Darnell Utrtnerttty ffiibcatfg Jltljata. 3to fork ALEXANDER GRAY MEMORIAL LIBRARY ELECTRICAL ENGINEERING THE GIFT OF The McGraw-Hill Book Co.. Inc. ENGINEERING LIBRARY Cornell University Library QA 304.B99 Elements of the differential calculus, wi 3 1924 004 644 492 Cornell University Library The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924004644492 ELEMENTS DIFFERENTIAL CALCULUS, EXAMPLES AND APPLICATIONS. A TEXT BOOK W. E. BYERLY, Ph.D., ASSISTANT PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY GINN & COMPANY BOSTON • NEW YORK • CHICAGO ■ LONDON Entered according to Act of Congress, in the year 1879, by W. E. BYERLY, in the office of the Librarian of Congress, ai Washington. 48.7 tgfte gttenaum 3@rtsg GINN & COMPANY . PRO- PRIETORS • BOSTON • U.S.A. PREFACE. The following book, which embodies the results of my own experience in teaching the Calculus at Cornell and Harvard Universities, is intended for a text-book, and not for an exhaustive treatise. Its peculiarities are the rigorous use of the Doctrine of Limits as a foundation of the subject, and as preliminary to the adoption of the more direct and practically convenient infinitesimal notation and nomenclature ; the early introduc- tion of a few simple formulas and methods for integrating ; a rather elaborate treatment of the use of infinitesimals in pure geometry ; and the attempt to excite and keep up the interest of the student by bringing in throughout the whole book, and not merely at the end, numerous applications to practical problems in geometry and mechanics. I am greatly indebted to Prof. J. M. Peirce, from whose lectures I have derived many suggestions, and to the works of Benjamin Peirce, Todhunter, Duhamel, and Bertrand, upon which I have drawn freely. W. E. BYERLY. Cambridge, October, 1879. TABLE OF CONTENTS. CHAPTER I. INTRODUCTION. Article. Page. 1. Definition of variable and constant 1 2. Definition of function and independent variable 1 3. Symbols by which functional dependence is expressed .... 2 4. Definition of increment. Notation for an increment. An in- crement may be positive or negative 2 5. Definition of the limit of a variable 3 6. Examples of limits in Algebra 3 7. Examples of limits in Geometry 4 8. The fundamental proposition in the Theory of Limits 5 9. Application to the proof of the theorem that the area of a circle is one-half the product of the circumference by the radius . . 5 10. Importance of the clear conception of a limit 6 11. The velocity of a moving body. Mean velocity ; actual velocity at any instant ; uniform velocity ; variable velocity 6 12. Actual velocity easily indicated by aid of the increment notation 7 13. Velocity of a falling body 7 14. The direction of the tangent at any point of a given curve. Definition of tangent as limiting case of secant 8 15. The inclination of a curve to the axis of X easily indicated by the aid of the increment notation 8 16. The inclination of a parabola to the axis of X 9 17. Fundamental object of the Differential Calculus ....... 10 CHAPTER II. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. 18. Definition of derivative. Derivative of a constant 11 19. General method of finding the derivative of any given function. General formula for a derivative. Examples 11 vi DIFFERENTIAL CALCTTLUS. Article. ' P»8 a 20. Classification of functions 12 21. Differentiation of the product of a constant and the variable; of a power of the variable, where the exponent is a positive integer 13 22. Derivative of a sum of functions 14 23. Derivative of a product of functions 15 24. Derivative of a quotient of functions. Examples 17 25. Derivative of a function of a function of the variable 18 26. Derivative of a power of the variable where the exponent is negative or fractional. Complete set of formulas for the differentiation of Algebraic functions. Examples 19 CHAPTER III. APPLICATIONS. Tangents and Normals. 27. Direction of tangent and normal to a plane curve 22 28. Equations of tangent and normal. Subtangent. Subnormal. Length of tangent. Length of normal. Examples 23 29. Derivative may sometimes be found by solving an equation. Examples 25 Indeterminate Forms. 30. Definition of infinite and infinitely great 26 31. Value of a function corresponding to an infinite value of the variable 26 32. Infinite value of a function corresponding to a particular value of the variable 27 33. The expressions g, 55, and OX 00, called indeterminate forms. When definite values can be attached to them 28 34. Treatment of the form g. Examples 1 28 35. Reduction of the forms % and X 00 to the form ■= 30 Maxima and Minima of a Continuous Function. 36. Continuous change. Continuous function 31 37. If a function increases with the increase of the variable, its derivative is positive ; if it decreases, negative 31 38. Value of derivative shows rate of increase of function .... 32 39. Definition of maximum and minimum values of a function . . 32 TABLE OF CONTENTS. Vll Article. Page. 40. Derivative zero at a maximum or a minimum 33 41. Geometrical illustration 33 42. Sign of derivative near a zero value shown by the value of its own derivative T . 34 43. Derivatives of different orders 34 44. Numerical example 34 45. Investigation of a minimum 35 46. Case where the third derivative must be used. Examples . . 35 47. General rule for discovering maxima and minima. Examples . 36 48. Use of auxiliary variables. Examples 38 49. Examples 39 Integration. 50. Statement of the problem of finding the distance traversed by a falling body, given the velocity 41 51. Statement of the problem of finding the area bounded by a given curve 41 52. Statement of the problem of finding the length of an arc of a given curve 42 53. Integration. Integral 44 54. Arbitrary constant in integration 44 55. Some formulas for direct integration 44 56. Solution of problem stated in Article 50 . 45 57. Example under problem stated in Article 51. Examples ... 46 58. Examples under problem stated in Article 52 48 CHAPTER IV. TRANSCENDENTAL FUNCTIONS. 59. Differentiation of log x requires the investigation of the limit • f ( 1+ s)" 49 60. Expansion of (l+-J by the Binomial Theorem 50 61. Proof that the limit in question is the sum of a wellrknown series 50 62. This series is taken as the base of the natural system of loga- rithms. Computation of its numerical value 52 63. Extension of the proof given above to the cases where m is not a positive integer 53 64. Differentiation of log a: completed 54 65. Differentiation of a x . Examples 55 Vlll DIFFERENTIAL CALCULUS. Trigonometric Functions.. Article. Page. 66. Circular measure of an angle. Reduction from degree to cir- cular measure. Value of the unit in circular measure ... 57 67. Differentiation of sin x requires the investigation of the limit sin Ax , 1 — cos Ax „_ — ; — and ; 57 Ax Ax 68. Investigation of these limits 58 69. Differentiation of the Trigonometric Functions. Examples . 59 70. Anti- or inverse Trigonometric Functions 60 71. Differentiation of the Anti-Trigonometric Functions. Examples 60 72. Anti- or inverse notation. Differentiation of anti- functions in general 61 73. The derivative of y with respect to x, and the derivative of x with respect to y, are reciprocals. Examples 62 CHAPTER V. INTEGRATION. 7i. Formulas for direct integration 65 75. Integration by substitution. Examples 66 76. If fx can be integrated, /(a + fix) can always be integrated. Ex- amples 67 J_ 78- L-n-rr^- Example 68 77 - f* .„„2_ v *v Examples 67 79. Integration by parts. Examples 69 80. /i sin" x. Examples 69 81. Use of integration by substitution and integration by parts in combination. Examples 70 82. Simplification by an algebraic transformation. Examples ... 71 Applications. 83. Area of a segment of a circle ; of an ellipse ; of an hyperbola . 72 8i. Length of an arc of a circle 74 85. Length of an arc of a parabola. Example ... 75 CHAPTER VI. CURVATURE. 86. Total curvature ; mean curvature ; actual curvature. Formula for actual curvature 77 TABLE OF CONTENTS. ix Article. Page. 87. To find actual curvature conveniently, an indirect method of differentiation must be used 77 88. The derivative of s with respect to y is the quotient of the derivative of z with respect to x by the derivative of y with respect to a; 78 89. Reduced formula for curvature. Examples 78 90. Osculating circle. Radius vf curvature. Centre of curvature . 81 91. Definition of evolute. Formulas for evolute 82 92. Evolute of a parabola 83 93. Reduced formulas for evolute. Example 85 94. Evolute of an ellipse. Example ,. 85 95. Every normal to a curve is tangent to the evolute 87 96. Length of an arc of evolute 88 97. Derivation of the name evolute. Involute 88 CHAPTER VII. THE CYCLOID. 98. Definition of the cycloid 90 99. Equations of cycloid referred to the base and a tangent at the lowest point as axes. Examples 90 100. Equations of the cycloid referred to vertex as origin. Exam- ples 92 101. Statement of properties of cycloid to be investigated .... 93 102. Direction of tangent and normal. Examples 93 103. Equations of tangent and normal. Example 94 104. Subtangent. Subnormal. Tangent. Normal .... .94 105. Curvature. Examples 95 106. Evolute of cycloid 96 107. Length of an arc of cyloid 97 108. Area of cycloid. Examples 98 109. Definition and equations of epicycloid and hypocycloid. Exam- ples 99 CHAPTER VIII. PROBLEMS IN MECHANICS. 110. Formula for velocity in terms of distance and time 102 111. Acceleration. Example. Differential equations of motion . .102 112. Two principles of mechanics taken for granted 103 113. Problem of a body falling freely near the earth's surface . . . 103 X DIFFERENTIAL CALCULUS. Article. Page. 114. Second method of integrating the equations of motion in the case of a falling body 104 115. Motion down an inclined plane 106 116. Motion of a body sliding down a chord of a vertical circle. Example 107 117. Problem of a body falling from a distance toward the earth. Velocity of fall. Limit of possible velocity. Time of fall. Examples 108 118. Motion down a smooth curve. Examples 112 119. Time of descent of a particle from any point of the arc of an inverted cycloid to the vertex. Cycloidal pendulum. Tau- tochrone 113 120. A problem for practice 116 CHAPTER IX. DEVELOPMENT IN SERIES. 121. Definition of series. Convergent series. Divergent series. (Sum of series 117 122. Example of a series 117 123. Function obtained by integration from one of its derivatives. Series suggested 118 124. Development of /(£„+ ft) into a series. Determination of the coefficients on the assumption that the development is pos- sible. Examples 119 125. An error committed in taking a given number of terms as equivalent to the function developed 122 126. Lemma 122 127. Error determined 123 128. Second form for remainder. Taylor's Theorem 125 129. Examples of use of expression for remainder. 'Test for the possibility of developing a function 126 130. Development of log (1 + a:) 127 131. The Binomial Theorem. Investigation of the cases in which the ordinary development holds for a negative of fractional value of the exponent. Example 129 132. Maclaurin's Theorem 132 133. Development of a 1 , e", and e 133 134. Development of sin x and cos x 134 135. Development of sin- 1 % and tan -1 x. Examples 134 136. The investigation of the remainder in Taylor's Theorem often omitted. Examples 136 TABLE OF CONTENTS. Xi Article. Page. 137. Leibnitz's Theorem for Derivatives of a Product 136 138. Development of tan x. Example 137 Indeterminate Forms. 139. Treatment of indeterminate forms by the aid of Taylor's The- orem. The form q. Example 138 140. Special consideration of the case where the form « occurs for an infinite value of the variable 139 141. The form ^ ; special consideration of the cases where its true value is zero or infinite 141 142. Reduction of the forms oo°, 1°°, 0° to forms already discussed. Examples 142 Maxima and Minima. 143. Treatment of maxima and minima by Taylor's Theorem . . . 145 145. Generalization of the investigation in the preceding article. Examples 146 CHAPTER X. INFINITESIMALS. 146. Definition of infinitesimal 149 147. Principal infinitesimal. Order of an infinitesimal. Examples . 149 148. Determination of the order of an infinitesimal. Examples . . 150 149. Infinitesimal increments of a function and of the variable ou which it depends are of the same order 151 150. Lemma. Expression for the the coordinates of points of a curve by the aid of an auxiliary variable 152 151. Lemma 153 152. Lemma 153 153. Geometrical example of an infinitesimal of the second order . 154 154. In determining a tangent the secant line can be replaced by a line infinitely near 155 155. Tangent at any point of the pedal of a given curve 156 156. The locus of the foot of a perpendicular let fall from the focus of an ellipse upon a tangent. Example 157 157. Tangent at any point of the locus of a point cutting off a given distance on the normal to a given curve 158 158. Tangent to the locus of the vertex of an angle of constant magnitude, circumscribed about a given curve. Example . 159 xii DIFFERENTIAL CALCULUS. Article. p «K e ' 159. The substitution of one infinitesimal for another . . . . . .160 160. Theorem concerning the limit of the ratio of two infinitesi- mals 160 161. Theorem concerning the limit of the sum of infinitesimals . . 161 162. If two infinitesimals differ from each other by an infinitesimal of higher order, the limit of their ratio is unity 162 163. Direction of a tangent to a parabola 163 164. Area of a sector of a parabola 164 165. The limit of the ratio of an infinitesimal arc to its chord is unity 165 166. Rough use of infinitesimals 166 •167. Tangent to an ellipse. Examples 167 168. The area of a segment of a parabola. Examples 168 169. New way of regarding the cycloid 169 170. Tangent to the cycloid 170 171. Area of the cycloid 171 172. Length of an arc of the cycloid . . . : 172 173. Radi us of curvature of the cycloid 174 174. Evolute of the cycloid 176 175. Examples 177 176. The brachistochrone, or curve of quickest descent a cycloid . . 177 CHAPTER XI. DIFFERENTIALS. 177. In obtaining a derivative the increment of the function may be replaced by a simpler infinitesimal. Application to the de- rivative of an area ; to the derivative of an arc 183 178. Definition of differential 185 179. Differential notation for a derivative. A derivative is the actual ratio of two differentials 185 180. Advantage of the differential notation 185 181. Formulas for differentials of functions. Examples 186 182. The differential notation especially convenient in dealing with problems in integration. Numerical example 187 183. Integral regarded as the limit of a sum of differentials. Defi- nite integral 188 184. An area regarded as the limit of a sum of infinitesimal rectan- gles. Example 189 185. Definition of centre of gravity. The centre of gravity of a parabola 190 TABLE OF CONTENTS. Xlll Differentials of Different Orders. Article. Page. 186. Definition of the order of a differential 192 187. Relations between differentials and derivatives of different orders. Assumption that the differential of the independent variable is constant 193 188. Expression for the second derivative in terms of differentials when no assumption is made concerning the differential of the independent variable 194 189. Differential expression for the radius of curvature 194 190. Finite differences or increments of different orders 194 191. Any infinitesimal increment differs from the differential of the same order by an infinitesimal of higher order 196 192. Lemma 197 193. Proof of statement in Artiele 191 197 CHAPTER XII. FUNCTIONS OF MORE THAN ONE VARIABLE. Partial Derivatives. 194. Illustration of a function oft wo variables 199 195. Definition of partial derivative of a function of several varia- bles. Illustration 199 196. Successive partial derivatives 200 197. In obtaining successive partial derivatives the order in which the differentiations occur is of no consequence 200 198. Complete differential of a function of two variables. Example . 202 199. Use of partial derivatives in obtaining ordinary or complete derivatives. Example 203 200. Special case of Article 199. Examples 204 201. Use of partial derivatives in finding successive complete deriv- atives. Example 205 202. Derivative of an implicit function. Examples .206 203. Second derivative of an implicit function. Examples .... 207 CHAPTER XIII. CHANGE OF VARIABLE. 204. If the independent variable is changed, differentials of higher orders than the first must be replaced by more general values 209 205. Example 209 XIV DIFFERENTIAL CALCULUS. Article. Page. 206. Example of the change of both dependent and independent variable at the same time. Example 211 207. Direction of a tangent to a curve in terms of polar coordi- nates. Examples 213 208. Treatment of the subject of change of variable without the use of differentials. Example 214 209. Change of variable when partial derivatives are employed. First Method. Examples 215 210. Second Method. Examples 217 211. Third Method. Examples 218 CHAPTER XIV. TANGENT LINES AND PLANES IN SPACE. 212. A curve in space is represented by a pair of simultaneous equations 220 213. Equations of tangent line to a curve in space. Equation of normal plane 220 214. Tangent and normal to helix 221 215. Expressions for equation of tangent line to curve in space in terms of partial derivatives. Examples 224 216. Osculating plane to a curve in space. Example 225 217. Tangent plane to a surface. Examples 226 CHAPTER XV. DEVELOPMENT OP A FUNCTION OP SEVERAL VARIABLES. 218. Taylor's and Maclaurin's Theorems for functions of two inde- pendent variables. Example 227 219. Taylor's Theorem for three variables. Example 231 220. Euler's Theorem for homogeneous functions. Example . . .232 CHAPTER XVI. MAXIMA AND MINIMA OP FUNCTIONS OP TWO OR MORE VARIABLES. 221. Essential conditions for the existence of a maximum or minimum _ 234 222. Tests for the detection of maxima and minima 235 223. Formulas for maxima and minima 236 224. Examples 23B 225. Examples # 239 TABLE OF CONTENTS. XV CHAPTER XVII. THEORY OF PLANE CURVES. Concavity and Convexity. Article. Page. 226. Tests for concavity and convexity of plane curves 240 227. Points of inflection 240 228. Application of Taylor's Theorem to the treatment of convexity and concavity and points of inflection 241 229. Examples 243 230. Singular points 245 231. Definition of multiple point ; osculating point ; cusp; conjugate point ; point d'arrit ; point saillant 245 232. Test for a multiple point 246 233. Detection of osculating points, cusps, conjugate points, points d'arret, and points saillant 247 234. Example of a double point • . . . . 247 235. Example of a cusp 249 236. Example of a conjugate point. Examples 250 Contact of Curves. 237. Orders of contact 251 238. Order of contact indicates closeness of contact 252 239. Osculating circle. Examples 253 Envelops. 240. An equation may represent a series of curves. Variable parameter. Envelop 255 241. Determination of the equation of an envelop 255 242. Example 257 243. An evolute the envelop of the normal. Examples 257 DIFFEKENTIAL CALCULUS. CHAPTER I. INTRODUCTION. 1. A variable quantity, or simply a variable, is a quantity which, under the conditions of the problem into which it enters, is susceptible of an indefinite number of values. A constant quantity, or simply a constant, is a quantity which has a fixed value. For example ; in the equation of a circle x 2 + y 2 = a 2 , x and y are variables, as they stand for the coordinates of any point of the circle, and so may have any values consistent with that fact ; that is, they may have an unlimited number of different values ; a is a constant, since it represents the radius of the circle, and has therefore a fixed value. Of course, any given number is a constant. 2. When one quantity depends upon another for its value, so that a change in the second produces a change in the first, the first is called a function of the second. If, as is generally the case, the two quantities in question are so related that a change in either produces a change in the other, either may be regarded as a function of the other. The one of which the other is considered a function is called the independent variable, or simply the variable. 2 DIFFERENTIAL CALCULUS. [Art. 3. For example ; if x and y are two variables connected by the relation we may regard x as the independent variable, and then y will be a function of x, for any change in x produces a corresponding change in its square ; or we may regard y as the independent variable, and then x will be a function of y, and from that point of view the relation would be more naturally written x= sfy. Again, suppose the relation is y — sin x, we may either regard y as a function of a;, in which case we should naturally write the relation as above, or we may regard x as a function of y, and then we should more naturally express the same relation by x = sin -1 ?/, i.e., x is equal to the angle whose sine is y. 3. Functional dependence is usually indicated by the letters /, F, and y. Thus we may indicate that y is a function of x by writing y =fx, or y = Fx, oxy= & c, where the ratio is — i . Here the limit of the sum as n increases indefinitely, is § . Let w start with the value 1 and increase ; when m = l, s = l, and is greater than the limit f ; when w = 2, s = £, and is less than §, but is nearer § than 1 was ; when n=3, s = £, and is greater than f ; when n = 4, s = £, and is less than f ; and as n increases, the values of s are alter- nately greater and less than the limit f , but each value of s is nearer § than the value before it. 7. It follows immediately from the definition of a limit, that the difference between a variable and its limit is itself a variable which has zero for its limit, and in order to prove that a given constant is the limit of a particular variable, it will always suf- fice to show that the difference between the two has the limit zero. For example ; it is shown in elementary geometry that the difference between the area of any circle and the area of the inscribed or circumscribed regular polygon can be made as small as we please by increasing the number of sides of the polygon, and this difference evidently can never become absolutely zero. The area of a circle is then the limit of the area of the regular inscribed or circumscribed polygon as the number of sides of the polygon is indefinitely increased. It is also shown in geometry, that the difference between the length of the circumference of a circle and the length of the perimeter of the regular inscribed or circumscribed polygon can be decreased indefinitely by increasing at pleasure the number of sides of the polygon, and this difference evidently can never Chap. I.] INTRODUCTION. 5 become zero. The length of the circumference of a circle is ' then the limit of the length of the perimeter of the regular in- scribed or circumscribed polygon as the number of sides of the latter is indefinitely increased. 8. The fundamental proposition in the theory of limits is the following Theorem. — If two variables are so related that as they change they keep always equal to each other, and each approaches a limit, their limits are absolutely equal. For two variables so related that they are always equal form but a single varying value, as at any instant of their change they are by hypothesis absolutely the same. A single varying value cannot be made to approach at the same time two different constant values as nearly as we please ; for, if it could, it could eventually be made to assume a value between the two constants ; and, after that, in approaching one it would recede from the other. 9. As an example of the use of this principle, let us prove that the area of a circle is one-half the product of the length of its radius by the length of its circumference. Circumscribe about the circle any regular polygon, and join its vertices with the centre of the circle, thus divid- ing it into a set of triangles, each having for its base a side of the polygon, and for its altitude the radius of the circle. The area of each triangle is one-half the product of its base by the radius. The sum of these areas, or the area of the polygon, is one-half the length of the radius by the sum of the lengths of the sides, that is, by the length of the perimeter of the polygon. If A' is the area, and P the perimeter of the polygon, and R the radius of the circle, we have A'=iBP; a relation that holds, no matter what the number of sides of 6 DIFFERENTIAL CALCULUS. [Art. 10. the polygon. A' and £BP evidently change as we change the number of sides of the polygon ; they are then two variables so related that, as they change, they keep always equal to each other. As the number of sides of the polygon is indefinitely increased, A' has the area of the circle as its limit ; P has the circumference of the circle as its limit. Let A be the area and C the circumference of the circle ; then the limit A'=A, and the limit %BP=$BC. By the Theorem of Limits these limits must be absolutely equal ; .-. A = £BC. Q.E.D. 10. It is of the utmost importance that the student should have a perfectly clear idea of a limit, as it is by the aid of this idea that many of the fundamental conceptions of mechanics and geometry can be most clearly realized in thought. 11. Let us consider briefly the subject of the velocity of a moving body. The mean velocity of a moving body, during any period of time considered, is the quotient obtained by dividing the dis- tance traversed by the body in the given period by the length of the period, the distance being expressed in terms of a unit of length, and the length of the period in terms of some unit of time. If, for example, a body travels 60 feet in 3 seconds, its mean velocity during that period is said to be $p-, or 20 ; and the body is said to move at the mean rate of 20 feet per second. The velocity of a moving body is uniform when its mean ve- locity is the same whatever the length of the period considered. The actual velocity of a moving body at any instant, is the limit which the body's mean velocity during the period imme< diately succeeding the instant in question approaches as the length of the period is indefinitely decreased. In the case of Chap. I.] INTRODUCTION. 7 uniform velocity, the actual velocity at any instant is obviously the same as the actual velocity at any other instant. If the actual velocity of a moving body is continually changing, the body is said to move with a variable velocity. 12. If the law governing the motion of a moving body can be formulated so as to express the distance traversed by the body in any given time as a function of the time, we can indicate the actual velocity at any instant very simply by the aid of the increment notation already explained. Represent the distance by s and the time by t. Then we have a= ft. Suppose we want to find the actual velocity at the end of t 9 seconds. Let At be any arbitrary period immediately succeeding the end of t seconds (it can fairly be considered an increment as we really increase the time during which the body is sup- posed to have moved by At seconds) , and let As be the distance traversed in that period. Then, by definition, the mean velocity As during the period At is — , and the actual velocity de,sired is the .limit approached by this ratio as At approaches zero. We shall indicate this by limit n^f] At=0\_Aty which is to be read "the limit of As divided by At, as At ap- proaches zero" ; the sign = standing for the word approaches. 13. Take a numerical example. In the case of a body falling freely in a vacuum near the surface of the earth, the relation connecting the distance fallen with the time is nearly s = 16« 2 , s being expressed in feet and t in seconds ; required, the actual velocity of a falling body at the end of t seconds. Let At seconds be an arbitrary period immediately after the end of t seconds, 8 DIFFERENTIAL CALCULUS. [Akt 14. then in k+At seconds the body would fall 16 (k + Atf feet, or 16 1 2 + 32 t At + 16 (At) 2 feet. In t seconds it falls 16 tj feet, so that in the period At in question, it would fall 16 i„ 2 + 32 1 At + 16 ( A) 2 - 16 * or 32« A+16(z(0 2 feet, which must therefore be As. If v be the required actual velocity, by Art. 12. _ limit v °-At=0 As At ^t At+ie(Aty =m+i6 ^ , , . , limit and obviously Jt—0 'As At ^32< . Hence v Q = 32 1 , the result required ; and in general t the velocity v at the end of t seconds is v = 32t. 14. Let us now consider a geometrical problem : To find the direction of the tangent at any point of a given curve. The tangent to a curve, at any given point, is the line with which the secant through the given poipt and any second point of the curve, tends tp coincide as the second point is brought indefinitely near the first. In other words, its position is the limiting position of the secant line as the second point of inter- section approaches the first, i.e., a position that the secant line can be made to approach as nearly as we please, but cannot actually assume. 15. Suppose we have the equation of a curve in rectangular coordinates, and wish to find the angle r that the tangent at a given point (x ,y v ) of the curve makes with the axis of X ; that is, what is called the inclination of the curve to the axis of X. Chap. 1.1 INTRODUCTION. The equation of the curve enables us to express y in terms of x, that is, as a function of x. We have then • y=fx. Let x -\-Ax be the abscissa of any second point P of the curve, and 2/o+ Jy the corresponding ordinate. If y is the angle which the secant through P and P makes with the axis of X, it is clear from the figure that tan

.(V£); bj- the general method. Ana. (1) 20; (2) 3x>; (3) --L; (4) [ as" w 2 V(«) 20. In order to deal readily with problems into which deriva- tives enter, it is desirable to work out a complete set of formu- las, or rules for finding the derivatives of ordinary functions ; and it will be well to begin by roughly classifying functions. The functions ordinarily considered are : — (1) Algebraic Functions : those in which the only operations performed upon the variable, are the ordinarj- algebraic opera- tions, namely : Addition, Subtraction, Multiplication, Division, Involution, and Evolution. Example. £x ! + 3-fy(x — l). (2) Logarithmic Functions: those involving a logarithm of the variable, or of a function of the variable. Examples. x log x ; log (x 2 — ax + b) . Chap. II.] DIFFERENTIATION. 13 (3) Exponential Functions: those in which the variable, or a function of the variable, appears as an exponent. Example. a*^ 2- ^. (4) Trigonometric Functions : Example. cos * — sin 2 a;. 21. We shall consider first, the differentiation* of Algebraic Functions of x. Required D x (ax) where a is a constant. By the general method (Art. 19), a (x + Ax) — ax ~ V>.vl—. = So Jx limit 'Jx=0 aJx ~ Jx limit r n = 43,-0 M=° ; . . D x (ax)=a. If a = 1 , this becomes D x x = l. Required D x x n where n is a positive integer. [1] [2] rn ^-i _ limit By the Binomial Theorem, (x + Jx) Jx J" (x +Jx)*= x Q » + nx?- 1 Jx+ n ( n ^ x n ~ 2 (Jx) 2 + . . + (Jx)" z (a; + /(a;)" — x n „_, n(n— 1) _ e . , ., . . i-2-! — -± 2. = nx n J -) i_ L x "~ 2 Jx + . . . + (Jx)"- 1 . Jx 2 Each term after the first contains Jx as a factor, and therefore has zero for its limit as Jx approaches zero, so that ~(x + JxY-xf limit Ax=Q Jx ■ nx " ■■■ [.D x x«-\ x=x =nx?-\ * To differentiate is to find the derivative. 14 DIFFERENTIA!; CALCULUS. [Art. 22. As x is any value of x, we may drop the subscript, and we have D x x" = naf - 1 . [3] 22. "We shall next consider complex functions composed of two or more functions connected \>y algebraic operations ; the sum of several functions, the product of functions, the quotient of functions. Required the derivative of u + v + w, where each of the quantities u, v, and w is a function of x. Let Ax be any increment given to x, and Au, Av, and Aw the corresponding increments of u, v, and w. Then, obviously, the increment of the sum u + v + w is equal to D x {u + v + w) = limit Ax=0 Au+Av+Aw, Au + Av + Aw~ limit 'Ax=0 Au Ax Ax . limit "zte=0 _ limit Ax=0 ' Av Ax + limit Ax=0 and we have Au ' Av Avf Ax Ax Ax Aw' Ax but, since Au and Ax are corresponding increments of the func- tion u and the independent variable x, in like manner limit Ax=Q limit Ax=0 'Au Ax Av" Ax and hence limit ["Jw" Ax = t) \_Ax =D x u; =D x v, = D x w; D x (u + v +v:) = D x u + D x v + D x w. [1] It is easily seen that the same proof in effect may be given, whatever the number of terms in the sum, and whether the connecting signs are plus or minus. So, using sum in the sense of algebraic sum, we can say. the derivative with respect to x Chap. II.] DIFFERENTIATION. 15 of the sum of a set of functions of x is equal to the sum of the derivatives of the separate functions. 23. Required, the derivative of the product uv, where u and v are functions of x. Let x , m , and v be corresponding values of x, u, and v ; let Ax be an increment given to x, and Au and Av the corresponding increments of u and v. Then, rn , Nn limit (m + Jm) Q + Av) — u v ~ Jx (w + Jm) (i! + Av) — u v = m Av + v Au + Au Av and [i»x(M«)] a;= x == jJ^o u Av + v Au Ju + Jw Av~\ te J limit : Ja;=0 Mo Ja; + limit Ax=0 v Au Ax + Jx limit Ax=0 'Au Av" Ax «o does not change as Ax changes, and „ = [A,*]» =eil ; limit Ax=0 Av Ax so that limit Ax=0 u Av Ax and in like manner limit Ax=0 Au' Ax : «o[A»], =Ih ! = »i[A«],= v may be written Au — or Av — . Let us consider Ax Ax Jx limit Ja;=0 Jm Jv 'Ax As Ax approaches 0, Jm, being the corresponding increment of Av Ax the function u, will also approach ; and the product Au Av will approach as its limit, if — approaches any definite value ; Ax 16 DIFFERENTIAL CALCULUS. [Art. 23- that is, if D x v has a definite value. It is, however, perfectly conceivable that — may increase indefinitely as Ax approaches Ax Av zero, instead of having a definite limit ; and, in that case, if — Ax should increase rapidly enough to make up for the simultaneous decrease in Au, the product Au — would not approach zero. Ax We shall see, however, as we investigate all ordinary functions, that their derivatives have in general fixed definite values for any given value of the independent variable ; but, until this is established, we can only say, that nit \ Au ^L\ ={ = 1_ Ax] JS-I "' I " when — or — has a definite limit, as zte=0 ; that is, when Ax Ax \.D x v-\ x=Xo or [D.«] a=a , o has a definite value. With this proviso, we can say, [A (w)L=* = v o [AuL^+Uo [■D.«]x=« b ; or, dropping subscripts, D x (uv) = uD x v + vD x u. [1] Divide through by uv, and we have the equivalent form, D x (uv) ^ D x u D,v_ r 2 -, uv u v If we have a product of three factors, as uvw, we can repre- sent the product of two of them, say vw, by z, and we have D* (uvw) D x (uz) D x u D x z But uvw uz ' u D x z D x (vw) _D x v D x w Z VW V "• w Chap. II.] DIFFERENTIATION. 17 D x (uvw) D x u D x v D x w UVW ~ U V w [3] This process may be extended to any number of factors, and we shall have the derivative of a product of functions divided by the product equal to the sum of the terms obtained by dividing the derivative of each function by the function itself. 24. Required the derivative of the quotient — , where u and v v are functions of x. Employing our usual notation, we have D x (-X] = limi1 limit M + Au v + Av Ax but m + Au u _v Au — u Av v + Av v v 2 + v B Av and dividing by Ax, we have D, . limit ' Ax=0 Au Av' v - u — Ax Ax _ vJ s + v Av «0 [A «]« = -,- Mo [I>.«] a v ( u ^\ — v ^x u ~ uD x v [1] Examples. Find (i)fl,^ + *-VW]; (2)A[^VW]; (8) A^?. Ans. (1) 8^+1-JL .; (2) -?£-. -+2xy/(x) ; (3) 2V(«) 2V(«) 2*VO0 18 DIFFERENTIAL CALCULUS. [Akt. 25 (4) Find, by Art. 24, [1],-D.AY (5) Deduce D x x n from last part of Art. 23. 25. If the quantity to be differentiated is a function of a function of x, it is always theoretically possible, by performing the indicated operations, to express it directly as a function of x, and then to find its derivative by the ordinary rules ; but it can usually be more easily treated by the aid of a formula which we shall proceed to establish. Required, D x fy, y being itself a function ofs.. Let x and y be corresponding values of x and y ; let Ax be any increment given to x, and Ay the corresponding increment of y ; then \PJy\ x => _ limit - x a Ax=0 and this can be written Ax ''} limit Ax=0 ~f{yo+Ay)-fyoAy Ay *y\ Ax]' As Ax and Ay are corresponding increments, they approach zero together ; hence is the same as limit Ax=0 limit Ay=0 ~ f(yo + Jy)-fy ~ Ay ~f(yo+Ay)- fy Ay '} which is equal to [Pyfy] y =y - ••• U>Jy\ x = x = {.D y fy-] y=yo .[D x y] x=x<> ; or, dropping subscripts, DJy = DJy.D x y. This gives immediately, as extensions of Art. 21, [1] and [3], D x (ay) = aD x y, D x y n = ny n ~' i D x y. [1] Chap. II.] DIFFEKENTIATTON. 19 26. Art. 21, [3] can now be readily extended to the case where n is any number positive or negative, whole or fractional. Let n be a negative whole number —m,m of course being a positive whole number. Let y = x n = x~ m , then we want D x y. Multiplying both members of y= x~ m by x m , we have x m y = 1. Since x m y is a constant, its derivative with respect to x must be zero; but by Art. 23, [1] and Art. 21, [3], D x [_x m y~\ = x m D x y + yD x x m = x m D x y + ma; m-1 y m being a positive integer ; .'. x m D x y + mx m ~ 1 y = 0, and D x y= — mx~ 1 y= — mx~ m ~ 1 = nx n ~ 1 . q.e.d. P Let n be any fraction - where p and q are integers either posi- tive or negative. As before, let p y = x n = *« ; required D x y. Clearing of radicals, we have y=x*; and since the two members are equal functions of x, their deriva- tives must be equal ; D m tf = D m &, or qy q - 1 D x y=px p - 1 , p x*~ x p x"- 1 p £-i and V * y -q y^-q (x l y -rq ™ ' «- E ' D 20 DIFFERENTIAL CALCULUS. [Art. 26. The formula,. D x x n = nx n ~% Art. 21, [3], holds, then, whatever the value of n. Example. Prove Art. 24, [1] by the aid of Art. 21, [3] and Art. 23, [1], regarding — as a product, namely uv~\ v By the aid of these formulas, D.o = 0; [1] D x ax = a; [2] D.x=li [3] D x x n = nx n ~ 1 ; [4] D x (u + v + w)=D x u+D x v+D x w; [5] D x (uv)=uD x v + vD x it; [6] u\ vD x u — uD.v D. [7] D x (fy)=D y (fy).D x y; [8] any algebraic function, no matter how complicated, may be dif- ferentiated. Examples. Find D x u in each of the following cases : — (1) u = m + nx. Ans. D x u = n. (2) u- (a + bx) a?. Ans. D x u = (4 bx + 3 a) 3?. Chap. II.] DIFFERENTIATION. 21 (3) M=V(^ + a 2 )- Solution: u = ^ (a? + a?) = (x* + a?)h. Let y = x 2 + a 2 , then u = 2/5. D x u = D x y* = D s yh.D x y by Art. 26, [8], D s y* = iy-'* by Art. 26, [4], D x r/=2a;; .-. D x ii = xy-i = x (x* +■ a 2 )~4 =- < 4 > Ht-J VO^+O 4ms. ZLw=- (l+a;) M+1 fK\ v? a n 2x(a — 2a?) (a+iB 3 ) 2 (a-j-a 3 ) 3 (6) u=(l + x)-s/(l-x). Ans. D x u = - 1 } 2V(l-«) (7) « = JL 4ws. JP.w= 2 f, + 1 -2x. 4ms. Dvit = —- (8) w / A-VW Y ■W + VOO/ 2(l + V5)V(*-^) 1 ; -^(1+^-7(1-^)' 22 DIFFERENTIAL CALCULUS. [Art. 27. CHAPTER III. APPLICATIONS. Tangents and Normals. 27. We have shown, in Art. 15, that the angle t, made with the axis of x by the tangent at any given point of a plane curve, when the equation y=f® of the curve, referred to rectangular axes, is known, may be found by the relation Ax tanr= limit Ax=Q where Jy and Jx are corresponding increments of y and x, the coordinates of a point of the curve. If the point be (x ,y ) , we have, then, tanr = [\D l2 /] . At any point (%,y) tan - r : D*y- [1] A line perpendicular to any tangent, and passing through the point of contact of the tangent with the curve, is called the nor- mal to the curve at that point. If v Q be the angle which the Chap. III.] APPLICATIONS. 23 normal at the point (% ,y ) makes with the axis of X, then it is evident from the figure, that "o=90° + r ,. and from trigonometry, 1 tan v = — cot r = — \D x y] !B=!Btt Of course, for any point (x,y) tan»> = — . [21 D.y ' 28. Since the tangent at (x ,y ) passes through (x ,y ), and makes an angle t with the axis of x, its equation will be, by analytic geometry, y — yo= tan t (x — x ) ; or, since tanr = [D x y~] x=a , o , y-y = [_D x y-\ x=Xo {x - x ). [1] In like manner, the equation of the normal at (x ,y ) is found to be y-y =-— — (x-x ). [2] The distance from the point of intersection of the tangent with the axis of X to the foot of the ordinate of the point of contact, is called the subtangent, and is denoted by t x . The distance from the foot of the ordinate of the point to the intersection of the normal with the axis of X, is called the sub- normal, and is denoted by n x . In the figure, TA and AN are respectively the subtangent ami subnormal, corresponding to the point (x ,y ) of the curve. Obviously |2.= tan r = [D x y] x , 24 DIFFERENTIAL CALCULUS. [ART. 28 and ?1 = tan (180° - n>) = - tan v t = — — i ; hence R>r the point (#o>2/o)j The distance from the intersection of the tangent with the axis of X to the point of contact is sometimes called the length of the tangent, and may be denoted by t. The distance from the point at which the normal is drawn to the point where the normal crosses the axis of X is sometimes called the length of the normal, and may be denoted by n. It is easily seen from trie figure, that *=VO/o 2 + 4 2 ), and n = V(2/o 2 + % 2 ) 5 hence * = y a [D.y]^ V(l + [0.»]U«») > and « = yo VC 1 + LAy]!^) • For any point (a;,?/) , our formulas become n x = yD z y; [4] *=2/[^2/]-V(l+[-Dx2/] 2 )5 [5] « = »V(i + [-D.y] 1 ). [6] Examples. (1) Show that the inclination of a straight line to the axis of X is the same at every point of the line ; i.e., prove tan t constant. ,m ,ii Chap. III.] APPLICATIONS. 25 (2) Show that the subnormal in a parabola y 2 = 2 mx , is constant, and that the subtangent is always twice the abscissa of the point of contact of the tangent. (3) Find what point of the parabola must be taken in order that the inclination of the tangent to the axis of X may be 45°. 29. If the equation of the curve cannot be readily thrown into the form y=fx, D x y may be found by differentiating both members with respect to x and solving the resulting equation algebraically, regarding D x y as the unknown quantity. For example ; required the equation of the tangent to a circle at the point (x ,y ) of the curve. The equation of a circle is x 2 + y 2 = r 2 , r being constant. Differentiating with respect to x, we have, by Art. 26, [8], 2x + 2yD x y = 0. Solving, D x y=--^=--. 2y y and by Art. 28, [1], the required equation is 2/ -2/o=-— (»-«») ; or, clearing of fractions, y y-y can be made greater than any assigned value. So we say tan

) ^ ^ ns ^ (7) z n — 1 ' at + Sx 3 — Ix 2 — 27a;— 18 " a; 4 - 3a 3 - 7^+ 27a;- 18 si- l + (a;- 1)1" (x>-l)l-x + l ~ xl-\ + (iB-l)i' . V(^-i) . l-V(l-a) Lv(i + «)-V(i + ^). VO 8 ) -V(«) +V(a:-o) ' V(^-a 2 ) a 3 — 3a; + 2 1 n ^.ns. 10. Ans. 2 .4ms. 0. Ans. 1. rt—Qtf + Sx — S V(2a) .4ns. oo. 35. If /a; and Fx both increase indefinitely as x approaches the value a, or, as we say for the sake of brevity, if Chap. III.] APPLICATIONS. 31 fa= oo and Fa = go, we can determine the true value of >" Fx ' 1 " by first throwing the fx Fx fraction i— into the equivalent form — , which assumes the form Fx 1 Fx T' - when x=a, and may be treated by the method just described. If fx = and Fx= when oj = a, the true value of [/a.-Fo;],—,, can be determined by throwing fx fx.Fx into the equivalent form '—, which assumes the form - Fx when x = a. Maxima and Minima of a Continuous Function. 36. A variable is said to change continuously from one value to another when it changes gradually from the first value to the second, passing through all the intermediate values. A function is said to be continuous between two given values of the variable, when it has a single finite value for every value of the variable between the given values, and changes gradually as the variable passes from the first value to the second. 37. If the function is increasing as the variable increases, the increment Ay, produced by adding to a; a positive increment Ax, will be positive ; -^ will therefore be positive, and "imt -^ 1 , Ax Ax=0 \_Ax] will also be positive ; that is, D x y will be positive. If a function decreases as the variable increases, the increment Ay, produced by giving x a positive increment Ax, will be nega- tive ; -1. will therefore be negative, and umlt -^ will also be Ax Ax=0 \_Axj negative ; that is, D x y will be negative. 32 DIFFERENTIAL CALCULUS. [Art. 38. Since D x y, being, as we have seen, itself a function of x, may happen to be positive for some values of x and negative for others, it would seem that the same function may be sometimes increasing and sometimes decreasing as the variable increases, and this is often obviously the case. For example ; sin

y=6x-2; [Z> l2 /] x=0 =0, [D x 2 2/] I= o=-2. Since D x 2 y is negative when x = 0, D x y must have been de- creasing as x passed through the value zero, and as [Z> x2 /] I=0 =0 Chap. III.] APPLICATIONS. 35 D x y must have been positive before a;=0, and negative after x — ; therefore, y must have been increasing before x = 0, and decreasing after x = 0, and must consequently have a maximum value when x=0. To confirm our conclusion, let us find the values of X s — x 2 when x = — .1, when x = 0, and when x = .1 : [a 3 -2y\ x =x i s negative, D/y must have been decreasing as x passed through the value x , and being when x = x Q , must 36 DIFFERENTIAL CALCULUS. [Art. 47. have been positive before and negative after. D x y therefore must have been increasing before x = x and decreasing after ; and as [AyL=«fc = o, it must have been negative both before and after x = x . The function y, then, must have been decreasing both before and after x = x , and y is neither a maximum nor a minimum. Examples. (1) Shov j that if and [-Dx 3 3/L=* >0, y is neither a maximum nor a minimum. (2) If [A 3 */L =Xo =o, and [£* 4 2/L=* <0, y is a maximum (3) If [A 2 2/L = * o = 0, and CA 4 y],=x >o, ?/o is a minimum, 47. The preceding investigation suggests the following method of finding the values of the variable corresponding to maximum or minimum values of the function. Differentiate the function and find what values of x will make the first derivative zero. This may, of course, be done by writing the derivative equal to zero, and solving the equation thus formed. Substitute for x, in turn, in the second derivative, the values of x thus obtained, and note the signs of the results. Those values of x which make the second derivative positive correspond to minimum values of Chap. III.] APPLICATIONS. 37 the function, and those that make the second derivative nega- tive, to maximum values of the function. If any make the second derivative zero, they must be substituted for x in the third derivative, and the result interpreted by the method of Art. 46. Examples. Find what values of x give maximum and minimum values of the following functions : — (1) w=2« 3 -21a 2 + 36a;-20. Ans. x=\, max. ; x= 6, min. (2) u = x 3 -dx 2 +15x-3. Ans. x = 1, max. ; x= 5, min. (3) M=3a^-125a^+ 2160a;. Avis. Max. when x = — 4 or 3 ; min. when x = — 3 or 4. (4) Showthat u = a?— 3^+ 6cc+ 7 has neither a maximum or a minimum value ; and that w = ar i — 5:^ + 5a 3 — 1 is neither a maximum nor a minimum when x = 0. (5) A person in a boat, three miles from the nearest point of the beach, wishes to reach, in the shortest possible time, a place b L five miles from that point, along the shore. Supposing he can walk five miles an hour, but can row only four miles an hour, required the point of the beach he must pull for. 38 DIFFERENTIAL CALCULUS. [Art. 48. With the notation in the figure, the distance rowed is ^/(x 2 + 9 ) miles, the distance walked is 5 — x miles, and u, the whole time taken, is evidently 7(^+9) 5- u = -f =— hours, and x must have a value that will make u a minimum. x 1 D x u = DJu = 47(^+9) 5' 9 4(^+9)1 S0lYinS 4 V (^+9) -H' we get x = ± 4 ; but, on substituting these values of x in turn in the expression for D x u, we see that a; = 4 is the only value which will make D x ii = 0, since we must take the positive value of ^/(a^+9), from the nature of the case, as it represents a distance traversed. Remembering this fact, we find L x Ja: - 4 500 and u then is a minimum when a; = 4, and the landing-place must be one mile above the point of destination. 48. In problems concerning maxima and minima, the func- tion u can often be most conveniently expressed in terms of two variables, x and y, which are themselves connected by some equation, so that either may be regarded as a function of the other. In this case, of course, u can, by elimination, be expressed in terms of either variable, and treated by the usual process. It is generally simpler, however, to differentiate u, regarding one of the variables, x, as the independent variable, and the other as a function of it, and then to substitute for D x y Chap. III.] APPLICATIONS. 39 its value obtained from the given equation between x and y by the process suggested in Art. 29. Examples. (1) Required the maximum rectangle of given perimeter. If a be the given perimeter, we have y X 2x+2y = a; (1) and the area u = xy. (2) Differentiate (1) with respect to x, and we have 2 + 2D x y=0, whence D x y = — 1 ; (3) D x u = xD x y + y=-) , + (Jy) , > Chap. III.] APPLICATIONS. P27= Ax . sec r. NQ=Ay- MN. MN= Ax . tan t. hence NQ = Ay—Ax tan t, and we have 43 \l{Axy+ (Jy) 2 X ?/) 2 ; hence 2o[ SeCr + ^~ tanr == Vl + (Z> I y)* + 2>.y-Ay, limit Ax= or =Vl + (D^) 2 . As As — lies always between two quantities which have the same Ax limit, Vl + (D x y) 2 , its limit must be Vl + {D x yY, and we have D x s=^l + (D x yy. D x y can be found from the given equation, and therefore 44 DIFFERENTIAL CALCULUS. [Art. h'o. Vl + (D x y) 2 can be determined. We can then regard D x s as given, and again we are required to obtain a function from its derivative. 53. To find a function from its derivative is to integrate, and the function is called the integral of the given derivative. Thus the integral of 2 a: is x* + C, where C is any constant, for D x {x^ + C) is 2x. In other words, if y is a function of x, that function of x which has y for its derivative is called the integral of y with respect to x, and is indicated by f x y, the symbol f„ standing for the words integral with respect to x. 54. Since the derivative of a constant is zero, we may add any constant to a function without affecting the derivative of the function ; so that if we know merely the value of the derivative, the function is not wholly determined, but may contain any arbitrary, i.e., undetermined, constant term. In special problems, there are usually sufficient additional data to enable us to determine this constant after effecting the integration. 55. Since integration is defined as the inverse of differentia- tion, we ought to be able to obtain a partial set of formulas for integrating by reversing the formulas we have already obtained for differentiating. Take the formulas — D.z=l; D x ax=a ; D x ay = aD x y; D x x" = rue"- 1 ; D x (u + v + w + &c.) = D x u +D x v + D x w + &c. ; and we get immediately — f x l = x+G; (1) J' x a = ax+C; (2) Chap. III.] APPLICATIONS. 45 f x aD x y = ay+C; (3) f,nx n ~ 1 = x" + C ; (4) f x (D x u +D x v +D x w +&c.)?=u + v + w + &c. + C; (5) where in each case is an arbitrary constant. The forms of the last three can be modified with advantage. In (3) , call D x y = u ; then y =f x u, and (3) becomes f x au = af x u + G. (6) By the aid of (6) , (4) can be written, nf x x"- 1 =x a + C. Change n into n + 1, and we get {n + \)f x x» = x n+x + C, or f m3 * = £±l + C, (7) w + 1 where C is any arbitrary constant, although, strictly speaking, different from the C just above. In (5) , let D x u = y, D x v = z, &c, then u—f x y,v=f x z,&c. and f x (y + z+&c.)=f x y+f x z+&c.+ C; or, the integral of a sum of terms is the sum of the integrals oj the terms. 56. We can now solve the problem stated in Art. 50. The velocity of a falling body at the end of t seconds is gt feet, g 46 DLFFEKBNTIAL CALCULUS. [Art. 57. being a constant number ; required the distance fallen in t sec- onds. We have seen that, if v, t, and s are the velocity, time, and distance respectively, v=D,s; hence s=f t v. Here s=f t gt + 0\ but by Art. 55, (6) and (7), f.gt = gf,t = 9*1+0 =igt* + C; and in this case we can readily determine C, for when the body has been falling no time, it has fallen no distance, :o s must equal zero when t = 0, and we have = ig(0y + C=0+C, and C=0; and our required result is s = £gt 2 . 57. Required the area intercepted by the curve y 2 = 4:X, the axis of X, and the ordinate through the focus. From the form of the equation we know that the curve is a parabola with its vertex at the origin and its focus at the point (1,0). The initial ordinate in this case is evidently the tangent at the vertex. If A is the required area, D x A = y, (Art. 51) , then A=f x y. y = 2\Jx = 2afr; hence A=f x 2ab = 2f x a* = — +C=-aA+0. § 3 A stands for the area terminated by the ordinate correspond- ing to any abscissa x. It is obvious from the figure that if we make x = 0, the ter- minating ordinate y will coincide with the initial ordinate through the origin, and A will equal zero. So we can readily determine C, Chap. III.] for we have so that and APPLICATIONS. 47 A=0ifx=0; = -0S + C = C, 3 A = -xl. If x = 1, as it must in order that y may pass through the focus, A = -, the required area. Examples. (1) Find the area bounded by the curve x s =4 : y, the axis of X, and the ordinates corresponding to the abscissas 2 and 8. Ans. 42. (2) Prove that the area cut off from a parabola by a double ordinate is two-thirds of the circumscribing rectangle. (3) Required the area intercepted between the curves ?/ 2 =4aa; and x 2 =Aay. Ans. 16a 2 (4) Find a formula for the area bounded by a curve x=fy, the axis of Y, and two lines parallel to the axis of abscissas. Ans. A=f s x + C. (5) Find a formula for the area intercepted by a curve y =fx, the axis of X, and two ordinates (oblique coordinates) . Ans. A = sin u>f x y + C, u> being the inclination of the axes. 48 DIFFERENTIAL CALCULUS. [Akt. 58. (6) Prove that the segment of a parabola cut off by any chord is two-thirds of the circumscribing parallelogram. 58. Required the length of the portion of the line 4x-3y+2 = (1) between the points having the abscissas 1 and 4. We have seen that D x s = Vl+(A2/) 2 Art. 52, where s is the length of an arc ; hence s=f x Vl + (D x y) 2 . From (1) we get 4-3 D,y = 0, ■sI\ + (D x yy=%; and therefore s=f x % = %x + C, where s stands for the length of the arc from the first point to any second point whose abscissa is a;. If we make x = 1 , the two points will coincide and s must equal ; then = $ + 0, o=-h and s = %(x — 1). To get the required distance, x must equal 4 and we get s= 5 Example. Find the length of the portion of the line Ax + By+C=0 between the points whose abscissas are x and a^. Ans. V^' + g) fo.^. Chap. IV.] TEANBCENDENTAL FUNCTIONS. 49 CHAPTER IV. TRANSCENDENTAL FUNCTIONS. 59. In order to complete our list of formulas for differentiat- ing, we must consider the transcendental forms, logo;, a x , since, &c. Let us differentiate log a;. By our fundamental method, we have 7-. , limit x 6 Jai=0 log (a; + Ax) - Ax ■ loga;"] J' log (a; + Ax) — log a; and r, , limit D ^°^ x = Ax=Q 1 , Vx + Ax~\ 1 , f, , Ax~\ - log — !— - = — log 1 + — , ix (_ x J Ax |_ a; J f Ax\ But as Ax approaches zero, log I 1 -| ) approaches log 1 zero, and — increases indefinitely ; so that it is by no means easy 1 / Ax\ to discover the limit of the product — log ( \ -) ) . , i.e., This product can be thrown into a simpler form by introduc- ing m = ■ Ax in place of Ax. Ax T- log 1 + — then becomes— log 1 + — , or - log 1+- m m Ax As Ax approaches zero, — or m increases indefinitely, and Z> c loga; = limit -log 1+ — x & V m 50 DIFFERENTIAL CALCULUS. IArt. 60. and the value we have to investigate is the value approached by M-j j as its limit as m increases indefinitely, which we in- m ,. , , limit dicate by m 60. Let us first suppose that m in its increase continues always a positive integer. Then we can expand M -| J by the Bi- nomial Theorem. / 1 Y* m/l\, m(m-l)/l V m(m— l)(m— 2)/ 1 1.2.3 \m + &c. to m + 1 terms = 1+ 1 , m V mj\ m 1 1.2 1.2.3 ..lY^lY^lj mj\ mj\ in J ,■.*.,„ + .... to m+1 terms. 1.2.3.4 Now, as m increases indefinitely, each of the first n terms of the series, n being any fixed number, approaches as its limit the corresponding term of the series 1 1.2 1.2.3 1.2.3.4 ■•"' so that we have reason to suppose that there is some simple rela- tion between this latter series and our required limit. 61. To investigate this question we shall divide the first series into two parts. The first part, consisting of the first n + 1 terms, where n is any fixed whole number less than m, we shall repre- sent by S ; the second part, consisting of the remaining m — n terms, we shall call B. Chap. IV.] TRANSCENDENTAL FUNCTIONS. 51 Then 1+-L )=S+B. m , 1 , 1-— (1 1 1.2 1.2.3 L )(l--) m/\ mj i-iYi-i- m/V m n-1 1.2.3. ...» As n is a fixed number, we have limit ro-i I , I , J_ , _J_ m=co lP -> ~ l ~l~ t ~ 1.2 1.2.3 + ....+ B (l- L )(l--)- 1-- n — 1 m 1.2.3 » 1.2.3. n ( n\f n+1 1 1 ) 1 — m \ inj\ m _n + l (n + l)(n+2) fi-^Yi-^Yi-^ 2 ) fi-^=i (n + l)(n+2)(n+3) m Since n is less than m, each numerator in the value of R is posi- tive and less than 1, and R< l 1.2.3 n 1 n+1 (n+1) 2 (n+1) 3 The sum of the decreasing geometrical series, (n+l)™-" + n+1 (n+1) 2 (n+1) is by algebra less than - ; therefore B< n(1.2.3....nV 52 DIFFERENTIAL CALCULUS. [>RT. 62. , limit rpl . 1 m=oo L J «(1.2.3 w) and we have at last, _ i4.I + _L + ^_ +_!_+.... + i ~ ^ 1^1.2 + 1.2.3 1.2.3.4 T ^1.2.3 n + something less than ,„ . „ , w(1.2.3 »i) n being any positive whole number. Thus we obtain the relation that the difference between our required value and the sum of the first n + 1 terms of the series 1 1 , 1 1+ l" 1 " 1.2 1.2.3 "*"'■" is less than n(1.2.3 n) The greater the value of n the less the value of — — ; ° n.l.2.3....n and by taking a value of n sufficiently great, we may make this difference as small as we please. Consequently, by Art. 7, our required value is the limit ap- proached by the sum of the first n terms of the series 1+ 1 1.2 1.2.3 H as n is indefinitely increased, or what is ordinarily called the sum of this series. 62. The series l-\ 1 1 h . . . . plays a very impor- 1 1 . — 1 . 'Z . o tant part in the theory of logarithms. It is generally represented by the letter e, and is taken as the base of the natural system of logarithms. Its numerical value can be readily computed to any required number of decimal places, since each term of the Chap. IV.] TRANSCENDENTAL FUNCTIONS. 53 series may be obtained by dividing the preceding one by the number of the term minus one. Carrying the approximation to six decimal places, we have 1. 1. 0.5 0.1666666 0.0416666 The error in the approximation is 0.0083333 less than one-eleventh of the last 0.0013888 term we have used, and therefore 0.0001984 cannot affect our sixth decimal place. 0.0000248 0.0000027 0.0000002 0.0000000 e=2.718281+, correct to six decimal places. 63. Let us now remove from m the restriction we placed upon it when we supposed it to have none but positive integral values, and suppose it to increase passing through all positive values. Let ix represent at any instant the integer next below m, then jj. +1 will be the integer next above m, and as m increases it will always be between /j. and ;j. + 1 , unless it happens to coincide with p. +1, as it sometimes will. We have, then, in general, fi and "»*(i + IY(l + I) = .xl-.; hence Mmit fl+— V=e. Again : let m be negative, and represent it by — r, - KH'-r=(^r=(^H^)' and limit fl + -Y m=oo y m/ - ( JEU (i+~i)'"'( 1+ ,4t)=« >— We see, then, that always 64. In Art. 59 we found that -~Alo ga! =^glo g (l + l)']. We have, then, D x loga; = _loge. If by logarwe mean, as we._shall always mean hereafter, naturai logarithm of a;, loge will equal 1, and Chap. IV.] TRANSCENDENTAL FUNCTIONS. 55 X y Exponential Functions. 65. Required D x a x , a being any constant. Let u = a x and take the log of each member, logw = »loga. Take D x of both members, Dm u = loga; D x u = u\oga, D x a'=a x \oga. [1] If a = e; since log e = 1 , we have D x e* = e. [2] Of course, D x a" = a" log ot-D^, and D x e" = e'D x y. Examples. Find D^w in each of the following cases : — (1) u=e x (\-x i ). a lg0 The circular unit is evidently the angle which has its arc equal to the radius, and its value in degrees is easily found. Let x represent the required value in degrees ; then x° r A 180° ■ = and x = . 360° 2w Hence, to reduce from circular to gradual measure, we have only 180 to multiply the circular value by 6 7 . Required T> x sins.. By our usual method, we have D x sinx= ]imit Ax=0 .sin(a; + Ax) — sinaf Ax sin (a; + Ax) — since _ sin a; cos Ax + cos a; sin Ax — sin a; Ax Ax _ cosa; sin Ax — sina;(l — cos Ax) Ax t\ • limit r si n dx . 1 — cos Ax~] D x smx= llmlu cosa; sma; Ja;=0 \_ Ax Ax [~ sinJa; ~]_ 3ma . limit r i-cosJa n ! Ax J Ax=Q Ax I = cosa; limit Ax=0\ But as Ax = 0, sin Ax=0 and cos Ax = 1 , so both of our limits, in their present form, are indeterminate, and require special investigation. 58 DIFFERENTIAL CALCULUS. [Art. 68. 68. Suppose an arc described from the vertex of the angle Ax, d :B with a radius equal to unity, then this arc measures the angle, and is equal to Ax, and the lengths of the lines, marked s and c in the figure, are the sin -da; and cos Ac, respec- tively. limit rj_1 and limit fklJ =o|_/ "We wish to find )|_ Ax J Ax=0\_Axj Ax=0 arc Ax"f >: 1 + c Ax 1+c + s and lilttit j -L | must be between 1 and limit [~— Ax=0\_AxJ zte=0|_l + since, as Ax = 0, s = 0, and c = 1, + c + c + s ; but Chap. IV.] TRANSCENDENTAL FUNCTIONS. 59 limit Ax therefore nit r_l±£_l = ? =1 . = o|_l+c + sJ 2 ■]., limit Ax=0 sin Ax Ax In like manner, 1=1 >1=± >(M0±^ or s s 2 Ax s(l+c) + s* >^>- «(l+f) J* s(l+e)+S*' \ s ^ 1 — c. s -> >: lim Ax imit (~ 1 — c~ [ a;=0 |_ Ja; J lit I" » i = 0|_l + c + sj 1 + c Ja; 1+c + s' lies between ^mit Jic=0 1 + c , or 0, and limit Ax therefore , or 0; limit Ax nit r i-cos'faa -T Q =o L &*. , J 69. Substituting these values in Art. 67, we have XL sin a: = cos a;. (1) Prove (2) Prove from the relations Examples. D x cosx= — sin a;. X> I tano5= sec 2 *, D x ctnx= — esc 3 a;, D x secx = tana;seca;, Z^csca; = — etna; esc a;, sin a; tana; =- cos a; 60 DIFFERENTIAL CALCULUS. 1 [AKT. 70. ctna; = - seca; = . (3) Given prove 4) Prove tana; 1 cos a; 1 csca; = . sins; vers a; = 1— cos a;, D x versx= sina:. D x log sin x = etna; ; Z> I logcosa; = — tana; ; 2> a: logtana;= secajcsea;; .Djlogctna; = — seca;csca! ; D x log sec x = tana; ; .D a .logcscx= —etna;. Anti- Trigonometric Functions. 70. In trigonometry, the angle which has a sine equal to x is called the inverse sine or the anti-sine of x, and is denoted by the symbol sin" 1 . Hence sin - 'a; means the angle which has x for its sine, and is to be read anti-sine of x. In the same way we speak of anti-cosine, anti-tangent, &c. 71. To differentiate sin -1 a;. Let y = sin _1 x ; then x = siny. Differentiate both members with respect to x. l=eosyD x y; Chap. IV.] TBANSCENDENTAL FUNCTIONS. 61 D x y = -1—. cosy It remains to express cosy in terms of a. s\ny = x, COS 2 y = l—X 2 , cosy = V(l-a^) 5 1 hence D x sin~ 1 x-. V(i-^) Examples. (1) Prove J.ooa-^g- . (2) Atan- 1 a; = — L~ 1 + ar (3) Z» I ctn- 1 a;= - j \+a? (4) Asec- ] a;= — — — — . ^VC* - — 1) (5) 2> x csc _1 a;= — (6) .D.vers- 1 ^: 1 V(2*-a^) 72. The emfa'-, or inverse, notation is not confined to trigono- metric functions. The number which has x for its logarithm is called the anti-logarithm of x, and is denoted by log -1 a; ; and, in general, if x is any function of y, y may be called the corre- sponding anti-function of x, and the relation of y to x will be indicated by the same functional symbol as that which expresses 62 DIFFERENTIAL CALCULUS. [Art. 73. the dependence of x upon y, except that it will be affected with a negative exponent, which, however, must not be confounded with a negative exponent in the algebraic sense. Thus, ifx=fy, we may write y =f~ x x. Any anti-function can be readily differentiated, if the direct function can be differentiated, and by the method we have em- ployed in the case of the anti-trigonometric functions above. Let y =f~ 1 x, then x=fy, differentiate, and 1= T> y fy .D x y. D x y- or D x f~ l x = DJy 1 DJy and it is only necessary to replace y in this result by its value in terms of x. 73. Since, in the formula above, fy = x, we have D x y = — — ; D y x a result so important that it is worth while to establish it by more elementarj- considerations. Suppose x and y connected by any relation, so that either may be regarded as a function of the other. Let Ax and Ay, be cor- responding increments of x and y. Then Ax may be regarded as having produced Ay, or as having been produced by Ay, ac- cording as we regard x or y as the independent variable ; and on either Irypothesis they will approach zero together. By definition, D x y _ limit \^f\ Ax=0 \_AxJ- > Chap. IV.] TBANSCENDENTAL FUNCTIONS. 63 and D v x= limit \— 1. * Ax=Q\_Ay] Ax Ax' Ay limit [ A y~\- 1 1 Ax=0 Ax limit. fAxl li mit. fAxl and mb=\j \_ax j zJa;=0 |_^2/J J?/=0 |_^yj since Ax and Jy approach together. Therefore D x y = — — . Examples. Find D x u in the following cases : — (1) w = sin 2 a;. Ans. D x u = 2smxcosx. (2) « = cosma;. Ans. "D x u = — m&mmx. (3) u = xe c °". Ans. D x u = e C0Ba: (l— aisina;). (4) m= cos(sina;). Ans. D x u= — cos a; sin (sin a;). (5) u= sin(loga;). Ans. D x u = -cos (logx). (6) w = ^2^ - tana; + x. Ans. D x u ^tsu^x (7) u'= (a 2 + a/Han- 1 -. Ans. D x u= 2a;tan~ 1 - -f a. v a a x (8) w = a;sin _1 a;. Ans. D x u=sirr 1 x + V(i-.^) (9) u = sin- 1 £±i . ,4ns. D r u = ^ ' V(2) r V(l-2«-^) 64 DIFFERENTIAL CALCULUS. [Art. 73. (10) u.^—*.-... Ans. ^-^j^- (11) w=sec~ 1 a — 7-. .4ms. D x u = (12) u=sin~ 1 ^J(sm%). Ans. D x u = £ -^(1 + esex). (13) w = tan -1 ;• Ans. D x u = -. 1 — ar 1 -far (14) « = tan-' l( 1 - cosx \, Ans. D x u = h Wl+cosa;/ 2 Chap. V.] INTEGRATION". 65 CHAPTER V. INTEGRATION. 74. We are now able to extend materially our list of formulas for direct integration (Art. 55) , one of which may be obtained from each of the derivative formulas in our last chapter. The following set contains the most important of these : — ZUoga; = - gives f x - = log». x x D x a x = a x logc " f x a x loga = a x . D x e x = e x " f x e x =e x . D x smx = cosx " / I cosa; = sina;. _D I cosa; = — sina: " f x (— sina;) = cos*. DJog sin x = ctn x " /* ctn x = log sin x. J>j.logcosa;= — tana; " f x ( — tana;) = log cos x. D * s ™- lx =^=r^ " x v(I^) = sin " la; - Atan-a;= I A_ .. /.^-W. ^'' ^(iM " / V(2i-^) =VerS " la; - The second, fifth, and seventh in the second group can be written in the more convenient forms, 66 DIFFERENTIAL CALCULUS. [Art. 75. M = ; — ; log a f x svax = —cos*; f x tanx = — logcosa;. 75. When the expression to be integrated does not come under any of the forms in the preceding list, it can often be prepared for integration by a suitable change of variable, the new variable, of course, being a function of the old. This method is called integration by substitution, and is based upon a formula easily deduced from D x (Fy)=D y Fy .D x y ; which gives immediately Fsf=f m (D,Fy.Djf). Let u=D t Fy, then Fy=f s u, and we have f u u =f x (uD x y) ; or, interchanging x and y, f,u=f,(uD,x). [1] For example, required f x {a + bx) n . Let 2 = a + bx, and then /.(a + bx)" =f x z» =/,(*» . D.x) , by [1] ; but x= z "b" a D z x = 1 V hence /«(« + 6*)" = -/,«"= 1 ft ftn + 1, Chap. V.] INTEGRATION. 67 Substituting for z its value, we have f,(a + to) = - -i — ! . JxK ' b n+1 Example. Find/, — — . Ans. hog(a + bx). a + bx o 76. If fx represents a function that can be integrated, f(a+bx) can always be integrated ; for, if z = a + b%, 1 then D„x = - b and X/(a + 6*) =f x fz =f.faD.x = i/Jfc. Examples. Find (1) X sin ax. Ans. cosoa; (2) f x cosax. Ans. -sinaa; (3) f x tan ax. (4) f x cinax. 77. Reared f x ^_^. 1 _1 1 v( "'- l!) *'\H£ T x ^ Let 2 = -, a then a = a«, D,a; = a, 68 l -L a 4}-m DIFFEKENTIAL CALCULUS. a Jx ^(\-z*) a J '^/(l-z 2 ) [Art. 78. D t x 1 . . i X z^f M — — = sm _i 2 = sin~ V(i-« 2 ) Examples. Find <"n c 1 ,* .4ns. -tan -1 - a a {) Jx a 2 +x 2 (' o ^ r X Ans. vers -1 - (X K ' } Jx ^/(2ax-x 2 ) 78. Required f x — -• Let z = x + sj{x i + a 2 ) ; then z — x = - s /(x 2 + a 2 ), z 2 — 2 zx + x 2 = x 2 + a 2 , 22a; = 2 2 — a 2 , 2 2 — a 2 2z ^/(x 2 + a?) = z — x = z- z 2 — a 2 _z 2 + a 2 2z 22 D,X: z 2 + a 2 2z 2 L z fx-z~. — , — f»-z—. — „ D x x V(a^ + a 2 ) 2 2 + a 2 ^V + a 2 =/. 22 2 2 + a 2 z 2 + a 2 2 2 2 =/*- = log2 = log(a; + vap + a 2 ) . Example. Find/, 1 VO^-a 2 ) Ans. log(x + Var* — a 2 ) . Chap. V.] INTEGRATION. 69 79. When the expression to be integrated can be factored, the required integral can often be obtained by the use of a formula deduced from D x (uv) = uD x v + vD x u , which gives uv =f x uD x v +/ x vD x u or f x uD x v = uv—f x vD x u. [1] This method is called integrating by parts. (a) For example, required /Jog*. log* can be regarded as the product of logo; by 1. Call log* = u and 1 = D x v, then D x u = -, x v = x; and we have f x logx =f x l log* =/ x uD x v = uv —f x vD x u = a; log a; —f x - = a; log a; — *. x Example. Find f x x log*. Suggestion : Let log* = u and * = D x v. Ans. -an log* 2 V 2 80. Required f x sin 2 *. Let u = sin* and D x v = sin*, then D x u = cosx, v= — cos*, /,sin 2 * = — sin*eos* +f x cos i x ; 70 DIFFERENTIAL CALCULUS. [Art. 8L but cos 2 a; = 1 — sin 2 *, so f x cos 2 x=X c l— f x sm 2 x = x — / x sin*« and / a an?x = x — sin&cosa;— - / x sin 2 a;. 2f x siv?x = x — sin a; cos*. Xsin 2 a; = \ {x — sinajcosas). Examples. (1) Find,/ x cos 2 a;. Ans. - (x + sin x cos a;). It (2) f x aiaxcoax. Ans. . , & 81. Very often both methods described above are required in the same integration. (a) Required f x sivr 1 x. Let sin _1 a; = y, then x=smy; D„x = cosy, J x si.vr l x=f x y =f v ycoay Let u = y and D y v = cosy ; then D y u = l, v = sin y, and f s ycoBy=ysmy—f l amy=ysmy+cosy=xsm~ 1 x+yf(\—x 2 ). Any inverse or anti-function can be integrated by this method if the direct function is integrable. (6) Thus, /J- 1 !»=f x y=f s yD,fy = yfy-f v fy where y=f~ 1 x. Chap. V.] INTEGRATION. 71 Examples. (1) FindXcos-'a;. Ans. xcog- 1 x — - s l(l—x*). (2) /.tan -1 *. Ans. £Ctan _1 a; — -log(l-j-a^). (3) f x vers~ 1 x. Ans. (x — l)vers~ 1 x + ^/(2x—x 2 ). 82. Sometimes an algebraic transformation, either alone or in combination with the preceding methods, is useful. (a) Required f x - -. x 2 — a? _i_ = W_i L_\ 3?— a 2 2a\x — a x + a/ and, by Art. 75 (Ex.), (6) Required f x A(^\. V\i-x) V(i- 1 . + ■ • x*) V(l-a^) V(l-^) /•I • -l f, — = sin l x. f x can be readily obtained by substituting y = (l — x s ), and is — V(l — ^) i hence ^-J(fz|) = sin_la; ~ V(l " «^) • (c) Required f x \j (a 2 — ic 2 ) . ,, , ^ _ « 2 — z 2 a 2 a? v - ; ~V(« 2 -* 2 ) - V(« 2 - a;2 ) V(« 2 -^)' 72 DIFFERENTIAL CALCULUS. [Art. 83. a,- 2 and Mp-^S.-^-/.--—, whence f,y/(a? -a?)=a? sin" 1 * _/. f , by Art. 77 i ft "\J [CI ~~ CuJ but f.y/(<* - a?) = Xy/(<* -<*)+/. V(« 2 -^) 6y integration by parts, if we let u = -J(a 2 — x 2 ) and D^v = 1. Adding our two equations, we have 2/xV(a 2 - *?) = ^VCa 2 - 3 s ) + a'sin- 1 - ; and •XV(« 2 - a;2 ) = |uVa 2 -« 2 + a 2 sin- 1 ^Y Examples. Find (1) /.V(^ + o«). 1 ^ ws - g O V (^ + « 2 ) + « 2 log(a; + Va* + a 2 ) J , (2) XVC^-a 2 ). 1 21 ^ ns - " [xy/{a? - a 2 ) - a 2 log(a; -f- V^-a 2 ) J, Applications. 83. To find the area of a segment of a circle. Let the equation of the circle be x 2 + y* = a 2 , and let the required segment be cut off by the double ordinates through (x ,y ) and (x,y) . Then the required area A=2f x y + 0. Chap. V.] INTEGRATION. 73 From the equation of the circle, y = ^j(a 2 -x 2 ), hence A=2/ x - s /(a 2 -x 2 )+C ; and therefore, bj- Art. 82 (c) , A = x^/(a 2 - o?) + a 2 sin- 1 - +C. As the area is measured from the ordinate y to the ordinate y, A = when x = x ; therefore = a^VO 2 -^ 2 ) + ^sin" 1 - + 0, O = — x V(« 2 — x 2 ) — a* sin -1 -A and we have A = x V(a 2 - a; 2 ) + a 2 sin-^ - a;„ V(a 2 -»o 2 ) - a 2 sin" 1 - If a? = 0, and the segment begins with the axis ofY, x A = x y/ (a? — x 2 ) + a 2 sin -1 — If, at the same time, x= a, the segment becomes a semicircle, and A = a y/(a? — a 2 ) + a 2 sin -1 - = — • The area of the whole circle is na 2 . 74 DIFFERENTIAL CALCULUS. [Art. 84. Examples. (1) Show that, in the case of an ellipse, 3? It 1 a 2 ^ 6- ' A = l the area of a segment beginning with any ordinate y is x V(« 2 - x 2 ) + ^sin- 1 - - x J(a 2 - x a 2 )~ a^in" 1 ^ a a That if the segment begins with the minor axis, x-sj{a i — x i ) + a 2 svcr l - . A= h - a That the area of the whole ellipse is xab. (2) The area of a segment of the hyperbola cr b- is A = - OVC^ 2 - « 2 ) ~ a 2 log(a; + V^T^ 8 ) - x V (a; 2 - a 2 ) 4- u 2 log(a: + Va; 2 — o 2 )]. If x = a, and the segment begins at the vertex, A =-[a;V(a! 2 - a 2 )-a z log(a;+V^^a 2 ) + a 2 loga]. 84. To find the length of any arc of a circle, the coordinates of its extremities being (x„,y ) and (x,y) . By Art. 52, s =/.V[l + (A*/) 2 ]- From the equation of the circle, a! 2 + y 2 =a 2 , Chap. V.] INTEGRATION. 75 we have 2x+2yD x y = 0, y y r When x=Xq, s = ; hence = a sin -1 - + C, G = — a sin 1 - , and s=«(sin _1 sin -1 — V a a . If a; = 0, and £7ie arc ;s measured from the highest point of the circle, s=a sin -1 -- a If the arc is a quadrant, x = a, s = asin l (l) = — , and the whole circumference = 2 -a. 85. To find the length of an arc of the parabola y 2 = 2mx. We have %yD x y = 2 m ; r, m D x y = —; y V[l + {D x yy] =y j(^+^ = I V(™ 2 + f) ; 76 DIFFERENTIAL CALCULUS. [Art. 85. »=/. V(^ 2 + 2/ 2 ) D.x = 1 iV(m 2 + f) 2>,« .2/ 2/ i>x2/ ™ by Art. 73 ; s = — /, Vm 2 + f = -L [y Vm 2 + , which is equal to - — - or Jr. The mean curvature is —I, and the actual curvature at P is As limit Js=0 At As = D,r. 87. To find D s r in any particular example, we must, in theory, begin by expressing t in terms of s by the aid of our old relations tanr =D x y, D x s = ^[\ + {D z y)^, 78 DIFFERENTIAL CALCULUS. [Art. 88. together with the equation of the given curve ; but, in practice, this part of the work may be avoided. By the aid of the rela- tions just referred to, r and s may be expressed in terms of a; ; and, consequently, we may regard them as functions of x, and can obtain their derivatives with respect to x ; and then the de- rivative of either with respect to the other may be found by the following principle. If y is a function of x, and 2 is a function of x, Dz=^. For limit ["ifl D x z Ax=0 \_Axj D 'V limit Ax- limit Ax=0 Az Ax Ax nit \*£\ =0 \_Axj limit [*f\=D v z; for Ax, Ay, and Az approach zero simultaneously. _ limit Ax=0 [1] 89. We have thus, if x represents the curvature at any point of the curve, Since but and D x s " tanr = D x y, sec?rD x T = D x 2 y, Dr- D *y- JJ x T — 5— > se^r = 1 + tan 2 r = 1 + (2) x y) s , D*r = D x *y i + (A?) 8 '' Chap. VI.] CURVATURE. 79 and, as D x s = V[l + (Ay)*], we have B?y ±[\ + {D x yy]l Either the positive or the negative value might be chosen as the normal one. For reasons that will be evident hereafter, it is customary to use the negative one ; and we have \l+(D x yf]l (a) For example, let it be required to find the curvature of a straigM line Ax +By + C=0 at any point. Differentiating with respect to x, we have A+BD x y = 0; 71 A D - y =-B' l + (D x y) - 2 _ A*+B\ B* ' D 'y ° - = 0; [\ + {D x yf]l /A*+B*\i B 2 a result which might have been anticipated (b) The curvature of a circle, x*-\-y 2 = a 2 . 2x.+ 2yD x y=Q; D x y=--\ y 80 DIFFERENTIAL CALCULUS. [Art. 89. a? _ y-xD x y V y x° + tf a? T f Hence the curvature of a circle is the same at every point, and is equal to the reciprocal of the radius. If a = l, and the unit of curvature is the curvature of the circle whose radius is unity. (c) The curvature of a parabola, y 2 = 2 mx. 2yD t y=2m; y D*y=-™ 2 D z y=-%; y r l + {Dx y)>="l+t; _ m 2 _ (m 2 + y 2 )i _ m 2 , ~ f^ f ~(m 2 + 2/ 2 )i ; and is a function of y, one of the coordinates of the point con- Chap. VI.] CTJKVATTJBE. 81 sidered. From the form of /., it is obvious that the curvature is greatest when y = ; that is, at the vertex of the curve ; that it decreases as y in- creases or decreases, and that it has equal values for values of y which are equal with opposite signs. Examples. (1) Required the curvature of the ellipse * + t = l Ans. at any point. a 4 6 4 (ftV + aV)* (2) Of the hyperbola _-£ = l. a 1 V Ans. a 4 & 4 (6V + aV)5 (3) Of the equilateral hyperbola a 2 " 2 Ans. a 2 O^ + y 2 )* Osculating Circle. 90. As the curvature of a circle has been found to be the reciprocal of its radius, a circle may be drawn which shall have any curvature required. A circle tangent to a curve at any point, and having the same curvature as the curve at that point, is called the osculating circle of the curve at the point in question. Its radius is called the radius of curvature of the curve at the point, and its centre is called the centre of curvature. From the definition of the radius of curvature, it is obviously 82 DIFFERENTIAL CALCULUS. [Art. 91. normal to the curve, and its length is the reciprocal of the curva- ture at the point. If p represents the radius of curvature, we have P = Of course, p is generally a function of the coordinates of the point of the curve, and changes its length as the position of the point in question is changed. Evolutes. 91. The locus of the centre of curvature of a given curve is the evolute of the curve. Problem. To find the equation of the evolute of a given curve y=fx. LetP, coordinates (x,y) , be any point of the curve, and P\ (x',y') the corresponding point of the evolute ; v the angle made by the normal with the axis of a;, and /> the radius of curvature at P. p and t can be found from the equation of the curve, and \ = t - 90°. We see from the figure, that X'=X — (OCOSX, y'=y — painv: Chap. VI. 1 CTTEVATTJEB. 83 p and v can be expressed in terms of x and y ; and then, with the given equation, y=fx, we shall have three equations connecting the four variables, x, y, x', and y'. We can eliminate x and y, and so obtain a single equation connecting x' and y', the variable coordinates of any point on the evolute ; and this will be the equation required. 92. For example : Let us find the evolute of the parabola y 2 =2 mx. tan- =D x y = — ; V tanv = tan (r — 90°) = — cotr = — 1. ; m sec 2 * = 1 + tan 2 v = — ^- ; . m cosv= ± V(m 2 + 2/ 2 )' sin v = ± . V(m 2 + 3/ 2 ) Since v is given by its tangent, it may always be taken less than 180° ; therefore we may take the positive value of sinv, and in that case, as tanu is negative, we must take cosv with the nega- tive sign : we have then V sm "= It 2i a\ ' V(m 2 + 2/ 2 ) — m cosv = ■ V(m 2 + 2/ 2 ) We have seen, Art. 89 (c) , that — m - (m 2 + y 2 )t , 84 DIFFERENTIAL CALCULUS. TArt. 92. hence />=- — „ ' ; mr x'= x -\ > m , (m 2 + V 2 ) V y =y- ± — / ,!> ; mr and these equations, together with y 2 —2 mx, are the equations of the evolute. Reducing, we have x' = m + 3x; whence X: X — m 3 ■ y' = - 2 ' mr and whence V=— (™ 2 y')'- Substituting in the equation of the parabola, we have (m?y')i= (x'—m). o Reducing, m 4 y' 2 = — m s (x'— m) 3 , y'* = -2-(x'-m) 3 ; 27 m or, dropping accents, y 2 = (x — m) 3 , * 27m v ' the required evolute ; a semi-cubical parabola. Chap. VI.] CURVATURE. 85 93. By expressing p and * in terms of x and y in the general equations of the evolute of y =fx, we can throw these equations into a rather more convenient form. We have the values P = _ £!+_( A%V) 2 ] 3 and Reducing D?y tanr = ~-D x y, tanj> = 1 D x y coty = -D x y. 1 X = X ■ [1 + (A2/) 2 ]*' D x y \\ + (D x yr-]V [\ + {D x yy]i D x y D*y [l + (D x yy-]V 2/ = y + — — ' J Z), 2 ]/ [l + (A2/) 2 ]^ y = y + i + (A.v)' u:-y [i] Example. Required the evolute of a circle. -4ws. *'= 0, y'= 0. 94. To find the evolute of an ellipse, a 2 "^& 2 ' 86 DIFFERENTIAL CALCULUS. [Akt. 94- ri, b* ay ,_ b 2 x b 4 x 2 + a 4 y 2 f « 2 ?A_ x(b 4 x t + a*f) b 4 x* + a 4 y 2 ( a 2 f\_ y(b 4 x 2 + a 4 y 2 ) ,r> f V b * J y=y+ ,;., [--w)=y- - a > b * o ./* Since -„ + £ ■ = 1 , a- &- tfx 2 + « 2 */ 2 = " 2 & 2 ; aV = a 2 & 2 (a 2 -.i: 2 ) ; and i 4 .r 2 =« 2 ?> 2 (?> 2 -i/ 2 ). fc 4 ^ + a 4 ;/ 2 = 6 2 (a 4 - a 2 ^ + & 2 a: 2 ) , or ar(b 4 -b 2 y 2 + a 2 y 2 ). x(a 4 -a?x 2 + b 2 x 2 ) a 2 -& 2 , y-n b 4 - b t y > O'X \i a- — 67 y \a 2 -b 2 ) Substituting in -. + \ = 1 , a 2 ft 2 Chap. VI.] CURVATURE. 87 we have ( ax ' V ■ / W V_i . Va 2 -& 2 y ^tf-67 ' or, dropping accents, (aa;)s + (ty)*=(a 2 -& 2 )S. ' Example. Find the evolute of the hyperbola — — U- = i a 2 6 2 Ans. (ax)S-(by)i = (a 2 +b !! )i. Properties of the Evolute. 95. "We have defined the evolute as the locus of the centre of curvature of the curve. It is also the envelop of the normals of the given curve, as may be readily shown ; that is, every normal to the curve is tangent to the evolute. Let v be the in- clination of the normal at the point (x,y) of the given curve to the axis of X, and r' the inclination of the tangent at the corre- sponding point (x',y') of the evolute. We have seen already that the normal at (x.y) passes through (x',y'), so it is only necessary to prove that -'= v. But tanr'= D^y' and tanv=— — — . D x y Hence we must show that D^ y'= — — — . D x y By Art. 88, D*>y'= ^', JJ x x since x' and y' may both be regarded as functions of x. x - x DFy ' DIFFERENTIAL CALCULUS. l + (^) 2 . [Art. 96. y = y + Dfy D v _i (A 8 ?/) 2 + HD*yY (A 2 y) 2 - D*y [i+ (Ay) 2 ] A 3 y (A 2 y) 2 Ay 1 3 D x y(D*yf - [1 + ( Ay) 2 ] A 3 y j , (A 2 */) 2 n,Ln, i •2P,.tWy)'-[i+(P.») r ]J,'y *y- *y-t- Wy) , _ 3J,y(Q. , j) , -[i + (Ay)W» ■ (A 2 y) 2 A* Ay Q.B.D. 96. A second important property of the evolute is that the length of any arc of the evolute is the difference, between the lengths of the radii of curvature of the given curve which pass through the extremities of the arc in question. Let (a>i,yi) and (a; 2 ',?/ 2 ') be the extremities of any arc of the evolute ; p^ and p s the radii of curvature drawn from these points to the curve ; s^ the length of the arc of the evolute measured from some fixed point on the evolute to (#i',2/i') ; and s 2 ' the length of an are measured from the same fixed point to (x 2 ', y 2 ') • Then we wish to prove that or or or ap-. or As'=l, where s' must be regarded as a function of p. S 2 — Si = /'s- - Pi Js'=Jp, Jp limit J P =0 _jp = 1 Chap. VI.] CURVATURE. 89 But D s'=M and D x s'= P^ = D x . s'. D x x'. D x . x i i> x y D x ,y'=--±, by Art. 95; hence D x ,s'= -L [l + (Zy/) 2 ]*. JJxV \ n s r_ [l + lP.y) 5 ]* \3D x ij(DJyy- [l+(D«y)*] A 3 yj by Art. 95. [1 + (D^) 2 ]?. /' = £, 2 2/ xP ~' (WW ' D„S'=^ = 1. Q.E.D. ■DxC 97. These two properties enable us to regard any curve as traced by the extremity of a stretched string unwound from the evolute, the string being always tangent to the evolute, and its free portion at any instant being the radius of curvature of the curve at the point traced at that instant. From this point of view, the curve itself is called the involute. 90 DIFFERENTIAL CALCULUS. [Art. 98. CHAPTER VII. SPECIAL EXAMPLES AND APPLICATIONS. The Cycloid. 98. The cycloid, a plane curve possessing very remarkable geometrical and mechanical properties, was first studied just before the invention of the Calculus, and has always been a favorite with mathematicians. It is the curve described in space by a fixed point in the rim of a wheel as the wheel rolls along in a straight line ; or, more strictly, it is the curve described by any fixed point in the cir- cumference of a circle, as the circle, keeping always in the same plane, rolls without sliding along a fixed straight line. The rolling circle is called the generating circle, and the fixed point the generating point, of the cycloid. The curve will evidently consist of an indefinite number of equal arches, and can be cut by a straight line in an unlimited number of points. Its equation, then, cannot be of a finite degree, and so cannot be an algebraic equation. The curve is a transcendeyital, as distinguished from an algebraic, curve. 99-. As the arches are all alike, it will do to consider a single one. Its base is obviously equal to the circumference, and its height to the diameter, of the generating circle, and its right and left hand halves are symmetrical. We can get its equation most easily with the aid of an auxil- iary angle. Take as axes the base of the cycloid, and a per* pendicular to the base through the lowest position of the generat- ing point, and represent by the angle made by the radius drawn to the generating point at any instant, with the radius drawn to the lowest point of the generating circle. The arc Chap. VII.] SPECIAL EXAMPLES AND APPLICATIONS. 91 joining the two points just mentioned is a0, by Art. 66, if a is the radius of the circle ; and this is therefore the length OT. If x and y are the coordinates of P, any point on the cycloid, x = aO — a sin y = a — a cos U) and these may be taken as the equations of the cycloid. Of course, may be eliminated between these equations, and a single equation obtained, containing x and y as the only varia- bles. We get COS0 = a — y •> a 1 — cos = - = vers 0-, hence y d = vers -1 - > a sin^ = V( 1 - cos2y )= ±aV( 2a 2/-2/ 2 ), and y x = a vers" 1 - ^ V( 2 a V ~ 2/ 2 ) > (B) where the upper sign before the radical is to be used for points corresponding to values of 0<-, that is, for points on the first half of the arch, and the lower sign for points on the second half of the curve. 92 DLFFEKENTIAL CALCULUS. [Art. 100. Examples. (1) Discuss completely the form of the cycloid from equations (^1), supposing to increase from to 2 -a. (2) Discuss the form of the cycloid from equation (J?), sup- posing y to increase from to 2 a. 100. If our axes are lines parallel and perpendicular to the base through the highest point of the curve, the equations have a slightly different form. Let be measured from the highest point of the generating circle. and OT=AB = ad Examples. (1) Obtain equations (C) from equations (^4) by transform- ation of coordinates, noting that the formulas required are x= a- + x', y=2a+y', = T!+0'. (2) Eliminate aD<3 obtain a single equation for the cycloid referred to its vertex as origin. Chap. VII.] SPECIAL EXAMPLES AND APPLICATIONS. 93 101 . The properties of the curve can be investigated from the equations involving or from the single equation. In the text we shall employ the former. We ought to be able to determine (1) the direction of the tangent and normal at any point of the curve ; (2) the equations of tangent and normal ; (3) the lengths of tangent, normal, subtangent, and subnormal ; (4) the curva- ture of the cycloid at any point ; (5) the evolute ; (6) the length of an arc of the curve ; (7) the area of a segment of the curve. (1) 102. x = all — a sin ft, y = a — acosB, Unr = D x y = ^ = °» in " = si »" = ^"^ = cotj , D g x a — neosfl 1— costf 2sin 2 | tan ■< = — cot r = — tan|. Since, as we have seen in Art. 99, sintf=- V(2«2/ — y 2 ) a and 1 — costf = ^, a tan t can be written =.. ( — ■ \\y V and tan/= — V(2a?/-2/ 2 ) Since tan>= — tan§, by trigonometry. In the figure (see next page) , PTO being formed by a tangent and a chord, is measured by half the arc PT, and therefore is equal to \. PTA, then, is equal to v, and the line PT is a normal. Hence the normal at any point on the cycloid passes through the lowest point of the generating circle. The tangent must, therefore, pass through the highest point of the generating circle. 94 DIFFERENTIAL CALCULUS. [Art. 103. Examples. ( 1 ) At what point of the curve is the tangent parallel to the base of the cycloid ? Perpendicular to the base ? Where does it make an angle of 45° with the base ? (2) Obtain the values of tan r and tan* from equation (B). (2) 103. The equation of the tangent at the point (x ,y ) is by Art. 28, y — y = cot f> (x — x ) , t> r y- y»=JC— - u( x - x <>) ; of the normal, is y — y„ = — tan |° (x — x ) , or y -y = ¥o V( 2 «2/o-#o 2 ) (a; — a; ) . Example. Show, from the equation of the normal, that it passes through the point (a0,O), the lowest point of the generating circle. (3) 104. We have the formulas, t - y D,y n, = yD t y, Chap. VII.] SPECIAL EXAMPLES AND APPLICATIONS. 95 D x y n = y V[l + {D,y)']; (Art. 28) ; a(l-cosO) 2asin 2 g hence t t = -i '- = — ■ = 2 a sin 2 £tanf • cot | cot I n x =a(l— cos0)cot| = 2asin 2 Scot.i = 2asinfcos|-= asia.8, t=2a sin 2 ftan| csef = 2 a sinf tanf . n = 2 a sin 2 ! cscf = 2 a sin \. Since D z!f = ^-iy the value of n may be expressed, n = ^/(2ay). (1) 105. * = ~ D *y • (Art. 89). W {l + {D x yy]V K 2 fl,(D. y ) = Zi^l = _ 1 csc fl * 2«sin 2 | 4a [l+(i) x y) 2 ]5 = csc 3 |, hence J_ 4a v and / o = I = 4asinf=2?i=2V(2a2/) ; and the radius of curvature at any point is equal to twice the normal drawn at the point. 96 DIFFERENTIAL CALCULUS. TArt. 106- Examples. ( 1 ) Find at what points of the curve the curvature is great- est ; at what least. (2) Obtain the expression for the curvature from the equa- tion (B). (5) 106. The equations of the evolute of a curve are y = y + i+(Ay) 2 D?y (Art. 93 [1]). Here cot|!csc 2 f x'= ad — asmd = a$ — asintf + 4asin|cosg Ta csc^ = a0 — asiaO+ 2asin0, = aO + asinO. csc'f y'= a — a cos -\ = a — a cos — 4 a sin 2 1 — t~ csc 4 ! 4a z = a(l — cos 9)— 2a(l— cos0) = — a + acosd; and we have, as the equations of the evolute, x'= a9 + asm0 1 y'= — a + acosO J but these (Art. 100) are the equations of an equal cycloid re- ferred to the tangent and normal at the vertex as axes. The Chap. VII.] SPECIAL EXAMPLES AND APPLICATIONS. 97 cycloid and its evolute would be situated as indicated by the figure. The property of the evolute established in Art. 96 enables us to obtain easily the length of the arc of an arch of the cycloid The length of the half-arch of the evolute is the difference be- tween the radii of curvature at the highest and the lowest points of the given curve ; that is, [/)]«=!- [|o]0=o= 4asin^— 4asin0 = 4a, and S, the whole arc, = 8a. (6) 107. The length of an arc of the cycloid can be found from the formula s = / x [l + (D x y) 2 ]* without using the evolute. We have D x y = cotg, hence s = f x esc % =/ $ esc | D g x : but D g x= 2 a sin 2 1, and s=2a/ 9 sin|. 98 DIFFERENTIAL CALCULUS. [Art. 108, Let z = \ ; then D z = 2, s= 2af g smz = 2af 2 sinzD z = — 4acosz + C, s — — 4 a cos I + C- If we measure the arc from the origin, s must equal when = 0. 0=— 4acosO+C, C=4a, and we have s = 4a(l — cos|). To get the whole arch, let 0=2-, s = 4a(l— cos-) = 8a. (7) 108. For the area of a segment of the arch, we have the formula A=f x y + C. SxV = af x (l—cos0) = af ti {\- cosO)D s x = a%{\ — costf) 2 = a 2 / e (l-2cos0 + cos 2 0) = a 2 (/ 9 l-2/ e cos0+/ e cos 2 6l), f 9 cos0= sinO, f e cos 2 — ^(f) + sm6cose) [see Art. 80, Ex. (1)] ; hence A = a? [8 — 2 sin + $(0 + sintfcos 0)~\ + C. If the segment is measured from the origin, A = when = 0; = a 2 [0-0 + £(0 + 0)] + C and O=0. Chap. VII.] SPECIAL EXAMPLES AND APPLICATIONS. 99 The area of the whole arch is obtained by making 0=2*. A = a?[2n — 2sin2* + J-(2tt + sin27rcos2jr)]= 3wa 2 , so that the area of the arch is three times the area of the gen- erating circle. Example. Find the length of an arc and the area of a segment from the equation (B). 109. If the generating circle rolls on the circumference of a fixed circle, instead of on a fixed line, the curve generated is called an epicycloid, if the rolling circle and the fixed circle are tangent externally, a hypocycloid, if they are tangent internally. The equations of these curves may be readily obtained. Let the figure represent the generation of an epicycloid, P being the generating point and E the starting point. Call AOB, 6 ; and PGA,

= D t s. 111. The acceleration of a moving body at any instant is the rate at which its velocity is changing at that instant. If the velocity is increasing, the acceleration is positive ; if diminish- ing, the acceleration is negative. We shall represent it by a, and it is evidently a function of t. Since the derivative of a function measures the rate at which its value is changing (Art. 38) , we shall have a=D t v=D t 2 s, since v=D,s. For example : in the case of a body falling freely near the sur- face of the earth, we have approximately the law s = l6t*. Here v=D t s=32t, and a=D t v=D, 2 s = 32, and the acceleration is constant and is equal to 32 feet a second ; that is, the velocity of the fall at any instant is 32 feet a second greater than it was a second before. The relations v=D,s and a=D l v=D, 2 s, Chap. VIII.] PROBLEMS IN MECHANICS. 103 and the corresponding formulas, v=f,* + C, s=f t v+C, obtained by integrating them, are of great importance in prob- lems concerning motion. 112. We shall assume the following principles of mechanics : (1) A force acting on a body in the line of its motion produces an acceleration proportional to the intensity of the force; and this acceleration is taken as the measure of the force. We speak of a force as a force producing an acceleration of so many feet a second; or, more briefly, as a force of so many feet a second. (2) The effect of a force in producing acceleration in any direc- tion not its own, is the product of the magnitude of the force by the cosine of the angle between the two directions; or, in other words, it is the projection of the line representing the force in direction and intensity upon the line of the direction in question. Problem. 113. The force exerted by the earth's attraction upon any particle of matter is constant at any given part of the earth's surface, and is nearly equal to 32 feet a second. Let g repre- sent the exact value of this force at any given point of the earth's surface, required the velocity of a falling body at the end of t seconds, and the distance fallen in t seconds. Here a is constant and equal to g. V=/,a=f,g = gt + C. If the body falls from rest, its velocity is when tis ; = gx0+O, C=0; and v = gt. 104 DIFFERENTIAL CALCULUS. [Art. 114. s =/,v =f,gt = gf,t = £gt* + C. * When £ is 0, the distance fallen must be ; = tgxO+C, C=0, and s = %gf. If the body, instead of being dropped, had started with an initial velocity i> , — for example, if it had been fired from a gun directly down or directly up, — we should have found a different value for C in the expression for the velocity, v = gt + G; for now, when t = 0, v = v ; hence v =g xO+O, C=v , and v = gt + v . s=f t v=f,(gt + v li ) = £gP + v t + C; but as s = when t = 0, C=0, and s = £gt 2 + v t. 114. The equation a = g or D?s = g can be integrated by a second method of considerable interest and generality. Multiply both members by 2D,s. 2D t sD t 2 s=2gD,s; Chap. VIII.] PROBLEMS IN MECHANICS. 105 but 2D t sD?s = D t (D t s) 2 ; hence f t 2D t sD t *s = (D,sy, and we have (D,sy=2gf t D l s = 2gs + C or v 2 =2gs + C. In the case of a falling bod}', when t = 0, v = 0, and s = ; hence C = and w 2 =2grs, v = y/(2ga), [1] or D ( s = 7(2^8). We cannot integrate directly here, for the first member is a function of t and the second member a function of s ; but since As = — -, by Art. 73, DJ=. l ! - V(2/(£ "We see the two velocities are identical ; that is, the velocity acquired by a body descending an inclined plane is precisely what it would have acquired falling through the vertical distance it has actually descended. PEOBLEMS IN MECHANICS. 107 is the mean velocity of the body during its descent, and Chap. VIII.] s H(¥) y_ \gy for the inclined plane, for the falling body. Hence the mean velocity of a body descending an inclined plane is equal to the mean velocity of a body which has fallen freely the same vertical distance. 116. Let the figure represent a vertical circle. The time of descent of a body sliding down any chord is by Art. 115. If a is the radius, s.sec(90°-c-)=2a 2ssec(90°-?)' ; -sl(£ and which is also the time a body would require to fall vertically the distance 2a. Therefore, the time of descent down a chord of a vertical circle from the highest point of the circle to any point of the circumference is constant, and is equal to the time it would take the body to fall from the highest to the lowest. point of the same circle. 108 DIFFERENTIAL CALCULUS. [Art. 117. Example. Show that the time of descent down a chord from any point of a vertical circle to the lowest point of the circle is constant. Problem. 117. To find the velocity acquired by a body falling from a distance toward the earth under the influence of the earth's attraction. Here we cannot regard the attracting force as constant, as we do in dealing with small distances near the surface of the earth, but must take it as inversely proportional to the square of the distance of the body from the centre of the earth. Let E be the radius of the earth ; r the distance from the centre of the earth to the point at which the bod} r started ; r the distance from the centre to the position of the falling body when the time t has elapsed. Let g be the force of the attraction of the earth at the earth's surface, and / the force exerted at P. f B 2 Then we have J - = — ? 9 r or f= y — r =a. r s, the distance fallen in the time t , equals r — r. D,s= — D t r = v, D?s=-D t 2 r=a; hence — D?r = ?—-., r Chap. VIII.] PROBLEMS IN MECHANICS. 109 Multiply by 2D t r ; 2D t r D t 2 r = ~ 2 9R 2 D,r^ r 2 Integrate : {D t r)> = -, 2g&f*>£^ -. 2gR*f}- = ^! + C ; ,-< r 2 r and v 2 = !^ 2 + o. ?■ When the body was on the point of starting, its velocity was zero ; hence, when r =r , v = ; 2 ^ 2 ~ and = — [-C, and v 2 =2gR 2 - \r r 0i When the body reaches the surface of the earth, r = R and v ?=2gR2(±-± Th3 greater the value of r in this result — that is, the greater the distance of the starting point from the centre of the earth — the nearer — comes to the value 0, and the nearer i> 2 approaches 2aR? to —2 — or to 2gR. In other words, the limiting value of the R velocity acquired by a bod}- falling from a distance to the surface of the earth under the influence of the earth's attraction, as the distance of the starting point is indefinitely increased, is \J(2gR) . Let us compute roughly the numerical value of this expression. g is about 32 feet per second ; and as we use the foot as a unit in one of our values, we must in all : therefore R must be ex- 110 DIFFERENTIAL CALCULUS. TART. 117. pressed in feet. R is about 4,000 miles, or 21,120,000 feet. V(J2) = 4,600, nearly. V(2fir) = V(64)=8. ->J(2gR) = 36,800 feet, or nearly seven miles ; and our required velocity is nearly seven miles a second ; and neglecting the re- sistance of the air, this is the velocity with which a projectile would have to be thrown from the surface of the earth to prevent its returning. We can easily go on and get an expression for the time of the fall by a second integration. We have (D t rY = 2gB 2 (- - -\= 2gB 2T °~ r ; \r r J r r - d -<=VGpM^) bJ ' Art • ,3 • n — r) \/{r r — r 2 ) ' an expression to which we can apply the method of integration by parts. Let u = r, then D r u = 1 ; 1 and let D r v = VCv-r 2 )' then ■y = vers- 1 ir by Art. 77, Ex. (2), n> f'—Ti ^T ==rvers Avers 1 — ■s/(r t> r — r i ) r r by Art. 79, [lj. Chap. VIII.] PROBLEMS IN MECHANICS. Ill Let „& then D r — -°. 2 f^ers- 1 — = £7; vers-' 1 z = r l [(z - 1) vers" 1 * + V(2z - z 2 )] by Art. 75, [1], and Art. 81, Ex. (3). Replacing z by its value, /..vers- 1 IT = (r - r A vers" 1 ^ + VCv - r 2 ) . '0 V 2/ r + G. Whence -^ Uy 'Jvers -1 — — V(»"o»' _^ ''0 > -r 2 ) When r = r , t = ; hence o=* IfAY^Hc, » d - (aB >|(8p»)K w,rl ^-' r )- 2v(r,r - ,J) } Examples. (1) The mean distance of the moon from the earth being 237,000 miles, find the velocity a body would acquire, and the time it would occupy, in falling from the moon to the earth's surface, neglecting the retarding effect of the moon's attraction. (2) The force of the sun's attraction at its own surface is 905.5 feet; find the velocity a body would acquire, and the time it would occupy, in falling from the earth to the sun. Earth's mean distance = 92,000,000 miles ; sun's diameter = 860,000 miles. 112 DIFFEEENTIAL CALCULUS. [Art. 118. (3) Find the limit of the velocity a body could acquire fall- ing from a distance to the sun. (4) How long would it take Saturn to fall to the sun, Saturn's mean distance being about. 88,0,(300,000 miles? Problem. 118. To find the velocity acquired under the influence of gravity by a body sliding without friction down a given curve, or in any way constrained to move in a fixed curve. Here the effective accelerating force is always tangent to the curve at the point the moving particle has reached. Suppose the origin of coordinates at the starting point, and let the direc- tion downward be the positive direction of the ordinates. Of course, this will amount to changing the sign of D x y ; that is, will make *• the supplement of its usual value. The acceleration a = gcos? = x */) 2 =sec 2 r, 1 i + (Ay) ,= COS'r, [i+(Z>^]i- smT ' Chap. VIII.] PBOBLBMS IN MECHANICS. 113 hence a =D, 2 s = 9_xV . [1 + {D x yyy but [l + (D x yy-\i=D x s; DtS = g-^-z=gD,y. JJ x s Multiply by 2D t s: 2D t sD?s = 2gD.yD t a = 2gD,y. Integrate with respect to t, and v*=(D,sy=2gy + C. If the particle started from rest at 0, v=0 when y = 0, and (7=0, « = V( 2 9'2/) ; but this is precisely the velocity it would have acquired in falling freely through the vertical distance y (Art. 114, [1]). So we are led to the remarkable result, that the velocity of a material par- ticle, sliding without friction down a curve, under the influence of gravity, is the same at any instant as if it had fallen freely to the same vertical distance below the starting point. A special case of this has already been noticed in Art. 115. Example. Prove, from the equation of a circle, and the equation of a chord through its highest point, that the time of descent is inde- pendent of the length of the chord. Problem. 119. To find the time of descent of a particle from any point of the arc of an inverted cycloid to the vertex of the curve. Taking the origin at the vertex of the curve, its equations are 114 DIFFERENTIAL CALCULUS. [ART. 119. x = aO + a sin 8 y = a — a cos j (Art. 100). Let y n be the ordinate of the starting point, and y the ordinate of the point reached after t seconds. Then the vertical distance fallen is y„ — y, and v = y/2g(y„ — y), by Art. 118. D,(s -s) = -D,s = ^2g(y -y) 1 -D,t=- -t=f. V'23(2/ -2/) V'2gr(2/ -2/) '" \/2g{y -y) D s s; £ lS = Vl+(A3/) 2 ; D fl a;=a + acosfl; De2/ = asin0; n ^— I) ex_ 1+cosfl , Dey sms cos0 = a — y Chap. VIII.] PROBLEMS IN MECHANICS. 115 2a — y 1 + cos B = — - — ; a 2 — a 2 + 2 ay — y 2 2ay — y i sin 2 0=l— cos 2 = ^ = — ^2 — ' sind = -^/(2ay-y i ); D 2a-y _ x [(2a~=W _ l f2a-y\ . » ■ yj(2ay-tf) \{2a~y)y \\ j j' 1 + (D,x)* = y'> J) If a by Ait. 77 (2) . When y = y ,t=0; V2^=^) ~\W Jy V(y.y - 2/ 2 ) ~\W veib % ^=/,^^4 = Jn/,-^-^ = J^Vers-^ + a hence = ( _ j vers" 1 (2) + C. vers -1 (2) is the angle which has the cosine — 1, that is, the angle jr. Hence, C= — T, and — i= |(_) [ vers -1 — — 7T ). When the particle reaches the vertex, y=o, vers -1 — = 0, and £=7r 116 DIFFERENTIAL CALCULUS. [Art. 120. As this expression is independent ofy , the ordinate of the starting point, the time of descent to the vertex will be the same for all points of the curve. If a pendulum were made to swing in a cycloid, this time ^^ |( - ) would be one-half the time of a com- plete vibration, which would therefore be independent of the length of the arc. On account of this property, the cycloid is called the tautochrone curve. Example. 120. It is shown in mechanics, that, if the earth were a per- fect and homogeneous sphere, and a cylindrical hole having its axis coincident with a diameter were bored through it, the at- traction exerted on any body within this opening would be pro- portional to its distance from the centre. Find the expression for the velocity of a body at any instant, supposing it to have been dropped into this hole, and the time it would take to reach any given point of its course. Compute (1) its velocity when half-way to the centre ; (2) when at the centre ; (3) the time it would take it to reach the centre, if dropped from the surface ; (4) if dropped from any point below the surface. Given g = 32; i2, the radius of the earth, = 4,000 miles. Chap. IX.] DEVELOPMENT IN SERIES. 117 CHAPTER IX. DEVELOPMENT IN SERIES. 121. A series is a sum composed of an unlimited number of terms which follow one another according to some law. If the terms of a series are real and finite, the sum of the first n terms is a definite value, no matter how great the value of n. If this sum approaches a definite limit as n is indefinitely increased, the series is convergent ; if not, it is divergent. The limit approached by the sum of the first n terms of a convergent series as n in- creases indefinitely, is called the sum of the series, or simply the series. Thus, we may express the result arrived at in Art. 6 by saying the sum of the series 1 + £ + £ + £-(- is 2; or, more briefly, the series 1+J- + I + J + = 2. Example. 122. Take the series \ + x + x i + X i + , ad infinitum . The series is a geometrical progression, and the sum of n terms can be found by the formula s = . J r-1 x n — 1 1 — as™ s = = . X—l 1 — X- Here and the sum of the series = , a definite value, and the series 1—x is, therefore, convergent. 118 DIFFERENTIAL CALCULUS. [Art. 123. If x> 1, x" = oo when n = oo , and the sum increases without limit as the number of terms in- creases indefinitely, and the series is divergent. The series \+x-\-x* + x z + can be obtained from by actual divi- \—x sion, but the fraction and the series are equal only when xl ; but, as we have seen, \-x ' the series in that case has not a definite sum. It is very unsafe to make use of divergent series, or to base any reasoning upon them, for, from their nature, they are wholly indefinite. Con- vergent series, on the other hand, are perfectly definite values. It is easily seen that the sum of the first n terms of a series cannot approach indefinitely a fixed value as n is increased, un- less, as we advance in the series, the terms eventually decrease; or, in other words, unless the ratio of the nth term to the one before it eventually becomes and remains less than unity as n is increased. This, however, affords only a negative test for the convergency of series, as a series may not be convergent even when each term is less than the term before it. 123. The series we have just considered is an example of a series arranged according to the ascending powers of a variable, and such series play an important part in the theory of functions. We are naturally led to the consideration of terms of such a series whenever we attempt to obtain a function from one of its derivatives. Suppose D„ n f(x + h) = z where h is a variable, x a a given value, and z, of course, a func- tion of h. Let p stand for J, &c. , so that /" =J"~'- Then DrV(x + h)=A^ +f h z, where A x is a constant ; i>r 2 /(Zo +h) =A, + A,h +f h *z, Chap. IX.] DEVELOPMENT IN SEEIES. 119 £r 3 /(Zo + h)=A 3 +A 2 h + %A 1 h i +/ s 3 z, DT 4 /(zo + h) =Ai +A s h + %AJ? + ^-Aih? +/»«*, /fa + h) =A n +A n _ 1 h + H-^ ! + 2^- A-.** + ■ Aft"- 1 +/»"*, 2.3 (n-1) and we have a set of terms arranged according to the ascending powers of h. Although, by increasing n indefinitely, we can make the second member above a true series, it does not by any means follow that every function can be developed into such a series. In the first place, it may not be possible to increase n indefinitely in the expression above, as the nth derivative of the function may become at last infinite or discontinuous, so that yi n 2 cannot be dealt with. Next, the series may be a diver- gent series, and then it could not be equal to the definite value f(x + h) . But the result is a remarkable one, and suggests the careful investigation of the development of functions in series. 124. Assuming, for the moment, that/(a: + /i) can be devel- oped into a convergent series arranged according to the ascend- ing powers of h, let us see what the coefficients of the series must be. Let f(x + h) = A +A 1 h +A 2 h 2 +A s h s + +A n h" + The function and the series are both functions of h, and may be differentiated relatively to h. D h f{x + h) =A 1 +2A 2 h + 3 A 3 h 2 +i A i 7v s + +n A n h n - 1 + We shall find it convenient to adopt the following notation : Let f'x stand forD x fx, f'x for DJ>fx, f™x for D x n fx. Let 120 DIFFERENTIAL CALCULUS. [Art. 124. / ,a; 0)/ (n) *o stand for the results obtained by substituting x for x in f'x,f in) x, where x may be a single term or any complicated function. Let n ! (which is to be read n admiration) stand for Ix2x3x4x X n. Call (x + 7t) = x, then D h f(x + h) = DJx = D x fxD h x =f'(x + h) D h (x + h) =f'(x + h). In like manner, we could show that D. , /(3b + A)=/ w (3b + *)i A" (a* + h) =/"" (x + h) , &c. /'(a; +/i)= A +2^+--+nA^ -1 +-- ; f"(x +h)= 2A 2 +....+n(n-l)4,A" -, +-- 5 /(»'(»o+/i)= «!i„+(n+l)n...'.2A +1 H- Let 7i = in these equations, and we have /* = A, /'"x„=3M 3 , fx =A u / Iv a; =4!A, /%=2^ 2 , /<»>*„ = n! ^„, hence -4 =/a; , A a = — /'" a*,, o ! 4 ! A 2 = ±f»x , A n =l-fWx , 2 n! Chap. IX.] DEVELOPMENT IN SERIES. 121 I, 2 lfi 7i 4 and /(<*„ + h) =fx +hf'x a + '^f"x + ^/'"x + ^/"x, + + -.f (n) x + , [1] n ! "- J if/C^o + 'O can be developed. Examples. (1) To develop (a + 7i)". Call (a + h) = x, then fx = x n , f'x = nx n_1 , f"x = n(n-l)x n -\ f"'x = n(n-l)(n-2)x n - s , &c. fa = a", fa^na"- 1 , /"a = n(w-l)a"- 2 , /'"a = w(w-l)(n-2)a"- 3 , &c. (a + h) n = a" + na"- 1 /^ + M" -1 ) a"" 2 ^ + W (n-l)(«-2) a n- 3/l 3 + t if (a -f h) n caw 6e developed. (2) To develop sin A. sinA = sin(0 + ft). Let x = 4- h. 122 DIFFERENTIAL CALCULUS. [Art. 125. /a; = sina;, yt) = sinO = 0, f'x=cosx, /'0 = cosO = l, f'x = - sina;, /"O = - sinO = 0, /"'»=- cosas, /"'0=-cos0=-l, f"x = sina;, &c. f"0 = sinO = 0, &c. sin(0 + 7i) = + 7i + 0.— - — + 0. — + — + , 2! 3! 4! 5! . , , h 3 , 7i 5 h T , h 9 , sinft = /t h , 3! 5! 7! 9! (/"sinh can be developed. (3) Assuming that cos7i can be developed, determine the series. 125. Let us find what error we are liable to commit if we take f(x + h) equal to n+1 terms of the series (Art. 124, [1]). Let R be the difference between f(x + h) and the sum of the first n+1 terms ; then f(x + ft) =fx + 7if'x + ¥-f'x + + —f M x +R, 2 ! n ! and we want to find the value of R. Lemma. 126. If a continuous function becomes equal to zero for two different values of the variable, there must be some value of the variable between the two for which the derivative of the function will equal zero. For, in passing from the first zero value to the second, the function must first increase and then decrease as the variable Chap. IX.] DEVELOPMENT IN SERIES. 123 increases, or first decrease and then increase. If it does the first, the derivative must at some point change from a positive to a negative value ; if the second, the derivative must change from a negative to a positive value, and in so doing it must, in either case, pass through the value zero. 127. To determine S. Let P = E- (n+1)!' 7,» + l then R = — P, (n + 1)! and f( Xo + h) = f x ab+ *" +1 P 2 ! n ! (?i + l) ! or f(x +h)-/x -hf'x -ff"x -^>o a . o __^Lp = . 2 ! n\ (n + 1) ! CaU fa + h) = X; then h= X — x , and we have 1 ! n. ! _ (X-Q" +1 p=0 J-, («+l)! L J Form arbitrarily the same function of a variable 2 that the first member of [1] is of x , and call it Fz. Fz=fX-fz- ( X - z) fz- i^L^f'z- _ (X-zy fM0 _ (X-s)" +1 p n\ J (n + 1)! 124 DIFFERENTIAL CALCULUS. [Art. 127. If z = x„, Fz becomes identical with, the first member of [1], and therefore = 0. If z=X,Fz=0, since each term disappears from containing a zero factor ; and we have succeeded in forming a function of z, which becomes equal to zero for two values, x and X of z. If Fz is continuous, there must be some value of z between x„ and X for which F'z = 0. Differentiating Fz, and remembering that P is constant, we have F'z = 0-f'z+fe- ( X ~ z ) f'z + (X ~ z) f'z - ( X - Z ^f'"z j -tv 1 ! 1 ! 2 ! ^ 2! y ^ («-l) ! _ (X-.)" (X-s)" p. n ! nl All the terms but the last two destroy one another, and Fz = _ ( J -')> -">» + ( Z ~ 8 )" P. ?i ! n ! But this must be equal to zero for some value of z between x Q and X. Such a value can be represented by x + 0{X— x a ) where is some positive fraction less than 1, i.e., O<0<1. Substituting this value, we have = _l X - X 0- (^-^)Tf(n + l)^ Xo+0 ^X- Xo )^ [X-x B -0(X-x )¥ p "•" n! "Whence P=/< n+a) [> + tf(X-a: )]. X — x = /*, P=/<« +1) (sb+«*), Chap. IX.] DEVELOPMENT IN SEKIES. 125 h h 2 whence f(x + h) =fx„ + — /'«„ + — f"x Q + + -.P n) xo+ r^jV-,/ ( " +1) to + eh), nl (n +1) ! where all that we know about is that it lies between and 1 . 128. The expression for the last term may be obtained in a different form by assuming at the start R = hP instead of i? = P. (n+1)! Making this assumption, show that /Oo + h) =fx + hf'x, + ^f"x + + —. f w *o + h ^ , > /<" + » (xo + Oh) . nl nl Since in each of these formulas x was any given value, we can represent it in the result just as well by a;, and the formulas may be written f(x + h)=fx+hf'x + ^f"x + + -.f ln, x + 7 ^-.f i " +1 \x + 0h); [1] n\ (n+1) ! f(x + h) =fx + hf'x + J £f"x + + £./(»> a; +_ V- ) fi»+»( x + 0h). [2] n ! n I and these formulas are known as Taylors Theorem. 126 DIFFERENTIAL CALCULUS. L A « T - 129 - Example. 129. To develop (2+1)*. Let us see what error we are liable to if we stop at the second term. fx=tf, /'"a;=24a:, /'* = 4ar J , . f"x =24, f"x=l2x 2 , fx = 0. (2+l) 4 = 2 4 +1.4.2 3 + — 12(2 + 0) s . If 0=0, the last term is 24. If = 1, the last term is 54. Hence, if we stop at the second term, our error lies between 24 and 54. In point of fact, it is 33. Suppose we stop with the third term. (2 + l) 4 =2 4 + 1.4.2 3 + |!l2.2 2 + |l 24(2+0). 2i \ 1 If 0=0, the last term is 8. If = 1, the last term is 12, and the error must be between 8 and 12. It is actually 9. Suppose we stop with the fourth term. (2 +1) 4 = 2 4 + 1.4.2 3 + — 12. 2 2 + — 24.2 + — 24. I 4 Here the error is precisely — 24 = 1 . Example. To find sin (0+1). Let n=l. fx = sin a;, f"x = sin a;, /'a; = cosa:, fx = cosx, f"x = — sinx, f n x = — sin*, &c. f"x = — cosa:, 'Chap. IX.] DEVELOPMENT IN SEKIES. 127 sin(0 + l)=l- — + — - — + — sin0. v ' 3 ! 5 ! 7 ! 8 ! If = 0, If = 1, sin# TV = 0. sin # _ sinl TT - 40320 %£ih is -within ^few of the true value of sin 1 . If in any development the general expression for the etror decreases indefinitely as we increase n, it follows that, as the number of terms of the series is indefinitely increased, the sum will approach as its limit the value of the function, which is therefore equal to a series of the form obtained, and is said to be developable. 130. Let ns consider some examples. To develop log(l + a;). Let 2= (1+a;). fz = \og8, f* = *-\ f"z=-tr°, f'"z = 2z- 3 , f TV z=-Z\ir i , /<*>*= (_i)»-i(n-l) !«-", fi»+i) g= . (_i)»n!r-»- 1 , /(1) = 0, /'0) = i, /"(1)=-1, /'"(I) = 2, X nl 128 DIFFERENTIAL CALCULUS. [Art. 130.' /"(l)=-3!, /<»>(l) = (-l)»-'(n-l)!, f {n+ v{\+(ix) = {-\yu\{\ + ox)- n - 1 . By Taylor's Theorem, io g (i+ a; )=^-| + | 3 -^ + + ( _i)-i£ f _1 1» r " + l + ^ ' , (1 + to)— 1 . n + 1 (-l)(n-l)x The ratio of the «th term to the term before it is , n or — [ 1 — ) a;. If a; is greater than 1 in absolute value, [1 — \ nj \ n will eventually become and remain greater than unity as n in- creases, and the series x- , X s x* , x f- 2 3 4 is divergent and cannot be equal to log(l + a;). So we need only investigate the expression for the error for the values of x between +1 and — 1. Suppose x is positive, and less than 1. x n + 1 Then (1+to) - " -1 approaches zero as its limit as n in- n + 1 creases indefinitely, for it may be thrown into the form — (— — V +I . Since x <1, — ^— fa' where is less than 1 ; 1 - ox' , limit fx' — 0x\ n hence ) = 0, n=#:= m(m — l) (m — n + l)^" -B , fin + »z — m ( m _1) ( m _ l,)z m -' — \ 130 DIFFERENTIAL CALCULUS. [Art. 131. /(1) = 1, /'(l)=m. /"(l) = m(m-l), f"(l) = m(m-l)(m-2), / w (l) = m(m-l) (m — n + 1), /c»+i> (i + to) = m(m - 1) (?n -n) (1 + to)"*— 1 . By Taylor's Theorem, If m is a positive whole number, /<"•>£ = m!, /<«•+«* = 0, and all succeeding derivatives are 0, so in that case (1 + x) m is equal to the sum of a finite number of terms, namely (m+1) terms. If m is negative or fractional, however, this is not the case. Let us see whether (1 +x) m is then developable. The ratio of the general term of the series to the one before it is "2L- x or I 1 J x. If a; is numerically greater than 1 , this ratio will eventually become and remain greater than 1 in absolute value as n increases, and the series is divergent and cannot be equal to the function. Hence we need examine the value of the error only for values of x between 1 and — 1 . The expression for the remainder after n + 1 terms is m(m-l) (m-n) +1 n . w_„_i or+iyi x {l+0x) which may be thrown into the form pm(m-l) (m-n) a , n+1 "| L (" + !)! J (l+0x) n + 1— m Chap. IX.] DEVELOPMENT IN SERIES. 131 As n increases, the limit approached by — — ; — is not greater than 1 . Increasing n by unity multiplies the quantity in parenthesis by m ~ w ~ x, which may be written m — 1 n \ n + 2 n + y and by taking n sufficiently large, this multiplier may be brought as near as we please to the value — as. If a; lies between and 1 , — x is numerically less than 1 ; and as n increases indefinitely, we multiply our parenthesis by an indefinite number of factors, each less than 1 , and so decrease the product indefinitely. Therefore, for values of x between and 1 , the expression for the error approaches zero as its limit as n increases indefinitely, and (1+ x) m is equal to the series l + ma; + m ( m - 1 ) g! 2 + m(m-l)(m-2) ^ ^ Example. Show, by considering the second form for the error, Art. 128, [2], that for values of x between and — 1, (\ + x) m is devel- opable. The Binomial Theorem follows easily from the development of (l + x) m . (x + h) m = x m (l-\-- and if h is less than x in absolute value, we have (x+7i)™=x m +mx m - 1 h + m ( m ~ 1 } x m ~ 2 h 2 + m(m-l)(m-2) ^-.^ + ? [2 , 3 ! no matter what the value of m. 132 DIFFERENTIAL CALCULUS. [ART. 132. Of course, if h is greater than x, we can write (a; + h) m in the (x\ m 1 -f- - ) , and shall then get as a true development, (h + x) m = h m -i-mh m - 1 x + Maclaurin's Theorem. 132. If, in Art. 128, [1] and [2], we let x = 0, we get /&=/(0) + hf(0) + 1^/"(0) + |l/'"(0) + In h n+1 + */w(0)+_!L_ /<•+«#*, /a =/(o) + hf '(0) + |^/"(o) + 1^/'"(0) + + *!/(«) (0) + ^ n+1 (l- g )" y(n+i) ^ It does not matter what letter we use for the variable in these formulas. Change h to x, and f x =f(0) + xf(0) + ^f"(0) + + J/ (n) (0) ./<» + !> to. [1] -K 2 - X" (»+i)r /* =/(0) + xf'(0) + ^/»(0) + •■ ■• + ^/ w <0) + a" +1 (l- <> ) n y(n+i) to- [-2] These results are called Maclaurin's Theorem, and they enable us to develop a function in a series arranged according to the ascending powers of the variable. Chap. IX.] DEVELOPMENT IN SERIES. 133 133. To develop a*. fx=a*, /(0) = a° = l. f'x = a x log a, /' (0) = log a, f"x = a*(\ogay, /"(0) = (loga) 2 , /<»)» = a*(loga) B , / w (0) = (log a)", /<"+ 1 >a;-_a I (loga) n+1 , /(»+«to = a fe (loga) n+1 . By Art. 132, [1], a* = l+aloga + |^(loga) 2 + ^(loga) 3 + + ^I(loga)" ,,n + l — (loga)" +1 ak a: n+1 (loga)" +1 9x _ 9x a; log a aloga xloga #loga aloga (»+l) ! " ' 1 2 3~~ n ' n + 1 ' No matter what value a; may have, after « has attained a cer- tain value in its increase, some of the factors of this product will approach the limit zero, and the whole product will there- fore have zero for its limit as n increases indefinitely, and a* = l + a;loga + !^loga) 2 +^(loga) 3 + [1] for all values of x. If a = e, logo, = 1, and e * = l + - + —+~ + — + [2] 1 2! 3! 4! Let x = 1 , and [2] becomes e=l + i + — + — + — + ; [3] 12! 3! 4! a result already established in Arts. 61 and 62. 134 DIFFERENTIAL CALCULUS. [Art. 13t 134. We can now test the accuracy of the provisional devel- opments of sine and cosine given in Art. 124, (2) and (3). -By Art. 132, [1], /V»3 /yi5 ,y»7 sva.x = x \- +-B, 3! 5! 7! /y.n + 1 ™n + 1 where E = — - /<" +1) to=± — sinto (» + l)! (» + l)I or ± cos to. (»+l) ! In either case, one factor sin to or cos to is between 1 and —1, and the other approaches zero as n increases indefinitely ; there- />i3 /yi5 ™7 ™9 fore, S mx = x - — + — - — +— - 3! 5! 7! 9! Example. Cu 3C Cu Cu Prove that cos x = l 1 1 2! 4! 6! 8'! 135. By the aid of the Binomial Theorem, tan -1 a; and sin -1 a: can be very easily developed. D x t&n~ 1 x=—}— i = (l+x 2 )- 1 . (Art. 71, Ex.) X -J— 3/ For values of x less than 1, (l + a^) _1 can be developed by Art. 131, [2], (l + x i )- 1 = l-x 2 + x i -x 6 + x 8 - Integrate both members. tan~ 1 a; = C+ a- -+---+-- 3 5 7 9 To determine our arbitrary constant C, let k=0; Chap. IX.] DEVELOPMENT IN SERIES. 135 then tan- 1 = C+0-- , 3 and (7=0. tan- 1 «= a;- - + ---+-- [1] 3 5 7 9 L J when x is less than 1 ; that is, when tan -1 * is less than -. 4 D uB m- 1 x= l = (l-x*)~h, by Art. 71. For values of a; less than 1, (1— a 2 ) - * can be developed by Art. 181, [2]. (1-^-4=1 + ^4-—^+— ^+ 1,8 ' 5,7 !C 8 + v ' 2 2.4 2.4.6 2.4.6.8 Integrating . _i r, , , 1 X s , 1.3 a; 5 , 1.3.5 a; 7 , sin 1 x = C + x-\ h 2 3 2.4 5 2.4.6 7 When a;=0, sin -1 a; = and C= 0. . _, , 1 X s , 1.3 ar 5 , 1.3.5 x 7 , ron sin l x = x-\ h |2 1 2 3 2.4 5 2.4.6 7 L J Examples. (1) Show that sin (a; + 7i) is equal to the series h h 2 h s h* . sinaH 'cosaj sina; — — cosa; + — sma)+ • (2) Show that e = l +TT + _-4.__ 136 DIFFERENTIAL CALCULUS. [Art. 136. 136. Although the strict proof that any given function is equal to the series obtained by Taylor's Theorem requires the investi- gation of the remainder after n+1 terms, it is often convenient to obtain terms of the series in cases where the expression for the remainder is too complicated to admit of the usual examina- tion. When such a series is employed, it is to be remembered that it is equal to the function in question only provided that the function is developable. Sometimes the possibility of de- velopment can be established by other considerations, and some- times in rough work no attempt is made to fill out the proof of the assumed equality. Examples. (1) Develop |-log(l + a;). 1-f-a; Ans. 2x--x 2 + -x>--x i + -x 5 + 2 3 4 5 (2) Obtain 4 terms of the development of log( 1 + e x ) . x , x' x Ans. log2 + £ + £- 2 2 3 2 3 .4 ! 137. In the work of successive differentiation required in applying Taylor's Theorem, a good deal of labor can often be saved by making use of Leibnitz's Theorem for the Derivatives of a Product. Let y and z be functions of x. Represent D x y, D?V, D*y by y', y", y(-> and D x s, D x *z, D x "s by 0\ z", *<»>. D x {ys) = y'z +yz', D x \yz) = y' , z + 2yV +yz", D x \yz) = y"'z + 3y"z'+ 3y B Let ACB represent any such path from A to B. Draw a normal to the plane at C and the perpendiculars p and q. Call DE = c, and let v x and v t be the velocities in the first and second media respectively. AC = psecd, CE=pta,n6, Chap. IX.] DEVELOPMENT IN SERIES. 147 BC = gsecsec0 is the time required to pass from A to C ; -BO _ g sec 0j is the time required to pass from G to B ; . psectf gsecflj f = ( 1>1 v 2 is the function we wish to make a minimum. and X are the only variables in t, and they are connected by the relation ptan0 + gtan #!=<;. _, p q D g t = — sec 0tan 6 -\ — sec 0jtan O^Ds B-y. Differentiate p tan + g tan 0j = c . psec?0 + qsec 2 6 1 D 9 X = 0, n , psec'fl g sec #! J9 g j3sec 2 2 1 gsec 2 e £ must equal zero in order that t may be a minimum. Ex- press everything in terms of sine and cosine. p sin q sin d x p cos 2 #i _ n v r cos 2 ^ cos 2 X q cos 2 148 DIFFERENTIAL CALCULUS. [Art. 145. smO sin0! z= 1>1 v 2 sin0 «i sin 1 v 2 By taking D#t and substituting sin d _ Vi sin*?! - Vz we should obtain a positive result ; so that this relation between the angles gives the path of quickest passage required. This result is the well-known law of the refraction of light, and our solution establishes the fact that a ray of light, in passing from a point in one medium to a point in another, takes the course that enables it to accomplish its journey in the least possible time. (2) What value of x will make sin 3 a;cosa; a maximum? Ans. x = — A (3) What value of a; will make sin x (1+ cos x) a maximum? Ans. x = — 3 (4) Show that a^ is a maximum when x = e. (5) A statue a feet high stands on a column b feet high ; how far from the foot of the column must an observer stand that the statue may subtend the greatest possible visual angle ? Ans. V6(a + 6) feet. (6) Required the shortest distance from the point (x ,y ) to 'i the line Ax + By + = 0. Ans. A ^o + Byo+C Chap. X.] INFINITESIMALS. 149 CHAPTER X. INFINITESIMALS. 146. An infinitesimal or infinitely small quantity is a variable which is supposed to decrease indefinitely ; in other words, it is a variable which approaches the limit zero. What we have called the increment of a variable has, in every case considered, been such a quantity ; and what we have called a derivative has been the limit of the ratio of infinitesimal incre- ments of function and variable. 147. When we have occasion to consider several infinitesimals connected by some law, we choose arbitrarily some one as the principal infinitesimal. Any infinitesimal such that the limit of its ratio to the princi- pal infinitesimal is finite, is called an infinitesimal of the first order. An infinitesimal such that the limit of its ratio to the square of the principal infinitesimal is finite, is called an infinitesimal of the second order. An infinitesimal such that the limit of its ratio to the nth power of the principal infinitesimal is finite, is called an infinitesimal of the nth order. Let a represent the principal infinitesimal, and a x any infini- tesimal of the first order, a 2 of the second order, a„ of the nth order. Then, by our definition, limit — = K, a a J£ being a finite quantity. a. 150 DIFFERENTIAL CALCULUS. [Art. 148. where e is an infinitesimal (Art. 7) , limit -j = K' ; a n =a. n {K ( - n) + £ (n) )- Examples. Show, by the aid of these expressions, that the limit of the ratio of any infinitesimal to one of the same order is finite ; to one of a lower order is zero ; to one of a higher order is infinite. That the order of the product of infinitesimals is the sum of the orders of the factors, and that the order of the quotient of infini- tesimals may be obtained \>y subtracting the order of the denomi- nator from the order of the numerator. Show that, if the limit of the ratio of two infinitesimals is unity, they differ by an infinitesimal of an order higher than their own. 148. The sine of an infinitesimal angje is infinitesimal ; for, as the angle approaches zero, the sine approaches zero as its limit. If we take the angle as our principal infinitesimal, the sine is an infinitesimal of the first order ; for we have seen that limit a=0 ina~| , — J = l, (Art. 68). The vers a is infinitesimal if a is infinitesimal, for vers a = 1 — cos a ; Chap. X.] INFINITESIMALS and as a = 0, COSa = l ; hence versa = 0. 151 It is an infinitesimal of a higher order than the first, for we have 1 — cos a~\ — J=0, (Art. 68). ., , limit seen that . n Let us see if it is of the second order ; that is, let us see if limit «=0 1 — COS a~\ is finite. s — assumes the form - when o? a = 0, and we can find our required limit by the method of Art. 139, which gives us - as the value sought. Therefore, when a is Id infinitesimal, versa is infinitesimal of the second order. Examples. Taking a as the principal infinitesimal, show that (1) tana is an infinitesimal of the first order. (2) a — sin a is an infinitesimal of the third order. (3) tana — a is of the third order. 149. Let y be am r function whatever of x, if we give x an infinitesimal increment Jx, the corresponding increment Ay of y will be an infinitesimal of the same order as Jx, unless for par- ticular single values of x. To establish this proposition, we must show that j _^o "Jr -j- cannot be zero, except for single values of a; ; for, suppose it could become and continue zero ; . _^ a J r is D x y, and we have seen (Art. 38) that D x y shows the rate at which y is changing as x changes. If D x y becomes and remains zero, y does not change at all as x changes ; and, therefore, is not a function of x, but a constant. „ ., limit is finite. Jx ^ Q 152 DIFFERENTIAL CALCULUS. [Art. 150. Ax=0 ~Ax cann °t become and continue infinite ; for, in that case, a v ^_(\ ~r~ would be zero, D y x would be zero, and x, regarded as a function of?/, would be constant. _Jx finite, and Ay and Jx are of the same ordej. c . limit Smce dx=0 can be neither zero nor infinite, it must be 150. If the coordinates of the points of a curve are expressed as functions of a third variable (/, the distance between tioo infi- nitely near points of the curve is an infinitesimal of the same order as the difference between the values of a to ivhich the points corre- spond. The ordinary equations of the cycloid, x = ad — a sin y = a — a cos 8 are a familiar example of the wa}- in which the coordinates of points of a curve may be expressed as functions of a third varia- ble. In the case of any curve, it is obvious that this may be done in a great variety of ways. Any two equations containing x, y, and a that will reduce on the elimination of a to the ordi- nary equation of a given curve, can be used as equations of that curve. For example : x = 2a ] > are equivalent to x— 2 ?/ + 4 = ; y=a+2 j x = a cos a 1 > are equivalent to a? + y- = a 2 ; y= a sin a J x = a cos a j 2 2 !• are equivalent to ^ + ^ = 1 ; y=bsina\ a ° Chap. X.J INFINITESIMALS x = a sec ii- ] y 153 x 2 are equivalent to 6 tan a I a ■t-1. The proof of our proposition is as follows : Let a and a+Aa be the two values of a in question, and (x,y) and (a; + Ax, y + Jy) be the two corresponding points. The distance D between these points will be, if we use rectangular coordinates, -\/(Axy-\-(Ay) 2 . \D' Aa We wish to prove that . . „ * Aa = {) is finite. p_ ^/(jxy + (j y y ^ lf^\\fM 2 Aa Aa \\Aa) + ' \AaJ ' and, by Art. 149, ^^ hence order as Aa. limit [JD~] . Aa±0\_Aay Ax' Ik and limit Aa=0 Ay Aa are both finite ; is finite, and D is an infinitesimal of the same 151. If two curves are so connected that the points of one cor- respond to the points of the other, so that when a point of the first curve is given, the corresponding point on the second is determined, the distance between two infinitely near points on the first curve is an infinitesimal of the same order as the distance between the corresponding points of the second curve. For, if we suppose the coordinates of the points of the first curve expressed as functions of some variable a, the coordinates of the points of the second curve can also be regarded as functions of a ; and, by Art. 150, each of the distances in question will be an infini- tesimal of the same order as Aa, and each will therefore be o( the same order as the other. 152. If a straight line moves in a plane according to some law, so that each of its positions corresponds to some value of a varia- ble a, the angle between two infinitely near positions of the line is an infinitesimal of the same order as the difference between the corresponding values of a. 154 DIFFERENTIAL CALCULUS. [Art. 153. Suppose lines drawn through a fixed point parallel to the moving line in its different positions. From 0, with the radius unity, describe an arc. Consider any two positions of the moving line, and the corresponding lines at 0, we wish to prove that; the angle

rovided that this limit is finite, any infinitesimal may be replaced by another so related to it that the limit of the ratio of the second to the first is unity. Let Proof. '-l + «2 + «S+ + a n be a sum of infinitesimals of such a nature that the number of the terms increases as each term decreases in absolute value, so that the limit of the sum is some finite quantity. Let ft, /9 2 , /9 3 , ,3 n be a set of infinitesimals so related to the first , h ' 3 - , K set that limit— = 1, limit— = 1, &c, limit — =1, «1 «2 «» then - = l + e„ - = l + £ ,, &c, ^ = l+£„, c], i2 , s„ being necessarily infinitesimal (Art. 7). Pi="i + "-\-u h— '-'2 + «2 £ 27 ft~ = «« -f- «« t' , /*! + ^ + ' ? 3 + + /*» = «1 + «2 + «S + + «n + « 1 e 1 + r/ 2 s 2 4 .z 3 c 3 + + "„s„. 162 DIFFERENTIAL CALCULUS. [Art. 162. Let r, be such a variable that at any instant it shall be equal to the greatest in absolute value of the quantities e^i £ n - Of course, since each of these approaches zero as its limit, vj must also approach zero as its limit ; i.e., j? is infinitesimal. «l £ l + «2 £ 2+ a 3 E 3+ + °n-n< r l( a l + a 2 + «3 + + «„)) hence ft + ft + ft + + ft - («i + « 2 + « 3 + + «„) / are infinitesimals of a higher order than the first, by Art. 147, Ex. PM+ P'M= PT+P'T+ r, + V, or the difference between the sum of the tangents and the chord is of a higher order than the first. The difference between the arc and the chord is less than this, therefore the limit of the ratio of an infinitesimal arc to its chord is unity. 166. It is customary to say roughly that lines which make with each other an infinitesimal angle, that is, lines which approach the same limiting position, coincide, and that finite values which differ by an infinitesimal or infinitesimal values which differ by an infinitesimal of a higher order, that is, values such that the limit of their ratio is unity, are equal; and this way of speaking is very convenient, especially for preliminary investigations. It is important, however, to be able to put a proof given in this form into the more exact language of limits. It is easily seen from what has just been said, that the line Chap. X.] INFINITESIMALS. 167 joining two infinitely near points of any curve, can, speaking roughly, be regarded at pleasure as chord, arc, or tangent, so that an infinitesimal arc can be treated as a straight line. 167. As an example of this loose form of proof, let us show that a tangent to an ellipse makes equal angles with the focal radii drawn to the point of contact. Let P and P' be two infinitely near points of the ellipse, then PP' is the tangent in question. From F and F' as centres, draw the arcs PA and P'B ; PA and P'B being infinitesimal arcs, are straight lines, and PAP' and P'BP are right angles, since the tangent to a circle is perpendicular to the radius drawn to the point of contact. F'P+PF=F'P'+P'F, by the definition of an ellipse. Take away from the first sum F'P + BF, and we have left PB ; take away from the second sum the equal amount F'A -f- P'F, and we have left PA ; .-. PB=P'A; and the right triangles PAP' and PBP' have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, and are equal ; and the angle FPP'= F'P'P ; but the lines F'P' and F'P coincide, so that the angle F'P'P is the same as the angle F'PT ; and .-. F'PT=FPP', and the tangent makes equal angles with the focal radii, q.e.d. 168 DIFFERENTIAL CALCULUS. [Art. 168. Example. Prove that a tangent to an hyperbola bisects the angle between the focal radii drawn to the point of contact. 168. To find the area of a segment of a parabola cut off by a line perpendicular to the axis. Compare the required area with the area of the circumscribing rectangle. We can regard the first as made up of the infinitesimal rectangles of which PMUS is a type, and the second of the corresponding rectangles of which QNPE is one. Draw the directrix. PF=SD and DO=OF, by the definition of the parabola ; but - PF=FT by Art. 163; .-. TO= OS. The triangles P'MP and PST are similar, and P'M = PM = PM . PS ST -2 OS' hence PM x PS = 2 OS x P'M= 2PRQN, or rectangle PU= 2PQ ; .-. ZPU=2ZPQ, Chap. X.] INFINITESIMALS. 169 and the segment in question is twice the external portion of the circumscribing rectangle, and, therefore, is two-thirds of the whole rectangle. Example. Prove the theorems of Arts. 167, 168, strictly, by the method of limits. 169. The properties of the cj'cloid can be very simply and neatly obtained by the aid of infinitesimals ; though, for this pur- pose, it is better to look at the curve from a new point of view. Let a fixed circle equal to the generating circle be drawn tan- gent to the base of the cycloid at its middle point ; through the generating point P, draw PQD parallel to the base. From the nature of the cycloid, the arc PN= ON and OB = ACB, PQ = NB=OB- 0N= ACB -QB = ACQ. Hence points of the cycloid can be obtained by erecting perpendicu- lars to a diameter of a fixed circle, and extending each until its external portion is equal to the distance along the arc of the circle from the perpendicular in question to a given end of the diameter. 170. The tangent to the cycloid passes through the highest point of the generating circle. 170 DIFFERENTIAL CALCULUS. T A, [Art. 170. Rough Proof. — Let P and P' be infinitely near, then PP' is the required tangent ; through P' draw an arc parallel and simi- lar to QQ'. This arc may be regarded as a straight line. The triangle PP'M is isosceles, since QP= ACQ and MQ = P'Q'= ACQ', hence PM=QQ'=MP'; .-.the angle PP'M= P'PM. P'M is parallel to the tangent at P to the generating circle, hence PP'M= TPT, and PT bisects the angle MPT', bisects the arc PTS, and con- sequently passes through the highest point of the generating circle. q.e.d. Strict Proof. — Draw the chord P'M, and regard PP' as a secant line ; in the triangle PP'M we have sin PP'M ^ PM sin'TPM P'M" ,. .. sinPP'M ,. .. PM .-. limit = limit The sin TPM P'M htcP'M=PM, and the chord P'M differs from the arc by an infinitesimal of a higher order than that of the chord. Chap. X.] INFINITESIMALS. 171 . ■ . limit PM limit PM = 1, chord P'M arc P'M hence limitPP'Jf = limit TPM. The limiting position of P'M is the tangent PT' ; . • . limit TPQ = limit TPT\ and the tangent passes through the highest point of the generat- ing circle. The Area of the Cycloid. 171. Rough Investigation. — Circumscribe a rectangle about the cycloid, and its area is evidently equal to the circumference of the generating circle multiplied by its diameter ; that is, to four times the area of the circle. The area of the cycloid is .N a b T A / r i d \^ J/ / c f J 1 T\ / <\ \^_ ^/ this area minus the area of the external portion of the rectangle. The external area ANO may be divided into trapezoids, of which abPP' is anj r one. The tangent PP' passes through the highest point of the generating circle, and is a diagonal of the rectangle TaPc, Tc being a diameter. From geometry, abP'P=cdP'P, which is equal to Qg ; therefore the sum of the trapezoids abP'P is equal to the sum of the corresponding rectangles Qg, or the external area ANO is equal to the semi-circle ACB : but ANO 172 DIFFERENTIAL CALCULUS. [Art. 172. is half of the external portion of the circumscribing rectangle ; consequently, the area of the cycloid is three times the area of the generating circle. Strict Proof. — The external area is the sum of the curvilinear quadilaterals of which abP'P is any one ; that is, area = labP'P = limit labP'P = limit labhP, for abhP - abP'P< eP'hP, which is of the second order. P'P" is of the second order, since Na b'b T A it is proportional to the distance from P' to the tangent at P (Art. 153); therefore bhh'b' is of the second order, and limit labhP = limit lab'h'P. ab'h'P=edcP= Qg, hence the external area = limit -Qg = area of ACB. Length of an Arc of the Cycloid. 172. Bough Proof. — The arc AP is equal to the sum of the infinitesimal chords of which PP' is one. The chord AQ is the sum of the differences between each chord and the one drawn to a point of the fixed circle above the point in question and in- finitely near it ; QS is such a difference, hence arc AP= IPP 1 and chord^lQ= IQS. Chap. X."| INFINITESIMALS. 173 PP' and QR are equal, Q'QR = Q'RQ, by Art. 170, and QQ'R is isosceles. Q'S, an infinitesimal arc described from A as a centre, may be regarded as a straight line perpendicular to QR, and therefore bisects QR, and PP'= 2QS, SPP'= 21QS. The Arc^lP=2chord^Q. arc^lO=2.-LB, and the whole arc of the cycloid is eight times the radius of the generating circle. Strict Proof— P'P", Q'T", and US are infinitesimals of the second order, each being proportional to the distance from a point VT of a curve to the tangent at a point infinitely near. Vv is also of second order, as it is the projection of Q'T" on AQ. 174 DIFFERENTIAL CALCULUS. The arcAP=limit2PP'=limit.TPP'', since, in triangle PP'P", PP'-PP", Chap. X.] and consequently, INFINITESIMALS. QB=bPX; 175 and the radius of curvature is twice PN, the portion of the normal within the generating circle. Strict Proof. — The centre of curvature is the limiting posi- tion of X. PP"X' is similar to QRB, hence PX' PP" , v ..PX' ,. ..PP" ——— = — r — and limit = limit . (1) QB QB QB QR K ' Let PP' be the principal infinitesimal, then P'P" is of the second order ; therefore, in (1) , PX can be substituted for PX'. RQ'S is similar to BQR, hence — — = ^ — . QR QB QR and QS are infinitesimal, QB is finite, RS is of the second order, and QS can be substituted for QR in (1), and r ..PX ,. ..PP" limit — — = limit : QB QS but, by Art. 172, limit PP" = 2; QS limit PX= 2QB= 2PN. Q.E.D. 176 DIFFERENTIAL CALCULUS. [Art. 174, Evolute of the Cycloid. 174. Extend the diameter TN to N', making NN'= TN and draw a circle on NN' as diameter. The centre of curva- ture X, corresponding to P, will lie on this circle, since PX=2PN. Draw a tangent to the second circle at N', drop a perpendicular from to this tangent, and lay off B'O' equal to one-half the circumference of the generating circle. The arcPJV= 0N= B'N ; .-. the arc N'X=N'0', and X lies on a cycloid equal to the given cycloid, having its origin at 0' and its highest point at 0, and this must be the evolute required. Chap. X.] INFINITESIMALS. 177 Examples. 175. (1) From a point situated in the plane of a plane curve, radii vectores are drawn to different points of the curve, and on each one a distance is laid off from inversely propor- tional to the length of the radius vector ; to determine the tan- gent at any point of the locus of the points thus obtained. (2) Take any two curves in the same plane, and consider as corresponding points those at which the tangents are parallel ; draw through a fixed point lines equal and parallel to those uniting corresponding points of the two curves. Prove that a tangent to the locus of the points thus obtained is parallel to the tangents at the corresponding points of the given curves, and that any arc of this curve is the sum or difference of those which correspond to it upon the given curves. (3) From a point radii vectores are drawn to a given curve, and each is extended beyond the curve by a constant length. Prove that the normal to the curve on which the extremities of the radii vectores lie, the normal at the corresponding point of the given curve, and the perpendicular through to the radius vector of the point, have common intersection. 176. To show the power of this method of infinitesimals, we shall give an investigation into the nature of what is called the Brachistochrone, or Curve of Quickest Descent. The problem is a famous one, and the solution below is in effect the one given by James Bernouilli, and is very much simpler and more ele- mentary than the usual analytical solution which requires the use of the Calculus of Variations. The problem is, given two points not in the same horizontal plane, nor in the same vertical line; to find the curve doivn which a particle moving without friction can slide in the least time from the upper point to the lower, the accelerating force being terrestrial gravitation. Let us first consider a simpler question : To find the path of quickest descent on the hypothesis that it is to consist of two 178 DIFFERENTIAL CALCULUS. [Art. 176. straight lines intersecting on a givenhorizontal plane, assuming that the particle moves down each line with a uniform velocity equal to the mean velocity with which it would actually descend the line in question. It is easily seen that both lines must lie in the vertical plane containing the two given points. Let PNP' and PMP' be two paths of equal time from P to P'. Then the required path must lie between them. If we suppose them to approach, continuing still paths of equal time, the required path of quickest descent will be the limiting posi- tion of either of them. Let v be the mean velocity of a particle sliding from P to M; then, by Art. 115. v will also be the mean velocity of a particle sliding from P to N. Let Vi be the mean velocity of a particle sliding from M to P\ supposing that the particle started from M with the velocity actually acquired by sliding down PM\ then v x is also the mean velocity of descent from iVto P', by Art. 115. As we are going to make the paths PMP' and PNP' approach indefinitely, MN is an infinitesimal. Draw the arcs NS' and MR' from P and P' as centres, and the perpendiculars NS and MR. On our hy- pothesis, the time of descent from P to S' equals time of descent from P to N, and time of descent from M to P' equals time of descent from R' to P'; hence, as time PMP' equals time PNP', the time of descent from S' to M equals time from N to R', S'M_ NR< V 1>! S'M v whence and NR' v 1 limit^=limit^; NR'- v, ' r ..S'M y . + SM limit • ; = limit , NR' NR' Chap. X.] infinitesimals. 179 SM _ cos PMN _ cosy . NB ~ cos P'MN ~ cos y x ' . „ . cosy , V nence limit = limit — COS

l, Call C= " V(2«) and assume y = a — a cos > and we have a s sin 2 0. 1 1 — COS 2 1 + cosfl (Dgx) 2 1 - cos o 1-cosfl 182 DIFFERENTIAL. CALCULUS. [Art. 176. (D ^.\2_ « 2 sin 2 g(l- cosg) _ a 2 (l- cos 2 Q) (1- cosO) 8 ' ~ 1+COS0 ~~ l+cos0 = a 2 (l-cos0) 2 , D g x = a(l— cos/9), x = af g (l— cosd) = aO — asin0 +C, when x = 0, y = 0, and = ; hence (7=0, and our equations are x — ad — asin'0 ' y = a — a cos ( the familiar equations of a cycloid ; and the brachistochrone is an inverted cycloid with its cusp at the higher of the given points. Chap. XI.] DIFFERENTIALS. 183 CHAPTER XL DIFFERENTIALS. 177. A derivative has, in effect, been defined as the limit of the ratio of infinitesimal increments of function and variable. Consequently, in getting a derivative, we can replace the incre- ment of the function by any quantity differing from it by an infinitesimal of a higher order. For example : in getting D x a?, we find A (a;) 2 = o? + 2xAx + (Ax) 2 - x 2 = 2xAx + {Ax) 2 . 2xAx differs from A (a?) by (Ax) 2 , which is of the second order if we take Ax as the principal infinitesimal, and 2x Ax may be substituted for A(a?) in getting D x v?, which then equals limit Ax=0 '2xAx Jx = limit [2 x -\=2x. In our old problem of getting the derivative of an area we can use this same principle. Take Ax as the principal infinitesimal, then AA and Ay are of the first order, by Art. 149. A A differs from the rectangle yAx y by V ^.^ / KA x X bx by less than the rectangle Ax Ay, which is of the second order, by Art. 147, Ex. ; and we have 184 DIFFERENTIAL CALCULUS. [Art. 178. D A= limit \^A_ 1_ limit Ax±0\_Ax] Ja;=0 yAx Ax 0-* Take the problem of Me derivative of an arc. 7 ^_^ 1 Ay -p= vl x & v Let Ax be the principal infinitesimal ; then As is of the first order. As differs from its chord V( Ax) 2 -\-( Ay) 2 by an infinitesi- mal of a higher order, by Art. 165. Hence we have D,s= limit Ax=0 As Ax . limit ' Ax=0 W(Axf+(Ayf Ax . limit Ja;=0 'M® A« = Vl + (D,y)». 178. In general, n ,. limit ~ f(x + Ax)-fx ~ Ax therefore f(x + Ax)—fx „ . . y ' — — = D x fx + e, Ax where e is an infinitesimal, by Art. 7. f(x + Ax) —fx = D x fx . Ax + eAx. But f(x + Ax)—fx is the actual increment of fx, caused by the increment Ax of x. eAx is of as high an order as the second, if we take Ax as our principal infinitesimal ; and we get the impor- tant result that D x fx . Ax differs from the actual increment of fx by an infinitesimal of a higher order, and may consequently be used in place of Afx in any case where we have to deal with the limit of the ratio or of the sum of such increments. This quan- Chap. XI.] DIFFERENTIALS. 185 tity) D x fx . Ax is called the differential of fx, and is denoted by dfx, d being a symbol for the word differential. By the definition of differential, dx = D x xJx = Jx. This definition may now be restated as follows : The differen- tial of the independent variable is the actual increment of that variable. The differential of a function is the derivative of the function multiplied by the differential of the independent variable; or formulating, dx = Jx, dy = D x y.dx, y being a function of x. It is to be noted that o differential is an infinitesimal, and that it differs from an infinitesimal increment by an infinitesimal of a higher order. 179. Since dy=D x y.dx, dy n dx =D *y- As, by Art. 73, D„x= , 3 Dj D v x=^. dy Consequently, if two quantities are so connected that either is a function of the other, the derivative of either with respect to the other is the actual ratio of the differential of the first to the differ- ential of the second. 180. The differential notation has the advantage over the derivative notation, that it is apparently simpler, and that the formulas in which it is used are more symmetrical than those in which the other notation is employed ; and although the differ- ential is defined by the aid of the derivative, and the formulas 186 DEETEBENTIAL CALCULUS. [Art. 181. for the differentials of functions are obtained from the formulas for the derivatives of the same functions, there is a practical advantage, after the formulas have once been obtained, in regard- ing the differential as the main thing, and looking at the derivative as the quotient of two differentials. 181. By multiplying each of our derivative formulas by dx, we get the following set of formulas for the differentials of functions. „^ . D^ is no longer necessary, as it gives us dfy = DJy .dy = Mdy = dfy, an identity. Examples. Work the examples in Chap. IV. by the differential formulas just given, remembering that D x y = ^. y dx 182. The differential notation is especially convenient in deal- ing with problems in integration, and leads to an entirely new way of looking at an integral. 188 DIFFERENTIAL CALCULUS. [Art. 183. Let y = x 2 , and suppose that x changes from the value 1 to the value 5 ; to find the whole change produced in y. Let x change by succes- sive increments, each of which may be called Ax ; then the whole change in y is the sum of the corresponding increments of y, which we will indicate by 2 Ay. The whole change in y is the i=i actual sum of these infinitesimal increments ; it is then the limit of their sum as Ax is indefinitely decreased, and each Ay decreases correspondingly ; that is, it is limit -2" Ay. But as we are deal- i=i ing with the limit of a sum of infinitesimals where the limit is, from the nature of the case, finite, each term may be replaced by any infinitesimal differing from it by an infinitesimal of a higher order (Art. 162) . Each Ay may then be replaced by the corresponding dy, and we get as the whole change produced in x=5 x=5 the value of y, limit 2 d(x 2 ) = limit 2 2xdx. i=i i=i As y=y?i this change must be [ +a '' A 2 y = dry + adx*. d 2 y is an infinitesimal of the second order, by Art. 187. adx 2 is of the third order ; consequent!} 1 , d 2 y may be used in place of A 2 y in problems concerning the limit of a ratio or of a sum. By similar reasoning, it can be shown that * A s y = d s y + adx 3 ; and, in general, that A n y = d"y + adx", when Jx is infinitesimal. Chap. XII.] FUNCTIONS, ETC. 199 CHAPTER XII. FUNCTIONS OF MORE THAN ONE VARIABLE. Partial Derivatives. 194. Dp to this time we have considered only functions of a single variable, but a complete treatment of our subject requires us to study functions of two or more independent variables. Plane Analytic Geometry has furnished us with numerous ex- amples of functions of the former kind ; Analytic Geometry of Three Dimensions introduces us to functions of the latter sort. The equation of a surface contains three variables, as, y, and z, and any one may be expressed as a function of the other two ; and when this is done, the one so expressed may be changed by changing either of the others, or by changing them both, as they are entirely independent. 195. The derivative of a function of several variables obtained on the hypothesis that only one of them changes, is called a par- tial derivative; and, as all the variables except one are, for the time being, treated as constants, a partial derivative can be ob- tained by the rules for differentiating a function of one variable. For example : D x x 2 y= 2xy, if x alone changes ; D y ^y = x 2 , if y alone changes. 2xy is the partial derivative of x 2 y with respect to ar, and a? is the partial derivative of x*y with respect to y. We shall represent partial derivatives by our old derivative notation, indicating ordinary or complete derivatives, when it is necessary to make any distinction between the two, by the ratio of two differentials. 200 DIFFERENTIAL CALCULUS. [ART. 19fi. 196. If a function contains two variables, its partial deriva- tive with respect to either will generally contain both variables, and may be differentiated again with respect to either of them. Take a?y>. D x x*f=2xtf; D?x*tf=2f; D y D x x 2 y 2 = lxy; D v x'y>=2x*y; D x D y %*tf = ±xy; D y 2 x 2 y 2 =2x 2 . Take a = x log//. D x u = logy; DJu = ; D 2 D x u=-\; y D ,m = -; D x D s u = -; 197. In both 'iiese examples we see that D x D !l u is the same as D t D x it, and in the second D 2 D x u = D x D?u. Chap. XII.] FUNCTIONS, ETC. 201 Let us see whether it is true in general that the order in which the differentiations are performed is immaterial. Let u=f(x,y)> To see if' D,D x u = D x D t u. D y u = limit * Jy=0 f(x,y+Ay)-f(x,y) Ay by Art. 7, where e is an infinitesimal and a function of x, y, and Ay. Similarly, D x u = A* + **,y) -/(»*) , £ <, Ja; where e' is an infinitesimal and a function of a;, y, and Ax. D x D y u is equal to limit U(n + An,y+Ay)—f(x + fa,y)- -f{x,y+Ay)+f{x,y)~ Ax=d [_ Ax Ay _ + D x e; [1] D y D x u is equal to limit [70 + Ax ^y + A y) -A x ,y + 4/) - -f(x + Ax,y)-hf(x,y)~ Ay=0 [_ Ax Ay _ + D,e'. [2] The second expression for D y u is absolutely true, whatever the value of Ay, and so is the expression for D x D y u. We may then suppose Ay to approach indefinitely near zero, and D x D y u will be equal to the limiting value approached by the second member of [1] . The limit of e as Ay approaches is ; therefore, by Art. 192, J^O^^ ' and D x D v u is equal to limit f f(x + Ax,y + Ay)-f(x+Ax,y)-f(x,y + Ay)+f(x,y) l |_ J« Jy J-' as both Ay and Ax approach 0. 202 DIFFERENTIAL CALCULUS. [Art. 198. By similar reasoning, it may be shown that D y D x u is this same limit, and hence that D x D y u=D y D x u. By applying this theorem at each step, we may prove that, in obtaining any successive partial derivatives, the order in which the differentiations occur is of no consequence. For example, let us show that D x -D y u=D y D 2 u; D x 2 D y u=D x {D x D y u) = D x {D y D x u) =D x D„D x u =D,D x D x u=D t D*u. 198. In a previous chapter, we saw that, while the increment of a function due to any increment of the variable is generally a very complex expression, the differential of the function, which differs from the true increment only by an infinitesimal of a higher order than the increment of function or variable when the latter is infinitesimal, is usually very much simpler, and yet can be used instead of the true increment in many important problems. It is worth while to see if we cannot get a simple expression capable of replacing the infinitesimal increment of a function of two or more variables in similar problems. A function of two independent variables may be changed by changing either of the variables alone, or by changing both. Suppose we give to each variable an infinitesimal increment of the same order. Let u =f(x,y) . Increase x by Ax and y by Ay, Au =f(x + Ax,y + Ay) -f(x,y) . Add and subtract f(x ,y-\-Ay), and we get Au =/(as + Ax,y + Ay) -f(x,-y + Ay) +f(x,y + Ay) -f(x,y) . Chap. XII.] FUNCTIONS, ETC. 203 f( x i1/ + Ay)—f(x,y) is the increment of f(x,y) produced by- changing y alone, and differs from D y f(x,y)Ay by an infini- tesimal of a higher order than Ay, by Art. 178. In like man- ner, we see that f(x + Ax,y + Ay)—f(x,y+Ay) differs from D x f(x,y + Ay)Ax by an infinitesimal of a higher order than Jx. D x f(x,y + Ay) is a new function of x and y, and any infinitesi- mal change in y will produce in it a change of the same order, by Art. 149. D x f(x,y + Ay), then, differs from D x f(x,y) by an infinitesimal of the same order as Ay, and D x f(x,y + Ay)Ax differs from D x f(x,y) Ax by an infinitesimal of the second order. D x f(x,y)Ax+D y f(x,y)Ay, or, using the differential notation and remembering that x and y are both independent, D x f(x,y) dx +D s f(x,y)dy differs from the true increment of u by an infini- tesimal of a higher order than dx and dy % and therefore may be used in place of Au whenever the limit of a ratio or the limit of a sum is sought. This is called the complete differential of u, and is indicated by du ; hence, when u =f(x,y) , du =D x udx +D y udy. Example. Prove that, if u =f(x,y,z) , du =D x udx +D s udy +D z udz. 199. Partial derivatives may very often be used with profit in obtaining ordinary or complete derivatives. Suppose that y =Fx and z =F 1 x and u =f(y,z) ; u is indirectly a function of x, and we can therefore speak of the complete derivative of u with respect to x, which we shall indi- , , du cate by — J dx 204 DIFFERENTIAL CALCULUS. [Art. 200. We wish to find the limit of the ratio — . In so doing, we Ax can replace Au by du, which equals D v uAy +D z uAz, since, as y and z are not independent variables, Ay and Az differ from dy and dz ; hence **= l^it \ D yU ^+D,u^\ dx zte=0|_ * Ax^ Ax] du r. dy r, dz or — =D y u-?-+D x u — . dx dx dx Example. To find dsin (y 2 ~ z ) , knowing that \ dx U = as 2 , Solution : D y sin (y 2 — z) = 2y cos (^ 2 — «) , IX, sin (y 2 — 2) = — cos(y 2 — «) . d l- dx 1 = — ) X dz _ dx 2x, <*nn(f-z) = 2yco S (y*-z) _ 2a , 2 _ g) dx x 2(y — x?) cos (y 2 — z) Confirm this result by expressing y and g in terms of x before differentiating. 200. If u=f(x,y) and y=Fx, the formula of the last article becomes dx dx Chap. XII.] FUNCTIONS, ETC. 205 Examples. (1) u = z 2 + tf + zy z = sina; v = e ^. , dw Find t . dx Ans. — = (Sy 2 +z)e' + (2z + y)cosx. dx (2) u=io g *r du y } Find -j-. dx y = sin x J (3) u = ta,n~ 1 (xy) y = e x „. , du Find -T-. da; .4ns. - — = etna;. da; a; da; 1 + x 2 y 2 (4) w = sin -1 ( - ) when z and y are functions of a;. Find — yj dx (5) u = ^| " -; — — „ ) when 2 and « are functions of x. Find — zr + y-y da; 201 . Higher derivatives of a function of functions of x can be obtained by an easy application of the method suggested by the formulas above. For example : u =f(y,z) , y=Fx, z=F 1 x, required d 2 u dx 2 ' d 2 u _ dx 2 ' d + dx D s D z u^+D 2 u D s 2 u^L+D y D z u dx d£ dx dy dx d V , ni,^' dy dx ' da; dy dz dz dx + ">(I)" + "W--4- -E + w(g'+A-8 +»■*& 206 DIFFERENTIAL CALCULUS. [Art. 202. In obtaining this formula, since y and z are given functions of x, -i and — are also explicit functions of x, and are therefore dx dx treated as constants in obtaining the partial derivatives with respect to y and z ; but now — is a function of (x,y and z) , hence dx w« must take also its partial derivative with respect to x. Example. Given u =f(x,y) , , , . d 2 u , d s u obtain — - and — ; . dar dx 3 V=Fx, Implicit Functions. 202. If, instead of having y given in terms of x, we have an equation connecting x and y, y is called an implicit ' function of • x, and — can be readily found by the aid of Partial Derivatives. dx Suppose /0,2/) = 0, to find — . Call f(x,y) u. dx Then u = ; hence — must also equal zero, dx ^=D x u+D,u*L dx dx dy = D x u dx D„u (1) ax m — ye ny = 0. Find dx Examples. dy Chap. XII.] Solution : or, as FUNCTIONS, ETC. D x u = max m ~ 1 , B y u— — e ny — nye ny , -D,u max m ~ 1 u + 2 Dx D v uf x+D> (^j + n y u^ 208 DIFFERENTIAL CALCULUS. [Art. 203. dy _ D x u dx~ D y u d 2 y_ D x i u{D y uy-2D x D y uD x uD v u+D s i u{D x u) i da? ~ (D v uy Examples. 11) f + a?-3axy=0. Find % Ans. ^ = - . f"' 8 * dx 2 da? (y 2 —ax) 3 (2) x l + 2aa?y-atf=0. Find ^ and ^|. da da? Chap. XIII.] CHANGE OF VARIABLE. 209 CHAPTER XIII. CHANGE OE VARIABLE. 204. If we use the differential notation, we have seen that there is no need of distinguishing carefully between function and independent variable, a single formula always giving a relation between the two differentials by which either can be expressed in terms of the other. This, however, is the case only when we are dealing with differentials of the first order. A differential of the second order or of a higher order has been defined by the aid of a derivative, which always implies the distinction between function and variable, and on the hypothesis of an important difference in the natures of the increments of function and varia- ble ; namely, that the increment of the independent variable is a constant magnitude, and that, consequently, its derivative and differential are zero. If, in any function involving differentials of a higher order than the first, we have occasion to change the independent va- riable, we can no longer assume that the differential of the old independent variable is constant, but must go back and replace all the differentials of higher order than the first by values ob- tained on the supposition that all the differentials are variable, before we attempt the introduction of the new variable, vide Arts. 187 and 188. 205. In any particular example in which it is necessary to change the variable, the method just described can be easily applied. Take the differential equation, , d 2 u du , A ar — - + x 1- u = 0, dx 2 dx 210 DIFFERENTIAL CALCULUS. [Art. 205. where x is the independent variable, and introduce y in place of x. Given y = \ogx. Our d 2 u here is the second differential of u taken on the assump- tion that x is the independent variable, and this can be indicated by writing it d 2 u, and we have dxd?u — dud i x >z <* — -* dy = dx = dx x ' ■■xdy, dx d?x = ■■ d (xdy) -- = xd 2 y + dxdy ; but d 2 y = 0, as y is to be the independent variable, hence d 2 x = dxdy, by Art. 188. , , 2 xdyd 2 u — xdudy 2 ™ j i and d x i u = — 2 i-=d'u — dudy: xdy l — afdx 1 — x 2 Given x = cost. Ans. — f-L«=0. d? ^ a (2) Change the variable from x to in the equation, + dy 2 )l dxd?y — dyd?x dx= —r sin d

+ @)j /° = - ! _,£T + 2^ V d? 2 dp .EXAMPLE. Find the radius of curvature of the circle r = cos x -2-— y r dx " and that ds? = dx 2 + dy 2 becomes d£=dr s + r 2 d

.*Wy-i>.yi>M t * . If x and y are given in terms of 2, we can calculate the values of D z x, D z y, D z x, and D z y, and substitute them in these form- ulas. Take Example (3), Art. 205. x= coat, {l-x*)D*y-xD x y=0, M,x p 2 D t xD?y-D t yD?x * y ~ (D t xf D t x= —sint, D, 2 x = — cosi, n D.y D x y= --^,, smt n2 -$\ntD?y + QostD t y — sin J £ 1 — a^= sin 2 *, . , -smtD?y + costD t y , costD t y _ n smt =lWt + sin* ~ °' D*y=0. 209. Suppose we have a function of two independent varia- bles, and its partial derivatives with respect to them, and wish 216 DIFFERENTIAL CALCULUS. [Art. 209. to introduce, in place of our old variables, two others connnected •with them by given relations. For example : let z be a function of x and y, and let it be required to introduce, instead of x and y, u and v, which are connected with x and y by given equations. If the equations can be readily solved so as to express u in terms of x and y, and v in terms of x and y, we may proceed as follows : — After the substitution, z is to be an explicit function of u and u. Suppose the substitution performed. As u and y are functions of x, z is indirectly a function of x. To get B x u, we suppose y constant, so that x is for the time being the only inde- pendent variable, and we can get D z z, by Art. 199, which gives us B x z = B u z B x u + B„ z B x v where all the derivatives are partial derivatives. In the same way, B y z=B u zB y u+B„zB v v. B x u, B x v, B y ii, and B y v are found from the values of u and v mentioned above, and are generally functions of x and y, and B x z and B s z are at first obtained in terms of u, v, x, and y. x and y must be replaced by u and v by the aid of the given equations, and B x z and B y z are then in terms of u and v alone. By extending the process, we can get B x z, B x B y z, Bj>z, &c, in terms of u and v. For example : introduce u and v in place of a; and y in the equation D x z=D y z. Given u — x + y v = x — y B x u = l, B x v = l, JV=1, B e v = -1, Chap. XIII.] CHANGE OF VARIABLE. 217 D.z=D u z + D,z, D v z=B u z — D v z, D x 2 z = D u h + 2D u D v z + D*z, D?z = D u *z -2D u D v z + DJ>z, D„ 2 z +2D u D v z + D*z = D a 2 z-2D u D„z + D*z, 4D„Z>„«=0. D u D v z = 0, the required equation. 210. If it is more convenient to express x and y in terms of u and v at the start, we can proceed thus : z is explicitly a func- tion of x and y, and if we regard v as constant for the time being, z is indirectly a function of the single variable w. Hence, D u z = D x zD u x + D s zD u y ; in like manner, D v z= D x zD v x + D s zD v y. D n x, D u y, D v x, and D v y are found in terms of u and v, and then by elimination between the equations, we get D x z and D v z in terms of u and v. Examples. (1) Given x = rcosD r z — smoDj,z) r D y z = - (rsm — ftp _ y — 2/0 _ z — zb • Ax Ay by Analytic Geometry ; or y-yo = -iyy x — x Ax Z — Z„ Az Az x — x Ax ■ Let Ax approach zero, and the secant line approaches the re- quired tangent as its limit, and this will have for its equations, y-yo x — x z — z X — Xn dy dxl x — Xa y- x Jx=Xf, or, writing them in a more symmetrical form, x — x _ y — yo _z — z dyo dx dz„ where, by -^?, we mean the value -^ has when x = x . dx dx A plane through the given point perpendicular to the tangent line is called the normal plane at the point in question. Prove that its equation is x-x + (y-y )^ + (z-z )p = 0. Example. 214. The helix is a curve traced on the surface of a cylinder of revolution by a point revolving about the axis of the cylinder 222 . DIFFERENTIAL CALCULUS. [Art. 214. at a uniform rate, and at the same time advancing with a uni- form velocity in the direction of the axis. We can easily express its equations by the aid of an auxiliary angle, the angle through which the point has rotated. Calling this angle 6 and the radius of the circle a, we readily see that x = acostf, y= asin.0. From the nature of the helix, z must be proportional to the z angle 6 ; hence ;=»*, e and z=M, The required equations are then x=a cos y=asaxd a constant, Chap. XIV.] TANGENT LINES AND PLANES. 223 To find the tangent line and normal plane at (x ,y ,z^) , dy = a cos 6d0, dx= —asmddd, ^=_ctn0 = dx X dz = kdff, dz _ k __ dx a sin 6 y The equations of the tangent are x — x = y — Vo _ z~ s o 1 x k Jo ~ V* or x-x _y-y _z-za [1] — y X k The normal plane is y (x-x )-x (y-y )-k(z-z o )=0. [2] The direction cosines of line [1] are, by Analytic Geometry, ~Vo COS a: or cosa = VOtf + ^ + fc 2 ) -Vo V(« 2 + ^)' rosr= VK+W 224 DIFFERENTIAL CALCULUS. [Art. 215 Cos j- is, then, not dependent on the position of the point P; therefore the helix has everywhere the same inclination to the axis of the cylinder ; or, in other words, it crosses all the ele- ments of the cylindrical surface at the same angle. If, then, this surface is unrolled into a plane surface, the helix will de- velop into a straight line. 215. The equations of the tangent line to the curve /(a;,y,«)=0, F(x,y,z) = 0, can be obtained in a very convenient form if we use partial derivatives. We have, by Art. 199, % = DJ+ DJ^L + DJ^- = dx dx dx OF dx dy ,dz = D x F+D tl F?Z + D,F'^ = dx dx (1) From these equations we can obtain the values of — and — ■ ■ dx dx Substituting these in the equations of Art. 213, and reducing, we get (x - x )D x J+ (y - y )D y J+ (z - z )D z J= (x - x )D x F+ (y - y )Dy F+ (z - z )D z „F= as the equations of the required tangent. The same result may be obtained much more easily by substituting in (1) the values of -f- and — given by the equations in Art. 213. u£c cms Chap. XIV.] TANGENT LINES AND PLANES. 225 Examples. (1) Given x 2 + y 2 — ax — ax + z 2 — a 2 = as the equations of a curve, find the tangent at (x ,y ,z ) . Ans. x x + y y + z z = a 2 a(x — x ) + 2 (z — z )z = (2) Given the circle x 2 + y 2 + z 2 = i x + z = a find the tangent at (x ,y ,z ) . Ans. x x + y B y + z a z = a 2 x + z= a 216. The osculating plane at a given point of a curve in space is the limiting position approached by a plane through the point and two other points of the curve as the latter approach indefi- nitely near the given point. If (x ,y ,z ) is the given point, and we regard x as our inde- pendent variable, we can represent two other points of the curve (Art. 190) by (x + Ax,y + Ay,z + Az) and (x +2Ax,y + 2Ay+A 2 y,z + 2Az + A 2 z). Forming the equation of the plane through these three points, dividing by Ax 2 , and taking the limiting values as Ax approaches zero, we shall get as the osculating plane, , , fdy cPz dz d 1 y\ , „ . (d 2 z\ , ,/j n Example. Obtain the osculating plane of the helix at (x ,y ,z ) . 226 DIFFERENTIAL CALCULUS. [Art. 217. 217. The tangent plane at a given point of any surface f(x,y,z) = can be found by the aid of the equations of Art. 215. Let F(x,y,z) = be any second surface whatever passing through (x ,y ,z ) . The tangent line to the curve of intersection of the two surfaces at the point (x ,y ,z ) , that is, to any curve through lx ,y ,z ) traced on the given surface, has for its equations (x - x a )D x J+ (y - y )DyJ+ (z - z )D z J= (a> - X )D X F+ (y - Vl> )DyF + (z - z a )D z F= It therefore lies in the plane represented by the first of these equations, which must then be the required tangent plane, (x - x )D x J+ (y - y )Dy o f+ (z - z )D Zo f= 0. Examples. (1) Find the tangent plane to a sphere. x 2 + y 2 + z 2 = a 2 . Ans. x x+y y + z z = a 2 . (2) Find the tangent plane to an ellipsoid. a 2 W c 2 Ans. Jg + ^ + M.!. The normal line at (x ,y ,z ) is easily seen to be x— x„ _ y — y _ z — z D,f D y J D.f' Chap. XV.] FUNCTION OF SEVERAL VARIABLES. 227 CHAPTER XV. DEVELOPMENT OP A FUNCTION OF SEVERAL VARIABLES. 218. To develop f(x + h,y + k) into a series arranged accord- ing to the powers of h and k, where h and k are any arbitrary increments that may be given. Let a be any variable, and call h , k , — = "i» — = "i> a a so that h = a/ij and k = a\. If now x and y are regarded as given values, f(x + h,y + k~) Is a function of h and k, which depend on a ; and hence f(x + h,y + k) can be considered a function of a. Call it Fa, and it may be developed by Maclaurin's Theorem, which gives Fa = F0 + aF'0 + — F"0 + —^F"'0 + + — F (n) + ■ "" +1 F< n+1 > 0a. n\ (n+1)! When a = 0, Fa or F(x + h,y + k) =f(x,y) . Call x+ah l = x' and y + ak 1 =y\ then Fa=f(x',y'), F , a = df(x>,y>) =Dxif(xW) §f + DtAafl tf f, da Cla aa by Art. 199 ; 228 DIFFERENTIA!. CALCULUS. £Akt. 218. dx' h aa — = «,, aa F'a = h,D x .f(x\y') + k^f^.y'), F'0 = hD x f{x,y) + k.DJ^y) , which we shall write 7^1),./+ k x D y f. F"a= ( **-2 = h x D x ,F'a + k,D^ F'a da = hfDJfix^y') + 2fcA D„ D s .f{x',y<) + ^D/J\x>,y<), F"'a = h?D/f{x\ y') + MfhDSDifix 1 , y<) + 3J h k^D x , D/f(x', y') + h*D/f(x\ y') . In F" a and F"' a the terms have a striking resemblance to the terms of the second and third powers of a binomial. Let us see whether this will hold for higher derivatives. Assume that it holds for the F (n) a, and see if it holds for the .F<" +1) «. If f«a = h 1 n D x . n f(x\y')+nh 1 n - 1 k 1 D x , n - i D !l ,f{x',y') + n{n ~ 1) K-*kSD x ,»-'D v :-f(;x',y<) + "(" -!)("- 2) Kr-*k»D^DSW, y') + o ! F<" +1 > a = \Bj F™ a + \D y , F«> a = ft 1 " +1 .D 1 ," +I /(s^y , ) + (n + WhDSDsf^y') Chap. XV.] FUNCTION OF SEVERAL VARIABLES. 229 If, then, the observed analogy holds for any derivative, it holds for the next higher. It does hold for the third ; it holds then for the fourth, and for all succeeding ones. F"0 = VZ» X 2 /+ 2h i k l D a DJ+ yfl,'/, F^O = KD x n f+ nhf-^Df-^DJ p(*+i) 6a _ h«+iD, n+1 f(x + dh,y + 6k) + (n + l)h 1 n k 1 D x n D„f(x + 6h,y + 6k) + By this notation we mean that x + 6h, y + 6k, are to be sub- stituted for x and y after the differentiations are performed. We have then, remembering that ahy = h and aTq = k, f(x + h,y + k)=f(x,y) + (hD x f+kD v f) + i (WD*/ + 2hhD.D,f+ k'Dff) + i- (h*DJ>f+ 3h?kD x *D y f+ SWD.D*f+l?D*f) + 3! +±-(h»D x y+nh»- 1 kD x «-WJ+?l?^h"-WD x »-'D v *f+ ) + , * , (h^D x ^f(x + 6h,y + 6k) (n + l)l \ + (n + l)h n kD x n D y f(x + 6h,y + 6k) + (n+\)n h —ij? D .-ij)»f(g, + eh,y + 0k) + Y If we use (JiD x + kD ll ) n f as an abbreviation for (h n D x n f J-nh n - 1 kD x n - 1 D l/ f+ ); that is, understanding Va.&t{hD,+kD y ) n 230 DrPPEKBNTIAL CALCULUS. [Art. 218. is to be expanded just as though it were a binomial, and then to have each term written before /, we can simplify the above ex- pression. f(x + h,y + k) =f(x,y) + (hD x + kD y )f(x,y) + jj (hD x +kD s yf(x,y) + ± (hD x + kD y )*f(x,y) + + ± ] (liD x + kD v yf(x,y) + . * , (W>. + kD y y^f(x + 6h,y + 6k) , which is Taylor's Theorem for two independent variables. If we let x = and y—0, we get f(h,k) =/(0,0) + (hD x + kD s )f (0,0) + ±(hD x + kD y ) i f(0,0) + ; or, changing h and k to x and y, A*,y) =/(0,0) + (xD x + yD y )f (0,0) + i (*D. + yi) y ) 2 /(0,0) + ± (xD x + yD s Yf (0,0) + + ± (xD x + yD y yf (0,0) + * (xD x + yD y y^f(6x,e y ) , which is Maclaurin's Theorem for two variables. Example. Transform Ax 2 + Bxy + Cy* + Dx + Ey + F=0 to (x ,y ) as a new origin, the formulas for transformation being x = x + x', y = y + y'. Chap. XV.] FUNCTION OF SEVERAL VARIABLES. 231 Call our given equation f(x,y) ; we wish to develop f(x + x',y + y'). f(x + x',y + y<) =f(x , yf> ) + (x'D Xn + y'Dy o )f(x ,y ) D x f(x,y) = 2Ax + By + D, D v f(x,y)=,Bx + 2Cy + E, D*f(x,y) = 2A, D x D y f(x,y) = B, D,*f(x,y)=2Ci all higher derivatives are 0. f{x + x', y + y') = AxQ under these same. circumstances, u is a minimum. By Taylor's Theorem, f(x + h,y + k) -f(x ,y ) = (hD x + kD s ) f(x ,y ) 2! + ^ (hD x + kD y Yf(x + 6h,y + 8k) . If we take the values of h and k sufficiently small, we can always make £ (hD x + kD v Yf{x + Oh,y + 6k) < (hD x + kB y ) f(x , y ) , and then the sign of the second member will be the sign of (hD x + kD y ) f(x ,y ); that is, of 7iD x u + kD y w , which evi- dently depends upon the signs of h and k. In order, then, that the sign of f(x + h,y + k) —f(x ,y ) should be constant, — that is, in order that for x ,y u should be either a maximum or a minimum, — the terms hD x u + kD y u must disappear, no matter what the values of h and k ; or, in other words, D x u and D y u 'Va must both equal 0. We get, then, as essential to the existence of either a maximum or a minimum, the conditions for the values of x and y in question. Chap. XVI.] MAXIMA AND MINIMA OF FUNCTIONS. 235 222. Carrying the development a step farther, and assuming that D x m and D y u are zero, /Oo + h,y + ft) -/Oo,2/o) = ^y (AD. + kD y ) 2 f(x ,y ) + ij (&Z>. + JcD y ) 3 f(x + 6h,y + 6k) . As before, it is evident that for small values of h and ft, the sign of the whole second member will be that of the terms % (h 2 D x 2 u a + 2hk D x D y u + k 2 D y 2 u ). Let us investigate this carefully. Let A = D x u , B = Dx a D y u , (7= Dy Q U , our parenthesis becomes Ah 2 + 2Bhk + Oft 2 ; and for a maximum or minimum the sign of this must be independent of the signs and values of h and ft. Ah 2 + 2Bhk + Cft 2 = i- (A 2 h 2 + 2ABhk + ACk 2 ) , Jx = - (A 2 h 2 + 2ABhk + B 2 k 2 - B 2 k 2 + ACk 2 ) , A = 1 [(Ah + Bk) 2 + (AC - JS 2 )ft 2 ] . (Ah + Bk) 2 and ft 2 are necessarily positive. If AC — B 2 is also positive, the sign of the whole expression will be independent of h and ft, and will be positive if A is positive, and negative if A is negative. If AC — B?=0, the result is the same ; but if AC— B 2 is negative, the sign of the parenthesis will de- pend upon the sign and relative values of h and ft, and we shall have neither a maximum nor a minimum. 236 DIFFERENTIAL CALCULUS. [Art. 223. 223. To sum up : — If D x u =0 D y u = Q Dx\ «o D y * Mo — (Dx Dy M ) 2 = or > If D x u = Dy o u =0 D*v*DJv t -(D m D,u 9 y= or>0 ■ m is a maximum. D x 2 u >0 u is a minimum. Examples. 224. (1) To find a point so situated that the sum of the squares of its distances from the three vertices of a given tri- angle shall be a minimum. Let (#1,2/1), (#2,2/3), (#812/3) be the given vertices, and (x,y) the required point. u = (x- xj 2 + (2/ - 2/0 2 + (x - x 2 ) 2 + (y - y 2 y + (x-x s y + (y-y B y is the function which we must make a minimum. D x u= 2(x — x{) + 2(x — x 2 ) + 2(x — x a ) , D,u=2(y-y 1 ) + 2(y-y i ) + 2(y-y s ), D*u = 2 + 2 + 2 = 6 = J., D x D s u = = B, D s 2 u = 2 + 2 + 2 = 6=0. "We must make D x u and D y u both equal to zero. Chap. XVI.] MAXIMA AND MINIMA OP FUNCTIONS. 237 2(x -x 1 ) + 2(x- x 2 ) + 2(x-x 3 ) = 0, x x l ~\~ x 2 ~f" x $ ~ 3 2 (y - Vi) + 2 (y - y t ) + 2 (y - y 3 ) = 0, y ~ 3 ' AC - B 2 = 36 -0>0, ^1=6>0. Hence u is a mininmm when afr + g. + g , and y! + y 2 + y 3 3 3 3 The required point is the centre of gravity of the triangle. (2) To inscribe in a circle a triangle of maximum perimeter. Join the centre with each vertex and with the middle point of each side. The angles between the three radii are bisected by the lines drawn to the middle points of the sides. Call these half-angles ft, 2 , 3 . - -=- r = sin ft, 2 a= 2r sin ft, b= 2rsin0 2 , c= 2rsin0 s , 238 DIFFERENTIAL CALCULUS. TAkt. 224. 20, + 20ft + 20 3 = 2n, 8i+0 2 +6 s = t:, (1) p = a + b + c= 2r(sin0 1 + sin0 2 + sin0 s ) is the function we are to make a maximum, and is a function of two independent variables, say 1 and 8 2 ; for we can regard 8 as depending on 0j and 2 through equation (1). As r is a con- stant, it will be enough to make u = sin0 1 + sin0 2 + sin0 3 a maximum. D^u = cos 0! + cos 03.D0J03 ; for, since X + 2 + g = t, changing 6 X without changing 2 will change g . -Ds 1 3 =-i; hence De t u = cos a — cos S . Dg^u = cos 2 — cos 3 , for Z>8 2 3 =-1, D^u = — sin 0! — sin S , De x Z>9 2 M= — sin0 3 , De^u = — sin 2 — sin S . Make Dg^u = and -Dfl 2 « = 0. cos X — cos 3 = ' COS0 2 — COS0 8 = Oj 0, = 2 = g . Chap. XVI.] MAXIMA AND MINIMA OF FUNCTIONS. 239 Substitute these values in Dg*u, &c, and D s \u= — 2sin0 1 = ^l, De 1 De^u = — sin 0j = B, Dfiu=— 2sin0,= C\ AQ—W = 4sin 2 ^ — sin 2 ^ = 3 sin 2 ^>0, A = — 2 sin ^ e curve is concave towards the axis of a;, if above the axis ; convex, if below. At any d 2 v point corresponding to a value of x that makes 753 >0, the curve is convex towards the axis of a;, if above the axis ; con- cave, if below. Any point corresponding to a value of x that makes zJ! = dx 2 is in general a point of inflection. We have seen that the curvature, _fy dx 2 k = - H It is easily seen that at a point of inflection this value changes sign. 228. These same tests for concavity, convexity, and inflec- tion can be very simply obtained by the aid of Taylor's Theorem. Let y =fx be the equation of a curve, and let it be required to discover whether the curve is concave or convex toward the axis of X at the point corresponding to the value x=a. Draw a tangent at the point in question, and erect ordinates to the curve and to the tangent near the point of contact. It is evident that the ordinate of a point in the curve minus the ordinate of the corresponding point of the tangent must be negative on both sides of the point of contact, if the curve is concave, and positive on both sides of the point of contact, if the 242 DIFFERENTIAL CALCULUS. [Art. 228. curve is convex. If the point is a point of inflection, this differ- ence will have opposite signs on different sides of the point. INFLECTION. The equation of the tangent at the point corresponding to x = a is y-fa=fa(x-a), by Art. 28, [1]. Let x = a + h in the equations of curve and tangent, and call the corresponding values of y, y x and y 2 ; then y 2 =fa + hf'a. Vl =fa + hfa + £ f"a + £ f"(a + eh) , by Taylor's Theorem. K* yi-y, = ^ ] f"a + ^- l f'"{a + eh). If f"a does not equal zero, h may be taken so small that the h 2 sign of y x — y% will be the sign of — f"a. — I Iff "a is positive this sign is positive whether h is positive or negative, and the curve is convex. If f"a is negative, y x — y 2 is negative both before and after x = a, and the curve is concave. If f"a = and f'"a does not vanish, the sign of y x — y 2 will change as the sign of h changes, and we shall have a point of inflection. Chap. XVII.] THEOKY OF PLANE CTTKVES. 243 229. For example, let us see whether the curve sb 2 -}- 2/3=25 is convex or concave towards the axis of X at the point (3,4). 2xdx + 2ydy=0. (1) 2dx 2 +2dy 2 +2yd 2 y = 0. (2) From(l) dy=-—. y Substitute in (2) , 2dx> + ^^ + 2yd 2 y = 0, y (x* + tf)dx> + tfd*y = 0, 25dx* + tfd 2 y=0. da? f 64 at the point (3,4); and the curve is concave. Again, let us see whether the curve y = x (x — a)* has points of inflection %=(x-aY + ±x{x-a)\ d?y da? = 36 (a; - a) 2 + 24iB(a; - o) , d*y ^ = 96 (a; -a) + 24®. Write g-o. 244 DIFFERENTIAL CALCULUS. [Art. 229. and we get 8 (x — a) 8 +12x{x.— a) 2 = or 2(x — a,y + 3x(x-ay — 0. One root is x = a ; divide by (a; — a) 2 , and 2x — 2a + 3x = 0. 2a . ., x = — is tne remaining root. 5 If x=% — f does not equal zero, and we get a point of inflection. dx 3 If x=a, ^ = dx 3 "' d*v — | does not equal zero, and the point is not a point of inflection. (1) If Examples. a 2 + ! there is a point of inflection at the origin, and also when a=±aV(3). « If NC"?)' 3a there is a point of inflection when a; = — .. 4 (3) If a* = logy, there is a point of inflection wnen x = 8. Chap. XVII.] THEORY OF PLANE CURVES. (4) If xy = a 2 log-, there is a point of inflection when x = ael. 245 Singular Points. 230. Singular points of a curve are points possessing some peculiarity independent of -the position of the axes. Such points are, — 1. Points of inflection (Art. 228); 2. Multiple points ; 3. Cusps ; 4. Conjugate points ; 5. Points d' arret ; 6. Points saillant. 231. (2) A multiple point is one through which two or more branches of the curve pass. If only two branches pass through DOUBLE POINT. OSCULATING POINT. CONJUGATE POINT. POINT D'ARRET. POINT SAILLANT. the point, it is a double point. A double point at which the branches of the curve are tangent is an osculating point. (3) An osculating point where both branches of the curve stop is a cusp. (4) An isolated point of a curve is a conjugate point. 246 DIFFEBENTIAL CALCULUS. [ART. 232. (5) A point at which a single branch of a curve suddenly stops is a point d'arrit. (6) A double point at which the two branches of the curve stop without being tangent to each other is a point saillant. Multiple Points. 232. At a multiple point, the curve will in general have more than one tangent, and therefore -^ will have more than one value. dx Let f = be the equation of the curve in rational algebraic form. dy _ —D x

and D y y will have each a definite value, as they are rational polynomials ; and -^ will have but one value, unless D x

a (Jx) n+1 p n+1) (a + 6Ax), ' (n+1)!' y 2 = Fa + AxF'a + (Ja ^ F"a + + !MI F™a (Jx) n+1 F(«+»( a + e'Ax). + (n + l)! V T ' If the curves have contact of the nth order, Fa =fa, F'a=f'a, etc., F™a=f™a. 2/i-2/2= £^r! lf n+1 >(a + 0Jx)- F*+»(a+6'Ax)], (n+1)! Chap. XVII.] THEORY OP PLANE CURVES. 253 which is infinitesimal of the (w+l)st order, if Ax is an infini- tesimal. It follows, then, that the order of contact indicates the closeness of the contact ; that is, the higher the order of contact of two curves, the less rapidly they recede from each other as they depart from the point of contact. 239. Let it be requhed to find the equation of the circle having contact of the second order with the curve y=fx at the point (x^yi). Let a and b be the coordinates of the centre, and r the radius of the required circle. Call (X,Y) any point of the required circle, then its equation is (X-a) 2 + (Y-b) 2 = r i . By our conditions, [-r^l =(-¥■] > J \dXjx=x 1 \dxJ a . =:Vl ' (d 2 Y\ = f i tfy) / r 2 V dy x-a= — \ /; d*y dx W/ substituting in (x — a) 2 + (y — &) 2 = r 2 , ufcc 2 / W/ *(£)'-+ (I)' [ 1+ (D r=± do 2 which is the familiar value of the radius of curvature of y=fx at the point (x,y). Hence, our osculating circle is that circle having contact of the second order with the given curve at the point in question. Chap. XVII.] THEORY OE PLANE CURVES. 255 Examples. (1) In the curve y = x* — Ax 3 — 18a; 2 , show that the radius of curvature at the origin is ^g-. (2) Find the parabola whose axis is parallel to the axis of ', which has the closest ] at the point where x = a. ™3 Y, which has the closest possible contact with the curve y ■= — a 2 Result, (x )=-(y ]. (3) Prove that, if the order of contact of two curves is even, they cross each other at the point of contact ; if odd, they do not cross. Envelops. 240. If the equation of a curve contain an undetermined constant, to different values of this constant will correspond different curves of a series. Such an equation is said to contain a variable parameter, the name being applied to a quantity which is constant for any one curve of a series, but varies in changing from one curve to another. For example : in the equation (x — a) 2 + y 2 = r 2 , let a be a variable parameter ; then the equation represents a series of circles, all having the radius r, and all having their centres on the axis of X. A curve tangent to each of such a series of curves is called an envelop. 241 . Two curves of such a series corresponding to two differ- ent values of the parameter will in general intersect. If they are made to approach each other indefinitely, by bringing the two values of the parameter nearer together, their point of intersec- tion will evidently approach the enveloping curve, which then may be regarded as the locus of the limiting position of a point 256 DIFFERENTIAL CALCULUS. [Art. 241. of intersection of any two curves of the series as the curves are made to indefinitely approach. From this point of view the equation of an envelop is easily obtained. Let /(z,2/,«) = (1) be the given equation of the series of curves, a being a variable parameter. f(x,y,a -f Aa) = (2) will be any second curve of the series. The equation f(x,y,a + Aa)-f{x,y,a)=0 (3) represents some curve passing through all the points of intersec- tion of (1) and (2) by the principle in Analytic Geometry : " If u = and v = are the equations of two curves, u + kv = rep- resents a curve containing all their points of intersection, and having no other point in common with them." f(x,y,a + A a )-f(x,y,a) Aa " is equivalent to (3) . If, now, Aa be decreased indefinitely, limit Ja=0 ~ f(x,y,a+Aa)-f(x,y,a) ~\_ t Aa D a f{x,y,a) = 0, (4) contains the limiting position of the point of intersection of (1) and (2). Let (x',y') be this point, and therefore any point of the required locus. Since (x',y') is on (4), and also on (1), D a f(x',y',a) = and f(x',y',a) = 0; we can eliminate a between these equations, and we shall have a single equation between x' and y', which will' be the equation of the required envelop. Chap. XVII.] THEORY OF PLANE CURVES. 257 242. For example : let us find the envelop of (x -ay + tf-r>=0, (1) u being a variable parameter. D a f=-2(x-a) = 0. X — a = 0. (2) Eliminating a between (1) and (2) , we get 2/ 2 - r 2 = 0, the equation of a pair of straight lines parallel to the axis of X, as the required envelop. 243. When dealing with the properties of evolutes, we proved that every normal to the original curve must be tangent to the evolute. We ought, then, to be able to find the evolute of any curve by treating it as the envelop of the normals of the curve. Let y —fx be the equation of the original curve y-y»=-(p) (v-vo) \dyjx=x is the equation of the normal, or (^y y -y ) + x-x =0. (1) x is the variable parameter, ^=§°^KS) 2 - 1=0 ' (2) \dx 0j d 2 y dx 2 258 DIFFERENTIAL CALCULUS. [Art. 243. %or 1+ M/oy~| x=x — ■ <* go dx 2 but 2/o=/*o, and we must eliminate x and 2/07 by the aid of these three equa- tions, to obtain the equation of the evolute. These equations are the ones obtained by a different method in Art. 93. Examples. (1) Find the envelop of all ellipses having constant area, the axes being coincident. Result. A pair of equilateral hyperbolas. (2) A straight line of given length moves with its extremities on the two axes, required its envelop. Result. xi + yi= ai. (3) Find the envelop of straight lines drawn perpendicular to the normals to a parabola y 2 = 4ax at the points where they cut the axis. Result, y 2 = 4a (2a — a;) . (4) Circles are described on the double ordinates of a parab- ola as diameters. Show that their envelop is an equal parabola. (5) Find the envelop of all ellipses having the same centre, and in which the straight line joining the ends of the axes is of constant length. Result. x±-y=±c. (6) Show that the envelop of a circle on the focal radius of an ellipse as diameter is the circle on the major axis.