jlMjt 7/Uy^^:-^ From the Library of Professor Arthur Hall donated by- Major Douglas Merkle 210 1^ CORNELL UNIVERSITY LIBRARY GIFT OF Professor Joe Bail Cornell University Library QA 303.T24 1902 Elements of the differential and Integra 3 1924 015 968 146 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924015968146 ELEMENTS OF THE DIFFEEENTIAL km INTEGRAL CALCULUS WITH Examples and Applications JAMES M. TAYLOR, A.M., LL.D PKOFESSOK OP MATHEMATICS COLGATE USIVEUSITY BOSTON, U.S.A. GINN & COMPANY, PUBLISHERS 1902 COPYBKiHT, 1884, 1891 By JAMES il. TAYLOR COPYKIGHT, 1898 By JAMES M. TAYLOR ALL r.TOHTS RF.SF.RVED PREFACE TO THE REVISED EDITION. oXKo In this revision an attempt has been made to present in their unity the three methods commonly used in the Calculus. The concept of Rates is essential to a statement of the prob- lems of the Calculus ; the principles of Limits make possible general solutions of these problems, and the laws of Infini- tesimals greatly abridge these solutions. The Method of Rates, generalized and simplified, does not involve "the foreign element of time." For in measuring and comparing the rates of variables, the rate of any variable may be selected as the unit of rates, dy jdx is the x-rate of y, or the ratio of the rate of y to that of x, according as the rate of X is or is not the unit of rates. The proofs of the principles of differentiation by the Method of Rates, and the numerous applications to geometry, mechanics, etc., found in Chapter II, render familiar the problem of rates before its solution by the Method of Limits or Infinitesimals is introduced. In Chapter III, by proving that It (Ay/Ax) = dy jdx, the problem of rates is reduced to the problem of finding the limit of the ratio of infinitesimals. The Theory of Infinitesimals is that part of the Theory of Limits which treats of variables having zero as their common limit. In approaching its limit an infinitesimal passes through a series of finitely small values before it reaches infinitely small values. Infinitesimals can be divided into orders, and their laws can be established and applied when iv PREFACE. they are finitely small as well as when they are infinitely small. Hence, in the study of infinitesimals it is not neces- sary to determine that indefinite boundary between the finitely small and the infinitely small. Any small quantity becomes an infinitesimal when it begins to approach zero as its limit, not when it reaches any- particular degree of smallness. A quantity, however small, which does not approach zero as its limit is not an infinitesimal. If it is recognized that the essence of infinitesimals lies in their having zero as their limit, rather than in their smallness, the study of them ceases to be mystical, obscure, and difficult. Again, the concept of a limit as a constant whose value the variable never attains removes the necessity of studying the anatomy of Bishop Berkeley's "ghosts of departed quanti- ties." Infinitesimals never equal zero and should not be denoted by the zero symbol. This distinction between infini- tesimals and zero involves that between infinites and a/0. The much-abused form 0/0 cannot arise in the Calculus or elsewhere from any principle of limits ; a distinctive service of the Theory of Limits is that it enables us to evaluate any determinate expression when it assumes this or any other indeterminate form. Those who prefer to study the Calculus by the Method of Limits or Infinitesimals alone can omit the few demonstra- tions in Chapter II, which involve rates, and substitute for them the proofs by limits or infinitesimals in Chapter III. To meet an increasing demand for a short course in diifer- ential equations, a chapter has been devoted to that subject. A table of integrals arranged for convenience of reference is appended. Throughout the work, as in previous editions, there are numerous practical problems from mechanics and other branches of applied mathematics which serve to exhibit the usefulness of the science, and to arouse and keep alive the interest of the student. PREFACE, V At the option of the teacher or reader, Chapters I and II of the Integral Calculus can be read after completing Chapters I and II of the Differential Calculus ; also many of the numerous examples and problems may be omitted. The author takes this opportunity of expressing his grati- tude to the friends who by encouragement and suggestions have aided him in this revision. James M. Taylok. Hamiltox, N. Y., 1898. OO^TEI^TS. PART I. — DIFFERENTIAL CALCULUS. 1,2 3-5 6,7 8 9 10 Chapter I. Functions. Bates. Differentials. Sections 1-5. Classification of variables and functions . . 6-8. Notation of functions. Continuity. Increments . 9, 10. Uniform cliange. Measure of rates IX Differentials . . ... 12, 1.3. dy /dx, a rate or a ratio of rates . . . 14. Definition of tlie Differential Calculus Chapter II. Differentiation. Applications. 15-28. Differentiation of algebraic, logarithmic, and exponential functions . . 11-20 29, 30. Derivatives. Velocity . . . .... 21-23 31-33. Tangent, dy/dx = slope ... .... 24 34, 35. Equations of tangents and normals . . . . 25 36. Values of subt., subn., tan., and norm. . 26-29 37-44. Differentiation of sin «, cos u, etc 30-32 45-52. Differentiation of sin-i'M, cos-'m, etc. . . 38-37 Miscellaneous examples 38, 39 VIH CONTENTS. Chapter III. Problem of Bates Solved by Limits. Sections Pages 53-55. Limits. Notation. Lt {Ay /Ax) = dy/dx . . 40-42 56,57. Lt (Ay/Ax), how found. Derivative as a limit . 4.3 58-60. Infinitesimals. Infinites. Lt (Ay / Ax) — slope 43, 44 61,62. Theorems concerning limits . 45 63. Subt., subn., tan., and norm., in polar curves 46,47 64-66. Orders of infinitesimals. Notation ... 48, 49 67-70. Theorems concerning infinitesimals . . 50 71. Rule for differentiating a function . ... 51-53 72,73. Orders of infinites. Qon.o = 54 74. The symbol ap ... 55 75, 76. Use of infinitesimals. Limits in position . .... 56 Chapter IV. Successive Differentiation. 77. Successive differentials . S7 78-80. Successive derivatives ... 58-61 81, 82. Leibnitz's theorem. Acceleration . 62-64 Chapter V. Indeterminate Forms. 83. Value of a function of a; for a; = a 84. Indeterminate form 0/0 .... 85. Indeterminate form ap/ap 86. Indeterminate forms d ■ ap, ap — ap . . 87. Indeterminate forms 0", c^", 1 - ), and d(-) = d{- ■ z^-- dz. 20. d(logaU) H m • du/u, where u is positive. [6] That is, the differential of the logarithm of a variable is the modulus of the system into the differential of tlie variable divided by the vnrhilile. For, n being a general constant, let u = ny. (1) . ■ . log^M = log„ra + log^y. (2) From (2), by [1], [2], and [3] we obtain dQog^u) = dQog^). (3) Differentiating (1) and dividing by (1), we obtain du/u = dy/y. (4) LOGARITPIMIC FUNCTIONS. 13 Dividing (3) by (4), we obtain d GogaW) ■.du/u = d (log^) -.dy/y. (5) It remains to prove that the equal ratios in (6) are constant. Let m denote the common ratio in (5) when y = y' ; then d (log„M) =^171 ■ du / u, (6) when u = ny'. But, as ?i is a general constant, ny' denotes any number ; hence (6), or [5], holds true for all values of u, m being a constant. The constant m is called the modulus of the system of logarithms whose base is a. The modulus of the common system of logarithms, obtained in § 97, is 0.434294 • • •. From the nature of logarithms we know that logaU changes the faster the smaller we take a. From [5] we learn that logaM changes at the rate of m/M to 1 of u. Hence, the modulus m varies with the base a. Ex. The number u changes how many times as fast as logiou when u - 2560 ? du u 2560 d (logaU) in 0.434^H ■ = 5895, nearly. That is, u changes nearly 5895 times as fast as Its common logarithm when u — 2560. 21. Natural logarithms. The system of logarithms whose modulus is unity is called the Naperian or natural system. The symbol for the base of the natural system is e. Hence, d(logeU) = du/u. [6] Natural logarithms are evidently the simplest and most natural for analytic purposes. Hereafter when no base is written, e is understood. In the natural system du = ud (log u) ; that is, the number u changes u times as fast as its natural logarithm. 14 DIFFERENTIAL CALCULUS. 22. d(uy)=ydu + udy. [7] When u and y are both, positive, we have log {uy) = log M + log y; _ d{uy) ^du ^ dy by [61 ' ' uy u y ' .' . d (uy) = ydu + udy. (1) Multiplying (1) by ab, by [4] we obtain d (au ■ by) =by -d (au) + au ■ d(by). (2) Since a and b are general constants, au and by denote any, variables, real or imaginary. Hence [7-] holds true for any real or imaginary values of u and y. 23. d(uvyz- • •) = (vyz • • •) du + (uyz • ■ •) dv + (uvz • • ■) dy H . [8] That is, the differential of the product of any number of variables is the sum of the products of the differential of each into all the rest. If in [7] we put vw for u, we obtain d (vwy') = ydiyw) + vwdy = wydv + vydw + vwdy. (1) By repeating this process the theorem is proved for any number of variables. liv= w = y, (1) becomes d{y^) = 3 y'^ily. 24. d(u/y)s(ydu-udy)/y2. [9] That is, the differetitial of a fraction is the denominator into the differential of the numerator minus the .numerator into the differential of the denominator, divided by the square of the denominator. EXPONENTIAL FUNCTIONS. 15 Let « = w/y ; then zy = u. .•.ydz + sdy = du ; by [1], [7] .■.dz = (du-zdy)/y, or ^ /m\ <^M - {u/y)dy ^ ydu - udy \yj y y^ Cor. d(a/y) = -ady/yl [10] 2/ r 2/ 25. Differentials of u^, b^, u". When u is positive, logaC^") = y logaW ; (^("m") du •'•'"''^;r = '^yi;;: + ^°SaU-dy; by [5] .-. d(uy) syuy-^du + u^ log^u dy/m. [11] Putting h for u in [11], by [2] we obtain J I d (by) = by log^b dy/m. [12] Putting n for y in [11], by [2] we obtain d(u°) = nu°-'du. [13] Multiplying [13] by c", by [4] we obtain d{cuf = n(ciCf--'^ dicu). (1) Since cm in (1) denotes any variable base, [13] holds true for any value of u, positive or negative, real or imaginary. The function m" is continuous only when u is positive ; hence [11] and [12] are limited to positive values of u and h. 16 DIFFERENTIAL CALCULUS. 26. Stating [13] in words, we have The differential of a variable base affected with a constant exponent is the product of the exponent, the base with its exponent diminished by one, and the differential of the base. CoK. dVu = du/2Vu. [14] Por d{v}''')=\u-^'''du = du/2-s/u. 27. Stating [12] in words, we have The differential of an exponential function with a positive constant base is the function itself into the logarithm of the base into the differential of the exponent, divided by the modulus of the system, of logarithms used. Cor. In the natural system, m = 1 ; hence, as log e = 1, from [12] we have d(by) = bylogbdy; [15] and d (e^) = e^ dy. [16] 28. Comparing [13] and [12] with [11], we see that The differential of an exponential function with a positive variable base can be obtained by first differentiating as though the exponent were constant, and then as though the base were constant. EXAMPLES. By one or more of the preceding formulas exclusive of [5], [6], [11], [12], [15], [16], differentiate 1. y = IS — 8 a; + 2 x2. d!/ = d (a;3 — 8x + 2 x2) \^j j-l"! = d(x3) + cJ(-8a;) + d(2x2) by [3] = 3 xHx — 8 dx ■+ 4 xdx. by [4] , [l.S] . . % /da; = 3 x^ — 8 + 4 X = the x-rate of y. § 12 algebEaic functions. 17 That is, y, or x' — 8 a; + 2 x^, changes at the rate of 3 a;^ — 8 + 4 a; to 1 of K ; or 2/ changes 3 a;^ — 8 + 4 x times as fast as x. When x = — 4, ^ is increasing at the rate of 24 to 1 of x ; ■when X = 0, y is decreasing at the rate of 8 to 1 of x ; when x = 5, y is increasing at the rate of 87 to 1 of x. 2. y = 3 ax^ — 5nx — Sm. dy = (6 ax — 5n) dx. Note. At first the student should give the meaning of each differ- ential equation which he obtains. 3. 2/ = 5 ax2 — 3 b^x' — ahx^. dy~(\S)ax — % V^^ — 4 abx^) dx. 4. 2/ = aS + 5 62a;s + 7 aH^. dy = {15 bV + 35 aH^) dx. 5. 2/ = ax3/2 + 6x1/2 + 0. (Z2//dx = (3ax + 6)/2Vx. 6. y = (b + ax^Y '*• dy = ^(b + ax'Y '* axdx. 7. 2/ = (ex + 6x3)* / 3. dy = I (ex + 6x8)1 / 3 (c 4- 3 bx^) dx. 8. 2/ = (1 +2x2) (1 + 4x3). d2/ = 4x(l + 3x + 10x3)dx. dy = (1 + 2x2) d(l + 4x3) + (1 + 4x3) d(l + 2x2). 9. y = (x + l)5(2x-l)3. (i2/=(16x+l)(x+l)4(2x-l)2dx. / , > / '^y a — 3x 10. 2/ = (a + x)Vci — X. — = — , dx 2va — X 11. y = (x^ + 1)^(2 x^ + x)K dy = {x^ + l) (2 x2 + x)2 (20 x3 + 7 x2 + 12 X + 3) dx. 12. 2/ = (1 - 3 x2 + 6 x*) (1 + x2)3. dy = 60 x6 (1 + x2)2 dx. 13. 2/ = x«(a + 2x)3(a-3x)2. d2/ = 6 x5 (a + 2 x)2 (a — 3 s) (a2 — ax — 11 x2) dx. ■r -i- ni 14. y- 15. y = x + a2 d?/ _ 6 - a2 x + 6 dx (X + 6)2 (x + 6)d(x + a2) -(x + a2)d(x + 6) (X + 6)2 2x* dy 8a2x3 — 4x5 a2-x2 dx (a? - x2)2 di/ _ 2 ax + 6 Vax2 + 6x + c. S^-l logX •dx. d2/ _ dx~ 2 a^ log a (a- + ly dy 1 dx X (1 + X2) dy 1 dx X logx d^^ dx =ero°^i- ')■ dy _ x2-2x -2 dx 2V1 -x^x2 + x4- 1)3'2 In exataple 20 and some that follow, pass to logarithms. _ X" dy _ nax"—'^ dx ~ {a + x)" + i dy_ _2a2-x2_ dx-^^la^-xY"'^ ' • %__J ')• (b) For example, the equation of the normal to the parabola y'^ = ^px at the point (x', y') is y — y' = — (y'/2p) (x - x'). 36. Subtangent, subnormal, tangent, normal. Let FT be the tangent at the point P{x', y'), and PS the normal. Draw the ordinate PM; then TM is called the subtangent, and MS the subnormal. Hence, TANGENTS AND NORMALS. 27 Subt. = TM=MP cot <^ = y'dx'/dy'. (1) Subn. =MS = MP tan <^ = y'dy'/dx'. (2) Tan. = TP =-\/mP'' + ^W = y'^1 + (dx' /dy'y. (3) Norm. = SP =\/"mF + 'MS^ = ij'-^Jl + {dy' / dx'f. (4) If the subtangent is reckoned from the point T, and the subnormal from M, each will be positive or negative accord- ing as it extends to the right or to the left. Note. The problem of tangents was foremost among the problems which led to the inventign of the Differential Calculus. EXAMPLES. The equations of the tangent and the normal to 1. The circle, x- + y'^ = r^, are yy' + xx' = r'^, and x'y = y'x. 2. The ellipse, x^/aS + ?/2/62 = 1, are xx'/a^ + VV' /^- = 1. and y — y'= {aY/b'^') (x — x'). 3. The hyperbola, x^/a^ — y^/b^ = 1, are xx' /a^ —yy' /b^= 1, and y — y'— — {aHi'/W'X') (x — %'). x^ , , ^x' (.3 a — x') , ,, 4. The cissoid, y"^ = ^a-x '^'^^ V - V = ± (2a-xy2 (* - »; ), and 2/ - 2/' = =F 'l"~'^'^''' (x - x'). § 155, fig. 3. V x' (.3 a — x') 5. The hyperbola, xy = m, are x'y + y'x — 2 m, and j/'y — x'x = y'^ — x"^. 6. The circle, x^ + y'' = 2 rx, are y — y' = (x — x') {r — x') / y', and . ^ , y — y'={x- x')y'l (x' — r). 7. Find the equations of the tangent and the normal to the curve 2/2 = 2 x^ — x', (1) at the point whose abscissa is 1, (2) at the point whose abscissa is — 2. 8. Find the slope of the curve y^ = x' + 2 x* at x = 2. 9. Find the slope of the curve y — x' — y''+ 1 a,t x = 2 ; x—1; x — 0; 28 DIFFERENTIAL CALCULUS. 10. At what point on j/'^ = 2 .c< is the slope 3 ? At what point is the curve parallel to the a;-axis ? Ans. (2, 4); (0, 0). 11. At what angle does y'^ = 8x intersect ix^ + 2y'^ = i8? The points of intersection are (2, 4) and (2, — 4) ; the slopes of the curves are 4/?/ and — 2x/y respectively, which for the point (2, 4) become 1 and — 1. Hence, the curves intersect at right angles at (2, 4). 12. At what angles does the line 3y — '2x — 8 = cut the parabola y^-Sx? Ans. tan- 10.2; tan- 10.125. 13. The cissoid y^ = x^/{2a — x) cuts its circle x^ + y'^ = 2ax at tan -12. 14. Find the subtangents and the subnormals of the conic sections and the cissoid. Ans. Parabola : subt. = 2x' ; subn. = 2p. Elhpse: subt. = (j:"- — a?)/x'\ subn. = — Is-x'/a-. Hyperbola: subt. = (x"^ — a?) / x' ; subn. = b'^x'/a^. _. ., ,. x'(2a — x') x'2(3a — x') Cissoid : subt. = — e —^ ; subn. = — tt -r^ • 6 a — x (2 a — x )2 Find the subtangent and the subnormal of 15. The hyperbola xy = m. 16. The semi-cubical parabola ay^ = x'. § 155, fig. 2. 17. Find the slope of the logarithmic curve x = log„?/. The slope \ arius as what ? In the curve x = log y the slope equals what ? 18. Find the equations of the tangent and the normal to x = log^j/. SliDW that the subt. = m, and find the subn. 19. Find the normal, subnormal, tangent, and subtangent of the catenary !/=.-, (e^'" + e-^'"). §197, fig. Ans. '^i 2(e2-/»-e-2-/''); - -"- ; ^ "•' ■ a 4' ' V2/2 _ a2 ' Vj/--! - a^ 20. The path of a point is an arc of the parabola y^= 4px, and its velocity is ; find its velocity in the direction of each axis. VELOCITIES. 29 Let s denote the length of the path measured from any point upon it; then ds/dt — V. t;. , . dy 2p dx Substituting these values in dx yv n dy 2pr> — — ana — ~ we obtain 2L Find the velocities required in example 20, when the path is, (i) an arc of the circle x- + y-= r", x^ w2 (ii) an arc of the ellipse -; + f; = 1, X^ w2 (iii) an arc of the hyperbola — — -^ = 1. 22. A comet's orbit is a parabola, and its velocity is v ; find its rate of approach to the sun, which is at the focus of its orbit. Let p denote the distance from the focus to any point on 2^ = 4px ; then p-x+p. .: dp/dt = dx/dt. (1) Hence, the comet approaches or recedes from the sun just as fast as it moves parallel to the axis of its orbit. dp^dx^ ^/ ^ Example20 dt dt Vy2 + 4p2 At the vertex y = 0; hence, at the vertex dp/dt is zero. When 3^ = 2i), dp/dt = (1/2) V2 B. Wheny = 3p, dp/dt ==(3/ 13) ^'13 1>. 23. The curves 2/ =/(z) a.ndy = F{x) have the point [a, /(a)] in com- mon ; show that these cuiTes intersect at this point at an angle whose . f'(a) — F'{a) 30 DIFFERENTIAL CALCULUS. Trigonometric Puflctions. 37. A radian is an angle which, when placed at the centre of a circle, intercepts an are a radius in length. It equals 180° /tt, or 57°.3 nearly. Let u denote the number of radians in any angle at the centre of a circle, r the number of units in the radius, and s the number in the intercepted arc ; then u = s jr ; or if r = 1, u = s. 38. d* sin u = cos u du. [17] d cos u = — sin u du. [18] Let the point P {x, y) move along the arc XPY of a unit circle. Denote the number of linear units in the arc XP by 5, and the number of radians in the angle XOP by u. We shall then have CX u = s, du = ds y = sm II, dy = d sin u, X = cos u. dx = d cos u. (1) Angle EDP equals u, and dx is negative ; hence, from the triangle EDP by § 33 we obtain dy = cos u ds, dx = — sin u ds. (2) Substituting for dy, ds, and dx in (2) their values as given in (1), we obtain [17] and [18]. 39. d tan u = sec^u du For [19] tan u = sin m/cos u. cos u d sin u — sin u d cos u . d tan u = 5 (ooahi + sin^iA du „ , = ^ ; — = sec'M du. * When not needed to avoid ambiguity tlie parentlieses after the sign d are often omitted. TRIGONOMETRIC FUNCTIONS. 31 40. d cot u = — csc^u du. [20] For cot M = tan (tt /2 - m). .-. dcotu = sec2(7r/2 — u) d (7r/2 — u) — — csc^M du. 41. d sec u = sec u tan u du. [21] For sec u = 1/cos u. , Binudu .: a sec u = 5— = sec u tan m du. 42. d CSC u s — CSC u cot u du. [22] For cscM = see(7r/2 — m). .-. c^ CSC M = sec (7r/2 — m) tan (7r/2 — u)d(Tr/2 — u) = — esc M cot u du. 43. d vers u = d(l — cosu) = sin u du. [23] 44. d covers u = cZ (1 — sin u) = — cos u du. [24] EXAMPLES. 1. y = sin 01. dy — a 00s ax • da;. 2. y =cos(x/a). % = — a-i sin(x/a)da;. 3. ^ ?/ = COS a;3. dy = — 3 a;^ sin x^da;. 4. f(e) = tan^e. /'(«) = m tan""-! e-sec^ff. 5. /(fl) = tan 3 6/ + sec 3 e. f'(e) - 3 sec^ 3 9 + 3 sec 3 e tan 3 «. 6. fix) — sin (log ctx). /'(x) = x-i cos (log ax). /(x) changes at the rate of /'(x) to 1 of x (§ 12). 7. /(x)= log (sinax). /'(x) = a cot ax. 8. /(e) = log (tan aS). f'W) = „ . ^'^ = -^ — ■' ^ ' * ^ ' ■^ ^ ' 2 sm aff cos a9 sm 2 ae 9. f(e) — log (cot ad). f'{e) = — 2 a/sin 2 a0. 32 DIFFEKENTIAL CALCULUS. ir. jy, ^ 1 —tan OX 10. f(x) — = cos ax — sm ax. ' sec ax 11. f{x) = X" e™ »^. f'(x) = x"-^ e«in ^ (n + a; cos x). 12. /(x) = sin nx • sin«x. f'(x) = ■" sin»-i x • sin (nx + x). 13. /(x) = e^ log sin X. /'(x) = e='{cotx + log sinx). 14. /(x)= tan(logx). 15. /(x) = log secx. 16. /(x)=cosx/2sin2x. 17. /(x) = 4 sin™ ax. /'(^) = 4 am sin""— i ax cos ax. 18. /(x) = xs™^. /'(x) = x=i°^{sinx/x + logx cos x). 19. f($) = (sin e)'"" 9. f'{e) = (sin «)»" 9 (1 + sec^ $ log sin «). 20. /(x)=tanai/^. /'(x) = — ai/a;sec2aVa:ioga/x2. 21. f(e)-i tanS 6» - tan 9 + 9. /'(8) = tan* e. 22. /(x) = e(«+ =')" sin x. /'(x) = eC« + ^y [2 (a + x) sin x + cos x]. 23. /(x) = e-"*^' cos rx. f'(x) = — er-<^=^{2 aH cos rx + r sin rx). I — (sec Vl — x)2 24. /(x) = tan Vl - x. /'(x) = ^ . -'- ■ ^ V J. — X 25. /(ff) = log Ji^i^ ■ /'(») = CSC e. \ 1 + COS 9 ^ ' By differentiation derive each of the following pairs of identities from the other : 26. sin 2^ = 2 sine cos 9, cos 2 S = cos^ e — sin^ e. __ . _„ 2tan9 „„ 1 — tan^e 27. sin2fl = --— — --. cos29Hr-; tt' l + tan^fl l + tan^e 28. sin 3 9 = 3 sin 9 — 4 sin^ d, cos 3 e = 4 cos^ e — 3 cos 9. 29. sin (to + n) 9 = sin mS cos mfl + cos mB sin nB, cos (m -\- n)9 = cos mfl cos n9 — sin mfl sin nB. 30. d/(a + x2) = /'(a + x^) 2 xdx. 31. d/(ax2) =f'(axP) 2axdx. 32. i-/(^) =/'(?!) 2^. (Jx \a/ \a/ a 33. d/(xy) = /'(XJ/) {xdy + j/dx). INVERSE-TEIGONOMETRIC FUNCTIONS. 33 Inverse-Trigonometric Functions, 45. d sin-i u = du/ Vi - u^.* [25] That is, the differential of an angle in terms of its sine is the differential of the sine divided by the square root of one minus the square of the sine. j . _, u. ytJic- i*.cL^ Let 2/ = sin-^M ; then sin y = m. VVv'--m^ , dti du .-.dy cos y Vl — sin^y du 46. dcos-»u = c?( ^-sin-^M ) = ^ ■ r261 .~ , du 47. dtan-^u=^^-p^,- [27] Lety = tan-iM; then tan y = m. JUjjL '^ v*^''*-^ _ du _ du _ du ^ Y + u, y ~ seeV ~ 1 + tan^y ~ 1+u^' 48. dcot-'u = cz[|^-tan-iM j = -^- [28] 49. dsec-^u = - — , [29"! Let y= sec^u ; then sec y = m. . ~' H. ^ vac**. - U'^f^ sec ?/ tan y ii^u^ — 1 * To avoid the ambiguity of the double sign ±, we shall in these for- mulas limit sin— ^M, cos— 'it, etc., to values between and 7t/2. 34 DIFFERENTIAL CALCULUS. 50. d CSC-' VL-df"^ scc-^ u\ - ^^ [30] \2 J uVu^ - ' I SI, d vers-' 11^-— ^^^ rsii Let 2/ = vers-^M ; then vers y = m. du du .-.dy- sin y Vl — cos^y du Vl — (1 — vers yf du du Vl - (1 - uf V2 M - iv" 52. dcovers-'u = + 1 : tan— i(?i tanx). , x+a _ i ■ 1 — ax : sin— ivsmx. ,&'—«-'' dy ^ 1 dx Vl — 2 X — x2 dx Vl — x2 (i2/ _ 2 ?ix" — '- dx~ x2 « + 1 dy n dx cos^x + n^sin^x dy . _i /sin-ix , logx \ dx V X VI— x2/ d2/^^^ dx 1 + x2 e^ + e-" d^ _ Vl + CSC X dx~^ 2 dy _ - 2 . dx e^ + e-^ , Vx + Va _-/ , /, d!/ 1 1— Vox ix 2Vx(H-x) = sec-i-^^ .^Si^X 1 + x dy _ dx 2V1 — x2 : cot- = cot- , i+vr+x2 .,,^^-, d2/_ 1 X - ^ dx 2 (1 + x2) V2 + X dy^ 1 dx li V2 — X — x2 -i_l^..;:Ux ^ = ^ 1 + X2 dx 1 + x2 '*^° 1-3x2 ^-^ '^ dx l + x2 2 x2 — 1 ^ dx VI — x^ l + x2 X -r-^^^Tx-l dy_ ^2_ + x2 36 DIFFERENTIAL CALCULUS. cm ,3 + 5C0SX 5 + 3 cos X 23. 2/ = tan-i 1 : . aX^ y- Vl-X2 dy 4 dx dy _ 5 + 3 cos a; 1 dx Vl - X2 24. A wheel whose radius is r rolls along a horizontal line with a velocity v ; find the velocity of any point P ui its rim, also the velocity of P horizontally and vertically. The path of P is a cycloid whose equations are x = r{e — sin 6), 1 2/ = r (1 — cos 6) = r vers 6, J where 6 denotes the variable angle DCP, and r the radius CD. Since the centre of the wheel is vertically over D, , V — the time-rate of OD = d(r0)/dt=r-de/dt. .-. d8/dt = v/r. Differentiatmg equations (1), by (2) we obtain dx/dt= !) • vers d = the velocity horizontally, dy/dt = v sine = the velocity vertically. ds dt (1) and (2) (3) ^ = u V2 (1 - cos e) = v V2y/r :; ^ /£*h £ (5) = the velocity of P aloiig its path. . i ^ „ /^ 1 <^^ dy ds At 0,6- 0, and — = ;rf = — : dt cJi dt At E, e = ; At X, e = IT, dx _ d^ _ dt ~ dt ~ '"' dx dt ■ ds di' : 2u, = 0. ds di dy dt = «V2. = 0. THE CYCLOID. 37 From (5) we obtain ds/dt : V — V2r-y : r. Hence, the velocity of P is to that of C as the chord DP is to the radius DC ; that is, P and C are momentarily moving about D with equal angular velocities' 25. Find the subnormal and the normal of the cycloid. Subn. = r sin fl = PH — ED. Thus the normal at P passes through the foot of the perpendicular to OM from C. Hence, to draw a tangent and normal at P, locate C, draw the perpendicular DCB equal to 2 r, and join P with B and D ; then PB and PD will be respectively the tangent and the normal at P. n Normal = DP = ^DB ■ DH = V2r^. = Xt ' ; In 26. Eliminating e tn equations (1) of example 24, find the equation of the cycloid in the form x=r vers-i (y/r) =p V2 ry — ifi. 27. The equation of the tangent to the cycloid is y-y'= V(2r-yO/y'(a;-a:')- 28. A vertical wheel whose circumference is 20 ft. makes 5 revolutions a second about a fixed axis. How fast is a point in its circumference moving horizontally when it is 30° iseffl- either extremity of the horizontal diameter? g-feo" ^-^/-.r-- ^^ 50 ft. a second. 29. What is the slope of the curve 2/ = sin a; ? Its inclination lies be- tween what values ? What is its inclination at s = ? What at a; = 7r/2? The slope = cos x ; hence, at any point, it must be something be- tween — 1 and -f- 1 inclusive. Hence, the inclination of the curve at any point is something between and 7r/4, or something between 37r/4 and w inclusive. 30. Find the equation of the tangent to the curve 2/ = sin s ; y = tan x ; y = sec X. 38 DIFFERENTIAL CALCULUS. MISCELLANEOUS EXAMPLES. 1. y = log tan— 1 x. 10. y = e'^ sin"'rx. 2. y- [x+ Vl — xA". 11. y = log {log (a + ftx")}. 3 y^^jlE^ 12 ^ Vl + x^ + Vl - g^ \(i + a;2)8' '" vr+T2-vr^ ^ ^ _ (sin MX)"' _ jjj example 12 rationalize the (cos mx)" denomkiator before differentiating. 5. 2/ = e=^ tan-ix. 3 + 2X 13. /(x) = (a2 + x2)tan-l^• 6. 2/ = sin-i — -^_ . Vis 7. /(x) = e(" + ^)'sinx. 14. 1/ = Vl ■ ■ x-' sm— 'x — X. xlogx,, ,, . 15. 2/ = tan-i5+ log J5_^. 8. 2/ = _ + log (1 — x). a \x + a a 1 g '^ ,c Vx2 + 1 + X 9. 2/ = sec-1 , • 16. 2/ = . Va2 — x2 Vx2 + 1 — X Vx2 + a2+ Vx2 + 62 IT. y= . = , Vx2 + a2 - Vx2 + 62 d^__2x_/2+ / x^ + g^ ^ f x2 + 62 \ tJx~a2-62\ \x2+62 \x2 + a2/' 18. 2/ = log ( Vl + x2 + Vl - x2). ^^l^i ^ y dx x^ Vl — x*-' 19. /(x) = (X- 3) e2=« + 4xe»^ + x + 3. 20. y = log. (2x — 1 + 2 Vx^ — x — 1) 21. . = log(i±-^) dy ^ 1 d^ Vx2 — X — 1 1 + X '^i^^ tan-ix dy _ x^ 2 • c?x "" 1 — X* ■T— ?/ 22. X = e » . (% _ X — 2/ dx X (log X + I) Here logx = -■ y MISCELLANEOUS EXAMPLES. 39 23. y = X2 1+ ■ 1 + Here y ■ a;2 1 + etc. to infinity. 1 + y' 24. y = log (X + Vx2 - a2) + sec-i - • 25. 2/ = log Vl — x2 + X V^ dy _1 / x + g dx ~ X \x — a ■\/2 26. 27. 2/ = log"\ Vl - x-' Vl + x2 + : d2/_ dx (Vl — x2 + x^/2)(l — x2) (J?/. Vl + x2 — X Here ^ = log( Vl + x^ + x). dx Vl + x2 2x2-2x+l , ^ 2x'^ + 2x+l + *^'^"^ • i/ = iog-^; 28. 2/ = cot-i- + logJ^^- X \x + a 79. y = sin-i I tanS Vg2 — x2 30. , = log{ex(^-^f^}. 2 X d2/ dx' 8x2 4x4 + 1 1- 2x2 dy 2 0X2 dx X*- -a* ^2/ g2 tan a2 — a e 1 dx .2 Vg2 — x2sec2 e ^_ X2- J. dx x2 — 4 CHAPTER III. PROBLEM OF KATES SOLVED BY LIMITS. 53. Limit. When according to its law of change a vari- able approaches indefinitely near and continually nearer a constant, 'JM!i*-'©aa-«ey6r-«eaafe»44>, the constant is called the limit of the variable. It is assumed that the reader is familiar with the elementary theorems of Limits ; but for convenience of reference we state them below : 1. If two variables are equal, their limits are equal. 2. The limit of the sum, or product, of a constant and a variable is the sum, or product, of the constant and the limit of the variable. 3. The limit of the variable sum, or product, of two or more variables ' is the sum, or product, of their limits. 4. The Umit of the variable quotient of two variables is the quotient of their limits, except when the limit&of the divisor is zero. 54. Notation. The sign = denotes "approaches as a limit." Thus, Aa; = is read " Ax approaches zero as its limit." The limit of a variable, as z, is often written It («). Limit fAf] ^^^^Ay Ax = \_Ax \, or It-^) denotes It ( -^ ) when Ax = 0. J Ax V^*/ Limit [Ayl _ dy. Let v', x\ and y' denote any corresponding values of v, x, and y, from which Av, Ax, and Ay are estimated. Let the rate of v be the unit of rates ; then evidently Ay _ J the y-rate of y at some value of y \ Av \ between y' and y' -j- Ay J ' ' -' LIMIT OF ^Y/^X. 41 .".It (Ay/Av) = the v-rate of y at the value y'. (2) Also It (Ax/Av) = the v-rate of x at the value x'. (3) Dividing (2) by (3), we obtain, in general, limit rAy"! _ the ■y-rate of y a^ Aa; =b l_Ax J the »-rate of x Comparing (4) with (1) of § 12, we obtain [33]. (4) To illustrate (1) and (2), let s denote the number of feet a falling body descends in t seconds. Let s' and f denote any corresponding values of s and t, from which As and At are reckoned ; then, evidently, f the time-rate of s at some value of s 1 I between s' and s' + As / As. At' .-. It {As/ At) = the time-rate of s at the value s'. EXAMPLES. By § 53 and [33] of § 55, prove 1. d(uy) = ydu + udy, or formula [7]. Let z = uy, and let x' represent any value of x, and u\ y\ and z' the corresponding values of u, y, and z, respectively ; then z' = uy. (1) When x = x' + Ax, u = u' + Au, y — y' + Ay, and z = z' + Az; hence, z' + Az = (u' + Am) {y' + Ay) = u'y' -)- 2/' Am -|- m'A?/ + AuAy. (2) Subtracting (1) from (2), we obtain Az = y'Au -\- ii'Ay + AuAy. (3) .-.lt^ = ,t(.'^)+lt[(M'+AM)g = y' It ^ -)- It iu' -)- Am) ■ It ^ • Ax Ax dz ,du ,dy , r??i 42 DIFFERENTIAL CALCULUS. Hence, as x' is any value of x, we have, in general, dz = d (uy) — ydu + udy, or [7]. If in [7] we put for y the constant a, we obtain d(au) = adu, or [4]. If in [7] we put vw for u, we obtain d(vwy) = wydv + vydw + vwdy, or [8]. .3. d{u + y + z+a) = du+dy + dz, or [3]. Let v — u + y + z + a; then Av — Au + Ay + Az. -^ Av ,^ Am , , Am , , Az .-. It " = It ~ + It -^ + It — • Ax Ax A.c Ax do _ du dy dz^ ' ' dx dx dx dx .-. dv = d{u + y + z + a) = du + dy + dz. 3. Ily = z, dy — dz, or [1]. If y = z, Ay = Az. • It — = It — • . # — ^ . Ax Aj; ' ' ■ dx dx 4. Aa = 0; .-. da = 0, or [2]. 5. d(u/y) — {ydu — udy)/ij^, or [9]. Let v = u/y, ., . m' + Au u' y'Au — ■u'A?/ then Av = -;-; — : ; = ,„ , ^ ■ ^ + A?/ 2/ !/'2 + y'Ay 6. Assuming the binomial theorem, prove : d(w>) = nu"-^du, or [13]. 56. By [33] of § 55 the problem of rates is reduced to one of limits. By theorems proved later in this chapter, proofs by limits are very much abbreviated. INFINITESIMALS. 43 The reader should note that we cannot write Ace ~ It (Arc) - ^^ For any proof of the principle applied in (1) fails when the limit of the divisor is zero. Moreover, the determinate expres- sion It (Ay /Ace) cannot be identical with the indeterminate expression 0/0. To find It (Ay/Acc), we find the limit of an equal variable, as in the examples of § 55 ; or we find the limit of some variable .which, though not equal to A^z/Ace, has the same limit, as in § 63. The theory of limits never gives rise to the form a/0, or to any of the indeterminate forms 0/0, etc. 57. Derivative as a limit. By § 29 and [33] we have It (Ay /Ace) = the derivative of y with respect to x. Or, since A/(a;) =f(x + Acb) —f(x), we have It (Ay/Acc), or dy jdx, is often denoted by Djy. 58, Infinitesimals. Zero is defined by the identity a — a = 0. An infinitesimal is a variable whose limit is zero. Hence, Ace = may be read " Ace is an infinitesimal." In approaching its limit zero, an infinitesimal becomes in- definitely small and continually smaller, but it never equals zero. Any small quantity becomes an infinitesimal when it begins to approach zero as its limit, not when it reaches any particular degree of smallness. A quantity, however small, which does not approach zero as its limit is not an infinitesi- mal. 44 DIFFERENTIAL CALCULUS. Infinitesimals are indefinite variables used as auxiliaries in the study of finite quantities. Ttieir essence and utility lie in their having zero as their limit, and not in their smallness. Tlie reader should not make the study of them difficult and obscure by thinking of them as mysteriously small. 59. An infinite is a variable wliich under its law of change can exceed all assignable values, however great. The general symbol for an infinite is oo . The reciprocal of an infinitesimal is an infinite, and conversely. For example, when x = 0, l/x = a); that is, when x is an infinitesi- mal, 1/x is an infinite. When e = 0, cot (± e) = ± 00 , and tan (7r/2 q: ff) = ± co . When xiO, logx= — qo. An infinite does not approach a limit ; in arithmetic value it increases without limit. 60. Geometric meaning of It (Ay /Ax). Let mn be the locus of 2/ =/(x), FP' a secant, and FD a tangent at P. Draw the ordinates MP and NP', also PC parallel to OX. Let 0M= X, and MN = Ax ; then MP = y, and CF' = Ay. Hence, Ay/Ax= CF'/ PC = the slope of the secant FP'. Conceive the secant FP' to re- volve about F so that arc FP' = ; then Aa; =0, Ay = 0, and the slope of the secant =^ the slope of the tangent at P. Hence, It (Ay /Ax) = the slope of the curve y = f(x) at the point (x, y). CoE. If when Ax = 0, Ay /Ax varies, the locus of y =f(x) is a curved line, and in general Ay / Ax approaches a limit ; but at a point where the locus is perpendicular to the x-axis Ay / Ax = 00 when Ax = 0, PRINCIPLES OF LIMITS. 45 61. The limit of the ratio of an infinitesimal are of any plane curve to its chord is unity. Let s represent the length of the arc mF (§ 60, fig.), and arc PaP' = As ; then PC = Aa;. Since s is a function of x, we have It (As / Ax) = ds / dx. (1) But It (chord PP'/Ax) = It (sec CPP') = sec PPD = ds/dx. (2) Dividing (1) by (2), we obtain limit r As Asd^OLchordPP' I "" ^^^ CoE. Let u and s have the same meanings as in § 38 ; then J- = 1. limit r M ~| _ limit f 2s ~[ _ limit r 2 ■ XP ~\ M = |_sin M J ~ s = 1_2 sin mJ ~ XP = 1_2 ■ CPJ That is, the limit of the ratio of an infinitesimal angle to its sine is unity. 62. The limit of the ratio of two variables is not changed when either is replaced by any other variable the limit of whose ratio to it is unity. Let a, tti, y8, and /81 be any four variables, so related that lt« = 1, It /8_ /81 1, and It' a a .A ft" a, /81 a ft. ■■■-r It a' It-^ •■'t = '■ "i (1) by(i) in which a is replaced by ai, and ^ by /81, without changing the limit. This principle often enables us to simplify a problem of limits by sub- stituting for an infinitesimal arc its chord, as in the following article. 46 DIFFERENTIAL CALCULUS. 63. Subtangent, subnormal, tangent, normal, in polar curves. Let arc mP = s, and arc Pi) = A.s ; then Z. POD = ^9, circular arc P3I = pA6, and MB = Ap. Draw the chords PM and PD, the tangents BPH and 2'PZ, and ZH perpendicular to PIT, Z being any point on the tangent PZ. When As d= 0, the limiting positions of the secants PJ/and PD are the tangents RPH and TPZ, respectively ; hence, It (Z PJID) = Z BPK= 77/2 = A PHZ, It (Z ODP) = Z OPT = ^ = Z ^-ZP, and It (Z ilfPX>) = Z HPZ. A\ Again, by §§ 61 and 62, and trigonometry, we have As :lt 3ID = lt sin MPD ^ = sin HPZ as chord PZ» " sin PMD' HZ PZ' (1) POLAK CURVES. 47 1^ pAg _ ^^ chord MP _ ^^ sin M DP Also, It ^^ - it ^j^^j.^ p^ - itgijipjfl^; .,P^ = 3ini.^P = f|. (2) From (1) and (2), it follows that, if ds = PZ, dp = HZ and pdO = HP. Draw OT perpendicular to OP, and P^ and OiV perpen- dicular to the tangent TP. Then the length PT is called the polar tangent; PA, the polar normal; OA, the polar subnor- mal; and OT, the polar subtangent. From the right-angled triangle HPZ we have d^ = dp'' + pW ; (3) 3,n^ = — , cos^ = -, tan^=— • (4) Polar subt. = OT = OP tan i/^ = p^de/dp. (5) Polar subn. = 0^1 = OP cot ■/. = dp/dd. (6) Polar tan. = PT = VOP^TOT' = pJl + p^'^ (7) Polar norm. = AP = ^OP' + 0^^ = ^p^ + ^^. (8) _p = C)ivr= OP sin ./^ = pW/ds „2 = , '^ (9) ^p^ + (dp/dey ,^ = ,/. -I- 61. _ (10) CoE. If P.Z' represents the velocity at P of (p, 0) along the line of its path, HZ and PH will represent its component velocities at P along the radius vector and a line perpen- dicular to it. 48 DIFFERENTIAL CALCULUS. EXAMPLES. 1. Find the subtangent, subnormal, tangent, normal, and p of the spiral of Archimedes p — ad. Ans. subn. = a ; subt. = p^/a; norm. = Vp^+cfi ; tan. = p VT+7V^ ; P = pV{p^ + a^)"^. 2. In the spiral of Archimedes show that tan \j/ = 6. 3. Find the subtangent, subnormal, tangent, and normal of the logarithmic spiral p — a^. Ans. subt. = p/loga; subn. =ploga; tan. = p Vl + (log(i)-2 ; norm. = p Vl + (loga)'^. 4. In the logarithmic spiral, show that \j/ is constant. li a = e, \p = Ti /i, subt. = subn., and tan. = norm. Since the logarithmic spiral cuts every radius vector at the same angle, it is often called the equiangular spiral. 5. Find the subtangent, subnormal, and p of the lemniscate of Bernouilli p"^ — a^ cos 2 S. Ans. subt. = — t^/a^ sin 2 0; subn. — — a^sin2e/p. p = pV Vp4 + a*sin22e = p^/a^. 6. In the lemniscate show that \p = 26 + 7r/2 and 4> = 3e+ Tr/2. 7. In the curve p = a sin' (S/3), show that (p = i\p. § 155, fig. 14. 0. In the parabola p = a sec^ (S/2), show that + ^ = ;r. 9. The velocity of a point along the spiral p = ad \sv; find its com- ponent velocities along the radius vector and a line perpendicular to it. 10. By the method of limits (§ 60, fig.) prove that, if PB = dx, BD = dy and PD — ds, where s = mP. 7 ' 64. Orders of infinitesimals. Any variable which is neither an infinitesimal nor an infinite is called a finite variable. Two infinitesimals are said to be of the same order when their ratio is a constant or a finite variable. ORDERS OF INFINITESIMAXS. 49 For example, when Ax = 0, 7 Ax and (5 a + 0)Ax are infinitesimals of the same order ; so also are 9 (Ax)2 and (8 x — 7) (Ax)2 ; so also are hy and Ax when It (Ay/Ax) is other than zero. Again, when e^^TC/i, 1 — sin and cos^S are infinitesimals of the same order ; for limit r l — sin9 ~| _ limit r 1 n _ 1 e = 7C/i\_ cos^e J ~ e=ic/2 Ll + sinflj ~~2 In order to classify the infinitesimals in any problem as belonging to different orders, we choose some one of them as the principal infinitesimal, and adopt the following definitions: An infinitesimal is of the first order when it is of the same order as the principal infinitesimal ; of the second order when it is of the same order as the square of the principal infinitesi- mal ; and so on. In general, an infinitesimal is of the wth order when it is of the same order as the nth power of the principal infinitesimal. Thus, when Ax is taken as the principal Infinitesimal and It (Ay /Ax) is other than zero, Ay is an infinitesimal of the first order ; 5 x (Ax)'^ and 7 y^yda, are infinitesimals of the second order ; 7 y (Ax)^ and 4 x (^xYAy are of the third order; and 2/(Ax)'' and x-(Ax)''-™ (Aj/)™ are of the nth order. 65. Notation. Let v^, v^, • • •, ■y„ represent finite variables or any constants except zero, and let i represent the principal infinitesimal ; then v^i, v^i?, ■ • ; vj," will represent respec- tively infinitesimals of the first, the second, • • •, the wth order. According to this notation, Ax = i is read " Ax is the principal infini- tesimal"; 5xAx = Di Is read "5xAx is an infinitesimal of the first order"; 1 x^xhy = v^i"^ is read "7xAxAi/ is an infuiitesimal of the second order"; and so on. The subscript need not be written with the first V which appears in any problem or discussion. A finite quantity may be regarded as an infinitesimal of the zero order ; for x = xi" and vi" = v. Ex. If 61 = i, 1 — cos fi = ui2. _, limit rl — cosfl-] limit rl — cosflT .. „ ^ limit [- 1 -1 ^1. 50 DITFERENTIAX CALCULUS. X) 66. Geometric illustration of in- „ finitesimals of different orders. -E Let CAB be a right angle inscribed in tlie semicircle CAB, BD a tangent at B, and AE a perpendicular to BD. From the similar triangles CAB, BAD, and AED we have AD ■.AB = AB-.AC; .■.AD=^ (1/AG)'ab'; (1) and DE : AD = AB : BC ; .: DE = {l/BG) -AD- AB. (2) Suppose A to approach B, and let AB = i ; then, from (1) and (2), AD-(l/^)-P = m'i, and DE - (1 /BC) ■vi'^-i- v^iK 67. Orders of products and quotients. The order of the product of two or more infinitesimals is equal to the sum, of the orders of the factors. For vi"- ■ ■y„,i'" = vvj,"- + '" The order of the quotient of any two infinitesimals is equal to the order of the dividend minus the order of the divisor. For vi''/rj"' = (v/i'„) i"-"". 68. If the limit of the ratio of one infinitesimal to another is zero, the first is of a hiijher order than the second; and con- versely. For It {vi^lvjf) = It \{y Iv^ i"-"'] = when, and only when, n > m. Cor. If the ratio of one infinitesimal to another is infinite, the first is of a lower order than the second. 69. If the limit of the ratio of two variables is imity, their difference is an infinitesimal of a higher order than either ; and conversely. LAWS OF INFINITESIMALS. 61 Let a= 13 + e; then It (a/(8) = 1 + It (c/^). If lt(a/j8) = l, lt(c/|8) = 0; hence, by § 68, e is of a higher order than /S. Conversely, if e is of a higher order than /3, lt(e/ 13) = 0; .■.lt(a//3) = l. Cor. If arc FF' = i, arc PF' = chord FF' + vi", where n>l. Also, if angle u = i, u = sin u + vi", where w > 1. § 61 70. From sums of infinitesimals of different orders, all infini- tesimals of the higher orders vanish in the limit of a ratio. For It :! + "r^+"f+--- = It = • § 62 vh + V jt + V 3I + • • • VI This principle of limits often greatly shortens the opera- tion of finding the limit of a ratio, and together with [33] of § 55 furnishes the following simple 71. Rule for differentiating a function. (1) Find the value of the increment of the function in terms of the increments of its variables. (2) Supposing the increments to he infinitesimals of the first order, in all sums drop the infinitesimals of the higher orders, and in the rem/iining terms substitute differentials for incre- ments. Tor by § 70 the infinitesimals of the higher orders in a sum will vanish in the limit, and by [33] of § 55 diiferentials will take the place of increments in the remaining terms. CoK. Anticipating, in step (1), the result of step (2) we need to express exactly only those terms of A/(7«) which are linear in Am. (See examples 7-11.) 62 DIFFERENTIAL CALCULUS. EXAMPLES. u, y, ■ ■ • being different functions of x, by § 71 prove 1. d (uy) = ydu + udy, or [7]. A ("2/) — {u + Au) (2/ + Ay) — uy § 8, example 3 = yAu + uAy + AuAy. Let Ax = i ; then AwA-i/ = Jii^^ Hence, dropping AuAy, and sub- stituting d{uy), du, and dj/i respectively, for A{uy), Au, and Aj/ in the remaining terms we obtain [7]. 2. d(u/y) = (ydu — udy)/y^, or [9]. a/'*\ _ m + Am _ « _ yAw — «A^ \y) y + Ay y y'^ + yAy 2/2 = i)p, and j/A?/ = Vii ; hence, in the sum y^ + 2/^2/1 ^y (2) of § 71, we drop yAy. Substituting differentials for increments in the remaining terms, we obtain [9] . 3. d{au) = adu, or [4]. 5. Formula [8]. 4. Formula [3]. 6. d(M«) = TOt"-idM, or [13]. A (W) = (u + Au)" — W — jiu»-i Au + " '". — - m"-2Am2 + • • •. 11 . .(J(tt") = nW—^du. 7. dsmu= eosudu, or [17]. A (sin u) — sin {u + Am) — sin u = cos u sin Au + sin u cos Au — sin u by Trig. = cos M (Au — ui") — (1 — cos Am) sin u. § 69, Cor. = cos u ■ Au — vi" ■ cos M — 1)2*^ sin u. § 65, example .-. (Jsintt= cosMdu. In obtaining the value of A(sinu) we express exactly only those terms which are linear in Am ; for by § 70 all the other terms vanish in the limit. 8. (Z cos M = — sin udu, or [18]. 9. ds = Vdx^ + dy^. Let As = arc PP' =i; § 60, fig. then As = c hord PP' + vi", where n > 1, § 69, Cor. = V ax^ 4- A2/2 + vi". .-. ds = Vdx^ + dJ/2. DIFFERENTIAL OF AREA. 63 10. Find the differential of the area between the x-axis, the curve RP, or y = f(x), the fixed ordinate HB, and the variable ordinate MP, or y. Let P be any point (x, y) on the curve. Conceive the area HRPM as gener- ated by the ordinate MP, or y, and denote this area by A. Let MB = Ai ; then A^ = DP', and AA = MBP'P = y£\x + PDF'. Let iiX=-i; then, since PI}P' = ap, i.e. when tan assumes the form cp, we know that is coterminal with */2, or 3;r/2 ; and conversely. When cot (p — ap,vre know that c/t is coterminal with or jr. If, when X = c,f(x)^0/a = 0, the reciprocal of /(x) as- sumes the form a/0, or cp, when x = e. 56 DIFFERENTIAL CALCULUS. 75. In this chapter, to obtain the ratio of differentials, or the ratio of rates, we employ infinitesimal increments of variables as auxiliary quantities. The division of infinitesi- mals into orders affords clear and brief statements of prin- ci]3les of limits which greatly abridge and simplify the work of finding the ratio of differentials. In this as in the previous chapters, diiferentials are regarded as finite quantities. 76. Limit in position. When according to its law of motion a point, or line, approaches indefinitely near and con- tinually nearer a fixed point, or line^-bufcean -nevetJEeadii*^ the fixed point, or line, is called the limit of the variable point, or line. For example, when in § 60 arc PP' = 0, the fixed point P is the limit of P', and the tangent PD is the limit of the secant PP". Whether the word limit has reference to magnitude or to position will always be evident from the context. EXAMPLES. 1. When X = c, a/(x — c) = ± oo ; when x = c, a/ (x — c) = ap. 2. ■Wlienx = 0, a/x=±cc; when x = 0, a/x = ap. 3. When = 7r/2, sec = ± oo ; when = Tt /i, sec (j> = ap. i. When == 0, esc = ± oo ; when = 0, osc = ap. a/0(x) ^/(x) a/f(x)- f"(^)> f"'i^)> r{^)> ■ ■ ■> /"(«=)■ Thus if f(x) = x*, f (x) = 4 x^, /" (x) = 12 x^, /'" (X) = 24 X, fy (x) = 24, f (X) = 0. Hence, if m = /(:-'■) and x is independent, du „,, ^ d^a „„, ^ d"u „ , ^ dx •''-■" dx' ■'"-■>' ' dx'' •' ^ ' Successive derivatives are often called successive differential coefficients. 79. The nth derivatives of some functions can be readily obtained by inspection. Ex.1. /(x) = e^; find/''(x). /'(x) = e^, f"(x) = (F, f"'(x) = e^, • • • ; . . /"(x) = e^. Ex.2. /(x) = a"^; find/»(x). /'(x) = log a ■ w, f"(x) = (log afa^; f"'(x) = (log aj'a", ■ • ■; .: /"(x) = (log a)''a''. SUCCESSIVE DERIVATIVES. 59 Ex. 3. /(z) = log (1 + x) ; find /"(i). r(x) = (1 + x)~^, f"{x) = (- 1)(1 + X)-'-, /'"(x) = (- l)2[2(H-x)-3, /.v(a:) = (_ l)3[3(l + x)-^ • • ■ ; ■■/"(x)=(-l)"-' |»-l (l +x)-" Ex. 4. /(9) = cos ae ; flnd/"{9). /'(9) = — a sin (i9 = a cos (a9 + 7r/2), f"(e) - — cfi sin {a9 + Tt/2) — a? cos (afl + 2 ■ 7r/2), /-"(6I) = — aS sin (afl + 7C) = a? cos (off + 3 • ;r/2), ■ • /"(*) = a" cos (a« + n - Tt/2). 80. Each of the successive derivatwes of f (x) equals thi; ^-rate of the preceding 'Icricative. For /"(x) = fZ/» - 1 ( .•■) / (Zx = the x-rate of /" - ' (x) . Cob. /""'(x) is an increasing or a decreasing function of X according as /"(x) is po£-live or negative ; and conversely. EXAMPLES. 1. /(x) = cx3 + ax2 + a. f"'(x) = 6 c, /■■'(x) = 0. Here/"(x) changes at the rate of 6 c to 1 of x (§ 80). 2. fix) = x« + 4x* + 3x + 2. /-"(x) = [6. 3. /(x) = xlogx. /"(x)=(— l)"-^ |?i — 2 x1-". Here/»-i(x) changes at the rate of {— l)°-^ |n — 2 x^-" to 1 of x. 4. /(x) = ax">. /"(x) = am(m — 1)(ot — 2)- • -(m — n + l)x'"-«. 6. /(x) = x3 1ogx. /'T(x) = 6i-i. dx3 (e^ + e-'')' 6. 2/ = log(e=^ + e-==). (b 7. /(x) = ^x3Gogx-5/6). /"(x) = (-1)"-< |b-4 x3-". 8. /(x) = (x2 - 3 X + 3) e2x. /'"(-j) = g x^eS^. 9. /(x) = X* log X. /^' W = — L* ^" 60 DIFFERENTIAL CALCULUS. 10. f(x) = a;^. f"(x) = x^{l + loga;)^ + x'-K __k3_ diy _ 24 ^ ^ " 1 - X ' dx* ~ (1 - a;)5 ' 12.f{x) = e^. f<'{x) — a''e'". 13. f(e) = sin a 9. /"(«) = a" sin (ad + n ■ tc/I). 14. /(x) = (1 + x)". /"(x) = 771 (m — 1) • • • (m — n + 1)(1 + x)"--". d"y_ (-l)"4"[>i :(4x + 2)-i 16. 2/ = 4x + 2 ^ ' / • (jj-n (4x4-2)''+i 2 - X d'-i/ _ (-1)"4[^ 17. y 2 + x dx» (2+x)"+i |-^ = - 1 + -^ = - 1 + 4 (2 + x)--'. 2+x 2+x ^ ' .6x-l (;„y_-5(-l)"3''[TO 3x + 2 dx» (3x + 2)'' + i ^ 4i-i-l dx» ^ ^ ^)^(2x-l)» + i (2x+l)''+iJ' (2x — l)-i — (2x + l)-i. 4x2 — 1 Prove each of tiie following differential equations 19. When ti =; Vseo 2 x, d?u/6ai^ = 3 «5 — ^, da sec 2 x tan 2 x 6x v sec 2 X Mtan2x. (1) .-. -T—„ = tan 2 X — + 2 a sec^ 2 x. ox^ ctx = M tan2 2 X + 2 M seo^ 2 x. by (1) = u (sec2 2 X — 1) + 2 a seo^ 2 x = 3 it^ — m. 20. Wlien M=e="sinx, ^-2=^ + 2u = 0. 21. When m = a sin (logx), x^ -^-z + x — + M = 0. SUCCESSIVE DERIVATIVES. 61 6 22. "Wheiij/ = e-'^^sinmo;, -^— 2c-^+ (c2 + m2)?/ = 0. dx2 dx ^ ' 23. When y = sin (sin x), -r^+ -^ ta,nx + y cos^x = 0. 24. When M = cos (a sin- 'x), (1 - x^) f^ - x ^ + a^u = 0. ^ ' dx^ dx 25. When « = (sin-ix)2, (1 - x^) f^' - x ^ = 2. dx2 dx 26. Wheny = ^(e-/» + e-/«),g = ^-- Of each of the following implicit functions ohtain that derivative Vfhich is given at its right : 27. 11y^ = 2xy — c, d^/dx^ = c/ (x — y)K d(y'^) = d(2xy-c); .-.^ = ~^— (1) ^ " dx y — X ' (-<^Kl-o_^-= dx'^ (^ — x)2 (y — x)2 From (1), (2), and the given equation, we obtain dh) _y^ — 2xy _ c _ ^2 "" (,y — x)3 ~ (X — y)3' 28. If 2/2 = 4px, d?y/dx^ = 24pV^*- d2//cix = 2p/?/; d2y _ —2p(dy/dx) _ _ 4 1)2 dx2 y2 2^3 d?y_ \2y'^p'^(dy/dx) _ 24 p» dx^ ?/* ?/5 ^3 29. If x2 + ^2 = ^2^ d'^y/dx^ = - r^/y^. 30. IE 2/3 = a2x, d^y/dx^ = — 2 aV9 2/^ 31. If x2/a2 + 2/2/62 = 1, dhj/dx'^= — ¥/a^\ 32. If x2/a2 — 2/2/62 = i^ d^y/dx^ = — b*/a'^K (2) 62 DIFFERENTIAL CALCULUS. 33. if^. = -^_,^ = ± 3^=.. 2 a - X da;2 Vx (2 a — x)* 34. If x2/3 + 2/2/3 = a2/3^ d'^/dx^ = a'^'^/5y^'^x*"'. 35. If 2/2 _ 2 oxj/ + x2 = c, d^ _ {a? — 1) (j/2 — 2 ax;/ + x^) _ c (g^ — 1) dx^ (2/ — ox)" (2/ — ax)8 36. If 2/3 + x3 = 3 ax2/, d%/dx2 = — 2 a^x?// (?/2 — ax)^. 61. Ite ^v-xy, ^^^- x2(2/-l)8 81. Leibnitz's theorem is a formula for the mth differential of the product of two variables. Let u and v be functions of x ; then d (uv) = du-v + udv. (1) In general, du and dv will be functions of x ; hence, c?^ (uv) = c?M . v + liw c^i; + cZm c^w + M rf^t; = d'u ■v + 2dudv + u d?v. (2) .-. d\uv) = d^u-v + 3 d'u dv + 3dud''v + u d'v. (3) The coefficients and the exponents of operation in (2) and (3) follow the laws of the coefficients and exponents in the Binomial Theorem. However far we continue the differen- tiation, these laws will evidently hold ; hence, we have d'-iuv) = d''u-v + nd''-'^udv+ ^ — ^ d^-^ud'v + • • • + n du d"-^ V + u d"v. (4) ACCELERATION. 63 EXAMPLES. 1. Emd(25(e'^x2). Here u = 6% dHi = a»e™ dx" ; and v — x\ dv = 2x dx, d% = 2 dx^, d^v = 0. Substituting tliese values in (4), we obtain, wlien n = 5, £j6 (eaa:x2) = (a^e'"^ • x^ + 5 • a^e"^ ■ 2 x + 10 • a^e"^ • 2) dx^ = a^e^ (a2x2 + 10 ox + 20) dx^. 2. Find ^''(x^sinax). Here it = sin as, d'^ = a'^ sin. {ax + n ■ 7i / 2) dx" ; and u = x2, (i!» = 2 x dx, d'h = 2 dx^, #u — 0. Substituting these values in (4), we obtain d" (x^ sin ax)/dx'' — x^a" sin (ax + n-7t/2) + 2mxa''-i sin [ax + {n — l);r/2] + ji (n — 1) a«-2 sin [ax + (ii — 2) jr/2]. 3. d" (xe^) = e='(x + n) dx". ifi 4. d" (x^e"^) /dx" = a''-^ gax [aQ^js -|- 2 amx + m (m — 1)]. 5. d" (x^a-T^) = a^ (log a)"-^ [(x log a + Ji)^ — n] dx". 6. d''(x2 logx) = 2(— 1)"-i |b — 3 x^-''dx". 82. Acceleration is the time-rate of the velocity v. Hence, if s = the distance, and a = the acceleration, _ds _ d ds _ dh '" ^ It '^ ~ dt dt~ df ^ ' EXAMPLES. I. A point moves along the arc of the parabola 2/2 = 4px with the constant velocity »' ; find its acceleration in the direction of each axis. From example 20 of § 36, we obtain dx _ yv' dy _ 2pv' dt Vy^ -I- 4^2 lit V2/2 + 4p2 d^ _ 8pH'^ dhj _ 4p^yv'^ ■'■ dP ~ (?/ + ipy' dt^ ~ (2/2 + 4^)2)2 ■ 64 DIFFERENTIAL CALCULUS. The velocities in the directions of the axes are the time-rates of x and y in the first quadrant. Since d'hi/dP' is positive, tlie velocity in the direction of the a-axis continually increases. Since for y positive dhj/dt?' is negative, the velocity in the direc- tion of the 2/-axis is constantly decreasing. 2. Find the accelerations required in example 1, vrhen the path of the point is (1) an arc of the circle x^ + y'' = r^, (2) an arc of the ellipse ~^ + T^ = li (3) an arc of the hyperbola — — — = 1. 3. If s denotes the number of feet a body falls In t seconds, and 9 = 32.17, s = gt^/2 is the law of falling bodies in a vacuum near the earth's surface ; find the velocity and the acceleration. Ans. v = ds/dt = gt; a = d'^s/dP = g, a. constant. 4. Given s — ct^'^; find v and a at the end of four seconds. CHAPTER V. INDETERMINATE FOBMS. 83. The value of a function of x for x = a usually means the result obtained by substituting a for x in the function. When, however, this substitution gives rise to any one of the indeterminate forms 0/0, aplap, 0-ap, ap - ap, 0", ap", W, the definition given above is inapplicable and must be enlarged as below : The value of a function for any particular value of its variable is the limit which the function approaches when the variable approaches this particular value as its limit. This definition is of general application, but it is practically useful only when the ordinary and simpler definition fails. /(x) is often written without the parentheses, as/x. The expression /a:]„ denotes the value of fx when x = a. EXAMPLES. By principles of limits prove that 1. (a;-a)i/V(x--a^)^'*]a = 0. ^'Vhenx = a this fraction assumes the indeterminate form 0/0. Hence, to evaluate it for x = a we must find its limit when x = a. Por value.s of x other than o, we have (x — ay ^ (x — a)*''^^ _ (x-ay'^ limit r (x-a)i/3 -| ^ umit r(x^-aV^-l ^ _0_ ^ " x = a L(x2 — a2)i/*J a: = a L (x + a)i/* -I V2a That is, the given fraction equals zero when x — a. 66 DIFFERENTIAL CALCULUS. x3 - ga -l ^ 3a x^ - I n ^ 5 P x2-a2j<, 2 x3_iJi 3 (g^ — x^)i/g+ (g-x) -1 _ V2a (g-x)i'2+ (a^-xyd" l + gVs 5. (1 — cose) /sin e]o—0. For values of 6 other than zero, we have 1-cosg ^ 2sm2(fl/2) _ ^0. sin 9 =2 sin (9/2) cos («/2) - ^°2 limit rl ~ cos mit rl — cosfln limit r en „ -oL^i^^J"«=oL'''°2j==°- 84. If lf^/<^x\ = 0/0, then [/r/<^x]„ =[/'cc/,^'a:]„. That is, (/" i/te ratio of two functions of x assumes the form 0/0 when X = a, i/iere i!Ae raiio of these functions when x = a /.■.• equal to the ratio of their derivatives when x = a. By [33] of § 55, the limit of the ratio of the increments of tviro variables is equal to the ratio of their differentials ; hence. limit r f{x + t^x) -fx l _ f'x ■ dx _ fx _ Aa; = |_ (x + Ax) — ^x\ (j>'x ■ dx 'x Substituting a for x and remembering that /a = <^a = 0, we obtain limit Aa; dt rf(a±Ax)l ^fa ^^ ,fif\ _ ,fxl = 0\_cl,(a + Ax)_\ 'a' '^xj, ^'a;J„ If fxl^'x\ also assumes the form 0/0, by the principle just proved we have fxl^'x\=f"xl^"x\; and so on, until we obtain a fraction which does not assume the form 0/0 when x = a. INDETERMINATE FORMS. 67 EXAMPLES 1. log(c/(a;-l)]i=l. log a; X 2. (1 — cosa;)/x2]o=l/2. 1 — cos XT 1 — cos X' n _0. . logx-1 _l/x-| _ 3. ^-1 x»- - COS x ~| _ 1 — cos x ~\ _ sing~l _ x2 Jo~0' ■'■ x2 Jo~ 2xJo~0' sinx ~| _ cosx ~| _ 1 1 — cos x ~| _ 1 ■■ 2x Jo~ 2 Jo~2' ■■ x2 Jo~2' Ln — Jl . q /sinmx\'""| _ iJi m ' \ X ) Jo""" ■ 4. — : I = 2. 10. — ap. sin x J 1 — X + log X J 1 ^ 5_ e^-e-^-2x -j ^^^ ^^ xlog(l+x) -| ^^ X — sin X J ' ' 1 — cos x J o 1^ , a ,- tanx — sinx"! 1 -J„='°sr- 12- sinBx J„=2- — 1 = - 1 ■ 13 ^~^ ~\ = - 2 Jo 6 ■ 1— x + logxJi X X — sin sin^x ax+i — i,x+\ -i _, a se c^x — 2 tan x ~l _1 x + 1 J_i~" ^^6' ^^ l+cos4x ]^/~2' fx I do fx I /* X I 85. When ''— assumes the form — > '- — = '^ ■ (1) ^kJ^ -^ ap xy, we obtain limit X -. limit r.^n ^ limit r/x"j ^^ B DIFFERENTIAL CALCULUS. EXAMPLES. 1. CSC 2 K/csc 5 a;]o = 5/2. CSC 2 X sin 5x CSC 5 s sin 2 X CSC 2 x"| sin 5 x~| 5 cos 5 x~| 5 when X = 0. §84 ]_ sm 5 x ~| _ 5 cos 5 x ~| _ ~ sin 2 X J 2 cos 2 x J o 2 Here we transform the given fraction into an identical fraction which assumes the form 0/0 when x = 0. 2. log x/cscx]o =: 0. log XT - cp logxl 1/x -[ ^^ esc X J Cp ' ' ' OSC X J — CSC X cot S J 1/x _ — sin^x 1/x ~l _ — sin% ~| _ — esc X cot X ~ X cos X ' ' ' — CSC xootxjo xcosxjo — sin^n — 2 sin X cos x' X cosx H :^2jmx^osxn p §84 J cos X — X sm X J Here we derive a new fraction by § 85, and transform that into an identical fraction which assumes the form 0/0 when x = 0. logx ' cotx. 1=0. 6.^5^1 =-3. Jo SeCiiXj,r/2 4 tanx -l ^3 ' ^ ]og{lj-xn ^^ \-^ tan3xJn-/2 sec(?fx/2)Ji g log(x-?r/2 )-l _^ g logUn2x-| __^ tanX j7r/2 ' " logtanxJ:r/2 g (e^-l)tan2x -1 ^ r e^ - 1 /taxy-] ^ j ^ ^ _ ^ x3 JoLxVx/Jo . tan X — x ~l _ 1 — sin X + cos x ~| _ ' X — .sinxjo ' ' sinx + cosx — lJ„/2~ 86. The forms ■ ap and ap — ap. A function of x which assumes the form () ■ ap qx ap — ap when a; = a is evaluated by- first transforming it into an identical fraction which will assume the form 0/0 or ap I ap when x = a. INDETERMINATE FORMS. 69 EXAMPLES. 1. (1 — a) taii(7ra;/2)]i = 2/7r. Taking the reciprocal of the infinite factor, we have (1 — x)tan — s— n TTTs =rwhenx = 1. 2 cot(7fx/2) 2. sec 3xoos7x],r/2 = 7/3. 5. sin x log x]o = 0. 3. (1 — tanx)sec2x]„/4 = 1. (i 6. xlogx]o = 0. 4. tanxlogsinx].,, = 0. 7. [-^ - -L^] ^= _ 1 . 2 1 1-x ^ ~~ = -z when X = 1. x2 — 1 X — l-x2 — 1 «■ [ii-i3|^]i=-l- !"• [=«tanx-|secx]^^ = -l. 9- r-V-^r-^— 1 =^ 11. xlogfl+«)l -a. Lsm^x 1 — cosxJo 2 ^\ x/J„ Limit r / a\"| log(l+az)-l , 1 87. The forms 0", ap", and I*'*'. When for a; = a, a func- tion of X assumes one of the forms 0°, ap", or I'^'V, the loga- rithm of the function will assume the form ±Q ap, and can be evaluated by § 86. From its logarithm the value of the given function can be obtained. EXAMPLES. 1. x^]o=l. log (x^) = X log x = —0 -op when x = 0, X log x]o = 0. § 86, example 6 .-. log x"^]o = ; .-. x^]o = 1. log('n-^~)'"] =xlog(l-l-^~)1 =a. § 86, example 11 70 DIFFERENTIAL CALCULUS. 3. x='°^]o = l. 8. (H-ax)i/^]o = e''. sa.-Ui'^') 4. (sina;)'i>''^],r/2 = L 9. (loga;)^]o = 1. 5. aM/(i-=:)]i = e-i. f'i 10. (e^ + a;)i"]o = e^. 6. (cosma;)'!'»:]o = 1. 11. (cos 2a;)i'^]o = e-^. 7. x^-i]o = l. 12. (logK)^-i]i = l. 88. Evaluation of derivatives of implicit functions. When y is an implicit function of x, its derivative, though containing both x and y, is a function of x. Hence, when the derivative assumes an indeterminate form for particular values of X and y, it can be evaluated by the previous methods. EXAMPLES. 1. Find the slope of a^?/ — a^x^ - x* = at (0, 0). „ dy 2a2x + 4x8 . Here -f- = — — -; = - > when x = w = 0. dx 2o^ %-| _ 2 g^x + 4 x3 -i _ 2 g^ + 12 x^ n _ 1 n ' dx\ 0,0 2 a'^y J o,o 2g2 ■ dy/dx\ o,o ~ d^/dxj o,o ' . . (d2//dx)2]o,o = 1, or dy/dx]o,a = ± 1. 2. Find the slope of y^ = gx^ — x> at (0, 0). TT ^"1 _ 2 gx — 3 x^ l _ 2a — 6x H dx\o,o~ Sy^ Jo,o 6?/-d2//dxJo,o' /d2/\2-i _ 2 a — 6 x "! _2a_ dy-l _ ■■\dx/Jo,o~ 6y Jo,o~ ""'"'' ""^ dxj o,o ~ ''^' 3. Find the slope of x^ — 3 oxj/ + y^ = o at (0, 0). Ans. dy/dx]o,o — or ap. 4. Find the slope of x* — g^xy + 6V = o at (0, 0). Ans. dy/dx]o,o = or a^/b\ 5. Find the slope of {y^ + x^)"- — 6 azy'' — 2 ox^ .+ aH^ = at (0, 0) and (o, 0). Ans. dy/dx']o,o =±ap; dy/dx]a,o = ± 1/2. CHAPTER VI. EXPANSION OF FUNCTIONS. 89. A series is a succession of terms whose values are all determined by any one law. A series is finite or infinite according as the number of its terms is limited or unlimited. The sum of a finite series is the sum of all its terms. The suTTh of an infinite series is the limit of the sum of its first n terms as n increases indefinitely. When such a limit exists, the series is said to be convergent; when no such limit exists, the series is divergent and has no sum. For example, the series 1 + 1/2 + 1/4 + 1/8 + • • • + l/2»-i + • • • is convergent ; for the sum of its first n terms == 2 when n = oo. The series l-l + l-l+--- + (- l)»-i + ■ • ■ is divergent ; for the sum of its first n terms does not approach a limit when ji = 00. 90. To expand a function is to find a series the sum of which shall equal the function. Hence, the expansion of a function is either a finite or a convergent infinite series. When the expansion is an infinite series, the difference between the function and the sum of the first n terms of the series is called the remainder after n terms. When re = co, this remainder must evidently approach zero as its limit. For example, by division we obtain '^ ;1 + x + a;2 + a;3+ . . . + a;»-i-).^_-51_. (i) 1— X 1 —X 72 DIFFERENTIAL CALCULUS. Here x"/(l — x) is the remainder after n terms. When «. = w and X > — 1 and < 1, this remainder i 0, and therefore the sum of n terms of the series = the function ; but when n = co and x > 1 or < — 1 , the remainder increases arithmetically, and therefore the sum of n terms of the series diverges more and more from the value of the function. Hence, the series in (1) is convergent and its sum equals the function only for values of x between — 1 and + 1. Some functions may be expanded by division, as above ; some by indeterminate coefficients ; others by the binomial theorem ; and so on. The binomial theorem, the logarithmic series, the exponential series, etc., are all particular cases of Taylor's theorem, which is stated and proved in § 92. Eor the proof of this theorem we need the following lemma : 91. Lemma. If tj>z and 's are each continuous betiveen a and a + h, and <^a = <^ (a + /j) = ; 'z must equal zero for at least one value of z hetiveen a and a + h ; that is, (f)'(a + 6h) = 0, where 6 is some positive proper fraction. For if <^s is continuous and ^a = <^ (a + 7t) = ; then, as z changes from a to a + h, '« must change from + to — or from — to + ; and therefore, if con- tinuous, it must pass through for some value of z between a and a + A. Denoting this value of « by a + dh where $ has some value between and + 1, we have tf>' (a + Oh) = 0. 92. Taylor's theorem. Whenfz,fz,f'z,- • ^fz are each continuous between x and x + h, f(x + h) = fx +f'x \+f"x I + • . • +/.-^ a^ -^!I^ +/" (x + eh)^, (A) where the last term is the remainder after n terms, and 6 is some positive proper fraction. TAYLOK'S THEOREM. 73 Let Ph''/\n denote the remainder after n terms when x = a; then we have t^ L>v^ -n-ts,^— h h^ h"—^ ' h" 1 [£ \n — l \n We proceed to find the value of F. Putting A = 6 — a in (1), and transposing, we obtain fh -fa -fa —^ /"a ^ > - f"a '- ' - ■ [2 -/-c.(^^^'-pfc^" = o, Let <^s represent the function of z obtained by substituting z for a in the first member of (2) ; then i-o A •^ l>t-l lw "^ -^ Differentiating (3) to obtain <^'z, we find that the terms of the second member destroy each other in pairs with the excep- tion of the last two, and obtain By hypothesis fz, f'z, • • ■, /"z are continuous between a and a + h; hence, from (3) and (4), it follows that z and ijt'z are continuous between a and a + A. Putting a for z in (3), by (2) we have <^a = 0. Putting b for z in (3), we have ^5 = 0, i.e. <^ (a + A) = 0. Hence, by § 91 we have — 1, /a; and all its successive derivatives are con- tinuous. Using the second form «f R^,, -we obtain J aUJL a.^<^ t-ff^f.- 1 + exj 1-6 The second factor in R^i^ is finite, and the first factor = when re = oo and a- > — 1 and < 1 or a; = 1. Therefore, the series in (1) is the expansion of log (1 + x) when a; > — 1 and < + 1 or x = 1. Putting X = 1 in (1), we obtain log,, 2. Putting X = 1/2, we obtain log„(.3/2), or log„3 — log„2; etc. CoE. 1. To deduce a series more rapidly convergent than that in (1), we put — x for x in (1) and obtain \og^{l-x)=m[-x-^---j+---y (2) Subtracting (2) from (1), we obtain T X 1 ,, 1+X S + 1 Let ^ = n — T^; *li6n = (4) 2z + l 1—x s ^ ^ Substituting in (3) the values in (4), we obtain log„t±i = 2™(^ + 3^^3 + ---)- (5) 78 DIFFERENTIAL CALCULUS. ■•■ ^°^« (^ + 1) = ^°^«^ + 2 r.(^^ + 3^2^ +••■)• (6) When s>0, 0 h arithmetically, and w. = oo, the product of the first and second factors = ; hence, -Ky = 0. Hence, the binomial theorem holds true when the first term is greater than the second arithmetically. When TO is a positive integer, we obtain ym+i^ = 0, /"•+2a; = 0, • • •; hence, in this case, the expansion in (1) is a finite series of m + 1 terms. 99. Failure of Maclaurin's theorem. The successive derivatives of log x are ap and discontinuous when x = ; hence, (B) fails to expand log x. For a like reason, (B) fails to expand cotx, cscx, vers~^x, a^^% sin(l/x), ■ • •. When (B) fails to expand fx, (A) will fail to expand fix + K) for X = 0. For example, when x = 0, (A) fails to ezpand log (a; + A), cot (x + A), vers-i(x + A), • • •. (A) may fail to expand fix + h) for other values than x = 0. The limits between which any expansion holds true should be carefully determined. 80 DIFFERENTIAL CALCULUS. EXAMPLES. 1. tan X = X + — + -TTz- + -^rr + • • ■. 3 15 315 x3 , x5 x' , x' 2. tan-ix = x-^ + --y+9 . Here /'x = (1 + x^)- 1 = i — x^ + x* — x^ + ■ • • ; .../"x = —2x + 4x3 — 6x5 + • •, etc. „ ■ , , 1 x3 ^ 1 • 3 x5 , 13-5 x^ , 3. Bm-ix = x + ---3+^-- + ^Tj:^-y+---. 4. From the series in example 3 find the value of ir. Putting X = 1 /2, we obtain '"' 2-0-2(^+^ + ^ + 7168+ )• .-. TV = 3.141592- • •. , , x'^, 5x< 61xf(a - h) and fa >f(a + h) ; that is, fa is a maximum of fx. § 100 If f"a is +, fa = is a maximum of xi/^. s : ^^V • ^ ^ ■ '-^ "> ^^*^ 18. 3v3/4 is a maximum of sin 5(1 + cose). tr rt \ 19. S/(l + 9 tanS) is a maximum vfhen = cosS. i^^-k- VWv^, u i-<-tv ^ -- ttt^ Examine the reciprocal of this function for maxima and minima. j 20. Show that 2 is a maximum ordinate and — 26 a minimum ordi- v . \ ' ^ nate of the curve w = x^ — 5 x^ + 5 x^ + 1. X / ; \ 21. Show that a/3 is a maximum ordinate and a minimum ordinate of the curve y = (2x — ay^ {x—a)^'^. ^ _ ^1 T-I' "mo^ I^- Vtvi-v\ 13 ' > 22. Show that sVs/lB is a maximum ordinate of the curve ^ y = sm^x cosx. v _ n TT ,i7t 1 1 "t Y - , 1^ +-^:- %. K V 88 DIFFERENTIAL CALCULUS. PROBLEMS IN MAXIMA AND MINIMA. 1. Eind the altitude of the maximum cylinder that can be inscrihed in a given right cone. Let DAB be a section through the axis of the cone and the inscribed cylinder. Let a = DC, b = AC, y = MC, x = IM, and V = the volume of the cylinder ; then V = jcxy^. From the similar triangles ADC and IDH, y = (b/a)(a — x); A M) C KB .■.V = Tt{b/aYx(a-xY. V will be a maximum vrhen x (a — x)^ is a maximum. Hence, let fx = x{a — x)^, etc. Ans. The altitude of the cylinder = 1/3 that of the cone. 2. Find the altitude of the maximum cone that can be inscribed in a sphere whose radius is r. Let A CD and A CB be the semicircle and the triangle which gener- ate the sphere and the cone, respectively. Let X = AB, y — BC, and V = the volume of the cone ; then y = i nzij^. y^ = AB ■ BD = x(2r — x); .. V = iax^{2r — x). V will be a maximum when x^{2r — x) is a maximum. Ans. The altitude of the cone =4/3 the radius of the sphere. 3. Find the altitude of the maximum cylinder that can be inscribed in a sphere whose radius is r. \E Let I = AB, and y = BE ; then F = 2 7rxy2 = 2 7rx()-2 — x2). Ans. Altitude = 2 r V3/3. 4. The capacity of a closed cylindrical vessel is c ; iind the ratio of its altitude to the diameter of its base when its entire inner surface is a minimum ; find its altitude. MAXIMA AND MINIMA. 89 Let u equal the radius of the base, x the altitude, aud S the entire inner surface ; then C = TCXU^, (1) and S = 2 «u2 + 2 xxu. (2) Trom (1), du/dx = —u/2x. (3) From (2), dS = i nudu + 2 axdu + 2 TCudx. (4) When S is a minimum, dS/dx — 0, or dS = 0^ hence, du/dx = —u/{2u + x). (5) Prom (3) and (5), we obtain 2 a; = 2 M + E, or X = 2 It. (6) Hence, as S evidently has a minimum value, it is a minimum when the altitude of the cylinder is equal to the diameter of its base. From (1) and (6), we find the altitude x = 2Vc/2;r. 5. Find the maximum rectangle which can be inscribed in the ellipse x^/a^ + y^/b^ = l. (1) Let (x, y) be the vertex of the rectangle in the iirst quadrant, and let u denote the area ; then u = 4 X2/. (2) Differentiating (1) and (2), and proceeding as in example 4, we find that the maximum area is 2 ab. 6. Find the maximum cylinder which can be inscribed in an oblate spheroid whose semi-axes are a and 6. The ellipse which generates the spheroid is xVa^ + y^/b^ = 1. (1) Let (x, y) be the vertex in the first quadrant of the rectangle which generates the inscribed cylinder ; then F = 2 TCyx^. (2) 7. The capacity of a cylindrical vessel with open top being constant, what is the ratio of its altitude to the radius of its base when its inner surface is a minimum ? %-^ u 8. A square piece of sheet lead has a square cut out at each comer ; find the side of the square cut out when the remainder of the sheet will form a vessel of maximum capacity. ^ — 4< ^i/ tu.'^ 90 DIFFERENTIAL CALCULUS. 9. The radius of a circular piece of paper is r ; find tlie arc of the sector which must be cut from it that the remaining sector may form tlie convex surface of a cone of maximum volume. .4ms. Arc = 2 ttj- (1 — V6/.3). Let X = the altitude of the cone ; then Y-Ttxif' — x'^)/Z. 10. A person, be*ng in a boat 3 miles from the nearest point of the beach, wishes to reach in the shortest time a place 5 miles from that point along the shore ; supposing lie can walk 5 miles an hour, but row only at the rate of 4 miles an hour, required the place where he must l^'Ud. Ans. 1 mile from the place to be reached. 11. Find the maximum right cone that can be inscribed in a given right cone, the vertex of the required cone being at the centre of the base of the given cone. ^^. The ratio of their altitudes is 1 : 3. 12. A Norman window consists of a rectangle surmounted by a semi- circle. Given the perimeter, required the height and the breadth of the window when the quantity of light admitted is a maximum. Ans. The radius of the semicircle = the height of the rectangle. 13. Prove that, of all circular sectors having the same perimeter c, the sector of maximum area is that in which the circular arc is double the radius. Let X — the radius of the sector ; a; (c — 2 a;) then area = — ^ — r 14. Find the maximum convex surface of a cylinder inscribed in a cone whose altitude is 6, and the radius of whose base is a. A-KS. Maximum surface = nab/ 2. 15. Find the altitude of the cylinder of maximum convex surface that can be inscribed in a given sphere whose radius is r. Ans. Altitude — rV2. 16. Find the altitude of the cone of maximum convex surface that can be inscribed in a given sphere whose radius is r. Ans. Altitude = 4 r/3. MAXIMA AND MINIMA. 91 17. A privateer has to pass between two lights, A and i', on opposite headlands. The intensity of each light is known, and also the distance between them. At what point must the privateer cross the line joining the lights so as to be in the light as little as possible 1 Let c = the distance AB, and X = the distance from A to any point P on AB. Let a and b be the intensities of the lights A and B, respectively, at a unit's distance. The intensity of a light at any point equals its intensity at a unit's distance divided by the square of the distance of the point from the light. Hence, the function whose minimum we seek is o/x2 + 6/(c — x)2. Ans. x = cai'8/(ai/3 + 6i/3). 18. The flame of a lamp is directly over the centre of a circle whose radius is r ; what is the distance of the flame above the centre when the circumference is illuminated as much as possible ? Let A be the flame, P any point on the circum- ference, and X = AC. The intensity of illumina- tion at P varies directly as sin CPA, and inversely as the square of FA. Hence, the function whose maximum is required is ax/ {r^ + x^)^/^, where r is the radius of the circle, and a is the intensity of illumination at a unit's distance from the flame. Ans. rV2/2. 19. On the line joining the centres of two spheres, find the point from which the maximum of spherical surface is visible. Let cp = r, CP = R, cC = a, and cA = X, A being any point _ on mM. From A draw the tan- gents Ap and AP ; then the sum of the zones whose alti- tudes are nm and NM, respec- tively, is the function whose maximum Is required. By geometry this function is L \x a — x/j Ans. X - ar^n/ (r^/^ -\- B3/2). 92 DIFFERENTIAL CALCULUS. 20. Assuming that the work of driving a steamer tlirough the water varies as the cube of her speed, show that her most economical rate per houi' against a current running c miles per hour is yc/2 miles per hour. Let V = the speed of the steamer in miles per hour. Then av' — the work per hour, a being a constant ; and V — c = the actual distance advanced per hour. Hence, av^/ (u — c) = the work per mile of actual advance. 21. The amount of fuel consumed by a certain ocean steamer varies as the cube of her speed. When her speed is 15 miles per hour- she con- sumes 4i tons of coal per hour at ^4 per ton. The other expenses are $12 per hour. Find her most economical speed and the minimum cost of a voyage of 2080 miles. ^„s. 10.4 miles per hour ; §3000. 22. Find the parabola of minimum area which shall circumscribe a given circle whose radius is r. Ans. y'^ — rx. 23. One dark night the captain of a man-of-war saw a privateersman crossing his path at right angles and at a distance ahead of c miles. The privateersman was making a miles an hour, whUe the man-of-war could make only 6 miles in the same time. The captain's only hope was to cross the track of the privateersman at as short a distance as possible under his stern, and to disable liim by one or two well-directed shots ; so the ship's lights were put out and her course altered so as to effect this. Show that the man-of-war crossed the privateersman' s track {c/b)\/{a^ — t^) miles astern of the latter. 24. The limited line AB lies without and is oblique to the indefinite line CD ; find the point P in CD so that the angle APB will b e a maxi - mum. Ans. If AB produced meets CD in C, PC — ^AC ■ BC. CHAPTEE, VIII. POINTS OF INFLEXION. CUKVATTJKE. EVOLTJTES. 108. A curve w concave upward or downward at any point (x, y) according as d-y/dx^ is positive or negative. When a curve, as ah, is concave upward, tan ^ or dy /dx evidently increases when x increases ; hence, by Cor. of § 80, d^y Ida? is positive. When a curve, as cd, is concave downward, tan ^ or dy/dx evidently decreases when x increases ; hence, d^y fdx^ is neg- ative. 109. A point of inflexion is a point, as P, where the tan- gent crosses the curve at the point of contact. On opposite sides of a point of inflex- ion, as P, the curve is concave in oppo- site directions, and dhj f dv? has opposite signs ; hence, at a point of inflexion dy [ dx has either a maximum or a mini- mum value (§ 101). Therefore, to examine a curve for points of inflexion, we examine its slope dy/dx for maxima and minima. The road or path whose grade is aPb is steepest at the point of inflexion, P. 94 DIFFERENTIAL CALCULUS. EXAMPLES. 1. Examine y = a + c(x + by for points of inflexion. Here d^/ /dx^ = ec{x + b). The root of 6c(x + b)-0 is — 6, and Qc(x + b) evidently changes from — to + when x passes through — 6 ; hence, (— 6, a) is a point of inflexion, or a point of minimum slope. To the right of (— b, a) the curve is concave upward. 2. Examine x^ — 3 bx^ + ah/ = for points of inflexion. Alls, (b, 2b^ /cC) is a point of inflexion, or of maximum slope, to the right of whicli tlie curve is concave downward. 3. Examine ?/ = x^ — .3 x^ — 9 x + 9 for points of inflexion. Ans. (1, — 2) is a point of inflexion, to the right of which the curve is concave upward. 4. Examine y = c sin (x/a) for points of inflexion. Aiis. (0, 0), (±a7i, 0), (±2 a*, 0), • • •. 5. Examine the witch of Agnesi y = ia^ / (x^ + i a^) for points of inflexion. Ana. (± 2aV3/3, 3a/2). 6. Examine the curve y = x^/(a- -\- x^) for points of inflexion. Ans. (0, 0), (aVs, 3aV3/4), (- aVI, -3aV3/4). 110. Polar curves. Prom tlie figure it is evident that when a polar curve, as ab, is concave toward the pole, p or OD increases as p increases; hence, dp j dp is positive. When a curve, as cd., is convex toward the pole, p decreases as p increases ; hence, dp j dp is negative. POINTS OE INFLEXION. 95 That is, a polar curve is concave or convex toward the pole according as dp /dp is positive or negative. At a point of inflexion on a polar curve, dp /dp changes its quality, and therefore j? is a maximum or a minimum ; and conversely. Hence, to examine a polar curve for points of inflexion, we examine p for maxima and minima. Ex. Examine the lituus p^e = a? for points of inflexion. Here P § 63, (9) Vp2 + (dp/ciey^ V4 a" + p4 dp _ 2 g^ (4 g^ — p*) '■ dp " (4g* + p'»)3/2 ' Hence, p = gV2 renders p a maximum; therefore, (gV2, 1/2) is a point of inflexion. In the logarithmic spiral p = g^, dp /dp is always positive ; hence, the curve, heing concave toward the pole at all points, has no point of inflexion. 111. The curvature of any curve, as APQ (§ 112, fig.), at any point, as P, is the s-rate at -which the curve bends at P, or the s-rate at which the tangent revolves, where s denotes the length of the variable arc AP. 112. The curvature of a curve at (x, y) is d<^/ds radians to a unit of a. Let AP = s, and let denote (in radians) the variable angle XMP as P moves along the curve APQ ; then, evidently, the curvature of APQ at P equals the s-rate of ^, or d I ds. 113. Curvature of a circle. Let APQ be the arc of a circle whose radius is r ; then the angle MEN, or A<^, will equal the angle subtended by PQ, or As, at its centre. 96 DIFFERENTIAL CALCULUS. Hence, by § 37, we have A^ = As/r; .-. d/ds = 1 /S, or JJ = ds/d(j>. If at P (§ 112, fig.) the direction of the path of (x, y) became constant, (X, y) would trace the tangent at P ; if at P the change of direction of the path became uniform "with respect to s, (x, y) would trace the circle of curvature at P. 115, To find R in terms ofx and y. ds/dx = [1 + (d^/dxyy '\ § 33, Cor. 1 (f> = tan~^ (dy/dx) ; § 33 dcf} _ d'y /dx' dx 1 + (dy/dxy . d^ ^[l±(dy/dx)^ ■ d dhjldx^ ^ > R will be positive or negative according as d^y fdx^ is posi- tive or negative ; that is, according as the curve is concave upward or downward. If we take the reciprocals of the members of (1), we obtain the curvature. RADII OF CURVATURE. 97 EXAMPLES. Find E and the curvature of each of the following curves : 1. The parabola y'^ — 4px. dy/dx = 2p/y, d'^/dx^= —4p^/yK Substituting these values in (1) of § 115, -we obtain \ y2 ) 4^,2 pl/2 We neglect the quality of E, since the quality of cPy /dx^ indicates whether the curve is concave upward or downward. At the vertex (0, 0), E — 2p, and the curvature is {l/2p) radian to a unit of arc. 2. The equilateral hyperbola 2 a;?/ = a2. E = {x^ + y'^)^'^/a^. 3. TheeIUpse| + S=l. S = (^i^^^f^W^ " The maximum cui'vature is a/b^, and the minimum b/a^. 4. Thecurve2/ = x* — 4x3 — 18x2at(0, 0). K = i/36. dtp _ my 5. The logarithmic curve y — a". 6. The cubical parabola y^ — a'h:. ds (m2 + yY'^ d _ 6 a*y ds ~ (9y* + a*)3/2" 7. The cycloid x = r vers- ' (y /r) if V2 rj/ — y^. E = 2V2 ry. /' '' > At the highest point iJ = 4 r, or the maximum of B is 4 r. 8. The catenary y = | i^'" + e-'^'"). 9. The hypocycloidx2/8 + 2^2/8 = a2/s. 10. The curve x^ '2 + ^^i /2 = ^i /2. 11. The curve 2/8 = 6 x^ + x'. d _ ds ~ a yl- E = ZiaxyY'K d ai/2 ds 2(X + 7/)3/2 d(p _ 8x^2/ ds [2/4 + (4x + x2)2]3/2 98 DIFFERENTIAL CALCULUS. 116. To find E. in terms of p and 6. From (4) and (10) of § 63, we have ^ = + \j/, i/f = tan""^ dp/de .d±_ d4 d^ _ {dp / dOf - p ■ dJ'p / dff" ■'■ d6~ de' dd~ p^ + (dp/ddy . d _ p^ + 2 (dp /doy -p-d^pjdS" " d6~ p^ + {dp/dey ds/do ^ fp' + (dp fdoyj'^ ' d^/dd p'' + 2 {dp / dey - p ■ d'p / ds'' § 63, (3) EXAMPLES. Find R in each of the following curves : 1. The spiral of Archimedes p = a0. Here dp/de = a, d^p/de^ = 0; p2 + 2 a2 2. The logarithmic spiral p = a'. 3. The lemniscate p^ — a? cos 2 0. 4. The cardioid p = a(l — cos S). 5. The curve p = a sec^ (9/2). 2 + 2 i? = /)Vl + (logo)2. iJ = aV3p. iJ = 2vTa^/3. fi = 2 a sec3 (9/2). 117. Co-ordinates of centre of curvature. Let P{x, y) be any point on the curve ab, and C{a, ji) the corresponding centre of curvature. Then PC equals R and is per- pendicular to the tangent PD. Hence, /_BCP = A XDP = <^. OA =^0E -JiP, AC = EP + BC; EVOLUTES AND INVOLUTES, that is, a = a; — ^ sin <^, ^8 = y + ^ cos <^ ; or a = X — B- Substituting in (2) the values of B and ds, we have dy dx wm 1+ P = y + - \dx/ 99 (1) (2) (3) d^y I da? If the point (x, y) moves (Py jdx^ 118. Evolutes and involutes. along the curve MN, by equa- tions (3) of § 117 the point y (a, yS) will trace some other curve, as AB. The curve AB, which is the locus of the cen- tres of curvature of MN, is called the evolute of MN. To express the inverse rela- tion, MN is called the involute of AB. 119. Properties of the involute and evolute. I. From Cor. 1 of § 33 and ds = Bdiji, we have dx = cos ds = B cos dtjj, (1) and dy = sin ds = B sin d. (2) Differentiating equations (1) of § 117, and using the rela- tions given in (1) and (2), we obtain da = dx — B cos dcft — sin (j> dB = — sin <^ dB, (3) dp = dy — B sin (f>d dB = cos ^ dB. (4) Dividing (4) by (3), we obtain dp I da = — cot x intersect in the points Pi, -=■'*■ Let OJ/i = a, and let h denote the distances from Ml to the ordinates :i.5_ MiPi,M,P„M,P„---; that is, let O M, Ml mTx h = 0, MiM^, MiM„ ■ ■ : Then, for each of these values of h, we have (a + h)=f(a + h), or =/(« + A) - <^(a + h). (1) Expanding the second member of (1) by Taylor's theorem, we have = (fa- 4>a) + (fa - 'a) h + {fa - <^"a) - + if '"a - "'a) I . + ...+ (/^-a - ^^a) - + • • • (2) If Pj coincides with Pj, that is, if two values of h are zero ; by the theory of equations, from (2) we have fa = a, fa = 4,'a. (3) Hence, (3) are the two conditions for contact of the first order. If three values of 7i are zero, from (2) we have fa = «, fa = 'a, fa = cj>"a. (4) Hence, (4) are the three conditions for contact of the second order. In general, if ^ + 1 values of h are zero, from (2) we have fa = ^a, fa = 'a, fa = (j>"a, • • •, fa = KJ^^a. (5) Hence, (5) are the k + 1 conditions for contact of the A;th order. OSCULATING CURVES. 109 Or, in the differential notation, if when x = a y, dy /dx, d^y jdo?, ■ ■ •, d^y j dx^ all have the same values in one of two equations as in the other, the loci of these equations have contact of the fcth order at (a, y). Ex. Find the values of a, 6, c when the curves y = ox^ + 6x + c and y = log (« — 3) have contact of the second order at the point (4, 0). From y = log (x — 3), we have when x = 4, 2/ = 0, dy/dx-1, dhj/dx'^ = — \. (1) From y = 0x2 + 6a; + c, we i-^^^^ when x = 4, y = l&a + ih + c, dy/dx = Sa + b, d^y/dx^ = 2a. (2) Equating the values of y, dy/dx, and d^y/dx^ m (1) and (2), -we obtain 16a + 46 + c = 0, Sa + b-l, 2a=-l. (3) Solving system (3), we obtain a = -l/2, & = 5, c=-12. Hence, the curve y — — x'^/2 + 5x — 12 has contact of the second order with the curve y — log(x — 3) at the point (4, 0). Only three independent conditions can be imposed on the three general constants a, 6, c ; hence, in general, the given curves cannot have an order of contact above the second. 127. Osculating curves. The straight line of closest contact (a tangent) has, in general, contact of the first order ; for two and only two independent conditions can be imposed upon the two arbitrary constants in the general linear equation, y = nix + c. The circle of closest contact, called the osculating circle, has, in general, contact of the second order ; for three and only three independent conditions can be imposed upon the three arbitrary constants in the general equation of the circle, {x - ay + (y - hf = r". 110 DIFFERENTIAL CALCULUS. The parabola of closest contact, called the osculating parabola, has, in general, contact of the third order ; for the general equation of the parabola has four arbitrary constants. The conic of closest contact, called the osculating conic, has, in general, contact of the fourth order ; for the general equa- tion of the conic has five arbitrary constants. It was necessary to qualify the above propositions by the words ' in general ' ; for at particular points the contact may be of a higher order than at points in general. For example, at a point of maximum or mini- mum slope the tangent has three coincident points in common with the curve, the contact is of the second order, and the tangent crosses the curve (§ 109). Again, at a point of maximum or minimum curvature, the circle of curvature, which does not cross the curve at the point of contact (§ 130), has contact of the third order at least. 128. The osculating circle is the circle of curvature. From §§ 115 and 126 it follows that any two curves which liave contact of the second order have the same curvature at their common point ; and conversely. 129. The circle of curvature, in general, crosses the curve at the pobit of contact. For its contact is, in general, of an even order. 130. *At a point of maxim.um or minimum curvature, the circle of curvntnre does not cross the curve. For on each side of a point of maximum curvature, the curve changes its direction more slowly than at this point ; hence, on each side of this point, the curve lies without the circle of curvature at this point, and therefore does not cross it. For a similar reason, the circle of curvature at a point of minimum curvature does not cross the curve. Since the circle of curvature at a point of maximum or minimum curvature does not cross the curve, the contact must be of an odd order (the third at least). * By a maximum or a minimum curvature is meant an arithmetic maximum or minimum, the quality of the curvature not being considered. ORDERS OF CONTACT. Ill EXAMPLES. 1. By drawing the figures, show that the four intersections of a circle and an ellipse coincide when the circle becomes the o.sculating circle to the ellipse at either end of the major or the minor axis. 2. Show that the curve ^y = ?>x'^ — x^ and the line 4 2/ = 3 a; — 1 have contact of the second order. 3. Show that the parabola 8 2/ = i^ — 8 and the circle x^ + ^2 = 6 2/ + 7 have contact of the third order. 4. Find the order of contact of the hyperbola xy — \ and the parabola (I -2)2 + (2^ — 2)2 = 21!/. 5. Find the value of a when the hyperbola xy = ?jX — \ and the parabola ;/ = a; + 1 + a (x — 1)^ have contact of the second order. .4)18. a = — 1. 6. Find the values of m and c when y = mx + c has contact of the second order with 2/ = x' — 3 1" — 9 x + 9. ^jjg m = — 12, c = 10. 7. Find the values of a, b, c when the curves y = x^ and y = ax^ + bx + c have contact of the second order at the point (1, 1). 8. Find the values of a, b, c when the curves y = smx and y — ax^ + 6x + c have contact of the third order at (ff /2, 1). Ans. a = - 1/2, 6 = )r/2, c = 1 - t('-/%. CHAPTER X. CHANGE OF THE INDEPENDENT VARIABLE. 131. Different forms of the successive derivatives of dy/dx. (i) When x is independent, by § 78 we have d dy _ dhj d d r/y _ d^y dx dx dx^ dx dx dx dx^ (ii) When neither x nor y is independent, dy/dx is a fraction having a variable numerator and a variable denomi- nator, and ddx = d^x, etc. ; hence, d dy _dxd?-y — dyd^x dx dx dx^ ^ ' d_ d_ dy _ dxhPy - dxdyd^x - 3 dxrPxdh^ + 3 dy(d''x) dx dx dx dx'' (2) (iii) When y is independent, dhj = 0, d^'y = 0, • • • ; hence, from (1) and (2) we obtain d dy dyd'x dx dx dx^ d d dy _3 dy (cPxy — dxdyd'x dx dx dx dx^ (1') (2') 132. Change of the independent variable. In the appli- cations of the Calculus it is sometimes necessary to make a differential equation depend on a new independent variable instead of the one which was originally selected ; that is, we need to change the independent variahle. CHANGE OF THE IXDEPEXDEXT VARIABLE. 113 When X = ^(z) and we wish to change the independent variable from x to 2, we substitute for cPy/(z). Ex. 1. Given x = cosS, change the independent variable from x to ein ^ — ^ ^ ■ y = m dx? I — x^ dx'^ \ — x^ ■ ^ ' Sabsdtating for dh//dz^ the second member of (1) in § 131, we have dxd?y — dyd?x __ x dy y _ .-, i(^)> ** is directly a function of x and indirectly a function of x through y and z. The differentiation of such functions is often simplified by using the formulas in the next article. 138. If u = f(x, y, z), y = <^(x), and z = <^i(x), du _ 6m 9m dy 8m dz dx dx dy dx dz dx ' where du/dx is the total derivative ofw. as a function ofx. du = ^dx + ^yhj + ^^dz. §135 Dividing by dx, we obtain (1). CoR. 1. If M =f{y, z), y = (x), and z = i(x), du _ Qu dy 9m dz dx dy dx dz dx ^ ■' COK.2. n-=/(3/)and, = <^(.),| = ||. EXAMPLES. 1. i( = z^ + 2/3 -f- zy, z = sin X, ?/ = e'' ; find du /dx. Here du/dy = oy'^ + z, du/dz = 2 z + y, dz/dx = cosx, dy/dx — eP'. Substituting these values in (2) of § 138, we have du/dx = (3y^ + z)&= + {2z + y)oosx = (3 e2»^ + sin x) &■ + (2 sin x + e') cos x = 3 e^^ + e^ (sin x + cos x) + sin 2 x. 2. M = tan-i (x?/), 2/ = e'. dtt/(to = 6^(1 + x) /(I + x%2^). 3. M = e»^(2/ — z), y = o sinx, z = cosx. du/dx = (a^ + l)e<" sinx. 4. M = tan-i^. x2 + 2/2 = ^2. ^= ^ X ax Vr2 — x2 5. M = sm - , z = e^, 2/ = x2. — == (x — 2) — cos — • ?/ ox ^ ' x^ x2 PARTIAL DIFFERENTIALS. 119 6. u = vx^ + 2/^1 y = mz + c. — = -^-;==^= • dx vx^ + (mx + c)2 7. u = sin— !(?/ — 2), 2/ = 3x, z = 4x8. du/dx = 3/ Vl — x^. 8. u = xV — a;%/2 + x*, 2/ = logx. dit/dx = xS[4(logx)2 + 7/2]. 139. Partial differentials and derivatives of higher orders. If we suppose only one of tlie independent variables to vary at the same time, by successive differentiations we obtain the successive partial differentials Q\u, d\u, Q^^u, Q^yU, • • ; Qhi Qhi &u Q^u ^^^' ^^^2/^ -^,dA ^d^f, •■-. For example, if it = x^ + xy'^ + y^, dxU = (2 X + 2/^) dx, Q^xU = 2 dx^, d^xU = ; a,jU = (2 xy + 2y) dy, Q\ii = (2 x + 2) dy\ Q\u - 0. If we differentiate u with respect to x, then this result with respect to y, we obtain the second partial differential, 0„,u, or ., , dxdy. " dxdy " For example, if it = x' + xHf'; QxU = (3 x^ + 2 xy^) dx, B'^xyU — 4 xydxdy. Similarly, the third partial differential 9'j,^2W, or ^ dydx', denotes the result obtained by differentiating u once with respect to y, then this result twice successively with respect to X. The symbols for the partial derivatives are Qhi ■Qhi 9^ 8% 8% d'l? dxdy dy' dx^ dyda? In iinding the successive partial differentials and deriva- tives of u, or f{x, 2/), we treat dx and dy as constants, since x and y are independent variables. 120 DIFFERENTIAL CALCULUS. 140. If u =f(x, y), ^ = ^> (1) dx'dy "" dxdydx ~ dydv? ' that is, if u is differentiated successively m times with respect to X and n tvm,es with respect to j, the result is independent of the order of these differentiations. Suppose X alone to vary in u =f(x, y); then \,u = fix + Ax, y) -f{x, y). (3) Supposing y alone to vary in (3), we obtain \i.^xU) =/(a;. + ^x,y + Ay) -f(x, y + Ay) -f(x + ^x, y)+f(x, y). In like manner we find ^xi^^y"") =/(a: + Ax, y + Ay) -fix + Ax, y) -f{x, y + Ay)+f{x, y). •■• \(.^xu) = A,(A„m), or \/A^A^A^/A, Ay \Ax J Ax\Ay 1} <*) Every term in Aji/ Ax which contains Ax vanishes in the A A u limit ; again, every term in —"- It —^ which contains Ay van- ishes in the limit ; hence, every term in -^ — ^- which con- ' ' .7 Ay Ax . tains either Ax or Ay vanishes in the limit. Likewise every term in — - —"— which contains either Ax ■' Ax Ay or Ay vanishes in the limit. Hence, by (4) we have SUCCESSIVE DERIVATIVES. 121 Differentiating (1) with respect to x, we obtain 9 / 3^M \ 9 / a^M \ 9^M _ 9°M ,g dx\dxdyj~dx\dydxj' dxdydx~ dydx^ ^' Applying the principle in (1) to Qu/dx, we have rfy dx\dx J ~ dx dy\dx J' d:i?dy'~ dxdydx ^ '^ From (5) and (6), we obtain (2); and so on. CoE. If, when Aa; = i and tvy = vi, we have A^j^M = <^ (x, t/) Ax\y + vi", where n7> 2; then 9^a:j,w = <^(x, y) dxdy. (7) EXAMPLES. Verify the identities (1) and (2) of § 140 in each of the four following functions : 1. ji = cos {x + y). 3. « = tan— 1 (y/x). 2. u = s^2 + a,y^_ 4. jt = sin (bx^ + ay^). ^ ' dx^ dxdy dx 6 jiu= "^^ X—+ 9^ ^2^"- x + y^ dx^ dxdy dx 7. If. = (.^ + ,^)i- .^|f + 2x,g + ,^5 = 0., 8. IfM = (.3 + ,3p^x=g + 2x.|| + .^5 = |«. 1 3^ 3^ 3^_ (x2 + 2/2 + z2)l/2' (Jx2 dy2 + ^^2 10. If M = e'«^ J^ ^^ = (1+ 3 xyz + x^j/ZzZ) m. . , , , 38m 1 + 2 x^y^z^ 11. If M = sm— 1 (x2^z), - ■ ■ = 7i „ „ „.,,„ ■ 122 DIFFERENTIAL CALCULUS. 141; To find the successive total differentials of a function of two independent variables in terms of its partial .differentials. Let u = f(x, y) ; then, by § 135, we have du = ^^ dx H — ;— dii. (1) dx dy . • . a'u = -r-T dx' + - — ;- dxdy + , ., rfyrfx + -r-r a?r, rfx^ rfiCfZe/ "^ dydx "^ dy' or ,V = g.x^ + 2^..., + ^:.^. (2) Differentiating (2), remembering that, in general, each term in the second member is a function of both x and y, and apply- ing the principle of § 140, we obtain By successive differentiations, we obtain d*u, d^u, etc. From the analogy between these results and the binomial theorem, the formula for d'^u is easily written out. 142. Expansion of f (x + h, y + k). Regarding x as the only variable, by Taylor's theorem we obtain f(x + h,y + k)=f(x, y + k) + h -^f{^, y + k) Regarding y as the only variable, we obtain k 9 W 9^ f(x, y + k) =f{x, y) + - -^yfi^, 2/) + jl ^/(^' V) + ' ' ■■ Substituting in (1) this value off(x, y + k"), we obtain f(x + h, y + k)=f(x, y) + h — f(x, y) + k—f{x, y) TAYLOR'S THEOREM. 123 A symbolic expression for this formula is where (h~ + k^) \ ax ay J is to be expanded by the binomial theorem and/(x, y) written after each of the resulting terms. If u =f{x, y), we may put u fov f(x, y) in (C) and (C). Compare (C) with (1) in § 144. CoK. 1. By a similar course of reasoning Taylor's theorem is extended to the expansion of functions of three or more independent variables, in series analogous to that in (C), or (C). CoE. 2. By Taylor's theorem we obtain the value of f(x + h)—f(x) in ascending powers of A; that is, Taylor's theorem expresses the increment of f{x) in ascending powers of the increment of x. Similarly, by the extension of Taylor's theorem, we express the increment of a function of two or more independent variables in ascending powers of the incre- ments of those variables. 143. Maxima and minima of f(x, y). f{a, b) is a maximum of f{x, y) when, for all small positive or negative values of h and k, f(a + h,b + k)-f(a,b)<0. f(a, b) is a minimum of f{x, y) when, for all small positive or negative values of A and k, f(a + h,b + k)-f(a,b)>0. 124 DIFFERENTIAL CALCULUS. 144. Conditions for maxima and minima of f(x, y). Putting u =f{x, y), from (C) of § 142, we obtain f{x + h,y + k) -fix, y) = ^''-^ + ^-^ When h and k are very small, the quality of f{x + h,y + k) -fix, y), or of the second member of (1), will evidently depend upon the quality of h and k unless . Qu /dx = 0, and Qu/ dy = 0. (a) But, by definition, when fix, y) reaches a maximum or a minimum value, the quality of fix + h, y + k) — fix, y) is independent of h and k. Hence, equations (a) express one condition for a maximum or a. minimum oi fix, y). Suppose a; = a, y = i to be one solution of system (a). Let A, B, C denote the values of Q^u/dx^, Q^u/dxdy, Q^u/dy^, respectively, when x = a and y = h; then from (1), we have fia + h,b + k)- fia, V) = .4 {Ah? + 2 Bhk + Ck^ ^ _ iAh + Bk)'' + iAC-B^)¥ ^ ^-^ - Z]2 +■■■• ^^> The quality of the second member of (2) is independent of h and k when and only when AC — B^ is positive or zero.* For when AC — B^ is negative, the numerator will be positive when k = 0, and negative when Ah + Bk = 0. Hence, a second condition for a maximum or a minimum of f(x, y) is * The limits of this treatise exclude the investigation of the exceptional case when AG = IP or when A = B— C = 0. MAXIMA AND MINIMA. 125 When condition (b) is satisfied by x = a, y = b; from (2) we see that f(a, b) will be a maximum or a minimum of f{x, y) according as A, or '&u J dx^\j,^ is negative or positive. CoK. 1. Condition (b) requires that 8^m/cZx^]„^j and Q^u / dy^^^j, have the same quality. Note. This discussion assumes that fix, y) and all its successive derivatives are continuous functions. CoE. 2. By a similar course of reasoning we may obtain the conditions for maxima and minima of functions of three or more variables. EXAMPLES. Examine for maxima and minima 1. u =f{x, y) — x'h/ + xy^ — axy. du/dx = {2x + y — a)y, du/dy = (2y + x — a)x, ahi,/dx^ = 2y, aH/dy^ = 2x, dhi/dxdy = 2x + 2y — a. Hence, condition (a) of § 144 is (2x + y-a)y = 0, (2y + x- a)x-0; (1) and condition (b) is 4:xy>(2x + 2y — a)^. (2) System (1) has the four solutions (0, 0), (a, 0), (0, a), (a/3, a/3). The last, and it only, satisfies (2) ; hence, /(a/3, a/3) is a maxi- mum or a minimum of /(x, y). If a is positive, d^/dx^ is isositive when y = a/3 ; hence, /(a/.3, a/.3), or - aV27, is a minimum. If a is negative, d^u/dx^ is negative when y = a/S; hence, — 0^/27 is a maximum. 2. M = a;3 -f- ^3 _ 3 axy. Ans. When a is + , — a^ is a min. ; when o is — , — a^ is a max. 3. u = x^ + xy + y^ — ax — 'by. Ans. {ab — a^ - 6^) /3 is a min. 4. u = %hf- (a — X — tj). Am. aV432 is a max. 126 DIFFERENTIAL CALCULUS. 5. « = M* + 2/* — 2 a;2 + 4 312/ — 2 2/2. j^^^ — 8 is a min. 6. u = (2ax — x^)(2by — y^). Ans. a^ft^ is a max. 7. Find the maximum of xyz subject to the condition xVa^ + y^/b'^ + zVc^ = 1- Since xyz is arithmetically a maximum when x^y- ■ z^/c^ is a maxi- mum, we put S2 2/2 N :.%2?, = XV(1-|.-|)- Here condition (a) of § 144 is and condition (b) is /, 6x2 yi., yp. 6y^\^,/ 2x2 2j^\\ V~ a^~ bOK a^ b^ )^V a^ b^ ) ^' The only solution of system (1) which satisfies (2) is x = a/V3, 2/ = 6/ Vs. 9^ o »/-, 0x2 y2. 862 , a b _ = 22/2(i-_^_-) = --^-, when x = ^, 2/ = ^- Hence, abc/sVs is a maximum of xyz. 8. Given the sum of the three edges of a rectangular parallelepiped ; find its form when its volume is a maximum. Ans. A cube. 9. Divide m into three parts x, y, z such that x''y''z'' may be a, maxi- nium. ^jjs_ x/a — y/b = z/c = m/{a + b + c). 10. Divide 24 into three such parts that the continued product of the first, the square of the second, and the cube of the third may be a maxi- mum. A.ns. 4, 8, 12. 11. The vertices of a triangle are (Xi, 2/1), (X2, 2/2)1 and (X3, 2/3) ; find the point the sum of the squares of whose distances from the vertices is a minimum. Ans. [(xi + X2 + X3)/3, (2/1 + 2/2 + 2/3)/ 3], or the centre of gravity of the triangle. CHAPTER XII. ASYMPTOTES. SINGTJLAB POINTS. CTTEVE TKACING. 145. An asymptote to a curve is a fixed line which is the limit of a tangent when the point of contact moves out along an infinite branch of the curve. 146. To obtain the equations of the asymptotes to the curve f{^, y) = 0, (1) where f (x, y) is of the nth degree in x and j. Let y = mx + 1 (2) be the equation of a tangent to (1). Substituting mx + I for y in (1), we obtain /Of, mx + 0=0. (3) Equations (2) and (3) form a system which is equivalent to the system (1) and (2); hence, the n roots of (3) are the abscissas of the n points common to the curve (1) and its tangent (2). Since (2) is a tangent to (1), two roots of (3) are equal. Conceive the point of contact to move out along an infinite branch ; then the two equal roots of (3) will become infinites. Therefore, by algebra,* the coefficients of a;" and a;"~^ in (3) will approach zero as their common limit. * Substituting 1/x for x in (1), and multiplying "by x", we obtain (2). X" + pia:"-^ + PiX''--^ + ■ • • + p„_2a;^ + A-iX + p„ = 0. (1) p„x" +p„_ix»-i+i)„_2a;"~''+ • • • + PiX^+PiX+ 1=0. (2) The n roots of (2) are the reciprocals of the n roots of (1). Whenpn = and^„_i = 0, two roots of (1) = 0; .-. two roots of (2)=oo. When jj„ = and Pn—i = 0, two roots of (1) = ; .-. two of the n roots of (2) assume the form ap, and (2) has, in reality, only n — 2 roots. 128 DIFFERENTIAL CALCULUS. Hence, by putting the coefficients of x" and a;"~^ equal to zero, and solving the resulting system of equations for m and I, we obtain the slope and the intercept of each asymptote. Ex. Examine for asymptotes the curve, 2/8 = ax2 — xs. (10 Let y = mx-^l (20 be the equation of an asymptote to (!'). Substituting mx + I for y in (1') and arranging the terms according to the powers of x, we have (ms + l)x3 + (ZmH - a)x2 + Smi^x + /» = 0. Putting the coefficients of x^ and x"^ equal to zero, we have to3 + 1 = 0, ZmH — a = Q. The only real solution of system (30 ism = — 1, Z = a/3. Substituting these values in (20, we obtain y=—x + a/3, which is the only real asymptote to curve (10. The locus of (r) is the curve nOPm, and that of (40 is the asymptote AB. (30 (40 CoE. 1. If we expand f(x, mx + Z) and arrange the result according to the descending powers of x, (3) will assume the form A^x" + J„_ia;"-i + J„_2X»-2 H + ^iX + A = 0. (4) ^„ will be of the reth degree in m, but will not contain I. ^„_i will be linear in I. EQUATIONS OF ASYMPTOTES. 129 Hence, the system, A^= and ^„_i = 0, (when determi- nate) has at most only n solutions ; whence a curve of the nth order cannot, in general, have more than n asymptotes. When, as is assumed in this article, equation (1) is of the wth degree in y, there will be no asymptote parallel to the y-axis ; and conversely (§ 148). CoE. 2. The n roots of equation (4) are the abscissas of the n points common to the curve (1) and the line (2). When J„ = and ^„_i = 0, (4) is, in reality, of the (n — 2)th degree and has only n — 2 roots, the other two of the n roots assume the form op. Hence, an (isymptote to the curve (1) cannot have more than n — 2 points in com.mon with the curve. For example, (1') and (4') form a system whicli is defective in two solutions; hence, the asymptote ^JS, or (4'), has only the one point P in common with the curve nOPm, or (!'). Any line parallel to an asymptote has a slope m which satisfies A„ = ; but when A„ = 0, (4) has only w — 1 roots. Hence, any line parallel to an asymptote to the curve (1) can- not have more than n — 1 points in common with the curve. For example, (1') and y — —x -\- c form a system which is defective in one solution ;. hence, any line parallel to AB cannot have more than two points in common with the curve nOPm. EXAMPLES. Examine for asymptotes 1. The folium of Descartes x^ + y^ — 3 axy. § 165, fig. 6 2. 2/3 = ax2 + a;8. y = x + a/Z. 3. The conic sections. ay — ± bx. 4. (y2 — 1) 2/ = (a;2 — 4) a;. y = x. 5. y* — x* + 2 ax^y = b-^x^ y = ±x — a/2. 130 DIFFERENTIAL CALCULUS. 6. X* — y* — a'kcy = Vh/'^. y = ±x. 7. ys _ g xy'^ + 11 x^y — 6x^ + x + y = 0. y = x, y = 2x,y = 3x. 8. x^ + Sx^y — xy^ — 3ys + x^ — 2xy + Sy'' + 4x + 5 = 0. y = -x/-a- 3/4, 2/ = X + 1/4, y=-x + S/2. 9. Prove that the asymptote y = — x lacks three points of intersec- tion with its curve a;^ + j/^ = a^. 147. Parallel asymptotes. In (4) of § 146 it sometimes happens that A,,_^ = does not determine I for one or more of the values of m given by -4„ = 0. If, in tliis case, I is determined by ^„_2 = 0, -which is a quadratic equation in I, we shall obtain for each value of in two values for I. This gives two parallel asymptotes, each of which lacks three points of intersection with its curve; for, in this case, (4) in § 146 has only w — 3 roots, the other three roots assume the form ap. If I is not determined by A„_2 = 0, but is determined by A„_^ = 0, we obtain three parallel asymptotes ; and so on. CoR. Any one of 7,' parallel asymptotes lacks ^ -f- 1 points of intersection with its curve ; and any line parallel to k par- allel asymptotes lacks 7c points of intersection with its curve. Ex. Examine y^ — xij- — z-y + x^ + x^ — y^ — 1 for asymptotes. Substituting mx + I for y, we obtain (m^ — m^ — m -f 1) a;^ -t- (3 m-l — m^ — 2 ml — I + 1) x^ + (3 mP — 2ml — P)x + {l" — P ~ 1)- 0. Equating the coefficients of x^ and x^ to zero, wei.liave ms — )n2 — m + 1 = 0, (1) -i 3mH — vi'^ — 2 ml — 1 + 1 = 0. (2) J From (1), m = — 1 or 1. When vt, = — 1, from (2) we obtain 1 = 0; hence, one asymptote is y = —x. ASYMPTOTES PARALLEL TO AN AXIS. 131 When m = 1, (2) does not determine I. This indicates that there are parallel asymptotes having the slope 1. To obtain the values of I for these asymptotes, we equate the coefficient of x to zero, and obtain 3mP — 2ml — P = 0. (3) When m — 1, from (3) v?e obtain l — O or 1; hence, the parallel asymptotes are y = x and y — x + 1. 148. Asymptotes parallel to either axis. In § 146, let the axes be revolved until the ^/-axis is parallel to an asymp- tote ; then one value of m will assume the form op ; hence, A„ = will be below the nth degree in m, and therefore f{x, y) = will be below the wth degree in y. Conversely, if f(x, y) = is below the rath degree in y, there will be one or more asymptotes parallel to the y-axis. Hence, if f(x, y) = is below the wth degree in either x or y, there wUl be one or more asymptotes parallel to a co-ordi- nate axis. To find the equations of these asymptotes, equate to zero the coefficients of the highest powers of x aiul y. The following example will make clear this principle : Ex. Eind the asymptotes of the curs'e, y'^^ — 3yx^ — 5xy^ + 2x^ + 6y^ + X + 7j + l = 0. (1) Arranging (1) in descending powers of x, we have {y2 — 3y + 2)x^ — {5y^ — l)x + 6y^ + y + l = 0. (2) Equating the coefficient of x'^ to zero, we obtain y'' — ?jy + 2 — 0; that is,y=l, y — 2. Substituting 1 for y, (2) becomes — K + 2 = 0. Hence, (2) and y — 1 form a system which is defective in three solu- tions ; that is, (2) and y = 1 have only one common point. Substituting 2 for y, (2) becomes — 19 x -|- 27 = ; hence, (2) and y = 2 form a system which is defective in three solutions. Therefore, 2/ = 1 and y = 2 are two parallel asymptotes, which are parallel to the x-axIs. Arranging (1) in descending powers of y, we have {x^ — 5x + e)y^ — (Sx'^ - ■l)y + 2x^ + x + I = 0. (3) From (3) we see that x = 2 and x = S are two parallel asymptotes, which are parallel to the y-axis. 132 DIFFERENTIAL CALCULUS. EXAMPLES. Find the asymptotes to 1. The cissoid of Diodes (2 a — x) j/" = ^s. See example 3 and flg. 3, § 155. 2. The strophoid x (x^ + y^) + a (x^ - y^) = 0. See example 7 and fig. 7, § 155. 3. xy^ + x^ = aK x — 0,y = 0. 4. 2/2(x2-a2) = x. 6. xh/^ = c^(x^ + y''). 5. xij — cy — bx = 0. 7. (x — 6)2 (j/ — c) = a'. S. {x — a)y'^ = x'+ ctxK x = a, y = ±x±a. 9. x'> + 2x^ + xy'^ — x^ — xy— —2. x = 0,y=—x,y=—x + l. 10. 2/8 _ -52,2 _ xSj, + x8 + x2 — 2/2 = 1. ^ = ± X, 2/ = X + 1. 11. x22/2 — x^!/ — X2/2 + 2 X + 3 ^ + 1 = 0. 2/ = 0, 2/ = 1, x = 0, x = l. 12. 2/8 — 5x2/2 + 8x22/ — 4x3 - 3 2/2 + 9x2/ — 6x2 + 2^ — 2x = 1. y — X, y = 2x + 1, y = 2x + 2. 149. Asymptotes to polar curves. When OP, or p, revolves about 0, it will become ap when, and only when, it is parallel to an asymptote, as SN. Hence, any value of 6 which makes p = ap gives the direction of an asymptote, and the corresponding value of the subtangent 08 gives the distance and direction of this asymptote from the pole 0. EXAMPLES. Examine for asymptotes 1. The hyperbolic spiral pB = a. When 9 = 0, p — ap, and the suht. = — a. Therefore, the line parallel to the polar axis and at the distance a above it is an asymptote to the spiral. EQUATIONS OF ASYMPTOTES. 133 2. The curve p cos 9 = a cos 2 9. When e = 7t/2, p = ap, and subt. = — a. Hence, the line perpendicular to the polar axis at the distance a to the left of the pole is an asymptote. 3. The curve p^ cos « = a^ gin 3 g^ 4. The curve p = a sec 2 9. There are four asj'mptotes forming the sides of a square, each being at the distance a/2 from the pole. 5. The polar axis is an asymptote to the Utuus pVe = a. Singular Points. 150. The singular points of a curve are those points which have some peculiar property independent of the position of the co-ordinate axes. Points of inflexion and multiple points are varieties of singular points. Points of inflexion have already been considered in § 109. 151. A multiple point is one through which two or more branches of a curve pass, or at which they meet. A multiple point is double when there are only two branches, triple when only three, and so on. A multiple point of intersection is a multiple point at which the branches cross each other (fig. a). An osculating point is a multiple point through which the branches pass and at which they are tangent to each other (figs, b and c). 134 DIPFERENTIAL CALCULUS. A cusp is a multiple point at -which the branches terminate and are tangent to each other (figs, d and e). A cusp or osculating point is of the first or the second species according as the two branches are on opposite sides (figs, b and d) or the same side (figs, c and e) of their common tangent. 152. To find the Tnultiple points of a curve. At a multiple point each branch has its own tangent ; hence, at such a point dy / dx has two or more values. Let f(x, 2/) = be the algebraic equation of the curve free from radicals and fractions ; then, by § 136, we have dy Qu/dx j>/ \ Since u contains neither radicals nor fractions, this expres- sion for dy j dx can have only one value at any given point unless it assumes the indeterminate form 0/0; hence, at a multiple point we have ^ujdx = 0, ^ujdy = 0. (1) Any solution of system (1) which satisfies the equation of the curve gives a point on the curve at which dy/dx = 0/0. (2) The form in (2) can be evaluated by the method of § 88. If, in (2), dy /dx has two or more unequal real values, the point is, in general, a tnultiple point of intersection. If, in (2), dy /dx has two equal real values, the point is, in general, an osculating point or a cusp. Ex. 1. Examine the curve x* + ax,h/ — ay^ — for multiple points. Here m = x* + ax'h/ — ay^ = ; (1) .-. Qu/dx = 4 x^ + 2 axy, 'du/dy — ox^ — 3 ay'': (2) Equating these partial derivatives to zero, we have X (2 x2 + ay) = 0, x^ — 3 2/" = o. (3) MULTIPLE POINTS. 135 The only solution of system (3) whioh will satisfy (1) is a; = 0, y = ; hence, th« only point to be examined is (0, 0). From (2) and § 136, we have dy 4:x^ + 2axy , (x = 0, 'T- = -T, — 5 J = n when i dx 3ay^ — ax^ 1 y = 0. Evaluating this fraction by the method of § 88, we find dy/dx'jojo = 0, +1, — 1. Hence, the origin is a triple point of intersection at which the inclina- tions of the branches are, respectively, 0, 71/4, Zit/i. The general form of the curve at the origin is shown in § 151, fig. a. Ex. 2. Examine the curve a*^2 = f^%jA _ j;6 for multiple points. Here u = a%2 _ ^H^ + ^jo = Q; (1) . . ait/da; = — 4 aH^ + 6 x^, du/dy = 2 a^y. (2) The only solution of the system, — 4 aH^ + 6 x5 = 0, 2a*y = 0, (3) which will satisfy (1) is a; = 0, y = 0. dy 4 aH^ — 6 x^ , [x = 0, Tx= 2a^y =0^'^"'^i2/ = 0. Evaluating this fraction, we have dy/dx\f„(,= ±0. An inspection of its equation shows that the curve passes through the origin and is symmetrical with respect to the ic-axis ; hence, the origin is an osculating point of the first species (§ 155, fig. 5). Ex. 3. Examine y^ = x (x + aY for multiple points. (x — —a, dy 3x^ + iax+ a" , Here v^ = ; = - when dx 2y \-y - Evaluating this fraction, we obtain dy/dx]-a,o = ± V^. When a is negative, (— a, 0) is a double point of intersection. When a is positive, the slopes of both the branches which pass through the point (— a, 0) are imaginary ; hence, these branches do not lie in the plane of the axes, but are a part of the imaginary locus. When a is positive, the multiple point (— a, 0) is isolated from the rest of the plane locus, and is called a conjugate point. 136 DOTERENTIAL CALCULUS. 153. A conjugate point is a multiple point which is formed by imaginary branches meeting or crossing each other in the plane of the axes. Such a point is, in general, entirely isolated from the plane locus; but in exceptional cases it may lie on it. At a conjugate point dy /dx is, in general, imaginary; but in excep- tional cases it may be real, since the tangents to the imaginary branches at such a point may lie in the plane of the axes. A shooting point is a multiple point at which two or more branches end but are not tangent to each other. A stop point is a point at which a single branch of a curve ends. As a shooting point or a stop point never occurs on an algebraic curve, they will not be further considered. EXAMPLES. Show that the curve 1. ay'^ = x^ has a cusp of the first pppcies at (0, 0). § 155, fig. 2. 2. 2/' = 2 oi^ — x^ has a cusp of the first species at (0, 0). § 155, fig. 4. 3. x^ + y' — S axy has a double point of intersection at (0, 0). § 155, fig. 6. 4. 2/2 — 5;2 (g2 — x^) has a double point of intersection at (0, 0). The form of this curve is similar to that of the curve in § 155, fig. 13. 5. X {x^ + y^) — a (j/^ — x'^) has a double point of intersection at (0, 0). § 155, fig. 7. 6. y" (a^ — x^) = X* has a point of osculation of the first species at (0, 0). 7. 2/2 = 2 x^y + x% — 2 X* has a conjugate point at (0, 0). The slope of each branch at (0, 0) is real ; but (0, 0) is an isolated point ; hence, it must be a conjugate point at which the tangents to the imaginary branches lie in the plane of the axes. 8. ay^ = (x — o)2 (x — 6) has at (o, 0) a conjugate point when a < 6, a double point of intersection when a > 6, and a cusp when a = b. 9. w'y^ — 2 abx^ = x^ has at (0, 0) a point of osculation and a point of inflexion on one branch. CURVE TRACING. 137 Curve Tracing. 164. Symmetry. The following principles of symmetry of loci are easily proved : A locus is symmetrical with respect to the a;-axis when its ecLuation contains only even powers of y. A locus is symmetrical with respect to the 2/-axis when its equation contains only even powers of x. A locus is symmetrical with respect to the origin when the terms in its equation are all of an odd or all of an even degree in X and y. 155. To trace the locus of an equation we first find its Axis or centre of symmetry, if any ; Intercepts on the axes, and its limits ; Maxima and minima ordinates, if any ; Asymptotes and singular points, if any. It is useful to remember also that an infinite branch is convex toward its asymptote. EXAMPLES. 1. Trace the cubical parabola v?y — xK Y Fio. 1. The origin is a centre of symmetry and a point of inflexion, to the right of which the curve is concave upward. The infinite branches are in the first and the third quadrant. When x = ± co, = */2. For the form of the curve, see fig. 1. 138 DIFFERENTIAL CALCULUS. Fig. 2. 2. Trace the semi-oubioal parabola a?-'^y = x^'^, or ay'^ = x^. The curve is symmetrical with respect to the K-axis, and lies to the right of the y-axis. The origin is a cusp of the first species. When a; = 00 , (^ = 7r/2. For the form of the curve, see flg. 2. The curve ai—iy = x" (1) is frequently called the parabola of the nth degree, n being greater than unity. When n — 2, (1) becomes ay = x% (2) the locus of which is the common parabola with its axis on the y-axis. When n is an even integer or a fraction having an even numerator and an odd denominator, the general form of curve (1) is that of the common parabola (2). When n is an odd integer or a fraction vrith an odd numerator and an odd denominator, the general form of curve (1) is that in flg. 1. When n is a fraction having an odd numer- ator and an even denominator, the general form of (1) is that in fig. 2. 3. Trace the cissoid of Diodes, 2/2 =x^/ (2 a — x). The X-axis is an axis of symmetry. The curve passes through (a, a) and [a, —a). It lies between x = and the asymptote x — 2a. The origin is a cusp of the first species. (See fig. 3.) To construct the cissoid geometrically, draw any line OR from to ^E, and take EP= OS; then P will be a point on the cissoid. Fig. 3. CURVE TRACING. 139 Fig. 4. 4. Trace the curve 2/' = 2 aa;2 _ ^s, y=—x + 2 a/3 is an asymptote. (2 a, 0) is a point of inflexion, to tlie right of which the curve is concave upvrard. Hence, the infinite branch in the fourth quadrant lies above the asymp- tote, and the one in the second below it. When a; = 4 a/3, ^ = 2aV4/3, a maximum ordinate. The origin is a cusp of the first species, the 2/-axis being tangent to both branches. 5. Trace the curve a'^y'^ = a^x* — x^. Each axis is an axis of sym- metry. The origin is an oscu- lating point of the first species. The curve is enclosed by the tangents x=±a,y = ±2aV3/9. When x-± aV6/3, y — 2 a v3/9, a max imum. When x = ± a V27 — 3 V33/6, (x, y) is a point of inflexion, fig. 6.) 6. Trace the folium of Descartes y^ — 3 axy + x^ — Q. y = — X — a is an asymptote. The infinite branches are concave upward, and hence lie above this asymptote. dy/dx = at (0, 0) and (aV^, aVi); dy/dx = ap at (0, 0) and (aV^, aV2). This indicates a double point of intersection at the origin and a loop tangent to the axes and the lines y — aVi and x = aVi. (See fig. 6.) (See fig. 4.) Fig. 5. (See 140 DIFFERENTIAL CALCULUS. 7. Trace the strophoid x (x'^ + y^) + a (x^ — y'^) = 0. (See fig. 7.) 8. Trace the hypocycloid of four cusps a;"" + j/^/a — (j2/s_ Each axis is an axis of symmetry. The limits of the locus are x = ± a, y=±a. dy/dx = — (2//x)i/8 = 0, at (- a, 0) or (a, 0), = ap, at (0, — a) or (0, a) ; hence, there is a cusp of the first species at each of these four points. d^y/dx^ = a2'V3x*'32/i/3; hence, the curve is concave upward in the first and second quadrants. (See fig. 8.) Y Fig. 7. In order to examine this curve for cusps by the method of § 152, it vfould be necessary first to rationalize its equation. This curve is traced by a point, P, in the circle PB (fig. 8) as it rolls within the fixed circle XBYX', whose radius is four times that of the circle PB. 9. Trace the evolute of the ellipse, (ax)2/8 + (6j/)2/8 = (a2e2)2/8^ where a and 6 denote the semi-axes, and e the excentricity of the ellipse. Each axis is an axis of sym- metry. The curve is enclosed by the lines x = ± ae^, y = ± a'^e^/b. dy/dx- — (a'^y/Vkoy^ — 0, at (— ae2, 0) or (ae2, 0); dy/dx — ofi, at (0, - a2eV6) or (0, a2e2/6). Fig. 9. .---r-' CURVE TRACING. 141 : Hence, there is a cusp of the first species at each of these four points. The curve is concave upward in the first and second quadrants. (See fig. 9.) 10. Trace the conchoid x'hp- = (6^ — y'^) (a + y)\ 11. Trace the witch of Agnesi (a;^ + 4 a^) ^ = 8 a^. 12. Trace the curve x^ + y^ = a^. 13. Trace the polar curve p — a cos 8 + 6. When 9 = 0, n/i, tc/2, cos-i(— 6/a), p = a + b, V2a/2 + 6, h, 0; when e = 37t/4, n, • ■ ', p=— V^o/2 + 6, — a + 6, ■ ■ •. These values of p will recur in reverse order as 8 increases from tc to 2 * ; hence, the locus is symmet- rical with respect to the polar axis. Locating these points for a = 3 and _l 6 = 2, we obtain the curve in fig. 10. When ix> 6, the point is a double point of intersection, as in the figure. When a = 6, O is a cusp. When a < &, there is a point of inflexion at B and at S, and is not on the curve. When a = 6, this curve is called Fie- l"- the cardioid. When a = 2 &, it is called the limaqon. 14. Trace the curve p = a sin 3 9. p = o, a maximum, when sin 3 9 = 1; that is, when 8 = ic/d, 5 7C/6, ■ • ■, p = — o, a minimum, when sin39 = — 1; that is, when 8= 7t/2, 77t/6, • ■ : When 8 = 0, «/6, j(/3, Tt/2, 2 7t/3, 5 7r/6, *, P = 0, a, 0, — a, 0, a, 0, The curve consists of three equal loops. (See fig. 11.) 142 DIFFERENTIAL CALCULUS. 15. Trace the curve p = a sin 2 0. The curve consists of four equal loops. (See fig. 12.) The locus oi p = a sin 9 is a circle, a curve of one loop. From the number of loops when m = 1, 2, 3, we infer that the locus of p = a sin ne consists of n loops when n is odd, and 2 n loops when n is even. Fig. 11. Fig. 12. Fig. 13. 16. Trace the lemniscate p^ = a^ cos 2 S. The lemniscate is the locus of the intersection of a tangent to the equi- lateral hyperbola and a perpendicu- lar to the tangent from the origin. (See fig. 13.) 17. Trace the curve p = a sin' (6/3). (See flg. 14.) Fig. 14. 18. Trace the curve p = a cos 8 cos 2 9. 19. Trace the logarithmic spiral p = e^. PART II. INTEGRAL CALCULUS. CHAPTEE I. STANDARD FOKMS. DIRECT INTEGBATIOW. 156. Integration. Having given the differential of a func- tion, integration is the operation of finding the function. In other words, having given the ratio of the rate of a function to that of its variable, integration is the operation of finding the function. A function is called an integral of its differential. Thus, fx is an integral of f'{x) dx. 157. The general or indefinite integral of any differential ^ (a;) dx is tlie most general function vi^hose differential is (j> (x) dx. The sign of indefinite integration is I . Thus, | in the expression | <^ (x) dx denotes the operation of finding the indefinite integral of c^ (x) dx, vsrhile the whole expression denotes the indefinite integral itself. For example, if C is a general constant, a^ + C is the most general function whose differential is 3 x'^dx ; that is, /' x^, x' — 1, a;^ + 8, ■ • • are particular integrals of Sx^dx. The signs d and j indicate inverse operations, and, in gen- eral, neutralize each other. 144 INTEGRAL CALCULUS. Thus, d I (x) dx = §^ ios;;7X7 = ^i°gj 6^x2 2 6c ^bx + c 2 6c * 6x — c dx „ /* 2 adx C ^'^ — n /* 2adx J ax2 + 6x + c "" J (2 aa: + 6)2 + 4 ac - 62 2 ^ , 2ax+6 ... : tan- 1 = 1 (1) VToc — 62 V4 ac — i 1 , 2 ox + 6 — V62 — ^ac V62 — 4 ac 2 ax + 6 + V62 - 4 c (2) We use the form in (1) or (2) according as 4 ac > or < 6^ ; that is, in any given case we take the form which is real. /^ da _ r Sdx _ 1 _j .3x + 2 J 3x2 + 4x + 7~J (3x + 2)2 + 17~ Vn Vl7 r 2X + 7. ^ f (2^ + 4)dx /• dx J x2 + 4x4-5 J x2 + 4x+5 J (x + 2)2 + 1 = log(x2 + 4x + 5) + 3 tan-i(x + 2). DIRECT INTEGRATION. 155 r {3x + 2)dx _ 3 n (4x + 4)dx _ r dx ■ J 2x-^ + 4x + 5~ 4J 2i2 + 4x + 5 J 2x= + 4x + 5 ^' = flog(2x2 + 4x4-5)- ^tan-i^^^- In like manner any differential of the form ' „ , . , — can be ... J ajy' + OX + C integrated. Note that the numerator of the first of the two fractions in the second member of (1) is the differential of its denoininator. ^^ f x^ + tx + 5 '^ " I '°^<''' + 4x + 5) - tan-i{x + 2). ,„ /» dx 1 /• bdx 1 . ,bx 17. I =: =T I , :^^= = -Sm-' J Va^c^ — 62a;2 6J v'(ac)2 - (6x)2 6 ^ 18. r , '^^ - = i log (bx + V62j2 ± a2c2). J V62x2 ± a2c2 & /• dx _ /• & . _, 2x + 1 ■ J Vl-x-¥2~J V5/4-(x + 1/2)2 ^"^ Vg 20. r "^ =^log(xVa + Vgx^-6). J Vox^ — b va 21. r ^-^ = log(x + a +Vx^ + 2 ox). J Vx2 + 2ax 22 f ^^!^ ^ 1 r ^^'"'^^ = 1 sin-ix/^- J V8-4x3 U V(2V2)2- (2x3/2)2 a \2 23 r 'fa' . =J- f J Vax2 4 6x + c V^J V(2 2adx ax + 6)2 + 4 ac - 62 = -^ log (2 ax + 6 + 2VaVax2 + 6x + c). Va 156 INTEGRAL CALCULUS. _, n ^ _J_ n 2afc J V— 0x2 + bo; + c V^J V4 ac + 52 — (2 ax — Vf 1 . , 2 ax — 6 :sm' 1-1 Va vTocTft^ -_ /• da; J_ /* 4 da _ _1_ . _ ^ 4 a— 3 J V4 + 3 X - 2 x2 V2 J v'41-(4x — 3)2 V2 V41 26. J V2x2 27. /-^=J^= = 4=log(4x + 3 + 2V2V2x2 + 3i + 4). V2x2 + 3x + 4 V^ r , '^ • 29. f-, J V3x-x2-2 J V2x2 + 2x + 3 34. / (?x gjj r* xdx Vl + x2 + X ' J Va* — x* / dx _ /* cdx J_ _i2i5 xVc2x2 — a262 J cxV(ca;)2 — {ahf <^ «* / dg gg r 5dx xV62a;2 _ a2 ' J xV3x2 — 5 / dx _ r V7dx _ 1 _,xV7 V7 x* — 5 x2 J X V7 ■ V7 x2 — 5 V5 V5 „, /• — dx 1 /* — cdx 1 , ex 35. I — = - I , = - covers- ' — ■ J V8 ex - c2x2 cj V8 • ex - {cx)2 c 4 36. f ^ ^^ . 37. C^M^- J V2abx — b'h:^ J V ox — x2 3g i- (2x + 3)dx ^ /» (2x + l)dx ^ „ /» dx ' J Vx2 + X + 1 J Vx2 + X + 1 J Vx2 + X + 1 = 2 Vx2 + X + 1 + 2 log (2 X + 1 + 2 Vx2+x+l). can In like manner any differential of the form ^ ' be integrated. Vax2 + 2 6x + c Note that the numerator of the first of the two fractions in the second member of (1) is the differential of the base in the denomi- nator. DIRECT INTEGRATION. 157 r —xdx /"c — 2x — c, 39. I = I „ , -dx J 'vcx — x'^ J ^ycx — x"^ I ; c , %x = V ex — x^ — - vers— ' — 2 c • (x^ — ay^^dx _ /• (x^-'a^)(?x 40. ^ J X Vx2 - a2 / xdx _ r a?dx Vx2 — a'^ J xVx2_ aa = Vx^ — a^ — a sec— 1 - • a -/ Vl+; dx = sin— ix — Vl — x^. 42, Vl-x !. r ^ f ^" dx = sec- 1 - + log (X + Vx2 - a2). •^ X vx — a •* CHAPTER II. DEFINITE INTEGHAES, APPLICATIONS. 164. Definite and corrected integrals. Let b> a; then the increment produced in the indefinite integral fx+Chj the increase of x from a to S is fb+C-(fa+C)=fb-fa. This increment of the indefinite integral of ^(x)dx is called "the definite integral of <\i(x)dx between the limits a and b," and is denoted by j ^{t)dx. <^ (x) dx=fb— fa. (1) in the expression } (x)dx from x = a to x = b. b is called the 'upper' or ' superior ' limit, and a the ' lower ' or ' inferior ' limit. The limits a and b must be so chosen that x will be finite, continuous, and of the same quality, from x = a to x = b. The expression /* —/a is often denoted hj fx . _ia If, in (1), we make the upper limit variable and put x in the place of b, we obtain (f>(x) dx = fx - fa, OT fx \ , (2) which is called "the corrected integral of (x)dx is meant the limit of I <^(x)dx when a; = oo. DEEINITE INTEGRALS. 159 EXAMPLES. ^T"_ X* 1 2. rnxdx = ~ a2j ^= I (62 - a^). 3- J^ (^-^-^^) = log(2x-x2)J^ = log(2x-x2). T" Xdx 1 , „T' TT 5. £6 x^dx = 24. 10. J'x^dx = ^!±ilZ*LL\ » + 1 6. rv-xvx=5- 11. r^^^g^ 8. I , =-• 13. I er-'^dx = -- Jo Va2 - x2 2 Jo a . /"'/■' sin ed9 r: - ,. /""^^ . „ » 1 15- I , =: = sin-1 — 1=- = sin-i — ;= sin-i — ;= - Jo V 5 — 2 X — x2 V6 J Ve Ve Jj 3 + 2X + X-' v^V V2 / J'^xa-e — 1 4. aa:— 1 1 „ x' + e^ cix = -log(x- + e^). 160 INTEGRAL CALCULUS. 165. Corresponding definite or corrected- integrals of equal differentials are equal. For the corresponding increments of variables which, change at equal rates are equal. ^^^^dx = 31og3+3- E^- r "J— 3^2 dx = 3 log 3 + Let z = X — 3 ; then x = z + 3, dx = dz, and 3x — 1 ^ 32 + 8_, , ax = ;; — dz. (X - 3)2 z2 When X = 4, z = 1, and when x = 6, z = 3 ; 3x-l , /•33Z + 8 J^'S 3x — 1 ^ r^ dz 6 165 Z2 = [31ogz-8/z1 = 3 log 3 + 16/3. This example and those which follow illustrate also how the introduction of a new variable often simplifies a given differential and renders it directly integrable. EXAMPLES. 1. I . _ „■ 3 dx = I 1 dz = log3 + 4, wherez = x — 3. „ C'^ x^dx /•«(z — l)dz 3,3/- 2- J^ (x2 + 1)^/3 -j^ A-^^ = g(V^ + 3),wherez = x^ + L /•" dx _ _ 1 r'__d^_ Ji X Va2 - x2 aj„ Vz2-1 = log [a + Va2 — 1) /a, where z = a/x. /•I e^^dx _ p + i (z — l)dz ■ J„ (e- + l)i/*~J, zi/* = zt[(3 e - 4)(e + l)3/i + Vs], where z = e^ + 1. _ /•« dx 1, a + Va2 + 1 5. I — , = - log jz Ji X Va2 + x2 « 1 + V2 In example 5 put x — a/z, in example 6 put S + a = z. 6. I 7T-; — - = (0 + Oleosa — Sin a iog(oos6/cosa). J_2„ oos(6> + a) ^ &\ / / AREAS OF CURVES. 161 166. Geometric meaning of j ^(x)dx, j ^(x)dx, j ^(x)dx. Let SPQ be the locus oi y = x. Conceive the variable ordinate y or NQ to trace the area between the x-axis and y = x, some undetermined fixed ordinate, as BS or B'S', and the moving ordinate NQ. Let dx = NN'; then dA = NQQ'N'; .• . dA = ydx = 4>(x) dx. (1) . j ydx = j <^(x) dx = A. (2) Let OM = a, and OM' = b ; then the increment produced in A by the increase of x from a to 6 is MPP'M'; J ydx = j (x)dx (3) r area bounded by the x-axis, \y= x, X = a, and the ordinate y } (4) In (2), A is indeterminate so long as the fixed ordinate RS, or B'S', is indeterminate. 162 INTEGRAL CALCULUS. From (1), by Cor. 1 of § 12, we know that the areas in (3) and (4) will be positive or negative according as x is posi- tive or negative from x = a to x = b. Hence, if a curve crosses the a^axis the area above, and the' area below, this axis must be obtained separately. For example, to find the area bounded hy y = z, the x-axis, and M'P\ we find the areas B"S'K and KPP'M' separately and take their arith- metical sum. From the geometrical meaning of an integral it follows that ff>(x)dx has an integral whenever (jjx is continvious. Ex. Give the geometric meaning of the definite integral in each of the examples on page 169. EXAMPLES. 1. Find the area hounded by the x-axis and y = x^ + ax^. The curve cuts the x-axis at {— a, 0) and (0, 0) ; hence, the limits are — a and 0. Here dA = ydx = (x^ + ax^) dx ; .-. area= C (x^ + ax^)dx = Vxi/i + ax^/s] by (3) = aV12. In each problem the reader should first gain a clear idea of the boundary of the figure whose area is required. 2. Find the area bounded by the x-axis, the parabola x^ 4- 4 j/ = 0, and the line X = 4. Ans. 16/3. Since the area lies below the x-axis, the formula gives a negative expression for it. 3. Find the area bounded by the x-axis, the curve y = x', and the lines X = — 2 and x = 2. ^^^ g Here we find the area below, and the area above, the x-axis sepa- rately, and take their arithmetical sum. 4. Find the area bounded by the curve 2/ = e^, the x-axis, the 2/-axis, and the line x = h. ^^g e'' — 1. AREAS OF CURVES. 163 5. Show that the area bounded by y = x — x' and y — is 1/2. 6. Find the area bounded by the parabola y^ = ipx and any chord perpendicular to its axis. Ans. 4 aw/3. Here the area required is twice the area bounded by the a;-axis, y = 2p^'^x^'^, and the ordinate y. . . area = 2 C%p^''^x^i'^dx, = (2/3)2x2/ ; by (4) ./o that is, the area is 2/3 the circumscribed rectangle. 7. Find the area bounded by the witch y(x^ + 4 a'^) = 8 a' and its asymptote y = 0. Area = 2 I „ , , „ = Saltan-' —- =i7ca^. J^ x2 + 4a2 2oJo Here the area between the curve and its asymptote is iinite. 8. Find the area bounded by the hyperbola x^ = 1, its asymptote y = 0, and the lines x = 1 and x = a. Ans. log a. Wben a = 00, log a = co ; hence, the area between the hyperbola and an asymptote is infinite. 9. Find the area bounded by the curve y = cos x, the x-axis, the y-asis, and x = it. ^,i5. 2. 10. Show that the area bounded by the curve y = tan x, the x-axis, and the asymptote x = jr/2 is infinite. 167. Formulas for accelerated motion. Let t denote a portion of time, s the distance traversed by a moving body, V the velocity, and a the acceleration ; then we have the fol- lowing formulas : (i) V = ds/dt; .'. s= I vdt, t= I ds/v. (ii) a = dv/dt; .'.v= j adt, t= i dv/a. 164 INTEGRAL CALCULUS. EXAMPLES. 1. To find the fundamental formulas for uniformly accelerated motion. Let Do and So denote, respectively, the initial velocity and distance ; that is, the values of v and s when t = ; and let ce" denote the con- stant aocelsration. We then have « = Do ■ ■ So + Jo J VI dt- ■ a't + «o, : a't^/2 + Dot + 1 If D = and s — when i = 0, (1) and (2) become D = a't, s = a:'t''/2; .-. t = V2s/a', D = V2 a's. (1) (2) (3) (4) The acceleration produced by gravity at the earth's surface is about 32.17 ft. a second, and is usually represented by g. Substitut- ing g for a' in equalities (3) and (4), we obtain the four formulas for the free fall of bodies in a vacuum near the earth's surface. 2. A rifle ball is projected from in the direction OF with a velocity of c feet a second. Find its path, knowing that its velocity acquired in t seconds along the action hne of gravity OX is gt feet a second. Let OX and OY be the co-ordinate axes ; dy/dt — c, dx/dt = gt; then -S: cdi = ct, x = gty2. (1) X Eliminating t between equations (1), we have 2/2 = 2c'kc/g. Hence, the path of the ball is an arc of a parabola. 3. A body starts from 0, and in f seconds its velocity in the direction of OX is 2 act, and in the direction of OY it is aH^ — c^ ; find its velocity along its path Onm, the distances in the direction of each axis and along the line of its path, and the equation of its path, the axes being rect- angular. ACCELERATED MOTION. 165 Let OX and OT be the axes, and let s denote the length of the path Onm ; then dx , . 11 = ''^'' dt ' ds ..x= i •2actdt = acP; (1) J-i (aH'^ — c^) dt = aH^ /Z — cH; (2) .-. s= r'{a2i2 + c2)di = a2iV3 + c2i. (4) Eliminating t between (1) and (2), we obtain _/ax „\ Is lac (5) which is the equation of the path Onm. 4. Given v =ft, to represent the time, the velocity, the distance, and the acceleration geometrically. Construct the locus of u = ft, t being represented by abscissas and V by ordinates ; then the distance will be represented by the area between v =ft and the x-axis, and the acceleration by the line rep- resenting dv when dt is represented by a unit length. 5. A body is projected upwards with a velocity of 80 feet per second ; find in what time it will return to the place of starting. Ans. 5 seconds, nearly. 6. Erom a balloon which is ascending at a uniform velocity of 20 feet per second two balls are dropped, one of them 3 seconds before the other ; find how far apart they vsdll be 5 seconds after the first one was dropped. Ans. 398 feet, nearly. 166 INTEGRAL CALCULUS. 168. Change of limits. The following formulas are readily proved from the definition of a definite integral : (x)dx = — I (x)dx; (i) for each member =fb —fa, if dfx = (x)dx. Xb /*c /»6 (x)dx= I tl}(x)dx+ I (l>(x)dx; (ii) for the second member =fc —fa +fb — fc =fh —fa. <^{x)dx = I <^{a — x)dx; (iii) for the second member = — j (^(a — x) d(a — x) Jo = -/(«-a^)]"=>-/0. Note. The Integral Calculus was invented to olDtain the areas of curvilinear figures, or for quadrature, as it is often called. CHAPTER III. INTEGRATION OF RATIONAL FRACTIONS. 169. Decomposition of fractions. When the numerator of a rational fraction is not of a lower degree than its denomi- nator, the fraction, said to be improper, should be reduced to a mixed expression before integration. For example, .,02 n dx = (x-~ 2) dx + , , „ „ ^ dx. x^ + 2x^ — X — 2 ^ ' x^ + 2x^ — X — 2 The numerator of this new rational fraction is of a lower degree than its denominator. Such a fraction, called & proper fraction, if not directly integrable, can be decomposed into partial real fractions which are integrable. These partial fractions will differ in form, according as the simplest real factors of the denominator of the given fraction are : I. Linear and unequal. II. Linear and equal. III. Quadratic and unequal. IV. Quadratic and equal. To present the subject in the simplest manner we solve below particular examples in each of the four cases. 170. Case I. To each of the unequal linear factors of the denominator as a; — a, there will correspond a partial fraction of the form A/(x — a). The denominator x^ + x^ — 2 x s x (x — 1) (x + 2). 168 INTEGRAL CALCULUS. Hence, by addition of fractions we know that the given fraction is the sum of three fractions wliose denominators are x, x — 1, and x + 2, respectively, and whose numerators do not involve x. We therefore assume 2x + 3 -A+ ^ A. C* x(x — l)(x + 2) xx-lx + 2' ^ ' in which A, B, and C are unknown constants. Clearing (1) of fractious, we obtain 2x + 3 = ^(x-l)(x + 2) + B(x + 2)x+ C(x — l)x (2) = (A + B+ C)x^ + {A+2B—C)x — 2A. (3) Equating the coefficients of like powers of x in (3), we have A + B+ C = 0,A + 2B — C = 2, -2A = 3. (4) Solving system (4), we obtain A ^-3/2, B = 5/3, C = -l/6. Substituting these values in (1), we obtain the identity, 2x + 3 _ 3 5 1 x3 + x2 — 2 X ~ 2 X 3 (X - 1) 6 (X + 2) ' / (2x + S)dx _ _ 3 /'dx 5 /^ dx _ 1 /• dx x3 + x2-2x~ 2J X 3jx-l ej x + 2 = - ?logx + |log(x - 1) - pog(x + 2). The values of A, B, and C may be obtained directly from (2) as follows : Making x = 0, (2) becomes 3 = — 2^; .. ^ = — 3/2, Making x=:l, (2) becomes 5 = 3B; .-. jB = 5/3. Making x = - 2, (2) becomes - 1 = 6 C ; .-. C = — 1 /6. EXAMPLES. , rx^--i, ,3, X 1- J ;;iTr7 («a; = x + j log x2 — 4 4 ^x + 2 x2 + X — 1 /rt-a _1_ ™ 1 ^8+\,-2_6^ 'is' = log [^' " (^ - 2)^ '^ (X + 3)1 /3]. 3- f.^4^^\s^^- 4. J- x2 + X — 2 /> (x^ + 1) cfa; ^ 1 . ^. (x + 1)6 (x - 1)2 ■ J (2x+ 1) (x-^-1) °° (2x + l)6 RATIONAL FRACTIONS. IgQ r xP + xi-8 3:3 3:2 a" (a: -2)5 „ r xdx 3 + v^, ^ 3 — V2 li /, n iBv, 3, X — 3— V2 - -l0S(x^ — 6x + 7) H ;=log- 8. 2 2V2 X — 3+V2 X (2x „ /* x^dx ;x + l)(x2 + x-6)' / xdx /^ (x-a)(x-6)' ^- J ^ 171. Case II. To r equal linear factors of the denominator as (x — by, there will correspond a series of r partial fractions of the form -— + ,_ + ■ • ■ + (x — by (x - by-^ x — b x2 — 7x + 6^ o, -._,.. .^ , X — 2 (X - 1)2 ■ Ex. 1. pjL_J^^ = 3 iog(^ _ 1) + Expressing the numerator in powers of (x — 1), we obtain 3x2 — 7x + 6-3(x-l)2 — (x-l) + 2; (j) for 3(x - 1)2 = 3x2 — 6x + 3 and — X + 3 = — (x — 1) + 2. Dividing both members of (1) by (x — 1)', we obtain 3x2-7x + 6_ 3 1 2 + /o. _ 1 \3 ' (2) (X — 1)3 -X — 1 (X— 1)2' (x-1) /3x2 — 7 x + 6 , „ /• c«x r dx , „ /• dx (x - 1)3 = 3 log (x - 1) + (X — l)-i — (X - l)-2- From (2) we see that to resolve the given fraction into partial fractions by undetermined ooeflficients, we should assume 3x2-7x + 6_ A B C (X - 1)3 (X - 1)3 (X - 1)2 X-1 and proceed as in § 170. 170 INTEGRAL CALCULUS. Here both Case I and Case II are involved ; hence, we assume (1) ^ - ^ +^ + ^- (2) (x — 1)2 (z + 1) (X - 1)2 X — 1 X + 1 Clearing (2) of fractions, equating the coefficients of the like powers of X, and solving the resulting system, we ohtain ^ = 1/2, £ = —1/4, C = 1/4. Substituting these values in (2) and integrating, we obtain (1). EXAMPLES. l-/jTi^'i- = 3 1og(x + 2) + ^- „ /•2x2-3x + 4, „, , „, 18X-41 2- J ,-.._a^a ''^ = 2 1og(x-3)- ^^^__^^, (X-3)8 — nV- / 2(X-3)2 4 x2 — 6 X + 7 , 1 , ,„ , o^ , 9 25 „ /»4 x2 — 6 X + 7 , 1 , ,„ , „, , » ^o 3- J (2x + 3)« ''^ = 2l°S(2^ + ^) + 2(2x + 3)-4(2x + 3)2 4 r (2x-5)(;x _ 7 ^11 2(x+i: Ji (X + 3) (X + 1)2 2 (X + 1) 4 "^ X + 3 , 2(^+1) 7 J"^ x^ „ X3 + 5X-2 1f87 + -4 = .-f! + "'-l' + «-^- »^ / '"•;y;rH-?,.'" "'-^['"^"(;T-.)'i-;' /» x'dx /♦ (X + 1) dx ^' J (X- 1)3(3; + 1)' J (X-l)2(X+2)2 172. Case III. To each of the unequal quadratic factors of the denominator as x^ + ^x + y, there will correspond a partial fraction of the form -5— —— • x'' + ^a; + 2' „ , /» (x2 + l)dx 1 ^ , X , 1 ^ , r^ ,^s Ex.1. ( TT-r-STTrrV^TT: - — 7=tan-i-7:H ptan-ixV2. (1) J (x2 + 2)(2x2 + l) 3^^ ^ 3V2 By addition of fractions we know that the given fraction is the sum of two real fractions whose denominators are x2 + 2 and 2 x2 + 1 respec- tively, and whose numerators cannot be above the first degree in x. RATIONAL FRACTIONS. 171 Hence, we assume a:^ + l _ Ax + B Cx + D (s2 + 2)(2x2 + l) a;2 + 2 "*" 2x2 + 1' ^^> Clearing (2) of fractions, we obtain x^ + l = (2A + C)x^+(2B + D)x^ + {A + 2C)x + B + 2D. .■.2A + C^0,2B + D = 1,A+2C^0,B + 2I) = 1. .: A =^C- 0, B-D = 1/S. Substituting these values in (2) and integrating, we obtain (1). T, „ /^ xdx 1, (x2+lU/2 1 ^"- '■ J (x + l)(x2 + l) = ^"'^^i+T- +2*^-^- (1) Here both Case I and Case III are involved ; hence, we assume (X + 1) (X2 + 1) X + 1 ^ X2 + 1 ^"'^ Proceeding as above, we find ^ = — 1/2, B = C = 1/2. Substituting in (2) and integrating, we obtain (1). EXAMPLES, x^cix 1, X — 1 , V2 , X /• x^dx /• x^dx J x* + 3x2 + 2" J l-x*' x* + x2-2 6^x + l 3 V2 x'dx x2(Zx 1 , 1, x2 — 2x+ 1 ■ + Tlog- (x - 1)2 (x2 + 1) 2 (X - 1) 4 ^ x2 + 1 dx 1, (x + l)2 , 1 ^ ,2x-l ^B- + T = 6l°ga;2-x + l + Vl '^"^ ~Vr ■ / dx _ 1 (^_dx_ _\ C (x — 2) cix X8 + I~3jx + 1 Sj X2-X + 1' / (x — 2) dx _ 1 /' (2x — l)dx 1 /»_ x^ — x + l~2J x2 — x + 1 2jx' .3(?x x + 1' dx 1, (X + l)2 , 1 ^ ,2x-l /dx 1 , x2 + x + 1 , 1 , ,2x + 1 172 INTEGRAL CALCULUS. J a;* + 8x2 -9' ^' J ^ r i :^^^. 8. r,-,^±^... J i^x^ — \)dx y. / (x2 + 3)(x2-2a; + 5) , x2-2x + 5,5, ,a; — 1 2, x 10. ' *" /? + x)2(l +2x + 4x2) ^1, (1 + x)" L_4._2_t -i 4x+l 3°^l + 2x + 4x2 3(l + x) 3V3 V3 "/^ x2 + 3a; + l 4 ,2x-l 2 ,2x + l - — dx = -^ tan-i — p p tan-i 1=- ■ + x2 + l V3 V3 V3 V3 ^^- /(x^ + a^Hx-^ ■ "> = ^"^^^^-^ *^"~' - + ^-^^'^ *^°"'' 1 _iX !_ 2 + 62) a (6-i - a2) ^" a 6 (a^ - 6^) ""■" ft 173. CAS'i; IV. To r equal quadratic factors of the denom- inator as (x^ + px + qy, there will correspond r partial frac- tions of the form A.T + B Cx + D Lx^ M {3? + px + 2')'' (x' +px + qy~'^ x^ -\-px + J _ /•2 x' + fc- -r .3 J. -r i , Ex. J ^^,-,-,,, dx '2x3 + x2 + .3x + 2 (X2 + 1)2 = log(x2 + l) + |tan-^x + ^^^- (1) Expressing the numerator in powers of (x2 -|- 1), we obtain 2 x3 -I- x2 + 3 X + 2 = (x2 -I- 1) (2 X -I- 1) -I- (X -I- 1). 2x3 + x2 + 2x-|-l x-f 1 /2x3 + x2-|-3x-F2 , /•2x-|-l^ , /- x + 1 , (2) : log(x^ + 1) + tan-x - ^^^^^ +/(S^. ■ (3) RATIONAL FRACTIONS. 173 By example 23 of § 183 we have ^+1)5 = 2(^ + 1) + 2*^^"'*- W From (3) and (4) we obtain (1). From (2) we see that to resolve the given fraction into partial frac- tions by undetermined coefficients we should assume 2x^ + x^ + 3x + 2 _ Ax + B Cx + D (X2 + 1)2 (X2 + 1)2 X2 + 1 Note. The solution of the following examples should be deferred until after reading Chapter V. EXAMPLES. r 2xdx 1, x2 + l ,lx — 1 ) (1 + xY 4 * (X + 1)2 2 x2 + 1 r dx _ 1 x2 1 1 J X (x2 + 1)3 2 ^^ 1 + x2 "^ 2 (1 + x2) "^ 4 (1 + x2)2 „ rx^ + x — l ^ 1 , ,„,„,, 2 — X -v^^ , X 3. J -(^^ + ^'^ = 2^og(x2 + 2) + j^^^- — tan-i-^. ^- J X2(x2 + l)2 ^^ = l°g X2 - 2x(x2+l) -2^"°"^^- dx da (x-l)2(; ;x2-+i)2 = ^°s (S)"2 - j^^r+I) " 4(^ + i *^'^"' ^• „ /• (4x2-8x)c?x 3x2 -X (x-l)2 , ^ CHAPTEE IV. INTEGRATION BY EATIONALIZATION, 174. Rationalization by substitution. Chapters I and III provide for tlie integration of any rational algebraic differen- tial whether it is entire or fractional in form. In some irrational differentials we can substitute a new variable so related to the old that the new differential will be rational and therefore integrable. 175. A differential containing no surd except a linear base, as a + bx, affected with fractional exponents, can be rational- ized by assuming a + bx = z°, where n is the lowest common denominator of the several fractional exponents. Por if a + te = «", x, dx, and the fractional powers oia -\-bx will each be rational in terms of z. Hence, the new function in z will be rational. J (a;-2)6'6 + (x-2)2/3 = 6 (x — 2)1/6 - 6 log [(x _ 2)1 '6 + 1]. Here the linear base is a; — 2, and n = 6 ; hence, we assume a; — 2 = 26. ..dx = Qz^dz, (X — 2)6/6 =z5^ (X— 2)2/8 = 24. dx /'' Gz^dz . n zdz dx r Gz^dz _ n (a;-2)6/6 + (x — 2)2/8~ J z6 + z4-°J ' z + 1 = 6[z-log(z + l)] = 6 (X - 2)1/6 -6 log [(X- 2)1/6 + Y^ RATIONALIZATION BY SUBSTITUTION. 175 Here the linear base is x, and n = 6; hence, we assume /a;l/2_a;2/3 ^ 2x1/0 '^^ = 3j (^^-^^)'i^ EXAMPLES. J 15 6'^ - f xdx 2 (2 a — 6x) / — — r- ^- X^l + x)^'-^ + (l+x)i/2 = 2 tan-i VfT^ - 1 ■ r^ xdx _ 2(2a + 6x) 2(2 + 6)Va • X (a + te)3/2 6-2y^qr^ 6Vl"T6 ' dx _ 1 , Va + 6x — Va 6. I — . = -p log J XV a + 6x Va + 6x Va Va + 6x + Va when a >• 0, 2 ^ , / a + 6x . , tan— 1 -» / 1 when a < 0. /^ \ -a ^ xi /3 (Jx 7. f— J Xl/2 + 2x2/3 = 3 2/3_la;l/2 + |xl/3_|xl/6 + Alog(i_)_2xl/6). 4 J o o lb XX da* 1/2+ 1^3 = 2x1/2 -3x1/8 + 6x1/6 -6 log (xl/6 + 1). /-)-l/6 4-1 fi 12 S77?+V4'^*=-^ + ^I7r2 + 2 1ogx-241og(xi/i2 + l). xdx (2x + 2)8/* + 4(2x+2)i/* = (2/5) (x+36) (2 x+2)i/^ - (4/3) (2x + 2)3'" - 28tan-i(^-Y^ V'*' 176 INTEGRAL CALCULUS. 176. A differential containing no surd except '\x^ + ax + b can be rationalized by assuming Vx^ + ax + b = z — x. For if "Vx^ + ax + b = s — x, ax + b = s' — 2 zx, z^-b , 2(z^ + az + b) , 2z + a (2z + ay /^— — g2 + as + h Vx'' + ax-\-b = z — x = — ^ ; I z + a Hence, x, dx, and Va;^ + ax + b are each, expressed ration- ally in terms of z. ^ f' dx _,__ Vx2 -l-x + 1 +x — 1 (1) J a: Va;2 + X + 1 ° Vxa + x + 1 + a; + 1 Assume Vx^ + x + \=z — x; then X + 1 = z2 _ 2 xz, (2) _ z^-l _ 2 (z2 + z + 1) dz ^""22 + 1' *^~" (2z + l)2 nr, — TT z2 + z + 1 Vx^ + x+l=z — X = — r r~-, — ' 2z -I- 1 / dx _ n dz _ z — 1 xVx-^ + X + 1 ~ J ^'^-i" °^^n- Substituting for z its value in (2), we obtain (1). 177. A differential containing no surd except V— x^ + ax +b can be rationalized by assuming V- x^ + ax + b = (^ - x) z, where j3 — x w one of the factors of — x^ + ax + b. Denoting the other factor oi — x'^ + ax + b hj x + y, assume -^-x' + ax + b = V(/3 -x)(x + y) = {p-x)z; (1) o^2 _ then x + y—{p — x)z^, x = ^ '- • z^ + 1 Hence, x and therefore dx andV — x'' + ax + b are expressed rationally in terms of z. Note. In § 176 the coefficient of x^ is + 1 ; in § 177 it Is — 1. RATIONALIZATION BY SUBSTITUTION. I77 EXAMPLES. — . = 2 tan-i (X + Vx^ + 2x - 1). / (ix _ V2 j^^ Vx2- x + 2 + X-V2 xVx^ — ~x"+^ 2 ° Vx^ — s + 2 + X + \^ /Vx^ + 2 X d. ~5 (ix = log (X + 1+ Vx2 + 2x) , ^ x + Vx2 + 2a 4 r tix _ '^ 1 v'^ + 2 X ~ V2 — X ■ J (2 + 3x)vi^r^2 8 °^VIT2^ + V2^:^' Assume V4 — x^ - V(2 — x) (2 + x) = (2 — x) z ; then 2 + X = (2 — X) 22, 2z2-2 _, 8z(iz X = „ , , 1 dx — z? + 1 (z2 + 1)2 V4 — x^ = (2 - X) z = 4z z^H- 1 z2-4 2 + .3x = ■ z'■ -r i dx _ 1 r <^2 "^1 v^z — 1 dx 1 /• dz (2 + .3x)V4 — x2~2j 2z^-l log- (2 + .3x)V4 — x2 2j2z^-l 8 V2z + 1 _ \^ V4 + 2X-V2-X 8 V4 + 2 X + V2 — X Z'" dx — "^ 1 "^2 + 2 X — V2 — X J2 xV2 + x — x2 2 V2 + 2X + V2-X „ r^Vax — x2 , /a — X „ /. a — X dx 2/2 — x\i/2 2a^ J"^ ^ 2/ 2 — x y/^ 2V „ (1 + x)V2 + x-x2 SVl + x/ 3 p ^ ^_ / x+Vx^ + x + 1 . 3+ V3 \ . Ji (1 +x)Vx2 + x+l °^\2 + x + Vx^ + x + l 1+VsJ r^ xdx _ 3 + x V3 (3 + 2x-x2)3/2 4V3 + 2X-: Assume V3 + 2 x — x^ = (3 — x) z. 178 INTEGRAL CALCULUS. 178. Irrational differentials of the form x" (a + bx")''" dx, where r and s.are integers and s is positive, can be rationalized by assuming, T ,1. 7 m + 1 . i. a + Dx° = z^, when is an integer or zero. TT ,1 7 m + 1 r . ii. a + Dx° = z^x", when 1 — is an integer or zero. n s " Assume, a + bx^ = s= ; then (a + bx''y = s^ (1) fz^-a\"' f^-aX'" z° — a\n (2) Multiplying (1), (2), and (3) together, we obtain (\ ™_±J_i ^-j^) " d^- (4) The second member of (4) is rational, and therefore in- tegrable, when (m + 1) /n is an integer or zero ; hence I. Assume a + bx"- = «'a;", or a;" = a (z' — b)"^ ; then (a + 6a;")'''» = (z'x^y/' = «''»'■/» («" — S)-''^", (I) a; = a'^"(a' — ^|)-'/", x" = a"/" (s» — 5)-™/», (2) S -1-1 dx= a'''",r'-'(2» — i) " (Zsr. (3) Multiplying (1), (2), and (3) together, we obtain r g m + l r ( m + l ,r \ a;™ (a + bx")^dx = a » » fs' — 6) v » » ^ z''+'-'^dz. The last member is rational, and therefore integrable, when h - is an integer or zero ; hence II. BINOMIAL DIFFERENTIALS. 179 EXAMPLES. Here = — - — = 2, and s = 2 ; hence, we assume n 3 ' a + 6x3 = z2; .. (a + 6a;8)-i'2 = z-i, (1) a;6 = ^^ "^ ; ... 6x^dx = ^ (z2 - a) zdz. (2) Multiplying (1) by (2) and dividing by 6, we obtain / x^dx _ 2 f,2_ ^^ -A. 3_ 2a. (a + 6a;y/2~3 62j ^^ ")''^-9 62« 3 62« = ^ (a + 6a;3)3 '^ _ |iL („ + ^3)1/2. 2. r-^=^= = ^log dx _ 1 . xVa^ + a;2 « "" vVTx^ + a „ m+l — 1+1 . , „. Here = = 0, and 8 = 2; hence, we assume a2 + x2 = z2 ; .■.xdx = z dz, x^ = z^ — aK / dx _ n dz _ 1 z — a a;V^TF2~J z2-a^~2a °^z + a 1 , Va^ + x2 — a = — log — ^ ■ 2a " Va2 + x2 + a 1, X = -log- a Va^^^^ + a ^ C dx 1 , X 3. I ; = - log— ;=^=^= J X Va2 - x2 a Va2 - x^ + i /'' ^ X' _ *• J„ (2-3 x2)3/2 ~ 9 V2-3x2 9 V2 '^ X3(fe _ 4 — 3 x2 Jo Va + 6x2 362 362 180 INTEGRAL CALCULUS. /• dx. _ (2xg-l)Vl + a:^ ™_i_i « — 4-1-1 1 Here -}- - = — ^-!— ^ — ^ = - 2, and 8 = 2; hence, we n s 2 2 assume 1 4-x2 = z2a;2; .•.x = (z2-l)-i^^ a;-4 = (22 - 1)2, (1) dx = -(z2-l)-s/2z(jz, (2) (1 + x2)-l/2 = z-lx-l = 2-l(z2 - 1)1/2. (3) Multiplying (1), (2), and (3) together, we obtain where z = vT+^/x. P " adx _ ox /•^ dx _ X (2 x2 + 3) . ■ J„ (l-|-a;2)6/2~3(l-|-iK2)8/2' r^ x2 dx _ x' J„ (a -I- 6x2)6/2 -3a(a + 6x2)3/2" r dx _ _ g + 2 6x2 _ J x2(a + 6x2)3/2 o2x (a + 6x2)1/2" /• dx ^ 3x8 + 2a ■ J x2(a + x3)6/3~ 2a2x(a + x3)2/3' CHAPTEE V. INTEGRATION BY PARTS. REDUCTION FORMULAS. 179. Integration by parts. By differentiation we have d (uv) = udv + vdu. Integrating both members and transposing, we obtain j udv = uv — I vdu. (1) The use of formula (1) is called integration hy parts. In applying (1) to particular examples the factors u and dv should be so chosen that dv is directly integrable and vdu is a known form or one easier to integrate than udv. xlogxdx = — logx — — • Let u = log X ; then dv = x dx, du — dx/x, V = x^/2. Substituting in (1), we obtain x^ /^x^ dx /x^ fx^ dx logx- xdx = logx--- I --• — = (x2/2)logx — xV4. (2) In (1) the second product can be obtained from the first by integrating its second factor, and the third product from the second by differentiating its first factor. The following examples will illustrate how we can by the use of this law abbreviate the applications of (1). 182 INTEGRAL CALCULUS. j X ■ cos xdx = X ■ sin X — j sir Ex. 1. 1 X ■ cos xdx = X ■ sin X — j sin xdx - X sin X + cos X. '. — x^Y'^xdx x-e^dx = x-- i a J a Ex. 3. Txs (a - x2)i / 2 dx = Tx^ • (a - x^)! / 2 = X2 [-|(a-x2)8/2] + ri(0-x2)"2.2x(fo = - ^ (a - x2)3'2 - ^ (a - x2)6 /2. EXAMPLES. 1. j log xdx = X (log X — 1). 2. ^xn log Xdx = ^ (log X - ^) ; J"! 1 x» log xdx = — . , .3 ■ § 86, example 6 (™ + i) 3. j X sin xdx = I — X cos x + sin x I = *. 4. I sin— ^ xdx = I x sin— i x + Vl — x" I = — — 1. 5. I tan-i xdx = x tan-i x — - log (1 + x") I 6. Jcos-.xdx = xcos-^x-Vr^. 7. j cot-i xdx = X cot-i X + - log (1 + x2). n."\l ' - * _ log 2 INTEGRATION BY PARTS. 183 r n, Sj a; Va^ — x^ , a? . .x ,,, 8. I Va2 — 3;2eja; = 1. gm-i-. n\ /Va2 — 12 . dx = Va^ — a;2 . ^ + J , J Va2 - x2 = xVa^ — x2 + I / ' dx J Va2 - 12 = xVa2 — x2 + a2 sin-i- _ J Va^ — x^dx. Transposing the last term and dividing by 2, we obtain the result in (I). 9. r Vx2 — a2 tfo; = I Vxs — a^ - ^ log (x + Vx2 — a^). 10. r Vx2 + a3dx= I Vx2 + a2 + ^log(x + Vx2 + a^). 11. I x^ sin X(Jx = 2 X sin X — (x2 — 2) cos i. (1) I x2 • sin x dx = x2 . (— cos x) + 2 | cos x ■ x (Jx (2) I X • cos X dx = X ■ sin x — j sin xdx = xsinx + cosx. (3) Substituting in (2) the value in (3), we obtain (1). 12. I x"" sin xdx = — x"> cosx + m | x"-! cos xdx. 13. j x2 cos X dx = 2 X cos x + (x^ — 2) sin x. 14. I x™ cos X dx = x"" sin X — m | x™-^ sin x dx. 15. j versin-i x dx = (x — 1) versin-i x + V2 x — xK xe^^dx = — (ex — 1). 184 INTEGRAL CALCULUS. 17. Cx^e'^ dx = y('x2-^ + |V 18. I x'"e™dx = I x">-ie<;^dx. 19. fx3ac.dx = ^7x3-^ + i4l^-^i^V 20 r»%- dx = -^ TxB - 5^ + ^-^-^ - i^^ii-I . J logaL log a (loga)2 (loga)'J 21. I = -\ I dx. J X"' m — 1 X'"- 1 m — lj x™-! 22. I e'=']ogxdx = ^ I — dx. J c cj X 24. I xcos-ix(Jx= - cos— 1 X — - Vl — x2 + - sin— i x. J 2 4 4 25. rx2sin-ixcix = ^sin-ix + ^(x2 + 2)Vl — x^. J o 9 /x^dx /I \ , , ^ tan-ix = ( X — - tan-ix jtan-ix — log v 1 + x^. 180. Additional standard formulas. For convenience of reference we write below the formulas obtained from examples 8, 9, and 10 of § 179, and examples 2 and 3 of § 178. f-y/a^ - u" du = '^ Va^ - ^^ + TT sin"! - + C. (1) I ^u^ ± a^ du = ^ Vtf^±a^±|-log(w+VM2d=a2)+ C. (2) /<^M 1 , M ^ — , =-log , + C. (3) REDUCTION FORMULAS. 185 181. Reduction formulas. A formula by which any inte- gral not directly obtainable is made to depend upon a standard form or on a form easier to integrate than the original func- tion is called a reduction formula. Thus, the formula for integration by parts is a general reduction formula. Many special formulas are obtained by applying this general formula to particular forms. 182. Reduction formulas for j x™(a + bx°)i'dx. j «"'(« + hx'^ydx = ,, ^, -TTT^ J, , XT; x-^—{a + hx^ydx, (np + in + 1) b{np + rn + l)J (A) ai^+Va + Sec")" , anp T ™/ , , „n„ i / /-on or ^ —^ H , ^ , . a;"(a + bx'y-^dx, (B) np + m + l np + m + IJ ^ ' ^ ' a;'"+i(a + te")"+i b(np + m + n + l) r ^,^ ^ ,,„.„, a (»i + 1) a (m + 1) ^ (C) ^^ _ x--^ (a + 5x")"^' _^ np + in + n + l T , :^^. ara ( » + 1 ) «« (fJ + 1) ^ (D) Formula (A) decreases m by n. Formula (B) decreases p by 1. Formula (C) increases m by n. Formula (D) increases p by 1. Formulas (A) and (B) fail when np + m + 1 = 0. Formula (C) fails when m -f 1 = 0. Formula (D) fails when p + i = 0. When (A), (B), or (C) fails, the method of § 178 is appli- cable. When (D) fails, p = — 1 and previous methods apply. 186 INTEGRAL CALCULUS. 183. Proof of formulas (A), (B), (C), and (D). Integrating by parts when u =x"'~"+^, we have Cx"'(a+bx")'>dx = , ) ,/ — — TTT I x"'—'(a + bx'')P-i-^dx. (1) nb(p + l) nb{p + V)J \ ' ' W Ta;"'-" (a + 6a;")^ + ^dx = C x""-" (a + bx'')^ (a + bx") dx = a Cx""-" (a + bx^ydx + b Cx'" (a + bx'^ydx. (2) Substituting the last member of (2) for its equal in (1), and solving for j x"" (a + bx^)''dx, we obtain (A). Solving (A) for i x'"~" (a + bx")'' dx, and substituting in the resulting identity m-\- n for m, we obtain (C). Ta;'" (a + bx")" dx = j x" (a + bx") (a + 5x")^-'c?x = a fx"" (a + Jx")^-'cZx + b Cx'" + '' (a + bx")''-'dx. (3) Substituting, in (A), m + n for m and p — 1 for p, we obtain Cx"'+''{a + bx")"-^dx x'^ + '^ia + bx")" a(m + l) C . . , ■. , , .,. = A/ \ VTT - w _L _L1N )»'"(« + bx^-^dx. (4) (wp + m + 1) (rep + m + 1)J ^ ' ^ ' Substituting in (3), and combining similar terms, we obtain (B). Solving (B) for j x" (a + 5x")p-'6Zx, and substituting^ + 1 for^, we obtain (D). REDUCTION FORMULAS. 187 EXAMPLES. X 1. i , = — - Va^ — x2 + — sin-i- J Va2— a;2 2 2 a Here m = 2, n = 2, p =; — 1/2, a = a^ 6 = — 1. Substituting these values in (A), we obtain 2 2 a We use formula (A), because decreasing m by ti in the given dif- ferential reduces it to a known form. x2 J Va2 — x2 \4 4-2/ 4-2 3. r ,^^— = |x^±"^T^log(x + V^±^). J Vx2 ± a2 2 -r 2 ^^ ' (Zx Va2 — x2 , 1 /» dx Va2 — x2 , 1 , X J x3 Va2 - x2 2 a2x2 2 a^ ^ Va^ - x^ + a Apply (C), and then use (3) of § 180. r dx _ _ Vx^ + a^ 1_ X J x8Vx2 + a?~ 2a2x2 2a3 °^ a + Vx^ + a? dx Va;2 — a^ , 1 ,x J x'Vx2 — o^ 2 0^x2 ' 2 0'""" a /» dx _ Vx2 ± g" J x^Vx^ ± a» ^ a^ dx Va2 — x2 9. rx2 Va2 - x2 dx = ^(2x2 - a2) Va2 - x2 + ^sin-i- • Apply (A), and then use (1) of § 180. 188 INTEGRAL CALCULUS. a* = ^ (2 x2 ± a2) Va;2 ± a^ - ^ log (x + Vx^ ± a^). o o Vx^ — a^ 11. i dx = Vx2 — o^ — a sec-i - • J X a Apply (B). 12. I dx = Vx^ + a^ + a log , • J X a+ Vx2 + a^ 13. p^°'~^' dx = Va2 - x2 + a log f Ja X a + V a^ — x2 14. r(a2 - x2)3/2dx = I (5 a2 -2x2)Va2-x2 + ^ sin-i-- Va2 — X* , Va2 — x2 . , x (tt = sm— 1 - • x^ X a Apply (C), and then use (1) of § 180. dx h log (x + vx2 ± a^). dx Apply (D). , dx _ ± X lo. / (X-2±a2)3/2 „2Vi^±^ 20. r , of'^a^a/-. = y "^ + log (X + V^^T^). J (x^±ay^ Vx2 ± a2 21. r(x2±a2)3/2£jx : I (2 x2 ± 5 a2) Vx2 ± a2 + ^ a* log (x + Vx^ ± a^). REDUCTION FORMULAS. 189 22. J (^' 2(x2 + o2)"'"2a3*^'^ ^a 23 f '^ = ■ J (x2 + a2)2 2 a' 24 /* (fe _ a: 3g 3 _^x ■ J (x2 + a2)3 4 (j2(x2 + a2)2 "^ 8 a* (a^ + x^) "^ 8 a^ a ' 25 r ^^ ■ J (x2 + a^)" _ 1 X 2?i — 3 Z' dx _ 2 (ji — 1) a2 {x2 + a2)»-i 2 (ji - 1) a2 J (x^ + a^)"-' ' 26. r V2 ax - x2 dx = ^-=^ V2 ox - x^ + ^ sin-i ^-^^ • ^ 2 2 a Reduce tlie expression to the form in (1), § 180. 27. 28. 29. 31. fx^V^^^^ax=fx^.^^^VI^xdx x'»-i(2ax — x2)8'2 (2nj4-l)a /• , /- -^ ^ :— ; h^ r-f- I X"'-1V2 ax — X^dx. m + 2 m + 2 J /x^dx _ x"— iv2 ax — x2 (2?n — l)a /* x™— i(Zx V2 ax - x2 ™ '™ J V2ox-x2 xdx /2 ax — x2 dx 30. / X2(?X _ •^2 ax — x2 X"' V2 ox — x^ V 2 ax — x2 m — 1 (2 m — 1) ax™ (2 m -l)aj x"-! dx v'2 ox — x2 CHAPTER VI. INTE6BAII0N OF TBIGONOMEIBIC FORMS. 184. j tan=uduor j cot°udu, n any integer. Wten re is a positive integer I tan" udu= | tan"~^ u (sea'u — 1) du tan"-'M r^ = • z I ta,n''~'' uau. n — 1 J By repeating this process the integration of tan" udu is made ultimately to depend upon the integration of tan u du or du according as n is odd or even. When w. is a negative integer, as — m, we have ta,n~'" udu = coV" udu, which may be integrated by a process similar to that used above. Ex. I tajLi—*-dx= I cot*-dx= I cot^-f esc''- — 1 j dx = -cots^- r(csc2^-l)dx = - cots (a;/3) + 3 cot {z/3) + x. TRIGONOMETRIC FORMS. 191 185. I sec°uduor j csc°udu, n even and positive. I sec'^ udu= I sec^^^M . see^Mi^M = ntSin^u + iy-^^'^-dtanu, (1) whicli can be expanded and integrated directly when n is even and positive. Ex. j sec" a;dx = | (tan^ x + l)^ -d tan x = tan^ x/5 + 2 tan^x/S + tan x. 186. j tan™usec°udu or j cot"" u CSC udu, m odd and positive or n even and positive. Ttan"" M sec" M <^M = j (sec^w — ly^^'^'^sec""'?* • c^secM, (1) or ftan™ u (tan'' u + iy-^^'^ ■ dtanu. (2) The form in (1) can be expanded and integrated when m is odd and positive ; the form in (2) can be when n is even and positive. Compare (2) with (1) of § 185. Ex. 1. j tan^ X sec^ xdx= ( (sec^ x — 1) sec* x ■ d sec x = sec' x/7 — sec^ x/5. Ex. 2. j cot6/2xcsc*xdx = — j cot6/2x(cot2x + 1) -d cotx = -^C0tll/2x- -^C0t'/2a;. EXAMPLES. 1. /» , , tan^x , , 1 tan^xcto = -^— + log cosx. 2. j tan* xdx = — ^ tan X + x. 192 INTEGRAL CALCULUS. 3. j t&n^xdx = —T ~ 1- log sec x. A C^ a J tan^x tan'x , 4. I ta.n^xdx=—p l-tanx — x. c /* s n J tan' 2 X , 3 tan^ 2 x , tan^ 2 x , tan 2 x 6. ( csc6|(Jx = -|cot55-f cot8|-2cot^ = cos^x/S — cos^x/S. Jtan^x. 3 6 3 8. I tan^x sec^/'xdx = — sec"/3j; _-_ secii'^j; + _ secs/sj;. 17 11 5 Aan' /2 X see* x dx = — tanis '2 x + - tan^ '^x. 13 ,„ .'secf'xdx ^ „ ^ cot'x 10. I — — -. — = tan X — 2 cot x — f' tan*x 3 cot^x cot^x 11. i cot^xcso^xdx = — D o 12. Jcot3xcsc'.^x.x. 13. Jcot-3.xcsc^x<^. /• 8 14. I (sec X + tan x)* dx = - (sec^ x + tan^ x) — 4 sec x + x. 15. j (tanx + ootx)3dx = -(tan^x — cot^x) + 2 log tan x. 187. I sin"" u cos° u du, m or n odd and positive. I sm^ u cos" u du = j sin""?* (1 — sin''M)~2~ d sinu, (1) COS" m(1 — cos^m) 2 dcosu. (2) TEIGONOIIETRIC FORMS. 193 The form in (1) can be expanded and integrated when n is odd and positive ; the form in (2) can be when m is odd and positive. Ex.1, j sin3/5a;cos8a;dx = j siii3'5a;(l — sm^i)^ sinx = - sin' '^ X — — sinis /^x. o 18 xilx _ /* (1 — .sin- x) d sin x _ 1 1 Ex.2. r52!!^= r J sin*x J sin*x sinx 3 sin^x 188. I sin" u cos" u du, m + n even and negative. //^sin™ u sin"" M cos" M \ / sin*" + 1 a; cos"- 1 re , n — 1 f . „ , or ; 1 ; — I sin'"xcos"-^xc?a;, (2) m + n m -\- nJ ^ ' / Gos''xdx _ cos" + 'a; m — n — 2 roos''xdx sin"'aj (m — V) sm"'~^ X m — l J siii'"~^a; ^ -^ / sin'"xdx sm™ + 'a; n — m — 2 ^sin'"xdx cos"x (m — 1) cos"'~^a: n — 1 J cos"-^a; ^ ' 191. Proof of formulas in § 190. Letting u = sin*"-' x, and integrating by parts, we have J sin™ X cos" x dx sin^-'a; cos"+^a; , m — 1 /^ . . ,„ , = -^ 1 —T- I sin"'"^^ cos" + ^a;da;. n + 1 n + IJ gjjjm-2 ^ (.Qgn+ 2 x = sin™'^ X COS" a; (1 — sin^ a;) ; .". I sin^-^a; cos""*"^ a;£?a; = j sin™"^ a: cos"xcZa; — | sin"a; cos"xrfa;. Substituting this value in the first identity, and solving for I sin™ a; cos"a;cZa;, we obtain (1). Letting u = cos"-^a;, and proceeding in a similar manner^ we obtain (2). Substituting 2 — m for m in (1), and solving for | — : > %j sin X we obtain (3). Substituting 2 — w for n in (2), and solving for | ; ^ COS o^ we obtain (4). TRIGONOMETRIC FORMS. 197 EXAMPLES. ^ r ■ , sin"!— ixcosx , m — \ p . „ , 1. I sm"'xax= 1 I sm"— ^xda;. J m m J Putting for n in (1) of § 190, we obtain tlie formula above. 2. By the formula in example 1 show that /* . . , sin^x oosx , 3 , (a) I sm*xdx= — ■ r h „ (x — smx oosx). 16 sinPxdx= —{ — o 1" T^sin^x + -sinx j + ■ /smxcos"— ix , n — 1 /• cos"xax= 1 I I n n J 4. j cos^xdx = — j-fcos^x + -cosx) + /^ dx cosx m — 2 Z' I sin^x (m — l)sin'"— ix m — 1 J 3x sm™— ■'x /» ^, cosx/1 , 3 \,3, ^ X I 4 \sm''x 2sm2x/ 8 2 P dx sinx m — 2 /» cfa; _ ■ J cos" X ~ (m — 1)008" -ix K — ij COS"— 2x ^ r - , sinx /I , 5 , 5\ 8. I sec'X(Xx = T; ;— (s 2 it;; ^ — r S ) I 2cos2x\3 cos*x 12cos2x 8/ + (5/ 16) log (sec X + tan x). /, , sinx/ cos^x , oos^x , cosxn , x sin2xoos*a;dx= -^ (^- -y- + -j^ + -y- j + j^- = dx = 1- sin^ X cos X — - (x — sin x cos x) 2x COSX 11. r^ sin'x 3, . , = (x — sm X cos x). COS X 2 ^ X , cosx ,„ „ , 3x -dx = — —-. — (3 — cos^x) — — ■ X 2 sm X ^ '2 /dx 1 cos X , 3 , ^ X 1 cos^ X cos X 2 sin2 x 2 ^ 2 198 INTEGRAL CALCULUS. du , r ivL 192. a + b cosu = a[ sin^^ + cos^ o ) + ^ I °°^^ f ~ ®"^^ 5 J a + b COS u J a + b sia u = (a-b) sin^ (■m/2) + (a + 6) cos' (t«/2)- / du 2 /^ Vg. — bd tan (u/2) a + b cos u~ Va — 6 J (« — *) t^^i^ (^/2) + (« + &) Va' - i'' /rfw _ — 2 /» V6 — ad tan (t< /2) 0. + 6 cos M V6 — aJ {b — a) tan' {u/2) — (b + a) _ 1 -y/b — a tan (m / 2) + V& + a ,„. ~V&'-a' °^ V6-atan(M/2)-VH^ , i • f ■ i'^ < 2"^ r Oi • " ^ a + 6 sm M = a ( sm' - + c.os^ - 1 + 2 6 sm - cos - ■ /du _ /^ sec^ (u/2) du a-\-b sin m J a tan' (u/2) + 2b tan (u/2) + a -/f ffl sec' (ic/2)du/2 a tan (m/2) + &]' + (a' - &^) _2^^^^_, c.tan(t^+5 ^ Va' - 6' Va' - &' 1 , a tan(M/2) + 5- V6'-a' or log ^ — - — (4 ) V5' - a' a tan (m/2) + b + Vfi' - a' If a > 5 arithmetically, we use the forms in (1) and (3) ; if a < 5 arithmetically, we nse the forms in (2) and (4) ; that is, in each case we use that form of the integral which is real. TRIGONOMETRIC FORMS. 199 EXAMPLES. 1 /• tig _ 1 /• d (2 X) 1 ^- J 5-3cos2a;-5j 5 -3 oos2cc = i**'' (^ '^''*)- '■/ ^ _1 _ , .5 tan X + 4 5 + 4sin2x~3 3 ^x _J_ tan(3x/2) + 2 3+ 5 cos 3 X 12 °^tan(3x/:i) — 2' ^ _1 2 tan (.3x72) + 1 4 + 5 sin 3 X ~ 9 °^ 2 tan (3 x/2) + 4 ' dx 1 , 2 tan X + 1 - ;log; ■ 5 cos 2 X 8 2 tan x — 1 193. Integration of trigonometric forms by substitution. Assume sin x = s; then cos X = (1 - zy, dx = (l- s^y^dz. .-. Csin'"zoos''xdx = Cz'" (1 — z^^-'^^'^dz. The last form is integrable for all integral values of m and n, positive or negative. We might have assumed cos x= z instead of sin x = z. This method is applicable to any rational trigonometric differential. EXAMPLES. /* . . „ , cosx/sin^x sin^x sinx\ , x 1. J smixcos!!xdx=-^(-^ ^ 8") + 16" Let sinx = z ; . . cos^x = 1 — z^, dx = (1 — !fi)^^'^dz. .-. j sin^xcos'^xdx = j z*(l — z^)^''^dz. _ {\—z^Y'^ (z^ 2° z\ sin-' ~ 2 V 3 12 8/ 16 Substituting sinx for z, we obtain the integral in (1). (1) 200 INTEGRAL CALCULUS. 2. j sec* xdx — —J— (sec' x + -secx) + - log (sec x + tan x). Let secx = z, or x = sec— ^z ; . . dx — Vz2-1 j sec^xdx = j z* (2- — 1)— 1^- (iz = i (z2 - 1)1/2 (z3 + 1 2) + |log (z + Vz2 - 1). Substituting sec x for z, we obtain the required integral, ■•cos* X (is cos'x , , , X ■ „ /'COS* X ax cos'x , , , x 3. I — : = — 1- cos X + log tan - • P smx 3 ^2 Assume cos x = z. /dx cos X , 1 , X . „ = — n . „ — h - log tan - sin^x 2sin2x 2 * 2 /.- cfe 1 ,- ^ X ^ + log tan : sinxcos^x oosx 2 ft /^ dx _ /" (sin^x + cos2x)(ix J sin X cos* X J sin x cos* x 'sinx(Zx , /• dx / smxdx . /• cos*x J sm X cos-" X 7 /• dx _ _\. cosx 3 x I sin^ X cos2 X cos X 2 sin^ x 2 2 °- I — rTl — ;; = „ . ,„ + „ . ,., log (a cos 9 + 6 sm $). J a + b tan 8 a^ + V a^ + b^ °^ ' Assume tan 6 — z. 9- j33^Ji^ = il°S'=°K4-0-2- TRIGONOMETRIC FORMS. 201 ia/i rp r i,^ ■ 1, T e''''(asinbx — bcosbx) ,.^ 194:. To prove I e''^ smbx dx = — =^ ; (1) J r .^ ^ 1 e*''(bsinbx + acosbx) and I e»== cos bx dx = — ^^ — — ^-- (2) »/ a"* + b^ ^ ■' Integrating e'^ sin bx dx by parts, first with u = sin ix and then with u = e"^, we obtain (3) and (4). /„^ • 7 , e"^ sin 6a; b f , , e"^ sm bx dx = I e'" cos 6a; t^a;, (3) /„.-,, 6°^ cos 6a; , a r , , e"^ sm bxdx = ~ — + - \ e'^ cos 6a; dx. (4) Subtracting (3) from (4), we obtain (2). Multiplying (3) by a/6 and (4) by b/a, and adding the results, we obtain (1). EXAMPLES. 1. t ef" (sin ax + cos ox) dx = e"™ sin ox/a. 2. j e3^(siii2x — cos2x)dx = e3^(sin2x — 5 cos 2 x)/ 13. /•sin X , sin X + cos x 3. i dx = — r J eF 2ef . r sin-ixdx ^ , 1 u -1 4. I T-. ittt:; = 2 tan z + log cos z, where « = sin- 1 x. ^ (1 — x''Y' /sin-ixdx r „ , ^ r^ J — ST7S = I z • sec^zda = z tanz — I tanzaz. (l-x2)3/2 J J 'X + sm X , . X _, , _ — ^ dx = X tan - • Put x = 2 z. L + cosx 2 195. Integration by expansion in series. When by any of the preceding methods we cannot integrate a given differ- ential exactly, we can expand the differential in a series, integrate its terms separately, and thus obtain the integral approximately between the limits of convergency of the series. 202 INTEGRAL CALCULUS. EXAMPLES. ' sing J _ _ ic^ , _x^ xj_ X '^^^ 3-[3''"5-[5 7-[7 ' 9-19 1. J5^dx = x-5^ + ^,-^ + 5^ • (1) x^ , x^ x' , x^ „ „, .mx = x-^ + ^-^+^ . §94 Multiplying hj dx/x and integrating, we obtain (1). x^ 2-|2 ' 4-[4 6-[6 ' 8-[8 a^^ g^x^ g^x* 2.[2"'"3-[3"'"4.[4 dx _ .. 1 • x5 1 ■ 3 ■ x8 _ 1 ■ 3 ■ 5 ■ xi3 (1) /•cosx , , X^ , X* x» , — dx = logx + ga; + £^ + i^^ + fnr + " *' J Vn- x* ~ " 2-5 ' 2-4-9 2-4-6-13 ^•fS|- = -(l. + 2^ + i + 4^+---)- (1 — x)-i = 1 + a; + x2 + x3 + • •, when — 1 < x < 1 ; J'^i ]ogx Z*^ _ <^^ — I (log X + X log X + x2 log X + • • ■) dx. (2) ^ ^ Jo Integrating each term in (2) by the definite integral given in example 2 of § 179, we obtain (1). 6. By integrating (1 + x'^)-'^dx directly and by series, prove that . , x3 x5 x'' , x8 tan-^x = x-- + --y + - , (1) (1 + X2)-1 = 1 - X2 + X* - X« + X8 — ■ • ■ ; „ ^ • , , x3 1 • 3 ■ x5 , 1 ■ 3 • 5 ■ x' , 7. Prove sm-ix = x + — + -^-^-^ + ^-^-^+---. Integrate dx/ Vl — x^ both directly and by series. 3J iC*^ CC^ iC^ 8. Prove log (g + x) = log g -I z— + r-^ — — — + Integrate dx/(g + x) both directly and by series. 1 ■ x' , 1 • 3 • x6 1 • 3 • 5 • x' n T> 1 / I , f, — i ;% 1 ■ X" , 1 • •:> • X" 1 • i5 • • X' , 9. Provelog(x + Vl+^) = x-^^ + ^^^^-^-^-^ + -' CHAPTEE VII. LENGTHS AND AREAS OF CURVES. SURFACES AND VOLUMES OF SOLIDS OF REVOLUTION. 196. Lengths of curves. Rectangular co-ordinates. Let s denote the length of the arc whose ends are tlie points (xo, y„) and (x, y) ; then from ds^ = dx'^ + dy^, by § 165, we have s= Cll + idy/dxf'l'-'^dx, (1) Jxf, or s=C [(dx/dyy + If'dy, (2) according as we express ds in terms of x or of y. In any given curve we use that formula which gives the simpler expression to integrate. EXAMPLES. 1. Find s of the semi-cubical parabola ay^ = x'. (dy/dx)^ = 9x/4:a; ,(■+1!) * "".('I =lf[(- !!)"■-(' +!?)■'"]■ » When (So, 2/o) is the origin, (1) becomes 8ar/, , 9a;\8/2 -i 2. Find s of the cycloid x — r vers-i (y/r)^ V2 ry — y^. (dx/dyY = y/('i.r-y); ... s= V27 \\2r-y)-^'^dy § 196, (2) = 2V2r[(2r-yo)i/2- (2r-2/)i^2]. 204 INTEGRAL CALCULUS. Putting 2/0 = and y = 2r and taking twice the result, -we find that the length of one arch is 8 r. 3. rind s of the parabola y'^ = ipx, (xo, j/o) being the origin. Am. g = -^V4i)^ + y^+plog ^"'" ^P^ + y^ 4. Find s of the circle x^ + 2/^ = r^, and the circumference. Ans. r sin— 1 (x/ r) — r sin— ^ (xo/r) ; 2 ?rr. 5. Find s of the hypocyoloid x^'^ -f- 5/2/8 = £i2/3^ and the length of the curve. . 3 ,,., ,,, om ^ 6. Find s of the ellipse y'^= {1 — e^) (a^ — x^), and the length of curve, e being the eccentricity. rlr ^y /-F-^ i (1) hence, the length of the elliptic quadrant s, is (a2-eH^)^n—^=. (2) V a2 — x2 The integrals in (1) and (2) cannot be obtained directly, but (o^ — e2j-)i/2 can be expanded by the binomial theorem, and the terms of the result can be integrated separately. Thus, J"" dx e^ r" x^dx e^ /•" x*dx Va2 - x2 2aJ„ Va^ - x^ ^"-^Jo Va^ - x^ _ *a / e' 3 ■ e* 3^ ■ 5 ■ e^ _ . . .\ ~ 2 V 22 22-42 22 • 42 • 62 ' ' '/ ' 7. Find s of the ellipse when given by the equations, X = a sin 0, y — b cos S, where $ denotes the complement of the eccentric angle. (fe2 = a2 cos2 edffi, dy^ = 62 sin2 ede2 ; .-. ds2 = (a2 cos2 6> + 62 sin2 fl) d&^ = a2 (1 — e2 sin2 e) dff\ ,. — e^sin^mi/2f ''so r = a r* (1 ■ ^so r = a T" ''(l--e2sin29--eisin4fl-— e^sinSff )d9. Jo ^ » lo THE CATENARY. 205 197. To find the length and the equation of the catenary. Let NOM be the curve in whicli a chain or flexible string hangs when suspended from two fixed points Jf and N; then NOM is a catenary. ^ Let w denote the weight of a unit length of the chain, and s the length of the arc whose ends are the lowest point (0, 0) and the point (a;, y'), or B; then the load suspended, or the vertical ^ -^ tension, at B is sw. Denote the horizontal tension, which is the same at all points, by aw. Let DA be a tangent at B ; then if c ■ BD represents the total tension of the chain at B, c ■ BE and c • ED will represent, respectively, its horizontal and its vertical tension at B. dy c- ED sw s _ ^ Hence, — = — 5^ = — = - ; (1) dx c- BE aw a ^ ' /A s y/ds^ — do? , ads - ; .' .dx = ■ a dx Va^ + . ds s+VaF+~? = alog (2) Va^ + s Solving (2) for s, we obtain as the length of OB S = ^(e^/a_e-x/a-) (3) Eliminating 5 between (1) and (3), we obtain .•.3/ + a = 1(6^/" + 6-^/") (4) is the equation of the catenary referred to the axes OX and Y. If O'O = a, and the curve be referred to the axes O'X' and O'Y, its equation will evidently be y = ^(e^/» + e-^/«). (5) 206 INTEGRAL CALCULUS. 198. To find the length and the equation of the tractrix. The characteristic property of the tractrix is that the length of its tangent FT is constant. Denote the constant length of the tangent FT by a. Let 'FM=ds; then — FN = dy, and NM = dx. ds _ PM _ a ' ' dy~ FN~ y' Hence, if s is measured from A, or (0, a), we have J"''dii , a -^ = a log - • Again, from the figure we have dy I dx = — y 1^ a? — y^. (1) (2) (3) Integrating (3), remembering that y = a when a; = 0, we have X = — Va^ — y'^ + a log [(a + Va'' — y"^) /y] as the equation of the tractrix. 199. Lengths of polar curves. Let s denote the length of the arc whose ends are (po, 0^ and (p, ff); then from ds =V^d¥Td^^ by § 165, we have s= CV^MO^l^^ or s= fV^WT^, (1) according as cZs is expressed in terms of 6 or p. CURVES m SPACE. 207 EXAMPLES. 1. Find s of the spiral of Archimedes p = a0. ..s = a 1 Vl + ff^de = ^ fflVi + 92 + log ($ + Vl + fl2)1 '■ ^ L J So Putting flo = and 9 = 2 tt, we obtain as the length of the first spire a [tt Vl + i7f^ + ilog (2 ff + Vl +4 7r2)]. 2. Find s of the logarithmic spiral p = fte^/". e/a = log{p/b) - log/) - log6; .-. pdB = adp. .: ds = Va^ + l dp. . . s = Va2+ 1 r''dp = Va2+i(p_p„). 3. Find s of the cardioid p = 2 a(l — cos 9). Ans. s = 8a[cos{eo/2) — cos(9/2)] ; entire length = 16 a. 4. The entire length of the curve p — a sin^ (fl/3) is .3 7ra/2. 200, Curves in space. Let s denote the length of an are of a curve in space whose ends are (a;,,, yo, Zg) and (x, y, z), and let As denote the length of an infinitesimal are whose ends are (x, y, z) and (x + Aaj, y + Ay, s + As) ; then As = A^AaM-Ap + As^ + wi", where »i > 1. § 69 ,-. ds = V^M^"^M^^. § 71 .-.5= pV5^M^^?T^'. 208 INTEGRAL CALCULUS. EXAMPLES. 1. Find the length of an arc of the helix, x = a cos 6, y = asinS, z = ke. Here dx = — a sin ed0, dy = a cos ede, dz = kdS ; . . ds = Va2 4- /c2 (jfl. .-. s = Va2 + k"- C de «/eo - Va2 + fc2 (9 _ 9o). 2. rind 8 of the curve y — x^/2a, z = x^/Q a^, s being reckoned from the origin. -S.V-&) ■■x + x^ 6a2' dx X + z. 3. Find s of the curve y = 2 Vox — x, z = x— (2/3) Vx^/a, s heing reckoned from the origin. =XXVi+Vf-o^=»+^-^ 201. Areas of curves. Let NBC be the locus of y = (^cc, , and JVZ>;SC that of y = fx. Denote their intersections, N by («o, 2/o) and C by (xj, i/i), and the variable area NBD by J. Let X = OM, dx = DE; then dA = DELE = { {a^ — x^) is ia^/H. 11. The area of one loop of (c'j/^ = b-x^ {a^ — x^) is 2 ah/3. 12. Solve the first examjjle by formula (2) in § 201. 13. Find the area between the curve y-x = 4 a- (2 a — x) and its asymptote x = 0. ^„s. 4 ^„2. 202. Areas of polar curves. Let B be any fixed point (p(i, 6u) and P any variable point (p, 6). Conceive the area ^ BOI' as generated by the radius vec- tor p, and denote it by A. Dyj With OP as a radius draw arc PZ>, and let d6 = A POP' ; then (^^ = OPD = (p72) (^^ ; 1 r* .•.^ = 2J^pm (1) For the proof of (1) by limits see § 71, example 11. AREAS OF POLAR CURVES. 211 EXAMPLES. 1. Find the area of the cardioid p = 2a(l — cos 6). X2ir (1 — COS 9)2 de — G 7ta?. 2. Find the area of the lemniscate p^ — a'' cos 2 8. Area = =2o2 r n^/4 cos 2$de=: a-. 3. Find the area between the first and the second spire of the spiral of Arcliimedes p = ai. Area = — / e^de — -~j (r^dB-S a^xO. ^ J-2-" ^Ja 4. The area generated by the radius vector of the logarithmic spiral p = e^ from e = to S = ;r/2 is (g"^ — 1) /4 a. 5. The area of one loop of the curve p = a sin 2 fl is ?ra^/8. 6. The area of one loop of the curve p = a sin 3 5 is jta^ /12. 7. The area of a sector of the spiral pff — a is(S — So) a^/2 ff0Q. 8. The area of a sector of the spiral p^ff = a^ is — log — • ^ do 9. The whole area of the curve p = a cos 2 9 is Tro?/2. 203. Areas of surfaces of revolution. Let ^ be a fixed point (xo, ?/o) ^"(i -f ^ variable point (x, y) on the curve ntAF. Let AP = s, and PP' = As. Draw PT and P'B each parallel to OX and equal to As. Let S denote the surface generated by the revolution of AP about the a;-axis ; then \S equals the surface generated hjPP'. Evidently that is. surface PT< A6f< surface P'B; 2 TryAs <^S<2'7^(1/ + ^y) As. 212 INTEGRAL CALCULUS. Let As = i ; then AS = 2 iri/As + vi", where m > 1. .". dS = 2 Tryds ; or /S = 2 tt / yds. Jo (1) In any particular example, yds is obtained in terms of x, y, or any other variable as may happen to be convenient. Similarly, if the y-axis is the axis of revolution, we have ,Sf = 2 7r r.rds. (2) 204. Volumes of solids of revolution. Let ^ be a fixed point (Kq, 2/o) and P a variable point (x, y). Let V denote the volume of the solid generated by revolving BAPMahovA OX. Conceive this solid as generated by a circle whose centre moves along OX, and whose variable radius is the ordinate y of the R /p- y n .£ 1 ^ --->' O B M N X curve AP. When X = OM let dx = MW ; then cZ r = cylinder MPDN = iry-dx. . ■ . V = IT I y^dx. §11 (1) Similarly, when the iz-axis is the axis of revolution, we have 7' = TT I X^(i(/. (2) The proofs of (1) and (2) by the method of limits are left as exercises for the reader. In the following examples, a segment of a solid of revolution means the portion included between two planes perpendicular to its axis ; and a zone means the convex surface of a segment. SURFACES OF REVOLUTION. 213 EXAMPLES. 1. Find the area of a zone of a sphere. Here yds = rdx. .: S — 2 X j yds — 2 7cr I dx Jo Jxf, — 2Ttr(x — Xo). The enth-e surface = ^ 2 rtrx = 4 xf^. 2. rind the area of a zone of the surface generated by the cycloid revolving about its base. yds = V^r {2r — y)- ^'^ydy. .: S = 27tV2r j"y(2r — y)-^'^dy = 2 « Wr f— I (4 »• + ?/) (2 r - 2/)i ^2"] ". L i Jy^ The entire surface = 4 TtVWr I — 7:{ir + y)(2r — yy^ I - 64 Ttr^/S. 3. Find the area of a zone of a prolate spheroid. The generating cuiTe is y^ = {1 — e") (a- — x"^) and yds = - Va" — e^x^ dx. .S = 2Tt- C Va^ — e?a;2dx a I = ;r - fx Va^ - 6^x2 + - sin- 1 -'] "• The entire surface — 'J,Tth\b -\- (a/e) sin- ' e]. 4. Find the area of a zone of the surface generated by the catenary revolving about the x-axis. Here yds = - (e'^/" + e- »/"')-dx. t(j2 ~lx 214 INTEGRAL CALCULUS. 5. rind the area of a zone of the surface generated by the tractrix revolving about the x-axis. ^„g. 2 ita {ya — y)- 6. rind the area of a zone of the paraboloid of revolution. 6p 7. The entire surface generated by revolving the hypocycloid 3;2/3 -|- 2^2/3 — (fi/s about the ir-axis is 12 ita? /^i. 8. The surface generated by revolving the catenary about the 2/-axis, from X = to a; = a, is 2 TCa"^ (1 — e— i). 9. Find the volume of a segment of the prolate spheroid. J Xf, J Xf, The entire volume = * , a?x — \r\ = tt jraft^, a^ L iA-a 3 which is tVFO-thirds of the circumscribed cylinder of revolution. Putting 6 = a we obtain the volume of a segment and the entire volume of a sphere vi'hose radius is a. 10. The volume of the oblate spheroid is two-thirds that of the circum- scribed cylinder of revolution. . > 11. The volume of the parabofoid is one-half the circumscribed cylinder of revolution. 12. The volume of the solid generated by revolving an arch of the cycloid about its base is flve-eighths of the circumscribed cylinder. Here V = '€^ V2ry — y'^ 13. Find the volume of the solid generated by the revolution of the tractrix about the x-axis. Volume = ?f / y'' 1. ' — K I (x — d J Vn — a)^dy. (1) The student should prove (1) by the method of § 204. 16. If the figure bounded by x = ct and the parabola y^ = ipx is revolved about the line a; = a as an axis, the volume of the solid gener- ated is .32 7ia^Vpa/16. 17. Find the volume of the solid generated by the revolution of the cissoid about its asymptote. Ans. 2 Tt^a^. 205. Let V denote the volume generated by any plane figure moving parallel to a fixed plane. Let x denote the distance of the generating figure from some fixed point, and let (x) ■ Ax and ^ (a; -|- Ax) ■ Ax. Hence, when Ax = i, A F = <^ (.r) ■ Ax -|- vi", where m. > 1. .-. r= C,i,{x)dx, (1) the limits being so chosen as to include the volume sought. EXAMPLES. 1. Find the volume of any pyramid or cone. Let B denote the area of the base and a the altitude. Let x denote the area of a section parallel to the base at the distance x from the vertex. Then by geometry we have 0x : JJ = x2 : a2 ; .-. x-pqx FT X2. = a'v^rx V = a I Vara = 7[r^a/2. -x'^dx 3. A rectangle moves parallel to and from a fixed plane, one side vai-y- ing as its distance from this plane, and the other as the cube of this dis- tance. At the distance of 3 feet the rectangle becomes a square of 4 feet. Find the volume then generated. Ans. 4. An isosceles triangle moves perpendicular to the plane of the ellipse x^/a^ + y^/V = 1, its base is the double ordinate of the ellipse, and its vertical angle 2 4 is constant. Find the volume generated by the triangle. Ans. i ab'^ cot A/ S. 5. A woodman fells a tree 2 feet in diameter, cutting halfway through from each side. The lower face of each cut is horizontal, and the upper face makes an angle of 45° with the horizontal. How much wood does the man cut out? ^,j^ 4 / 3 cubic feet. 6. Obtain formula (1) in § 205 by the method of proof employed in § 204. CHAPTER VIII. DOUBLE AND TRIPLE INTEGRATION. APPLICATIONS. 206. Double and triple integrals. If we reverse the oper- ations represented by dxdy, we obtain the function u. That is, ,=Jj|^^,rfy, (1) which indicates two successive integrations, the iirst with reference to y, x and dx being regarded as constants, and the second with reference to x, y being regarded as a constant. In (1) the right-hand sign of integration is used with the variable y; that is, the signs of integration are taken from right to left in the same order as the differentials. Let u' denote the definite integral when the limits for x are Xq and Xi, and those for y are y^ and y^ ; then u'= C' C"'-p^dxdy. (2) •Jxo ^VQ dxdy Ex. 1. C" C xv{x-y) dxdy = C^xdx f^ - ^-1 = aW(a-6)/6. Oftentimes the limits of the first integration are functions of the variable of the second. -^Vi- , . . , , r^V ?/* , y^ Sy''\ , 676^ 20 J {x + y)dydx=j^ (^+62)^^ = The second member of (1) denotes what is called an indefi- nite double integral, and the second member of (2) a definite 218 INTEGRAL CALCULUS. double integral. Similarly, we have indefinite and definite triple and multiple integrals. I I %yz d.x,dyd,z= \ | zy dx dy I zdz ) = -^- Jo "* 3. I I xydxdy — -r- J^b fpib 7 62 pdpde^ — - 6/2 i/O ^ /"S /-Zj, 11 (,4 5. I I xydydx = -^-- ^0 x/iz — b I pd0dp = ^(a2-62). *y2bcos4' COS"f> ■>■„.„,,„ 65 _ £,3 (cos /3 — cos 7y. 8. I I (x — a)(y — b) dy dx = a^b r • ■ Xa f-a /•a Ifi I I (?/ + s2) fc dx dy dx = Y ^a^- X6 /'a /'aft 1 1 r r r »/o •/o »/o 11. I II e'^ + v + 'dxdydz — — T '^ ^'^- AREAS BY DOUBLE INTEGRATION. 219 Rectangular co-ordi- Jc 207. Areas by double integration. nate. Let NBC be the locus of 2/ = x — /x) rfx, which is the differential of the area NDB. Integrating {<^x — fx) dx between the limits Xa and o-x, we obtain the area NBCD, or A. O M Hence, ^ ♦^/^ dxdy. (1) When y and % are constants, PGQF, or dydx, will be the differential of the area HPGK. Hence, integrating dydx between the limits H'H and H'S, we obtain the area HSRK, which is the differential of the area NHS. Integrating this between the limits M'N and XC, we obtain the area NBCD, as before. Hence, ■- [ jdydx. (2) the limits being taken so as to include the required area. The order of integration, therefore, is indifferent, provided the limits assigned in each case be such as to include the area sought. CoR. 9, A = dxdy and 9,,^. A ■ dydx. 220 INTEGRAL CALCULUS. Ex. 1. Find the area bounded by the parabolas y^ — iax and x- = i ay. The parabolas intersect at the points (0, 0) and (4 a, 4 a). Hence, if we use formula (1), the constant limits for x will be and 4a, and the variable limits for y will be x^/i a and v4ax. I Using formula (2), we obtain ^•""■^ ^ 16 a2 ax ay = ^ — J-40 r^*"!/^ _, 16 a2 I dydx = -^ J.^/^„ ^ V/4 Ex. 2. Find the area between the parabola y'^ = ax and the circle 7j^ — 2ax — x^. 1.oj: — :^ Area - ^ ^ Tca^ 4a2 dxdy = -r —■ 208. Areas of polar curves by double integration. Let NBG be the locus oi p = e, and HDE that of p =/(9. Let Z XON= e„, and Z A'OG = 6,. Denote the area HEGN by A. Let P be any jjoint (p, 6) in this area, p and 6 being independent. Let PJIf = dp a.n6. ZPOS=de; then arc PS = pdd. Construct the rectangle PC AM where PC = PS = pdO; then triangle POC= sector PO>S. AREAS BY DOUBLE INTEGRATION. 221 When 6 and dO are constant, PCAM, or pdOdp, will be the differential of the triangle POC, or of its equal, POS. Hence, integrating pdddp between the limits OD and OB, or fd and ,f>e, we obtain the area BBB'U, or \ \_{^fff — (fdf] dO, which is the differential of the area HDBN. Integrating this differential between the limits 6^ and 0^, we obtain the area HEGN, or A. Hence, A = T^ C^ dd dp. (1) Cor. dopA = pdOdp. EXAMPLES. 1. Find the area between the two tangent circles p = 2 a cos 6 and p = 2 6 cos 6, where a > 6. J^n/2 /^2acos9 I pdedp ^ 2b cos 9 X.r/2 2. Find the area, (1) between the first and the second spire of tlie spiral of Arcliimede.s p = afl ; (2) between any two consecutive spires. 3. By double integration find the area, (1) of a rectangle ; (2) of a parallelogram ; (.3) of a triangle. 4. Find the whole area of the curve (y — mx — c)^ = a'^ — x^. Ans. Tta^. 209. Area of any surface by double integration. On the surface « =f(x, y), let P be any point {x, y, z), and Q the point (.r + Aa;, y + A?/, z + As), x and y being independent ; then P'N= \x and P'M=\y. Conceive a tangent plane at P, not shown in the figure. The planes through P and Q parallel to the co-ordinate planes A'.^ and YZ will cut a curved quadrilateral PQ from the sur- face «=/(«, y), and a parallelogram Py from the tangent plane. 222 INTEGRAL CALCULUS. Let Ax = i and Ay = vi ; then area FQ = area Pq + vi", where n > 2 = &Tea, F'Q' -sec y + vi", (1) where y is the angle which the tangent plane at P makes with the plane XY. From (1) we have A^jS = Ax Ay sec y + vi". ■ . d^„S = sec y • f/.r dy. § 140, Cor. From analytic geometry we have --[-(iy-d)"]" .-. S: =//['-(!)'+ (1)7 -*<^) the limits being so chosen as to include the required surface. Let S denote that part of the surface x =f(x, y), z being a one-valued function, which is included by the cylindrical surfaces y = <^|,a;, y = oX, y = ^x ; and the planes x = a, x = b ; I I dxdydz. (2) Cor. 3x112^= dxdydz, Qy^^V= dydzdx, ■ ■ ■. Ex. Find the volume of the ellipsoid xVa^ + 2/V6^ + zVc'^ = 1- The entire volume is eight times that in the first octant, where the limits are 2 = 0, z = cV] — a;2/a2 _ y-i/lfi-^ y~^t y =^ by/ 1 — x- / a'^ ; X = 0, x = a. EXAMPLES. 1. Find the volume bounded by the plane x — a and the surface Z-/C + !/'/b-2x. The entire volume is four times that in the first octant ; .. l' = 4 f" r^"^ r'''-''^-'^^'"dxdydz = rra^^e. 2. Find the volume bounded by the surfaces, ^2 -\- yi = (.2^ x^ + y^ — «.r, 2 = 0. Ic Ans. 2 r C'-'' r''-'''>/^axaydz = '-^ Jo Jo Jo "-' 3. Find the volume bounded by the cylinder x^ + ?/2 = r'^ and the planes 2 = and z = mx. ^,jj^ 4 mr^/3. 4. Find the volume bounded by the surface x^z"^ + cfly" — c^x^ and the planes x = and x = a. Ana. it€-a/-2 VOLUMES BY DOUBLE INTEGRATION. 225 5. Find the area of the zone of the sphere, x2 + 2/2 + z2 = r\ (1) included between tlie planes x = a and x = 6. From (1), Qz/dz = — x/z, dz/dy = — y/z. '9z\2 , /9z\2i"/2 ^ H-. v^m^mr-^ ■ X2 — 2/2 The area required is four times that in the first octant, where the limits are x = a, x = 6, 2/ = 0, ?/ = Vr2 — x2 ; •'b |^v,^_ja rdxdy "S'S Vr2 — x2 — 2/2 2ffr(6-a). §209, (2) 6. Find the surface of the cylinder x2 + z2 = ^2 intercepted by the cylinder x2 + 2/2 = r2. ^„g 8r2. 211. Solids of revolution. Let F be any point (x, y) in the area NBCD (§ 207, fig.), x and y being independent. Let PF = Aa; and P(? = Ay. Conceive NBCD to revolve through ^ radians about OX as an axis ; then By -AxAyK a',, V 2. .■- 9^r=%(/a;(?2/. I 2/^«<^2/- (1) Putting 6 = 2 TT, we obtain the volume generated by a com- plete revolution of the area. CoK. If the a;-axis cuts the area, formula (1) will give the difference between the volumes generated by the two parts. Hence, F = when these two parts generate equal volumes. 212. The moment of a force about an axis perpendic- ular to its line of direction is the product of its magnitude by the perpendicular distance of its line of action from the axis. 226 INTEGRAL CALCULUS. and measures the tendency of the force to produce rotation about the axis. The force exerted by gravity on a body varies as the mass of the body, and may be measured by the mass. The centre of mass of a body is a point so situated that the force of gravity produces no tendency in the body to rotate about any axis passing through this point. The mass of any homogeneous body is the product of its volume by its density. 213. To find the centre of mass of a hodij. Let the points of the body be referred to the rectangular axes OX, OY, OZ, the plane XY being horizontal. Let m denote the mass of the body, and M the moment of the force of gravity on m about an axis parallel to OY and passing through C, (x, y, z). Let P be any point (x, y, z) in the body, and Q the point (x + Aa;, y + Ay, .~ + As). Let Aw equal the mass of the parallelopiped PQ; then {x — x) Am < AJf < (a; + Ax — x) A?/4. Am = point P ; Ax = i, Am = v^W, AM= (x — x) Am + vi", where m > 3. .'. dM = (x — x)dm; .' . M = I xdm — x I dm. (1) When (x, y, z) is the centre of mass, M=0; hence from (1) x= i xdnij I din. [1] Let then, if and CENTRE OF MASS. 227 In like manner we obtain - fydm - ^ /zdm y Jdm ' " /dm ^ -^ To obtain z place the «-axis horizontal. Whether the body is homogeneous or not, dm denotes the mass of a homogeneous solid whose density is that of the body at the point P (x, y, z), (§ 11). Hence, denoting the volume of dm by dv and the density of the body at P by k, we have dm = kdv. Substituting Mv for dm in [1] and [2], we have Jxlidv - fijkdv - fzl; dv "^ ^jkd^' y ^ fkdv ' ^ ^ fkdv ' ^^ When the body is not homogeneous, k is some function of the coordinates of the point (x, y, s), and dv = 'd^jsV= dxdydz. When the body is homogeneous, k is constant, and formulas [3] become _ fxdv - fi/di' _ fzdv ^,_, fdv ^ fdv fdv "- -■ CoE. In formulas [4] dv may equal Q^^ V, 9^ V, or d V. For when the body is homogeneous, all points in the plane ARP have the same moment. Hence, to prove [4] we may let Am equal the mass of the body between the planes ARP and XNQ, or an increment of this mass. In the first case dv wDl equal dV, and in the second it will equal 9^,^ or Q^V. 214. Centre of mass of right cylinders and areas. Let c denote the altitude of the right cylinder whose convex sur- face is made up of the cylindrical surfaces y=fx, y = (f>x, and the planes x = x^, x = x^, the plane XY being midway between and parallel to the bases. 228 INTEGRAL CALCULUS. Evidently 5 = 0. To find X and y we have du = 9^ V — cdx dy. Hence, from [4] of § 213, we have - _ ffx dx dy ffdxdy y ^ ffydxdy ^ ffdxdy [5] the limits for x being a;„ and Xi, and for y, fx and ^x. As the values of x and y depend solely on the plane area bounded by the plane curves y=fx, y ^= x, and the lines X = x„, X = Xx\ for convenience the point (x, y) is called the mass-centre of this area. Cor. 1. If a plane area be revolved about a line through its mass-centre, tlie two parts will generate equal volumes. For let this line coincide with the a;-axis ; then y = 0, and from [5] we have y dx dy = 0. (1) '//= Hence, by Cor. of § 211, the two volumes are equal. Cor. 2. If an area is symmetrical with respect to the a;-axis, y = 0, and x is the same for one of the symmetrical halves as for the whole area. 215. Centre of mass of rods and curves. Suppose the mass-centre of the plane figure CD to move along the curve HPB, its plane being always perpendicular to the curve. Let A denote the constant area CD; V, the volume of the rod generated by CD ; s, the arc HP ; and As, PP' Through P' draw X in the plane C"D" the line P'n parallel to tliQ plane CD'. On P'n as an axis revolve C"D" until it becomes parallel to CD'. CENTRE OF MASS. 229 Then, by Cor. 1 of § 214, we have ^V = A ■ As + vi", where n > 1. .■.dV=Ads. (1) Substituting Ads for dv in [4] of § 213, we obtain fxds - fyds - fzds ""^fd^'y^jdi'^^jd^- [6] As the values of x, y, and i depend solely on the curve HB, for convenience (x, y, z) is often called the mass-centre of the curve HB. If the curve is in the plane XY, 5 = 0; if in addition it is symmetrical with respect to the x-axis, y = 0, and x is the same for one of the symmetrical halves as for the whole curve. CoK. From (1) we obtain V= As. (2) That is, the volume of the rod CBP equals the area of CD into the length of the arc traced by the mass-centre of CD. The proofs of equations (1) and (2) fail when CD outs the evolute of the curve BB. EXAMPLES. 1. Find the centre of mass of the area bounded by the parabola y^ = ipx and a double ordinate. From the symmetry of the curve, ^ = 0, and I xdxdij/ I I dxdy = 3x/b. 2. Find the centre of mass of the area bounded by the semicubical parabola 02/^ = I' and a double ordinate. Ans. x = 5x/7. 3. Find the centre of mass of the area bounded by the ^-axis and the curve X2/2 = b^{a-x). j^ns. x = a/4. 4. Find the centre of mass of the area of the first quadrant of the ellipse x-/a'^ + y^/b^ = 1. ^„s. i = 4a/37i, y = i 6/3 k. 230 INTEGRAL CALCULUS. 5. Find the centre of mass E of any circular arc BOD. D Let OB = OB = s, being the origin. The equation of BOD is y'^ = 2rz — x'^, r being the radius. From the symmetry of tlie curve y = o, and xdx _ fo'xds _ r . fo'ds sJo V2 rx — a;2 — r — ry/s — OE. Hence, CE = 2ry/2s = r chord BD/arc BOD. 6. Find the centre of mass of the arc in the first quadrant of the curve X2/3 + 2/2/3 = «2/3. ^^^ i = y = 2a/b. 1. Find the centre of mass of the arc of a catenary cut off by any horizontal chord. Aiis. y — {ax + ys) /2 s, where 2 s is the length of the arc. 8. Find the centre of mass of the curve xy^ = b'' (a — x). Ans. X = a/4. 9. The axis of a homogeneous solid of revolution is the i-axis ; show that y =~z = 0, and : j j xydxdy/ j j ydxdy. §211 10. Find the volume of the ring generated by the revolution of an ellipse about an external axis in its own plane, the distance of the centre of the ellipse from the axis being r. Ans. 2 TC^abr. § 215, Cor. 11. If an arc of a plane curve revolve through e radians about an external axis in its own plane, the area of the surface generated will be equal to the length of the revolving arc, multiplied by the length of the path described by the mass-centre of this arc. From [6] of § 215 we obtain iy-s = 8 j yds, or the theorem. §203 MOMENT OF INERTIA. 231 12. Find the surface of the ring generated by tlie revolution of a circle (radius a) about an external axis, the distance of the centre of the circle from the axis being r. ^„g_ 4^„^_ 216. The moment of inertia of a plane area about a given point in its plane is the limit of the sum of the products obtained by multiplying the area of each infinitesimal portion by the square of its distance from the given point. Denote by M.I. the moment of inertia of the area NBCD, or A, about (§ 207, fig.). Let P be any point (a;, y) in this area, x and y being independent ; then 'oF = x' + 2/1 Let PF= Aa; = i, and PG = Ay = vi ; then A^(j)/.7.) = {x^ + if) Ax Ay + vi", where n>2. ■■• 9w(^V-f-) = («' + if)dxdy. .-. M.I. = r C {.I? + if)dxdy, the limits being taken so as to include the required area. Ex. 1. Find the moment of inertia about the origin, of the circle x2 + 2/2 = a?-. M.I.= \ j "'"-^ (x2 + 2/2) dxdy = ^- Ex. 2. Find the moment of inertia about the origin, of the smaller area hounded by the s-axis, the parabola y^ = iax, and the line x + y = 3a. Jr-2a /».3o-» 314 a* I (x^ + y^)dydx- x/y^/ia 35 CHAPTER IX. DEFINITE INTEGRAL AS A LIMIT. INTRINSIC EQUATIONS OF CTTKVES. 217. Definite integral as a limit. Heretofore we have considered differentials as finite; in this article we shall regard them as infinitesimal. A definite integral has been defined as an increment of an indefinite integral. We pro- ceed to show that a definite intcfjral eqiidls the limit of the sum, of an infinite nuviher of infinitesimal differentials. To make the theorem and its proof as clear as possible, let us consider the area MiPiBX, which we will denote by A. Let Olfi = a, OX = h, and P^B be the locus of // = <^.r. Divide JfjA' into n equal parts, JfiJlfa, M^M^, ■ ■ ; M„X, M, M, M. M„ X and divide the area A as in the figure. Let dx = JfiJfj = -3^2-3^3 = • • ■ = M^X, then M^Qi = (a + dx^) {h - dx) dx. .'. A = <^(a) dx + tfiia, + dx) dx + <\>(a + 2 dx) dx + ■ • ■ + (b-dx)dx + T, (1) where T is the sum of the triangles PiQiP,, P^Q.P„ ■ ■ ; PnQuB- By the notation of sums, (1) is written DEFINITE INTEGRAL AS A LIMIT. 233 Evidently Tbydx. limit yV(<^.r)=f7x = r= C'7r{,l>.rfdx. § 203 dx = 0- EXAMPLES. 1. The effect of gravity in making a body tend to rotate about any given axis is the same as if its mass were concentrated at its centre of mass. From [1] of § 213, by integration we have /•"■ , limit w-v™ , xm = I xdm = , . „ > xam. (1) The given axis being Oy, xm is what would be the moment of the force of gravity on m if m were concentrated at its centre of mass. The last member of (1) is the limit of the sum of the moments of the force of gravity on all the material points (dm) of m when dm = 0. Hence, (1) proves the theorem. 2. Show that the area of a polar curve is the limit of the sum of an infinite number of infinitesimal differentials. 234 INTEGRAL CALCULUS. 3. Using the figure in § 207 show that V dzdy = (0x —fx) dx = I dx dy, ^^fx Jfx limit /.^-\^i and dx 2^ ('P^ —fx)dx = A= I {(px—fx) dx. ^Xq »yx(i (1) (2) (1) holds true whether dy is finite or infinitesimal ; for dx being constant the sum is constant. In (2) tlie sum varies with dx, and dx must he infinitesimal to cause this sum to approach its limit A. 4. Using the figure in § 210, show that dxdy n dz = NRdxdy dxdy dz ; (1) limit -v^-f-s r^'fi „ >, NRdxdy- AMB-dx- I NRdxdy; (2) = -^^o %/(! dy j™^^""4M2?.dx = 0Zr-X mit -^oj = 0-^0 AMB ■ dx. (3) In (1) dx and dy are constants, and dz may he either a constant or an infinitesimal. In (2) dx is a constant, but dy = 0. 218. Intrinsic equation of a curve. Let s denote the arc between a fixed point, Q, and a variable point, P, of the curve QP, and T the angle ABP included between the tangents at Q and P ; then the equation which expresses the relation between the variables s and t is called the intrinsic equa^ tion of the curve. Ex. 1. Eind the intrinsic equation of the circle. Let QP, or s, be an arc of a circle whose radius is r. Let C denote the centre of this circle ; then T = Z. ABP = Z. QCP = s/r. Hence, s = rr is the intrinsic equation of the circle. INTRINSIC EQUATIONS. 235 Ex. 2. Find the intrinsic equation of tlie catenary. In § 197 let OB = s; tlien t = Z. ^^B ; and tan t = dy/dx = s/a. § 197, (1) Hence, s = a tan t is tlie intrinsic equation of the catenary. Ex. 3. Find tlie intrinsic equation of the tractrix. In § 198 let AP = s ; then sec r = sec EPT= a/y. Hence, s = a log sec t § 198, (2) is the intrinsic equation of the tractrix. 219. To obtain the intrinsic equation of a curve from its rectangular or polar equation, we find the values of s and t and eliminate the other variables between these equations. Ex. Find the intrinsic equation of the cycloid. When s is reckoned from the cusp (§ 196, example 2), we have -2//V2r) and (1) s — ir{\ — Var - COST = dy/ds— Var — y/ v2r. . . s — ir(l — cost). When s is reckoned from the vertex, we have -V2r I {2r — y)-^'^dy = 'lr- V2r — y/V2r, and sinT — —dy/ds = V2r — y/^/2r. .-. s = 4 r sin t. (2) 220. If the intrinsic equation of the involute QP is s = fr, the intrinsic equation of the evolute QiPi is S = fr - f'O. (1) The curvature of QP is dr /ds. ■.R = ds/dr=fT. (2) s, = P,P-Q,Q ' §119 = f'r -f'O by (2) = Ax -f'O, (3) Ti = T. Omitting the subscripts in (3), we have (1). 230 INTEGRAL CALCULUS. Ex. 1. The evolute of the traotrix s — a log sec t is d(log seoT)T' s = a ^ , = o tan t, which, by example 2 in § 218, is the catenary. Ex. 2. The evolute of the cycloid s = 4 j- (1 — cos t) is , dll —cost)"!'" , s = ir -^ ; I = 4 r sin T, dr Jo •which, by the example in § 219, is an equal cycloid. EXAMPLES. 1. Find the evolute of the catenary s = a tan t. 2. Find the intrinsic equation of x'"^ + ?/-" = a^/' and of its evolute. Ans. s = (3a/2) sin^r ; s = {■ia/2) sin2T-. 3. Find the intrinsic equation of the logarithmic spiral p = bei"^ and of its evolute. When s is measured from the point (6, 0) where the spiral crosses the initial line, we have s = 6 Vl + a? (e9/« — 1). § 199, example 2 Since ^ is constant, t = 9. .-. s = 6Vl + a2(e'''>-l). CHAPTEE X. ORDINABY DIFFERENTIAL EQUATIONS. 221. A differential equation is an equation which involves one or more differentials or derivatives. An ordinary differential equation is one which involves only- one independent variable. For example, dy = cosxdx, (1) (J22//(Zx2 + 2/ = 0, (2) and y — X- dy/dx + rVl + (dy/dx)'-, (3) are ordinary differential equations. The order of a differential equation is the order of the highest differential or derivative which it contains. The degree of a differential equation is that of the highest power to which the highest differential or derivative which it contains is raised, after the equation is freed from fractions and radicals. Thus equation (1) is of the first order and first degree, (2) is of the second order and first degree, while (3) is of the first order and second degree. 222. The general solution of a differential equation is the most general equation free from differentials or derivatives, from which the former equation may be derived by differen- tiation. The general solution of equation (1) in § 221 is y — sm.x+ C, where C is the constant of integration. y = sin I, y — sinx + ?,••', are particular solutions of (1), which are included in its general solution. 238 INTEGRAL CALCULUS. The general solution may not include all possible solutions. A solu- tion not included in the general solution is called a singular solution. For a discussion of singular solutions the reader will consult some treatise on differential equations. The general solution of a differential equation of the wth order contains n arbitrary constants of integration. It is often called the cow,plete integral or jrrimitive of the differential equation. 223. In the foregoing chapters it was our object to obtain the general solution of differential equations of the form dy = (x) dx. In this chapter we shall extend the process of integration to differential equations of the more general form Mdx + Ndy = 0, (1) where Tlf and iVare functions of x and y. The variables in Mdx + Ndy are said to be separated when M, or the coefllcient of dx, contains x only, and iV contains y only. When Mdx + Ndy is the total differential of some function of x and y, it is called an exact differential, and (1) is called an exact differential equation. For example, xdy + ydx is the exact diffei-ential of xy ; hence, the general solution of the exact differential equation, xdy + ydx = 0, is xy = C. 224. Equations of the form <^i (x) dx + <^2 (y) dy = 0. (1) When, as in (1), the variables are separated, a differential equation is solved by integrating its terms separately. For example, the general solution of the differential equation e^dx + Zy^dy-fi, is e"' + 2/3 = C. VARIABLES SEPARATED. 239 When an equation is, or may be, written in the form <^i (x) ■ -i (y) dx + <^3 (x) ■ <^4 (y) dy = 0, the variables may be separated by dividing both members by Ex. 1. Solve (1 - x) d?/ - (1+ 2/) (fc = 0. (1) Dividing by (1 — k) (1 + y) to separate the variables, and integrating, we obtain log (\+y)+ log (1 - X) = log C. (2) . . (1 + y) (1 - X) = C. (3) Equations (2) and (3) are two equally correct ways of expressing the general solution of (1). Solution (3) could be obtained without separating the variables in (1) by noting that (1 — x) d?/ — (1 + i/) dx is the exact differential of (l-x)(\ + y). Ex. 2. Solve (x2 + 1) dy = (y^ + 1) dx. X + c Here tan— ' y — tan- ' x + tan- ' c = tan ■ ■y = 1 — ex X + c 1 — ex EXAMPLES. Solve each of the following differential equations : 1. dy _x^ + x + l dx y^ + y + 1 X? — y3 x^ — y^ . 3 2 -^ ' 2. y + l ^^ where E, R, and L are given constants. Find the value of C, having given that C = when t = 0. Ans. C = E{l~e-Xi/L) /jj. 225. Equations homogeneous in x and y. After being divided by x" (n being the degree of each term in x and y), any equation homogeneous in x and y can be put in the form dy=.f{y/3-)dx. (1) Putting y = vx in (1), we obtain V dx + xdv = f(v) dx. (2) The variables in (2) are easily separated ; hence, its solu- tion is found by § 224. Ex. 1. Solve (x2 -I- y^^)dx = 2xydy. (1) Putting 2/ = DX and dividing by x^, we obtain (1 +«2)(ix = 2J)(X(^^)^- udx). (2) HOMOGENEOUS EQUATIONS. 241 Separating the variables and integrating, we liave log[a;(l-!)2)] = logC. Putting y /x for i), the solution becomes a;2 - 2,2 = Cx. Ex. 2. Solve (x2 + xp) dy — xydx,. (1) Putting y — vx and dividing by x^, we obtain (1 + V-) {x dv + V dx) = V dx. (2) Separating tlie variables and integrating, we have logX + logC = 1)2/2 — logM, or Cy — e,^''^^. EXAMPLES. Solve each of the differential equations : 1. x''-dy — y-dx = xydx. logx + x/y = C. 2. {2V^-x)dy + ydx = 0. y = Ce~^^. 3. yHx + xHy — xy dy. y — C&i^. 4. xdy = (y + Vx^ + y-^)dx. x^ = C^ + 2 Cy. 5. (X + y)dy = (y- x) dx. log (x2 + ^2) + 2 tan-i (y/x) = C. 6. x^dy = yMx. y — x— Cxy. 7. (8?/ + 10x)dx + (52/+ 7x)d2/ = 0. {y + xY{y + 2xY - C. 8. Find the system of curves at any point of which, as (x, y), the sub- tangent is equal to the sum of x and y. Ans. y — C^'". 226. Non-homogeneous equations of the first degree in X and y. These equations are of the form {ax + by + c)dx + (a'x + b'y + c')dy = 0. (1) Putting x' + h for x and y' + k for y in (1), we obtain (ax' + by' + ah + bk + c) dx + (a'x' + b'y' + a'h + b'k + c') dy = 0. (2) 242 INTEGRAL CALCULUS. Giving to h and k the values determined by the system, ah + bk + e = 0, a'h + b'k + c' = 0, (3) equation (2) becomes {ax' + bij') dx + {a'x' + hhj) dij = 0, (4) which is homogeneous in x' and y', and can therefore be solved by the method of § 225. This method fails when a' / a = h' /b ; for then h and k in system (3) are infinite or indeterminate. In this case assume a' /a = b' /b = in, or a' = ma, b' = mb. Equation (1) then assumes the form (ax + bi/ + c) dx + [m (ax + by) + c'] dy = 0. (5) Let ax + hy = v; then dy = (dv — adx) /b. Substituting these values in (5), we obtain [b(v + c) — a (mv + c')] dx + (mv + c') dv = 0, where the variables are readily separated. EXAMPLES. 1. Solve (2x + 3y — 8)dx - {X + y — 3)dy = 0. (1) Putting x' + k for X, and y' + k for y, we obtain (2x' + Zy' + 2h + Zk — ?,)dx' = (x' + y' + h + k — Z) dy'. (2) Assume the system 2;i + 3 7i; — 8 = 0, ^ + fc — 3 = 0; or 7i = l, fc = 2. Equation (2) then becomes (2 X' + 3 y') dx' - (x' + y') dy'. (3) Putting y' = vx', (3) becomes (2 + "j v) dx' = (1 + u) {vdx' + x'dv), NON-HOMOGENEOUS EQUATIONS. 243 dxf B + 1 " U-l)2-3 + Vl(«-l-V3 ~ «-i+V3)J "^"^ .-. - logx' = \ log{(» - 1)2 - 3} + -^^log ""^" I- + C. ^ V 3 » — 1 + V3 where x' = x — 1 and d = ^^ X — 1 2. (3 2/ — 7 X + 7) dx + (7 2/ - 3 X + 3) d?/ = 0. vlns. (2/ - X + 1)2 (2/ + X - 1)6 = C. 3. (2x-|-2/ + l)dx + (4x + 22/-l)*y = 0. ^BS. X + 2 2/ + log (2 X + 2/ — 1) = C. 4. (22/ + X+ l)dx = (2x + 42/ + 3)Ay. ^?is. 4 X — 8 2/ = log (4x + 8 2/ + 5) + C. 5. (72/ + x + 2)cZx = (3x + 52/ + 6)d2/. vlns. x + 52/+2 = C(x — 2/ + 2)^. 227. Exact differential equations. The condition that Mdx + Ndy = (1) may he an exact differential equation is 8M/cly = aN/dx. (2) Comparing (1) with the exact differential equation du = —— dx + -;— dii = 0, (3) dx dy ^ ' we obtain M=Qu/dx, N=Qu/dy, (4) as the conditions that (1) be exact. From conditions (4) by differentiation, we obtain BM ■&U QN dy dydx dx or (2). 244 INTEGRAL CALCULUS. Condition (2) is called Euler's Criterion of Integrability. When condition (2) is satisfied, (1) may be solved by regard- ing y as constant and putting u=^Mdx+fy, (5) and then determining fy so that Qu/dy = N. (6) Or regarding x as constant, we may put ^c = JNdy+fx, and so determine fx that Qu/dx = M. Equations (5) and (6) involve the conditions in (4). Ex. Solve x(x + 2v)dx + (x^ — y^) dy = 0. (1) Here M=x(x + 2y), N = x^ — y^. (2) . . dM/dy = 2x- 3N/dx ; hence, condition (2) of § 227 is fulfilled. Eegarding y as constant, we put =/ x(x + 2y)dx +fy = x'>/3 + yx^+fy. (3) To determine fy, from (3) and (2) we have du/dy — x^ + fy = N = x^ — y^. ■■■fy = — y^, OT fy~ — 2/3/3. (4) Prom (3) and (4) we obtain u = x^/H + yx^ — y^/3. .: x^ + Zyx^ — y^= C is a solution of (I). INTEGRATING FACTOR. 245 EXAMPLES. Solve each of the following differential equations : 1. (6zy-y^)dx + (3x^-2xv)dy = 0. 3x^~y''x=G. 2. (x3 + 3 xy2) dx + (y^ + S x^y) dy = 0. x« + 6 xV + 2/* = C. 3. {x^ + y^)dx + 2xydy = 0. z^ + Sxy"^- C. 4. (x2 — 4 xy — 2 2/2) dx + (y^ — ixy — 2 x^) dy = 0. x" — 6x2j/ — 6x2/2 + ys= C. 5. (1 +2/Vx2)(Jx-(2?//x)d!/ = 0. x2-2/2=Cx. 6. xdx + ydy + "'''^7^/"' = 0. x^ + ^2 + 2 tan-i^ = C. x^ + y^ X In example 6 divide both terms of the fraction by x^. 7. e^(x2 + 2/2 + 2 X) dx + 2 ye^'dy = 0. e^ (x2 + 2/2) = C. 8. (2ax + by + g) dx + i2cy + bx + e) dy = 0. 228. . Integrating factor. When the equation, Mdx + Ndy = 0, is not exact, it may sometimes be made exact by multiplying it by a factor called an integrating factor. Sometimes an integrating factor may be found by inspec- tion, as in the examples below : Ex.1. Solve ^dx — xd?/ = 0. (1) Equation (1) is not exact, but when multiplied by y~^, it becomes ydx — xdy _ 2/2 which is exact, and which has for its solution x/y = G. (2) Multiplied by 1 /xy, (1) becomes dx/x — dy /y = 0, which is exact, and has for its solution log(x/2/)=logC. (3) Solution (2) is readily obtained from (3). Multiplying (1) by x— 2, we obtain the solution y/x = Ci. 246 INTEGRAL CALCULUS. Ex.2. Solve (1 + XT/) ydx + (l- xi/)xdy = 0. (1) From (1), ydx + xdy + xij^dx — xhjdy = 0, or d{xy) + xy'^dx — x^ydy = 0. (2) Dividing (2) by x^'^, we obtain d(xy) , ^ ' ■^ ) "^ °' ^^^ that is, equations of the nth order not containing y directly. ^ dy ,, cPy dp d"ii d''~^p Put V = -T-\ then -^ = ^, • ■ • — ^ = i- ■ -' dx' dx^ dx ' dx" dx"-^ Substituting these values in (c), we obtain which is an equation of the (n — l)th order between p and x. Ex. Solve d^y/dx'^ = a? + V' (dy/dx)K (1) Putting p = dy /dx, (1) becomes dp/dx = a2 + 62p2. . . tan-i (bp/a) — ah{x + Ci), or bp = a tan [a6 (x + Cj)]. .-. ft^d)/ — tan [a6 (x + Ci)] aftda;. .-. 622/ = log CSC [06 (X + Ci)] + C2. (d°v dv N j-^' ■ ■ •> ^' y ) = 0. (d) Put „_^. then^-z,^' ^ = „2^ + ./^^Y etc 254 INTEGRAL CALCULUS. Substituting these values in (d), we obtain an equation of the (n — l)th order between p and y. Ex. Solve dHj/dx?' + a{dy/dxY = 0. (1) Putting p = dy /dx, (1) becomes -^ + ap = 0, or - = — ady. dy ^ ' p .-. p= Cie-"", or e"Jdy = Cidx. .-. e"j = c'l ax + Co. EXAMPLES. Solve each of the following equations : 1. dhi = xeFdx^. y — x&: — Ze' + Cix^ + dx + C,. 2. d*y = x''-iU^. 3. X d^y = 2 dx^. y = x^ log x + CiX^ + c^x + Cs, where fi= Ci/2 -.3/2. 4. dhj = a^y dx'^. ax = log {y + V?/2 + a) + Ca, where a'^Ci = Ci. 5. 2/3 (Jij/ = f( (Jx2. Cl7/2 - ci2 (a; + C2)2 + d. 6. Va2/d% = d!x2. .Sx = 2 ai'< (j//^ _ 2 pj) (,/i/2 + cj)i/2 + g2_ 7. xd^y/dx^ + dy/dx = 0. y = ci\ogx + C2. 8. a2 (diy/dx'>- )2 = 1 + (dy/dxy. 2a-^y = cie^'" + Ci-'e-^/o + C2. 9. (1 + x-^) • dhj/dx"^ + 1 + (dy/dxY = 0. ?/ = Cix + (Ci2 + 1) log (ci — a;) + Ca. 10. (\—x'^)-d:hi/dx'^ — x-dy/dx = 2. y = Cisin-'x + (sin-i.r)- + C2. 11. y ■ d'hi / dx'^ + (dy / dxY ~ I. ?/ = j:'^ + CiX + Co. 12. The acceleration of a body moving toward a centre of attraction , (7, varies directly as its distance from that centre ; determine the velocity and the time. APPLICATIONS TO MECHANICS. 255 Let a = the acceleration at a unit's distance from C ; X = tlie varying, and c tlie initial, distance of the body from C; then xa = the acceleration at the distance x. Here s = c — x ; .-. v = ds/dt = — dx/dt ; (1) ..xa = d^/dt!' - — dH/dt\ (2) Since v — — dx/dt = when x — c,hy integrating (2) we have (dx/d<)2 = ac2-aa;2. .-. I) = — dx/dt — Va (c2 — a;2). (3) Since S = when x = c, from (3) we have i = a-i/2cos-i(x/c). (4) Putting X = in (3) and (4), we obtain D = cVa, the velocity at the centre of force, C. and t = (l/2)ffa-i/2, (3/2)7ra-i/2, (5/2)7ra-i'2, • ■ .. (6) Hence the motion is periodic, the time-period being jt/Va, which is entirely independent of the initial distance. The acceleration due to gravity at the earth's surface is 32.17 feet per second, and below the surface it varies as the distance from the centre. Hence, a particular case of the periodic motion considered above would be»that of a body which could pass freely through the earth. Such a body would vibrate through the centre from surface to surface. Calling the diameter of the earth 20919360 feet, we would have in this case = 32.17/20919360; .-. period = *a-i'2 = 3.1416 V20919360/32.17 sec. = 42 min. 13.4 sec. 13. Assuming that the acceleration of a falling body above the surface of the earth varies inversely as the square of its distance from the earth's centre, And the velocity and time. Let X = the varying, and c the initial, distance of the body from the earth's centre ; r = the radius of the earth ; g = the acceleration due to gravity at its sui-face ; a = the acceleration due to gravity at the distance x. Here 3 = c — x, and from the law of fall a : g = r^ : x^; or a: = gr^/x^. ..-d-^x/dl-^ = gr^/x^ (1) 256 INTEGRAL CALCULUS. Since v = when x = c, from (1) we have Since t = when x = c, from (2) we obtain 14. Assuming that r, the radius of the earth, is 3962 miles ; that the smi is 24,000 r distant from the earth ; and tliat the moon is 60 r distant ; find the time that it would take a body to fall from the moon to the earth, and the velocity, at the earth's surface, of a body falling from the sun. The attraction of the moon and the sun, and the resistance of any medium, are not to be considered. 15. A body falls in the air by the force of gravity, the resistance of the air varying as the square of the velocity ; determine the velocity on the hypothesis that the force of gravity is constant. Let 6 = the resistance when the velocity is unity ; and t = the time of falling through the distance s. Then 6 (ds/cU)'^ = the resistance of the air for any velocity ; and g = the acceleration downward due to gravity alone. Hence, g — h(da/dt)'^ = the actual acceleration downward ; that is, (Xh/dt^ = g — h (ds/dt)^. (1) Integrating (1) and solving for ds/dt, we obtain _ ds _ jg e'^tVbj; — 1 ^ dt \be^''^^ + l As t increases, v rapidly approaches the constant value 'Vg/b. 16. A body is projected with a velocity, Vo, into a medium which resists as the square of the velocity ; determine the velocity and the dis- tance after t seconds. Let 6 = the resistance of the medium when the velocity is unity ; then b (ds/dt)'^ = the resistance for any velocity. Hence, d2s/d<2= —6((te/dS)2. (1) Integrating (1) and solving for ds/dt, we obtain V = ds/dt — Vo/^'. (2) APPLICATIONS TO MECHANICS. 257 Integrating (2) and solving for s, we obtain s-log(bvot+ l)/6. The velocity decreases rapidly and ik when s = oo. 17. A body slides without.f riction down any curve, mn. The accelera- tion caused by gravity at any point, P, is g cos DPA, PA being a tangent. m ■ PD = dy. Find the velocity of the body. Let PA = ds ; then - .-. d^s/dt^ = g cos DP A = —gdy/ds. (1) Let yo be the ordinate of the start- ing point on the curve ; then u = when y — yo. Integrating (1), we obtain V = da/dt = ^:ig(yo — y)- From (2) it follows that if a body falls from the line y = yoto the line y = b, the velocity acquired is the same for all curves of descent. 18. A body falls from the point P along the arc of a cycloid, PO; find the time of descent. From (2) of example 17 we have V = ds/dt = V2g{yo — y). The equation of the cycloid referred to OX and OF Is (1) x = r vers-i (y/r) + V2n/- -y. Hence, ds = —^2r/ydy. (2) Eliminating ds between (1) and (2) and Integrating, we obtain t = VT/g [it — vers- 1 (2 V/Vo)]- .: t = aVr/g, when y = 0. Hence, if a pendulum swings i n th e arc of a cycloid, the time required for one oscillation is 2 TtVr/g. The time of an oscillation being independent of the length of the arc, the cycloidal pendulum is isochronal. APPENDIX. oXKo FOBMVLAS FOB SEFEBENCE. Standard Forms. wdu = — — -• 3. I a^du = . • n + 1 J log a 2. I — =logM. 4. I e"dw = e". J « J 5. Jsin... = -cos«,orvers.. 6. Jcos..u = sin.,or-covers„. 7. I sec^MciM = tanit. 11. j tan a du = log sec M. 8. I csc^Mdtt = — cotM. 12. I cotudM = log sin«. 9. j sec u tan udu = sec u. 13. j esc udu = log tan - ■ j cscw cotMdM= — csc«. 14. j seottdM = log tanf - + — y ,^/'(Jm 1. ,M 1,,M 15- i "IT"; — ;=— tan-i-! or cot-i-- J w + a^ a a a a 10 260 APPENDIX. n du 1, u — a 1, a — u 17. 1 =sm— 1-1 or — cos— !-• J Va2 — 142 a a 18. T— J^= = log (« + Vm2 ± a2). , _ /* (Jm 1 , M 1 ,U 19. I — =-sec— '-' or cso— i-- I u Vit2 (j2 a a a a = = vers—' - 1 or — covers— ' - • ■ V 2 au — m2 a a 22 23. Elementary Principles and Formulas. 21. I T FORMS INVOLVING a + bu^. 30 r_i?5__ = 1 _ 1 , g + fett J u(a +6it)2 a(a + 6«) a^ ^^ m Forms Involving a + hu?. /du 1 , [b ^+ 6u2 = Vl *^"" " Va' ''•''° a > and 5 > ; 261 32, 33. = — . log —p r= 1 when a > and 6 < 0. ■ J (a+6tt2)2-2a(a + 6u2) "'"2aJ a + ftM^' 35 /• t^M _ 1 M . 2?- — 1 /* CJM J (a + 6tt2)r + i^2ra (a + 6u2)'- 2m J (o + 6m2)'-' - r_i^dii_ _u _a P du ■ J a + 6m2 ^ 6 bj a + bu?' „_ /^__Jt^dM___ _ — M 1_ /* d " J {a + lnfiY + '^~ 2rb(a + lm?'Y 2rbJ (a + 30 I ■ J M (a + 6m2) 2 a '"* a + 6^2 6u2)'- dtt 1 , u^ + 6u2 39 f <^ = _i_of_ ^ m2 (a + hu^) an aj a ^. /• du _1 /^ da 6 Z' aw ■ J M^ ((J + 6„2)r + '~ aJ u^{a + bu^ a J (a + bu^y + 1 ' 262 APPENDIX. Forms Involving a + bw. 41. j M"'(a + 6M")J'(iu y^m-n + iia + bu")P + ^ aim — n + l)/^ , ,,,, 6(»ip + m + l) 6(np + m + l)^ ^ ' 42. or \ —r- H , , ■■ I tt"'(a + 6M")p-idM: np + m + 1 np + )7i + 1^ 43. or ^- 7-—^ ' -^ , — ' I «"'+''(« + 6M'')i'd«: a (m + 1) a (m + 1) ^ ^ ' ' M" + 1 (a + 6m»)p + imp + m + n. + l/' , ,, ^ ,,, 44. or ^"7 — t 1- -^ ; r--r I M™ (a + &U")P + 1 d«. Forms Involving au^ + 6m + c. .- n dii 2 ^ 2 aw + 6 45. I — r— — r — ; — = , ==rtan-t- J «**' + 6« + c Vi ac — 62 V4ac — 62' .- 1 2aM + 6— V62 — 4tic 46. = , = log , V 62 — 4 ac 2 au + 6 + v 62 — 4 ac "''■ /s;?ffT-c = 2^'°S(-^ + ^'' + ^)-fa/: att2 + 6u + c Forms Involving Va + 6u. 48. f„V^T^c^u = -^i^^^lMJ^±M!l^ ^ 15 62 49. Cu- V^ + ^ du = ' <^ ''^ - ^^ """ + ^^ '''^'^ (" + "'^^^^^ ■ J 105 63 ttd» _ 2 (2 tt — 6m) Vg + 6« . 50. r_jig^^_ /^ M"dM _ 2M"Va + 6M 2wa /♦ w^-^du ' J V^TTm' (2n+l)b (2^ + l)6j V^ + 6m rOEMS mVOLVING Vo^"^^. 263 E„ r ^ 1 , Va + 6m— Va , ^ ^ 52. I — . = -ji log ^ ) when a > ; ^ «va + 6it Va va+ &it + va 53. = tan-i-y/ ' whena<0. V^ \ -a ^ J «"Vo + 6it~ (7i-l)au''-i (2n-2)aJ u-'-iVa + bit + bu Forms Involving Va^ — u^. 56. I , : = sm— 1- • , = sin— 1 - • V a2 — u2 a __ /* dM 1 , tt 57. I — , --log = J It Va2 — ^2 o a + Vci2 — , du _ Va2 _ y2 _ 58. f- .2Va2 - u2 59. CVa^ -u^du = '^ Va2 - m2 + 60. — sm-i-- 2 a a* 8 Va2- -w2 + sir a + vV — lt2 log It -sin- ■1- . a 61. r ^°' '^'' du = VS^"=ir2 _ a J u -„ rVa2 — m2 , Vo2 - m2 62. I ;; du = J V? u .„ /• M2(fy u ^— a? , u 63. I , = — - V a2 _ „2 + _ sm-i - • J Va2 - m2 2 2 a 64 r ^^^ = " J (a2-M2)3/2 a2Va2-;i2 264 APPENDIX. 65. f(a^ - u2)3/2 dM = I (5 a2 - 2 u^) Va^ - u^ + ^ sin-i - • P o ad : — sin— 1 - Forms Involving Vm^ ± a^. 67. /^ (2u = log (u + Vm2 ± a2). 68. I — , = - sec-1 - 69. ( ; = - log u 70. mVu^ + cfi "■ ° a + Vm,2 + d^ da Vm2 ± a2 a-'M J u2Vtt2±a2 J M3Vjt3_a2~ 2aV^ 2a-5 / 72, (Ju 73. r Vit2 ± a? du = " Vu2 ± a2 -t- ^ log (u + V^J±~^). 74. Tu^ Vii2 ± a2 d« = ^ (2 m2 ± a^) VitS ± o2 - 1- log (a + Vu^ ± a2). 75. /Vm2 - a2 , /-T ; , a du — vm2 _ (j2 — o cos-1 - • u u „„ rVM2 + a2 ^— - — - 76. I du — Vm2 + a2 — a J u log a + Vu2 + a2 'VM2±a2 , Vm2 ± (,2 77. I :; du = h log (u + vm2 ± a^). FORMS INVOLVING V2o«-u2. 265 78. r-^^ = |'\4^±T2 3:^log(tt + V^;iT^). 79. Tt^ '^^ - *" 80. /u^dM — M , , , , rir~, — s^ (u2±a2)3/2 Vll2 ± a2 81. C{U^ ± o2)3/2du = ^(2«2 ±5a2)VM2±a2 + 5|^ log(it + Vu2 ± a^). 82. / Forms Involving V2 au — u^. du , u v^ au — u' f W'du _ _ u"'— 'V2aM — m2 (2 m — 1) ct /• * u™-^du J V'2 a« - m2 to "1 J V2 au — u^ J M'"V2 au — 1*2 _ _ V2 gu — m2 m — 1 /• du (2 »i — !)««"' (2 TO — 1) a I ifn— iV2^ 85. r V2 ou - u2 (Zu = ^^-—^ V2au-u^ + ^ sin- 1 I 2 2 a 86. r*M"'V2au — ^2^^ / V2 au — u2 du M"— '(2au — u2)3/2 , (2m + l)a /» , /- 5, = ^ ; h- 1-77— I u"'-W2au — u^du. TO + 2 m + 2 87. f M" (2aM — ugp^2 in —3 r ^2au — v?du _ '(2to — 3)au"' (2 TO — 3) a J m'»-i 266 APPENDIX. Forms Involving V± au^ + bu + c, -where a > 0. 88. C ^^ = -^ ^"S (^ "" + ft + i^ V"- V"M^ + ft"- + (■)■ J Vmj? + 6u + c V a 89. C^av? + bu + cciu 4a 8a J Vaw^ + bu + c f du 1 . , 2 au — 6 90. 1 , = -^sin-i . J V — aifi + bu + c Va V 62 + iac 91. I V— (iit2 + 6it + c du 2au — b I j-r-; — ; — , 6^ + 4 ac ; ^ V — avk'' + 6u + c H ;; 92. / - 4ae n dw 4a ' "'"■-"•■■'■ 8a J V- au^ + 6u + c V± au2 + 6u + c V ± (im2 + bu + c 6 /• (Jm ± a "^ 2 a J V± avi?' + bu + c 93. rj(V±au2 + 6u + cdu = ^ '- — 3: ^ I v± av} + fiii + can. Za ^2 a J Forms Involving Transcendental Functions. /w 1 sin^ttdti = - — 7 sin 2 m. 2 4 /u 1 cos2 tt da = - + - sin 2 M. 96. j sin^M cos^MoJu = -(m — -sin4« J • 97. I seoitoscMdM= I -: = logtan«. J J sw-u 00s u - COtM. TRANSCENDENTAL DIFFERENTIALS. 267 98. I sec^tt csc^itiJu = i -r-^ r— = tanM — J J sm^ u cos^ u 99. I sin™ u cos" u if Sinm-ly cogn + l^ m— 1 1 . 1 1 ; — I sm"»-2u cos"Md«; m + n m + n ,„„ siii™ + iu cos^-'tt , n, — 1 /- . „ , 100. = — 1 1 — I sm^M cos''-2Mdu. , 1 ; — i s m + n m + nj 101. I sm"'MdM = 1 i smp^-^udu. J m m J -inn /* „ J sinMcos"-iM , n — 1 /» , , 102. I coS^udu = 1 I cos^-^udu. J " ^ J 103. i du = — : 1 ; i J cos^u (n — l)cos"— !« n — 1 J cosf-^u ,.. /•cos"M, cos" + iu , m — n — 2 fco&^udu 104. I du = —-, TT-. -, 1 ; — I -. TT-- J sm™M ()7i — 1) sin™— 1m m — 1 J sin™— ^m J. _ /* du cosu m — 2 /* du J sin™M (m — l)sin™— 'm m — \J sin™— ^j^ lOe C-^HL- = sinw ^ TO — 2 /• J cos"M (n — l)cos"— ^M n — 1 J c I — 1 I cos"— ^M tan"udM = ^Tj I tan"-2udit. /cot** — ^ M /^ cofudM = — j j C0t»-2Md«. 109. f— f^=-^J=tan-i(J^tan^),ifa^>6^; J a + 6 cos M Va2 _ 52 V\o + 6 2/ ^ , Vft — o tan - + V6 + a 110. = , log = ,ifo2<62. idu. 268 APPENDIX. 111. j u"'sinudu = — M^cosu + m j «"'-icosu 112. j Mi«costtdM = if^sinM — m j u^-ioosudtt. ,,„ /•sina , ifS , m5 «' m" 113. J^r'^" = "-37^ + 5T^~r^ + 94l'"' 114. r?i^du= L^^ + ^C^au. /'cos u , , u'^ , u'^ u^ , u^ rcoau , 1 cos It 1 /'sinu , 116. I du= r I rdu. I W" m — lu"-! m— 1 f M"'-i 117. j M sin-i udu = t [{2 »- — 1) sin-' u + mVi — m^]. 118. I xt" sin- ludu = r— I , J 71 + 1 n + lj Vl -u2 ,„ /• , , W's + l COS-I tt , 1 /•«» + !(?« 119. I tt"cos— iMatt = r~-, 1 r~; I , J n + 1 «+ IJ Vl — m2 ,„„ /» , , M« + itan-i« 1 /•u" + idtt 120. 1 M" tan-l uau = r-r r-r I ^r— ; — 5- • f n + 1 n + l^l + «2 121. J«" log«du = „»+ ' [i^ - ^^] ■ a aj 123. I — du — • r -i r I : du. J -UP- K— 1 M"-l 1—1 J ""^"^ ,„./', , eoi'loeu 1 /»en« 124. I e''"logMd« = I — du. TRANSCENDEXTAL DIFFERENTIALS. 269 e"" (g sin nu — n cos nu) _ 125. I e»" sin nudu— , , ., a^ + n^ j e"«si e°" (g cos nu + n sin mu) /""' 126. I e»"cosnudu= „ , „ a^ + n" Miscellaneous Forms. '■Urn 127. |A/?^(i" = V(a + M)(6 + M) + (a - 6) log ( Va + u + Vb + u). To prove formula 127 let 6 + u = z\ 128. fyjf^ du = V(a - «)(6 + u) + (a + 6) sin-^^~±^ ■ 129. I — =9p.nt.-i-. / = -2sm-^\h J V(u-a)(6-w) \M-a \b-a ,„^ r du 1 , V«'' + M" — g 130. I — , = — log = J u^w + a^ "" vg^ + an -I- a 131. f /!^ = SHORT COURSE IN THE CALCULUS. To those who wish to give in Taylor's Calculus a short course including the fundamental principles, problems, methods, and applications of the Calculus, the following suggestions may prove helpful : PAET I. Chapter I. Take it all thoroughly. Chapter II. Omit exs. 28-34, pp. 18, 19 ; exs. 6, 8, 12, 13, 14, 25-30, pp. 19, 20 ; exs. 8-10, pp. 21, 22 ; exs. 19-23, pp. 28, 29 ; exs. 23-29, p. 32 ; exs. 18-23, pp. 35, 36 ; exs. 1-30, pp. 38, 39. Chapter III. Omit all after p. 44 except § 74. Chapter IV. Omit all after ex. 15, p. 60, except the defini- tion in § 82. Chapter V. Omit exs. 9-14, p. 67 ; exs. 8-11, p. 68 ; all the chapter after ex. 9, p. 69. Chapter VI. Read the proof of Taylor's theorem with the class, but do not require its reproduction. Omit Cors. 2^, p. 74; §§ 93, 96, 99; the proofs of convergency in §§ 94, 95, 97, 98 ; exs. 11-20, p. 81. Chapter VII. Omit exs. 12-19 and 22, p. 87 ; exs. 14-16, p. 90 ; exs. 19-24, pp. 91, 92. Chapter VIII. Omit § 110 ; exs. 8-11, p. 97 ; § 116 and examples ; exs. 7-9, p. 102. Chapter IX. Omit exs. 8-13, pp. 106, 107 ; exs. 1-8, p. 111. Chapter X. Omit it all. 270 SHORT COURSE IN THE CALCULUS 271 Chapter XI. Omit pp. 118-126. Chapter XII. Omit exs. 7-9, p. 130 ; exs. 9-12, p. 132 ; § 149 and examples ; § 153 and examples. Read curve tracing with the class so that the most important curves are made familiar. PART II. Chapter I. Omit exs. 39-48, p. 160 ; exs. 57-59, p. 151 ; exs. 22-24, p. 153 ; exs. 19-24, pp. 155-156. Chapter II. Omit exs. 4-6, p. 160 ; § 168, p. 166. Chapter III. Omit exs. 7-9, p. 169 ; exs. 6-8, p. 170 ; exs. 8-12, p. 172 ; exs. 3-6, p. 173. Chapter IV. Omit exs. 6, 10, p. 175 ; exs. 7-9, p. 177 ; exs. 8-11, p. 180. Chapter V. Omit exs. 17-26, p. 184; exs. 10, 14, 19-21, 25-31, pp. 188, 189. Chapter VI. Omit exs. 11-15, p. 192 ; exs. 10-13, p. 194 ; exs. 5-7, p. 195 ; §§ 190, 191, and examples, pp. 196, 197 ; § 192, and examples, pp. 198, 199 ; exs. 5-9, p. 200 ; § 194 and examples, p. 201 ; exs. 7-9, p. 202. Chapter VII. Omit exs. 5, 7, p. 204 ; § 198 ; exs. 3, 4, p. 207 ; § 200 and examples ; exs. 7-13, p. 210 ; exs. 6-9, p. 211 ; exs. 5-8 and 13-17, pp. 214, 216 ; § 205 and examples. Chapter VIII. Omit all after ex. 3, p. 221. To gain an idea of limits and infinitesimals as used in the Calculus, read Chapter III in Part I and Chapter IX in Part II. If a course still shorter than that outlined above is desired, omit all of Chapters VII-XII in Part I, all of Chapter VIII in Part II, and more examples in the other chapters.