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Digitized by tine Internet Arcliive in 2007 witli funding from IVIicrosoft Corporation littp://www.arcliive.org/details/differentialcalcOOpliiluoft d^j^^ ft: ^9 WORKS OF H. B. PHILLIPS, PH.D. PUBLISHED BY JOHN WILEY & SONS, Inc. Differential Equations. Second Eklition, Rewritten. vi+116 pages. 5 by 7M. Illustrated. Cloth, $1.50 net. Analytic Geometry. vii + 197 pages. $1 75 net. Differential Calculus. V + 162 pages. $1.60 net. Integral Calculus V + 194 $1.75 ne Y + 194 pages. f. 5 by 7M- Illustrated. Cloth, 5 by lyi. Illustrated. Cloth, 5 by 7H- Illustrated. Cloth, Differential and Integral Calculus. In one volume. $2.50 net. ,) ,oA^ ^^j^ J-^i^ >'? cyK k) IFFERENTIAL CALCULUS BY H. B. PHILLIPS, Ph.D. Associate Professor of Mathematics in the Massachusetts Institute of Technology NEW YORK JOHN WILEY & SONS, Inc. London: CHAPMAN & HALL, Lixitbd Copyright, 1916^ BY H. B. PHILLIPS 79996^ Stanbope ipress K H. GILSON COMPANY BOSTON, U.S.A. PREFACE In this text on differential calculus I have continued the )lan adopted for my Analytic Geometry, wherein a few cen- ral methods are expounded and appUed to a large variety »f examples to the end that the student may learn principles nd gain power. In this way the differential calculus makes nly a brief text suitable for a term's work and leaves for the ntegral calculus, which in many respects is far more impor- ant, a greater proportion of time than is ordinarily devoted o it. As material for review and to provide problems for which nswers are not given, a supplementary list, containing about alf as many exercises as occur in the text, is placed at the nd of the book. I wish to acknowledge my indebtedness to Professor H. W. 'yler and Professor E. B. Wilson for advice and criticism nd to Dr. Joseph Lipka for valuable assistance in preparing tie manuscript and revising the proof. H. B. PHILLIPS. Boston, Mass., August, 1916. CONTENTS ^APTEB Pages I. Introduction 1- 9 II. Derivative and Differential 10- 18 III. Differentiation of Algebraic Functions 19- 31 IV. Rates 32-38 V. Maxima and Minima 39- 48 VI. Differentiation of Transcendental Functions. ' 49- 62 VII. Geometrical Applications 63- 84 /'III. Velocity and Acceleration in a Curved Path . 85- 93 IX. Rolle's Theorem and Indeterminate Forms 94-100 X. Series and Approximations 101-112 XI. Partl\l Differentiation 113-139 Supplementary Exercises 140-153 Answers 154-160 Index 161-162 DIFFERENTIAL CALCULUS CHAPTER I INTRODUCTION 1. Definition of Function. — A quantity y is called a function of a quantity x if values of y are determined by values of X. Thus, U y = 1 — x^, y is a function of x; for a value of x^ determines a value of y. Similarly, the area of a circle is a function of its radius; for, the radius being given, the area is determined. It is not necessary that only one value of the function correspond to a value of the variable. Several values may be determined. Thus, if x and y satisfy the equation x^ — 2xy-\-y^ = x, then y is a function of x. To each value of x correspond two values of y found by solving the equation for y. A quantity u is called a function of several variables if u is determined when values are assigned to those variables. Thus, if z = x- + y^, then 2 is a function of x and y; for, values being given to x and y, a value of 2 is determined. Similarly, the volume of a cone is a function of its altitude and radius of base; for the radms and altitude being assigned, the volume is determined. 2. Kinds of Functions. — An expression containing variables is called an explicit function of those variables. Thus Vx + y is an explicit function of x and y. Similarly, if y = Vx-^ 1, y is an explicit function of x. DIFFERENTIAL CALCULUS Chap. I. A quantity determined by an equation not solved for that quantity is called an implicit function. Thus, if x^ — 2 xy -\- y^ = X, \y is an implicit function of x. Also x is an implicit function of y. Explicit and implicit do not denote properties of the func- tion but of the way it is expressed. An implicit function is rendered explicit by solving. For example, the above equa- tion is equivalent to y = X ± Vx, in which y appears as an explicit function of x. A rational function is one representable by an algebraic expression containing no fractional powers of variable quanti- ties. For example, xV5 + 3 x'^ -\-2x is a rational function of x. An irrational function is one represented by an algebraic expression which cannot be reduced to rational form. Thus V X + y is an irrational function of x and y. A function is called algebraic if it can be represented by an algebraic expression or is the solution of an algebraic equa- tion. All the functions previously mentioned are algebraic. Functions that are not algebraic are called transcendental. For example, sin x and log x are transcendental functions of x. 3. Independent and Dependent Variables. — In most problems there occur a number of variable quantities con- nected by equations. Arbitrary values can be assigned to some of these quantities and the others are then determined. Those taking arbitrary values are called independent vari- ables; those determined are called dependent variables. Which variables are taken as independent and which as de- pendent is usually a matter of convenience. The number of independent variables is, however, determined by the equa- tions. Chap. I. INTRODUCTION 3 For example, in plotting the curve y = 3^ + X, values are assigned to x and values of y are calculated. The independent variable is x and the dependent variable y. We might assign values to y and calculate values of x but that would be much more difficult. 4. Notation. — A particular function of x is often repre- sented by the notation / (x) , which should be read, function of X, or / of X, not / times x. For example, fix) = V^TT means that/ (ar) is a symbol for Vx^ -\. i. Similarly, y=f(x) means that y is some definite (though perhaps unknown) function of x. If it is necessarj' to consider several functions in the same discussion, they are distinguished by subscripts or accents or by the use of different letters. Thus, /i (x), f-z (x), /' (x), /" (x) , g (x) (read /-one of x, /-two of x, /-prime of x, /-second of X, g of x) represent (presumably) different functions of x. Functions of several variables are expressed by WTiting commas between the variables. For example, v=f{r,h) expresses that t; is a function of r and h and V = f{a, 6, c) expresses that y is a function of a, 6, c The / in the symbol of a function should be considered as representing an operation to be performed on the variable or variables. Thus, if f(x) = V^^~+l, f represents the operation of squaring the variable, adding 1, and extracting the square root of the result. If x is replaced 4 DIFFERENTIAL CALCULUS Chap. I. by any other quantity, the same operation is to be performed on that quantity. For example, / (2) = V22+ 1 = Vs. f{y+l) = V(y +1)2+1 = \/i/2 + 2y + 2. Similarly, if / {x, y) = x^ + xy - y\ then / (1, 2) = 12 + 1 . 2 - 22 = -1. If / (x, y, 0) = a:2 + i/2 + ^^ then f (2, -3, 1) = 22 + (-3)2 + 1 = 14. EXERCISES 3 3 3 1. Given x + y = a , express y as an explicit function of x. 2. Given logio (x) = sin y, express x as an explicit function of y. Also express y as an explicit function of x. 3. If / (x) = x2 - 3 X + 2, show that / (1) = / (2) = 0. 4. If F (x) = x* + 2 x2 + 3, show that F (-a) = F (a). 5. li F (x) = x+ -, find F {x + 1). Also find F (x) + 1. 6. If <!> (x) = Vx2 - 1, find <> (2 X). Also find 2 </. (x). 7. If^W=2T^3,find^(^). Also find ^. 8. If /i (x) = 2% /2 (X) = xS find /i !/2 (y)]. Also find /^ [/i (y)]. 9. If / (x, ?/) = X - -, show that/ (2, 1) = 2/ (1, 2) = 1. 10. Given / (x, y) = x^ + xy, find / (y, x). 11. On how many independent variables docs the volume of a right circular cylinder depend? 12. Three numbers x, y, z satisfy two equations X2 + ?/2 + 22 = 5, X ■\- y -\rz ==1. How many of these numbers can be taken as independent variables? j 6. Limit. — If in any process a variable quantity ap- proaches a constant one in such a way that the difference of the two becomes and remains as small as you please, the con- stant is said to be the limit of the variable. The use of limits is well illustrated by the incommensurable Chap. I. INTRODUCTION 5 cases of geometry and the determination of the area of a circle or the volume of a cone or sphere. 6. Limit of a Function. — As a variable approaches a limit a function of that variable may approach a limit. Thus, as X approaches 1, a:^ + 1 approacnes 2. We shall express that a variable x approaches a limit a by the notation X = a. The symbol = thus means "approaches as a limit." Let / {x) approach the limit A as x approaches a; this is expressed by lim/(x) = A, which should be read, " the limit of / (x), as x approaches a, is A." Example 1. Find the value of lim sHi)- As X approaches 1, the quantity x + - approaches 1 + t or 2, Hence 1s(^+i) = 2. Ex. 2. Find the value of sin 6 lim ;e=0 1 + COS d As 9 approaches zero, the function given approaches Hence ,. sin d _ hm :r— = 0. 9=0 1 + cos 7. Properties of Limits. — In finding the limits of func- tions frequent use is made of certain simple properties that follow almost immediately from the definition. 6 DIFFERENTIAL CALCULUS Chap. I. 1. The limit of the sum of a finite number of functions is equal to the sum of their limits. Suppose, for example, X, Y, Z are three functions ap- proaching the limits A, B, C respectively. Then X-\-Y-{-Z is approaching A -{- B -\- C. Consequently, lim (X + F + Z) = ^ + 5 + C = lim Z + lim 7 + lim Z. 2. The limit of the 'product of a finite number of functions is equal to the product of their limits. If, for example, X, Y, Z approach A, B, C respectively, then XYZ approaches ABC, that is, lim XYZ = ABC = lim X lim Y lim Z. 3. // the limit of the denominator is not zero, the limit of the ratio of two functions is equal to the ratio of their limits. Let X, Y approach the limits A, B and suppose B is not X A zero. Then y approaches ^ , that is, ,. X A limX ^''^Y=B=V;^' A If B is zero and A is not zero, t, will be infinite. Then ts X A X =7 cannot approach ^ ^sa limit; for, however large ^ may become, the difference of t^ and infinity will not become small. 8. The Form ^. — When x is replaced by a particular value, a function sometimes takes the form r- Although this symbol does not represent a definite value, the function may have a definite limit. This is usually made evident by writ- ing the function in a different form. Example 1. Find the value of ,. x'-l lim r Chap. I. INTRODUCTION 7 When X is replaced by 1, the function takes the form 1-1 1-10* Since, however, a^-1 , , ^^=^ + 1' the function approaches 1 + 1 or 2. Therefore 3:2—1 Hm^ ^ = 2. 1=5=1 a: — 1 Ex. 2. Find the value of ,. (VTTx - 1) lim -• x=0 X When X = the given function becomes 1-1 ^0 o" Multiplying numerator and denominator by Vl -f a: + 1, Vl + x- 1 ^ X 1 X ~ X (Vl+x + 1) ~ Vl+x + 1 * As X approaches 0, the last expression approaches ^. Hence 9. Infinitesimal, -r- A variable approaching zero as a limit is called an infiriiiesimal. Let a and /3 be two infinitesimals. If lim- is finite and not zero, a and /3 are said to be infinitesimals of the same order. If the limit is zero, a is of higher order than /3. If the ratio - approaches infinity, /3 is of higher order than a. Roughly speaking, the higher the order, the smaller the infinitesimal. 8 DIFFERENTIAL CALCULUS Chap. I. For example, let x approach zero. The quantities X/>«2 /y^ /y^ Ck^ n are infinitesimals arranged in ascending order. Thus x* is of higher order than x^; for x^ lim — = lim x^ = 0. x=0X^ 1=0 Similarly, a^ is of lower order than a^, since X* X approaches infinity when x approaches zero. As X approaches x , cos x and cot x are infinitesimals of the same order; for ,. cos a; ,. . Iim — 7 — = lim sm x = 1, .,cota; i-fl^ which is finite and not zero. EXERCISES Find the values of the following limits: ... a;*- 2 a; + 3 ... Vl - x» - Vl + x' 1. hm ^ 4. Iim z ■ x=Q X — 5 x=o X* _ ,. sin 9 4- cosd , sin 2 + cos 2 9 6. hm ^-^• 6=2 edzoi&n9 ^ ,. x^ - 3x + 2 - ,. sin » 3. hm r-! 6. lim xAi X — 1 9=0 sin 29 7. By the use of a table of natural sines find the value of ! ,. sin X lim x=o a; 8. Define as a limit the area within a closed curve. 9. Define as a limit the volume within a closed surface. 10. Define V2. 11. On the segment PQ (Fig. 9a) construct a series of equilateral tri- angles reaching from P to Q. As the number of triangles is increased, Chap. I. INTRODUCTION their bases approaching zero, the polygonal line PABC, etc., approaches PQ. Does its length approach that of PQl \ / Fig. 9a. 12. Inscribe a series of cylinders in a cone as shown in Fig. 9b. As the number of cyl- inders increases indefinitely, their altitudes approaching zero, does the sum of the vol- umes of the cylinders approach that of the cone? Does the sum of the lateral areas of the cylinders approach the lateral area of the cone? Fig. 9b. 1 13. Show that when x approaches zero, tan - does not approach a limit. 14. As X approaches 1, which of the infinitesimals 1 — a; and Vl — x is of higher order? V 16. As the radius of a sphere approaches zero, show that its volume is an infinitesimal of higher order than the area of its surface and of the same order as the volume of the circumscribing cylinder. CHAPTER II DERIVATIVE AND DIFFERENTIAL 10. Increment. — When a variable changes value, the algebraic increase (new value minus old) is called its in- crement and is represented by the symbol A written before the variable. Thus, if X changes from 2 to 4, its increment is Ax = 4 - 2 = 2. If X changes from 2 to —1, Ax = -1-2= -3. When the increment is positive there is an increase in value, when negative a decrease. Let 2/ be a function of x. When x receives an increment Ax, an increment A?/ will be determined. The increments of X and y thus correspond. To illustrate this graphically let X and y be the rectangular coordinates of a point P. An equation y=f{x) represents a curve. When x changes, the point P changes to some other position Q on the curve. The increments of X and y are Ax = PR, Ay = RQ. (10) 11. Continuous Function. — A function is called con^ tinuous if the increment of the function approaches zero as the increment of the variable approaches zero. 10 Chap. II. DERIVATIVE AND DIFFERENTIAL 11 In Fig. 10, y is a continuous function of x; for, as Ax approaches zero, Q approaches P and so Ay approaches zero. In Figs. 11a and lib are shown two ways that a function can be discontinuous. In Fig. 11a the curve has a break at T I X / if Fig. 11a. Fig. lib. P. As Q approaches P', Ax = PR approaches zero, but Ay = RQ does not. In Fig. lib the ordinate at x = a is infinite. The increment Ay occurring in the change from a: = a to any neighboring value is infinite. 12. Slope of a Curve. — As Q moves along a continuous curve toward P, the line PQ turns about P and usually approaches a limiting posi- tion PT. This line PT is called the tangent to the curve at P. The slope of PQ is RQ Ay PR Ax' As Q approaches P, Ax ap- proaches zero and the slope of PQ approaches that of PT. Fig. 12a. Therefore Slope of the tangent = tan = lim -r^ • (12) 12 DIFFERENTIAL CALCULUS Chap. II. The slope of the tangent at P is called the slope of the curve at P. Exam-pie. Find the slope of the parabola y = x'^ aX the point (1, 1). Let the coordinates of P \yQ X, y. Those of Q are x -\- Aa:, y + Ly, Since P and Q are both on the curve, = 'r2 y =x'^ and ?/ + Ay = (a; + Aa;)2 = a^2 + 2 a; Ax + (Ax)2. Subtracting these equations, we get ^y = 2x^x-\- (Ax)2. Dividing by Ax, A?/ Ax = 2 X + Ax. As Ax approaches zero, this approaches Slope at P = 2 X. This is the slope at the point with abscissa x. The slope at (1, 1) is then 2-1 = 2. 13. Derivative. — Let 2/ be a function of x. If ^y Ax approaches a limit as Ax approaches zero, that limit is called the derivative of y with respect to x. It is represented by the notation D^y, that is, A-. (13a) D^y = lim -^• Ax=0 Ax If a function is represented by /(x), its derivative with respect to x is often represented by/' (x). Thus A/(x) /' (x) = lim Az=0 Ax = D^(x). (13b) Chap. II. DERIVATIVE AXD DIFFERENTIAL 13 Fig. 13. In Art. 12 we found that this Umit represents the slope of the curve y = Six). The derivative is, in fact, a function of X whose value is the slope of the curve at the point with ab- scissa X. The derivative, being the limit Aw of ^, is approximately equal to a small change in y divided by the corresponding small change in x. It is then large or small according as the small in- crement of y is large or small in comparison with that of x. If small increments of x and Aw y have the same sign -r^ and Ax its limit Dxy are positive. If they have opposite signs D^y is negative. Therefore D^y is positive when x and y increase and decrease together and negative when one increases as the other decreases. Example. y = 3? — Zx-\-2. Let X receive an increment Ax, The new value of x is x + Ax. The new value of j/ is y -f Ay. Since these satisfy the equation, 2/ + Ay = (x + Ax)3 - 3 (x + Ax) + 2. Subtracting the equation y = x3-3x-t-2, we get Ay = 3 x2 Ax + 3 X (Ax)^ -J- (Ax)' - 3 Ax. Dividing by Ax, Aw ^=3x2 + 3xAx4- (Ax)2 - 3. As Ax approaches zero this approaches the limit D^y = 3 x2 - 3. 14 DIFFERENTIAL CALCULUS Chap. II. The graph is shown in Fig. 13. At A (where x = \) y = and DxV = 3 • 1 — 3 = 0. The curve is thus tangent to the a;-axis at A. The slope is also zero at B (where x = —1). This is the highest point on the arc AC. On the right of A and on the left of B, the slope DxV is positive and x and y in- crease and decrease together. Between A and B the slope is negative and y decreases as x increases. EXERCISES 1. Given y = Vx, find the increment of y when x changes from X = 2 to X = 1.9. Show that the increments approximately satisfy the equation Ay ^ 1 Ax 2 Vx 2. Given y = logio x, find the increments of y when x changes from 50 to 51 and from 100 to 101. Show that the second increment is ap- proximately half the first. 3. The equation of a certain line is y = 2 x + Z. Find its slope by Au calculating the limit of — • .- — ^ 4. Construct the parabola y = x^ — 2 x. Show that its slope at the point with abscissa x is 2 (x — 1). Find its slope at (4, 8). At what point is the slope equal to 2? 6. Construct the curve represented by the equation y = x* — 2 x*. Show that its slope at the point with abscissa x is 4 x (x" — 1). At what points are the tangents parallel to the x-axis? Indicate where the slope is positive and where negative. In each of the following exercises show that the derivative has the value given. Also find the slope of the corresponding curve at x = — 1. 6. y = (x + 1) (x + 2), Dx2/ = 2 X + 3. 7. y = X*, Dzy = 4x'. 8. 2/ = x» - x2, Dj^y = 3x^ -2 x. 9. t/ = -. ^'y=-^^- 10. If X is an acute angle, is Dx cos x positive or negative? 11. For what angles is Dx sin x positive and for what angles negative? 14. Approximate Value of the Increment of a Function. — Let y be a function of x and represent by e a quantity such that ^y Chap. II. DERIVATIVE AND DIFFERENTIAL 15 As Ax approaches zero, -r^ approaches D^y and so « ap- proaches zero. The increment of y is Ay = Bxy Ax -f cAx. The part D,y Ax (14) is called the -principal part of Ay. It differs from Ay by an amount cAx. As Ax approaches zero, e approaches zero, and so eAx becomes an indefinitely small fraction of Ax. It is an infinitesimal of higher order than Ax. If then the principal part is used as an approximation for Ay, the error will be only a small fraction of Ax when Ax is sufficiently small. Example. When x changes from 2 to 2.1 find an approxi- mate value for the change in y = -• X In exercise 9, page 14, the derivative of - was found to X be 1. Hence the principal part of Ay is -Aax= -j(.l) = -0.0250. -J.2 4 The exact increment is ^^ = (21)5 - ^^ = -0-«232. The principal part represents Ay with an error less than 002 which is 2% of Ax. 15. Differentials. — Let x be the independent variable and let y be a function of x. The principal part of Ay is called the differential of y and is denoted by dy; that is, dy = D^y Ax. (15a) This equation defines the differential of any ftmction y of x. In particular, if y = x, Dzy = 1, and so dx = Ax, (15b) that is, the differential of the independent variable is equal to 16 DIFFERENTIAL CALCULUS Chap. II. its increment and the differential of any function y is equal to the product of its derivative and the increment of the independent variable. Combining 15a and 15b, we get whence dy = Dxy dx, dy dx = D^y, (15c) (15d) dy that is, the quotient ^ is equnl to the derivative of y with respect to X. Since D^y is the slope of the curve y =f(x), equations 15b and 15c express that dy and dx are the sides of the right tri- angle PRT (Fig. 15) with hypotenuse PT extending along the tangent at P. On this diagram, Ax and Ay are the increments Ax = PR, Ay = RQ, occurring in the change from P to Q. The differen- tials are dx = PR, dy = RT. A point describing the curve is moving when it passes through P in the direction of the tangent PT. The differential dy is then the amount y would increase when x changes to x + Ax if the direction of motion did not change. In general the direction of motion does change and so the actual increase Ay = RQ is different from dy. If the in- crements are small the change in direction will be small and so Ay and dy will be approximately equal. Equation 15c was obtained under the assumption that x was the independent variable. It is still valid if x and y are continuous functions of an independent variable t. For then dx = DtX At, dy = Dty At. T y X P^ J ^ dx Ay ^/[H -Ax---> R x! Fig. 15. Chap. II. DERIVATIVE AND DIFFERENTIAL 17 The identity Ay _ Ay Ax At Ax ' At gives in the limit Dty = D^y • DtX. Hence Dty At = D^y • Dtx At, that is, dy = D^y dx. X 4- I Example 1. Given y = , find dy. In this case ,, + Ax+l x+1 Ax ^ x + Ax X X (x + Ax) Consequently, Ay 1 Ax X (x + Ax) As Ax approaches zero, this approaches ^= -i. dx 3?' Therefore , dx Ex. 2. Given x = f*, y = <3, find ^• The differentials of x and y are found to be dx = 2tdl, dy = '^f^dt. Division then gives, dx~r" Ex. 3. An error of 1% is made in measuring the side of a square. Find approximately the error in the calculated area. Let X be the correct measure of the side and x + Ax the value found by measurement. Then dx = Ax = ±0.01 x. 18 DIFFERENTIAL CALCULUS Chap. II, The error in the area is approximately dA =d (a;2) =2xdx= ±0.02 x^ = ±0.02 A, which is 2% of the area. EXERCISES 1. Let n be a positive integer and y = x". Expand Ay = {x + Ax)" — x" by using the binomial theorem. Show that -^ = nx"-». ax What is the principal part of Ay? 2. Using the results of Ex. 1, find an approximate value for the in- crement of x« when x changes from 1.1 to 1.2. Express the error as a percentage of Ax. dA 3. If A is the area of a circle of radius r, show that -y- is equal to the ar ■circumference. 4. If the radius of a circle is measured and its area calculated by using the result, show that an error of 1% in the measurement of the radius will lead to an error of about 2% in the area. 5. If V is the volume of a sphere with radius r, show that -p is equal ar to the area of its surface. 6. Let V be the volume of a cylinder with radius r and altitude h. Show that if r is constant -rr is equal to the area of the base of cylinder and if A is constant -j- is equal to the lateral area. 7. li y — f (x) and for all variations in x, dx = Ax, dy = Ay, show that the graph of y = / (x) is a straight line. 8. If y is the independent variable and x = f (y), make a diagram showing dx, dy, Ax, and Ay. 9. If the y-axis is vertical, the x-axis horizontal, a body thrown hori- zontally from the origin with a velocity of 50 ft. per second will in / seconds reach the point ' X = 50t, y = -16 fi. I Find the slope of its path at that point. 10. A line turning about a fixed point P intersects the x-axis at A and the y-axis at B. If Ki and K2 are the areas of the triangles OP A (and OPB, show that dKi^PA^ ■ dK, P&' CHAPTER III DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 16. The process of finding derivatives and differentials is called differentiation. Instead of applying the direct method of the last chapter, differentiation is usually per- formed by means of certain formulas derived by that method. In this work we use the letter d for the operation of taking the differential and the symbol -r- for the operation of taking the derivative with respect to x. Thus d {u -\- v) = differential of {u-\-v), -j-{u -j- v) = derivative of (m + v) with respect to x. To obtain the derivative with respect to x we proceed as in finding the differential except that d is everywhere replaced ^^dx' ^^ 17. Formulas. — Let u, v, w be continuous functions of a single variable x, and c, n constants.* I. dc = 0. n. d {u + v) = du + dv. in. d {cu) = c du. IV. d (vv) = udv -\- V du. du — u dv VI. d (m") = nw'*-^ du. * It is assumed that the functions w, v, w have derivatives. There exist continuous functions, " = / (x), 19 20 DIFFERENTIAL CALCULUS Chap. III. 18. Proof of I. — The differential of a constant is zero. When a variable x takes an increment Ax, a constant does Ac not vary. Consequently, Ac = 0, ^ = 0, and in the limit dc ~r- == 0. Clearing of fractions, dc = dx ' = 0. 19. Proof of- n. — The differential of the sum of a finite number of functions is equal to the sum of their differentials. Let y = u-\- V. When X takes an increment Ax, u will change to m + Aw, v to y + Ay, and yioy -]- Ay. Consequently y + Ay = u-i- Au-i-v + Av. Subtraction of the two equations gives Ay = Au + Av, whence Ay _ Au Av Ax ~ Ax Ax As A^ approaches zero, ^. ^, ^^ approach |, |, ^° respectively. Therefore dy _du dv dx dx dx' and so dy = du-\- dv. By the same method we can prove I d(w±y±iy± • • ' ) = du^dv zLdw ± • ••, audi that Au I Ax does not approach a limit as Ax approaches zero. Such a function has no derivative DxU and therefore no differential du = Dzudx. Chap. III. ALGEBRAIC FUNCTIONS 21 20. Proof of in. — The differential of a constant times a function is equal to the constant times the differential of the function. Let y = cu. H*J^^ ^'^ Then y + Ay = c (m + Am) m +z\ij -- ^^x ♦ Ax) (. and so Ay = c Am, ^'j^ fC^^^^)-fC^l Ay Am AV- /Yil^i^M^ Ai = 'Ai- ^^ ^^ As Ax approaches zero, -r-^ and c -r- approach 3^ and c -t- • Therefore ofif. /.W, f^jc+Ax)- ^^c— ^^ A^^c' — ^^ dx dx* whence dy = c du. Fractions with a constant denominator should be differen- tiated by this formula. Thus ^(c)=^(^)=^'^^- 21. Proof of IV. — The differential of the product of two functions is equal to the first times the differential of the second plus the second times the differential of the first. Let y = uv. Then y + Ay = (m + Am) (y + Aw) = uv -{- V Am + (m + Am) Ap. Subtraction ^ves Ay = f Am + (m + Am) Ar, whence Ay Am , , , . . Ap Since m is a continuous function, Am approaches zero as Ax approaches zero. Therefore, in the limit, dy _ du, dv dx dx dx' 22 DIFFERENTIAL CALCULUS Chap. III. and so dy = V du-\- u dv. In the same way we can show that d (uvw) = uv dw -\- uw dv + vw du. 22. Proof of V. — The differential of a fraction is equal to the denominator times the differential of ike numerator minus the numerator times the differential of the denominator ^ all divided hy the square of the denominator. Let u Then and Ly = u + Am u V Am — m Ay V + Ay V V {v -\- Ay) Dividing by Ax, . Am Ay Am y-T M-r- -r^ = Ax Ax Ax — -, 1 — 7-T"* V {v -\- Av) Since y is a continuous function of x, Av approaches zero as Ax approaches zero. Therefore du dv dy _ dx dx dx v^ whence , V du — udv dy = — "— i 23. Proof of VI. — The differential of a variable raised to a constant power is equal to the product of the exponent, the variable raised to a power one less, and the differential of the variable. We consider three cases depending on whether the exponent is a positive whole number, a positive fraction, or a negative number. For the case of irrational exponent, see Ex. 25, page 6J . Chap. III. AL^tEBRAIC FUNCTIONS 2^ (1) Let n be a positive, integer and y = w". Then nin — 1) and Ay = nw"-» Am + ^ ^^, ^^ W-^ (Am)^ + = nM"~^ Dividing by Au, Aw , , n (n — 1) , /* X , -r^ = nu"-^ H ^ — - M"-2 (Am) + As Am approaches zero, this approaches dy du Consequently, dy = nti"~^ du, V TO ~ (2) Let n be a positive fraction - and y = u''= u^. Then y« = M". Since p and g are both positive integers, we can differentiate both sides of this equation by the formula just proved^ Therefore gy«~^ dy = pw^"* du. p Solving for dy and substituting m' for y, we get P-i f"^ dy = ^- — — du = - u du = nM"~^ du. qu « (3) Let n be a negative number —m. Then y = M" = M""* = ^ M" Since 7?i is positive, we can find d (u") by the formulas proved above. Therefore, by V, J M"'d(l) — 1 d(M'") —mu'"-^du , , _, , dy = ' ., — - = ^- • =v-«fnM~"*~^aM = nu'^^ du^ 24 DIFFERENTIAL CALCULUS Chap. III. Therefore, whether n is an integer or fraction, positive or negative, d (w") = nw"~^ du. If the numerator of a fraction is constant, this formula can be used instead of V. Thus d{~\= d {cu~^) = —cu~^ du. Example 1. y = Aa^. Using formulas III and VI, dy = Ad{x?) = 4 • 3 a;2 dx = 12 a:2 dx. Ex.2, y = Vx-\--^+3. Vx This can be written y = x^ -}- x~^ + 3. Consequently, by II and VI, dy ^d (x^) ^ d (x'h ^ d (3) dx dx dx dx 1 -idx 1 -sdx , ^ 2 dx 2 dx^ 1 1 2V^ 2V^ Ex. 3. y=(x + a) (x^-b^). Using IV, with u = x + a, v = x^ — h^, = (x + a) (2x-0) + (x2-62) (1+0) = 3 x2 + 2 ax - 62. Ex.^.y = ^-^' Chap. III. ALGEBRAIC FUNCTIONS 25 Using V, with w = a^ + 1, y = x- — 1, (x» - 1) d (x2 + 1) - (x* 4- 1) d (x* - 1) dy = (X^ _ 1)2 (x^- l)2xdx- (x' + l)2xcte (x2 - 1)2 4xdx Ex. 5. t/ = Vx2- 1. Using VI, with w = x^ — 1, = i(x«-l)-i(2x) = X Vx2-1 Ex. 6. x^ + xy — y^ = 1. We can consider y a function of x determined by the equa- tion. Then d (x2) + d (xy) - d (y') = d (1) = 0, that is, 2 X dx -\- X dy -^ y dx — 2 y dy = 0, {2x-\-y)dx + ix-2y)dy = 0. Consequently, dy ^ 2x + y dx 2y — X Ex.7. x = t-{-j, y = t--. In this case dx = dt-rp, dy = dt-{--^* Consequently, dy ^ ^ t^ ^ f-hl dx 1 ^ - 1 ' Ex. 8. Find an approximate value of y = ( 1 when 26 DIFFERENTIAL CALCULUS Cbap. III. When X = 0, y = 1. Also , 2dx dy = - 3 {I - x)i (1 + x)^ When X = this becomes dy = —I dx. If we assume that dy is approximately equal to Ay, the change in y when x changes from to 0.2 is approximately dy = -f (0.2) = -0.13. The required value is then y = l- 0.13 = .87. EXERCISES In the following exercises show that the differentials and derivatives have the values given: l/t^y = 3x* + 4x^ - 6x^ + 5, di ■= 12 {x^ + x^ - x) dx. x3 - x2 + 1 dy 3 x^ - 2 X ^ ~ 5 ' dx 5 4. ?/ = (x + 2 a) (x - a)2, dy = Z (x^ - a«) dx. 5. 1/ = X (2 X - 1) (3 X + 2), ^^ = 18 x^ + 2 X - 2. 1 , -2xdx 2 X + 3 -22 (to rf 1 V ^ITi-g d 1 - 2x _ 2x d g' - 2 s'' ^' dx (X - 1)2 - (X - 1)'* ^^' ds « ^«' - «' = v^rrr^i- ax \x / x^ Va^ - x* 14 ± d't^ = ^^^7=- 16. ^xY = 2xi/^ + 2x-^j/^. ^ ' dxVx^+1 i£^ + l)Vx*-l dx " ^ "dx C2+3x«)S 20 , . xdxi-'ydu 16. d^^^^ = - :^ (2+3x<')Sdx. 17. d Vx^ + r/^ = W^FP^' Chap. III. ALGEBRAIC FUNCTIONS 27 18. y = (i+l)(2-3i)»(2x-3)', ^ = (24 + 13x -36i«) (2- 3z) (2x - 3)». ax _ a:* dy _ mcuf^~^ 19- y- m» ^- n. • (a + fti")" (a + 6x»)«+* * 3z* ' dx x4 Vx* + 1 (x + VTT70'^^ , (x + vrr^)"-' ^^ ^ = — ^+1 — + — -^r^n — ' dj/ =2(i + v^l +xO"(ic- 22. X* + y* = a*, 24. 2x2-3xy + 4t/» = 3x, ax dy _ 4x - 3y - 3 dx ~ 3x -8y 25. - + ?^ = 1, ydx-xdy = 0. y X ' 1 dx do " 26. y=-, + , = 27. i/=" + x*"*/" = x**", mydx = nx dy. t 2t + S dy , 28. x=^^, y=-^— ^^^ = 0. 29. x=<-Vi2-l, y = <+V<2-l, xdy + ydx=0. - 3a< ^ 3a<^ dy ^ 2t -t* 1 + Z'' ^ ~ 1 4- i'' dx ~ 1 - 2t^' X 31. Given y = / ^ „ , find an approximate value for y when X = 4.2. 32. Find an approximate value of \/l x + 1 x^ + x + 1 when X = .3. 33. Given y = a^, find dy and At/ when x changes from 3 to 3.1. Is d(/ a satisfactory approximation for Ay? Express the difference as a percentage of Ay. 34. Find the slope of the curve J/ = X (x» + 31)^ at the point x = 1. 35. Find the points on the parabola y^ = 4 ax where the tangent is inclined at an angle of 45° to the x-axis. 28 DIFFERENTIAL CALCULUS Chap. III. . 36. Given y = (a + a?) Va — x, for what values of x does y increase as X increases and for what values does y decrease as x increases? 37. Find the points P (x, y) on the curve ,1 where the tangent is perpendicular to the line joining P to the origin. 38. Find the angle at which the circle x2 + 2/2 = 2x-32/ intersects the x-axis at the origin. 39. A line through the point (1,2) cuts the z-a^ at (x, 0) and the 2,-axis at (0, y). Find g- " ^ -- ^^ ( X -^O 40. If x'* — X + 2 = 0, why is the equation not satisfied? ^ 41. The distances x, x' of a point and its image from a lens are con- nected bv the equation x^x' /' / being constant. If L is the length of a small object extending along the axis perpendicular to the lens and L' is the length of its image, show that L \x] approximately, x and x' being the distances of the object and its image from the lens. 24. Higher Derivatives. — The first derivative -i^ is a function of x. Its derivative with respect to x, wntten -r-^ » is called the second derivative of y with respect to x. That is, dx^ dx \dx) Similarly, ty = ±('^\ d3? dx W/ Chap. III. ALGEBRAIC FUNCTIONS 29 The derivatives of / (x) with respect to x are often written /' (x), /" (x), r (x), etc. Thus, if y = / (x). |=n.,g=r(x). g=r'(x).etc. Example 1. y = x^. Differentiation with respect to x gives S = #(6) = 0. All higher derivatives are zero. Ex. 2. x2 + xy + y^ = 1. Differentiating with respect to x, whence dy ^ 2x + y rfx X + 2 y The second derivative is ^J ^ _± ( 2x + y \ ^^Tx' ^y dx^ dx\x + 2y) (x + 2 y)^ dy Replacing -^ by its value in terms of x and y and reducing, cPy 6 (x2 + xy + y2) 6 dx2 (x + 2y)3 (x + 2y)' The last expression is obtained by using the equation of the curve x^ + xy + y2 = i gy differentiating this second derivative we could find the third derivative, etc. 30 DIFFERENTIAL CALCULUS Chap. IH. 25. Change of Variable. — We have represented the second derivative by -— . This can be regarded as the quo- tient obtained by dividing a second differential d^y = d (dy) by (dxy. The value of d^y will however depend on the vari- able with respect to which y is differentiated. Thus, suppose y = x"^, x = t^. Then -j^= ^ and so dx^ d'y = 2 (dxY = 2 (3 f2 dty = 18 f^ {dt)\ If, however, we differentiate with respect to t, since y = i^, g=30f^and d'y = 30 t^ (dty, which is not equal to the value obtained when we differen- tiated y with respect to x. For this reason we shall not use differentials of the second or higher orders except in the numerators of derivatives. Two derivatives like --t^ and ■— must not be combined like fractions because cPy does not have the same value in the two cases. If we have derivatives with respect to t and wish to find derivatives with respect to x, they can be found by using the identical relation du d _ du dt_ _ dt ,nK\ dx dt dx dx dt For example, <Py d_ /dy\ ^ d_ /dy\ di^^de_^ dx\dtj dt\dt) dx dx^ dt Chap. IV. ALGEBRAIC FUNCTIONS approxima _. ^1 . , 1 c , , . , e. Given x = t — - , y = t-\--:, find velocity a t ^ t case / This e would m A^ a rule ,, >. . ^ luently, dy ^ t^ t^-l dx ,, , dt f -\-l dt + j. dl'y _ d (t^-\\ At dt At 1 31 dry dx^' 4<» dx2 dx \t^ + 1/ (f + 1)2 dx {p + 1)' . , 1 {^ + If EXERCISES Find ^ and -r4, in each of the followong exercises: 1. y = _ • 6. x^ + y^ = a^ 2. J/ = V^"ir^. 6. x2 - 2 2/! = 1. 3. y = {x - ly (x + 2)*. 7. xy = 2 +y. 4. j/! = 4 X. 8. x' + y' = a'. 9. If a and 6 are constant and y = ax'' + bx, show that dx* ax 10. If a, b, c, d are constant and y = ax^ -\- bx^ + ex + d, show that d*y dx^ «• 11. Show that 12. Show that d^ I .d^ _ \_ . fPx di\dt ^l~ dfl' dxVdx^ '^'^ dx^^^^'dx ^y) ''dx* 13. Given x = t^ + t', ?/ = f - <', find ^ and ^■ dx' dj^ 14. By differentiating the equation dx ^ dy with respect to x, find j^ in terms of derivatives of x with respect to y. ■ Chap. III. Vited the CHAPTi^R IV 1 the quo- RATES 26. Rate of Change. — If the change in a qua'/ proportional to the time in which it occurs, z is sai*-' /the vari- at a constant rate. If Az is the change occurri 9/ in- terval of time A^, the rate of change of z is If the rate of change of z is not constant, it will be nearly Az constant if the interval At is very short. Then -^ is ap- proximately the rate of change, the approximation becoming greater as the increments become less. The exact rate of change at the time t is consequently defined as limff = |, (26) At=o At at that is, the rate of change of any quantity is its derivative with respect to the time. If the quantity is increasing, its rate of change is positive; if decreasing, the rate is negative. 27. Velocity Along a Straight Line. — Let a particle P move along a straight line (Fig. 27). Let s = OP be con- 8 As ,^ ^ ■^■■■■■■^ p Fig. 27. sidered positive on one side of 0, negative on the other. If the particle moves with constant velocity the distance As in the time At, its velocity is As ) At' ^ If the velocity is not constant, it will be nearly so when At As is very short. Therefore -rr is approximately the velocity, the 32 Chap. IV. RATES 33 approximation becoming greater as A< becomes less. The velocity at the time t is therefore defined as This equation shows that ds is the distance the particle would move in a time dt if the velocity remained constant. As a rule the velocity will not be constant and so ds will be different from the distance the particle does move in the time dt. When s is increasing, the velocity is positive; when s is decreasing, the velocity is negative. Example. A body starting from rest falls approximately s=mf feet in t seconds. Find its velocity at the end of 10 seconds. The velocity at any time t is f = ^ = 32 « ft./sec.* dt At the end of 10 seconds it is y = 320 ft./sec. 28. Acceleration Along a Straight Line. — The accelera- tion of a particle moving along a straight line is defined as the rate of change of its velocity. That is This equation shows that dv is the amount v would increase in the time dt if the acceleration remained constant. The acceleration is positive v/hen the velocity is increasing, negative when it is decreasing. Example. At the end of t seconds the vertical height of a ball thrown upward with a velocity of 100 ft./sec. is /i = 100 < - 16 ^. Find its velocity and acceleration. Also find when it is rising, when falling, and when it reaches the highest point. * The notation ft./sec. means feet per second. Similarly, ft./sec.*, used for acceleration, means feet per second per second. -p Fig. 29. 34 DIFFERENTIAL CALCULUS Chap. IV. The velocity and acceleration are v = j^ = (100 - 32 t) ft./sec, a = ^= -32ft./sec.2. at The ball will be rising while v is positive, that is, until t = -^ = 3i. It will be falling after ^ = 3^. It will be at the highest point when < = 3|. 29. Angular Velocity and Acceleration. — Consider a body rotating about a fixed axis. Let 6 be the angle turned through at time t. The angular veloc- ity is defined as the rate of change of 6, that is, angular velocity = w = -r- • at The angular acceleration is the rate of change of angular velocity, that is, , IX- do} cPd angular acceleration =« = —- = —-. at dt^ Exam-pie 1. A wheel is turning 100 revolutions per minute about its axis. Find its angular velocity. The angle turned through in one minute will be CO = 100 • 2 TT = 200 IT radians/min. Ex. 2. A v^^heel, starting from rest under the action of a constant moment (or tv/ist) about its axis, will turn in t seconds through an angle e = kt^, h being constant. Find its angular velocity and acceleration at time t. By definition w = 37 = 2 /e/ rad./sec, dt a = -TT = 2k rad./sec.'. ai ' Chap. IV. / RATES 35 30. Related Rates. — In many cases the rates of change of certain variables are known and the rates of others are to be calculated. This is done by expressing the quantities whose rates are wanted in terms of those whose rates are known and taking the derivatives with respect to t. Example 1. The radius of a cylinder is increasing 2 ft, /sec. and its altitude decreasing 3 ft./sec. Find the rate of change of its volume. Let r be the radius and h the altitude. Then V = irr^h. The derivative with respect to t is By hypothesis Hence dv _« d^ , o u^'"' dr _ dh _ '^~^' dt~ '^' $ = 4 Trr/i - 3 7rr2. at This is the rate of increase when the radius is r and altitude h. If r = 10 ft. and /i = 6 ft., -77= — 60 T cu. ft./sec. at Ex. 2. A ship B sailing south at 16 miles per hour is north- west of a ship A sailing east at 10 miles per hour. At what rate are the ships approaching? Let X and y be the distances of the ships A and B from the point where their paths cross. The distance between the ships is then s = Vx2 + y\ This distance is changing at the rate dx dy ds J^di'^^'dU dt Vx2 + w2 36 DIFFERENTIAL CALCULUS Chap. IV. By hypothesis, 1=10 *= -16, ^= = ^^ = cos45° = ^. at at Vx^ + ?/2 Vx^ + 2/2 v2 Therefore ds 10 - 16 dt V2 = -3V2 mi./hr. The negative sign shows that s is decreasing, that is, the ships are approaching. EXERCISES 1. From the roof of a house 50 ft. above the street a ball is thrown upward with a speed of 100 ft. per second. Its height above the ground t seconds later will be h = 50+ loot - U fi. Find its velocity and acceleration when I = 2. How long df^3s it con- tinue to rise? What is the highest point reached? » 2. A body moves in a straight line according to the law s = i /< - 4 ^3 + 16 ^2. Find its velocity and acceleration. During what interval is the velocity decreasing? When is it moving backward? 3. If V is the velocity and a the acceleration of a particle moving along the a:-axis, show that adx = V dv. 4. If a particle moves along a line with the velocity v^ = 2 gs, where g is constant and s the distance from a fixed point in the line, show that the acceleration is constant. 5. When a particle moves with constant speed around a circle with center at the origin, its shadow on the a;-axi3 moves with velocity v satisfying the equation 1,2 4- n^x^ = nVS n and r being constant. Show that the acceleration of the shadow is proportional to its distance from the origin. 6. A wheel is turning 500 revolutions per minute. What is its angular velocity? If the wheel is 4 ft. in diameter, with what speed does it drive a belt? Chap. IV. RATES 37 7. A rotating wheel is brought to rest by a brake. Assuming the friction between brake and wheel to be constant, the angle turned through in a time t will be = a + bt — cP, a, b, c being constants. Find the angular velocity and acceleration. When will the wheel come to rest? 8. A wheel revolves according to the law w = 30 / + ^, where « b the speed in radians per minute and t the time since the wheel started- A second wheel turns according to the law 6 = iC*, where / is the time in seconds and the angle in degrees through which it has turned. Which wheel is turning faster at the end of one minute and how much? 9. A wheel of radius r rolls along a line. If r is the velocity and a the acceleration ot its center, a> the angular velocity and a the angular acceleration about its axis, show that V = roj, a = Ta. ^ 10. The depth of water in a cylindrical tank, 6 feet in diameter, is increasing 1 foot per minute. Find the rate at which the water is flow- ing in. 11. A stone dropped into a pond sends out a series of concentric ripples. If the radius of the outer ripple increases steadily at the rate of 6 ft. /sec., how rapidly is the area of disturbed water increasing at the end of 2 seconds? 12. At a certain instant the altitude of a cone is 7 ft. and the radiiis of its base 3 ft. If the altitude is increasing 2 ft. /sec. and the radius of its base decreasing 1 ft. /sec., how fast is the volume increasing or de- creasing? 13. The top of a ladder 20 feet long sUdes down a vertical wall. Find the ratio of the speeds of the top and bottom when the ladder makes an angle of 30 degrees with the ground. 14. The cross section of a trough 10 ft. long is an equilateral triangle. If water flows in at the rate of 10 cu. ft. /sec., find the rate at which the depth is increasing when the water is 18 inches deep. 15. A man 6 feet tall walks at the rate of 5 feet per second away from a lamp 10 feet from the groimd. When he is 20 feet from the lamp post, find the rate at which the end of his shadow is moving and the rate at which his shadow is growing. 16. A boat moving 8 miles per hour is laying a cable. Ass umin g that the water is 1000 ft. deep, the cable is attached to the bottom and stretches in a straight line to the stem of the boat, at what rate is the cable leaving the boat when 2000 ft. have been paid out? 17. Sand when poured from a height on a level siurface forms a cone with constant angle /3 at the vertex, depending on the material. If the / / 38 DIFFERENTIAL CALCULUS Chap. [V. sand is poured at the rate of c cu. ft. /sec, at what rate is the radius in- creasing when it equals a? 18. Two straight railway tracks intersect at an angle of 60 degrees. On one a train is 8 miles from the junction and moving toward it at the rate of 40 miles per hour. On the other a train is 12 miles from the junction and moving from it at the rate of 10 miles per hour. Find the rate at which the trains are approaching or separating. 'K-C/t^tA-^ 19. An elevated car nmning at a constant elevation of 50 ft. above the street passes over a surface car, the tracks crossing at right angles. If the speed of the elevated car is 16 miles per hour and that of the sur- face car 8 miles, at what rate are the cars separating 10 seconds after they meet? ^7 f ? ' j ; i_^ 20. The rays of the sun make an angle of 30 degrees with the hori- zontal. A ball drops from a height of 64 feet. How fast is its shadow moving just before the ball hits the ground? I CHAPTER V MAXIMA AND MINIMA 31. A function of a; is said to have a maximum at x = a, if when x = a the function is greater than for any other value in the immediate neighborhood of a. It has a minimum if when X = a the function is less than for any other value of x sufficiently near a. If we represent the function by y and plot the curve y = / (^)> ^ maximum occurs at the top, a minimum at the bottom of a wave. If the derivative is continuous, as in Fig. 31a, the tangent is horizontal at the highest and lowest points of a wave and the slope is zero. Hence in determining maxima and minima of a function / (x) we first look for values of x such that d dx /(x)=/'(x)=0. If a is a root of this equation, / (a) may be a maximum, a minimum, or neither. Fig. 31a. If the slope is positive on the left of the point and negative on the right, as at A, the curve falls on both sides and the ordinate is a maximum. That is, / (x) has a maximum value 39 40 DIFFERENTIAL CALCTijUS Chap. V. atx = a, iff (x) is positive for values >f xa little less and nego/- live for values a little greater than a. If the slope is negative on the left and positive on the right, as at B, the curve rises on both sides and the ordinate is a minimum. That is, / (x) has a minimum at x = a, if f (x) is negative for values of x a little less and positive for values a little greater than a. If the slope has the same sign on both sides, as at C, the curve rises on one side and falls on the other and the ordinate is neither a maximum nor a minimum. That is, / (re) has neither a maximum nor a minimum at x = a if f (x) has the same sign on both sides of a. Example 1. The sum of two numbers is 5. Find the maxi- mum value of their product. Let one of the numbers be x. The other is then 5 — x. The value of x is to be found such that the product y = X (5 — x) = 5 X — x^ is a maximum. The derivative is ^ = 5-20:. ax This is zero when x = |. If x is less than f , the derivative is positive. If x is greater than f the derivative is negative. Near X = ^ the graph then has the shape shown in Fig. 31b. At re = f the function has its greatest value f (5 - f ) = \^ Ex. 2. Find the shape of the pint cup which requires for its construction the least amount of tin. Let the radius of base be r and the depth h. The area of tin used is A=rr^ + 2 rrh. Let V be the number of cubic inches in a pint. Then V = n^h. I Chap. V MAXIMA AND MINIMA 41 Consequently, and h = irr» A 2 . 2f r Since r and v are constants, dA dr = 27rr -■7^=K^> This is zero if irr^ = v. If there is a maximum or minimum it must then occur when r = v/-; dA for, if r has any other value, -j- will have the same sign on both sides of that value and A will be neither a maximum nor a minimum. Since the amount of tin used cannot be zero there must be a least amount. This must then be the value of A when v = irr^. Also v = rr^h. We therefore conclude that r = h. The cup requiring the least tin thus has a depth equal to the radius of its base. Ex. 3. The strength of a rec- tangular beam is proportional to the product of its width by the square of its depth. Find the strongest beam that can be cut from a circular log 24 inches in diameter. In Fig. 31c is shown a section of the log and beam. Let x be the breadth and y the depth of the beam. Then X2 + J/2 = (24)2. The strength of the beam is S = kxy^ = kx (242 _ 3.2)^ Fig. 31c. 42 DIFFERENTIAL CALCULUS Chap. V. k being constant. The derivative of *S is dS dx = k (242 - 3 x2). If this is zero, x = ±8 V3. Since x is the breadth of the beam, it cannot be negative. Hence a: = 8 a/3 is the only solution. Since the log cannot be infinitely strong, there must be a strongest beam. Since no other value can give either a maximum or a minimum, a; = 8 Vs must be the width of the strongest beam. The corresponding depth is y = 8 Vq. Ex. 4. Find the dimensions of the largest right circular cylinder inscribed in a given right circular cone. Let r be the radius and h the altitude of the cone. Let X be the radius and y the altitude of an inscribed cylinder (Fig. 31d). From the similar triangles DEC and ABC, DE^AB EC EC ' that is. y ^h^ r — X r The volume of the cylinder is y = - (r - x). Fig. 31d. V = irx^y = — (rx^ — X?). Equating its derivative to zero, we get 2 rx - 3 a;2 = 0. Hence x = or x = | not give the maximum, radius must then be x The value x = obviously does Since there is a largest cylinder, its I r. By substitution its altitude is then found to be ?/ = \h. 32. Method of Finding Maxima and Minima. — The method used in solving these problems involves the following steps: Chap. V. MAXIMA AND MINIMA 43 (1) Decide what is to be a maximum or minimum. Let it be y. (2) Express y in terms of a single variable. Let it be x. It may be convenient to express y temporarily in terms of several variable quantities. If the problem can be solved by our present methods, there will be relations enough to elimi- nate all but one of these. dv (3) Calculate -^ and find for what values of x it is zero. (4) It is usually easy to decide from the problem itself whether the corresponding values of y are maxima or minima. du If not, determine the signs of -j^ when x is a little less and ax a little greater than the values in question and apply the criteria given in Art. 3L EXERCISES Find the maximum and minimum values of the following functions: 1. 2 x» - 5 X + 7. 3. X* -2x^ + 6. \/2. 6 + 12i -i». A. —=£=' V a* — X* Show that the following functions have no maxima or minima: 6. x». 7. 6x5 - 15 X* + 10x3. \/ 6. x' + 4 X. 8. X Va» + x*. 9. Show that x A — has a maximum and a minimum and that the X maximum is less than the minimum. 10. The sum of the square and the reciprocal of a number is a mini- mum. Find the number. 11. Show that the largest rectangle with a given perimeter is a square. ?/ 12. Show that the largest rectangle that can be inscribed in a given circle is a square. , 13. Find the altitude of the largest cylinder that can be inscribed in a sphere of radius a. , ■^ 14. A rectangular box with square base and open at the top is to be " made out of a given amount of material. If no allowance is made for thickness of material or waste in construction, what are the dimensions of the largest box that can/be made? 44 DIFFERENTIAL CALCULUS Chap. V. ; ,16. A cylindrical tin can closed at both ends is to have a given capacity. Show that the amount of tin used will be a minimum when the height equals the diameter. 16. What are the most economical proportions for an open cylindrical water tank if the cost of the sides per square foot is two-thirds the cost of the bottom per square foot? 17. The top, bottom, and lateral surface of a closed tin can are to be cut from rectangles of tin, the scraps being a total loss. Find the most economical proportions for a can of given capacity. 18. Find the volume of the largest right cone that can be generated by revolving a right triangle of hypotenuse 2 ft. about one of its sides. 19. Four successive measurements of a distance gave Ci, 02, as, cu as results. By the theory of least squares the most probable value of the distance is that which makes the sum of the squares of the four errors a minimum. What is that value? 20. If the sum of the length and girth of a parcel post package must not exceed 72 inches, find the dimensions of the largest cylindrical jug that can be sent by parcel post. 21. A circular filter paper of radius 6 inches is to be folded into a conical filter. Find the radius of the base of the filter if it has the maximum capacity. 22. Assuming that the intensity of light is inversely proportional to the square of the distance from the source, find the point on the line joining two sources, one of wliich is twice as intense as the other, at which the illumination is a minimum. 23. The sides of a trough of triangular section are planks 12 inches wide. Find the width at the top if the trough has the maximum capacity. 24. A fence 6 feet high runs parallel to and 5 feet from a wall. Find the shortest ladder that will reach from the ground over the fence to the wall. 26. A log has the form of a frustum of a cone 29 ft. long, the diameters of its ends being 2 ft. and 1 ft. \ beam of square section is to be cut from the log. Find its length if the volume of the beam is a maximum. 26. A window has the form of a rectangle surmounted by a semi- circle. If the perimeter is 30 ft;, find the dimensions so that the greatest amount of fight may be admitted. 27. A piece of wire 6 ft. long is to be cut into 6 pieces, two of one length and four of another. The two former are bent into circles which are held in parallel planes and fastened together by the four remaining pieces. The whole forms e, model of a right cylinder. Calculate the lengths into which the wire must be divided to produce the cylinder of greatest volutue. Chap. V. MAXIMA AND MINIMA 45 28. Among all circular sectors with a given perimeter, find the one which has the greatest area. 29. A ship B is 75 miles due east of a ship A. IS B sails west at 12 miles per hour and A south at 9 miles, find when the ships will be closest together. 30. A man on one side of a river J mile wide wishes to reach a point on the opposite side 5 miles further along the bank. If he can walk 4 miles an hour and swim 2 miles an hour, find the route he should take to make the trip in the least time. 81. Find the length of the shortest line which will divide an equi- lateral triangle into parts of equal area. 32. A triangle is inscribed in an oval curve. If the area of the tri- angle is a maximum, show graphically that the tangents at the vertices of the triangle are parallel to the opposite sides. 33. A and C are points on the same side of a plane mirror. A ray of light passes from A to C by way of a point B on the mirror. Show that the length of the path ABC will be a minimum when the lines AB; CB make equal angles with the perpendicular to the mirror. 34. Let the velocity of Ught in air be t'l and in water ij. The path of a raj' of Ught from a point A in the air to a point C below the surface of the water is bent at B where it enters the water. If di and fit are the angles made by AB and BC with the perpendicular to the surface, show that the time required for light to pass from A to C \st11 be least if £ is so placed that singi _ Vi sin di Vi 35. The cost per hour of propelling a steamer is proportional to the cube of her speed through the water. Find the speed at which a boat should be run against a current of 5 miles per hour in order to make a given trip at least cost. 36. If the cost per hour for fuel required to run a steamer is propor- tional to the cube of her speed and is S20 per hour for a speed of 10. knots, and if the other expenses amount to $ ICO per hour, find the most econona- ical speed in still water. 33. other Types of Maxima and Minima. — The method given in Art. 31 is sufficient to determine maxima and minima if the function and its derivative are one-valued and continu- ous. In Figs. 33a and 33b are shown some tj'pes of maxima and minima that do not satisfy these conditions. ^At B and C, Fig. 33a, the tangent is vertical and the de- 46 DIFFERENTIAL CALCULUS Chap. V. that on the right. The derivative is discontinuous. At A and E the curve ends. This happens in problems where values beyond a certain range are impossible. According to Fig. 33a. our definition, y has maxima &t A, B, D and minima at C and E. If more than one value of the function corresponds to a single value of the vari- able, points like A and B, Fig. 33b, may occur. At such points two values of y coincide. These figures show that in determining max- ima and minima special attention must be given to places where the de- FiG. 33b. rivative is discontinuous, the function ceases to exist, or two values of the function coincide. Example 1. Find the maximum and minimum ordinates on the curve y* = a^. In this case, y = x^ and dx 3^ • No finite value of x makes' the derivative zero, but x = Chap. V. MAXIMA AND MINIMA 47 makes it infinite. Since y is never negative, the value is a mipitrmm (Fig. 33c). Ex. 2. A man on qne side of a river ^ mile wide wishes to reach a point on the opposite side 2 miles down the river. If he can row 6 miles an hour and walk 4, find the route he should take to make the trip in the least time. Fig. 33d. Let A (Fig. 33d) be the starting point and B the destina- tion. Suppose he rows to C, x miles down the river. The time of rowing will be ^ VxM-~l and the time of walking i (2 — x). The total time is then t = i V^Ti + i (2 - x). Equating the derivative to zero, we get 6\/z* + i 4 which reduces to 5 x* + | = 0. This has no real solution. 48 DIFFERENTIAL CALCULUS Chap. V. The trouble is that J (2 — x) is the time of walking only if C is above B. If C is below B, the time is \{x — 2). The complete value for t is then the sign being + if x < 2 and — if x > 2. The graph of the equation connecting x and t is shown in Fig. 33e. At X = 2 the derivative is discontinuous. Since he rows faster than he walks, the minimum obviously occurs when he rows all the way, that is, x = 2. EXERCISES Find the maximum and minimum values of y on the following curves: 1.- x' 4- y' = o'. \J 3. y* = a:* - 1. 2. ^ = x* (a; - 1). 4. X = l^ + i\ y = (^ - i*. 6. Find the rectangle of least area having a given perimeter. 6. Find the point on the parabola j/^ = 4x nearest the point (-1,0). 7. A wire of length I is cut into two pieces, one of which is bent to form a circle, the other a square. Find the lengths of the pieces when the sum of the areas of the square and circle is greatest. 8. Find a point P on the line segment AB such that PA^ + PB^ is a maximum. 9. If the work per hour of moving a car along a horizontal track is proportional to the square of the velocity, what is the least work re- quired to move the car one mile? 10. If 120 cells of electromotive force E volts and internal resistance 2 ohms are arranged in parallel rows with x cells in series in each row, the current which the resulting battery will send through an external resistance of \ ohm is ^ 60xE ^ x* + 20* How many cells should be placed in each row to give the maximum current? CHAPTER VI DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS 34. Formulas for Differentiating Trigonometric Func- tions. — Let u be the circular measure of an angle. Vn. d sin w = cos ti du. 4^1]d^ = cosv CK^ . , Vin. <2 cos u = — smwdu. dc£i^ . -^r*^-/ IX. d tan u = sec"* u du. j +<u>'^ , d^c '^ - X. d cot M = — CSC** It rfu. <*£^- XI. d sec u = sec u tan u di/! ^4^ ^ ^^^ >= ^" ^ Xn. d CSC M = —CSC u cot M da. cU^l^^ - -cos^Ci'^^ "^ The negative sign occurs in the differentials of all co- functions. 35. The Sine of a Small Angle. — Inspection of a table of natural sines will show that the sine of a small angle is very nearly equal to the circular meas- ure of the angle. Thus sin l** = 0.017452, r 180 = 0.017453. We should then expect that ,. sin , hm— -= 1. (35) Fig. 35. To show this graphically, let 9 = AOP (Fig. 35). Draw PM p)erpendicular to OA. The circular measure of the angle is defined by the equation arc arc AP 6 = rad. OP 49 60 DIFFERENTIAL CALCULUS Chap. VI. ,, . ^ MP „ Also sin 6 = jyp . Hence sing^ MP ^ chord QP d arc AP arc QP As 6 approaches zero, the ratio of the arc to the chord ap* proaches 1 (Art. 53). Therefore the limit of — r— is 1. 36. Proof of VII, the Differential of the Sine. — Let J/ = sin w. ■ Then • fcj--5»n3^ J/ + Ay = sin (w + Aw) ' - 3(^^ 3 and so Ay = sin (w + Aw) — sin w. It is shown in trigonometry that sin A - sin 5 = 2 cos § (^ + ^) sin ^ {A - B), If then A = w + Aw, B = u, ^ Ay = 2 cos (w + § Aw) sin \ Aw, whence ^''" ^ ^ ■^**'^ -^^1^^%^^)^^ ii^^j^ 4- c ^ = 2 cos ^ +4 Aw) j"2u ^ =1^0^ tw +% Aw) ^^1^ . As Aw approaches zero *^( sin I Aw _ sing ~^5Au~~ B approaches 1 and cos (w + i Aw) approaches cos w. There- fore dy 3^ = cos w. aw Ck)nsequently, *^L " -^ ^^ dy = cos w dw. 37. Proof of Vm, the Differential of the Cosine. —By trigonometry cos w = sinf^ — wj- , Chap. VI. TRANSCENDENTAL FUNCTIONS 53 22. Find the points on the curve y = x + sin 2 x where the tangent is parallerfto the line y = 2 x + S. 23. A weight supported by a spring hangs at rest at the origin. If the weight is lifted a distance A and let fall, its height at any subsequent time ( will be y = A cos (2 mt), n being constant. Find its velocity and acceleration as it passes the origin. WTiere is the velocity greatest? Where is the acceleration greatest? 24. A revolving light 5 miles from a straight shore makes one com- plete revolution per minute. Find the velocity along the shore of the beam of light when it makes an angle of 60 degrees with the shore line._ ^ 26. In Ex. 24 with what velocity would the light be rotating if the spot of light is moving along the shore 15 miles per hour when the beam makes with the shore line an angle of 60 degrees? 26. Given that two sides and the included angle of a triangle have at a certain instant the values 6 ft., 10 ft., and 30 degrees resf)ectiv6ly, and that these quantities are changing at the rates of-S ft., —2 it., and 10 degrees per second, how fasf is the area of the/triangle ch^ging? 27. OA is a crank and AB a connecting nxi attached to a piston B moving along a line through 0. If OA re*x>l\W about with angular velocity w, prove that thp velocity of fi is wOC/where C is the point in which the line BA cuts/he line through perpendicular to OB. 3/ Gcoa rims ^eam that 28. An alley 8 ft /What is the longest street into the alley? 29. A needle rest^'ith one end needle will sink to a p>osition in If the length of them^edle eq the position of equmbrium? 30. A rope with a ring at o horizontal line and held taut the rope slips freely, tUe wei the angle formed at th^ bott* dicubir to a street 27 ft. wide, mitved horixontally along the a smAOth hemispherical bowl. The hich tpe center is as low as possible. meter of the bowl, what will be is looped over two pegs in the same weight fastened to the free end. If descend as far as f>ossible. Find of the loop. 31. Find the angle at the bottom of the loop in Ex. 30 if the rope is looped over a circular pulley instead of the two pegs. 32. A gutter is to b^ made by bending into shape a strip of copper so that the cross section sl^all be an arc of a circle. If the width of the strip is a, find the radius of the cross section when the carrying capacity is a maximum. 33. A spoke in the front wheel of a bicycle is at a certain instant per- pendicular to one in the rear wheel. If the bicycle rolls straight ahead, an what position will the outer ends of the two ^x)kes be closest together? 64 DIFFERENTIAL CALCULUS Chap. VI. 39. Inverse Trigonometric Functions. — The symbol sin~* X is used to represent the angle whose sine is x. Thus y = sin~^ X, x = sin y are equivalent equations. Similar definitions apply to cos~^ x tan~^ x, cot~* x, sec~^ x, and csc~^ x. Since supplementary angles and those differing by mul- tiples of 2 7r have the same sine, an indefinite number of angles are represented by the same symbol sin~^ x. The algebraic sign of the derivative depends on the angle dif- ferentiated. In the formulas given below it is assumed that sin-i u and csc~^ u are angles in the first or fourth quadrant, cos~^ u and sec"^ u angles in the first or second quadrant. If angles in other quadrants are differentiated, the opposite sign must be used. The formulas for tan~^ u and cot~^ u are valid in all quadrants. ^ 40. Formulas for Differentiating Inverse Trigonometric Functions. — Xm. dsin-^u = XIV. d cos-^u = - XV. d tan-* u = XVI. d cot-* u = - XVII. dsec-^u = XVIII. dcsc-'u = fl' • d[a>^^^.-_l-^ y = sin~* u. 41. Proof of the Formulas. — Let Then sin y = u. Differentiation gives cos y dy = du, Chap. VI. TRANSCENDENTAL FUNCTIONS 55 whence ' But dy = du cosy cosy = =fc Vl - sin* y = zfc Vl - u». If 2/ is an angle in the first or fourth quadrant, cos y is positive. Hence cos y = Vl — u* and so , du dy = , The other formulas are proved in a similar way. EXERCISES \J L y = Bin-'(3x-l), 2. y=coe-(l-?). 5. y = 8ec-» V4 1 + 1, ft 1 -. 3 . V 7. y = tan-i' dy = dy = Zdx v6x - 9i» rfx V2ax-x* dy _ -2 dx x» + l' (4 1 + 1) Vi dy 1 X + a 8. X = csc~i (sec 6), s/.9. y = r- ^ 10. y = sec" Vtf -x» 1 Vl -I* <^ V2 + 2x-4x» dx~ 3? +a*' d9 ^• ^^ ^ <** (o» - x») Va» - 2 1»* dy 1 11. y = §V^^^ + ^'Bin-»-, ^ = < o i -I 4 sin I 12. y = tan-i r-— — , 3 + 5 cosx dx Vl -z» Va»-x». 4 13. y = S€< ■» 2x»-l' dx 5 + 3cosx ^= 2 56 DIFFERENTIAL CALCULUS Chap.VL \/l4. y - a8in-»5+ V^rr^, ^ - V^^. " o ' dx ^ a -{■ X 16. y = 2 (3 X + 1)* + 4 cot-i ^^^-±^, rfy^ i dx (3x + l)* + 4(3x4-l)** 1 _, 3x dy 1 -x" 16. I/-gtan 2 + 2x»' dx 4x* + 17x2+4' „ _,x + 1 2 _, 2x dy X + 1 17. « = cos ' — s :^co8 ^5 , j^ = . =L' 2 V3 3-x dx (3_a;)V3-2x-x» 1ft _ ^x' - g* 1_ _jX dy 1 "• ^~ 2a^x« 2a»^'' o' dx ~ -^a v^TTT^j* 19 , ,x + Vi2 + 4x-4 '^I/. I. y = tan-i ^ 1 ^Z - 2 <te xVx2+4x-4 20. y = X sin-i x + Vl - x^, ^ = , ^ dx2 Vl -x' 21. y = x2 sec-i I - 2 Vx^^Tl^ ^ = 2xsec-i|- Z uX i2 22. Let 8 he the arc from the x-axis to the point (x, y) on the circle x2 + y2 = a*. Show that * = -?, |.|, *. = ^. + .^. V 23. Let A be the area bounded by the circle x* + y" = o*, the y-axis and the vertical line through (x, y). Show that A = xy + a^ tan~* - , dA = 2ydx. 24. The end of a string wound on a pulley moves with velocity v il along a line perpendicular to the axis of the pulley. Find the angular velocity with which the pulley turns. ^ 25. A tablet 8 ft. high is placed on a wall with its base 20 ft. above the level of an observer's eye. How far from the wall should the observer stand that the angle of vision subtended by the tablet be a maximum? 42. Exponential and Logarithmic Functions. — If a is a positive constant and u a variable, a" is called an exponential function. If w is a fraction, it is understood that o" is the positive root. If y = a", then u is called the logarithm of i/ to base a. That is, y = a", ?' = logaV ' " r *. .^^^SCENDENTAL FUNCTIONS 57 Elimination of u are by definition equivalent equations gives the important identity aio«.v = y. (42b) This expresses symbolically that the logarithm is the power to which the base must be raised to equal the number. 43. The Curves y = a". — Let a be greater than 1. The graph of y = a' has the general appearance of Fig. 43. increment h, the increment in y is At/ = 0*+* — a' = a' (a'' — 1). This increases as x increases. If then X increases by successive amounts h, the increments in y form steps of increasing height. The curve is thus concave upward and the arc lies below its chord. The slope of the chord AP join- ing A (0, 1) and P (x, y) is a' — \ If X receives a small Fig. 43. As Pi moves toward A the slope of APx increases. As P2 moves toward A the slope of APj decreases. Furthermore the slopes of AP2 and APi approach equality; for and a~* approaches 1 when k approaches zero. Now if two numbers, one always increasing, the other always decreasing, approach equality, they approach a common limit. Call this limit m. Then lim^^ = m. (43) bb DIFFERENTUL CALCUI.Ao jpp. m^^^^Wii^-^- This is the slope of the curve 2/ = a* at the point where it crosses the ^/-axis. 44. Definition of e. — We shall now show that there is a number e such that ^-^ - (44) In fact, let •■ 1 1 lim where m is the slope found in Art. 43. Then ,. e* - 1 ,. o*" - 1 1 ,. a*" - 1 1 hm = lim = — hm = — • m = 1. x=o a: 1^0 X m x^o x_ m m The curves y = a' all pass through the point A (0, 1). Equation (44) expresses that when a = e the slope of the curve at A is 1. If a > e the slope is greater than 1. If a < e, the slope is less than 1 . Y 1 1 . / a>e, 1 / i I / ^ /a<e ■ A I/-'-' — :::SS5^ X- Fia. 44. We shall find later that e = 2.7183 approximately. Logarithms to base e are called natural logarithms. In this book we shall represent natural log- arithms by the abbreviation In. Thus In u means the natural logarithm of u. Chap. VI. TRANSCENDENTAL FUNCTIONS 59 45. Differentials of Exponential and Logarithmic Func- tions. — XIX. de~ = c- dw. - e = I -*--L 4 -L-(- U. L? XX. da" = a" In a du. XXL d\nu = ^-^}i^'Jk cJe^.e"" 46. Proof of XIX, the Differential of e". — Let Then ^ whence di<2^ yy^^ Ay = e«+^» - e« = e" (e^" - 1) ZHT ^ land ^ Au Au As Au approaches zero, by (44), C e^" - 1 Au approaches 1. Consequently, \ 3^ = e", dy = e" dw. 47. Proof of XX, the Differential of a". — The identity a = e'°" gives a" = e"*"". Consequently, da" = e" •" ° d (w In a) = e" •" ° In a dw = a" Ina du. 48. Proof of XXI and XXII, the Differential of a Log- arithm. — Let y = Inw. Then e" = u. Differentiating, ^ dy = du. 60 Therefore DIFFERENTIAL CALCULUS du du Chap. VI. The derivative of loga u is found in a similar way. Example 1. y = Iri (sec^ x). d sec^ X 2 sec x (sec x tan x dx) ^ sec^ X Ex. 2. y = 2*^'^"'^ sec^ X dy = 2^"""' In 2d (tan-^ x) = — T+l^ — = 2 tan X dx. 2tan-iiin2da; EXERCISES I. 2/ = ea;, \/ 2. y = a*^'^^*^ z-l 3. 2/ = e=^+S ^ . 5. 1/ = x" + n*, \j e. y = a'sf, 7. y = In (3 x2 + 5 X + 1), 8. 2/ = In sec* x, 9. y = In (x + Va;2 - a?), ^10. y = In (sec ax + tan ax), II. y = In (o* + &*), dx X* dx dy = 2 a**'' ^ "^ In o sec* 2 a;. 2 — dx (x + 1)2 dy_( 2 Y dx Ve* + e-^/ — = wx*~^ + n'' In n. dx ^ = a='x°-Mo + a;lno). dx dy dx 6x4-5 3x2 + 5x + l ^ = 2 tan X. dx dy L_ -^ = a sec ax dx ', 12. y = In sin X + 5 cos* x, 1 2 IS. 2/ = 9l^*«'°2~2wH^' dx Vx* — o* a sec ax. _aMno+_b^4n6 ^ - a» + 6* dy _ cos' x ^ dx ~ sin X dy_ 1 , dx sin* z Chap. VI. TRANSCENDENTAL FUNCTIONS 61 ,. _1, X' 1_ dy ^ 8 14. y - ^in^, _^ a:«-4' dx x(i»- 4)>* 16. 1/ = - In , > :7- = , « o + Va» -I* <« X V o» - X* 16. y = iB ( VI+3 + vT+2) + V(x + 3) (I + 2), ^ = y |^ 17. y = In (VT+^ + V^, ^ = •ii^ 2 Vx» + ax 18. y = X tan-i --% In (x«+a»), ^ = tan"' - • a Z ax a V 19. y = ef" (sin ax — cos ax), -^ = 2 ae" sin ax. ax \l 20. y = i tan* x — 5 tan* z — In cos x, -p = tan' x. 22. X = e* + e-', y = e< - e"', = - ^• 23. y=llnx, g=^(21nx-3). 24. y = x^, g=(x + n)e'. 26. By taking logarithms of both sides of the equation y = x" and differentiating, show that the formula --x" = nx"-* dx is true even when n is irrational. 26. Find the slope of the catenary y = |Ve^+e'-) at X = 0. 27. Find the points on the curve y = e** sin x where the tangent 18 parallel to the i-axis. y 28. If y = Ae"* + Be""*, where A and B axe constant, show that 29. If y = 2e"'*, where z is Miy function of x, show that di*^*'<ix^^^ * dx» SO. For what values of x does y = 5 hi (x - 2) + 3hi (x + 2) + 4» Bcrease as x increases? 62 DIFFERENTIAL CALCULUS Chap. VI. 31. From equation (44) show that e = lim (1 + x) *• 32. If the space described by a point is s = ae' + &e""', show that the acceleration is equal to the space passed over. 33. Assuming the resistance encountered by a body sinking in water to be proportional to the velocity, the distance it descends in a time t is g and k being constants. Show that the velocity v and acceleration a satisfy the equation a = g — kv. Also show that for large values of t the velocity is approximately con- stant. 34. Assuming the resistance of air proportional to the square of the velocity, a body starting from rest will fall a distance s=^ln /e« + e-*^'^ in a time t. Show that the velocity and acceleration satisfy the equa- tion k^i^ a = g Also show that the velocity approaches a constant value. CHAPTER VII GEOMETRICAL APPLICATIONS 49. Tangent Line and Normal. — Let mi be the slope of a given curve at Pi (xi, yi). It is shown in analytic geometry that a line through (xi, yi) with slope nil is represented by the equation y — yi = miix - xi). This equation then rep- resents the tangent at (,^1, yi) where the slope of the curve is mi. The line PiN perpen- dicular to the tangent at its point of contact is called the normal to the curve at Pi. Since the slope of the tangent is mi, the slope of a perpendicular line is and so TWl y-yi= - — (^-xi) 7/1 1 is the equation of the normal at (xi, yi). Example 1. Find the equations of the tangent and normal to the eUipse x^ + 2 ^^ = 9 at the point (1,2). The slope at any point of the curve is dy _ X dx 2y At (1, 2) the slope is then mi = —I. ,The equation of the tangent is 2/-2 = -i(x-l). 63 Fig. 49. 64 DIFFERENTIAL CALCULUS Chap. VII. and the equation of the normal is y-2 = A(x-l). Ex. 2. Find the equation of the tangent to x^ — y^ = a? at the point (xi, y\). Xi ' The slope at (xi, yi) is — . The equation of the tangent yi is then Xi y-yi = -(.^- ^i) 2/1 ^ -r.Z — which reduces to XiX - yiy = xi^ - yi' Since {xi, yi) is on the curve, Xi^ — y-^ = a^. The equation of the tangent can therefore be reduced to the form XiX - yiy = a\ 50. Angle between Two Curves. — By the angle be- tween two curves at a point of intersection we mean the angle between their tangents at that point. Let mi and nii be the slopes of two curves at a point of intersection. It is shown in analytic geometry that the angle /3 from a line with slope mi to one with slope Fig. 50a. m2 satisfies the equation taujS = m2 — mi 1 +mim2 (50) This equation thus gives the angle /3 from a curve with slope mi to one with slope nh, the angle being considered positive^ when measured in the counter-clockwise direction. I I Chap. VII. GEOMETRICAL APPLICATIONS 65 Example. Find the angles determined by the line y = x and the parabola y = x^. Solving the equations simultaneously, we find that the line and parabola intersect at (1,- 1) and (0, 0). The slope i r of the line is 1. The slope at any point of the parabola is ax At (1, 1) the slope of the parabola is then 2 and the angle from the line to the parabola is then given by Fig. oOb. tan /3i = 2-1 whence 1 3' 1+2 /3i = tan-4 = 18° 26'. At (0, 0) the slope of the parabola is and so the angle from the line to the parabola is given by the equation tan Pi = = — 1. ^ 1+0 ' whence ft = -45°. The negative sign signifies that the angle is measured in the clockwise direction from the line to the parabola. EXERCISES Find the tangent and normal to each of the following curves at the point indicated: 1. The circle 3^ + f = 5 at (-1, 2). 2. The hj-perbola xt/ = 4 at (1, 4). 3. The parabola y* = ax at i = a. 4. The exponential curve y = oif at x = 0. 6. The sine curve i/ = 3 sin x at x = ^• D X* IT 6. The eUipse -j + ^ = 1, at (x,, y,). 66 DIFFERENTIAL CALCULUS Chap. VIL / 7. The hyperbola x^ -\- xy — y^ = 2 x, at (2, 0). 8. The semicubical parabola y^ = x^, at ( — 8, 4). 9. Find the equation of the normal to the cycloid X = a {4> — sin <^), y = a (1 — cos <^) at the point ^ = 4>\. Show that it passes through the point where the rolling circle touches the x-axis. Find the angles at which the following pairs of curves intersect : 10. 2/2 = 4 x, x^ = 4 2/. 13. y = sin x, y — cos x. y/ 11. x^ + ?/2 = 9, x^ + y'^ — 6 x = 9. 14. y = logio x, y = In x. 12. x2 -+ 2/2 + 2 X = 7, 2/2 = 4 X. 15. y = \ (e^ + e"^), 2/ = 2 e*. 1^ 16. Show that for all values of the constants a and h the curves x^ — y^ = a*, xy = l^ intersect at right angles. 17. Show that the curves y = 6°'", y = e"^ sin (&x + c) are tangent at each point of intersection. ^ 18. Show that the part of the tangent to the hyperbola xy = a? in- tercepted between the coordinate axes is bisected at the point of tan- gency. ^ 19. Let the normal to the parabola y^ = ax at P cut the x-axis at A'^. Show that the projection of PN on the x-axis has a constant length. 20. The focus F of the parabola y^ = ax is the point (l a,0). Show that the tangent at any point P of the parabola makes equal angles with FP and the line through P parallel to the axis. /' 21. The foci of the ellipse ^ + P=1' «>^ are the points F' ( - Va^ - b^, O) and F (Va* - 6^^ q) . Show that the^ tangent at any point P of the ellipse makes equal angles with FP and F'P. 22. Let P be any point on the catenary y = ^\f' +e ° ) , M the projection of P on the x-axis, and A'^ the projection of M on the tangent at P. Show that MA'' is constant in length, 23. Show that the portion of the tangent to the tractrix a. /a + y/a^ - xA ,-- -, intercepted between the 2/-axis and the point of tangency is constant in iength. ft Fig. 51. Chap. VII. GEOMETRICAL APPLICATIONS 67 24. Show that the angle between the tangent at any point P and the line joining P to the origin is the same at all points of the curve hi Vx- + y2 = k tan-i -• ^ X 26. A point at a constant distance along the normal from a given curve generates a curve which is called parallel to the first. Find the parametric equations of the parallel curve generated by the point at distance h along the normal drawn inside of the ellipse X = a cos 4>, y = b sin <t>. 61. Direction of Curvature. — A curve is said to be con- cave upward at a point P if the part of the curve near P lies above the tangent at P. It is concave downward at Q if the part near Q lies below the tangent at Q. A • T dhj . At 'points where -p^ is pos- d^n itive, the curve is concave upward; where -7-^ is negative, the curve is concave downward. For d^y _ d (dy\ dx^ dx \dx/ (Py dv If then -T-| is positive, by Art. 13, -r- • the slope, increases as x increases and decreases as x decreases. The curve therefore rises above the tangent on both sides of the point. If, however, -j^ is negative, the slope decreases as x increases dx^ and increases as x decreases, and so the curve falls below the tangent. 52. Point of Inflection. — A point like A (Fig. 52a), on one side of which the curve is concave upward, on the other concave downward, is called a point of inflection. It is assumed that there is a definite tangent at the point of in- flection. A point like B is not called a point of inflection. 68 DIFFERENTIAL CALCULUS Chap. VII. The second derivative is positive on one side of a point of inflection and negative on the other. Ordinary functions change sign only by passing through zero or infinity. Hence to find points of inflection we find where r^ is zero or infinite. -t Fig. 52a. If the second derivative changes sign at such a point, it is a point of inflection. If the second derivative has the same sign on both sides, it is not a point of inflection. Fig. 52b. Fig. 52c. Example 1. Examine the curve 3y = x* — Qx^ for direc- tion of curvature and points of inflection. The second derivative is (Py da^ = 4 (x2 - 1). Chap. Vn. GEOMETRICAL APPLICATIONS 69 This is zero at x = ± 1. It is positive and the curve con- cave upward on the left of x = —1 and on the right of X = -\-l. It is negative and the curve concave downward between x = —1 and x =-\- I. The second derivative changes sign at A (—1, — f) and B (+1, — §), which are therefore points of inflection (Fig. o2b), Ex. 2. Ejcamine the curve y = x* for points of inflection. In this case the second derivative is This is zero when x is zero but is positive for all other values of X. The second derivative does not change sign and there is consequently no point of inflection (Fig. o2c). Ex. 3. If X > 0, show that sin x > x — 5-: •* o! Let ^3 y = sm X - X + ^j We are to show that y > 0. Differentiation gives dy , , a^2 cPy -y- = cos X — 1 + ?ri ' T^ = — sm x + x. dx 2! dx2 When X is positive, sin x is less than x and so -7^ is positive. Therefore -^ increases with x. Since -^ is zero when x is ax ax du zero, j^ is then positive when x > 0, and so y increases with X. Since y = when x = 0, y is therefore positive when X > 0, which was to be proved. * If n is any p>ositive integer n ! represents the product <rf the integers frwn 1 to n. Thus 3! = l-2-3 = 6. 70 DIFFERENTIAL CALCULUS Chap. VII. EXERCISES Examine the following curves for direction of curvature and points of inflection: 1. 2/ = a;3 - 3 z + 3. 6. ?/ = xe*. 2. 2/ = 2 x3 - 3 a;2 - 6 X + 1. 6. y = e"^. 3. y = x^ - 4 x» + 6 x2 + 12 X. 7. x^y - 4 x + 3 y = 0. 4. 2/3 = X - 1. 8. X = sin <, 2/ = ^ sin 3 t. Prove the following inequalities: 9. X In X - X - ^' + I > 0, if < X < L 10. (x - 1) e^ + 1 > 0, if X > 0. 11. 6== < 1 + X + ^ e°, if < X < a. 12. hisecx>|, if _|<a;<|. 13. According to Van der Waal's equation, the pressure p and volume y of a gas at constant temperature T are connected by the equa- tion RT a ^ m{v -h) v^' a, b, m, and 12 being constants. If p is taken as ordinate and v as ab- scissa, the curve represented by this equation has a point of inflection. The value of T for which the tangent at the point of inflection is hori' zontal is called the critical temperature. Show that the critical tem- perature is „ _ 8 am ~ 27 lib' 63. Length of a Curve. — The length of an arc PQ of a curve is defined as the hmit (if there is a hmit) approached by the length of a broken line with vertices on PQ as the number of its sides increases indefinitelj^ their lengths ap- proaching zero. We shall now show that if the slope of a curve is continu- ous the ratio of a chord to the arc it subtends approaches 1 as the chord approaches zero. In the arc PQ (Fig. 53) inscribe a broken line PABQ. Projecting on PQ, we get PQ = proj. PA -f proj. AB -{■ proj. BQ. GEOMETRICAL APPLICATIONS 71 The projection of a chord, such as AB, is equal to the prod- uct of its length by the cosine of the angle it makes with PQ. On the arc AB is a tangent RS parallel to AB. Let a be the argest angle that any tangent on the arc PQ makes with the Fig. 53. chord PQ. The angle between RS and PQ is not greater than a. Consequently, the angle between AB and PQ is not greater than a. Therefore proj. AB = AB cos a. Similarly, proj. PA = PA cos a, proj. BQ = BQ cos a. Adding these equations, we get PQ = (PA +AB + BQ) cos a. It is evident that this result can be extended to a broken ine with any number of sides. As the number of sides in- sreases indefinitely, the expression in parenthesis approaches ihe length of the arc PQ. Therefore PQ = arc PQ cos a, ;liat is, chord PQ ^pc- = cos a. arc PQ If the slope of the curve is continuous, the angle a ap- proaches zero as Q approaches P. Hence cos a approaches L and ,. chord PQ , lim 5;^ = 1. Q-P arc PQ ^ 72 DIFFERENTIAL CALCULUS Chap. VII. Since the chord is always less than the arc, the limit cannot be greater than 1. Therefore, finally, ,. chord PQ . lim ^^7^ = 1. (53) Q^P arc PQ 54. Differential of Arc. — Let s be the distance measured along a curve from a fixed point A to a variable point P. Then s is a function of the coordinates of P. Let </> be the angle from the positive direction of the x-axis to the tangent PT drawn in the direction of increasing s. T r Q / .^^^'' /^^ • w ^ B 1 S \ A ~o Fig. 54a. Fig. 54b. If P moves to a neighboring position Q, the increments in X, y, and s are Ax = PR, Ay = RQ, As = arc PQ. From the figure it is seen that Ax Ax As cos (RPQ) = PQ = ^ PQ' . /r.T>/^N Ay Ay As As Q approaches P, RPQ approaches and As _ arc PQ PQ ~ chord PQ approaches 1. The above equations then give in the limit dx dy cos<^ = ^. 8m</, = ^ (54a) Chap. VII. GEOMETRICAL APPLICATIONS 73 These equations express that dx and dy are the sides of a right triangle with hj-potenuse ds extending along the tangent (Fig. 54b). All the equations connecting dx, dy, ds, and <f> can be read off this triangle. One of particular importance is ds^ = dx^-\- dy\ (54b) 55. Curvature. — If an arc is everywhere concave toward its chord, the amount it is bent can be measured by the angle /3 between the tangent-s at its ends. The ratio /3 <{>' - <t> _ A<l> ~ As arc PP' As The is the average bending per unit length along PP'. limit as P' approaches P, ,. A<f> d<t> A«^o As ds is called the curvature at P. It is greater where the curve bends more sharph% less where it is more nearly straight. Fig. 55a. In case of a circle (Fig. 55b) Fig. 55b. <i> = d-\- s = aB. Consequently, d<t> _ dB 1 ds add a* 74 DIFFERENTIAL CALCULUS Chap. VII. that is, the curvature of a circle is constant and equal to the reciprocal of its radius. 66. Radius of Curvature. — We have j ust seen that the radius of a circle is the reciprocal of its curvature. The radius of curvature of any curve is defined as the reciprocal of its curvature, that is, ds radius of curvature = p = jr ' (56a) It is the radius of the circle which has the same curvature as the given curve at the given point. To express p in terms of x and y we note that Consequently, * = tan-.|. ^dx , 1 , /dy\ _ dx^ Also ds = Vdx^ + dyK Substituting these values for ds and d4>, we get dPy dx" (56b) If the radical in the numerator is taken positive, p will have d^v the same sign as -r^ , that is, the radius will be positive when the curve is concave upward. If merely the numerical value is wanted, the sign can be omitted. By a similar proof we could show that h(l)T dy' (56c) Chap. VII. GEOMETRICAL APPLICATIONS 75 Example 1. Find the radius of curvature of the parabola y2 = 4 x at the point (4, 4). At the point (4, 4) the derivatives have the values dy^2^1 ^^_4 1^ dx y 2' dx^ y^ ~ IQ Therefore P - ^ - _j_ - - 10 V5. dx" 16 The negative sign shows that the curve is concave downward. The length of the radius is 10 Vo. Ex. 2. Find the radius of curvature oi the curve repre- sented by the polar equation r = a cos 6. The expressions for x and y in terms of 6 are X = r cos 5 = a cos 6 cos 6 = a cos^ ^, y = r sin ^ = a cos dsmd. Consequently, dy a (cos* B — sin* 6) a cos 2 6 . « „ ■T- = ^ a ■ a = ^-^ = -cot 2 d, dx —2a cos 6 smd — asm 2d m ^ = V^^/ = _ 2 CSC* 2 Odd ^ 2 dx* dx asm2d dd a [1 + cot* 20]^ (csc*2 0)i = — a a 2 ,^, 2 esc' 2 2 esc' 2 a The radius is thus constant. The curve is in fact a circle. 67. Center and Circle of Curvature. — At each point of a curve is a circle on the concave side tangent at the point with radius equal to the radius of curvature. This circle is called the circle of curvature. Its center is called the center of curvature. Since the circle and curve are tangent at P, they have the 76 DIFFERENTIAL CALCULUS Chap. VII. dv same slope -^ at P. Since they have the same radius of curvature, the second derivatives will also be equal at P. Fig. 57. The circle of curvature is thus the circle through P such that dv d/^v ~ and 74 have the same values for the circle as for the curve dx dx^ at P. EXERCISES 1. The length of arc measured from a fixed point on a certain curve is s = x^ + X. Find the slope of the curve at x = 2. 2. Can X = cos s,y = sin s, represent a curve on which s is the length of arc measured from a fixed point ? Can x = sec s, y = tan s, represent such a curve? Find the radius of curvature on each of the following curves at the point indicated: 3. ^2 + p=l. at (0,6). 5. r = e^, at = ^• 4. x^ + xy + y^ = 3, at (1, 1). 6. r = a (1 + cos 0), at = 0. Find an expression for the radius of curvature at any point of each of the following curves: 7. y =l\^ + e »). 9. X =■ iy* — ihxy. 8. X = In sec y. 10. r = a sec* f 9. 11. Show that the radius of curvature at a point of inflection is infinite. Chap. VII. GEOMETRICAL APPLICATIONS 77 12. A point on the circumference of a circle rolling along the x-axis generates the cycloid X = a (ip — sm<f>), J/ = o (1 — COS 4>)t a being the radius of the roUing circle and <f> the angle through which it has turned. Show that the radius of the circle of curvature is bisected by the point where the rolling circle touches the x-axis. 13. A string held taut is unwound from a fixed circle. The end of the string generates a curve with parametric equations X = a cos + ad sin 9, y = asind — aO cos 9, a being the radius of the circle and 9 the angle subtended at the center by the arc unwound. Show that the center of curvatiu"e corresponding to any point of this path is the point where the string is tangent to the circle. 14. Show that the radius of curvature at any point (x, y) of the hypo- cycloid X* + 2/' = a' is three times the perpendicular from the origin to the tangent at (x, y). 1 cos x 58. Limit of It is shown in trigonometry that Consequently, 1 — cos X = 2 sin' r 1 2sin2| 1 — cos X 2.x = = sin - X X 2 /. x\ sin- X \ 2 / As X approaches zero, . X X 2 approaches 1. Therefore iimi^:i^^ = 0.1 = 0. 1=0 X 69. Derivatives of Arc in Polar Coordinates. — The angle from the outward drawn radius to. the tangent drawn Fthe direction of Increasing s is usually represented by the tter ^. 78 DIFFERENTIAL CALCULUS Chap. VIL Let r, 6 be the polar coordinates of P, and r + Ar, d + M those of Q (Fig. 59a). Draw QR perpendicular to PR and let As = arc PQ. Then RQ _ (y+Ar) sin A9 _ ^ , ^ >, sin ^^ ^^ As 'As'PQ' sin (/2PQ) = cos (RPQ) = PQ PQ PR _ (r + Ar) cos A0 - r PQ~ Ad Af = cos {^9) p^ - PQ r (1 — cos A9) PQ = cos {^^) Ar As r (1 — cos A0) A0 As As PQ Ad As PQ Fig. 59a. Fig. 59b. As Ad approaches zero, 1- /no^x , 1- sinA0 , ,. 1-cosA^ \im (RPQ) =rl/, lim-^^=l, hm ^ The above equations then give in the limit, = 0, lim As PQ 1. sin }p = rdd ds cos\f/ = dr ds (59a) These equations show that dr and rdd are the sides of a right triangle with hypotenuse ds and base angle ^. From this triangle all the equations connecting dr, dd, ds, and \p can be obtained. The most important of these are . , rdB tan^ = dr ds^ = dr^ + r2 dd\ (59]j) Chap. VII. GEOMETRICAL APPLICATIONS 79 Example. The logarithmic spiral r = as^. In this case, dr = ae^ dd and so The angle ^ is therefore constant and equal to 45 degrees. The equation , dr 1 ^^ = di = ^ dr . shows that -r is also constant and so r and s increase propor- tionally. EXERCISES Find the angle ^ at the point indicated on each of the following curves: 1. The spiral r = ad, a.t 6 = ^ • o 2. The circle r = a and at 6 =-• 4 3. The straight line r = a sec 6, at 9 = ^ • o 4. The eUipse r (2 — cos d) = k, at = ^• 6. The lemniscate r* = 2 a* cos 2 0, at = f x. 6. Show that the curves r = ae^, r = ae~^ are perpendicular at each of their points of intersection. 7. Find the angles at which the curves r = a cos 0, r = a sin 2 fl intersect. 8. Find the points on the cardioid r = o (1 — cos 9) where the tan- gent is parallel to the initial line. 9. Let P (r, 9) be a point on the hyperbola r^ sin 2 S = c. Show that the triangle formed by the radius OP, the tangent at P, and the X-axis is isosceles. 10. Find the slope of the curve r = e** at the point where 9 =%• 4 60. Angle between Two Directed Lines in Space. — A directed line is one along which a positive direction is assigned. This direction is usually indicated by an arrow. 80 DIFFERENTIAL CALCULUS Chap. VII. An angle between two directed lines is one along the sides of which the arrows point away from the vertex. There are two such angles less than 360 degrees, their sum being 360 degrees (Fig. 60). They have the same cosine. If the lines do not intersect, the angle between them is de- fined as that between intersecting lines respectively parallel to the given lines. Fig. 60. Fig. 61. 61. Direction Cosines. — It is shown in analytic geome- try* that the angles a, /3, 7 between the coordinate axes and the line P1P2 (directed from Pi to P2) satisfy the equations zi — Zi COS a = p p , cos /3 = ^p p^ , cos 7 = -5-B- ' (61a) rir2 -Lii^i i^U2 These cosines are called the direction cosines of the line. They satisfy the identity cos^ a + cos^ /3 + cos^ 7 = 1. (61b) If the direction cosines of two lines are cos ai, cos j8i, cos 71 and cos 0:2, cos /32, cos 72, the angle 6 between the lines is given by the equation cos 6 = cos ai cos az + cos ^2 cos ^2 + cos 71 cos 72. (61c) In particular, if the lines are perpendicular, the angle d is 90 degrees and = cos ai cos a2 + cos jSi cos 02 + cos 71 cos 72. (6 Id) * Cf. H. B. Phillips, Analytic Geometry, Art. 64, et seq. Chap. VII. GEOMETRICAL APPLICATIONS 81 62. Direction of the Tangent Line to a Curve. — The tangent line at a point P of a curve is defined as the limiting position PT approached by the secant PQ as Q approaches P along the curve. Let s be the arc of the curve measured from some fixed point and cos a, cos /3, cos 7 the direction cosines of the tangent drawn in the direction of increasing s. If X, y, z are the coordinates of P, ^^" ^^^' X + Ax, y + At/, z + Az, those of Q, the direction cosines of PQare Arc Ay Az PQ' PQ' PQ' As Q approaches P, these approach the direction cosines of the tangent at P. Hence Ax ,. Ax As cos a = lim 757: = lim -r— 757. • Q=P PQ As PQ On the curve, x, y, z are functions of s. Hence ,. Ax dx ,. As ,. arc ^ * ^^°^A^=d^' ^^°^PQ = ^^°^c-hord = ^- Therefore Similarly, cos a = 3- • (62a) as cos^ = ^, cosT = |. (62a) These equations show that if a distance ds is measured along the tangent, dx, dy, dz are its projections on the coordi- nate axes (Fig. 62b). Since the square on the diagonal of a * The proof that the limit of arc/chord is 1 waa given in Art. 53 for the case of plane curves with continuous slope. A similar proof can be given for any curve, plane or space, that is continuous in direction. 82 DIFFERENTIAL CALCULUS Chap. VII. rectangular parallelopiped is equal to the sum of the squares of its three edges, c?.s2 = dx^ + di/ + dz\ (62b) Fig. 62b. Example. Find the direction cosines of the tangent to the parabola X = at, y = bt, z = \ cP at the point where t = 2. At i = 2 the differentials are dx = a dt, dy = b dt, dz = \ctdt = c dt, ds = zt Vdx^ + dy^ + dz^ = ± Va^ + b^ -{- c^ dt. There are two algebraic signs depending on the direction s is measured along the curve. If we take the positive sign, the direction cosines are dx _ a dy _ b ds ~ Va^ + 62 _j_ c2 ' ds ~ Va^ + fe^-fc^' dz c ds ~ Va2 + 62 + c2 63. Equations of the Tangent Line. — It is shown in analytic geometry that the equations of a straight line Chap. VII. GEOMETRICAL APPLICATIONS 83 through a point Pi (xi, yi, Zi) with direction cosines propor- tional to A, B, C are x-xi _ y -yi _ z-zi ,^„v The direction cosines of the tangent line are proportional to dx, dy, dz. If then we replace A, B, C hy numbers pro- portional to the values of dx, dy, dz at Pi, (63) will represent the tangent line at Pi. Example 1. Find the equations of the tangent to the curve x = t, y = f, z = i^ at the point where t = I. The point of tangency is t — 1, Xi = 1, yi = 1, Zi = 1. At this point the differentials are dx : dy : dz = dt :2tdt iSe dt = I : 2 : 3. The equations of the tangent line are then X — 1 _y — 1 _ 2—1 1 2 ~~3 Ex. 2. Find the angle between the curve Sx -\- 2y — 2 z = 3, 4 x^ + y^ = 2 2^ and the line joining the origin to (1, 2, 2). The curve and line intersect at (1, 2, 2). Along the curve y and z can be considered functions of x. The differentials satisfj^ the equations Zdx + 2dy -2dz = 0, Sxdx-\-2y dy = 4:zdz. At the point of intersection these equations become 3dx-\-2dy-2dz = 0, Sdx -{- ^dy = Sdz. Solving for dx and dy in terms of dz, we get dx = 2dz, dy = —2 dz. Consequently, ds = Vdx^ -\- dy^ + dz^ = ddz and dx 2 a dy -2 dz 1 cos a = -5- = ;r, cos /3 = V- = -5- , COS 7 = -r = 5. ds 3 ds 3 ' ds 3 84 DIFFERENTIAL CALCULUS Chap. VII. The line joining the origin and (1, 2, 2) has direction cosines equal to 12 2 3» 3> 3' The angle 6 between the line and curve satisfies the equation 2-4 + 2 cos e = - — ^^^^ = 0. The line and curve intersect at right angles. EXERCISES Find the equations of the tangent lines to the following curves at the points indicated: 1. a; = sec t, y = tan t, z = at, at t = -:• 2. X = e*, y = e~*, z = fi, at < = 1. 3. X = e^ sin /, y = e* cos t, z = kt, at < = 5* 4. On the circle X = a cos 6, y = a cos (0 + 0^), z = a cos ( ^ + 5 "" J show that ds is proportional to dd. 5. Find the angle at which the helix X = a cos 6, y = a sin 6, z = kd cuts the generators of the cylinder x^ + y^ = a^ on which it lies. 6. Find the angle at which the conical helix X = t cos t, y = t sin t, z = t cuts the generators of the cone x^ + y^ = z^ on which it lies. 7. Find the angle between the two circles' cut from the sphere x'^ + y^ + z^ = 14 by the planes x — y + z = and x + y — z = 2. CHAPTER Vin VELOCITY AND ACCELERATION IN A CURVED PATH 64. Speed of a Particle. — When a particle moves along a curve, its speed is the rate of change of distance along the path. Let a particle P move along the cm^e AB, Fig. 64. Let s be the arc from a fixed point A to P. The speed of the par- ticle is then Fig. 64. Fig. 65a. 66. Velocity of a Particle. — The velocity of a particle at the point P in its path is defined as the vector* PT tangent to the path at P, drawn in the direction of motion with length equal to the speed at P. To specify the velocity we must then give the speed and direction of motion. * A vector is a quantity having length and direction. The direction is usually indicated by an arrow. Two vectors are called equal when " they extend along the same line or along parallel lines and have the i same length and direction. 85 86 DIFFERENTIAL CALCULUS Chap. VIII. The particle can be considered as moving instantaneously in the direction of the tangent. The velocity indicates in magnitude and direction the distance it would move in a unit of time if the speed and direc- tion of motion did not change. Example. A wheel 4 ft. in di- ameter rotates at the rate of 500 revolutions per minute. Find the speed and velocity of a point on its Let OA be a fixed line through the center of the wheel and s the distance along the wheel from OA to a moving point P. Then s = 2dh. The speed of P is ^ = 2^=2 (500) 2 TT = 20007r ft./mm. at at Its velocity is 2000 tt ft./min. in the direction of the tangent at P. The speeds of all points on the rim are the same. Their velocities differ in direction. 66. Components of Velocity in a Plane. — To specify a velocity in a plane it is customary to give its components, that is, its projections on the coordinate axes. If PT is the velocity at P (Fig. 66), the x-component is „^ „^ ds , ds dx dx PQ = Prcos<^=^cos0 = ^^ = ^, and the ^/-component is ^^ ^,r, . ds . ds dy dy The components are thus the rates of change of the coordinates. Since PT^ = PQ2 + QT\ Chap. VIII. VELOCITY AND ACCELERATION 87 the speed 'S expressed in terms of the components by the equation (dsV ^ (dxV (dyV \dt) \dtl "^ \dtj Fig. 66. Fig. 67. 67. Components in Space. — If a particle is moving along a space curve, the projections of its velocity on the three coordinate axes are called components. Thus, if PT (Fig. 67) represents the velocity of a point, its components are ds dx _ dx dt ds dt * PQ = PT cos a dt ds dt RT = PT cosy = ^^^ = ^- dt ds dt Smce PT^ = PQ^ + QR^ -\- RT^, the speed and compo- lents are connected by the equation (f=(tr+(fj+(fj 88 DIFFERENTIAL CALCULUS Chap. VIII. 68. Notation. — In this book we shall indicate a vector with given components by placing the components in brack- ets. Thus to indicate that a velocity has an a;-component equal to 3 and a ^/-component equal to —2, we shall simply say that the velocity is [3, —2]. Similarly, a vector in space with x-component a, ^/-component h, and 2-component c will be represented by the symbol [a, h, c]. Example 1. Neglecting the resistance of the air a bullet fired with a velocity of 1000 ft. per second at an angle of 30 degrees with the horizontal plane will move a horizontal distance x = 500t Vd and a vertical distance y = 500t- 16.1 1^ in t seconds. Find its velocity and speed at the end of 10 seconds. The components of velocity are ^ = 500 V3, § = 500 - 32.2 t at at At the end of 10 seconds the velocity is then V= [500 V3, 178] and the speed is j^ = V(500V3)' + (178)2 = 884 ft./sec." Ex. 2. A point on the thread of a screw which is turned into a fixed nut describes a helix with equations X = r cos 6, y = rsind, z = kd, S being the angle through which the screw has turned, r the radius, and k the pitch of the screw. Find the velocity and speed of the point. I Chap. VIII. VELOCITY AND ACCELERATION 89 The components of velocity are dx . ^dd dy „dd dz J dd Since -77 is the angular velocity « with which the screw is rotating, the velocity of the moving px)int is V = [— rw sin 6, rw cos d, ha] \ and its speed is ds dt = VrV gjn2 e^r^f^ cos" d + kW = « Vr^ + k^, which is constant. /:5i Fig. 69a. Fig. 69b. 69. Composition of Velocities. — By the sum of two velocities Vi and V2 is meant the velocity Vi + V2 whose components are obtained by adding corresponding compo- nents of Vi and Fo. Similarly, the difference V2 — Vi is the velocity whose components are obtained by subtracting the components of Vi from the corresponding ones of V2. Thus, if Vi = [ai, bi], Vi = [aa, 62], Fi + F2= [01 + 02,61 + 62], Vz-Vt= [02-01,62-6,]. If Fi and F2 extend from the same point (Fig. 69a), Fi + F2 is one diagonal of the parallelogram with Fi and F2 as adjacent sides and F2 — Fi is the other. In this case F2 — Fi extends from the end of Fi to the end of Fj. 90 DIFFERENTIAL CALCULUS Chap. VIII. By the product wF of a vector by a number is meant a vector m times as long as V and extending in the same direc- tion if m is positive but the opposite direction if m is negative. It is evident from Fig. 69b that the components of mF are m times those of F. F 1 The quotient — can be considered as a product — F. m m Its A = Hm . At=0 i^t components are obtained by dividing those of F by m. 70. Acceleration. — The acceleration of a particle mov- ing along a curved path is the rate of change of its velocity AF dV dt ' In this equation AF is a vector and -TT At ing the byA^ Let the particle move from the point P where the velocity is F to an adjacent point P' where the velocity is F + A F. is obtained by divid- components of AF Fig. 70a. The components of velocity will change from dx dt |t° dx .dx dt^ di dy. .dy dt'^ dt Consequently, _ Vdx dy] ^ ~ Idt ' Tt\ Subtraction and division by At give F + AF =[' dx ^^ f. dx dt'^^dt dy,Kdy' dt'^ dt . ,, r . dx . dyl A F dx . dy dt At dt At As At approaches zero, the last equation approaches ._dV_VdH d^l Chap. VIII. VELOCITY AND ACCELERATION 91 In the same way the acceleration of a particle moving in space is found to be (70b) Equations 70a and 70b express that the components of the acceleration of a moving particle are the second derivatives of Us coordinates with respect to the time. Example. A particle moves with a constant speed V around a circle of radius r. Find its velocity and acceleration at each point of the path. Let = AOP. The co- ordinates of P are x== r cos Qy The velocity of P is -[- r sm -t: , r cos Q dt „. ^ ds dd Smce s = r^, y: = V = r -fi , dt dt The velocity can therefore be written V=[—vsmd, wcos^]. Since v is constant, the acceleration is . dV Vd , . ^s d , ,,1 = |_-.cos0^. -.sm^^J. Replacing ;77 by - , this reduces to [t'- v^ ~\ v^ COS0, sua© = -[— COS0, — sin^l. r r J r Now [— COS0, — sin0] is a vector of unit length directed 92 DIFFERENTIAL CALCULUS Chap. VIIL along PO toward the center. Hence the acceleration of P is directed toward the center of the circle and has a magnitude equal to — . EXERCISES 1. A point P moves with constant speed v along the straight line y=a. Find the speed with which the line joining P to the origin rotates. 2. A rod of length a sUdes with its ends in the x- and y-axes. If the end in the x-axis moves with constant speed v, find the velocity and speed of the middle point of the rod. 3. A wheel of radius a rotates about its center with angular speed « while the center moves along the x-axis with velocity v. Find the velocity and speed of a point on the perimeter of the wheel. 4. Two particles Pi (xi, yi) and Pt (xz, 2/2) move in such a way that xi = 1 + 2 <, r/i = 2 - 3 <2, X2 = 3 + 2 <2, y^= - 4 i». Find the two velocities and show that they are always parallel. 5. Two particles Pi ( xi, yi, Zi) and P2 (x2, 2/2, Z2) move in such a way that Xi = o cos 6, yi = a cos (6 + i ir), Zi = a cos (^ + § jt), Xi = a sin d, yz = a sin {9 + i ir) , Zi = a ain {9 -{■ i it) . Find the two velocities and show that they are always at right angles. 6. A man can row 3 miles per hour and walk 4. He wishes to cross a river and arrive at a point 6 miles further up the river. If the river is If miles wide and the current flows 2 miles per hour, find the course he shall take to reach his destination in the least time. 7. Neglecting the resistance of the air a projectile fired with velocity [a, b, c] moves in t seconds to a position X = at, y = bt, z = ct — \ gfi. Find its speed, velocity, and acceleration. 8. A particle moves along the parabola 3? = ay in such a way that ■J- is constant. Show that its acceleration is constant. dt 9. When a wheel rolls along a straight line, a point on its circum- ference describes a cycloid with parametric equations X = o (^ — sin 0), y = a (I — cos <t>), a being the radius of the wheel and </> the angle through which it hab rotated. Find the speed, velocity, and acceleration of the moving point. Chap. Vni. VELOCITY AND ACCELERATION 93 10. Find the acceleration of a particle moving with constant speed v along the cardioid r = a {1 — cob 6). 11. If a string is held taut while it is unwoxind from a fixed circle, its end describes the curve X = a cos d + a d an 0, y = a tan d — a 9 coe d, e being the angle subtended at the center by the arc unwound. Show that the end moves at each instant with the same velocity it would have dd if the straight part of the string rotated with angular velocity -t- about the point where it meets the fixed circle. 12. A piece of mechanism consists of a rod rotating in a plane with constant angular velocity w about one end and a ring sUding along the rod with constant speed v. (1) If when t = the ring is at the center of rotation, find its position, velocity, and acceleration as functions of the time. (2) Find the velocity and acceleration immediately after t = ti, if at that instant the rod ceases to rotate but the ring continues sliding with unchanged speed along the rod. (3) Find the velocity and acceler- ation immediately after / = /i if at that instant the ring ceases sliding but the rod continues rotating. (4) How are the three velocities re- lated? How are the three accelerations related? 13. Two rods AB, BC are hinged at B and lie in a plane. A is fixed, AB rotates with angular speed « about A, and BC rotates with angular speed 2 u about B. (1) If when t = 0, C lies on AB produced, find the path, velocity, and acceleration of C. (2) Find the velocities and accelerations immediately after < = /i if at that instant one of the rotations ceases. (3) How are the actual velocity and acceleration related to these partial velocities and accelerations? 14. A hoop of radius a rolls with angular velocity on along a horizon- tal line, while an insect crawls along the rim with speed cuat. If when t = the insect is at the bottom of the hoop, find its path, velocity, and acceleration. The motion of the insect results from three simultaneous actions, the advance of the center of the hoop with speed cu^, the rota- tion of the hoop about its center with angular speed a>i, and the crawl of the insect advancing its radius with angular speed wi. Find the three velocities and accelerations which result if at the time t = ti two of these actions cease, the third continuing unchanged. How are the actual velocity and acceleration related to these partial velocities and accelera- tions? CHAPTER IX ROLLE'S THEOREM AND INDETERMINATE FORMS 71. Rolle's Theorem. — // /' (x) is continuous, there is at least one real root of f (x) = between each pair of real roots off{x) =0. To show this consider the curve y=fix)- Let / (x) be zero at X = a and x = b. Be- tween a and b there must be one or more ^^^- ^^^- points P at maximum distance from the x-axis. horizontal and so \t such a point the tangent is | = /'(x) = 0. That this theorem may not hold if /' (x) is discontinuous is shown in Figs. 71b and 71c. In both cases the curve Y Fig. 71b. Fig. 71c. crosses the x-axis at a and b but there is no intermediate point where the slope is zero. n Chap. IX. ROLLE'S THEOREM 95 Example. Show that the equation a:' + 3x - 6 = cannot have more than one real root. Let Then /' (x) = 3 a;2 + 3 = 3 (x^ + 1). Since /' (x) does not vanish for any real value of x, / (x) =0 cannot have more than one real root ; for if there were two there would be a root of /' (x) = between them. 72. Indeterminate Forms. — The expressions ^, -, 0.x, X - 00, 1-, 0^ 00° are called indeterminate forms. No definite values can be assigned to them. If when X = a a function / (x) assumes an indeterminate form, there may however be a definite limit lim/(x). x=a In such cases this limit is usually taken as the value of the function at x = a. For example, when x = the function 2x _0 X "0* It is e\'ident, however, that Urn — = lim (2) = 2. x=0 X This example shows that an indeterminate form can often be made definite b}' an algebraic change of form. aa 73. The Forms -r and - . — We shall now show that, if 00 ' f (x) for a particular value of the variable a fraction „\ \ assumes F(x) the form - or — , numerator and denominator can be replaced 96 DIFFERENTIAL CALCULUS Chap. IX. by their derivatives without changing the value of the limit approached by the fraction as x approaches a. 1. Let /' {x) and F' (x) be continuous between a and 6 Iff («) = 0, F (a) = 0, and F (6) is not zero, there is a number Xi between a and b such that fib)_f(x^) Fib) F'(xi) To show this let ^^ = R. Then Fib) (73a) fib) -RFib)=0, Consider the function / ix) - RF ix). This function vanishes when x = b. Since / (a) = 0, F (a) = 0, it also vanishes when x = a. By RoUe's Theorem there is then a value Xi between a and b such that /' ixO - RF' ixO = 0. Consequently, Fib) "" F'ix^)' which was to be proved. 2. Let/' ix) and F' ix) be continuous near a. /// (a) = and F (a) = 0, then (73b) r f(x) ,. fix) :,=„ i^ (a:) x=« F' ix) For, if we replace b by x, (73a) becomes fix) ^fix,) Fix) F'ix,y Xi being between a and x. Since Xi approaches a as x ap- proaches a, iS F (x) i;E F' ixO S F' (x) 3. In the neighborhood of x = a, let /' ix) and F' ix) be Chap. IX. ROLLE'S THEOREM 97 continuous at all points except x = a. If f (x) and F (x) approach infinity as x approaches a, "™ F (X) "2 F' (I) To show this let c be near a and on the same side as x. Since f (x) — f (c) and F {x) — F (c) are zero when x = c, by Theorem 1, f{xr)^ f{x)-f(c) ^ /(x)' fix) F'(xi) F (x) - F(c) F(x) F(c) ' where Xi is between x and c. As x approaches a, / (x) and F (x) increase indefinitely. The quantities / (c)/f (x) and F (c)/F (x) approach zero. The right side of this equation therefore approaches Since Xi is between c and a, by taking c sufficiently near to a the left side of the equation can be made to approach .2 F' (x) Since the two sides are always equal, we therefore conclude that "2 F (x) 'iS f ' (x) sin X Example 1. Find the value approached by as x approaches zero. Since the numerator and denominator are zero when x = 0, we can apply Theorem 2 and so get ,. sin X ,. cosx lim = lim — - — = 1. x=0 ^ 1=0 1 Ex. 2. Find the value of lim -. r^ • x=, (tt — x)* 98 DIFFERENTIAL CALCULUS Chap. IX. When X = T the numerator and denominator are both zero. Hence ,. l + cosx ,. (— sinx) hm -, Tj- = hm -^^. ^ = -• x=^ (tt - xf x=T -2 {tt — x) Since this is indeterminate we apply the method a second time and so obtain ,. sin X _ y cosx _ 1 x=x ^ [tt X) x=ic ■^ ^ The value required is therefore \. + Q fl ^ O* Ex. 3. Find the value approached by — as x ap- tan X proaches x • id TT When X approaches - the numerator and denominator of this fraction approach oo. Therefore, by Theorem 3, lim tan 3 a: ,. 3 sec^ 3 a; ,. 3 cos^ x . T -7 = lim ^ = lim — ^-^r- ' a;=2 tanx sec^x cos^Sx TT 0- When X is replaced by ^ the last expression takes the form j: • Therefore ,. 3 cos^ aj ,. 6 cos x sin x hm — TTir- = hm cos^ 3 a; 6 cos 3 x sin 3 x cos^ X — sin^ X 1 = lim 3 (cos2 3 X - sin2 3 x) 3 74. The Forms • oo and oo — oo. — By transforming e expression t^ For example, the expression to a fraction it will take the form r: or — a; In a; Chap. DC. ROLLE'S THEOREM 99 has the form • oo when x = 0. It can, however, be written , In a; I xmx = -r— » 00 which has the form — 00 The expression sec X — tan x has the form x — oo when x = ^- It can, however, be written 1 sin a; 1 — sin x sec X — tan x = = , cos X cos x cos X which becomes ^ when a: = ^ • 75. The Forms 0**, 1* , oo°. — The logarithm of the given function has the form • x. From the limit of the log- arithm the limit of the function can be determined. 1 Example. Find the limit of (1 + x)' as x approaches zero. 1 Let i/ = (l+x)'. Then 1 1 1 /I I \ In (1 + x) \ny =-ln(l + x) =^ '- TNTien x is zero this last expression becomes ^ . Therefore ^^!n(l+£) _ ,.„ 1 ^ J x-0 a: 1 + X The Umit of In y being 1, the limit of y is e.] 100 DIFFERENTIAL CALCULUS Chap. IX. EXERCISES 1. Show by RoUe's Theorem that the equation X* — 4x — 1 = cannot have more than two real roots. Determme the values of the following limits: 3^-1 'ITxio - l' x" - 1 'ir X — 1 1 — cos X x=o sin X 2. Lira 3. Lim a;=l 4. Lim x=0 6. Lim X — a . Lim : • 1=0 a; — sm a; _ T • x^ cos X 7. Lim -. x=0 cos X — 1 8. Lim^5j£^. x=3 X —3 rt T • In cos a; 9. Lim • x=0 X 10. Lim 11. Lim =^2 (x - 2)2 1 + cos X — sin a; cos X (2 sin x — 1) 12. Lim ^"gio (s^" ^ ~ sin a) x=a logio (tan X — tan a) 13. Li^6sinx-6x + x3^ a;=0 a;2 14. Lim ^^^^ <A - 2 tan <^ ^ 1 + cos 4 </> 16. Limi^. xdro cot .r 16. Lim^. 17. Lim sec S X — X If sec X — X 18. Liml±^^. x=^|sec(x + ^j 19. Lim a; cot x. 20. Lim tan x cos 3 x. 21. Lim (x + a)lnfl+-V X = CI0 \ X/ 22. Lim (x - 3) cot (ttx). a;=3 23. Limn ffx + ^)~f(x)' n=oo L V n/ _ f 1 24. Lim x^e**. x=0 26. Lim ). a:=0 \a; ^ - 1/ 26. Lim (cotx — Inx). x=0 27. Lim tanx — : r-r— | .tL smx— sm*a;J ^~9 28. Lira X*. x=0 29. Lim(sinx)**°*. -=^1 1 30. Lim (1 + ax)*. x=0 1 81. T<im (x"* — a*") in «. Z»ao CHAPTER X SERIES AND APPROXIMATIONS 76. Mean Value Theorem. — // / (x) and /' (x) are can- tinuousfram x = a to x = b, there is a value X\ between a and b such thai fib)-f(a) b — a = /' (xi). (76T To show this consider the curve y = f(x). Since f (a) and / (6) are the ordinates at x = a and x = b, f(b)-f(a) ^ ^ ^j ^^^^^ ^^ b — a On the arc AB let Pi be a point at maximum distance from Fig. 76. the chord. The tangent at Pi will be parallel to the chord and so its slope/' (xi) will equal that of the chord. Therefore b — a which was to be proved. Replacing 6 by x and solving for/ (x), equation (76) becomes Six) =/(a)4-(x-a)/'(xi). 101 102 DIFFERENTIAL CALCULUS Chap. X. Xi being between a and x. This is a special case of a more general theorem which we shall now prove. 77. Taylor's Theorem. — If f (x) and all its derivatives used are continuous from a to x, there is a value Xi between a and x such that f (x) =f{a) + {x- a)r (a) +^^^r (a) + ^^^ /'" («)+••• +^^^ /" (^>). To prove this let <i>{x) =f{x)-f{a)-{x-a)f'{a) _ {x - aY . ^ _ . . . _ {x - ay-' . 2! ^ ^""^ (n- 1)! ^ ^"^* It is easily seen that <t> (a) = 0, <l>' (a) = 0, <t>" (a) = 0, . . . r-' (a) = 0, 0" (x) = /» (x). When x = a the function <l>(x) {x - a)* therefore assumes the form ^. By Art. 73 there is then a value 2i between a and x such that <f> (x) <t>' (zi) (x — aY n {zi — a)"'^ This new expression becomes ^ when z, = a. There is con- sequently a value Zi between Zi and a (and so between x and a) such that <f>'(zi) ^ <i>" (z,) n {zi — a)"-i n (n - 1) (22 - a)""" ^ A continuation of this argument gives finally <i>{x) ^ 0" (Zn) ^ /" (Zn) V (x — a)" n! n! * ' Chap. X. SERIES AND APPROXIMATIONS 103 2» being between x and a. If Xi = 2„ we then have Equating this to the original value of <t> (x) and solving for / (x), we get f{x)=f{a) + (x-a)r(ay + (^V'(a)+...+^V"(xO. which was to be proved. Example. Prove hix=(x-l) 2~ + ~~3 4^7"* where Xi is between 1 and x. When X = 1 the values of In x and its derivatives are /(x)=bi(x). /(1)=0. /' (X) = l^ r (1) = 1, r(x) = -i, r(i) = -1, r'(x) = |. r'(i) = 2, r"(x) = -^, r"(xo = -(!),• Taking a = 1, Taylor's Theorem gives hix = + l(x-l)-i(x-l)2 + ?(x-l)3-^^^^\ which is the result required. 78. Approximate Values of Functions. — The last term in Taylor's formula (x — a)" , , , _ 104 DIFFERENTIAL CALCULUS Chap. X. is called the remainder. If this is small, an approximate value of the function is S(x)=fia) + ix-a)ria) (x - ay „ , . (x - g)"-^ . , , . + -^j / (a) + • • • + -(^j-JTiyr /" («)' the error in the approximation being equal to the remainder. To compute/ (x) by this formula, we must know the values of/ (a), /' (a), etc. We must then assign a value to a such that /(^)» /' (^)> ^^c., are known. Furthermore, a should be as close as possible to the value x at which f (x) is wanted. For, the smaller x — a, the fewer terms (x — a)^, (x — aY, etc., need be computed to give a required approximation. Example 1. Find tan 46° to four decimals. The value closest to 46° for which tan x and its derivatives are known is 45". Therefore we let a = t' 4 /(a;) = tana;, "^W^^' . f (x)=sec«x, ^'(S)^^' f" (x) = 2 sec2 X tan x, f" (|) = 4, f" (x) = 2 sec* X -\- A. sec^ x tan^ x. Using these values in Taylor's formula, we get tan. = l + 2(.-|) + i(x-.)%/^(.-|; and tan46« = l + 2(j|5) + 2(j|5)'= 1.0355 approximately. Since Xi is between 45° and 46°, /'" (xi) does not differ much from f" (45°) = 8 + 8 = 16. Chap. X. SERIES AND APPROXIMATIONS 105 The error in the above approximation is thus very nearly 6" (l^) < 3750? ^ 40:000 = 0-0^25. It is therefore correct to 4 decimals. Ex. 2. Fmd the value of e to four decimals. The only value of x for which e* and its derivatives are known is a; = 0. We therefore let a be zero. fix) = e-, / (x) = e*, r{x)=^, , f- (x) = e*, /(O) = 1, /' (0) = 1, /" (0) = 1, , /» (xx) = e^. By Taylor's Theorem, ^ = i + x + 2i + 3i+ • • • + (;rriy! + ^- Letting x = 1, this becomes 6 = 1+1+^ + 57+ • • • + 7 TV-i + — 1- 2! 3! (n — 1)! n! In particular, if n = 2, e = 2 + I e*'. Since Xi is between and 1. e is then between 2| and 2 + ^ e, and therefore between 2^ and 4. To get a better ap- proximation let n = 9. Then e = l + l + ^ + Jr+--- +3^2-7183 approximately, the error being 9! ^ 9] ^ 9] ^ .00002. The value 2.7183 is therefore correct to four decimals. EXERCISES Determine the values of the following functions correct to four decimals: 1. sin 5". 5. sec (10°). 2. cos 32°. 6. hi (t?5). 3. cot 43°. 7. VI. 4. tan 58°. 8. tan"! i^). 9. Given In 3 = 1.0986, hi 5 = 1.6094, find hi 17. 106 DIFFERENTIAL CALCULUS Chap. X. 79. Taylor's and Maclaurin's Series. — As w increases indefinitely, the remainder in Taylor's formula Rn = ■^^— ^/" (Xx) n often approaches zero. In that case S{x)=\\m\f{a) + {x-a)r{a)+ • • • + ^T'^lXj '"' («)| This is usually written Six) =/ (a) + (X ~ a)r (a) +^^^V" («) 3! the dots at the end signifying the limit of the sum as the number of terms is indefinitely increased. Such an infinite sum is called an infinite series. This one is called Taylor's Series. In particular, if a = 0, Taylor's Series becomes /(a;)=/(0)+a;f(0)+|r(0)+|^r'(0)+ .... This is called Maclaurin's Series. Example. Show that cos x is represented by the series x^ , X* x* , cosx=l-2| + 4j-^+ • • • . The series given contains powers of x. This happens when a = 0, that is, when Taylor's Series reduces to Maclaurin's. / (x) = cos X, / (0) = 1, /' (x)= -sinx, /(0)=0, r(x) = -cosx, /"(0)=-l, /'" (x) = sin X, f" (0) = 0, f" (x) = cos X, J"" (0) = 1. These values give cos X = 1 - 2j + 41 - • • • ± -, /" (a;i). Chap. X. SERIES AND APPROXIMATIONS 107 The nth derivative of cos x is ±cos x or isin x, depending on whether n is even or odd. Since sin x and cos x are never greater than !,/„ (xi) is not greater than 1. Furthermore ^ _ X X X X n! ~ l'2'3* ' ' ' n can be made as small as you please by taking n sufl&ciently large. Hence the remainder approaches zero and so , X2 , X* x^ , cosx = l-2! + 4!~6!"^ ' ' * ' which was to be proved. EXERCISES 1. smx = x-3J + ^-yj + - • • . - - , ^ 2x» 4x« 4x6 ,8x^, 6. (a + x)" = a» + no»-»x + " %7 ^^ """^ + • • • , if |xl*< |o|. 8. lnx = ln3+^-^^^' + ^^^ .iflx-3I<l. 9. ln(x + 5)-ln6+^^-^|^'+^|^' ,if|x-l|<l. 80. Convergence and Divergence of Series. — An in- finite series is said to converge if the sum of the first n terms approaches a Umit as n increases indefinitely. If this sum does not approach a limit, the series is said to diverge. The series for sin x and cos x converge for all values of x. The geometrical series a -\- ar -{- ar^ -\- ar^ -\- ar* -\- ■ • • * The symbol |x| is used to represent the numerical value of x with- out its algebraic sign. Thus, | — 3 1 = | 3 | = 3. 108 DIFFERENTIAL CALCULUS Chap. X converges when r is numerically less than 1. For the sum of the first n terms is Sn = a -]r ar -\- ar^ + ...-{- ar""-^ = a \ ~ ^" • 1 — r If r is numerically less than 1, r" approaches zero and Sn approaches 1 — r as n increases indefinitely. The series 1-1 + 1-1 + 1-1+.. . is divergent, for the sum oscillates between and 1 and does not approach a limit. The geometrical series 1 + 2 + 4 + 8+16+ • • • diverges because the sum increases indefinitely and so does not approach a limit. 81. Tests for Convergence. — The convergence of a series can often be determined from the problem in which it occurs. Thus the series 1 — ^ 4- ^ _ ^^ _j_ 2!"^ 4! 6"!"^ ' ' * converges because the sum of n terms approaches cos aj as n increases indefinitely. The terms near the beginning of a series (if they are all finite) have no influence on the convergence or divergence of the series. This is determined by terms indefinitely far out in the series. 82. General Test. — Fcrr the series Wl + W2 + W3 + • • • + Wn + • • • to converge it is necessary and sufficient that the sum of terms beyond Un approach zero as n increases indefinitely. For, if the series converges, the sum of n terms must ap- proach a limit and so the sum of terms beyond the nth must approach zero. Chap. X. SERIES AND APPROXIMATIONS 109 83. Comparison Test. — A series is convergent if beyond a certain point its terms are in numerical value respectively less than those of a convergent series whose terms are all positive. For, if a series converges, the sum of terms beyond the nth will approach zero as n increases indefinitely. If then another series has lesser corresponding terms, their smn will approach zero and the series will converge. 84. Ratio Test. — If the ratio -^ of consecutive terms ^ Un ^ approaches a limit r as n increases indefinitely, the series ^1 + 1^2 + ^3+ • • • +W„ + Un+l + • • • IS convergent if r is numerically less than 1 and divergent if r is numerically greater than 1 . Since the limit is r, by taking n sufficiently large the ratio of consecutive terms can be made as nearly r as we please. If r < 1, let ri be a fixed number between r and 1. We can take n so large that the ratio of consecutive terms is less than ri. Then Wn+l < riMn, Mn+2 < ^Mn+i < r^U,,, etC, Beyond w„ the terms of the given series are therefore less than those of the geometrical progression Un + riUn + rihin + • • • which converges since rx is numerically less than 1. Con- sequently the given series converges. If, however, r is greater than 1, the terms of the series must ultimately increase. The terms do not then approach zero and their sum cannot approach a limit. Example. Find for what values of x the series X + 2 x2 ^ 3 3^ ^ 4 3^ _^ , . . converges. The ratio of consecutive terms is Wn+i (n + 1) x"+i Un nx" -(}-¥■ 110 DIFFERENTIAL CALCULUS Chap. X. The limit of this ratio is r = lim [I + -] X = X. n=oo \ n/ The series will converge if x is numerically less than 1. 85. Power Series. — A series of powers of {x — a) of the form P (x) = oo + ai (x — a) + 02 (x — a)2 + a3 (x — a)' + • • • , where a, ao, ai, a^, etc., are constants, is called a power series. If a power series converges when x = b, it will converge for all values of x nearer to a than b is, that is, such that \x — a\ < \b — a\. In fact, if the series converges when x = b, each term of Oo + ai (6 - a) + 02 (6 - o)2 + as (6 - a)3 -I- • • . will be less than a maximum value M, that is, |a„ {b - o)"| < M. Consequently, The terms of the series oo + oi (x — o) + 02 (a; — 0)2 -{- as (x — a)^ + • • • are then respectively less than those of the geometrical series in which the ratio is \x — a\ \b-a\ If then |x — o|<|6 — a|, the progression and consequently the given series will converge. // o power series diverges when x = b, it will diverge for all valu£s of X further from a than b is, that is, such that jr — o| > |b — to|, Chap. X. SERIES AND APPROXIMATIONS 111 For it could not converge beyond b, since by the proof just given it would then converge at b. This theorem shows in certain cases why a Taylor's Series is not convergent. Take, for example, the series X^ 3^ X^ ln(l + a;) =^-2!~^3]~4]'*" ' ' * * As X approaches — 1, In (1 + x) approaches infinity. Since a convergent series cannot have an infinite value, we should expect the series to diverge when x = — 1. It must then diverge when x is at a distance greater than 1 from a = 0. The series in fact converges between x = — 1 and x = 1 and diverges for values of x numerically greater than 1. 86. Operations with Power Series. — It is shown in more advanced treatises that convergent series can be added, subtracted, multiplied and divided like polynomials. In case of division, however, the resulting series will not usually converge beyond a point where the denominator is zero. Example. Express tan x as a series in powers of x. We could use Maclaurin's series with / (x) = tan x. It is easier, however, to expand sin x and cos x and divide the one by the other to get tan x. Thus x^ x° smx ^~6"'~120~'** ,x3,2x«, tanx= = z -T =^+-5-+^rT-+ • * * • cos X i_^_i_^_ 315 EXERCISES 1. Show that l°(l-^) = l^(l+^)-l^(l-^)=2(x+| + | + y+ • •• y and that the series converges when \x\ < 1. 2. By expanding cos 2 x, show that . , 1 — cos 2 X - X* _, ar» , -. !« sin3^=— ^ . = 22;-23^ + 2«gj . Prove that the series converges for all values of x. 112 DIFFERENTIAL CALCULUS Chap. X. 3. Show that and that the series converges for all values of x. 4. Given / (x) = sm"i x, show that Expand this by the binomial theorem and determine /" (z), etc., by differentiating the result. Hence show that , 1 x» 1 3a:5 1 3 5a;7 8m-x=x + -^ + -.--^+^.^.gy+... and that the series converges when |x| < L 6. By a method similar to that used in Ex. 4, show that tan-i x = x--5-+g — y+-'« and that the series converges when |x| < L 6. Prove sec X = = l+-^+-^x*+-««. cos X 2 24 For what values of x do you think the series converges? I CHAPTER XI PARTIAL DIFFERENTIATION 87. Functions of Two or More Variables. — A quantity u is called a function of two independent variables x and y, u =fix,y), if u is determined when arbitrarj^ values (or values arbitrary within certain limits) are assigned to x and y. For example, U = VI — X2 _ y2 is a function of x and y. If w is to be real, x and y must be so chosen that x^ + y^ is not greater than 1. Within that limit, however, x and y can be chosen independently and a value of u will then be determined. In a similar way we define a function of three or more in- dependent variables. An illustration of a function of vari- ables that are not independent is furnished by the area of a triangle. It is a function of the sides a, b, c and angles A , B, C of the triangle, but is not a function of these six quantities considered as independent variables; for, if values not be- longing to the same triangle are given to them, no triangle and consequently no area will be determined. The increment of a function of several variables is its in- crease when all the variables change. Thus, if u=f{x, y), u + Au = f{x -\- Ax,y -h Ay) and so Aw = / (x + Ax, y + Ay) - / (x, y). A function is called continuous if its increment approaches zero when all the increments of the variables approach zero. 113 114 DIFFERENTIAL CALCULUS Chap. XI. 88. Partial Derivatives. — Let u =f(x, y) be a function of two independent variables x and y. If we keep y constant, w is a function of x. The derivative of this function with respect to x is called the 'partial derivative of u with respect to x and is denoted by fx °' ^'^^^y^' Similarly, if we differentiate with respect to y with x con- stant, we get the partial derivative with respect to y denoted by — or fy{x,y). For example, if u = x^ -\- xy — y"^, then du ^ . du - a^=2x + y, Yy^^-^y- Likewise, if w is a function of any number of independent variables, the partial derivative with respect to one of them is obtained by differentiating with the others constant. 89. Higher Derivatives. — The first partial derivatives are functions of the variables. By differentiating these functions partially, we get higher partial derivatives. For example, the derivatives of — with respect to x and y are Similarly, dx \dx/ dx^' dy \dx) dydx d /du\ _ d^u d /du\ _ d^u dx \dy) ~ dxdy' dy \dy) ~ dy^ It can be shown that dhi _ dhi dxdy dydx* Chap. XI. PARTIAL DIFFERENTIATION 115 if both derivatives are continuous, that is, partial derivatives are independent of the order in which the differentiations are 'performed* Example, u — xhf + xy^. ^''^ = A (x2 + 2xy) =2x + 2y, ^ = ^(x2 + 2a^)=2x. dxdy dx ' ' "" "' dy^ dy 90. Dependent Variables. — It often happens that some of the variables are functions of others. For example, let w = x^ + ?/' + z* and let 2 be a function of x and y. When y is constant, z will be a function of x and the partial derivative of u with respect to X will be --=2x4-2 z— • dx dx Similarly, the pari;ial derivative with respect to y with x con- stant is — = 2y + 2z — • dy '^ dy If, however, we consider z constant, the partial derivatives are du ^ du ^ — = 2 X. — = 2 V. dx ' dy ^ The value of a partial derivative thus depends on what quantities are kept constant during the differentiation. The quantities kept constant are sometimes indicated by subscripts. Thus, in the above example dxly,^ ' \dx/y dx V^x/a ^ dx * For a proof see Wilson, Advanced Calculus,^ 50. 116 DIFFERENTIAL CALCULUS Chap. XI. Then — will repre- It will usually be clear from the context what independent variables u is considered a function of. sent the derivative with all those variables except x constant. Example. If a is a side and A the opposite angle of a right ^ triangle with hypotenuse c, find (—] • \dcjA From the triangle it is seen that c a = c sin ^. Fig. 90. Differentiating with A constant, we get da . . — = sm A, dc which is the value required. 91. Geometrical Representation. — Let z = f (x, y) be the equation of a surface. The points with constant y- coordinate form the curve AB (Fig. 91a) in which the plane y = constant intersects the surface. In this plane z is the vertical and x the horizontal coordinate. Consequently, dz dx is the slope of the curve AB at P. Similarly, the locus of points with given x is the curve CD and dy is the slope of this curve at P. Example. Find the lowest point on the paraboloid 2 = x2 + i/2 — 2a: — 4y + 6. At the lowest point, the curves AB and CD (Fig. 91b) will have horizontal tangents. Hence |^=2x-2 = 0, dx 1 = 2.-4 = 0. Chap. XI. PARTIAL DIFFERENTIATION 117 Consequently, x = I, y = 2. These values substituted in the equation of the surface give z = I. The point required is then (1, 2, 1). That this is really the lowest point is shown by the graph. Fig. 91a. Fig. 91b. EXERCISES In each of the following exercises show that the partial derivatives satisfy the equation given: 1. u = x' + y^ x + y' du , du ' dx dy 2. e = (x + a)(, + 6). g | = z. 3. z = (x' + j/»)", dz dz ^^~ dy dx 4. M = In (x« H- xy + ?/«), x--- + y—- = 2. dx ay z X y' 6. u = tan"' 7. u = (^)' 1 du . du , du _ dx» ^ dy^ S^u . cShx dhi. Vx* + y^ + 2*' <^* ^y* ^2* In each of the following exercises verify that dx dy dy dx ■jo [dxjy 118 DIFFERENTIAL CALCULUS Chap. XL 8. u = -. 10. u ^ sin (x + y). 9. M = In (x^ + 2/2)^ 11, ^ _ ^yg^ 12. Given i; = V^~+^i'+^, verify that 9^ _ dh) dx dy dz dz dy dx Prove the following relations assuming that 2 is a function of x and y: 15. . = (. + .).^, | + |.(l+. + ,)(l+| + |)e«^. .. /aw aw\ / dz dz\ U.u = xyz, z^x--y-) = u\x~-y-y ^ ' dxdy \dxdy^ dx dyj ^ .- a / dtt 3z\ d^w a^z 16. • — l Z — — U — \ = Z — : — U • dx\ dx dxj dx^ dx^ 17. If X = r cos 8, y = r sin d, show that fdx^ 18. Let a and b be the sides of a right triangle with hypotenuse e and opposite angles A and B, Let p be the perpendicular from the vertex of the right angle to the hypotenuse. Show that ') --■ ijA C 19. If K is the area of a triangle, a side and two adjacent angles of which are c, A, B, show that ldK\ ^h2 ldK\ ^a^ 20. If K is the area of a triangle with sides a, b, c, show that If) -loot A. \dajb,c 2 21. Find the lOwest point on the surface z = 2x^ + y'^ + Sx -2y + 9. 22. Find the highest point on the surface z = 2 y - x^ + 2 xy - 2 y^ + 1. 92. Increment. — Let u = f (x, y) be a function of two independent variables x and y. When x changes to a; + Aa; and y to y -\- Ay, the increment of u is ^u=f{x + ^x,y + Ay) - f (x, y). (92a) [dajb c^' [daj Chap. XI. PARTIAL DIFFERENTIATION 119 By the mean value theorem, Art. 76, / (x + Ax, y + ^y) =fix,y + Ay) + Az/, (xi, y + Ay), Xi lying between x and x + Ax. Similarly / (x, y + Ay) = / (x, y) + lyfy (x, yi), yi being between y and y + Ay. Using these values in (92a), we get Aw = Ax/x (xi, y + Ay) + Ay/„ (x, yi). (92b) As Ax and Ay approach zero, Xi approaches x and yi ap- proaches y. If /x (x, y) and /„ (x, y) are continuous, a.. /. (iCi, y + Ay) = /, (x, y) + ci = — + ei, a-. /v (a;, yi) = /» (a;, y) + €2 = ^ + ^, ci and €2 approaching zero as Ax and Ay approach zero. These values substituted in (92b) give Aw = ^ Ax + ^ Ay + ei Ax + €2 Ay. (92c) The quantity V- Ax + — Ay dx dy is called the principal part of Aw. It differs from Au by an amount ei Ax + ez Ay. As Ax and Ay approach zero, ci and e» approach zero and so this difference becomes an indefinitely small fraction of the larger of the increments Ax and Ay. We express this by sajdng the principal part differs from Aw by an infinitesimal of higher order than Ax and Ay (Art. 9). When Ax and Ay are sufficiently small this principal part then gives a satisfactory approximation for Aw. Analogous results can be obtained for any number of in- dependent variables. For example, if there are three inde- pendent variables x, y, z, the principal part of Aw is dw . , du . , du . — Ax+— Ay + — Az. dx dy dz In each case, if the partial derivatives are continuous, the 120 DIFFERENTIAL CALCULUS Chap. XI. principal part differs from Am by an amount which becomes indefinitely small in comparison with the largest of the in- crements of the independent variables as those increments all approach zero. Example. Find the change in the volume of a cylinder when its length increases from 6 ft. to 6 ft. 1 in. and its diam- eter decreases from 2 ft. to 23 in. Since the volume is y = irr^, the exact change is Av = TT (1 - ^y (6 + tV) - x. p. 6 = -0.413 TTCU. ft. The principal part of this increment is I A. + I AA = 2 . rt (- 1) + . r^ (j|) = -0.417 . cu', ft. 93. Total Differential. — If m is a function of two inde- pendent variables x and y, the total differential of u is the principal part of Aw, that is, This definition applies to any function of x and y. The particular values u = x and u = y give dx = Ax, dy = Ay, (93b) that is, the differentials of the independent variables are equal lo their increments. Combining (93a) and (93b), we get <^w = ^ fix + ^ dy. (93c) We shall show later (Art. 97) that this equation is valid even if x and y are not the independent variables. The quantities , du , , du , dxU = r- dx, dyU = -r- dy dx dy ^ are called partial differentials. Equation (93c) expresses tha t the total differential of a function is equal to the sum of the partial differentials obtained by letting the variables change one at a time. Chap. XI. PARTIAL DIFFERENTIATION 121 Similar results can be obtained for functions of any number of variables. For instance, if m is a function of three inde- pendent variables x, y, z, du = ^- Ax + ■— Ay -\- —- Az. dx dy dz The particular values u = x, u = y,u = z give dx = Ax, dy = Ay, dz = Az, The pre\aous equation can then be written du=pdx-\-pdy+^dz (93d) dx dy dz and in this form it can be proved valid even when x, y, z are not the independent variables. Example 1. Find the total differential of the function w = x^ + xy^. By equation (93c) , du . du J du = -r~dx + --dy dx dy = (2x2/ + y') dx+(x'-{-2xy) dy. Ex. 2. Find the error in the volume of a rectangular box due to small errors in its three edges. Let the edges be 5, y, z. The volume is then V = xyz. The error in v, due to small errors Ax, Ay, Az in x, y, z, is Av. If the increments are sufficiently small, this will be approxi- mately dv = yz dx -\- xz dy + xy dz. Dividing by v, we get dv _ yz dx -{- xz dy -\- xy dz V xyz dx , dy , dz = h — H xyz dx Now — expresses the error (2x as a fraction or percentage of x. 122 DIFFERENTIAL CALCULUS Chap. XI. The equation just obtained expresses that the percentage error in the volume is equal to the sum of the percentage errors in the edges. If, for example, the error in each edge is not more than one per cent, the error in the volume is not more than three per cent. 94. Calculation of Differentials. — In proving the formu- las of differentiation it was assumed that u, v, etc., were functions of a single variable. It is easy to show that the same formulas are valid when those quantities are functions of two or more variables and du, dv, etc., are their total differentials. Take, for example, the differential of uv. By (93c) the result is d (uv) = -r- (uv) du -\--r- (uv) dv = V du -\- u dv, du dv which is the formula IV of Art. 17. Example, u = ye^ -\- ze". Differentiating term by term, we get du = y^ dx -\- ^ dy -}- ze" dy + e" dz. We obtain the same result by using (93d) ; for that formula gives ^-1/ Silt Sxt du = -;- dx -^ -T- dy + -r- dz = yei' dx -\- (e' + ze^) dy + e« dz. dx dy dz 95. Partial Derivatives as Ratios of Differentials. — The equation dxU = -r- dx dx shows that the partial derivative — is the ratio of two dif- ferentials dxU and dx. Now dxU is the value of du when the same quantities are kept constant that are constant in the calculation of ^r-. Therefore, the partial derivative ^r- is the dx dx Chap. XI. PARTIAL DIFFERENTIATION 123 value to which -r- reduces when du and dx are determined with dx the same quantities constant that are constant in the calculation . du Example. Given u = x^ -\- y^ -{- z^, v = xyz, find (^) * Dififerentiating the two equations with v and z constant, we get du = 2 X dx -\- 2 y dy, = yz dx -\- xz dy. Eliminating dy, du = 2xdx-2^ dx = 2 f ^' ~ ^' ) dx. X \ X / Under the given conditions the ratio of du to dx is then du ^ 2 (x^ - y^y dx X Since i; and z were kept constant, this ratio represents f — j ; that is, lbu\ ^ 2 (x^ - y») W/r,z ^ EXERCISES 1. One side of a right triangle increases from 5 to 5.2 while the other decreases from 12 to 11.75. Find the increment of the hypotenuse and its principal part. 2. A closed box, 12 in. long, S in. wide, and 6 in. deep, is made of material \ inch thick. Find approximately the volume of material used. 3. Two sides and the included angle of a triangle are h = 20, c = 30, and A = 45°. By using the formula a* = 6^ + c^ — 2 6c cos A, find approximately the change in a when b increases 1 unit, c decreases 5 unit, and A increases 1 degree. 4. The period of a simple pendulum is r = 2xv'-. Find the error in T due to small errors in I and g. A 124 DIFFERENTIAL CALCULUS Chap. XI. 6. If J7 is computed by the formula, s = h gi\ find the error in g due to small errors in s and t. 6. The area of a triangle is determined by the formula K = ^ ah smC. Find the error in K due to small errors in a, b, C. Find the total differentials of the following functions: 7. xyV. 9. - + ^ + -. y z X 8. xyainix + y). 10. tan-i - + tan-i -• X y 11. The pressure, volume, and temperature of a perfect gas are con- nected by the equation -pv = kt, k being constant. Find dp in terms of dv and dt. 12. If X, y are rectangular and r, d polar coordinates of the same point, show that xdy - ydx = r^djd, dx^ + dy"^ = dr^ + r^ dfi. 13. li X = u — V, y = u^ -\- v^, find ( — I ' \dV jy 14. li u = xy + yz -\- zx, x^ + z'^ = 2 yz, find I -r- ) • 16. li yz = ux + v^, vx = uy + z^, find ( — ) \dzju,x 16. A variable triangle with sides a, h, c and opposite angles A, B, C is inscribed in a fixed circle. Show that cos A cos B cos C 96. Derivative of a Function of Several Variables. — Let u = f (x, y) and let x and y be functions of two variables ^ and t. When t changes to < + Af, x and y will change to :c + Ax and y + Ay. The resulting increment in u will be Aw = — Ax + — Ai/ + d Ax + C2 Ay. Consequently, Am _ du Lx du ^y Ax Ay A<~dxA7"^dyAf"^''A«'^''A< As A< approaches zero, Ax and Ay will approach zero and so Chap. XI. PARTIAL DIFFERENTIATION 125 ci and 62 will approach zero. Taking the limit of both sides, dt dx dt dy dt dx If a: or y is a function of t only, the partial derivative — at or ^ is replaced by a total derivative -r: or -^ . If both x dt at at and y are functions of <, w is a function of t with total deriva- tive du _du dx du ^ /OAK> dt~didt'^dydt' ^^^^ Likewise, if t( is a function of three variables x, y, z, that depend on t, du _ du dx du dy du dz (Qf^\ m ~ dx It '^ dy It '^ dz W ^ ^ As before, if a variable is a function of t only, its partial de- rivative is replaced by a total one. Similar results hold for any number of variables. The term du dx dx ~di is the result of differentiating u with respect to t, leaving all the variables in u except x constant. Equations (96a) and (96c) express that if u is a function of several variable quanti- dll ties, — can he obtained by differentiating with respect to t as if only one of those quantities were variable at a time and adding the results. Example 1. Given y = x', find -p- The function x* can be considered a function of two vari- ables, the lower x and the upper x. If the upper x is held constant and the lower allowed to vary, the derivative (as in case of X") is X . x*-i = «*, 126 DIFFERENTIAL CALCULUS Chap. XI. If the lower x is held constant while the upper varies, the derivative (as in case of a") is af In a;. The actual derivative of y is then the sum -^ = 0:=" -h af In x. Ex. 2. Given u = f (x,y, z), y and z being functions of x, n J du find 3-- ax By equation (96c) the result is du _ du du dy du dz dx dx dy dx dz dx In this equation there are two derivatives of u with respect to X. If y and z are replaced by their values in terms of x, u "will be a function of x only. The derivative of that function is -T- . liy and z are replaced by constants, u will be a second du function of x. Its derivative is t- • ax Ex. 3. Given u = f {x,y, z), z being a function of x and y. Find the partial derivative of u with respect to x. It is understood that y is to be constant in this partial differentiation. Equation (96c) then gives du _ du du dz dx dx dz dx In this equation appear two partial derivatives of u with respect to x. If z is replaced by its value in terms of x and y, u will be expressed as a function of x and y only. Its partial derivative is the one on the left side of the equation. If z is kept constant, u is again a function of x and y. Its partial derivative appears on the right side of the equation. We must not of course use the same symbol for both of these derivatives. A way to avoid the confusion is to use the Chap. XI. PARTIAL DIFFERENTIATION 127 letter / instead of u on the right side of the equation. It then becomes du _ df^.dl dz ' dx dx dz dx It is understood that/ (x, y, z) is a definite function of x, y, z and that -r=^ is the derivative obtained with all the variables dx but x constant. 97. Change of Variable. — If m is a function of x and y we have said that the equation , du , , du J du = -r- dx ->r -— ay dx dy is true whether x and y are the independent variables or not. To show this let s and t be the independent variables and x and y functions of them. Then, by definition, du = -r- ds + -^ dt. ds dt Since u is a function of x and y which are functions of s and t, by equation (96a), du _ du d3C du dy du _ du dx du dy ds dx ds dy ds ' dt dx dt dy dt Consequently, J /du dx , du dy\ J , fdu dx , dudy\ ,^ du/dXj , dx ,A , du/dy J , du jA du , , du , = rx{a^''' + M'^) + ry[fa'^+mV = rx''^ + Ty'''<' which was to be proved. A similar proof can be given in case of three or more variables. 98. Implicit Functions. — If two or more variables are connected by an equation, a differential relation can be ob- tained by equating the total differentials of the two sides of the equation. 128 DIFFERENTIAL CALCULUS Chap. XI. Example 1. f (x, y) = 0. In this case d'fix,y) =^dx+^dy = d*0 = 0.\ Consequently, dy dx Ex. 2. / (x, y, z) = 0. Differentiation gives df df dx df ^dx+^dy+^dz^^a, dx dy dz If z is considered a function of x and y, its partial derivative with respect to x is found by keeping y constant. Titen dy = and d£ _ _ ax dx ~ dj_' dz Similarly, if a; is constant, dx = and dz dy_ dy~ df dz Ex. 3. /i {x, y, z) = 0, /2 {x, y, z) = 0. We have two differential relations df: dfi dfi f^ax+f^dy^f^dz^O. fUx-^l^dy^fUz^O, We could eliminate y from the two equations /i = 0, fz = 0. We should then obtain s as a function of x. The total de- Chap. XI. PARTIAL DIFFERENTIATION 129 rivative of this function is found by eliminating dy and solving dz for the ratio -r- . The result is dx dz dfidf2_ dy dx _dhdh dx dy dx dfidh_ dz dy _dfidf2 dy dz 99. Directional Derivative. — Let u = f {x, y). At each point P (x, y) in the xy-plane, u has a definite value. If we move away from P in any definite direction PQ, x and y will Fig. 99. be functions of the distance moved. Thfc derivative of u with respect to s is du du dx , dii dy dii , du . ds dx ds dy ds dx dy This is called the derivative of u in the direction PQ. The partial derivatives -r- and -r- are special values of -r- which ^ dx dy ds result when PQ is drawn in the direction of OX or OY. Similarly, if w = / (x, y, z), du du dx , dudy , du dz du , du ^ , du T- = T~T~ + ^Tr + ^j~ = "E~ cos a-f^- cos /3 + -T- cos y ds dx ds dy ds dz ds dx dy dz is the rate of change of u w4th respect to s as we move along a line with direction cosines cos a, cos /3, cos y. The partial 130 DIFFERENTIAL CALCULUS Chap. XI. du derivatives of u are the values to which -r- reduces when s is ds measured in the direction of a coordinate axis. Example. Find the derivative of x^ + y^ in the direction <t> = 45° at the point (1, 2). The result is ■^(x^-hy') = 2x^+2y^==2xcos<f>-\-2ysm<f> OS OS OS = 2~ + 4-4== 3V2. 100. Exact Differentials. — If P and Q are functions of two independent variables x and y, Pdx + Qdy may or may not be the total differential of a function w of a; and y. If it is the total differential of such a function, P dx -\- Q dy = du = -r- dx -\- -T- dy. dx dy Since dx and dy are arbitrary, this requires du dx dy Consequently, dP dH dQ dhi dy ~ dydx dx dx dy Since the two second derivatives of u with respect to x and y are equal, An expression P dx -\- Qdy \s called an exact differential if it is the total differential of a function of x and y. We have just shown that (100a) must then be satisfied. Conversely, it can be shown that if this equation is satisfied P dx ■{• Qdy is an exact differential.* ^ See Wilson, Advanced Calculus, § 92. (100b) Chap. XI. PARTIAL DIFFERENTIATION 131 Similarly, if Pdx + Qdy + Rdz is the differential of a function u of x, y, z, dP^dQ dQ^dR dR^dP dy dx ' dz dy' dx dz and conversely. Example 1. Show that (x2 -h2xy)dx+ (x2 + y') dy is an exact differential. In this case The two partial derivatives being equal, the expression is exact. Ex. 2. In thermodynamics it is shown that dU = T dS — p dv, V being the internal energy, T the absolute temperature, S the entropy, p the pressure, and v the volume of a homogene- ous substance. Any two of these five quantities can be assigned independently and the others are then determined. Show that \dp)s \ds). The result to be proved expresses that TdS + v dp is an exact differential. That such is the case is shown by replacing T dS by its value dU + pdv. We thus get TdS + vdp = dU -\- pdv-^vdp = d(U + pv). EXERCISES 1. If w = / (x, y), y = <f> (x), find ^• 2. If « = / {x, y, z),z=4> (x), find [^ 132 DIFFERENTIAL CALCULUS Chap. XI. 3. l!u =f{x, y, z), z = <l> (x, y), y = ^ (x), find ^. 4. If M = / (x, y), y = 4> {x, r), r = ^ (x, s), find f ^ j , f |^j , and 5. U/ (X, J,, 2) = 0, z = F (X, y), find^- 6. If F (x, y, z) = 0, show that dx dy i dy dZ dx dx dy dz^ . 7. If M = x/ (z), z = -, show that x - — f- V t— = «• X dx " dy 8. If w =/ (r, s), r = x+a<, s = y + 6<, show that -rr = a t- + & t — 3i dx 5j/ 9. If z = / (x + ay), show that -r- = o — ' dy dx 10. li u = f (x, y), x = r cos 9, y — r sin 0, show that (5r+(^sy=(s)'+(s)' 11. The position of a pair of rectangular axes moving in a plane is determined by the coordinates h, k of the moving origin and the angle <t> between the moving x-axis and a fixed one. A variable point P has co- ordinates x', y' with respect to the moving axes and x, y with respect to the fixed ones. Then X =f (x', y', h, k, <i>), y = F (x', y', h, k, 0). Find the velocity of P. Show that it is the sum of two parts, one repre- senting the velocity the point would have if it were rigidly connected with the moving axes, the other representing its velocity with respect to those axes conceived as fixed. 12. Find the directional derivatives of the rectangular coordinates X, y and the polar coordinates r, ^ of a point in a plane. Show that they are identical with the derivatives with respect to s given in Arts. 54 and 69. 13. Find the derivative of x^ — y' 'm the direction <f> = 30° at the point (3, 4). 14. At a distance r in space the potential due to an electric charge e is F = - . Find its directional derivative. r 16. Show that the derivative of xy along the normal at any point of the curve x^ — y^ = a^ is zero. Chap. XI. PARTIAL DIFFERENTIATION 133 16. Given u = f {x, y), show that if «i and 82 are measvu*ed along i>erpendiciilar directions. Determine which of the following expressions are exact differentials: y dx — X dy. {2x + y)dx + (x-2y) dy. exdx+ eydy + (x + J/) asdz. yzdx — xzdy + y^dz. Under the conditions of Ex. 2, page 131, show that \dT)p \dp)T \dT}v KdvJT In case of a perfect gas, pv = kT. Using this and the equation dU = TdS -pdv. 17. 18. 19. 20. 21. 22. show that dp Since U is always a function of p and T, this last equation expresses that U is & function of T only. 101. Direction of the Normal at a Point of a Surface. — Let the equation of a surface be F (x, y, z) = 0. Differentiation gives dF , .dF, ,dF, ^ dx dy " dz (101a) Let PN be the Une through P (x, y, z) with direction cosines propor- tional to dF^dF^dF dx' dy ' dz If P moves along a curve on the surface, the direc- tion cosines of its tangent PT are proportional to dx : dy : dz. Equation (101a) expresses that PN and PT are perpendicu- lar to each other (Art. 61). Consequently PN is perpendicu- 134 DIFFERENTIAL CALCULUS Chap. XI. lar to all the tangent lines through P. This is expressed by- saying PA'" is the normal to the surface at P. We conclude that the normal to the surface F (x, y, z) = at P {x, y, z) has direction cosines proportional to ^:^:^. (101) dx dy dz 102. Equations of the Normal at Pi {xi, yi, zi). — Let A, B, C be proportional to the direction cosines of the normal to a surface at Pi (xi, yi, zi). The equations of the normal are (Art. 63) x- xi _ y-yr _ z- Zi A ~ B ~ C ' ^^^^^ 103. Equation of the Tangent Plane at Pi (cci, j/i, si). — All the tangent lines at Pi on the surface are perpendicular Fig. 103. to the normal at that point. All these lines therefore lie in a plane perpendicular to the normal, called the tangent plane at Pi. It is shown in analytical geometry that \i A, B, C are pro- portional to the direction cosines of the normal to a plane passing through (xi, y\, Zi), the equation of the plane is A{x-Xx)^B{y- y{) + C (z - z,) = 0.* (103) * See Phillips, Analytic Geometry, Art. 68. Ch.\p. XI. P.\RTL\L DIFFERENTL\TIOX 135 If A, B, C are proportional to the direction cosines of the normal to a surface at Pi, this is then the equation of the tangent plane at Pi. Example. Find the equations of the normal line and tan- gent plane at the point (1, —1, 2) of the ellipsoid x2 + 2t/2 + 3z» = 3x+12. The equation given is equivalent to x2 + 2 j/2 + 3 22 _ 3 X - 12 = 0. The direction cosines of its normal are proportional to the partial derivatives 2 z - 3 : 4 y : 6 2. At the point (1, —1, 2), these are proportional to A :B :C = -1 : -4 : 12 = 1 : 4 : -12. The equations of the normal are X - 1 _ y 4- 1 2-2 1 ~ 4 ~ -12 ' The equation of the tangent plane is X - 1 + 4 (y + 1) - 12 (0 - 2) = 0. EXERCISES Find the equations of the normal and tangent plane to each of the following surfaces at the point indicated: 1. Sphere, z» + y» + 2' = 9, at (1, 2, 2). 2. Cylinder, x» + ly + !/» = 7, at (2, -3, 3). 3. Cone, 2^ = x» + t/*, at (3, 4, 5). 4. Hj'perbohc paraboloid, xy = 3 z — 4, at (5, 1, 3). 6. EUiptic paraboloid, x = 2 y* -|- 3 2*, at (5, 1, 1). 6. Find the locxis of points on the cylinder (x + 2)* + (y - 2)« = 4 where the normal is parallel to the xy-plane. 7. Show that the normal at anj- point P (x, y, z) of the surface J/* 4- 2* = 4 X makes equal angles with the x-axis and the line joining P and A (1, 0, 0). 8. Show that the normal to the spheroid 9 "^25 at P (x, y, z) determines equal angles with the lines joining P with A' (0, -4, 0) and A (0, 4, 0). 136 DIFFERENTIAL CALCULUS Chap. XI. 104. Maxima and Minima of Functions of Several Variables. — A maximum value of a function w is a value greater than any given by neighboring values of the variables. In passing from a maximum to a neighboring value, the func- tion decreases, that is Aw < 0. (104a) A minimum value is a value less than any given by neigh- boring values of the variables. In passing from a minimum to a neighboring value Aw > 0. (104b) If the condition (104a) or (104b) is satisfied for all small changes of the variables, it must be satisfied when a single variable changes. If then all the independent variables but X are kept constant, u must be a maximum or minimum in x. du If — is continuous, by Art. 31, i = 0. (1040 Therefore, if the first 'partial derivatives of u with respect to the independent variables are continuous, those derivatives must be zero when u is a maximum or minimum. When the partial derivatives are zero, the total differential is zero. For example, if x and y are the independent vari- ables, du =^dx +^ dy = ' dx + ' dy = 0. (104d) Therefore, if the first partial derivatives are continuous, the total differential of u is zero when u is either a maximum or a minimum. To find the maximum and minimum values of a function, we equate its differential or the partial derivatives with re- spect to the independent variables to zero and solve the resulting equations. It is usually possible to decide from the problem whether a value thus found is a maximum, minimum, or neither. Chap. XI. PARTIAL DIFFERENTIATION 137 Example 1. Show that the maximuin rectangular paralielo- piped with a given area of surface is a cube. Let X, y, zhe the edges of the parallelepiped. If F" is the volume and A the area of its surface V = xyz, A = 2 xy ■\- 2 xz -\- 2 yz. Two of the variables x, y, z are independent. Let them be X, y. Then A -2xy z — Therefore V = 2{x-\-y) xy (A -2 xy) 2{x + y) &V^t P -2x^-4xy l ^ dx 2L {x + yy J "' dV ^x^T A -2y^ -4 xy 1 dy 2L (x-\-yr J "• The values x = 0, y = cannot give maxima. Hence A - 2x^ - 4:xy = 0, A - 2y^ - 4:xy = 0. Solving these equations simultaneously with A = 2 xy -\- 2 xz •\- 2 yz, we get We know there is a maximum. Since the equations give only one solution it must be the maximum. Ex. 2. Find the point in the plane x + 2i/ + 30= 14 nearest to the origin. The distance from any point (x, y, z) of the plane to the origin is D = a/x2 + 2/2 + 2?. 138 DIFFERENTIAL CALCULUS Chap. XI. If this is a minimum d.j) ^ ^dx-\-ydy-\-zdz ^ ^ Va;2 + 7/2 + 22 that is, X dx + y dy -{- z dz = 0. (104e) From the equation of the plane we get dx -\- 2 dy -{- 3 dz = 0. (104f) The only equation connecting x, y, z is that of the plane. Consequently, dx, dy, dz can have any values satisfying this last equation. If x, y, z are so chosen that Z) is a minimum (104e) must be satisfied by all of these values. If two linear equations have the same solutions, one is a multiple of the other. Corresponding coefficients are proportional. The coefficients of dx, dy, dz in (104e) are x, y, z. Those in (104f) are 1, 2, 3. Hence X _y _ z I~2~3" Solving these simultaneously with the equation of the plane, we get a; = 1, 2/ = 2, 2 = 3. There is a minimum. Since we get only one solution, it is the minimum. EXERCISES 1. An open rectangular box is to have a given capacity. Find the dimensions of the box requiring the least material. 2. A tent having the form of a cylinder surmounted by a cone is to contain a given volume. Find its dimensions if the canvas required is a minimum. 3. When an electric current of strength / flows through a wire of resistance R the heat produced is proportional to PR. Two terminals are connected by three wires of resistances Ri, Ri, Rz respectively. A given current flowing between the terminals will divide between the wires in such a way that the heat produced is a minimum. Show that the currents Ii, h, h in the three wires will satisfy the equations IiRi = TiRi = /jRj. 4. A particle attracted toward each of three points A, B, C with a force proportional to the distance will be in equilibrium when the sum Chap. XI. PARTUL DIFFERENTUTION 139 of the squares of the distances from the pwints ib least. Find the posi- tion of equilibrium. 6. Show that the triangle of greatest area with a given perimeter is equilateral. 6. Two adjacent sides of a room are plane mirrors. A ray of light starting at P strikes one of the mirrors at Q, is reflected to a point R on the second mirror, and b there reflected to 5. If P and S are in the same horizontal plane find the positions of Q and R so that the path PQRS may be as short as possible. 7. A table has four legs attached to the top at the comers Ai, Ai, A 3, Ai of a square. A weight W placed upon the table at a point of the diagonal A1A3, two-thirds of the way from Ai to A3, will cause the legs to shorten the amounts Si, sj, sj, S4, while the weight itself sinks a dis- tance h. The increase in potential energy due to the contraction of a leg is /:«*, where A; is constant and s the contraction. The decrease in potential energy due to the sinking of the weight is Wh. The whole system \s"ill settle to a position such that the potential energy is a mini- mum. Assuming that the top of the table remains plane, find the ratios of Sj, Sj, Sj, S4. SUPPLEMENTARY EXERCISES CHAPTER III Find the differentials of the following functions: ^- ^ • 6. X (a2 + a;2) Va^ - x\ 2 , "" 7 (2x+l)(2x + 7)^ 6 Vox* + & (2x + 5)3 3 2Vax'' + bx (x + 2)« (x + 4)^ 6x * ' (x + 1)2 (x + 3)6" 4, 2ax + b , (2x»-l)V^Hn: Vox- + &a; + c x' (ax + ft)'H-2 6(ax+b)"+i „_, ^- o2 (n + 2) a^ (n + 1) ' 10- a; (x" + n) » ' Find -T- in each of the following cases: 11. 2x2-4x2/4-3^2 = 6x-4y + 18. 12. x^ + 3 x«7/ = 2/3. 13. x = Sy^ + 2yK 14. (x2 + 2/2)2 = 2a2(x2-2/«). 15. X = 1 + -^, y = 2t- ^ i-1' ^ " (<-l)« < 1 16. X = / ' y = , V 1 + <2 '^ Vl - /2 17. X = < (<2 + a2)4 y = t{t^-\- a2)5. 18. X = z2 + 2 8, 2 = 2/2 + 22/. 19. x2 + 22 = a2, 2/2 = &2. 20. The volume elasticity of a fluid is e = —v-j-. according to Boyle's law, p» = constant, show that e = p. 21. When a gas expands without receiving or giving out heat, the pressure, volume, and temperature satisfy the equations pv = RT, pv" = C, R, n, and C being constants. Find -^ and ^« 140 SUPPLEMENTARY EXERCISES 141 22. If 9 is the volume of a spherical segment of altitude h, show that jT is equal to the area of the circle forming the plane face of the segment. an 23. If a polynomial equation /(x)=0 has two roots equal to r, / (x) has (x — r)* as a factor, that is, / (x) = (X - r)Vi (x), where /i (x) is a polynomial in x. Hence show that r is a root of f (x) = 0, where/' (x) is the derivative of/ (x). Show by the method of Ex. 23 that each of the following equations has a double root and find it: 24. x^ - 3 x» + 4 = 0. 25. x3 - x^ - 5 X - 3 = 0. 26. 4x»-8x* -3x + 9 =0. 27. 4x^ - 12x» + x* + 12x + 4 =0. Find -^ and -j-^ in each of the following cases. <ir ox* 28. w = X Vo* - X*. 31. ax + by + c = Q. ^ 32. X = 2 + 3<, y = 4-5/. ''•'-(?+T?- 33. x = ^., „. " 30. XT/ =a^ * + !' ^ + 1 34. If J/ = x^, find g and 0- 35. Given x* — y* = 1, verify that d^y _ _ (Pxfdy^ dbfi ~ dy^ \dxj ' 36. If ra is a positive integer, show that -5-: x" = constant, dx" 37. If « andi; are functions of x, show that d* , . ihi , .dhi dv , „ dhi <Pv , . du cPv , d*v ^(u.) -=^.. + 4^.^ +6^.^ + 4^.^ + u^. CJompare this with the binomial expansion for (u + »)*. 38. If / (x) = (x — r)3/i (x), where /i (x) is a polynomial, show that /'(r)=/"(r)=0. ^ 142 DIFFERENTIAL CALCULUS CHAPTER IV 39. A particle moves along a straight line the distance s = 4 <3 - 21 i2 + 36 < + 1 feet in t seconds. Find its velocity and acceleration. When is the particle moving forward? When backward? When is the velocity increasing? When decreasing? 40. Two trains start from different points and move along the same track in the same direction. If the train in front moves a distance 6 t^ in t hours and the rear one 12 t^, how fast will they be approaching or separating at the end of one hour? At the end of two hours? When will they be closest together? 41. If s = Vt, show that the acceleration is negative and propor- tional to the cube of the velocity. 42. The velocity of a particle moving along a straight line is V = 2P -3t. Find its acceleration when t = 2. k 43. li V- = -, where k is constant, find the acceleration. s ' 44. Two wheels, diameters 3 and 5 ft., are connected by a belt. What is the ratio of their angular velocities and which is greater? What is the ratio of their angular accelerations? 45. Find the angular velocity of the earth about its axis assuming that there are 365| days in a year. 46. A wheel rolls down an inclined plane, its center moving the distance s = 5P in t seconds. Show that the acceleration of the wheel about its axis is constant. 47. An amount of money is drawing interest at 6 per cent. If the interest is immediately added to the principal, what is the rate of change of the principal? 48. If water flows from a conical funnel at a rate proportional to the square root of the depth, at what rate does the depth change? 49. A kite is 300 ft. high and there are 300 ft. of cord out. If the kite moves horizontally at the rate of 5 miles an hour directly away from the person flying it, how fast is the cord being paid out? 50. A particle moves along the parabola 100 y = 16x2 in such a way that its abscissa changes at the rate of 10 ft./ sec. Find the velocity and acceleration of its projection on the j/-axis. 51. The side of an equilateral triangle is increasing at the rate of 10 ft. per minute and its area at the rate of 100 sq. ft. per minute. How large is the triangle? SUPPLEMENTARY EXERCISES 143 CHAPTER V 52. The velocity of waves of length X in deep water is proportional to V^ when a is a constant. Show that the velocity is a minimum when X = a. 53. The sum of the surfaces of a sphere and cube is given. Show that the sum of the volumes is least when the diameter of the sphere equals the edge of the cube. 54. A box is to be made out of a piece of cardboard, 6 inches square, by cutting equal squares from the comers and turning up the sides. Find the dimensions of the largest box that can be made in this way. 55. A gutter of trapezoidal section is made by joining 3 pieces of material each 4 inches wide, the middle one being horizontal. How wide should the gutter be at the top to have the maximum capacity? 56. A gutter of rectangular section is to be made by bending into shape a strip of copper. Show that the capacity of the gutter wOl be greatest if its width is twice its depth. 57. If the top and bottom margins of a printed page are each of width a, the side margins of width b, and the text covers an area c, what should be the dimensions of the page to use the least paper? 58. Find the dimensions of the largest cone that can be inscribed in a sphere of radius a. 59. Find the dimensions of the smallest cone that can contain a sphere of radius a. 60. To reduce the friction of a Uquid against the walls of a channel, the channel should be so designed that the area of wetted svu^ace is as small as possible. Show that the best form for an open rectangular channel with given cross section is that in which the width equals twice the depth. 61. Find the dimensions of the best trapezoidal chaimel, the banks .iiaking an angle 6 with the vertical. 62. Find the least area of canvas that can be used to make a conical tent of 1000 cu. ft. capacity. 63. Find the maximum capacity of a conical tent made of 100 sq. ft. of canvas. 64. Find the height of a light above the center of a table of radius a, so as best to illuminate a point at the edge of the table; assimaing that the illumination varies inversely as the square of the distance from the Ught and directly as the sine of the angle between the rays and the surface of the table. 144 DIFFERENTIAL CALCULUS 65. A weight of 100 lbs., hanging 2 ft. from one end of a lever, is to be raised by an upward force appUed at the other end. If the lever weighs 3 lbs. to the foot, find its length so that the force may be a minimum. 66. A vertical telegraph pole at a bend in the line is to be supported from tipping over by a stay 40 ft. long fastened to the pole and to a stake in the grovmd. How far from the pole should the stake be driven to make the tension in the stay as small as possible? 67. The lower corner of a leaf of a book is folded over so as just to reach the inner edge of the page. If the width of the page is 6 inches, find the width of the part folded over when the length of the crease is a minimum. 68. If the cost of fuel for running a train is proportional to the square of the speed and $10 per hom* for a speed of 12 mi./hr., and the fixed charges on $90 per hour, find the most economical speed. 69. If the cost of fuel for running a steamboat is proportional to the cube of the speed and $10 per hour for a speed of 10 mi./hr., and the fixed charges are $14 per hour, find the most economical speed against a current of 2 mi./hr. CHAPTER VI Differentiate the following fimctions: sin X 76. sec^ x — tan^ x. 70. 71. 72. . „ 77, sin'- sec =• smd X 3 1 — cos TO i ^ , , „ 78. tan- 1 + cos B 1 — X sin e ,__ 2 tan x 73. sin ax cos ax. ' 1 — tan^ x „, ,9 80. 5 sec^ - 7 sec* 9. 74. COtTT — CSCtt- o, - , 2 2 81. sec X CSC a; — 2 cot X. 75. tan 2 x — cot 2 x. Differentiate both sides of each of the following equations and show that the resulting derivatives are equal. 82. sec^ x + csc-x = sec'' x csc^ x. 83. sin2x = 2 sin x cos x. 84. sin 3 X = 3 sin X — 4 sin' x. 85. sin (x + a) = sin x cos a + cos x sin a. 86. sec'^ X = 1 + tan* x. 87. sin X + sin o = 2 sin J (x + a) cos ^ (x — a). SUPPLEMENTARY EXERCISES 145 88. coso - cosx = 2sm Ha; + o) sin Ha; - o). Find ^ and $^ in each of the following cases: 89. x = a cos" 9, y = a sin- 6. 90. X = a cos* e, y = a sin* d. 91. X = tan9 -e, y = cosB. 92. X = sec* e, y = tan* e. 93. X =8ec9, y = tanO. 94. X = CSC — cot ff, y = esc + cot 6. Differentiate the following functions: 95. Bin-V/^- ^^2. ac8C-^ + V^^^. 96. cos- (^). 103. pq-.-cot- 97. tan tan (^)- 104. Vl — X sin-i X — "n/x- ,-. ,x + 1 , . .z — 1 - ^ , , lOo. sec-i r+sin * — --: • 2 2x + l x-1 x + 1 VS"^ ^^ 106. sm-^ ;+^^^^ - _ + a cos X 99. cos-^^^py- jQ. 1 os-x + ^vT:^ V5 ^ 100. CSC ' 2x-l ' 108. Vi^-a* -asec-»-- 101 . sec -'K^-^y- 109. e^. lis. -tan-'-+^hi(a* + x»). 110. Ve*. a a ' 2 119. e-«cos(o+W). 121. (. + l),„(x+l)-x-i. 113. 7=^. 114. a^lnx. ,-_ , V.T + a + Vx — a ,,, 1 • n 122. In , = . llo. lnsin"x. Vi 4- a — Va- — a 116. hilnx. 123. tan"! H^* + e"*). 117. Infi^y 124. hi(Vi + V7T2). 125. (x + l)hi(x* + 2x + 5)+|tan-i^^. 126. sec§xtan|x + ln(sec|x + tan|x). 146 DIFFERENTIAL CALCULUS 127. x sec-i I /^x + -] - In (x2 + 1) . 128. I In f I x2 + 1 ) - I a; + tan-i |x. CHAPTER VII Find the equations of the tangent and normal to each of the follow- ing curves at the point indicated: 129. y^ = 2x + y,at (1, 2). 130. x2 - 2/2 = 5, at (3, 2). 131. x^ + y^ = x + 3y,at i-1,1). 132. x* + ?/* = 2, at (1, 1). 133. y = \nx, at (1, 0). 134. x2 (x + y) = a" (x - y), at (0, 0). 135. X = 2 cos 0, 2/ = 3 sin 0, at 6 = 5- 136. r = o (1 + cosd), at (9 = ^• Find the angles at which the following pairs of curves intersect: 137. x2 + 2/2 = 8 X, 2/2 (2 - x) = x^ 138. 2/^ = 2 ox + a^ x2 = 2 62/ + 62. 139. x2 = 4: ay, (x^ + 'ia^)y = 8a\ 140. y^ = &x, x2 + ^2 = 16, 141. 2/ = He^ + e-^ ), 2/ = 1. 142. 2/ = sin X, 2/ = sin 2 x. 143. Show that all the curves obtained by giving different values to n in the equation are tangent at (a, 6). 144. Show that for all values of a and 6 the curves x' — 3 X2/2 = 0, 3 x^y — y^ = h, intersect at right angles. Examine each of the following curves for direction of curvature and points of inflection : i^c 1 —X 145. y = rr^- 146. 2/ = tan x. 147. x = 62/2-2i/». 1 SUPPLEMENTARY EXERCISES 147 148. x = 2<-i, ^/ = 2f + |• 149. Clausius's equation connecting the pressure, volume, and tem- perature of a gas is RT c ^ v-a r(t; + b)'' R, a, b, c being constants. If T is constant and p, v the coordinates of a point, this equation represents an isothermal. Find the value of T for which the tangent at the point of inflection is horizontal. 150. If two curves y = f (x), y = F (x) intersect at x = a, and /' (a) = F' (a), but /" (a) is not equal to F" (a), show that the curves are tangent and do not cross at x = a. Apply to the curves y = 3? and y = x* at X = 0. 151. If two curves y = f (x), y = F (x) intersect at x = a, and r (o) = F' (a), /" (a) = F" (a), but /'" (o) is not equal to F'" (a), show that the curves are tangent and cross at x = a. Apply to the ciuTres y = X* and y = x* + (x — 1)' at x = 1. Find the radius of ciu^rature on each of the following curves at the point indicated: 152. Parabola y* = ax at its vertex. X* w* . 153. Ellipse — + rr = 1 at its vertices. a- b- 154. Hyperbola ^ - ^ = 1 at x = Va* + 6». 155. 1/ = Incscx, at f ^, )• 156. x = §siny — jln (sec y + tan y), at any point (x, y). 157. X = a cos's, y = a sin' 6, at any point. 158. Find the center of curvature of y = In (x — 2) at (3, 0). Find the angle ^ at the point indicated on each of the following curves: 159. r = 2*, at fl = 0. 160. r = a + 6 cos d, at = ^• 161. r(l -cose) =k,ate = |- 162. r = a sin 20, at » = |- Find the angles at which the following pairs of curves intersect: 163. r (1 — cosd) = a, r = o (1 — cosd). 164. r = asec*s> r = 6csc*=' 148 DIFFERENTIAL CALCULUS 165. r = acosd, r = a cos 2 d. 166. r = asecd, r = 2 a sin d. Find the equations of the tangent Unes to the following curves at the points indicated: 2 167. X = 2 1, y = - z = f^, at t = 2. 168. X = sint, y = cos t, z = sec t, at t = 0. 169. x^ + y^+z^ = 6, x + y + z = 2, at (1, 2, -1). 170. z=x^ + y%z'' = 2x- 2y, at (1, -1, 2). CHAPTER VIII 171. A point describes a circle with constant speed. Show that its projection on a fixed diameter moves with a speed proportional to the distance of the point from that diameter. 172. The motion of a point (x, y) is given by the equations x = ^ V^^rT2 4- |%in-i 1 , ^ ^ Of y = \ V^2Tp72 -j. ^ In + v^rqr^). Show that its speed is constant. Find the speed, velocity, and acceleration in each of the following cases: 173. x = 2 + 3<, 2/ = 4-9<. 174. X = a cos {o)t + a), y = a sin (ost + a). 175. X = a + at, y = b + fit, z = c -\- yt. 176. a; = e* sin t, y = e^ cos t, z = kt. 177. The motion of a point P (x, y) is determined by the equations X = a cos {nt + «), 2/ = & sin {nt + a). Show that its acceleration is directed toward the origin and has a magnitude proportional to the distance from the origin. 178. A particle moves with constant acceleration along the parab- ola y^ = 2 ex. Show that the acceleration is parallel to the x-axis. 179. A particle moves with acceleration [a, o] along the parabola 2/2 = 2 ex. Find its velocity. -T-^ , -T-| extends along the normal at (x, y) and is in magnitude equal to the curvature at (x, y) , SUPPLEMENTARY EXERCISES 149 CHAPTER IX 181. Show that the function xi-1 vanishes at x = — 1 and x = 1, but that its derivative does not vanish between these values. Is this an exception to Rolle's theorem? 182. Show that the equation x*-5x + 4 = has only two distinct real roots. 183. Show that x*sm- Lim — : = 0, a: =5= sm X but that this value cannot be found by the methods of Art. 73. Explain. 184. Show that _ . 1 — cos X „ Lim = 0. x:S.O cos X Why cannot this result be obtained by the methods of Art. 73? Find the values of the following limits: 185. Lim ,^'-,^ . 189. Lim f i - ^- x = ol-cos2x z = 0\X* X / ,„^ ^. vTx - Vl2 -X 190. Limx"e-'*. 186. Lim x = 32x -3 Vl9 -ox z = 00 .«, _. xlnx TX 191. Lim — 187. Lim tan^ ■ x = osin2x - xcotx ^ 1 1 + esc (x — 1) 192. Lim (.sec x) *»• 188. Lim^(^-^^ ■0 ; A I cot (irx) 193. The area of a regular polygon of n sides inscribed in a circle of radius a is na'sm - cos — n n Show that this approaches the area of the circle when n increases indefinitely. 194. Show that the curve x^ + y^ = Zxy is tangent to both coordinate axes at the origin. 150 DIFFERENTIAL CALCULUS CHAPTER X Determine the values of the following functions correct to four decimals: 195. cos 62°. 198. vTl. 196. sin 33°. 199. tan-i (t^)- 197. hi (1.2). 200. esc (31°). 201. Calculate tt by expanding tan~* x and using the formula \ = tan-i (1). 202. Given hi 5 = 1.6094, calculate hi 24. 203. Prove that _ D = V^h is an approximate formula for the distance of the horizon, D being the distance in miles and h the altitude of the observer in feet. Prove the following expansions indicating if possible the values of X for which they converge: 204. In (1 + a;2) = hi 10 + f (x - 3) - 2^ (x - 3)2 + • • • . 205. ln(e^ + e-)=f-g + g+ .... 206. In (1 + siax) = X - h x^ + ^ ^ - ^ X* + ' • ' . 207. e^secx = 1 + x + x^ + ^x^ + h x* + • . . . 208. In (x + Vr+^0 =^-|f+^f+---- 209. ln^ = 2p + J-3 + ^ + ...]. X — 1 La;3x^5x* J 210. In tan X = hi X + ^ + ^ + • • • . 211. e«-- = l+x + |-j-^ . 212. eta°- = l+x + |j+^' + ^+ • • • . Determine the values of x for which the following series converge: 213. 1+3^ + 1+1+5+ '■ ' • 2H. (.-,) + (^'+fe^* + (1^111'+.... 215. 1 +2x4-3x2 + 4x3+ ... . X + 2 (X + 2)' (X + 2)» _ ^ 216. 2 + -jT2-+-2T3~ + ~3T4~+ • SUPPLEMENTARY EXERCISES 151 CHAPTER XI In each of the following exercises show that the partial derivatives satisfy the equation given: 217. u = xy + y^z\ x - + z - = y -■ 218. 2 = x'-2xV + y*, J/?+^§=0- ax ay 219.„.(x + ,)ln^. x(g-g)..f. 221. u = xy + -, 222. z = ln(^ + y^), g + ^ = 0. 223. u = y-±^, y -z axay "^ dj/az "^ dz* di»* Prove the following relations assuming that z is a function oK x and y^ 224. u = ix + y- zy, du _du _ dudz du dz 3y dx dy dx dx dy 225. u = z + e^, du du dz dz X V — = X — — 1/ — • dx ^ dy dx " dy 226. u = 2 (a? - J/*), ^ai + ^a^^^^-^a^a^+'^ai;) 227. If X = I (e* + e"*), y ="^{6^ - e"*), show that Var/e \dxjy 228. lixyz = a', show that 152 DIFFERENTIAL CALCULUS In each of the following exercises find A2 and its principal part, assuming that x and y are the independent variables. When Ax and Ay approach zero, show that the difference of Az and its principal part is an infinitesimal of higher order than Ax and Ay. 229. z = xy. 232. z = V^'+^z. 230. z = x2 -2/2 + 2X. 231. z = —^ x^ + 1 Find the total differentials of the following functions: 233. ax* + bxY + cy*. 234. hi{x^ + y^+z^). 235. x* tan-i ^ - y^ tan-i -• X " y 236. yzfF + zxe" + xye'. 237. If w = x»/ (2), z = ^, show that 238. Ku =f(r,s), r = x + y, s = x — y, show that 5m . du _ ^du dx dy ~ dr' 239. If M =/(r, s, 0, r = -, s = ^, < = -, show that y z X du , du , du „ 240. If a is the angle between the x-axis and the Hne OP from the ' origin to P (x, y, z), find the derivatives of a in the directions parallel to the coordinate axes. 241. Show that (cot y — ysecx tan x) dx — (x esc* y + sec x) dy is an exact differential. Find the equations of the normal and tangent plane to each of the following surfaces at the point indicated: 242. x2 + 2y2 - z2 = 16, at (3, 2, -1). 243. 2x + 3?/ -4z = 4, at (1, 2, 1). 244. z2 = 8xy, at (2, 1, -4). 245. 2/ = z2 - x2 + 1, at (3, 1, -3). 246. Show that the largest rectangular parallelopiped with a givai surface is a cube. SUPPLEMENTARY EXERCISES 153 247. An open rectangular box is to be constructed of a given amount of material. Find the dimensions if the capacity is a maximum. 248. A body has the shape of a hollow cylinder with conical ends. Find the dimensions of the lai^est body that can be constructed from a given amoimt of material. 249. Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid. 250. Show that the triangle of greatest area inscribed in a given circle is equilateral. 251. Find the p)oint so situated that the sum of its distances from the vertices of an acute angled triangle is a minimum. 252. At the point (x, y, z) of space find the direction along which a given function F (x, y, z) has the largest directional derivative. I i ANSWERS TO EXERCISES Page 8 1. -f 4. -1. 2. V2. 6. 1. 3. -1. 6. i. Page 14 3. 2. 5. The tangents are parallel to the x-axis at (—1, —1), (0, 0), and (1, —1). The slope is positive between ( — 1, —1) and (0, 0) and on the right of (1, —1). 10. Negative. 11. Positive in 1st and 4th quadrants, negative in 2nd and 3rd. Pages 27, 28 31. When x = 4, t/ = | and dy = 0.072 dx. When x changes to 4.2, dx = 0.2 and an approximate value for y is y + dy = 0.814. This agrees to 3 decimals with the exact value. 32. When x = 0, the function is equal to 1 and its differential is — dx. When x = 0.3, an approximate value is then 1 — dx = 0.7. The exact value is 0.754. 34. 18. 35. (a, 2 a). 36. Increases when x < 5, decreases when x > ■=• o o 37. x=±^. 39. -(^:n)5 2 38. tan-i|. Page 31 1 2 4 9 ^ «' {x-iy (x-l)» • V^rr^j* (a2-x2)3 3. (x - 1)2 (x + 2)2 (7x + 2), (x - 1) (x + 2)2 (42x2 + 24x - 12). 4. 2. _4 7 1-^ 2 y' y' x-r (x-l)3 8. ^ y^ x* 3xij/S X 1 6 2y' 42/» 154 ANSWERS TO EXERCISES 155 13. ^ 12 gj _ 12 dl» ~ U2 + 30'' rfy* U2-30*' Pages 3&-38 1. r = 100 — 32 f, a = — 32. Rises until t = 3|. Highest point h = 206.25. 2. V = P ~ 12P + S2t, a = SC- - 241 + 32. Velocity decreasing between t = 1.691 and t = 6.309. Moving backward when t is negative or between 4 and 8. 7. CO = 6 — 2 d, a = —2 c. 'SVTieel comes to rest when < = tj-- J c 10. 9xcu. ft./min. 15. 12^ ft./sec., 7| ft./sec. 11. 144 X sq. ft./sec. 16. 4 VS mi.,/hr. 12. Decreasing 8tcu. ft./sec. ^_ ctan/3, , ^ 17. z- it./aec. 13. %^ : 1. TO* ' 14. ^ V3 in. /sec. 18. Neither approaching nor separating. 19. 25.8 ft. sec. 20. 64 \^ ft./sec. Pages 43-45 1. Minimum 3|. 2. Minimum — 10, maximum 22. 3. Maximimi at x = 0, minima at x = — 1 and x = +1. 4. Minimum when x = 0. 13. ' a V3. 10. fv^2. 14. Length of base equals twice the depth of the box. 16. Radius of base equals two-thirds of the altitude. 4 17. Altitude equals - times diameter of base. X 16 X V3 20. Girth equals twice length. 27 21. Radius equals 2 Ve inches. 19. i(ai + «i + aj + 04). 22. The distance from the more intense source is v^2 times the dis- tance from the other source. 23. 12 V2. 25. 19ift. 24. [5' 4- 6*]'. 26. Radius of semicircle equals height of rectangle. 27. 4 pieces 6 inches long and 2 pieces 2 ft. long. 28. The angle of the sector is two radians. 29. At the end of 4 hours. 166, DIFFERENTIAL CALCULUS 30. He should land 4.71 miles from his destination. a V2 3L — -^ , a being the length of side. 35. 2|mi./hr. 36. 13.6 knots. Page 48 1. Maximum = a, minimum = —a. 2. Maximum = 0, minimum = — {^)K 3. Minimum = — 1. 4. Minimum = 0, maximum = ^. 10. Either 4 or 5. Pages 62, 53 19. A = 3. „„ IT , . 20 ^ = _ 7 B = -y? 22. WTT ± g, n bemg any mteger. 21. V3 -|. 23. Velocity = —2irnA, acceleration = 0. 24. —^ miles per minute. ^"- 2+b^tV3. or o V ,. 28. 13 Vl3. 25. I radians per hour. 29. The needle will be incUned to the horizontal at an angle of about 32° 30' 30. 120°. a 31. 120°. '*'^- TT* 33. If the spokes are extended outward, they will form the sides of an isosceles triangle. Page 56 V 24. w = - cos <p, r being the radius of pulley and <t> the angle formed T by the string and line along which its end moves. 25. 4V35. Page 61 27. X = rnr + cot~i 2, n being any integer. 30. a;<-3, x>2, or -2 < x < 1. Pages 65, 66 1. 2y-x = 5, y + 2x = 0. 2. 7/ + 4x = 8, 4y-x = 15. 3. 2y=F X = d:a, y ±2x = ±Sa. ANSWERS TO EXERCISES 157 4. y = aix\nb + l), x + ayhxb = a^hab. a^ vr 7. x + y = 2, 2-y = 2. 12. 90°. 8. x + 3i/ = 4, y-3x = 28. 13. tan-»2V2. 9. y -\-xX&a.\4>x = a<iniaiih<h- _, In 10 - 1 10. 90°, tan-if. ^^' ^^ In 10 + 1' 11- 45°. 15. tan-i3V3. Page 70 1. Point of inflection (0, 3). Concave upward on the right of this point, downward on the left. 2. Point of inflection (|, —\). Concave upward on the right of this point, downward on the left. 3. The curve is everjTV'here concave upward. There is no point of inflection. 4. Point of inflection (1, 0). Concave on the left of this point, down- ward on the right. 6. Point of inflection! —2, — -j)- Concave upward on the right, downward on the left. 6. Points of inflection at x = ± — ^^ . Concave downward between V2 the points of inflection, upward outside. 7. Points of inflection (0, 0), (±3, ±1). Concave upward when -3 < X < or X > 3. 8. Point of inflection at the origin. Concave upward on the left of the origin. Pages 76, 77 1. ±2V6. -7 t 7. ^. a ^- b' 8. aecy. 4. 3V2. 9. ^^ + ^>'. 4y ^•^'^' 10. 2a8ec«^- 6. ta. 2 158 DIFFERENTIAL CALCULUS Page 79 There are two angles \}/ depending on the direction in which s is measured along the curve. In the following answers only one of these angles is given. 2 "L. 7. 0°, 90°, and tan-» 3 Vs. '4 s. d = dziT. 3. |. 10- 3. o Page 84 1. X ,_!^ 5. tan-?. •v/o 1 2 — J- 5. tan"' 7 • — ^'^ _ y — 1 ^ 4_ k V2 2 a ■ ft X -1 « 6. tan '■——' - X — e , 2 — 1 V2 2. = 1 — we = — -^ — • e ^ 2 7. 69° 29'. 3. irfc Pages 92, 93 1. The angular speed is ^ ^ , where x is the abscissa of the moving point. 2. If Xi is the abscissa of the end in the x-axis and yi the ordinate ct' the end in the y-axis, the velocity of the middle point is the upper signs being used if the end in the x-axis moves to the right, the lower signs if it moves to the left. The speed is ^ — • 3. The velocity is [v — aw sin 0, aw cos 6], where is the angle from the x-axis to the radius through the moving point. The speed is y/i^ -\- aW — 2 omv sin 6. 6. The boat should be pointed 30° up the river. 7. Velocity = [o, h, c — gl], Acceleration = [0, 0, —g\, Speed = Va2 -\-¥-^{e- gt)K ANSWERS TO EXERCISES 169 9. Velocity = [oko (1 — cos<^), aw sin <^], Speed = aw ^2 — 2 cos <> = 2 aw sin ^ ^, Acceleration = \au?€\a.<i>, aoj^cosi^]. ^ r 3r^ sinf g ZrP- cos f g ~| L 4 a sin ^ ' 4 o sin 5 9 J 12. X = vi cos w/, y = vt sin w/. The velocity is the sum of the par- tial velocities, but the acceleration is not. 13. X = a cos cot +b cos 2 at, y = a sin w< + & sin 2 ut. The velocity is the sum of the partial velocities and the acceleration the sum of the partial accelerations. 14. X = ao)it — a sin {ui + oh) t, y = a cos («i + Wi) t. The velocity is the sum of the partial velocities and the acceleration the sum of the partial accelerations. Page 100 2. A. 3. n. 4. 0. 5. e». 6. 2. 7. -2. 8. 1. 9. 0. 10. ir2. 11. 1. 12. 1. 13. 0. 14. -i- 15. 0. 16. 0. 17. -i. 1. 0.0872. 2. 0.8480. 3. 1.0724. 4. 1.6003. 5. 1.0154. 21. (-2,1,0) 18. h 19. 1. 20. -3. 21. a. 1 22. ir 23. f'{x)dx. 24. 00. 25. 1 2- 26. 00. 27. 00. 28. 1. 29. 1. 30. a. 31. gm Page 106 6. 0.1054. 7. 1.6487. 8. 0.0997. 9. 2.833. Page 118 22. (1, 1, 2) 160 DIFFERENTIAL CALCULUS Pages 123, 124 1. Increment = —0.151, principal part = —0.154. . dT l(dl dg\ ^. ,, , J ^ ., ~T ~ 2\~l a J' '^"^^® "' ^^^ ^ ™^y "^ either positive or negative, the percentage error in T may be I the sum of the percentage errors in I and g. 6. The percentage error in g may be as great as that in s plus twice that in T. 13. -"^-tl. 15, 2z-^-uy ^ u zx — 2uv 14. hx^ + y^ + xy - z-'). Pages 131, 132 J du ^df_ dl_d6^ ^ /du\ ^df_ df^d^^ ' dx dx~^ dy dx' "■ \dx/y dx "*" dz dx' ' dx dx dy dx dz \dx dy dx*' dldF_dldF 13. 3V3-4. dx ^,dldF ' 14. --3(xcosa+2/cos^+zcoS7). dy "^ dz dy p dz dy dx dx dy p 5. - = -^T TT-^TT- 14. _£ Page 135 j_ ^l^y_-2^£-^^ (a;_i)+2(i/-2)+2(z-2)=0. 2. Normal, y + 4 x = 5, z = 3. Tangent plane, x — 4 j/ — 14 = 0. a; - 3 2/ - 4 g - 5 01^ c n 3. — g— = ^ = -3^ , 3x + 42/ — 5z = 0. a; — 5 ?/ — 1 z — 3 ,^ „ , ^ 4. "l— = "^-5— = ^^. a: + 52/-32-l =0. a;-5 7/-lz-l cicn 5. ^rr'"~^ = ~6~' a;-42/-6z + 5=0. 6. x + z = y-z = db V2. Pages 138, 139 1. The box should have a square base with side equal to twice the depth. 2. The cylinder and cone have volumes in the ratio 3 : 2 and lateral surfaces in the ratio 2 : 3. 4. The center of gravity of the triangle ABC, INDEX The numbers refer to the pages. Acceleration, along a straight line, 33. angular, 34. in a curved path, 90, 91. Angle, between directed lines in space, 79, 80. between two plane curves, 64. Approximate value, of the incre- ment of a function, 14, 15, 118-120. Arc, differential of, 72. Continuous function, 10, 113. Convergence of infinite series, 107- 111. Curvature, 73. center and circle of, 75. direction of, 67. radius of, 74. Curve, length of, 70. slope of, 11. Dependent variables, 2, 115. Derivative, 12. directional, 129. higher, 28, 29, 114. of a function of several vari- ables, 124-127. partial, 114. Differential, of arc, 72. of a constant, 20. of a fraction, 22. of an nth power, 22. of a product, 21. of a sum, 20. total, 120, 121. Differentials, 15. exact, 130, 131. of algebraic functions, 19-31. of transcendental functions, 49- 62. partial, 120. Differentiation, of algebraic func- tions, 19-31 Differentiation, of transcendoital functions, 49-62. partial, 113-139. Directional derivative, 129. Direction cosines, 80, 81. Direction of curvatiu^, 67. Divergence of infinite series, 107- 111. Exact differentials, 130, 131. Exponential fimctions, 56-62. Function, continuous, 10, 113. discontinuous, 10. explicit, 1. imphcit, 2, 127, 128. irrational, 2. of one variable, 1. of several variables, 113. rational, 2. Functions, algebraic, 2, 19-31. exfKinential, 56-62. inverse trigonometric, 54-56. logarithmic, 56-62. transcendental, 2, 49-62. trigonometric, 49-53. Ftmctional notation, 3. Geometrical apphcations, 63-84. ImpUcit functions, 2, 127. Increment, 10. of a function, 14, 15, 118, 119. Independent variable, 2. Indeterminate forms, 95-100. Infinitesimal, 7. Infinite series, 106-112. convergence and divergence of, 107-111. Maclaurin's, 106. Taylor's, 106. Inflection, 67. 161 162 INDEX Length of a curve, 70. Limit, of a function, 5. .sin 5 .„ of -^,49. Limits, 4-9. properties of, 5, 6. Logarithms, 56, 58. natural, 58. Maclaurin's series, 106. Maxima and minima, exceptional types, 45, 46. method of finding, 42, 43. one variable, 39-48. several variables, 136-138. Mean value theorem, 101. Natural logarithm, 58. Normal, to a plane curve, 63. to a surface, 133, 134. Partial derivative, 114. geometrical representation of, 116, 117. Partial, differentiation, 113-139. differential, 120. Plane, tangent, 134. Point of inflection, 67, 68. Polar coordinates, 77-79. Power series, 110, 111. operations with, 111. Rate of change, 32. Rates, 32-38. related, 35. Related rates, 35. RoUe's theorem, 94. Series, 106-112. convergence and divergence of, 107-111. Maclaurin's, 106. power, 110, 111. Taylor's, 106. Sine of a small angle, 49. Slope of a curve, 11. Speed, 85. Tangent plane, 134. Tangent, to a plane curve, 63. to a space curve, 81-83. Taylor's, theorem, 102. series, 106. Total differential, 120, 121. Variables, change of, 30, 127. dependent, 2, 115. independent, 2. Vector, 85. notation, 88. Velocities, composition of, 89, 90. Velocity, components of, 86, 87. along a curve, 85-89. along a straight line, 32. angular, 34. \ V r WORKS OF H. B. PHILLIPS, PH.D. PUBLISHED BY JOHN WILEY & SONS, Inc. Differential Equations. Second Edition, Rewritten. vi+116 pages. $1.50 net. Analytic Geometry. vii + 197 pages. $1-75 net. Differential Calculus. V + 162 pages. $1.50 net. Integral Calculus. V + 194 pages. $1.75 net. 5 by 7}4. Illustrated. Cloth, 5 by 7M. Illustrated. Cloth, 5 by 7H. Illustrated. Cloth, 5 by 7M. Illustrated. Cloth, Differential and Integral Calculus. In one volume. $2.50 net. INTEGRAL CALCULUS BY H. B. PHILLIPS, Ph.D. AssocicUe Professor of Malhematics in the MassachttseUt Institxde of Technology r- NEW YORK JOHN WILEY & SONS, Isc Lo(ax>N: CHAPMAN & HALL, LmtTBD Copyright, 1917, BY H. B. PHILLIPS Stanbope ]press f, H.GILSON COMPANY BOSTON, U.S.A. 5-25 PREFACE This text on Integral Calculus completes the course in mathematics begun in the Analytic Geometry and continued in the Differential Calculus. Throughout this course I have endeavored to encourage individual work and to this end have presented the detailed methods and formulas rather as suggestions than as rules necessarily to be followed. The book contains more exercises than are ordinarily needed. As material for review, however, a supplementary list of exercises is placed at the end of the text. The appendix contains a short table of integrals which includes most of the forms occurring in the exercises. Through the courtesy of Prof. R. G. Hudson I have taken a two-page table of natural logarithms from his Engineers' Manual. I am indebted to Professors H. W. Tyler, C. L. E. Moore, and Joseph Lipka for suggestions and assistance in preparing the manuscript. H. B. PHILLIPS. Cambridge, Mass. June, 1917. CONTENTS PAOsa HAFTXB 1-13 ^ I. Integration II. Formulas and Methods of Integration 14-34 III. Definite Integrals 35-46 IV. Simple Areas and Volumes 4/-59 V. Other Geometrical Applications 60-69 VI. Mechanical and Phtsical Appucations 70-89 VII. Approximate Methods 90-96 VIII. Double Integration 97-111 IX. Triple Integration 112-125 X. Differential Equations 126-lo6 Supplementary Exercises 157-170 Answers 171-185 Table of Integrals 186-189 Table of Natural Logarithms 190-191 Index. 193-194 INTEGRAL CALCULUS CHAPTER I INTEGRATION 1. Integral. — A function F ix) whose differential is equal to / (x) dx is called an integral of / {x) dx. Such a function is represented by the notation / / (x) dx. Thus F (x) = Jf (x) dx, dF (x) = / (x) dx, are by definition equivalent equations. The process of finding an integral of a given differential is called integration. For example, since d (x-) = 2 x dx, /^ 2 X dx = x^. Similarly, I cos xdx = sin x, j e' dx = e'. The test of integration is to differentiate the integral. If it is correct, its differential must be the expression integrated. 2. Constant of Integration. — If C is any constant, d[F{x) +C] =dF{x). If then F (x) is one integral of a given differential, F {x) + C is another. For example, j 2xdx = x^ -\- C, / cos X dx = sin X + C, "where C is any constant. 2 Integration Chap. 1 We shall now prove that, if two continuous junctions of one variable have the same differential, their difference is constant. Suppose Fi {x) and F^ {x) are functions having the same differ- ential. Then dFi {x) = dFi (x). I Let y = Fi (x) - Fi {x) and plot ^ =• the locus representing y as a X function of x. The slope of this Fig. 2. locus is dy-^ dF^jx) -dF,{x) ^Q dx dx Since the slope is everywhere zero, the locus is a horizontal line. The equation of such a line \s, y = C. Therefore, Fi (x) - F, (x) = C, which was to be proved. If then F (x) is one continuous integral of / (x) dx, diTiy other continuous integral has the form Jf{x)dx = F{x) + C. (2> Any value can be assigned to C. It is called an arbitrary constant. 3. Formulas. — Let a and n be constants, u, v, w, variables. I. I du db dv zt dw = I du ± j dv zt j dw. n. fa du = a I du. u»^' III. / u" da = ^ , , + C, if n is not -1. J n + 1 IV. fu-^du=f^ = lnu + C. Art. 3 Formulas 3 These formulas are proved by showing that the differential of the right member is equal to the expression under the integral sign. Thus to prove III we differentiate the right side and so obtain \n+ I / n + 1 Formula I expresses that the integral of an algebraic sum of differentials is obtained by integrating them separately and adding the results. Formula II expresses that a constant factor can be trans- ferred from one side of the symbol I to the other without changing the result. A variable cannot be transferred in this way. Thus it is not correct to write I xdx = X I dx = xi^. Example 1. / x* dx. Apply Formula III, letting u = x and n = 5. Then dx = du and / Ex.2. CzVidx. By Formula II we have fs Vx dx = 3 fxs dx = ^+C = 2x^ + C. Ex. 3. Ax - 1) (x + 2) dx. We expand and integrate term by term. J{x - 1) (x + 2) dx = J {7^ + X - 2) dx = ix3 + Ax2-2x + C. 4 Integration Chap. 1 Ex. 4. I dx. Dividing by a^ and using negative exponents, we get In a: + 2a;-i-^a:-2 + C 2 1 a; 2x2 — '• •V.J /V2 ^x. 5. / V2 X + 1 dx. li u = 2 X -{- I, du = 2 dx. We therefore place a factor 2 before dx and | outside the integral sign to compensate for it. rV2 X + 1 dx = ^ r(2x + l)^2dx = ^ fu^ du Apply IV with w = x^ + 1. Then du = 2xdx and /xdx 1 r2xdx 1 Tdw 1, , /^ i ^/ , ■ v , ^ By division, we find 4x4-2 ^ , 4 2x- 1 ' 2x-l Therefore Art. 4 Motion of a Particle fi EXERCISES Find the values of the following integrals: Vi (x2 + 2x + 1) dx. / ^"' J x^+ax + ft*^- (2 X + a) dx J^. /(V^- v^)'dx. l/21- /: Vx- + ax + 6 7. fx (x + a) (x + b) dx. «^' f <'d/ •^ ^^^* J 1 — a a/* "di. ^8. /2£±3dx. ^. jK.-,0« j (x» + i)y-2) ^^^ • //,;_i_x.^d^. ^-L-^- .-./(,-i)'f. X^ C X dx A,y3 o 4. Motion of a Particle. — Let the acceleration of a particle moving along a straight line be a, the velocity v, and the distance passed over s. Then, dv ds a = Consequently, " = 5' " = * dv = adt, ds = vdt. •6 Integration Chap. 1 If then a is a known function of the time or a constant, v= iadt + Ci, s = iv dt + C2. (4) If the particle moves along a curve and the components of velocity or acceleration are known, each coordinate can be found in a similar way. Example 1. A body falls from rest under the constant acceleration of gravity g. Find its velocity and the distance traversed as functions of the time t. In this case dv Hence v=Jgdt = gt + C. Since the body starts from rest, v = when t = 0. These values of v and t must satisfy the equation v = gt -\- C. Hence = g-0 + C, ds whence C = and v = gt. Since v = -r., ds = gtdt and = Jgtdt + C = hQt' + C. When f = 0, s = 0. Consequently, C = and s = \ gf. Ex. 2. A projectile is fired with a velocity Vo in a direction making an angle a with the horizontal plane. Neglecting the resistance of the air, find its motion. Pass a vertical plane through the line along which the particle starts. In this plane take the starting point as origin, the horizontal line as x-axis, and the vertical line as y-axis. The only acceleration is that of gravity acting downward and equal to g. Hence ^ = ^=-g. di^ ' dt^ ^ Art. 6 CtJBVEs WITH A Given- Slope Integration gives, dx di = Ci |=-^ + c,. dx dii When ^ = 0> "j: ^^^ ~jI are the components of t'o. Hence C\ = t'o cos a, Cj = Vo sin a, and dx -J- = Vo cos a, at dy -^ = Vosma — gt. Integrating again, we get x = v4 cos a, y = Vot sin a — I gf-, Fig. 4. the constants being zero because x and y are zero when t = 0. 5. Curves with a Given Slope. — If the slope of a curve Is a given function of x, l=/w. then dy =f (x) dx and y = //(x) dx-[-C is the equation of the curve. Since the constant can have any value, there are an infinite number of curves having the given slope. If the curve is required to pass through a ^ven point P, the value of C can be found by substituting the coordinates of P in the equation after integration. Fig. 5. 8 Integration Chap. 1 Example 1. Find the curve passing through (1, 2) with slope equal to 2 x. In this case ax Hence y = j2xdx = x'^ + C. Since the curve passes through (1, 2), the values x = 1, y = 2 must satisfy the equation, that is 2 = 1 + C. Consequently, C = 1 and y = x^ -\- 1 is the equation of the curve. Ex. 2. On a certain curve d^y — - = r dx^ ^• If the curve passes through (— 2, 1) and has at that point the slope — 2, find its equation. By integration we get At (- 2, 1), x = -2 and ^ = - 2. Hence - 2 = 2 + C, or C = — 4. Consequently, y =J{h x2 - 4) dx = i rr^ - 4x + C. Since the curve passes through (—2, 1), 1 = - I + 8 + C. Consequently, C = — 5f , and y = \a^ — 4:X — bl is the equation of the curve. Arte Separation of the VARLtBLES 6. Separation of the Variables. — The integration formu- las contain only one variable. If a differential contains two or more variables, it must be reduced to a form in which each term contains a single variable. If this cannot be done, we cannot integrate the differential by our present methods. Example 1. Find the curves such that the part of the tangent included between the coordinate axes is bisected at the point of tangency. Let P (x, y) be the point at which AB \& tangent to the curve. Since P is the middle point oi AB, OA =2y, OB = 2x. The slope of the curve at P is dy dx OA OB This can be written y X Since each term contains a single variable, we can integrate and so get In y + In X = C. This is equivalent to Inxy = C. Hence xy = e^ = k. C, and consequently k, can have any value. The curves are rectangular hyperbolas with the coordinate axes as asymptotes. Ex. 2. According to Newton's law of cooling, when k is constant, a the temperature of the air, and 6 the temperature at the time t of a body cooling in the air. Find ^ as a function of t. 10 Integbation Chap. 1 Multiplying by dt and dividing by 6 — a, Newton's equation becomes d — a Integrating both sides, we get \n{d- a) = -kt-\- C. Hence e - a = e-*'+<^ = e^e-*'. When t = 0,\ete = do. Then do— a = e^e" = e^, and so d - a= {do- a) e-*' is the equation required. Ex. 3. The retarding effect of fluid friction on a rotating disk is proportional to the angular speed co. Find co as a function of the time t. The statement means that the rate of change of w is pro- portional to oj, that is, do) J , It = ^'"' where k is constant. Separating the variables, we get — = kdt, 0} whenco \n 0} = kt -\- C , and CO = e*^'+<^ = e^g*'. Let Wo be the value of w when t = 0. Then 050 = e^'^e^ = e^. Replacing e^ by wq, the previous equation becomes CO = OJoC*', which is the resuJt required. Art. 6 Separation of the Variables 11 Ex. 4. A cylindrical tank full of water has a leak at the bottom. Assuming that the water escapes at a rate pro- portional to the depth and that ^ of it escapes the first day, how long will it take to half empty? Let the radius of the tank be a, its height h and the depth of the water after t days x. The volume of the water at any time is ira^x and its rate of change , dx This is assumed to be proportional to x, that is, Va^ "77 =KX, at where k is constant. Separating the variables, irar dx , -. = kat. X Integration gives 7ra2 In X = kt + C. When t = the tank is full and x = h. Hence TraHnh = C. Subtracting this from the preceding equation, we get xa^ In T = kt. n When t = 1, X = -^jj h. Consequently, xa' In ^ji = k. When X = \h, to,^ In -r , , . n, In^ « c- 1 < = ^^ = j^ = 6.5/ days. 12 Integration Chap. 1 -y EXERCISES V 1. If the velocity of a body moving along a line ia v = 2 1 + 3 fl, fiiid the distance traversed between t = 2 and i = 5. /I. Find the distance a body started vertically downward with a velocity of 30 ft. /sec. will fall in the time t, ./ 3. From a point 60 ft. above the street a ball is thrown vertically upward with a speed of 100 ft. /sec. Find its height as a function of the time. Also find the highest point reached. . ^ 4. A rifle ball is fired through a 3-inch plank the resistance of which causes a negative constant acceleration. If its velocity on entering the plank is 1000 ft. /sec. and on leaving it 500 ft. /sec, how long does it take the ball to pass through? \^\, 6. A particle starts at (1, 2). After i seconds the component of its '' velocity parallel to the x-axis is 2 < — 1 and that parallel to the t/-axis is 1 — <. Find its coordinates as functions of the time. Also find the. equation of its path. y' 6 A bullet is fired at a velocity of 3000 ft. /sec. at an angle of 45° from a point 100 ft. above the ground. Neglecting the resistance of the air, find where the bullet will strike the ground. 7. Find the motion of a particle started from the origin with velocity Vo in the vertical direction, if its acceleration is a constant X in a direc- tion making 30° with the horizontal plane. t/ 8. Find the equation of the curve with slope 2 — x passing through (1, 0). y^"9. Find the equation of the curve with slope equal to y passing through (0, 1). ^ 10. On a certain curve 1 = ^^+3. If the curve passes through (1, 2), find its lowest point. 11. On a certain curve ■J- = X— 1. / If the curve passes through (— 1, 1) and has at that point the slope 2, find its equation. / 12. On a certain curve v/ If the slope is — 1 at x = 0, find the difference of the ordinates at x = 3 and X = 4. L' Art. 6 Separation- of the Vablables 13 13. The pressure of the air p and altitude above sea level h are con- noted by the equation dh '"P' where k is constant. Show that p = po^"**, when pjis the pressure at sea level. 14. Radium decomposes at a rate proportional to the amount present. If half the original quantity disappears in 1800 years, what percentage disappears in 100 years? 15. When bacteria grow in the presence of imlimited food, they increase at a rate proportional to the number present. Express that number as a function of the time. 16. Cane sugar is decomposed into other substances through the presence of acids. The rate at which the process takes place is propor- tional to the mass x of sugar still unchanged. Show that x = cc~^. What does c represent? 17. The rate at which water flows from a small opening at the bottom of a tank is proportional to the square root of the depth of the water. If half the water flows from a cylindrical tank in 5 minutes, find the time required to empty the tank. 18. Solve Ex. 17, when the cylindrical tank is replaced by a conical funnel. 19. A sum of money is placed at compound interest at 6 per cent per annum, the interest being added to the principal at each instant. Ilovr many years will be required for the sum to double? 20. The amount of light absorbed in penetrating a thin sheet cf water is proportional to the amount falling on the surface and approxi- mately proportional to the thickness of the sheet, the approximation increasing as the thickness approaches zero. Show that the rate of change of illumination is proportional to the depth and so find the illumination as a function of the depth. cL CHAPTER II FORMULAS AND METHODS OF INTEGRATION 7. Formulas. — The following is a short list of integra- tion formulas. In these u is any variable or function of a single variable and du is its differential. The constant is omitted but it should be added to each function determined by integration. A more extended list of formulas is given in the Appendix. u" du = — r^ , if n is not — 1. n + 1 TT r*^" 1 n. / — = In u. J u in. / cos udu = sin u. IV. / sin udu = — cos u. \/ ^ V. / sec** udu = tan u. VI. / esc** udu = — cot u. v/ ^ Vn. / sec u tan udu = sec u. VIII. / CSC u cot u du = — CSC u.^ €> IX. I tan a du =4 — In cos u. X. / cot u du = In sin u. v_^ \ XI. / sec u du = In (sec u + tan u). 14 .\rt 8 Integration by StTBJsTrrmox' 17 Xn. Jcscuda = ln(cscu-'^^^™^- Letu=xV3, xm. r_±= = sin-^-* /flit 1 U y du u'* + a** a a — w- XV. / =^^= = - sec ^ — \^ I r XVI • ^" r di J II \ ir r-74i= = In (u + VII^T^. xvn. rT±i-^ = -Lin!l^. \ J u'^ - a" 2a u -{-a \ XVm. Tc- du = e*. Any one of these formulas can be proved by showing that the differential of the right member is equal to the expression under the integral sign. Thus to show that / sec M dw = In (sec u + tan u), we note that J , / , ^ . (sec u tan u + sec' u) du , d In (sec u + tan u) = ;-— = sec m du. sec u -\- tan u 8. Integration by Substitution. — When some function of the variable is taken as u, a given differential may assume the form of the differential in one of the integration formulas or differ from such form onlj'^ by a constant factor. Inte- gration accomplished in this way is called integration by substitution. Ekch differential is the product of a function of u by du. More errors result from failing to pay attention to the du * In Formulas XIII and XV it is assumed that sin-* - is an angle in the 1st or 4th qvAdrant, wid sec-* - an angle in the 1st or 2nd quadrant. In other cases the algebraic sign of the result must be changed. I Methods of Integration Chap. U 4ie cause. Thus the student may m Formula III that the integral of a iO write / cos 2 X dx = sin 2 x. ' sm^ xdx _, ?cf-^- + cos ^ X lA. FORMULAS 7. Formt'' ^^ let 2 x = u, dx is not du but \ du and so tion form- ^ C i . i • o . , , cos 2 X arc = 4 / cos m aw = ^ sm w = f sm 2 x. smgle Y< ''J , omitt*' r ^^ by ia'xampZe 1. / sin' x cos x dx. ;*;;. If we let u = sin x, du = cos x dx and / sin' X cos xdx — j u^du = \u* + C = j sin* x + C. Ex.2. Jf We observe that sin ^ x dx differs only by a constant fa tor from the differential of 1 + cos ^ x. Hence we let w = 1 + cos 3 X. Then dw = — I sin ^ X dx, sin ^ x dx = — 3 dw, , r sin ^ xdx ^ rdw „, , ^ and / -— r-^ — r— = — 3/ — = -31nw + C J 1 + cos f X J u = - 3 In (1 + cos ^ x) + C. Ex. 3. I (tan x + sec x) sec x dx. Expanding we get / (tan X + sec x) sec x dx = / tan x sec x dx + / sec^ x dx = sec X + tan x + C n ^ r 3dx ■ JEx, 4. / , J V2-3x» Art 8 Integration by Substitutiox' 17 This resembles the integral in formula XIII. Let u =x Vs, « = V2. Then du = Vsdx and r 3 — r 3dx ^ I V3 ^ /- r du ■ J V2-Zx^ J Va2 - u^ ^J Va2 - u^ = VSsin-i - + C = V3 sin-i^-^ + C. o V2 Ex. 0. / * f V4 <2 - 9 This suggests the integral in formula XV. Let w =2i, a = 3. Then / rf^ _ /^ 2dt _ r du t V4 f 2 - 9 ~ J 2 f V4 <2 - 9 ~ J w Vw2 _ a- = - sec~^ - + C = 7, sec~^ tt + C'- a a 3 3 J V2 2:2 + 1 + This may suggest formula XVI. If, however, we let u = X V2, du = V2 dx, which is not a constant times x dx. We should let M = 2x2 + 1. Then xdx = I du and r xdx 1 rdu 1 r . J = |V^4-C = ^V2x2 + l+C. Ex. 7. / e**° ' sec2 x dx. If M = tan X, by formula XVIII / c**°' sec2 X dx = / e" du = e" + C = e**° ' 4- C. 18 . Formulas and Methods of Integration Chap. 2 EXERCISES Determine the values of the following integrals: 1. J (sin 2 X - cos 3 x) dx. 21. J cos« x sin a; dx. 3. Jsin (n< 4- a) d<. i 23. ^~ 4. fsec^^edd. 24. r «ec^ (ax) dx •^ -^ 1 + tan (ax) 6. ycsc ^ cot ^ dd. I 25. r ^^^ sec" X dx + 2 tan X cos 2 X dx sin 2x' V3— 2x2 6. Jcos^sin^de. 26 f— ?i^ J 3 X- + / dx cos" X 27. I — - J "^ X + 4 dx V3 x" — 4 ■ J sin"x ' 29. r ^i^ — sin X dx -^ V 7 x" + 1 cos* X 30. r__zE_ sin" X 29. 11. J f CSC 2 — cot- j CSC -de 12. J'cos(x2— l)xdx. 31. 32. / dx VTx^ C- J X Va"x" / / dx 3- 4x" dx 13. r+ff ^ dx. cos"3x gg /' (3x- 2)dx 14. r(8ecx- 1)2 dx. ' -^ ^4-x" /• 2x + 3 . „ rl — sin X J 1 34. j dx. 15. I dx. -^ V x" + 4 ^ cosx . ' /x -p 4 4 X" - 5 '^* 17 r (co8x + sinx)" , 36. f ^^~^ // ntj f COS . sin" X cos X dx. K ^ '* J ^2^ dx. cos X dx sin" X sin X cos x dx *n r ■, oo /•^''in X C( 19. I tan' X sec* x da?, JttJ. I — ;== J ) J V 2 — sin" X 20. r8ec"xtanxdxl , 89. r_22££i^. ^ ^ ^ J 1 + sin" X Art 9 Integrals Co.vtai.vi.vg a Qctadratic Expression 19 sec-xdx , -_ r^'dx 40. / sec-xdx ^ ^ rj tanx Vtan^x- l' * -^ 1 + e*' Vl-cosd" 60- /« '^• • /x[4-(lnx)^r 6L j^^ r sin X cos x ax •^ v'cos- X — siii^ X 52. /*_^_«£ 45. r-^ e-»^ x^ CO r g°'' <^J^ I k2^ 63. , dx 46. /e-..dx. ^^^--,g ^^ 47. /(."+e-)«dx. 64. /e^ + e- 9. Integrals Containing ax^ + 6x + c. — Integrals con- taining a quadratic expression ax- -\-hx -\- C can often be reduced to manageable form by completing the square of 03? + hx. Example 1. / ^ ., , f — j-^- Completing the square, we get 3 x2 + 6 a; + 5 = 3 (x2 + 2 X + 1) + 2 = 3(x + 1)- + 2. If then M = (x + 1) Vd, du f dx ^ r rf (x + 1) ^ _i_ r_( J 3x2 + 6x + 5 ~J 3(x + l)2 + 2 V3 J w- + 2 = 1 tan-(£±lI^ + C. Ve V2 J V2 - 3 X - The coefficient of x- being negative, we place the terms x* and 3 X in a parenthesis preceded by a minus sign. Thus 2 - 3 X - x2 = 2 - (x2 + 3 x) = V- - (a; + iY- 20 Formulas and Methods of Integration Chap. 2: If then, w = x + I, we have r 2dx ^ 2 r J V2-dx-x^ J Ex.3. ^ (2^-1)^^- , = 2sin-i f-7^ + C. /; V4 x2 + 4 a: + 2 Since the numerator contains the first power of x, we resolve the integral into two parts, dx r (2x-l)dx ^1 r (Sx-\-4:)dx ^ C J V4rr2 -4-4 3:4-2 4 J V4rr2 4-4 3:4-2 J ^ V4x2-f4a:+2 4«^ V 4x^+4 x-h2 ^ V4x2+4x+2 In the first integral on the right the numerator is taken equal to the differential of 4 a;^ + 4 x + 2. In the second the numerator is dx. The outside factors \ and — 2 are chosen so that the two sides of the equation are equal. The first integral has the form The second integral is evaluated by completing the square.. The final result is J* {'2,x—\)dx 1 / . „ , . r-x >/4x2+4x + 2 2 - In (2x -f l-f V4x2 + 4x + 2) + C. EXERCISES 1 r__— ^£__ 7 r (2 X -f- 5) dx ' Jx2-f6x4-13" J4x«— 4x— 2* J V9 -1- 4 -r J. tJ (2 X - 1) dx ^- •^V 2-H4x-4x i' -^Vax^-ex + i 3^ C dx ^- J 3 x2 + 2 X -h 2 >•/ X dx j-2x (2x rf:c "" J (2x 4-1)^4x^-1- 4 x-1 VS x2 + 4 X -h 2 f (2 X -1- 3) dx. /; VI+6X-5X-' 11. r .f'^r^V'^,. - J (x^— 2x 4- 3)1 •^ (x- 3) V2x2- 12x4-15 ^2. J y- 6 --dx. dx ^^ r e'dx • J (i-|-o)(x-f 6)" ^^' J (i-|-o)(x4-6) c/ 2e*^4-3e*— 1 •• J Integrals of Trigonometric Fdnctioks 21 10. Integrals of Trigonometric Functions. — A power of a trigonometric function multiplied bj' its differential can be integrated by Formula I. Thus, if w = tan x, I tan^ X • sec- xdx = I u* du = ^ tan^ x -\- C. Differentials can often be reduced to the above form by trigonometric transformations. This is illustrated by the following examples. Example 1 . I sin* x cos* x dx. If we take cos xdx as du and use the relation cos^ x = 1 — sin^ X, the other factors can be expressed in terms of sin X without introducing radicals. Thus / sin* X cos' xdx = j sin* x cos* x • cos x dx = I sin* a; (1 — sin* x) dsinx = } sin* a; — | sin'^ a: + C Ex. 2. / tan' x sec* x dx. If we take sec* x dx as du and use the relation sec* x = 1 + tan- x, the other factors can be expressed in terms rf u = tan x A^nthout introducing radicals. Thus I tan' X sec* xdx = j tan' x • sec* x • sec* x dx = j tan' X (1 + tan* x) d tan x = itan*x+itan«x + C. Ex. 3. I tan^xsec'xrfx. ^2 Formulas and Methods of Integration Chap. i\ If we take tan x sec x dx = d sec x as du, and use the rela- tion tan^ X = sec^ x — 1, the integral takes the form / tan^ X sec' xdx= j tan^ x • sec^ x • tan x sec x dx = j (sec^ a; — 1) sec^ X'dsecx = i sec* X — I sec' x-{- C. Ex. 4. I sin 2 a; cos 3 x dx. This is the product of the sine of one angle and the cosine of another. This product can be resolved into a sum or difference by the formula sin AcosB = ^ [sin (A + 5) + sin (A - B)]. Thus ~~^ sin 2 X cos Sx = ^ [sin 5 a; + sin (— x)] = I [sin 5 re — sin x] Consequently, / sin 2 X cos dxdx = ^ j (sin 5 a; — sin x) dx = — tV cos 5 X + ^ cos x + C Ex. 5. / tan* x dx. If we replace tan^ x by sec^ a; — 1, the integral becomes / tan* a: da; = I tan' a; (sec- a; — l)dx = jtan^x — / tan^xdx. The integral is thus made to depend on a simpler one / tan' X dx. Similarly, / tan' xdx = j tail x (sec^ a; — 1) da: = ^ tan^ a: + In cos x. Hence finally I tan* xdx = I tan* x — | tan^ x — In cos x + C. Art. 12 Trigonometric SuBSTmrrioxs 23 11. Even Powers of Sines and Cosines. — Integrals of the f^rm sin" X cos" X dx, f^ where m or n is odd can be evaluated by the methods of Art. 10. If both m and n are even, however, those methods fail. In that case we can evaluate the integral by the use of the formulas 1 — cos 2 U sin^u = cos'^u = sin u cos u = 1 + cos 2 u 2 ' sin 2u 2 (11) dx Example 1. / co^xdx. By the above formulas I cos^ xdx = I (cos^ xy dx = j I ^ j = r(i + |cos2x + icos22a;) = f [\ + ^ cos2x + 1(1 -\- cos^x)]dx — f a; + 4 sin 2 X + 3^5 sin 4 X + C. Ex. 2. / cos^ X sin- x dx. I cos* X sin* xdx = I J sin* 2xdx= j 1(1 — cos Ax) dx = I 2: — ^V sin 4 X + C. 12. Trigonometric Substitutions. — Differentials con- taining Va- — X-, Va2 4- X-, or Vx- — a*, which are not 24 Formulas and Methods of Integration Chap. 2 reduced to manageable form by taking the radical as a new variable, can often be integrated by one of the folbwing substitutions: For Va^ — x^, let x = a sin Q. For Va^ + x^, let a; = a tan Q. For V'a:^ — a^, let x = a sec Q. Example 1. l Va^ — x^dx. Let X = a sin 6. Then V a^ — x"^ = a cos 0, dx = a cos c?^. Consequently, fVa''- x^dx = a2 Ccos^ddd = ^(^ + ^sin 2 ^W C. Since a: = a sin d, • , a^ 1 • « « • « /, a; Va^ — x"^ 6 = sm~^ - > - sm 2 = sm cos 6 = — - — 5 • a 2 a^ . Hence finally fV^^^r^^dx = ^sin-i-+^Va2-x2 + C. J 2 a 2 ^^•2- J ^^2^aY If we let a; = a tan ^, 3:^ + 0^ = a^ sec^ 6, dx = a sec^ d^j and /(^^ = i/s"S-0 = ^/^^^'^^^ = iT^ (^ + sin cos e) + C. Since a; = otan0, = tan-^-, sin cos = -t-i — :.■ ' a a^ -\- x^ Hence / dx _ 11" _i 5 , «a; "I ^ (a;2 + a2)2-2a3L a "^a^ ^^^^J "^ ^' Art 12 it6 : 77 Trigonometric ScBSTrrunoNis 1. J sin' X dx. 2. J cos'xdz. 3. J (cos z + sin x)» dx, 4. J cos* z sin* X dx. 5. J sin* § I cos* \xdx. 6. rsin«3»coe»3ddff. 7. JCcos* — sin* 9) sin ad0 g /• cos^ X dx J 1 — sin X g /• cos- zdx J sin X 10 r sin* g dg J cos g 11. I sec*xdx. 12. jcsc^ydy. 13. j tan*xdx. 14. pec3g + tan»g J sec + tan fl EXERCISES 21. J sin* ax di. 22. J cos* ax dx. 23. Jcoe* X sin* X dx. 24. Jcos* i X sin* ^ rdx. 26. I sin'xdx. 26. j^_^. . 16. Jtan ^ X sec* ^ x dx. 16. j tan* 2 x sec' 2 x dx. 17. Jcot'xdx. 18. I tan'xdx. j^g /• cos*xdx J sin* X y 20. J sec' X C8C X dx. sinx 27. ( - •/ 1 + cos X 28. j/l +Bined0. 29. J*Vx*— a«di. 80. fV^^+^'dx. 31. f. ^'^ . 32. f ^-^ ■ •^ (x*- o*)* (a*— x2)i 36. fx^V3? + a*dx. 37. r-^^=. -/ X* V I* + a* 88. f^x'— 4x + 5dx, 39. r ^-^---^^^^ •^ V2— 2x— 4x» 26 Formulas and Methods of Integration Chap, 2 13. Integration of Rational Fractions. — A fraction, such as x^ -2x-S whose numerator and denominator are polynomials is called a rational fraction. If the degree of the numerator is equal to or greater than that of the denominator, the fraction should be reduced by division. Thus = X + 2 ' x'-2x-Z ' ' x2-2a;+3 A fraction with numerator of lower degree than its denomi- nator can be resolved into a sum of partial fractions with denominators that are factors of the original denominator. Thus lOx + G ^ lOx + 6 ^ 9 , 1 x^-2x-^ (a;-3)(x+l) x-3"^x + l' These fractions can often be found by trial. If not, pro- ceed as in the following examples. Case 1. Factors of the denominator all of the first degree ^id none repeated. „ , rx* + 2x + 6. Dividing numerator by denominator, we get x* + 2x-\-Q ^^_^ 3x2 + 6 ^ x^ -\- x^ — 2x 7? -\-x^ — 2x 3x2 + 6 = x-l + X (x - 1) (x + 2) Assume 3 x2 + 6 ^A^. ^ JL. ^ X (x - 1) (x + 2) X ' X - 1 ' X + 2 The two sides of this equation are merely different ways of Art. 13 Integration of Rational Fractions 27 writing the same function. If then we clear of fractions, the two sides of the resulting equation 3 x2 + 6 = A (x - 1) (x + 2) + fix (x + 2) + Cz (a; - 1) = {A-{-B-\-C)x^-\-{A-\-2B-C)x-2A are identical. That is A+B + C = 3, A-\-2B-C = 0, -2A = Q. Solving these equations, we get A = - 3, B = S, C = 3. Conversely, if A, B, C, have these values, the above equa- tions are identically satisfied. Therefore rx^ + 2x + 6 , n 1 3 , 3 , 3 \ , J^ +a:^-2x ^=Jl^-^-x+^^ + ^^^ = |x2-x- 31nx + 31n(x- 1) +31n(x + 2)+C z X The constants can often be determined more easily by substituting particular values for x on the two sides of the equation. Thus, the equation above, 3 x2 + 6 = A (x - 1) (x + 2) + fix (x + 2) + Cx (x - 1) is an identity, that is, it is satisfied by all values of x. In particular, if x = 0, it becomes 6= -2A, tihence A = — 3. Similarly, by substituting x = 1 and X = — 2, we get 9 = 35, 18 = 6 C, whence fi = 3, C = 3. Case 2. Factors of the denominator all of first degree but some repeated. * „ r (8x^ + 7)dx J (x + l)(2x + ] 28 Formulas and Methods of Integration Chap. 2 Assume Corresponding to the repeated factor {2 x -\- ly, we thus introduce fractions with (2 x + 1)' and all lower powers as denominators. Clearing and solving as before, we find A = 1, B = 12, C = - C, D = 0. Hence Case 3. Denominator containing factors of the second degree but none repeated. r Ex.3. /"*-^^t£ + idx. The factors of the denominator are x — 1 and x^ -\- x -\- 1. Assume 4 x^ + x+ 1 ^ A Bx-\-C X^ -\ X - l^ x^ + X -\- l With the quadratic denominator x^ + x + 1, we thus use a numerator that is not a single constant but a linear function Ex + C. Clearing fractions and solving for A, B, C, we find A =2, B = 2, C = 1. Therefore J x^ — 1 J \x — I x^ + X 4- 1/ = 21n (x - 1) + In (x2 + X 4- 1) + C. Case 4. Denominator containing factors of the second degree, some being repeated. Ex.4:. C-f^A'^.dx. (X2+ 1)2' Art 14 I^'TEGRAI;s 29 Assume x' + 1 ^A Bx-hC Dx-\-E X (X^ + 1)2 X ' (X2 + 1)2 ' X- + 1 Corresponding to the repeated second degree factor (x^ + 1)^, we introduce partial fractions having as denominators (x^ + ly and all lower powers of x^ + 1, the numerators being all of first degree. Clearing fractions and solving for A, B, C,D, E, we find A = l, B=-l, C=-l, D=-l, E = l. Hence J X (X2 + 1)2 "^^ J Ix (X2 +1)2 X2 + 1 '^ = In , + ^ tan-i x — „ , , , ,. + C Vx2+ 1 2 2 (x2 + 1) ' 14. Integrals Containing {ax + b)'. — Integrals contain- p ing (ox + 6)9 can be rationalized by the substitution ax-\-h = z'^. If several fractional powers of the same linear function ax + 6 occur, the substitution ax + & = 2" may be used, n being so chosen that all the roots can be extracted. Examvlel. f ^^■ J l + Vi Let X = 2^. Then dx = 2 z dz and = 2 2 - 2 In (1 + 2) + C = 2 Vi - 2 In (l + Vx) + C. Ex.2. rJ2£r3)^^x J r2r- (2x-3)^+l 30 Formulas and Methods of Integration Chap. 2 To rationalize both (2 a; — 3)* and (2 a; — 3)», let 2x-Z = z\ Then J (2a; -3^ + 1 J 2' + 1 J\ 2'+! dz = 3(y - f + 1 - 2 + tan-i2) + C = f (2 a; - 3)^ - § (2 X - 3)^ + (2 a: - 3) ^ - 3 (2a; - 3)« + tan-^ (2a: - 3)« + C. 1. 2. 3. 4. 6. 6. 7. 8. 9. 10. 11. 12. 13. X2- J x' X3 + X2 3a; + 2 2x + 3 dx. EXERCISES L4. x^ + 1 {x'' - 1) x3- 1 dx. dx. xdx r x^ + l ^^^ J X {x^ — "^ r x^— ] J 4x»- J (X + 1) (x + 3) (X + 5) 16xdx (2x-l)(2x-3)(2x-5)' f4^.dx. J x^ — x^ x'dx (x + l)(x- 1)** J dx /(f^y- / dx x^— X* / x'dx (x*— 4)*' xdx / xdx (x* - 4)> / x*dx X*- 1' /dx x3 + 1* x^dx x^ + l' dx X* — X* + x^ — 1 2 x2 + X — 2 /• 2x-= + ; J (x2- / 20. 21. (X2— 1)2 ■'^^" X* + 24 x^ - 8 X (x + D* dx dx. J X i_ ^i /x* — X' x» + 1 dx. 22. /x^/^JT6dx. 23. 24. 26. vT+^- 1 r vx- »/ X / / + 3 dx dx. (X* - 1) (X* + 1) dx y/x+l - Vx— 1 Art. 15 Integration by Parts 31 15. Integration by Parts. — From the formula d (uv) = udv -\- udv we get f^d** udv = d {uv) — V du, whence j udv = uv — j V du. (15) If j vdu is known this gives / v du. Integration by the use of this formula is called integration by parts. Example 1. jlnxdx. dx Let u = \n X, dv = dx. Then du = — , v = x, and X l]nxdx = \nx'X— j x- — ' = xQnx- 1) +C. Ex. 2. / x^ sin x dx. Let u = x^ and dv = sin x dx. Then du = 2 x dx, t; =» — cos x, and / x^ sin xdx = —x^ cos x + I ! ^ sin xdx = —x^ cos x -\- j 2x cos x dx. A second integration by parts with u = 2 x, dv = cosx dx^ gives I 2 X cos xdx = 2xsmx — j 2 sin xdx = 2 X sin a; + 2 cos x + C Hence finally / x* sin xdx = — x^ cos x + 2 x sin x + 2 cos x + C. 32 Formulas and Methods of Lvtegration Chap. 2 The method of integration by parts applies particularly to functions that are simplified by differentiation, like In x, or to products of functions of different classes, like x sin x. In applying the method the given differential must be re- solved into a product u • dv. The part called dv must have a known integral and the part called u should usually be simplified by differentiation. Sometimes after integration by parts a multiple of the original differential appears on the right side of the equation. It can be transposed to the other side and the integral can be solved for algebraically. This is shown in the following •examples. Ex. 3. / Va2 - x^dx /v^ Integrating by parts with u = Va^ — x^, dv = dx, we get — x^dx /v a^ — x^dx = X Va^ — x^ — / , J Va2- Adding a^ to the numerator of the integral and subtracting ^n equivalent integral, this becomes /Va^ — x^dx = x Va^ — x"^— I , dx-^-d? / . J Vo2-x2 J Va'-x^ = X Va^ - x^ - fVa'-x^dx + a^ f /^ - J J Va^-x^ Transposing / ^a^ - ^^ dx and dividing by 2, we get / Va2 -x^dx = ^ Va2 - x^ + ^ sin"^ ^ + C. Ex. 4. I e"* cos hx dx. Integrating by parts with u = e^', dv = cos bx dx, we get /. , e"^ sin 6a; a C n. • v j e'"' cos bx dx = — r r / e"^ sm bx dx. Art. 16 Reduction Formulas 3S Integrating by parts again with u = ef', do = sin hx dx, this becomes /, , e^^sinftx a[ e"'cos6x , a /*_, • t j T €f"cos,hxdx= T r T hr I (f'smhxdxX (b sin hx -\- a cos hx\ a^ P . . , " ^' i P j " ^ J e"sm6x(ix. Transposing the last integral and dividing by 1 + -^, this gives fh sin hx -\- a cos hx\ j e" cos 6a; rfx = e*"' f a2 + 62 / 16. Reduction Formulas. — Integration by parts is often used to make an integral depend on a simpler one and so to obtain a formula by repeated appUcation of which the given integral can be determined. To illustrate this take the integral / sin" X dx, wnere n is a positive integer. Integrating by parts with u = sin*"^ X, dv = sin x dx, we get /sin"xdx= — sin"~^xcosx+ / (n — 1) sin*~'xcos'a;dx = — sin""^ x cos X + (n— 1) I sin""- x (1 — sin' x) dx = —sin""' X cos X -{- (n— 1) / sin""' x dx — (n— 1) / sin'xdx. Transposing the last integral and dividing by n, we get /. „ J sin"~' X cos X , n — I C . , , sin" xdx = 1 I sin"~' x dx. n n J By successive application of this formula we can make / sin" X dx depend on j dx or j sin x dx according as n IS I even or odd. I 34 Formulas and Methods of Integration Chap. 2 Example. j sin^ x dx. By the formula just proved /. « J sin^ X cos X , b C • A 1 sin^ xdx = ^ 1" fi / ^^^ ^ ^^ sin^ x cos x , 5 r sin^ a; cos re , 3 T . „ , "I = g + g|_ ^ + jjsin^a:dxj sin^a;cosa; 5 . , 5 . . 5 . ^ = a H7 sin^ X cos a; — 7^ sin a; cos X + 77; a; + (7. o 24 Id Id EXERCISES 1. \ X cos 2 X dx. 11. i X? e-* dx. 2. Jinx- xdx. 12. J* (x — 1)2 sin (2 x) dx. 3. Jsin-ixdx. 13_ JVi^^T^dx. 4. I X tan-^ X dx. >. , •^ 14. J Va^ + x^dx. 5. fin (x + Va-'' + x'O dx. /, ^ j^ 16. I e^ sin 3 x dx. In X dx J Vx- 1 ^ - ^T 16. I (f cos X dx. 7. J In (In x)^. -^ 8. Jx^ sec- xdx. "• /^-^«i'^2xdx. 9. re-*ln(c* + l)dx. 18. j'sec'^edd. 10. (x^e^dx.. 19. I sin 2 X cos 3 xdx. 20. Prove the formula /_ , V , sec"-'' X tan x , n — 2 /• «_» / s , sec" (x) dx = 1 r I sec"-2 (x) dx. n — 1 n— IJ and xise it to integrate J sec* x dx. 21. Prove the formula and use it to integrate j (o* — x*)' dx. CHAPTER III DEFINITE INTEGRALS 17. Summation. — Between x = a and a: = 6 let f{x) be a continuous function of x. Di\'ide the interval between a and h into any number of equal parts Ax and let X\, Xz, . . , Xn, be the points of di\'ision. Form the sum /(a)Aa:+/(x,)Ax+/(x2)Aa;+ • • • +/(x,)Ax. This sum is represented by the notation Since / (a), /(xi), /(xj), etc., are the ordinates of the curve y = f (x) at X = Xi, X2, etc., the terms / (a) Ax, / (xi) Ax, / (xa) Ax, Fig. 17a. etc., represent the areas of the rectangles in Fig. 17a, and 2^^f (x) Ax is the sum of those rectangles. Example 1. Find the value of V' x^ Ax when Ax = |, The inter%'al between 1 and 2 is divided into parts of length Ax = ^. The points of division are 1{, U, If. Therefore 2^'x2Ax= P.Ax+(|)2Ax-h (1)2 Ax + (1)2 Ax = -«/Ax = ^3.i = 1.97. Ex. 2. Find approximately the area bounded by the X-axis, the curve y = V^, and the ordinates x = 2, x = 4. From Fig. 176 it appears that a fairly good approxima- 35 Fig. 176. 36 Definite Integrals Chap. 3 tion will be obtained by dividing the interval between 2 and 4 into 10 parts each of length 0.2. The value of the area thus obtained is ^Wx Ax={V2-\-V2^+\/2A-\- ' • • +V3^) (0.2) =3.39. The area correct to two decimals (given by the method of Art. 20) is 3.45. 18. Definite and Indefinite Integrals. — If we increase indefinitely the number of parts into which 6 — a is divided, f (x) Ax usually approaches a limit. This limit is called the definite integral of / (x) dx between x = a and x = b. It is represented by / f{x)dx. That is f'f (x) dx = lim 2 V (^) Aa;. (18) the notation The number a is called the lower limit, b the upper limit of the integral. In contradistinction to the definite integral (which has a definite value), the integral that we have previously used (which contains an undetermined constant) is called an in- definite integral. The connection between the two integrals will be shown in Art. 21. 19. Geometrical Representation.— If the curve y = f (x) lies above the a:-axis and a < 6, as in Fig. 17a, / / (x) dx represents the limit approached by the sum of the inscribed rectangles and that limit is the area between x = a and x = b bounded by the curve and the ic-axis. Fig. 19o. Art. 19 Geometrical Representation 37 At a point below the a;-axis the ordinate / {x) is negative and so the product / (x) Ax is the negative of the area of the corresponding rectangle. Therefore (Fig. 19a) ^ / (x) Ax = (sum of rectangles above OX) — (sum of rectangles below OX), and in the limit i: f (x) dx = (area above OX) — (area below OX) (19a) Fig. 196. If, however, a > 6, as in Fig. 196, x decreases as we pass from o to 6, Ax is negative and instead of the above equation we have / (x) dx = (area below OX) — (area above OX). (19b) Example 1. Show graphically that / sin'xdx = 0. The curve y = sin^ x is shown in Fig. 19c. Be- tween X = and x = 2ir the areas above and below the X-axis are equal. Hence 27r r sin'xdx = Ai — ^2 = 0. Fig. 19c. Ex. 2. Show that / €-''■ dx = 2 / e-^' dx. 38 Definite Integrals Chap. 3 The curve y = e~^' is shown in Fig. 19a. It is symmetrical with respect to the ^-axis. The area between x = —1 and a: = is therefore equal to that between x = and X = 1. Consequently Fig. 19d. Je'' dx = Ai-\-A2^ 2 At = 2 fe-^'dx. Jo 2.x: Ax = 1. EXERCISES Find the values of the following sums: 1. ^ xAx, Ax = \. 10 Ax X 3. V]_ VxAx, Ax = \. 4. Show that V 7 J sin X Ax = 1 — cos - approximately. Use a table of natural sines and take Ax = -^^ 5. Calculate x approximately by the formula >i Ax '-s:r + X2 Ax = 0.1. 6. Find correct to one decimal the area bounded by the parabola y = X*, the X-axis, and the ordinates x = 0, x = 2. The exact area is f . 7. Find correct to one decimal the area of the circle x^ + y^ = 4. By representing the integrals as areas prove graphically the following equations : 8. f'sin {2x)dx = 0. Jo J-2t cos^ X dx = 0. ir 10. I sin' xdx = 2 \ sin* x dx. Jo Ja Art 20 Debit ATTV£ of Ab£a 39 It +» xdx 1 +x* = 0. dx •X " J_l 1 + X* Jo l+x* 13. J" fix)dx =j"' fia- x)dx. 20. Derivative of Area. — The area A bounded by a curve y=f{x), a fixed ordinate x = a, and a movable ordinate MP, is a function of the abscissa x of the movable ordinate. Let X change to x + Ax. y The increment of area is AA = MPQN. Construct the rectangle MP'Q'N equal in area to MPQN. If some of the points of the arc PQ are above P'Q', others must be below to make MPQN and MP'Q'N equal. Hence P'Q' intersects PQ at some point R. Let y' be the ordinate of R. Then y' is the altitude of MP'Q'N and so ^A = MPQN = MP'Q'N = y' Ax. Consequently Fig. 20. ^A Ax = y' When Ax approaches zero, if the curve is continuous, y' approaches y. Therefore in the hmit dA . . Let the indefinite integral of / (x) dx be f{x)dx = Fix)-\-C. (20a) I' 40 Definite Integrals Chap. 3 From equation (20a) we then have A = Cf{x)dx = F{x)+C. The area is zero when x = a. Consequently = F(a)+C, whence C = — F (a) and A=F{x) -F(a). This is the area from x = a to the ordinate MP with abscissa X. The area between a: = a&ndx = 6 is then A = F (b) - F (a). (20b) The difference F (h) — F (a) is often represented by the notation F {x) , that is, Fix) = Fib)-F (a). (20c) 21. Relation of the Definite and Indefinite Integrals. — The definite integral I / (x) dx is equal to the area bounded by the curve y = f (x), the x-axis, and the ordinates x = a, x = b. If '*fix)dx = F(x) + C, f^ by equation (20b) this area is F (b) — F (a). We therefore conclude that £ f{x)dx^F{x) ^F{b)-F{a), (21) that is, to find the value of the definite integral j f (x) dx, substitute x = a, and x = bin the indefinite integral J f (x) dx and subtract the former from the latter result. Art 22 Properties of Definite Integrals 41 Example. Find the value of the integral '1 dx r Jo 1 +x» The value required is n dx Jo 1+X2 = tan~^ X I = tan~* 1 tan-iO = 7- 4 22. Properties of Definite Integrals. — A definite inte- gral has the foUowing simple properties: I. rfix)dx= - rfix)dx. n. r / (x) dx = ff ix) dx-^i'f (x) dx. m. /V W dx^ih- a)f(xd, a = x^ = h. The first of these is due to the fact that if Ax is positive when X varies from a to 6, it is negative when x varies from h to a. The two integrals thus represent the same area with different algebraic signs. r Fig. 22a. Fig. 22&. The second property expresses that the area from a to c is equal to the sum of the areas from a to 6 and b to c. This is the ease not only when h is between a and c, as in Fig. 22a, but also when 6 is beyond c, as in Fig. 226. In the latter case sum I f (x) dx is negative and the rf(x)dx-\- rf{x)dx da t/5 is equal to the difference of the two areas. 42 Definite Integrai^ Chap. 3 Equation III expresses that the area PQMN is equal to that of a rectangle P'Q'MN with altitude between MP and NQ. 23. Infinite Limits. — It has been assumed that the limits a and h were finite. If the integral 'J a dx Fig. 22c. approaches a limit when b increases indefinitely, that limit / (x) dx. That is, a f f (x) dx = lim fV (x) dx. (23) t/r. 5=30 ^a If the indefinite integral Jf{x)dx = F{x) approaches a limit when x increases indefinitely, r /(re) dx = lim [F (6) - F (a)] = F (oo) - F (a). »/a 6 = 00 The value is thus obtained by equation (21) just as if the limits were finite. ..00 Example 1. / — »/o 1 dx + x^ The indefinite integral is f dx ^ J l-\-x' tan~^ X. When X approaches infinity, this approaches jr . Hence J/100 r dx = tan~^a; J Art. 24 Infixtte Values op the PuNcnoN iZ r*aa Ex. 2. / cosxdx. £' The indefinite integral sin a: does not approach a limit when X increases indefinitely. Hence r cosxdx has no definite value. 24. Infinite Values of the Function. — If the function / (x) becomes infinite when x = 6, j f{x) dxis defined as the limit Jf (x) dx = lim / / (x) dx, a z=b *J a z being between a and 6. Similarly, if / (a) is infinite, J/ (x) dx = lim / / (x) dx, a z = a *J z z being between a and 6. If the function becomes infinite at a point c between o and b, I f (x) dx is defined by the equation rf{x)dx= rf{x)dx+ rf(x)dx. (24) t/a i/a »/c Example 1. / dx When X = 0, -77= is infinite. We therefore divide the vx integral into two parts: J -I V X J_i V X Jo dx ^ 3 3^ v^ 2 2 44 Definite Integrals Cha|i. 3 dx Ex.2 £ If we use equation (21), we get '^dx^_l '-1 x^ X x: 1 = -2. -1 Since the integral is obviously positive, the result — 2 is absurd. This is due to the fact that -5 becomes infinite x^ when X = Q. Resolving the integral into two parts, we get ri dx ndx , n dx 25. Change of Variable. — If a change of variable is made in evaluating an integral, the limits can be replaced by the corresponding values of the new variable. To see this, suppose that when x is expressed in terms of t, s> fix)dx = Fix) is changed into ' <i>{t)dt = ^{t). f< If to, tif are the values of t, corresponding to Xo, Xi, F (xo) = $ (to), F (x,) = $ (<i), and so F{xr)-F{xo)=^{U)-^{k), that is rf(x)dx= f <f>{t) dt. If more than one value of t corresponds to the same value of X, care should be taken to see that when t varies from k to ti, X varies from Xo to Xi, and that for all intermediate values, / (x) dx = 4> (t) dt. Example. j Va^ — x^ dx. Art. 26 Change of Variables 45 Substituting x = a sind, we find When X = a, sin0 = 1, and = K' When x = —a, sind = — 1 and d = —-. Therefore T r J'' V^^^^dx = a^r cos^ Odd =^ld-\-^sm 2e]' = ^• ~2 "2 Since sin^Tr = —1, it might seem that we could use |t as the lower limit. We should then get "'I cos'dde= -^ s This is not correct because in passing from |t to ^x, crosses the third and second quadrants. There cos 6 is negative and Va^ — 7? dx = {— a cos d) cos Odd, and not o' cos' d d6 as assumed above. EXERCISES Find the values of the following definite integrab: /•4 6. I xlnxdx. 1. 1 sec* a; dr. ■'- J a Va- — 8. J tanxdx. 3. f {x-l)^dx. . (^ xdx - ra\a2( £ _£\ ^-4 Vj2 + 144 Jo >• ^ ^ sin' do. 10. J^ ;^- 46 Definite Integrals Chap. 3 J 13. f e-k'x^xdx. 11. J csc^xdx. •'<' 2 12 »/i X y/x^ — 1 Evaluate the following definite integrals by making the change of variable indicated: r* Vr — 1 16. -^ -dx, x-\=z\ X dz _^ ^ 1^ 18. I ^^-. —-; r-T-f Sin 0=2. Jo 6 — 5 sm + sin^ 9 Jo a^+x^ ^ CHAPTER IV SIMPLE AREAS AND VOLUMES 26. Area Bounded by a Plane Curve. Rectangular Coordinates. — The area bounded by the curve y = / (x), the X-axis and two ordinates x = a, x = b, is the limit approached by the sum of rectangles y Ax. That is, Fig. 26a. Fig. 266. A = hm V y Ax = / ydx= I f{x)dx. (26a) Ai=0 ^^ a *J a *J a Similarly, the area bounded by a curve, the abscissas y = a,y = b, and the y-axis is A = lim 2 X A?/ = / xdy. Ay=0 %J a (26b) Example 1. Find the area bounded by the curve x = 2 -|- y — y- and the i/-axis. 47 48 Simple Areas and Volumes The curve (Fig. 26c) crosses the 2/-axis at y = y = 2. The area required is, therefore, Chap. 4 • 1 and Fig. 26c A = J\dy=£^ {2 + y-y^)dy = 2y + ^-y^ = 4i Ex. 2. Find the area within the circle x^ -{■ y^ = 16 and parabola x^ = 6 ?/. Solving the equations simultaneously, we find that the parabola and circle intersect at P (— 2V3, 2) and Q (2\/3, 2). The area MPQN (Fig. 2M) under the circle is ydx= I \/l6-x2da: = ^x + 4V3. The area MPO + OQN under the parabola is x 2v/3 3.2 S ,- -2V3 6 The area between the curves is the difference MPQN - MPO - OQN = J/tt + J Vs. Ex. 3. Find the area within the hypocycloid x = a sin^ 0, y = a cos^ 4>. Art. 26 Rectangular Coordinates The area OAB in the first quadrant is Jydx= / a CDs' <^ • 3 a sin^ <f> cos <t> d<t> X = 3 a^ I cos* sin'^ 4>d<l) = g\ x a^ 49 Fig. 26c. The entire area is then ^•OAB = lira\ EXERCISES 1. Find the area bounded by the Hne 2 y— S x— 5 =0, the x-axis, and the ordinates x = 1, x = 3. .^. 2. Find the area bounded by the parabola y = 3 x-, the y-axis, and r the abscissas y = 2, y = 4. 3. Find the area bounded hy y' = x, the line y = — 2, and the ordinates x = 0, x = 3. 4. Find the area bounded by the parabola y = 2 x — x- and the X-axis. .y 6. Find the area bounded by y = In x, the x-axis, and the ordinates 'z = 2, X = 8. 6. Find the area enclosed by the ellipse ^ + g = l. a- cr 7. Find the area bounded by the coordinate axes and the curve x* + j/i = a*. 50 Simple Areas and Volumes Chap. 4 8. Find the area within a loop of the curve x^ = y^ (i— y^). y 9. Find the area within the loop of the curve y"^ = (x— I) {x— 2)2. / 10. Show that the area bounded by an arc of the hyperbola xy = k^, the X-axis and the ordinates at its ends, is equal to the area bounded by the same arc, the y-axis and the abscissas at its ends. tX 11. Find the area bounded by the curves y^ = 4. ax, x^ = A ay.~\ y- 12. Find the area bounded by the parabola y = 2x— x"^ and the iline y = — X. / 13. Find the areas of the two parts into which the circle x^ + r/^ = g is divided by the parabola y^ = 2 x. / 14. Find the area within the parabola x^ = 4 y + 4 and the circle a;2 4-^ = 16. 5. Find the area bounded by y* = 4 x, x^ = 4 y, and x'' + y^ = 5. 16. Find the area of a circle by using the parametric equations X = a cos 0, y = a sin 6. fn^ Find the area bounded by the x-axis and one arch of the cycloid. X = a (4) — sin </>), y = a (1 — cos 0). </ 18. Find the area within the cardioid X = a cos d {1 — cos 6), y = a sin 0(1 — cos 6). 19. Find the area bounded by an arch of the trochoid, X = cuj) — b sin </>, y = a — b cos <(>, and the tangent at the lowest points of the curve. 20. Find the area of the ellipse x" — xy + y^ = 3. 21. Find the area bounded by the curve y"^ = ^ — — — and its as- ymptote X = 2 a. 22. Find the area within the curve l+(f)' = '- 27. Area Bounded by a Plane Curve. Polar Coordi- nates. — To find the area of the sector POQ bounded by two radii OP, OQ and the arc PQ of a given curve. Divide the angle POQ into any number of equal parts A^ and construct the circular sectors shown in Fig. 27a. One of these sectors ORS has the area i OR^ A0 = ^ r2 A^. Art 27 Polar Coordinates 51 If a and /S are the limiting values of 6, the sum of all the sectors is then As A^ approaches zero, this sum approaches the area A of the sector POQ. Therefore A= lim V §r2A5= / ^r^ dd. (27) Fig. 27a. Fig. 276. In this equation r must be replaced by its value in terms of d from the equation of the curve. Example. Find the area of one loop of the curve r ^ a sin 2 (Fig. 276). TT A loop of the curve extends from d = to 6 =-. Its area is A = r ^r^ dd = r ^sm^2e) dd =fra- cos 4:d)dd = xa*= 52 Simple Areas and Volumes Chap. 4 EXERCISES 1. Find the area of the circle r = a. 2. Find the area of the circle r = a cos Q. 3. Find the area bounded by the coordinate axes and the line r = asecfe— -^y 4. Find the area bounded by the initial line and the first turn of the spiral r = ae^- \b^ Find the area of one loop of the curve r^ = o^ cos 2 6. 6. Find the area enclosed by the curve r = cos fl + 2. TT) Find the area within the cardioid r = a (1 + cos 0). 8. Find the area bounded by the parabola r — a sec^ -^ and the y-axis. 9. Find the area bounded by the parabola 2a f = . 1 — cos 6 and the radii 9 = -:, 9 = ■^- 4 2 10. Find the area bounded by the initial line and the second and third tiu-ns of the spiral r = aB. 11. Find the area of the curve r = 2 a cos 3 9 outside the circle r = a. 12. Show that the area of the sector bounded by any two radii of the spiral rd = am proportional to the difference of those radii. 13. Find the area common to the two circles r = a cos 9, r — a cos 6 + asin 9. n 14. Find the entire area enclosed by the curve r = a cos' 5 • 16. Find the area within the curve (r — aY = a? {I — ff^). 16. Through a point within a closed curve a chord is drawn. Show that, if either of the areas determined by the chord and curve is a maxi- mum or minimum, the chord is bisected by the fixed point. 28. Volume of a Solid of Revolution. — To find the volume generated by revolving the area ABCD about the X-axis. Inscribe in the area a series of rectangles as shown in Fig. 28a. One of these rectangles PQSR generates a circular Art 28 Volume of a Solid of Revolution' 53 cylinder with radius y and altitude Ax. The volume of this cylinder is Ty-^x. D Fig. 28a. If a and b are the limiting values of x, the sum of the cylinders IS 2 TZ/2AX. The volume generated by the area is the limit of this sum x?/-Ax= I iry-dx. (28) If the area does not reach the axis, as in Fig. 286, let ?/i and t/2 be the distances from the axis to the bottom and top r Fig. 286. of the rectangle PQRS. When revolved about the axis^ it generates a hollow cy Under, or washer, of volume T (2/2^ - Vi^) Ax. 54 Simple Areas and Volumes Chap. 4 The volume generated by the area is then b /»6 V = lim y TT (?/2^ - yi^) Ax = I TT (yi^ - y^) dx. Ai=0 ^^ a tJ a If the area is revolved about some other axis, y in these formulas must be replaced by the perpendicular from a point of the curve to the axis and x by the distance along the axis to that perpendicular. Example 1. Find the volume generated by re- volving the ellipse O Ax ax Fig. 28c. about the x-axis. From the equation of the curve we get The volume required is, therefore, J-'o ^52 fa 4 Tty^dx = -Y J {a- — x^) dx = -Tah\ Ex. 2. A circle of radius a is revolved about an axis in its plane at the distance b (greater than a) from its D center. Find the volume generated. Revolve the circle, Fig. 28d, about the line CD. The rectangle MN generates a washer with radii R^ = b-x=b- Va^-y^, R^=b-^x=b-\- Va^-y\ Fig. 28d. The volume of the washer is T {Ri" - Ri") =4x6 Va'-y-'Ay. Art. 28 Volume of a Solid of Revolution 55 The volume required is then V = r 4x6 Va2 _ y2dy = 2Tr^a%. Ex. 3. Find the volume generated by revolving the circle r = a sin 6 about the x-axis. In this case y = rsind = a sin- 6, X = rcosd = acosflsin^. The volume required is V = / Ty^ dx = I ira? sin* (cos- d — sin- 6) d9 = x-a-' The reason for using x as the lower limit and as the upper is to make dx positive along the upper part ABC of the curve. Fig. 28e. As 6 varies from x to 0, the point P describes the path OABCO. Along OA and CO dx is negative. The integral thus gives the volume generated by MABCN minus that generated by 0AM and OCN. EXERCISES 1. Find the volume of a sphere by integration. 2. Find the volume of a right cone by integration. 3. Find the volume generated by revolving about the z-axis the area X)unded by the i-axis and the parabola y = 2 x — x*. 56 Simple Areas and Volumes Chap. 4 4. Find the volume generated by revolving about OY the area bounded by the coordinate axes and the parabola x' + 2/ = a • 5. Find the volume generated by revolving about the x-axis the area bounded by the catenary y = -x ye" -j- e"y , the x-axis and the lines X = ± a. 6. Find the volume generated by revolving one arch of the sine curve y = sin X about OX. 1/ 7. A cone has its vertex on the surface of a sphere and its axis coincides with a diameter of the sphere. Find the common volume. V 8. Find the volume generated by revolving about the y-axis, the part of the parabola y"^ = "iax cut off by the line x = a. iX 9. Find the volume generated by revolving about x = a the part of thp parabola y^ = 4tax cut off by the line x = a. \/ 10. Find the volume generated by revolving about y = — 2 a the part of the parabola r/^ = 4 ax cut off by the line x — a. 11. Find the volume generated by revolving one arch of the cycloid X = a ((^ — sin <^), y = a (1 — cos <^) about the x-axis. 12. Find the volume generated by revolving the curve X = o cos' 0, 2/ = o sin' <> about the y-axis. 13. Find the volume generated by revolving the cardioid r = a (1 -|- cos (?) about the initial line. 14. Find the volume generated by revolving the cardioid r = o (1 + cos 0) about the line x = — j* 16. Find the volume generated by revolving the eUipse x^ -h xy -h 2/2 = 3 \t^^ ^* about the x-axis. 16. Find the volume generated by revolving about the line y = x the part of the parabola x* -H 2/* = "* cut off by the line x -{■ y = a. 29. Volume of a Solid with Given Area of Section. — Divide the solid into slices by parallel planes. Let X be the area of section at distance x from a fixed point. The plate PQRS with lateral surface perpendicular to PQR has the volume PQR 'Ax = X Ax. Art. 29 Volume of a Solid with Given Area of Section 57 If a and h are the limiting values of x, the sum of such plates is Xx^x. The volume required is the limit of this sum v = \\my\XAx=rXdx. (29) Ai=0 Example 1. Find the volume of the ellipsoid ?! _L ^' 4_ !! = 1 ^2 -r ^2 "f- ^2 ^• Fig. 296. The section perpendicular to the x-axis at the distance x from the center is an ellipse ^ -J- ?! = 1 _ ^ 52-1-^2 A o2* 58 Simple Areas and Volumes Chap. 4 The semi-axes of this ellipse are MP = cv/l-|, MQ = 6\/l-^. By exercise 6, page 49, the area of this ellipse is TT . MP . MQ = Tbc ll - ~) • The volume of the ellipsoid is, therefore, I Trbc (1 ^jdx = ^ irabc. Ex. 2. The axes of two equal right circular cylinders intersect at right angles. Find the common volume. Fig. 29c. In Fig. 29c, the axes of the cylinders are OX and OZ and OABC is I of the common volume. The section of OABC by a plane perpendicular to OF is a square of side MP = MQ = Va^ - y\ I Art. 29 Volume of a Solid with Given Abea of Section 59 The area of the section is therefore MP • MQ = a2 - y*, and the required volume is EXERCISES 1. Find the volume of a pjTamid by integration. 2. A wedge is cut from the base of a right circular cylinder by a plane passing through a diameter of the base and inclined at an angle a to the base. Find the volume of the wedge. 3. Two circles have a diameter in common and lie in perpendicular planes. A square moves in such a way that its plane is perpendicular to the common diameter and its di^onals are chords of the circles. Find the volimae generated. 4. The plane of a moving circle is perpendicular to that of an ellipse and the radius of the circle is an ordinate of the ellip)se. Find the vol- ume generated when the circle moves from one vertex of the eUipse to the other. 5. The plane of a moving triangle is perpendicular to a fixed diam- eter of a circle, its base is a chord of the circle, and its vertex Ues on a line parallel to the fixed diameter at distance h from the plane of the circle. Find the volume generated by the triangle in moving from one end of the diameter to the other. 6. A triangle of constant area A rotates about a line perpendicular to its plane while advancing along the line. Find the volume swept out in advancing a distance h. 7. Show that if two soUds are so related that every plane parallel to a fixed plane cuts from them sections of equal area, the volumes of the soUds are equal. 8. A cj-lindrical siu^ace passes through two great circles of a sphere ■which are at right angles. Find the volume within the cylindrical surface and sphere. 9. Two cylinders of equal altitude h have a common upper base and their lower bases are tangent. Find the volume common to the two cj-linders. 10. .\ circle moves with its center on the 2-axis and its plane parallel to a fixed plane inclined at 45° to the 2-axis. If the radius of the circle is always r = Va- — z*, where 2 is the coordinate of its center, find the volimae described. CHAPTER V OTHER GEOMETRICAL APPLICATIONS 30. Infinitesimals of Higher Order. — In the applica- tions of the definite integral that we have previously made, the quantity desired has in each case been a limit of the form lim V' fix) Ax. Ai=0 ^a We shall now consider cases involving limits of the form lim V Fix, Ax) when F ix, Ax) is only approximately expressible in the form / (x) Ax. Such cases are usually handled by neglecting infinitesimals of higher order than Ax. That such neglect does not change the limit is indicated by the following theorem : // for values of x between a and b, F ix, Ax) differs from f ix) Ax by an infinitesimal of higher order than Ax, lim Y Fix, Ax) =lim V fix) Ax. To show this let e be a number so chosen that F ix. Ax) = fix) Ax -\-e Ax. If F ix. Ax) and / ix) Ax differ by an infinitesimal of higher order than Ax, e Ax is of higher order than Ax and so e ap- proaches zero as Ax approaches zero (Differential Calculus, Art. 9). The difference y" Fix, Ax) - V'fix) Ax = V%Ax 60 Art. 31 Rectangtjlar Coordinates 61 Am kO. is graphically represented by a sum of rectangles (Fig. 30), whose altitudes are the various values of c. Since all these values approach zero * with Aa;, the total area approaches zero and so limVV(x,Aa;)=limV /(x)Ax, - which was to be proved. ^^^- ^^• 31. Length of a Curve. Rectangular Coordinates. — In the arc AB of a curve inscribe a series of chords. The length of one of these chords PQ is VA^M^A^ = y/l + (^)'Aa:, Y Ay B A f" c 1 i X Fig. 31a. and the sura of their lengths is The length of the arc AB is defined as the limit approached by this sum when the number of chords is increased indefi- nitely, their lengths approaching zero. * For the discussion to be strictly accurate it vaaai be shown that there is a number larger than any of the e's which approaches zero. In the language of higher mathematics, the approach to the limit must be uniform. In ordinary cases that certainly would be true. A similar remark applies to all the applications of the above theorem. 62 Other Geometrical Applications Chap. 5 The quantity V 1 +(t^) is not a function of x alone. \^x) When Ao; approaches zero, however, the difference of es zero. If then we V/^ + {^ ^^^ v/l + (^J approach replace V1+(t^) Aa;byyi+ f -^ J Ax, the error is an infinitesimal of higher order than \x. Therefore the length of arc is dv In applying this formula ~ must be determined from the equation of the curve. The result can also be written s= r Vd^+~dy^. (31) In this formula, y may be expressed in terms of x, or x in terms of y, or both may be expressed in terms of a parameter. In any case the limits are the values at A and B of the variable that remains. Example 1. Find the length of the arc of the parabola y^ = 4:X between x = and x = 1. -\ dx V In this case t- = ^ . The limiting values of y are and 2. dy 2 Hence s = J'\/l+(^Jdy = £lV^^^dy = V2 + \nil + V2). Ex. 2. Find the perimeter of the curve x = a cos^ </>, y = a sin^ </». In this case ds = Vdx^ + dy^ = Vo a^ cos* sin^ 0+ 9 a^ sin" <^ cos^ <f> d<l> = 3 a cos (f) sin <j) d<i>. Art. 32 Polar Coordinates 63 One-fourth of the curve is described when <f> varies from to jz. Hence the perimeter is = 4 rSa Jo cos <f> sin <f>d<f> = 6 a. EXERCISES 1. Find the circumference of a circle by integration. 2. Find the length of y^ = x' between (0, 0) and (4, 8). 3. Find the length of x = In sec y between y = and y = -^y 4. Find the length ofx = \jf— |lny between y = 1 and y = 2. 5. Find the length of y = e^ between (0, 1) and (1, e). 6. Find the perimeter of the curve x' +2/ = tt • 7. Find the length of the catenary y = -M + e 9 between x = — a and x = a. 8. Find the length of one arch of the cycloid x — a {<(> — sin <f>), y = a {1 — cos (^). 9. Find the length of the involute of the circle X = a (cos +dsia6), y = a (sin 6—6 cos 6), between ^ = and 6 = 2v. 10. Find the length of an arc of the cycloid X = a{d + siD.6), y = a{l—cosd). If s is the length of arc between the origin and any point (x, y) of the same arch, show that s2 = 8 ay. 32. Length of a Curve. Polar Coordinates. — The differential of arc of a curve is (Differential Calculus, Arts.. 54,59) ds = Vdx^ + di/2 = Vdr2 + r2 dd\ 64 Other Geometrical Applications Equation (31) is, therefore, equivalent to Chap. 5 (32) r i M ^•^ X Fig. 32. In this case dr = add and In using this formula, r must be expressed in terms of or ^ in terms of r from the equation of the curve. The limits are the values at A and B of the variable that re- ■^ mains. Example. Find the length of the first turn of the spiral r = ad. Jo 2 dd^ + a2^2 (IQ2 Jo + 02, = 7ra Vl +4x-+^ln(2x + Vl+4x2). EXERCISES 1. Find the circumference of the circle r — a. 2. Find the circumference of the circle r = 2 a cos Q. 3. Find the length of the spiral r = e"* between = and Q — -> a 4. Find the distance along the straight line r = a sec ( ^ ~ 5 ) from = to (9 = ^• 6. Find the arc of the parabola r = a sec* J 6 cut off by the ^-axis. 6. Find the length of one loop of the curve r = a cos* 7' 4 7. Find the perimeter of the cardioid r = a (1 + cos6). 8 8. Find the complete perimeter of the curve r = a sin' = . o Art. 33 Area of a Surface of Revolution 05 33. Area of a Surface of Revolution. — To find the area generated by revolving the arc AB about the x-axis. Join A and Z^ by a broken line with vertices on the arc. Let X, y be the coor- dinates of P and X + q_ Ax, y -\- Ay those of Q. The chord PQ generates a frustum of a cone whose area is 7r(2y + Ay) PQ = ir(2y + A?/) VAx2+A2/2. Fig. 33a. The area generated by the broken line is then 2) 7r(2 2/ + Ay)VAx2 + Ai/2. The area S generated by the arc AB is the limit ap- proached by this sum when Ax and Ay approach zero. Neg- lecting infinitesimals of higher order, (2 i/ + Ay) VAx^ + Ay^ can be replaced by 2 y Vdx^ + dy^ = 2y ds. Hence the area generated is S= \2Tryds. (33a) In this formula y and ds must be calculated from the equation of the curve. The limits are the values at A and B of the variable in terms of which they are expressed. Similarly, the area generated by revolving about the 2/-axis is S =i>^ xds. (33b) Example. Find the area of the surface generated by revohing about the y-axis the part of the curve y = \ — x* above the x-axis. 66 Other Geometrical Applications In this case Chap. 5 ds = y 1 + (^Jdx = Vl + 4x''dx. The area required is generated by the part AB of the curve between x = and x = 1. Hence Fig. 336. S = f 2t xds= j 2txVi-\-4x'' D v) dx EXERCISES 1. Find the area of the surface of a sphere. 2. Find the area of the surface of a right circular cone. 3. Find the area of the spheroid generated by revolving an ellipse about its major axis. 4. Find the area generated by revolving the curve x' + 2/ = o about the 2/-axis. 6. Find the area generated by revolving about OX, the part of the catenary between x = — a and x = a. 6. Find the area generated by revolving one arch of the cycloid X = a (<t> — sin <^), y = a (1 — cos (f>) about OX. Art 34 Unconventional Methods 67 7. Find the area generated by revolving the cardioid r = a (1 + cos 0) about the initial line. / 8. The arc of the circle «* + y* = a' between (a, 0) and (0, a) is revolved about the line x -{- y = a. Find the area of the surface generated. 9. The arc of the parabola ^ = 4 x between x = and x = 1 is I revolved about the line y = — 2. Find the area generated. 10. Find the area of the surface generated by revolving the lemnis- ! cate r^ = 2 a' cos 2 about the line fl = 7- 4 j 34. Unconventional Methods. — The methods that have I been given for finding lengths, areas, and volumes are the I ones most generally applicable. In particular cases other methods may give the results more easily. To solve a problem by integra- tion, it is merely necessary to ex- press the required quantity in any way as a limit of the form used in defining the definite integral. Example 1. When a string held taut is unwound from a fixed circle, its end describes a curve called the involute of the circle. Find the length of the part de- scribed when the first turn of the string is unwound. Let the string begin to unwind at A. When the end reaches P the part unwound QP is equal to the arc AQ. Hence QP = AQ = ad. When P moves to R the arc PR is approximately the arc of a circle with center at Q and central angle M. Hence Fig. 34a. PR = ad Id 68 Other Geometrical Applications Chap. 6 approximately. The length of the curve described when d varies from to 2 tt is then s = lim 2i ad Ad Ae=o Jo addd = 2ira\ Ex. 2. Find the volume generated by rotating about the y-axis the area bounded by the parabola x^ = y — 1, the X-axis, and the ordinates a: = ±1. Resolve the area into slices by ordinates at distances Aar apart. When revolved about the y-axis, the rectangle PM between the ordinates x, x -\- Ax generates a hollow cylinder whose volume is TT (x -\- Ax)- y — TTX^y = 2 irxy Ax + Ty {AxY. Fig. 346. Fig. 34c. Since Try {AxY is an infinitesimal of higher order than Ax, the required volume is lim V 2 TTxy Ax = I 2 ttx (1 + x^) dx = %Tr. Ex. 3. Find the area of the cylinder x^ -\- y^ = ax within the sphere x^ -\- y^ -\- z^ = a^. Fig. 34c shows one-fourth of the required area. Divide the circle OA into equal arcs As. The generators through Art. 34 Unconventional Methods 6^ the points of division cut the surface of the cylinder into rips. Neglecting infinitesimals of higher order, the area the strip MPQ is MP • As. If r, d are the polar coordinates M, r = a cos d and As = a A0, MP = Va- — r- = a sin 6. The required area is therefore given by — = hm 4 Ad- im T^a^sinflA^ = Ja-smddd. Consequently S = 4a' / sindde = Aa\ Jo EXERCISES 1. Find the area swept over by the string in example 1, page 67. 2. Find the area of surface cut from a right circular cyUnder by a plane passing through a diameter of the base and incUned 45° to the base. 3. The axes of two right circular cylinders of equal radius intersect at right angles. Find the area of the soUd coounon to the two cylinders (Fig. 29c). 4. An equilateral triangle of side a is revolved about a line parallel to the base at distance b below the base. Find the volume generated. 5. The area bounded by the hyperbola i* — y^ = a- and the lines y = ±a is revolved about the x-axis. Find the volume generated. 6. The vertex of a cone of vertical angle 2 o is the center of a sphere of radius a. Find the volume common to the cone and sphere. 7. The axis of a cone of altitude h and radius of base 2 a is a gen- erator of a cylinder of radius a. Find the area of the surface of the cyhnder within the cone. * 8. Find the area of the surface of the cone in Ex. 7 within the cylinder. 9. Find the volume of the cylinder in Ex. 7 within the cone. CHAPTER VI MECHANICAL AND PHYSICAL APPLICATIONS 35. Pressure. — The pressure of a liquid upon a hori- zontal area is equal to the weight of a vertical column of the liquid having the area as base and reaching to the surface. By the pressure at a point P in the Hquid is meant the pressure upon a horizontal surface of unit area at that point. The 1 Fig. 35a. Fig. 356. volume of a column of unit section and height h is h. the pressure at depth h is p = wh, Hence (35a) w being the weight of a cubic unit of the liquid. To find the pressure upon a vertical plane area (Fig. 356), we make use of the fact that the pressure at a point is the same in all directions. The pressure upon the strip AB parallel to the surface is then approximately pAA, p being the pressure at any point of the strip and AA its area. The reason for this not being exact is that the pressure 70 Art. 35 Pressitbe 71 at the top of the strip is a little less than at the bottom. This difference is, however, infinitesimal, and, since it multi- plies ilA, the error is an infinitesimal of higher order than Ai4. The total pressure is, therefore, P = lim VplA = fpdA = w ChdA. (35b) Before integration dA must be expressed in terms of h. The limits are the values of h at the top and bottom of the submerged area. In case of water the value of w is about 62.5 lbs. per cubic foot. Example. Find the water pressure upon a s^-nicircle of Fig. 3oc. radius 5 ft., if its plane is vertical and its diameter in the surface of the water. In this case the element of area is dA = 2 V25 - h^ dh. Hence P = w jhdA =2w jhV2o-h^dh = i5 o. ^ = 2.^0. (62.5) = 5208.3 lbs. EXERCISES 1. Find the pressure sustained by a rectangular floodgate 10 ft. broad and 12 ft. deep, the upper edge being in the surface of the water. 2. Find the pressure on the lower half of the floodgate in the pre- ceding problem. 3. Find the pressure on a triangle of base 6 and altitude h, sub- merged so that its vertex is in the surface of the water, and its altitude vertical. 4. Find the pressure upon a triangle of base b and altitude h, sub- merged so that its base is in the surface of the liquid and its altitude vertical. 72 Mechanical and Physical Applications Chap. 6 5. Find the pressure upon a semi-ellipse submerged with one axis in the surface of the liquid and the other vertical. 6. A vertical masonry dam in the form of a trapezoid is 200 ft. long at the surface of the water, 150 ft. long at the bottom, and 60 ft. high. What pressure must it withstand? 7. One end of a water main, 2 ft. in diameter, is closed by a vertical bulkhead. Find the pressure on the bulkhead if its center is 40 ft. below the surface of the water. 8. A rectangular tank is filled with equal parts of water and oil. If the oil is half as heavy as water, show that the pressure on the sides is one-fourth greater than it would be if the tank were filled with oil. 36. Moment. — Divide a plane area or length into small parts such that the points of each part differ only infinitesi- mally in distance from a given axis. Multiply each part by the distance of one of its points from the axis, the distance being considered positive for points on one side of the axis and negative for points on the other. The limit approached by the sum of these products when the parts are taken smaller and smaller is called the moment of the area or length with respect to the axis. Similarly, to find the moment of a length, area, volume, or mass in space with respect to a plane, we divide it into elements whose points differ only infinitesimally in distance from the plane and multiply each element by the distance of one of its points from the plane (considered positive for points on one side of the plane and negative on the other). The moment with respect to the plane is the limit approached by the sum of these products when the elements are taken smaller and smaller. Example. Find the moment of a rectangle about an axis parallel to one of its sides at distance c. Divide the rectangle into strips parallel to the axis (Fig. 36). Let y be the distance from the axis to a strip. The area of the strip is b Ay. Hence the moment is ybAy== J bydy = ab[c-\--]. Art. 37 Center of Gravity of a Length or Area in a Plane 73 Since ab is the area of the rectangle and c + x is the distance from the axis to its center, the moment is equal to the product of the area and the distance from the axis to the center of the rectangle. b - r-*~ a ; 1 ' Fig. 36. Fig. 37a. 37. The Center of Gravity of a Length or Area in a Plane. — The center of gravity of a length or area in a plane is the point at which it could be concentrated without changing its moment with respect to any axis in the plane. Let C (x, y) be the center of gravity of the arc AB (Fig. 37a), and let s be the length of the arc. The moment of AB with respect to the x-axis is /; yds. If the length s were concentrated at C, its moment would be sy. By the definition of center of gravity sy whence J yds, Similarly, - 1: xds X = 74 Mechanical and Physical Applications Chap. 6 The limits are the values at A and B of the variable in terms of which the integral is expressed. Let C {x, y) be the center of gravity of an area (Figs. 376, 37c). Divide the area into strips dA and let {x, y) be the center of gravity of the strip dA . The moment of the area with respect to the X-axis is /^ dA. Fig. 376. If the area were con- centrated at C, the moment would be Ay, where A is the total area. Hence Fig. 37c The strip is usually taken parallel to a coordinate axis. The area can, however, be divided into strips of any other kind if convenient. Example 1. Find the center of gravity of a quadrant of the circle x"^ -\- y^ = a^. In this case ~^ i ds = Vdx^ -j- dy"^ = -dx \J f\ Art. 37 Center OF Gravity OF A Length OR Area IN A Plane 75 and j yds = j y ' -dx = a^. The length of the arc is s = - (2 7ra) = -a. Hence y = I yds 2_a IT T ^^ is X ^ , Tig. 37d. It is evident from the symmetry of the figure that x has the same value. Ex. 2. Find the center of gravity of the area of a semi- circle. From symmetry it is evident that the center of gravity is in the y-axis (Fig. 37e). Take the element of area parallel to OX. Then dA = 2xdy and j ydA = I 2xydy = 2 j y Va^ — y^ dy = ia\ The area is A = jr a^. Hence CydA 4 o Fig. 37e. Fig. 37/. Ex. 3. Find the center of gravity of the area bounded by the X-axis and the parabola y = 2x — x^. Take the element of area perpendicular to OX. If {x, y) 76 Mechanical and Physical Applications Chap. 6 are the coordinates of the top of the strip, its center of gravity is (a^, I). Hence its moment with respect to the x-axis is The moment of the whole area about OX is then i2 15 The area is A = I ydx = I (2X — X^) dx =;: Hence y = %. Similarly, \xdA r{2x^-x^) dx X = = 1. 38. Center of Gravity of a Length, Area, Volume, or Mass in Space. — The center of gravity is defined as the point at which the mass, area, length, or volume can be con- centrated without changing its moment with respect to any plane. Thus to find the cen- ter of gravity of a solid mass (Fig. 38a) cut it into slices of mass Am. If {x, y, z) is the cen- ter of gravity of the slice, its moment with respect to the a;y-plane is z Am and the moment of the whole mass is lim V z Am = I z dm. Am*0 ^^ t/ Fig. 38o. Art 38 Center of Gravitt 77 If the whole mass M were concentrated at its center of gravity {x, y, z), the moment with respect to the xy-plane ! would be zM. Hence zM or = 12 dm, J zdm Similarly, / X dm j y dm ^ = ^M-' y = -M-' (^^> The mass of a unit volume is called the density. If then dv is the volume of the element dm and p its density, dm = pdv. To find the center of gravity of a length, area, or volume it is merely necessary to replace M in these formulas by s, S, or V. Example 1. Find the center of gravity of the volume of an octant of a sphere of radius a. The volume of the slice (Fig. 38a) is dv = I irx^ dz = \w (a^ — z^) dz. Hence fzdv = £l.ia^-z^)zdz = f-^a\ The volume of an octant of a sphere is ^ ird^. Hence - /^^" i^^' 3 V x , 8 6" From symmetry it is evident that x and y have the same value. 78 MECHANICAL AND PHYSICAL APPLICATIONS Chap. 6 Ex. 2. Find the center of gravity of a right circular cone whose density is proportional to the distance from its base. Cut the cone into slices parallel to the base. Let y be the distance of a slice from the base. Except for in- finitesimals of higher order, its volume n is TX^ dy, and its density is ky where K is constant. Hence its mass is Am = kwx^y dy. Fig. 38fe. By similar triangles x = rih — ?/). Hence M Therefore, finally, h' = J dm = J -i^iji-yfydy = h' 12 / ydm y = —n— 2, M 5 EXERCISES 1. The wind produces a uniform pressure upon a rectangular door. Find the moment tending to turn the door on its hinges. 2. Find the moment of the pressure upon a rectangular floodgate about a horizontal line through its center, when the water is level with the top of the gate. 3. A triangle of base h and altitude h is submerged with its base horizontal, altitude vertical, and vertex c feet below the surface of the water. Find the moment of the pressure upon the triangle about a horizontal line through the vertex. 4. Find the center of gravity of the area of a triangle, 6. Find the center of gravity of the segment of the parabola y^ = ax, cut ofT by the line a; = a. Art 38 Center of Gravity 7^ 6. FLid the center of gravity of the area of a quadrant of the ellipse a* ^ 6'- 7. Find the center of gravity of the area bounded by the coordinate axes and the parabola x' + y* = a'. 8. Find the center of gravity of the area above the x-axis bounded by the curve x* + y = a*. 9. Find the center of gravity of the area bounded by the x-axis and one arch of the curve y = sin x. 10. Find the center of gravity of the area bounded by the two parab- olas y- = ax, x^ = ay. 11. Find the center of gravity of the area of the upper half of the cardioid r = o (1 -f- cosfl). 12. Find the center of gravity of the area bounded by the x-axis and one arch of the cycloid, X = a (<^— sin0), y = a (1— co8<^). 13. Find the center of gravity of the area within a loop uf the lemnis- cate r- = a^ cos 2 d. 14. Find the center of gravity of the arc of a semicircle of radius a. 16. Find the center of gravity of the arc of the catenary y = ^(^ + e"°) between x = — a and x = a. 16. Find the center of gravity of the arc of the curve x* + y* = a* in the first quadrant. 17. Find the center of gravity of the arc of the curve X = \y^— h\ny between y = 1 and y = 2. 18. Find the center of gravity of an arch of the cycloid X = a{<i>— sm<i>), y — a{\— cos4>). 19. Find the center of gravity of a right circular cone of constant density. 20. Find the center of gravity of a hemisphere of constant density. 21. Find the center of gravity of the solid generated by revolving about OX the area bounded by the parabola y- = 4 x and the line x = 4. 22. Find the center of gravity of a hemisphere whose density is proportional to the distance from the plane face. 23. Find the center cf gravity of the soUd generated by rotating a sector of a circle about one of its bounding radii. so Mechanical and Physical Applications Chap. 6 24. Find the center of gravity of the solid generated by revolving the cardioid r = a (1 + cos e) about the initial line. 26. Find the center of gravity of the wedge cut from a right circular cylinder by a plane passing through a diameter of the base and making with the base the angle a. 26. Find the center of gravity of a hemispherical surface. 27. Show that the center of gravity of a zone of a sphere is midway between the bases of the zone. 28. The segment of the parabola' i/^ = 2ax cut off by the line x = a is revolved about the x-axis. Find the center of gravity of the surface generated. 39. Theorems of Pappus. Theorem I. — If the arc of a plane curve is revolved about an axis in its plane, and not crossing the arc, the area generated is equal to the product of the length of the arc and the length of the path described by its center of gravity. Theorem II. If a plane area is revolved about an axis in its plane and not crossing the area, the volume generated is equal to the product of the area and the length of the path described by its center of gravity. To prove the first theorem, let the arc be rotated about the X-axis. The ordinate of its center of gravity is I' ^ yds whence 27r / yds = 2irys. The left side of this equation represents the area of the surface generated. Also 2Try is the length of the path described by the center of gravity. This equation, therefore, expresses the result to be proved. To prove the second theorem let the area be revolved about the X-axis. From the equation JydA Art. 39 Theorems of Pappus 81 we get 2ir I ydA = 2iryA. Since 2t f y dA is the volume generated, this equation is equivalent to theorem II. Example 1. Find the area of the torus generated by- revolving a circle of radius a about an axis in its plane at distance b (greater than a) from its center. Since the circumference of the circle is 2 7ra and the length of the path de- scribed by its center 2 wb, the area gen- erated is S = 27ra-2 7r6 = 4:ir-ab. Fig. 39a. Ex. 2. Find the center of gravity of the area of a semi- circle by using Pappus's theorems. When a semicircle of radius a is revolved about its diameter, the volume of the sphere generated is J xa'. If y is the distance of the center of gravity of the semicircle from this diameter, by the second theorem of Pappus, ^ Tra^ = 2 Try A = 2 x^ • i- xa^, whence $xa' y = TT^a- 4a 3x" Ex. 3. Find the volume gen- erated by revolving the cardioid r = a (1 + cos 6) about the initial line. The area of the triangle OPQ is approximately Fig. 396. 1 r^ Ad, and its center of gravity is f of the distance from the vertex to the base. Hence y = f r sin 0. 82 Mechanical and Physical Applications Chap, ft By the second theorem of Pappus, the volume generated by OPQ is then approximately 2 TT?/ A A = f 7rr^ sin 6 M. The entire volume is therefore V = j lTr^smddd = ^Tra^ J {1 + coseysmddd Jo t/o 2 -(l + eos^)-'|- 8 , O 4 I o EXERCISES V 1. By using Pappus's theorems find the lateral area and the volume of a right circular cone. Y 2. Find the volume of the torus generated by revolving a circle of radius a about an axis in its plane at distance b (greater than a) from its center. J 3. A groove with cross-section an equilateral triangle of side i inch is cut around a cylindrical shaft 6 inches in diameter. Find the volume of material cut away. </ 4. A steel band is placed around a cylindrical boiler 48 inches in diameter. A cross-section of the band is a semi-ellipse, its axes being 6 and Vq inches, respectively, the greater being parallel to the axis of the boiler. What is the volume of the band? l^y^. The length of an arch of the cycloid X = a {<j> — sin <^), y = a (I — cos <j>) is 8 a, and the area generated by revolving it about the x-axis is ^ iro*. Find the area generated by revolving the arch about the tangent at its highest point. i/^&. By the method of Ex. 3, page 81, find the volume generated by revolving the lemniscate r^ = 2 a^ cos 2 about the x-axis. 7. Obtain a formula for the volume generated by revolving the polar element of area about the line x = — a. Apply this formula to obtain the volume generated by revolving about x = — a the sector of the circle r = a bounded by the radii d = — a, 6 = -\- a. 8. A variable circle revolves about an axis in its plane. If tho distance from the center of the circle to the axis is 2 a and its radius is a sin 6, where d is the angle of rotation, find the volume of the horn- shaped solid that is generated. 9. Can the area of the surface in Ex. 8 be found in a similar way? Art. 40 Moment of Inertia 83 10. The vertex of a right circular cone is on the surface of a right circular cylinder and its axis cuts the axis of the cyhnder at right angles. Find the volume common to the cyhnder and cone (use sections deter- mined by planes through the vertex of the cone and the generators of the cylinder). 40. Moment of Inertia. — The moment of inertia of a particle about an axis is the product of its mass and the square of its distance from the axis. To find the moment of inertia of a continuous mass, we divide it into parts such that the points of each differ only infinitesimally in distance from the axis. Let Aw be such a part and R the distance of one of its points from the axis. Ebccept for infinitesimals of higher order, the moment of inertia of \m about the axis is R- Aw. The moment of inertia of the entire mass is therefore 7 = UmX^^^^= fR'dm. Am=0 ^ J (40) By the moment of inertia of a length, area, or volume, we mean the value obtained by using the differential of length, area, or volume in place of dm in equation (40). Example 1. Find the mo- ment of inertia of a right cir- cular cone of constant density about its axis. Let p be the density, h the altitude, and a the radius of the base of the cone. Di\nde it into hollow cyhndrical sUces by means of cylindrical sur- faces having the same axis as the cone. By similar triangles the altitude y of the cylin- drical surface of radius r is Fig. 40a. y=-(a-r). 84 Mechanical and Physical Applications Chap. C Neglecting infinitesimals of higher order, the volume between the cylinders of radii r and r + Ar is then Ay = 2 irry Ar = r (a — r) dr. a The moment of inertia is therefore 2Trhp a I = I r^ dm = I r^p dv = The mass of the cone is t/O r^(a — r) dr = Tpha* 10 M = pv = I wpa^h. I = Hence Ex. 2. Find the moment of in- ertia of the area of a circle about a diameter of the circle. Let the radius be a and let the x- axis be the diameter about which the moment of inertia is taken. Divide the area into strips by lines parallel to the x-axis. Neglecting infinitesimals of higher order, the area of such a strip is 2 a; Ay and its moment of inertia 2 xy^ Ay. The moment of inertia of the entire area is therefore Fig. 406. / = j2xy^dy = 2 j Va-' - y^y^ dy = ira* EXERCISES 1. Find the moment of inertia of the area of a rectangle about one of its edges. 2. Find the moment of inertia of a triangle about its base. 3. Find the moment of inertia of a triangle about an axis through its vertex parallel to its base. 4. Find the moment of inertia about the r/-axis of the area bounded by the parabola y^ = iax and the line x = a. 5. Find the moment of inertia of the area in Ex. 4 about the line X = a. 6. Find the moment of inertia of the area of a circle about the axis perpendicular to its plane at the center. (Divide the area into rings with centers at the center of the circle.) Art. 41 Work Done by a Forge 8^ 7. Find the moment of inertia of a cylinder of mass M and radius a. about its axis. 8. Find the moment of inertia of a sphere of mass M and radius a. about a diameter. 9. An ellipsoid is generated by revolving the eUipse ^-ul^= 1 about the x-axis. Find its moment of inertia about the x-axis. 10. Find the moment of inertia of a hemispherical shell of constant density about the diameter perpendicular to its plane face. 11. Prove that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the center of gravity plus the product of the mass and the square of the distance between the two axes. 12. Use the answer to Ex. 6, and the theorem of Ex. 11 to determine the moment of inertia of a circular area about an axis, perpendicular to- ils plane at a point of the circumference. 41. Work Done by a Force. — Let a force be applied to a body at a fixed point. Wtien the body moves work is done by the force. If the force is constant, the work is defined as the product of the force and the distance the point of appli- cation moves in the direction of the force. That is, W = Fs, (41a) where W is the work, F the force, and s the distance moved in the direction of the force. If the direction of motion does not coincide with that of the force, the work done is the product of the ^^^^ I ^j . force and the projection of the dis- Fig. 41a. placement on the force. Thus when the body moves from. A to 5 (Fig. 41a) the work done by the force F is W = Fs cos e. (41b> If the force is variable, we divide the path into parts As. 'Tn moving the distance As, the force is nearly constant and. so the work done is approximately FcosflAs. As the m Mechanical and Physical Applications Chap. 6 intervals As are taken shorter and shorter, this approximation becomes more and more accurate. The exact work is then the Hmit W = ]imy.FcosdAs= jFcoseds. (41c) As=0 ^ J To determine the value of W, we express F cos 6 and ds in terms of a single variable. The limits of integration are the values of this variable at the two ends of the path. If the displacement is in the direction of the force, 6 = 0, cos 6=1 and W -P ds. (41d) -B vwvvww-^- Fig. 416. Fig. 41c. Example 1. The amount a helical spring is stretched is proportional to the force applied. If a force of 100 lbs. is required to stretch the spring 1 inch, find the work done in stretching it 4 inches. Let s be the number of inches the spring is stretched. The force is then F = ks, k being constant. When s = 1, F = 100 lbs. Hence k = 100 and F =lOOs. The work done in stretching the spring 4 inches is JFds = I 100 s ds = 800 inch pounds = 66§ foot pounds. *Jo Art. 41 Work Done by a Force 87 Ex. 2. A gas is confined in a cylinder with a movable piston. Assuming Boyle's law pv = k, find the work done by the pressure of the gas in pushing out the piston (Fig. 41rf). Let V be the volume of gas in the cylinder and p the pressure per unit area of the piston. If A is the area of the piston, pA is the total pressure of the gas upon it. If s is the distance the piston moves, the work done is W = JpAds. But A ds = dv. Hence W= f^pdv= r'^dv = k\n^ is the work done when the volume expands from I'l to i'2. Fig. Aid. Fig. 41e. Ex. 3. The force with which an electric charge Ci repels a charge e^ at distance r is ke-iCz where k is constant. Find the work done by this force when the charge e^ moves from r = a to r = 6, Ci remaining fixed. Let the charge d move from A to B along any path AB (Fig. 41e). The work done by the force of repulsion is W = jFcosdds = ffdr = r^dr The work depends only on the end points A and B and not on the path connecting them. Mechanical and Physical Applications Chap. 6 EXERCISES 1. According to Hooke's law the force required to stretch a bar from the length a to the length a -{- x ia kx where K is constant. Find the work done in stretching the bar from the length a to the length b. 2. Supposing the force of gravity to vary inversely as the square of the distance from the earth's center, find the work done by gravity on a meteor of weight w lbs., when it comes from an indefinitely great distance to the earth's surface. 3. If a gas expands without change of temperature, according to van der Waal's equation, ^ >, P V — b a, b, c being constant. Find the work done when the gas expands from the volume vi to the volume t'2. 4. The work in foot pounds required to move a body from one altitude to another is equal to the product of its weight in pounds and the height in feet 41J. ^j^^^ j^ jg raised. Find the work required to pump the water out of a cylindrical cistern of diameter 4 ft. and depth 8 ft. 5. A vertical shaft is supported by a flat step bearing (Fig. 41/). The frictional force between a small part of the shaft and the bearing is HP, where p is the pressure between the two and m is a constant. If the pressure per unit area is the same at all points of the supporting surface, and the weight of the shaft and its load is P, find the work of the fric- tional forces during each revolution of the shaft. 6. When an electric current flows a distance x through a homo- geneous conductor of cross-section A, the resistance is kx A' where K is a constant depending on the material. Find the resistance when the current flows from the inner to the outer surface of a hollow cylinder, the two radii being a and b. Art 41 Work Doxe by a Force 89 7. Find the resistance when the current flows from the inner to the outer surface of a hollow sphere. 8. Find the resistance when the cmrent flows from one base of a tnmcated cone to the other. 9. When an electric current i flows an in- finitesimal distance AB (Fig. 41^) it produces at any point O a magnetic force (perpendicular to the paper) equal to ide r ' where r is the distance between AB and 0. Find the force at the center of a circle due to a current i flowing around it. 10. Find the magnetic force at the distance c from an infinite straight line along which a current i is flowing. CHAPTER VII APPROXIMATE METHODS 42. The Prismoidal Formula. — Let yi, ys, be two ordinates of a curve at distance h apart, and let 2/2 be the ordinate midway between them. The area bounded by the a:-axis, the curve, and the two ordinates is given approximately by the for- mula Fig. 42a. A= lh(yi-\- 4:yo+ ys). (42a) This is called the prismoidal formula because of its similarity to the formula for the volume of a prismoid. If the equation of the curve is y = a -i- bx -i- cx~ -\- dx?, (42b) where a, 6, c, d, are constants (some of which may be zero), the prismoidal formula gives the exact area. To prove this let k be the abscissa of the middle ordinate and t the dis- tance of any other ordinate from it (Fig. 42a). Then X = k -[-t. [f we substitute this value for x, (42b) takes the form y = A+Bt + Ct^-{- Dt\ where A, B, C, D are constants. The ordinates yi, 1/2, ys are Hence h h obtained by substituting t = — k, 0,^. 90 Art. 42 The Prismoidal Forjiula 91 Also the area is -2 12 This is equivalent to which was to be proved. If the equation of the curve does not have the form (42b), it may be approximately equivalent to one of that type and so the prismoidal formula may give an approximate value for the area. While we have illustrated the prismoidal formula by the area under a curve, it may be used equally well to determine a length or volume or any other quantity represented by a definite integral, V (x) dx. £ Since such an integral represents the area under the curve y = / (^)> its value can be found by replacing h in (42a) by 6 - a and t/i, y^, yz by / (a), / f ^ ^ j , / (6) respectively. Exam-pie 1. Find the area bounded by the X-axis, the curve y = e~^\ and the ordinates X = 0, X = 2. The integral / e-'^dx Fig. 426. cannot be expressed in terms of elementary functions. Therefore we cannot obtain the area by the methods that we 92 Approximate Methods Chap. 7 have previously used. The ordinates yi, y^, yz, in this case are yi = 1, y2 = e-S yz = e-^. The prismoidal formula, therefore, gives A = |(l +^+^1=0.869. The answer correct to 3 decimals (obtained from a table) is 0.882. Ex. 2. Find the length of the parabola ?/- = 4 x from a; = 1 to X = 5. The length is given by the formula =xv^-^ dx. By integration we find s = 4.726. To apply the prismoidal formula, let y -^' Then /i = 4, and Vl = V2, 7/2 = Vl, 7/3 = Vf, s = 4 ( V2 + 4 v^ + Vf ) = 4.752. Ex. 3. Find the vol- ume of the spheroid generated by revolving the ellipse a- 0- about the x-axis. The section of the Fig. 42c. spheroid perpendicular to OX has the area A = Try'- = irh- ll - ^y Art. 43 Simpson's Rule 93 Its volume is =£■ Adx. Since A is a polynomial of the second degree in x (a sp)ecial case of a third degree polynomial), the prismoidal formula gives the exact volume. The three cross-sections corre- sponding to x = —a, X = 0, X = a, are Ai = 0, Ao = irlr, Ai = 0. Hence y = ^[Ai + 4A2 + A3]=| ra62. 43. Simpson's Rule. — Divide the area between a curve and the x-axis into any even number of parts by means of equidistant ordinates yi, y^, 1/3, ... , y„. (An odd number of ordinates will be needed.) Simpson's rule for determining approximately the area between yi and y„ is ^l/, + 4y2 + 2i/, + 4i/4 + 2i/3+ • • • 4- A 1 + -i + 2 + 4 -f 2 + + 1 jby (43) h being the distance between the ordinates yi and ?/„, In the numerator the end coefficients are 1. The others are alternately 4 and 2. The denominator is the sum of the coefficients in the num- erator. This formula is obtained by applying the prismoidal formula to the strips taken two at a time and adding the results. Thus if the area Y / 'Jx V. y» "4 ^ "s £ Fig. 43. is divided into four strips by the ordinates r/i, 1/2, ys, J/4, y h the part between r/i and 1/3 has a base equal to area as given by the prismoidal formula is 2' Its 94 Approximate Methods Chap. 7 ' Similarly the area between ys and y^ is The sum of the two is A^h ( ^^' + ^ ^2 + ^ ^3 + 4 ^4 + ys N By using a sufficiently large number of ordinates in Simpson's formula, the result can be made as accurate as desired. Example. Find In 5 by Simpson's rule. Since in 5 ndx 1 . we take y = - m Simpson's formula. Dividing the interval into 4 parts we get In 5 = 4 /i + 4»^ + 2.^ + 4-i + ^ 12 = 1.622. If we divide the interval into 8 parts, we get In 5 = ^\- (1 + I + I + I + f + f + I + I + i) = 1.6108. The value correct to 4 decimals is In 5 = 1.6094. 44. Integration in Series. — In calculating integrals it is sometimes convenient to expand a function in infinite series and then integrate the series. This is particularly the case when the integral contains constants for which numerical values are not assigned. For the process to be valid all series used should converge. Example. Find the length of a quadrant of the ellipse ^^t.= 1 a^ "^ 62 ^' Art 44 Integration in Series 95 Let a be greater than 6. Introduce a parameter <t> by the equation X = a sin <t>. Substituting this value in the equation of the ellipse, we find y = b cos 4>. Using these values of x and y we get r s= fVdx^ + dy- = jVa- - (a^ - b^) sin^ <f, d<t>. This is an eUiptic integral. It cannot be represented by an expression containing only a finite number of elementary functions. We therefore express it as an infinite series. By the binomial theorem Va- — (a- — b~) sin^ (f> = «L^ - 2 -^^^^^ '^-2li-^j ^^ '^ • • -J- Since we find by integrating term by term *~"L2 8 a' 128V a^ j ' gj ^Trgf g^-b^ 3 /a- - ¥V 1 ~ 2 L 4a2 64V a- )''']' If a and 6 are nearly equal, the value of s can be calculated very rapidly from the series. EXERCISES 1. Show that the prismoidal formula gives the correct volume in each of the following cases: (o) sphere, (6) cone, (r) cyhnder, (d) P3'ramid, (e) segment of a sphere, (/) truncated cone or pyramid. 2. Find the error when the value of the integral j i*dx is found by the prismoidal formula. •96 Approximate Methods Chap. 7 In each of the following cases compare the value given by the pris- moidal formula with the exact value determined by integration. 3. Area bounded hy y = Vx, y = 0, x = 1, x = 3. 4. Arc of the curve y = 3^ between x=— 2, x = +2. 5. Volume generated by revolving about OX one arch of the sine ■curve y = sin x. 6. Area of the surface of a hemisphere. Compute each of the following by Simpson's rule using 4 intervals: dx 4 Jo 1 +x^ dx '•r Vi+x» 9. Length of the curve y = Inx from a; = 1 to a; = 5. 10. Surface of the spheroid generated by rotating the ellipse x* + ■4y^ = 4 about the x-axis. 11. Volume of the solid generated by revolving about the x-axis the ^rea bounded hy y = 0, y = ^ , ^ , x = — 2, x = 2. 12. Find the value of by expanding in series. 13. Express J cos (x*) dx, x ' sin (Xx) dx as a series in powers of X. 14. Find the length of a quadrant of the ellipse x* + ly^ = 2. CHAPTER VIII DOUBLE INTEGRATION 45. Double Integrals. — The notation / / f{x,y)dxdy is used to represent the result of integrating first with respect to y (leaving x constant) between the limits c, d and then with respect to x between the limits a, b. As here defined the first integration is with respect to the variable whose differential stands last and its limits are attached to the last integral sign. Some writers integrate in a different order. In reading an article it is therefore necessary to know what convention the author uses. Example. Find the value of the double integral Jo J -i (x- -H y^) dx dy. We integrate first with respect to y between the limits —x, X, then with respect to x between the limits 0, 1. The result is f j\x-'+y'')dxdy = j^dx{xhf + \y'y_^ = J\x'dx = l. 46. Area as a Double Integral. — Divide the area be- tween two curves y = f {x), y = F (x) into strips of width Ax. Let P be the point {x, y) and Q the point (x + Ax, y + Ay). The area of the rectangle PQ is Ax Ay. The area of the rectangle RS (Fig. 46a) is Ax V Ay = Ax / dy. ^f(z) J fix) ^ 97 98 Double Integration Chap. 8 The area bounded by the ordinates x = a, x = h is then. Sb nF(x) nb PF(x) Ax j dy = j j dx dy. ^-y, a Jf(x) J a JS(.i) If it is simpler to cut the area into strips parallel to the a;-axis, the area is = J I dy dx, i the limits in the first integration being the values of x at the ends of a variable strip; those in the second integration, the values of y giving the limiting strips. Example. Find the area bounded by the parabola y"^ = 4 ox + 4 a^ and the straight line y = 2a — x (Fig. 466). Fig. 46a. Fig. 46&. Solving simultaneously, we find that the parabola and the line intersect at A (0, 2 a) and B (Sa, —6a). Draw the strips parallel to the rc-axis. The area is y2 /2a f2a-y po / / dydx= I 2( 4a'\ , 64 , The limits in the first integration are the values of x at i? and S, the ends of the variable strip. The limits in the second integration are the values oi y at B and A, corre- sponding to the outside strips. Art 47 Volume by Double Integratiox 99 47. Volume by Double Integration. — To find the volume under a surface z = / (x, y) and over a given region in the xy-plane. The volume of the prism TQ standing on the base Ax Ay (Fig. 47a) is z Ax Ay. The volume of the plate RT is then Ax Ay = Ax / 2 dy, a»=u R J fix) / (x), F (x) being the values of y at R, S. The entire volume is the limit of the sum of such plates Ax / zdy = I I zdxdy, ui=u a Jf(.x) J a Jf(x) a, b being the values of x corresponding to the outside plates. Example. Find the volume bounded by the surface az = a^ — x^ — 4 y2 and the xy-plane. z Fig. 47a. Fig. 476. Fig. 476 shows one-fourth of the required volume, y = 0. At iS, 2 = and so y = ^ Vo« - x2. Ati2, 100 Double Integration Chap. 8 The limiting values of x at and A are and a. Therefore v = ^\ I zdxdy = il / -id'-x''-4y^)dxdy 3 a Jo 4 48. The Double Integral as the Limit of a Double Summation. — Divide a plane area by lines parallel to the coordinate axes into rectangles with sides Ax and Ay. Let (a:, y) be any point within one of these rectangles. Form the product / {x, y) Ax Ay. This product is equal to the volume of the prism standing on the rectangle as base and reaching the surface z = f {x, y) at some point over the base. Take the sum of such products Fig. 48a. for all the rectangles that lie entirely within the area. We represent this sum by the notation 2) ^J{x,y)AxAy. When Ax and Ay are taken smaller and smaller, this sum approaches as limit the double integral // / {x, y) dx dy, Art 48 The Limit of a Double Summation lot with the limits determined by the given area; for it approaches the volume over the area and that volume is equal to the double integral. Whenever then a quantity is a limit of a sum of the form. %^f(x,y)AxAy its value can be found by double integration. Furthermore, in the formation of this sum, infinitesimals of higher order than Ax Ay can be neglected without changing the Umit. For, if e Az A?/ is such an infinitesimal, the sum of the errors thus made is %%€^x Ay. When Ax and Ay approach zero, e approaches zero. The sum of the errors approaches zero, since it is represented by a volume whose thick- ness approaches zero. Example 1. An area is bounded by the parabola y' = 4 ax and the line X = a. Find its moment of inertia about the axis perpendicular to its plane at the origin. Divide the area into rectangles Ax Ay. The distance of any point P (x, y) fro m the ax is perpendicular to the plane at is i2 = OP = Vx- + y^. If then (x, y) is a point within one of the rectangles, the moment of inertia of that rectangle is Rr ^x \y = (x^ -\- y«) Ax Ay, approximately. That the result is approximate and not exact is due to the fact that different points in the rectangle differ slightly in distance from the axis. This difference is. Fig. 486. 102 Double Integkation Chap. 8 however, infinitesimal and, since R^ is multiplied by Ax Ay, the resulting error is of higher order than Ax A?/. Hence in the limit J-2\^z 344 (x^ -\- y^) dx dy = j^a*- Ex. 2. Find the center of grav- ity of the area bounded by the parabolas y^ = 4:X -j- 4:,y^ = — 2x + 4. By symmetry the center of gravity is seen to be on the X-axis. Its abscissa is - / xdA X = If we wish to use double inte- gration we have merely to replace dA by dx dy or dy dx. From the figure it is seen that the first integration should be with respect to x. Hence Fig. 48c. X2 /'i(4-»2) I xdy dx _ 2 Ji(yi-4) 11 dydx «/-2 t/ J (1^2-4) 16 8 EXERCISES Find the values of the following double integrals: dxdy ^' J, Ji {x + yy Ja rdedr. -i /•1V3 3. I I xydxdy. J -2 IT /•'« I e-kr^rdddr. •'0 6. C C {x^ + y^)dydx. Jo Jy I dy dx. *'0 Art. 49 Double Integratiox. Polar Coordinates 103 7. Find the area bounded by the parabola y^ = 1x and the line X = y. 8. Find the area bounded by the parabola j/* = 4 ax, the line X + y = 3 a, and the x-axis. 9. Find the area enclosed by the ellipse {y - xT- + x^ = l. j 1 10. Find the volume under the paraboloid z = -i — x^ — y^ and over the square bounded by the lines x = ±l,i/ = ±lin the xy-plane. 11. Find the volume bounded by the x(/-plane, the cyhnder x- + y^ = 1, and the plane x + y + z = 3. 12. Find the volume in the first octant bounded by the cylinder (x — 1)- + (i/ — 1)^ = 1 and the paraboloid xy = z. 13. Find the moment of inertia of the triangle bounded by the coordinate axes and the line x + y = 1 about the line perpendicular to its plane at the origin. 14. Find the moment of inertia of a square of side a about the axis perpendicular to its plane at one comer. 15. Find the moment of inertia of the triangle bounded by the lines x + y = 2, X = 2, y = 2 about the x-axis. 16. Find the moment of inertia of the area bounded by the parab- ola y- = ax and the line x = a about the line y = — a. 17. Find the moment of inertia of the area bounded by the hyper- bola xy = 4 and the line x -{- y = 5 about the line y = x. 18. Find the moment of inertia of a cube about an edge. 19. A wedge is cut from a cylinder by a plane passing through a diameter of the base and inclined 45° to the base. Find its moment of inertia about the axis of the cylinder. 20. Find the center of gravity of the triangle formed by the lines x = y, x + y = 4, X — 2y = 4:. 21. Find the center of gravity of the area bounded by the parabola J/* = 4 ax -f 4 a* and the line y = 2a — x. 49. Double Integration. Polar Coordinates. — Pass through the origin a series of lines making with each other equal angles Ad. Construct a series of circles with centers "at the origin and radii differing by Ar. The hues and circles di\ide the plane into curved quadrilaterals (Fig. 49a). Let r, 6 be the coordinates of P, r + Ar, 4- A0 those of Q. Since PR is the arc of a circle of radius r and subtends the angle Ad at the center, PR = r Ad. Also RQ = Ar. 104 Double Integration Chap. 8 When Ar and \d are very small PRQ will be approximately a rectangle with area PR'RQ = rAd Ar. Fig. 49a. It is very easy to show that the error is an infinitesimal of higher order than A^ Ar. (See Ex. 5, page 107.) Hence the sum ^^rAeAr, taken for all the rectangles within a curve, gives in the limit the area of the curve in the form A = j jrdddr. (49a) The limits in the first integration are the values of r at the ends A, B of the strip across the area. The limits in the second integration are the values of d giving the outside strips. If it is more convenient the first integration may be with respect to 6. The area is then = // r dr dd. Art. 49 Double Integration. Polar Coordinates 105 The first limits are the values of 6 at the ends of a strip between two concentric circles (Fig. 496). The second hmite are the extreme values of r. Fig. 496. Fig. 49c. The element of area in polar coordinates is dA = rdddr. (49b) We can use this in place of dA in finding moments of inertia, volumes, centers of gravity, or any other quantities expressed by integrals of the form / fir,d)dA. Example 1. Change the double integral {x'^ + y^)dxdy Jo Jo to polar coordinates. The integral is taken over the area of the semicircle y = V2 ax — x^ (Fig. 49c). In polar coordinates the equation of this circle is r = 2 a cos 6. The element of area 106 Double Integration Chap. 8 dx dy can be replaced by r cW dr.* Also x^ -\- y^ = r^. Hence J^2a n V2ai-i2 ni /»2 COS 9 / (x^ -\- y'^) dx dy = \ I r^ • r dd dr. Jo Jo J The limits for r are the ends of the sector OP. The limits for 6 give the extreme sectors 6 = 0, 9 = -• Ex. 2. Find the moment of in- ertia of the area of the cardioid r = a (1 + cos 6) about the axis -X perpendicular to its plane at the origin. The distance from any point P (r, d) (Fig. 49d) to the axis of rota- tion is OP = r. Fjg. 49d. Hence the moment of inertia is na(l + cosO) „4 /V OK r^-rdedr= - (I -\- cosdYdd = ^ira*. iJo 10 Ex. 3. Find the center of gravity of the cardioid in the preceding problem. The ordinate of the center of gravity is evidently zero. Its abscissa is xdA 2 / / r cosd ' rdddr Jo Jo X = J'm If rdddr 5 Ex. 4. Find the volume common to a sphere of radius] 2 a and a cylinder of radius a, the center of the sphere being] on the surface of the cylinder. * This does not mean that dx dy = rdd dr, but merely that the sum of all the rectangular elements in the circle is equal to the sum of all the polar elements. Art. 49 Double Integration. Polar Coordinates 107' Fig. 49e shows one-fourth of the required volume. Take a system of polar coordinates in the xy-plane. On the element of area r dd dr stands a prism of height z= V4 a2 - r', Fig. 49e. The volume of the prism is z-rdddr and the entire volume is 2acoe9 dB I V^a'-r^.rdddr = 4:l ^ X = ^ r (1 - sin'^) de = ^a' (3t - 4). EXERCISES Find the values of the following integrals by changing to polar coordinates: (i2 + y2)dy(ir. 3. J J e-^'*+*''> dxdy. 2. J^ J^ dxdy. ^ J J ^d^-x'-y^dxdy. 6. Find the area bounded by two circles of radii a, a + £i.a and two lines through the origin, making with the initial line the angles a, ■ j a + ^a, respectively. Show that when Aa and Aa approach zero, the ' result differs from a Act Aa by an infinitesimal of higher order than Aa Ao. 108 Double Integration Chap. 8 6. The central angle of a circular sector is 2 a, Find the moment of inertia of its area about the bisector of the angle. 7. An area is bounded by the circle r = a V2 and the straight line r = asec id — j] . Find its moment of inertia about the axis per- pendicular to its plane at the origin. 8. Find tha center of gravity of the area in Ex. 6. 9. The center of a circle of radius 2 a lies on a circle of radius a. Find the moment of inertia of the area between them about the common tangent. 10. Find the moment of inertia of the area of the lemniscate r^ = 2 a^ cos 2 about the axis perpendicular to its plane at the origin. 11. Find the moment of inertia of the area of the circle r = 2 a outside the parabola, r = a sec* ^ about the axis perpendicular to its plane at the origin. 12. Find the moment of inertia about the y-axis of the area within the cu-cle {x - aY + (y - o)* = 2 a?. 13. The density of a square lamina is proportional to the distance from one corner. Find its moment of inertia about an edge passing through that corner. 14. Find the moment of inertia of a cylinder about a generator. 15. Find the moment of inertia of a cone about its axis. 16. Find the volume under the spherical surface x^ -{- y^ ■\- z^ = a? and over the lemniscate r* = a^ cos 2 5 in the xy-plane. 17. Find the volume bounded by the xy-plane, the paraboloid az = X* + y'^ and the cylinder x* + y* = 2 ax. 18. Find the moment of inertia of a sphere of density p about a diameter. 19. Find the volume generated by revolving one loop of the curve 1 r = a cos 2 d about the initial line. 50. Area of a Surface. — Let an area A in one plane projected upon another plane. The area of the projection A' = A cos 0, when <f) is the angle between the planes. To show this divide A into rectangles by two sets of lines respectively parallel and perpendicular to the intersection MN of the two planes. Let a and h be the sides of one of Art. 60 Area of a Sttrface 109 these rectangles, a being parallel to MN. The projection of this rectangle will be a rectangle with sides a' = a, h' — h cos ^, and area a'h' = ab cos <{>. The sum of the projections of all the rectangles is ^a'b' = ^ab cos 4>. As the rectangles are taken smaller and smaller this approaches as limit A' = A cos <t>, which was to be proved. Fig. 50a. To find the area of a curved surface, resolve it into elements whose projections on a coordinate plane are equal to the differential of area dA in that plane. The element of surface can be considered as lying approximately in a tangent plane. Its area is, therefore, approximately dA cos<f> where is the angle between the tangent plane and the coordinate plane on which the area is projected. The area of the surface is the limit J cos </) 110 Double Integration Chap. 8 The angle between two planes is equal to that between the perpendiculars to the planes. Therefore ()> is equal to z Fig. 50&. the angle between the normal to the surface and the co- ordinate axis perpendicular to the plane on which we project. If the equation of the surface is F (x, y, z) = 0, the cosine of the angle between its normal and the 2-axis is (Differential Calculus, Art. 101) dF dz cos 7 = v/fj dxj'^\dy)'^[dzj py The cosines of the angles between the normal and the x-axis or w-axis are obtained by replacing ^r- by 7— or t— . In dz "^ dx dy finding areas the algebraic sign is assumed to be positive. Example 1. Find the area of the sphere x^ + y^ -{- z^ = a^ within the cylinder x^ -}- y^ = ax. Project on the x^z-plane. The angle <f) is then the angle 7 between the normal to the sphere and the 2!-axis. Its cosine is z z cos 7 = Vx2 + 2/2 + 22 a J Art. 60 Area of a Surface 111 Using polar coordinates in the xj/-plane, z = Va- — x^ — if = Va2 — j^. Hence the area of the surface is S = fj" = 4 r r°-^£iL = 2aHx - 2). J COS 7 Jo Jo V a^ — r^ Ex. 2. Find the area of the surface of the cone y^ -h z^ = x2 in the first octant bounded by the plane y -\- z = a. Project on the yz-plane. Then <f> = a and X X 1_ Vx2 +1/2+22 ~ V2x2 ~ V2 The area on the cone is therefore S= r r " V2dydz = Jo Jo EXERCISES 1. Find the area of the triangle cut from the plane x + 2y + 3z = 6 by the coordinate planes. 2. Find the area of the surface of the cylinder x- -\- y- = a* between the planes z = 0, 2 = mx. 3. Find the area of the surface of the cone x- -\- y- = z* cut out by the cylinder i^ + t/^ = 2 ax. 4. Find the area of the plane x + y + z = 2a in the first octant bounded by the cyUnder x^ + y^ = a*. 6. Find the area of the surface ^ = 2xy above the zy-plane bounded by the planes j/ = 1, x = 2. 6. Find the area of the surface of the cylinder x^ + y^ = 2ax between the ly-plane and the cone x- -\- y- = z-. 7. Find the area of the surface of the paraboloid y- -\- z- = 2 ox, intercepted by the parabolic cyUnder rf- = ax and the plane x = o. 8. Find the area intercepted on the cylinder in Ex. 4. 9. A square hole of side a is cut through a sphere of radius a. If the axis of the hole is a diameter of the sphere, find the area of the surface cut out. CHAPTER IX TRIPLE INTEGRATION 51. Triple Integrals. — The notation I I fix, y, z) dx dy dz is used to represent the result of integrating first with respect to z (leaving x and y constant) between the limits Zi and Za, then with respect to y (leaving x constant) between the limits 2/1 and y^, and finally with respect to x between the limits Xi and Xg- Fig. 52a. 52. Rectangular Coordinates. — Divide a solid into rectangular parallelepipeds of volume Ax Ai/ A2 by planes parallel to the coordinate planes. To find the volume of 112 'Art. 62 Rectangitlar Coobdinates 113 the solid, first take the sum of the parallelepipeds in a vertical column PQ. The result is V Ax \y A2 = Ax Ay / dz, Zi and Zt being the values of z at the ends of the colunm. Then sum these columns along a base MN and so obtain the volume of the plate MNR. The result is Um V Ax Ay / dz = \x J I dy dz, t/i and ya being the Umiting values of y in the plate. Finally, take the sum of these plates. The result is the triple integral V = Um X -^"^ / / dy dz = I J I dxdydz, Xi, X2 being the limiting values of x. It may be more convenient to begin by integrating with respect to x or y. In any case the limits can be obtained from the consideration that the first integration is a summa- tion of parallelepipeds to form a prism, the second a summa- tion of prisms to form a plate, and the third integration a summation of plates. Let (x, y, 2) be any point of the parallelepiped Ax Ay \z. Multiply Ax Ay Az by / (x, y, z) and form the sum 2 2 2-^ ^^' ^' ^^ ^* ^^ ^^ taken for all parallelepipeds in the soUd. When Ax, Ay, and Az approach zero, this sum approaches the triple integral III' f (x, y, 2) dx dy dz as limit. It can be shown that terms of higher order than Ax Ay Az can be neglected in the sum without changing the limit. 114 Triple Integration Chap. 9 The differential of volume in rectangular coordinates is dv = dx dy dz. This can be used in the formulas for moment of inertia, center of gravity, etc., those quantities being then determined by triple integration. Example 1. Find the volume of the ellipsoid c^ y^ & Fig. 52a shows one-eighth of the required volume. Therefore V = S I I I dxdydz. The limits in the first int egration are the values z = at P and z = cyl i^fl ^^ ^- '^^^ Umits in the second integration are the values of y a t M a nd N. At M, y = I x^ and at iV, 2 = 0, whence y = hy \ ^. Finally, the limits for X are and a. Therefore V = S I I I dx dy dz = ^ irdbc. Fig. 52&. Ex. 2. Find the center of gravity of the solid bounded by the paraboloid y^ -j- 2 z^ = 4 x and the plane x = 2. Art. 62 Rectangular Coordinates 115 By symmetry y and z are zero. The x-coordinate is xdv 4 1/ / xdzdydx . Jo Jo Jiiyi+2^) _4 X = Jd. 111'^'^ dx The limits for x are the values x = j (y^ + 2 2*) at P and X = 2 at Q. At S, X = 2 and y = V4 x - 2 2^ = V8 - 2 2*. The Umits for y are, therefore, y = at ^ and y = VS - 2 2^ at S. The limits for 2 are 2 = at'A and 2 = 2 atB. Ex. 3. Find the moment of inertia of a cube about an edge. Fig. 52c. Place the cube as shown in Fig. 52c and determine its moment of inertia about the 2-axis. The distance of any point (x, y, z) from the 2-axis is R = Vx2 + i/2. Hence the moment of inertia is I / (3^ + y')dxdydz=la\ »/o «/0 where a is the edge of the cube. 1 116 Triple Integration Chao. 9 EXERCISES 1. Find by triple integration the volume of the pyramid determined by the coordinate planes and the plane x + y -\- z = 1. 2. Find the moment of inertia of the pyramid in Ex. 1 about the a>axis. 3. A wedge is cut from a cylinder of radius a by a plane passing through a diameter of the base and inclined 45° to the base. Find its center of gravity. 4. Find the volume bounded by the paraboloid f; + -; = 2 - and b^ c' a the plane x = a. 6. Express the volume of the cone (z - 1)2 = a;2 + 2/2 in the first octant as a triple integral in 6 ways by integrating with dx, dy, dz, arranged in all possible orders. 6. Find the volume bounded by the surfaces j/^ = 4 a" — 3 ax, y^ = ax,z = ±/i. 7. Find the volume bounded by the cylinder ^ = \ — x — y and the coordinate planes. 53. Cylindrical Coordinates. — Let M be the projection of P on the a;?/-plane. Let r, Q be the polar coordinates of M in the xy-plane. The cyhndrical coordinates of P are r, Q, z. From Fig. 53a it is evident that x = r cos Q, y = T sin d. By using these equations we can change any rectangular into a cylindrical equation. The element of volume in cy- lindrical coordinates is the volume PQ, Fig. 536, bounded by two cylindrical surfaces of radii r, r + Ar, two horizontal planes z, z -\- Az, and two planes through the 2-axis making angles ^, ^ + A^ witli OX. The base of PQ is equal to the polar element MN in the xy~ plane. Its altitude PR is Az. Hence dv = rdd dr dz. (53) Fig. 53a. Art. 63 Cttjxdrical Coordixates 117 This value of dv can be used in the formulas for volume, center of gravity, moment of inertia, etc. In problems con- z Fig. 536. Fig. 53c. lected with cylinders, cones, and spheres, the resulting itegrations are usually much easier in cylindrical than in jtangular coordinates. 118 Triple Integration Chap. 9 Example 1. Find the moment of inertia of a cylinder about a diameter of its base. Let the moment of inertia be taken about the x-axis, Fig. 53c. The square of the distance from the element PQ to the X-axis is i^a = ^2 _|_ 22 = ^2 sin2 4- z\ The moment of inertia is therefore R^dv= I I I {r^smH-\-z^)rdddzdr Jo Jo Jo IT (3a2+4/i2). The first integration is a summation for elements in the wedge RS, the second a summation for wedges in the slice OMN, the third a summation for all such slices. Ex. 2. Find the volume bounded by the xy-plane, the cylinder x^ -\- y^ = ax, and the sphere x^ -{- y^ -\- z^ = c?. z Fig. 53d. In cylindrical coordinates, the equations of the cylinder and sphere are r = a cos B and r^ -\- z^ = a^. The volume j required is therefore v P2 r a cose f*^ai>-f v = 2 I I I rd9drdz= ^a^Zir -^). Jo Jo Jo Art. 64 Sphekical Coordinates 119 54. Spherical Coordinates. — The spherical coordinates of the point P (Fig. 54a) are r = OP and the two angles 6 and <i>. From the diagram it is easily seen that a; = r sin cos 6, i/ = r sin sin Q, z = r cos (j>. The locus r = const, is a sphere with center at 0; 6 = const, is the plane through OZ making the angle 6 with OX; <f> = const, is the cone gener- ated by lines through making the angle ^ with OZ. The element of volume is the volume PQRS bounded Fig. 54a. Fig. 545. by the spheres r,r -\- Ar, the planes 6,9 + Ad, and the cones </>, + A0. When Ar, A^, and A0 are very small this is 120 Triple Integration Chap. 9 approximately a rectangular parallelepiped. Since OP = r and POR = A<^, PR = r A</). Also OM = OP sin <^ and the arc PS is approximately equal to its projection MN, whence PS = MN = rsm(j> Ad, approximately. Consequently Av = PR'PS'PQ = r" sin (/> A^ • A0 • Ar, approximately. When the increments are taken smaller and smaller, the result becomes more and more accurate. Therefore dv = r^ sin (j>ddd<j) dr. (54) Spherical coordinates work best in problems connected with spheres. They are also very useful in problems where the distance from a fixed point plays an important role. Fig. 54c. Example. If the density of a solid hemisphere varies as the distance from the center, find its center of gravity. Take the center of the sphere as origin and let the 2-axis be perpendicular to the plane face of the hemisphere. By symmetry it is evident that x and y are zero. The density Art. 55 Attraction 121 is p = kr, where A; is constant. Also z = r cos <p. Hence z = fzdrn^f_ krzdv rr r2 Pa r* cos <t> sin <l>dd d<t) dr Jr»2T /»2 na »/0 t/0 -5"- r^ sin <t> dd d<t> dr EXERCISES 1. Find the volume bounded by the sphere a? + y* + 2^ = 4 and the paraboloid x* + y* = 3 z. 2. A right cone is scooped out of a right cylinder of the same height and base. Find the distance of the center of gravity of the remainder from the vertex. 3. Find the volume bounded by the surface z = e— (^H-i*) and the xy-plane. 4. Find the moment of inertia of a cone about a diameter of its base. 5. Find the volvune of the cylinder i* + i/^ = 2 ax intercepted between the paraboloid x^ + y^ = 2az and the xy-plane. 6. Find the center of gravity of the volume common to a sphere of radius a and a cone of vertical angle 2 a, the vertex of the cone being at the center of the sphere. 7. Find the center of gravity of the volume boimded by a spherical surface of radius a and two planes passing through its center and in- cluding an angle of 60°. 8. The vertex of a cone of vertical angle - is on the surface of a sphere of radius a. If the axis of the cone is a diameter of the sphere, find the moment of inertia of the volume common to the cone and sphere about this axis. 55. Attraction. — Two particles of masses mi, mz, sepa- rated by a distance r, attract each other with a force kmiiiv. 122 v Triple Integration Chap. 9 where fc is a constant depending on the units of mass, dis- tance, and force used. A similar law expresses the attraction or repulsion between electric charges. To find the attraction due to a continuous mass, resolve it into elements. Each of these attracts with a force given by the above law. Since the forces do not all act in the same direction we cannot ob- tain the total attraction by merely adding the magnitudes of the forces due to the sev- ^^°- ^^"• eral elements. The forces must be added geometrically. For this purpose we calculate the sum of the components along each coordinate axis. The force having these sums as components is the resultant attraction. If dm is the mass of an element at P, r its distance from 0, and 6 the angle between OX and OP, the attraction between this element and a unit particle at is , 1 • dm _ k dm This force acts along OP. Its component along OX is cos 6 • k dm The component along OX of the total attraction is then '*k cos 6 dm = / The calculation of this integral may involve single, double, or triple integration, depending on the form of the attracting mass. Example 1. Find the attraction of a uniform wire of length 2 I, and mass M on a particle of unit mass at distance | c along the perpendicular at the center of the wire. Take the origin at the unit particle and the a:-axis perpen- dicular to the wire. Since particles below OX attract down- Art. 66 Attraction 123 ward just as much as those above OX attract upward, the vertical component of the total attraction is zero. The component along OX is, therefore, the total attraction. The mass of the length dy of the wire is Mdy Hence X = 21 kM ^cos 6 dy 21 r- For simplicity of integration it is better to use 6 as variable. Then y = c tan 6, dy = c sec^ 6 dd, and X ~ 21 J-„ Fig. 556. cos • c sec2 d dd kM where a is the angle XOA X = & sec^ Q cl In terms of I this is kM sma, Ex. 2. Find the attraction of a homogeneous cylinder of mass M upon a particle of unit mass on the axis at distance c from the end of the cyUnder. By symmetry it is clear that the total attraction will act along the axis of the cyUnder. Take the origin at the attract- ing particle and let the y-ends be the axis of the cyhnder. 124 Triple Integration Chap. 9 Divide the cylinder into rings generated by rotating the elements dx dy about the y-axis. The volume of such a ring is 2 irx dx dy and its mass is M 2M dm = — 57 '2-Kxdxdy = —zj-xdx dy. Since all points of this ring are at the same distance from O and the joining Unes make the same angle 6 with OY, the vertical component of attraction is , fcosOJm _ , rydm _ 2 Mk A+* A xydydx J r" ~ J H ~ a^h Jc Jo (r^+y2)f = ^ Ih + V^T^ - Va'-\-(c-hh)q . aril Art. 56 Attraction 125 EXERCISES 1. Find the attraction of a imiform wire of mass M and length I on a particle of unit mass situated in the line of the wire at distance c from its end. 2. Find the attraction of a wire of mass M bent in the form of a semicircle of radius a on a unit particle at its center. 3. Find the attraction of a flat disk of mass M and radius a on a unit particle at the distance c in the perpendicular at the center of the disk. 4. Find the attraction of a homogeneoxis cone upon a unit particle situated at its vertex. 6. Show that, if a sphere is concentrated at its center, its attraction upon an outside particle will not be changed. 6. Find the attraction of a homogeneous cube upon a particle at one comer. CHAPTER X DIFFERENTIAL EQUATIONS 56. Definitions. — A differential equation is an equation containing differentials or derivatives. Thus (x^ + y^) dx -\-2xydy = 0, ^^_dy^ 2 dx"^ dx are differential equations. A solution of a differential equation is an equation connect- ing the variables such that the derivatives or differentials calculated from that equation satisfy the differential equa- tion. Thus y = x^ — 2 xisa solution of the second equation above; for when a;^ — 2 a; is substituted for y the equation is satisfied. A differential equation containing only a single independent variable, and so containing only total derivatives, is called an ordinary differential equation. An equation containing partial derivatives is called a partial differential equation. We shall consider only ordinary differential equations in this book. The order of a differential equation is the order of the highest derivative occurring in it. 57. Illustrations of Differential Equations. — Whenever an equation connecting derivatives or differentials is known, the equation connecting the variables can be determined by solving the differential equation. A number of simple cases were treated in Chapter I. The fundamental problem of integral calculus is to find the function = jf{x)dx, y 126 Arts? Illustrations of Differential EkjUATioNS 127 when / (x) is given. This is equivalent to solving the differ- ential equation dy = f (x) dx. Often the slope of a curve is known as a function of x and y. The equation of the curve can be found by solving the differential equation. In mechanical problems the velocity or acceleration of a particle may be known in terms of the distance s the particle has moved and the time t. ds dt='^ df = a. The position s can be determined as a function of the time by solving the differential equation. In physical or chemical problems the rates of change of the variables may be known as functions of the variables, and the time. The values of those variables at any time can be found by solving the differential equations. Example. Find the curve in which the cable of a suspension bridge hangs. Let the bridge be the x-axis and let the i/-axis pass through the center of the cable. The portion of the cable AP is in I 128 Differential Equations Chap. 10 equilibrium under three forces, a horizontal tension fl" at A, a tension FT in the direction of the cable at P, and the weight of the portion of the bridge between A and P. The weight of the cable, being very small in comparison with that of the bridge, is neglected. The weight of the part of the bridge between A and P is proportional to x. Let it be Kx. Since the vertical com- ponents of force must be in equilibrium r sin = Kx. Similarly, from the equilibrium of horizontal components, we have Tcos<l> = H. Dividing the former equation by this, we get tan <f> = y^x. £1 But tan <j) = -^ . Hence dx H ' The solution of this equation is The curve is therefore a parabola. 58. Constants of Integration. Particular and General Solutions. — To solve the equation we integrate once and so obtain an equation with one arbi- trary constant, y = jf(x)dx + c. Art. 58 Constants of Integration 129 To solve the equation we integrate twice. The result y = j jf{x)dx^+ CiX + Cj contains two arbitrary constants. Similarly, the integral of the equation contains n arbitrary constants. These illustrations belong to a special type. The rule indicated is, however, general. The complete, or general, solution of a differential equation of the nth order in two vari- ables contains n arbitrary constants. If particular values are assigned to any or all of these constants, the result is still a solution. Such a solution is called a particular solution. In most problems leading to differential equations the result desired is a particular solution. To find this we usually find the general solution and then determine the constants from some extra information contained in the statement of the problem. Example L Show that aJf-f_y2_ 2cx = is the general solution of the differential equation y^-x^-2xy^ = 0. Differentiating x^ -\- y- — 2cx = 0, we get whence dy _ c — X dx u 130 Differential Equations Chap. 10 Substituting this value in the differential equation, it becomes yi — x^ — 2xy -^ = y~ — x^ — 2x{c — x) = if -]-x^ — 2cx = 0. Hence x^ -{- y^ — 2 ex = Q \s, a solution. Since it contains one constant and the differential equation is one of the first, order, it is the general solution. Ex. 2. Find the differential equation of which y = Cie"" + C2e^^ is the general solution. Since the given equation contains two constants, the differential equation is one of the second order. We there- fore differentiate twice and so obtain d'y dx" = C\e- ' + 4 cae^*. EUminating Cl, we get d:'y dx" dy _ dx = 2c2e2^, dy _ dx ■y = de"'. Hence diy_dy^^/dy_\ dx^ dx \dx / or p^-3^ + 2y=0. ax- ax This is an equation of the second order having y = Cie" -\- C2fi^' as solution. It is the differential equation required. EXERCISES In each of the following exercises, show that the equation given is a solution of the differential equation and state whether it is the general or a particular solution. Art. 59 The Fikst Obder ix Two Variables 131 2.^-yS=cx, ix' + y')dx-2xydy = 0. 3^ 1/ = 06* sin X, g_2| + 2y = 0. 4. y =ci+ctsin(x+c), ^ + df "^ ®- Find the differential equation of which each of the following equa* tions is the general solution: _ cj 7. y = Ci sin X + ci coe z. 5. y-cix + -' ^ x*y = ci+c4hxz + cvr». 6. y = cxe'. 9. i^ + Cixy + Cjy* = 0. 59. Differential Equations of the First Order in Two dv Variables. — By solving for -p an equation of the first order in two variables x and y can be reduced to the form To solve this equation is equivalent to finding the curves with slope equal to f(x,y). The solution contains one Fig. 59. arbitrarj' constant. There is consequently an infinite num- ber of such curves, usually one through each point of the plane. We cannot always solve even this simple tj'pe of equa- tion. In the follo\sing articles some cases will be discussed 132 Differential Equations Chap. 10 which frequently occur and for which general methods of solution are known. 60. Variables Separable. — A differential equation of the form Mdx + Ndy = is called separable if each of the coefficients M and N «;on- tains only one of the variables or is the product of a function of X and a function of y. By division the x's and dx can be brought together in the first term, the y's and dy in the second. The two terms can then be integrated separately and the sum of the integrals equated to a constant. Example 1. (I -^ x^) dy — xy dx = 0. Dividing by (1 + x^) y, this becomes dy _ xdx y ~ 1 +a;2' whence ln^ = iln(l+a;2)+c. If c = In k, this is equivalent to \ny = lnVT+^+ln/c = lnfc VT+i", and so y = k Vl + x^, where k is an arbitrary constant. Ex. 2. Find the curve in which the area bounded by thej curve, coordinate axes, and a variable ordinate is proportions to the arc forming part of the boundary Let A be the area and s the length of arc. Then A = ks. Differentiating with respect to x, dA _ yds dx dx' or ''-v/'+dj Art. 61 Exact Differential Equations [13a Solving for ■—, dy_ Vy^ - k^ dx~ k ' whence dy dx Vt/2 - fc2 k' The solution of this is In (2/ + Vi/2 _ A-2) =^ + c. -+e Therefore y + V?/2 — ^-2 = e* = e*e* = Cie* , where Ci is a new constant. Transposing y and squaring, we get Hence, finally, y' k" = [cie^j - 2 cie*2/ + y^. k^ -I y = 2' -^27.' k. 61. Exact Differential Equations. — An equation du — 0, obtained by equating to zero the total differential of a func- tion u of X and y, is called an exact differential equation. The solution of such an equation is u = c. The condition that M dx -\- N dy be an exact differential is (Diff. Cal., Art. 100) dy dx This equation, therefore, expresses the condition that M dx + N dy = be an exact differential equation. (61) 134 Differential Equations Chap. 10 An exact equation can often be solved by inspection. To find u it is merely necessary to obtain a function whose total differential is M dx -\- N dy. If this cannot be found by inspection, it can be determined from the fact that du = M dx -{-N dy and so — = M. dx By integrating with y constant, we therefore get w = / Mdx-\-j{y). Since y is constant in the integration, the constant of inte- gration may be a function of y. This function can be found by equating the total differential of u to M dx -\-'N dy. Since df (y) gives terms containing y only, / (y) can usually be found hy integrating the terms in N dy that do not contain x. In exceptional cases this may not give the correct result. The answer should, therefore, be tested by differentiation. Example 1. (2x — y) dx + (iy — x) dy = 0. The equation is equivalent to 2xdx + 4:ydy— (ydx+xdy) = d {x^ -\- 2 y"^ — xy) =0. It is therefore exact and its solution is x^ -\- 2 y^ — xy = c. Ex.2. (\iiy-2x)dx + {^-2'^dy=Q. In this case -r- =-^{\ny -2x) =-) dy dy^ "^ ' y dx dx\y V y' Art 62 Integrating Factors 135 These derivatives being equal, the equation is exact. Its solution is x]ny— 3^ — y' = c. The part x In y — x^ is obtained by integrating (In 2/ — 2 x) dx with y constant. The term — y^ is the integral oi — 2 y dy, which is the only term in f ^yjdy that does not contain x. 62. Integrating Factors. — If an equation of the form M dx + N df/ = is not exact it can be made exact by multiplj-ing b}' a proper factor. Such a multipher is called an iiUegrating factor. For example, the equation xdy — ydx = is not exact. But if it is multiplied by -j* it takes the form xdy— ydx -0-^ which is exact. It also becomes exact when multiplied by -5 or — . The functions -;»-;? — are all integrating factors y- xy x^ y^ xy oi xdy — y dx = 0. While an equation of the form M dx -\- N dy = always has integrating factors, there is no general method of finding them. Example 1. y (I -\- xy) dx — x dy = 0. This equation can be written y dx— xdy -\- xy^dx= 0. Dividing by y^, ydx — X dii y' + X dx = 0. Both terms of this equation are exact differentials. The solution is ^ 1 1 2 y 2 136 Differential Equations Chap. 10 Ex. 2. (i/2 + 2 xy) dx + {2x^ + 3 xy) dy = 0. Tliis is equivalent to y'^ dx + Z xy dy -\- 2 xy dx ■\- 2 x^ dy = 0. Multiplying by y, it becomes y^dx + 3 xy^ dy + 2 xy^ dx + 2x^ydy = d {xy^ + x V) = 0. Hence xy^ + x^y^ = c. 63. Linear Equations. — A differential equation of the form % + Py = Q> (63a) where P and Q are functions of x or constants, is called liJiear. The linear equation is one of the first degree in one of the variables (y in this case) and its derivative. Any functions of the other variable can occur. If the linear equation is written in the form (63a), fPdx e is an integrating factor; for when multiplied by this factor the equation becomes The left side is the derivative of Hence y/^'^^ J/^'^Qdx + c (63b) is the solution. Example 1. ^ + ^ ^ = ^* i Art. 64 Equations Reducible to Linear Form 137 In this case P'^-f -dx = 2hix = hix^. Henc3 fpdx lnx» „ The integrating factor is, therefore, x^. Multiplying by x* and changing to differentials, the equation becomes j^ dy -\- 2 xy dx = a^ dx. The integral is x^y = I x« + c. Ex. 2. a+y')dx-ixy-\-y+y')dy = 0. This is an equation of the first degree in x and dx. Divid- ing by (1 + t/2) dy^ it becomes dx y _ di'TTy^'^-y- P is here a function of y and fpdy 1 vTT? Multiplying by the integrating factor, the equation becomes dx xydy _ ydy whence = vr+7 + r-i-c and X = 1+ 2/2 -I- c Vl +?/2. 64. Equations Reducible to Linear Form. — An equation of the form ^4-Pi/ = Q2/», (64) 138 Differential Equations Chap. 10 where P and Q are functions of x, can be made linear by a change of variable. Dividing by ?/", it becomes If we take I as a new variable, the equation takes the form 1 du 1 — ndx which is linear. Examvle. f^ + lv =%^ + Pu = Q, Division by y^ gives -3 _ dx ^ x^ x^ .dy . 2 „ 1 Let Then u = y~^. dx dx' whence -3 ^ — —1 — ^ dx~ 2dx' Substituting these values, we get 2dx'^x'^ x'' and so dw_4 ^ _2 dx X x^ This is a linear equation with solution 3x2 w = t:;:2 + ex*' or, since u = y~^, y2-3x2 + ^^' Art. 65 HoMOGEXEOTJS Equations 139 65. Homogeneous Equations. — A function / (x, y) is said to be a homogeneous function of the nth degree if fitx,ty) =t-f{x,y). Thus Vx^ -f- y^ is a homogeneous function of the first degree; for Vx¥T^ = t Vx2 + y2. ( 7\ It is easily seen that a polynomial whos^ terms are all of tnfe nth degree is a homogeneous function of the nth degree. The differential equation Mdx+Ndy = is called homogeneous if M and N are homogeneous functions of the same degree. To solve a homogeneous equation substitute y = vx. The new equation will be separable. dy Example 1. x-^ — y = Vx^ + y^. This is a homogeneous equation of the first degree. Sub- stituting y = vx, it becomes whence (v+^j-)- vx = Vx^+v^ .| = v/IT7. This is a separable equation with solution X = civ + Vl-\-v^). Replacing t; by -, transposing, squaring, etc., the equation becomes x2 - 2 q/ = c*. "140 Differential Equations Chap. 10 dy dx Ex.2. ym\2x%-y = (i. Solving for j-, we get dy^ _ — X ± Vx^ + y^ dx~ y or ydy+xdx= ± Vx^ + y^ dx. This is a homogeneous equation of t he first degree. It is much easier, however, to divide by Vx^ + y^ and integrate at once. The result is xd x-\-ydy ^_^^^^ v x^ + y^ whence VxH-^ = cztx and 2/2 = c2 ± 2 ex. Since c may be either positive or negative, the answer can be written y2 = c2 + 2 crc. 66. Change of Variable. — We have solved the homo- geneous equation by taking as new variable X It may be possible to reduce any equation to a simpler form by taking some function w of a; and y as a new variable or by taking two functions u and v as new variables. Such func- tions are often suggested by the equation. In other cases they may be indicated by the problem in the solution of which the equation occurs. Example, (x ~ VY^ =" ^** Art. 66 Change of Variable 14 J Let X— y = u. Then dy du dx dx and the differential equation becomes whence 2 2 2<^" u^ — a^ = w* . dx The variables are separable. The solution is X =u +7^ In—— he 2 u -\- a or , a, X — y — a , = x — 7/+?: In ^— he, 2 a; — 1/ + o a x — y — a 2/ = - In ^— h e. ^ 2 X — 2/ + a EXERCISES Solve the following differential equations: L x'dy -jfdx = 0. 2. tan X sin* y dx + cos* x cot ydy = 0. 3. (xy* + x) dx + (i/ - x*y) dy = 0. 4. (xy* + x) dx + (x*j/ -y)dy = 0. 6. (3x* + 2xy-y*)dx+(x»-2xy-3t/*)di/ = 0. 6. ;c^-t/ = y». 12. (2x2/»-y)dx + xrfy = 0. 7. xdx + ydy = a(x» + y»)dy. 13. (i_a^)^+2xi/ = (l-x*)». , 14. tan x— — y = a. #v dy fc_ dx 9. -f- — ay = er'. 10. Jf?-2xv = 3. W.x|-3,+xV.O. dx 11. x*g-2xy = 3t,. 16. % + y = xy». 142 Differential Equations Chap. 10 17. (x2 - 1)3 dy + (x^ + 3 xy VjnTi) dx = 0. 18. X dx -{- {x + y) dy = 0. 19. (a;2 + i/2) dx-2xydy = 0. 20. 2/da; + (a: + ?/)dy = 0. 21. (x3 _ 3 x2y) dx + (2/3 - a;') dy = 0. 22. 2/e'' dx = {f + 2 xe^) dy. 23. I xye^ + j/^j dx - x^e^ dy = 0. 24. {x + y - 1) dx + (2x + 2 y - 3) dy = 0. 25. 3y2^-y3 = a;. 30. The differential equation for the charge 5 of a condenser having a capacity C connected in series with a circuit of resistance R is ^dl + c-^' where E is the electromotive force. Find v as a function of i if ^ is constant and g = when t = 0. 31. The differential equation for the current induced by an electro- motive force E sin at in a circuit having the inductance L and resist- ance R is di L-r, + Ri = ^ sin at. dt Solve for i and determine the constants so that i = I when t = 0. Let PT be the tangent and PN the normal to a plane curve at P (x, y) (Fig. 66a). Determine the curve or curves in each of the following cases' 32. The subtangent TM = 3 and the curve passes through (2, 2). 33. The subnormal MN = a and the curve passes through (0, 0). 34. The intercept OT of the tangent on the a>axis is one-half the abscissa OM. 35. The length PT of the tangent is a constant a. 36. The length PN of the normal is a constant a. 37. The perpendicular from M to PT is a constant o. Art. 67 Certain Equations of the Second Ordeb 143 Using polar coordinates (Fig. 666), find the curve or curves in each of the following cases: Fig. 66a. Fig. 666. 38. The curve passes through (1, 0) and makes with OP a constant angle lA = ^• 39. The angles i^ and d are equal. 40. The distance from O to the tangent is a constant a. 41. The projection of OP on the tangent at P is a constant a. 42. Find the curve passing through the origin in which the area bounded by the curve, x-axis, a fixed, and a variable ordinate is pro- portional to that ordinate. 43. Find the curve in which the length of arc is proportional to the angle between the tangents at its end. 44. Find the curve in which the length of arc is proportional to the difference of the abscissas at its ends. 46. Find the curve in which the length of any arc is proportional to the angle it subtends at a fixed point. 46. Find the curve in which the length of arc is prop>ortional to the difference of the distances of its ends from a fixed point. 47. Oxj'gen flows through one tube into a liter flask filled with air while the mixture of oxygen and air escapes through another. If the action is so slow that the mixture in the flask may be considered uniform, what percentage of oxygen will the flask contain after 10 liters of gas have passed through? (Assume that air contains 21 per cent by volume of oxygen.) 67. Certain Equations of the Second Order. — There are two forms of the second order differential equation that 144 Differential Equations Chap. 10 occur in mechanical problems so frequently that they de- serve special attention. These are The peculiarity of these equations is that one of the vari- ables (y in the first, x in the second) does not appear directly in the equation. They are both reduced to equations of the first order by the substitution dy This substitution reduces the first equation to the form This is a first order equation whose solution has the form p = F{x, ci), dy or, smce p = -f-^ This is again an equation of the first order. Its solution is the result required. In case of an equation of the second type, write the second derivative in the form ^ _ dp _dp dy _ dp dx^ dx dy dx dy The differential equation then becomes p| = /(y,p). Art. 67 Certain Equations of the Second Ordeb 145 Solve this for p and proceed as before. Example 1. (1 + x^) g + 1 + [^J = 0. Substituting p for --^, we get (l+x2)g + 14-p^ = 0. This is a separable equation with solution Ci — X whence dy = T^- dx. The integral of this is By a change of constants this becomes 2/ = ex + (1+ c2) In (c - x) + c'. Substituting dy ^ _ _ ^P we get dx ^' dx2 ^dj/' The solution of this is y2p2 = t/2 _|_ c^_ dv Replacing p by -^ and solving again, we get i/2 + c, = (a; + C2)2. 146 Differential Equations Chap. 10 Ex. 3. Under the action of gravitation the acceleration of k a falhng body is -^ , where k is constant and r the distance from the center of the earth. Find the time required for the body to fall to the earth from a distance equal to that of the moon. Let Ti be the radius of the earth (about 4000 miles), r2 the distance from the center of the earth to the moon (about 240,000 miles) and g the acceleration of gravity at the surface of the earth (about 32 feet per second). At the surface of the earth r = ri and k a = — 2= — g. The negative sign is used because the acceleration is toward the origin (r = 0). Hence k = — gr-^ and the general value of the acceleration is ar r where v is the velocity. The solution of this equation is r When r = r2, v = 0. Consequently, C=-24 and The time of falling is therefore =XV: ^ dr = 116 hours. 2 gri' {1-2 — r) This result is obtained by using the numerical values of n and Ti and reducing g to miles per hour. Art. 68 Constant Coefficiknts 147 68. Linear Differential Equations with Constant Coeffi- cients. — A differential equation of the form d'^y , d"~^y , d"~^y , , r/ \ /^o \ where ai, ch, . . . an are constants, is called a linear differen- tial equation with constant coefficients. For practical appUca- tions this is the most important type of differential equation. In discussing these equations we shall find it convenient to represent the operation — by D. Then ^ = Dt, ^ = D^ etc dx ^' da^ ^^, etc. Equation (68a) can be written (D- + aiD"-i + a^D"-'- + • • • + a,)y = / (x). (68b) This signifies that if the operation Z)» + aiZ)"-i + aiD"-^ + • • • + a„ (68c) is performed on y, the result will be /(x). The operation consists in differentiating y, n times, n — 1 times, n — 2 times, etc., multiplying the results by 1, Oi, as, etc., and adding. With the differential equation is associated an algebraic equation r" + air"-i + a^r''-'^ + • • • + On = 0. If the roots of this auxiliary equation are ri, r^, . . , r„, the polynomial (68c) can be factored in the form (D - n) (D - r^) . . . (D - r,). (68d) If we operate on y with D — a, we get {D-a)y = ^-ay. 148 Differential Equations Chap. 10 If we operate on this with D — 6, we get ^D - b) ' (D - a)y = {D - b)(^^- ay^ The same result is obtained by operating on y with iD-a){D-h) =D^- (a-i-b)D+ ab. Similarly, if we operate in succession with the factors of (68d), we get the same result that we should get by operating- directly with the product (68c). 69. Equation with Right Hand Member Zero. — To solve the equation (D» + aiZ)"-i + (hD^-^ + • • • +an)y = (69a) factor the symbolic operator and so reduce the equation to the form (D - ri) (D - ra) . . . (D - Vn) y = 0. The value y = Cie^^^ is a solution; for (D— n) Cie'"'^ = Cirie""!^ — riCie'"'^ = and the equation can be written (D-r^) • • • (Z)-r„) . {D-n)y=iD-r,) ■ • • (D-r„).0 = 0. Similarly, y = C2e'"'^ y = c^e^^, etc., are solutions. Finally y = cie^>^+ 026"'^ + • • • + c„e'""^ (69b) is a solution; for the result of operating on y is the sum of the results of operating on Cie"^^, c^d^^, etc., each of which is zero. If the roots Vi, 7*2, . . . , /•„ are all different, (69b) contains n constants and so is the complete solution of (69a). If, however, two roots ri and r^ are equal CiC*' + C2e''»^ = (ci + Co) e*"" Art. 69 Equatiox vrira Right Hand Member Zero 149 contains only one abitrary constant Ci + C2 and (69b) con- tains only n — 1 arbitrary' constants. In this case, however, xe"^^ is also a solution; for {D — n) xe^^ = Tixe^^ -\- e'*' — nxe^^ = e'** and so (D - ri)2 xe*-!' = (D - rO e^ = 0. If then two roots r^ and rt are equal, the part of the solu« tion corresponding to these roots is (Ci + CiX)e^. More generally, if m roots n, r2, . . . r^ are equal, the part of the solution corresponding to them is (ci + CoX + C3X2 + • • . + c^x'^^)e^^. (69c) If the coefl5cients Oi, Oo, . . . On, are real, imaginary roots occur in pairs ri = a + /3 V- 1, r2 = a - /3 V^T. The terms Cie*"'', C2e''»^ are imaginary' but they can be replaced by two other terms that are real. Using these values of r^ and r2, we have {D-T,){D-r^) = {I)-aY + ^. By performing the differentiations it can easily be verified that [{D - a)2 + /32] . e«^ sin ^x = 0, [(D - a)2 + /32] . e« cos /3x = 0. Therefore ef^ [ci sin /3x + C2 cos |3x] (69d) is a solution. This function, in which a and /3 are real, can, therefore, be used as the part of the solution corresponding^ to two imaginary' roots r = a ± /3 V— 1. To solve the differential equation (D- + aiZ)»-i + aj)--^ + • • • + a„) y = 0, 160 Differential Equations Chao. 10 let ri, r2, . . . , r„ be the roots of the auxiliary eqvxition r" + air"-i + a^r""-^ + • • • + an = 0. If these roots are all real and different, the solution of the ^qvMion is y = cie^>^ + c^e''^ + • • • + c„e'"''^ // m of the roots ri, rz, . . . , rm are equal, the corresponding part of the solution is (ci + CiX + C3x'^ + ' ' ■ + CmX"^-^) e'-'^ The part of the solution corresponding to two imaginary roots r = ado^ V— 1 is e°^ [ci sin ^x + Cz cos ^x], This is equivalent to {D^-D-2)y = 0. The roots of the auxiUary equation r2 - r - 2 = a,re — 1 and 2. Hence the solution is y = Cie~' + de^', Ex.2. f^ + *|-5^ + 3s, = 0. ■ dx^ dx^ dx ^ The roots of the auxihary equation are 1,1, — 3. The part of the solution corresponding to the two roots equal to 1 is (ci + c^x) e". Hence 2/ = (ci + Gjx) e" + Cie~^'. Art. 70 Right Hand Member a Function of x 151 Ex.Z. (D2 + 2D + 2)i/ = 0. The roots of the auxiliary equation are - izb v^ni. Therefore a = — 1, /3 = 1 in (69d) and y = e~' [ci sin a: + Co cos x]. 70. Equation with Right Hand Member a Function of x. — Let y = uhe the general solution of the equation (I>» + aiD"-i + a2D"-2 + • • • +a„) y = and let y = t; be any solution of the equation (D" + axZ)»-i + a^D--' + - - - an) y = f (x). (70) Then y = u + v is a solution of (70) ; for the operation D" + ai2)»-i + azD'^^ + • • • + a» when performed on u gives zero and when performed on v gives / (x). Furthermore, u -\- v contains n arbitrary' con- stants. Hence it is the general solution of (70). The part u is called the complementary function, v the 'particular integral. To solve an equation of the form (70), first solve the equation with right hand member zero and then add to the result any solution of (70). A particular integral can often be found bj' inspection. If not, the general form of the integral can usually be deter- mined by the following rules: 1. If fix) = ax" + flix"-! + • • • + a„, assume y = Ax" + Aix"-i + • • • + A,. But, if occurs m times as a root in the auxiliary equation, assume 1/ = x™ [Ax" + Aix"-i + • • • + A«]. 2. If / (x) = ce", assume y = Ae^. 152 Differential Equations Chap. 10 But, if a occurs m times as a root of the auxiliary equation, assume y = Ax"'e'". 3. If f (x) = a cos ^x -\-b sin /Sx, assume y = A cos ^x -\- B sin /Sx. But, if cos fix and sin fix occur in the complementary function, assume y = x[A cos fix -^ B sin fix]. 4. If / (x) = ae°^ cos fix + 6e"^ sin fix, assume y = Ae"^ cos /3a; + /3e°^ sin ySa;. But, if e"^ cos jSx and e"^ sin /3rr occur in the complementary function, assume y = xe"^ [A cos fix -\- B sin j8a;]. If / (x) contains terms of different types, take for y the sum of the corresponding expressions. Substitute the value of y in the differential equation and determine the constants so that the equation is satisfied. Example 1. -^^ -\- 4 y = 2 x + S. A particular solution is evidently y = i{2x-{-3). Hence the complete solution is y = ci cos 2 X + C2 sin 2 a; + J (2 x + 3). Ex.2. (D2 + 3 Z) + 2) 1/ = 2 + e\ Substituting y = A -\- Be'', we get 2A -^QBe' = 2-{-e'. Hence and 2A = 2, QB = 1 y = 1 + i e^ + cie-' + C2e-2*. Ex.3. d'y . <i'y _ ^2 Art. 71 Simultaneous Equations 153 The roots of the auxihary equation are 0, 0, — 1. Since is twice a root, we assume y = x'{A3^ + Bx-{-C)=Ax*-\-Bx^-\- Cx'. Substituting this value, 12 Ax2 + (24 A + 6 5) z + 6 B + 2 C = x2. Consequently, 12.4 = 1, 24A+6B = 0, 6B + 2C = 0, whence A=^\, B=-l C = l. The solution is y = j\x^ - Ix^ +x~ -\-Ci +C2PC -{■ c^-'. 71. Simultaneous Equations. — We consider only linear equations with constant coefficients containing one independ- ent variable and as many dependent variables as equations. All but one of the dependent variables can be eliminated by a process analogous to that used in soh-ing linear algebraic equations. The one remaining dependent variable is the solution of a linear equation. Its value can be found and the other functions can then be determined by substituting this value in the previous equations. Example. -z--\-2x — 3y = t, Using D f or -r., these equations can be written {D-\-2)x-3y = t, {D-\-2)y-3x = ^'. To ehminate y, multipU' the first equation by Z) + 2 and the second by 3. The result is (D + 2)2x-3(Z) + 2)i/= 1+2^, 3(D+2)i/-9x = 3€2'. 154 Differential Equations Chap. 10 Adding, we get [(D 4-2)2 -9]a: = l+2« + 3e2'. The solution of this equation is x= -M-H + ye^'+Cie' + Cae-^'. Substituting this value in the first equation, we find y = i (^ + 2) X - i « = - f < - if + ^ e^ ' + cie' - C2e-". EXERCISES Solve the following equations: 2. (, + i)g_(. + 2)|+x + 2 = 0. d<2 ~ S^" rfx3 ■"dx^'^'^dx 12. g+j,=0. 25. g-a=!, = e". 14. ^=V. 27. g -,-..- >.. «-S-^| + ^''='>- 28.g-4|+3v = e...„.. Art. 71 SufCTLTANEOTrs Equations 155 29. ^-9y = e»«co8x. 31. ^ + 4y = co82x. ax -y = e-*. dy = X -y + 2 dx dt + 2/ = cos<. 33. f+. = e., 34. g=x-2y + l, „- . dx dy , n ■ s 3^- ■*di-i + ^^ = ^"^'' 37. Solve the equation ^+f^V = i do^^\dx] and determine the constants so that y = and -^ = \ when x = 0. 38. Solve j^ = 3 v^ under the hypothesis that y = 1 and 3^ = 2 when X = 0. 39. When a body sinks slowly in a Uquid, its acceleration and velocity approximately satisfy the equation a = g — ko, g and A; being constants. Find the distance passed over as a function of the time if the body starts from rest. 40. The acceleration and velocity of a body falling in the air approxi- mately satisfy the equation a = g — kv^, g and A; being constants. Find the distance traversed as a fimction of the time if the body falls from rest. 41. A weight supported by a spiral spring is lifted a distance b and let faU. Its acceleration is given by the equation a = — fc*s, fc being constant and s the displacement from the position of equiUbriima. Find s in terms of the time t. 42. Find the velocity with which a meteor strikes the earth, assimi- ing that it starts from rest at an indefinitely great distance and moves toward the earth with an acceleration inversely proportional to the square of its distance from the center. 43. A body falling in a hole through the center of the earth would have an acceleration toward the center proportional to its distance from the center. If the body starts from rest at the surface, find the time reqxiired to fall through. 156 Differential Equations Chap. 10 44. A chain 5 feet long starts with one foot of its length hanging over the edge of a smooth table. The acceleration of the chain will be proportional to the amount over the edge. Find the time required to slide off. 45. A chain hangs over a smooth peg, 8 feet of its length being on one side and 10 on the other. Its acceleration will be proportional to the difference in length of the two sides. Find the time required to slide off. SUPPLEMENTARY EXERCISES xdx J a + bi^' J{a + hx)*dx. a -\-hx dx. J p + qx 4. fxV2 -3i*dr. 5 f (x + a)dx J Vx» + 2 ax + 6 7. J*(i-l)(x»-2x)'dr. /x- 1 dx X*' CHAPTER II 18. 19. 20. 21. 22. 23. 8. r. J SI dx 0. 1. sin ax ■in ox dx. COS* ax cos2x sin2x sec^xtanxdx /cos 2.., \ r-^T- dx. 1 — su /sej a + bsec*x / dx secx cot xdx /i / / / sinx sin ax cos ax dx dx. cos X — sin X dx sec X — tan x dx sin' ax COS* ax cosxdx cos X + sin X dx Vtan' X + 2 dx X Vx» - 1 dx. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. cot X dx Vl + sin' X dx / / / / / dx sin* X — cos* / ^ (2x-l) V4i*-4x fxe-^dx. /• f^dx J 6 + ce"' j sec* xe**" * d!x. jah^dx. e - j tanxlncoexdx. / / e' + l dx_ e" dx a*x* + 2a6x + 6** xdx Vx2 _ 2x + 3 (2x + 3)dx (i - 1) Vx* -2x' 167 158 Supplementary Exercises 35. r — f'^ • 56. fx3 Va2 - x^ dx. •^ X V 2 X + 3 J •^ x2 Vax2 + 6* -^ ^^2 - x^ 37. j {a? - x')'^ dx. 58. /^^r^^ 38 49. x^dx IP dx 1 f ^^ 69. f ^^• J X Vax^ + 6x -^ («' - ^')* 39. JV3-2x-x^dx. 60. /^^^- 40. r(a3-x?)3dx. 61. Je^ + in*dx. 41. fcosSxsm^xdx. 62. J e'^sinftxtte. 42. r(l+cosx)^dx. 63. J^ d^- 43. rtan2xsec6 2xda;. 64. Tx In (ex + &) da;. 44. jTcot^xdx. 65. Jll^^^da: 45. fi ^r — r- 66. Cxcoi-^xdx. J tan X + cot X J 46. /(secx+tanx)^dx, 67. j^^^—0^^-^, tan X — 1 ,_ „„ r x" dx 47 ._ rcos^xdx ^^ r 48. I —^-, 69. I tan X + 1 ■ "°- J (r> + 1) (x* - 2) • cos'xdx „„ r xdx x^ + 1 /sin 2 X cos^ X dx. 70. C - — ^--r- dx. J (x — 1)* 50. JVI + cos2 X sin 2 X dx. 71, fx^cosixdx. 51. C dx ^ f 2x^ + 3x •^ X Va^x + 62 '^- J (x-l)(x-2)(x4-3) ^2 f ^^ 70 f (3x-5)dx ■ ^ x^ V^3-2 "^- J X (x + 3)2 * -^ (x - 1) Vx + 2 J ^-^ 64. Jx(ax + 6)«dx. 75. /jj^^* 65. (^^M.dx. 76. f ^^^ , . J Vax + 6 •' (1 -2x2)» StIPPLEMENTART EXERCISES 159 x2 - 3j + 2 dx. 77. faec^xtaxi^xdx. 84. f- , J J Vj^ _ 4 X + 3 78. I sin Sxcos 4xdx. a- CKrLUJl dx. J y a + X 79. j'sin2xcos2idx. 86. J(sinx - co8x)»(ir. 80. fsinxsinSxdx. 87 f ^^ — . rco8 2xco6 3xdx. 88 r log (x + Vx^ - l) J Vx2- 1 . J (cotx + cscx)^dx. 89 Jgec^^dx. /(3x — l)c?x /. , -===• 90. /(x«+««)idx. cir. 83. CHAPTER IV 91. Find the area bounded by the x-axis and the parabola y = x* -4x + 5. 92. Find the area bounded by the curves y = 3?, y^ = x. 93. Find the area bounded by the parabola y* = 2 x and the witch 1 y- + 1 94. Find the area within a loop of the curve j/* = x= — x*. 95. Find the area of one of the sectors boimded by the hyperbola X* — y* = 3 and the lines x = =fc 2y. 96. Find the area bounded by the parabolas y' = 2 ax + a-, ^ -\- 2ax = 0. 97. Find the area within the loop of the curve x = ; — ■. r, y = 1 + wi 3aw^ 98. Find the area bounded by the parabola x = a cos 2<t>, y = o sin (^ and the line x = —a. 99. Find the area inclosed by the curve x = a cos^ <l>, y = b sin* 4>. 100. Find the area bounded by the curve x = asin^, y=a cos* 0. 101. Find the area of one loop of the curve r = a cos nd. 102. Find the area of a loop of the cxu^'e r = a (1 — 2 cos 0). 103. Find the area between the curves r = a (cos -{- 2), r = a. 104. Find the total area inclosed by the cxu-ve r — a sin \ 0. 105. Find the area of the part of one loop of the curve 1^ = 0^ sin 3 & outside the curve r^ = a sin 0. 100 Supplementary Exercises 106. By changing to polar coordinates find the area within one loop of the curve (x^ + ifY = a?xy. 107. By changing to polar coordinates find the area of one of the regions between the circle x"^ -\- y^ = la? and hyperbola x^ — y* = a^. 108. Find the area of one of the regions bounded by d = sin r and the line e = \. 109. Find the volume generated by revolving an ellipse about the tangent at one of its vertices. 110. Find the volume generated by revolving about the y-axis the area bounded by the curve y^ = 3? and the line x = 4. 111. Find the volume generated by rotating about the y-axis the area between the x-axis and one arch of the cycloid x = o (<^ — sin ^), y = o (1 — cos<^). 112. Find the volume generated by rotating the area of the preced- ing problem about the tangent at the highest point of the cycloid. 113. Find the volume generated by revolving about the x-axia the part of the ellipse ^ — xy -\- y^ = I'va. the first quadrant. 114. Find the volume generated by revolving about 9 —-^ the area enclosed by the curve r" = (^ sin B. 115. The ends of an ellipse move along the parabolas z* = ox, J/'' = ax and its plane is perpendicular to the x-axis. Find the volume swept out between x = and x = c. 116. The ends of a helical spring lie in parallel planes at distance h apart and the area of a cross section of the spring perpendicular to its axis is A. Find the volume of the spring. 117. The axes of two right circular cylinders of equal radius intersect at an angle a. Find the common volume. 118. A rectangle moves from a fixed point, one side varying as the <iistance from the point, and the other as the square of this distance. At the distance of 10 feet the rectangle becomes a square of side 4 ft. What is the volume then generated? 119. A cylindrical bucket filled with oil is tipped until half the bot- tom is exposed; if the radius is 4 inches and the altitude 12 inches find the amount of oil poured out. 120. Two equal ellipses with semi-axes 5 and 6 inches have the same major axis and lie in perpendicular planes. A square moves with its •center in the common axis and its diagonals chords of the ellipses. Find the volume generated. 121. Find the volume bounded by the paraboloid 12 3 = 3 x'^ -f y* And the plane 3 = 4. StIPPLEMENTARY ElXERCISES 161 CHAPTER V 122. Find the length of the arc of the curve y = i X Vx2 - 1 — ^ In (x + Vi* - l) between i = 1 and x = 3. 123. Find the arc of the curve 9 !/- = (2 x — 1)' cut off by the line X = 5. 124. Find the perimeter of the loop of the curve 9^ = i2y-l)(y-2r-. 125. Find the length of the curve x=P+t, y = t^— t below the a>axis. 126. Find the length of an arch of the curve z = aV3(2</. — sin2</>), y = ^ (1 — co8 3<^). 127. Find the length of one quadrant of the curve X = a cos* 0, y = b sin* ^. 128. Find the circumference of the circle r = 2 sin » + 3 cos fl. 129. Find the perimeter of one loop of the curve (I) 130. Find the area of the surface generated by revolving the arc of the curve 9 y* = (2 x — 1)* between x = and x = 2 about the y-axis. 131. Find the area of the surface generated by revolving one arch of the cycloid x = a {<f> — sin <f>), y = a (1 — cos (/>) about the tangent at its highest point. 132. Find the area of the surface generated by rotating the curve t" = a^ sin 2 about the x-axis. 133. Find the area generated by revolving the loop of the curve 9 X* = (2 J/ - 1) (i/ - 2)2 about the x-axis. 131. Find the volume generated by revolving the area within the ciu^e j/2 = X* (1 — x^) about the y-axis. 135. The vertical angle of a cone is 90°, its vertex is on a sphere of radius a, and its axis is tangent to the sphere. Find the area of the cone within the sphere. 136. A cylinder with radius b intersects and is tangent to a sphere of radius a, greater than b. Find the area of the surface of the cylinder within the sphere. 137. A plane passes through the center of the base of a right circular cone and Ls parallel to an element of the cone. Find the areas of the two parts into which it cuts the lateral surface. 162 Supplementary Exercises CHAPTER VI 138. Find the pressure on a square of side 4 feet if one diagonal ia vertical and has its upper end in the surface. 139. Find the pressure on a segment of a parabola of base 2 b and altitude h, if the vertex is at the surface and the axis of the parabola is vertical. 140. Find the pressure on the parabolic segment of the preceding problem if the vertex is submerged and the base of the segment is in the surface. 141. Find the pressure on the ends of a cylindrical tank 4 feet in diameter, if the axis is horizontal and the tank is filled with water under a pressure of 10 lbs. per square inch at the top of the tank. 142. A barrel 3 ft. in diameter is filled with equal parts of water and oil. If the axis is horizontal and the weight of oil half that of water,. find the pressure on one end. 143. Find the moment of the pressure in Ex. 138 about the other diagonal of the square. 144. Weights of 1, 2, and 3 pounds are placed at the points (0, 0),. (2, 1), (4, — 3). Find their center of gravity. 145. A trapezoid is formed by connecting one vertex of a rectangle to the middle point of the opposite side. Find its center of gravity. 146. Find the center of gravity of a sector of a circle with radius a and central angle 2 a. 147. Find the center of gravity of the area within a loop of the curve yi = x^ — x^. 148. Find the center of gravity of the area boimded by the curve y^ = ^ and its asymptote x = 2a. 149. Find the center of gravity of the area within one loop of the curve r^ = a^ sin 6. 150. Find the center of gravity of the area of the curve x = a sin' </», y = b sin' 4> above the x-axis. 151. Find the center of gravity of the arc of the curve 9 y^ = (2x — l)3cutoff by thelinex = 5. 152. Find the center of gravity of the arc that forms the loop of the curve 9 2/2 = (2 X - 1) (x - 2)2. 153. Find the center of gravity of the arc of the curve x = P + 1, y = t? — t below the x-axis. 154. Show that the center of gravity of a pyramid of constant density is on the line joining the vertex to the center of gravity of the base, J of the way from the vertex to the base. StrPPLEMENTART EXERCISES 163 155. Find the center of gravity of the surface of a right circular cone. 156. Show that the distance from the base to the center of gravity of the surface of an obhque cone is j of the altitude. Is it on the line joining the vertex to the center of the base? 157. Find the center of gravity of the soUd generated by rotating about the line x = i, the area above the 3>-axis bounded by the parab- ola y* = 4 X and the Une x = 4. 158. The arc of the curve x* + y* = a* above the x-axis is rotated about the j/-axis. Find the center of gravity of the volume and that of the area generated. 159. Assuming that the specific gravity of sea water at depth h in miles is find the center of gravity of a section of the water with vertical sides five miles deep. 160. By using Pappus's theorems, find the center of gravity of the arc of a semicircle. 161. The eUip>se is rotated about a tangent inchned 45° to its axis. Find the volume generated. 162. The volume of the ellipsoid ^+^+- =1 is ^vabc. Use this to find the center of gravity of a quadrant of the elhpse -5 + ^ = 1. 163. Find the volume generated by revolving one loop of the curve r = a sin d about the initial line. 164. A semicircle of radius a rotates about its bounding diameter whUe the diameter sUdes along the line in which it hes. Find the volume generated in one revolution. 165. The plane of a moving square is perpendicular to that of a fixed circle. One comer of the square is kept fixed at a point of the circle while the opposite comer moves around the circle. Find the volume generated. 166. Find the moment of inertia about the x-axis of the area bounded by the x-axis and the curve y = 4 — x*. 167. Show that the moment of inertia of a plane area about an axis perpendicular to its plane at the origin is equal to the siun of its moments of inertia about the coordinate axes. Use this to find the moment of 164 Supplementary Exercises inertia of the ellipse ~5 + rj = 1 about the axis perpendicular to its plane at its center. 168. Find the moment of inertia of the surface of a right circular cone about its axis. 169. The area bounded by the x-axis and the parabola ^^ = 4 ax — x- is revolved about the x-axis. Find the moment of inertia about the X-axis of the volume thus generated. 170. From a right circular cylinder a right cone with the same base and altitude is cut. Find the moment of inertia of the remaining volume about the axis of the cylinder. 171. A torus is generated by rotating a circle of radius a about an axis in its plane at distance h, greater than a, from the center. Find the momeat of inertia of the volume of the torus about its axis. 172. Find the moment of inertia of the area of the tonos about its axis. 173. The kinetic energy of a moving mass is j 5 y- dm, where v is the velocity of the element of mass dm. Show that the kinetic energy of a homogeneous cylinder of mass M and radius a rotating with angular velocity w about its axis is \ Mw'^a'^. 174. Show that the kinetic energy of a uniform sphere of mass M and radius a rotating with angular velocity co about a diameter is ^ Mu^a'. 175. When a gas expands without receiving or giving out heat, its pressure and volume are connected by the equation pv^ = k where y and k are constant. Find the work done in expanding from the volume vi to the volume V2 . 176. The work done by an electric current of i amperes and E volts is iE joules per second. If E — Eo cos (Jit, i = h cos {wt + a), where Eo, h, w are constants, find the work done in one cycle. 177. When water is pumped from one vessel into another at a higher level, show that the work in foot poimds required is equal to the product of the total weight of water in pounds and the distance in feet its center of gravity is raised. CHAPTER VII 178. Find the volume of an ellipsoid by using the prismoidal formula. 179. A wedge is cut from a right circular cylinder by a plane which passes through the center of the base and makes with the base an angle a. Find the volume of the wedge by the prismoidal formula. StrPPLEMENTABY ExERCISES 16& 180. Find approximately the volume of a barrel 30 inches long if its diameter at the ends is 20 inches and at the middle 24 inches. 181. The width of an irregular piece of land was measured at inter- vals of 10 yards, the measurements being 52, 56, 67, 49, 45, 53, and 62 yards. Find its area approximately by using Simpson's rule. Find the values of the following int^rals approximately by Simpson's rule: i2. f Vj-5 + 1 dx. 53. t -ilnxdx. Jl X* + J* 186. Find approximately the length of an arch of the curve j/ = sin x. 187. Find approximately the area bounded by the a!>-axis, the curve y = , and the ordinates x = 0, x = r. X CHAPTER VIII E.xpress the following quantities as double integrals and determine the limits: 188. Area bounded by the parabola y = x- — 2 x + 3 and the line y = 2x. 189. Area bounded by the circle x- + y^ = 2 a- and the curve y- =:r: — :• 190. Moment of inertia about the x-axis of the area within the circles x*-fy* = 5, x' + y' — 2x — 'iy = 0. 191. Moment of inertia of the area within the loop of the curve y* = x^ — x* about the axis perpendicular to its plane at the origin. 192. Volume bounded by the xy-plane the paraboloid z = x- -{- if and the cylinder x* + y^ = 4. 193. Volmne boimded by the xy-plane the paraboloid z = x- + y^ and the plane z — 2 x -\- 2y. 194. Center of gravity of the soUd bounded by the xz-plane, the cylinder x- -\- z- = o*, and the plane x •\- y -{- z = \ a. 195. Volume generated by rotating about the x-axis one of the areas bounded by the circle x^ -\- y- = b a^ and the parabola y^ = A ax. In each of the following cases determine the region over which the integral is taken, interchange dx and dy, determine the new limi ts, and so find the value of the integral: 166 Supplementary Exercises , ix+y)dydx. 199. | I Vx^ + xy dydz. Express the following quantities as double integrals using polar ■coordinates: 200. Area within the cardioid r = a (1 + cos d) and outside the circle r = | a. 201. Center of gravity of the area within the circle r = a and outside the circle r = 2 a sin 9. 202. Moment of inertia of the area cut from the parabola 2a T = 1 — COS 9 by the line y = x, about the x-axis. 203. Volume within the cylinder r = 2 a sin B and the sphere x^ + 2/2 + 2= = 4 a-. 204. Moment of inertia of a sphere about a tangent line. 205. Volume bounded by the paraboloid z = x- -{■ y^ and the plane z = 2x + 2y. 206. Find the area cut from the cone x^ + y^ = z"^ by the plane x = 2z-Z. 207. Find the area cut from the plane by the cone in Ex. 206. 208. Find the area of the surface z^ + (x + yY = a^ in the first octant. 209. Determine the area of the surface z^ = 2 x cut out by the planes 3/ = 0, 2/ = X, X = 1. CHAPTER IX Express the following quantities as triple integrals: 210. Volume of an octant of a sphere of radius o. 211. Moment of inertia of the volume in the first octant bounded by the plane - + ^ + - = 1 about the x-axis. a c 212. Center of gravity of the region in the first octant boimded by the paraboloid z = xy and the cylinder x"^ ■]- y^ = o^. 213. Moment of inertia about the z-axis of the volume boimded by the paraboloid z = x^ + y"^ and the plane z = 2 x + 3. 214. Volume bounded by the cone x"^ = y^ + 2z'^ and the plane 3z + j/ = 6. Express the following quantities as triple integrals in rectangular, cylindrical, and spherical coordinates, and evaluate one of the integrals; SuPPLEBfENTAKY ExERCISES 167 215. Moment of inertia of a right circular cylinder about a line tangent to its base. 216. Moment of inertia of a segment cut from a sphere by a plane, about a diameter parallel to that plane. 217. Center of gravity of a right circular cone whose density varies as the distance from the center of the base. 218. Volume bounded by the xy-plane, the cylinder 3? -\- y- = 2 ax and the cone z^ = x* + j/*. 219. Find the attraction of a uniform wire of length I and mass M on a particle of unit mass at distance c from the wire in the perpen- dicular at one end. 220. Find the attraction of a right circular cylinder on a particle at the middle of its base. 221. Show that the attraction of a homogeneous shell bounded by two concentric spherical surfaces on a particle in the enclosed space is zero. CHAPTER X Solve the following differential equations: 222. ydx + {x-xy)dy =0. 223. sin X sin y dx + cos x cos ydy = 0. 224. (2x2/-y2+6x«)dx + (3y*+x*-2xy)di/ = 0, 225. x^+y = xh/. 226. X :r^ + y = cot X. ax 227. xdy - \y + e') dx = 0. 228. (1 + x*) dy + (xy + x) dx = 0. 229. xdx -\- ydy = x dy — y dx. 230. (sin X -\- y) dy -\- (y cos x — x*) dx = 0. 231. y (e* + 2) dx + (e* + 2x) dy = 0. 232. (xy* - x) dx + (y + xy) dy = 0. 233. (l+x2)^+xy = 2y. 234. xdy — ydx = Vi* + y2 dx. 235. (x - y) dx + X dy = 0. 236. xdy -ydx = x^3^-ry^dx. 237. c*^-^' dy + (1 + e*) dx = 0. 238. (2x + 3y- l)dx + (4x + 6y-5)dy =0. 239. (3y« + 3xy + x»)dx = (x* + 2xy)dy. 240. (1+ x») dj^ + (xy - x») dx = 0. 168 Supplementary Exercises 241. {x'y + t/*) dx - (x3 + 2 xy^) dy = 0. 243. 2^ +y + x2/» = 0. ax 244. ydx = (y^ — x) dy. 245. y -1^ + t/2 cot X = cos x. 246. (x2 - y2) (dx + dy) = (x« + r/^) (dy - dx). 247 x-'^ 4-^ = 1 253. f2-!^.e". dx^ dx 258. j^ + aV = sin ox. 260. ^-^-2y = e-*sin2x. dx'^ dx 261. j^ + 9 y = 2 cos 3 X - 3 cos 2a;. 262. g + 6^+5y = (e- + l)^ Supplementary EIxercises 169 263. ^ - y = xe''. dx- " 265. ^+2x = sm/, ^-2y = co6<. at at 267. According to Newton's Law, the rate at which a substance cools in air is proportional to the difference of the temperature of the sub- stance and the temperature of air. If the temperature of air is 20° C. and the substance cook from 100° to 60° in 20 minutes, when will its temperature become 30°? 268. A particle moves in a straight line from a distance a towards a point with an acceleration which at distance r from the point is k f^i. If the particle starts from rest, how long will be the time before it reaches the point? 269. A substance is undergoing transformation into another at a rate proportional to the amount of the substance remaining untrans- formed. If that amount is 34.2 when t = 1 hour and 11.6 when t = 3 hours, determine the amount at the start, t = 0, and find how many hours will elapse before only one per cent will remain. 270. Determine the shape of a reflector so that all the rays of light coming from a fixed point will be reflected in the same direction. 271. Find the curve in which a chain hangs when its ends are sup- ported at two points and it is allowed to hang under its own weight. (See the example solved in Art. 57.) 272. By Hooke's Law the amount an elastic string of natural length I stretches under a force F is JdF, k being constant. If the string is held vertical and allowed to elongate under its own weight w, show that the elongation is 5 kwl. 273. Assuming that the resistance of the air produces a negative acceleration equal to k times the square of the velocity, show that a projectile fired upward with a vdocity vi will return to its starting point with the velocity ^-V' g + kvr' g being the acceleration of gravity. 274. Assuming that the density of sea water imder a pressure of p pounds per square inch is p = 1 + 0.000003 p. 170 SUPPLEMKNTART EXEKCISES show that the surface of an ocean 5 miles deep is about 465 feet lower than it would be if water were incompressible. (A cubic foot of sea water weighs about 64 pounds.) 275. Show that when a liquid rotating with constant velocity is in equilibrium, its surface is a paraboloid of revolution. 276. Find the path described by a particle moving in a plane, if its acceleration is directed toward a fixed point and is proportional to the distance from the point. ANSWERS TO EXERCISES Page 6 2. |x3 + l+C. 3. |x' + 2x* + C. 4. i v^(2I-3)+C.- ~6. a' X - 2 ox' + f a* x^ - I x' + C. 8. 2x + 31nx + C. 9. ^^ + 4y + 41nj/ + C. 10. xiCAx^-^x^-e) +C. 11. In (x + 1) + C. 23. - i (a* - P)» + C. 12. - ■:r^ + C. 24. X + 3 In (x - 2) + C. 13. V2X + 1+C. 25. 5 In (2x^ + 1) + C. x + 1 27+1 14. i In (x^ + 2) + C. . " 1 15. Vx^ - 1 + C. ~ 4 (2 X* + 1) ^^- - 46(a+6xy +^- 26. ^ (l - 1]' + C. 17. - 1 (o^ - x*)' + C. 1 18. iln(a3 + x3)+C. 27. "„ („_i) (^n_^o)»-i + C 20 ln(x2 + ax + 6)+C. 28. ^'^ -^^ ^"^ +(7. 21. 2 Vx« + ax + 6 4- C. 1 29. ix3-|ln(x3 + 2)+C. ^ in (1 - a^) + C. 30. i.i:8_|a;5^ia^^C'^ Pages 12, 13 1. 138. 2. |^ + 30<. 3. A = - ^ ff^ + 100 < + 60. It reaches the highest point when t = 3.1 sec, h = 215.3 ft. 4- Tis^ffT sec. 171 172 Answers to Exercises 5. X = (^ — t + I, y = t — ^ t^ + 2. These are parametric equa- tions of the path. The rectangular equation is x^ + 4 xy + 4 y^ _ 12x-22t/ + 31 =0. 6. About 53 mUes. 10. (-i -^) 11. 6y = 12. -12 7. x = jV3fi,y = ^fi + Vot. 11. 6i/ = x3-3x2 + 3x + 13. 4 ■ "-.^ 4' 8. J/ = 2 X - -^ x2 - f . 9. y = e*. 14. About 4 per cent. 15. X = xoe**, where x is the number at time t, xo the number at time t = 0, and k is constant. 17. 17 minutes. 19. 11.6 years. 18. 11.4 minutes. Pages 18, 19 1. 2. 3. 4. - (1 cos 2 x+l sin 2 x) + C. 5 • f2x-3\ 6. 7. 8. 9. \ sin2 e + C. tan X + C. - ^ cot 2 X + C, — CSC X + C - i cos (nt + a) +C. n 5. -4csc^ + C. 10. J sec* X + C. 11. 2 (esc I - cot 1^ +C. 12. i sin (x« - 1) + C. 13. I (tan 3 X + sec 3 x) + C. 14. tan X + X — 2 In (sec x + tan x) + C. 15. hi (1 + sin x) + C. 16. e + cos2 + C. 17. sin X + hi (esc x — cot x) + C. 18. ism'x + C. 19. ltan*x + C. 20. Itan^x + C. 21. -icos«x + C. 22. Hn(l +2tanx) + C. 23. -Jhi(l -sin2x) + C. 24. i In (1 + tan ax) + C a _. 1 . . x V2 . ^ 25. — = sin-' — -=7- + C. V2 V3 Answers to Exebcisbs 173 27. isec-i^ + C! 28. itan-i2y + C. 29. -^ In (x V7 + V7 1* + l) +C. v7 30. |sec-'Y+C. ^^ 1 , 2J + V3 , ^ 31. 7= m 7= + ^- 4 V3 2 X - V 3 32. In (2z + V4x^-3) + C. 33. -3 VT^T^ - 2sm-i | + C. 34. 2 V:?T4 + 3ln (x + Vi« + 4) + C. 35. 5ln(4x2-5)+-^ln|^-=-^ + C. 8 ' V5 2 X + V 5 36. J V3x»-9 - -|= In (x + V?"=^) + C. V3 37. ein-.(^-^)+C. 46. -i.->^ + e. 38. -V2-sin«x + C'- 47. ^ (e*« - e"*-*) + 2 x + C 39. tan-i (sin x)+C. 48. i In (1 + e^*) + C. 40. sec-i (tan x) + C. ^^ 49. In (e* + e"') + C. 41. In (sec x + v'sec' x + 1). _1 Art n / _ "0. e "T C. 42. 2 VT^-^e + C. 51^ t^-, (^) ^ c. ._ 1, 2 + lnx , ^ 1 — e* 43- 4^2 ^n^ + ^- 52. In f:^ + C. 44. - ^ Vcos^x-8in«x + C. 53 1 ^^_, ^^^ ^ ^ 1 _ X* a 45. 2 sin ' ^ + C. 54 tan-i (e*) + C. Page 20 ^l*-i^ + 3.„ -l._, 2x-l,- 3. -^ In (3 X + 2 + V9x* + 12x + 6) + C. V3 . 1 . _, (2x- 1) V5 . ^ . 1 _, (x-3)V6,_ ^- VB'^" 3 +^- 5- Vl""" 3 + ^- 174 Answers to Exercises 6. 1 in^ + C. — a X + 7. -ln(4x^-4a:-2)+ — In^^-^-p-^ + C. 8. I V3x2 -6x + 1 + 4= In [3 (a; - 1) + VQx^ - 18x+3] + C. 3 Va 9. 1 In (3 x2 4- 2 X + 2) + -^ tan-i ^^^ + C. 6 ^ 3 V 5 V 5 10. 4= sec-i ^^ .t^ + ^ln (2x + 1 + V4x'^ + 4x-l) + C. V2 V2 2 11. -— =J=+C. Vx2 - 2 X + 3 12. Vx2 - X - 2 + I In (2 X - 1 + V4x2 -4x-8) + C. 13. JL,„(iil±lii^Uc. Vl7 V4e* + 3 + Vl7/ Page 26 1. — cos X + 3 cos' X + C. 2. sin X — I sin^ x + ^ sin^ x + C. 3. sin X — f cos' x + f sin' x — cos x ■{• C. 4.-3 cos' X + 5 cos^ X + C. 5. I sin^ 1 X - f sin'' ^ X + I sin' I X + C. 6. -Jy sin« 3 (9 - ^ sinS 3 + C. 7. - f cos'0 + cos0 + C. 8. sin X + 5 sin^ x + C. 9. cos X + In (esc x — cot x) + C. 10. cos^ d — I cos* - In cos + C 11. tan X + i tan' x + C. 12. - (cot y + i cot' 2/ + I cots y _|. ^ cot^ y + i cot» y) + C. 13. tan X — X + C'. 14. 2 tan - sec - tf + C. 15. |sec'|x + C. 16. t'j sec^ 2 X - i sec^ 2 x + i sec' 2x + C. 17. — I csc^ X — In sin X + C. 18. I sec* X — f sec* x + | sec'' x + In cos x + C. 19. - i cot« X - i cot' x + C. 20. ^tan^x + lntanx + C. 21. |-^ sin (2 ax) + C. 22. 1 + j^sin(2ox) + C. Answers to Exercises 17a 23. t^ X - sV sin 4 a; - tV sin' 2 X + C. 24. xV^- V2sin2x + ism»x + C. 25. ^ X — I sin 2 X + ^\ sin 4 z + -^^ sin' 2 x + C. 26. tan x + sec x + C*. 27. tan ^ X + C. 28. 2 f sin ^ - cos I j + C. 29. I Vx2 -oT- - |- In (x + ^3? - a") + C. 30. I Vx2 + d^ + |. In (x + Vx* + a?) + C. 31. I Vx2 + a^ - ^' In (x + Vx* + a?) + C. 32. -^ +C. a^ V x^ — a- 33. -In - ^ +C. « O + V o2 — X* 34. -:^I^Zi + C. . ax 35. . ^ +C. Vo* -x* 36. i (x2 + a2)« - I (x^! + a^)* + C. 37. -^^7^ + C. a-x 38. X - 2 \/x2-4x + 5 + iln(x-2 + V'x*-4x+5) + C. 2 -^ ^-^ ' " ' 2 39. ^^ - ^^ V 2 - -.x - 4x^ + ^ sin-^ ^^ + ^ + C 32 ^•^^64 3 ^^' Page 30 1. ^ + 4x-2In(x- 1) + 121n(x-2) +C. 2. 31nx -ln(x + 1) 4-C. 3 j^(x_-i^(x+2)+c. 4. l + lnx- jgln(2x-l)-^ln(2x + l)+C. 5. f In (X + 3) - i In (X + 1) - f In (X + 5) + C. 6. ^ln(2x- 1) -31n(2x-3) +|ln(2x-5) + C. I . X -\ h In h C. J> X 376 Answers to Exercises 8. lln(x + l)+|ln(x-l)-2^^j+C. »--Ur4^+'"l^)+- 10. x-81n(x + l) 3(^ + 1)3 +g- 13- 2(4^ + ^- 14. x + ^ln^^-Kan-iz + C. 4 X + 1 2 „ 1, J + 1 1 , ,2x-l . - 15. 5 In , ^: H — tan-i 7= h C, 3 Vx2 - X + 1 Vs Va 16. i In (x3 + 1) + C. 17 1 . 1, X- 1 2V3^^ , 2x-l . ^ 6(x + l) 4 x + l 9 \/3 19. --^+lln— "-=^ + 1 tan-x^ + C. a^ - 8 6 Vx2 + 2 X + 4 2 V3 V3 20. 3 (x + D* + In [(x 4- D* - 1] - ^^ tan-i ^ (x + 1 ) +1 ^ ^^^ v3 21. -1 A+1 i_l i_:,A+llnl±£5+c. 1-x^ 5 22. 3|,(ax + 6)i»-|A(a, + ,)i + C. 23. 2 Vx + 2 - In (x + 3) - 2 tan-i V^T2 + C. 24. 4x* + 21n (x* - l) + In (x* + l) - 2tan-» x* + C. 25. Hx + l)* + i(x-l)« + C. Page 34 1 X 1. 2[ cos 2 X + 2 sin 2 X + C 3, x sin"! x + Vl - r* + C. 2. |lnx-| + C. 4. ^-|^tan-ix-|x + C. 6. X In (x + Va» + 3?) - Vo« + x« + C. Answebs to Exercises 177 6. 2 VF^^ In X - 4 Vj - 1 + 4 taii-i Vx - 1 + C. 7. Inxln(lnx) -Inx + C. 8. ^ x»8ec-ix - I V^^n _ 1 In (x 4. V^^TT) + C. 600 9. X - (1 + e-') In (1 +6*) + C. 10. (x» - 2 X + 2) e* + C. 11. - (x» + 3x» + 6x + 6)e^ + C. 12. ^^sm2x-^^^^^^^^^cos2x + C. 13. ^ Vx'-a^ - ^ In (x + V^i:^») + C. 14. I Va* + x* +|ln (x + V^*T^) + C. 15. ^(28in3x-3coe3x) +C. lo 16. -^ (cob X + sin x) + C'. 17. -?(sin2x + 2cos2x) +C. o 18. H9ec0tan0 + ln(sec» + tan0)] + C. 19. 5 cos X — T^jj cos 5 X + C. 1- f 2. 2.829. 1. 1 - i V3. 2. 3' 3. -20. 4. 2. 5. 2 3- 6. 1.807. 7. 0.2877 8. 0. 9. \a. 10. 2. 11. 00. Page 38 3. -0.630. Pages 45, 46 12. 2* 13. 1 2fc» 14. 0.5493. 15. r + 1. 4^2 16. 1.786. 17. 0.4055. 18. 0.2877. 19. f (l-ln2) 178 Answers to Exercises Pages 49, 50 1. 11. 13. 27r + |, 67r-f 2. |V3(4-V2). 14. 4V3 4-Y'r. 4. i 5. 9.248 6. irab. 7. ia?. 8. 51. 9. A. 11. ^aK 12. f. 2. >2. 4 3. fa^Vs. 4. ^ (e"- - 1). 5 ^■ ^- 2 17. 18. 19. 20. 21. 22. 37ra2. lira". Tr{¥ + 2ab). X V3. 3ira\ f 7ro6. Page 52 9. 2a2(l+|V2). 10. 167r3a2. 11. ^'{i+H- 13. f(.-i). 14. (lO» + 9V3)g. 15. m'. Pages 56, 66 5. ^^'+4-4V 6. Itt. 7. iira^. 8. |a2. 3. iU- . irO' 4- -6"' 6. 1x2. 7. f xa'(l — cos* a), where a is the radius of the sphere and 2 a the vertical angle of the cone 8. \^xa». 13. fxa'. 9. Mxa». 14. Jji^x^a^. 10- Vto^. 15^ 8xV3. 11. 5ir^aK ^gz 12. ^ira\ 16. "16 V2. Page 69 2. f a' tan a. 3. la". ^• 4. ixo62. 9. ta^A. 5. ^ira%. j^ 4 ^ 3 6. .4A. • 3 v'2 '^^ ' l-'O+i) Answers to Exebcises 179 Page 63 2. ^(lOVlO-l). 6. 6a S. ln(2 + V3). ^- "(^-ij* 4. f + ^ln2. 8. 8a. 5. 2.003. 9. 2x2a. Page 64 3. ^^^^ (e - 1). 5. 2a [V2 + In (l + ^l . ±a 6. i^a. • Vs' 7. 8a.~ 8. ixa. Pages 66, 67 \ V a^ - 6* a / 4. V^a^. 7. V'o*- ''■2^ 6== + ^; 9. 8x[V2+ln(l + v^)]. 6. V»-a*. 10. 4xa». Page 69 1. fT»o». 6. fxa'a -cos a). 2- 2a«. 7. 2a/i(x-2). ^- 1^°*- _ 8. §xaVA» + 4a^ 4. ixaKa + 26V3). g ia«A(9x-16). 5. |xoK2 V2 - l). Pages 71, 72 1. 45,000 lbs. 3. i«*A». 2. 33,750 lbs. 4. |u'6/j«. 5. § ipofc*, where a is the semi-axis in the surface and 6 the vertical semi-axLs. 6. 300,000 IP. 7. 40 xw. Pages 78-80 1. i pd'b, where p is the pressure per unit area, a the width, and 6 the height of the door. 2. ^wa^b. 4. The intersection of the 3. ^wbh* {4:C + 3h). medians. 180 Answers to Exercises 5. (!a,0). _ 4a _ 6. x= 77—, y Stt' " 7. (a a\ 8. / 256a\ P 3157rj _ IT 9. ^=8' 10. (^ a, ^ a) 11. (5 16a' 46 12. y = ia. 13. X = Itt V2a. 14. At distance — from the TT bounding diameter. 1^- y- 4e(e^-l) 16. (^a, fa). 17. (0.399, 1.520). 18. y = ia. 19. On the axis i of the distance from the base to the vertex. 20. At distance f a from the plane face of the hemisphere, where a is the radius, 21. (iO). 22. Its distance from the plane face is ^ of the radius. 23. On the axis at distance f a (1 + cos a) from he vertex, a being; the radius and a the angle of the sector. 24. (|a,0). 25. The dista'nce of the center of gravity from the base of the cylinder is ^j ira tan a. 26. At the middle of the radius perpendicular to the plane face. 28. x.:«^^. 15 v^ - 5 Pages 82, 83 2. 2 x^a^b. 6. J xa» [3 hi (1 + V2) - V2). 3. ^(12V3-1). 7. §7ra3(3a + 2sina). 4. 7r(36x + V6)V6. 8- '^'«'- 5. "/Tra^. 10. § TTo' (4 sin a — sin' a) tan a, where a is the radius of the cylinder and 2 a the vertical angle of the cone. Pages 84, 85 1. ^ a%, where b is the edge about which the rectangle is revolved- 2. T^ bh^, where b is the base and h the altitude. 3. i b¥, where b is the base and h the altitude. 4. ^a*. 6. ^^a*. 6. ^ ira*. Answers to Exercises 181 7. i Ma^h, where h is the altitude. 8. iMa\ 9. — — : — , where p is the density. LO 10. I 3/a*, where M is the mass and a the radius. 12. I TO* Pages 83, 83 k (b - a)» 2a 2. au; ft. lbs., where a is the radius of the earth in feet. „ , Vt — h , a a 3. c In J H • Vi — Vt Vi 4. 25,133 ft. lbs. 5. ^ TTfiPa, where a is the radius of the shaft. 6. 2~T ^ - > where A is the altitude of the cylinder. 7. -T— ( ^ ~ T ) > where a and b are the inner and outer radii. kh 8. —7 , where a and b are the radii of the ends and h the altitude t>. irab' 9. 2xi r 2. 8.5. 7. 0.785392. 8. 1.26. 9. 4.38. 10. 21.48. 1. In If. 2. h^a\ 3. ^. 5. -1. „ TO* 6- -^- 10. 2i. c Pages 95, 96 11. 4.27. 12. 0.9045. 13. aX - 3L3 5L5"*" * * * 14. 1.91. Pages 102, 103 7. f. 15. 4. 8. ^a\ 16. |a*. 9. X. 17. 16 hi 2 --V 10. 13§. 18. fa*. 11. 3x. 19. |o*. 12. X. 20. (i -f). 13. h 14. ia*. 21. (Va, -2o) 182 Answers to Exercises Pages 107, 108 , va* . ica^ 2^ I^. 6. J aM2 a - sin 2 a). ^ 7. i^aaMeTr-S). 8. On the bisector at the distance — x from the center. o a 9. ^xo*. 11. aU4ir -If). 10. \T,a\ 12. 3xa^ 15. X5 ilfa^, M being the mass and a the radius of the base. 16. ia3(3^_4). 18, ^^Trpa\ i7. l-a3. jg (8V2-9)4^. lUo Page 111 1. 3 Vii. 2. There are two areas between the planes each equal to 2 mofi. 3. Two areas are determined each equal to ira^ V2. 4. ixa2 V3. 7. ^iva' (3 Vs - l). 5. 4. ' 8. a? (tt - 2). €• 8 a'- 9. 8anan-iiV2. Page lie 1. \. 4. wahc. 2. A. 6. ^^a^A. 3. Its distance from the base 7. ^. is -/j TTO. Page 121 1- ^s^"-- 4. I^ (2/i2 + 3a«>. 2. f A, where A is the altitude. * 60 3. IT. 5. fTra'. 6. On the axis of the cone at the distance f o (I + cos a) from the vertex. 7. If the two planes are = ± ?, the spherical coordinates of the 9 T ■center of gravity are r = T^a, 0=0, <^=k' 10 ^ 8. Httos. Answers to Exercises 183 Page 125 kM 2 kMc ri 1 ~ | 2. 2M. xa- 4. 2 xArpA (1 — cos a), where p is the density, A the altitude, and 2 a the vertical angle of the cone. 6. The components along the edge through the comer are each equal to 2Mk fx , , 2 + >/2l Pages 130, 131 dx^ dx " d^ d3? dx Q. X dy — y (x + 1) dx = 0. — 4 y = 0. 7. ^ + j/ = 0. 9. ydx =xdy. Pages 141-143 10. u=cx* • 2. tan' X — cot* y = c. x 3. 1^ + 1 = c (x2 - 1). -? .,,,,, 11. u = cx*e *. 4. x-(^ + i» - ys = c. 19 L, _ ^ . „, 5. x^ + x-^/-x^/*-y3=c. jg ^ = (i _ ^) (;, + c). 6. y^ =cx^{y^ + 1). 14 ^ ^ csinx - a. 7. x' + ^ = ce*"*. 15. 7 1» = y (x' + c). 8. xy = c (y - 1). 1 , 1 , .^ 16. — = X + - + «?»* ■^ ^h-a^' 17. x< + 4y(x»-l)*=c 18. hi (x» + xy + y*) + A tan'i ^-^^ = c. V3 xVs 19. X* - y* = ex. 25. ys = ce* - X - 1. 20. y* + 2xy=c. 26. e» = |e* + ce"*. 21. x*-4x»y + y* = c. „ c x» 27. y =--— . 22. 4 = c - e-". "^ ^'^ y 28. y = ^ X* + c, or y = og». f 29. y* = 2 ex + c*. 23. e" + In X = c. / , ^ 24. x + 2y + ln(x + y-2) =c. 30. q = Ec\\ - f Re] , 31. i =/e~l' + ^i-q^^^ /i!sina/-Lafcosa/-e"iM . 184 Answers to Exercises 32. 2/3 = 8 6=^-2. 34. y = cifi. 33. 2/2 = 2 ax. 35. X = o In ^ + Vo" - 1/2 + c. a + Va2 - j/2 36. y2 + (x - c)2 = a2. 37. 2/ = |e^ + |3e--. 38. r = e". 39. r = c sin 0. 40. r = a sec (0 + c). 41. o0 = Vr2 - a2 - o sec"! - + c. a 42. y = e*. 43. A circle. 44. A straight line. 45. A circle with the fixed point on its circumference or at its center. 46. The logarithmic spirals r = ce^o. 47. 0.999964. Pages 164^156 1. y = Ci In X — I a:2 + C2. 2. y — X + Cixe^ + C2. 3. y = cie'^ + C2e-°*. 4. y = Ci sin ax -\- d cos ax. 6. s = ^ In (cie°*< - g-""") + cj. a^ ^- J/ = l^'-2^1nx + c. 8. y = 5^ [e<=i*+<'2 + e-i'^'^t)]. 9. y = ci + c-ifi^'. 10. 2/ = Cie*^ + Cae"*. 11. y = (ci + cix) e'^. 12. J/ = Ci cos X + C2 sin x. 13. y = Ci + Cje"* + cse'*. 14. y = cic^ + C2e~* + Cs cos x + C4 sin x. 15. y = e* [ci cos (x V2) 4- Ci sin (x V2)]. 1ft -i«r xVs , . xVsl 16. 2/ = e »* ci cos — 2 f- Ci sm — g— • Answers to EIxercises 186 17. y = ae' + cie-' + Cje"^ -{- Cxer' ^. 18. y = (ci + cjx + Car*) e*. 19. !/ = X + 3 + Ci cos X + cj sin X. 20. y = cic^i + C2€-^* -he'. 21. y = cie^^ + c^e-^"^ - 5 -c' - tt^ - xk- • ^ 22. J/ = ce* — J (sin x + cos x). .^ ' ^ 23. 2/ = Ci + ce*' - i X* + X. y /K/^ \ 24. y = Cxe-' + (^-^* + ix-A + ^e''. ^ J (^Z ^ 25. y = cie" + cje"" + #- e". */ '^ r a;V3 xVsl 1 *^^ 26. y = e»* Ci cos — ^ h cj sin , I — y^ (2 sin 2 x + 3 cos 2 x). [X v's . X V^~| Cx cos — X H cj sin — ^ I — x* + x* — 6. 28. y = cie* + ce»'- ^e»*sinx. 29. y = Ci(?' + Cj€~" + jV ^' (6 sin X — cos x). 30. y = Ci + Cjx + cjx^ + 046"^ + x^y (4 cos 4 X — sin 4 x). 31. y = Ci cos 2 X + (ci + J x) sin 2 X. 32. y = e-' (ci + cx + ^x*) + ie*. 33. X = Ci cos <H-Cj8inf + |(e* — e~*)» y = Ci sin < — cj cos < + 3 (e* — e~'). 34. y — Cx cos < + Cj sin < — 1, X = (ci + C2) cos ^ + (cj — Ci) sin < — 3. 35. X = CiC* + C3€~", y = Cie~' + 3 Cje~" + cos I. 36. X = Cic' + C26~* + cj cos < 4- C4 sin f, y = Cie' + CjC"' — C3 cos < — C4 sin t. 37. y = X. 38. 2 y* = X + 2. 39. « = f<+|(e-*'-l). ^•«=jk^l 2 j- 41. s = 6 cos {kt). 42. About 7 miles per second. 43. About 42? minutes. 44. < = V-ln (5 + ^^). 45. < = -^ In (9 + 4 V5). TABLE OF INTEGRALS u" du = — -— , if n is not — 1. n + 1 2. r^ = lnu. J u r du \ u 3. I -;;— — = - tan 1 - • J V? -\- a^ a a . r du \ , u — a 4- I —^ -o = ^7-1^^ — \ J w — c^ 2 a u -\r a 5. { e^ du = c". 6. Ca'^du In a Integrals of Trigonometric Functions 7. I sin udu = — costt. 8. I sin^ udu = lu — lsm2u = I (u — sinu cos m). 9. 1 sin* w dw = I If — J sin 2 M + ^^j sin 4 u. 10. j sin^ udu = f'^u — ls\n2u + :^g sin' 2 u + ?? sin 4 u, 11. J costtdw = sin u. 12. J cos^ « du = § M + -J- sin 2 M = 5 (w + sin u cos u). 13. j cos* M (iu = I u + i sin 2 M + -g^j sin 4 u. 14. j cos^ udu — ^^u + \sm2u — ^g sin' 2 u + ^'j sin 4 u. 15. I tan udu = —In costt. 16. j cot udu = In sin u. 186 17. 18. 19. 20 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. Table of Integrals Jsec udu = In (sec u + tan u) = In tan ( o "I" I )* I sec* u du = tan u. j.'sec' udu = ^ sec M tan t* + i In (sec t* + tan u). esc u du = In (esc u — cot «) = In tan -x* I CSC* udu = — cotu. jcsc^ udu = — ^cscucotu + |ln (esc u — cot u). 187 Integrals containing Va* — u* JV^^^rV* du = I Va«-tt» + ^ sin-i ^• fu^ V^i^r;r« du = 3 (2 u* - a») Vo* - u* + ^* sin"! -• /Va- - u- J ^/-z ;. , I o - "^q ' - «* du = Va* — u* + a In u u /u' du u /— , a* . , u .- -, = - :5 ^«* - "' + 2 «^"" n ' va- — u- -^ z a r du _ 1 , a - V^s^T^ J u v'a* - u- " In u / du _ V g- — fid' - u*)' du = '^ (5 a* - 2 u*) V^^^TII^ + ^ sin-» H «/ o 5 a du dh (n- — (.0- — u*)* a* Va* — u* Integrals containing Vu* — a* J* ViiT^^i du = I Vu« - o* - I" In (u + Vu^-a*). J*u* Vt^^T^ du = I (2 u»-a*) Vi^TT^ ~ ^ ^ ^^+ Vu»-rf). 188 Table of Integrals 34 r Vu^ - g^ ^„ ^ ViF^r^^ - a sec"! -• J u a 35. J Vl/2 — rt2 Vyi - o2 36. ^-^l^^ = HV^73T2 + ^^in(t* + V^ir^^. -' V M* — a- '^ ^ 37. r_^^_. = 1 sec- H. •^ w V «« _ a2 a a dw V ^2 _ q2 S-, 39. J(m« - a2)*du =|(2u2-5a2) V«2 _ ^2 4.§|*i„ (u+ Vu^-a^). 40 • '^^ Integrals containing V^* + a* 41. J V^^Srp^ '^^ = I Viii"4r^2 + ^ In (u + Vu^+a^). 42. Ji/2 Vm2 + a^ dtt = I (2 m2 4- a2) Vm2 + «« _ |'ln (u + V^iH^). 43. r^^:Z±^d. = V^I-F^ + aln^^^^+^^«. J u u 44. f— JL= = ln(«+V^lHr^2). *' V U^ -|- ^2 ^ ^ 46. f /^ ^l,^V^^ + a^-a •^ w Vm2 -[- o2 a du Vm2 -[- o2 a w 48. J(m* + 0^)3 du = |(2 u2+5 a^) v^To^ +^'ln U+ ^vH^^)' 49. f ^^ = ^ . Table of Integrals 189 Other Integrals 60. l\/?^±|dx ^ J \ ax + b = - V{ax + b)(j)x + q) - ^7^ 1n(Vp(qx + 6) + Va(px + g)) a Vap = ^ V (ax + b) (px + g) - ^-^^ tan-i :^ -«Pjg^ ^\ ° aV—ap oVpx + 5 . et r n* • 1. J e" (a sin 6x — b cos bx) ^51. Je"sm6xdx^ ^-^ 1. -62. fe-cosbxdx = ^"^ (^ ^in bx + a cos 6x) . 190 Natural Logarithms 0-609 N 1 2 3 4 6 6 7 8 9 0.0000 0.6931 1.0986 1.3863 1.6094 1.7918 1.9459 2.0794 2.1972 1 2.3026 2.3979 2.4849 2.5649 2.6391 2.7081 2.7726 2.8332 2.8904 2.9444 2 9957 3.0445 3.0910 3.1355 3.1781 3.2189 3.2581 3.2958 3.3322 3.3673 3 3.4012 4340 4657 4965 5264 5553 5835 6109 6376 6636 4 6889 7136 7377 7612 7842 8067 8286 8501 8712 8918 5 9120 9318 9512 9703 9890 4.0073 4.0254 4.0431 4.0604 4.0775 6 4.0943 4.1109 4.1271 4.1431 4.1589 1744 1897 2047 2195 2341 7 2485 2627 2767 2905 3041 3175 3307 3438 3567 3694 8 3820 3944 4067 4188 4308 4427 4543 4659 4773 4886 9 4998 5109 5218 5326 5433 5539 5643 5747 5850 5951 10 6052 6151 6250 6347 6444 6540 6634 6728 6821 6913 11 7005 7095 7185 7274 7362 7449 7536 7622 7707 7791 12 7875 7958 8040 8122 8203 8283 8363 8442 8520 8598 13 8675 8752 8828 8903 8978 9053 9127 9200 9273 9345 14 9416 9488 9558 9628 9698 9767 9836 9904 9972 5 0039 15 5.0106 5.0173 5.0239 5.0304 5.0370 5.0434 5.0499 5.0562 5.0626 0689 16 0752 0814 0876 0938 0999 1059 1120 1180 1240 1299 17 1358 1417 1475 1533 1.591 1648 1705 1761 1818 1874 18 1930 1985 2040 2095 2149 2204 2257 2311 2364 2417 19 2470 2523 2575 2627 2679 2730 2781 2832 2883 2933 20 2983 3033 3083 3132 3181 3230 3279 3327 3375 3423 21 3171 3519 3566 3613 3660 3706 3753 3799 3845 3891 22 3936 3982 4027 4072 4116 4101 4205 4250 4293 4337 23 4381 4424 4467 4510 4553 4596 4638 4681 4723 4765 24 4806 4848 4889 4931 4972 5013 5053 5094 5134 5175 25 5215 5255 5294 5334 5373 5413 5452 5491 5530 5568 26 5607 5645 5683 5722 5759 5797 5835 5872 5910 5947 1' 27 5984 6021 6058 6095 6131 6168 6204 6240 6276 6312 28 6348 6384 6419 6454 6490 6525 6560 6595 6630 6664 29 6699 6733 6708 6802 6836 6870 6904 6937 6971 7004 30 7038 7071 7104 7137 7170 7203 7236 7268 7301 7333 31 7366 7398 7430 7462 7494 7526 7557 7589 7621 7652 32 7683 7714 7746 7777 7807 7838 7869 7900 7930 7961 33 7991 8021 8051 8081 8111 8141 8171 8201 8230 8260 34 8289 8319 8348 8377 8406 8435 8464 8493 8522 8551 35 8579 8608 8636 8665 8693 8721 8749 8777 8805 8833 36 8861 8889 8916 8944 8972 8999 9026 9054 9081 9108 37 9135 9162 9189 9216 9243 9269 9296 9322 9349 9375 38 9402 9428 9454 9480 9506 9532 9558 9584 9610 9636 39 9661 9687 9713 9738 9764 9789 9814 9839 9865 9890 40 9915 9940 9965 9989 6.0014 6.0039 6.0064 6.0088 6.0113 6.0137 41 6,0162 6.0186 6.0210 6.0234 0259 0283 0307 0331 0355 0379 42 0403 0426 0450 0474 0497 0521 0544 0568 0591 0615 43 0638 0661 0684 0707 0730 0753 0776 0799 0822 0845 44 0868 0890 0913 0936 0958 0981 1003 1026 1048 1070 45 1092 1115 1137 1159 1181 1203 1225 1247 1269 1291 46 1312 1334 1366 1377 1399 1420 1442 1463 1485 1506 47 1527 1549 1570 1591 1612 1633 1654 1675 1696 1717 48 1738 1759 1779 1800 1821 1841 1862 1883 1903 1924 49 1944 1964 1985 2005 2025 2046 2066 2086 2106 2126 60 2146 2166 2186 2206 2226 2246 2265 2285 2305 2324 N 1 2 3 4 6 6 7 8 9 Natural Logarithms 191 60O-10O9 N 1 2 3 4 5 6 7 8 9 50 6.2146 6.2166 6.2186 6.2206 6.2226 6 2246 6.2265 6 2285 6.23a5 6 2324 51 2344 2364 2383 2403 2422 2442 2461 2480 2500 2519 52 2538 2558 2577 2596 2615 2634 2653 2672 2691 2710 53 2729 2748 2766 2785 2804 2823 2841 2860 2879 2897 54 2916 2934 2953 2971 2989 3008 3026 3044 3063 3081 55 3099 3117 3135 3154 3172 3190 3208 3226 3244 3261 56 3279 3297 3315 3333 3351 3368 3386 3404 3421 3439 57 3456 3474 3491 3509 3526 3544 3561 3578 3596 3613 58 3630 3648 3665 3682 3699 3716 3733 3750 3767 3784 59 3801 3818 3835 3852 3869 3886 3902 3919 3936 3953. 60 3969 3986 4003 4019 4036 4052 4069 4085 4102 4118 61 4135 4151 4167 4184 4200 4216 4232 4249 4265 4281 62 4297 4313 4329 4345 4362 4378 4394 4409 4425 4441 63 4457 4473 4489 4505 4520 4536 4552 4568 4583 4599 64 4615 4630 4616 4661 4677 4693 4708 4723 4739 4754 65 4770 4785 4800 4816 4831 4»46 4862 4877 4892 4907 66 4922 4938 4953 4968 4983 4998 5013 5028 5043 5053 67 5073 5088 5103 5117 5132 5147 5162 5177 5191 5206 68 5221 5236 52.50 5265 5280 5294 5309 5323 5338 5352 69 5367 5381 5396 5410 5425 5439 M53 5468 5482 5497 70 5511 5525 5539 5554 5568 5582 5596 5610 5624 5639 71 5653 5667 5681 5695 5709 5723 5737 5751 5765 5779 72 5793 5806 5820 5834 5*48 5862 5876 5889 5903 5917 73 5930 5944 5958 5971 5985 5999 6012 6026 6039 6053 74 6067 6080 6093 6107 6120 6134 6147 6161 6174 6187 75 6201 6214 6227 6241 6254 6267 6280 6294 6307 6320 76 6333 6346 6359 6373 6386 6399 6412 6425 fr438 6451 77 6464 6477 6490 6503 6516 6529 6542 6554 6567 6580 78 6503 6006 6619 6631 6644 66.57 6670 6682 6695 6708 79 6720 6733 6746 6758 6771 6783 6796 6809 6821 6834 80 6846 6859 6871 6884 6896 6908 6921 6933 6946 6958 81 6970 6983 6995 7007 7020 7a32 7044 7056 7069 7081 82 7093 7105 7117 7130 7142 7154 7166 7178 7190 7202 83 7214 7226 7238 7250 7262 7274 72S6 7298 7310 7322 84 7334 7346 7358 7370 7382 7393 7405 7417 7429 7441 85 7452 7464 7476 7488 7499 7511 7523 7534 7546 7558 86 7569 75S1 7593 7604 7616 7627 7639 7650 7662 7673 87 7685 7696 7708 7719 7731 7742 7754 7765 7776 7788 88 7799 7811 7822 7833 7845 7856 7867 7878 7890 7901 89 7912 7923 7935 7946 7957 7968 7979 7991 8002 8013 90 8024 8035 8046 8057 8068 8079 8090 8101 8112 8123 91 8134 8145 8156 8167 8178 8189 8200 8211 8222 8233 92 8244 8255 8265 8276 8287 8298 8309 8320 8330 8341 93 8352 8363 8373 8384 8395 ^405 8416 8427 8437 8448 94 8159 8469 8180 8491 8501 8512 8522 8533 8544 8554 95 8565 8575 8586 8596 8607 8617 8628 8638 8648 8659 96 8669 8680 8690 8701 8711 8721 8732 8742 8752 8763 97 8773 8783 8794 8804 8814 8824 8835 8845 8855 8865 98 8876 8886 8896 8906 8916 8926 8937 8947 8957 8967 99 8977 8987 8997 9007 9017 9027 9037 9048 9058 9068 100 9078 9088 9098 9108 9117 9127 9137 9147 9157 9167 N 1 2 3 4 6 6 7 8 9 i INDEX The numbers refer to the pages. Approximate methods, 90-96. Area, by double integration, 97. bounded by a plane curve, 47-52. derivative of, 39. of a surface of revolution, 65. of any surface, 108. Attraction, 121. Center of gravity, 73-80. Change of variable, 44. Constants of integration, 1, 128. Curves with a given slope, 7. Cylindrical coordinates, 116. Definite integrals, 36. properties of, 41. Differential equations, 126-156. exact, 133. homogeneoas, 139. linear, 136, 147. of the second order, 143. reducible to linear, 137. simultaneous, 153. solutions of, 128. with variables separable, 132. Exact differential equations, 133. Formulas of integration, 14. Homogeneous differential equa- tions, 139. Infinite limits, 42. Infinite values of the inteerand, 43. Integral, definite, 36. definition of, 1. double, 97. indefinite, 36. triple, 112. Integrals, containing ax- + 6x -f- c, 19. p containing (ax + &) « , 29. of rational fractions, 26-29. of trigonometric functions, 21— 23. relation of definite and indefinite, 40. Integrating factors, 135 Integration, 1. by substitution, 15-19. constant of, 1. formulas of, 14. geometrical representation of, 36. in series, 94. of rational fractions, 23-29. Length of a curve, 61, 63. Limits of integration, 36, 42. Linear differential equations, 136, 147. Mechanical and physical applica> tions, 70-89. Moment, 72. of inertia, 83. Motion of a particle, 5. 193 194 Index Order of a differential equation, 126. Pappus's theorems, 80. Physical and mechanical applica- tions, 70-89. Pressure, 70. Prismoidal formula, 90. Polar coordinates, 50, 63, 103. national fractions, integration of, 26-29. Reduction formulas, 33. Separation of the variables, 9, 132. Simpson's rule, 93. Spherical coordinates, 119. Summation, 35. double, 100. triple, 113. Trigonometric functions, integrals of, 21-23. Trigonometric substitutions, 23. Variables, separation of, 9, 132. Volume, by double integration, 99. of a solid of revolution, 52. of a soUd with given area of sec- tion, 56. Work, 85. ^<;q'^~V WNOING SECT. AUG 2 6 1971 QA P5 Phillips, Henry Bayard Differential calculus Applier' Physic PLEASE DO NOT REMOVE CARDS OR aiPS FROM THIS POCKET UNIVERSITY OF TORONTO LIBRARY