BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF Henrg W. Sage 1891 ..l^^jsix^-.Q, 3.uci..m\- °mnm»mLm!iJ!}*^^^' calculus for l 3 1924 031 219 318 The original of tliis 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/cu31924031219318 DIFFERENTIAL AND INTEGRAL CALCULUS ■s? :^^^ DIFFERENTIAL AND INTEGRAL CALCULUS FOR TECHNICAL SCHOOLS AND COLLEGES BY P. A. LAMBERT, M.A. ASSISTANT PROFESSOK OF MATHEMATICS, LEHIGH UNIVERSITY Wetoiotk ■ THE MACMILLAN COMPANY LONDON : MACMILLAN & CO., Ltd. 1898 All rights reserved ( COPTEIGHT, 1898, ^UJ-CtJ By the MACMILLAN COMPANY. J. S. CuBhing & Co. — Berwick & Smith Norwood Mass. U.S.A. PREFACE This text-book on the Differential and Integral Calcu- lus is intended for students who have a working know- ledge of Elementary Geometry, Algebra, Trigonometry, and Analytic Geometry. The object of the text-book is threefold : By a logical presentation of principles to inspire confi- dence in the methods of infinitesimal analysis. By numerous problems to aid in acquiring facility in applying these methods. By applications to problems in Physics and Engineer- ing, and other branches of Mathematics, to show the prac- tical value of the Calculus. The division of the subject-matter according to classes of functions, makes it possible to introduce these applica- tions from the start, and thereby arouse the interest of the student. The simultaneous treatment of differentiation and inte- gration, and the use of trigonometric substitution to simplify integration, economize the time and effort of the student. P. A. LAMBERT. TABLE OF CONTENTS CHAPTER I On rnNCTiONS ARTICLE PAGE 1. Definition of a Function 1 2. The Indefinitely Large and the Indefinitely Small ... 2 3. Limits 3 4. Corresponding Differences of Function and Variable . 5 5. Classification of Functions 6 CHAPTER II The Limit or the Ratio and the Limit of the SnM 6. Direction of a Curve 9 7. Velocity 12 8. Rate of Change . . 14 9. The Limit of the Sum .15 10. General Theory of Limits 17 11. Continuity . . 18 CHAPTER III Differentiation and Integration of Algebraic Fdnctions 12. Differentiation 21 13. Integration 29 14. Definite Integrals 34 15. Evaluation of the Limit of the Sum 36 16. Infinitesimals and Differentials . .... 37 CHAPTER IV Applications of Algebraic Differentiation and Integration 17. Tangents and Normals 43 18. Length of a Plane Curve 45 vii viii ' CONTENTS ARTICLE PAGE 19. Area of a Plane Surface 48 20. Area of a Surface of Revolution 50 21. Volume of a Solid of Revolution 52 22. Solids generated by the Motion of a Plane Figure ... 53 CHAPTER V ScccESSivE Algebraic DirFERENTiATioN and Integration 2.3. Tlie Second Derivative . . 57 24. Maxima and Minima 63 25. Derivatives of Higlier Orders 74 26. Evaluation of the Indeterminate Form - 75 CHAPTER VI Partial Differentiation and Integration of Algebraic Functions 27. Partial Differentiation 78 28. Partial Integration 82 29. Differentiation of Implicit Functions 83 30. Successive Partial Differentiation and Integration ... 86 31. Area of Any Curved Surface 89 32. Volume of Any Solid 91 33. Total Differentials . . 92 34. Differentiation of Indirect Functions 96 35. Envelopes 97 CHAPTER VII ClRCULAE AND INVERSE CIRCULAR FdKCTIONS 30. Differentiation of Circular Functions . . . 100 37. Evaluation of the Forms oo • 0, oo — oo, ^ . . . 105 38. Integration of Circular Functions ...... 107 39. Integration by Trigonometric Substitution .... Ill 40. Polar Curves 113 41. Volume of a Solid by Polar Space Coordinates . . . 117 42. Differentiation of Inverse Circular Functions . . 118 43. Integration by Inverse Circular Functions .... 120 44. Radius of Curvature 123 CONTENTS IX CHAPTER VIII LOGAHITHMIC AND EXPONENTIAL FUNCTIONS ARTICLE PAGE 45. The Limit of ( 1 + - j wlien Limit 2 = oo 128 46. 47. 48. 49. 50. 51. Differentiation of Logarithmic Functions .... Integration by Logarithmic Functions .... Integration by Partial Fractions Integration by Parts ... . . Integration by Rationalization .... Evaluation of Forms 1"", 00", 0° 131 1.33 135 139 142 144 52. 53. 54. Differentiation of Exponential Functions . Integration of Exponential Functions .... The Hyperbolic Functions 145 147 148 55. Tlie Definite Integral ^"'""e-^^ dx 150 56. 57. Differentiation of a Definite Integral Mean Value 151 152 CHAPTER IX Center of Mass and Moment of Inertia 58. Center of Mass 154 59. Center of Mass of Lines 155 60. Center of Mass of Surfaces 157 61. Center of Mass of Solids . . . ' 161 62. Theorems of Pappus 162 63. Moment of Inertia 165 04. Moment of Inertia of Lines and Surfaces 167 65. Moment of Inertia of Solids 170 CHAPTER X Expansions 66. Convergent Power Series 173 67. Taylor's and Maolaurin's Series 175 68. Euler's Formulas for Sine and Cosine 184 69. Differentiation and Integration of Power Series . . . 186 70. Expansion of ais/(x + ft, J/ + A;) 192 CONTENTS CHAPTER XI Applications of Taylok's Series ARTICLE PAOB 71. Maxima and Minima by Expansion 194 72. Contact o£ Plane Curves 198 73. Singular Points of Plane Curves 201 CHAPTER XII Oedinary Differential Equations of First Order 74. Eormation of Differential Equations 205 75. Solution of Eirst Order Differential Equations of First Degree . 209 76. Equations of First Order and Higlier Degrees .... 218 77. Ordinary Equations in Three Variables 222 CHAPTER XIII Ordinary Differential Equations of Higher Order 78. Equations of Higher Order and First Degree .... 225 79. Symbolic Integration 230 80. Symbolic Solution of Linear Equations 234 81. Systems of Simultaneous Differential Equations . . . 237 CHAPTER XIV Partial Differential Equations 82. Formation of Partial Differential Equations .... 239 83. Partial Difierential Equations of First Order .... 240 84. Linear Equations of Higher Order 244 DIFFERENTIAL AND INTEGRAL CALCULUS* CHAPTER I ON PUNOTIONS Aet. 1. — Definition of a Function A constant is a quantity which retains the same value throughout a discussion. A variable is a quantity which has different successive values in the same discussion. If two variables are so related that to every value of the first there correspond one or more determinate values of the second, the second is called a function of the first. Denote the variables by a; and y, and let the relation between them be expressed by the equation y = mx + n, where m and n are constants. If in a particular discussion m = 3 and n = 5, the equation becomes y = 3x + 5. Arbitrary values may be assigned to x, and the corresponding values of y calcu- lated. If in another discussion .m = — 2 and n = 6, the equa- * The Differential and Integral Calculus was invented independently by Newton and Leibnitz, Newton antedating Leibnitz by several years. Leibnitz published his method in the "Acta Eruditorum" of Leipzig, in 1684 ; Newton published his method in his "Natural Philosophy," in 1687. B 1 2 DIFFERENTIAL AND INTEGRAL CALCULUS tion becomes y = — 2x + 6. Again, arbitrary values may be assigned to x, and the corresponding values of y calculated. The variable x, to which arbitrary values are assigned, is called the independent variable. The variable y, whose value depends on x, is called the dependent variable or function. In the equation y = 3x + 5,ii x increases, y increases ; if x decreases, y decreases. This fact is expressed by calling y an increasing function of x. In the equation y = — ^— -, if x increases, y decreases ; if x decreases, y increases. This fact is expressed by calling y a decreasing function of x. Aet. 2. — The Indefinitely Large and Indefinitely Small The term of the geometric progression 1, 2, 2^, 2', 2*, 2", ••• continually increases, and becomes larger than any number that can be assigned when the progression is extended suffi- ciently far. The term of the geometric progression 1, |, i, i, i, 1,..: continually decreases and becomes smaller than any number that can be assigned when the progression is extended sufB.- ciently far. A variable quantity whose numerical value continually increases and becomes larger than any quantity that can be assigned is said to become indefinitely large. A variable quantity whose numerical value continually decreases and becomes smaller than any quantity that can be assigned, is said to become indefinitely small. 3 If y = ~, and X is positive and becomes indefinitely large, the corresponding value of y is positive, and becomes ON FUNCTIONS 3 indefinitely small. If x is negative, and starting from zero continually approaches — 6 in such, a manner that the differ- ence between the value assigned to x and — 5 becomes indefi- nitely small, the corresponding value of y is positive and becomes indefinitely large. If x continues to decrease beyond — 5,y becomes negative and its numerical value decreases, becoming indefinitely small when the numerical value of x becomes indefinitely large. All quantities which lie between those which are indefi- nitely small and those which are indefinitely large are called finite. If £ denotes an indefinitely small quantity, and n is finite, the product n-t must also be indefinitely small. Tor, if n-€ = m, and m is finite, e = — , the ratio of two finite quanti- n ties. Hence e would be a quantity whose value can be as- signed, which is contrary to the hypothesis. Since y(x + 5)=3 is always true ii y = -, and when x becomes indefinitely large, y becomes indefinitely small, it fol- lows that the product of an indefinitely small quantity and an indefinitely large quantity may be finite. The sum of a finite number of indefinitely small quantities is indefinitely small. For if e is the largest of the n indefi- nitely small quantities C], e^ ^3> ^ts "■) 'n) their sum ei + ea + es + ^iH H «„ cannot be greater than n ■ e, which is indefinitely small when n is finite. Akt. 3. — Limits If one quantity continually approaches a second quantity in such a manner that the difference between the two becomes 4 DIFFERENTIAL AND INTEGRAL CALCULUS indefinitely small, the second quantity is called tlie limit of the first. For example, if y = 5 + - when x becomes indefinitely large, 3 ^ - becomes indefinitely small, and by the definition 5 is the X limit of y. The limit of a quantity which becomes indefinitely small is zero. The limit of a quantity which becomes indefinitely large is called infinity, and is denoted by the symbol oo. This conception of a limit is used in elementary geometry when the circumference of a circle is proved to be the limit of the perimeter of the inscribed regular polygon when the num- ber of sides becomes indefinitely large ; when the area of a circle is proved to be the limit of the area of the inscribed regular polygon when the number of sides becomes indefinitely large ; when the volume of a triangular pyramid is proved to be the limit of the sum of the volumes of inscribed triangular prisms of equal altitude and with bases parallel to the base of the pyramid when the number of prisms becomes indefi- nitely large. In elementary algebra, the sum of n terms of the geometric progression a, a-r, a-r^, a-r^, a-r*, •■■ is proved to be a — a-r" a a- r' s„ = - 1 — r 1 —r 1 — r If r is numerically less than unity and a is finite, ^'^ be- 1 — r comes indefinitely small when n becomes indefinitely large. Hence when n becomes indefinitely large, the limit of s„ is a ON FUNCTIONS 5 If r is positive, the successive values of s„ as n increases all lie on the same side of the limit ; if r is negative, the succes- sive values of s„ oscillate from one side of the limit to the other. Por example, the limit of the sum of an indefinitely- large number of terms of the geometric progression 1, — -|-, ^, 11 1 1 1 1 1 _L ... ^o 2 TllO flTof ten successive approximations are Sj = 1, S2 = ^, Sg = f , 84 = |, *5 = To! *6 == ¥?) ®7 ~ ¥¥' ®8 = ^F) ^9 = m> *io = jjj- The approximation s„ is larger than the limit when n is odd, smaller than the limit when n is even. x' — 1 If y = —, y has a determinate value for every value of x — 1 X, except for x=l. When x = l, y takes the indeterminate form -, which may have any value whatever. The true value of y when a; = 1 is defined as the limit of the values of y cor- responding to values of x whose limit is 1. Eor example, when a; = 1.1, 1.01, 1.001, 1.0001, 1.00001, 1.000001, •••, 2/ = 2.1, 2.01, 2.001, 2.0001, 2.00001, 2.000001, -. Hence 2 is the true value of y when a; = 1, and for all values a^ — 1 of X, including x=l, y = = x + l. x—1 Art. 4. — Coeebsponding Diffeebnces of Function AND Vaeiable If 2/ = a;^, when a; = - 4, -3, -2, -1, 0, 1, 2, 3, 4, 2/= 16, 9, 4, 1, 0, 1, 4, 9, 16. Starting from a; = 0, a difference of +1 in the value of x causes a difference of +1 in the value of y ; starting from 6 DIFFERENTIAL AND INTEGRAL CALCULUS x = + l, a difference of + 1 in the value of x causes a differ- ence of + 3 in the value of y ; starting from a; = — 3, a dif- ference of + 1 in the value of a; causes a difference of — 5 in the value of y. Observe that the same change in the value of the variable in different parts of the function causes different changes in the value of the function. In general, if y = a^, and starting from any value of x the corresponding differences of y and x are denoted by Ay and Aa;, so that x-\- Ax and y + Ay must satisfy the equation y = ixF, by subtracting the equations y + Ay = (x + Axy and y = ay', there results Ay = 2x-Ax+ (Axy. The difference in the value of y corresponding to a difference of Ax in the value of X is seen to depend on x and on Aa;. Akt. 5. — Classification or Functions A function is called algebraic if the relation between func- tion and variable can be expressed by means of a finite number of the fundamental operations of algebra, addition, subtrac- tion, multiplication, division, and involution and evolution with constant indices. For example, y =3? — 7x + 7 explicitly defines y as an integral, rational, one-valued, algebraic function of x. The equation a^ + y^^Q implicitly defines y as a two-valued alge- braic function of x. The relation y = defines y as a 1 + a;* fractional, irrational, one-valued function of x. All functions not algebraic are called transcendental. The expression of transcendental functions by means of the fundar mental operations of algebra is possible only in the form of the sum of an indefinitely large number of terms, or in the form of the product of an indefinitely large number of factors. ON FU2SICTI0N8 7 The elementary transcendeiital functions are : The exponential function y=a' and its inverse the loga- rithmic function x — log^y. The trigonometric or circular functions y = sin x, y = tan x, y = sec X, together with their complementary functions ; and the inverse functions x = sin~^y, x=ta.n~^y, x=sec~^y. In general, the fact that y is an explicit function of x, with- out specifying the nature of the function, is denoted by writing y=f(x), or y = F(x), or y = ^(x); the fact that y is an im- plicit function of x is denoted by writing /(«, y) = 0, or F(x,y)=0, or (x,y)=0. PROBLEMS 1 . Determine the values of the function y = 3ai'— 5 cor- responding to a; = 0, 1, 2, 3, 4, 5. a^ 4 2. Find the true value of y = — when x = 2. x — 2 a? — 1 3. Find the true value of « = — when x = l. x — 1 4. Starting from x = 3, calculate the difference in y cor- responding to a difference of 2 in the value of x if y = !)f-5x + 12. 5. Starting from a; = 2, calculate the difference in y corre- sponding to a difference of — 2 in the value of a; if i/ = 7 a; — 3 a^. 6. Find the limit of the sum of an indefinitely large num- ber of terms of the series 1 4- i -I- 1 + tt + tt H ■ 7. Find the limit of the sum of an indefinitely large num- ber of terms of the series ^-^ + ^h + tsts + loooo + ■•■• 8 DIFFERENTIAL AND INTEGRAL CALCULUS 8. Show that any difference in the abscissa of the straight line whose'equation is y = mx + n causes a difference m times as large in the ordinate. 9. Show that the ordinate of the straight line whose equa- tion is y — 2/o = m (a; — x^) changes m times as fast as the abscissa. 10. li y = 30^ — 7 X, determine the change Ay in the value of y corresponding to a change of Aa; in the value of x. 11. Compare the values of Ay corresponding to Aa!=l, starting from a; = 0, 1, 2, 3, if 2/ = a^ - 3 a; + 10. 12. If 2/ = ax' + bx + c, where a, b, and c are constants and when a; = a;o, y = y„, determine the change Ay in the value of y corresponding to a change of Aa; in the value of x, starting from X = Xff. CHAPTER II THE LIMIT OP THE EATIO AND THE LIMIT OP THE SUM Art. 6. — Dibection of a Cubve Let {x„,yo) be any point of the curve whose equation is y = x'. Let (a;„ + \x, y^ + Ay) be any other point of the curve, Aa; and At/ representing corresponding differences in abscissa and ordinate. The ratio — ^ is the slope of the secant line Ax through {xa, y^ and {x^ + Aa;, 2/0 + ^y), that is, the tangent of the angle of inclination of the secant to the X-axis. The equa- tion of the secant is y- A?// -x„). rio.i. This is true whatever may be the magnitude of the corre- sponding differences Ax and Ay. Now the tangent to a curve is defined as the limiting ,posi- tion of the secant whose two points of intersection are made to continually approach each other. If the point {Xf, + Ax, 2/0 + ^y) continually approaches the point (xq, y^, the limit of Aaj is zero, and the corresponding limit of the ratio — ^ 9 10 DIFFERENTIAL AND INTEGRAL CALCULUS is the slope of the tangent to the curve at («„, y„). This limit of the ratio is to be determined. By hypothesis, 2/0 + ^2/ = {^0 + Aa;)^, 2/0 = ^o^ i whence Ax Ax The limit of this ratio, when the limit of Ax is zero, cannot be found by placing Aa; = in this value of the ratio. Eor this makes the ratio take the indeterminate form - as it ought to, for the two points are made coincident, and through one point an infinite number of straight lines may be drawn. By performing the operations, indicated in the numerator of the value of the ratio — ^ and then dividing out the factor Ao; com- Ax ^ men to numerator and denominator, there results --^ = 2a!o + A% Aa; If now Aa; becomes indefinitely small, that is, if the limit of Aa; is zero, the limit of the ratio — ^ is 2 Xq. Ax Denoting by a the angle of inejination to the X-axis of the tangent to y = siy' at* (a;„, j^o). tana = 2a;o and the equation of the tangent is y — yo = 2xQ(x — Xo). At the point (2,4), tan a = 4 and a = 75° 58'. The tangent makes an angle of 45° with the X-axis if 2a;o = tan45° = l. Solving the equations 2 a;„ = 1 and 2/0 = a^o^ the point of tangency is found to be ^0 = "Z"? 2/0 == T' If the curve whose equation is y =a? is generated by the continuous motion of a point, when the generating point passes the point (a;„, ?/„) of the curve it tends at that instant to move along the tangent y — yo=2xo(x — x„) at the point (xo,yo)- THE LIMIT OF THE BATIO 11 Hence the direction of the tangent to the curve at any point is called the direction of the curve at that point. If a point moves along the straight line y~ya = 2xo{x — x^, the ordinate changes 2 aio times as fast as the abscissa. When 2 Wo is positive, the ordinate is an increasing function of the abscissa; when 2a;o is negative, the ordinate is a decreasing function of the abscissa; when 2x0 = 0, the line is parallel to the X-axis and a change in the abscissa causes no change in the ordinate. Hence, when the point generating the curve y^a? passes the point (ajo, y^, the ordinate of the curve is at that instant changing value 2a;o times as fast as the abscissa changes. At the point (|-, ■^) ordinate and abscissa are changing value at the same rate ; at the point (2, 4) the ordinate is increasing 4 times as fast as the abscissa increases ; at the point (— 2, 4) the ordinate decreases 4 times as fast as the abscissa increases. By precisely the same analysis it is proved that the slope of the tangent to the curve whose equation is y =/(») at any point (x, y) of the curve is tan a = limit — ^ = limit -^ — -^ — "^-^j ^ "'^ Ax Ax when the limit of Ax is zero, and that the limit of this ratio measures the rate of change of ordinate and abscissa at (x, y). It is essential to remember that in the calculation of this limit Ax must start from some finite value and then be made to approach the limit zero.* PROBLEMS 1. Find the slope of the tangent to y = a;^ — 3 a; at a; = 5. 2. Find the direction in which the point generating the graph of y = 3a? — x tends to move, when a; = 1. *The tangent problem prepared the way for the invention of the Differential Calculus, in the seventeenth century. 12 DIFFERENTIAL AND INTEGRAL CALCULUS 3. Find the rate of change of ordinate and abscissa of- y = 3a^ — X at x = l. 4. Find at what point of the curve whose equation is y — ix^ the tangent makes with the X-axis an angle of 45°. 5. Find the equation of the tangent to y = 2si!' — 5x at x=3. 6. Find where the ordinate of y = 3x — ia^ decreases 6 times as fast as x increases. Aet. 7. — Velocity Suppose a locomotive to start at station A, to pass station B distant So miles from A after tf, hours, and station G distant s miles from A after t hours. The average velocity per hour -So r c .-S — J to tl Fio. i. from B to G, that is, the uniform number of miles per hour the locomotive must run from .B to O to cover the distance s — So miles in t —to hours, is ■ ~ ° . Calling the difference of t — f distance As and the difference of time At, the average velocity is — . The equal ratios — = ~ " determine the average At At t — to velocity of the locomotive during the interval of time At = t — to, whatever may be the magnitude of this interval of time. Now if station G is taken nearer and nearer station B, the interval of time At becomes indefinitely small and has zero THE LIMIT OF THE BATIO ' 13 for limit. The average velocity from B to C continually approaches the actual velocity at B, since the interval of time during which a change of velocity might take place continu- ally decreases. Hence the limit of the ratio — when the At limit of Ai is zero is the actual velocity of the locomotive at 5. This analysis shows that if the relation between distance s and time t of the motion of a body is expressed by the equBr tion s =f(t), and the velocity at any time t is denoted by v, V = limit •'^ — '' ~-'^ ' when the limit of AJ is zero. At For example, in the case of a freely falling body, starting from rest, s = 16.08 1^, where s is distance measured in feet, and t is time measured in seconds. Here At = limit 16.08 (2 < + At) = 32.16 *, when the limit of At is zero. Hence the velocity at the end of the third second is 96.48 feet per second. Velocity is seen to be the rate of change of distance per unit of time. PROBLEMS 1. If s z^^gt", where g is a. constant, determine the velocity at time t. 2. If s = lot + 16.08 1', calculate the velocity at time t. 3. If s = Mt —jgt% where u and g are constants, determine the velocity at time t. 4. If s = Mt + is't^, where u and g are constants, determine the velocity at time t. 14 DIFFERENTIAL AND INTEGRAL CALCULUS Art. 8. — Rate of Change In the function y = oi^ — 5x let x„, y^ and X(, + Ax,yo + ^y be two sets of corresponding values of variable and function. That is, starting from Xq, y^ to a change of Aa; in the value of the variable there corresponds a change of Ay in the value of the function. The ratio of the corresponding changes of function and variable — ^ determines the average rate of change Ax of the function throughout the interval Ao; ; that is, the uni- form change of the function for change of the variable by unity which in the interval Aa; causes a change of A3/ in the value of the function. This is true for all values of the interval Aa;. Now, if Aa; becomes smaller and smaller, the ratio — ^ con- Aa; tinually approaches the actual rate of change of the function at Xf,, since the interval Aa; during which the rate of change might vary continually decreases. Hence the actual rate of change of the function y = x^ — 5x at a;o, 2/0 is limit ^ = limit (^'o + Atc)^ - 5 (a!o + Aa;) - (xp' - 5 Xp) Aa; Aa; = limit (2xp — 5 + Aa;) = 2a;o — 5, when the limit of Aa; is zero. The function y = a? — 5x increases 3 times as fast as x increases when 2 ajo — 5 = 3, that is, when a;o =^ 4 ; the function decreases 5 times as fast as x increases when 2a;o — 5 = — 5, that is, when Xp = Q; the function is stationary, that is, it is neither increasing nor decreasing, when 2 Xq — 6 = 0. This analysis shows that for any value of x the limit of f(n' _i_ A^ . f(x) ■'^ ^-^-!- when the limit of Aa; is zero, measures the Aa; THE LIMIT OF THE SUM 15 rate of change of f(x) foi- that value of x. If this limit is positive, f(x) is an increasing function of x; if this limit is negative, fix) is a decreasing function of x; if this limit is zero, f{x) is stationary. The calculation of the limit of the ratio f{^+^^)-f{«) A* when limit Aa; = for all functions f(x) is the fundamental problem of the Differential Calculus. PROBLEMS 1. Calculate the rate of change of 4 a; + 7. 2. Find the rate of change of 3/ = «^ + 3 a;. 3. Calculate the rate of change of '6x — a? at a; = 1. 4. rind where the function a;^ — 2 a; increases twice as fast as X increases. 5. rind where the function a;^ — 2a; decreases twice as fast as X increases. 6. Find where the function a;^ — 2 a; is stationary. Aet. 9. — The Limit of the Sum Let y =f(x) be the equation of the given curve. Denote by A the area of the surface bounded by the curve, the X-axis, and the lines x = a, x = b. Divide the portion of the X-axis from x = a to x = b into any number, say 6, of equal parts, and call each part Aa;. Constructing rectangles on each Aa;, as indicated in the figure, and denoting by A' the 16 DIFFERENTIAL AND INTEGRAL CALCULUS sum of the areas of the rectangles, A'=f{a) • Aa; +/(a + Aa;) • Aa; +/(a + 2 A*) • Aa; +f(b - 2 Aa;) . Aa; +f{b - Aa;) ■ Aa;, which may be written A' = S/(a;) • Aa;. Now, as the number x = a of equal parts into which b — a is divided is indefinitely in- creased, Aa; becomes indefinitely small, and A' continually approaches A. Hence A = limit of S/(a;) • Aa; when the limit x = a of Aa; is zero. The calculation of this limit of the sum is one of the fundamental problems of the Integral Calculus.* In some simple problems the limit of the sum may be calculated by means of the formula for the sum of n terms of an arithmetic progression a + (a + d) + (a + 2d) + — + Q-2d) + (l-d) + l, namely, s = (a + l)^- Fig. 4. For example, let it be required to calculate the area of the surface bounded by the straight lines y = x + 2, x = l, a; = 5, and the X-axis. Divid- ing the distance from a; = 1 to a; = 5 into n equal parts and calling each part Ax, Aa; = -, and n * Historically the calculation of the areas of surfaces bounded by curved lines led to the invention of the Integral Calculus. THE LIMIT OF THE SUM 17 1 = 4 A = limit %f{x) Ak = limit{3 + (3 + ^) + (3H-^) + (3 + l^) + ... = limit|6 + («-l)H^i = limit 2(^10 --V 20 (. n) 2 11 \ nj when n is indefinitely increased. This agrees with the result obtained by elementary geometry. Aet. 10. — General Theory op Limits Let u denote a function of x whose limit is U when the limit of X is a. This relation may be denoted by the equation W(a;=a±{) = f7± «) where e must become indefinitely small when S becomes indefinitely small. Limit of the sum. — Suppose that when the limit of x is a, limit u^ = Ui, limit Mj = Ui, limit «3 = XJ3. The hypothesis is equivalent to whence (% + u^ — i«3)ci=a ± s) = f^i + ^2 — f^s ± «i ± 62 T «3- Since when S becomes indefinitely small, tj, e^t «3 each become indefi- nitely small, .± £1 ± €2 T £3 for all combinations of signs also becomes indefinitely small. Hence when the limit of x is a, limit (mi + Mj — U3) = C/i + CTj — J/g = limit Uy -j- limit «j — limit M3. That is, the limit of the algebraic sum of a finite number of quantities is the like algebraic sum of their limits. Limit of the product. — If, when the limit of x is a, limit Ui = Ui and limit Wj = ZZj, i«i(«=a±s) = C^i ± ^i and M2(«=o±6) = 1/2 ± £2. By multiplication, («1 • M2)(« = a± 6) = f^l • f72 ± C^2 • fi ± C7i • £j. 18 DIFFERENTIAL AND INTEGRAL CALCULUS Hence if Ui and U^ are finite, wlien the limit of x is a, limit (iti ■ Us) = Vi-U2 = limit Ml • limit u^. That is, the limit of the product is the product of the limits. Limit of the quotient. — If, when the limit of x is a, limit Ml = CTj and limit M2 = t^2) Wi(«=a±s) = t^ii-^i) "2(i=a+s) = C^2±«2- By division, /mj\ ^ ?7i±ei _Ui ^ p-i±€i Ui \W2/(, = a±{) f/2±«2 t/2 (72 ± «2 U2 _^Ul . ± ?72 • t2 T g7i • £2 U2 C4(U2±£2) Hence if CTi and U2 are finite, when the limit of x is a, limit ^ = -^ = ii5!Hii^. That is, the limit of the quotient is Ma U2 limit M2 the quotient of the limits.* Art. 11. — Continuity The function y=f(x) is said to be continuous at x — Xi^ if the limit of the difference /(x^ ± Ax) —f(x^ is zero when the limit of Aa; is zero. This definition may also be written [/(«o + ^^) — /(a'o)]A»=±ii = ± «) where e. must become indefi- nitely small when 8 becomes indefinitely small. The function is said to be discontinuous 2A, x = Xo if t does not become indefinitely small when 8 becomes indefinitely small. For example, if the curve in the figure is the graph of y=f(x),f(x) is continuous at all points except at X(, = l and * Jordan, in his Gours d'analyse, Paris, 1893, perhaps the most com- plete treatise on the Calculus ever written, says : " Arithmetic and Algebra employ four fundamental operations, addition, subtraction, multiplica- tion, and division. A fifth can be conceived of, consisting in replacing a variable quantity by its limit. It is the introduction of this new opera^ tion that characterizes the Calculus." THE LIMIT OF THE SUM 19 x„ = — 3. Starting from (1, — 2), f(x) is continuous for in- creasing values of x ; starting from (1, 2), /(x) is continuous for decreasing values of x. Y Starting from (1, 2), limit [/(I + Ax) -/(I)] =4 when limit Ax = 0. Starting from (1,-2), limit[/(l-Aa;)-/(l)] = -4^ when limit Ax = 0. Starting from (—3, 00 ), limit[/(-3 + Aa;)-/(-3)] = — oo Fio. 5. when limit Aa; = 0. Starting from (—3, — oo ), [/(- 3 — Aa;) — /(— 3)] = + 00 when limit Aa; = 0. Starting from (1, 2), limit limit =^IlziM_zZ(ll= tan 60° — Aa; Aa; when limit Aa; = ; when limit Aa; = 0. Starting from (1, — 2), limit /(I + ^^) -fO-) = tan 330° when limit Aa; = ; Aa; limit /(1-Aa;)-/(1) . • Aa; when limit Aa; = 0. Hence it is evident that, at points of discontinuity of f(x), the limit of the ratio /(a'o + Aa;) -f{xa) ^^^^ ^. ^^-^ Aa; = Aa; is not independent of the algebraic sign of Aa;. At points of continuity, the limit of this ratio generally is independent of the sign of Aa;. 20 DIFFERENTIAL AND INTEGRAL CALCULUS Example. — Show that log a; is a continuous function of x. 1 ± — 1 = 0, when limit Ax = 0, except for a; = 0. PROBLEMS Show that the following are continuous functions of x : 1. 3a^-5» + 2. 2. sin a;. 3. e'. Find the points of discontinuity of 5x-7 CHAPTER III DiriEEENTIATION AND INTEGRATION OF ALGEBEAIO rUNOTIONS Abt. 12. — Differentiation The function of x which is the limit of the ratio f{x + Ax)-f{x) Ax when limit Aa; = 0, is called the first derivative of f(x) with respect to x, and is denoted by the symbol — f(x), or f'{x). If f(x) is denoted by y, the first derivative is denoted by -^• ctx The operation of forming the first derivative, denoted by the symbol — , is called differentiation. General rules for the dx differentiation of algebraic functions are to be established. I. Let u represent any continuous function of x. Represent by Am the change in the value of lo corresponding to a change of Ax in the value of x. By definition, — = limit — - when dx Ax limit Aa; = 0. *Tlie notation ^ was invented by Leibnitz (1646-1716). Newton (1642-1727) denotes — by s, a notation still used in mechanics. La,- grange (1736-1813) denotes the first derivative of f(x) hyf(x), Cauchy (1789-1857) by Df(x). 21 22 DIFFERENTIAL AND INTEGRAL CALCULUS II. Let M = c, where c is a constant, that is, a change in the value of X does not cause a change in the value of c. Then *^ = ^ = limit — = when limit Ax = 0. Conversely, if da; dx Ax — = 0, that is, identically zero, u is independent of x. For dx — = means that a change in x causes no change in u. dx Hence u is independent of x. III. Let u = x. Then — = — = limit — = 1 when limit dx dx Ax Ax = 0. IV. Let f{x) = c • M, where c is a constant and u represents a continuous function of x. Forming the first derivative, ^f{x) = limit c-(u + Au)-c.u ^ j.^.^ c . ^ = c . *^ dx Ax Ax dx when limit Ax = 0. Hence the first derivative of a function multiplied by a constant is the constant times the first deriva- tive of the function. V. Let f(x) = u + V — w, where u, v, w represent continuous functions of x. Denoting by Au, Av, and Aw the changes in the values of u, v, and w, corresponding to a change Ax in the value of x, A f(x) = limit ^ + ^^ + ^ + A-» — w — Aw — (m + -» — w) dx-^^ ' Ax = limit ^ + limit ^ - limit ^ = *f + ^ _ ^ Ax Ax Ax dx dx dx when limit Ax = 0. That is, the first derivative of the alge- braic sum of a finite number of functions is the like algebraic sum of the first derivatives of the functions. If two functions /(x) and ^ (a;) have the same first deriva- BIFFERENTIATION AND INTEGRATION 23 tive, their difference f{x) — 4, (x) is a constant. By hypothesis, |./(.)-|<^(.) = |i/(.)-^(.)J.O, hence f(x) — t^ (x) = c. VI. Let f(x) = u-v, where u and v represent continuous functions of x. ±f(x) = ±(u.v) = limit (" + A^)-(^ + At>)-«-7; dx-'^' dx^ ' Aa; = liniit(« + A«)-^ + liniit'«.^' = «.^ + i;.^ Aa; Aa; dx dx when limit Aa; = 0. Hence the first derivative of the product of two functions is the first function times the first derivative of the second function plus the second function times the first derivative of the first function. In like manner it is proved that d , V dw , dv , du — (u-v-w) =u-v \-u-w \-v -w dx dx dx dx VII. Let f(x) = - , where u and v are continuous functions of a;. d_ dx" . /(^) ^±(^\ = limit ^ + ^^ ^ r ^ ' dx\vj Aa; Aw Ai; du dv V u V u — • .. ., Aa; Aa; dx dx = limit — 2- = 2 V^ + V- ^V V when limit Aa; = 0. That is, the first derivative of the quo- tient of two functions is the divisor times the first derivative of the dividend minus the dividend times the first derivative of the divisor, divided by the square of the divisor. 24 DIFFERENTIAL AND INTEGRAL CALCULUS dv — c • If f(x)= -, where c is a constant, -=-( - 1 = 5 Hence ■^ ^ '' V dx\vj v^ the first derivative of a fraction whose numerator is constant and whose denominator is a function of x, is the numerator with its sign changed times the first derivative of the denomi- nator divided by the square of the denominator. VIII. Let f{x) = u", where u is a continuous function of x, and n is any finite, positive integer. Af(x) = A„« = limit (" + ^^)"-^^" dx dx Ax = limit I n ■ u"-^ + ^ ^" ~ -*-) • m-" • Am + ... + n-u. (Am)»-2 + (Aji)"-! I ^ ) Ax .1 du = n ■ ?t" ' — , dx when limit Ax = 0. Hence the first derivative of a function affected by a finite, positive, integral exponent is the product of the exponent, the functicSn with its exponent diminished by unity, and the first derivative of the function. r If f{x) = v = u', where r and s are finite, positive integers, V' = u'. Forming the first derivatives of both sides of this equation, s-v'-'- ■ — =r-u'-^.~. Solving for ^, there re- dx dx ^ dx suits *^ = r.?^.*^ = r.?^\*^ = r.J->.*f. Hence dx s v'~^ dx s ^r-i dx s dx dx s dx DIFFERENTIATION AND INTEGRATION 25 that is, the first derivative of a function affected, by a finite, positive, fractional exponent is the product of the exponent, the function with its exponent diminished by unity, and the first derivative of the function. If /(») = M~", where n is finite and either integral or frac- tional, fix) = — , and d „. ■. d fl\ dx ___, du dx-'^ '' dxXu") u^' dx Hence the first derivative of a continuous function affected by any finite constant exponent is the product of the 6xpo- nent, the function with its exponent diminished by unity, and the first derivative of the function. If u = X, — a;" = m ■ x"~'^. dx If 2/ is a continuous function of x, x is also a continuous function of y. Trom the equation y=fi(x) an equation of the form x = fJy) is obtained. Differentiation gives -^ and , dx — . The relation between these derivatives is to be found. The equation — — = 1 is true for all values of Ax, hence ^ Ax Ay ' limit ^ . ^ = limit^ . limit^ ^dydx^j^^ Ax Ay Ax Ay dx dy when limit Aa; = and limit Ay — 0. There results -^ = — ; dx dx dy that is, the first derivative of y with respect to x is the recip- rocal of the first derivative of x with respect to y. If 2/ is a continous function of z, y =/i (2:), and z is a contin- uous function of x, z =fs{x), y is also a continuous function of X. The derivative -^ is to be calculated. dx 26 DIFFEBENTIAL AND INTEGRAL CALCULUS The equation — ^ = — e . — is true for all values of Ax. Ax Az Aa; When limit Aa; = 0, limit — ^ = limit — • limit — ; that is, , , Aa; Aa Aa; dy _ dy dz^ dx dz dx The rules of this article are sufficient for the differentiation of all algebraic functions. Example I. — Form the first derivative of 5a?-7a^ + 12x-15. — (5 a;' - 7 a;''' + 12 a; - 15) = 5 — a!« - 7 — a;2 ^ 12 — a; - ^ 15 dx dx dx dx dx = 15x'^Ux + 12. Example II. — Eorm the first derivative of (2 — 5 a;^^. This expression has the form u" whose derivative is d „ „_i du dx dx In the problem m = 2 — 5 a;^, w = f . Hence — (2 - 5af)^ = ^(2 - 5x')i --^(2 - 5x^ =-15x(2 -53f)K dx dx Example III. — Form the first derivative — of the implicit dx function a^ + y^ = 9. Forming the first derivative of both sides of this equation, 2x + 2y^=0, whence ^ = -5 and ^ = -^. dx dx y dy x Example IV. — If ^ = 1 — and x^ = z'' + 1, form ^^ dz /j,2 + 1)J ^^ Prom aP = z^ + 1, — = - • Hence dx z DIFFERENTIATION AND INTEGRATION 27 dy_dy dz _ ■^ x_ z'-x _ {x'—l)-x _ 1 dx dz dx /g2 _^ ^\i z /g2 _|_ -^i x' X 1 — X Example V. — Form the first derivative oi y = - . (1 + x')^ Applying the rule for differentiating the quotient of two func- tions, (1 + a;^)* . A(l -x)-(l- x).^(l + x^i dy dx dx dx~ 1+0? ^ _ (1 + x^)^ - a;(l - x) (1 + a?)'^ ~ 1 + x^ — l — a? — x + x^_ 1+x (1 + of)i (1 + a!=)^ PROBLEMS Form the first derivative of, 1. 3 a; + 5. 6. 1 l-x" 2. 2a;^ + 7a ; + 3. 7. 1 + a? 1-a? 3. 3a=-8. 8. {l-xf. 1 + a;. 5 1 + ^ 9. 10. (1-aO^. (1-0.)- ^- 1-x 11. (l-a?)i 12. (3 + 5 a;) ^ 13. \l + xj 14. (5a;-7a^)l 15. (l-x + x'^i. 16. (a + te")"*. 17. (a + bx")". 1 — a? 2 18. Form the first derivatives of and -, and find the difference of the functions. 28 DIFFERENTIAL AND INTEGRAL CALCULUS In the following equations y is an implicit function of x. Form ^. dx 19- ^ + fi = l- 20. -,-|J = l. 21. 2/»-2pa; = 0. a' ¥ a' o' 22. a?y + y — x = 0. 23. a^ - 3 o^ + 2/" = 0. Determine the rate of change of function and variable of 24. /(a;) = 2 a - 05^. 26. /(a;) = (1 - a;^)^ at a; = f 25. f{x) = —^. 27. f(x) = (4 - a;'')^ at a; =2. 28. Pind at what point of the circle x' + y'' = 9 the ordinate increases twice as fast as the abscissa. 29. Discuss the rate of change of ordinate and abscissa of — 1-^=1 for different points of the ellipse, a. 6^ The rate of change of ordinate and abscissa is measured by fill liOf* • -^ = When X and y have like signs the ordinate is a dx ary decreasing function of x; when x and y have unlike signs the ordinate is an increasing function of x ; when a; = 0, the ordi- nate is not changing value ; when 3/ = 0, the ordinate changes infinitely more rapidly than the abscissa. These results agree with the results obtained by examining the ellipse. 30. Determine the rate of change of area and side of an equilateral triangle. Area = — — a^, x being a side. 31. Find the rate of change of area and radius of a circle when the radius is 10. 32. A man walks on level ground towards a tower 80 feet high. When 60 feet from the foot of the tower find the rela- tive rate of approaching the top and foot of the tower. DIFFERENTIATION AND INTEGRATION 29 Calling the man's distance from the top of the tower y, his distance from the foot of the tower x, y' = x^ + 6400. Find the slopes of the tangents to the following curves at the point (x, y) of the curve : 33. v = 4=a;l 34. y = Qx — x\ 35. — — ^=1. 36. Find the slope of the tangent to y = 8 a; — a^ at a; = 1. 37. Find where the point generating the circle »^ + 2/^=4 tends to move parallel to the X-axis. Where parallel to the y-axis. Determine the velocity at time t supposing the relation between s the distance and t the time to be expressed by the following equations, where a and u are constants : 38. s = \ at". 39. s = ut + l af. 40. s = ut—^ ai\ 41. If ^ = 2^(1 + z)l and 1 + « = ar', find ^. dz dx 42. If ^ = / and 1 + 2 = A find ^• Aet. 13. — Integration The difference between two functions which have the same first derivative is constant. Hence, if the first derivative of a function is known, the function itself is known, except for an additive arbitrary constant. The process of obtaining a func- tion f(x) from its first derivative -—f(x) is called integration, /ux The operations denoted by the symbols — and | neutralize each other ; that is, /^/C*) =/(«=)+(? and £//(«')=/(«:), 30 DIFFERENTIAL AND INTEGRAL CALCULUS where C is au arbitrary constant. Hence the rules of integra- tion are at once inferred from the rules of differentiation. The result of integration is called an integral, and the integral is correct if the derivative of the integral is the expression which was integrated. The rules of algebraic integration, obtained from the rules of algebraic differentiation of Art. 12, are expressed by the formulas : I. C—=:u + a. III. Cl = x+C. J dx J II. Co = C. IV. Ca- — = a-u + G. J J dx -IT r/du , dv dw\ , , ri V. {[-- + — -—] = u + v-w+C. J \dx dx dxj VI. C(u.^ + v.^\=u.v+C. J \ dx dx) du dv V ■ u J V' V VIII. Cu"- — = ^!^^ + a When !t = a;, this formula be- J dx n + 1 x" = + C When n = — 1, this result is n + 1 absurd. The rule of formula VIII. may be stated, the integral of any function affected by an exponent other than — 1, multiplied by the first derivative of the function, is the function with its exponent increased by unity divided by the increased expo- nent, plus an arbitrary constant. DIFFERENTIATION AND INTEGRATION 31 Since | a- — = a-M+(7 and a\ — = a-u-\-C, a constant J dx J dx factor may be shifted from one side of the sign of integration to the other, without affecting the result. By formula V. the integral of the algebraic sum of a finite number of functions is the like algebraic sum of the integrals of the functions. Example I. — Find the function of x whose first derivative is 4a!^ — 5x. Denoting the function by fix), — f{x) = 4a;^ — 5a; and f{x) = \ {i:X^ — 5x) = A: i x^ — 5 \ x = ^a? — ^a? + C, where the arbitrary constant C is called the constant of integration. This integral is called the indefinite integral. If f(x) is re- quired to have a given value for a given value of x, for example, if f{x) = 10 when x = l, the value of C is found from the equation f{x)=^o? — ^x^+C to be 11|^. Hence under the given conditions f{x)=^a? — ^a? + ll\. This re- sult is called the corrected integral. Example II. — Eind the function of x whose first derivative is a; (1 + a?Y. Denoting the integral by f{x), f(x)=Jx{l + x^i This integration can be performed by means of the formula /m" • — = — — - + C if a; (1 -f- x')^ can be separated into factors dx n + 1 V I / f of the form u" and — Placing u = 1 + a?, — = 2 a;. Hence dx dx fi<^) = */(l + a^^*£ (1 + aO = i (1+ x^^ + C. 32 DIFFERENTIAL AND INTEGRAL CALCULUS Example III. — If ^ = 10 (a; - 2) (a^ - 4 x)-l, find y. Placiiigrt = a;=-4a;, — = 2a;-4, dx and 2/ = 10 r(a;-2)(a^-4a;)-i Example IV. — Determine the equation of a curve such that the slope of the tangent to the curve at any point {x, y) is the negative ratio of the abscissa to the ordinate. The condition of the problem is expressed by the equation — = , whence y^ = — x. Integrating, -y" = a^ + (7 or dx y dx 2 2 a? + y^ = 2G. This equation represents any circle with center at the origin of coordinates. If the circle is required to con- tain the point (3, 4), the value of 2 C must be 26, and the problem has the determinate solution a? + y^ = 25. PROBLEMS Find the f{x) whose first derivative is, 1. 2 + X, knowing that /(«) = 7 when x=2. 2. 3 — 5ar^, knowing that /(a;) = 20 when x=5. 3. l + x + aP + a^, knowing that /(a;) = 12 when x = l. 4. x^. 5. Sx'^. 6. x^ + 5. 7. 2xi — x^. 8. (1 — a;) (2 + a;^. Multiply out and integrate term by term. 9. (3 a; - 5) (2 a;- 3 a;^. 12. (2 a; + 3) (a;'' + 3 a;)*. 10. (l+af)(3a^ + 2). 13. (3 a;^ - 10 a;) (a^ - 5 a^i 11. (4 a; - 5) (2 a^- 5 a;)*. 14. (2a! + 6) (a;' + 5a;- 7)*. DIFFERENTIATION AND INTEGBATION 33 15. (1 + 9 a;)*. 17. 15(3a^ -8x) {3«? -12a^)i. 16. (a + bx)l 18. (a^ + 3x-5)(2a^ + 9ay'-30x)K Find tlie equation y =/(«) of a curve sucli that the slope of the tangent at any point {x, y) is, 19. 3 a; — 7. 20. 2x + 5, the curve passing through (5, 0). 21. a^ + Bx, the curve passing through (0, 0). 22. — „•-, the curve passing through (a, 0). a^ y 23. For the cable of a suspension bridge with load uni- formly distributed over the horizontal, -^ = ~-x, where w dx io is the uniform load per hori- zontal linear unit and to is the tension at the lowest point. Find equation of curve assumed by cable. 24. Find the function of x whose rate of change is 2 x — 6. Find the relation between s and <, knowing that s = when i = 0, and that the velocity is, 25. u + at. 26. u — at. Fig. 6. Find the relation between x and y, knowing that, dy_s^^ gjj dy _ 1 —3x + 5a^ ' y ' dx y — 4:y^ . l + a: . 31_ dy^ 1 dx 1 — y dx 29. ^=-^. dx 1+2/ dx 28. ^-- a?f 32. 0?^ = ^. dx y 34 DIFFERENTIAL AND INTEGRAL CALCULUS Find the values of the following integrals : 33. C /-^ • 34. f ^. 35. fci-a^', -' VlT^ "^ (1 - a;^^ •^ 36. /C-^' »'• /o (2 ax - a^)^ Multiplying numerator and denominator by (a;-^^ = x'^, the problem becomes f ^I^ = _ -L ("(2 ax-^ - l)-f A(2 ax-i - 1) (2aa;-i-l)^ = ^(2 aa;-i - l)-i + 0= — — L= + 0. <* aV2 aa; — a;'' 38. r V2 aa; - a;" . 39, T 1 40. f. 3^ '^ (a^ + a;')^ -^ Vl + i Calling the integral ?/, the problem is, given -^, to find y. (XX By the substitution 1 + a; = 2^, dy^dy dx_ x 22 = "^^^^^ "^ —I'y? -I. dz dx dz t/1 _|_ X z 2 By integration y = ~i^ — 2z + C, o hence 2/ = |(H-a;)l - 2(1 + a;)i + 0. 41 ■ - - " ■ ^ ra^(H-a;)i 42. f- Art. 14. — Definite Integrals If -f F(x) =f(x), ff(x) = F(x) + C- If F(x) + C is dx J evaluated for a; = & and x = a and the second result subtracted from the first, the final result i?'(6) — F(a) is called the defi- DIFFERENTIATION AND INTEGRATION 35 nite integral of f(x) between the limits x = a and x = b. This operation is denoted by writing f(^) = Fix) + C = F(b)-F{a), where a is called the lower limit and 6 the upper limit of the definite integral. From this definition it follows that and also that, if 6 — a = (6 — c) + (c — a), a, \Jc %/ a Example. — Find the value of the definite integral r(3 x2 - 6 £B + 7). (3a^-5a;+7): = p- far' + 7a;+C = 63. PROBLEMS Calculate the definite integrals, 1. r(3a!^-5). 6. Cx{^-v?)^- 2. r(2a;2 + 4a;-6). 7. ^\{l-o^^. 3. j a;(4-ar^. 8. r°a;(a2-a^i. 4- Ja + ^)(i-a^- 9. C'x^dx. 5. f (1-x^). 10. r 3a5*da!. 36 DIFFERENTIAL AND INTEGRAL CALCULUS 11. Cx'Vl + x. Write — = x'VT+x and determine — Jo '^^ /I /I ^ '^'' when l+x = z\ There results 5E = ^ . 25 = 2 zHz" - 1)\ dz dx dz Since 2 = 1 when a; = 0, and » = 2 when a; = 3, M = rVVT+a = r'2 «^(«2 _ 1)2 = 16.1. 12. If s is the distance in feet and t the time in seconds of a body's motion, find the distance the body moves from the end of the third to the end of the fifth second, knowing that - = 32.16 «. dt Akt. 15. — Evaluation of the Limit of the Sum The value of >S = limit "^.fix) • Aa; when limit Aa; = 0, where 1=6 x=a %f(x) • Aa; stands for x = a f{a) • Aa;+/(a+Aa;) • Aa;H 1-/(6-2 Aa;) • Aa;4-/(6-Aa;) • Aa;, is to be determined. Suppose Fix) to represent a continuous func- tion of X whose first derivative is f{x), that is, — F{x) =f(x). This means that — ^-^ — ^^^ — ^=/(a;)±e, where e be- Aa; comes indefinitely small when Aa; becomes indefinitely small. From the last equation, f(x) • ^x = F(x + Aa;) — F(x) ± e • Aa;. Whence by substituting for x successively a, a + Ax, a + 2 Ax, • ••, b — 2Aa;, 6 — Ax, f(a) -Ax =F(a + Ax)—F (a) ± tj • Aa;, /(a + Aa;) • Aa; = F(a + 2 Aa;) — F{a + Ax) ± t^ ■ Ax, f(a + 2 Aa;) ■ Aa; = F(a + 3Aa;) — F(a + 2Aa;) ± £3- Aa;, f(b - 2 Aa;) • Aa; = F(b - Aa;) - F(b - 2 Aa;) ± e„_i ■ Ax, /(b - Ax) ■ Ax = F(b) - F(b - Ax) ± e„ • Aa;. DIFFERENTIATION AND INTEGRATION 37 By addition, ^f(x) ■ ^x = F(b) - F{a) ± %e ■ Ax. Since F(x) is continuous, each e becomes indefinitely small when Ax be- comes indefinitely small. Denoting by E the numerically largest e, Se-Aa; ^E 'S,Ax = E (b — a), which becomes indefi- nitely small when E becomes indefinitely small, since 6 — a is supposed to be finite. Hence S = limit's/(a;) • Ax = F(b) - F(a), when limit Ax = 0. But FQj) — F(a) is the value of the defi- f(x), for by hypothesis — F(x)=f(x). Hence, " dx finally, if ~F{x)=f(x), S = limitlf(x) Ax= ffix) when dx x=:a \Ja limit Ax = 0. If the summation limits are a and x, the result obtained becomes 'S' = I f(x) = F(x) — F(a). Forming the derivative oiS,f- = ^F(x) = fix). dx dx Akt. 16. — Infistitesimals and Differentials A quantity which becomes indefinitely small is called an infinitesimal. If several infinitesimals occur in the same problem, any one may be chosen as the principal infinitesimal. Denote the principal infinitesimal by a. Infinitesimals whose ratio to a is finite are said to be of the first order. An infinitesimal /S whose ratio to the Kth power of a is finite is said to be of the nth order. If n is positive .and larger than unity, ^8 is said to be of a higher order than a. Denoting the finite ratio of /3 to a" by »"; -^ = - • ——i = '>'> 38 DIFFERENTIAL AND INTEGRAL CALCULUS whence " = 7- • a"~\ an infinitesimal, and - = :, an indefi- nitely large quantity. Hence, if /8 is an infinitesimal of a higher order than a, p = e.-a, where c is an infinitesimal. The laws governing the use of infinitesimals are contained in the following two propositions. I. In the limit of the ratio of infinitesimals any infinitesimal may be replaced by another infinitesimal differing from it by infinitesimals of higher orders. Let o and /3 represent any two infinitesimals, and consider the ratio l3Ay + Br^ + C,-l3' + Drp'+-' where the coefiicients B, C, D, ••• and Bi, Cy, Di, ••■ are finite. Let M be the numerically largest of the coefficients B, C,D, ■■• , Ml the numerically largest of the coefficients Bi, Ci, Z)j, •••. ThenB ■ a+C ■ a?+ D ■ a^ + ■■■> M(a + a? + a?+ -) =M-^^, 1 — a an infinitesimal, and another infinitesimal. Hence in the limit .^« A-[-B■a+C^a^ + D■n^^ ^Aa II. In the limit of the sum of infinitesimals, provided this limit is finite, any infinitesimal may be replaced by another differing from it by an infinitesimal of a higher order. Suppose limit (% + aj + «3 H ) = c, a finite quantity, and let ^1 = «i + ci ■ «!, ^2 = "2 + £2 ■ «2) /?3 = «3 + £3 • "j) •••) whcro DIFFERE/fTIATION AND INTEGRATION 39 ci, £2, «3, "•• are infinitesimals of which e is the numerically largest. Adding 08i + ft + A+-) = («1 + «2 + "3 -I ) + («! ■ «1 + «2 • «2 + ^3 • «3 + •••)• Now ei •«! + £2 • «2 + «3 •'"3 +•••>«(«!+ «2 + «2+ •••) ==^*<'j ^^ infinitesimal. Hence in the limit A + ft + ft H = «1 + «2 + «3 H • This proposition is true even when c is indefinitely large, pro- vided £ • c is an infinitesimal. If the limit of the ratio of two infinitesimals is unity, the infinitesimals can differ only by infinitesimals of higher orders. For if limit - = 1, — =1 ±£ and B = u ± e • a, where e is an a a ' infinitesimal. Hence the rules governing the use of infinitesi- mals may be stated, in the limit of the ratio and in the limit of the sum any infinitesimal may be replaced by another infinitesimal, provided the limit of the ratio of the two infini- tesimals is unity. Since infinitesimals of higher orders disappear from the limit of the ratio and from the limit of the sum, the solution of problems involving the limit of the ratio or the limit of the sum may be simplified by dropping infinitesimals of higher orders at the start. If y =f(x) is a continuous function of x whose first deriva- tive is f'x, -^ = /' (x) + £, whence Ay = /'a; • Aa; -f- £ • Aa;, where Aa; A.y is the difference in the value of the function corresponding to a difference of Aa; in the value of the variable, and e be- comes an infinitesimal when Aa; becomes an infinitesimal. When the difference Ax becomes an infinitesimal it is denoted by dx, which is read differential a;. The change in the value 40 DIFFERENTIAL AND INTEGRAL CALCULUS of y corresponding to dx is f (x) •dx + e-dx. Defining dy, read differential y, by tlie equation dy =/' (a;) • dx, dy differs from the change in the value of y corresponding to a change of dx in the value of x by e • dx, an infinitesimal of a higher order than dx. Hence in problems involving the limit of the ratio or the limit of the sum the actual change in y may be replaced by dy =/' (x) ■ dx. The first derivative of a function is therefore the factor by which the differential of the variable must be multiplied to obtain the corresponding differential of the function. This explains why the first derivative of a function is frequently called the differential coefficient of the function. The operation of finding the differential of a function corre- sponding to the differential of the variable is called differentiar tion. It will be observed that differentiation as here defined is performed by the rules established for differentiation as defined in Art. 12. For example, ii y = a? — 1 a? + 15 x + IQ, d?/ = A (x3 _ 7 a^ + 15 a; -f 10) da; = (3 a^ - 14 a; + 15) dx. If F(x) is a continuous function of x whose first derivative is f(x), AF(x) = F(x + Aa;) — F{x) =f(x) • Aa; + e • Aa;, where AF(x) and t become infinitesimals when Ax becomes an infini- tesimal. The sum of the elements AF(x) of F(x) from a; = a x = z to 35 = a; is F{x) — F{a) = S [/(a;) • Aa; + e . Aa;], and the sum of the elements Aa; of x from a; = a to a; =x is x — a. This is true for all magnitudes of Aa;. "When the element Aa; becomes the infinitesimal element dx, F(x) - F(a) = ^\f(x) -dx + e- dx}. DIFFEUENTIATION AND INTEGRATION 41 Supposing X — a to be finite, in the limit F(x) -F(a) = % fix) ■ dx, since £ • dx is an infinitesimal of a higher order than dx. Calling the operation of finding the limit of the sum integrar tion and denoting it by the symbol j , and indicating that the sum is to extend from a; = a to a; = a; by writing I , the result f(x)-dx = F(x) — F(a). Now F(x) — F(a) was found in Art. 14 to be the value of the defi- X" d f(x), provided — F(x) =f(x). dx In general, if y=f(x) and -^=:f'(x), in the notation of differentials dy=f(x)-dx and y= \ f (x) ■ dx=f(x) + G. Differentiation and integration as defined in this article are again inverse operations. It will be observed that integration as here defined is per- formed by the rules established for integration in Art. 13. For example, if dy= (Zx^~5x + W)dx, y = 3? -^x' + lQx + C; the sum of the. elements of y from ce = to x = x is CiSay'- 5x + 10)dx=[x'~ fx^ + lOa; + o]] =a?-^se' + Wx; the sum of the elements of y from a; = to x = i is r*(3 a;^ - 5 aj + 10) da; = [a;« - f a^ + 10 a; + o]] = 64. While the method of differentials is based on the method of limits, the method of differentials has two decided advantages : first, calculations are simplified by droyjping differentials of higher order at the start ; secondly, the successive steps in the 42 DIFFERENTIAL AND INTEGBAL CALCULUS application of the method of differentials, especially in sum- mation problems, are more directly intelligible than is the case when using the method of limits. Hereafter the derivative and differential notations are used interchangeably. PROBLEMS Form the differentials of the functions, 1. y = 3a^-5. 3. 2/ = ^- sr + l 2. y={l-x^i. 4. 2/=(j:^/ Find the functions whose differentials are, 5. d7j=(2a? — 5x)dx. 6. dy={l+a^dx. 7. dy=(l + xydx. Evaluate the definite integrals, 8. C\4:X-5)dx. 9. C\2x>-6x)dx. 10. C (x'-3x + 5)dx. 11. C a^ia^ + iy^dx. Placing ay' + l = z'', dx = - dx, and Jo X 3?(!i? + Vf^ dx = 3? -z-^ ---dz = a? ■ z-'' -dz X = (z^-l)-z-^-dz = dz--- T While X takes all values from to 3, 2 takes all values from 1 to 10. Hence 12. ( '3^(^-\-x)^dx. CHAPTER IV APPLIOATIONS OF ALGEBRAIO DIFFERENTIATIOIT AND lUTEaRATIOir Aet. 17. — Tangents and Normals The slope of the tangent to the curve whose equation is y =f{x) at any poiiit (xq, y^ is the first derivative of y =f(x') evaluated for x — Xq, y = y^. This is denoted by writing dxi) tana : The equation of the tangent TT' to y=f(x) at .(^2/0 (xo, 2/0) is 2/-2/0 =^°(»-a'o)- The intercept of the tangent on the X-axis, found by placing y = in the equar tion of the tangent and solv- ing for X, is AT=: x^—y^ dx„ Ay a the intercept on the F-axis is ^r' = 2/„-a;, Tia. 7. #0. dXf, The portion of the tangent bounded by the point of tan- gency and the point of intersection of the tangent with the X-axis is called the length of the tangent. Prom the figure 43 44 DIFFERENTIAL AND INTEGRAL CALCULUS the length of the tangent is PT = y^-Jl + f^Y. The projec- tion of the tangent PT on the X-axis is called the subtangent. dx From the figure the subtangent is DT = Vq—-"- dyo The equation of the normal NN' to y=f{x) at (xg, y^) is y — yg = ^(x — Xq). The intercept of the normal on the dyo X-axis is AN=X(, + yo^; the intercept of the normal on dx„ the y-axis is AN' = 2/0 + '"o — -• dyo The portion of the normal bounded by the point (x^, 2/0) and the intersection of the normal with the X-axis is called the length of the normal. From the figure the length of the normal is PN = yoyjl + ('^\ The projection of PN on the X-axis is called the subnormal. From the figure the sub- normal is DN = yo—- dxo Example. — Find the equations of tangent and normal to si^ + 2y^ — 2xy — x = at the point (1, 1). Also the length of tangent and subtangent, and of normal and subnormal. Differentiating a^+2y^ — 2xy— x = with respect to x, 2x + iy^-2y-2x^-l = 0, whence dy^ l+2y-2x _ dx dx dxAy — 2x At the point of tangency Xo = 1, 2/0 = 1, — = -x, — - = 2. dxo 2 dyo Hence the equation of the tangent is y — l = ^(x—l), reducing to y=:^x + ^; the equation of the normal y — 1 = — 2 {x — T), reducing to y = — 2x + 3; the length of the tangent is VS ! the length of the subtangent 2; the length of the normal ^ VS ; the length of the subnormal ^- APPLICATIONS 45 PROBLEMS 1. Find the equations of tangents and normals to x'y + y — x = at x = + l and at a; = — 1. 2. Find the equations of tangent and normal to xy = 4: at x = 2. 3. Find equation of tangent to x^ + y'^ = a^ at (xo, yo). 4. Show that the sum of the intercepts of the tangent to x^ + y^ = a^ is the same for all positions of the point of tangency. 5. Find the subtangent of the ellipse — + ^ = 1. 6. Find the subnormal of the parabola y^ = 2px. 7. Find the length of the normal to 4a^ + 16?/^ = 100 at x = 3. 8. Find the length of the tangent to ix^ + 16 y^ = 100 at x = 3. 9. Find the length of the subnormal to a^ + y^ = 25 at (3, 4). 10. Determine the curves whose subnormal is constant. The hypothesis is expressed by the equation y-^ = a, where a is the constant length of the subtangent. From this equa- tion — = ^ or adx = ydy. By integration ax = ^y' + C ay Oi or y'^ = 2ax — 2C. This equation represents a system of parabolas. Art. 18. — Length of a Plane Curve Denote by s the length from x = a to x=h of the plane curve whose equation is the continuous function y=f{x). 46 DIFFERENTIAL AND INTEGRAL CALCULUS k n A/ i X AV B Divide ab into any number of equal parts Ax, and draw ordi- nates at the points of division of ab. Draw the chords of the arcs As into which these Y ordinates divide the curve AB. At the ends of any one of these arcs, such as mn, draw tangents to the curve AB. Assume as axiomatic that chord mln < arc mn < broken line mkn. From the right triangles Mm and Mn, Fig. 8. km = Im ■ sec hnl. Ten = In • sec knl. These relations are true for all magnitudes of the chord mn. As the length of the chord mn is diminished, the angles kml and knl approach zero and their secants approach unity. Hence km = Zm (1 + ti) and kn = Zn (1 + tj), where ej and tj approach zero when mn approaches zero. Adding, km + A;re = (Im + In) — (ti -Im -\- e^- In). When the chord mln becomes an infinitesimal, Im, In, cj, and e^ become infinitesimals. It follows that broken line mkn — chord mln = tj • ?m + £2 • In, an infinitesimal of a higher order than the chord mln. Hence the difference between the infinitesimal arc mn and its chord mln is an infinitesimal of a higher order than the chord, and in problems involving the limit of the ratio or the limit of the sum, the infinitesimal arc may be replaced by its chord. It follows at once that the length of the curve is the limit of the length of the inscribed broken line when the number of APPLICATIONS 47 parts into which ab is divided is indefinitely increased, which is equivalent to saying when limit Aa; = 0. That is, s = limit'i\Aa;2 + A?/')* = limit'^sYl + — "^^ • ^^ x=a x=c\ Aajy when limit Aa = 0. Since -^ = tan a, where a is the angle of inclination to the dx X-axis of the tangent to the curve y =/(») at (x, y), — = (1 + tan' ay = sec a, whence cos a = -^■ dx ds Since sin a = tan a • cos «, sina = — ^ — = -2. These results, dx ds ds obtained by the method of limits, may be roughly inferred from the figure. As Ax approaches zero, the element of curve As approaches equality with its chord and becomes a part of the tangent. Hence in the limit, when Ax-, Ay, and As become the infinitesimals dx, dy, and ds, d^ = dx' + df = fl+^\dx', ds = (l+^'.dx, . dy . dy dx tan a = -2., sm a = ^i-, cos a = — dx ds ds Example. — Pind the length of the semi-cubic parabola y^ = Ax? from x = 5 to a; = 10. From the equation of the curve, y = 2 a;^ and ~ = 3x^. dx ••10 = 31.07. Hence, s = | (l+9x)^ ■dx = The length of the curve from the origin (0, 0) to any point (a;, y) is s = r\l ■+ 9x)i -dx^ ^(1 + 9x)^ - ^. 48 DIFFERENTIAL AND INTEGRAL CALCULUS The length of the curve between any two points x = a and a; = 6 is PROBLEMS 1 . Find the length of y = 2x from x = 5 to a; = 10. 2. Find the length of y = 3x + 5 from x = 5 to a; = 20. 3. Find the length of 9 j/^ = 4 a^ from (0, 0) to {x, y). 4. Find the length of 9 y^ = 4 .r' from a; = to a; = 10. 5. Find the length of 9y^ = iii? from a; = 5 to a; = 15. Art. 19. — Abea of a Plane Surface Denote by A the area of the surface bounded by the continu- ous curve whose equation is y=f{x), the lines x = a, x = b, and the X-axis. Divide the portion of the X-axis from x = a to x= b into any number of equal parts Ax. Construct rec- tangles on each Aa; and the adjar cent ordinates as indicated in the figure. Denote by AA the portion of A included between two successive ordinates. Then AA = y-Ax + 6-Ax- Ay, where f>o- 9. is less than unity. This is true for all magnitudes of Aa;. When Aa; becomes an infinitesimal, Ay also becomes an infinitesimal. Hence the infinitesimal element of area differs from the area of the infinitesimal rec- tangle by an infinitesimal of a higher order, and it follows that 1 = 5 /^h A = limit %y-Ax= I y-dx when limit Aa; = 0. APPLICATIONS 49 Example. — Find tlie area of the surface bounded by the parabola y^ = 4:X, the X-axis, and the lines a; = 4, a; = 9. For this curve Y y = 2 a;* and ^ = 2 f xi-dx=* x^ =28. The area from the vertex to the point (a;, y) is A = 2 ry.dta;=|a;*. The area from x = a to x= b is A = 2 Cx^ ■ da; = 1(6^ - a'). The area of the surface bounded by the parabola y^ = i x, the y-axis, and the lines y = 4, ?/ = 6 is /*6 /^6 A = J^ xdy = \l y'--dy = r = 12.67.* * The areas of surfaces bounded by curves whose equations are not known may be found mechanically by means of an instrument called the planimeter. E 50 DIFFERENTIAL AND INTEGRAL CALCULUS PROBLEMS 1. Find area bounded by y = 3x, the lines x = 0, x = 8 and the X-axis. 2. Find area bounded by y = 5x, the lines x=l, a; = 4 and the X-axis. 3. Find area bounded by y=mx+n, the lines x = a, x = b and the X-axis. 4. Find area bounded by y = mx + n, the lines y = a, y = b and the Y-axis. 5. Find area bounded by y^=9x, x — 0, a; = 4 and the X-axis. 6. Find area bounded by the parabola y' = 2px, the or- dinate of the point (x, y) of the parabola and the X-axis. 7. Find area bounded by the parabola y^ = 2px, the ab- scissa of the point (a;, y) of the parabola and the Y-axis. 8. Find area bounded by x^ + y^ = a' and the coordinate axes. 9. Find area bounded hj y''=9x, y = x and a; = 4. 10. Find area bounded hj y^ = 9 x, y = 2x and y = 6. 1 1 . Find area bounded hy y' = ix and y=^x. 12. Find area bounded by the parabolas y'= 4 x and x'= 4 y. Akt. 20. — Area of a Surface of Ebvolution Denote by A the area of the surface generated by the revo- lution of the continuous curve y = f{x) from a; = a to x= b about the X-axis. A is the limit, virhen limit Aa; = 0, of the sum of the areas of the frustums of cones of revolution APPLICATIONS 51 generated by the revolution of the parts of the broken line ANB. Hence A = limit'i'2 .^-iMi:^^ A + ^]i Ax x = a 'iX'+i/- when limit Ax = 0. Example. — Find the area of the surface generated by the revolution of the line y = 2x about the X-axis and bounded by planes perpen- dicular to the X-axis at a; = 3 and a? = 8. Here ^ = 2 TT ^2 a;(l + 4)i . do; = 4 VS TT Pa; da; = = 2 V5 TT • 55 = 110 V5 TT = 765.592. PROBLEMS 1. Find the area of the surface generated by the revolution of the line y = 3 a; -1- 5 about the X-axis and included by the planes perpendicular to the X-axis at a; = 0, x = 5. 2. Find the area of the surface bounded by the revolution of the line y=c about the X-axis and bounded by planes perpendicular to the X-axis at a; = a, a; = &. 3. Find the area of the surface of revolution generated by the line y = mx + n revolving about the X-axis and included by planes perpendicular to the X-axis lA x = a, x = b. 4. Find the area of the surface generated by the revolution of x^ + y^ = a' about the X-axis. 52 DIFFERENTIAL AND INTEGRAL CALCULUS Aet. 21. — Volume op a Solid of Revolution Denote by V the volume of the solid bounded by the surface generated by the Ur-Ax--", TT^ revolution of the contin- uous curve y =/(») about the X-axis and planes per- pendicular to the X-axis at x= a, x — b. Denoting by AFthe vol- ume of the part of the solid included by planes perpendicular to the X-axis at X and x + Ax, Fig. 12. and 7r2/=. Aa; < AF< t(2/ -f- Ayf ■ Ax, AY- Trf-Ax = eTr(2y-Ay + Ay^Ax, where 6 is less than unity. When Ax becomes an infinitesi- mal, Ay also becomes an infinitesimal. Hence in the limit when limit Ace = 0, AV=iry^Ax, and V = 1 Try' Ax = ir \ y'' dx. Example. — Find the volume of the prolate spheroid. The prolate spheroid is gener- ated by the revolution of the ellipse — + ^ = 1 about or ¥ the X-axis. The entire spheroid is comprised be- tween x= + a and x = — a. Hence, fm. is. •/-a (X APPLICATIONS 63 6^ a- a^x — ^a^ ■■^vob". PROBLEMS 1. Eind the volume of the solid generated by the revolution of the ellipse — + ^ = 1 about the F-axis. This is called the a- 0^ oblate spheroid. 2. Find the volume of the part of the sphere si^ + y'' + z' = 25 between x = 2 and x = 4. 3. Find the volume of the solid bounded by the surface generated by the revolution of the parabola y' = 2px about the X-axis and the plane x = a- 4. Find the volume of the solid bounded by the surface generated by the revolution of — — i- = 1 about the X-axis and the planes x = c, x = d, where c and d are greater than a. 5. Find the volume of the solid bounded by the surface generated by the revolution of y = 3x + 2 about the X-axis and planes perpendicular to the X-axis at x — 2, x = 7. 6. Find the volume of the solid bounded by the surface generated by the revolution of x^ + y^ = a^ about the X-axis and the planes a; = 0, x= a. Art. 22. — Solids Generated by the Motion of a Plane Figure Example. — The ellipsoid — \- — -\ — = 1 may be generated a^ r & by an ellipse whose center moves on the X-axis, whose plane 54 DIFFERENTIAL AND INTEGRAL CALCULUS is perpendicular to the X-axis, and whose axes in any position are the intersections of the plane of the generating ellipse with the fixed ellipses, The area of the generating ellipse in any position is IT rs • rt. Since (x, rs, 0) and (a;, 0, rt) are points in the ellipsoid ^ + ^' + ?-' = !, rs = -(a^ -x^^ a.iid rt = -(a'-x^K c? h^ AF> (X - AX) Aw, AF=XAa;-e-AX-Aa;, be, where AX represents the change of X = ir —„(a' — x') corre- al sponding to a change of Ax in the value of x and 6 is less than unity. Since AX becomes an infinitesimal when A* becomes an infinitesimal, in the limit, when limit Ax = 0, the volume of the ellipsoid is : limit 2 X-Ax= | i=— o */ — X-dx bo r^' = '^-; I (a^ — x^ dx = 4 T abc. a'J-a APPLICATIONS 55 PROBLEMS 1. Find the volume of the part of the elliptic paraboloid ,,2 ~1 ''--{ — =2x included between the planes a; = and x= 5. 9 4 ^ 2. Find the volume of the solid bounded by the hyperboloid of one sheet — + ^ — 2^ = 1 and the planes « = 3, « = 5. 3. Two equal semi-circles x' = 2 rz — z^, -f = 2rz — z^ lie in the perpendicular planes X^, YZ. The solid generated by the square whose center moves on the Z-axis, whose plane is per- pendicular to the ^axis, and whose dimensions in any posi- tion are the chords of intersec- tion of the plane of the square with the fixed semi-circles, is called a semi-circular groin. Find the volume of this groin. Notice that the groin might be defined as the solid which the two cylinders x^ = 2 r« ■ common. and 'if = 2rz — z^ have in 4. Two semi-circles a? = 2riZ — z', y^ = 2r2Z — z^ ot unequal radii lie in the perpendicular planes XZ, YZ. Find the volume of the groin generated by the rectangle whose plane is perpendicular to the Z-axis and whose dimensions are the chords of intersection of the plane of the rectangle with the given semi-circles. Take depth of groin d. 5. The two parabolas a?=2piZ, y''-=-2'p^z lie in the per- pendicular planes XZ, YZ. Find the volume of the groin 56 DIFFERENTIAL AND INTEGRAL CALCULUS generated by the rectangle whose plane is perpendicular to the ^axis and whose dimensions are the chords of intersection of the plane of the rectangle with the given parab- olas. Take depth of groin d. 6. Two circular cylinders have equal bases and equal altitudes. Their lower bases are tangent to each other, their upper bases coincident. Find the vol- ume of the solid common to the two cylinders. CHAPTER V SUOOESSIVE ALGEBEAIO DIPFEEENTIATION AND INTEGEATIOU , is denoted by —j^fi^) orf"(x), and is Art. 23. — The Second Derivative The first derivative of /(a;), denoted by — f(x) or f'(x), is in general a fanction_of x. The first derivative of the first derivative, — — f(x) dx\_dx called the second derivative of f(x). If f(x) is denoted by y, the first and second derivatives are written — ^ and — %• dx dx^ For example, if f(x) = x^-7x + 7, f'(x) = 3x^-7, f"{x) = 6 x. Geometrically -J- measures the slope of the tangent to the curve whose equation is y =/(x) at (x, y). Hence ■ — | — ) = — ,, J. J, , dxKdxJ dar measures the rate of change ,, ^ ' of the slope of the tangent. When -^ is positive, -^ daP ^ ' dx or the slope of the tangent increases as x increases. From the figure this is seen to be the case for the part cde of the curve, that is, when the curve is concave FiQ. 17. 57 58 DIFFERENTIAL AND INTEGRAL CALCULUS upward. When — ^ is negative, the slope of the tangent de- dx- creases as x increases. This is the case for the part abc of the curve; that is, when the curve is convex upward. When — ^ = 0, the tangent is stationary. A point of the curve where da? the tangent is stationary is called a point of inflection, since at such a point the curve changes its direction of curvature. If the relation between the distance passed over and the time of the motion of a body is expressed by the equation ds s = f(t), — is the velocity at any time t. If the velocity has different values for different values of t, represent the velocity at time to by v^, at time t by v. The average rate of change of velocity in the interval of time t — to is ~ ' • Calling the interval of time A^, the change in velocity Av, the ratio — At measures the average rate of change of velocity in the interval of time A^ This is true whatever may be the magnitude of the interval of time At. The limit of this ratio when limit A* = is the actual rate of change of velocity at time .t, for the interval of time during which the rate of change of velocity might vary continually decreases. The actual rate of change of velocity is called acceleration. Hence acceleration T ., AV dv d ds d's ,, i -j? ^^/^o.^ = limit — = — = = — -• For example, if s = 16.08 i. At dt dt dt df ^ ' ' denoting the acceleration by a, a = 32.16. As a direct consequence of the definition d^ f( s _ d ^|/(.)] it follows that/|l/(.) =|/(.) + 0. For example, let it be required to find the function whose second derivative is 2 a; — 6. Denoting the function by f{x), — f(x)=2x + 5. Integrating, —f(x) = x'-5x+G,. Inte- dor' dx SUCCESSIVE DIFFERENTIATION 59 grating again, f(x) = \3? ~ ^3? + dx + C^. Oj and Oj are two arbitrary constants of integration, whose values become known if the function and its first derivative are required to take given values for given values of x. For instance, if it is re- quired that f{x) = 10 when a; = 2 and — f{x) = 1 when a; = 0, Ci = 1 and C-i = 15 1. Under these conditions, if ^J(x) = 2x-5,fix) = i^-^x' + x + 15i. PROBLEMS Find the first and second derivatives of, 1. 3»2 2. 4.3^-5. 3. 2a^-7a?. .1 -1 „ X 4. — 5. X 1 —X 1 + a^ Find between what values of x the curves which represent the following equations are convex upward, concave upward, and find the points of inflection, 7. y = a^-7x + 7. 8. 2/ = — ^„- 9. 2/ = ^-a^ + 2. l + x' 3 Determine the first and second derivatives of y with respect to x in the implicit functions, 10. a? + f= 7^. The first derivative is ^ = - -• dx y Differentiating both sides of this equation with respect to x, d^_ dx da? ~ y^ Substituting ^ = _^, there results ^ = -t.. dx y da? ]^ a? X ''■ ^ + ^='- ''■ i-h = '- ''■ ^^ = 2p.. 60 DIFFERENTIAL AND INTEGRAL CALCULUS Find — from the following equations, dx' 14. ( — J =y^. Differentiating both sides of this equation with respect to a;, 2 ^ ^ = 3 y'% whence ^ = f y\ dxd'j? dx da? 15. ^ = y\ 16. ^ = xy. 17. ^ = ^. dx dx dx X 20. The equation f(x, y) = defines y as a continuous func- tion of X. Consequently it also defines a; as a continuous function of v. It has been shown that -^ = — - Prove that ^ dx dx ^ dy d'y _ dy' dx'~~7M?' A body moves in such a manner that the relation between s and t is expressed by the following equations. Find velocity and acceleration at time t. a and u are constants. 21. s = u-t + ^a-t'. 23. s = u-t' + a- f. 22. s = u-t — ^a-tK 24. s=u-t'-a-f. Determine the functions whose second derivatives are, 25. 3 a; + 2, knowing that when x=l, f(x)=10; when x = 0, ^f(x) = 3. dx 26. 5 a; — 7 a;", knowing that when a; = 0,/(a;) = 0, ^/(^) = 1- 27. Find the curve through the origin and making an angle of 45° with the X-axis at the origin, knowing that — ^ = 2 a; -f- 6. SUCCESSIVE DIFFERENTIATION 61 28. Find the curve through (6, 7) and parallel to the X-axis at a; = 3, knowing that — ^ = 2 k^ — 4 a;. dxr 29. A body starts moving with a velocity u and has a uniform acceleration a. Find the relation between s and t. The conditions of the problem are expressed by — ^ = a, and ds • • "^^ when f = 0, s = 0, — = m. By integration ctv di ds Since s = and — = u when < = 0, Ci = u and C2 = 0. Hence the resu.lt is s = ut + ^ af. 30. A body starts moving with a velocity u and has a uniform acceleration — a. Find the relation between s and f. 31. A horizontal beam has one end fixed. The bending of the beam due to a weight P causes an elongation of the upper surface of the beam and a com- pression of the lower surface. -^-^ Some intermediate surface re- __1____Z Z1 -— -^_._ mains unchanged in length and 1 , 1 , | -2/t^I!!;;i3^ is called the neutral surface. 1 .i <■ 1- • The figure shows a vertical f?\ longitudinal section of the ^*^ , , , Fis. 18. upper, lower, and neutral sur- faces of the beam. This section of the neutral surface is called the elastic curve. In text-books on Strength of Mate- rials it is proved that for any point (x, y) in the elastic curve, EI —^ = — P-x, where jE is a constant depending on the mate- dor rial of the beam, / a constant depending on the cross-section of the beam. By the nature of the problem ?/ = when a; = 0, 62 DIFFERENTIAL AND INTEGRAL CALCULUS and -^=0 when x= I. Determine the equation of the elastic dx ^ curve and the maximum deflection of the beam. 32. Suppose the load uniformly distributed over the beam of Problem 31. If the load per linear unit is w and the origin . -| of coordinates is taken at the , I I free end of the beam, it is I _l__^jyQ QQQQ[Vy,^ proved in Strength of Materials .1 . I I ~--^^^^^-<^Qd, ^^^^ ^^^ ^^y point (^) y) ii" the :^^ 2?t^::^ a'y -wa? _i— ^ elastic curve EI—^ = - -I 1 dx^ 2 Fia- 19- By the nature of the problem -^ = when x=l, and 2/ = A when x = l, where A represents dx the deflection of the free end. Determine the equation of the elastic curve. 33. A horizontal beam rests on two supports the distance between which is Z. If a load P is placed at the middle of the beam and the origin of coordinates is taken at the left support, it is proved in Strength of Materials that for any point (x, y) in the elastic curve EI—^ = —-. By the nature of the prob- lem _i = when x = Xl, and v = when a; = 0. Find the dx ^ > n equation of the elastic curve and the maximum deflection. 34. The beam of Problem 33 supports a uniform load. If w is the load per linear unit, it is proved in strength of mate- rials that for all points in the elastic curve ■p-rd^y _ wlx W3? By the nature of the problem -^ = when x = \l, and y = SUCCESSIVE DIFFERENTIATION 63 when a; = 0. Find the equation of the elastic curve and the maximum deflection. 35. A circular disc whose weight is W is free to turn about a horizontal axis through the center of the disc and perpen- dicular to the plane of the disc. A cord wound around the circumference of the disc has a weight P attached to its free end. In Mechan- ics it is proved that d'g^ P-r-g dt^ W-k' + P-T^' where g is the earth's gravity constant, r is the radius of the disc in feet, k is a constant depending on the size of the disc, and 6 is the angle in radians through which the disc turns in t seconds. Find the relation between 6 and t. In the following problems find -^• dx 36. — ^= y^- Multiply both sides of the given equation by -^ and integrate. There results LJ Fig. 20. dx da? dx ^^^ KIJ=^^+^- 37. g = 2^-f 0^. 38. ^=2y^-5y. aar dx dx' Art. 24. — Maxima and Minima A continuous function f(x) increases as x increases when its first derivative — fix) is positive, and decreases as x increases dx when its first derivative is negative. If f{x) changes from an 64 DIFFERENTIAL AND INTEGRAL CALCULUS increasing to a decreasiag function when x = x^, the value of f{x) is greater when x = Xa than it is just before x reaches x^, and also greater than it is just after x passes x„. This is expressed by saying that f{x) has a maximum value when X = Xq. At a maximum value of f(x) therefore the first derivative — f(x) changes sign from + to — for increasing dx values of x. If f{x) changes from a decreasing to an increasing function when X = Xg, the value of f{x) is less when x= x,, than it is just before x reaches a;, and also less than it is just after x passes Xo- This is expressed by saying that f(x) has a mini- mum value when x = x^. If f{x) changes from a decreasing to an increasing function, the first derivative must change sign from — to +. Hence at a minimum value of f{x) the first derivative —fix) changes sign from — to + for increasing dx values of x. An algebraic function can change sign only by passing through zero or by passing through infinity. Hence when f{x) has a maximum or a minimum value, — /(a;)=0 or —f{x)=ao. ■J O/OC QiiC -J If — f(x) = and f(x) has a maximum value, — f(x) changes sign from + to — by passing through zero. Hence — f(x) is a decreasing function, and its first derivative, which d" is the second derivative of f(x), that is — ■,/(*)) must be negative. ^ If — f(x) = and f(x) has a minimum value, — f(x) dx dx changes sign from — to + by passing through zero; hence — f(x) is an increasing function, and its first derivative, which dx is the second derivative oif(x), must be positive. These results, obtained analytically, may also be directly SUCCESSIVE DIFFERENTIATION 65 obtained from the graph of y =/(x). For increasing values of X the point generating the continuous curve which represents the equation y =f(x) can cease moving away from the X-axis and start moving towards the X-axis only in one of two ways, either by tending to move parallel to the X-axis, as at Fi, Y R .P' which requires that -^ = 0, or by tending to move parallel to ax J the F-axis, as at P^, which requires that -^=00. At P., , dx while -^ = 0, 2/ is neither a maximum nor a minimum. At P4, dx while -^=00, y is neither a maximum nor a minimum. An dx examination of the figure shows that in all cases for increasing values of x the slope of the tangent, that is — , changes sign dx from + to — at a maximum, from — to + at a minimum. At a maximum, when -^ = 0, the curve is convex upward, and hence — ^ is negative ; at a minimum, when ^ = 0, the curve di? « dx is concave upward, and — ^ is positive. From the figure it is evident that maximum and minimum values must occur alter- nately and that a minimum may be greater than a maximum. From this investigation is inferred the following method of examining a function of one variable f(x) for maximum and 66 DIFFERENTIAL AND INTEGRAL CALCULUS minimum values. Find the roots of the equations — f(x) = and — f(x) = 00. If, for increasing values of a;, — / (a;) cix ax changes sign from + to — when x passes one of these roots, this root makes /(k) a maximum; if — f(x) changes sign from — to + when x passes one of these roots, this root makes fix) a minimum ; if — f(x\ does not change sign when x dx passes through one of these roots, this root makes f(x) neither a maximum nor a minimum. The roots of the equation , = which make ■^^^^ nega- ^ dx da? ^ tive make fix) a maximum ; the roots of ^^' = which make ■' \^ ' positive make fix) a minimum. (lit Example I. — Examine ar* — 7 a; + 7 for maximum and minimum values. Here /(«) = a;' — 7 a; + 7, "^^ ' = 3a^ — 7, -^— = 6 a;. Equating the first derivative to zero, Sa? = 'I and a; = ± V J. x = + V|- makes ,^ positive and /(a;) /- ., dYix) = — .145, a minimum, a; = — V^ makes , » negative and fix) = 14.145, a maximum. ix 4- 2V Example II. — Examine y = '' ^ — l- for maxima and (a; - 3)'' minima. Here -^ = V^ + ^ '■^ "~ ' ) . The first derivative ■^ da; (a; — 3)= equated to zero gives x = — 2, a; = + 13 ; the first derivative equated to infinity gives a; = + 3. When x is just less than — 2, the signs of the three factors of -^ are "' and -^ is dx — dx positive ; when x is just greater than — 2, the signs of the factors of — are — — and -i is still positive. Since — dx — dx dx SUCCESSIVE DIFFERENTIATION 67 does not change sign when x passes through —2,yis neither a maximum nor a minimum when x = — 2. When X is just less than +13, the signs of the factors of _i are — — and — ? is negative ; when a; is j ust greater than dx + dx tj M J- tJ + 13, the signs of the factors of -i are '" and -i is posi- dv dx + dx tive. Since -i changes sign from — to + when x passes dx through 13, ?/ is a minimum when x = 13. This minimum is y = 33|. When X is just less than + 3, the signs of -^ are ■ and dv ~ -^ IS positive ; when x is just greater than + 3, the signs of -/- are and -^ is negative. Since -^ changes sign from dx + dx ° dx ° ° + to — when x passes through +3, ?/ is a maximum when a; = + 3. This maximum is y = oo. PROBLEMS Examine the following functions for maximum and mini- mum values : 1. x^-Zx + 5. 9. X-1 2. a:^-4ci; + 7. (x + 2y 3. 2«?-5x. 10. {x~iy{x + 2Y. 4. a?-2x\ 11. y = 2x-a?. 5. ix'-^x^-^: x^ + 2x. 12. y = 2Rx — a?. 6. a^_5a^_10a ; + 4. 13. 2/ = ^/2aa;-af), 7. (a; -1)2 {x + lf X 14. IK a 2/ = a;5 + 3a; — 5. l + x^ (2 a; -3)''' 68 DIFFERENTIAL AND INTEGRAL CALCULUS 16. 2/ = 10 V8 X — x'. Suggestion. 10 VSa; — a^ is a maxi- mum when 8 a; — »^ is a maximum, a minimum when 8x — a? is a minimum. Hence constant factors may be dropped and the radical sign removed before forming the first derivative. 17. V3a^-10x + 6. 18. -^J- i- 19. SVSa^-lOw^. ^ X -\-3 20. Show that x^ — Bos' + 6x lias neither a maximum nor a minimum value. 21. Show that — — — is a maximum value of ax^ + 2bx + c a when a is negative, a minimum value when a is positive. 22. Divide a into two parts whose sum is a minimum. 23. Divide a into two parts such that the sum of their squares is a minimum. 24. Divide a into two parts such that their product is a maximum. 25. Divide a into two parts such that the sum of their square roots is a maximum. 26. From a square sheet of tin 18 inches on a side equal squares are cut at the four corners. From the remainder of the sheet of tin a vessel with open top is formed by bending up the sides. Find side of small squares when the vessel holds the greatest quantity of water. 27. From a rectangular sheet of tin 3 feet by 2 feet equal squares are cut at the four corners and a box with open top formed by turning up the sides. Find sides of squares cut off when contents of box are greatest. 28. A box, square base and open top, is to be constructed to contain 108 cubic inches. What must be its dimensions to require the least material ? SUCCESSIVE DIFFERENriATION 69 29. A circular cylindric standpipe is to be built to hold 10,000 cubic feet of water. Find altitude and diameter of base whicli require least material. 30. Find the shortest distance from (3, 6) to the line 2 3 31. Find the shortest distance from (7, 8) to the parabola 2/2 = 4 a;. 32. Find the shortest line that can be drawn through (a, b) meeting the rectangular axes. 33. A Norman window is composed of a rectangle sur- mounted by a semicircle. Find the dimensions of the window so that with a given perimeter the window admits the greatest amount of light. 34. Two trains are running, the one due east at 30 miles per hour, the other due north at 40 miles per hour. When the first train is 30 miles from the intersection of the tracks the second is 20 miles from this point. Find the least dis- tance between the trains. 35. A person 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 beach. If he can walk 6 miles an hour, but row only 4 miles an hour, where must he land ? 36. The strength of a rectangular beam varies as the product of the breadth and the square of the depth. What are the dimensions of the strongest beam that can be cut from a log whose cross-section is a circle 18 inches in diameter ? Strength pre- vents breaking. PraT^. 70 DIFFERENTIAL AND INTEGRAL CALCULUS 37. The stiffness of a rectangular beam varies as the product of the breadth and the cube of the depth. Find the dimen- sions of the stiffest beam that can be cut from a log whose cross-section is a circle 18 inches in diameter. Stiffness pre- vents bending. 38. The bending moment of a simple beam whose length is I when the uniform load per ( nt I '*' " I linear unit is w is "H Ft"'"' at the point whose distance ^'°' ^^' from the left support is x. Find where the bending moment is a maximum. 39. The distance between two lights A and B is d. The intensity of A at unit's distance is a, of B is 6. If the inten- sity of a light varies inversely as the square of the distance, find the points between the lights of maximum and minimum illuminabion. 40. If the illumination varies as the sine of the angle under which light strikes the illuminated surface divided by the square of the distance from the source of light, find the height of an electric light directly over the center of a circle of radius r when the illumination of the circumference is greatest. 41. If c^r+- is the total waste per mile going on in an r electric conductor, r resistance in ohms per mile of conductor, c the current in amperes, t a constant depending on interest on investment and depreciation of plant, find the relation between resistance and current when the waste is a minimum. 42. If v is the velocity of an ocean current in knots per hour, X the velocity of a ship in still water, and if the quantity of fuel burnt per hour is proportional to a?, find the value of x SUCCESSIVE DIFFERENTIATION 71 which makes the consumptioa of fuel a minimum for a run of s miles. 43. Given n voltaic cells of E. M. F. e and internal resist- ance r, to find the way in which the cells should be arranged to send a maximum current through a given external resist- ance R. Let X cells be placed in series, then the current / = xe ay'r + B 44. The equation of the path of a projectile is y = x tan 6 " — 2 w^ cos^^' where 6 is the angle of projection, u the velocity of projection, g = 32.16. The range, the value of X when y = 0, is u'sin{26) a. Find the greatest height. 6. Find the angle of projec- tion which gives the greatest height for a given velocity. '^'®' ^' c. Find the angle of projection which gives the greatest range for a given velocity. 45. Find the dimensions of the isosceles triangle of maxi- mum area that can be inscribed in a circle of radius r. Calling altitude of triangle x, base 2y, t/^ = 2 ra; — a^, and area A= x V2 rx — x''. ^ is a maximum when ^' = 2 ra;^ — cB'' is a maximum. = 6 ra^ — 4 x', dx which is zero when a; = f r. The first derivative changes sign from -|- to — when X passes through ^r. Therefore 72 DIFFERENTIAL AND INTEGRAL CALCULUS the area of the inscribed triangle is a maximum when its alti- tude is f r. 46. Find the dimensions of the rectangle of maximum area that can be inscribed in the ellipse — + -^ = 1. The area of y the rectangle is A = A,xy, which is a maximum when A' = a!2/ is a maximum. dA' dx dii , ax Fio. 26. '* y = — -, and the area of the maximum V2 From the equation of the di/ 1) Sj ellipse -^ = — . By sub- da; a'y fj A' nil f) ^r stitution, = —^—z — '—. Hence A' is a maximum when dx ary a'y^ — bV = 0. Combining this equation with the equation of the ellipse, x ■- rectangle is 2ab. 47. Find the area of the maximum rectangle that can be inscribed in the parabola y^ = 2px whose limiting coordinates ^ are a, b. 48. Find the dimensions of the cylinder of revolution of maximum volume that can be in- scribed in a sphere of radius r. Calling the radius of the base of the cylinder x, the altitude 2 y, (x, y) is a point of the circle x'-\-y'^=r^ which generates the Fio. 2t. Sphere. Calling the volume of SUCCESSIVE DIFFERENTIATION 73 the cylinder V, V=2 ■iryx' = 2iry (r^ — y'). F is a maximum when -— = 2 Trr^ — diry^ = 0, that is when y = -^, since this value of y makes — - = — 12 ttm negative. The maximum is 4 dy' F= -irr^. The ratio of the volume of the maximum 3V3 inscribed cylinder to the volume of the sphere is ^^• V3 49. Find the cone of maximum volume that can be inscribed in a sphere of radius r. 50. Find the cylinder of maximum volume that can be inscribed in a cone, altitude h, radius of base r. 51. Find the maximum cylinder that can be inscribed in the prolate spheroid. 52. Find the maximum cylinder that can be inscribed in the oblate spheroid. 53. Find the cone of maximum volume that can be in- scribed in a paraboloid of revolution, the limiting point of the generating parabola y^ = 2px being (a, 6), the vertex of the cone at intersection of axis of paraboloid with base of paraboloid. 54. Of all right circular cones whose convex surface is the same find the dimensions of that whose volume is greatest. V=^Try\ — = \T,\2xy-^ + oA and iryV^'+f = constant. a^y^ + y* = constant, and ^ = "'"^ , - A Hence §E=i./^2||-^^Y Equating the /f\ first derivative to zero, x = y V2, which / ^_i \ makes V a maximum since — changes sign (^ i—y^J^ dx \^,_ ^ from -|- to — when x passes through y V2. pie. 28. 74 DIFFERENTIAL AND INTEGRAL CALCULUS 55. Of all right circular cylinders whose convex surface plus the surface of the lower base is constant, find the dimen- sions of that whose volume is greatest. 56. The distance between the centers of two spheres of radii r and B is d. Find on the line joining the centers the point from which the greatest amount of spherical surface is visible. 57. Find the axis of the parabola of maximum area that can be cut from a right circular cone, radius of base r, altitude h. Art. 25. — Derivatives of Higher Orders It has been found convenient to call the first derivative of the first derivative of f{x) the second derivative of /(»), and to denote it by — ^/(a;) or /"(»). In like manner it is found convenient to call the first derivative of the second derivative the third derivative of /(«), and to denote it by the symbol -—f(x) or f"'{x). By an extension of the same notation /"(a')) ri^)j •■■> f\^) denote the fourth, fifth, •••, nth derivar tives of f{x). For example, if fix) =a^ + 3a?-7x'-27x-18, /'(») =ia? + 9xP-Ux-27, f"{x) =12a^ + 18a;-14, /"'(a;) = 24x + 18, r{x)=2l, r(.x) =0. It follows immediately from the definition that the integral of any derivative is the next lower derivative. SUCCESSiri! DIFFEBENTIATION 75 For example, if /"(a;) = 24 a; -18, /"'(a;) = 12a^-18a; + (7i, /"(a;) =^3?-93?+Ci-x+C^, fix) =x*-3a?+\Ci-3?+G.,-x + C^, f(x) =^x'-^x^ + ^G,-a!' + iO,-x'+C,-x+C,. The arbitrary constants Ci, Cj, Og, C4 become known if the function f(x) and its first three derivatives /'(a;), f"{x), f"'(x) are required to take given values for given values of x. The successive derivatives f\x), f'\x), f"'(x), •••, are called derivatives of the first, second, third, • • •, orders. If /"(a;) is given, f{x) is found by n successive integrations, and the general expression for f(x) must therefore contain n arbitrary constants.* PROBLEMS Form the derivatives of the first three orders of, 1. a:=-6a;^ + 15. 2. Sx' + ix-l. 3 1-x Find f(x) when, 4. f"{x) ^x'-Sx. 6. f"(x) = 15 a;^ + 7. 5. /"(a;)=l + a;. T.f"{x) = 10. Aet. 26. — Evaluation of the Indeterminate Form - The ratio - may have any value whatever; that is, the value of the ratio is indeterminate. If, however, the ratio of * In Lagrange's notation tlie successive derivatives of f(x) are denoted by /'W, /"W, /'"(«)•••; in Cauchy's notation, by Df(,x), D^f{x), D'^fipc) .... Newton denotes ^ by s. 76 DIFFERENTIAL AND INTEGRAL CALCULUS two functions -^-^ takes the form - for some special value of X, such as x = a, the true value of ,, (. when a; = a is defined as the limit of the ratio J^ — — ^ when limit Ax = 0. <^(a + Aa;) Since by hypothesis /(a) = and (/> (a) = 0, f(a + Aa;) -f(a) liniit %h^ = limit -^-^^——=m Aa; when limit Aa; = 0. Hence by the definition the true value of i-^ when a; = a is ^-yz-i, where f(a) and '(x) when a is substituted for x. If ^ ,,' = r,the same analysis shows that the true value of ^'(a) ^ is the , , — same order. rid) /(a) ^/M =/M. In general, the true value of =^ ,^(a) (x) x'-a^-x + l ' f"(x) 6 a; 3 , . ^ = 6^:1:2 = 2' ^^^"^ = 1- /j3 3a; -1-2 Hence 8^ is the true value of — — ~ — when x = l. af — x'—x + l SUCCESSIVE DIFFERENTIATION 77 PROBLEMS Find the true values of, 1. — 5- — wlien x = 2. a;2-14x + 24 x^-16 , , 2. when a; = 4. x' + x-20 „ ^-1 3?-l when X ■■ ^^— when x = l. 7a^-9x + 2 x*-2a? + '2x-l , 1 ■ when x = l. a;6- 15*2 + 24 a; -10 ,^ + 3c^-7a;^-27.-18 ^hena, = 3. x'^-3ii?-7x' + 27x-18 when x = l. X— i. X" — g" a; — a when X ■- (k — 1)^—1 ,. ■yfx — yfa + Vx — a 1, 10. ' when x — a. y/v? — a' 11. '^'*' ~ '^'^ when 0! = a. Here -^^ = - when a; = a for (a - a;)* ^"(a!) all values of n and the true value of the ratio cannot be found by the method of derivatives. The removal of the factor (a — a?)* from numerator and denominator leads to a deter- minate result. CHAPTER VI PAETIAL DirFEEENTIATION AND INTEGEATIOIT OP ALGEBEAIO FUNCTIONS Art. 27. — Partial Differentiation If to pairs of arbitrarily assigned values of x and y there correspond one or more determinate values of z, z is called a function of the two independent variables x and y. This is denoted by writing z =f{x, y). The function z = f(x, y) is said to be continuous at a%, y^ if limit f(xo + Sy ?/o + 82) — /{xq, yo) = when limit Sj = and limit 82 = 0. It follows that if f(x, y) is a continuous func- tion of X and y, it is also a continuous function of x and y separately. The converse is not necessarily true. The equation z = f{x, y), where x and y are independent variables and « is a continuous function of both x and y, when interpreted in rectangular space coordinates represents a curved surface. Let (xg, y^, z^ be any point in the surface. If x re- tains the fixed value a:,,, the equation z =f(xQ, y) represents the projection on the .^F-plane of the intersection of the plane x = Xq with the surface 78 Fig. 29. PARTIAL DIFFERENTIATION 79 % = f(x, y). Hz— f{x„, y) is differentiated with respect to y, the result is called the partial derivative of z =/(», y) with dz respect to y, and is denoted by the symbol — . Denoting by - -, dy ^ — 5 the value of — ■ when x — Xn, y = y^jZ = z^, — - is the slope Syo ■ Sy dy„ of the tangent to z =/(x„, y) at (yo, z^) and measures the rate of change at (x„, y^, Zg) of z when the point (x, y, z) moves along the curve of intersection of a; = Kq and z —fix, y). If y retains the fixed value i/o, the equation z = f(x, y^) rep- resents the projection on the ZX-plane of the intersection of the plane 2/ = 2/o with the surface z =f(x, y). It z = fix, y$j is differentiated with respect to x, the result is called the partial derivative of z=^fix, y) with respect to x, and is de- o2 nV ciz noted by the symbol — . Denoting by — - the value of — ax ~, dxa ax when x = Xa, y =:yQ, z = Zq, — - is the slope of the tangent to dxo z=fix, 2/o) at (»o, a^o) ^'^^ measures the rate of change at (xo, ya, Zo) of z when the point (a;, y, z) moves along the curve of intersection of y = yo and z =f{x, y)* The equations of the straight line tangent to the curve of intersection of the plane x = Xq and the surface z =f(x, y) at (a;,,, y„, «„) are x = Xq and » — »o = — ^ (2/ — 2/o) j the equations of dyo the straight line tangent to the curve of intersection of j/ = 2/0 dz and z =f{x, y) at (% y^, z^) are y = ?/„ and z - ^o = ^ (a: — ^a)- The plane containing these two tangent lines is the tangent plane to the surface z=f(x, y) at (% y„, «(,). In the analytic geometry of three dimensions it is proved that the plane Aix — x^ + Bijj — 2/0) + O (z — 2o) = through the point (a;o, 2/0) ^0) contains the line x — XQ = a{z — z^y — y„='biz — z^, *The use of 9 to denote partial differentiation was introduced by Jaoobi (1804-1851). 80 DIFFERENTIAL AND INTEGRAL CALCULUS when Aa + Bb + C = Q. Hence the equation of the tangent plane is found to be « — «„= — (x — x„) +—(y— Vo) • dxa dyo Denoting by Z the angle made by the tangent plane with the XF-plane, cos Z-- The normal to the curved surface z =f(x, y) at («;„, y^, «o) is the line through {x,,, yo, «o) perpendicular to the tangent plane at this point. Hence the equations of the normal are X-X,= -^{Z-Z,), y-y, = -^(z-Zo). dxo dyo For example, let it be required to find the equation of the tangent plane and the equations of the normal to the sphere a^ + y^ + z' = 14: at (1, 2, 3). Differentiating, regarding y con- dz staut, x + z — =0. Differentiating, regarding x constant, dz ^* y + z — =0. At the point of tangency dy rr —A ,/ — 2 « — ^ 32o — _ 1 ^^0 _ _ 2 Xq — i-, y V) ± ^ where e de- ^ ■'^ dx Aa; ' pends on x, y, and \x, and approaches zero with Aa;. It is convenient to write e = ei/i(a;, y), where ej depends only on Aa; and approaches zero with Aa; while /i (x, y) must be finite. Similarly 5% _ f(x+^x, y +Ay) — f(x, y + Ay)— /(a; + Ag;, y) +f(x, y) dydx Ax Ay Ay where limit ./K^) y + ^y) -/i("'» 2/) when limit Aw = 0, must Ay 88 DIFFERENTIAL AND INTEGRAL CALCULUS be finite, e^ must vanish with Aj/, and /s (a;, y) must be finite by hypothesis. In like manner ^= f{'^>y + ^y)-f(^>y) ± e^fs (x, y) and dy Ay d'e _ (x + Ax,y + Ay) — f(x + Zia;, y) — f(x, y + Ay) + f(x, y) dxdy Ay Ax ± ,,M-±^^.M}^iMml ± .,/,(«,, y), Ax where £3 must vanish with Ay, £4 must vanish with Ax, limit f3{x + Ax,y)-Mx,y') ^^^^ j.^^^^ Ao; = must be finite, Ax and f^{x, y) must be finite. It follows that when limit Ax = Q and limit Ay = 0, lij^j^. f{x +Ax,y + Ay) -f(x + Ax, y) -f(x, y + Ay) +f(x, y) Ay Ax ^ d'z ^ ff'z dxdy dydx PROBLEMS '^"^ asc' aa:^' ayaa;' dy dy^' dxdy ' 1. 2 = a;^+y^. 2. 2 = a^-^y. 3. 2 = ^±^- 4. 2 = a;(y-2). CB— y 5. 2 = cB+y 6. 2 = aV. 7. 2 = x^y^. 8. 2 = a^y" Integrate 9. — = a^y + 3a; — 5y + 2. Integrating, eon- dx^ sidering y constant, — = ^ a^y + f a;^ — 5 ya; + 2 a; + /i (y). Integrating again, considering y constant. The arbitrary functions /i(y) and /^{y) become known if 2 and — are known for some value of x. Suppose that when a; = 0, aa; PARTIAL DIFFEBENTIATION 89 z = y+3,^=f + S. Then My) = f + 5, My) = y + 3, and the result becomes z = J J a;''!/ + -|- a;' + I ya^ + a^ + xy'' + 5x + y + 3. 10. dccdy = xy + 2x — 7y+5, knowing that, when x = 0, dz z = y — 2 and when y = 0, — =2oiP. dx 11. —— = 3a;?/l oyax 12. ^=xy'- 13.g = a.-^. 14- ^ = (3' -3) (2/ -2). 5a;ay ^ ^^^ ^ 15. _5^ dydx = x(y-5). Aet. 31. — Area of Any Curved Surface Let the equation of the curved surface be z=f{x, y) con- tinuous in X and y. Planes perpendicular to the X-axis at intervals of Aa; divide the given surface into strips of surface CDEF. Planes perpendicular to the y-axis at intervals of Ay divide these strips into elements of surface abed. The V = BC strip CDEF = 2 abed, and the given surface s,=o x=OA x = OA y = BO A = -S, CDEF =-2. 2 abed. a: = 1 = y = This summation holds what- ever be the magnitude of Aa; and Aj/. The projection of the element of surface abed on the XT^plane is the rec- tangle ai&iCidi, whose area is Ax Ay. The plane through ad and the line aV parallel to ttifti intersects the prism which 90 DIFFERENTIAL AND INTEGliAL CALCULUS projects abed on the XF-plane in the parallelogram ab'c'd. Denoting by 6', the angle made by the plane of ab'c'd with the XY"-plane, area ab'c'd = f- When Aa; and Aw ap- proach zero, the plane of the parallelogram ab'c'd approaches the tangent plane to the curved surface at a, and the paral- lelogram ab'c'd approaches the surface element abed. Hence, when limit Aa; = and limit Ay = 0, = 0A y^BO x=OA y = BC x = OA y = BC A = l 2 abed = limit S S ab'c'd = limit S 2 ^^^ 1=0 j=o 1=0 j=o 1=0 v=o cos 6', J--x=OA /» 1 = t/t/-- y = BC/ a«2 flyZNl Example. — Find the area of the groin formed by the inter- section of the semicircular cylinders a^ + z^ = r', y^ +z^ = r'. One-eighth of the surface of the groin is bounded by the cylinder ay'+z^='r^, the .ZX-plane, the Xy-plane, and the intersec- tion of the cylinders, which intersection lies in the plane y=x. Hence for this part of the surface of the groin 92 _ _ a; dz dx and = 0, Fia. 81. i/i=0 c/j=0 dxdy =0 (f-a?)^ z' dy 1=0 ^} = \^ i Xi=>- xdx =0 (f— a!2)i 'dxdy PARTIAL DIFFERENTIATION 91 Aet. 32. — Volume op Any Solid Denote by V the volume of the solid bounded by the con- tinuous surface z—f(x,y) and the coordinate planes XY, XZ, YZ. Planes perpendicular to the X-axis at intervals Ax divide the solid into laminae LL'; planes perpendicular to the y-axis at intervals A?/ divide these laminae into prisms FP'; planes perpendicular to the Z-axis at intervals Aa divide these prisms into elements of volume aa'. The volume of aa' = Ax- Ay ■ Az; the volume of PP' = S Ax ■ Ay ■ Az—6 ■ Ax ■ Ay • Az, where 6 is less than unity. When Ax, Ay, Az approach zero, volume of PP' = 2 dx-dy-dz — 6-dx-dy-dz. Since 6 - dx • dy -dz is an infinitesimal of a higher order than 2 dx-dy •dz = PD • dx ■ dy, in the limit of the sum volume of PP =£ dx- dy ■ dz. The volume of the lamina LL' = 2 PP' ; the volume of the solid is ' x=OA x=OA y = BO F = 2 iL' = 2 2 PP'- X=0 95 = 1^ = Hence in the limit I I dx dy dz. 92 DIFFERENTIAL AND INTEGRAL CALCULUS Example. — Find the volume of the solid bounded by the 2 ^^ ZX-plane, the XF-plane, and the planes x=a,y=b, z=mx. In this problem I j dxdydz = m\ xdxdy Fio. 33. : mb £ xdx = ^mba'. Akt. 33. — Total Differentials Let M =f(x, y) represent a continuous function of the inde- pendent variables x and y. Let Aj.m denote the change in the value of u corresponding to a change of Aa; in the value of x when y remains unchanged; Aj,?^ the change in the value of u corresponding to a change of Ay in the value of y when x remains unchanged ; Am the change in the value of u corre- sponding to a change of both x and y by Ax and Ay respectively. Then (1) A^u = /(^ + Aa;,y)-/(a;,y) Ax (2) ^^^^ f(^,y + ^y)-f(<^,y) ^y^ (3) Am =f(x + Ax,y + Ay) -fix, y) ^ f{x + Ax,y + Ay) -f{x, y + Ay) ^^ Ax I f(^,y + ^y)-f(!«,y) ^^y Ay ^ f(x + Ax, y) -f(x, y)±i ^^ Ax ■ _l_ f{x, y + Ay) - f{x, y) ^y^ Ay PARTIAL DIFFERENTIATION 93 where e vanishes when Ay approaches zero. If Aa; and A?/ are indefinitely decreased, becoming dx and dy, and the correspond- ing values of A^m, \u, Am are denoted by djt, d^u, du respec- tively, equations (1), (2), and (3) become d^u = — dx, du = — dy, du = — dx-\ dw = d^u + dji. dx dy dx dy The quantity d^u is called the partial differential of u with respect to x ; d^u the partial differential of u with respect to y ; du the total differential of u. The equation du = d^u + d,/u expresses the fact that the total differential equals the sum of the partial differentials. Tor example, if u = x' + y' — xy, d^u = —-dx=(2x — y)dx; d,u = ydy={2y- x) dy; du oit du = — dx-\ dy = {2x — y)dx+(2y — x') dy. The differential expression (4) du = Pdx + Qdy, where F and Q are functions of the independent variables x and y, is said to be exact if it can be obtained by differentiating some function (6) u=f(x,y). The differential of (5) is (6) du = pdx + ^dy,a.nd if (6) is identical with (4), (7)— = P, (8) ^ = Q. The partial y-derivative of (7) is -^ = ^: the dy dy dx dy partial c»-derivative of (8) is _^=^. since ^'" - ^''"■ dxdy dx dydx dxdy' the hypothesis that (4) is exact leads to the condition -J-=°M_ Qp QQ dy dx Conversely, if -- = -^, the differential expression dy dx ' du = Pdx + Qdy can be integrated. All the terms of u which 94 DIFFERENTIAL AND INTEGRAL CALCULUS contain x are found by integrating Pdx, considering y constant. That is, (1) M = I Pdx + Y, where Y stands for the terms of u which do not contain x. To find Y from the 2/-derivative of (1), -r- = — I Pdx H . If the expression du = Pdx + Qdv dy dyJ dy can be integrated, — = Q. Hence — j Pdx ^ = Q, and -T~ ~ ^ ~ a^ I P^'^- Y can be found by integration if Pdx is independent of x. Forming the as-derivative «-i/^ by, of this expression, dx dyJ J dx ox oyj ^sQ_±±rpax^dQ_dp dx By dxj dx dy by hypothesis. Hence Y can be found by integration and the integral of the given differential expression becomes known. Consider, for example, the differential expression du = (3 xy' -a^)dx-(l + 6y^-3 x'y) dy. This can be integrated since — =6xy = —^. Integrating the dy dx first term of du, considering y constant, m = | x'y^ — ^a^ + Y. Differentiating partially with respect to y, ^ = 3x'y + ^=-l-6f + 3x'y. dy dy Whence '~^=-l-6y^ a,nA Y=-y -2 f + 0. Finally « = | ^y^ — \!)? — y — 2i/+0. PABTIAL DIFFERENTIATION 95 PROBLEMS Form the total differential of 1. u = xy. 2. M=-. 3. u = a?y. 4. m=— . y y 5. The pressure of a gas on the containing vessel varies directly as the temperature and inversely as the volume ; that is, p = C-, where c is a constant. Find the change of 1} pressure when the temperature remains constant ; the change of pressure when the volume remains constant ; the change of pressure when temperature and volume both change. Integrate the differential expressions : 6. dw = (3 a;^ + 2 ax) dx + {ax^ + ^y^ dy. 7. du=(o? + 3xy'^dx + {f + da?y)dy. 8. du=(a? — 'Lxy — 2y^)dx + {y^ — ^xy — 2oiF) dy. (ill nil 9. u = ax'y'. Show that x-— + y-- = 5u. ox ay 10. M = aoi?'^ + hxy^. Show that x—- + y— = 5 m. 11. u = F(x, y), where F(x, y) is homogeneous of degree n. Show that X f-?/— - = n-M. ox oy The terms of F(x, y) are of the general form m, = Ax^^^'^y'. dit wtj/ For all terms of this form x-~ + y~ = n ■ u„ and the truth ax ay of the proposition follows directly. This is Euler's theorem on homogeneous functions; 12. If du = /i (x, y) dx + /a {x, y) dy is exact and homogeneous of order n — 1, show that n • u = xfi (x, y) + y/j (x, y) + C. Denoting the integral of the given difEerential expression by M = f(x, y), where f(x, y) must be homogeneous of degree 96 DIFFERENTIAL AND INTEGRAL CALCULUS n, since differentiation diminishes by unity the degree of an expression, aw = — dx-\-—-dy and n-u = x h y — In the ax ay ax ay given expression -^ = fi(x,y), -^ = /j (a;, y). Hence ax ay n-u = xf^ (x, y) + 2//2 {x, y). Integrate the following expressions : 13. du = {2y^x + Sf)dx+ (2 x'y + 9a;/ + ^f) dy. 14. du = (f + Qxy)dx+{2xy + Za?)dy. 15 . dw = (^ x~^ + 6 2/*) da; + (a;^ + 3 a;y~^) d?/. Art. 34. — Differentiation of Indirect Functions If z =f(x, y) and y = F{x), z is said to be a function of x directly and also indirectly through y. Denoting by Ax, A?/, and Az the corresponding changes in the values of x, y, and z, A« =f(x + Aa;, 2/ + A?/) —f{x, y) and A?/ = i?'(a; + Aa;) -F{x), whence Ag_ /(a; + Aa;, y+Ay)-/(a;,y+Ay) ^ f(x,y+Ay)-f(x,y) Ay Aa; Aa; At/ Aa; and ^==^L^+.^^IziZM. Passing to the limit when Ax Aa; limit Aa; = 0, ^ = ^ + ^^ and ^=F'{x). dx ax ay dx dx Example. — A point moves aloiig the intersection of the paraboloid z — Zo?-^5y'^ and the plane y = 2x. Find the rate of change of z and x. Here — = 6 a;, — = 10 «, and ^ = 2. ^ dx ' dy ^' dx Hence — = 6a; + 20w. dx ^ PARTIAL DIFFERENTIATION 97 PROBLEMS dz Determine — when dx X. x' + y' + z' = 25 and y = 2x + 3. 2. x' + y'' + «' = and 2x-3y = 0. 3. ^ + |5 + -' = 1 andx= + 2/^ = r=. a" ¥ r 4. ^ + ^' + -' = 1 and2/'-3a;2/ + x' = 0. a'' 0'' r 5. Determine — and -- whenM = »' and z = a! + y. OK ay Akt. 35. — Envelopes. The equation f{x, y,d) — represents a curve whatever the value of a. By assigning to a all real values an infinite system of curves is obtained. The locus to which every curve of this system is tangent is called the envelope of the system of curves. The equation of the envelope is to be determined. Let (1) f(x, y,a) = Q and (2) f(x, y,a + Aa) = represent any two curves of the system. Their points of intersection approach the points of tangency of f(x, y,a)—0 > with the envelope when Aa approaches zero. Hence the envelope may also be defined as the locus of the ultimate intersections of f(x, y,a)=0 and f(x, y,a + Aa) = when Aa approaches zero. Fig. 84. The points of intersection of (1) and 98 DIFFERENTIAL AND INTEGRAL CALCULUS (2) satisfy f(x,y,a)=0 and /(^^^ 2/, « + Aa) -/(a;, ?/, a) ^ q_ Aa These equations, when Aa approaches zero, become f(x, y,a) = and g-f(x, y, a) = 0. The elimination of a from the last pair of equations gives the equation of the envelope. In like manner the envelope of the singly infinite system of surfaces f(x, y, «, a) = is found by eliminating a from f{x, y, z, a) = and ^/C*, V, «, a) = 0. The envelope of the doubly infinite system of surfaces f(x, y, z, a, 6) = is found by eliminating a and b from f{x, y, 2, a, b) = 0, ^/(a:, V, «, «, &) = 0, j^f{x, y, z, a, h) = 0. Example. — Find the envelope of the system of circles with centers on the X-axis and radii one-third of the distance of center from the origin. a? The equation of the system of circles is (x — of -\-y' = ~, where a is the distance of center from origin. Differentiating with respect to a, —2(x — a) = — . Eliminating a from the equations (x-ay + y^ = ^ and -2(x-a)=—, y'' = \a?, the equation of the envelope. This equation represents the two straight lines y = — —x and y = —x. 2V2 ^ 2V2 PROBLEMS 1. Eind the envelope of the system of lines ^-1-^ = 1 when a + b = c, where c is a fixed constant. PARTIAL DIFFERENTIATION 99 2. Show that the envelope of the normals to the parabola y' = 2px is a semi-cubic parabola. 3. Find the envelope of the system of ellipses —4-^=1 for which the area nab is constant. 4. Find the envelope of the system of ellipses ^ + '^ = 1 for which a+b = c, where c is a fixed constant. 5. Find the envelope of the system of planes - + ^-{-- = 1 when ahc = m, where m is a fixed constant. 6. Find the envelope of the system of spheres whose centers lie in the XF-plane and whose radii vary as the dis- tance from origin to center. CHAPTER VII OIROULAE AND INVEESE OIROULAE PUNOTIONS Art. 36. — Diffekentiation op Circular Functions Denote by x the circular measure of an angle less than a right angle. From the figure triangle OPD < sector OPA < triangle OTA. Hence •^ r" • sin X • cos a;<-^r'''a!<^r*' tana; X ^ 1 and cos x < -^^^^ < sin X cos X This in- equality is true for all values of x less than ^- When x approaches zero, cos x approaches unity. Since -: — lies between two numbers, sma; -, whose common limit is unity when x ap- cosa; preaches zero, limit cos a; and - = 1 when limit a; = 0. This limit is sma; fundamental in this chapter. Denote by u a continuous function of x which changes by Am when x changes by Aa;. Then T d ■ T ., sin(?i + Am) — sinw Am I. — sin M = limit — 5^ — ■ 1- dx Am Aa; _,. ■, 2sin^ Am-cos^(2m + A?f) Am Am Aa; 100 CIBCULAR FUNCTIONS 101 Am TT ' de 33. ^ = «2 (2 ax - a?)^. Find ^ when a; = a (1 - cos ff). dx ^ ' de \- ; 34. Form -^ and -4 for circle a; = i2.cos^, v = B-sine. dx dx" ' ^ Since x and y are continuous functions of Q, if Aw, A^/, and Afl denote corresponding changes in x, y, and e, limit Aa; = and limit A?/ = when limit A9 = 0. For all values of A^, Ay dy ^ = ^. Hence when limit A^ = 0, ^ = ^- Aa; Ax dx dx A0 dS 104 DIFFERENTIAL AND INTEGRAL CALCULUS 35. Form ^ and ^ for cycloid x = Be-Iisme, y = B—E cos 6. 36. Form p. and ^ for cycloid x = B-Bcoae, CtX QiQj y = Be + Bsm 6. 37. Show that 2tana; — tan^K has a maximum value when 4 38. Show that tan 0! + 3 cot a; has a minimum value when 39. Find the maximum radii vectores of r = a sin (3 0). 40. Examine sin a; cos' a; for maxima and minima. 41. Examine sin a; + cos a; for maxima and minima. Fig. 86. 42. Find the length of the arc of the sector which must be cut from a circular piece of sheet iron so that the remainder may form a conical vessel of maximum capacity. V=^i^ sin^a; cos x. o 43. A steamer whose speed is 8 knots per hour and course due north sights another steamer directly ahead whose speed is 10 knots and course due west. What course must the first CUiCULAli FUNCTIONS 105 steamer take to cross the track of the second steamer at the least possible distance from her ? Eind the true values of, .. 1— cosx , n An a — sin a; cos a; i « 44. when a; =0. 47. ■whena; = 0. af 9? 45. ~ when a; = 0. 48. '^^^— when a; = 0. a? X — sm a; . „ sin a; , n ,. « a; sin x , „ 46. when a; = 0. 49. : — when a; =0. X x — 2 sm X 50. Show that the radius of curvature of the cycloid x = B9 — B sin 6, y = R — B cos 6 is twice the normal. Aet. 37. — Evaluation of the Forms oo • 0, oo — cio, — CO If fix) • (x) takes the form oo — oo when x = a, _1 i_ /(.)-.#.(.) =4--4- = i(f-4(^ = 5 when x = a, and again the method of Art. 26 may be applied. The reduc- tion of the form oo — oo to the form - can. frequently be effected more directly. 106 DIFFERENTIAL AND INTEGRAL CALCULUS 1 If y = /M = ?^ wliena; = a, j, = ^ = ^ when a; = a, and the method of Art. 26 gives i/(a')r Hence, y = f^^ and 2/ = /M = /M when x = a. That is, the form — is evaluated in precisely the same manner as the form -• Example I. — Evaluate (a' — a?) tan^ when x = a. 2a (a? — Qi?) tan— = • oo when x = a. (a? - x") tan^ = ^^^^ = 5 when x = a. 2a Hence, (a= - x^ tan^ = =-^^ = — when x = a. 2 a _^cosec='— "" 2a 2a Example II. — Evaluate sec x — tan x when a? = ^• sec X — tan a; = oo — oo when « = ^• Li sec a; — tan a; = = - when a; = — • cos X K) 2 Hence, sec x — tan x = : — = when a; = -• — sin X 2 CIRCULAR FUNCTIONS 107 PROBLEMS Evaluate, 1. (1 — a;) tan ^ when x = l. 2. Ax when a; = 0. 3. cusin^— when a; = oo- ,irX X cot- 4. a; cot a; when a; = 0. 5 . sec (3 x) cos (5 a;) when a; = ~ 6. 2 a; tan a; — IT sec a; when a; = ^• 7 . (1 — tan a;) sec (2 x) when a^ = j* Art. 38. — Integration of Circular Functions The eight formulas of Art. 36 may be written, I. I cos M ■ — = sin M + C, III. I sec^M ■ — = tan u + C. J dx J dx II. I sinw- — = — C0SM + C, IV. j cosec^w- — = — cotw+C, J dx J dx V. I tan u • sec u — = sec u + C, J dx VI. jcotw-cosecM — = — cosecM+C, J dx VII. I sin u — = vers u-\- C, J dx VIII. I cos u • — = — covers u + C. J dx Example I. — Integrate -^ = a; • sin (2 a^). This derivative has the general form sin u ■ — . Placing du ^^ u = 2x^, — = 4 a;, and the given derivative may be written dx dx 108 DIFFERENTIAL AND INTEGRAL CALCULUS -^ = i • sin (2 x^ • 4 x. Hence, by the formula, /sinw-— = — cosM+(7, w = — i cos (2 a;^) + C dx Example II. — Integrate — = sin° a; • cos x. dx Since —sin a; = cos a;, — =sin°9; — sina;, and w=fsin''a!-|-C. dx dx dx ■ PROBLEMS Integrate, 1. ^= cos (4 a;). dx ' 7. ^-i.cos-. dx x^ X 2. ^= cos (2 a; -6). dx 8. |=3sin(5.-7), 3. ^=a;^-c6s(a^ + 2). dx 9. — = sin^a;-cosa;. dx 4. ^=sec2(3a; + 5). dx 10. -^= sinta;'COsa;. aa; 5. ^=sec^('^Y dx \2) 11. -^= cos'a;'Sina;. da; 6. ^= cos (fa;) -cot (fa;). 12. -^ = tan'a;-sec^a;. da; Frequently an expression containing circular functions, when not directly integrable, may be transformed into an expression which can be integrated by means of the trigonometric rela- tions sin^a; + cos^a; = 1, sec^a; = 1 + tan^a;, sin^a; = \ — \ cos (2 x), cos^a; =\-\-\ cos (2 x), sin x cos x = \ sin (2 a;). The last three relations are special cases of the formulas sin a; sin 2/ = -^ cos {x — y)—\ cos (x + y), cos X cos y = \ cos (a; — 2/) + ^ cos (a; + y), sin a; cos y = \ sin (a; + 2/) + |- sin (a; — 2/), which are frequently useful. CIEGULAB FUNCTIONS 109 Example I. — Integrate -^ = sin^o; • cos^a;. do; Write -^ = sin' a; ■ cos^ a; = sin* a; • cos'^ a; ■ sin x dx = —(1 — cos^ajV • cos^a; — cos x dx = — 008" a? — cosa; + 2 DOS'* a; — cos a; — cos' a; — cos a;, da; dx dx Integrating term by term, ^ = — •!■ cos^ a; + 1 cos' a; — | cos" a; + C. Example II. — Integrate ^=tan*a;. da; "Write ^ = tan* x = tan^ x (sec^ a; - 1) dx = tan^x ■ sec^a; — tan^a; = tan^a; • sec^a; — sec^a; + 1. Integrating term by term, y = ^ tan^ x — tan a; + a; + C Example III. — Integrate -^ = sin* x. dx Write ^ = sin*a; = J^-^cos(2a;)}2 CtiX) = |:-|cos(2a;) + ^cos''(2a;) = i-|cos(2a;)+i{i + icos(4a;)| = f — "I cos (2 a;) + ^ cos (4a;). Integrating term by term, 2/ = |- a; — -^ sin (2 a;) + -^ cos (4 a;) + C. Example IV. — Integrate ^ = ?E:i^. da; cos* x Write ^ = 5iB!^ = ?^._l^ = tan2a;.sec2a;. Integrating, dx cos* a; cos^x cos a; y = \ta,ii^x + G. ilO DIFFEllENTIAL AND INTEGRAL CALCULUS Example V. — Integrate -^ = sin'' a; • cos'' x. dx Write -^ = sin" x • cos" a; = -l- sin" (2 x) dx = i li - T cos (4 ce) I = i - ^ cos (4 a;). Integrating, y = ix — -^j sin (4 a;) + C. Example VI. — Integrate -^ = sin x cos 4 x. dx Write -^ = sin a; cos 4 a; = ^ sin 5 a; — ^ sin 3 a;. Integrating, 2/ = — ^ cos 5 a; + ^ cos 3 a; + C PROBLEMS Integrate, 1. ^=sin"(3a;)-cos^(3a;). 8. ^ = sin' a; • cos' a;. dx dx 2. ^ = sin' (A a;). 9. ^ = sin*a!.cos*a!. dx ^^ ' dx 3. ^ = cos" (2 a;). 10. ^ = tan''a;. dx dx 4. ^ = tan' (2 a;). 11. ^ = cos" (3 a;). dx ^ ' dx ^ ' 5. ^ = 3cos'a;. 12. ^ = 3a;.sin"(a!").cos'(a^. dx dx 6. ^ = 5sin"(Aa;). 13. ^ = a;" • sec" (3 a^). dx ^^ ^ dx ^ ' 7. ^= sin' a;, cos" x. 14. ^ = i^^. dx dx cos" a; Show that, m and n being positive integers, sin (wx) • cZa; = 0. 16. I cos (ma;) • da; = 0. CIRCULAR FUNCTIONS 111 17. I sin (mx) • sin (nx) • dx = 0, m^n. X2ir COS (ma;) • cos (nx) •dx = 0, m + n. X2ir COS (mx) ■ sin (nx) ■ dx = 0. cos^ (mx) ■dx = w. 21. I sin'* (mx) • da; = tt. Aet. 39. — Integkation by Tbigonometkic Substitution First derivatives with respect to x involving -Va" — x' may be transformed into trigonometric derivatives by the substitu- tion x = a- cos 6 ; those involving Va^ + x^ by the substitution x — a- tan ; those involving Va;^ — a' by the substitution x = a-seo6; those involving ■v'2 ax — x' by the substitution x=a(X — cos 6). Example. — Integrate -^ = '^^ x\s?-ay Placing a; = a • sec 6, — = a . sec d • tan 5, (a;''— a?)l = a • tan 6. dO Integrating, y = — { ^ fl + i sin (2 0) | + C Erom a; = a ■ sec ^, e = sec-^-, sin (2 6) = 2 sin^ cos 6» = 2^-fl - ^Y = ^(a;''-a')*- a x\ x^J or Finally,, = ^,.sec-^ + ^(.^-aO^+a PROBLEMS Integrate, 1. :^ = ± — -. 2. ^ f'^' (a^-x^^ ^^ ar'(l + a;'^^ 112 DIFFERENTIAL AND INTEGRAL CALCULUS 3. dy_ 1 d^ a^iaP-l)^ Q dy _ a^ ■ dx (i_a^i 4 dy _ a? ^ dy _ a? dx y 1 _ a,2 dx (i_a^4 5. dy 3 '^^ (2 + a?y 8 #_ a^ 9. dy^ dx -^{1. - a;2)i. 10. dy_ 1 dx {a? + aFf Substitute a; = a • tan 6. 11 dy_ 1 dx (1 - af)2 12. ^=-y|^^ — ^- Substitute a!* = sin ft 13. Find tbe area of the circle x^ + y^ = i^. A = 2 I (r^ — oiF)i • da;. Substituting a; = r • cos 6, and noticing that when x runs through all values from + r to — r, 6 runs from + ^ to — ^, there results ^ = j-='r 'cos2e•d(9 = ,r•r2- 14. Find the area of the ellipse ^ + ^, = 1. a'' ¥ 15. Find the area of the hypocycloid x^ + y^ = a* 16. Find the volume of the solid generated by the revolu- 8 a? tion of w = ; about the X-axis. 17. Find the area bounded by the curve w = and the 2a — X line a; = 2 a. -4 = ) Substitute a; = 2 a • sin'' &. -'» (2a-a;)» CIRCULAR FUNCTIONS 113 Aet. 40. — PoLAK Curves Tangents and Normals. — Let r = f(ff) be the equation of a continuous plane curve, P(r, 6) any point in the curve, and. P' (r + Ar, + ^6) any other point in the curve. The ratio — ^ measures the average rate of change of r in the interval a6, and — = limit — when limit A^ = measures the actual ' de AB rate of change of r at the point (»•, ff). The secant through (r, 6), (r + Ar, 6 + Aff) approaches the tangent to the curve at (r, ff) as Ad approaches zero. Hence the angle AP'S ap- proaches the angle APT = <^ included by the tangent at (r, 6), and the radius vector to the point (r, 6) when A^ approaches zero. Drawing a perpendicular from (r, ff) to AP', tan ^ = limit tan AP'S = limit ' sin AO r + Ar — r • cos A6 = limit sinAfl AB AB rsin — 2 . A6 , Ar -Ae-''''-2+AB = r • — when limit AB = 0. dr 114 DIFFERENTIAL AND INTEGRAL CALCULUS The distance AT from the pole to the tangent measured on the perpendicular to the radius vector to the point of tangency is called the polar subtangent; the distance AJff from the pole to the normal measured on the same per- pendicular is called the polar sub- normal. From the figure subnormal AN=—; normal PiV" = r' + dr\i dr' subtangent AT = r' dicular from pole to tangent d6'^ perpen- AD=2J = - ^ + Asymptotes. — When in a polar curve r=f(ff), for some finite value oi 6 = 6^ r becomes infinite and the subtangent is finite, the tangent to the curve at infinity passes at a finite distance from the pole and is called an asymptote. If the sub- tangent is positive, lay it off to the right of the radius vector looking towards the infinitely distant point of the curve ; if neg- ative, lay off the subtangent to the left of this radius vector. Example. — Examine r = - for asymptotes. For = 0, r = oo. The subtangent = r^ • — - = - dr ■ a. Hence for = 0, subtangent =—a, and the asymptote is obtained by laying off on the perpendicular to the polar axis at the origin to the left when facing in the direc- tion 6 = the distance a and drawing the perpendicular FT. CIRCULAR FUNCTIONS 115 PROBLEMS 1. Find the subtangent of r = a- 6. 2. Find the subtangent of r^ = aP- cos (2 6). 3. Examine r = — for asymptotes.. 4. Find the angle under which the radius vector cuts P r = i . 1 — cos 6 5. Show that the radius vector cuts r = a' under a constant angle. Length. — Denoting the length of the curve r=f{6) from 6 = 60 to 6 = 6u by s fl=«i s = limit 2 V?-^ . sin= A0 + (r + Ar - r • cos l^fff = 00 = limit^iVr'rgH^^Y+ i '•(I - '^'f) +^'J\6 r2 + ^ . de, when limit AO = 0. 116 DIFFERENTIAL AND INTEGRAL CALCULUS Area. — Denoting the area bounded by the curve r —/(&) and the radii vectores to the points (?"o, 60), (rj, Oi) by A, A = limit S i r" . Ae = I C'r" ■ 06. 8=90 i/flo Example. — Find the length and the area of the cardioid ?• = «(! + cos^). The length s = 2j^ Yr^ + ^y . de = 2 J^" j (a + a ■ cos e)^ + a^ sin' e} *d(9 The area A = 2. :^ C''r'.de = a' C (l+cosef -dO = a'C (1 + 2 cos ^ + cos'5) • d$ = a'C (I + 2 cos e + |cos 2 6) dO =a?he + 25me + \sm2ey = ^7ra\ PROBLEMS 1. Find the area of the lemniscate r^ = a^- cos 2 6. 2. Find the area of r = 2a- sin 6. 3. Find the length of r = a- sin'-. 4. Find the area of one loop of r = a • sin (2 6). 5. Find the area of r = a • sec^ - from 6 = to ^ = ^■ 2 2 CIBCULAR FUNCTIONS 117 Fia. 41. Art. 41. — Volume of a Solid by Polae Space cooedinates Tlie polar space coordmates of a point are r, <^, and 6. The conical surfaces corresponding to 6 and 6 + ^6 include a conical wedge of the volume to be determined. The planes corresponding to and + A<^ cut from this wedge a solid of the nature of a pyramid. The spherical surfaces corresponding to r and r + Ar cut from this pyramid an element of solid ■which, when Ar, A(f>, and A6 are indefinitely decreased, ap- proaches as its limit the rec- tangular parallelopiped whose dimensions are a6 = Ar, ad =r- cos 6 • A<^, ac = r • A6. Hence in the limit the element of volume ='i^ • cos 6 -dr • d- dd, r'-cos6 -dr-dtfi- dd, I r' • COS 6 -dr-dil)- dO, and the entire volume I j 7^ ■ cos 6 -dr-dtiy -de. =0 «/(>=0 i/r = For example, let it be required to find the volume of the trirectangular spherical pyramid, the radius of the sphere being a. In this problem r extends from to a, 6 from to ^, <^ from to ^- Hence, 118 DIFFERENTIAL AND INTEGRAL CALCULUS F= I M M r'-cose-dr-d.m . =^ j M ''cQse-d^-d6 3»/fl=o »/0=o 6 »/8=o 6 Aet. 42. — Differentiation of Inveese Circular Functions To differentiate y = sin"' u, where m is a continuous function of X, write sin y = u, and differentiate with respect to x. du There results cos « • — = — , whence -^ = : and, since dx dx dx cos 2/ cos du y = Vl — sin^w = Vl —u\ -^ = -— sin~' u = - ^ dx dx -y/j _ y2 du dM In like manner — cos 'm = ; --tan '■u=: dw dw d ._, dx d _, da; — cot^M = — =— — 5; -— sec'w = — ; da; 1+u" dx mVm^-I _dw dw d _i da; d _i da; • — cosec^M = ; — vers 'm =■ , ^) da; mVm^-1 <^» V2 m - m" dw d 1 dx — covers"' M: da; V2 M— M' CIRCULAR FUNCTIONS 119 Example. — Differentiate tan _i 2 a; d, 1 2a! _ dccl — a;^ _ (1 — x-y dx l-a;^~j^ / 2 a! Y ]^7 T^ .l-a!7 (1-xy ^ 2(l-a;^) + 4a!^ ^ 2(1 +a;') ^ 2 (l-a!2)^+4a!2 l+2a!^ + a!* 1 + a;^' It appears that — tau~' ; = 2 — taii~^a:, which is as it da; 1 — a;'* dx 2x ought to be, since tan~'- — = 2 tan~' a; by trigonometry. X — Oj Differentiate, PROBLEMS 1. sin-i(3a!). .. .^-v^. 7. sec-'(a;^). 2. 3sin-'^. 5. 5 vers '-• 6 8. cot-'g). 3. sin-' (3 a; +5). 6. 3a;2 + 2cos-'a;. 10. cot->^~'^- 9. Stan-'?- X 1+a; 11 . Show that — sin-i (3 a; - 4 a!^) = 3 — sin"* x. dx dx 12. Form -^ and — ^ of equation of cycloid dx die a; = r vers"'- — ■\/2 ry—if. r Here ^ = ^ ; hence ^ = -v/^E^. t'j/ ■\/2ry-y' dx ^ y Squaring, ^ = ^I _ i. Differentiating, 2^^ = -^^- dar y dx dor y' dx Whence, ^ = -^. dar 2/^ 120 DIFFERENTIAL AND INTEGRAL CALCULUS 13. Form — and — ^ of cycloid y=r- vers"'- + V2 rx — a?- dx dor r 14. Find length of cycloid y = r • vers"'- + V2 rx — o?. 15. On a pedestal 25 feet high stands a statue 11 feet high. Find the distance from the base of the pedestal of the point in the horizontal plane through the base at which the statue subtends the greatest angle. Abt. 43. — Integration by Inverse Ciecular Functions The results of the preceding article when - is substituted for u may be written in the form, " du du I. I , =sin-'-+C; III. I ^- — 5 = -tan-'-+C'; _du _dM II. I , =p,os-'- + <7; IV. 1-2- — 5 = -cot-'- + C: du V. I = -sec"'- + a- _du ^^^ r dx 1 _i« , /7 VI. I — r=^ = -cosec '- + C; du VII. r-=^= = vers"'-+C; du VIII. r^^J^= covers"'- +0. CIRCULAR FUNCTIONS 121 1 Example. — Integrate —=^-^=z. This derivative has dx V2-4a^ du the general fnnn — — '■ Placing M^ = 4a^, u = 2x^ and <^M Q i w •.- dy x^ 1 dx -— = 3x\ Writing -f- = — = q ~7= , da; da; V2-4ar' 3^2- 4a;* 2/ = |sin-i(V2-a;t) + 0. PROBLEMS Integrate, 1. dy^ dx 3 4 + 9a;=' 2. dy^ dx 2 a;V3a;='-5 5. dx : ^* . Eedu( l+x' 6. dy_ dx X x* + A 7. dy^ dx 1 V6a; — ar* 8. dy _ dx X VI -a;* 12. dy^ dx 1 ar'-eaj + ll y = —-ta.n V2 _ia;-3 ^_ V2 13. dy _ 1 3. ^ = _^_. "da; 1 + a;* 4. ^ = - -^ da; V6 a;* - 2 a;^ Eeduce improper fraction to mixed number. a d?/ _ tan~^ X dx l + ar" in dy _ sin~' x dx~ ^TZT^ n dy _ sec~^ x dx XyJyS._1 — (a;-3) Since ^=. 1 '^^ — da; 2+(a;-3)2 2+(a;-3/ ^'^ Vl+3a;-ar' 122 DIFFERENTIAL AND INTEGBAL CALCULUS 14. The time of descent of a body down the arc of the vertical frictionless cycloid a; = r • vers~'-- Y - + V2 ry — y", from Fig. 42. y=hto the vertex, is i = r\i r- Jo - dy Show that the KdJ -'» V% - y^ time is the same for all positions of the starting point. 15. A body is suspended on a frictionless horizonal axis, turned through a small angle 6^ and then left free under the action of gravity. This constitutes a compound pendulum. The relation be- tween 6, the angle the pendulum in any position makes with the vertical and the time t measured in seconds after the pendulum is started, is expressed by the equation —^■■ ■gh 6, where g is Fig. 43. complete oscillation. de the acceleration of gravity, h and fcj are constants depending on the shape and material of the pendulum and the posi- tion of the axis. Find the time of a In this problem, when { = 0, 6 = do, dt = 0. CIBCULAE FUNCTIONS 123 16. In strength of materials it is proved that for any point (x, y) of the elastic cvirve of a long column da? ■P-y, where E and I are constants depending on the material and cross-section of the column; P is the load. Calling the maximum deflection A, when ?/ = A, -^ = 0, and when x = Q, y = 0. Find dx the equation of the elastic curve. Aet. 44. — Radius of Cukvatuee If a point moves in the circumference of a circle, the tangent to the circle at this point changes its direction. Sup- pose the point to start from A. When it reaches B, the tangent has turned through the angle T'ST=AOB. The ratio of the angle AOB to the distance AB the point has moved along the circle is called the rate of curvature of the circle, and equals -^ = i- That is, •• • $ r Fig. 46. the rate of curvature of the circle is the reciprocal of the radius of the circle. If a point moves along any curve y=f(x), the ratio of the angle through which the tangent turns to the distance the point moves is called the average rate of curvature of the curve for the distance the point moves. Denoting by A(/) 124 DIFFERENTIAL AND INTEGRAL CALCULUS tlie angle through which the tangent turns while the point moves a distance As along the curve, the average rate of cur- vature is — • The actual rate of curvature at any point {x, y) of the curve is the value at (a;, y) of limit — = -^ when ,. . . ^s ds limit As = 0. The reciprocal of the rate of curvature at any point {x, y) of the curve is the radius of the circle which has the same rate of curvature as the curve at the point {x, y). The reciprocal of the rate of curvature is called the radius of curvature, and is denoted by p, so that p = — . Now A = tan"'-^ hence _ds _ds dx _dx \ da?) \ dx^J ^ ~ d~ dx d~' d±~ d^y ~ d^y dx dx^ dx? da? The analysis supposes that the curve is such that y is a continuous function of x, and a continuous function of s. If the equation of the curve is given in the form x=fi(t), dy y =A{t), dy _ dx = _, and dx dt d'y_ da? dy ' dxdx dt dy d d^^ ' dt dx' dt d?y dx d'x 1 dt^ dt df dx~ da? dt df dy dt Hence P = [dt" ' d^y da de ' dt dy\^ ^df) ; d'x dy df dt CIBCULAR FUNCTIONS 125 For a polar curve r =f(ff), a; = r • cos ^, y = r- sin 6. Hence dx dd ■r-sin0 + cose-^ ■2-^-| + -^-||' dv A , • n dr J=r.cose+sme.-, g=-r.sin. + 2sin..|+sin..|^^. Hence for a polar curve p = --1-5' r2 + 2*f-r^.^ A circle of radius p, placed so that the circle and curve y=f{x) have a common tangent at (x, y) and lie on the same side of the common tangent, is called the circle of curvar ture of y = f(x) at {x, y); the center of the circle is called the center of curva- ture of y—f(x) at {x, y). Denoting the coordinates of the center of curvature by a and /?, (1) a = a; — p • sin <^ dx V dx^ d^ do? Fis. 46. 1 + (2) ^ = 2/ + P • cos (^ = 2/ + - da? da? 126 DIFFERENTIAL AND INTEGRAL CALCULUS The locus of the center of curvature as the point (x, y) traces the curve y=if(x) is called the evolute of y = f{x). The equation of the evolute is found by solving (1) and (2) for x and y in terms of a and fi and substituting in y =f(x). Example I. — Find the radius of curvature, coordinates of center of curvature, and evolute of the parabola y^ = 2px. Here^ = £, § = _£, hence p = -(^^; « = 3x+p, da; y aar ^ p^ j8 = — ^ ; the evolute is p' = -^ {a —p)', a semi-cubic parabola. The radius of curvature to the parabola at the vertex (0, 0) is p; at the extremity of the latus rectum {^p, p) the radius of curvature is 2 V2 • p. Example II. — Eind the radius of curvature of a;?/ = 4 at the point (1, 4). Eor this curve ^ = - 4 ^ = J- At (1, 4), ^= -4, dx of dx^ of dx g=8. Hence p = i(2)f. PROBLEMS Eind the radius of curvature of 1. 2/=^ = 4 a, at (1,2); (0,0); (4,4). 2. | + |' = lat(3,0); (0,2). 3 ^4-t-l 4 t-t-1 or ¥ a? V y I 5. x = r • vers"' V2 ry — y'. 6. x = v-t, y = \g-t^. 7. a; = a cos <^, 2/ = 6 sin (^. 8. X = R • e — B • sine, y = B — B • coa$. CIRCULAR FUNCTIONS 127 9. 2xy = a\ 12. r = a(l — cosS). 10. a;* + 2/^ = ai 13. i^ = a' ■ cos {2 6). 11. 2/ = sma;. 14. r = a*. CHAPTER VIII LOGAEITHMIO AND EXPONENTIAL PUNOTIONS Art. 45. — The Limit of ( 1 + -) When Limit z= (-J)-^ Assume first that z takes only positiye integral values m. By the binomial formula, \ m.J m 1 • 2 fn? . m (m — 1) (m — 2) 1_ 1-2.3 tn?'^'" . m (m — 1) (m — 2) • ■ • (m — w + 1) 1 l-2-3"-n m" . m(m — l)(m — 2) ■■■(m — w + l)(m — ?t) 1 . 1.2.3.4..-w(n + l) m*^^ . ?)i (ot — 1) (m — 2) • • • f m — (m — 1) j 1 1.2-3-4"-m m" =1 ,14. m V myy my - "*" ^ 1-2 "^ 1.2-3 "^ ^ V m)\ m)"\ ~ m ) 1_2\ A n-l\ 1.2T3...nr 128 LOGARITHMIC AND EXPONENTIAL FUNCTIONS 129 1-2 -S-n 171+1+ {n+l)(n+2) '^"' V mj\ m j \ m J "^ (ji + l)(n + 2)".m This expansion is true for positive integral values of m how- ever large m may be taken. Denote by 8 the sum of the first w + 1 terms of the expansion, by B the sum of the remaining terms. Then the equation limit ( 1 + - ) = limit S + limit B t(l + ^J=limi is always true. When m is indefinitely increased, limit/S = l + l+-i-+-J— - + T-7rir^ + - "*" 1-2 1-2-3 1-2-3-4: 1.2.3.4-n '^°<=^^i < s ^ m + 1. m + 1 When limit s = ao, limit m = oo. Hence when limit s=oo, limit ( 1 + - ) lies between two quantities whose common limit is e. Consequently limit fH — j =e when limit s = oo. Lastly assume that z = —r, and that limit r — ao. Now limit (l-l)-'=limit(?:^)-' = limit(-f3j = liniit(l + -i_J = limit(l+-^J-\(l+-A_) = e when limit r = oo. It appears that in every case, when limit « = oo, limit /'l + -Y = e = 2.718281828. This limit is fundamental in this chapter. LOGABITHMIC AND EXPONENTIAL FUNCTIONS 131 Akt. 46. — Differentiation of Logarithmic Functions Let u represent a continuous function of x, and denote by Am and Ax corresponding changes in u and a;. Then, if a is the base of the system of logarithms used, d ^ , lijnit log.(w + AM)-log„M _ Aw Ax log, Au u + Aw = limit '■ Am w Am Ax = limit - ■ log„ 1 4- A^A^ Aw M / Ax = limit - • log„ du dx Aw aS_ Aw Aw = log„e when limit Ax = 0. For by the nature of logarithms the difference of the loga- rithms of two numbers is the logarithm of the quotient of the numbers, and the logarithm of a number affected by an ex- ponent is the exponent times the logarithm of the number. Since m is a continuous function of x, limit Aw = when limit Ax = 0. Writing — = z, when limit Am = 0, limit « = oo. : 0. Writing — = z, when limit Am : Aw Hence limit Aw when limit Ax : : equals limit fl-|- when limit 2 = co • Calling, as is customary, log„e the modulus of the sys- tem of logarithms whose base is a, and denoting it by M, 132 DIFFERENTIAL AND INTEGRAL CALCULUS du — log„it = Jf— ; in words, the derivative of the logarithm of dx u a function of x is the modulus of the system of logarithms times the derivative of the function divided by the function. In the Napierian system of logarithms — log„M = -• dx u dx Unless otherwise specified, the Napierian system is to be used. h — X Example. — Differentiate log -\/ . ^ ^1 + x '■\l-x 1 + x] 1-x' = h PROBLEMS Differentiate, 1. loga^ 4. log(a!^. 7. log tan a;. 2. logl 5. log^a;. «• log VI -a^. '^ Q 1 I-'"' 3. log (3 a; -6). 6. log sin ». ^- ^^ 1 + x'' 10. logtanice. ^3_ j^^ 11. log^ — cos a; VTT^ + cosa! 14. log(a;+ Vl+a^. cos a;' 15. log (x + Var' + a')- 16. log (a; + Va:^ - aF). Find the true value of, 17. , ^°Sx ^hen x = l. 18. ^^ when x = l. (1-x)^ * - 1 LOOABITHMIC AND EXPONENTIAL FUNCTIONS 133 19. -^ when a; = 00. 22. a;'»log"a; when x = 0. 20. ^^ when x = 0. 23. _^_-— - whena; = l. cot a; X — ± iog x loetan('2a;) 1 n „^ loajsino; , tt 21. ^ ^^ — ^whena;=0. 24. ^ ° ^ — -whena; = -. log tan X {■!r — 2xy 2 Art. 47. — Integeation by Logaeithmic Functions The result of the preceding article may be written dit fdx r , du , , „ Example. — Integrate -^ = x dx 1 -\- a? The first derivative of the denominator is 2 x, hence — (1+0!=) PROBLEMS Integrate, J dy ^ l + 3a; ^ dy^ 2x + 3 ' dx 2x + 3oa'' ' Ux a^ + 3x 2. ^=3^. 5. ^ = tana;. dx of + 7 dx 3_ dy^nxdx ^ ^ = cotx. dec a^ + o;^ dx r. ^=J_. Writing ^ = J- = 3^ dx sin a; da; sin x 2 sin ^ a; • cos ^ x secHa;. — (ia;) -^tania; ^ |secH«^ ^ dx ^dx ,2/ = iogtania; + a tan-l^a; tan^a; tania; 134 DIFFERENTIAL AND INTEGRAL CALCULUS 8. ^ = - 10. ^ = dx dx cos X dy_ L dy^ 1 da; a;(l + a!^" Substitute x = tan ft vV + (? Write Va^ + a^ = 2 — a;, ■whence , /rF~; — 2 da' z ~x dy dy dx dz z dz dx dz z■ y=log.z+C=log,.(x + vs^ + aF) + a 1 z — x 1 ; and 11. dy _ 1 12. dx ■V^+2x d Writing Va;'+2a; V(x+1)'-1 y = \og\{x + l)+ \/x'+2x\ + a dy dx' dx (x + 1) V(x + 1)" - 1 =• and 13 dy^ 2 + 3x ' dx 1 + a? term by term. Write ^ = -2-, da; 1 + a;= + 3a; and integrate 14. dy _ m -\- n x dx a^ + X- 15 dy _ log^x dx da; a; 17. '^y= ^ ' dx x' +1 Reduce a^ to a mixed number. Fig. 47. 18. Find the area of xy = l from a; = to x = x'. 19. If A is the cross-section of a bar of uniform strength at a distance y from the lower end, — = '^l^-iA, where w is the weight dy S of the bar per cubic foot, and /S is a constant depending on the material of the bar and the area of the lower end. Find the relation between A and y. LOGAniTBMIC AND EXPONENTIAL FUNCTIONS 135 Art. 48. — Integeation by Partial Fractions F(x) Tlie function of x defined by > > ^ is called a rational frac- cl>(x) tion when x does not occur affected by a fractional exponent or under a radical sign. If the numerator of the rational fraction — )-l is not of (x) lower degree than the denominator, the given fraction may always be transformed by division into the sum of an integral function P(x) and a rational fraction •' ^ ■' , whose numerator is of lower degree than its denominator. For example, Consider the rational fraction ^i-V^ whose numerator is of {x) lower degree in x than the denominator. Suppose the denomi- nator (x) can be resolved into real factors of the first and second degrees. Let {x) = (x-a)(x-by{(x-cy + dm(x-ey+f\'- It is proposed to break up the fraction ■' ^ ■' into the sum of par- tial fractions whose denominators are the factors of ^ (x), and in every partial fraction the numerator is to be of lower degree than the denominator. 136 DIFFERENTIAL AND INTEGRAL CALCULUS Assume f(x)_ A B, B, B. Cx + D {x). Multiplying both members of the identical equation by <^ (a;), the right hand member will be of degree m — 1 in a;, while, by hypothesis, f(x) cannot be of a higher degree than n — 1. Collecting the terms of like powers of x in the right hand member, the co- efficients of the resulting n terms are linear in the n undeter- mined constants. Hence, by equating the coefficients of corresponding terms of both members of the identity, there result 11 equations linear in the n assumed constants. These equations determine the assumed constants uniquely. Hence the fraction l-^ can be broken up into partial frac- tions of the form assumed and in only one way. A few examples will explain the process of breaking up a fraction and show the importance of partial fractions in integration. dv 3/ 1 Example I. — Integrate -f- = — j- dx ar — 4 j^ ^ gS -^ 4aj— 1 Transforming — to a mixed number, — = x + — - — - ■ a;2 — 4 05^ — 4 x'—i , ix-1 A . B Assume —z = ^ + ar'-4 a! + 2a;-2' whence ix-l={A + B)x+ (-2A + 2B). LOGARITHMIC AND EXPONENTIAL FUNCTIONS 137 Equating the coefficients of like powers of x, A + B = 2, -2A + 2B = -1, and^ = |, B=i- -rr dy x^-1 ,4a;-l ,91 ,71 Hence, -f=-. — T = a;H — -, — T-^ + l—T^ + l o' do: a? — i x' — 4: 4a; + 2 4a; — 2 Integrating, y = | + 9 log (a; + 2) + 1 log (a; - 2) + C. Example II. — Integrate -^= „ ^ "*" — -• dx af + x — z Assume — ; — — — - = -| -, x^ + x-2 a;-l a; + 2' whence 5x + l = A{x + 2) + B(x — T). This identity is true for all values of x. When x = l, A=2; when a; = — 2', B = 3. Hence, -^ = -\ -• Integrat- dx X — 1 x + 2 ing, 2/ = 2 log (x — V) + 3 log {x + 2) + G. This result may also be written, y = log (a; + 1)" + log (x + 2)= + log c = log \c(x - iy(x + 2yi. af + Ax-2 Example III. — Integrate -^ = - dx 1 + x + a^ + aP Assume o^' + i^-^ - ^ , Bx + C 1+x + a^ + ai' 1 + x 1 + sa' whence x' + 4:X-2= (A + B)x^ + (B + C)x + A. Equating the coefficients of like powers of x, A = — 2, B+ C=i, A + B = l, whence A = - 2, B = 3, 0=1. Hence dy_ -2 3a; 1 . and 2/ = -21og(l+a;) + f log(l+a;^) + tan-'a;+ 0. 138 DIFFERENTIAL AND INTEGRAL CALCULUS Example IV.— Integrate ^ = 4±i^^. Assume ^ + ^-l^A^ + B Gx+D^ {a? + 2f {x' + 2f^ 3? + 2 By clearing of fractions and equating the coefficients of like powers of x, it is found that A = —l, B = — l, 0=1, D = 0. Hence '^^ = "'""^ 1 ^ ^ -'^ 1 ^ ?^. ' dx (a? + 2y x'' + 2 (x2 + 2)'i^ar' + 2 ix' + 2y The first two terms are directly integrable. To integrate the last term substitute a; = V2 • tan 6. PROBLEMS Integrate, 1. dy^ 2 jB + 3 dx a? -\-a? — 2x 2. ^y^ « ■ dx ^—a? 3 %^ 2-3a^ dx (a; + 2)2 ^ dy^ a? ' dx x* + a^-2 da; a;' — a;^ + 2 05 ■ 6. dy__ dx a!^ + 1 13. Show that I -— - = -— log— ^^ — |- O, u being a con- J w' — or 2 a u + a tinuous function of x. This result is very useful. 7. dy^ dx 1 a!*-l Q dy_ a; + l dx a!(l + aO Q dy^ dx X a^-x'-2 10. dy _ dx 1 a;(l+a!y n. dy^ dx 3a! + l 12. dy^ dx a;* !)?-l ^^ -«4 - CI. 11, hfiins LOGARITHMIC AND EXPONENTIAL FUNCTIONS 139 — (a; -3) 14. Integrate ^ = -^4 ?■ Writing ^^ /^^ „,, . , ^ dx x'-Gx + B dx {x-Sy-i byProb.l3, , = xiog|^|^+C=ilog|5f+C. 15. Integrate -^ = d!a; da; ai^ + Saj + l Akt. 49. — Integeation BY Parts Prom — (u-v) = u- — + V-— is obtained by integration dx dx dx /u- — = u-v— (■y-— , which is called the formula for inte- dx J dx gration by parts. The following examples will show the application of this formula. . Example I. — Integrate -^ = x- log x. dx Writing u = log x, — = x, whence — = -, v=ix' + Ci, the dx dx X application of the formula gives y=j'x.\ogx={^x'+G,).\ogx-J(ix'' + C,)-^ = ^x'-logx+ Ci-logx^^x'— Oi-\ogx+ C = ^»^-loga; — ^»^+ C Gi, the constant of the integration | — , always eliminates as J dx in this example. It may therefore be neglected. Example II. — Integrate -^ = a; • sin a;. dx Writing ic = x, — = sin x, whence — = 1 and v = — cos a;, dx dx the application of the formula gives y = J a; • sin a? = — a; ■ cos a; + j cos a; = — a; • cos a; + sin a; + C 140 DIFFERENTIAL AND INTEGRAL CALCULUS Example III. — Integrate -^ = a; • sin~' x. dx Writing M=sin~'a;, — = x, whence — = — -^^^ and dx dx -^1 _ ^ V = ^a^, the application of the formula gives y=jx- sin-^a; =lx' ■ sin-^a; - ^j—==- Write -— = ■ and substitute x = sin 0. There results ax -^1 _ a^ dd dx d9 T ^ ^ ;> and y' = ^e-lsin(2e) + G = ^6 - ^sind ■ cosO + C. Substituting sin 6 = x, cos 6 = Vl — x', 6 = sin"' x, y' = ^sm~^x — ^aj'Vl — a^ + 0, and 2/ = ^ar'-sin"'a!4-^sin~'a!— -Ja;- Vl — a^ + C- Example IV. — Integrate -^ = sec'a;. dx Writing ?t = sec x, — = sec^a;, whence — = sec x • tan x and dx dx V = tan X, the application of the formula gives y= I sec^a;= seca;- tana; — | secaj-tan^a; = sec a; -tana;— j sec^a; + j J J cos a! Hence 2 j sec^a; = seca;-tana;+ | J J cos X = seca;- tana; — log tan (- — -)+ C V4 2j = sec a; • tan x + log (sec x — tan x). LOGABITHMIC AND EXPONENTIAL FUNCTIONS 141 Example V. — Integrate ^— Va^ + a^. Writing u = -y/oF+l^, — = 1, whence — = — =^ and dx dx -y/aP + 3^ v = x, the application of the formula gives (1) 2/ = «-V^+^-/v^^^- dy a' + x^ a" , "^ Also -7- = , „ = , „ + ■ hence (2) 2/ = a^ . log (a, + V^M^") + / V^=^- Adding (1) and (2) and solving for y, 2/ = - 1 a; . V^T^ + 1 log (a; + VoM^) + C. PROBLEMS Integrate, 1. ^=a;^.loga;. 6. ^J^t:^. dx dx " x^ 2. dy _ dx "• dx = ar^-cosa;. e. ^=loga;. 3. ^ = tan->a,. 7. ^ = _^}£S^. dx dx (^2 ^ ^4 . cJw i 2 o % a^tan~^a; 4. ^^ = a;. tan^a;. 8. -^ = -— — . da; aa; 1 + x^ 9. rind the area of the cycloid, x = r ■ 6 — r ■ sin 0, y = r — r • cos 6. 10. Find the area of the cycloid, x = r — r • cos 0, y=r-6 + r- sin 0. 11. Find the volume of the solid generated by the revolu- tion of the cycloid a; = ?• • ^ — r • sin 5, y = r — r • cos 9, about its base. 142 DIFFERENTIAL AND INTEGRAL CALCULUS 12. Find the volume of the solid generated by the revolu- tion of the cycloid x = r — r-cosO, y = r- + r- sin 6, about its axis. 13. Find the length of the parabola y^ = 2px from x = to a; = x'. 14. Find the length of the spiral r = a- $ from r = to r = r'. 15. Find the area bounded by the hyperbola a^ — y^ = a?, the X-axis, and the ordinate to the point {x, y) of the hyperbola. 16. Show that the area bounded by the hyperbola 3?—y^=a^, the X-axis, and the line from (0, 0) to the point {x, y) of the hyperbola is ^ log. ^ "*" ^ ^ Art. 50. — Integeation by Rationalization A derivative -^ containing the binomial a + hx affected by fractional exponents may be transformed into a rational de- rivative -^ by the substitution a + bx = «", where n is the dz least common multiple of the denominators of the fractional exponents. Example I. — Integrate -^ = — Substituting « = .«, - = 6 ^, ^ = -|. - = -J-—, = — ^ = 62»-|-6«H6«+6+-^- «— 1 Integrating, y = ^z!" + 2^ + Sz^ + &z + Qlo^iz -V) + G ; whence, y = ^x^ + 2x^ + 3x^+Qx^ + Q\os(x^ — l) + C. LOGABITHMIC AND EXPONENTIAL FUNCTIONS 143 Example II. — Integrate -^= -. r- "'^ (1 + xy + (1 + xy 2 Integrating, y = 2tan~^z+ C; whence, 2/ = 2tan"^(l+a;)*+ (7. dy A derivative -^ containing only the surd Va + bx + 0^ may dx , be transformed into a rational derivative -^ by the substitu- ■ dz •' tion Va + &a; + or" = » — a;. A derivative — containing only the surd Va + bx ~a^ may be transformed into a rational derivative -^ by the substitu- .___ dz ■' tion Va + 6x — a^ = V(a; — rj) (r-j — x) = (x — j-i) • «. Example III. — Integrate ^ = ^^'" + '^ - da; a;'' Writing /o — T"— 2 2^ da; 2 »^ + 4 2 / + 2 « V2a; + ar' = « — a;, a;= -, — = — — -—-, z — x = —^ ' 22+2' dz 22 + 2' 22 + 2 Hence, dy^dy da;^ z'+4g + 4 ^ g" ■ 4(2 + 1) _ 1 4 d2~da;'d2 2'(2+l) 2^(2 + 1) 22(2 + 1)^2 + 1 z^' Integrating, y=log(2+l)-^ + C'=log(a!+l+V2^+^) i__, + C. 2 a;+V2a;+a;^ Example IV. — Integrate -^ = f^a: a;V2 + a; - a;'' Writing V2+a;-a;2 ^ V(2 - a;) (1 + a;) = (2 - a;) • 2, a; = 2^1^ ^ = _6^ and ^ = — 2__ 2=^ + 1' dz (z'^ + iy dz 2z'-l 144 DIFFEBENTIAL AND INTEGRAL CALCULUS By partial fractions, — = — = d« zV2-l 2V2 + I Integrating, 2/ = -1- log ( V2 -z-l) L log ( V2 . » + 1) + C. V2 a/2 PROBLEMS Integrate, 1. ^ = x'.il+x)i. 6. ^ = -^- ax "*' a;' + 1 2. ^ = _^! (^y^ 1 3. ^ = 8. ^ = - dx X 4. i^ 9. ^ = - dx a;.vr+^ ■ dx (2 + 3 a; -2 a;')*' da; a; + Va;-1 ' c^a; xy/a? + 2 a; - 1 Akt. 51. — Evaluation of Fokms 1°°, 00°, 0° If y =f(xy^'^ = 1°° when x = a, lege?/ = <^(a;) • log,/(a;) = 00 • when a? = a. If y =/(a;)*<'' = 00° when x = a, log^y =cf)(x)- log, /(a;) = • 00 when x = a. If 2/ =/(a;)*(" = 0° when x = a, log,y = whena;=0. 2. x^~' when x = l. 3. (sin a;)"-' when a; = 0. "^^ {~j "^^^^ '^ = ^- 4. (sin aj)"""' when a; = J- o / o n4 v n ^ ■' 2 8. (cos 2 a;)^ when a; = 0. 9. (log x)'-^ when x = l. Art. 52. — Differentiation op Exponential Functions The logarithm of the exponential function a" is ' log^a'' = M • log.a. Differentiating, 1 . ^a" = log.a • ^ ; whence, -^a" = a" ■ log.a • ^• a" dx dx dx dx That is, the derivative of an exponential function with a con- 146 DIFFERENTIAL AND INTEGRAL CALCULUS stant base is the product of the exponential function, the logarithm of the base, and the derivative of the exponent. Md „ ,, du a = e, — e" = e" dx dx Example. — Differentiate, y = of. Here -^=a'^ log, a ■ -r^a? = 2x • a"^ ■ log, a. dx ^' dx ^' PROBLEMS Differentiate, 1. 2/ = a». 2. y = ai-'. 3. y = a'""'- 4. 2/ = a'"""''. 6. y = e". 6. y = e^^+'^K 7. 2/ = e"""''. 8. Show that — e'"' = a" • e". dx" Differentiate, 9. y = af. Here log y = a; • log », and by dif- ferentiation - ■ ^ = 1 + log a;. Hence, ^ = af ■ (1 + log a;). y dx dx 10. y = ^. lX.y = e'. 12. y = iif'. Find the true value of, I 13. (e* + 1)^ when x = 0. - - - 16. !!=i2=2^whena;=0. X — sin X 14.. — — — when a; = 0. 1 - «' X o'" ono 17. — when a; = 00. 15. when x = a. (a; _ a)' 18. «: • sin a; when a; = 0. 19. ^ . ^ when a; = 0. sin a; 1 20. Show that af is a minimum when x= e. 21. Find the least value of ae*^ + be-"'. LOGARITHMIC AND EXPONENTIAL FUNCTIONS 147 Art. 53. — Intbgbation of Exponential Functions The results of the preceding article may be written fa". ^ = ^^+(7, fe".— = e"+C. J dx logjtt J dx Example. — Integrate -^ = e^~^. dx Writing u = 3x-2, — = 3 and ^ = i^-' .-^(3x -2). ^ dx dx ^ dx^ ' Integrating, y = \- ^'~'' + C. Integrate, PROBLEMS 1. ^ = ef. dx 4. da; 7. ^ = a- dec 2. ^ = e-. dx 5. dx 8. ^ = a'». da; 3. ^ = e- dx 6. ^ = x-^.ei dx 9. ^ = a2'^+'. da; 10. ^ = a3-''. dx 11. ^ = dx : e" ■ sin a;. Applying the formula for integration by parts by writing M = sin a;, ■ — = 6", y = ( e* ■ sin a; = e"' • sin a; — | e' ■ cos x. ax J J Applying the formula to t e"- cos x by writing u = cos x, , \ e" • cos a; = — e* • cos x+ j ^- sin a;, Hence | e'' • sin a; = e* • sin x + e'- cos a; — j e* • sin x, and y = I e'' • sin a; = |- e'' • sin x + ^e"- cos a; + C. d« da; 148 DIFFERENTIAL AND INTEGRAL CALCULUS 12. ^ = e'-cosa;. 13. ^=e'-x'. 14. ^ = a'-a^. dx dx dx 15. ^=e'-x''. 16. ^ = e-'.x'. dx dx Art. 54. — The Hyperbolic Functions The functions ^ (e* + e'') and ^ (e' — e~') are called the hyperbolic cosine of a; and the hyperbolic sine of x respectively, and are denoted by the symbols cosh x and sinh x. Hence by definition cosh a; = ^ (e"' + e~'), sinh x = ^(e' — e~'). It follows at once that (cosh xy — (sinh xy = 1. The inverse hyperbolic functions are denoted by cosh~' x and sinh"' X, so that y = cosh~' x and " y = sinh~' x are equivalent to a; = coshy and a; = sinh2/ respectively. The hyperbolic functions have been calculated and tabulated. Example I. — Find the derivative ot y = sinh x. Differentiating, ^ = — sinh x= ~Ue'- e"') = A (e' + e'') = cosh x. dx dx dx^^ J Y\ -r ; Example II. — Differentiate y = sinh~^a;. Writing x = sinh y and differentiating, 1 = cosh y ■ -^, whence -~-=- dx dx coshy Vl + siah'y Vl + a^ Hence — sinh~^a; = dx Vl + a^. PROBLEMS 1 . Show that -^ cosh x = sinh x. dx d _:_^ iX _ 1 Show that — sinh dx a -y/a? + x^ LOGABITHMIC AND EXPONENTIAL FUNCTIONS 149 3. Show that — cosh~^- = ax a ^x" — a' 4. ITind the minimum value of cosh a;. 5. An inextensible, perfectly flexible, homogeneous string fastened at two points in the same horizontal is acted on by gravity only. It is shown in mechanics that for any point (cc, y) in the position of equi- librium -^ = -, where s is the dx c length of string from the low- est point to (x, y), and c is the length of string whose weight equals the tension at the low- est point. Pind equation of position of equilibrium. The a^derivative of ~^ = - dx c ds IS da? c c » da?' whence Fro. 48. d dy dx dx 4 1 + df da? Integrating, sinh-i^ = - + Q. dx c When X ■- : 0,^ = 0, hence da; Oi = 0, and -^ = sinh-. Integrating again, y -. dx X : c-cosh-+ Cj. c " c In the figure, for a; = 0, y = c, hence Cj = 0, and X y = c- cosh-, the equation of the catenary. finally Using exponential functions, 2/ = |(e^ + eO- 6. Pmd the length of the catenary y = c- cosh - from x ■■ c to a; = x'. 7. Find the radius of curvature of the catenary. 150 DlFFEIiENTIAL AND INTEGRAL CALCULUS Art. 55. — The Definite Integral | e ^ • dx tX— 00 e-' • dx IS very important in mathematical physics and the theory of probability. Its value may be determined by the following analysis, due to Poisson. Denoting the value of the integral by ^, -4 = j e~°^ • dx, e"*" • dy, where x and y are assumed to be indepen- ■QO dent variables. Hence, e-'.da;. f e->''.dy=\ C e-<''+''> -dxdy. .QO ft/— 00 ^i/~~ = e~'■^ Hence the volume of the solid is 2 ir j e~'' •rdr=7r. Therefore, A-' = tt e""' • dx = -^/^^. g-c«2+2ax) ,^^^ write x^-\-2ax=(x-\-ay—a^', -00 whence, f ^ " e-^"' + ^ ""' • da; = e»' f ^ " e-^' + »>" d(x+a)=^- e\ Aet. 56. — Differentiation of a Definite Intbgbal Denoting the definite integral ) a? ■ a? ■ dx hj A, A = ia? dA . ..•-''' and — = 6 a. Differentiating a? ■ a? • dx with respect to a, da and integrating the result with respect to x between the limits a;=0, a; = 3, | — (a?o?dx)=\ 2aa?dx=6a. Hence for this Jo da Jo d C^ r^ d special definite integral, — i a^ • a? • dx — \ — (aVda;). da Jo Jo da ■ That is, to differentiate the definite integral with respect to a parameter, differentiate the function under the sign of integra- tion with respect to the parameter. To prove this proposition in general, consider the definite integral ^ = ( f(x, a) die. If f(x, a) is a continuous function /»6 Jc of a, I f(x, a) dx is also a continuous function of a. For, denoting by A J. and Aa the corresponding changes of A and a, ]f{x, a + Aa) —f(x, a)\dx=c\ dx=t(b — c), where c ^ c £ is a quantity which approaches zero when Aa approaches zero. Hence -4 is a continuous function of a. The derivar tive of A with respect to a is 152 DIFFERENTIAL AND INTEGRAL CALCULUS dA da Jc Aa Jc da-^ ^ ' '^ ' which proves the general proposition. By this proposition the values of definite integrals of a general form may be found from known special definite integrals. e-"' . dx= — a Differentiating with respect to a, J" X • e-"' • dx = —^, C" x' ■ e-" ■ dx = ^^, raP-e-'-dx = ^-^^,--, f'°x''.e-'^.dx= ^'^'^'""^ . Jo a* Jo a"+i Example II. — The definite integral j"° '^^ '' ^ aP + a' 2 a Differentiating with respect to a, dx IT 1 1 r'° dx TT 1 1 3 J dx _ir 11 C" (x' + a')'' 2'a'2' Jo ( (q? + ay 2 a 2' Jo (a^ + d^f 2 a 2 4' ' dx irll3 2n-l s: {a^ + a'y+^ 2 a 2 4 2n Art. 57. — Mean Value Let it be required to determine the mean or average value of the continuous function f(x) from x=ato x=b. This is equivalent to finding the mean ordinate of the curve y=f(x) from x = a to x = b. Diidde the portion of the X-axis from a; = a to a; = & into n equal parts Aa;, and draw an ordinate of y=f(x) at the end of each Aa; nearest the origin. Denot- ing the sum of these n ordinates by Sj/, their mean value is zy_^yj^ — 5. This is true for all values of n. When n is n n- Ax LOGABITHMIC AND EXPONENTIAL FUNCTIONS 153 indefinitely increased Ax becomes dx, %y becomes the sum of all the ordinates of y —f{x) from a; = « to a; = 6, and the mean ydx I ydx n • Aa; = 6 — a for all values of n. Example. — Find the mean ordinate of the sine curve y = sin X from a; = to a; = tt. I sin X • dx n Here mean ordinate = ^^ = — ir IT PROBLEMS 1. Find mean ordinate of circle ocP + y'' = t^ from x = + i to x = — r. 2. Find mean ordinate of ellipse ^ + ^ = 1 from x = + a , a- W to a; = — a. 3. Find mean value of sin^a; from a; = to a; = ■;r. CHAPTER IX OENTEE OF MASS AND MOMENT OP INEETIA Art. 58. — Center of Mass If the mass of a body is divided into infinitesimal elements of mass dm and the coordinates of dm, are x, y, %, the center of mass of the body is the point {x, y, z) so situated that x multiplied by the entire mass of the body is equal to the sum of the products of each element of mass dm by the distance of this element of mass from the FZ^plane, with like defini- tions for y and z. Hence the coordinates of the center of I xdm I ydm | zdm mass of a body are x='<^— , y = J-— , z='J-— , the I dm I dm I dm integration extending over the entire mass of the body.* Representing the magnitude of' the element of mass dm by dM, its density by Z), dm = £> • dM, and the coordinates of the center of mass become * In this chapter differentials are used directly. Lagrange says in the preface to his Mecanique Analytique (1811), "When we ^ave properly conceived the spirit of the infinitesimal method, and are convinced of the exactness of its results by the geometrical method of prime and ultimate ratios, or hy the analytical method of derived functions, we may em- ploy infinitely small quantities as a sure and valuable means of abridging and simplifying our demonstrations." 154 CENTER OF MASS AND MOMENT OF INERTIA 155 CxDdM _ CyDdM _ Ci fodif r T>dM ( ^zDdM ^DdM the integration extending over the entire magnitude of the body. If the body is homogeneous, D is constant, and ixdM iydM iz -^ , iy='i , z = ^ — dM idM ic The center of mass now becomes the center of figure. If the FZ-plane is a plane of symmetry of the body, to every term + x dM of ix dM there corresponds a term — x dM. Hence x = ; that is, the center of mass lies in the plane of symmetry of the body. In like manner it is shown that the center of mass lies in the axis of symmetry of the body and at the center of symmetry of the body. Unless otherwise specified, the density is assumed uniform ; that is, the body is homogeneous. In mechanics it is proved the center of gravity of a body coincides with its center of mass. Aet. 59. — Center of Mass of Lines Example I. — Find the center of jnass of a straight line of length I, whose density varies as the distance from one end. Denoting by da; an element of the length of the line, by x the dis- tance of dx from the end A,\>y Tc the density of the line at unit's dis- tance from A, whence .- a,-— A .- AX >^ 156 DIFFERENTIAL AND INTEGRAL CALCULUS r, \ kx- xdx 7, X = I kx-dx — 2 7 Example II. — Find the center of mass of the arc of a circle. Take as X-axis the axis of ^AS symmetry of the arc. The ele- ment of magnitude is From the equation of the cir- cle 3? + f = R\ FiQ. 51. R dy X Hence ds= — dy, and x- »x; »+l chord B\ dy -i chord arc radius x chord arc PROBLEMS 1. Find the center of mass of the straight line of length I, whose density varies as the square of the distance from one end. 2. Find the center of mass of the entire cycloidal arc. 3. Find the center of mass of the length of one loop of the lemniscate r^ = a^ cos (2 6), calling the length of the loop 2 1. Here dM=ds = (i^ + ^^ • d^ and a; = r • cos 6. 4. Find the center of mass of the quarter of the curve 3! + 2/ — o' included by the coordinate axes. CENTER OF MASS AND MOMENT OF INERTIA 157 5. Find the center of mass of the helix x = a y = a- cos , from <^ = to '. sin , '=ds = C. V d' d^ dy 6. Find the center of mass of a triangle. Break up the triangle into infinitesimal strips by lines parallel to the base at intervals dx measured on the median to the base, and call the distance from the vertex to any strip meas- ured on the median x. The median is an axis of symmetry of the triangle, and the center of mass must lie in the median. Concentrate the magni- tude of each strip on the median, and the median becomes a line whose density varies as the distance from the vertex. Hence, fc representing the density at unit's dis- rkx ■ xdx Fig. 5a. tance from the vertex, x ■■ s: Jcx • dx — 2 7 Art. 60. — Cbntek of Mass of Surfaces Y Example I. — Find the center of mass of the sur- face bounded by the parabola y^ = 2px, the r^axis, and the abscissa to the point (x„, y^ of the parabola. The surface is broken up into strips of breadth dy by lines parallel to the (xo.y^)^ ^y AM ^■--''■^ 1 ^ X A Tio. 63. 158 DIFFERENTIAL AND INTEGBAL CALCULUS ^-axis; each strip is broken up into elements of area dA by lines parallel to the F-axis at intervals dx. Hence dM= dA = dxdy and I 'rxdxdy r" r^'^^dxdy J° j ^''ydxdy ■ = Aa!o, dxdy = 12/0, since the object of the a^integration is to sum up the products ydxdy for the elements of area dxdy forming the strip, and the object of the ^-integration is to sum up the strips forming the given surface. Example II. — Find the center of mass of the surface of the cycloid x—r — r- cos 0, y = r-6 + r- sin 6. The center of mass lies on the X-axis and ■Xa; = X2f* 2xydx J-2r 1^ P(l -COS 6) (e + sin 6) sin 6 ■ dO r^ C'(e + sme)sine-de Fig. 54. Example III. — Find the center of mass of the circular sector -whose angle is 2 6„ and whose density varies as the square of the distance from the center. CENTER OF MASS AND MOMENT OF INERTIA 159 Here it is advisable to use polar coordinates. Drawing radii at angular intervals d9, the sector is broken up into infinitesi- mal sectors. Call distances measured out from the center p and draw circles concentric at A at intervals dp. Each infinitesimal sector is divided into elements of area dA = pdO-dp and x = p • cos 0. Denoting the density at unit's distance from center by k, fig. 65. :^ = e!andZ>: Tc 1 Hence X+flo /•« Jc- p" • p cos d • p • dOdp •So Jo k- p'- p ■ dQdp 4 -B • sin g 5 e Example IV. — Find the center of mass of the eighth of Tig. 56. the surface of the sphere a? + y^ + e' = B?' bounded by the coordinate planes. Jo Jo ^""""y 2 r^'--)^ hence x 160 DIFFERENTIAL AND INTEGRAL CALCULUS Here dA = (l+^ + ^dxdy =^dxdy, xdxdy i^rB^ ~TrBj Jo -y/':^— 7?-f ^_2x_ C"^dx=iB. 2TrRJo ^ By the nature of the problem x = y=:z. PROBLEMS 1. Pind the center of raass of the circular sector. 2. Find the center of mass of the quarter of the circle x' + y^= B^ included by the coordinate axes. 3. Find the center of mass of the quarter of the ellipse ^ + ^ = 1 included by the coordinate axes. or ¥ 4. Find the center of mass of the surface bounded by the parabola y^ = 2px and the double ordinate to the point {x, y). 5. Find the center of mass of the circular segment bounded by 2/^ = 2 Bx — a^ and the double ordinate through (x, y). 6. Find the center of mass of the surface bounded by the circle y'' = 2 Bx — a?, the y-axis, and the abscissa to the point (x, y). This surface is called the circular spandrel. 7. Find the center of mass of the part of a circular annulus bounded by the circles r — B,r=B', and the radii vectores 5 = — ^0) = + $(,. 8. Find the center of mass of the surface bounded by gfi j^yt — aj and the positive coordinate axes. 9. Find the center of mass of the area of one loop of the lemniscate. 10. Find the center of mass of a zone. CENTER OF MASS AND MOMENT OF INERTIA 161 Abt. 61. — Center op Mass of Solids Example I. — Find the center of mass of the half of the ellipsoid - + 1-^ + J = 1 lying to the right of the Z Emplane. The X-axis is an axis of symmetry, and dM= irrs -rt-dx a' ■ x^dx. Hence r— - I (a' — 3r)dx Fra. 57. ■abc = f' Example II. — Find the center of mass of the eighth of the sphere a? -\- 'if -\- e' = B'^ included by the coordinate planes, the density varying as the square of the distance from the center. Here it is advisable to use polar coordinates. Passing planes through the Z-axis at angular intervals d, the solid is divided into spherical wedges. Passing conical surfaces with vertex at 0, and whose elements Z make angles with the XF-plane increasing by d6, each wedge is divided into pyramids. Passing spherical surfaces concentric at 0, and whose radii increase by dp, each pyramid is divided into elementary parallelepi- peds, whose dimensions are dp, p cos 6 d<\>, p dd. Hence, V dM = p^ cos 6 dp d4> dd, x = p cos 6 cos ^, D = kp% M 162 DIFFERENTIAL AND INTEGRAL CALCULUS where k is the density at unit's distance from the center. Finally, Jr»2 /*2 ^*J Jo Jo fe • p' • cos^fl ■ COS i) • d(l) ■ dO ■ dp y^2 /»2 /^l Jo Jo Jo ■ 5 J? ■JJ.K. k- p* • cos 6 ■d r. 2. Find the area of the surface generated by the revolution of a semi-circumference of radius r about a tangent at its middle point. Multiplying both sides of the equation, y-- ffydxdy ^Cf-dx --^-^ = — by 2^.4, 1iTy-A = i.\fdx. j j dxdy A ^ 164 DIFFERENTIAL AND INTEGRAL CALCULUS The right-hand member of this equation represents the volume of the solid generated by the revolution of the area A about the X-axis; the left-hand member is the area A multiplied by the circumference described by the center of mass of A. This furnishes a convenient determination of the center of mass of the area when the volume generated is known, or of the volume when the center of mass of the area is known. Z Fig. 61. Example I. — Find the volume generated by the revolution about the X-axis of the ellipse whose axes are 2 a and 2 6, distance of center from X-axis c. F = 2 irc • irab = 2 ir'abc. Example II. — Find the center of mass of a circular sector. The volume generated by the circular sector whose angle is 2 6o, radius B, revolving about a diameter parallel to the chord of the sector is 2 wB • 2 R sin Of,-\B, the area of the sector is R'' • On- Hence, y = 2^^m^. CENTER OF MASS AND MOMENT OF INERTIA 165 PROBLEMS 1. Find the volume generated by revolving the cycloid y = r- vers ^- + V2 rx — x' about its base. 2. Find the distance of the center of mass of half the ellipse -5 + ^=1 bounded by a; = 0, x = a, from the center of the ellipse. Aet. 63. — Moment of Inertia If the mass of a body is divided into infinitesimal elements of mass, and each element is multiplied by the square of its distance from a fixed line, the sum of all these products is called the moment of inertia of the body with respect to the straight line as axis. Denoting the infinitesimal element of mass by dm, its distance from the axis by y, and the moment of inertia by J, 1= \ y^ • dm, the integration extending over the entire mass of the body. Representing an infinitesimal element of the magnitude of the body by dM, the density of this element by D, dm=D ■ dM, and the moment of inertia becomes J= I y^ ■ JD • dM, the in- tegration extending over the entire magnitude of the body. The moment of inertia occurs very frequently in Strength of Materials and in the Theory of Eotation of Bodies. For example, in Problems 31, 32, 33 of Article 23, J is the moment of inertia of the cross-section of the beam. If the material of the body is of uniform unit density, which is always assumed unless the contrary is stated, D • dM = dM, and the moment of inertia depends only on the shape of the body. The moment of inertia is one of the elements used in select- ing shapes for engineering structures. 166 DIFFERENTIAL AND INTEGRAL CALCULUS The moment of inertia divided by the mass of the body is called the square of the radius of gyration of the body for the moment axis used and is denoted by F. Hence fc^ I y^dm k is the m Fia. 68. distance from the moment axis to the point where the mass of the body must be concentrated so that the moment of inertia of the concen- trated mass shall equal the moment of inertia of the mass distributed throughout the body. In Problem 35 of Article 23 and in Problem 13 of Article 43, fe" is the square of the radius of gyration of the turning body. Denote by 7„ the moment of inertia of a body for any moment axis AA, by I, the moment of inertia of the same body for a parallel moment axis through the center of mass of the body. Through C, the center of mass of the body, pass a plane RS perpendicular to the line AG. Denote by D the distance between the two axes, by r the distance of dM from the axis AA, by r' the distance of dM from the axis CO. By I^=ir'^-dM. From the triangle Hence definition /„ = \i^- dM, AGP, i^^r'^ + jy + ^T'Davud. (1) Ci^dM= Cr"'dM+D''CdM+2DCr' sin 6 -dM. CENTER OF MASS AND MOMENT OF INERTIA 167 Now r' sinO = PE=y and j r' sm9-dM= ( ydM. CydM But y = — -T^ — = 0, since y is the distance of the center of mass of the body from the plane RS, and this plane is drawn through C perpendicular to the line QA. Hence (1) becomes 7„ = /„ + D^ • M, which may be written I^ = I^ — D-- M, and is known as the reduction formula. Stated in words, this formula reads the moment of inertia of a body for any moment axis equals the moment of inertia for a parallel moment axis through the center of mass of the body plus the square of the distance between the two axes into the mass of the body. Denoting the radii of gyration for the axis AA and the axis GG by /f„ and K^ respectively, I^ = M- K^, I^ = M • K^-, and by the reduction formula IQ = K^ + D". Aet. 64. — Moment of Inertia of Lines and Suefaces Example I. — Find the moment of inertia of a straight line of length I for moment axis perpendicular to line through one end of line. Denoting an element of the line by dy, the distance of this element ^ from the moment axis by y. The radius of gyration is found from A The moment of inertia for a parallel axis through the center of mass of the line is I^ = ^P — ^P = J^ P. The radius of gyration for this axis is found from ifc/ = ^ f. —-y- — »! -I 168 DIFFERENTIAL AND INTEGRAL CALCULUS Example II. — Find the moment of inertia of a triangle for moment axis through vertex parallel to base of triangle. Breaking up the triangle into strips by lines parallel to the moment axis at intervals dy, calling the length of any strip x, and its distance from the moment axis y, dM = x dy, and from simi- Hence, lar triangles - = - d Fig. 67. y The radius of gyration is found from The moment of inertia for a parallel axis through the center of mass of triangle is J. = i 6d' - 1 6d (I d)2 = ^ bd' and kj' = j\ d\ The moment of inertia for the base of triangle as axis is I. = ^V b<^ + i &^ a ^y = ^ biP and kj' = ^ d'. Example III. — Eind the moment of inertia of a circle for axis through center perpendicular to plane of circle. Here it is advisable to use polar coordinates. dM = pd6dp, and the distance of dM from the axis AP is p. Hence, I. J p'dOdp = iTrB*, and kj' .\R'. Fig. 69. Ip is called the polar moment of inertia of the circle. Denoting by I^ the moment of CENTER OF MASS AND MOMENT OF INERTIA 169 inertia of the circle for axis AX, by /„ the moment of inertia for axis AT, I. = Cy'dM, /j, = CaJ'dM, and I. + Iy= f(3^ + y')dM= Cp^dM = /,. Since the circle is placed in exactly the same manner with respect to each of the diameters AX and AT, I^ = /„. Hence, 2 /. = /, = -^ 7rB\ and J, = ^ ^B*, K" = \ R\ PROBLEMS 1. Find moment of inertia of a rectangle base b, altitude d for base of rectangle as axis. 2. From Problem 1 find by reduction formula the moment of inertia of rectangle for axis through center of rectangle parallel to base. 3. Find moment of inertia of rectangle for axis through center perpendicular to plane of rectangle. 4. Find moment of inertia of isosceles triangle for axis of symmetry as moment axis. 5. Find moment of inertia of ellipse for major diameter as moment axis. 6. Find moment of inertia of ellipse for minor diameter as moment axis. 7. Find moment of inertia of ellipse for axis through center of ellipse perpendicular to plane of ellipse. 8. Find moment of inertia of parabolic segment for axis of parabola as moment axis. 9. Find moment of inertia of circular spandril for diameter as moment axis. 170 DIFFERENTIAL AND INTEGRAL CALCULUS Akt 65. — Moment of Inektia of Solids Example I. — Find the moment of inertia of the rectangular parallelopiped whose dimensions are b, d, h for axis through centers of two opposite faces. Break up the solid into laminae by planes perpendicular to the axis at intervals dx and call the distance of any lamina /u X- / }^ / ".L I'lO. 69. from one of the faces x. The lamina may be broken up into elements dxdydz, and the moment of inertia of the lamina = dx I i p^dydz = moment of inertia of base of lamina mul- tiplied by the thickness of the lamina. Hence the moment of inertia of the lamina is ■^{y+d'')bddx, and the moment of inertia of the parallelopiped is 7= J^ (62 + d^) bdf" dx = -^ {V + d^) bdh. The radius of gyration is found from ^'= bdh =tV(6' + ^0- Example II. — Find the moment of inertia of a cone of revolution for axis through center of mass of cone parallel to base of cone. CENTER OF MASS AND MOMENT OF INERTIA 171 Break up the cone into laminae by planes parallel to the base at intervals dy. Call the distance of any lamina h-om the vertex y, the radius of the base of the i lamina x. The mo- ment of inertia of the lamina for axis X' X' through the center of the lamina and paral- lel to the axis XX is ^irx^dy, the distance h-omX'X' to XX is f-ff — 2/. Hence by the reduction formula the moment of inertia of the lamina for axis XX is ^Trx^dy + way' (^ H —yY dy, and the moment of inertia of the entire cone is I=Jjl7rx'dy + w^i^H-yfdyl . By similar triangles - = — .hence y H I=£'>^j^,tdy + w^M^-yrdy^=i,^R'H{R^ + \H^. Fig. 70. PROBLEMS 1. Find moment of inertia of the sphere for diameter as axis. 2. Find moment of inertia of cone of revolution for axis of symmetry as moment axis. 3. Find moment of inertia of cylinder of revolution for axis of symmetry as moment axis. 4. Find moment of inertia of cylinder of revolution for axis through center parallel to base. 172 DIFFERENTIAL AND INTEGRAL CALCULUS 5. Find moment of inertia of spherical cap for axis of symmetry as moment axis. 6. Find moment of inertia of ellipsoid for longest diameter as moment axis. 7. Find moment of inertia of the segment of a paraboloid of revolution for axis of symmetry as moment axis. CHAPTER X EXPANSIONS Abt. 66. — Convergent Powek Series The identical equation (1 — xy= 1 — 2 a; + x^ is true for all values of x. Expanding the fraction into a series by division, there 1 ^ X results the identical equation ^^^l+x+x'+a?+x*+ ■■■ x''-'^+af+x"^^+x'^^+x''+^-] , 1—x where the number of terms in the series is infinite. This identity is not true for all values of x. For example, if x = 2, the fraction equals — 1, the series equals infinity. To 1 — X determine for what values of x the identity is true, denote the sum of the first n terms of the series by s„, the sum of the remaining terms by r„. Then = s„ + ?•„, and if r„ can be 1—x made less than any assigned quantity, however small, by tak- ing n sufficiently large, = limit s„ when limit n = 1 — x and the series is said to be convergent. Now r„ = a;"(l +x + x' + x' + x*'-\ ). When x, since a„ and a; are by hypothesis not infinite and r is assumed less than unity. Hence the sum of the series s = s„ + limit s„ when limit n = oo; the series is convergent and Cauchy's test is proved. For example, in the series l + x + oi^ + a^+ •■■x" + a;"+^ -\ , the ratio of each term to the preceding term is x, and by Cauchy's test the series is convergent when — 1 < a; < 1.* Art. 67. — Tatlok's and Maclaurin's Series The values of explicit algebraic functions can be directly calculated for arbitrarily assigned values of the independent variable. For example, ity = x?—7x + 7, any value may be assigned to x, and the corresponding value of y becomes known. * Euler seems to have been the first to call attention to the fact that infinite series can be safely used only when convergent. 176 DIFFERENTIAL AND INTEGSAL CALCULUS Consider the function f(re) = x", when m is assumed to be a positive integer. When x=b=a + h, whence h = b — a, 1,2 /(6) = (a+h)" = a'' + m- a""^ ■ 7t + m (m — 1) a""^^^ + m(m-l)(m-2)a"'-3^ + ... oi by the binomial formula. This result may be written, (l)f{b)=/(a + h)=/{a)+f'(a)-h^ +/"(«)•!+/'"(«) •f,+ - or (2)/(6)=/(a)+/'(a).(6-a) where f(a), /'(a), f"(a), f"'(a), •■-, are the values of the suc- cessive derivatives of f(x) = a;" when x= a. The symbol n ! stands for 1-2-3-4-5- ••• n, and is read factorial n. It is proposed to derive a general series of the same form as series (1) and (2) for the approximate calculation of any func- tion f(x) for arbitrarily assigned values of the independent variable. The problem may be thus formulated : Let f{x) and its successive derivatives f'(x), f"{x), f'"(x), ■■•, be finite and continuous from x = a to x=b, and denote by f{a), f'ip), f"(a), f"'(a), ■••, the values of these functions when x = a. The value oif{b) is to be found in terms of f(a), f'{a), /"(«)» f"'(a), •■•, and the powers of b — a. The investigation is based on the following proposition, known as Rolle's theorem : If the function <^ (x) and its first derivative <^'(a5) are finite and continuous from x = atox=b and c^ (a) = and (6) = 0, the first derivative ^'(a;) must vanish for some value of x between a and b. EXPANSIONS 177 If (f>' (^) is identiually zero, the truth of the proposition is evident. If <^' (x) is not identically zero, ^' (a;) must change sign between x = a and x= b, for otherwise (x) would either continually increase or continually decrease from a; = a to X = b, and in neither case could i^ (a) and ^ (&) both be zero. Hence <^' (x) must change sign between x— a and x = b. Since ^' {x) is assumed to be finite and continuous from a; = a to a; = &, <^' (a;) can change sign only by passing through zero. Therefore (j>' (x) must vanish for some value of x between x = a and x = b. Denoting this value of a; by a + $(b — a), where 6 must be less than unity, i(x)=f(b) — f(x)—ki{b — x) and form the first derivation ^i'(a;)s— /'(a;)+fci. Since by hypothesis (^i(a;) and 4>-i{x) are finite and continuous from x=a to x=b and ^i(a)=0 and ,{x) = f{b)-fix)-f'(x) . (}, -x)-k,- ^^ ~f^' and form the first derivative <^2'(^) s — (6 — x)f"(x) + k^ib — x). Since by hypothesis <^2(^) S'ld 4>i(p) 3,re finite and continuous from x = a to x = b and <^2(«)=0 and t^2(&)=0) S^is'C^) must vanish for some value of x between a and b. Denoting this value of X by a+ diib — a), k.2=f'\a+6^{b — a)\. Hence /(6) = /(a)+/'(a) . (& - a)+r\a + 6,ib - a)\ ■^~^- 178 DIFFERENTIAL AND INTEGRAL CALCULUS Assume /(6) =/(«)+/'(«) . (6 - «)+/"(«) .(^^ + fc3 .(^^'. Write U=>')=f{b)-f{x) -f'{x) .(b-i>) J \ J 2! ^ 3! and form the first derivative Since i(x) and <^3'(a;) are by hypothesis finite and continu- ous from a; = a to x=b and <^3(a) = and (^3(6) = 0, 3{x) must vanish for some value of x between a and b. Denoting this value of a; by a + O^Q) — a), Aij = /" ' j a 4- 63(6 — a) | • Hence /(&)=/(«)+/'(«) • (6 - «) + /"(«) -^^^^ 2! +/"'Ja + e3(6-a)i.i^^' 3! Eepeating this operation n times, there results (1) m ^f(a) +f(a) .(b-a) +f"(a) ■ ^^ + /'"(a) • ^^ ■^ ^ ^ 4 ! •' ^ ^ (n - 1) ! +/"fa + g(6-a)|- (^~")" n ! where < 6 < 1. The error committed by placing f{b) equal to the sum of the first n terms of series (1) is r„=f''la + 6(p — a) j *■ ~ ^ • If this error can be made indefinitely small by taking n suffi- ciently large, the series is convergent and can be used to cal- culate the value of f(b) to any required degree of accuracy, provided /(a), /'(a), f"(a), /'"(*))•" ^'^e known. When n EXPANSIONS 179 becomes indefinitely large, the series becomes an infinite series and the region of convergence is most readily determined by Cauchy's test. In (1) place b — a = h, whence b = a + h. There results (2) f(b) ^fia + h) ^ f(a) + f'(d) . h +/"(a).|^ +/'"(«) -^^ This is Taylor's series. In (2) place a = 0. There results, (3) fQi) s/(0) +/'(0) • h +/"(0) ~+f"'(0) -^ (n — 1) ! n ! This is Maclaurin's series. The only restrictions on a and h in these series are that f(x), f'(x), f"(x), f"'(x),---f''-^(x) must be finite and continuous from x = a to x = a + h and that /" (a + 6/).) — must become less than any assignable quantity when n is indefinitely increased. Since the quantities represented by a and h are not fixed, they may be denoted by X and y respectively, when Taylor's and Maclaurin's series become f(x + y)=f{x)+f(x).y+r(x).yl+r{x).l.^ + ... f{y) s/(0) +/' (0) . y +/" (0) .|! +/'" (0) -t.^ + ... (?i — 1) ! n ! *Taylor (1685-17.S1) published his series in his " Methodus inoremen- torura." Maclaurin published his series in his "Treatise of Flexions" 1742. The expansions effected by these series had been previously obtained by laborious processes. 180 DIFFERENTIAL AND INTEGRAL CALCULUS Taylor's series expands a function of the sum of two varia- bles in the ascending powers of one of the variables; Mac- laurin's formnla expands a function of one variable in the ascending powers of that variable. Example I. — Expand log„(l+2/). This expansion is effected by Maclaurin's series since log, (1 + y) is a function of one variable. Forming the succes- sive derivatives, 2! (1 + 2// -3! /(2/) = log,(l+2/), f"'(y) ^'^^^=1 + .' r{y) f"(^)=7^.' riy) a+yy' (1+2/)" hence /(O) = 0, /'(0) = 1, /"{0)=-l, /"'(0) = 2!, /'-(0)=-3!, ... /»(0) = ±(n-l)!. Substituting in Maclaurin's series, \ogAl+y) = y-^+^-^+^-^+^-- , 2/""^ -jir 1 (n-1)! n\{l + eyY By Cauchy's test this series is convergent for values of y numerically less than unity. Hence this series may be used to calculate the Napierian logarithms of numbers from to 2. -I +1 Pig. 72. Taking y=.5 and w = 13, log, 1.6 = .40546914 with an error between .000000053 and .0000093. EXPANSIONS 181 Tlie region of convergence of this expansion of log^ (1 + y) is indicated graphically by the heavy line of the figure. Erom this expansion of log^ (1 + y) an expansion of loge (1 + ») with an enlarged region of convergence is obtained by the following analysis : log,{l + y)-y 2^3 4+5 q + j g + - for - 1 < 2/ < 1. y^ f y* f f f f log.(l-2/)^-r 2 3 4 5 6 7 8 for - 1< 2/ < 1. By subtraction, log,(l + 2/)-log,(l-2/) = log,^s2('2/+| + | + ^+..A for — 1 < 2/ < 1. Substituting y = —^, 1±I = ?-±1, and when -l0. 2z + l 1—y z Hence log, (!+«) = log^a I o/ 1 I 1 I 1 I 1 I \2z-\-l Z{2z + \Y 5{2z + lf 1 {2z +iy for 2 > ; a convenient formula for the calculation of the Napierian logarithms of numbers. By Maclaurin's series log,(l + 2/) = log,e{2/-|'+|-J + |-|+...|, that is, log, (l + y) = log. e • log, (1 + y). Placing l + y = a, log„a=log„e-logea, whence log.e =- and log„(l+2/) =- log,(l+?/). log, a log, a The factor :; , by which the Napierian logarithm of a loge« number must be multiplied to obtain the logarithm of the same 182 DIFFERENTIAL AND INTEGRAL CALCULUS number in the system whose base is a, is called the modulus of the system of logarithms whose base is a. In the common system a = 10, and -^—=.43429448. ■^ log. 10 Hence logio (1 + 2/) = .43429448 log. (1 + y). Example II. — Expand log. y. Here f{y) = log.y, /' {y) = \ f" (y) = - i, /'" (y) = ^, - ... r{y) = ± (^^^- Hence, /(O) = -oc, /'(O) =a>, /"(O) = -«,, /"'(0) = oo, •••, /"(0) = ±oo. The function log.y cannot be expanded by Maclaurin's series into a power series in y. However, writing logy = log 51 + (2/ — 1)|, Example I. gives logy=(y-l)- ^^ 2 + 3 4 "'' 5 ■"' a power series iny — 1 convergent when < y < 2. Example III. — Expand (a; + y)", where m represents any finite number, positive or negative, integral or fractional. This expansion is effected by Taylor's series. Here f{x) = O!", /' (x) = mx'"-\ f" (x) = m (m - 1) ar-% ■ ■ ■ /»-' (a;) = m (m — 1) • • • (m — n + 2) a;'"-"+S /"{x) = m (m — 1) ■■■ (m — n + 2) (m-n + l)a;'"-'', ••• Substituting in Taylor's series, {x + yY = x^+m ■ x"-' ■ y _[_ ^ ("^ - 1) a;"- Y I w(w-l)(m-2) ^„_3 , , ___ 3! m(m - 1) ... (m - ?i + 2) +i +i («-l)! ^ m(m - 1) ... (??i — n + 2) (m - w + l) ^„-n „ _j_ ..._ EXPANSIONS 183 The ratio of the nth term to the {n — l)th term is r = 1 — — 1 )-. Since m is finite, the factor — — 1 \ n Jx n approaches unity when n is indefinitely increased. Hence if - < 1, the ratio r becomes less than unity when the series is sufB.ciently extended, and the series is convergent by Cauchy's test. This proves the binomial theorem for all finite ex- y ponents, provided — 1 < - < 1. Example IV. — Expand {2 — x+y)~^ in ascending powers of y. Substituting v = 2 — x, (2 — x + y)~ ^ becomes (v + yy ^. By Taylor's series, yv-ryj - •^ -r^^V ^ '^ ^ dv'^ ^ 2! dw'=^ ^31 = v-i -iv-i-y + iv-i-y^-^\v-^-y''+-; whence, (2—x+y)~^ = (2 — x)~^ — -|.(2 — a;)"^ • y + i(2-x)-i.f-^\(2-x)-^.f+..., y convergent when — 1 < „ _ < 1- PROBLEMS Expand and determine the region of convergence, 1. sin 2/. 5. sm.{x — y). 9. log (a; + 2/). 2. cosy. 6. cos (a; — 2/). 10. tan~'a;. 3. sin (a; + 2/). 7. e". 11. sin^'a;. 4. cos (a; + 2/). 8. e"*. 12. log \x+ VI +x''\. 13. sinh a; s ^ (e* — e-"'). 14. cosha!= |-(e»^4-e-"'). 15. e'^sina;. 18. tana;. 21. (1 — a; + 2/)^. 16. e-""~'^ 19. logcosa;. 22. (a;^ — 2/')"'. 17. (1— sin^a;)"*. 20. cot 2/. 23. log (k' — 2/^). 184 DIFFERENTIAL AND INTEGRAL CALCULUS 10° 24. Compute sm 10°. Here a; = -g-j-^- 25. Compute sinh .2. 26. Compute logj2 and logio2. 27. Expand e"*'. Aet. 68. — Euler's Fobmulas foe Sine and Cosine 2^ gS ~4 g« .6 Let the series e'= 1 +« + |- + |: + J- + |: + |- + ..., which is the development by Maclaurin's formula of e' when » is real, be adopted as the definition of e' for all values of z, real and complex. Placing 2 = ix, where i stands for V— 1, whence i*" = 1, i^"+^ = i, t*"+^ = — 1, i*^'^^ = — i for all integral values of n, ^ ' 2! 3! 4! 5! 6! 7! In like manner, placing 2 = — ia;, ^ '^ 2! 3! 4! 5! 6! 7! Taking half the sum of (1) and (2), "2 ~ 2! 4! 6! 8! /o\ e H- e H a;- , a; a;- , ar (3) — — =1 -"scosa;; ^ ' 9, 9! 4! fil 8 ' dividing half the difference of (1) and (2) by 2i, (4) ^^ — 3-, + ^,-,-j + 9-y---sina,. In general, sin {nx) = — \ e'*" — e-*"" 5 > c°s {nx) = \ \ e*'" + e"'"* \ . These results are known as Euler's formulas for sine and cosine. These formulas are useful in trigonometric transformations EXPANSIONS 186 and in integrating derivatives involving trigonometric and exponential functions. Since the i in u = ix is of the nature of a constant factor, du _ . dx Example I. — Find the value of cos^a; in terms of the functions of the multiples of x. = ■jVI^'^'' + 5 e'''^ + 10 e'^ + 10 e'" + 5 e-<^ + e-<='| = -^{cos (5 a;) +5 COS (3 a;) + 10 COS a;}. Example II. — Integrate -^ = sin^a; • cos^ x. dx Substituting for sin a; = — (e** — e"**) for cos x = \{e''' + e"*"), ^ = -^ (e^" + e-^" - 2) = - 1 cos (4 a;) + 1. Hence dx y = - i^ sin (4:X)+ \x + a Example III. — Integrate -^ = e~^' sin (pf), where P and p are constants. Substituting sin (pt)= ^(e'** — e"*"); ^y L \d-p+ip)t — gi-p-ip)i > dt~2V *' hence y = — 2i 1 _e<--p+'W< H L_e(-i"-.i.)« I 4- c — P+ip P+ip 2iP^+p - -pl^ \P sin {pt)+p cos (j,t) ] + G 186 DIFFERENTIAL AND INTEGRAL CALCULUS PROBLEMS Find in terms of the functions of the multiples of x, 1. sin*;?. 2. sin'o;. 3. sin'' a; • cos^ (2 a;). Integrate, 4. dy dx~ : sin' X. 5. dy_ dx sin* X. 6. ^=COS^ dx X. 7. dy _ dx~ sin^ X. 8. dy_ dx e^. • sin^ X. 9. Show that e*" = cos x + i sin x, e~'' = cos x — i sin x. AkT. 69. DlFFEKENTIATION AND INTEGRATION OF PoWEB Sekies Consider the power series, (1) /(x) = ao + ai • a; + Wj • a;2 _| and denote the sum of the first n terms by s„ (a;), the sum of the remaining terms by r„ (a;). Absolute Convergence. — Denote the numerical or absolute value of any quantity z by the notation \z\. So that |— 6 1 = 5, 12 1 I 1 Suppose in the power series (1), denoting | a„ | by A„, that ^Xo" < M, where JHf is a finite quantity, for all values of n. -Xo -fXo Fig. 78. M Take | a; | = X < Xq ; then, since A„ < — ^ by hypothesis, X, \xj ^ \xj ^ Since by hypothesis M is finite and — < 1, by Cauchy's test EXPANSIONS 187 ■the right-hand member of the inequality, and consequently the left-hand member, is a convergent series. That is, the sum of the absolute values of the terms of the original power series is convergent for | a; | < Xq if A„X„'' < M. This is expressed by saying that the given power series is absolutely convergent within the region — Xo < a; < Xq. Example. — In the expansion, when x = l, each term is less than 2. Hence, the series is absolutely convergent for — 1 < a; < 1. Uniform Convergence. — Writing Ao+AX+A,X'+ ■■.A„X''+A„+,X-+'+...=S„{X)+B„(X), --(D"- • for 1 K 1 = X < Xo. "■-(f.r- ■-'iiJiir x„ Now, n may be taken so large that, for all values of X < Xq, (X\" 1 -^ J ^ for the same value of n becomes less than e, however small e may be assumed. Consequently, since '■„ (x) >■ iJ„ (X), for all values of x between — X„ and -f Xo the expression ?•„ (a;) can be made less than e for one and the same value of n. This fact is expressed by saying that the power series is uniformly convergent in the region - Xo < a; < Xo. Example. — The power series l+a;+a;^+a^H x"+x''-'^-\ is absolutely convergent for | a; | < 1. Assuming e = .000001 , 188 DIFFERENTIAL AND INTEGRAL CALCULUS determine n so that for this value of n r„ (x) < e for all values of X between — ^ and + \. Since r„ = , by the conditious of the problem r„< (^)"~^ The conditions of the problem are satisfied when (^)"~' = .000001 , that is, when n = 22. Continuity. — Denote by x and X(, any two quantities numeri- cally less than Xq. Since f(x) = s„ (a;) + ?•„ (x), f(x) -/(a;„) = {s„ (x) - s„ (a;„) } + f r„ («) - r„ (sBo) j • Since n, however large, is supposed to be finite, x may be taken sufficiently near x^ to make | s„ (x) — s„ (a!o) [ < e, however small £ may be assumed ; and since by hypothesis x and Xg are within the region of uniform convergence of the power series, n may be taken so large that | r„ (x) | and | r„ (x^) \ each become less than c. Consequently | f{x) — /(a!o) | < 3 e, and the function defined by the power series is continuous in the region of uniform convergence. Integration. — To show that between limits within the region of uniform convergence the limit of the sum of the integrals of the terms of a power series is the integral of the limit of the power series, write Jf(x)dx= I aD-dx+ I ai^-xdx+ I a2-x'dx+ ••• +J r„(x)-dx, where a and /3 lie within the region of uniform convergence of f(x) = ao + ai-x + a2-oc' + a3-a^+ ••• Now j r„ (a;) • da; < £ I dx=:e(J3 — a), where c is a quantity which approaches zero as n approaches infinity. This proves the proposition. EXPANSIONS 189 The series f{x) = ao-xj/{x) + ai-il/{x) -fj} (x) +a2-\l/(x)- [<^ {x)Y + — may be integrated term by term between the limits a; = « and x = p provided a and j8 lie within the region of uniform con- vergence of the series ao + % • ^ (x) + ctj • [<^ (a;)]^ + ■ • ■ and kj/ (x) is finite from x = a to x — ^. Example I. — Expand tan^^o; into a power series by inte- gration. — tan-^a; = — ^„ = 1 - a^ + a* - a!« + a;' - a;!" + •••, dx 1 -f ar a power series uniformly convergent for | a; | < 1. Hence term by term integration gives a valid result, and tan-ia; = a,-| + |-^-F|-g+... forla^Kl. From this expansion is obtained Euler's series for the calcu- lation of 17. Writing tan u = ^ and tan v = ^, tan (u + v) =1 = tan^. Hence j = m + v = tan~'-|- + tan"'^, ° 4 1^2 3-2'"''5-2' '"] \3 3-3' 5.3» '" ) 1,2 3^ 3^2' 3V 51,2=^37 Example II. — Find the value of t when f = ^ r ^y — . h<2r. -^2g'J^ -,/(li-y){2ry-y^ Here t is the time of vibration of a simple pendulum of length r, the bob starting at a distance h above the horizontal through its lowest position. 190 DIFFERENTIAL AND INTEGRAL CALCULUS Substituting y = h- sin^ 0, ir r^ 2 71-sine-cose-de J Jo V2^*^" a/A • cos e • sin 6 V2 rh - li' sin^ 6 ,|.||.(^J.3..,,...]., HH-l)"(^)"+-} If /t is small compared with 2r, t= 27r\- is an approxima- 9 tion sufficiently accurate for most purposes. Difierentiation. — To find under what conditions the sum of the derivatives of the terms of a power series, f(x) = ao + ai ■ a; + Oj • ar" + «3 • 9^ H a„-of+ ■-, uniformly convergent for | a; | < Xq, is the derivative of the function defined by the power series, write (^ (a;) = Oj + 2 02 • a; H « • a„ • x"-^+ (n +.1) ■ a,^i ■ x" -] . By hypothesis the series defining f(x) is uniformly convergent for I a; I < Xo- If from and after some fixed term the ratio of the corresponding terms of (x) and f(x) is not greater than unity, 4> (x) is uniformly convergent when f{x) is uniformly EXPANSIONS I'Jl convergent. The ratio of the (n + l)th terms of {x) and f(x) is «(— ) — , where xj, and x, denote respectively the variables of the and / series. Hence <^ (a;) is uniformly con- >\x^\. If vergent when nf — | — > 1, that is, when n -^ \X^J X^ Xf wj, is taken less than x,, limit «( — 1 =0 when limit n = oo. Hence (jc) is uniformly convergent for | a; | < X, when X lies within the region of uniform convergence of fix). For these values of x, (a;) may be integrated term by term, and fix)— I when a and ;8 lie within the region of uniform convergence. Differ- entiating this result, f'i^ — 4>ix) = 0; hence fix) = tti + 2 a^ • x + ••• + n ■ a„ • x"'^ + ••-. That is, a uniformly convergent power series may be differ- entiated term by term as long as x is within the region of uniform convergence of the power series. Example. — The expansion (1) (l-a;V* = l+|a;2+^.|a;^+|.|.|a;»-fi.f.|.|a;«+... is uniformly convergent for | a; | < 1. Obtain the expansion of (1 — a;^"^ and (1 — af)~^ by differentiation. Differentiating (1) and dividing by x, (2) il-^)-i = l+ia^ + i-^ai' + ^x'+-; differentiating (2) and dividing by 3 x, (3) (l-af)-i=l+^x' + ^ai'+-. PROBLEMS Expand by integration, 1. log(l-|-a;). 2. log(l — a;). 3. sin-'a;. 192 DIFFERENTIAL AND INTEGRAL CALCULUS From the expansion of (1 — a;)~' obtain by differentiation the expansion of, 4. (l-a;)-'- 5. (1 - »)-'. 6. (1 - a)-*- 7. Find the length of the ellipse x = a cos , y = b sin <^. If the arc is measured from the end of the major axis, s=a\ VI — e^ svD? d^, where e is the eccentricity of the ellipse. The entire length is 4 a I VI — e''sin^ ( ) ; Mi — Mo is positive, and. «o SiBo S2/o KdnHoSyoJ a minimum, wlien — - and — - are both positive, and d'F d^F / d^F V , dyo' Example. — Find the dimensions of the rectangular parallel- opiped of maximum volume, sides parallel to the coordinate axes, that can be inscribed in the ellipsoid — + f^ + -^ = 1 The volume of the parallelepiped is b' Fio. T4. V= 8 xyz = 8 cxy (-g- If F is a maximum, Vi = a^'y" — — f J^ is a maximum, and vice versa. Forming the piartial derivatives of Fi, dx 5Fi_o^. 2A 4:x'f dy "a" W ' 5a^ ^ a^ j2' APPLICATIONS OF TAYLOR'S SERIES 197 dx dy a^ b^ The conditions ^ = 0, ^^ = make x=-^, y=-^. dx dy V3 ^3 These values of x and y make ax^ 9 ' 52/2 9 ' accd?/ 9 ' Since ^and ^ are both negative and ^.^■^>(^^\\ dx" 8y' ^ dx' dy^ \dxdyj' Vi is a maximum. Hence the dimensions of the maximum parallelepiped are - — , — , — . PROBLEMS 1. A box with open top in the form of a rectangular paral- lelopiped contains 108 cubic inches. What must be its dimen- sions to require the least material in construction ? 2. Eind the point (x, y, z) the sum of the squares of whose distances from (oi, &i, Cj), (ajj hj "2), (as, 63, C3) is a minimum. 3. The sum of the three dimensions of a rectangular parallelepiped is a. Mnd the dimensions when the volume is a maximum. 4. Find the dimensions of the cistern of maximum capacity that can be built out of 3000 square feet of sheet iron, the cistern being of the form of a rectangular parallelopiped and without lid. 5. The sum of the three dimensions of a rectangular parallelopiped is a. Find the dimensions when the surface is a maximum. 198 DIFFERENTIAL AND INTEGRAL CALCULUS Aet. 72. — Contact of Plane Curves Plot the curves representing the equations F= F(X) and y = f{x) to the same coordinate axes, and denote by Yq and y^ the values of Y and y corresponding to X=x^, x=Xa, by Fj and y-^ the values of Fandy corresponding to X=x,j±h, x = Xq±Ji. By Taylor's series, yr _Y .dY,. J. d'Y,i±hl d''Yoi±]il,d'Yoi±Kl, ^' - ^°+dX„^^''^+ dX? 2! +dZ?"3! +dX;r^ 4! +■■■' y -y I dno( J. d\(±hy d?yo(±hY d%,(±hy Hence dxo^ 2! ' dxo' 31 /'dYo_d}h\ LYo dajn 2! + d^Fo d%\ (±Jiy dXo^ dx„V 3 ! + ■ The curves Y=F(X), y=f(x) have a common point if Yo = yo when X(,=a^. The difference between the ordinates • Y Fio. 75. corresponding to Xq ± h when h approaches zero is of the first dY dv degree in h if — - =?t — ^ : of the second degree in li and the dXo d Xa curves are said to have contact of the fi.rst order, if d Fo _ d2/„ , d^ Fo dVo . dX, dx. dXi dxn' APFL1CATI0N8 OF TAYLOR'S SERIES 199 of the third degree in h and the curves are sjiid to have con- tact of the second order if — - = -^, \ = — ^ and 'TjFs'^lfi' In general, the difference between tlie ordinates is of the {n + l)th degree in h and the curves are said to have contact of the nth order when the first pair of corresponding derivatives not equal are of order n + 1. If the contact of the two curves Y=F(X) and y=f(x) at (^o, 1/0) is of an even order 2 in, Y-v = (^""'^^» d'-^%\(±hY'«+^ Hence Yi — y-i changes sign with 7t and the curves intersect. If the contact is of an odd order, the curves do not intersect. Suppose the equation y = f{x) to be completely determined, while the equation Y=F{X) involves arbitrary constants, that is, parameters. The condition necessary for the inter- section of the curves represented by the equations Y=^ F(X) and y =/(») when X^ = x^ namely Yq = t/o, determines one of the parameters of Y=F(X); the conditions for contact of dY dv the first order when X^ = Wo, namely Fq = ^o, t^ = > aXo dx^ determine two parameters of Y= F(X) ; the conditions for contact of the second order Fq = Wq, -— ^ = --^, — — ^ = —^ dXo dxf, dX„ dx^ determine three parameters of Y= F(X). In general, con- tact of the mth order determines n+1 parameters of Y=F(X). It is also evident that the highest order of contact Y= F(X) can in general have with another curve y=f(x) is one less than the number of parameters of Y= F{X). Example I. — Determine the order of contact of 4 y = a;^ — 4 and a^ + y" — 2 y = 3. 200 DIFFERENTIAL AND INTEGRAL CALCULUS The common point is (0, — 1). Tor this point the first equar tioa gives 1 = 0, g=l, g-_0, g = 0, ..., the seeond •^""'»° S=». S4 &=»' S-i- «-''™ is contact of the third order. Example II. — Tlie equation y=f(x) represents a fixed curve, (X — mf + (Y — rif = R^ is the equation of any circle. The parameters m, n, R are to be so determined that the circle has contact of the second order with y = f(x) at the point (»o, 2/o)- The problem requires that m Y =v ^ = ^ d?Y^^d?ya ^ ' " ^'" dXo dxo dXi dxo^ when Xq = Xg. Differentiating the equation of the circle twice in succession, (X-my + (Y- nf = B% (X- m) + (F- n)|^ = 0, By the conditions (1), these equations become (x„ - my + (y„ - ny = R', (xo - m) + (y„ - n)^ = 0, dXa dxo" dxo" Whence V da;„7 I dx^jdxo , drR,^ dojo^ d^o^ dojo^ APPLICATIONS OF TAYLOR'S SERIES 201 This circle is called the osculating circle at the point (x^, y^) of the curve y =f(x), and by comparison with the results of Art. 44, this circle is seen to be identical with the circle of curvature. PROBLEMS Determine the order of contact of 1 . y^=4tx and a^ + y' = Ax. 2. x^y + y — x = and y=0. 3. ^ + 2/2 = 1 and a^ + 2/2 + 61/ - 7 = 0. 4. Show that at a point of inflection the tangent y = mx + n has contact of the second order with y =f(x). 5. Show that at a point of maximum or minimum curvature of 2/ =/(«) the osculating circle has contact of the third order. Art. 73. — Singular Points of Plane Curves Let (a;,,, 2/0) be any point of the plane curve F(x, y) = 0, (xg + A, 2/0 + ^) Einy other point. By Taylor's series Fix, + h,yo + k) = ^.7i+^.-k dxo dya \dXo^ dxodyo 92/o / Denoting the point {x^ + h, y„ + Jc) by (x, y), this series becomes + ^ {§(. - x.y + 2^(. - .„) (2/ - 2/0)+ 1|(. - .0)^ 1 + ... = 0. 202 DIFFERENTIAL AND INTEGRAL CALCULUS If either or both — , — differ from zero, wheu (x, y) in- definitely approaches {xo, i/q), (1) approaches -—(a; - Xo)+--{y - yo)= 0, dx„ dyo the tangent to F(x, 3/)= at (x^, y^). If = and — = 0, (1) approaches dxo dyo — ^(a; - Xo)' + 2 (x - x„) (y - y„) + —-^(y - y^y = 0. This equation is homogeneous of the second degree in (x—x^) and (y—yo), and therefore represents two straight lines through (»„, 2/o)- Tiiis means that at the point (xq, y^) of the curve F(x, y)=0 two tangents can be drawn to the curve. If the curve stops at (a^, 2/0) a-nd the tangents are real and distinct, the curve is said to have a salient point at (Xf,, y^ ; if the tan- gents are real and coincident, the curve is said to have a cusp APPLICATIONS OF TAYLOR'S SERIES 203 at (a!o, 2/0)) of the first species when the two branches of the curve lie on different sides of the common tangent, of the second species when both curves lie on the same side of the common tangent. If the curve does not stop at (% y^ and the tangents are real and distinct, (wo, 3/0) is ' a point where the curve crosses itself, called a node. If the tangents are real and coincident, two branches of the curve are tangent to each other at (Xn, y^, and (xq, 2/0) is called a tac-node. If the tangents at (ko, y^) are imaginary, (a;,,, 2/0) is an isolated point, called a conjugate point of the curve. An ordinary or regular point of a curve is a point (xq, y^ for which — and — are not both zero; all other points are singular. The points of inflection of a curve, where — ^ = 0, are also classed as singular points. Example I. — Examine 3? — Zxy + 'j^ = Q for singular points. Here — =3s^-3y, ~- = -3x+3y', -—=6x, _. =61/, — — = -3. ox ay axr dy' dxoy The conditions — =0, — =0 dx dy determine the singular point (0, 0). Eor the point (0, 0), dx^ dy' dx dy Hence the two tangents at (0, 0) are a; = and y = 0, the coordinate axes. Plotting the curve in the neighborhood of (0, 0), it is seen that the origin, is a node. , fig. 77. 204 DIFFERENTIAL AND INTEGRAL CALCULUS Example II. — Examine y^ — a? + 2a^ = for singular points. Here — = — 3ar + 4a;, — = 2y, — - = — 6 a; + 4, — - = 2, = 0. dx dy da? dy" oxdy The conditions — =0, — = determine the singular point 5a; dy 32pi fl2p a2p (0, 0). For this point |4 = 4, |4 = 2, f^ = 0. Hence dx' dy' dxdy the tangents at (0, 0), represented by i a? + 2 y" = 0, are y = ± V— 2 ■ X, and the point (0, 0) is an isolated point of the curve. Example III. — Examine y^ = ar' for singular points. „ dF o^ SF r, d^F c, d^F „ S'F ^ Here — = — 3ar, — = 22/, — r = — ox, — - = 2, = 0. 9a; dy dar dy dxdy The conditions — = 0, — =0 determine the singular point 5a; dy (0, 0). For this point ^= 0, ^= 2, -^ = 0. Hence the dx^ dy' dxdy two tangents at (0, 0), represented by 2y^= 0, coincide with the X-axis. Plotting the curve in the neighborhood of (0, 0), it is seen that the origin is a cusp of the first species. PROBLEMS Examine for singular points, 1. 3?+Za?-y^+5x+i.y-l=0. 4. y'^ = x^ + 2oi?. 2. (y — aff—o?. 5. y^ = a^ + oi?. 3. 2/' = ^- 6. (x' + yy = a\x'-y'^. 7. {xy + 1)' + {x- Ifix - 2) = 0. 8. y^ = 3^ — a^. CHAPTER XII OEDINAKY DirrEEENTIAL EQUATIONS OP PIEST OEDEE Aet. 74. — Formation or Diffbeential Equations An equation containing ordinary derivatives is called an ordinary differential equation. An equation containing partial derivatives \ ' ' ' dx dy da?' dy^J is called a partial differential equation. The order of a differential equation is tlie order of the highest order derivative occurring in the equation. The degree of a differential equation is the greatest expo- nent of the highest order derivative when the exponents of all derivatives in the equation are positive integers. Example I. — The equation (» — c)^ + j/^ = ^c^ represents all circles with center in the X-axis and with radius ^ the abscissa of the center. Differentiating this equation with respect to », a; — c -|- y^ — 0. Eliminating c from this result dv' dv and the given equation, 3y'-^^ — 2xy-^ + 4Ly'' — x'=0, the differential equation of the given system of circles. Solving 205 206 DIFFERENTIAL AND INTEGRAL CALCULUS the differential equation for ^, dy ^2xy±VW^lZWl_ dx dx y' Hence for every point (a;, y) of tlie plane where 16 0:^—48 y*>0, the differential equation determines two unequal values for — ; where 16 xPy' — 4^8 y* = 0, the two values of -^ are equal; dx dy dx where 16 x'y^ — iS y* < 0, the two values of y are imaginary. Geometrically these results mean that through points (x, y) for which — no circles of the system pass, o If X, y, -^ satisfy the differential equation, [x, y, -^ ) de- da; \ dxj notes a point in the circumference of one of the circles of the system and moving along the circumference. If (x, y, -^ ) \ dxj moves along in obedience to the differential equation, changing its direction continuously, it describes the circumference on which it started. ORDINARr DIFFERENTIAL EQUATIONS 207 If, however, [x.y, -^ ) denotes a point in either of the lines \ dxj y = ± — =, it may move along in obedience to the differential equation without a discontinuous change of direction, and describe the straight lines y = ± . These straight lines are the envelopes of the system of circles (x — Gy + y^ = ^ c'. The equation (x — cy +y^ = 1 c^ is called the general solu- tion of the differential equation ^ dx? ^dx " The solution obtained by assigning to the arbitrary constant in the general solution some particular value is called a par- ticular solution of the differential equation and represents a particular circle of the system. The solution y = ± ^-, which cannot be obtained from the V3 general solution by assigning a particular value to the arbi- trary constant, is called a singular solution of the differential equation. ' If the differential equation is written F{x, y, p)= 0, where p = ^, the preceding analysis shows that the singular dx solution is the ^-envelope of the equation. Example II. — Form the differential equation of the sys- tem of circles x' + y^ — 2ax — 2by + c = 0. Differentiating three terms in succession, ^ ■' ^ dx dx ' dx^ 208 DIFFERENTIAL AND INTEGRAL CALCULUS (3) 3^f^Y-g-(^Yg=0. ^ ' dxydx'J da? \dxj daf Observe that in Examples I. and II. the order of the differ- ential equation is the same as the number of arbitrary con- stants in the general solution. This is always the case. The solution of a differential equation is also called the primitive of the equation. The differential equation is ob- tained from its primitive either directly by differentiation or by the elimination of constants from the primitive and its derivatives. The process of finding the primitive of a differ- ential equation is called solving the equation.* PROBLEMS Form the differential equations of the following primitives, 1. y = cx + G — , y, Ci)/2 (a, 2/, c) ■ • •/„ {x, y, c) = 0, where c is an arbitrary constant. This last product is the general solution of the given equation. Example I. — Solve ^„-aa; = 0. Write the equation (-^ + a^x^\l^—a^x^\=0, and solve the equations ^ + a^x^ = 0, ^ - aM = 0. The product of dx dx these solutions, (2/ + |o^a;^ + c) (2/ — fa^a;^ + c) = 0, reducing to (2/ + c)^ — ■|aie' = 0, is the general solution of the given equation. Example II. — Solve y^^ + 2x^-y = 0. daf dx Solving for ^, ^ = - ^ ± ^^^±^ whence ^^^±1^ = dx. dx dx y y ±Vi^+f Integrating, ± (x^ + y''y = x + c. Hence the general solution is l(x + c) + {a^ + 2/2)*J j (a; + c) — (x^ + y^)^ = 0, reducing to y' = 2 ex + c'. Case II. — Suppose the equation (1) f(x, y, p) = 0, where p stands for ^, can be put into the form (2) y — F{x,p). dx Forming the w-derivative of (2) gives an equation of the form 220 DIFFERENTIAL AND INTEGRAL CALCULUS (3) p = F^ (x, p, ^\ If the primitive of (3) is (4)/i (x, p, c) =0, the elimination of p from (1) and (4) is the solution of (1). In like manner if (1) f{x, y,p) = can be put into the form (2) x=F{y,p), the y-derivative of (2) is (3) - = Fjy,p,f If the primitive of (3) is (4) /i {y, p, c) = 0, the elimination of p from (1) and (4) is the primitive of (1). Example. — Solve y =p^ + 2p\ The a;-derivative of this equation is p = 2p^ + 6 »" -^, dx dx ■whence c + x = 2p + dp^. Hence x = 2p + 3p^ — c and y =p'' + 2p^ for every value of p determine a pair of corre- sponding values of function and variable of the solution of the given equation. Case III. — If the equation F(x, p)=0 cannot be readily- solved for X or p, try the substitution p = xz. Example. — Solve a^ + -^ — ax-^ = 0. dor dx The substitution p = xz gives a? + 3?e^ — axh = 0, whence az -V dy dy dz X = -• jSI ow p = -^ = -^ — = xz, 1 4- r dx dz dx , dy _ az^ d_ f az \ _ a/'(z'^ — 2^) ""^ ~dz~l+?'dz\^T^r {1+^f ' and by integration y = ^ " %i , ^y + i "^YT^ "*" ^' Function y and variable x are now expressed in terms of the same quantity z. Case IV. — If the equation F{x, y, p)=0 is homogeneous in X and y, substitute y = xz. ORDINARY DIFFERENTIAL EQUATIONS 221 Example. — Solve {2p + l)x^y = a;V + 2 2/i The substitution y = zx, -^ = » + x — gives — ± = 0, dx dx X g _ gi whence ex + (z^ — 1)^ = 0, and ex + (z^ — 1)"^ = or cv? + (2/^ _ a!*)2 = and c + (y* - x^)-'' = 0. Case V. — Clairault's equation, y =px+/{p). The K-derivative of this equation is » = « + x^ +f'(p)—> dx dx WD which reduces to (1) ^{x +f'(p) j = 0. Equation (1) is satis- fied by (2) ^ = or (3) x +f'(p)=0. From (2) p = c and CtOG the general primitive of Clairault's equation is y = ex +f(c). The elimination of p from the given equation and (3) gives a singular solution of Clairault's equation. Example I. — Solve y=px-[-p—p^. The general primitive, found by substituting p = c, is y = ex + c — c'. Example II. — Solve «?(%/ — px) = yp^. Multiply the given equation by y and substitute u = y^, *^ = 2 2/^. There results ui? - 1 a?— = i— '• Substituting dx dx dx dor a? = v, whence — = 2x — , u = v 1 -, a Clairault's equa- dx dv dv dw tion whose primitive is m = cv + -^-^-T^ ^^^^ ox dy oy ox ox oz oz ox ay az oz ay to the identities ORDINARY DIFFERENTIAL EQUATIONS 223 Multiply (3) by B, (4) by P, (6) by Q, and add the products. There results the condition under which (1) can be solved. Example I. — Solve (y + z)dx + dy + dz = 0. Here P = y + z, Q = l, B = l, ^=1 ^=1 iQ=o ^ = ^=0 — =0 dy ' dz ' dx ' dz ' dx ' dy and condition (6) is satisfied. Considering x constant, the given equation becomes dy + dz = 0, whence y + z + X = 0, where X must be so determined that the avderivative of y + z + X is y + z. Hence ^=y + z = -X, 1^=-1, logX=c-a;, X=e'-'. dx ' X dx Finally, y + z + e''" = 0, the solution required. Example II. — Solve 2 (y + z)dx + (x + 3y + 2z)dy + {x + y)dz = 0. Here P=2(y + z), Q = x + 3y + 2z, R = x + y, dP^2 ^=2 ^ = 1 ^=1 ^=1 ^=1 dy ' dz ' dx ' dz dx dy ' and condition (6) is satisfied. 224 DIFFEBENTIAL AND INTEGRAL CALCULUS Considering y constant, the given equation becomes 2 dx dz 2(y+z)dx+(x+y)dz = 0, or + = 0. Integrating, log (x + yy + log (y + z) = log T, whence (x + y)" (y + z)= T, where 7" must be so determined that the y-derivative of (x + yy(y + z)-Y is (x + 3y + 2z)(x + y). Hence 2(x + y)iy+z) + (x+yf - ^= {x+Sy+2z)(x+y), ^= 0, dy dy T=G. Finally, the required solution is {x + yy{y + z)= C. PROBLEMS Solve, 1. (y + a)''dx + zdy — {y-\-a)dz = 0. 2. dx + dy+{x + y + z + r)dz = 0. 3. yzdx + zxdy + xydz = 0. 4. (y + z)dx + (z + x) dy + {x + y) dz = 0. 5. zydx — zxdy + y^dz = 0. 6. {y^ + yz) dx + (xz + z^dy+ {f -xy)dz = 0. CHAPTER XIII ORDIITAET DIFFERENTIAL EQUATIONS OF HIGHER ORDER Art. 78. — Equations of Higher Oedek and First Degree Standard I. — The primitive of an equation of the form — ^ =f(x) is found by n successive integrations ; the primitive of an equation of the form — ^ =/(?/) is found by multiplying di/ ' doc both sides by -^ and integrating, then solving for — and ax dy integrating again. Example. — Solve -^^ = 11. ,dy Multiplying by -t-> -r" -3— = y-r-- ^ J ^ ■' dx dx dx ^dx Integrating, l(^)' = ^y^+ C„ vrhence ^ = Vf + 2~C,. u \(xxj I otic o 1 • dx dx 1 Solving or dy dy ■y/y^ + 2Ci Integrating, x = logly + -y/y^ + 2 dl + Q. Standard II. — Equations of the form — ^ +/i (x) -^ =/2 (x) daf dx become linear by the substitution — =^3, — ^ = — • da; dx' dx Q 225 226 DIFFERENTIAL AND INTEGRAL CALCULUS Example. —Solve (1 - a^ ^ - a;^ = 0. dx^ dx Substituting ^=p, ^ = ^, the given equation becomes dx dar dx ldp^_x_ Integrating, ^ = d (1 - a;^)-^ Integrating pdx l — x" dx again, y = Ci sin"^a; + Cj. Standard III. — The equation linear in y and its derivatives, the coefficients Pq, Pi, Pi,--'Q being independent of y, is called the linear equation of order n. Suppose the coefficients Pq, Pj, Pi,---P,^ constant and Q = 0. dv ~^~x When n = 1, the equation becom es Po -^ + Pj 2/ = and y=e ''" , dx which has the general form y = e""- Substituting y = e"" in (1) p„|:|+P:f3+"-P"2/=o dx" do;"^ there results (2) ?„ m" + Pj m"-^ + P^ m"-^ + • • • P„ = 0, which shows that y = e°" is a solution of (1) for the n values of m which are roots of (2). Eepresenting these n roots by mi, m^, ma, m4, • ■ • m„, 2/1 = e"^', 2/2 = e""', 2/3 = e"^', • • • 2/„ = e""' are solutions of (1). Consequently (3) y=Ci- e"«' + C2 • e™^ + C3 • e"^ + ••• C„ • e"*"', where Ci, C2, Gs,---Gn ^'^^ arbitrary constants, is a solution of (1). Since (3) contains n arbitrary constants, it is the gen- eral solution of (1). The values of these n constants become known if the values of y and its first n — 1 derivatives are known for some value of x. DIFFERENTIAL EQUATIONS OF HIGHER ORVER 227 If the roots of equation (2) are real and unequal, (3) is a satisfactory form of the solution of (1). If equation (2) has pairs of conjugate imaginary roots, mi = a + & V— 1, m2 = a — 6 V— 1, or wij = a + ib, mj = a — ib, the corresponding terms of (3) are = e'" I Oi (cos bx + i sin 6a;) + Q (cos bx — i sin 6a;) \ by Problem 9, Art. 68. This result may be written e" \ (Ci + Oj) cos 6a; + i (Cj - Q sin 6a;|, which, by placing G-^ — \{A — iB), 02=\{A-\- iB), becomes e^iA cos 6a; +5 sin 6a;). Finally, placing ^= A'sin k, B=Kcos Jc, the result becomes Ke"'' sin {k + bx), where K and k are arbitrary constants. If equation (2) has equal roots mi = m^, (3) contains less than n arbitrary constants and is no longer the general primi- tive. In this case write mj = Wi + ft and the problem reduces itself to determining the form of the general primitive 2/ = C'l ■ e^" + Ci ■ e'^i+W"' + Og • e"'"' H 0„ • ""«", when 7i approaches zero. Now Ci • 6'"^'+ C2 • e<""+W' = e^^Xd + C2 • e"') =e-^=^|Oi+(72(l+7ia;+^+...)} by Maclaurin's series. When h approaches zero, this becomes e'"^'{(Ci+ C2) + C2hxl= e'"'^'{A + Bx), such values being as- signed to the arbitrary constants Cj and Cj that d -|- C2 = A, Cji = B when h approaches zero. If equation (2) has three equal roots, mi = mj = m^, a like analysis shows that the corresponding part of the general primitive is ^^^'{A -\-Bx+ Co?). 228 DIFFERENTIAL AND INTEOIiAL CALCULUS If equation (2) has equal imaginary roots, for example mi = mo = (a + ib) and mj = m4 = a — ib, the corresponding terms of the general primitive are e-^l {A + Bx) cos bx + {C + Dx) sin bx \ . Equation (2) is called the auxiliary equation of the differ- ential equation (1). Example I. — Solve ^, + 6 ^ + 13 y = 0. dar dx Substituting y = e"*, the auxiliary equation is found to be m^ + 6 m + 13 = 0, -whence mj = — 3 + 2 1, mj = — 3 — 2 1 and y = e~^ { J. cos 2 a; + 5 sin 2 a; j. Example II. — Solve ^-3^ + 4?/ = 0. dar dx^ The roots of the auxiliary equation »«,' — 3m^ + 4 = are — 1, 2, 2. Hence the general primitive is y = G-e-'+{A+Bx)e''. Standaed IV. — Linear equations with second member not zero. Eorm the successive a^-derivatives of P^.^ + P^.^L^Jf- ...P„.«=X ° dx' ' da;»-i " ^ until either the second derivative becomes zero or the elimina- tion of X from the given equation and the derivative becomes possible. Example. — Solve —4 + a^V = sin bx. da? ^ The second a^derivative of this equation is —. + ci^ ■ —^= — b' ■ sin&a;. dor dor DIFFERENTIAL EQUATIONS OF HIGHER ORDER 229 Eliminating sin hx from this derivative and the given equa- tion there results — ^ + (a^ -)- 6^)— ^ + a?b^y = 0. The auxiliary equation of this linear differential equation is m* + (a^ + b')m' + aV = 0, whence mi = ai, mj = — ai, m^ = hi, m^ = — bi. The general primitive is 2/= Oj cos {ax) + Oj sin (aa;) + C3 cos (hx) + C4 sin (bx). This value of 2/ is a solution of the given second order differ- ential equation if C^d^h — Ctd'h" = 0, — C^b^ — CJ) = 1, whence (73 = -— ^2, Ci = - ^ • The required solution is, therefore, y = d cos (ax) -|- C2 sin (ax) — — cos (6a;) — — — sin (6a;) . Standard V. — Change of the independent variable. In the equation ^ -|- P-^ -|- Qy = 0, where F and Q are daf dx functions of x, change the independent variable from a; to 2 by the relations ^ = ^.^, ^ = ^-^(^^Y + ^.^. There dx dz ax dx' W \dxj dz dor -»■"• » S(f+l(£^^l)+*=»- "" ^-- mine z so that ^-i-P^=0, whence (2) z= Ce-!'''".dx. dx^ dx J The elimination of x from (1) and (2) gives an equation whose solution leads to the solution of the given equation. Example. — Solve ^ -f- -^^ ^ -|- ^ = 0. dx' 1 + x' dx (1 + af)^ 2xdx e"!"^"^ ■ dx = I e ' i+«' ■ dx = tan~^ x, whence X = tan z. The transformed equation is — ^ -f- y = 0, whence y — Ci- cos z -\- Oi- sin z. Substituting for 2, y=Ci- cos 2: -f- C2 • sin 2 = Ci — == -|- Oi- Vl + a^ Vl+a^ 230 BIFFEHENTIAL AND INTEGRAL CALCULUS PROBLEMS Solve, 1. S^sin'a;. 12. 6^ = ^+v da? dx '' aa? dx^y ^^ 3. a?p- = l. 14. ^/_^_„ , 1 5- S + 6f^ + 92/ = 0. 16. ^-3^ + 2v-a!.e' f'*' «^» ■ dx" dx^ y~ ^^ '■ S-''^'2/ = 0. 17. g+32/ = sin(ma,). 7. g=a=^-sina,. is. g+42/ = cos(«4 - s-^s+^^=«- 21. ^=^^Y+i dx"" \dxj ^ 11. dx^ 22. «'^ + ^=a=". Abt. 79. — Symbolic Integration If M, v, w are any symbols whatever obeying the funda- mental laws of ordinary algebra, namely : DIFFERENTIAL EQUATIONS OF HIOUES OBDEB 231 the associative law, u + v + w ^ (u+v)+w, uvw = {uv)w ; tlie commutative law, u-\-v+w^^v-\-u-\-w, uvw s vuw ; the distributive law, (u + v)w = uw + vw\ the law of indices, «*"«" = «*"+" ; any algebraic transformation of expressions involving u, v, w gives a valid result. Denoting the operation of forming the first derivative by D, that of forming the second derivative by D", that of forming the third derivative by IP, ■•-, so that Dy = ~, D'-y = — ^, D^y = -^, •••; and defining the symbol D~^ by the equation D{D~^y) = y, whence D' is equivalent to the symbol of integration j , it follows that : I. D . Dy= D'y, D ■ D'y = l^y, I)-' ■ D'y = Dy, D~^ • Dy = D~^y ; that is, the law of indices of algebra holds for D affected by integral exponents. II. (a + bD- cDP) y = l(a + bD) - cD^ y ; that is, the associative law of algebra holds for D combined with constants. III. {aD + bD^y = D(a + bD)y; that is, the distributive law of algebra holds for D combined with constants. IV. (a + bD)y = (bD + a)y, aDy=Day; that is, the com- mutative law of algebra holds for D combined with constants. Since the operator D obeys the fundamental laws of algebra, the result of any algebraic transformation of expressions con- taining D and constants is valid. For example, by Maclaurin's series, f{D) = Ao + A,-D + A,.D'+-.A,.D'' + - = -^A„.D''. Hence f(D) y = ^A„. Dy. If f{D)y = X,y =^ X. This is the definition of the operator inverse to f(D). 232 DIFFERENTIAL AND INTEGRAL CALCULUS For instance, (I)^-3D + l)(2a^-5x' + 7x)=2a?-23a^ + 4:9x-31. Hence (2ocP -23x' + 49 x-31) = 2a? - Bx' + lx. Example I. — Show that (J)"^ - D-2)y={D+V){p-2)y. (Z>+l)(i)-2),.(i>+l)(|-2,).g-2| + |-2, d':^ dx Example II. — Show that /(D) e" s/(a) e'". f(D) = 2 A-D" and A-D-e"^ = A„a''e'''^ Hence f(D)e'"=f(a)e'". Inversely _^e" = -4-e°". Example III. — Show that — - — = y = ( — ^ —5—1 y. D^-D-2^ \D+1 0-2)" Performing the operation D^ — D — 2 on both sides of the assumed identity, (£)2 _ Z) _ 2)-^-^^ y =(iy-D- 2)f— i ^V or y= \ — \D-\-\ + \D + \\y = y, which proves the assumed identity correct. Example lY. — {iy-2D + 2)y = &x-93? + 2K?. Find?/. y = — — i -(6a;-9a;2 + 2a^) "0^-30 + 2^ ' - ^ (6a!-9a^ + 2a!^--^(6a!-9a^ + 2a!«) 1-D' ' 2-D s (1 + Z> + Z)2 + D^ 4- D" + • • •) (6 « - 9 a^ + 2 0^) DIFFEBENTIAL EQUATIONS OF HIGHER OEDER 233 This is a particular value of y since it does not contain arbi- trary constants. Example V. — y = — — - [0]. Find y. U — ^ This is equivalent to {D — 2)y = 0, which is the same as the linear equation -^ — 2y = 0. Whence y = e^. PROBLEMS 1. Show that /(D)[e'«X] = e°'./(D + a)[X], and conse- quently -^ [e'"X] = e" [Xl. f{D) ^ f{D + ar ^ 2. {D' + D + l)y = e'a:^. Showthat y = e'f~-x^ + ^x + ~\ 3. Show that /(D^ sin (ma;) =/(—m^ sin (ma;) and conse- quently — — — sin (mx) = — — sin (mx). 4. Show that /(i)'') cos (ma;) =/(—m^) cos (ma;) and conse- (l^^^^^J j7^ cos (^i^) = j., ^^ cos (mx). 1 or' 5. Show that -—r— — — a; = — — — a;. D(D-1) 2 6 . Show that -— — e^ sin a; = — e'' sin x. 7. Show that — =ri— - — re'' = t^. jy + D^ + D + l 15 8 . Show that — - — - e" sin x = e" (sin x — cos a;). 9. (D-m)y = 0. Show that y = --i — [0] = e"". 10. Show that y = ,^ ^ ,, [0] = e"« (Oi + Qx). (i> — my 234 DIFFERENTIAL AND INTEGRAL CALCULUS Art. 80. — Symbolic Solution of Linear Equations* Writing the linear equation P(,^+ Pi*^+ ... T'^y = with constant coefficients and second member zero in the form {P,D- + PiD"-' + P,D"-' + ••• P„) 2/ = 0, V = roi. Factoring, PoD" + PiD""' + P^D"-^ + —P„ = iD- mi) (D - ms) (D - wis) - (D - m„). Decomposing into partial fractions, D—mi D—mi D — rrhs D—m^ hence, y = GjeT'-'' + G^"^ + C^e'^ ^ 1- C„e""". Q If m-y = mj, a partial fraction of the form occurs. The value of - — ^^^-— -JO] is (Ci + C^)e'^\ {D — m])^ In general, if mi = m2 = mi= ••• m„ a fraction of the form occurs. The value of — fOl is (D - m^y (D - miY If m,i = m2 = a + U, ms = mi = a — hi, the corresponding terms of the solution of the differential equation are reducing to {A-^x + B^ cos (hx) + {A^ + B^) sin (fea;). The solution of (1) P„*!^ + Pi^!l^+ ... +P„w=X, with da;" dx"-^ *Maclaurin introduced the symbolic solution of differential equations. DIFFEBENTIAL EQUATIONS OF HIGHER ORDER 235 coefficients constant and right-hand member a function of x, is y = Y + u, where Y, called the complementary solution of (1), is the solution of P„^ + pS-^, + •■• + P^ = 0, and daf dx"~^ u is any particular solution of (1). For substituting y = Y+u -W'(^«l?+^^£?+-+^»^ +(^°S+^S+-^'.«)=^' dx" dx' which is a true equation by the hypothesis. d*v Example I. — Solve — ^ — « = k*. Writing the complementary equation (Z)* — 1) F = 0, or (D+1){D-1)(D+V^)(D-V^)Y=0, the complemen- tary solution is found to be Y— Cie"'+ C72e~"'+ C3 sina;-f- C4 cos x. The particular solution is u = --—ai" = {- 1 - I)* - D" )a;^ = -a^-24 Hence the general solution of the given equation is y = Cie" + 0.fi~' + C3 sin a; -f G, cos a; — a^ — 24. Example II. — Solve ^-2^ + y = x'^. dor dx The solution of the complementary equation (D'-2D + 1)Y=0 is Y= e'(Gi+C^). The particular solution is 1 i)2 _ 2 Z> -fl 1 (x'e^) -a^ = e^ (D + 3y-2{D + 3) + l {2+By- 236 DIFFERENTIAL AND INTEGRAL CALCULUS The general solution is y =e'(Oi + G^) + ^e^ {2 a? - ^x + 3). Example III. — Solve ^, + ^ + 2/ = sin (2 a;). da? dx The solution of the complementary equation (jy + D + l)T=0 is F=e-l/'cicos^a;+C2sin^a!y The particular solution is u = — sin (2 a;) = — — -sin 2x = --i— sin 2 x D- + D + 1 ^ ' D-3 I)'-9 = -^(D + 3)sin2x= - J^ (2 cos 2 a; + 3 sin 2 a;). Hence the general solution is _x/ -^/3 -^/3 \ y = e 2( C'jC0S-^a;+ Casin-^a; J — y'^ (2 cos 2 a; + 3 sin 2 a;). PROBLEMS Solve, 1. ^^ + y = l+x + a?. 6. ^^^2^+4:y=e'cosx. dor dor dx dar dx dor T 3. —^ + 2/ = cos a;. 8. — ^+ 42/ = cos(na;). dar dar 4. ^_3^ + 2w = a^2. g_ ^ + y = a.sin2a;. da? dx da? 5. ^ + 42/ = a;sin2a;. 10. ^, + 4 « = 2 a;' sin^ a;, da^ dar 11. ^ + 2^+22/ = e''sina! + cosa!. dar dx 12. ^ + 2^+2/ = a^cosa!. DIFFERENTIAL EQUATIONS OF HIGHER ORDER 237 Akt. 81. — Systems of Simultaneous Difpbeential Equations fjfj (Jv Let P-^=Q, Pi — = Qi, where x is the independent dx dx variable and y and z are the dependent variables, and P, Q, Pi, Qi are functions of x, y, z. It may be possible by combin- ing the given equations with their derivatives to obtain an equation in which one dependent variable and its derivatives do not appear. Example. — Solve —-7x + y = 0, ^-2x-6y=0. dt " ' dt ^ Eorm the f-derivative of the first equation, d^ _„dx.dy_ r, df dt dt ~ and eliminate y and -^ from this equation and the two given equations. There results the linear equation dt^ dt ' whence x — e^'{Ci • cos i + C2 • sin t). Substituting this value of X in the first of the given equations, y = e'" KO, - C) cos < + (Oi + C) sinf}. If the given equations can be written in the form dx __dy _ dz p~ q~B' each of these equal ratios equals ^dx + mdy -\-ndz ^^^ ^^^ ^ ^ IP+mQ + nR ' last ratio implies the same relation between x, y, and z as the three given ratios. If I, m, n can be so determined that 238 DIFFERENTIAL AND INTEGRAL CALCULUS IP + mQ + nli =0, it follows that Idx + mdy + ndz = 0. The integral of the last equation is an integral of the given system of equations. Example. -Solve ^ = ^ = ^. rfz xz y' TTT •. xdx _dy _dz _ lxdx -\- mdy -^ ndz yh xz y^ lyh + mxz + ny^ Placing 1 — — X, m = j/'^, n = 0, lyh + mxz + ny" = 0, whence — x'dx + y^dy = and y^ — sc^— d- Placing l = — l, m = 0, n = z, lyh + mxz + ny'^ = 0, whence — xdx + zdz^O and z' — a? = CV PROBLEMS Solve, 1. ^ + ^+2a; + 2/=0, ^+5a; + 32/ = 0. dt dt dt 2. — + 5x-2y = ^,^-x + %y=e''. dt " ' dt " „ dx_dy _ dz «? y^ xy 4. -dx= dy ^ dz ^ 3y + 4:Z 2y + 5z 5- ^-Sx + 4.y + 3 = 0,^+x + y + 5 = 0. df dr df dt dt'' dt " CHAPTER XIV PAETIAL DIPPEEENTIAL EQUATIONS Akt. 82. — FoEMATiON OF Pabtial Differential Equations Example I. — Form the partial differential equation of the system of spheres (1) (x — of +{y —hy ^z^ = R'^, whose centers (a, h, 0) lie in the XF-plane and whose radius is constant. Consider z the dependent, x and y the independent variables, and denote — by », — by q. dx ^ dy ■^ ^ Differentiating (1) partially with respect to x, (2) {x-a) + zp = 0; differentiating (1) partially with respect to y, (3) iy-b)+zq = 0. Eliminating a and b from (1), (2), (3), z\l +p' + q^= R% the partial differential equation required. Example II. — Eorm the partial differential equation of the expression z = ^^{y + ax) + tftj^j — ax), where <^i and 2 are arbitrary functions. Writing the given expression z = <^i(y-^+ ^iiyi), which requires that Vi = y + ax, v^^y — ax, and differentiating 239 240 DIFFERENTIAL AND INTEGRAL CALCULUS twice in succession partially with respect to x and also with respect to y, 1= *'.(.,)+ ^>.), g ^= \(v,)+ \{v,), ^= <^":K)+ '^"aW. d z ciZ By division, -^-^=0^ •■^^j the partial differential equation required. Observe that partial differential equations result either from the elimination of arbitrary constants from a function and its successive partial derivatives, or from the elimination of arbitrary functions from an expression and its successive derivatives. PROBLEMS Form the partial differential equations of the following expressions : 1. z = ax+--\-h. Z. z = ax-\-by + ab. 2. % = ax -\- a^y^ + h. 4. z — {y + mx). 5. y —bz = (x — a2). 6. z + ay + hx = ^\{x-af+(jj-pf + z% Akt. 83. — Partial Differential Equations of First Order Standard I. — Equations of the form F{p, q)=0. Try a solution of the form z= ax + by + c. Since p= a and q = b, z = ax + by + c is a solution of F(p, q)=0 if F(a, b)= 0. Denoting by f(a) the value of 6 obtained from PARTIAL DIFFERENTIAL EQUATIONS 241 the equation F(a,b)=0, the solution of F(p,q) = is z = ax+f(a)-y + c. Example. — Solve p'+q'' = m^. Here a/' + b^ = m', whence 6 = s/m' — a", and the solution is z = ax + Vm^ — a^ ■ y + c. Standard II. — Equations of the form F(z, p, q) = 0. Try a solution of the form z = ^(x-\- ay). Writing _ , / s , dz dv dz J dz dv dz z=4,{y), v = x+ay, p = — — = — and g= — — = a— . do ax dv dv ay dv By substituting these values of p and q the given equation becomes Fiz, — , a — )=0, an ordinary differential equation V dv dvj whose solution leads to the solutiion of the given partial differ- ential equation. Example. — Solve 9 (jfz + g^ = 4. Writing » = <^ (v), v = x+ay, whence p = — and g = a — , dv dv the given equation becomes „/ dz'^ , odz^\ J. ;, dz 2 1 9 «:^+a'3rJ=4, and dv^ dv^j dv 3 -y^g _|_ a? Integrating, |(v + c) =f (2 + a^)^ or {x + ay + cY={z + aFf. Standard III. — Equations of the form Fi(x,p) =Fi{y, q). Assume (1) F^ (x, p) = a, (2) F^ (y, q) = a, where a is an arbitrary constant. Integrating (1), « =/i (x, a) + Y, where Y represents the terms of z which do not contain x ; integrating - (2), z=f2 {y, a) + X, where X represents the terms of z which do not contain y. Hence z=fi(x,a)+f2(y,a) + G is the required solution. 242 DIFFERENTIAL AND INTEGRAL CALCULUS Example. — Solve p^ + q^ = x -\- y. Write this equation p^ — x = y — q^ = a, whence p = j- = (a+x)^ and g = -^ = (y - a)K Integrating, 2= |(a + a;)^ + F, z = |(2/- a)^ + X, and z = i{a + x)^ + i{y-a)^+0, the required solution. Standard IV. — The analogue of Clairault's equation z=px + qy + (j)(p,q). The solution of this equation is z = ax + by + '*') = 0> where / repre- P (4 R sents an arbitrary function, is also a solution of (1). For pl^f{u,v) + Qf^f(u,v) + Bl-nu,v) " ^5^^("' ") • a^ + ^a^-^^"' "-) • 9^ + ^a^-^^'*' ^) • Ty , n d ^/ N 5« , D 9 ,/ , 9m , T) 9 ./ ^ dv + ^ 9^-^(''' "> ■ 9^ + ^a^-^^'^' '^^ • 9^ + ^ 9^-^("' '^^ • 9^ + I;-^(«'^H''S+ ^l^+^lh*^ by hypothesis. The solution f(u, v) = may be written u = ^ (v), where <^ represents an arbitrary function. Example. — Solve xz ■ -— + yz ■ —- = oey. ax ay The system of equations — = — = — is satisfied by - = a •' xz yz xy "^ y and xy — z'= b. Hence the general solution of the given equation is xy — z^= fl>i-)- PROBLEMS Solve, 1. pq = k. 2. q = X2'>+p'. 3. z = px + qy + pq. 244 DIFFERENTIAL AND INTEGRAL CALCULUS dz , dz + q'^ = Zx. »■ 2 5. p' + q' = npq. ^5«,,52 4. pi + qi = 2x. 9. ''£ + 2'f^ = 3'- 6. p' = z'{l—pq). 10. x— + z— + y = 0. ox ay 11. g^^-xy— + f = 0. 7. pii+q) = qz- ^^ °y 8. « = pa; + 92/ + Sp^gi 12. a^ ^. + 2/^^ = «"• Akt. 84. — LiNEAB Equations of Higher Ordeb Standard I. — All derivatives of the same order. di^ dx dy dy^ Assume z = (i/ + mx), where represents an arbitrary function. Writing z= (v) and v = y + mx, — = m"(v) = 0. Hence z = (l>{y + mx) is a solution of |^ + 3-^ + 2— = oar dx dy dy' if m^ + 3 m + 2 = 0, that is if m = — 1, m = — 2. The required solution is therefore z = -''\ -where n, A„, and B„ are arbitrary constants is a solution of the given equation. Since SAe"'""^"' and S-B„e''<''-*' are arbitrary functions of x-\-y and y — X respectively, this solution may be written z = i(x + 2/) +6^(^2(2/ - »)• PROBLEMS Solve, 1.^ + 5-^ + 6^ = 0. 3. ^=a^^. flar 3a; 3?/ ^y'^ Sf dx^ „ 5^z Bh , dz dz n A 3^2 5^« , ^2 n dx^ dxdy dx dy dx^ dxdy dy dx^ dx dy dy' 5. 2^-3-^-2^ = 0. liiiWiKiSPlilj