integrate dy = 
 
 Vl + x + a,* 
 
 Assume vl + a; + ^ = 2— », and we have, as in Case I, 
 
 z 2 — 1 7 2 (s 2 + s + 1) dz 
 
 x = 
 
 1 + 2s 
 
 _ 2 (z 2 + z + 1) <fe (1 + 2s) (1 + 2s) _ 2<fe 
 ,# - rf y = (i + 2s) 2 (s 2 + z + 1) (* 2 - 1) ~~ 2 2 - X 
 
270 BINOMIAL DIFFERENTIALS, 
 
 " ^ xVl + x + 4 ~ J z-l~ J z + 1 
 
 m log z —\ = log *-i + v pSg 
 
 . 3z 
 
 - °S 2 + * + tyl + *'+*" 
 
 145. Binomial Differentials.- Expressions of the form 
 
 dy = x m (a + bx n ) p dx, 
 
 in which m, n, p denote any numbers, positive, negative, or 
 fractional, are called binomial differentials. 
 
 This expression can always be reduced to another, in 
 which m and n are integers and n positive. 
 
 1st. For if m and n are fractional, and the binomial of the 
 form 
 
 ar$(a + bx^f dx, 
 
 we may substitute for x another variable whose exponent is 
 equal to the least common multiple of the denominators of 
 the exponents of x, as in Art. 141. We shall then have an 
 expression in which the exponents are whole numbers. 
 Thus, if we put x = z% we have 
 
 ar*(a + bx^f dx = Gz~*(a + bz*fdz, 
 
 in which the exponents of z are whole numbers, and the 
 exponent of z within the parenthesis is positive. 
 
 2d. If n be negative, or the binomial of the form 
 x m (a + bx- n ) p dx } 
 we may put x = - , and obtain 
 
 x m (a + bx-^Y dx = — z-' H ~ 2 (a + bz 1l Y dz, 
 
 in which the exponents of z are whole numbers, and the 
 one within the parenthesis is positive. 
 
CONDITIONS FOR RATIONALIZATION. 271 
 
 3d. If x be in both terms, or the binomial is of the form 
 x m {ax t + bx n ydx, 
 we may take x 1 out of the parenthesis, and we shall have 
 X m+ P t ( a _j_ fa*-ty dx, 
 
 in which only one of the terms within the parenthesis con- 
 tains the variable. 
 
 146. The Conditions under which the General 
 Form p 
 
 dy = x m (a + hX 11 )? dx, 
 
 can be rationalized, any or all of the exponents being frac- 
 tional. 
 
 (1.) Assume a + bx n = z q . 
 
 2 
 
 Then (a + bx n ) q = z*. (1) 
 
 41 (z q — a\ k 
 
 Also x = [—t—j , 
 
 , m 
 
 and ar = (—J- (3) 
 
 Multiplying (1), (2), and (3) together, we have 
 
 m+l 
 
 dy = x"} (a + bx H )i dx = -~ z? +q - } ( — =-- ) dz, (4) 
 
 an expression which is rational when is an integer, 
 
 orO. n 
 
 (2.) Assume a + bx n = z q x n . 
 
 Then x n — a (** — b)~\ (1) 
 
278 CUX;>lTlo.\s OV i.\ri-:<;u ABILITY. 
 
 .-. a: = a* (* — &)-*. (2) 
 
 in m 
 
 A #" = w< (^ — *)"», (3) 
 
 tfc = — 2 a^ (2? — J)-^ _1 3?- 1 <fe. (4) 
 
 Multiplying (1) by b, adding a, and taking - power, we 
 have * 
 
 P P £ 
 
 (a + h&)* = a q (s* — b) q z?. (5) 
 
 Multiplying (3), (4), and (5) together, we have 
 
 a* (« + bx") q dx = — £ a Kn q B V — b) v " « V + «- ! <fe, 
 
 an expression which is rational when 1- - is an in- 
 
 teger, or 0.* 
 
 Therefore there are two cases in which the general bino- 
 mial differential can be rationalized f 
 
 1st. When the exponent of the variable without the 
 parenthesis increased by unity, is exactly divisible by 
 the exponent of the variable within the parenthesis. 
 
 2d. When the fraction thus formed, increased by 
 the exponent of the parenthesis, is an integer. 
 
 Rem.— These two cases are called the conditions of integrability of 
 binomial differentials^ and when either of them is fulfilled, the inte- 
 gration may be effected. If, in the former case, — 1 is a positive 
 
 m + 1 v 
 
 integer or 0, or in the latter case, + - + 1 is a negative integer 
 
 n q 
 
 * The student will observe that Art. 143 is a particular case of this Article, re 
 suiting from making m an odd positive integer, and n = 2. 
 
 t These are the only cases of the general form which, in the present state of 
 analysis, can be made rational. When neither of these conditions is satisfied, the 
 
 P 
 
 expression, if he a fractional index, is, in general, incapable of integration in a 
 
 finite number of terms. 
 
EXAMPLES. 279 
 
 or 0, the binomial ^ — a) or {p — b) will have a positive integral ex- 
 ponent, and hence can be expanded by the Binomial Theorem, and 
 
 each term integrated separately. But if, in the former case, 1 
 
 is a negative integer, or in the latter, — - 1- - + 1 is a positive inte- 
 ger, the exponent of the binomial (z? — b) will be negative, and the 
 form will be reduced to a rational fraction whose denominator is a 
 binomial, and hence the integration may be performed by means of 
 Chapter II. But as the integration by this method usually gives com- 
 plicated results, it is expedient generally not to rationalize in such 
 cases, but to integrate by the reduction formulae given in the next 
 Chapter. 
 
 1. Integrate dy = x 5 (a + x % )* dx. 
 
 Here — — 1 = 2, a positive integer, and therefore it 
 
 can be integrated by the first method. 
 Let (a + x>) = z 3 . 
 
 Then (a + x*)* = z. (1) 
 
 tf> = (z 3 — of, 
 afidx = l(# — a)*#dz. . (2) 
 
 Multiplying (1) and (2) together, we have 
 dy = x 5 (a + xrf dx = f (z 3 — dfzHz. 
 
 3 A n *\ „ , 7 3 /z 10 2az? «V\ 
 
 = *(« + * 2 )- - y (a + x>)i + 3 -£(a + x^. 
 
 2. Integrate dy = — r = x~*(l + x*)^dx. 
 
 n q 2 2 
 
 ative integer, and hence it can be integrated by the second 
 method. 
 
380 
 
 10 KXAMPLE8. 
 
 
 Let (1 + a- 2 ) = 2% 
 
 
 Then *• = (*- l)-t. 
 
 
 s = (z 2 -l)-i; 
 
 
 ar« = (* 8 -l) 2 . 
 
 (1) 
 
 (l + a*) = l-f-(s 2 -l)-i 
 
 
 = « 2 (^-l)-J. 
 
 
 (1 + a 8 )-* = a- 1 (s 2 - 1)*. 
 
 (2) 
 
 dx = — (2 2 — l)~$zdz. 
 
 (3) 
 
 Multiplying (1), (2), (3) together, we have 
 
 
 dy = ar 4 (1 + a 2 )-* da; = — (f — l)dz. 
 
 
 _ (i + *)* *(i + *)* _ (i + «2* ~> r2 n 
 
 3a? 
 
 EXAMPLES. 
 
 (2a;* — 3a*)rfa; ,. . ,,,. 
 
 1. % = S —L — (Art. 141.) 
 
 ox* 
 
 y = &%* — f^*. 
 
 z* __ 2a;* , 
 
 2. dy = — *?a;. 
 
 1 + x* 
 
 y = %x$ — 2x — fe$ + 3a** + 2a;* — 6^ — 6a;* 
 + 6 log (a;* + 1) + 6 fcur*a& 
 
 2x$ — 3a;* , 
 
 3. d# = — 5 dx. 
 
 3a* + a* 
 y = 12 (fas* — fa* + ^ T?¥ ~ 9a;*) 
 + 1908 [jtfA— Ja£+3a£— - ^ ' + 81^—243 log (a;^ + 3)]. 
 
EXAMPLES. 281 
 
 4. dy = -rdx. 
 
 y = 12 j IT t 8 - T ~ 4 + g+ilog(l +a *) 
 
 ^ L W + 8W + 1/ \i - xv. 
 
 5. ^--^j. (Art. 142.) 
 (1 + x)* 
 
 L Vi + d 
 
 6. <fy 
 
 a?V « + to 
 
 2 , V « + bx — \/a 
 
 V = ~p log 
 
 7. ^ 
 
 Va Vix 
 
 dx 
 
 (1 + *)* + (% + x)i 
 
 y = 2 tan" 1 (1 + ar)*. 
 
 8. dy = 4 (a; + Vz + 3 + \ / x + 3)dx. 
 
 y = 2 (^ + 3)2-12(^ + 3) + 4 (a?+3)* + 3 (*+3)i 
 
 9. dy=-^t=-. (Art. 143.) 
 
 Vi + x 2 ; 
 
 2/ = i(l + x*)$ - | (1 + a*)* + (1 + a?)*. 
 10. dy = . y = _ 
 
 (1 + a«)f 3 (1 + a*)* 
 
 1 -, 7 aMc £ 2 + 2 
 
 (1 + a*)i Vl + £ 2 
 
282 EXAMPLls. 
 
 12. dy = , a ==• (Art. 144.) 
 
 y = log(l + 2x + 2 Vl -f z 4- z 2 ). 
 (See Art. 144, Ex. 1.) 
 dx 
 
 13. tfy = 
 
 ^/xi _ a; — 1 
 y = log(2a; — 1 + 2 V^ 2 — a? — l). 
 
 14. dy = _ ^ — = • V = — 2 tan-U/-^^- 
 
 1E , dx 
 
 15. dy = 
 
 ^1 + x — a? 
 
 y=-2tan-V ijzi --^ 
 
 16. dy = 
 
 I Assume \ / ™ + a? = z — #, etc. ) 
 y = | log (to + V« 2 + ^ 2 ) 
 
 1 . (a+ V¥+^\ 
 
 18 . n m V + *** 
 
 y = log (a; + 1 + V&b + a; 2 ) — 
 
 a; + V2a; + a: 2 
 
19. dy = 
 
 (1 + &) Vl - x 2 
 
 EXAMPLES. 283 
 
 dx 
 
 y = -— tan -1 ( — =z=) 
 9 V2 Wl — aV 
 
 20. <fy 
 
 V2 Wl 
 
 adx 
 
 V%ax + a: 2 
 2/ = « log (a; + « + \/%ax + a?). 
 
 21 - rfy = vg— 7~ ' ?y = iiog^ + V2^ r 7)- 
 
 (Compare with Ex. 16.) 
 
 (Compare with Ex. 20.) 
 
 23. <fy = a? (2 + 3a*)* dfc. (Art. 146.) 
 
 * = A[^- - W+** + m#*} 
 
 24. ^ = X s (a + fos 2 ) * (fe. 
 
 25. dy = z?(a + bx 2 )? dx. 
 
 26. dy = x<> (a — a?)-b dx. 
 
 27, ^^T+W ' = .-CT6W 
 
 28. <fy = a (1 + a; 2 )-* da. y 
 
 aa; 
 
 (1 + a: 2 )*' 
 
2S-1 \MPIiB8. 
 
 29. dy = x~* (1 - 2a 8 )-* dx. 
 
 * = ~<i#(x-**>*. 
 
 30. ^ = (1+^^. 
 
 _ (3s*-2)(l+3 *)S 
 9 15 
 
 31. <fy = x-* (a + a«)-f efcr. 
 
 {Zx* + %a) 
 
 y " ~ 2a*x (a + x*)%' 
 
 32. <fy = a 3 (a* + x*)l dx. 
 
 V = A (« 2 + **)* (^ 2 - 3fl«). 
 
 33. J# = x 5 (a + fa 2 )t dx. 
 
 y ~~ 2^ 111 4 + 5 / ' 
 in which z = (a + bx 2 )$. 
 
 34. tfy =(* + **>* **■ 
 
 y = | (« + &z)f + 2« (« + &c)$ 
 
 + a* log __Z_ — . 
 
 Va + ox -f y <r 
 
 35. tf# = («2 + z 2 )i dx. 
 
 xia* + x*)* a\ r , . ,..i- 
 
CHAPTER IV. 
 
 INTEGRATION BY SUCCESSIVE REDUCTIONS. 
 
 147. Formulae of Reduction. — When a binomial dif- 
 ferential satisfies either of the conditions of integrability, it 
 can be rationalized and integrated, as in the last chapter. 
 But, instead of rationalizing the integral directly, it may 
 be reduced to others of a simpler kind, and finally be made 
 to depend upon forms whose integrals are fundamental, or 
 have already been determined. This method is called 
 integration by successive reduction, and is the process 
 which in practice is generally the most convenient. It is 
 effected by formulas of reduction. These formulas are 
 obtained by applying another, known as the formula for 
 integration by parts, and which is deduced directly as 
 follows : 
 
 Since d (uv) = udv + vdu, (Art. 16) 
 
 we have uv = / udv + / vdu ; 
 
 / udv = uv — / vdu ; 
 
 a formula in which the integral of udv depends upon 
 that of vdu. 
 
 148. To find a formula for diminishing the expo- 
 nent of x without the parenthesis by the exponent 
 of x within, in the general binomial form 
 
 fx m (a + bx n )i> dx. 
 
28G SPECIAL WORXVUM OP REDUCTION* 
 
 Let y =■ j #"' (« + h&Y dx = J udv 
 
 = uv —J vilu ; (1) 
 
 and put dv = a"" 1 (a -f bx")* dx and u == z m — H 
 
 Then — &£?& 
 
 and du = (m — n + 1) x m ~ n dx. 
 
 Substituting these values of w, w, d?/, d/*, in (1), we have 
 
 v = y *• (a + fo-)» * = - -^nriy 1 - 
 
 - %S TT? /*'"' ( " + ^ fc < 2 > 
 
 This formula diminishes the exponent m by w as was 
 desired, but it increases the exponent ;; by 1, which is 
 generally an objection. We must therefore change the 
 last term in (2) into an expression in which p shall not be 
 increased. 
 
 Now x m -" (a + bx n )p +1 — x m ~ n (a + bx n )p {a -f bx 11 ) 
 
 = ax m ~ n (a + bx n )p + bx m (a + btf 1 )* ; 
 
 which in (2) gives 
 
 - ^7** + * /V (a + fofWz. 
 rc (jo + 1) */ v ' 
 
 Transposing the last term to the first member and redu- 
 cing, we have 
 
SPECIAL FORMULJE OF REDUCTION. 287 
 
 Therefore we have 
 
 y = / x m (a + bx n Y dx 
 ^ m _„ + i ^ + fitfiy+i _( m —n+ l)a Cx m ~ n (a + bx n )Hx 
 
 b (np + w + 1) 
 ft^'e^ 2*s the formula required. 
 
 ;M) 
 
 149. To find a formula for increasing the exponent 
 of x without the parenthesis by the exponent of x 
 within, in the general binomial form 
 
 y = I x~ m (a + bx n )p dx. 
 
 Clearing (^4) of fractions, transposing the first member 
 to the second, and the last term of the second to the first, 
 and dividing by (m — n -+- 1) a, we have 
 
 Jx m ~ n (a + bx n Y dx 
 x m ^ + \a + bx n Y^--b[np-\-m-{-l) fx m (a + bx n ydx 
 
 a (m — n + 1) ^ ' 
 
 Writing -- m for m — n, and therefore — m -f n for m, 
 (1) becomes 
 
 y = I x~ m {a + bx n )p dx 
 x~ m+ \a + bx n Y +l + b(m — np — n—1) f x~ m+n (a + bx n ) p dx 
 
 — a (m — 1] 
 
 which is the formula required. 
 
 ■m 
 
SPECIAL FOHMULJE Of REDUCTION. 
 
 150. To find a formula for diminishing the expo- 
 nent of the parenthesis by 1, in the general bino- 
 mial form 
 
 y = Jx m (a + bx n )p dx. 
 j x'" (a + bx")p dx = J x m (a + bat 1 )**- 1 (a + bx n ) dx 
 
 = a j x m (a -f- bx")?- 1 dx + b j x m+n (a + bx n )p~ l dx. (1) 
 
 By formula (^1), we have for the last term of (1), by writ- 
 ing m + n for m and p — 1 for ^, 
 
 Cx m+n (a + bxry- 1 dx 
 
 x m+l (a + bx n )p — (m + 1) a C x m (a + bx n )p~ l dx 
 b [ n (P — !) + m + n + !] 
 which in (1) gives 
 
 y = j x" 1 (a -f ^ n ) p dx = aj x m (a + bx n )p~ l dx 
 #" +1 (a + for*)" — (fw -f 1) a Cx m {a + &e M )*'- 1 rfo 
 
 + 
 
 (rcjo + m + 1) 
 
 Therefore, uniting the first and third terms of the second 
 member, we have 
 
 y = J x m (a + bx n Y dx 
 of 1 * 1 (a -f bx")* + flr;z;; C x m (a + to")*- 1 $» 
 
 MjO -+- 7W + 1 
 
 which is the formula required. 
 
 (C) 
 
SPECIAL FORMULAE OF REDUCTION. 289 
 
 151. To find a formula for increasing the exponent 
 of the parenthesis by 1, in the general binomial form 
 
 y — J x m (a + b.M' n )-p dx. 
 
 By transposing and reducing (6 y ), as we did (A) to find 
 (/>), we have 
 
 fx m (a + bz")p~ l dx 
 x m+l (a -f- hx n Y — (np -f m + 1) I x m (a + bx 7, y dx 
 
 anp 
 
 a) 
 
 Writing — p for p — 1, and therefore — p + 1 for jt>, 
 (1) becomes 
 
 ij — i x m (a + bx n )~p dx 
 x m+1 (a + bx n )-v+ l - (m + n + 1 — w^) / x m (a + bx n )~p +] dx 
 
 = j — rv^ ;^> 
 
 #tt (/j — 1) 
 ?«7iic/i is the required formula. 
 
 Remark. — A careful examination of the process of reduction by 
 these formulae, will give a clearer insight into the method than can be 
 given by any general rules. We therefore proceed at once to exam- 
 ples for illustration, and shall then leave it to the industry and inge- 
 nuity of the student to apply the method to the different cases that 
 he may meet with. 
 
 EXAMPLES. 
 
 x m dx 
 
 1. Integrate dy - 
 
 Here y = J x m (a 2 — x 2 )-? dx, 
 
 a form which corresponds to 
 
 fx m (a -f- bx^y dx. 
 13 
 
290 EXAMPLES. 
 
 We BM that by applying formula (J) wo may diminish 
 m by v, and by continued applications of this formula, \w 
 can reduce w to or l according as it iseven or odd, so 
 that the integral will finally depend upon 
 
 ydx . ^ x , 
 
 — ===. = sm l - , when m is even ; 
 V« 2 - x 2 « 
 
 y» ^^fo i 
 
 — = — (a 2 — z 2 ) 2, when wi is odd. 
 V« 2 -^ 
 
 Making ??* = m, a = a 2 , b = — 1, w = 2, p = — J, 
 we have from formula (^4), 
 
 y = I x m (a 2 — x 2 )'? dx 
 
 z>«-2+i ( a 2 _ a?)* _ a % ( m _ 2 + 1) fx m ~ 2 (a 2 — x*y? dx 
 - [^ (- 1) + m + 11 
 
 m 
 
 (wi — 1) a 2 /V-2 (a 2 — x*)~i 
 
 dx 
 
 + *-= (i) 
 
 When m = 2, (1) becomes 
 
 When »i = 3, (1) becomes 
 
 y = f-~= = -h°? (« 2 - *»)* - i« a <« 2 - *)* 
 
 ^ va 2 — x 2 
 
 = _ J (fl8 _ **)* (^ + 2« 2 ). 
 
 When m = 6, (1) becomes, by applying (^4) twice in 
 succession, 
 
EXAMPLES, 291 
 
 Mx 
 
 = /: 
 
 Va 2 — x* 
 
 (which the student may show.) 
 
 x m dx 
 
 2. Integrate dy = 
 
 ya 2 + x 2 
 
 Here $f = / x m (a 2 + x 2 )~? dx. 
 
 Making m = wi, a = a 2 , 5 = 1, « = 2, ^? = — ^, 
 we have from (^1), 
 
 y = j x m (a 2 + £ 2 )~* $b 
 
 = i i — / x 11 2 (rt 2 + a,- 2 ) * <£c. (1) 
 
 m m v 
 
 By continued applications of this formula, the integral 
 will finally depend on 
 
 /— = log (x + V« 2 + x 2 ), when m is even, 
 A//?2 I /y>2 
 
 Va 2 + z 2 
 
 Va 2 + x 2 
 
 /xdx i 
 
 — — = (a 2 + x 2 Y, when wi is odd. 
 a / si2 _i_ ^-2 
 
 dx 
 
 3. Integrate <:??/ = -. 
 
 *■ (a 2 - x 2 )? 
 
 Here y = / a; - "' (a 2 — x 2 ) - ^ cfe, 
 
 from which we see that by applying (/?) we may increase m 
 by 2, and by continued applications of (B), we may reduce 
 m to or 1, according as it is even or odd, making the 
 integral finally depend on a known form. 
 
292 APPLICATIONS OF FORMULA. 
 
 Making m — m> a = a 2 , b = — 1, n = 2, p s ^ |, 
 
 (#) gives us 
 
 y = J x- w [a 2 — x 2 )'^ dx 
 
 x «fi («2 _ ^» _ ( w + i __ 2 — 1) /V'"+ 2 (rt 2 — z*) "i efe 
 
 = — o2(m-l) 
 
 _ _^.-^)l__ (in- 3)_ /* & 
 
 (»J - 1) rtV»" 1 "*" (/« - 1) d* J zm-2 ( a 2 _ X 2)h K ' 
 
 When m = 2, (1) becomes 
 
 _ f dx - _ (a 2 -* 2 )*. 
 
 (since the last term disappears.) 
 When m = 3, (1) becomes 
 
 dx 
 
 = r dx - _ ( q8 - * 2 )* 4 j_ /■ * 
 
 & 
 
 _ V« 2 — a? 1 , a— Va 2 — x 2 
 
 + 0^3 l0 S 
 
 2rt 2 z 2 ' 2a 3 & a; 
 
 (Ex. 17 of Art. 146.) 
 4. Integrate ^# = (a 2 — x 2 )? dx, when w is odd. 
 Here we see that by applying (C) we may diminish 
 
 - by 1, and by continued applications of (6') we can reduce 
 
 - to — £, making the integral depend finally upon a 
 
 lit • 
 
 known form. 
 
 IV) 
 
 Making m = 0, a = a 2 , I = — 1, n = 2, p = ^ , ( 0) 
 gives us 
 
 y = J(a* — a?)*dx 
 
 x (a 2 - x*f + wa« /V - a'T -1 dx 
 
APPLICATIONS OF FORMULAE. 293 
 
 When n = 1, (1) becomes 
 
 d . ox 1 7 z(« 2 — z 2 )* , « 2 • 
 «/nJ (a 2 — x 2 )* dx = — ^ — - — '- + ^ sin" 
 
 ■l _, 
 
 die 
 
 5. Integrate dy = - , when n is odd. 
 
 (a 2 — x' 2 ) 2 
 Here y =s y (a 8 - s*)-i <&, 
 
 from which we see that by applying (D) we may increase 
 the exponent - by 1, and by continued applications of {!)) 
 ft 
 
 we can reduce : to — J, making the integral depend 
 
 finally on a known form. 
 
 Maki] 
 gives us 
 
 71 
 
 Making m = 0, a = a 2 , b = — 1, n = 2, p = -, (D) 
 
 y — I (a 2 — x 2 ) F dx 
 ( a % _ x 2yV x _(3 - n) J (a 2 - x 2 ) 
 
 ~- +x dx 
 
 2a 2 
 
 (i->) 
 
 . • + -jl=3 r « (1) 
 
 (n - 2) a 2 (a 2 - x 2 )^ 1 ( n ~ 2) a 2 «/ ^ _ ^-i 
 When w = 3, (1) becomes 
 
 -/: 
 
 dx 
 
 (a 2 - x 2 )? a 2 (a 2 — x 2 )% 
 
 x m dx 
 
 6. Integrate dy — - 1' =• 
 
 v2ax — x 2 
 
 Here y = / af* (2#£ — a?)"*i dx = I x m ~^(2a—x)~idx, 
 
 which may be reduced by (A) to a known form. 
 
294 APPLICATIONS OF FORMULJE. 
 
 Making m = m — $, a = 2a, b = — 1, n = 1, p = —J, 
 (A) gives us 
 
 /x™dx 
 V2ax — x % 
 
 x m -h (2a — x)l — 2a (m — \) Cx m ~% (2a — x)~l dx 
 — m 
 
 (2m — l)a p x m ~hlx 
 
 When m = 2, (1) becomes 
 
 x*dx 
 V2ax — a? 
 
 a"- 1 /5 9 , (2m — 1) a P x m ~\lx ,* 
 
 V2ax — x 2 + i — / -— =— (x) 
 
 »» « J V2ax-x* 
 
 I* X*dx X + 3a vr -3 
 
 * «/ a/0/>/» ~2 2 
 
 d V2ax 
 
 x + 3« /~ 1 , „ „ .a? 
 
 = s — v2## — a^ + fa 2 vers * -• 
 
 7. Integrate dy = — 
 
 9 Vl - S 1 
 
 /x 5 1 • 5 , 13 5 \ « , L3-5. , 
 
 y = - (.6 + 4^^ + 2^70^)^-^ + 2-4 .1 ^^ 
 
 8. Integrate d?/ r= — 
 
 J \ 3«a? x 3«W T 
 
 9. Integrate dy = (I — x^ dx. 
 
 7/ = \x (1 — a*)* -f fa? (1 -* a; 2 )* -f | sin" 1 X. 
 
 dx 
 
 10. Integrate dy = 
 
 (1 + rf)8 
 
 + 5 • t^ 9\ + o tan * rr. 
 
 4 (1 + a; 2 ) 2 ' 8 . (1 + x 2 ) ' 8 
 
LOGARITHMIC FUNCTIONS. 295 
 
 x^dx 
 
 11, Integrate dy — - t -• 
 
 & * ^/2ax - x* 
 
 y = - (|- 2 + 1 1 « + f • t« 2 ) Vto^~=& + | • f« 3 vers-il 
 
 clx 
 
 12. Integrate cfo/ = 
 
 a* Vl — z 2 
 
 i + vr 
 
 These integrals might be determined by one or other of 
 the methods of Chapter III, but the process of integration 
 by reduction leads to a result more convenient and better 
 suited in most cases for finding the definite integrals.* 
 
 LOGARITHMIC FUNCTIONS. 
 
 152. Reduction of the Form / X (log x) n dx, in 
 whieh X is an Algebraic Function of x. 
 
 Put Xdx = dv and log 71 x = u. 
 
 /dx 
 Xdx and du = n log" -1 x — 
 x 
 
 Substituting in / udv = uv — J vdu, (Art. 147) 
 
 we have y = J X log" xdx 
 
 = log" x J* Xdx - fn log"-* x —f{Xdx) ; 
 or by making / {Xdx) as Xj , 
 
 we have y = J X log n zt/z 
 
 — ■ log" -1 £6& ; 
 
 * For a discussion of definife integrals, see Chap. V. 
 
29G EXAMPLES. 
 
 which diminishes the exponent of log x by 1, wherever it 
 is possible to integrate the form / Xdx. By continued 
 
 applications of this formula, when n is a positive int < 
 we can reduce n to so that the integral will finally 
 depend on 
 
 J x 
 
 Sch. — A useful case of this general form is that in which 
 X = x m , the form then being 
 
 y = ) x m log n xdx ; 
 
 and the formula of reduction becomes 
 
 of 1 log" xdx = ^^-j log" x 
 
 n 
 
 m + 
 
 - j x m Jog n " 
 
 xdx. 
 
 by means of which the final integral, when n is a positive 
 integer, becomes. 
 
 //£»l + l 
 x m dx = — — . 
 m -\- 1 
 
 EXAMPLES. 
 
 1. Integrate dy = x* log 2 xdx. 
 
 Making m = 4, and n = 2, we have 
 
 y = I x* log 2 xdx 
 
 Making m = 4 and n ss 1, we have 
 
 /*• log* <fc = ^p - J/aAfc (= fj, 
 
LOGARITHMIC FUNCTIONS. 297 
 
 which substituted in (1) gives us 
 
 9 = J*l 0( ?xdx = —f-— 6 (_JL_-._) 
 
 2. Integrate dy 
 
 = 5 Cog 2 *-! logz + ^). 
 
 x log # efo 
 Va 2 + # 2 
 
 Put = <Zv and log a; = u : 
 
 V« 2 + a? 
 
 then y = a/« 2 + x 2 and <?w = — 
 
 x 
 
 Vat + % 2 
 
 ... y = f^A^. = (flP + ^i log * _ /V*: 
 
 <£* 
 
 flftfa; /* zefce 
 
 = te* + 38)* logz _ /* g<fa? __ _ T 
 
 = (a 2 + z 2 )* l oga j + « log ( fl + ^ fl8 + ^ _ y'S+S. 
 (See Ex. 17, Art. 146.) 
 3. Integrate dy = j^^yf 
 
 V = y^c log X ~ log ^ + ^ 
 
 153. Reduction of y 
 
 log** 05 
 
 Put #•» = «, j-l-Jp*, 
 
 then dw = (m + 1) a m tfa; 
 
 and v ss 
 
 (ft — 1) log" -1 & 
 
898 LOGARITHMIC FUNCTIONS. 
 
 f&dx _ _ x" l + l m + 1 r afdx 
 
 V ~ J \og"x~ (n — l)\og"- l x + (n — \)J log" -1 a' 
 
 by means of which the final integral, when n is a positive 
 integer, becomes 
 
 / x m d x 
 log x' 
 
 beyond which the reduction cannot be carried, for when 
 r. = 1 the formula ceases to apply. We may, however, ex- 
 press this final integral in a simpler form ; thus, 
 
 Put z = x'» +] ; 
 
 then dz = (w + 1) x m dx and log z = (m + 1) log x. 
 
 r x m dx _ r dz 
 J loaf x ~ J 
 
 log x J log z' 
 
 an expression which, simple as it appears, has never yet 
 been integrated, except by series, which gives only an 
 approximate result. 
 
 Ex. 1. Integrate dy = ; — ^ " * 
 
 Here m = 4 and n = 2; therefore the formula gives us 
 
 /x i dx x 5 px 4 dx 
 
 log 2 x ~ log x J log x 
 
 Put z = x 5 ; then dz = bx*dx and log z = 5 log x; 
 
 therefore f**? = f * 
 
 t/ log a; «/ log 
 
 Now put log z — u ; then 2 = e" and dz — e u du. 
 
 / dz Pe n du 
 
 log z ~~ J u 
 
 = /(l + B + £ + ^ + ete .)£ ( A rt .6 8 ) 
 
EXPONENTIAL FUNCTIONS. 299 
 
 = log u + u + ^ + ^ + etc. 
 
 = log (log z 5 ) + logo: 5 + £- 2 log 2 :* 5 + -^-^ +etc. 
 
 Pz 4 dx 
 ~ J log 2 X 
 
 X 5 
 
 i +5 
 
 log SB 
 
 l02f 2 ^ 5 
 
 log (log x 5 ) + log £ 5 + — 5 * _ 
 
 log 3 £ 5 
 + -| 32 + etc. 
 
 ) 
 
 (See Strong's Calculus, p. 392 ; also, Young's Int. Cal., 
 pp. 52 and 53.) 
 
 EXPONENTIAL FUNCTIONS. 
 
 154. Reduction of the Form j a mx oc n dx. 
 
 Put a mx dx = dv and x n = u ; 
 
 a mx 
 
 then v =» — i and du = nx n ~* dx. 
 
 m log a 
 
 a mx x n dx =s — ; : / a mx x n ~ x dx. 
 
 m log a m log a <J 
 
 By successive applications of this formula, when n is a 
 positive integer, it can be finally reduced to 0, and the in- 
 
 /a mx 
 a mx dx = — i 
 m log a 
 
 Only a very few of the logarithmic and exponential func- 
 tions can be integrated by any general method at present 
 known, except by the method of series, which furnishes 
 only an approximation, and should therefore be resorted to 
 only when exact methods fail. 
 
300 EXPONENTIAL Fl'SVTlONS. 
 
 EXAMPLES. 
 
 1. Integrate dy = cP&dx. 
 
 Here m = 1 and n = 3 ; therefore, from the above for- 
 mula, we have 
 
 y = / a x x*dx 
 
 a x x* 3 f* 
 = , j / a^dx; (by repeating the process) 
 
 „ ** !_ /^ _ ^_ r a * xdx \ . 
 
 log a log a \log a log a ^ / ' 
 
 (by repeating the process) 
 _ a x x* _ 3a^ 2 _6_ / a x x 1 \ 
 
 ~ log « log 2 a log 2 a llog a log 2 « / 
 
 ~" log « \ log a log 2 a log 3 a/ 
 
 2. Integrate tfy = zPe^dx. 
 
 I& '5x* 6x 6\ 
 
 — — , when jm is a posi- 
 tive Integer. 
 
 Put x~ m dx = dv, a x = u; 
 
 x~ mJrl 
 
 then v = and du = a x log a dx. 
 
 m — 1 b 
 
 /a x dx _ a x log a pa x dx 
 
 x m ~ (m — 1) a^ -1 + m^-~l •/ a?"'" 1 ; 
 
 by means of which the final integral becomes 
 
 *a x dx 
 x 
 
 /- 
 
TRIGONOMETRIC FUNCTIONS. 301 
 
 which does not admit of integration in finite terms, but 
 may be expressed in a series, and each term integrated sepa- 
 rately. (See Lacroix, Calcul Integral, Vol. II, p. 91.) 
 
 Ex. 1. Integrate dy = — ^~- 
 
 By the formula just found, we have 
 /*e x dx _ e x re x dx 
 
 = -f + /[ 1 + a; +^+s +et0 -]?( Art - 63 ) 
 
 <P , x 2 x* 
 
 = --+log* + z + - 2 + -- 2 + etc. 
 
 TRIGONOMETRIC FUNCTIONS. 
 
 156. Cases in which sin m 6 cos n OdO is immediately 
 Integrable. — The value of this integral can be found im- 
 mediately when either m or n, or both, are odd positive 
 integers ; and also when m + n is an even negative 
 integer. 
 
 1st. Let m = 2r + 1 ; then 
 
 ysin w cos" dd = J* (sin 6) 2r+1 cos" 6 dd 
 
 = J* (I — cos 2 ey cos n sin dB 
 
 = — y (1 — cos 2 6) r cos" 6 dcos 0, 
 
 an expression in which the binomial (1 — cos 2 0) r can be 
 expanded, and each term integrated immediately. In like 
 manner, if the exponent of cos 6 be an odd integer, we may 
 assume n = 2r -f 1, etc. 
 
302 EXAMPLES. 
 
 2d. Let m + n = — 2r ; then 
 y ? sin m cos" dd = y\an m d (cos0) n+ffl d& 
 = f tan w (sec 0) 2r dd 
 
 = f tan'" 0(1 + tan 2 0) ,wl tf-tanfl, 
 
 each term of which, after expansion, pan be immediately 
 integrated. 
 
 exam ples. 
 
 1. dy = sin 2 cos 3 dd. 
 
 Here y = faitfe cos 8 dd 
 
 = y*sin 2 (9(1 — sin 2 d) d- sin 
 
 — J S i n 3 0_ £ Sm 5 0. 
 
 2. dy = — -7-z dd. 
 
 u cos 4 d 
 
 Here y = f ~^ </0 = Aan 2 d sec 2 <?0 
 = \ tan 3 0. 
 
 3. dy = sin 3 cos 4 </0. # = — £ cos 5 + \ cos 7 0. 
 
 4. ^ = sin 5 cos 5 dd. 
 
 y = — T V cos 6 (sin 4 + | sin 2 + J). 
 
 5. dy = sin 3 cos 7 J0. y = ^ cos lo — J cos 8 0. 
 
 sin 2 
 
 6. dy = Mffid ddm # = Jtan 3 + ^tan 5 0. 
 
 7. <fy 
 
 sin cos 5 
 
FORMULA OF REDUCTION 303 
 
 - C s ec 4 fl</0 A 1 + tan 2 (9) 8 se^fl dd 
 
 Here y - J tan6l cog ^ - J tan — 
 
 = log (tan 6) + tan 2 + £ tan 4 0. 
 
 8. % = — i 1 — 
 
 sin* cos* (9 
 
 Let x = tan 0; 
 
 1 x 
 
 then cos = , , sin = 
 
 
 and ^=i + ^ 
 
 /» rfg _ Al + a? ) <fa 
 ^ "" ^ sin*0cos*0 ~ J x§ 
 
 = *a* — 
 
 f tan*0 
 
 * 2 tan*0 
 
 CTITl3 /5 
 
 9. dy = =rzdO. y = \ tan 4 0. 
 u cos 5 * 4 
 
 10. dy = —j-q. y = tan0 + f tan*0 4 £tan 5 0. 
 
 157. Formulae of Reduction for 
 
 sii\ m d cos* 1 d do. 
 
 /• 
 
 When neither of the above mentioned conditions as to w* 
 and » is fulfilled, the integration of this expression can be 
 obtained only by aid of successive reduction. 
 
 We might produce formulae for reducing the expression 
 sin" 1 cos" directly ;* but, as it would carry us beyond 
 the limits of this book, we prefer to eifect the integration 
 •by transforming the given expression into an equivalent 
 algebraic form, and then reducing by one or more of the 
 
 * See Price, Lacroix, Williamson, Todhunter, Courtenay, etc. 
 
304 EXAMPLES. 
 
 formulae (A), (B), (C), (/)). Thus, put sin = x, then 
 sin'" = x'", cos = (1 — z 2 )-, cos" = (1 — a 2 )?, ami 
 r/0 = (1 — a?)"* cte. 
 
 .-. */ = /sin™ cos n 0^/0 = J x" 1 (1 — u^y^'dx. 
 
 or we may put cos 6 = x, and get 
 
 # = J sin m cos" 0^0 = y — re" (1 — x 2 f*~dx ; 
 
 either of which may be reduced by the above formulae. 
 
 This process will always effect the integration when m 
 and n are either positive or negative integers, and often 
 when they are fractions. The method is exhibited by the 
 following examples. 
 
 
 EXAM PLES. 
 
 1. 
 
 dy = sin 6 Odd. 
 
 Put 
 
 sin = x, 
 
 then 
 
 dd = (1 — x*)-% dx. 
 
 .-. y = y*sin 6 0^0 = J X s (1 — x z )~* dx 
 
 = " V6 + 4^ + 2^ V < X " *>* + 2.TT6 «T* * 
 
 (by Ex. 7, Art. 151); 
 cos0 , . .„ . 5 ..... „ . 53 . „, 5-3 
 
 = 6 ^ 
 
 n" a -+- j »in » -h j— ~ sin "/ " 
 
 !" 2-4-6 
 
 2. 
 
 . rf0 
 
 
 Put 
 
 sin 6 = x, 
 
 
 then 
 
 dd = (1 -z2)-£^. 
 
 
INTEGRATION. 305 
 
 cos I 1 3 
 
 Vsin 4 0^2 sin 2 0/ 
 
 , . , 1 + cos . 
 
 (since — log : — - — = log 
 
 v ° sin b 
 
 ^ftr 1 ** 
 
 1 — a; 2 
 
 2-4 10g 2 
 
 
 (by Ex. 12, 
 
 Art. 151); 
 
 1*3 
 
 + 2T4 log tan |(9. 
 
 
 sin ^ 1 , * ~ > 
 
 -— s = log tan id.) 
 
 1 + cos & 3 J 
 
 3. dy = sin 4 0^0. 
 
 + I sin 0) + |i 
 
 (See Ex. 10, Art. 135.) 
 
 cos 
 y = 7- ( sin3 ^ + ! sin 0) + f 
 
 4. % = cos 4 0^0. 
 
 sin cos 3 . *.'•*' 
 
 y = — h 4 sm cos + f 0. 
 
 (See Ex. 9, Art. 135.) 
 
 158. Integration of sin m cos w (16 in terms of 
 the sines and cosines of the multiple arcs, when m 
 and n are positive integers. 
 
 The above integrations have been effected in terms of the 
 powers of the trigonometric functions. When m and n 
 are positive integers, the integration may be effected with- 
 out introducing any powers of the trigonometric functions 
 by converting the powers of sines, cosines, etc., into the 
 sines and cosines of multiple arcs, before the integration is 
 performed. The numerical results obtained by this pro- 
 cess are more easily calculated than from the powers. 
 
 Three transformations can always be made by the use of 
 the three trigonometric formula). 
 
306 EXAMPLES, 
 
 (1.) sin a sin b = J cos (a — I) — \ cos (a + J). 
 (#.) sin « cos # = £ sin (a -}- £) + £ sin (a — b). 
 (8.) cos a cos b = i cos (a + b) + J cos (a — b). 
 
 EXAM PLES. 
 
 1. dy = sin 8 cos 2 6d0. 
 
 Here sin 8 cos 2 = sin (sin cos 0) 2 
 
 = sin (i sin 20) 2 [by (2)] 
 
 = J sin (sin 2 20) 
 
 . . _ /l — cos 40\ „ ..,.- 
 
 = i sin * ( ^ / [by (1)] 
 
 = | sin — £ sin cos 40 
 
 = i sin — J- (i sin 50 — J sin 30) 
 
 [by (2)] 
 = J sin — ^g- sin 50 + A sin 30 
 
 A y = Am 8 cos 2 0^0 
 
 = J* (| sin 0d0 — -jV sin 50^0 -f ^ sin 30d0) 
 = — -£ cos + ^ cos 50 — -fa cos- 3ft. 
 
 2. ofy = sin 8 cos 8 0d0. 
 
 # = — -fo cos 20 + rh cos 6 & 
 
 3. <fy = sin 3 0^0. 
 
 y = -^ cos 30 — J cos 0. 
 
 4. 6fy = cos 3 Odd. 
 
 y = Jg sin 30 + f sin 0. 
 
FORMULAE OF REDUCTION, 307 
 
 159. Reduction of the Form 
 
 / x n cos ax dx. 
 
 Put 
 
 u = x n , 
 
 and 
 
 dv = cos ax dx ; 
 
 fchei 
 
 du = nx n ~ l dx, 
 
 and 
 
 1 . 
 
 v = - sin ax, 
 a 
 
 .-.?/—/ x n cos axdx 
 
 = - x n sin ax - — / x 71 - 1 sin ax dx, 
 a a*J 
 
 Again, put 
 
 u — x 71 - 1 , 
 
 and 
 
 dv =. sin ax dx ; 
 
 then 
 
 du = (n — V^^dx, 
 
 and 
 
 1 
 
 V = COS #«. 
 
 a 
 
 / x n ~ l sin axdx = a 71-1 cos ax 
 
 u a 
 
 -\ / «*~* cos ax dx, 
 
 a v 
 
 ,\ y =z I x" cos axdx = - # n sin «# 
 
 ( x n - 1 cos az H ^— / x n ~ 2 cos a# dx) 
 
 a \ a a v / 
 
 of 1 ' 1 (ax sin &# + n cos «#) w (n — 1) /* „ _ 
 a 1 a 2 «/ 
 
 The formula of reduction for / x >l sin axdx can be 
 obtained in like manner. 
 
 EXAMPLE. 
 
 1. dy = x* cos a? dx. 
 > y = z 3 sin x -J- 3^-' cos # — to sin $ — 6 cos £. 
 
308 FORMULAS Of UF.hVCTION. 
 
 160. Reduction of the Form 
 
 / e™ cos n x dx. 
 
 Put U = COS" X, 
 
 md dv = e^ dx; 
 
 then du = — n cos" -1 x sin x dx, 
 
 and V = — • 
 
 \ y = J er* cos" xdx = 
 
 e™ cos" a; 
 
 + - J e ax cos" -1 # sin a; f?a\ (1) 
 
 Again, 
 
 put 
 
 
 u = cos" -1 a: sin a-, 
 
 and 
 
 
 
 $y = e a *dx; 
 
 then 
 
 
 
 du = — (w — 1) cos" -8 a; sin 2 a; c?a; 
 
 •f cos" x dx, 
 
 and 
 
 
 
 v = — 
 a 
 
 
 
 • 
 • • 
 
 / e M cos* -1 x sin #$r 
 
 = - e" cos* -1 
 
 a; sin a: / e ax [— (n — 1) cos n ~ 2 x sin 2 a; 
 
 + cos" x] dx 
 
 1 ^* « 1 • . (W — 1) /* M „ O 7 
 
 = - e** cos w_1 a: sin a; + * £ / e"* cos n_2 a:6?a; 
 
 a a v 
 
 / e a * cos" x dx. (Since sin 2 x = 1 — cos 2 x.) 
 
 Substituting in (1), and transposing and solving for 
 / &* cos" x dx, we get 
 
 /, e"* cos n_1 x (a cos x + n tin x) 
 e a * cos" xdx = r~- ' 
 a 1 + n* 
 
 ■ + -rr 3 -^ A" cos "" 2 * <& J (2) 
 
 a 2 4- M< »' v ' 
 
EXAMPLES. 309 
 
 which diminishes the exponent of cos x by 2. By con- 
 tinued applications of this formula, we can reduce n to 
 or 1, so that the integral will depend finally on 
 
 /e ax 
 (T dx = — , when n is even ; 
 
 or / e* 8 cos xdx s when n is odd. 
 
 (2) gives the value of / e a * cos x dx without an integra- 
 tion, since the last term then contains the factor n — 1 
 = 1 — 1 = 0, and therefore that term disappears. 
 
 The reduction of / e ax svtfxdx can be obtained in like 
 manner. 
 
 EXAM PLES. 
 
 1. dy = (F cos x dx. 
 
 y = a 1 {a cos a; -J- sin x). 
 
 = e aa cos 2 x dx. 
 
 e°* cos #(« cos x -f 2 sin #) 
 
 + 
 
 * 4 + ft 2 ' 4: + a? a 
 
 161. Integration of the Forms 
 
 f(x) sin -1 as *fc#, /(ac) tan -1 x dx 9 etc. 
 
 Integrals of these forms must be determined by the 
 formula for integration by parts (Art. 147) ; the method 
 is best explained by examples. 
 
 EXAMPLES. 
 
 1. dy = sin" 1 x dx. 
 
 
 Put dv = dx, 
 
 and u = sin -1 x\ 
 
 ien v = a\ 
 
 and r/?/ = 
 
 Vl-x 2 
 
310 EXAMPLES. 
 
 •\ y = J sin -1 x dx 
 
 
 . , P xdx 
 = x sin -1 x — I — 
 
 J VI —a* 
 
 
 = x sin" 1 x + (1 — a?)i 
 
 2. dy = 
 
 x 2 tan -1 # dfa; 
 1 +x* 
 
 Put 
 
 
 and 
 
 w = tan -1 a; ; 
 
 then 
 
 t; = a? — tan -1 a;, 
 
 and 
 
 , rfa; 
 
 . - /. 1 xo /7 %dx tan _1 a^a;\ 
 
 ... y = x tan *x - (tan- xf - J [—^ - T+lF) 
 
 = a; tan" 1 a; — (tan" 1 a:) 2 — £ log (1 + a: 2 ) + J- (tan" 1 a;) 2 
 = a; tan" 1 a? — J (tan" 1 xf — £ log (1 -f z 2 ). 
 
 3. e?y = a 2 sin -1 x dx. 
 
 X s 
 
 y = % sin" 1 x + £ (a: 2 + 2) Vl — a*. 
 
 4. <fy = sin" 1 x ^ y = i (sin- xj. 
 
 162. Integration of tft/ 
 
 a + b cos 
 
 -/ 
 
 ~ t/ « + h cos 
 
 d0 
 
 a (cos 2 ? + sin 2 1) + 8 (cos 2 1 - sin 2 1) 
 
-/• 
 
 EXAMPLES. 
 
 dd 
 
 311 
 
 -/ 
 
 (a + fycos^ + (a -b) sin* ^ 
 
 n 
 
 sec 2 ^ d6 
 
 4* 
 
 a + b + (a—b) tan 2 
 
 2 
 
 When a > b, 
 
 d- tan 
 
 
 
 a + b + (a — b) tan 2 - 
 
 a) 
 
 tan" 
 
 .£+!) tan fl 
 
 V a 2 - 6 2 
 
 (by Ex. 3, Art. 132.) 
 
 When a <b, we have, from (1), 
 
 ».-/ 
 
 <Z0 
 
 a + b cosO 
 
 I: 
 
 d-tan 
 
 2 
 
 b + a — (b — a) tan 2 
 
 
 
 
 
 y #> - <# 
 
 Vb + a -\- Vb — a tan „ 
 log i 
 
 a 
 
 Vb + a — Vb — a tan - 
 
 The integral of 
 
 (by Ex. 5, Art. 137). 
 dd 
 
 to be 
 
 a + 5 sin 
 
 can be found in like manner 
 
 • e i * d 
 a sin ■= + o cos - 
 
 i -1 — — - , when a > 5 ; 
 
 (« 2 -^ (« 2 -^ 
 
 cos 
 
 
 
312 EXAMPLES. 
 
 1 a tan | + b - (P - o>)* 
 and = r log , 
 
 v ^ ; a tan ^ + b + (P - « 2 )* 
 
 when a < J. 
 
 There are other forms which can be integrated by the 
 application of the formula for integration by parts (Art. 
 147). Those which we have given are among the most 
 important, and which occur the most frequently in the 
 practical applications of the Calculus. The student who 
 has studied the preceding formulae carefully should find no 
 difficulty in applying the methods to the solution of any 
 expression that he may meet with, that is not too compli- 
 cated. 
 
 The most suitable method of integration in every case 
 can be arrived at only after considerable practice and famil- 
 iarity with the processes of integration. 
 
 EXAMPLES. 
 
 1. tf y== -J*k y = _j(rf + 2)(i_a*)*. 
 
 2. dy = 
 
 VI— x* 
 x 4 dx 
 
 3. dy = 
 
 Vl — x 2 
 
 (a? 1-3 \ /= = 1-3 . , 
 
 V = * \I + 2.1*7 Vl " X + 2l " * 
 x 7 dx 
 
 (ofi 1-6 . 1-4-6 , l-2.4.6\ /T 
 
 . , x 5 dx 
 
 4. dy = -— 
 
 \/a + bx 2 
 1 /_ . 4rf 8«2\ . — __ 
 
5. dy = 
 
 EXAMPLES. 313 
 
 dx 
 
 Vl - x 2 
 
 / 1 Jj5_ 1-3-5 \ « 2 
 
 'W 4 4.6^ + 2.4.6W V1 ~^ 
 
 1-3-5, 1 + V1 — a 8 
 2^6 l0 S S 
 
 6. % = r— - 
 
 Vl + * , , 
 
 ? = i log 
 
 / yT+"g-i \ 
 
 Wl HM + 1/ 
 
 (See Ex. 1, Art. 142.) 
 
 7. 6??/ = (« 2 — x 2 )§ dx. 
 
 y = \x (a 2 - 2*)! + ~j a*x (a 2 - a*)§ 
 
 5-3 - \ « i 5-3 . a? 
 
 8. dy = £ 3 (1 + z 2 )* <&. ?/ = ^^T (1 + ^ f# 
 
 9. dy = (1— x 2 )§ dx. 
 
 y — \x (1 — z 2 )t + fa; (1 — x 2 f* + f siir 1 ^ 
 dx 
 
 10. % 
 
 y = \a~+bx* + a) 
 
 11. 4 tfy = 
 
 3aVtf + to 2 
 
 (« + foe 2 )* 
 y - \_(a + bx 2 ) 2 + da (a + bx 2 ) + 3a 2 J 5 «v^+^' 
 
 14 
 
514 
 
 EXAMPLES. 
 
 12. 
 
 rhlr 
 V ~ (1 _ 3»)l 
 
 
 z 3 — 3z , . 
 ii = & sin - 
 
 J Vl-a* 
 
 13. dy = . 
 
 V2arc - a* 
 
 /a* , 7a* , 7-5a; 7-5.3 8 \ . 
 
 14. dy = log a; da?, y = x (log a: — 1). 
 
 15. dy = x* log 8 a; dx. y = |a*(log 2 a; — $ log a; + f). 
 
 dx 
 16# dy = x~l^x' y = l0g (1 ° g X) ' 
 
 17. dy = a; log 3 a; d#. 
 
 a; 2 , „ 3a* 2 , 3a*, , 2 
 
 18. dy = x* log a; dr. ?/ = j- log a; — — • 
 
 ■» dx 1 
 
 19. d# = — y— 5 — y — — \ 
 
 ^ a; log 2 x * log a; 
 
 20. <?y = !?££^. y = _ 1 (log 2 a: + 4 loga: + 8). 
 
 21. dy = 
 
 x* x* 
 
 x*dx 
 log 8 x 
 
 x? 5x* 25 
 
 2 log 2 £ 2 log x + 2 
 
 [log (loga-) 
 
 + log a* + i log 2 & + ~ log 3 aA 
 
EXAMPLES. B15 
 
 22. dy — eFMx. y = ^(^—4^ + 12^—24^ + 24). 
 
 a x / 1 \ 
 
 23. dy = xa x dx. y = -, [x — -. )• 
 
 u * log a \ log a) 
 
 24. dy = x 2 e x dx. y = e x (x* — Zx + 2). 
 
 25. dy = ^- y = - e~ x (x* + 2x + 2). 
 na , a x dx 
 
 + ilog2«.| 2 + etc). 
 
 27. tfy= (6 flte+ Y • y = log (a« + an-). 
 
 28. ^ = eftFdx. y = e 6 *. 
 
 __ 1 e x xdx eF 
 
 29. tty = ,_, , > v y = — 
 
 * (1 + xf u 1 + x 
 
 [Put (1 -f x) = z ; then # = 2 — 1, <&«== ^, etc.] 
 
 7 sin 5 <#? ,j , 1M . 
 
 31. dy = 5-3— (Art. 156.) 
 
 * cos 2 v 
 
 ^ = sec + 2 cos — -J- cos 3 0. 
 
 32. ^ = sin* cos 3 dd. y = f sin* — f sin* 0. 
 
 7 sin 3 d$ 5 _ _ 1 n 
 
 33. dy = : y = I cos* (9 — 2 cos* (9. 
 
 cos*0 
 
 c . cos 3 <f<9 . i . 7 - 
 
 34. efy = — — i y = 3 sin* — | sins 0. 
 
 sin » 
 
316 EXAMPLES. 
 
 sin 5 e (IB 
 
 35. du = 
 
 J . cos 2 
 
 y = - u 1 (sin 4 + 4 sin 2 — 8). 
 
 3 cos0 
 
 36. dy = -7— AO 
 
 3 sin 4 cos 2 
 
 1 4 cos 8 cos 
 
 y ~~ cos sin 3 ^ _ 3 sin 8 ~~ 3 sinT 
 
 37. d y = d6 ? . y = 2 tan* 0(1 + i tan 2 0). 
 
 sin* cos* 
 
 sin* dS o - » z, 
 
 38. <fy = T-- y = 4 ton* 0. 
 
 cos*0 
 
 39. <fy = . .f g 4fl - y = - 8 cot 20 - } cot 3 20. 
 
 ^ sin 4 cos 4 * 
 
 40. % = sin 4 cos 4 «W. (Art. 157.) 
 
 y = — ^ (cos 3 0-H cos 0) - ?|~ (sin 3 + | sin 0) 
 
 + 
 
 41. d# = -r-s rz" y = sec + log tan -< 
 
 ^ sin cos 2 J 2 
 
 42. dy = -r-. 
 
 sin cos 4 
 
 ^3^s^0 + ci0 + lo ^ tan | 
 
 43. ^ =s sin 8 cos 6 d0. 
 
 14 
 
 cos' 
 y = — - TF - (sin 7 + 1 %sin 5 + ^sin 3 + -^sin 0) 
 
 + ^ (cos= + J cos'0 + Ycos 6) + ~ 
 
EXAMPLES. 311 
 
 44. dy = sin 4 dd. (Art. 158.) 
 
 y = fa sin 4(9 — J sin 20 + f0. 
 
 45. tf# = cos 4 dd. 
 
 y = fa sin 40 + J sin 20 + |0. 
 
 46. dy = sin 6 Odd. 
 
 y ~ fa (_ i. sin 60 + f sin 40 — - 1 / sin 20 + 100). 
 
 47. ## = :i 4 sin x dx. (Art. 159.) 
 
 y = —x 4 cos # -f 4a; 3 sin x + 12a? cos a; 
 
 — 24a; sin x — 24 cos x. 
 
 48. dy = e** sin 2 a; ^. (Art. 160.) 
 
 e ax sin#, _ x , %<f* 
 
 y == — ~ (« sm a; - 2 cos a;) + 
 
 4 + «* v ■ a (4 + a 2 ) 
 
 49. % = e* sin 3 a; dx. 
 
 y = fae x (sin 3 x + 3 cos 3 a? + 3 sin a — 6 cos x). 
 
 _ . 7 7 sin &£ + & cos foe 
 o0. % = e~ ax sm foe &?. y = U^Wf^~~" 
 
 _ , aMr sin -1 x iK .« +\ 
 
 51. rfy = • (Art. 161.) 
 
 V 1 — a: 2 
 
 Put oV == — and m = sin -1 a:; then 
 
 Vl — x 2 
 
 v — _ £ (38 + 2) VT^ 2 (by Ex. 1), etc. 
 
 ^3 
 
 ^ — _ j (32 + 2) Vl — a? • sin" 1 H g - + ?»■ 
 
 52. #V = . sin -1 x. 
 
 Vi - * 2 
 
318 EXAMPLES, 
 
 rlr 
 
 53. dy = - , ■ (Art. 162.) 
 
 3 2 + cos a; v ' 
 
 v = — r= tan -1 — — tan s • 
 
 s V3 La/3 2 J 
 
 X UC 
 
 54. dw = (See Formula 43, p. 345.) 
 
 + 1 log [x - 1 + (2 - 2x + *»)*]. 
 
CHAPTER V 
 
 INTEGRATION BY SERIES SUCCESSIVE INTEGRA- 
 TION — INTEGRATION OF FUNCTIONS OF TWO 
 VARIABLES — AND DEFINITE INTEGRALS. 
 
 163. Integration by Series.— The number of differ- 
 ential expressions which can be integrated in finite terms is 
 very small ; the great majority of differentials can be inte- 
 grated only by the aid of infinite series.. When a differen- 
 tial can be developed into an infinite series, each term may 
 be integrated separately. If the result is a converging 
 series, the value of the integral may be found with sufficient 
 accuracy for practical purposes by summing a finite number 
 of terms ; and sometimes the law of the series is such that 
 its exact value can be found, even though the series is infi- 
 nite. This method is not only a last resort when the 
 methods of exact integration fail, but it may often be em 
 ployed with advantage when an exact integration would 
 lead to a function of complicated form ; and the two methods 
 may be used together to discover the form of the developed 
 integral. 
 
 EXAMPLES. 
 
 1. Integrate dy = — ■ — in a series. 
 * a -f- x 
 
 By division, 
 
 1 1 x X 2 iC 8 
 
 + -o A 4- etc. 
 
 a + x a a 2 
 
 f* dx /71 xx 2 z 3 , , \ , 
 
 '. y = j — ■ — .= / ( + + etc. ) dx 
 
 u J a + x t/ \a a 2 a B « 4 / 
 
*20 EXAMPLES. 
 
 X X 2 X* X* 
 
 Bufc flTfi = log (a + x) ' [Art * 130 ' (4) ' ] 
 
 ... i g(« + .) = -__ + ___ + e tc. 
 
 2. % = a* (1 - re 2 )* <fc. 
 
 Expanding (1 — re 2 )* by the Binomial Theorem, we have 
 M 9X i ^ x 2 x* x* 5x* 
 
 /• i /, x 2 x 4 - x 6 5x* . \ . 
 = £z* — *»* — &a# — nb^*" — tAtf*^ — etc. 
 
 » ~ /r?^ = tan_1 * [Art m > (16) ] 
 
 X s x 5 x 1 x 9 
 = a! -3+5-7 +9- etC - 
 
 [Art. 144, (1)] 
 j* 3s 5 3 -5a? 
 
 " * ~ 2-3 + 2^5 " 24^7 + etC * 
 
 5. dy = af* (re — l)t ^. 
 
 ^ = f re* — 4z* -f -j^rr* + yJ^'V- — etc. 
 
SUCCESSIVE INTEGRATION. 321 
 
 164. Successive Integration. — By applying the rules 
 previously demonstrated for integration, we may obtain 
 the original function from which second, third or n ih dif- 
 ferentials, containing a single variable, may have been 
 derived. 
 
 If the second derivative ~ = X be given, when X is 
 
 any function of x, two successive integrations will be 
 required to determine the original function y in terms of x. 
 Thus, multiplying by dx, we have 
 
 p. = Xdx; 
 dx 
 
 or d^£j = Xdx. 
 
 Integrating, we get 
 
 ^ = f Xdx = X x + Ci. 
 dx J 
 
 Multiplying again by dx and integrating, we get 
 y = fx { dx + fc x dx = X 2 + G x x + C*. 
 
 Similarly, if we had -^ = X, three successive integra- 
 tions would give 
 
 x 2 
 y = X, + C, - + G,x + G z , and so on. 
 
 Generally, let there be the n th derivative 
 
 dx n 
 
 
 Xdx; 
 
322 success n/: INTEGRATION. 
 
 hence, by integrating we have 
 
 ££ = Jxdx = X, + 0> 
 Again, we get from this last equation, 
 
 and by integrating, 
 d n ~hi 
 
 fpj = jr, + ax + c r 
 
 Also from this we obtain 
 
 yidJ = X^dx + C x x dx -i- Ctdv, 
 and integrating, 
 
 _ =:X,+ fly + c& + c z . 
 
 And continuing the process we get, after n integrations, 
 J fry = J Xdx" 
 
 Y _i_ P _ i p 
 
 - -** -h i/, l#gf8 _ ^ ^ ^ _ x) + - 2 L2.3 . . . ( W _ 2 ) 
 
 + . . . . CU* + (t (i) 
 
 The symbol / Xdx n is called the n m integral of Xdx n , 
 
 and denotes that n successive integrations are required. 
 The first term X n of the second member is the n th integral 
 of Xdx", without the arbitrary constants ; the remaining 
 part of the series is the result of introducing at each 
 integration, an arbitrary constant. 
 
DEVELOPMENT OF INTEGRALS. 323 
 
 165. To Develop the n th Integral J Xdx tl into 
 a Series. — By Maclaurin's theorem, we have 
 
 f' l Xdx n = (f n Xdx n \ + (f^Xdx*-^*^ 
 
 + ( f Xdx ) T^B^ + « ItfE ~, i 
 
 (dX\ x n+l 
 
 + \dx/l-2...(n + 1) 
 
 /cPX\ x n +* 
 
 + V&)l.*...ini%)+^ (1) 
 
 in which the brackets 
 
 (f l Xdx n Y (f'^Xdx"- 1 } .... (fxdx), 
 
 are the arbitrary constants 
 
 G n , (7„_i , . . . . Ci, 
 
 for that is what these expressions become respectively, 
 when x = 0. 
 
 By Maclaurin's theorem, we have 
 
 ,_, (dX\x /d?X\x 2 /d 3 X\ x* 
 
 X" 
 
 which may be converted into (1) by substituting for 
 x°, x 1 , x 2 , x 3 , etc., in (2), the quantities 
 
 x n x n+l x n+i 
 
 -^, etc., 
 
 1-2... rc' 2-3... (» + !)' 3-4. .. (n + 2) 5 
 
 Since ^ /* Xrix* = f Xdx n - 1 . 
 
I VI' LBS. 
 
 ind prefixing the terms containing the arbitrary constants 
 as above shown, viz., 
 
 p p z p n — 
 
 c„, o-.p u^ x %i 0l 1.2-3 ... 0* — 1) 
 
 Lacroix, Calcul Integral, Vol. II, pp. 154 and 155.) 
 
 •EXAMPLES. 
 * drf 
 
 1. Develop / 
 
 J a/i - & 
 
 Here X = (1 - x*)~t 
 
 , a 1-3 . 1-3-5 . 
 
 = 1 + i* 2 + y.\* + 2^6^ + etc ' 
 
 Substituting in this series for x°, x 2 , x 4 , xP, etc., the 
 quantities 
 
 x* x* zP z 10 
 
 PTP * 
 
 1-2.3. 4' 3.4.5.6' 5-6.7-8' 7.8.9.10' ' 
 
 and prefixing 
 
 P P x P p 
 
 1' 2 l-2' x l.W 
 we get 
 
 / 4 dx * -p + p x ip x% ±c * 
 
 5 4.5.6 ^ 
 
 + etc. 
 
 2.3.4 ' 2.3.4.5.6 ' 2.4.5.6.7-8 
 1.3.5a: 10 
 
 1 2.4-6.7-8-9.10 
 2. Integrate d?y = 6a dz?. 
 Dividing by dx 2 we have 
 
 g = *»<** or rf(g) = 6 8( fc. 
 
EXAMPLES. 32u 
 
 ••• .A® -A*' 
 
 or g = 6o* + ^. 
 
 Multiplying by dx and integrating again, we have 
 
 % = 3aa? + (7,a; + ft. 
 da; 
 
 Multiplying again by da; and integrating, we have 
 y = aa; 3 + ft | + ftz + ft. 
 
 3. Integrate d 2 y = sin x cos 2 a? da£. 
 Dividing by dx, we have 
 
 J = sin x cos 2 a; dx. 
 dx 
 
 Integrating, we have 
 
 g« -*«*«+ o,. 
 
 Multiplying by da; and integrating again, we have 
 
 sin 8 a; sin a; , ~ , ■« 
 2/ = -9 3- + C i x + C 2« 
 
 4. Integrate d 3 y = ax 2 dx\ 
 
 5. Integrate d 8 ?/ == %x~Hx\ 
 
 y = \0gX + ^+ax+C t . 
 
 6. Integrate d 4 ?/ = cos a? da; 4 . 
 
 , fta; 3 C,x\ n ,' 
 ^ = cos a; + -g- + -g- + ft^ + ft. 
 
fflM I. \T KG RATION OF FUNCTIONS. 
 
 166. Integration of Functions of Two or More 
 Variables. — Differentia] functions of two or more vari- 
 ables arc either partial or total (Art. 80). When partial, 
 they are obtained from the original function either by 
 differentiating with respect to one variable only, or by 
 differentiating first with respect to one variable, regarding 
 the others as constant ; then the result differentiated witl 
 respect to a second variable, regarding the rest as constant, 
 and so on (Art. 83). For example, 
 
 J LP™ st 
 
 and iETy=u x >y ) 
 
 are differential functions -of the first and second kinds 
 respectively, in which u is a function of the independent 
 variables x and y. From the manner in which the 
 expression d?u „. 
 
 3* ■=/«*»> 
 
 was obtained (Art. 83), it is evident that the value of u 
 may be found by integrating twice with respect to x, as in 
 Art. 165, regarding y as constant; care being taken, at 
 each integration, to add an arbitrary function of y, instead 
 of a constant. 
 
 167. Integration of ^^ =f(x, y). 
 This equation may be written 
 
 du 
 It is evident that -z- must be a function such that if we 
 ax 
 
 differentiate it with respect to y, regarding x as constant. 
 
 the result will be f(x, y). 
 
 Therefore we may write 
 
 5 =//(**> 4* 
 
INTEGRATION OF PARTIAL DIFFERENTIALS. 327 
 
 Here, also, it is evident that u must be such a function 
 that if we differentiate it with respect to x, regarding y as 
 constant, the result will be the function 
 
 ff(x,y)dy. 
 Hence, u = I I f(x, y)dy \dx. 
 
 Therefore, we first integrate with respect to y, regarding 
 x as constant,* and then integrate the result with respect to 
 x, regarding y as constant,* which is exactly reversing the 
 process of differentiation. (Art. 83.) 
 
 The above expression for u may be abbreviated into 
 
 J J f (*. y) d v dx or j J f (* y) dx d y- 
 
 We shall use the latter form ; f that is, when we perform 
 the ^-integration before the ^-integration, we shall write ily 
 to the right of dx. 
 
 It is immaterial whether we first integrate with respect to 
 y and then with respect to x> or first with respect to x and 
 then with respect to y. (See Art. 84.) 
 
 In integrating with respect to y, care must be taken to 
 add an arbitrary function of x, and in integrating with 
 respect to x to add an arbitrary function of y. 
 
 In a similar manner, it may be shown that to find the 
 value of u in the equation 
 
 d 3 u .. x 
 
 dx dy dz 
 we may write it 
 
 u ~ J J j f-ift y> z ) dx dy dz > 
 
 * Called the y-integration and ^-integration, respectively. 
 t On this point of notation writers are not quite uniform. SeeTodhnnter'sOal;, 
 p. 78 ; also Price's Cal., Vol. II, p. 881. 
 
328 EXAMPLES. 
 
 which moans that we first integrate with respect to z, regard- 
 ing X and y as constant ; then this result with respect to y, 
 regarding x and % as constant; then this last result with 
 respect to x, regarding y and z as constant, adding with the 
 -integration arbitrary functions of x and y, with the 
 •/-integration arbitrary functions of x and z, and witli the 
 p-integration arbitrary functions of y and z. (See Lacroix, 
 Calcul Integral, Vol. II, p. 20G.) 
 
 EXAMPLES. 
 
 1. Integrate d?u = l&ydx? 
 
 Here ^\J~) = b&ydx. 
 
 du 
 
 •*• d X - =zfh: * ydx = ^y+f { y ) - 
 
 du = \bx*ydx +f(y)dx. 
 A u = ftbaty + f{y) x + (y). 
 
 2. Integrate d?u = 2x 2 ydxdy. 
 
 Here d(^j = 2x 2 ydy. 
 
 /. ^ = f**ydy = x*tf + 1>(z). 
 
 *?w = x 2 y 2 dx + (a;) <:/#. 
 .-. u = ix*y* + f(t>(x)dx+f(y). 
 
 3. Integrate d% = Sxy 3 dxdy. 
 
 u = %xY + f<l>(x)dx+f(y). 
 
 4. Integrate dhi — ax?y 2 dx dy. 
 
 u = ^y* + f<f>(z)dx+f(y). 
 

 INTEGRATION OF TOTAL DIFFERENTIALS. 329 
 
 168. Integration of Total Differentials of the First 
 Order. 
 
 If u = f{x, y), 
 
 
 we have (Art. 81), 
 
 
 
 , du 
 
 du = -y-< 
 dx 
 
 **+S* 
 
 
 du du 
 in which -5- dx and } dy are the partial differentials of u ; 
 
 also, we have (Art. 84), 
 
 
 
 <&u 
 
 d 2 ^ 
 
 
 dx dy 
 
 " e?# rfo;' 
 
 
 d (du\ 
 dy \dxl 
 
 6? /rf«A 
 
 (1) 
 
 Therefore, if an expression 
 
 of the form 
 
 
 du = Pdx + Qdy 
 
 (2) 
 
 be a total differential of u, we must have 
 
 
 du p 
 35' ' 
 
 du n 
 dy-V' 
 
 
 and hence, from (1), we must have the condition 
 
 dP _dQ (0 , 
 
 dy dx' K °> 
 
 which is called Euler's Criterion of Integr ability. When 
 this is satisfied, (2) is the differential of a function of x and 
 y, and we shall obtain the function itself by integrating 
 either term ; thus, 
 
 u = /Pdx+f(y), (4) 
 
 in which f(y) must be determined so as to satisfy the con- 
 dition 
 
EXAMPLES. 
 
 Remark. — Since the differentia] with respect to x of every term of 
 u which involves x must contain dx, therefore the integral of Pd.v will 
 give all the terms of u which involve x. The differential with respect 
 to y of those terms of u which involve y and not ./ . will he found only 
 in the expression Qdy. Hence, if we integrate those terms of Qdy 
 which do not involve x, we shall have the terms of u which involve y 
 only. This will be the value of f(y), which added with an arbitrary 
 constant to J'Pdx will give the entire integral. Of course, if every 
 term of the given differential contain x or dx, f(y) will be constant. 
 (See Church's Calculus, p. 274.) 
 
 EXAMPLES. 
 
 1. du = ±x*yWx -f 3x 4 y 2 dy. 
 
 Here P = 4ay, Q = dx*f. 
 
 ... ^ = i2aty» and ^ = 12ay. 
 
 Therefore (3) is satisfied, and since each term contains l 
 or dx, we have from (4), 
 
 u = fkx*y z dx = x*y 3 + C. 
 (3) is satisfied, therefore from (4) we have 
 
 » = /f +/<*>= J +/<y). 
 
 Since the term 2ydy does not contain x, we must have, 
 from the above Remark, f(y) = f2ydy = y 2 , which must 
 
 x 
 be added to - , giving for the entire integral, 
 
 « = | + ? + a 
 
 3. du = ydx 4- xdy. u = xy + C. 
 
 4. du = (Gxy — f) dx + (3z 2 — 2xy) dy. 
 
 u = 3a?y — y 2 x -f- C. 
 
DEFINITE INTEGRALS. 331 
 
 5, du = (2axy — 3bx 2 y) dx + {ax 2 — bx 3 ) dy. 
 
 u = ax 2 y — byx 3 + C. 
 
 The limits of this work preclude us from going further in 
 this most interesting branch of the Calculus. The student 
 who wishes to pursue the subject further is referred to 
 Gregory's Examples; Price's Calculus, Vol. II; Lacroix's 
 Calcul Integral. Vol. II ; and Boole's Differential Equations, 
 where the subject is specially investigated. 
 
 169. Definite Integrals.- It was shown in Art. 130 
 that, to complete each integral, an arbitrary constant C 
 must be added. While the value of this constant C remains 
 unknown, the integral expression is called an indefinite in- 
 tegral ; such are all the integrals that have been found by 
 the methods hitherto explained. 
 
 When two different values of the variable have been sub- 
 stituted in the indefinite integral, and the difference between 
 the two results is taken, the integral is said to be taken 
 between limits. 
 
 In the application of the Calculus to the solution of real 
 problems, the nature of the question will always require 
 that the integral be taken between given limits. When an 
 integral is taken between limits, it is called a definite 
 integral.* 
 
 The symbol for a definite integral is 
 
 / 
 
 b 
 f(x) dx, 
 
 a 
 
 which means that the expression f(x) dx is first to be inte- 
 grated ; then in this result b and a are to be substituted 
 successively for x, and the latter result is to be subtracted 
 from the former ; b and a are called the limits of integra- 
 tion, the former being the superior, and the latter the 
 inferior limit. Whatever may be the value of the integral 
 
 * In the Integral Calculus, it is often the most difficult part of the work to pass 
 from the indefinite to the definite integral. 
 
nrn.MTE iNTEti i;.\i.s. 
 
 at the inferior limit, that value is included in the value of 
 i he Integral up to the superior limit. Hence, to find the 
 integral between the limits, take the difference between the 
 values of the integral at the limits. 
 
 In the preceding we assume that the function is continu- 
 ous between the limits a and b, i.e., that it does not become 
 Imaginary or infinite for any value of x between a and b. 
 
 Suppose u to be a function of x represented by the equa- 
 tion 
 
 u = f(x); 
 
 then du = f (x) dx. 
 
 Now if we wish the integral between the limits a and b, 
 we have 
 
 « = ff'^)dx=f(b)-f(a). 
 
 v a 
 
 If there is anything in the nature of the problem under 
 consideration from which we can know the value of the 
 integral for a particular value of the variable, the constant 
 C can be found by substituting this value in the indefinite 
 integral. Thus, if we have 
 
 du = (abx — bx 2 )$ (ab — %bx) dx, 
 
 and know that the integral must reduce to m when x = a, 
 we can find the definite integral as follows: 
 
 Integrating by known rules, we have 
 
 u =r | (abx - bx*)% + C, 
 
 which is the indefinite integral; and since u = m when 
 z = a, we have 
 
 m = + C; .-. C = m; 
 
 which substituted in the value of u gives 
 
 u — | (abx — bx 2 j* + m. 
 
EXAMPLES. 
 
 333 
 
 EXAMPLES. 
 
 ^l. Find the definite integral of clu = (1 + $ax)$ dx, on 
 the hypothesis that u = when # = 0. 
 
 The indefinite integral is 
 
 Since when jc = 0, w = 0, we have 
 
 .-. = - 
 
 8 
 
 ° = m + (7 ' 
 
 
 which substituted in the indefinite integral, gives 
 
 for the definite integral required. 
 2. Integrate 6?w = Qxhlx between the limits 3 and 0. 
 
 Here 
 
 §Mx 
 
 54. 
 
 1 _ 1_ 
 
 o ~ n + V 
 
 Jo 
 
 3. u = / x"dx = 
 
 Jo [n + 1 
 
 /»oo ~| qo 
 
 4. u = J e~ x dx z=\ —e~* \ = - (0 — 1) = 1. 
 
 r«> dx ir .a?" 
 
 5. u = / - F - — ^ = - tan" 1 - 
 
 t/o « 2 4- a 2 # [_ a 
 
 r a dx ir, t x m 
 
 6. U = -t-t— 1 2 — " tan 
 
 77 
 
 « _ 7T 
 
 * This notation signifies that the integral is to be taken between the limits 3 
 and 0. 
 
CBANGM OF limits. 
 
 «/_ao fl 2 + «* «L ^J-o 
 
 = - rtau -1 oo — tan" 1 (— oo )1 = - 
 
 p a dx r. t a;"| a 
 
 Remark. — It should be observed here that the value of the infini- 
 tesimal element corresponding to the superior limit is excluded, while 
 that corresponding to the inferior limit is included in the definite in- 
 tegral ; for, were this not the case, as — zr^r— becomes equal to oo 
 
 when x — a, the integral of Ex. 8 between the limits a and would 
 not be correct ; but as the limit a, being the superior limit in Ex. 8, 
 and that which renders infinite the infinitesimal element, is not 
 included, the definite integral is correct. (See Price's Calculus, 
 Vol. II, p. 89.) 
 
 9. u = f a (a? — xrfdx. (See Ex. 4, Art. 151.] 
 
 10. u = f l f^L =. (See Ex. 7, Art. 151.) 
 t/ Vl-z 2 
 
 _ 1-3-5-7T 
 
 ~~ 2-4.6.2° 
 
 4* 
 
 11. u = I sin 7 x cos 4 x dx 
 
 3.5.7.11 
 
 170. Change of Limits. — It is not necessary that the 
 increment dx should be regarded as positive, for we may 
 consider x as decreasing by infinitesimal elements, as well as 
 increasing. Therefore, we have 
 
CHANGE OF LIMITS. 335 
 
 A*' (x) dx = (a) - (b) = - [0 (*) - («)] 
 
 = — / 0' (<r) rfa;. 
 
 That is, i/ we interchange the limits, we change the 
 sign of the definite integral. 
 
 Also, it is obvious from the nature of integration (Art. 129), that 
 
 pc pb pc 
 
 j <p (x) dx = / <$>{x)dx + I <p (x) dx, 
 
 and so on. Hence, 
 
 /»ir /»$"" pin /a fir 
 
 / cos a? <fa; = / cos x dx + I cos x dx + / cos a; efa; 
 
 t/0 t/0 t/iir J[|» 
 
 + / cos a; dx 
 
 J\n 
 
 pin pit 
 
 — I cos a; dx + I cos x dx — 0. 
 
 Jo Jin 
 
 Also, /^ V (*) dx = C f ( x ) dx + ^f ( x ) dx - (!) 
 
 J -a J -a t/0 
 
 Let x = — x; then <£c = — dx, and the limits and — a become 
 and + a ; therefore we have 
 
 p0 />0 pa 
 
 / f'(x)dx=- / f'(-x)dx= / f'(-x)dx, 
 J -a Ja t/0 
 
 which in (1) gives 
 
 pa pa pa 
 
 / f'(x)dx = / /'(_aj)<te+ / /'(«)<&> 
 ./-a t/0 »/0 
 
 = ^ [/' H») *8 + /' (*) *bJ. (2) 
 
 Now if /' (—x) — — f (x), (2) becomes, 
 
 yV (»)<& = a 
 
But if /' (-x) -f {x\ we have 
 
 j*J* (») dx = j™f (x) dx, (3) 
 
 pa pa 
 
 which in ( 1 ) gives / f'(x)dx = 2 / f (x) dx. (4) 
 
 J-a Jo 
 
 The following are examples of these principles. 
 
 1. u = / cos a; dx. 
 
 Here /' (— x) = f (x) = cos x. 
 
 /♦in- p\it 
 
 cos x dx = 2 I cos x dx = 2. 
 
 2. u= sin«^ = 0. [Since/' (—«) = —/'(.r).] 
 
 3. if = /"V - a^*rfj» = 2 /'"(a 2 - a*)*<& = ^ 
 
 *' -a vq Z 
 
 pit p\lt pit 
 
 4. u = / sin a; eta; = / smxdx-j- / sinxdx 
 
 = 2 / sin a; da;. 
 
 Since / sin a; da; = / * sin a; da; = 2. 
 
 pit p\it pit 
 
 5. u = / cos x dx = I cos x dx -\- I cos a; da; = 0. 
 
 Since / cos x dx = — I cosxdx. 
 
 <-'\it ^o 
 
 «/-i Vl — a; 2 t7 o Vl — z 2 
 
 (See Ex. 2, Art. 162.) 
 
EXAMPLES. 33? 
 
 E X AM PLES, 
 
 1. Integrate du = -— — =^= by series. 
 V« 8 — x 2 
 
 W " a + 1-2. 3a 8 + 2^. 5a* + 2.4.G-W + 
 But f * X = sin- 1 - (by Ex. 14 of Art. 131) ; 
 
 therefore, 
 
 a a; z 3 l-3z 5 l-3-5a; 7 
 
 aa; 
 
 Sm * a " a + 2«3« 3 + 2.4.5a* + 2.4-6-7a 7 + etc * 
 
 2. Integrate (Zm = — -__ 
 
 _ £ l^ 5 1-30 9 _ l-3- 5.r 13 
 W ~ I ~ 2T5" + 2^T§ ~~ 2^6713 + C * 
 
 dx 1 
 
 3. Integrate du — — - (1 — e 2 z 2 )^. 
 
 a/1 — a* 
 By the Binomial Theorem, 
 
 VT-1P = l - yy> -~~- j^ - etc. 
 
 Multiplying by —--—=_- , and integrating each term sep- 
 
 vi 
 
 X 
 
 arately (see Ex. 1, Art. 151), we have 
 
 u = f- ~— (1 - eW)i 
 J VI -^ 2 
 
 sm -1 a; 
 
 15 
 
 + \e 2 | Vl - x 2 - } sin" 1 aH 
 
 e 4 T/.r 3 1-3. «\ 7l r 1-3 . , ~1 
 
 + 2-4 \k + it) Vl - x - n sm x \ 
 
333 EXAMPLES 
 
 . l-S.^r/a 8 , 1-5* 3 . 1.3.5z\ ,- -3 
 
 1.3.5 . 
 
 27476 81 
 
 o 
 
 4. Given d*y = -gdz 3 , to find #. (Art. 164.) 
 
 # = logo; + 1(7,3* + fts + <7 3 . 
 
 5. tfty = — x^dx 4 . 
 
 y = $\ogx + idx* + i6' 2 ^ + C\x + ft. 
 C. rf»y = *W. y = yfg-^ 4- Jfta 2 + ft* + ft> 
 
 7. <ty = Sxhlx*. y = JzSx* + C,x + ft. 
 
 8. d 2 y = cos 2: sin 2 x dx 2 . 
 
 y = Icos 8 ^ — -Jcos# +Cj x + C 2 . 
 
 9. d*y = cos * rfz 4 . 
 
 y = cos re + £C,^ + \C&* + fts + C 4 . 
 10. d 3 */ = esffa 3 . y = e a + J (7,^ + fts + C 3 . 
 
 11 dfy = (1 + «*)-*<&* (Art. 165.) 
 
 r 2 /v3 ^4 
 
 y= ft+ftB+ft| + ft^ + 
 
 ' '2-3 ' 2.3.4 
 a 6 1-3^ 
 
 2.3.4.5.6 ' 2.4-5. 6-7. 8 
 1.3. 5x 10 
 
 4- etc. 
 
 2.4.6.7.8.9.10 
 
 12. cPu = axlyWxdy. 
 
 u = ^x*y* + f<(>(x)dx +f(y). 
 
 13. du = (2xy* + 9afy + 82*) rfar + {%x*y + 3s 8 ) %. 
 
 u = tfy* + 3^ + 2z* + G 
 
EXAMPLES. 839 
 
 14. u = f a (a 2 — z 2 )f dx — % f*( a * — z 2 )f dx. 
 
 (See Art. 170.) 
 u = ^g « 6 tt. (See Ex. 7 of Art. 162.) 
 
 i&. u 
 
 {**" > x*dx 
 Jo */2ax — x 2 
 
 7-5-3 . 
 
 16. u = f l «*(l-x)§dx 
 
 o 3.7.11.13' 
 
 dx 
 
 17. « ^ ; --■-— (i _ e 2^)l (See Ex. 3.) 
 
 «/o Vl — Z 2 
 
 W - 2 4' 2 ~ 22.4^ 2 ~ 2M^6 2 2 "" etC ' 
 
 18. u= f" f XX -^-% (Art. 167.) 
 
 We first perform the ^/-integration, regarding x as con- 
 stant, and then the ^-integration. 
 
 .-. u = jT H tan- 1 |~f» <*& (Art. 132, Ex. 3.) 
 
 /^a /># /»y 
 
 19. «« = / / / xyzdxdydz 
 
 pa nx xi ,z . , /™z 5 7 « 6 
 
 = y o y t dxd y = J r* = 4r 
 
 20. tt = / / / da; rfy tfz = £• 
 
340 formula; of integration. 
 
 f o •'o 
 
 /\* ,-*%a cos 6 
 / r*dO dr = fa 4 *. 
 
 For the convenience of the student, the preceding for* 
 mulse are summed up in the following table. 
 
 TABLE OF INTEGRALS. 
 CHAPTER I. 
 
 Elementary Forms. (Page 23 Si) 
 
 1. J (dv + dy — dz) = v + y — z. (130)* 
 
 2. /'ax»dx = -^. (131) 
 
 /W# a 
 
 J af == (» — 1) z"- 1- 
 
 4. y — - = a log & 
 
 5. / a 1 log adx — a x . 
 
 6. y e 35 ^ = e x . 
 
 7. J cos xdx = sin a?. 
 
 8. / sec 2 xdx = tan #. 
 
 9. / sec a? tan xdx = sec .r. 
 
 * The arbitrary constant is understood. (See Art. 131.) 
 
FORMULAE OF INTEGRATION. 541 
 
 10. / *?_1, in-.* (133) 
 
 J A /ni _ hit* o a 
 
 1, / 
 
 dx 1 , _j #a; 
 
 a 2 + ^ai* ~~-aJ a 
 
 (fa 1 , &e 
 
 = - sec - 
 
 ^^2 _ d z a a 
 
 P dx 1 .bx 
 
 13. / — -=i= = t vers -1 — « 
 «/ ^2abx — b 2 x 2 b ® 
 
 14. / tan a: tfa = log sec a;. 
 
 15. / - — = log tan ix. 
 
 t/ sm x ° 
 
 CHAPTER II. 
 
 Rational Fractions. (Page 256.) 
 /y(aO<fa _ f> Adx Bdx_ fJ^L nm 
 
 17 /* /(*)*? 
 17 - / 0(.r)" 
 
 /» ^4(fa , /» £(fa /» Zkfa , . 
 
 =y(5=^+y^^p + --v^^' (i38) 
 
 is rf{x)dx _ r (Ax + .g ) (fa 
 
 lb ' ,/ (a;) ~ */ [(a; ± a) 2 + 6 2 ]" 
 + J [(x ± af + J 2 ]- 1 + ' ' ' «/ (a ± «) 2 + & K 
 
 /* «(fa . [x — a 
 
148 FORMULJE OF I.\Ti:<; RATION. 
 
 CHAPTER III. 
 Irrational Functions. (Page 269.) 
 
 [where z = (a + to)*]. (142; 
 
 Pj^dx_ P (# — a)* <fc 
 
 ^ (a + &*)* " ' * n+1 ' 
 
 23. 
 
 (a + to 2 )* 
 [where z = (a + to 2 )*]. (143) 
 
 f/ v « + to + z 2 V/5 7 
 
 d Va + to — z 2 V a — «' 
 
 [where a + to — z 2 = (x — a) (0 — x). 
 
 25. fx^a-\-bx n )ulx=^J*z^ + ^(p^j ' <fe,(146) 
 
 (where ^ = « + to n ) ; 
 or s — I aU + f + ^y (#_g)V-=-V7 s</>+*-» (fo, 
 
 (where a:V = a + to w ). 
 86 . f-¥=== * log (v^+^-V^ . Ex . 6 , (ltf) 
 
 27 - ^v«S^ = ^ log (te+ v ^* ) - Bx - l6 ' 
 
FORMULAE OF INTEGRATION. 343 
 
 29. P _____ ___ = a log (x + a + <s/%ax + &). Ex. 20. 
 
 J V%ax + x* 
 
 30. f(a* + x*)?dx 
 
 = | (« 2 + z 2 ) 1 + 1 8 lo g l> + (« 2 + *)*]• Ex * 35 ' 
 
 CHAPTER IV. 
 
 Successive Reduction. (Page 285.) 
 
 31. Judv = uv — fvdu. (147) 
 
 32. fx m (a + bx n f dx (148) 
 x m - n+l (a + bx n ) p+1 — (m—n + 1) a fx m - n (a + bx n ydx 
 
 . K . (j\\ 
 
 b {np + m + 1) v ' 
 
 33. yV"» l (a + 6z n )* „» (149) 
 
 x~ m+] (a + te' 1 )^ 1 + b(m — np—n — 1) C'x- m+n (a + bx n ) p dx 
 — « (m — ij" ~~ 
 
 34. J x m (a + to w )" dx (150) 
 
 a*M-i (« 4- f)x n y 4- a^ /V* (a + bx")*- 1 dx 
 _ (C) 
 
 /> 
 
 35. J x m (a + bx n )~P dx (151) 
 
 #" l+1 (« + te n )-* +1 -(m + n + 1- np) fx m (a 4- Js*)-** 1 ^ 
 
 an(2 ) — 1 ) 
 36. f x m {a* — x*)-? dx 
 
 ^~ x / 2 _\4 . (w— 1) 
 
 (£) 
 
 _ <___ (a2_^2)i + _______ ^ ^m-2 (-» _ -*)-* fa 
 
;;il FORMULA Of INTEGRATION. 
 
 37. far (a 2 + z 2 )"^ dx 
 
 = ^ ( rt ? + aj)i _ fSLzH a 2 far~ % (a 2 + aM & 
 
 88 
 
 • ■/ 
 
 £fo 
 
 X >n ( rt 2 _ ^)i 
 
 ~ (m -. IJaftr-' + (m — 1) a 2 </ ^-2^2 _ ^ 
 39. f(a*-x*Ydx 
 
 x (a 2 — xrf + ??« 2 /V - a?Y~ X dx 
 
 4o 
 
 /; 
 
 »+ 1 
 
 dx 
 
 ( fl « _ z 2 )£ 
 _ a; w— 3 /* da? 
 
 r aTdx 
 
 V%ax — x 2 
 
 x m ~ l ,= - q (2m — l)a r x m ~ A dx 
 
 V %ax — x 2 + ± — / — 
 
 ™ m J V^ax-x 2 
 
 x n dx 
 Va -f bx + ex 2 
 
 = x n ~ x Va + bx + ex 2 n — la f* x n ~Hx 
 
 — 1 a r 
 
 n ' °J Va 
 
 n c v y/ a + bx + CX 2 
 2n—l b r x n ~ x dx * 
 
 r af_ 
 
 J Va~+ 
 
 2n cJ ^/ a + j )X + cx i 
 
 * See Price's Calculus, Vol. II, p. 
 
43 
 
 •7 
 
 FORMULA OF INTEGRATION. ; / 345 
 
 .T 2 6?£ 
 
 a/« + bx + &£ 2 
 
 x 36 \ . /3£ 2 a\ /» dx 
 
 44. A 
 
 ^ V a + fa + cz 2 
 
 ^a + bx + cx* b C dx 
 
 Va + bx + c# 2 
 
 #r/£ 
 
 
 fa + ex 2 
 
 45. y af* log" adk 
 
 sa --r log" a; / x m log"" 1 :wfc. (152) 
 
 m + 1 ° w -f 1 «/ ° v ' 
 
 46. /££ 
 
 */ log"£ 
 
 47. /* 
 
 ff ,n+ 1 m + 1 r x m dx 
 
 (n - 1) log"- 1 x + n — 1 «/ log"- 1 a;' * 15d ' 
 
 log ^ 
 
 = log (log z) + log z + p log 2 2 + ^-p log 3 z -fete. 
 
 flW<& = — >-- r^ — / (^x—'dx. (154) 
 
 m Jog « wz log # */ 
 
 49 f^x = _ <*> x , iog_« /»fl»<fo , 
 
 ' «/ £»> ( m _ i) aj»n-i ^ m — \J afi-i v ; 
 
 50. fs^d cos? ddO 
 
 = — y*(l — cos 2 0)'- cos n 0^ cos 0, (156) 
 
 * See Price's Calculus, Vol. II, p. 63. 
 
:}40 FORMULA OF INTEGRATION* 
 
 (when in = 2/- -f 1) ; 
 or = /'tan'" (1 + tan* 0)'-W tan 0, 
 
 (when m -f ft = — 8r) ; 
 or = /*" (1 — a?)~*~dx, (when a; = sin 0). (157) 
 
 51. / x" cos aa; da; 
 
 = ^-g-te sin aa; + w cos ax) - — ^— - / x n ~ 2 cos aa; </a;.(159) 
 
 p , e" x cos" -1 a; (« cos a: + w sin a:) 
 
 52. / e"* cos" xdx = 3— — 5 
 
 «/ a 2 -f »* 
 
 w ( W _ i) /^ cogn2 ^ ^ 
 « 2 + w 2 t/ 
 
 53. ysin- 1 xdx = x sin" 1 x + (1 - a; 2 )i (161) 
 
 54 /■_*— 
 
 «/ « + cos 
 
 = vfa *■*" [fefO 1 fl (when a>b) - {m) 
 
 1 rvo + tf + vo — # tan ri , . ... 
 
 = . log . : ; „ , (when a<b). 
 
 c 2 . M x n ~ l 
 
 -1) 
 
 55. /^*ft+CUj+CUjS+..« I5 ^ 
 
 ^ v ; 1-2-3... n^ \dxll.%.3,..(n + l) 
 
 (d?X\ af +2 
 
 w) i-a.3...(«H ^) + etc - (165) 
 
CHAPTER VI. 
 
 LENGTHS OF CURVES. 
 
 171. Length of Plane Curves referred to Rectan- 
 gular Axes. — Let P and Q be two consecutive points on 
 the curve AB, and let (x, y) be the 
 point P ; let s denote the length of 
 the curve AP measured from a fixed 
 point A tip to P. Then 
 
 PQ = ds, PR = dx, RQ = dy, 
 
 Therefore, from the right-angled 
 triangle PRQ we have 
 
 hence, 
 
 Vdx 2 + dy 
 
 *-/( 
 
 *#* 
 
 To apply this formula to any particular curve, we find 
 
 the value of -~ in terms of x from the equation of the 
 
 curve, and then by integration between proper limits s 
 becomes known. 
 
 The process of finding the length of an arc of a curve is 
 called the rectification of the curve. 
 
 It is evident that if y be considered the independent 
 variable, we shall have 
 
 -yvsy* 
 
 The curves whose lengths can be obtained in finite terms 
 are very limited in number. We proceed to consider some 
 of the simplest applications : 
 
RECTIFICATION OF TEM PARABOLA. 
 
 172. The Parabola. — The equation of the parabola Is 
 y* = 2px ; 
 
 hence, -j- = -• 
 
 ax y 
 
 or * = - /'(//> + y 2 )*<fy, (which, by Ex. 35. Art. 140> 
 
 - «^t£ + 1 log (y + Vf + f) + a • (i) 
 
 If we estimate the arc from the vertex, then s = 0, 
 y = 0, and we have 
 
 0=pogp+O; .: 0=-|lo gi >, 
 
 which in (1) gives 
 
 which is the length of the curve from the vertex to the 
 point which has any ordinate y. If, for example, we wish 
 to find the length of the curve between the vertex and one 
 extremity of the latus-rectum, y = p, we substitute p for y 
 in (2), and get 
 
 s = lpV2+£\og(l + V2) 
 for the required length. 
 
 We have here found the value of the constant C by the 
 second method given in Art. 1G9. We might have found 
 the definite integral at once by integrating between the 
 limits and p, as explained in the first method of Art. 109, 
 and as illustrated in the examples of that Article. Hence, 
 
RECTIFICATION OF THE CIRCLE. 349 
 
 we need not take any notice of the constant C, but write 
 our result 
 
 s = - f P (p 2 + y 2 )* dy, (see Art. 169) 
 
 and integrate between these limits. 
 
 173. Semi-Cubical Parabola.* — The equation of this 
 curve is of the form y 2 = ax 3 . (See Fig. 39. ) 
 
 Hence, -— == \^fax and -^ = \ax. 
 
 A s = J*(\ + %ax)^ 
 
 If we wish to find the length of the curve from A to P, 
 we must integrate between the limits and dp (see Art. 
 L28, Ex. 9) ; hence, ' 
 
 i = J o (1 + \ax)*dx = -- (1 + lax)i 
 = ^[(l4-¥^) f -l] =?(3t-l), 
 
 Q 
 
 by substituting -— for a. (See Art. 125, Ex. 1. Compare 
 Ex. 10, Art. 128.) 
 
 174. The Circle. — From x 2 -f- y 2 = v 2 , we have 
 
 dy _ x 
 dx ~ y 
 
 * ''This was the first curve which was rectified. The author was William Neil, 
 who was led to the discovery, about 1660, by a remark of Wallis, in his Arithmetics 
 lufinitorum. See Gregory's Examples, p. 420. 
 
RECTIFICATION OF THE ELLIPSE. 
 
 Benoe, for the length of a quadrant, we have (since the 
 limits are and /), 
 
 Jo \ y 2 / Jo A / r 2_ iC 2- 
 
 = r sin -1 - = Ittt, 
 
 which involves a circular arc, the very quantity we wish to 
 determine. The circle is therefore not a rectifiable curve ; 
 Ian the above integral may be developed iuto a series, and 
 an approximate result obtained. 
 By Ex. 1, Art. 170, we have 
 
 f (x , a* 1.33* 1.3.5a? , Y] r 
 
 s = [ r ^ + a 7 * 5 + *^ + STerw + etc ')j 
 
 * ^ = 1 + 2^ + S + ^li7 + etC - 
 
 By taking a sufficient number of terms, reducing each to 
 a decimal, and adding, we have n = 3. 141 592653589793 + . 
 For the approximation usually employed in practice, ?r is 
 taken as 3.1416, and for still ruder approximations as 3|. 
 
 175. The Ellipse.— From y 2 = (1— e 2 ) (a 2 — x 2 ), we have 
 
 *f _ -^/1_*Y?_ _^i 
 
 (1-^)= = 
 
 To find the length of a quadrant, we must integrate be 
 tween the limits and a ; hence, 
 
 C a /" , x*(l — e*) . /* Id^-ltx 2 ' 
 
lifiCTIFICATION OF THE CYCLOID. 
 
 351 
 
 This integration cannot be effected in finite terms, but 
 may be obtained by series. 
 
 X 
 
 Put - = z; then dx = adz. When x = a, z = l, and 
 when x = 0, z = ; therefore the above integral becomes 
 
 dz 
 
 s = a f\l-(W)i 
 
 Jo Vl-z 2 
 
 * {, <o 1-3 t l-3 2 -5 . h \ 
 = a-{l-le*- ^ - ¥7¥ ^^ - etc.), 
 
 (by Ex. 17, Art. 170), which is the length of a quadrant of 
 the ellipse whose semi-major axis is a and eccentricity e. 
 
 y 
 
 176. The Cycloid.— From x = rvers -1 - — \'2ry—y\ 
 
 we have 
 
 dx _ y 
 
 dy ^2ry — y* 
 
 r*2r j 
 
 .% * = V2r / (2r — y)~* dy 
 
 = _ 2 (2r)* (2r - y)H = 4r, 
 
 which is J the cycloidal arc; 
 hence the whole arc of the cy- 
 cloid is 8r or 4 times the diam- 
 eter of the generating circle. 
 
 If we integrate the above ex- 
 pression between y and 2r, we get 
 
 s = V2r f \lr — y)-* dy = 2 (2r)* (2r — y)* 
 
 == 2 V2r (2r — y) == arc BP. 
 
 A 
 
 Fig. 44. 
 
 But BD = VBAxBC = V2r{2r-y); 
 
 .: arc BP = 2 times chord BD.* 
 
 * This rectification was discovered by Wreu. Sec Gregory's Examples, p. 421. 
 
352 
 
 INVOLUTE OF A CIRCLE. 
 
 177. The Catenary.— A catenary is the curve assumed 
 in- m perfectly il«'.\ii>U' Btring, when 
 it- ends are fastened ;ii two points, 
 A and B, nearer together than the 
 length of the Btring. [te equation is 
 
 y = l(r +*-•)■ 
 
 Hence, 
 
 Fig. 45. 
 
 g = i (;- _ e -~y ... ds = I (;- + *-•') a*. 
 
 If 8 be measured from the lowest point V, to any point P 
 (./■. //), we have 
 
 178. The Involute of a Circle.— (See Art. 124.) Let 
 C be the centre of the circle, 
 whose radius is r ; APR is a 
 portion of the involute, T and 
 T' are two consecutive points 
 of the circle, P and Q two 
 consecutive points of the in- 
 volute, and the angle ACT. 
 Then TCT' = PTQ = d<}> 9 
 and PT = AT = r<p. 
 
 /. ds = PQ = nffi* ; Fi S- 46 - 
 
 A s = rf((>d<f) = lr<p + C. 
 If the curve be estimated from A, C = 0, and we have 
 
 For one circumference, (p = 2n ; .\ 8 = \r (2tt) 8 = 2r-n 2 . 
 For n circumferences, <f) = %m\ .-. & = ^r^mrf = 2rn 2 rrK 
 
THE CARD10IDE. 353 
 
 179. Rectification in Polar Co-ordinates. — If the 
 
 curve be referred to polar co-ordinates, we have (Art. 102), 
 
 ds 2 = r 2 dd 2 + dr 2 ', 
 hence we get s = / (r 2 + -==) J0, 
 
 180. The Spiral of Archimedes. — From r = ad, we 
 
 have 
 
 <& __ 1 
 
 dr ~ a 
 
 s = - f T (r 2 + a 2 )? dr 
 
 )ctf-±Jf±£), 
 
 r (a 2 + r 2 )\ a ,_ /r -f- Va 2 + 
 
 2a ' 2 
 
 (see Art. 172), from which it follows that the length of any 
 arc of the Spiral of Archimedes, measured from the pole, is 
 equal to that of a parabola measured from its vertex, r and 
 a having the same numerical values as y and p. 
 
 181. The Cardioide. — The equation of this curve is 
 
 r = a (1 + cos 6). 
 
 dr 
 Here iz = — a sin 6, 
 
 atf 
 
 and hence s = f[a 2 (1 + cos Of -f a 2 sin 2 d]?dS 
 = af{2 + 2 cosd)?dd 
 
 cos - JO = 4a sin - + G, 
 
 2 2 
 
354 
 
 LENQ THS OF CI ft l WS / . V sr \> r. 
 
 If we estimate the arc 
 from the point A, for which 
 = 0, we have 
 
 s = 0; 
 
 O=0. 
 
 Making = n for the 
 Baperior limit, we have 
 
 s = 4a sin 
 
 -> 
 
 4a, 
 
 Fig. 47. 
 
 which is the length of the arc ABO; hence the whole 
 perimeter is 8a. 
 
 182. Lengths of Curves in Space.— The length of 
 an infinitesimal element of a curve in space, whether plane 
 or of double curvature, from the principles of Solid Geom- 
 etry (see Anal. Geom., Art. 169) is easily seen to be 
 
 Vdx 2 + d\f + dz** 
 
 Hence, if s denote the length of the curve, measured from 
 some fixed point up to any point P (x, y, z), we have 
 
 Vdx 2 + dy 2 + dz 2 
 
 / 
 
 -/[*♦<#+©? 
 
 dx. 
 
 If the equations of the curve are given in the form 
 y=f(x) and z = (f>(x), 
 
 we may find the values of ~ and y- in terms of x, and 
 then by integration s is known in terms of x. 
 
 * The student who wants further demonstration of this, is referred to Price's 
 Cal., Vol. I, Art. 341, and Vol. II, Art. 164; De Morgan's Dif. and Integral Cal., 
 p. 444 ; and Ilomersham Cox'b Integral Cal., p. 95. 
 
EXAMPLES. 355 
 
 183. The Intersection of Cycloidal and Parabolic 
 Cylinders. — To find the length of the curve formed by the 
 intersection of two right cylinders, of which one has its 
 generating lines parallel to the axis of z and stands on a 
 parabola in the plane of xy, and the other has its generating 
 lines parallel to the axis of y and stands on a cycloid in the 
 plane of xz, the equations of the curve of intersection being 
 
 y 2 = 4px, z = a vers -1 - + V%ax — x 2 . 
 
 TT dy In , dz l%a — x 
 
 Here di = yi and 7n = \/-^-' 
 
 ... * = (i + 1 -+•«? _ i)» ax =ip + 2a)i % 
 
 \ X X I As/ / X 
 
 Estimating the curve from the origin to any point P, wo 
 have 
 
 dx 
 x$ 
 
 = f X (p 4- 2a)i cI ? = 2(p + 2a)* V^ 
 
 EXAMPLES. 
 
 1. Rectify the hypocycloid whose equation is 
 
 #i -f- yi = «t. 
 Am. The whole length of the curve is 6a. 
 
 X 
 
 2. Rectify the logarithmic curve y = be". 
 
 Ans. s = a log ===== + V«M- y 2 + C. 
 
 e x -f- 1 
 
 3. Rectify the curve e y = * between the limits 
 
 x = 1 and a? = 2. 
 
 ^s. 5 = log (e + e _1 ). 
 
i UI'LES. 
 
 4. Rectify toe evolute of the ellipse, its equation being 
 
 1. 
 
 er+tf 
 
 Put x = a cos 3 0, y = j3 sin 3 ; 
 
 then eb =3 — 3« cos 2 sin ^/0, 
 
 dy = 3j3sin 2 0cos0tf0; 
 
 = 3 y "(« 2 cos 2 + j3 2 sin 2 0)* sin cos r/0 
 
 _ qS — fla 
 ~"« 2 -i3 2; 
 
 a? — 3 s 
 therefore the whole length is 4 —^ -5- 
 
 If j3 = a, this result becomes 6«, which agrees with that 
 given in Ex. 1. (See Pi-ice's Calculus, Vol. II, p. 203.) 
 
 "). Find the length of the arc of the parabola x^-\-y\ = a? 
 between the co-ordinate axes. 
 
 Put x = a cos 4 0, y = a sin 4 ; 
 
 ••• 5 = 4rt /(cos 4 + sin 4 0)* sin cos d0 
 
 = — -2= /^"(l + cos 2 20)* d cos 20 
 
 = a + -"- log (a/2 + 1). 
 V2 
 
 G. Find the length, measured from the origin, of the 
 3urve 
 
 ic 2 == a 2 (l —e\ 
 
 Ans. s = a loir ( - J — x, 
 
 n Vrt — x' 
 
EXAMPLES. 
 
 357 
 
 7. Rectify the logarithmic spiral log r — 6 between the 
 
 limits 7'o and 
 
 Ans. s = (1 -f m*)l(ri — n), 
 
 8. If 100 yards of cord be wound in a single coil upon an 
 upright post an inch in diameter, what time will it take a 
 man to unwind it, by holding one end in his hand and 
 traveling around the post so as to keep the cord continually 
 tight, supposing he walks 4 miles per hour ; and what is the 
 length of the path that the man walks over ? 
 
 Ans. Time = 51^ hours ; distance = 204 T 6 T miles. 
 
 9. Find the length of the tractrix or equitangential 
 curve. 
 
 If AB is a curve such that PT, 
 the length of the intercepted 
 tangent between the point of 
 contact and the axis of x, is 
 always equal to OA, then the 
 locus of P is the equitangential 
 curve. 
 
 Let P and Q be two consecu- 
 tive points on the curve ; let 
 (a?, y) be the point P, and OA = PT = a. Then 
 
 Fig. 48. 
 
 
 
 PQ 
 PR - 
 
 a 
 
 y 
 
 
 
 
 
 ds _ 
 
 
 a 
 
 
 
 
 dy ~ 
 
 
 y 
 
 
 (the minus 
 
 sign 
 
 being taken since 
 
 y is a decreasing 
 
 function 
 
 of s or x). 
 
 
 
 
 
 
 Hence, 
 
 s 
 
 = - a r dJ[ 
 
 J a y 
 
 ss 
 
 
 
VXAMPLB8. . 
 
 This example furnishes an instance of our being able to 
 determine bbe length of a curve from a geometric property 
 
 of the curve, without previously finding its equation. 
 
 The equation of the tractrix may be found as follows - 
 
 PR_ PM 
 RQ ~ MT ; 
 
 hence -f = -—- ; 
 
 ax *Ja% _ yi 
 
 py Va 2 — if j 
 .: x = — / 2- dy 
 
 J a y 
 
 J a y^tfl^y 2 J(l Vat — f 
 
 = «log _ *._(**_ f)i, 
 
 (See Ex. 17, Art. 146.) 
 
 This curve is sometimes considered as generated by attaching- one 
 end of a string of constant length (= a) to a weight at A, and by 
 moving the other end of the string along OX ; the weight is supposed 
 to trace out the curve, and hence arises the name Tractrix or Tractory. 
 This mode of generation is incorrect, unless we also suppose the fric- 
 tion produced by traction to be infinitely great, so that the weight 
 momentum which is caused by its motion may be instantly destroyed. 
 Price's Calculus, Vol. I, p. 315. 
 
 10. A fox started from a certain point and ran due east 
 300 yards, when it was overtaken by a hound that started 
 from a point 100 yards due north of the fox's starting-point, 
 and ran directly towards the fox throughout the race. Find 
 the length of the curve described by the hound, both having 
 started at the same instant, and running with a uniform 
 velocity. Ans. 354.1381 yards. 
 
 This example, like the preceding, may be solved without finding 
 the equation of the curve. 
 
EXAMPLES. 
 
 359 
 
 11. Find the length of the helix, estimating it from the 
 plane xy, its equations being 
 
 x = a cos <p, y = a sin <f>, % = c<f>. 
 
 Ans. s = (a 2 + erf <f>. 
 
 12. Find the length, measured from = 0, of the curve 
 which is represented by the equations 
 
 x = (2a — b) sin <f> — (a — b) sin 3 <f>, 
 
 y = (25 — a) cos <p — (b — a) cos 3 </>. 
 
 ^tws. 5 = i (a + #) ^ + } (a — #) sin <p cos 0. 
 
 13. Find the length of the curve of intersection of the 
 elliptic cylinder a 2 y 2 + b 2 x 2 = a 2 &, with the sphere 
 
 a 2 + y 2 + # = a\ 
 
 14.4- 
 
 -4w$. 27ra. 
 
 
CHAPTER VII 
 
 AREAS OF PLANE CURVES. 
 
 184. Areas of Curves. — Let PM and QN be two con- 
 secutive ordinates of the curve AB, and let (x, y) be the 
 point P ; let ^4 denote the area included 
 between the curve, the axis of x, and 
 two ordinates at a finite distance apart. 
 Then the area of the trapezoid MPQN 
 is an infinitesimal element whose breadth 
 is dx and wlio.se parallel sides are y and 
 y -f dy ; therefore we have 
 
 dA= l±lM+M dx = ydx> 
 
 Fig. 49. 
 
 since the last term, being a differential of the second 
 order, must be dropped. 
 
 .\ A 
 
 Jydx, 
 
 the integration being taken within proper limits. If, for 
 example, Ave want the area between the two ordinates 
 whose abscissas are a and b, where a > b, we have 
 
 A= Tydx. 
 
 (1) 
 
 In like manner, if the area were included between the 
 jurve, the axis of y, and two abscissas at a finite distance 
 apart, we would have 
 
 A = J x dy, 
 where c and d are the y-limits. 
 
 (2) 
 
QUADRATURE OF THE CIRCLE. 
 
 301 
 
 185. Area between Two Curves. — If the area were 
 included between the two curves AB and ab, whose equa- 
 tions are respectively y = f(x) and y 
 y = (p {x), and two ordi nates CD and 
 EH, where OD = b and OH = a, 
 we should find by a similar course of 
 reasoning, 
 
 /a 
 [f{x) —^{x)]dx. Fig. 50. 
 
 The determination of the area of a curve is called its 
 Quadrature. 
 
 186. The Circle. — The equation of the circle referred to 
 its centre as origin, is y 2 = a 2 — x 2 ; therefore the area of 
 a quadrant is represented by 
 
 A = P(a 2 — x 2 )? dx 
 
 rx (a? — x 2 )s a 2 . , xi a /c . „ , . , „„.,. 
 
 = — ^ — - f - + - sin" 1 - (See Ex. 4, Art. 151.) 
 
 L/« & aJQ 
 
 therefore the area of the circle = na 2 . 
 
 Also, if OM = x, the area of OBDM 
 becomes 
 
 A = f\a 2 — x 2 )* dx 
 rx (a 2 — x 2 )? a 2 . , xi x 
 
 =1-2- - + 2 sm d„ 
 
 x (a 2 — x 2 )* a 2 . « x 
 = — — - — 4- - sin -1 -• 
 
 2 +2- s,n " 
 
 vtf-jr 
 
 Fig. 51. 
 
 This result is also evident from geometric considerations 
 16 
 

 QUADRATURE OF THE PARABOLA. 
 
 X 
 
 for the area of the triangle OMD = ' (a 2 — a 8 )*, and bhe 
 
 area of the sector ODB = - sin -1 -• 
 
 Z a 
 
 Remark. — The student will perceive that in integrating between 
 the limits x = and .r = a, we take in every elementary slice PQB£i 
 in the quadrant ADBO; also integrating between the limits x = 
 and x = x = OM, we take in every elementary slice between OB 
 and MD.* 
 
 187. The Parabola.— From y 2 — 2px, 
 we have 
 
 y = \/2p%. 
 
 Hence, for the area of the part OPM, 
 we have 
 
 A = V%p I x^dx = $ V%p %% ; *. e., \xy. 
 
 Fig. 52. 
 
 Therefore the area of the segment POP', 
 
 cut off by a chord perpendicular to the axis, is § of the 
 
 rectangle PHH'P'. 
 
 188. The Cycloid. — From the equation 
 
 x = r vers -1 - — V%ry — y 2 , 
 we have 
 
 dx= iM=. 
 
 V%ry — y 2 
 
 -/ 
 
 y 2 dy 
 
 V%ry — y 2 
 
 (See Ex. 6, Art. 151) = | the area of the cycloid. Since 
 integrating between the limits includes half the area of 
 the figure. 
 
 * The student should pay close attention in every case to the limits of the 
 integration. 
 
AREA BETWEEN PARABOLA AND CIRCLE. 363 
 
 Therefore the whole area = 37rr 2 , or three times the area 
 cf the generating circle.* 
 
 189. The Ellipse.— The equation of the ellipse referred 
 to its centre as origin, is 
 
 a 2 y 2 + ft 2 x 2 = a 2 ft 2 ; 
 
 therefore the area of a quadrant is represented by 
 
 A = 
 
 ft a 2 n 
 a~T 
 
 ft P a i 
 
 - / (a 2 — x 2 )* dx 
 aJ Q v ' 
 
 (See Art. 186) = \aftv. 
 
 Therefore the area of the entire ellipse is naft. 
 
 190. The Area between the Parabola y 2 = ax 
 and the Circle y 2 = 2ax — x 2 . — These curves 
 through the origin, and also intersect at A ^ 
 
 the points A and B, whose common abscissa 
 is a. Hence, to find the area included 
 between the two curves on the positive side ^ 
 of the axis of x, we must integrate between 
 the limits x = and x = a. Therefore, 
 by Art. 185, we have 
 
 A = f\(2ax — x 2 )? — (ax)*] dx 
 
 Fig. 53. 
 
 -na 
 
 _^L _ !«2 ; (See Ex. 6, Art. 151.) 
 
 which is the area of OPAP'. 
 
 * This quadrature was first discovered by Roberval, one of the most distin- 
 guished geometers of his day. Galileo, having failed in obtaining the quadrature 
 by geometric methods, attempted to solve the problem by weighing the area of the 
 curve against that of the generating circle, and arrived at the conclusion that the 
 former area was nearly, but not exactly, three times the latter. About 1628, 
 Roberval attacked it, but failed to solve it. After studying the ancient Geometry 
 for six years, he renewed the attack and effected a solution in 1634. (See Salmon's 
 Higher Plane Curves, p. 266.) 
 
564 
 
 TEE SPIRAL OF ARCHIMEDES. 
 
 191. Area in Polar Co-ordinates.— Let the curve be 
 referred to polar co-ordinates, being the pole, and lei 
 
 OP and OQ be consecutive 
 
 radii-vectoivs and PR an 
 
 arc of a circle described with 
 
 as centre ; let (r, 0) be the 
 
 point 1*. Then the area of the 
 
 intiniU'sinial element OPQ 
 
 = OPR + PRQ ; but PRQ is 
 
 an infinitesimal of the second order in comparison with 
 
 OPR, when P and Q are infinitely near points ; conse- 
 
 r 2 dd 
 quently the elementary area OPQ = area OPR = -— 
 
 <i 
 
 Hence if A represents the area included between the curve, 
 the radius-vector OP, and the radius- vector OB drawn to 
 some fixed point B, we have 
 
 Fig. 54 
 
 -*/ 
 
 If j3 and a are the values of 6 corresponding to the points 
 B and C respectively, we have 
 
 1 A-m 
 
 192. The Spiral of Archimedes. — Let r = ^- be 
 
 KiTT 
 
 its equation. Then 
 
 A — tt I r z dr = ^nr* + C. 
 
 If we estimate the area from the pole, we have -4 = 
 when r = 0, and .\ C = ; hence, 
 
 A = ^r 3 , 
 
 which is the value of the area passed over by the radius- 
 vector in its revolution from its starting at c to any value, 
 as r. 
 
EXAMPLES. 3 G5 
 
 If we made 6 = 2n, we have r = 1 ; therefore 
 
 A = Jtt, 
 
 which is the area described by one revolution of the radius- 
 vector. Hence the area of the first spire is equal to one- 
 third the area of the measuring* circle. 
 If we make 6 = 2 (2tt), r = 2 ; therefore 
 
 A = f 71, 
 
 which is the whole area described by the radius-vector 
 during two revolutions, and evidently includes twice the 
 first spire -f the second. Hence the area of the first two 
 spires = f rr — %n = -Jtt, and so on. 
 
 EXAM PLES. 
 
 1. Find the area of y = x — x % between the curve and 
 the axis of x. Ans. £. 
 
 The limits will be found to be x = 0, x — + 1 ; also x = 0, 
 = ■- l.f 
 
 2. Find the area of y = x 3 — b 2 x between the curve 
 and the axis of x. Ans. \¥. 
 
 3. Find the area of y = x B — ax 2 between the curve and 
 the axis of x. Ans. -^a\ 
 
 4. Find the whole area of the two loops of dhP 
 
 — X 2 ( a % _ a?). Ans. frt 2 . 
 
 5. Find the area of xy 2 = a? between the limits y = b 
 
 and y — c. A , b — c 
 
 v Ans. 2a 6 — ? — 
 
 ■ — - — be 
 
 6. Find the whole area of the two loops of aHf 
 
 — a 2 b 2 x 2 — VW. Ans. §ab. 
 
 * See Anai. Geom., Art. 158. 
 
 t The student should draw the figure in every case, and determine the limits of 
 the integrations. 
 
3CG \Ml'LES. 
 
 7. Find the whole area of a 2 y* = z* (2a — #). (Sc 
 Arts. 150 and 188.) Ans. rra* 
 
 8. Find the whole area between the Cissoid y 2 = - 
 
 and its asymptote. (See Art. 103.) '*'" x 
 
 J l y } Ans. 8ffrt 
 
 !). The equation of the hyperbola is a 2 y 2 — Px 2 = — aVfl ; 
 
 iind the area included between the curve, the axis of a:, and 
 
 an ordinate. xy ah < x + y^ _ ^ 
 
 ^4ws. -^ 
 
 - ¥ Iog( — -) 
 
 10. The equation of the Witch of Agnesi is 
 
 a*y = 4« 3 (2a — y) ; 
 
 find the area included between the curve and its asymptote. 
 
 Ans. 4a 2 rr. 
 
 11. Find the area of the catenary VPMO, Fig. 45. 
 Ans. =- f e° — e~«) = a (y 2 — a 2 )^. 
 
 12. Find the area of the oval of the parabola of the third 
 degree whose equation is cy 2 = (x — a) (x — b) 2 . (See 
 Art. 142.) . 8 ,, N s 
 
 Ans 'u^ {h ~ a) ' 
 
 13. Find the area of one loop of the curve 
 
 ay 2 = x 2 (a 2 — x 2 )%. 
 
 Ans. %a 2 . 
 
 14. Find the whole area between the curve 
 
 x 2 y 2 + aW = « Y 
 and its asymptotes. Ans. 2nab, 
 
 15. Find the whole area of the curve 
 
 Ans. $rrab. 
 
EXAMPLES. 367 
 
 16. Find the area included between the parabola y 2 = 2px 
 and the right line y = ax. 
 
 These two loci intersect at the origin and at the point whose ab- 
 scissa is — t ; hence the z-limits are and —. ; therefore, Art. 185, 
 a' 2 a 2 
 
 = J (y\/2px — ax)dx = ~g , Ans. 
 
 17. Find (i) the area included between the parabola 
 
 y 2 = 2px, 
 
 the right line passing through the focus and inclined at 45° 
 to the axis of x, and the left-hand double ordinate of inter- 
 section. (See Art. 185.) 
 Also find (£) the whole area between the line and parabola. 
 
 (1.) Here the ^-limits are found tobe^ (<\/2 + l) 2 and '- (\/2— l) 2 ; 
 hence we have 
 
 — / [\/%P ^ dx — {x — lp) dxj 
 
 f (1/3-1) 2 
 
 [ 
 
 W^px^-^ + lpx 
 
 = w - W*p* + VV = P* (¥ - 2 V»). 4iw. 
 
 (2.) vim !p 9 V2- 
 
 18. Find the whole area included between the four infi- 
 nite branches of the tractrix. Ans. mi 2 . 
 
 19. Find the area of the Naperian logarithmic spiral. 
 
 Ans. \r z . 
 
 20. Find the whole area of the Lemniscate r 2 = a 2 cos 20. 
 
 Ans. a 2 . 
 
[MPUS8. 
 
 21. Find the whole area of the curve 
 
 r = a (cos 20 -f sin 20). 
 
 Ans. na 2 , 
 
 22. Find the area of the Cardioide. (See Art. 181.) 
 
 Ans. f 7ra 2 « 
 
 23. Find the area of a loop of the curve r = a cos nO. 
 
 . va % 
 Ans. — < 
 
 24. Find the area of a loop of the curve 
 
 r = a cos nB + b sin nd. 
 
 A a* + W n 
 
 Ans. — -. • 
 
 4 n 
 
 25. Find the area of the three loops of the curve 
 
 r = a sin 30. (See Fig. 33.) 
 
 Ans. — — 
 4 
 
 26. Find the area included between the involute and the 
 e volute in Fig. 46, when the string has made one revolution. 
 
 Ans. Jr 2 ^, 
 
CHAPTER VIII 
 
 AREAS OF CURVED SURFACES. 
 
 193. Surfaces of Revolution. — If any plane be sup- 
 posed to revolve around a fixed line in it, every point in the 
 plane will describe a circle, and any curve lying in the plane 
 will generate a surface. Such a surface is called a surface 
 of revolution ; and the fixed line, round which the revolu- 
 tion takes place, is called the axis of revolution. 
 
 Let P and Q be two consecutive 
 points on the curve AB ; let (x, y) be 
 the point P, and s the length of the 
 curve AP measured from a fixed point 
 A to any point P. Then MP = y, 
 NQ = y + dy, and PQ = ds. 
 
 Denote by 8 the area of the surface 
 generated by the revolution of AP 
 around the axis OX; then the surface generated by the 
 revolution of PQ around the axis of x is an infinitesimal 
 element of the whole surface, and is the convex surface of 
 the frustum of a cone, the circumferences of whose bases 
 arc Irry and %it (y + dy), and whose slant height is PQ = ds ; 
 therefore we have 
 
 dS = **+*l<*+*) PQ = 2 rryds, 
 
 since the last term, being an infinitesimal of the second 
 order, must be dropped. Therefore, for the whole surface, 
 we have 
 
 8 = 2njyds = 2tt J y\/dx* + dy 2 , 
 
 Fig. 55. 
 
570 !/>/;.! 77 /,■/•; OF THE Sl'll, 
 
 the integral being taken between proper limits. If for 
 example, ire want the surface generated by the curve be- 
 tween the two ordinates whose ebscissas are a and 6, where 
 a> b, we have 
 
 In like manner it may be shown that to find the surface 
 generated by revolving the curve round the axis of y, we 
 have 
 
 8 = 2tt Jxcls. 
 
 194. The Sphere. — From the equation of the gener- 
 ating curve, x 2 + y 2 = r 2 , we have 
 
 • fif = 2nfy{l + ^)**» = %*frdx = 2frnr + (7. 
 
 Hence, the surface of the zone included between two 
 planes corresponding to the abscissas a and b is 
 
 S =2n f a rdx = %-nr (a — b); 
 
 *J b 
 
 that is, the area of the zone is the product of the circum- 
 ference of a great circle by the height of the zone. 
 
 To find the surface of the whole sphere, we integrate 
 between + r and — r for the a-limits ; hence we have 
 
 S = 2-rr jdx = 2ttt [r — (— r)] = 4nr 2 ; 
 
 that is, the whole surface of the sphere is four times the 
 area of a great circle. 
 
 Remark. — If a cylinder be circumscribed about a sphere, its convex 
 surface is equal to 2nr x 2r = 4-rrr' 2 , which is the same as the surface 
 of the sphere. If we add 2n-r 2 to this, which is the sum of the areaa 
 of the two bases, we shall have for the whole surface of the cylinder 
 
QUADRATURE OF PARABOLOID OF REVOLUTION. 371 
 
 6nr -2 . Hence the whole surface of the cylinder is to the surface of the 
 sphere as 3 is to 2. This relation between the surfaces of these two 
 bodies, and also the same relation between the volumes, was discovered 
 by Archimedes, who thought so much of the discovery that he ex- 
 pressed a wish to have for the device on his tombstone, a sphere 
 inscribed in a cylinder. Archimedes was killed by the soldiers of 
 Marcellus, b. C. 212, though contrary to the orders of that general. 
 The great geometer was buried with honors by Marcellus, and the 
 device of the sphere and cylinder was executed upon the tomb. 140 
 years afterward, when Cicero was questor in Sicily, he found the 
 monument of Archimedes, in the shape of a small pillar, and showed 
 it to the Syracusans, who did not know it was in being , he says it was 
 marked with the figure of a sphere inscribed in a cylinder. The 
 sepulchre was almost overrun with thorns and briars. See article 
 " Marcellus," in Plutarch's Lives, Vol. Ill, p. 120. 
 
 195. The Paraboloid of Revolution. — From the 
 equation of the generating curve y 2 = 2px, we have 
 
 y=V2Jx, and g = > ffi 
 
 [f * Vp (p + 2x)f\ = f tt Vp [(p + 2x)^ - pi] , (1) 
 
 which is the surface generated by the revolution of the 
 part of the parabola between its vertex and the point 
 
 & y)- 
 
 We might have found the surface in terms of y instead 
 of x, as follows : 
 
 dy p 
 
ill I PROLATE 8PMER0JD. 
 
 which result agrees with (1), as the student can easily 
 
 u-rify. 
 
 196. The Prolate Spheroid (See Anal. Geom., Art 
 J 01). — From the equation of the generating curve 
 
 y* - (1 _ 6 2) {a 2 _ X 2 )f 
 
 we have 
 
 2tt yds = 2nVl — e* Va 2 — a? 
 
 = 2tt VI - e 2 Va* — eh* dx (Art. 175.) 
 
 therefore for half the surface of the ellipsoid, since the 
 a-limits are a and 0, we have 
 
 
 ~ la* 
 
 2tt^ 
 a 
 
 + tr-5 sin l 
 
 2 ' 2e 2 a 
 
 (See Ex. 4, Art. 151.) 
 Tiber /aZ — aWYk « 2 . , n 
 
 = 7T6 2 -j sin * e. 
 
 e 
 
 \/ 197. The Catenary.— From the equation of the gen 
 erating curve, 
 
 y = ~{ea + e "a), 
 
SURFACE GENERATED BY THE CYCLOID. 373 
 
 we have for the surface of revolution around the axis of x 
 between the limits x and 0, 
 
 S = 2tt I yds == rra I (e« + e «) ds 
 = \-na f X (e« + e~ajdx (by Art. 177) 
 = — / le<* -\- 2 + e ajdx 
 
 = ?[!(—) + -] 
 
 = 7r -le«-e « I -f ax\ 
 
 = n (ys + ##), (where s = VP ? Fig. 45.) 
 
 £/ 198. The Surface generated by the Cycloid when 
 it revolves around its axis. — From its equation 
 
 y = r vers -1 - + \/%rx — x 2 , (1) 
 
 we have 
 
 , S = ( 1 + g)^ = V /f,, (3) 
 
 A S=%*,fya S -^2nfy^dx. (4) 
 
 Put u = y, dv = \ / — dx ; 
 
 .-. dw SB dy, and v = 2 \/2r£; 
 therefore (by Art. 147) we have 
 
374 POLAR CO-ORDINATES. 
 
 fy y~ dx = 2y ^** x "■ 2 v'as- yVs d v 
 
 = 2y y/2rx — 2 \/2rjV%r — x dx [by (2)] 
 
 = 2 ^/Jrx (r vers" 1 - + V%rx — ~x*\ + $ V2r (2r — a:)* 
 
 [by (1) and integrating.] 
 
 which in (4) gives 
 
 # = S^r 2 — -^Trr 2 = 8t-t» (tt — f). 
 
 199. Surfaces of Revolution in Polar Co-ordi- 
 nates. — If the surface is generated by a curve referred to 
 polar co-ordinates, its area may be determined as follows : 
 Let the axis of revolution be the initial line OX, see* 
 Fig. 54, and from P (r, 0) draw PM perpendicular to OX. 
 Then PM = r sin 0, and the infinitesimal element PQ 
 = ds will, in its revolution round OX, generate an infini- 
 tesimal element of the whole surface, whose breadth = ds 
 and whose circumference = 2nr sin 0. Hence, 
 
 S = /W sin ds* = %*fr sin (r 2 + ^fdS, 
 
 (Art. 179) 
 the integral being taken between proper limits. 
 
 / 200. The Cardioide. — From Art. 181, we have 
 
 a 
 
 ds = a (2 + 2 cos 0)* dd = 2a cos 5 dd. 
 
 * This expression might have been obtained at once by substituting in Art. 193, 
 for y, its value r sin 0. 
 
DOUBLE INTEGRATION. 
 
 8?5 
 
 For the surface of revolution of the whole curve about 
 the initial line, we have n and for the limits of 0, there- 
 fore we have 
 
 8 
 
 Jo 
 
 2nr sin 6 ds 
 
 4rra 2 / (1 + cos 0) cos - sin 6 dd 
 
 P n 6 
 = 16rf / cos 4 - sin - dd 
 
 t/o <0 & 
 
 — T_ ijwa* cos 5 17= *£na 2 . 
 
 201. Any Curved Surfaces. — Double Integration. — 
 
 Let (x, y, z) and (x -f dx, y -j- dy, z + dz) be two consecu- 
 tive points p and q on the sur- 
 face. Through p let planes be 
 drawn parallel to the two planes 
 xz and yz ; also through q let 
 two other planes be drawn par- 
 allel respectively to the first. 
 These planes will intercept an 
 infinitesimal element pq of the 
 curved surface, and the projec- 
 tion of this element on the 
 plane of xy will be the infini- 
 tesimal rectangle PQ, which = dx dy. 
 
 Let -8 represent the required area of the whole surface, 
 and dS the area of the infinitesimal element pq, and 
 denote by a, /3, y, the direction angles* of the normal at 
 p (x, y, z). Then, since the projection of dS on the 
 plane of xy is the rectangle PQ = dx dy, we have by Anal. 
 Geom., Art. 168, 
 
 dx dy = dS cos y. (l) 
 
 Fig. 56. 
 
 * See Anal. Geoni., Art. 170. 
 
SURFACE OF A SPUE in:. 
 
 Simila ly, if dS is projected on the planes yz and 
 We luu' 
 
 dy dz = dS cos « ; (2) 
 
 dz dx = dS cos 0. (3) 
 
 •Squaring (1), (2) and (3), and adding, and extracting 
 Bquare root, we have 
 
 rf£ ss (<fo% 2 + %W + dz 2 dx 2 )^ 
 
 (since cos 2 a + cos 2 (3 -f- cos 2 y = 1, 
 
 Anal. Geom., Art. 170). 
 
 .-. 8 = ff(dx 2 dy 2 + dtfd* + dz 2 dx 2 )\ 
 
 the limits of the integration depending upon the portion 
 of the surface considered. 
 
 1/ 202. The Surface of the Eighth Part of a Sphere.— 
 
 Let the surface represented in Fig. 56 be that of the 
 octant of a sphere ; then being its centre, its equation is 
 
 a? + yi + z 2 = a 2 . 
 
 tt dz x dz y 
 
 Hence, -j- = , -y- = — *■• 
 
 dx z dy z 
 
 p p adxdy 
 ~ J J i/ a * — x 2 — y 2 
 
 Now since pq is the element of the surface, the effect 
 of a ^-integration, x being constant, will be to sum up 
 all the elements similar to pq from H to I ; that is, 
 from y = to y = hi = y x = Va 2 — x 2 ; and the aggre- 
 
EXAMPLES. 377 
 
 gate of these elements is the strip llpl. The effect of a 
 subsequent ^-integration will be to sum all these elemental 
 strips that are comprised in the surface of which OAB 
 is the projection, and the limits of this latter integra- 
 tion must be x = and x = OA == a. Therefore, 
 we have 
 
 " Jo Jo \/t& — a? - 
 
 Jo d 
 
 = / \adx sin -1 — 
 Jo L #,J 
 
 = / 
 
 Va 2 — x 2 — y 2 
 
 a dx dy 
 vV — y 2 
 
 7 na 4 
 
 i-na ax = -— < 
 2 
 
 E X A M P LE S. 
 
 jS 1. Find the convex surface of a right circular cone, 
 whose generating line is ay — hx = 0. 
 
 Ans. nb V 'a 2 + £ 2 . 
 
 Remark. — It is evident that the projection of the convex sur- 
 face of a right circular cone on the plane of its base, is equal 
 to the base ; hence it follows (Anal. Geom., Art. 168) that the 
 convex surface of a right circular cone is equal to the area of its 
 base multiplied by the secant of the angle between the slant 
 height and the base. Thus, calling this angle a, we have in 
 the above example, 
 
 S = :r& 2 sec a = 7T& 2 ^4-— = ri \/'cfTb\ 
 o v 
 
 which agrees with the answer. 
 
 2. Find the area of the surface generated by the revolu- 
 tion of a logarithmic curve, y = e x , about the axis of x, 
 between the ^/-limits and y. 
 
 Ans. n\y(l+ tf)$ + log [y + (1 + */ 2 )*] f- 
 
378 IMP&X& 
 
 8. Find the area of the surface generated by the revolu- 
 tion of the cycloid (1) about its base, and ( .') about the 
 tangent at tile highest point. 
 
 Ans. (1) *£na 2 ; (2) S£-na 2 . 
 
 4. Find the area of the surface generated by the revolu 
 tion of the catenary about the axis of y, between tin 
 ic-limits and x. Ans. 2n [xs — a (y — a)]. 
 
 .\ S = 2ir I xds = 2n \xs — I sdx , 
 
 from which we soon obtain the answer. 
 
 ]/ 5. Find the area of the surface of a spherical sector, the 
 vertical angle being 2a and the radius of the sphere = r. 
 
 Ans. 4:ttt 2 (sin |) • 
 
 6. Find the area of the surface generated by the revolu- 
 tion of a loop of the lemniscate about its axis, the equation 
 being r 2 = a 2 cos 28. Ans% na 2 ( 2 _ 2^). * 
 
 Here find rds — a?dO : .*. etc. 
 
 i/ 
 
 7. Find the area of the surface generated by the revolu- 
 tion of a loop of the lemniscate about its axis, the equation 
 being r 2 = a 2 sin 2d. Ans. 2na 2 . 
 
 I / 8. A sphere is cut by a right circular cylinder, the radius 
 of whose base is half that of the sphere, and one of whose 
 edges passes through the centre of the sphere._ Fin<J the 
 area of the surface of the sphere intercepted by the cylinder. 
 
 Let the cylinder be perpendicular to the plane of xy ; 
 then the equations of the cylinder and the sphere are 
 respectively y 2 = ax — x 2 and x 2 + y 2 + z 2 = a 2 . It is 
 easily seen that the y-limits are and \/ax — x 2 = y x , and 
 the z-limits are and a. Therefore, Art. 201, we have 
 
a a ny, a ( j x 
 
 Jo t/o Vft 2 — s 
 = ft / si 
 
 EXAMPLES. 37 J 
 
 ft ffo ft*y 
 
 x 2 — y 2 
 
 % (ax — a£)* , 
 
 sin _1 ^ aa? 
 
 (ft 2 - z 2 )* 
 
 = a sin -1 (-— — )*«'(« + a;) 
 
 t/o \ft + xl 
 
 sx ft |(ft + a?) sin" 1 /_^-Y — v'flS (Art. 14 ?) 
 
 = -(!-')■ 
 
 Therefore, the whole surface == 2ft 2 (n — 2). (In Price's 
 Calculus, Vol. II, p. 326, the answer to this example is 
 #2(77 _ 2), which is evidently only half of what it should 
 be.) 
 
 9. In the last example, find the area of the surface of the 
 cylinder intercepted by the sphere. 
 
 Eliminating y, we have i — Va 2 — ax for the equation 
 of the projection on the plane xz of the intersection of the 
 sphere and the cylinder. Therefore the ^-limits are and 
 2, = Vft 2 — ax, and the ^-limits are and a ; hence, Art. 
 201, we have 
 
 dx dz 
 
 [dtfidf + dtfdz> + dz*dx*]i = [l + (fj.+ (J)' 
 
 ft dx dz 
 s= — : for an element of the surface of the cylinder. 
 
 ** 2 Vox — x 2 
 
 dxdz _ ft* r a dx 
 Vox — x 2 %Jox^ 
 therefore the whole area of the intercepted surface of tho 
 cylinder is 4a 3 . (Bee Gregory's Examples, p. 436.) 
 
 \j 10. The axes of two equal right circular cylinders inter- 
 sect at right angles ; find the area of the one which is inter- 
 cepted by the other. Ans. 8ft 2 . 
 
 ft f a /»*i dxdz a* r a dx . 
 
 *^o * 7 o Vax — x 2 4tJ o x^ 
 
l.rt the axes of tin* two cylinders be taken as the axes of // and :\ 
 
 and let a = the radius of each cylinder. Then the equations are 
 
 2** + z* = a 8 , x° + y 9 s a '•'. 
 
 11. A sphere is pierced perpendicularly to the plane of 
 ■me of its givaf circles by two right cylinders, of which the 
 diameters arc equal to the radius of the sphere and the axes 
 
 through the middle points of two radii that compose a 
 diameter of this great circle. Find the surface of that por- 
 tion of the sphere not included within the cylinders. 
 
 Ans. Twice the square of the diameter of the sphere. 
 
 12. Find the area of the surface generated by the revolu- 
 tion of the tractrix round the axis otx. Ans. A.mi\ 
 
 13. If a right .circular cone stand on an ellipse, show that 
 the convex surface of the cone is 
 
 5 (OA + OA') (OA.OA')* sin «, 
 
 where is the vertex of the cone, A and A' the extremities 
 of the major axis of the ellipse, and « is the semi-angle of 
 the cone at the vertex. (See Remark to Ex. 1.) 
 
CHAPTER IX. 
 
 VOLUMES OF SOLIDS. 
 
 203. Solids of Revolution. — Let the curve AB, Fig. 55, 
 revolve round the axis of x, and let V denote the volume 
 of the solid bounded by the surface generated by the curve 
 and by two planes perpendicular to the axis of x, one 
 through A and the other through P ; then as MP and NQ 
 are consecutive ordinates, the volume generated by the revo- 
 lution of MPQN round the axis of x is an infinitesimal 
 element of the whole volume, and is the frustum of a cone, 
 the circumferences of whose bases are %-ny and 2n (y + dy), 
 and whose altitude is MN = dx ; therefore we have 
 
 f 
 
 3 
 
 by omitting infinitesimals of the second order. Hence, for 
 the whole volume generated by the area between the two 
 ordinates whose abscissas are a and b, where a > #, we 
 
 have 
 
 pa 
 
 V = I Try 2 dx. 
 
 In like manner, it may be shown that to find the volume 
 generated by revolving the arc round the axis of y, we have 
 
 V = 7T / x 2 dy. 
 
 204. The Sphere. — Taking the origin at the centre of 
 the sphere, we have y 2 = a 2 — x 2 ; therefore we have 
 
 dV= Ky 2 + n(y + dy) 2 + ny(y + dy) dx = ^^ 
 
 V = nf a (a 2 — x 2 ) dx = [~n (a 2 x — Jz 8 ) 
 for the whole volume of the sphere. 
 
 $77«3 
 
889 VOLUME GENERATED BY CYCLOID. 
 
 Cor. 1. — To Bud the jolume of a spherical segment be- 
 tween two parallel pkuiee, let b and c represent the distanced 
 of these planes from the centre; then we have 
 
 V = *f\a* - x*)dx = n [a 2 (b - c) - \ (b* - c 3 )]. 
 
 Cor. 2. — To find the volume of a spherical segment with 
 one base, let h be the altitude of the segment; then b = a 
 and c = a — h, and we have 
 
 V = n J^ a (a 2 _ X 2 ) dx = 7T7/2 (ci - |). 
 
 Cor. 3. $tt« 3 = § of 7ra 2 x 2a = § of the circumscribed 
 cylinder. (See Art. 194, Remark.) 
 
 205. The Volume generated by the Revolution of 
 the Cycloid about its Base. 
 
 Here dx = &* (Art. 176) ; 
 
 V 2ry — y 2 
 
 and integrating between the limits y = and y = 2r, we 
 find for the whole volume 
 
 = 2Tr|r f --^M— (by Ex. 6, Art. 151) 
 * ^o v^n, -y 2 ' 
 
 = i^ r7r ( j^tt) (by Ex. 6, Art. 151) 
 
 We have S^r 8 = -Jt (2r) 2 x 2nr. 
 
 Hence, £7te volume generated by the revolution of 
 the cycloid about its base is equal to five-eighths the 
 circumscribing cylinder. 
 
SOLIDS BOUNDED BY ANY CURVED SURFACE. 383 
 
 206. The Cissoid when it revolves round its 
 Asymptote.— Here 0M = x, MP = y, 
 
 0A = 2a, MA = 2a - x, HD = dy ; 
 hence an infinitesimal element of the 
 whole volume is generated by the revo- 
 lution of PQDH about AT, and is 
 represented by n (2a — xf dy. 
 The equation of the Cissoid is 
 
 x? 
 
 Fig. 57. 
 
 f = 
 
 2a 
 
 " a y ~ (2a-~xf dx > 
 
 hence, between the limits x = and x = 2a, we have 
 (2a — xjdy = 2n 
 
 /»2a p2a , 
 
 V = 2n / (2« - a;)2rfy = 2rr / (3a— x) (2ax—x^ dx 
 
 v «/ 
 
 rt /' 2 « Ga 2 x — 5ax* + a 3 
 
 2 77 / ; 6?« = 27T 2 « 3 
 
 V2ax — x 2 
 
 (by Ex. 6, Art. 151). 
 
 207. Volume of Solids bounded by any Curved 
 Surface. — Let (x, y, z) and 
 
 (x + dx, y + dy, z-\-dz) be two 
 consecutive points E and F 
 within the space whose volume 
 j s to be found. Through E 
 pass three planes parallel to 
 the co-ordinate planes xy, yz, 
 and zx; also through F pass 
 three planes parallel to the 
 first. The solid included by 
 these six planes is an infinitesi- 
 mal rectangular parallelopipe- 
 don, of which E ami F are two opposite angles, and the 
 volume is dxdy dz ; the aggregate of all these solids between 
 
TRIPLE TNTXQRATtC \. 
 
 the limit* assigned by the problem is the required volume. 
 Hence, if V denote the required volume, we have 
 
 V "= J J \fdxdydx, 
 
 the integral being taken between proper limits. 
 
 In considering the effects of these successive integrations* 
 let as suppose that we want the volume in Fig. 58 contained 
 within the three co-ordinate planes. 
 
 The effect of the ^-integration, x and y remaining con- 
 stant, is the determination of the volume of an infinitesimal 
 prismatic column, whose base is dxdy, and whose altitude 
 is given by the equations of the bounding surfaces ; thus, in 
 Fig. 58, if the equation of the surface is z=f(x,y), the 
 limits of the ^-integration are/(«, y) and 0, and the volume 
 of the prismatic column whose height is Vp is f(x, y) dx dy ; 
 hence the integral expressing the volume is now a double 
 integral and of the form 
 
 V = fff(x, y) dx dy. 
 
 If we now integrate with respect to y, x remaining con- 
 stant, we sum up the prismatic columns which form the 
 elemental slice TLphnq, contained between two planes per- 
 pendicular to the axis of x, and at an infinitesimal distance 
 (dx) apart. The limits of y are 12 and 0, 12 being the y to 
 the trace of the surface on the plane of xy, and which may 
 therefore be found in terms of x by putting z = in the 
 equation of the surface; or, if the volume is included be- 
 tween two planes parallel to that of xz, and at distances //„ 
 and y x from it, y and y x being constants, they are in that 
 case the limits of y ; in the same way we find the limits if 
 the hounding surface is a cylinder whose generating lines 
 ;ire parallel to the axis of z. In each of these cases the 
 result of the ^-integration is the volume of a slice included 
 between two planes at an infinitesimal distance apart, the 
 length of which, measured parallel to the axis of y, is a 
 
EXAMPLES. 385 
 
 function of its distance from the plane of yz ; thus the limits 
 of the #- integration may be functions of x, and we shall 
 have 
 
 V = fff(*> y) dx dy = J F{x) dx, 
 
 where F(x) dx is the infinitesimal slice perpendicular to 
 the axis of x at a distance x from the origin, and the sum 
 of all such infinitesimal slices taken between the assigned 
 limits is the volume. Thus, if the volume in Fig. 58 be- 
 tween the three co-ordinate planes is required, and OA = a, 
 then the a-limits are a and 0. If the volume contained 
 between two planes at distances x and x x from the plane of 
 yz is required, then the ^-limits are x and x x . 
 
 EXAMPLES. 
 
 1. The ellipsoid whose equation is 
 - u- t 4- - — 1 
 
 (& + £2 I" $ - U 
 
 1 g — j-A 
 
 and 0, which call z, and 0; the limits of y are hi = 
 1 — —J" and 0, which call y x and ; the ^-limits are a 
 and 0. 
 
 First integrate with respect to z, and we obtain the infini- 
 tesimal prismatic column whose base is PQ, Fig. 58, and 
 whose height is Pp. Then we integrate with respect to y, 
 and. obtain the sum of all the columns which form the 
 elemental slice Hplmq. Then integrating with respect to x, 
 we obtain the sum of all the slices included m the solid 
 OABC. 
 
 pa py x r»z x 
 
 V=8 / / dxdydz 
 
 17 
 
386 EXAMPLES, 
 
 = — g- / (a 2 — x^dx = $nabc. 
 
 2. The volume of a right elliptic cylinder whose axis 
 coincides with the axis of x and whose altitude = 2a, the 
 equation of the base being 
 
 &tf + b 2 z 2 = W&. 
 Here the z-limits are 7 (b 2 — y 2 )? and 0, which call % x and 
 ; the ^-limits are b and ; the z-limits are a and 0. 
 
 r»a pb nz x 
 
 .-. V=8 / / dxdydz 
 
 t/o Jq Jq 
 
 ft 
 
 dx 
 
 = 8-— I dx = 2abcn. 
 
 (See Price's Calculus, Vol. II, p. 356.) 
 
 3. The volume of the solid cut from the cylinder x 2 -f- y i 
 = a 2 by the planes 2 = and 2 = x tan «. 
 
 Here the 2-limits are x tan « and 0, or z, and ; the 
 ^-limits are (a z — x 2 )? and — (a 2 — x 2 )?, or y, and — y x ; 
 the 2>limite are « and 0. 
 
EXAMPLES. 
 
 387 
 
 /»« nyi r>«x 
 ,\ V= / / / dxdydi 
 
 pa /»y, 
 
 = / /* (2; tan a) afo tf# 
 
 = 2 tan a f a x (a 2 — z 2 )* dx = f a? tan «. 
 
 4. The volume of the solid common to the ellipsoid 
 
 —• 4- Is +--s = l and the cylinder # 2 + «/ 2 = £ ? - 
 a 2 ^ &^ c l J * 
 
 (x 2 y 2 \h 
 1 — jA 
 
 and 0, or z x and ; the limits of the ^--integration are 
 (b 2 — y 2 )^ and 0, or x x and 0; the ^/-limits are and h.* 
 
 .-. V = 8 / / dydxdz 
 
 ^T't-S-S]^^ 
 
 — ® c C 
 
 aJn 
 
 * 
 
 
 «*■ 
 
 ^ + 
 
 02 
 
 sin" 
 
 •('-?)* 
 
 dy 
 
 
 4c /»* 
 a 1 
 
 
 * In this example, this order of integration is simpler than it would be to take 
 it with respect to y and then x. 
 
m 
 
 EXAMPLES* 
 
 (See Mathematical Visitor, 1878, p. '26.) 
 
 208. Mixed System of Co-ordinates.— I listen 1 <>!' 
 
 dividing a solid into columns standing on rectangular 
 
 bases, bo that zdxdy is the 
 
 volume of the infinitesimal 
 
 column, it is sometimes more 
 
 convenient to divide it into 
 
 infinitesimal columns standing 
 
 on the polar element of area 
 
 abed = r dr dO, in which case 
 
 the corresponding parallelopipedon is represented by 
 
 zr dr dd, and the expression for V becomes 
 
 Fig. 59. 
 
 V = J fzrdrdB, 
 
 taken between proper limits. From the equation of the 
 surface, z must be expressed as a function of r and 0. 
 
 EXAMPLES 
 
 1. Find the volume included between the plane z = 0, 
 and the surfaces x 2 + y 2 = 4az and y 2 = 2cx — x 2 . 
 
 Here z 
 
 x 2 + y 2 r 2 
 
 4a 
 
 ; hence the z-limits are j- and 0. 
 
 The equation of the circle y 2 = 2cx — x 2 , in polar co-or- 
 dinates, is r = 2c cos ; hence the r-limits are and 
 
 2c cos 0, or and r, ; and the 0-limits are and 5- 
 
 ,: V=2 / i-dddr 
 
 Jq Jq 4a 
 
 9/4 /Utt 9y4 
 
 = T J C0S4 ° dd = a **' ^ EX * 4> Art ' 157o) 
 
 3tT£4 
 
 8 a' 
 
EXAMPLES. 389 
 
 2. The axis of a right circular cylinder of radius b, 
 passes through the centre of a sphere of radius a, when 
 a > b ; find the volume of the solid common to both 
 surfaces.* 
 
 Take the centre of the sphere as origin, and the axis of 
 the cylinder as the axis of %\ then the equations o* the 
 surfaces are x 2 + y 2 + z 2 = a 2 and x 2 -f- y 2 = b 2 ; 01% in 
 terms 'of polar co-ordinates, the equation of the cylinder 
 is r = b. 
 
 Hence for the volume in the first octant, the z-limits are 
 
 y a « _ x 2 — y 2 or V« 2 — r 2 and ; the r-limits are b 
 and ; the 0-limits are - and 0. 
 
 .«. V = 8 / zrdr dd 
 
 Jo Jo 
 
 = 8 f*" f r (a 2 — r 2 )i dd dr 
 Jo Jo 
 
 (See Gregory's Examples, p. 428.) 
 
 209. The polar element of plane area is r dr dd (Art. 
 208). Let this element revolve round the initial line 
 through the angle 2n, it will generate a solid ring whose 
 volume is %nr sin Or dr dd, since 2nr sin 6 is the circumfer- 
 ence of the circle described by the point (r, 6). Let <p 
 denote the angle which the plane of the element in any 
 position makes with the initial position of the plane ; 
 then d(f> is the angle which the plane in any position makes 
 
 * This example, as well as the preceding one, might he integrated directly in 
 terms of x and y by the method of Art. 207, bat the operation would be more com 
 plex than the one adopted. 
 
390 IMPLB&. 
 
 with its consecutive position. The part of the solid ring 
 which is intercepted between fche reyolving plane in these 
 two consecutive positions, is to the whole ring in the same 
 proportion as d<f> is to 2^. Hence the volume of this 
 intercepted part is 
 
 r 2 sin d<f> d0 dr, 
 
 which is therefore an expression in polar co-ordinates for 
 an infinitesimal element of any solid. Hence, for the 
 volume of the whole solid we have 
 
 V = f'jfr 2 sin 6 dd> dd dr, 
 
 in which the limits of the integration must, be so taken 
 as to include all the elements of the proposed solid. In 
 this formula r denotes the distance of any point from the 
 origin, denotes the angle which this distance makes with 
 some fixed right line through the origin (the initial line), 
 and <p denotes the angle which the plane passing through 
 this distance and the initial line makes with some fixed 
 plane passing through the initial line. (See Lacroix Cal- 
 cul Integral, Vol. II, p. 209.) 
 
 The order in which the integrations are to be effected is 
 theoretically arbitrary, but in most cases the form of the 
 equations of surfaces makes it most convenient to integrate 
 lirst with respect to r ; but the order in which the 0- and 
 0-integrations are effected is arbitrary. 
 
 EXAMPLES. 
 
 1. The volume of the octant of a sphere. Let a = the 
 radius of the sphere ; then the limits of r are and a ; 
 hence, 
 
 V = ff* sin 6 d<p dd. 
 
 In thus integrating with respect to r, we collect all the 
 elements like r 2 sin 6 d<p dd dr which compose a pyramidal 
 
EXAMPLES. 39] 
 
 solid, having its vertex at the centre of the sphere, and for 
 its base the curvilinear clement of spherical surface which 
 is denoted by a 2 sin d<$> dd. 
 
 Integrating next with respect to between the limits 
 
 and 5 , we have 
 
 V = /£[(- cos *)]> = / J#. 
 
 In thus integrating with respect to 0, we collect all the 
 
 pyramids similar to - sin d<f> dd, which form a wedge- 
 o 
 
 shaped slice of the solid contained between two consecutive 
 
 planes through the initial line. 
 
 Lastly, integrating with respect to $ from to -, 
 we have 
 
 V = ™ 3 . (See Todhunter's Int. Cal., p. 183.) 
 
 In this example the order of the integrations is imma- 
 terial. 
 
 2. The volume of the solid common to a sphere of 
 radius a, and the right circular cone whose vertical angle 
 is 2a and whose vertex is at the centre of the sphere. 
 
 Here the r-limits are and a, the 0-limits are and «, 
 the 0-limits are and 2n. 
 
 p2n pa pa 
 
 .% V = / / / r 2 sin d(f> dd dr 
 J ^~ smdd4> dd 
 
 />2tt^S 
 
 =y 3 ( i -° os «)# 
 
 = fnfl 8 (1 — COS «). 
 
MM EXAMPLES. 
 
 EXAMPLES. 
 
 1. Find the volume of a paraboloid of revolution whose 
 altitude = a and the radius of whose base = b. 
 
 Ans. -afr. 
 
 2. Find the volume of the prolate spheroid. Also 
 the oblate spheroid. Ans. The prolate spheroid = |n<zd*. 
 
 The oblate spheroid = fyurb. 
 
 3. Find the volume of the solid generated by the revolu- 
 tion of y = a x about the axis of x, between the limits x 
 
 <md — oo, where a > 1. . n _,, . ; 
 
 Ans. - a 21 (log a)" 1 . 
 
 4. Find the volume of the solid generated by the revolu- 
 tion of y = a log x about the axis of x, between the 
 limits x and 0. Ans. ncPx (log 2 x — 2 log x -f 2). 
 
 5. Find the volume of the solid generated by the 
 revolution of the tractrix round the axis of x. Ans. f "« 3 . 
 
 6. Find the volume of the solid generated by the 
 revolution of the catenary round the axis of x. 
 
 Ans. x a (ys + ax). (Compare with Art. 197.) 
 
 7. Find the volume generated by the revolution of a 
 parabola about its base 2b, the height being //.* (See 
 Art. 206.) Ans. \%»bh*. 
 
 8. The equation of the Witch of Agnesi being 
 
 y 
 
 find the volume of the solid generated by its revolution 
 round the asymptote. Ans. ±n 2 a\ 
 
 * This solid is called a parabolic spindle. 
 
EXAMPLES. 393 
 
 t. Find the volume of a rectangular parallelopipedon. 
 three of whose edges meeting at a point are #, b } c. (See 
 Art. 207.) Arts. abc. 
 
 10. Find the volume contained within the surface of an 
 elliptic paraboloid * whose equation is 
 
 a o 
 and a plane parallel to that of yz, and at a distance c from it. 
 
 Ans. nc 2 (ab)?. 
 
 11. The axes of two equal right circular cylinders inter- 
 sect at right angles, their equations being x 2 + z 2 = a 2 and 
 x 2 -f y 2 = a 2 ; find the volume of the solid common to both. 
 
 Ans. ±£-a s . 
 
 12. A paraboloid of revolution is pierced by a right cir- 
 cular cylinder, the axis of which passes through the focus 
 and cuts the axis of the paraboloid at right angles, the 
 radius of the cylinder being one-fourth the latus-rectum of 
 the generating parabola ; find the volume of the solid com- 
 mon to the two surfaces. A - (% . w\ 
 
 Here the equations of the surfaces are 
 
 y 2 + 2 2 = 2px and z 2 + y 2 = px. 
 
 13. Find the volume of the solid cut from the cylinder 
 x 2 + y 2 = 2ax by the planes z = x tan a and z — x tan (3. 
 
 TTdfi 
 
 Ans. 2 (tan j3 — tan a) -g~ • 
 
 14. Find the volume of the solid common to both sur- 
 faces in Ex. 8 of Art. 202. (See Art. 208.) 
 
 Ans. |(3tt — 4c) a*. 
 
 15. Find the volume of the part of the hemisphere in the 
 last example, which is not comprised in the cylinder. 
 
 Ans. f« 3 . 
 
 * Called elliptic paraboloid because the sections made by planes parallel to the 
 planes of xy and xz are parabolas, while those parallel to the plane of yz are 
 ellipses. (Salmon's Anal. Geom. of Three Dimensions, p. 58.) 
 
094 EXAMPLES. 
 
 16. Find the volume of the solid intercepted between the 
 concave surface of the sphere and the convex surface of t lie 
 cylinder in Art. 208, Ex. 2. . Ans. fr (a* — P)i 
 
 17. Find the volume of the solid comprised between the 
 
 •urtace z = ae~~ c% and the plane of xy. Ans. •nacK 
 
 Here the r-limits are and oo ; and the 0-limits are and 2*. 
 
 18. Find the volume of the solid generated by the revo 
 hit ion of the cardioide r = a (1 + cos 0) about the initial 
 line. 
 
 pit pin r»a(l + COB6) 
 
 Here V= J J J r 2 sm0 d0 d<f> dr = etc. 
 
 (See Art 209.) . 8tt«s 
 
 v Ans. — — 
 
 19. Find the volume of the solid generated by the revo- 
 lution of the Spiral of Archimedes, r = ad, about the initial 
 line between the limits 6 = n and = 0. 
 
 AnS. -f7r%3 (tt2 _ c). 
 
 20. A right circular cone whose vertical angle = 2«, has 
 its vertex on the surface of a sphere of radius a, and its 
 axis coincident with the diameter of the sphere ; find the 
 volume common to the cone and the sphere. 
 
 Ans. —^- (1 — cos 4 «). 
 
 21. Find the volume of a chip cut at an angle of 45° to 
 the centre of a round log with radius r. (Mathematical 
 Visitor, 1880, p. 100.) Ans. |r 3 . 
 
 22. Find the volume bounded by the. surface 
 
 and the positive sides of the three co-ordinate planes. 
 
 . abc 
 
 Ans - w 
 
EXAMPLES. 395 
 
 23. Find the volume of the solid bounded by the tnree 
 surfaces x 2 + y 2 = cz, x 2 + y 2 = ax, and z = 0. 
 
 3ira 4 
 
 Ans ' aST 
 
 24. A paraboloid of revolution and a right circular cone 
 have the same base, axis, and vertex, and a sphere is 
 described upon this axis as diameter. Show that the volume 
 intercepted between the paraboloid and cone bears the same 
 ratio to the volume of the sphere that the latus-rectum of 
 the parabola bears to the diameter of the sphere. 
 
 25. Find the volume included between a right circular 
 cone whose vertical angle is 60° and a sphere of radius r 
 touching it along a circle, by the formula 
 
 V=fffdxdydz. 
 
 . 7T7* 3 
 
 Ans. -r-i 
 b 
 
 26. In the right circular cone given in Ex. 13 of Art. 
 202, prove that its volume is represented by 
 
 g(OA.OA')*sin2«cos«. 
 
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