Jt^ata, New Qntk BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT' OF HENRY W. SAGE 1891 arV19570 Differential calculus Cornell University Library 3 1924 031 254 042 olin.anx The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031254042 WORKS OF H. B. PHILLIPS, PH.D. PUBLISHED BY JOHN WILEY & SONS, Inc. Analytic Qeometry. Tii+197 pages. 5 by 7}^. Illustrated. Cloth, $1.50 net. Differential Calculus. v+162 pages. 5 by 7Ji. Illustrated. Cloth, $1.25 net. Integral Calculus. $1.25 net. Ready Spring, 1917. Differential and Integral Calculus, In one volume. $2.00 net. Ready Spring, 1917. DIFFERENTIAL CALCULUS BY H. B. PHILLIPS, Ph.D. AssisioTii Professor of Mathematics in the Massachusetts Institute of Technology FIRST EDITION FIRST THOUSAND NEW YORK JOHN WILEY & SONS, Inc. London: CHAPMAN & HALL, Limited 1916 Copyright, 1916, BY H. B. PHILLIPS StanDope pcex F H.GILSON COMPANY BOSTON, U.S.A. PREFACE In this text on differential calculus I have continued the plan adopted for my Analytic Geometry, wherein a few cen- tral methods are expounded and appUed to a la,rge variety of examples to the end that the student may learn principles, and gain power. In this way the differential calculus makes only a brief text suitable for a term's work and leaves for the integral calculus, which in many respects is far more impor- tant, a greater proportion of time than is ordinarily devoted to it. As material for review and to provide problems for which answers are not given, a supplementary list, containing about haK as many exercises as occur in the text, is placed at the end of the book. I wish to acknowledge my indebtedness to Professor H. W. Tyler and Professor E. B. Wilson for advice and criticism and to Dr. Joseph Lipka for valuable assistance in preparing the manuscript and revising the proof. H. B. PHILLIPS. Boston, Mass., August, 1916.' CONTENTS Chapter Pages I. Introduction 1- 9 II. Derivative and Dipfbrbntial 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 VIII. 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. Partial 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, if 2/ = 1 — x^, 2/ 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 3? — 2 xy ■]- 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 2 = x^ + y^, then 2 is a function of x and y; for, values being given to x and y, a value of z is determined. Similarly, the volume of a cone is a function of its altitude and radius of base; for the radius 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 + 2/ is an expUcit function of x and y. Similarly, if y = Vx-j- 1, y is an explicit function of x. 1 2 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 imphcit 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, a; V5 + 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 Vx + 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 = 31? + 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, / (x) = Vx^+ 1 means that/ (x) is a symbol for Vx^ + 1. Similarly, 2/ = / (aj) means that y is some definite (though perhaps unknown) function of x. If it is necessary 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), /a (x), /' (x), f" (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 writing commas between the variables. For example, V =f{r, h) expresses that «; is a function of r and h and V = f{a, b, c) expresses that « is a function of a, b, 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) = Vx^ + 1, /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(2/ +1)2+1 = Vf + 2y + 2. Similarly, if / (x, y) = x^ + xy - 2/2, then / (1, 2) = P + 1 . 2 - 22 = -1. If / {x, y, z) = x^ + y^ + z^, then f (2, -3, 1) = 2^ + (-3)2 + 1 = 14. EXERCISES 3 0^ 1. Given x + y = a , express !/ as an explicit function of x. 2. Given logio (x) = sin y, express x as an explicit function of y. Also express 2/ as an explicit function of x. 3. If / (x) = x^ - 3 X + 2, show that/ (1) = / (2) = 0. 4. If F (x) = X* + 2 x^ + 3, show that F (-a) = F (a). 5. If F (x) = X + i, find F (x + l). Also find F (x) + 1. X 6. If .* (x) = Vx2 - 1, find (2 x). Also find 2 .^ (x). 7. If^(x)=^3,find^(i). Alsofind^. 8. If /i (x) = 2% /2 (x) = x^ find /i !/2 (j/)]. Also find f, [/i (^)]. 9. If/ (x, J/) = X - i, show that/ (2, 1) = 2/ (1, 2) = 1. 10. Given / (x, j/) = x^ + xj/, find / {y, x) . 11. On how many independent variables does the volume of a right circular cylinder depend? 12. Three numbers x,y, z satisfy two equations X2 + 2,2 + 32 = 5, X + y +z =1. How many of these numbers can be taken as independent variables? 5. 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 volmne 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, x^ + 1 approaches 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, x=a which should be read, " the limit of / (x), as x approaches a, is A." Example 1. Find the value of lim (x + -)• x=l V xl As x approaches 1, the quantity x+ - approaches 1 +t X 1 or 2. Hence lim Ex. 2. Find the value of ,. sin 6 hm ■e=o 1 + cos 9 As d approaches zero, the function given approaches Hence ,. sin e - hm — = 0. 6=5=0 1 + cos 6 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 + Y + Z) = A + B + C = \imX + limY + \im 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 = lira X lim Y lim Z. 3. If 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 ^ approaches „ , that is, ,. X A limX If B is zero and A is not zero, -?, will be infinite. Then X A X ■y cannot approach „ as a limit; for, however large y. may become, the difference of -^ and infinity will not become small. 8, The Form jr , — When x is replaced by a particular value, a function sometimes takes the form ^ • Although this symbol does nob 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^-1 iim r- x=l X-l Chap. I. INTRODUCTION 7 When X is replaced by 1, the function takes the form 1-1 1-1 0' Since, however, T =X+l, X — 1 the function approaches 1 + 1 or 2. Therefore lim— ^ = 2. x=l x— 1 Ex. 2. Find the value of ,. (\/r+^ - 1) lim a:=0 X When X = the given function becomes 1-1 ^0 "O" Multiplying numerator and denominator by Vl + a; + 1, Vl+x- 1 ^ X 1 X ~ X {Vl + x + 1) ~ Vl + x + 1 As X approaches 0, the last expression approaches |. Hence ^■^(vr+^-i)^i x=o X 2 9. Infinitesimal. — A variable approaching zero as a limit is called an infinitesimal. Let a and fi be two infinitesimals. If lim^ is finite and not zero, a and ^ are said to be infinitesimals of the same order. If the hmit is zero, a is of higher order than |8. 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, 0?, 3?, 3^, etc. are infinitesimals arranged in ascending order. Thus x* is of higher order than x^; for lim -; = lim x^ — 0. x=oX'' 1=0 Similarly, 3? is of lower order than 3^, since 3^ X approaches infinity when x approaches zero. As x approaches ^ , cos x and cot x are infinitesimals of the same order; for ,. cos x ,. . lim — - — = lim sin cc = 1, T, cot X 1=0 which is finite and not zero. EXERCISES Find the values of the following limits: . ,. a;2-2x + 3 ... Vl - s^ - Vl + a;" 1. hm ^ 4. hm — ;=D X- 5 x=0 ^ „ ,. sin 9 + cos 9 2- >^"^.sin29 + cos2e- 5. Iim^i5_£. «= J 9=0 tan 9 « ,. a' — 3a; + 2 ... sin 3. hm r-' 6. hm ■ x=\ X — \ s=osin 2 9 7. By the use of a table of natural sines find the value of / ,. sinx Imi a:=0 ^ 8. Define as a Umit 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 9 their bases approaching zero, the polygonal line PABC, etc., approaches PQ. Does its length approach that of PQ? A \/ \/ 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. 13. Show that when x approaches zero, tan - does not approach a limit. 14. As X approaches 1, which of the infinitesimals 1 — x and Vl — x is of higher order? 15. 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 y he a, function of x. When x receives an increment Ax, an increment Ay 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 =fix) 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, 2/ 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 Y / <^ X / X Fig. 11a. Fig. lib. P. As Q approaches P', Ax = PR approaches zero, but Ay = RQ does not. In Fig. lib the ordinate at a; = a is infinite. The increment Ay occurring in the change from a; = a to any neigbboring 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 hue 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 Ay 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. Example. Find the slope of the parabola y = x^ at the point (1, 1). Let the coordinates of P be X, y. Those of Q are a; + Ax, y + Ay. Since P and Q are both on the curve, y =x^ and y + Ay=-{x + Axy = x2 + 2 a; Ax + (Aa;)^. Subtracting these equations, we get Ay = 2xAx+ {Axy. Dividing by Aa;, ^^2a=+A.T. Aa; As Aa; 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 w be a function of x. If -r^ 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 Dxy, that is, Ay D^y = lim ^ Ai=o Ax (13a) If a function is represented by / (x), its derivative with respect to x is often represented by/' (x). Thus f (x) = lim^^ = Z)^(x). ^=0 iiX (13b) Chap. II. DERIVATIVE AND DIFFERENTIAL 13 In Art. 12 we found that this Umit represents the slope of the curve y = fix). 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 -T-^ , 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 Fig. 13. its limit Dxy are positive. If they have opposite signs D^y is negative. Therefore Dxy is positive when x and y increase and decrease together and negative when one increases as the other decreases. Example. y = x? — 3x + 2. Let x receive an increment Ax. The new value of x is X + Ax. The new value oiy isy + Ay. Since these satisfy the equation, y + Ay={x + Axy -3{x + Ax)+ 2. Subtracting the equation y = oi^-3x + 2, we get Ay = 3x^Ax + 3x (Ax)^ + (Ax)' - 3 Ax. Dividing by Ax, Ay Ax = 3x^ + 3xAx + {Axy - 3. As fyx approaches zero this approaches the limit Dxy = 3 a;2 - 3. 14 DIFFERENTIAL CALCULUS Chap. U. The graph is shown in Fig. 13. At A (where x = 1) y = and Dxy = 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 D^y 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 I = 2 to X = 1.9. Show that the increments approximately satisfy the equation A^ 1_. Ax 2 VS 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 iay = 2 x + .3. Find its slope by Aw calculating the limit of -p- 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? 5. 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), Dxy = 2x + 3. 7. y = x*, Dxy = 4 X'. 8. y = x^ - x2, Dxy = 3x^ -2 X. 9. 1 y = -' o.y = -|- 10. If X is an acute angle, is Dx cos X positive or negative? 11. For what angles is Dx i sin X positive and for what angles negative? 14. Approximate Value of the Increment of a Function. — Let J/ be a function of x and represent by « a quantity such that ^ Chap. II. DERIVATIVE AND DIFFERENTIAL 15 As A.X approaches zero, -i— approaches Dxy and so « ap- proaches zero. The increment of y is Ay = DxV Ax + eAa;. The part D^y Ax (14) is called the principal part of Ay. It differs from Ay by an amount eAx. 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 = -■ In exercise 9, page 14, the derivative of - was found to be 1- Hence the principal part of Ay is -\Ax= -7(.1) = -0.0250. x^ 4 The exact increment is The principal part represents Ay with an error less than 0.002 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 function y of x. In particular, ii y = x, D^y = 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 dy = Dxy dx, (15c) whence X = "^"V' (1^^) dy that is, the quotient -^ is equal to the derivative ofy vnth 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, Aa; 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 ixi 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. Chap. II. DERIVATIVE AND DIFFERENTIAL 17 The identity Ay _ Ay Ax Ai~Ax"Ai gives in the limit Dty = D^y • DfX. Jieiice D,y At = Dx2/ • Dtx At, that is, dy = Dxy dx. X -\- 1 Example 1. Given y = , find dy. X In this case x+Ax+l x+l_ Ax x + Ax X X (x + Ax) Consequently, A2/ . . 1 Aa; x (x + Ax) As Ax approaches zero, this approaches ^= -i. dx y?' Therefore , dx Ex. 2. Given x = t\ y = i^ find ^• The differentials of x and y are found to be dx = 2tdt, dy = 3 f dt. Division then gives, dy^3 dx 2 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 = a;". Expand A?/ = (a; + Ax)" — x" by using the binomial theorem. Show that ^ = nx-K ax What is the principal part of Ay? 2. Using the results of Ex. 1, find an approximate value for the in- crement of a^ 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 -j- is equal to the 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 -y- is equal to the area of its surface. 6. Let V be the volume of a cylinder with radius r and altitude h. dv Show that if r is constant -jr is equal to the area of the base of cylinder an dv and if h is constant -7- is equal to the lateral area. 7. If 2/ = / (x) and for all variations ia x, dx = Ax, dy = Ay, show that the graph oi y = f (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 2/-axis is vertical, the a;-axis horizontal, a body thrown hori- zontally from the origin with a velocity of 50 ft. per second will in t seconds reach the point X = 50t, y = -16 tK 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-swa at B. If Ki and K2 are the areas of the triangles OP A and OPB, show that dKi^PA^ dKi PB^' 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-jior the operation of taking the derivative with respect to x. Thus d {u + v) = differential of (m + w), J- (m + w) = derivative of {u + 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 ""^dTx 17. Formulas. — Let u, v, w be continuous functions of a single variable x, and c, n constants.* I. dc = 0. II. d {u-\- v) = du + dv. III. d {cu) = c du. IV. d (tfv) = u dv -\- V du. fu\ V du — ti dv V. d[-) = 3 VI. d (it") = nu"'-'- du. * It is assumed that the functions u, v, w have derivatives. There exist continuous functions, u =f{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, -r— = 0, and in the limit dc -^ = 0. Clearing of fractions, dc = dx-0 = 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 w + Am, v to «; + Ay, and y to j/ + Ay. Consequently y + Ay = u + Au + v-j- Av. Subtraction of the two equations gives Ay = Au + Av, whence Ay _ Aw Ay Aa; ~ Ax Ax . . , Ay Au Av , dy du dv As Ax approaches zero, ^, ^, ^^ approach ^, ^, ^ respectively. Therefore dy _du dv dx dx dx' and so dy = du + dv. By the same method we can prove d{u±v±w dc • • • ) = du±dv±dw± • • • . such that Au Ax does not approach a limit as Ax approaches zero. Such a function has no derivative DxU and therefore no differential du = DxU dx. 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. Then y + Ay = c(u + Am) and so Ay = c~"Am, Ay _ Am Ax "" Ax A A 1 Ay , Am , dy , du As Ax approaches zero, -r-.and c-ir- approach j- and c-j-- Therefore whence dy = c du. Fractions with a constant denominator should be differen- tiated by this formula. Thus dy _ du dx dx' '^©='^(^)=^"- 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 = (u + Am) (v + Av) = uv -\-v Am + (m + Am) Av. Subtraction gives Ay = V Am + (m + Am) Av, whence Aw Am , , , . , Aw Since m is a continuous function, Am approaches zero as Ax approaches zero. Therefore, in the limit, dy _ du do 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 the numerator minus the numerator times the differential of the denominator, all- divided by the square of the denominator. Let u y = -■ Then and , w + Am u V Am — m At; Ay - - Dividing by Ax, V -\- Av V V {v + Av) . Am At; At/ V-: M-7- -r^ = Ax Ax Ax V (v + Av) Since t; 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 dv- ^ 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 61. Chap. III. ALGEBRAIC FUNCTIONS 23 (1) Let n be a positive integer and y = u". Then and Ly = nu"-^ Am + ^ ^\~ ^^ m"-^ (Am)^ + • ■ • . Dividing by Am, Ah , , n (n — 1) „ , , , , -^ =nM"-iH ^^"2 — ^M"-2 (Am) + • ■ • . As Am approaches zero, this approaches -^ = nM"-!. du Consequently, dy = nu"~''- du. V (2) Let n be a positive fraction - and y — u''= u". Then 2/« = M". Since p and g are both positive integers, we can differentiate both sides of this equation by the formula just proved. Therefore qy^~^ dy = ■pu'^~'^ du. p Solving for dy and substituting m' for y, we get VU^ V ^ dy = du = -u du = nu"~^ du. « p n P-- q qu * (3) Let n be a negative number — m. Then W = M" = M""" = " M" Since m is positive, we can find d (m") by the formulas proved above. Therefore, by V, , u"'d(l) — ld{u'^) —7rmr~Hu „,, ,, dy = ^ „^„ — - = ^ = —mu-'^-^du = nu"-^ du. 24 DIFFERENTIAL CALCULUS Chap. III. . Therefore, whether n is an integer or fraction, positive or negative, d (m") = nw"""' du. If the numerator of a fraction is constant, this formula can be used instead of V. Thus d ( - j = d icu~^) == —cu~^ du. Example 1. y = 43^. Using formulas III and VI, dy = 4 d (x^) = 4 • 3 a;2 dx = 12 a;2 da;. Ex.2, y = Vl + ~ + 3. Vx This can be written y = x^ + x 2+3. Consequently, by II and VI, dy _d (xi) d (x-i) d (3) dx dx dx dx 1 —idx 1 —sdx , n = 2^^d^-2^^di + 2VZ 2V'^ Ex. 3. y = (x + a) (x^ - b^). Using IV, with u = x + a, v = x^ — V, = (x + a)(2x-0) + (3?- ¥) (1 + 0) = 3 a;2 + 2 ax - 62. Ex.4.. y = $^- Chap. III. ALGEBRAIC FUNCTIONS 25 Using V, with u = x^ + 1, v = x^ — 1, , (x" -l)d (x' + 1) - (x^ +l)d {x^ - 1) '^'^ ~ {?? - ly ^ {x^-\)2xdx- jx" + l)2xdx _ Axdx '~ ~ (a;2 -1)2' Ex. 5. y = Va;2 - 1. Using VI, with u = x^ — 1, = i (a;2 - 1)-^ (2 a;) = Vx2- 1 Ex. 6. cc^ + a;2/ — 2/^ = 1. We can consider y a function of x determined by the equa- tion. Then d ix') + d (xy) - d (y2) = d (1) = 0, that is, 2 X dx + X dy + y dx ~ 2 y dy = 0, (2x + y)dx+{x-2y)dy = 0. Consequently, d^ ^ 2x-\-y dx 2y — x Ex.7. a; = < + -, y = t - -. In this case dx = dt — ^, dy = dt + ^- Consequently, % ^ ^ t^ ^ l±l dx 1 t^-l f /I _ x\i Ex. 8. Find an approximate value of 2/ = ( . 1 when X = 0.2. 26 DIFFERENTIAL CALCULUS Chap. III. When X = 0, y = 1. Also 2dx ^~ 3(l-a;)^(l+a;)*' When a; = this becomes dy = -f da;. 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 : 1. y = 3x* + ix' -6x' + 5, di ^ 11 {x" + x^ - x) dx. 3 3, dy Zx^-2 g3 - x^ + 1 dy 3 x' - 2 a : ^- y ~ 5 ' dx 5 ■ 4. y = (,x + 2a){x- ay, dy = 3 (y? - a") dx. 5. J/ = a; (2a;- 1) (3a; + 2), ^ = ISa;^ + 2a; - 2. 1 _ -2a;cfa; 6- 2/ = ^?:pi- ''■y - (a;2 + if _ 2 X + 3 -22dx (i 1 Ve^-l-e '•^"4x-5'''^ (4x-5J^ ^"- d9 0+V^2Tri= ^/W^^l ' d \ -2x _ 2x ^ a2-2s2 (ix (x - I)'' (x - 1)' ds« va- - s> = y^, _ ^j / {l-t)di J y a?dy dx Vx / x' Vo^ - x^ Chap. III. ALGEBRAIC FUNCTIONS 27 ^ = (24 + 13a; -36a;2) (2- 3a;) (2x - 3)=. 18. 2/ = (x+l)(2-3a;)2(2x-3P, 1-1 19 = a;" dy _ maa:"'-' * "^ ' da; ~ "^ _i_ 1 * (a + ia;")"" (a-|-6x")«+' 20. . = ^^V.-^. t 3 ar* ' dx jfi Va:2 + 1 ' (x + vr+T')"+^ (x + vrr^o"-' ' ij ,« I 1 ' 22. .x2 + 2/2 = a2, « + 1 K - 1 d?/ = 2 (x + VrTx'2)"dx. d^ _ X dx V 23. x3 + ,3 = 3 ax.. | = P^^- 24. 2 x^ - 3 X2/ + 4 2/2 = 3 X, d^ _ 4x — 3)/ — 3 dx ~ 3 X — 8 2/ 25. 1 + 1 = h ydx-xdy = 0. n~ 1 dx da 26. 2/ = -1 I " - - n a;' Vl+x'+Vr+a*" 27. 2/'" + x'"2/" = x^™, mydx =nx dy. I 2 J + 3 d2/ ^ 29. x=i-V<2-l, y=t+^e-l, xdy + ydx = 0. X 31. Given y = .^ ^ „ , find an approximate value for ?/ when X = 4.2. 32. Find an approximate value of i/5 T X ■X + 1 X^ -|- X + 1 when X = .3. 33. Given y = x^ find d2/ and Ay when x changes from 3 to 3.1. Is dy a satisfactory approximation for Ay? Express the difference as a percentage of Ay. 34. Find the slope of the curve 2/ = X (x5 + 31)^ at the point x = 1. 35. Find the points on the parabola y^ = iax where the tangent is incUned at an angle of 45° to the x-axis. 28 DIFFERENTIAL CALCULUS Chap. III. 36. Given y = (a + x) Va — x, for what values of x does y increaae as X increases and for what values does y decrease' as x increases? 37. Find the points P {x, y) on the curve ,1 y = x + - where the tangent is perpendicular to the line joining P to the origin, 38. Find the angle at which the circle < x^ + y'^ = 2x-Zy intersects the a;-axis at the origin. 39. A line through the point (1, 2) outs the x-axis at {x, 0) and the 2/-axis at (0, y). Find -—■■ 40. If a;2 — a; + 2 = 0, why is the equation :^ (x2 - a; + 2) = ax not satisfied? 41. The distances x, x' of a point and its image from a lens are con- nected bv the equation - + - = - x^x' r / 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 . MIT approximately, x and x' being the distances of the object and its image from the lens. 24. Higher Derivatives. — The first derivative ^^ is a function of x. Its derivative with respect to x, written ^-^ . is called the second derivative of y with respect to x. That is, ^ = -^ /%Y dx^ dx \dx) Similarly, d^^d^ l^\ db? dx \dxy' ^ = 1W etc Chap. III. ALGEBRAIC FUNCTIONS 29 The derivatives of / (x) with respect to x are often written /' (a=), /" (.x),f"' (x), etc. Thus, ii y = f (x), | = /'(x). g = /"(x), g = r(x),etc. Exafnple 1. y = x^. Differentiation with respect to x gives All higher derivatives are zero. , , \ JEx. 2. x^ + xt/ + 2/' = 1. ■^ '^^ Differentiating with respect to x, 2x + ,+x|+2,|=0, whence dy ^ 2x + j/ dx X + 2 J/ The second derivative is 3x^-Sm ^= __d /2xJ-_^\ ^_2dx__^ dx2 dx U + 2 2// (x + 2 y)' dv Y by its value in terms of x and y (Py 6 (x^ + x?/ + 2/^) 6 Replacing -~ by its value in terms of x and y and reducing, dx2 (x + 2 yy (x + 2 2/)' The last expression is obtained by using the equation of the curve x^ + xy -\- y^ = 1. By differentiating this second derivative we could find the third derivative, etc. 30 DIFFERENTIAL CALCULUS Chap. III. 25. Change of Variable. — We have represented the second derivative by -r^ . 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. d^y Thus, suppose y = x^, x = t'. Then -r^ = 2 and so d^ = 2 (dxY = 2 (3 <2 dty = 18 1^ {dty. If, however, we differentiate with respect td t, since y = t^, ^ = 30 ^« and df 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. ' dj^ti d^v Two derivatives like -^ and -r^ must not be combined like dr ax' fractions because d^y 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 ,„ V dx dt dx dx Tt For example, dx\dtj dt\di) dx dx Chap. III. ALGEBRAIC FUNCTIONS 31 1 1 d'y Example. Given x = t — - , y = t + -, find -r| ■ In this case dy ^ f_ ^ f - 1 dx ■,, , dt f + l' dl + -. Consequently, d^y _ d ff - 1\_ it dt _ 4t 1 _ 4f dx \f + 1/ {(' ■ EXERCISES Find -r and -r4 in each of the following exercises: ax ax' 1. y = r- 5. x^ + 2/" = a?. X — 1 2. y = Va2 - x2. 6. x'' -2y^ = 1. 3. y = (x - 1)3 (x + 2)<. 7. X2/ = a + ?/. 4. !/^ = 4 X. 8. x^ + t/» = ai 9. If a and b are constant and y = ax^ + bx, show that dx' dx 10. If a, b, c, d are constant and y = ax^ + bx' + ex + d, show that 11. 12. Show that d 1 dx \ dH dt\dt ^j dP' Show that ^L3J!^^_3..^ + 6x^-6,Ux3g. dx V dx3 dx' dx " j dx^ 13. Given x = f? + t\ y = t' - P, find ^ and ^• 14. By differentiating the equation dy^J_ dx dx dy with respect to x, find t^ in terms of derivatives of x with respect to y. CHAPTER IV RATES 26. Rate of Change. — If the change in a quantity z is proportional to the time in which it occurs, z is said to change at a constant rate. If Az is the change occurring in an in- terval of time Ai, the rate of change of z is Az Af If the rate of change of z is not constant, it will be nearly Az constant if the interval Ai is very short. Then -r- 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 ^fr-%' 0^' 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 hne (Fig. 27). Let s = OP be con- s 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 Ai, its velocity is As At' If the velocity is not constant, it will be nearly so when A< As is very short. Therefore -rr is approximately the velocity, the 32 Chap. IV. RATES 33 approximation becoming greater as At becomes less. The velocity at the time t is therefore defined as „=liin^ = g. (27) A(=o At dt ^ ' 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 approxinaately s = 16 i2 feet in t seconds. Find its velocity at the end of 10 seconds. The velocity at any time t is dv t, = ^ = 32 « ft./sec* dt At the end of 10 seconds it is V •= 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 « = 5=g- (^' This equation shows that Uv is the amount v would increase in the time dt if the acceleration remained constant. The acceleration is positive when 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 h = 100t-lQ f. 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. 34 DIFFERENTIAL CALCULUS Chap. IV. The velocity and acceleration are v = j^ (100 - 32 ft./sec, a = ^ = -32 ft./sec.2. The ball will be rising while v is positive, that is, until t = -^ = 3|. It will be falling after i = 3|. It will be at the highest point when < = 3|. 29. Angular Velocity and Acceleration. — Consider a body rotating about a fixed axis. Let d be the angle turned ^P through at time t. The angular veloc- ity is defined as the rate of change of 9, that is, angular velocity = to = -^ • at Fig '^9 The angular acceleration is the rate of change of angular velocity, that is, , , .. do} (P9 angular acceleration = a = -tt = ttt • dt dt^ Example 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 T radians/min. Ex. 2. A wheel, starting from rest under the action of a constant moment (or tv/ist) about its axis, will turn in t seconds through an angle 8 = kt\ k being constant. Find its angular velocity and acceleration at time t. By definition 0) = jT = 2 ftf rad./sec, a = -jT = 2k rad./sec*. at Chap. IV. RATES 35 30. Related Rates. — In matny 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. Exam,ple 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'-ll. The derivative with respect to t is By hypothesis dv ,dh,^ , dr dr _ _ dh _ di~^' dt~ ~^- Hence -r- = 4 irrh — 3 TT^- dt This is the rate of increase when the radius is r and altitude h. If r = 10 ft. and /i = 6 ft«, -77= — 60 TT cu. ft./sec. dt 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 = Vx^ + 2/2. This distance is changing at the rate dx dy ds ^dt^^di ^ Va;2 + 9 36 DIFFERENTIAL CALCULUS Chap. IV. By hypothesis, ;t7 = 10, -| = -16, , = , " = cos45° = — ^• at dt Vx' + y^ VX2 + 2/2 V2 Therefore ds 10 - 16 dt V2 = -3\/2mi./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 A = 50 + 100 < - 16 fi. Find its velocity and acceleration when t = 2. How long does it con- tinue to rise? What is the highest point reached? p4 / 2. A body moves in a straight line according to the law s = it* - il' + 16P. 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 tht acceleration of a particle jnoving along the a;-axis, show that adx = V dv. 4. If a particle moves along a line with the velocity t)2 = 2 gs, where g is constant and s the distance from a fixed point in the line, show that the acceleration is constant. 6. When a particle moves with constant speed around a circle with center at the origin, its shadow on the a^-axis moves with velocity v satisfying the equation 2,2 4- „2j;2 = ^2^2^ 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 e = a + bt - cfi, a, b, c being constants. Find the angular velocity and acceleration. When wiU the wheel come to rest? 8. A wheel revolves according to the law w = 30 < + <', where m is the speed in radians per minute and t the time since the wheel started. A second wheel turns according to the law 9 = Ifi, where t is the time in seconds and 9 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 v is the velocity and a the acceleration of its center, a the angular velocity and a the angular acceleration about its axis, show that V = Tec, a = ra. 10. The depth of water in a cylindrical tank, 6 feet in diameter, ia 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 radius 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 slides 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 ground. WTien 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. Assuming that the water is 1000 ft. deep, the cable is attached to the bottom and stretches in a straight line to the stern 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 surface forms a cone with constant angle ^ at the vertex, depending on the material. If the 38 DIFFERENTIAL CALCULUS Chap. IV. 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. 19. An elevated car running 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 mUes, at what rate are the cars separating 10 seconds after they meet? 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? CHAPTER V MAXIMA AND MINIMA 31. -A function of x is said to have a maximum a,t 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 = / Wi ^ 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 / (a;) we first look for values of x such that lfix)=nx) = o. If a is a root of this equation, / (a) may be a maximum, a minimum, or neither. T A C^ / / / /^ K^B ^ ^ X Fig. 31a. If the slope is positive on the left of the point and negative on the right, as at 4, the curve falls on both sides and the ordinate is a maximum. That is, / (x) has a maximum value 39 40 DIFFERENTIAL CALCULUS Chap. V. atx = a, iff (x) is positive for values of xa little less and negor tive 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, iff (x) is negative for values of xa 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, / (x) has neither a maximum nor a minimum at x = a if f (x) /los 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 ib — x) = 5 X — x^ is a maximum. The derivative is f = 5-2x. ax This is zero when x = f . If x is less than Fig. 31b. ti the derivative is positive. If x is greater than f the derivative is negative. Near a; = J the graph then has the shape shown in Fig. 31b. At a; = 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 = TTT^ + 2 irrA. Let V be the number of cubic inches in a pint. Then v = irr^h. Chap. V MAXIMA AND MINIMA 41 Consequently, and h = irr' 4 = Trr^ + 2v Since t and v are constants, dA dr 2 — — = 2 ( ^ ~ " V J.2 ^ ^2 y This is zero if tit' = v. If there is a maximum or minimum it must then occur when = V!; dA for, if r has any other value, -p 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 = in^. Also v = irr%. 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 a;2 + 2/2 = (24)2. The strength of. the beam is S = kxy^ = kx (242 _ a.2)^ Fig. 31c. 42 DIFFERENTIAL CALCULUS Chap. V. k being constant. The derivative of S is dS dx = k (24^ - 3 x2). If this is zero, x = ±8 Vs. Since x is the breadth of the beam, it cannot be negative. Hence a; = 8^3 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 v3 must be the width of the strongest beam. The corresponding depth is 2/ = 8 Ve. 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 ABCj DE^AB EC BC' that is, y h r y = - (r - x). Fig. 31d. r — X The volume of the cylinder is V = irx'^y = — {rx^ — x^) . Equating its derivative to zero, we get 2 rx - 3 a;^ = 0. Hence x = or a; = | r. The value a; = obviously does not give the maximum. Since there is a largest cylinder, its radius must then be a; = | r. By substitution its altitude is then found to be 2/ = ^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. (3) Calculate -p 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. If not, determine the signs of -j^ when a; is a little less and a little greater than the values in question and apply the criteria given in Art. 31. EXERCISES Find the maximum and minimum values of the following functions: 1. 2 a;2 _ 5 x + 7. 3. x* -2x^ + Q. 2. 6 + 12 a; - x\ 4. ^^ • Va? - x' Show that the following functions have no maxima or minima: 5. x\ 7. 6 a;* - 15 a^ + 10 x\ 6. a? + 4 X. 8. X Va^ + a;". 9. Show that x +- 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. 15. A cylindrical tin can closed at both ends is to have a given capacity. Show that the amount of tin used wiE 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 ai, 02, Oa, 04 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 which 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. A 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 Ught 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 a model of a right cylinder. Calculate the lengths into which the wire must be divided to produce the cylinder of greatest volume. 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. If 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. 31. Find the length of the shortest Une 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 Ught 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 light in air be Vi and in water Vi. The path of a ray of light 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 fli and 0i 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 wiU be least if B is so placed that sin $1 _ vi sin 62 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 $20 per hour for a speed of 10 knots, and if the other expenses amount to $100 per hour, find the most econom- ical speed in atill 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 types of maxima and minima that do not satisfy these conditions. At B and C, Fig. 33a, the tangent is vertical and the de- rivative infinite. At D the slope on the left is different from 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 at 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, Pig. 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- rivative is discontinuous, the function ceases to exist, or two values of the function coincide. Exatnyle 1. Find the maximum and minimum ordinates on the curve y^ = 7?. In this case, y = x^ and dy 2 _i dx 3 ■ Fig. 33b. No finite value of x makes the derivative zero, but a; = Chap. V. MAXIMA AND MINIMA 47 makes it infinite. Since y is never negative, the value is a minimum (Fig. 33c). Y Ex. 2. A man on one side of a river J 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. i_ 2 ■ Fig. 33d. Fig. 33e. 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 ^ Va;^ + J and the time of walking J (2 — a;). The total time is then t = i ViH=l + H2 - x). Equating the derivative to zero, we get '6 Vx^ + i 4 which reduces to 5 a;^ + f = 0. This has no real solution. 48 DIFFERENTIAL CALCULUS Chap. V. The trouble is that J (2 — a;) is the time of walking only if C is above B. If C is below B, the time is i (r — 2). The complete value for t is then t = i v^n ± i (2 - x), the sign being + if x < 2 and — if a; > 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' + 2/' = a^. 3. 2/3 = X* - 1. 2. y^ = x'{x - 1). 4. X = t' + i^, y = t' - tK 6. Find the rectangle of least area having a given perimeter. 6. Find the point on the parabola y^ — ix 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 J ohm is 60 xE C = s2 + 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. d sin u = cos u du. d cos v, — — sin u du. d tan u = sec^ u du. d cot u = — csc^ u du. d sec u = sec u tan u du. d CSC u = — CSC u cot M du. VII. VIII. IX. X. XL XII. 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 1° = 0.017452, 180 = 0.017453. We should then expect that ,. sin 6 lim-^= 1. 9=0 O (35) Fig. 35. To show this graphically, let d = AOP (Fig. 35). Draw PM perpendicular to OA. The circular measure of the angle is defined by the equation arc arc AP e = rad. OP 49 50 DIFFERENTIAL CALCULUS Chap. VI. Also sin d = jyp . Hence sing _ MP _ chord QP e ~ 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 y = sinu. Then y + Ay = sin (u -\- Am) and so Ay = sin (w + Am) — sin u. It is shown in trigonometry that ; sin A - sin 5 = 2 cos I (A + B) sin | (A - B). If then A = u+ Am, B = u, ~ Ay = 2cos {u + i Am) sin § Au, whence Am „ / , 1 . , sin A Am , , , . , sin i Am ^ = 2cos(m + i Am) ^^ = cos (m + |Am) ^ ^^ . As Am approaches zero sin § Am _ sin 8 I Am ~ ~e~ approaches 1 and cos (m + § Am) approaches cos u. There- fore dy -rr = cos M. au Consequently, dy = cos u du. 37. Proof of Vm, the Differential of the Cosine. — By trigonometry cosM = sin(^ — MJ- Chap. VI. TRANSCENDENTAL FUNCTIONS 51 Using the formula just proved, dcosu = d sin (^ — wj = cosf^ — u]dl- — u]= —sin u du. 38. Proof of IX, X, XI, and XII. — Differentiating both sides of the equation sin u tan u = COSM and using the formulas just proved for the differentials of sin u and cos u, cos M d sin M — sin w d cos u cos^ udu + sin^ u du d tan u = 5 = „^ . „. COS^ U COS'' u = sec^ u du. By differentiating both sides of the equations COS M 1 1 cot u = —. , sec u = , CSC u = sin u cos u sin M ' and using the formulas for the differentials of sin u and cos u, we obtain the differentials of cot u, sec u and esc u. Example 1. y = sin^ (a;^ + 3). Since sin2 (a;2 + 3) = [sin (x^ + 3)]^, we use the formula for u^ and so get dy=-2 sin (a;^ + 3)d sin (x^ + 3) = 2 sin (a;2 + 3) cos {x^ + 3) d {x^ + 3) = 4 X sin (a;2 + 3) cos {x^ + 3) dx. Ex. 2. y = sec2x tan 2 x. -r = sec 2 a; T- tan 2x + tan 2 a; -5- sec 2 a; dx ax dx = sec 2 a; sec^ 2 a; (2) + tan 2 x sec 2 a; tan 2 x (2) = 2 sec 2 a; (sec^ 2 a; + tan^ 2 x) . EXERCISES In the following exercises show that the derivatives and differentials have the values given: 1. 2/ = 2 sin 3 a; + 3 cos 2 a;, -p = Q (cos 3 x — sin 2 x). 52 DIFFERENTIAL CALCULUS Chap. VI. 2. y = sin' - , dy = sin ^ cos ^ dx. 3. 2/ = 2 cos a; sin 2 a;— sin a; cos 2 s, dy =Z cos x cos 2 x dx. , 1 — cos ix dy X — cos \ x 4» V ==^ — , — — ^ — • sin i a; dx Z sin' J a; dy 6. 2/ = tan 2 a; + sec 2 a;, -3^ = 2 sec 2 x (sec 2 a; + tan 2 a;). e ,,a; ,a; d?/ .a; .xt ,x , ,,x\ 6. y = cot'^csc'^, ^ = -cot2Csc'2(csc'2 + cot'-j- 7. X = a cos t, y = asin^ /, -^ = —3 sin t cos i. da; 8. a; = a ( — sin 0), « = a (1 — cos ), 3^ = cot ^• ax i d?v 1 9. .-c = cos i + i sin <, y = sint — t cos t, -rK = r-- . ax' t COS'' c ' 10. 2/ =|cot'i— icot'x+cotx+x, dy = —cot'xdx. J,' 11. y = i tfirf s + f tan' x+ tan x, dy = sec' x dx. 12. M = ^ sec' 9 — f sec' 9 +^ sec' 9, du = tan" 9 sec' 9 d9. 13 /cos'x \,1., ,2. -J J '• V = x\ — q— —cos ^ I + Q sin' X + ^ sm X, dy = x sm' x dx. ^. cosx/1 .. ,5., ,5. \,5 , -Bj 14. y = =— I ^sm'x + YoSin'x + gSinx j + T^x, dy = sm' x dx. ., _ 1 + sin X dy 2 cos x 15. y = -^r-^ — : — , -r- = ,ri -■ rz- 1 — sin X dx (1 — sin x)' ^„ _ sec X — tan x dy _ 2 sec x (tan x — sec x) sec X + tan x ' dx ~ sec x + tan x 17. y = (cotx — 3tanx) Vcotx, -,- = 5— • "^ 2cot^x 18. II y = A cos (?ix) + B sin (rax), where A and B are constant, show that 19. Find a constant A such that j/ = A sin 2 x satisfies the equation g + 52/ = 3sin2x. 20. Find constants A and B such that y = A sin 6 x + 5 cos 6 x satisfies the equation S+4x+«^=^^'"^- 21. Find the slope of the curve 2/ = 2 sin a; + 3 cos a; at the point TT Chap. VI. TRANSCENDENTAL FUNCTIONS 53 22. Find the points on the curve y = x + sin2x where the tangent is parallel to the line y = 2 x + 3. 23. A weight supported by a spring hangs at rest at the origin. If the weight is lifted a distance A and let f aU, its height at any subsequent time t will be y = A cos {2imt), n being constant. Find its velocity and acceleration as it passes the origin. Where is the velocity greatest? Where is the acceleration 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 respectively, and that these quantities are changing at the rates of 3 ft., —2 ft., and 10 degrees per second, how fast is the area of the triangle changing? 27. OA is a crank and AB a connecting rod attached to a piston B moving along a line through 0. If OA revolves about with angular velocity. fc), prove that the velocity of B is MC, where C is the point in which the Une BA cuts the line through O perpendicular to OB. 28. An alley 8 ft. wide runs perpendicular to a street 27 ft. wide. What is the longest beam that can be moved horizontally along the street into the alley? 29. A needle rests with one end in a smooth hemispherical bowl. The needle will sink to a position in which the center is as low as possible. If the length of the needle equals the diameter of the bowl, what will be the position of equilibrium? 30. A rope with a ring at one end is looped over two pegs in the same horizontal line and held taut by a weight fastened to the free end. If the rope slips freely, the weight will descend as far as possible. Find the angle formed at the bottom 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 be made by bending into shape a strip of copper so that the cross section shall 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, in what position will the outer ends of the two spokes be closest together? 54 DIFFERENTIAL CALCULUS Chap. VI. 39. Inverse Trigonometric Functions. — The symbol sin~i 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 sam'e 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~' 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-'w and cot~'M are valid in all quadrants. 40. Formulas for Differentiating Inverse Trigonometric Functions. — du Xm. d sin-' XIV. XV. XVI. dcot-'tt = XVII. d sec-' u = xvni. vT w du Vl-u^ du- l + u"' du 1+1?' du u Vw^ _ i' du u Vm^ _ 1 41. Proof of the Formulas. — Let y = sin-i u. Then siny = u. Differentiation gives cos y dy = du, Chap. VI. TRANSCENDENTAL FUNCTIONS 55 whence dy = ' '^ cosy But cos 2/ = ± Vl — sin^ 2/ = ± Vl — w'. If y is an angle in the first or fourth quadrant, cos y is positive. Hence cos y = Vl — u^ and so , du dy = Vl -m2 The other formulas are proved in a similar way. 1. ^ = sin-i {Sx - 1), EXERCISES 3dx V6x - 9a;2 V2 ax - x2 dx 3. 2/ = tan-'|^, - 6dx 4. 2/ = cof -(i-rj. dv_ -2 5. 2/ = sec~^ V4x + 1, d^/ dx x^ + l dx (4 X + 1) Vx 1,3 dy 1 6. 2/=^csc-" 1. y = tan 2 4x-l dx V2 + 2x-4x2 X — a dy _ a X + a dx x' -\- a? dx 8. X = csc~' (seo0), Hfi ~ ~^' o ■ -1 ^ ^2/ 9. 2/ = sin 1- Va? - x2 d^; (a2 _ 3;2) Vo^ - 2 x^ 10. y = sec-i , ^ , ^ = ^ . V 1 — x^ dx Vl — x^ 11. 2/ = I v^^:^^ + 1 sin-i^, g ^ v^^r^s. HO i _, 4sinx rf,. 4 12. 2/ = tan^ „ . , , 5» = 1 3 + 5COSX ckc 5 + 3COSX 13.2/ = sec-^,^, g=__^. 56 DIFFERENTIAL CALCULUS Chap. VI. 14. y = asin-i- + V^TT^, e the slope is greater than 1. If a < e, the slope is less than 1. Fig. 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. <«e" = e" du. XX. d!a" = a" In a du. XXI. dhiu = —- u V VTT ^ !.,„ log„e du XXII. d lOga'U' = — ^ ttl 46. Proof of XIX, the Differential of e». — Let y = e". Then y + Ay = e"+^", whence Ay = 6"+^" — e" = e" (e'^" — 1) and Am. Am As Am approaches zero, by (44), gAu _ 1 Am approaches 1. Consequently, -^ = e", dy = e" du. 47. Proof of XX, the Differential of a". — The identity ^ _ glna gives a" = e"'"". Consequently, da" = e" '""" d (m In a) = e" ■" " In a dw = a" In a dit. 48. Proof of XXI and XXn, the Differential of a Log- arithm. — Let y = Inu. Then e" = u. Differentiating, e^ dy = dM. 60 DIFFERENTIAL CALCULUS Chap. VI. Therefore J du du The derivative of logo u is found in a similar way. Example 1. y = ]n (sec'' a;). J d sec^ X 2 sec X (sec x tan x dx) „ ^ , dy = ^— — ^^ = 2 tan x dx. sec' x sec^ x Ex. 2. y = 2t='°"'^ 2""'"'Mn2cia; dy = 2'»°"'»' In 2 d (tan-i x) = 1 + x^ EXERCISES - dy 1 '- 2. y = a*^2^, g = 2 a*»°^^ln a sec^^ 2 x. x-l x-l e* -e" 4. 2/ = V + e-^' 5. 2/ = I" + re% ^ = »ia;"-i + n^ln n. 6. 2/ = a^2«, ^ = a^x»-i(a + a;lna). 7. 2/ = In (3 a;2 + 5 X + 1), dx = a-; 6l + 5 dz 3i2 + 5a; + i 8. y = ln sec^ x, -y- = "2 tan x. 9. 2/ = In (x + Vx2 — a?), dx dy 1 10. y = \n (sec ax + tan ax), -?- = a sec ax. ox 11.2/ = ln(a^ + ^.), g, a-lna + 5^1n. . u?/ COS X 12. 3/ = In sin X + i cos" x, -;^ = —. ax sin X ,n 1 , . X 1 cosx dy 1 13. 2/ = o *° t^'^ 2 2 2sin2x dx sin^x Chap. VI. TRANSCENDENTAL FUNCTIONS 61 J . ^ 1 , _^ 1_ dy ^ 8 "• ^ 4*^3:2-4 a;2-4' dx a; (a;^ - 4)2' ,.1, X dy 1 15. 2/ = - m - — — " a + Va^ — x2 dx ^j Va^ ■ X'' 16. 2/=ln(V^T3 + Vx + 2)+V(x + 3)(x + 2), J = V|-^- 17. 2/ = In (VxTa + V^), !^^ = ^ "^ 2 Vx2 + ax 18. 2/ =xtan-i--^ln(x'i+a^), ^ = tan-i-- a z ax a 19. y = ef^ (sin ax — cos ax), j^ = 2 ae<" sin ax. 20. 2/ = i tan* X— J tan* x — In cos x, -3^ — tan^ x. 21.x = aln,2. = i(.+ i} i=i(-J)- 22. X = e< + e-'. , 2/ = «' -e-', dx' y^' 23. 1, 2,=-lnx, S=^(^'--^)- 24. y = xe', i^.-i- + n)e^. 26. By taking logarithms of both sides of the equation y = x" and differentiating, show that the formula —- x" = nx"-i dx is true even when n is irrational. 26. Find the slope of the catenary y = lW + e"^l at X = 0. 27. Find the points on the curve y = e"^ sin x where the tangent is parallel to the x-axis. 28. If 2/ = Ae"^ + Be-^, where A and B are constant, show that 29. If 2/ = ze~'"', where z is any function of x, show that dx^" ^ •" dx + '^ ^ ^ dx' 30. For what values of x does 2/ = 51n(x -2) +31n(x + 2) +4a; increase as x increases? 62 DIFFERENTIAL CALCULUS Chap. VI. Sl.^From equation (44) show that e = lim (1 + »)»• x=o 32. If the space described by a point is s = ae' + fee"*, show that the acceleration is equal to the space passed over. 33. Assumfng 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 _ g ." eM + e-^ N fc2 ™\, 2 / in a time t. Show that the velocity and acceleration satisfy the equa- tion kV 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{xuyi). It is shown in analytic geometry that a line through (xi, yi) with slope rui is represented by the equation y — yi = mi(x - xi). This equation then rep- resents the tangent at (xi, 2/1) where the slope of the curve is mi. The line PiN perpen- dicular to the tangent at its point of contact is caUed the normal to the curve at Pi. Since the slope of the tangent is mi, the slope of a perpendicular line is and so is the equation of the normal at (xi, yi). Example 1. Find the equations of the tangent and normal to the ellipse x^ + 2y^ = 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 = -|. The equation of the tangent is 2/ - 2 = -i (a; - 1), 63 Fig. 49. 64 DIFFERENTIAL CALCULUS Chap. VII. and the equation. of the normal is 2/ - 2 = 4 (a; !)• Ex. 2. Find the equation of the tangent to x^ — y^ = a^ at the point {xi, yi). Xi The slope at (xi, j/i) is is then 2/1 The equation of the tangent 2/ - 2/1 = - (a; - a;i) which reduces to ^\x - y-oj = x^ - y^. Since (xi, yi) is on the curve, x^ — y^ = a!'. The equation of the tangent can therefore be reduced to the form Xix - yiy = a^. 50. Angle between Two Curves Fig. 60a. wia satisfies the equation 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 ma be the slopes of two curves at a point of intersection. It is shown in analytic geometry that the angle S from a line with slope mi to one with slope tan/3 = 1 + miTTh (50) This equation thus gives the angle /S from a curve with slope mi to one with slope rrh, the angle being considered positive when measured in the counter-clockwise direction. Chap. VII. GEOMETRICAL APPLICATIONS 65 Example. Find the angles determined by the Une y = x and the parabola y = x^. Solving the equations simultaneously, we find that the fine and parabola intersect at (1, 1) and (0, 0). The slope \ y of the line is 1. The slope at any point of the parabola is ^ = 2x. 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. 50b. tan |3i 1 3' whence 1 + 2 l3i = tan-ii = 18°26'. At (0, 0) the slope of the parabola is and so the angle from the fine to the parabola is given by the equation 0-1 whence tan j32 = ft 1+0 : -45' = -1, 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 x^ + y^ ='5 at (-1, 2). 2." The hyperbola xy = i&t (1, 4). 3. The parabola y' = ax a.t x = a. 4. The exponential curve y = aVa,tx = 0. 5. The sine curve 2/ = 3 sin a; at a; = ^■ x^ V^ 6. The ellipse -j + fe = 1, at (xi, yi). 66 DIFFERENTIAL CALCULUS Chap. VII. 7. The hyperbola x' + xy - y" - 2 x, a,t (2,0). 8. The semicubical parabola y' = x', at ( — 8, 4). 9. Find the equation of the normal to the cycloid x = a ( — sin — i' Show that it passes through the point where the rolling circle touches the z-axis. Find the angles at which the following pairs of curves intersect: 10. 3/* = 4 a;, x' = iy. IZ. y = sin x, y = cos x. 11. x^ + y' = 9, x^ + y^— 6 X = 9. 14. y = logw x, y = hix. 12. x'' + y^ + 2x = 7, 2/2 = 4 x. 15. y = i (e^ + e^), 2/ = 2 e*. 16. Show that for all values of the constants a and 6 the curves x^ — y^ = o'', xy = V intersect at right angles. 17. Show that the curves y = e°*, ^ = e"^ sin (bx + 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 N. Show that the projection of PN on the x-axis has a constant length. 20. The focus F of the parabola j/^ = ax is the point (i 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 are the points F' ( - Va^ - 6^, o) and F (Va^ - b\ O) . Show that the tangent at any point P of the ellipse makes equal angles with FP and F'P. I X _^x\ 22. Let P be any point on the catenary 2/ = sU^ + e "), Af the projection of P on the x-axis, and N the projection of M on the tangent at P. Show that MN is constant in length. 23. Show that the portion of the tangent to the tractrix " 2 \a- vV - xV intercepted between the y-axis and the point of tangency is constant in length. Chap. VII. GEOMETRICAL APPLICATIONS 67 24. Show that the angle between the tangent at any point P and the Une joining P to the origin is the same at all points of the curve In Vx' + y' = k tan-'-- " X 25. A point at a constant distance along the normal from a given curve generates a curve which is called parallel to the first. Fiad 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 (t>, y = b sin 0, show that sin x > a; — ^ ■* Let y = smx- x + —^- We are to show that y > 0. Differentiation gives -p- = cos a; — 1 + pT, , -r%= —sin a; + a;. dx 2! dx^ When X is positive, sin x is less than x and so -r^ is positive. Therefore -rr increases with x. Since -^ is zero when x is da; dx zero, -p is then positive when a; > 0, and so y increases with X. Since y = when x = 0, y is therefore positive when a; > 0, which was to be proved. * If n is any positive integer ra! represents the pi:oduct of the integers from 1 to ra. Thus 3! = l-2-3 = 6. 70 DIFFERENTIAL CALCULUS Chap. VII. EXERCISES Examine the following curves for direction of curvature and poiats of inflection: 1. y =a^-3x + 3. 2. 2/- = 2x' -3x2 -6z + l. 3. y = x* - 4 x3 + 6 1^ + 12 1. 4. 2/3 = X - 1. 5. y = xe 6. 2/ = 6-^. 7. x^i/ - 4 X + 3 2/ = 0. 8. X = sin t, ^ = 1 sin 3 <. Prove the following inequalities: 9. xhii -x-^ + l >0, ifO 0, ifx '>0. 11. e- -^ , if - -2<^<2- 13. According to Van der Waal's equation, the pressure p and volume u of a gas at constant temperature T are connected by the equa- tion RT a ^ m(v-b) v" a, b, m, and R 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 J, _ Sam 53. Length of a Curve. — The length of an arc PQ of a curve is defined as the limit (if there is a limit) approached by the length of a broken line with vertices on PQ as the number of its sides increases indefinitely, their lengths ap- proaching zero. We shaU 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 + proj. AB + proj. BQ. Chap. VII. 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 largest 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 line with any number of sides. As the number of sides in- creases indefinitely, the expression in parenthesis approaches the length of the arc PQ. Therefore PQ = arc PQ cos a, that is, chord PQ If the slope of the curve is continuous, the angle a ap- proaches zero as Q approaches P- Hence cos a approaches land chord PQ , hm 57P =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 jqTT^ = 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. FiQ. 54a. Fig. 54b. If P moves to a neighboring position Q, the increments in X, y, and s are Aa; = PR, Ay = RQ, As = arc PQ. From the figure it is seen that rr>T>r,\ Ao; Ax As cos (RPQ) = _ = -—, sin (RPO) - ^ - ^ Ai . As Q approaches P, RPQ approaches and As _ arc PQ PQ ~ chord PQ approaches 1. The above equations then give in the Hmit cos^ = dx ds' • , dy as (54a) Chap. VII. GEOMETRICAL APPLICATIONS 73 These equations express that dx and dy are the sides of a right triangle with hypotenuse ds extending along the tangent (Fig. 54b). All the equations connecting dx, dy, ds, and 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 tangents at its ends. The ratio . j3 _ _ Atj} arc PP' ~ As ~ As is the average bending per unit length along PP'. The limit as P' approaches P, A _d4 ds' is called the curvature at P. It is greater where the curve bends more sharply, less where it is more nearly straight. lim . As=o As Fig. 55a. In case of a circle (Fig. 55b) Fig. 55b. = B + Consequently, 2' s = ad. d _ dd 1 ds ado a ' 74 DIFFERENTIAL CALCULUS Chap. VII. that is, the curvature of a circle is constant and equal to the reciprocal of its radius. 56. Radius of Curvature. — We have just 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 = yr- (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 <^ = tan-|. Consequently, ^^dx 1 ldy\ dx^ Also ds = Vdx^ + dy^. Substituting these values for ds and d, we get " cPy dx^ If the radical in the numerator is taken positive, p will have 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 dy' (56b) Chap. VII. GEOMETRICAL APPLICATIONS 75 Example 1. Find the radius of curvature of the parabola y^ = Ax at the point (4, 4). At the point (4, 4) the derivatives have the values dy 2-1 dx y 2' Therefore U + &)1 \dx/ _ dx^ y^ f / nl (1+4-) 1 16 -10^ d^y dx^ 1 16 The negative sign shows that the curve is concave downward. The length of the radius is 10 Vs. Ex. 2. Find the radius of curvature of the curve repre- sented by the polar equation r = a cos d. The expressions for x and y in terms of 6 are X = r cos 6 = a cos d cos 6 y = r sin e = a cos 6 sin 6. Consequently, dy _a (cos^ d — sin^ 6) _ a cos 2 6 dx —2a cos dsinO — a sin 2 S a cos'' 6, = -cot 2 6, dx 2 csc^ 2ddd ' a sin 2 Odd esc' 2 9. a [l-|-cot2 29]^ '' " 2 - - esc' 2 e a = — a (csc^ 2e)i 2 csc^ 2 e a '2 The radius is thus constant. The curve is in fact a circle. 57. 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. same slope -r- at P. Since they have the same radius of curvature, the second derivatives will also be equal at P. T Fig. 57. The circle of curvature is thus the circle through P such that -rr and to have the same values for the circle as for the curve ax ax' at P. EXERCISES 1. The length pf 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: at (0, b). 5. r = e'. 3 1+^-1 ^- o^ + 62 ~ ^' at9=2- 4. x^ + xy + y' = 3, at (1, 1). 6. r = a (1 + cos 6), at S = 0. Find an expression for the radius of ciuvature at any point of each of the following curves: 7. y = l[^ + e ^)- 9. X = ly^ — ilny. 8. X = In sec y. 10. r = a sec^ \ 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-axia generates the cycloid X = a { — sm), y — a (1 — cos ), a being the radius of the rolling circle and ij> 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 6 -{- aB sin 8, y = asmB — aS cos 8, a being the radius of the circle and d the angle subtended at the center by the arc unwound. Show that the center of curvature 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 hjrpo- cycloid x^ + y^ = o' 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 a; = 2 sin" ^ ■ 2 sm^ ^ I sm - z 1 — COS X 2.x = = sm 5 X X 2 X \ 2 / . X sm- As X approaches zero, approaches 1. Therefore ,. 1 — cosa; hm = 0-1 = 0. 1=0 X 59. Derivatives of Arc in Polar Coordinates. — The angle from the outward drawn radius to the tangent drawn in the direction of increasing s is usually represented by the letter ^. 78 DIFFERENTIAL CALCULUS Chap. VII. Let r, B be the polar coordinates of P, and r -\- Lr,d ■\- M those of Q (Fig. 59a). Draw QR perpendicular to PR and let As = arc PQ. Then RQ _ (r+ Ar) sin ^e _.. . . sin Ag A9 _As PQ~ PQ -^^+^^) AO ' As' PQ- PR _ (r + At-) cos A9 - r PQ~ sin (/2FQ) = cos (,RPQ) = Ar = cos (Ae) p^ - r (1 — cos Afl) PQ /• (1 — cos Ad) Ag As ^°'(^^)ii-^- - A0 As PQ Fig. 59a. Fig. 59b. As Ad approaches zero, sin A& lim(i2FQ)=^, lim- A6I , ,. l-cosA0 As , = 1, hm ^g ^0, limp^=l. The above equations then give in the limit, rdd . dr sm^ = ds COSl^ ds (59a) These equations show that dr and rcW are the sides of a right triangle with hypotenuse ds and base angle ^. From this triangle all the equations connecting dr, dd, ds, and \l/ can be obtained. The most important of these are tan^ = ^. d^ = dr^ + r^ dS^. (59b) Chap. VII. GEOMETRICAL APPLICATIONS 79 Example. The logarithmic spiral r = ae^- In this case, dr = aeP dd and so tan yL = — ;— = 1. dr The angle ^ is therefore constant and equal to 45 degrees. The equation , dr 1 ''''l' = ds = V=2 shows that -j- is also constant and so r and s increase propor- tionally. EXERCISES Find the angle i^ at the point indicated on each of the following curves: 1. The spiral r = ad, a,t 6 = ^^ o 2. The circle r = asinflate = -:■ 4 3. The straight line r = a sec '9, at 9 = ^• 4. The ellipse r (2 - cos 6) =k,a.td = |- 6. The lemniscate r'' = 2 a? cos 2 9, at fl = | ir. 6. Show that the curves r = oe», r = ae~« are perpendicular at each of their points of intersection. 7. FLad the angles at which the curves r = a cos B, r = a sin 2 9 intersect. 8. Find the points on the cardioid r = a (1 — cos 6) where the tan- gent is parallel to the initial line. 9. Let P (r, 8) be a point on the hyperbola r-^ sin 2 9 = c. Show that the triangle formed by the radius OP, the tangent at P, and the a;-axis is isosceles. 10. Find the slope of the curve r = e^* at the point where 9=2" 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. i^&iC- M Fig. 60. Fig. 61. 61. Direction Cosines. — It is shown in analytic geome- try* that the angles a, fi, y between the coordinate axes and the line P1P2 (directed from Pi to P2) satisfy the equations cos a = 0^2 — Xl cos /3 = -yi cos 7 22 — Zi (61a) P1P2 ' ^""'' PlP2 ' """ ' P1P2 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 /3i, cos 71 and cos a^, cos j32, cos 72, the angle 6 between the lines is given by the equation cos 6 = cos ai cos a.2 + cos 182 cos ^2 + cos 71 cos 72. (61c) In particular, if the lines are perpendicular, the angle B is 90 degrees and = cos ai cos a2 + cos |8i cos 1S2 + cos 7i cos 72. (61d) * 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 j8, 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 + Aaj, y + A?/, s + Lz, those of Q, the direction cosines of PQare Lx 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 7^7; = am -r— 7^7, • Q=P PQ As PQ On the curve, x, y, z are functions of s. Hence ,. Ax dx ,. As ,. arc , ^ lim T- — -r ■ hm r-^ = hm -j 7 = 1.* As ds ■ PQ chord Therefore Similarly, cos a = -7- • (62a) a ^y d^ fan \ 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 was given ia 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 parallelepiped is equal to the sum of the squares of its three edges, ds2 = dx^ + dif + dz\ (62b) Fig. 62b. Example. Find the direction cosines of the tangent to the parabola X = at, y = bt, z = \ cf at the point where t = 2. At t = 2 the differentials are dx = a dt, dy = b dt, dz = \ctdt = c dt, ds = ± Vdx^ + dy^ + dz^ = ± Va^ + ¥ + 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 dy ds Va^ + ¥ + c^' ds Va^ + fe^ + c"' ds c ds~ Vcf+W+~^ 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, Z\) with direction cosines propor- tional to A, B,C are x-^i _ y-yi _ z-zi ,„ . The direction cosines of the tangent hne 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 = t^, z = i^ at the point where t = 1. The point of tangency is i = 1, a;i = 1, 2/1 = 1, Zi = 1. At this point the differentials are dx : dy : dz = dt ■.2tdt -.Sf dt = 1 :2 -.3. The equations of the tangent line are then X — 1 _ y — 1 _ g — 1 1 2 ~~3 Ex. 2. Find the angle between the curve Sx + 2y — 2z = 3, 4 x^ + j/2 = 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 S dx + 2 dy - 2 dz = 0, 8xdx + 2y dy ^ izdz. At the point of intersection these equations become 3dx + 2dy -2dz = 0, 8dx + 4:dy = 8dz. Solving for dx and dy in terms of dz, we get dx = 2 dz, dy = —2 dz. Consequently, ds = Vdx^ + */ + dz^ = S dz and dx 2 „ dy —2 dz 1 '=°^" = d-s=3' '°''^ = d^ = -3-' '^°'^ = ds = 3- 84 DIFFERENTIAL CALCULUS Chap. VII. The line joining the origin and (1, 2, 2) has direction cosines equal to 12 2 31 31 3- The angle d between the line and curve satisfies the equation . 2-4+2 - cos 6 = ^ = 0. The line and curve intersect at right angles. EXERCISES Find the equations of the tangent hues to the following curves at the points indicated: TT 1. X = sec /, y = tan t, z = at, a,tt = -• 2. X = e', 2/ = e-', z = P, at « = 1. 3. X = e' sin (, y = e' cos t, z = kt, at i = ^• 4. On the circle a; = ocos9, ^ = acos(e + o'^)i z = acos(9 + ^7rJ show that As is proportional to dS. 6. Find the angle at which the heUx X — a cos e, y = aeai.6, z = kO cuts the generators of the cylinder x' + y^ = a^ on which it lies. 6. Find the angle at which the conical heUx I = t cos t, y = tsint, 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 3? + y^ + ^ = liby the planes x — y + z = and x + y — z = 2. CHAPTER VIII 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 curve AB, Fig. 64. Let s be the arc from a fixed point A to P The speed of the par- ticle is then ds '"=dt- (64) Fig. 64. Fig. 65a. 65. 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 inotion. * 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 hnes and have the 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 rim. 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 = 2dit. The speed of P is ^ = 2^=2 (500) 2 X = 2000 x ft./min. at at Its velocity is 2000 t 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 a;-component is FQ = Prcosm ■ , rfs . , ds dv dv The components are thus the rates of change of the coordinates. Since pya = PQ2 -I- QT\ Chap. VIII. VELOCITY AND ACCELERATION 87 the speed is expressed in terms of the components by the equation (dsY ^ (dxV (dyf [dtj [dt) '^[dtj FiQ. 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 PQ = PT cos a = -77 T- = -77 . dt ds dt Since PT? = PQ^ + QR^ + RT^ the speed and compo- nents are connected by the equation (dsV ^ /dx\ (dyV (dzV \dt) \dt) '^[dtj '^[dtj 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 a;-component a, y-component b, and 2-component c will be represented by the symbol [a, b, 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 a; = 500 « Vs and a vertical distance y = 500 <- 16.1 «2 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 7= [500 V3, 178] and the speed is j^ = V (500 V3)' + (178)2 = 384 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 d, y = rsind, z = k9, 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. Chap. VIII. VELOCITY AND ACCELERATION 89 The components of velocity are dx . „dd dy „dd dz , dd at dt dt dt dt dt d8 Since -37 is the angular velocity &> with which the screw is rotating, the velocity of the moving point is V = [—na sin 6, rco cos 9, kia] and its speed is ds — = y/rW sin2 d + r^oj^ cos^ 6 + fcW = co Vr^ + k^, which is constant. X^ m,vy ^^ Fig. 69a. Fig. 69b. 69. Composition of Velocities. — By the sum of two velocities F] and V2 is meant the velocity Vi + F2 whose components are obtained by adding corresponding compo- nents of Vi and V2. Similarly, the difference V2 — Vi is the velocity whose components are obtained by subtracting the components of Vi from the corresponding ones of Vi. Thus, if Fi = [ai, 61], Vi = [02, 62], Fi + Fa = [ai + 02, h + h], F2 — Fi = [02— ai, 62 — &i]. If Vi 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 Fg — Vi extends from the end of Vi to the end of F2. 90 DIFFERENTIAL CALCULUS Chap. VIII. By the product mV 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 mV are m times those of V. V 1 The quotient — can be considered as a product — V. Its components are obtained by dividing those of V by m. 70. Acceleration. — The acceleration of a particle mov- ing along a curved path is the rate of change of its velocity AV dV dt ' A = lim . . M=0 At In this equation AF is a vector AV and -jrr is obtained by divid- ing the components of AF by A*. Let the particle move from the point P where the velocity is V to an adjacent point P' where the velocity is 7 + AF. The components of velocity will change from dx dy It' dt to da; .dx di'^ di' dy, .dy dt'^ dt Consequently, Subtraction and division by At give ^^'b%^t] AV At A dx ^1 At ' At As At approaches zero, the last equation approaches , dV [d'x d^yl ,,„ , 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 its 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 e = AOP. The co- ordinates of P are a; = r cos 0, The velocity of P is ■ ^dd .del ds Since s = rd, -y. = v = r -j- . dt dt 'df ^"" dtj The velocity can therefore be written V = [— «;sin5, «;cose]. Since v is constant, the acceleration is r .dd . .del Replacing 37 by - , this reduces to COS0, sm0 = -[ — cosS, — sin9]. Now [—cos 9, —sin 9] is a vector of unit length directed 92 DIFFERENTIAL CALCULUS Chap. VIII. 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 shdes with its ends in the x- and j/-axes. If the end in the s-axis moves with constant speed v, find the velocity and speed of the middle point of the rod. 3. A wheel of radius o rotates about its center with angular speed u 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 Pi (xa, 2/2) move in such a way that xi = 1+ 2 <, 2/1 = 2 - 3 «2, % = 3 + 2<2, 2/2= -4<'. Find the two velocities and show that they are always parallel. 6. Two particles Pi ( xi, yi, zi) and Pi (xa, y^, Zj) move in such a way that xi = a cos e, yi = a cos (5 + 4 jt), Si = o cos {0 + % ir), X2 = o sin 9, 2/2 = o sin (9 + J tt), 32 = a sin (9 + f ir). Fiud 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 — \ gf'. Find its speed, velocity, and acceleration. 8. A particle moves along the parabola 3? = ay in such a way that ■yr is constant. Show that its acceleration is constant. at 9. When a wheel rolls along a straight line, a point on its circum- ference describes a cycloid with parametric equations X = a (<^ — sin 0), y = a {1 — cos ), a being the radius of the wheel and rotated. Find the speed, velocity, and acceleration of the moving point. Chap. VIII. VELOCITY AND ACCELERATION 93 10. Find the acceleration of a particle moving with constant speed v along the cardioid r = o (1 — cos 9). 11. If a string is held taut while it is unwound from a fixed circle, its end describes the curve X — a cos 9 -{■ a 8 Bin d, y = a sin d — a ff cos 9, 9 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 if the straight part of the string rotated with angular velocity -w 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 / = U, 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 t = ii ii 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 w about A, and BC rotates with angular speed 2 a 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 = ii 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 oua. 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 aun, the rota- tion of the hoop about its center with angular speed wi, and the crawl of the insect advancing its radius with angular speed £02. Find the three velocities and accelerations which result if at the time t = h 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. — If /' ix) 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 =f(x). 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 Fig. 71a. distance from the a;-axis. horizontal and so dy dx That this theorem may not hold if /' {x) is discontinuous is shown in Figs. 71b and 71c. In both cases the curve \t such a point the tangent is = f (x) = 0. Fig. 71b. Fig. 71c. cro.sses the x-axis at a and b but there is no intermediate point where the slope is zero. 94 Chap. IX. ROLLE'S THEOREM 95 Example. Show that the equation a;S + 3a;-6 = cannot have more than one real root. Let f{x)=x' + 3x-&. Then fix) =3x' + S = 3(x2+l). 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 ^, -, O-oo, 00-00, r, 0°, 00° 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). In such cases this limit is usually taken as the value of the function at x = a. For example, when x = the function 2x0 X ~0" It is evident, however, that lim — = lim (2) = 2. x=0 X This example shows that an indeterminate form can often be made definite by an algebraic change of form. 73. The Forms -r and — . — We shall now show that, if U 00 f (x) for a particular value of the variable a fraction . , assumes the form t; 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 b. Jff W = 0, F (a) = 0, and F (b) is not zero, there is a number Xi between a and b such thai f(b)_f(x^) Fib) F'{x,) To show this let |^ = R. Then t (p) (73a) / (6) - RF (6) = 0. Consider the function / ix) - RF (a;). This function vanishes when x = h. Since / (a) = 0, F (a) = 0, it also vanishes when x = a. By Rolle's Theorem there is then a value Xi between a and b such that /' (xi) - RF' (xi) = 0. Consequently, fib) jf^rM F{b) "- F'ixO' which was to be proved. 2. Let/' (x) and F' (x) be continuous near a. Iff (a) = and F (a) = 0, then For, if we replace b by x, (73a) becomes fix) ^f'ixQ Fix) F'ix,)' xi being between a and x. Since xi approaches a as x ap- proaches a, iS F ix) '^a F' (Xi) Ta F' (X) 3. In the neighborhood of x = a, let /' (x) and F' (x) be Chap. IX. ROLLE'S THEOREM 97 continuous at all points except x = a. If J {x) and F (x) approach infinity as x approaches a, ,• /(^) ,■ fix) ^^a F (X) x=a F' (X) To show this let c be near a and on the same side as x. Since / (x) — / (c) and F {x) — F (c) are zero when x = c, by Theorem 1, 1 He) IM = /W-/(c) = /M /W F' (xi) F (x) - F (c) i^ (x) F_(^ ' F(x) where Xi is between x and c. As x approaches a, / (x) and F (x) increase indefinitely. The quantities / (c)// (x) -and F {c)/F (x) approach zero. The right side of this equation therefore approaches x=a F (x) 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 ™ F' (x) Since the two sides are always equal, we therefore conclude that limm = lim^^- F (x) .r. F' (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 ,. sinx ,. cosx lim = lim— — = 1. Ex. 2. Find the value of Um-^ r^- • xi^ (x — x)2 98 DIFFERENTIAL CALCULUS Chap. IX. When X = IT the numerator and denominator are both zero. Hence ,. l + cosx ,. (—sin a;) Iml 7 TV = Iim -^^r-j r = -• i=, (tt - xy x=T —2 (tt — X) Since this is indeterminate we apply the method a second time and so obtain ,. sin a; ,. cos a; 1 X=T 2 (TT — X) x=„ —2 2, The value required is therefore \. Ex. 3. Find the value approached by — as x ap- tan X proaches ^- TT When X approaches -^ the numerator and denominator of this fraction approach oo. Therefore, by Theorem 3, Iim tan 3 a; ,. 3sec^3a; ,. 3cos^a; TT —. = lim 5 = hm — 5-;r— • a:=2 tan X sec^ x cos^ 3 x When X is replaced by ^ the last expression takes the form 7: • Therefore ,. 3 cos^ a; ,. 6 cos x sin x hm — TTTT- = hm cos^ 3 a; 6 cos 3 x sin 3 x eos^ X — sin^ x 1 lim 3 (cos2 3 X - sin2 3 x) 3 74. The Forms • 00 and 00 — 00. — By transforming e expression t< For example, the expression to a fraction it will take the form tt or — 00 xlnx Chap. IX. ROLLE'S THEOREM 99 has the form • oo when a; = 0. It can, however, be written , In a; a; In a; = -j— i X which has the form — ■ 00 The expression sec X — tan x IT has the form oo — 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 x = -• 75. The Forms 0°, 1"=^, oo°. — The logarithm of the given function has the form • oo. From the limit of the log- arithm the limit of the function can be determined. 1 Example. Find the limit of (1 + xY as x approaches zero. 1 Let y = {1+ xf- Then , 1 1 /I I N In (1 + x) In 2/ =-ln(l + a;) = '- When X is zero this last expression becomes ^ . Therefore ,. ln(l + a;) ,. 1 bm — ^^ '- = hm = 1. 1=0 X 1 -|- a; The limit of lay being 1, the hmit 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. Determine the values of the following limits: „ T • ^' ~ 1 {x)=f(x)-f{a)-(x-a)f'(a) _ jx - ay .^s _ . . . _ (X - a)"-i ■ 2! ■^ ^''^ (n- 1)! ^ * ^■ It is easily seen that (a) = 0, 0' (a) = 0, (t)" (a) = 0, . . . r-' (a) = 0, r (x) = /" (x). When X = a the function 4>{x) (x — a)" therefore assumes the form zz. By Art. 73 there is then a ■^ value 3i between a and x such that 0(a;) _ 4,'(zi) (x — a)" 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 <^^(zi) _ ct>" (z,) n (zi — a)"-i 71 (n — 1) (Z2 — a)""^ A continuation of this argument gives finally (t>(x) ^ («)+---+^"/"(^0. which was to be proved. Example. Prove (x - ly {x - ly {x - ly 2 "^ 3 4x1* ■ In X = (x - 1) - ^ , , ^ ^ , where xi is between 1 and x. When X = 1 the values of In x and its derivatives are /(x)=ln(x), /(i) = o, /' (X) = l /' (1) = 1, f"(-)--h' /"(!) = -1, r{x) = l, /'" (1) = 2, r'w = -|. ^""^-) = -(xV Taking a = 1, Taylor's Theorem gives In X ='0 + 1 (X - 1) - i (X - 1)^ + ? (X - 1)3 - ^ ^^^*. which is the result required. 78. Approximate Values of Functions. — The last term in Taylor's formula 104 DIFFERENTIAL. CALCULUS Chap. X. is called the remainder. If this is small, an approximate value of the function is /W -/(o) + (l-a)f(o) the error in the approximation being equal to the remainder. To compute/ (a;) by this formula, we must know the values of/ (a), /' (o), etc. We must then assign a value to a such that /(«). /' («)i ete., are knovm. Furthermore, a should be as close as possible to the value x at which / (x) is wanted. For, the smaller x — a, the fewer terms {x — a)^, (x — a)', 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 = -7 • 4 /(a;)=tana;, ^(l)'" ■^' f'{x)=eec^x, •^'(i)=2. /" (x) = 2 sec2 x tan x, f" (^ = 4, /'" (x) = 2 sec* x + 4: sec^ x tan^ x. Using these values in Taylor's formula, we get and tan46° = l + 2(^) + 2(^)^=1.0355 approximately. Since xi is between 45° and 46°, /'" (x{) does not differ much from /'" (46°) = 8 + 8 = 16. Chap. X. SERIES AND APPROXIMATIONS 105 The error in the above approximation is thus very nearly f(lioJ<3|o)-3<4oko = 0-««««25. ' It is therefore correct to 4 decimals. Ex. 2. Find 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. / {x) = e-, /' (x) = e, f'{x)=e-, , /" {x) = e^, /(O) = 1, /' (0) = 1, /" (0) = 1, , /" (xi) = e\ By Taylor's Theorem, , x^ , x^ , , a;"-^ , a;"e^i Letting x = 1, this becomes 11 1 e'^ 6 = 1 + 1 + 21 + 3! + • • • +(;r^i)! + ^r In particular, if n = 2, e = 2 + i e^'. Since xi is between and 1, e is then between 2^ and 2 + J e, and therefore between 2^ and 4. To get a better ap- proximation let n = 9. Then e = l + l + ^ + ^+--- +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. ln(T%). 3. cot 43°. V. Ve. 4. tan 58°. 8. tan"' (^j). 9. Given In 3 = 1.0986, In 5 = 1.6094, find In 17. 106 DIFFERENTIAL CALCULUS Chap. X. 79. Taylor's and Maclaurin's Series. — As n increases indefinitely, the remainder in Taylor's formula n often approaches zero. In that case /(x)=lim[/(a) + (a;-a)/'(a)+ • • • + ''"^ /""Ha)]- This is usually written f{x)=f (a) + {x- a)f' (a) + ^^—^ f" (a) ■ , (^ - a)l^„, (^) + . . . ^ 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 f{x)=fiO)+xf'{0)+^f"{0)+f^f"iO)+ • • ■ . This is called Maclaurin's Series. Example. Show that cos x is represented by the series x'' , x^ x^ , cosx=l-2| + 4;-gj+ • • • . 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, f (0) = 1, f'{x) = -smx, /'(0)=0, f"ix) = -cosx, /"(0) = -l, /'" {x) = sm x, f" (0) = 0, /"" (x) = cos X, f"" (0) = 1. These values give cosa;=l-2] + 4j- ■ • ■ ±-,/"(xi). Chap. X. SERIES AND APPROXIMATIONS 107 The nth. derivative of cos x is =tcos x or ±sin x, depending on whether n is even or odd. Since sin x and cos x are never greater than 1, /„ (xi) is not greater than 1. Furthermore /viTC I* 'y* '>• f n\^l'2'3' ' ' 'n can be made as small as you please by taking n sufficiently large. Hence the remainder approaches zero and so cosa; = l-2; + 4|-g;+ • • • . which was to be proved. EXERCISES 1. „ 1 1 / 7r\ 1 / ttV , 1 ( irV , 3. 2^ = l+xln2 + (^^ + Mi2)!+. .. ryl '1-3 /r5 T'G X T' ,. ,, 2a;3 4s* 4x5, 8 a;' 5. e-cosx = l+x-^-^--gj- + -^+ ... 6. (o + x)" = a" + na"-'x + ^'^gT^ '*''"^^' + • • • , if |x|*< |o|. 7. V-x = 2 + — i-^+i^j^ ,if|x-4| |6 — a\. 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 ^2 /wo ^4 ln(l + x)=x-^ + ^-^+ • • • . 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 a; is at a distance greater than 1 from a = 0. The series in fact converges between a; = — 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 poljoiomials. 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 a:' x^ sin a; ^ ~ 6 ^ 120 ~ ' ' ' , a;^ 2 a;^ tana;= = z -j =a;+ir +-t?-+ ■ • • . cos a; 1 _ ^ I ^ _ 3 15j 2 "^ 24 EXERCISES 1. Show that and that the series converges when |a;| < 1. 2. By expanding cos 2 x, show that 1 — cos 2x _ a;2 „„ a;* , „, a? sin'' X = ■ 2 21-2' 4! + "' 61- 2 2! 4! ' 6! Prove that the series converges for all values of x. 112 DIFFERENTIAL CALCULUS Chap. X. 3. Show that (l+e^)2 = l+2e* + e=* = 4 + 4a; + 3a^+ix'+fx*+--- and that the series converges for all values of x. 4. Given / (x) = sin"' x, show that /' (X) = 1— = (1 - x^)-\. VI — x^ Expand this by the binomial theorem and determine /" {x), etc., by differentiating the result. Hence show that , 1 a^ , 1 .3 a? , 1 3 5x' and that the series converges when |a;| < 1. 6. By a method similar to that used in Ex. 4, show that *i^3 -lO ^7 U.n-^x = x-^+j-j+--- and that the series converges when \x\ < 1. 6. Prove cosx 2 24 For what values of x do you think the series converges? 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 =f{x, y), if u is determined when arbitrary values (or values arbitrary within certain limits) are assigned to x and y. For example, M = Vl - a;2 - 2/2 is a function of x and y. If u 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 + ^x,y + Ay) and so Au=fix + Ax,y + Ay) - f {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 g or fAx,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 „ ' _=2x + j/, -^-x-2y. Likewise, if m 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 d ldu\ dhi d ldu\ dhi tdu\ ^dhi d^ tdu\ _ dhb \dx) ~ dx^ ' dy \dx) ~ dy dx dx \dx/ dx^ dy \dxj dydx Similarly, d_ /du\ ^ d^u d^ /du\ ^ 6^ dx \dy/ dx dy ' dy \dy/ dy^ It can be shown that dhc _ 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 = x^y + xy^. — = 2 xy + y", — = x^ + 2 xy, af| = ^(^^ + 2^^)=2- + 2,,g=A(,. + 2.,)=2.. 90. Dependent Variables. — It often happens that some of the variables, are functions of others. F6r example, let u = x^ + y^ -\- f and let 3 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 — = 2 X -\- 2 z — dx dx Similarly, the partial derivative with respect.to y with x con- stant is P=2y + 2zp. dy " dy If, however, we consider z constant, the partial derivatives are du du -^ = 2x, — = 2 2/. 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 * For a proof see Wilson, Advanced Calculiis,^ 50. 116 DIFFERENTIAL CALCULUS Chap. XI. It will usually be clear from the context what independent variables u is considered a function of, Then -r— will repre- ox 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 (— j • From the triangle it is seen that P a = c sin A. Differentiating with A constant, we get da dc = sin A, 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 3 = a;2 + 2/2-2x-4j/ + 6. At the lowest point, the curves AB and CD (Fig. 91b) will have horizontal tangents. Hence f = 2a;-2 = 0, p-=2y-i = 0. dx dy Chap. XL PARTIAL DIFFERENTIATION 117 Consequently, x = 1, y = 2. These values substituted in the equation of the surface give 2 = 1. 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^ + i x + y 2. z = (.x+a}(.y + b), 4. u =\n {x^ -\-xy -\- y'^), x —- + y ■— = 2. ox dy du , du ""Tx-^^Vy- - u dz dz dx dy ~ ^' dz dz yYx^^'Ty- Z X 6. M = tan"' (!)■ du , du , du - dy ^"• dx dz dhl . dhl _ . dx' "*" dy^ ~ dx' "*" a?/8 "^ dz^ ~ 7. u = . =, Vx' + y' + z' In each of the following exercises verify that dhl _ d^u dx dy dy dx 18 DIFFERENTIAL CALCULUS Chap. XI, 8. u = ..y X 10. a = sin (x + y). 9. u = = hi(a;» + 2/'). 11. u = xyz. 12. Given v = Vs^ + j f + 8^ verify that a% an. dx dy dz dz dy dx Prove the following relations assuming that z is a function of x and y: 13. u = -ix + z) e''+^ du du dx'^ dy = {l+x- * ■'('+! +S)'" 14. u = -- xyz, \^Tx du\ ( dz az\ 16. u = = 6" + e« + dx If a; or 2/ is a function of t only, the partial derivative -r-. at du dx du or T7 is replaced by a total derivative -77 or -^ . If both x dt dt at and y are functions of i, w is a function of t with total deriva- tive du _du dx du dy (q(\w\ dt ~ dx dt'^ dy dt ' ^ ^ Likewise, if w is a function of three variables x, y, z, that depend on t, du _ du dx du dy du dz , , dt ~ dx 'di ^ dy ^ '^ 'dz Tt' *■ ^^ 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 dt 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- ties, — can be obtained by differentiating with respect to t as if only one of those quantities were variable at a time and adding the results. du Example 1. Given y = x", find -^- 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 oj") is X • cc^~i = a?. 126 DIFFERENTIAL CALCULUS Chap. XI. If the lower x is held constant while the upper varies, the derivative (as in case of a?') is iU'lnx. The actual derivative of y is then the sum -^ = x^ + a?' In a;. ax Ex. 2. Given u = f {x,y, z), y and z being functions of a;, ~ , du find ^-• 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 J- . If 2/ and z are replaced by constants, u will be a secopd function of x. Its derivative is -r- • dx 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 dx dx dz dx It is understood that / {x, y, z) is a definite function of x, y, z and that ^ is the derivative obtained with all the variables dx but X constant. 97. Change of Variable. — If w is a function of x and y we have said that the equation , 5m , , du I du == —- dx + ^- 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 , , du ,, du = -r- as + -rr dt. dS Of Since m is a function of x and j/" which are functions of s and t, by equation (96a), du _ du dx du dy du _ du dx du dy ds ~ dx ds dy ds ' Tt ^ dx 'di dy 'di\ Consequently, , /du dx , du dy\ , , /du dx , du dy\ ,. '^''=[di-d-s+d^-d-sr+[d^^+dimr du /dx , , dx ,\ . du /dy , , dy , A du , , du , = d-x[d-s'^ + -dt'^V + d^[ds^'+mV = d-x'^'' + d-y'^y' 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. / (x, y) = 0. In this case d-f{x,y)=^dx + ^dy = d-0 = 0.\ Consequently, df dy dx dx df dy Ex. 2. / (x, y, 2) = 0. Differentiation gives fdx+^dy+^dz = (i. 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. Then dy = and dl . dz _ dx dx~ ~W' dz Similarly, if x is constant, dx = and dl dz _ dy d^~ ~W dz Ex. 3. /i {x, y, z) = 0, U (x, y, z) = 0. We have two differential relations fdx + ^^dy+fdz = 0, dx dy dz fdx + ^^dy+fdz = 0. dx dy dz We could eliminate y from the two equations /i = 0, /z = 0. We should then obtain 3 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 dx' The result is dz dx dfidf2 dy dx dfidh dx dy dfidf2 dz dy dfidf, dy dz 99. Directional Derivative. — Let u = f {x, y). At each point P (x, y) in the aitz-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. The derivative of u with respect to s is du du dx , du dy du , , du . , ds dx ds dy ds dx dy This is called the derivative of u in the direction PQ. The partial derivatives — and — are special values of — which result when PQ is drawn in the direction of OX or OY. Similarly, if m = / (x, y, z), du du dx , dudy , du dz du , du „ , du •r- = T-j- + T--f + T-j- = ^COSa + -T-COS|8 + -r-COS7 ds dx ds dy ds ds as dx dy dz is the rate of change of u with 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. 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 a;^ + y^ in the direction = 45° at the point (1, 2). The result is ^(x^ + yi) =2x^+2y^ = 2xcos + 2ysm = 2~ + 4~= 3V2. V2 V2 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 = -T- dx + -r- dy. ax ay Since dx and dy are arbitrary. this requires P du Q- du ^ dy' Consequently, dP dhx dQ dhi dy dydx' dx ' dxdy Since the two second derivatives of u with respect to x and y are equal. An expression P dx + Qdy is 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 Caladus, § 92. Chap. XI. PARTIAL DIFFERENTIATION 131 Similarly, if Pdx + Qdy + Rdz is the differential of a function u of x, y, z, dPdQ dQ_dR dRdP ,.„„,. ^-'di' 'di-^' 'di'dz' ^^""^^ and conversely. Example 1. Show that (x2 + 2 xy) dx + (x^ + 2/2) 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 = TdS -pdv, U 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 /an ^ /dv\ \dp/s \dSjj, The result to be proved expresses that TdS + vdp 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 M = / (x, y), y == (t> (x), find ^• 2. If M = / (x, y, z), z = (x), find f ^j 132 DIFFERENTIAL CALCULUS Chap. XI. 3. If u = / (a;, y, z), z = (x, y), y = <{' (x), find ~ 4. and If « = / {x, y), y = (x, r), r = ^ (x, s), find [J^j , \J^j^' \ax), 5. If/ fe y,z)=Q,z = F {X, y), find^- 6. If F (a;, y, z) = 0, show that Sx ay Bz^ _ dy dz dx~ 7. If w = a;/ (z), z = -, show that x — + y — = u. 8. If u =/ (r, s), r = a;+), y = F (x', y', h, k, 4>). 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, 9 of a point in a plane. Show that they are identical with the derivatives with respect to s given in Arts. 54 and 59. 13. Find the derivative of x'' — y'' 'va. the direction ij> = 30° at the point (3, 4). 14. At a distance r in space the potential due to an electric charge e is y = - . Find its directional derivative, r 15. Show that the derivative of xy along the normal at any point of the curve 3? — y^ = c^ ia zero. Chap. XI. PARTIAL DIFFERENTIATION 133 16. Given u = f {x,y), show that if Si and sj are measured along perpendicular directions. Determine which of the following expressions are exact differentials: 17. y dx — X dy. 18. (2x + y)dx + {x -2y) dy. 19. exdx-\- eydy + (x + y) ezdz. 20. yz dx — xzdy + y'^dz. 21. Under the conditions of Ex. 2, page 131, show that \dT)p Kdph' {dTJv \av]T 22. In case of a perfect gas, pw = hT. Using this and the equation dU = TdS -p dv, show that 1^ = 0. dp Since U is always a function of p and T, this last equation expresses that J7 is a function of T only. 101. Direction of the Normal at a Poiat of a Surface. — Let the equation of a surface be F (x, y, z) = 0. Differentiation gives dF, .dF, ,dF, _ -T-dx +^-dy + -r-dz = 0. dx dy dz (101a) Let PN be the line through P (x, y, z) with direction cosines propor- tional to dF^SF^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. Fig. 101. 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 PN 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 f:f:f- (101) dx dy dz ^ ^ 102. Equations of the Normal at Pi (asi, yi, z^.' — 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-yi _ z - z, ..„„, ^4~~~B~~ C ■ ^^^^> 103. Equation of the Tangent Plane at Pi (xi, yi, «i). — 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 ii A, B, C are pro- portional ,to the direction cosines of the normal to a plane passing through (xi, yi, Zi), the equation of the plane is A(x-xO+B(y- yi) + C (z - z^) = 0* (103) * See Phillips, Analytic Geometry, Art. 68. Chap. XI. PARTIAL DIFFERENTIATION 135 It 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 a;2 + 2 2/2 + 3 z2 = 3 a; + 12. The equation given is equivalent to a;2 + 2 1/2 + 3 ^2 - 3 x - 12 = 0. The direction cosines of its normal are proportional to the partial derivatives 2x - 3 -.iy :6z. 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 - I _ y + 1 _ g - 2 1 ~ 4 ~ - 12 ■ The equation of the tangent plane is X - 1 + 4 (?/ + 1) - 12 (2 - 2) = 0. EXERCISES Find the equations of the normal and tangent plane to each of the following surfaces at the point indicated: 1. Sphere, t' + y^ + z'' = 9, at (1, 2, 2). 2. Cylinder, x^ + xy + y' = 7, at (2, -3, 3). 3. Cone, z' = x' + y', at (3, 4, 5). 4. Hyperbolic paraboloid, sj/ = 3 z — 4, at (5, 1, 3). 5. Elliptic paraboloid, x = 22/^ + 3 z^, at (5, 1, 1). 6. Find the locus of points on the cylinder {x + zy + {y-z)' = i where the normal is parallel to the iiz-plane. 7. Show that the normal at any point P (x, y, z) of the surface j^2 _]_ 22 = 4 X makes equal angles with the x-axis and the line joining PandA (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 m 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 Am < 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. rilj If -r- is continuous, by Art. 31, ox g=o. (104c) 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) ax ay 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 maximum rectangular parallele- piped with a given area of surface is a cube. Let X, y, z be the edges of the parallelopiped. If V 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 ^~2{x + y)' Therefore y ^xy{A-2 xy) 2{x + y) dV^f F A -2x^ -4 xy l dz 2L {x + yy J-"' dV ^^ V A -2y^ -Axy '\ dy~ 2l {x + yy J " "■ The values x = 0, y = cannot give maxima. Hence A - 2x^ - 4xy = 0, A - 2y^ - Axy = 0. Solving these equations simultaneously with A = 2 xy + 2 xz + 2 yz, we get x^y = z =V-g- 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 + 2y + iz= 14 nearest to the origin. The distance from any point {x, y, z) of the plane to the origin is D = Vx^ + y^ + z^. 138 DIFFERENTIAL CALCULUS Chap. XI. If this is a minimum , j^ ^ xdx + ydy + zdz ^ ^ Va;2 + 2/2 + 2^ that is, xdx + ydy + zdz = 0. (104e) From the equation of the plane we get dx + 2dy + Zdz = 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 D 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 1 ~2~3' Solving these simultaneously with the equation of the plane, we get X = 1, y = 2, z = 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 I^R. Two terminals are connected by three wires of resistances Ri, Ri, Ra 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 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. PARTLAi DIFFERENTIATION 139 of the squares of the distances from the points is least. Find the posi- tion of equilibrium. 5. 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 B on the second mirror, and is there reflected to S. If P and S are in the same horizontal pline find the positions of Q and B so that the path PQBS may be as short as possible. 7. A table has four legs attached to the top at the comers Ai, Ai, Ai, Ai of a square. A weight W placed upon the table at a point of the diagonal AiAs, two-thirds of the way from Ai to As, will cause the legs to shorten the amoxmts Si, Si, Sj, St, while the weight itself sinks a dis- tance h. The increase in potential energy due to the contraction of a leg is fcs^, where k is constant and s the contraction. The decrease in potential energy due to the sinking of the weight is Wh. The whole system will 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 Si, Si, Ss, St, SUPPLEMENTARY EXERCISES CHAPTER III Find the difierentials of the following functions: 2 ^_ 7 (2 a; + 1) (2 a; + 7)' ■ b^ax?+h (2a; + 5)» , 2 Vaa:' + hx „ (a + 2)' (a; + 4)' 6a; ■ (a; + l)na: + 3)«' ^ 2o3; + 6 _ _ (2x''-l) V^HH; Vax' + 6i + c ■ a^ . (ax + 6)'H^ fa(ai+&)'H-i ,„, ,5=1. 5- aHn + 2) ~ aUn + 1) ' ^0. a;(a;" + 7i) » Find -y in each of the following cases: 11. 2a;='-4xj/ + 3j/2 = 6x-42/ + 18. 12. x' + Zx'y = y\ 13. 2 = 32/'' + 22/'. 14. (x2 + 2/2)2 = 2o2(x'-2/''). 15. x = « + r^, y = 2t- ^ t-V " {t-iy 16. X = /:;--— ' y = Vl+ t2 ^ Vl - <2 17. X = < 02 + o2)* 2/ = < «2 + a2)*. 18. X = z' + 2z, 2 = 2/2 + 2^. 19. x" + 2^ = aS yz = V. 20. The volume elasticity of a &uid iae = —v-^. If a gas expands according to Boyle's law, pv = 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 = BT, pv" = C, B, n, and C being constants. Find ^ and ^- ^ 140 SUPPLEMENTARY EXERCISES 141 22. If « is the volume of a spherical segment of altitude h, show that yr is equal to the area of the circle forming the plane face of the segment. 23. If a polynomial equation / W = has two roots equal to r, / (x) has {x — r)^ as a factor, that is, / (x) = (x - ryf, (x), where /i (x) is a polynomial in x. Hence show that r is a root of /' W = 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. x3 - 3x2 + 4 = 0. 25. x= - x^ - 5 X - 3 = 0. 26. 4x'-8x'i-3x + 9 =0. 27. 4x4 - 12x3+ a;2 + i2x + 4 =0. Find '^ and j^ in each of the following eases. 28. y = x Va? - x\ 31. ax + by + c = 0. 3.2 32. X = 2 + 3(, 2/ = 4-5<. ^ ^^ 33. x = — — T y = 7:ri' 30. xj/ = a2. ' + 1 * + 1 34. If, = x^findgandg. 35. Given x^ — y' = 1, verify that dx^ ~ d!/2 \,dx/ ■ 36. If « is a positive integer, show that d" -r^ x" = constant. dx" 37. If M and v are functions of x, show that d^ , , d% , .d'u dv , -dhi cPv . . du dh) , d^ ^(w) = ^-« + 4^-^^ +6^.^,+4^.^ + «^. Compare this with the binomial expansion for (4t + w)*. 38. If / (x) = (x — rYfi (x), where /i (x) is a polsmomial, show that /'W=/"«=0. 142 DIFFERENTIAL CALCULUS CHAPTER IV 39. A particle moves along a straight line the distance s = it>-21P + 3&t + l 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 1' in t hours and the rear one 12 P, how fast will they be approaching or separating at the end of one hour? At the end of two hours? When wUl 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 / = 2. k 43. li tf' = -, 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 365j days in a year. 46. A wheel roUs.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. (rf cord out. If the kite moves horizontally at the rate of 5 nules an hour directly away from the person fls^ng it, how fast is the cord being paid out? 50. A particle moves along the parabola 100 y = 16a;2 in such a way that its abscissa changes at the rate of lO 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 a! 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 \^ X a a'^X when o is a constant. Show that the velocity ia 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 will 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 6, 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 liquid against the walls of a channel, the channel should be so designed that the area of wetted surface 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 channel, the banks making an angle B 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; assuming 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 minimiiTn . 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 ground. How far from the pole should the stake be driven to make the tension in the stay as small as possible? 67. The lower comer 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 hour 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 functions: sin X 76. sec^ x — tatf x. 70. 71. 72. X --..?> X . „ 77. sm'-secs" sm X A 1 — cos 9 78. tan: 1 + cos 9 ■ 1 — X sin 9 __ 2 tan x 73. sin ax cos ax. ' 1 — tan' x ^.^0 80. 5 sec' 9 - 7 setf 9. 74. cot K — CSC ;;• ., „ . 2 2 81. sec I esc I — 2 cot x. 75. tan 2x — cot 2 x. Differentiate both sides of each of the following equations and show that the resulting derivatives are equal. 82. sec' X + csc'a; = sec' x esc' x. 83. sin 2 X = 2 sin a; cos x. 84. sin 3 a; = 3 sins — 4 sin' X. 85. sin (x + o) = sin s cos a + cos x sin a. 86. sec'x = 1 + tan'x. 87. ainx + maa == 2mn^ (x + a) cos i (x — a). SUPPLEMENTARY EXERCISES 145 88. cos a — cos s = 2 sin 5 (a; + o) sin |(s — a). Find ■—- and -3-^ in each of the following cases: 89. X = a cos^ e, 90. X = a cos* e, 91. X = ta.ne — e, 92. X = sec2e, 93. X = sec e, y = a sin* fl. 2/ = a sitf 8. y = cos 9. y = tan* 9. 2/ = tan 8. 94. a; = esc 8 — cot 8, y = csc8 + cot 0. Differentiate the following functions: 95. sin-i 96. cos-' 97. tan-i 2 98. :7|tan- , 2x + l V3 ■ 99. cos-i 100. CSC-' 101. sec-i Vx^ + 1 V5 2a; -1' 109. e^. 110. V^. 111. (Vi)^ 112. 5thxt. 1 113. 7^. 114. a^lna;. 115. In sin" x. 116. hihi.-!;. U7. .{1=^. 102. a csc-i - + Va* - a;«- 103. ■ cot-^a;- l+a:^ 104. Vl — X sin-i .r - Vi- 105. sec->^i4 + siii"'^^- a; — 1 a; + 1 106. sm 1 r— ^ +a cos a; 10' 108. 118. ,- 1 2 cos-i X + 2 VT Vx* — a* — a sec" X*- itan-ii + ihi(a2 + x2)- 119. e-*' cos (a + 6/). 120. hi(a + Va' + x*). 121. (x + yin(x + y-x-L 122. ln^^ + ° + ^^^zJ. vx + a ~ V X — a 123. tan-i i (e^ + e-"^). 124. hi(Vx + ^^1^2). 125. (x + l)hi(x* + 2x + 5)+|tan-i-^^. 126. secixtanjx + ln (secix +tan Jx). 146 DIFFERENTIAL CALCULUS |(.+i)-M 127. xsec->^|a; + -j-In(a;2 + l) -z^[l^' + ')- 128. iln (HX^ + l)-|a; + tan-'|a;. 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,a.t{l,2). 130. x^-yi = 5, at (3, 2). 131. x' + j/^ = x + 3y,Sit{-l,l). 132.. I* + 2/* = 2, at (1, 1). 133. 2/ = hix, at (1, 0). 134. a^{x + y) = a" {x - y), at (0, 0). 135. X = 2 cos 9, 2/ = 3 sin 9, at 9 = 2 136. r = a (1 + cose), at 9 = 2" Find the angles at which the following pairs of curves intersect: 137. x^->ry'' = Sx, y^ (2 - x) = x'. 138. y'- = 2ax + a?, x^ = 2hy + ¥. 139. x' = Aay, (x^ + 4a2) i/ = Sa^. 140. 2/2 = 6x, x^ + y' = 16. 141. y = He' + €-"), y = i. 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 b the curves x' - 3 xi/' = a, 3 x^ -2/3 = 6, intersect at right angles. Examine each of the following curves for direction of curvature and points of inflection: i.ir 1 —X 146. y = tan x. 147. X = 6y' -22/3. SUPPLEMENTARY EXERCISES 147 148. x = 2«-i, 2/ = 2« + i- 149. Clausius's equation connecting the pressure, volume, and tem- perature of a gas is RT c ^ v-a T(.v + by' B, 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' (o), but /" (a) is not equal to F" (o), show that the curves are tangent and do not cross at x = a. Apply to the curves y = 7? and 2/ = s' at X = 0. 151. If two curves y = f {x), y = F (x) intersect at x = a, and f (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 curves y = x^ and y = x^ + (x — ly at x = 1. Find the radius of curvature on each of the following curves at the point indicated: 152. Parabola y^ = ax at its vertex. x^ i/^ 153. Ellipse -; + r: = 1 at its vertices. a-' 0' 154. Hyperbola ^' - |' = 1 at x = Va^ + ¥. 155. 2/ = In CSC x, at ( s , I • 156. x = Jsin2/ — iln (sec y + tan y), at any point (x, y). 157. X = a cos' 6, y = a sin' 6, at any point. 158. Find the center of curvature of j/ = In (x — 2) at (3, 0). Find the angle 2 -J^ 'y6 205. ln(e^ + 0=|-g + 5+ ■•■• 206. hi(l +sinx) =x - ix2+Ji= -t^eX<+ • • • . 207. e^secx = l+x + x2 + |xs + Jx<+ • ■ ■ . 208. In (x + VT+^') =x-i| + if|J+--.. 209. In 5^ = 2g + 3^ + ^ + ...]. X* 7x* 210. In tan x = In x + -5- + -^ + • • • . 211. e»-- = l+x+|-,-^ . 212. «*'°" = l+^+|+^' + ^+--- • Determine the values of x for which the following series converge: rt»2 1^ -M 213. 1+x + l+f + j + • • • . 214. (x-l) + (^V(^V^^+.... 215. 1 +2x + 3x2 + 4x'+ • • • . 01R » I a: + 2 (x + 2)' (x + 2)» 216. 2 + -Y-2-+^73-+-3TT-+' • 'Z -r— • dz SUPPLEMENTARY EXERCISES 151 CHAPTER XI In each of the following exercises show that the partial derivatives satisfy the equation given: «,n , , , du , du du 2n.u = xy + yV, x~ + z~=y-- 218. z = x« - 2 a;V + 2/*, ^ ^ + ^ E " °- / , M ^3m du\ du 219.^ = (x + 2/)lnxz, ^^_--j=^-. 220.. = (.+ i)tan-.(.-J), i+.^g=.^ 221. u = xy + -, ^'^ + 2..z— ^ + z^^ = 2/^-,. 222. z = hi (a:^ + t), ^^ + ^^ = °- 223. ,* = ^ + 5, y —z Shi Shi^ _L ^ = ^. dxdy dydz dz' dx^' Prove the following relations assuming that z iS'a function of x and y: 224. u = {x + y - zf, du du du dz du dz dy dx ~ dy dx dx dy 225. u = z+e^, du du _ dz _ dz ""to y'di~''dx y dy 226. u = z{x^ - if), du , 9m 't — hx -T- dx dy y"£+-'£ = (^-y'){y%+-'i) 227. If a; = ^ (e* + e"'), 2/ = | (e* - e''), show that ' Xdrje \dxjy 228. If xyz = o', show that /dy\ /dj\ /dA ^ \dx)i\dylx\dz]y 152 DIFFERENTIAL CALCULUS In each of the following exercises find As and its principal part, assuming that x and y are the independent variables. When Aa; and A^ 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 - Vx'' + y^. 230. 2 = a;2 -y^ + 2x. 231. ^ = ,-^- Find the total differentials of the following functions: 233. ax^ + bxV + cy*. 234. hi (a? + 2/2 +22). 235. a'tan-'^ -^Han-'-- X " y 236. yze' + zxe" + xye'. 237. If u = x"f (z), z = ^, show that 238. li u — f{r,s), r = x + y, s = x — y, show that aw aw _ - aw dxdy~ dr' 239. If M = / (r, s, 0, »• = -, s = ^, « = -, show that du , du , du . a;-:; — \-y t. r z r- = 0. dx " dy dz 240. If a is the angle between the a>axis and the line 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 — y sec x 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. x^ + 2y' -z'' = 16, at (3, 2, -1). 243. 22; + 32/ -42 = 4, at (1, 2, 1). 244. z' = 8x2/, at (2, 1, -4). 245. y = z^-x'' + l, at (3, 1, -3). 246. Show that the largest rectangular paraUelopiped with a given 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 largest body that can be constructed from a given amount of material. 249. Find the volume of the largest rectangular parallelepiped that can be inscribed in the eUipsoid. a2 "^ 62 "•" & 250. Show that the triangle of greatest area inscribed in a given circle is equilateral. 251. Find the point so situated that the sum of its distances from the vertices of an acute angled triangle is a minimum. 252. At the point (a;, y, z) of space find the direction along which a given function F (x, y, z) has the largest directional derivative. ANSWERS TO EXERCISES Page 8 1. -i 4. -1. 2. V2. 5. 1. 3. -1. 6. |. 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 = A, y = ^ and dy = 0.072 da;. 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 a; = 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 < s. decreases when x > -x- 37.x = ±^. ^^--(^r^riy^ 2 38. tan-i|. Page 31 , 2 _J: 2 - ' (x-iy (x-i)» ■ v^;rz^2' ^a'-x'^^ 3. {x-iy(.x + 2y(7x + 2), (a; -l)(x + 2)2(42x2 + 24 a;- 12). 4 2 _4. 7 l^Zjy _2_. y' y x-r {.x-iy X a' „ y* rii x_ _J_ 2y' At Sx^y 154 ANSWERS TO EXERCISES 155 1Q (ft/ _ 12 tPa; _ 12 di' <(2 + 3 0'' 2, or -2 < z < 1. Pages 66, 66 1. 2y -x = 5, y + 2x = G. 2. 3/ + 4a = 8, iy — x = \b. 3. 2 2/ =F a; = ±a, ^ ± 2 2 = ±3 a. ANSWERS TO EXERCISES 157 4. 2/ = a (x In 6 + 1), x -\- ay]nb = a?]nb. 5. y — 2 2 ^(--i)' ^-i+i^Ki)=«- 6. ^ + ^' = 1. fc'aJiy - a^ViX = (6^ - tf) ajij/i. 7. a + 2/ = 2, a; - 2/ = 2. 12. 90°. 8. a; + 3 2/ = 4, j/-3a; = 28. 13. tan-i2V2. 9. 2/ + a: tan hi = CKt>i tan ^ 0i. _ In 10 — 1 10. 90°, tan-'f. 1^. tan ij^^^-j-^- 11- 4^°- 15. tan-' 3 Vs. Page 70 1. Point of inflection (0, 3). Concave upward on the right of this point, downward on the left. 2. Point of inflection (J, — f). Concave upward on the right of thia point, downward on the left. 3. The curve is everywhere 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. 5. Point of inflection I —2, — ^)- Concave upward on the right, downward on the left. 6. Points of inflection at a; = ±— ;=. Concave downward between V2 the points of inflection, upward outside. 7. Points of inflection (0, 0), (±3, ±1). Concave upward when -3 < X < or a; > 3. 8. Point of inflection at the origin. Concave upward on the left of the origin. 1. ±2V6. 3 ^■ 4. 3V2. 5. e^ v'2' 8. fo. Pages 76,' n 7. a 8. secy. 9. (2/2 + 1)' iy 10. 2asec'=- 1158 DIFFERENTIAL CALCULUS Page 79 There are two angles 4/ depending on the direction in which s is measured along the curve. In the following answers only one of these angles is given. '■ --(I)- t r •-* 2 1. 7. 0°, 90°,andtaii-i3V3. ■ 4 8. fl= ±ijr. 3 Z. 10. 3. ^- 3 Page 84 1. I -V2 v-l ^"T ^- *^"' _ y — 1 4__ ■ k 6. tan '— =• X — e _ ^ ^ g — 1 V2 e 2/« 2 ■ 7. 69° 29'. 3. irfc '.-e^ y 2 Pages 92, 93 1. The angular speed is ^ ^ , where a; is the abscissa of the moving point. 2. If xi is the abscissa of the end in the a>axis and yi the ordinate of the end in the j/-axis, the velocity of the middle point is [*". '£]. 2yi_ the upper signs being used if the end in the avaxis moves to the right, the lower signs if it moves to the left. The speed is ^ — 2j/i 3. The velocity is [v — axis to the radius through the moving point. The speed is Vt^ + aW — 2 ooi!) sin 0. 6. The boat should be pointed 30° up the river. 7. Velocity = [a, h, c — gt], Acceleration = [0, 0, —g]. Speed = Vo" + 62 + (e - gt)^. ANSWERS TO EXERCISES 159 Velocity = [oa (1 — cos<^), ousin^], Speed = ao> Vz — 2 cos (^ = 2 aw sin J (^, Acceleration = [a«^ sin <^, aa'cos<^]. 10 r 3 1)' sin 1 9 St!" coaf q L 4a 6in|fl' 4ajSinJeJ 12. X = «J cos ait, y = vt sin aj<. The velocity is the sum of the par- tial velocities, but the acceleration is not. 13. X = a cos ait -\-b cos 2 at, j/ = a sin ui + 6 sin 2 at. The velocity is the sum of the partial velocities and the acceleration the sum of the partial accelerations. 14. X = (Mit — a sin (ui + uj) <, y = a cos (coj + aj) t. The velocity is the sum of the partial velocities and the acceleration the sum of the partial accelerations. Page 100 2. A. IS. i. 3. n. 19. 1. 4. 0. 20. -3. 5. e». 21. a. 6. 2. 22. 1 7. -2. TT 8. 1. 23. /' {x) dx. 9. 0. 24. 00. 10. •r'. 25. i. 11. 1. 26. 00. 12. 1. 27. 00. 13. 0. 28. 1. 14. -J. 29. 1. 15. 0. 30. a. 16. 0. 31. e" 17. -i. Page 106 1. 0.0872. 6. 0.1054. 2. 0.8480. 7. 1.6487. 3. 1.0724. 8. 0.0997. 4. 1.6003. 9. 2.833. 5. 1.0154. Page 118 21. (-2,1,0). 22. (1, 1, 2) 160 DIFFERENTIAL CALCULUS Pages 123, 124 1. Increment = —0.151, principal part = —0.154. 4. -ip=n{-j — — )■ Since dl and dg may be either positive or negative, the percentage error in T may be | the sum of the percentage errors in I and g. 5. The percentage error in g may be as great as that in s plus twice that in T. 13. -'^+^. 15. 2''-"^ zx — 2uv 14. -{x' + y' + xy-z'^). Pages 131, 132 dx dx dy dx "' \dxjy dx dz dx dx dx Bydx dz\dx dydx' df_aF_aldF 13. 3V3-4. _ dz _ dy dx dx dy g ^' d'x ^ WW" 14, --3(a:co8Q:+2/oos/3+zcos7). dy dz dy Page 135 1. £^1 = ^ = 1:^2, (a; _ 1) +2(2/ -2) +2(2- 2) =0. 2. Normal, y + ix = H, z = 3. Tangent plane, a; — 4 2/ — 14 = 0. „x — 3 2/ — 4 z — 5 „ ,. - - 3. — 3~= 4^ = ^15" • 3x + 42/-5z = 0. .X — 5 y — 1 z — 3 4. ^—=5 Z3-' x + 5y -Zz-l =0. . X — 5y— IZ— 1 a, en 5. —^ = '^-^' g— • X -42/-6Z + 5 =0. 6. x + z = 2/-2=± "^2. 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 fvmction, 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 transcendental functions, 49-62. partial, 113-139. Directional derivative, 129. Direction cosines, 80, 81. Direction of curvature, 67. Divergence of infinite series, 107- 111. Exact differentials, 130, 131. Exponential functions, 56-62. Function, continuous, 10, 113 discontinuous, 10. explicit, 1. implicit, 2, 127, 128. irrational, 2. of one variable, 1. of several variables, 113. rational, 2. Functions, algebraic, 2, 19-31. exponential, 56-62. inverse trigonometric, 54-56. logarithmic, 56-62. transcendental, 2, 49-62. trigonometric, 49-53. Functional notation, 3. Geometrical applications, 63-84. Implicit 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. , sinS .„ of— ,49. Limits, 4-9. properties of, 5, 6. Loganthms, 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. Natm-al 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. Rolle'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 strai^t line, 32. angular, 34.