wm Digitized by the Internet Archive in 2008 with funding from Microsoft Corporation http://www.archive.org/details/elementarytreatiOObowsrich 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«. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. REC'D LP m^ '6 4 -i pm tfEi-nss^ 8 FE B S'68-2PK t REC'D LD 3UI 1 5 7\ 7 py ? Q General Library 797%o •' •• QA THE UNIVERSITY OF CALIFORNIA LIBRARY