The MARTIN ROWAN CHAFFIN Collection of Pablic School By his grandchildren in honor of M. R. Chaffin, who taught public school in Davie and Yadkin counties for a number of years beginning in 1850, and in honor of his father, William Owen Chaffin, who first taught a North Carolina public school in 1843, in Yadkin county. For the especial use of the Department of Education and of the Durham county and city teachers. Text-Books PRESENTED TO DA TE , 3 > Digitized by the Internet Archive in 2016 with funding from Duke University Libraries https://archive.org/details/elementarycalcul01smit ELEMENTARY CALCULUS A TEXT-BOOK FOR THE USE OF STUDENTS IN GENERAL SCIENCE BY PERCEY F. SMITH, Ph.D. PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY NEW YORK- : • CINCINNATI • : • CHICAGO AMERICAN BOOK COMPANY Copyright, 1902 and 1903, by PERCEY F. SMITH. EL. CALC SMITH. W. P. 4 £ * £ * si 7 S bTL £ \ PREFACE This volume has been written in response to the un- mistakable and growing demand for a text-book on the Calculus which shall present in a course of from thirty- five to forty exercises the fundamental notions of this branch of mathematics. In American technical schools students pursuing courses distinct from engineering branches usually terminate their mathematical studies with Plane Analytic Geometry. But in view of the recent remarkable development of certain of the general sciences along mathematical lines, such a course can no longer be regarded as adequate. Moreover, there can be no differ- ence of opinion as to the relative advantage to the student of a knowledge of more than the mere elements of Ana- lytic Geometry and an introductory acquaintance with the Calculus. It is, I think, the experience of every teacher that the average student first realizes the power and use of mathematics when taught to solve problems in maxima and minima by means of the methods of the Differential Calculus. Certainly no stronger argument can be adduced in favor of an adjustment of the curriculum which shall include this branch of mathematics. Such a change has been effected in the Sheffield Scientific School, and results abundantly justify the step. For the general student in our colleges who elects a year’s work in mathematics beyond the usually required 3 4 PREFACE Trigonometry, the most satisfactory course would seem to be one in which the time is equally divided between Plane Analytic Geometry and Calculus. In writing this book I have everywhere emphasized the possibility of applications. The examples have been care- fully selected with this end in view. The first chapter may seem long, but the notion of limit certainly demands adequate treatment. While an elementary text-book offers no excuse for employment of the refinements of modern rigor, I have endeavored to avoid positive inaccuracies and have carefully distinguished between demonstration and illustration. I am indebted to my colleague, Dr. W. A. Granville, for many helpful suggestions. PERCEY F. SMITH. Sheffield Scientific School. CONTENTS CHAPTER PAGE I. Functions and Limits . 7 II. Differentiation ........ 23 III. Applications 51 IV. Integration 70 V. Partial Derivatives . 84 VI. Additional Examples ....... 90 5 349462 ELEMENTARY CALCULUS CHAPTER I FUNCTIONS AND LIMITS 1. Continuous Variation. In this book we are concerned with real numbers only. Geometrically, such numbers may be conveniently represented by points of a scale (Fig. i). etc. -5 — i —8 Eg r£ o 1 2 3 t 5 6 7 etc. Fig. i Then to every real number corresponds one point of the scale, and only one ; conversely, every point of the scale represents a real number. Any segment of the scale, however small, represents indefinitely many numbers. We speak indifferently of the number a and the point a of the scale. A variable x is said to vary contmuously between the numbers a and b when it assumes values corresponding to every point of the segment ab. 2. Functions. The problems arising in Elementary Calculus involve in general two variables in such a way that the value of one variable can be calculated as soon as a value is assumed for the other. Thus, in Geometry, the student has an illustration in the area and radius of a circle, two variables such that the area A can be calculated when we know the radius r from the formula A — n tA. 7 8 FUNCTIONS AND LIMITS Definition. A variable is said to be a function of a second variable when its value depends upon the value of the sec- ond variable and can be calculated when the value of the second variable is assumed. The first variable is called the dependent variable , and the second the independent variable. For example, the equations y = -r 2 , y = sinx, y = log 10 (.r 2 - i) state that y is a function of x. In the first two cases, y may be calculated for any value of x ; in the last case, how- ever, x is restricted to values numerically greater than I, since the logarithms of negative numbers cannot be found. In the first two cases, then, we say that the dependent variable (or the function) is defined for every value of x, and in the last the function is defined only when x exceeds i numerically. A function is defined for a value of the variable when its value can be calculated for that value of the variable. Elementary Functions. Power Function: x m , m any positive integer. Logarithmic Function: log a ;r, a> o; this function is defined only for x>o. Exponential Function : a x , a > o, He. the exponent is a variable, the number a being a constant. Circular Functions :* sin x, cos jr, tan .r, etc., i.e. involving the six trigonometric functions. * So called from the use of the circle in their definition, e.g. in Fig. 2, sin A OP = FL = .\fp jf ft js the unit of R linear measure. Hereafter, angles will always be meas- ured in circular measure , i.e. x = arc — = 'LL. In the radius R FlG. 2 unit circle, R = I, x = arc AP. FUNCTIONS AND LIMITS 9 Inverse Circular Functions: arc sin x, arc tan x, etc., i.e. the “arc whose sine is x,” “arc whose tangent is x,” etc. In the unit circle (see Fig. 2) R = 1 ; if x=MP, then arc sin x = arc AP. One thing is peculiar here. Assuming any value of x not exceeding 1 numerically , arc sin x may be calculated, but the number of answers is always indefinitely great. For not only is arc sin MP = arc AP, but also equal to any number of circumferences + arc AP, i.e. arc sin MP = arc AP + 2 ml, where n is any integer. For this reason the inverse circular functions are called many-valued functions. For definiteness we may always take the least positive arc. 3. Functional Notation. As general symbols for func- tions of variables we use the notation f(x), 0(y), (f>(r), etc., (read f function of x, theta function of y, phi function of r, etc.). We mean by this that f(x) is a variable whose value de- pends upon x, and can be found when a value is assumed for .r. The notation is extremely convenient, for it enables us to indicate the value of the function corresponding to any value of the variable for which the function is defined. Thus f(a) represents the value of f(x) for x = a, 6(0) the value of 6 ( y) for y = o, <£(^) the value of (r ) for r = 1 , etc. IO FUNCTIONS AND LIMITS EXERCISE 1 1. For what values of the variable are the following functions defined ? (a) A?is. For every value except ;r = o, since - cannot be x calculated.* 0 (b) vhr 2 — 6x. Since x- — 6 x or x(x — 6) must not be negative, x and x — 6 must always have the same signs. Ans. For every value except those between o and 6. 0) y/y - y 2 ; (d) vTo; ( e ) arc sin x; (/) arc sec x; (g) sin Vi + x; ( h ) log tan x. 2. Given f(x) = x 3 — 7 x 2 + 16 x— 12, show that f(j 2 .) =0, /($) —o. Does f(x) vanish for any other value of x ? 3. Given f(x) = logx; show that /O') +/OO =/(*/)• 4. Given $ ( x ) = a z ; show that O') <£(/) = <£ O' + /)• 5. Given 0 (x) = cosjt; then 6 (x) + 6 ( y ) = cos x + cos y. From Trigonometry, we know that cos .r + cos/ = 2 cos £ (x + /) cos £ (jr — /) ; 4. Graph of a Function. After determining for what values of the variable a given function is defined, it is im- portant to know in what manner the value of the function * The student should observe that the four fundamental operations of arithmetic, addition, subtraction, multiplication, and division, when performed with real num- bers, give real numbers, with the single exception that division by zero is cxchided. FUNCTIONS AND LIMITS II changes with the variable. Geometrically this is accom- plished by drawing the graph of the function , which is thus defined : The graph of a function is the curve passing through all points whose abscissas are the values of the variable and ordinates the corresponding values of the function. In the language of Analytic Geometry the graph of a function fix) is the locus of the equation y =/(*)• — l- - Fig. 3 By carefully drawing the graph of a func- tion a good idea is obtained of the behavior of the function as the variable changes. For ex- ample, the graph of log 3 -r, i.e. the locus of the equation * y = log 3 x, is drawn in Fig. 3. Here we see the following facts clearly pictured to the eye. (a) For x= i, log 3 ^= log 3 1=0. ( b ) For x> 1, logger is positive and increases as jv in- creases. * The values ofy are found from the formula proven in the theory of logarithms, logio x logs * = logio 3 12 FUNCTIONS AND LIMITS ( c ) For x< i, logg.tr is negative and increases indefi- nitely in numerical value as x diminishes. ( d ) For .r = o, log 3 .tr is not defined, since the logarithm of zero cannot be calculated. The graph of the general logarithmic function log a ;tr may be drawn by merely changing the ordinates in Fig. 3 in the constant ratio 1 -5- log 3 a. Graphs: (a) Of x 2 . (b) Of x*. Y The graph of x m has the appearance of (a) or (b) according as m is even or odd. 00 Of log 3 a-. (d) Of 3 1 . 1 / T I 0 / r (T \-i- 1 1 3 jLd-)- 1 1 “/ F Trl Since if we set y = log 3 a-, then x=y, the graph in ( c ) has the same relation to XX' and YY' as (d) to YY' and XX' . FUNCTIONS AND LIMITS 13 The graphs of the circular functions have the appearance of a curve repeated' over and over as the variable increases or decreases. As in (c) and {df if we revolve (e) and (/ )~around XX and interchange XX' and YY', we shall have the graphs of arc sin x and arc tan.tr respectively. 5. Limits. For the study of the Calculus it is absolutely essential that the student should understand perfectly the fundamental notion of a limit. He is already familiar with simple examples of limits from Geometry, such as the limit of the perimeter of an inscribed regular polygon as the number of sides is indefinitely increased is the circum- ference, and the limit of the area of the polygon is the area of the circle. These are examples of variables ap- proaching limits , the variable being in the first case the perimeter, and in the second the area of the regular polygon. The following definition states the matter gen- erally. Definition. A variable is said to approach a number A as a limit when the values of the variable ultimately differ 14 FUNCTIONS AND LIMITS from A by a member whose numerical value is less than any assignable positive number. If we represent the values of the variable by the infinite sequence ^l> ^2’ ^3» ’"*> ^ni fi'n+b “"> then on the scale (Fig. i) the points corresponding to a v a 2 , a 3 , •••, a n , a n+l , •••, etc., will ultimately approach nearer the point A than any assignable length, that is, will “heap up” at the point A. The definition interpreted A a x a 2 |< — h-> * — h — *| Fig. 4 geometrically means, then, that no length h (Fig. 4), however small , can be laid off from the point A, but that points of the sequence will fall within the segment. •We write Limit (#„) = A, or, also, if we denote the varia- ble whose values are a v a 2 , etc., by x, Limit (x)>= A. 6. Limiting Value of a Function. Continuous Function. Consider the elementary function \og a x (Fig. 3). Take any sequence ^ 2 * ^ 3 * * **> of positive numbers whose limit is some positive number A. For example, the sequence i-3. I -33> 1-333, — , the limit of which is f. Consider now the sequence of numbers logo («i), log a (tf 2 )> lo g« Os), •"> and draw their ordinates in Fig. 3. Then the student FUNCTIONS AND LIMITS 15 will see that this last sequence has the limit log a (^d); that is, When the variable x approaches a limit A greater than zero , the logarithmic function log a x approaches the limit log a A. We express this important fact by writing Limit (}oga x )x=A = lo g a A. The general relation brought out by this example is the following : When the values assumed by the variable x approach * a limiting value A, then the corresponding values of the function will also approach a limiting value ; and if the function is defined for the value A, then the limiting value of the function is its value for x= A. Or, in symbols, if f (A) is a number, then Limit (f(x)') x=A =f(A). For example, since cos 0=1, Limit (cos x)^ = 1. The property above described is that of continuity; i.e. a con- tinuous function is such that Limit f(x) = /(Limit x). For the purposes of the Calculus it is essential that a function should be continuous. The elementary functions of § 2 possess this property. 7 . Infinity. If the points on the scale of Fig. 1 cor- responding to the sequence of values of the variable x * The variable x may approach the limit A in any manner consistent with the definition of the function. In the above illustration the geometrical sequence 1, 1 + 1 , 1 -f - 1 + ie, 1 + * + is + 6 s, etc., whose limit is f, might also have been taken. i6 FUNCTIONS AND LIMITS ultimately advance to the right without limit, we say, “ x increases without limit,” or also, “ x approaches the limit positive infinity,” and we write Limit x = + oo. If under the same conditions the points advance to the left without limit, we say, “.r decreases without limit,” and write Limit x = — oo. Finally, if the points advance both to the right and left without limit, we write 4 Limit x = oo. The student should disabuse his mind of any previous notions of infinity not agreeing with the above definitions. The symbols + oo, — oo, cc, must be used always in the sense above described. 8. Fundamental Theorems on Limits. The student is asked to accept the following theorems as true : Given a number of variables whose limits are known ; then I. The limit of an algebraic sum of any finite number of variables equals the same algebraic sum of their respective limits. II. The limit of the product of any finite 7iumber of variables equals the product of their respective limits. III. The limit of a quotie7it of two variables equals the quotie7it of their respective limits when the limit of the denominator is 7iot zero. 1 7 Fig. s FUNCTIONS AND LIMITS 9. Two Important Limits. To prove * Limit = 1 . L a? J*=o In Fig. 5 let x= arc A T =-. arc A S, the radius OA being taken equal to unity. Then sin x = MT = SM, tanar = TQ = QS. Now ST 1 + I • 2 1-2 1 - 2-3 not defined for x = o. EL. CALC. — 2 i8 FUNCTIONS AND LIMITS Representing the successive terms by a v a 2 , a 3 , •••, we have — U a 2 = 1 + Y’ a Z ~ 1 + ~ + a i — 1 + ~ + a n — I +~ + I I • 2’ I I • 2 I I • 2 + + 1 - 2-3 I 1 - 2-3 + = I = 2 = 2.5 = 2.666 - 4 - , — - — *, etc. \n — i The numbers of this sequence continually increase. We may show, however, that any term is less than 3. If r> 3, then |r>2 r , and therefore ~ 1 I I I I I 2" a n I 2 2 i 2 3 2" 1 I I 2 since 1 1 1 I d 1 n d 5 + 2 2" 2 d + I 2 n— 1 is a geometrical progression and its sum may be imme- diately written by the usual formula. Hence a„ < 3 and taking n= 1, 2, 3, etc. ijifi- nitum, every term of the sequence is seen to be less than 3. c CTl do Q3 5 1 S h_ Fig. 6 The points, then, corresponding to the sequence (Fig. 6) must heap up at some point to the left of 3 ; that is, the sequence must have a limit. * The symbol \n — t , read " factorial n — i,” means the product of all integers from 1 to n — i inclusive. FUNCTIONS AND LIMITS 19 The calculation of this limit to any number of decimal places is a matter of no difficulty, as the following compu- tation to five decimal places will show. Write down 1. 000000 ( = 1). Divide by 2 ) 1 .000000 ^ 4) 3). 500000 ^ II 4). 166667 ( -5) 5 >.041667 ( -b) 6). 008333 ( -b) 7). 001388 ( =|) 8 >.000198 ^ 9 >.000025 ^ :-D 10 >.000003 ( :=s) Adding, 2.71828 neglecting the figure in the sixth decimal place, of which we cannot be sure. In fact, it can be easily shown that 20 FUNCTIONS AND LIMITS 2.71828 is the limit of the sequence correct to five deci- mal places. Writing the limit of the sequence in the form of an infi- nite series and denoting this limit by e, we have The number e is called in the Theory of Logarithms the Napierian base or natural base, and is a number of prime importance in mathematics. The expression for e in the form of an infinite series should be remembered and also its value to five decimal places. 11. To prove A rigorous proof of this very important limit is beyond the scope of this volume. We may perhaps best illustrate the meaning of the theorem by drawing the graph of the function for positive values of z. e = 2.71828 .... Limit Tl + -1 = e. L ZJz=OD Setting and for any value of z greater than zero p may be approxi- mately calculated, as for example in the accompanying table, which gives y to five decimal places. FUNCTIONS AND LIMITS 21 z y .01 1.04723 .1 1.27098 I. 2. IO 2.59374 IOO 2.70481 IOOO 2.71692 10,000 2.71815 100,000 2.71827 1,000,000 2.71828 The figure illustrates the theorem in showing that the graph approaches the line y = e as z increases indefinitely. When z diminishes toward zero, y approaches unity. 22 FUNCTIONS AND LIMITS EXERCISE 2 [The graph of the function considered must be drawn in every case . ] 1. Prove Limit | - — ~ 3 ^ + 4 1 _ 2 , L x-i We have merely to substitute 2 for x. 2. Prove Limit [ '* — — 1 =— 2 a. L x+a We cannot substitute directly, for we should get a meaningless 2 2 O expression. But ' = .r — a, and we may now substitute. x+a 3. Prove the following in which a is any number greater than zero ■ Limit 1~— 1 = + co ; Limit f-1 = oo ; Ur 2 J a-o La-J^ Limit ax — oo ; Limit — = o. L LjrJ^ The last three results are often written a a - = oo, a- oo=oo, — = o, o 00 but the student must remember that such equations are merely abbrevia- tions of the preceding. ~ Vx + h — Vjr J _ i 4. Prove Limit *[’ h = 0 2\ / X Hint. Multiply numerator and denominator by Vx + h + Vx. 5. Show that Limit [ tan '* 1 =i; Limit Ttan^rl = oo ; LsinxJx^ L J*=f Limit Qog e .rJ = — oo ; Limit = o. CHAPTER II DIFFERENTIATION 12. Increments. In order to understand the manner of variation of a function as the variable varies, it is essen- tial to know how great a change in value occurs in the function for a given change in value of the variable. Change in value is termed increment ; i.e. the increment of the function is the change in value of the function corre- sponding to a given change in value or increment of the variable. The problem now arises : To calculate the increment of a given function. Let fix ) be defined for all values of x from x to f{x+h)-f{x) x + h ( Fig _ g ) Nqw for x + h the value of the func- tion is fix + h), hence the increment of the function fix) corresponding to an x~ increme7it h in the variable x Fig. 8 x+h x is fix + h) -fix). We shall represent the increment of any variable by the letter A (read “ delta ”) prefixed to that variable, thus If Ax = h, then A fix) = fix + ft) — fix') . 23 24 DIFFERENTIATION Rule. To find the increment of a function , ca'culate the new value of the function by replacing x by x + h and sub- tract the old value of the function from the new value. EXERCISE 3 1 . Find Ax 2 . 2. Find A a)- Ax 2 = (x + h ) 2 — x 2 — 2 hx -f h 2 . Ans. A / 1 \ I I —k A ( — ) = — — t = — — Ans. \x ) x+h x x(x+h) 3 . Prove A \fx=- h dx-\-h + Vx 4. Find A lo zx. (Ex. 4, page 22.) A log x — log (x -f- //) — log x = log j =log^i + Ans. 5. Find Asin.tr. A sin x = sin (x + h) — sin x = 2 cos (x+ ^ h ) sin \ h, from Trigonometry. Ans. 6. Find Ae z . Ae* = — e* = — i). Ans. 7. Find A cos 2 .r. A cos 2 x = — 2 sin (2 .r + h) sin A. Ans. , h a 8. Find AVi + x , == Ans. vi+r+/i + Vi+r 13. The Increment Quotient. While the increment of a function as found in the preceding article is of impor- tance, still more essential in any investigation is the rate of change of the function , that is, the change in the function per unit change in the variable. If we form the quotient DIFFERENTIATION 25 we obtain the average rate of change of the function while the variable changes from x X.o> x h. For example, the “ law of falling bodies,” given in Mechanics, asserts that the distance s traversed by such a body falling freely from rest in a vacuum varies as the square of the time t , that is, s = 16. 1 t 2 , the constant 16. 1 being determined experimen- tally when is measured in feet and t in seconds. Therefore As = 16.1 (t + Jif — 16. 1 t 2 , or — = 16.1(2 t + h), since At = h. At v ’ For example, the average velocity throughout /\ c the third second is given by setting in — , t — 2, h—i, and is 80.5 feet per second. EXAMPLES 1. From Physics we learn that for a perfect gas at constant tem- perature the product of the pressure p and volume v is constant, or , . , c , , \t> c = a constant c, i.e. t> = - : show that = , ; — v tsv v 2 + vtSv 2. Show from Ohm’s law, viz. current strength C equals electro- motive force E divided by the resistance R , that for constant R the change of current strength per unit change of electromotive force is constantly equal to 1 ~ R. 14 . Derivative of a Function. In the illustration taken from the law of falling bodies given in § 13, let us propose to ourselves to find the velocity at the end of two seconds. Making t= 2, we have t = 3 Fig. 9 t= 0 f= 3 26 DIFFERENTIATION which gives us the average velocity throughout any time h after two seconds of falling. Our notion of velocity shows us, however, that by the velocity at the end of two seconds we do not mean the average velocity during one second after that moment, or even during or ^oVo of a second after that moment, but, in fact, we mean the limit of the average velocity when h diminishes toward zero ; that is, the velocity at the end of two seconds is 64.4 feet per second. Thus, even the everyday notion of velocity involves mathematically the notion of a limit, or, in our notacion, Thus, after t seconds have elapsed, the velocity of a falling body is 32.2 t feet per second. Again, let it be required to find the slope of the tangent at any point P of a plane curve whose equation is given in rectangular coordinates x and y. The tangent at P is constructed as follows (Fig. 10): Through P and any point P' on the curve near P draw the secant AB. Let the point P' move along the curve Velocity = Limit DIFFERENTIATION V toward P, the secant AB meanwhile turning around P. Then when P' coincides with P, the secant AB becomes the tangent TP. Now if P is (x, y) and P ' ( x + Ax, y + Ay), the slope of AB is tan 6 = SP' = Ay PS Ax As P’ approaches P as above described, Ax will ap- proach zero as a limit, while 6 approaches the angle PTO or 7 ; hence, at the limit, tan -y = Limit j ^ = slope of tangent at any point P. For example, the slope of the tangent at any point of the parabola y = x 2 + 3 is 2 x. Law of Linear Expansion. If l 0 is the length of a rod at o° Centi- grade, and l the length at t° on the same scale, then experiment estab- lishes the law of expansion / = /q - f- at bE, a and b being constants. The coefficient of linear expansion at any tem- perature t is the increase in length per unit change in temperature, i.e. coefficient of expansion = Limit ( — ’] \ A/ / £,=0 We easily find, then, that coefficient of expansion = a + 2 bt, and .\ a — the coefficient of expansion at o°. Specific Heat of a Substance. The specific heat of any substance is the quantity of heat necessary to raise a unit mass of the substance one degree in temperature. If Q is the measure of the quantity of heat in unit mass, and t the corresponding temperature, then by definition, specific heat — Limit 28 DIFFERENTIATION These examples show that we obtain an important new function of the variable if we can find the limit of the Increment Quotient when the increment of the variable approaches zero. This function is called the derivative of the function. Definition. The derivative of a function is the limit of the quotient of the increment of the function and the increment of the variable when the latter increment approaches the limit zero. The step of finding the limit of when Ax ap- Ax proaches o is indicated by changing the A’s to ordinary d’s, so that d f - x l = Limit ^ , or also, if dX \ AX / Aar=0 Ax = h, = Limit [ +h)-f(x)l h J/i=0 The symbol c ^S: x .l is read, “ derivative of f(x) with dx respect to x.” This, being a new function of x , is often written f (x), so that also, 4f(*) dx =/'(*> Thus in the illustrations given, velocity = — , J dt i.e. velocity is the derivative of the space traversed in the time t with respect to the time. Slope of tangent = dx DIFFERENTIATION 29 i.e. equals the derivative of the ordinate of the point with respect to its abscissa. Coefficient of linear expansion = — , dt or the derivative of the length with respect to the temperature. Specific heat = dt that is, equals the derivative of the quantity of heat in unit mass with respect to the temperature. Many more illustrations of physical magnitudes might be given which take the form of a derivative. We call — the sign of differentiation , so that the pre- dx fixing of — to any function of x means that the following dx process is to be carried through : General Rule of Differentiation. 1°. Calculate the quotient of the increment of the function and the increment of the variable (i.e. the increment quotient). 2°. Find the limit of this quotient when the increment of the variable approaches the limit zero.* It must be emphasized here that the characteristic thing in differentiation is finding the limit of a quotient. From the standpoint of the Differential Calculus a function is of no interest if the limit mentioned does not exist. Func- tions possessing derivatives are said to be differentiable , and it is of prime importance to show, for example, that the elementary functions of § 2 are differentiable. * The student must notice that the limit of the increment quotient cannot be found by Theorem III, § 8, since the limit of the denominator is zero. 30 DIFFERENTIATION 15 . Differentiation of the Elementary Functions x m , sin x, log a x. (a) To prove —x m = rnx m ~ 1 . dx Now A(x m ) = (x + Ji) m — x m if Ax = h. But (x + li) m = x m + inx m ~ 1 h H \- h m , the terms not written containing powers of h. A(x m ) = mx m ~ l h 1- h m ; = vix m ~ x -} f A m ~\ Ax where again the terms not written contain powers of h. Putting h — o, we find (0 d_ dx (x m ) = mx‘ .771 — 1 (b) To prove — sin ;tr = cos ;r. dx Since A sin x — sin (x + A) — sin x = 2 cos ( x + J Ji) sin \h (§ 12, Ex. 5), we find A sin x _ 2 cos (x + ^ h) sin \ h Ax h = cos (x + j h) • — 2— 2 But Limit (— y-f— ^ =I (§ 9 ). \ \ ll Jh = 0 and Limit (cos (x + ^/i )) h==0 = cos (since cos x is continuous, § 6), so that we may apply the theorem II, § 8, and we have (2) - sin x = cos x. dx DIFFERENTIATION 3 * 00 To prove log a x = log a e ~ From § 12, Ex. 4, we have A 10g a X Ax log a ^i IT =- 1Q g* k\ h I _l — I* Xj since the introduction of the exponent — is, by the princi- ples of logarithms, equivalent to multiplying the logarithm itself by that exponent. / /A* Now ( 1 + - r is the expression of § 1 1 if we write in that expression z = — h Also Limit A=0 hence Limit rr.Af — Limit [ 7 . + ±Yl _V x) _ h= 0 Lv z) J = e. Limit b cr a 1 H — ) ft Xj h = 0 = log a e. and we have ( 3 ) (since logx is a continuous function), 4- lo g « x — lo &« e ax x Formula (3) becomes most simple when a — e, for then d , 1 — log e x = ~ dx x Logarithms to the base e are called natural logarithms or Napierian logarithms (§ 10), and the factor log a e in (3) is called the modulus of the system whose base is i.e. the member by which natural logarithms enust be multiplied in order to obtain logarithms to any given base a. We write 32 DIFFERENTIATION For example, the modulus of the common system of base io is log ;o e, and logio e = 0.43429 to five decimal places. If in (1), (2), and (3) we write u for jr, we have (4) u m = mu m ~ 1 ; sin u = cos v ; ~~ log a u = log a e 1 • 7 du du du u EXERCISE 4 1 . Differentiate with respect to x. ( a ) x 2 + 3 ax+ b. (, b ) — — . (r) Vir (cf. Ex. 4, p. 22). 2. Prove — cos .r = — sin x. dx 3. Prove — Vi + x= 1 dx 2 V 1 + x 4 . Prove d fa\_ a dv\v we found (§ 14) or Prove What does ds dt = 32.2 1, velocity = v = 32.2 1. dv „ „ — = 32.2. dt J — = Limit dt Ans. 2 3 a. A us. (x+ b )' 2 Ans. 2Vx 5. Prove — {Cu') = C, if C is any constant. du 6. From the law of falling bodies s = 16. 1 1 2 , AzA A/ y ^^=0 represent in Mechanics ? Acceleration. Ans. DIFFERENTIATION 33 7 . Find the velocity and acceleration of the motion defined by (1) s = at + \gt 2 \ Ans. V=a+gt ; accel. =g. (2) s = at — \gt 2 -, Ans. V = a — gt ; accel. = —g. 8 . Find the slope of the tangent to y = 6 x — x 2 at the origin. Ans. 6. 16. Certain General Rules. We prove in this section several important rules for differentiation of a general character. Let the variables u, v, zv, etc., be functions of the vari- able ;r. I. To differentiate any algebraic sum of these variables. For example, to find — (u + v — w). dx Now A(u + v — w) = u + A u + v + Av — (zv + Azv) — {tt + v — zv), = An + Av — Azv, A(zt + v — zv) _ Ate Av Azv Ax Ax Ax Ax Since Limit Au = — , Limit Av _Ax_ A3'=o dx _Ax_ _ dv ^^=0 dx Limit Azv A x _ dzv — 0 dx we may apply I, § 8, and we have ( 5 ) d , , x du , dv — (u + v-zv)= — + — - dx dx dx dzv dx II. To differentiate a product. For example, to get ~ (uv). dx EL. CALC. — 3 34 DIFFERENTIATION Now A {uv) — {u + A u)(v + Av) — uv, = u Av + v A u + A u Av, A (uv) Av . An . A Av •• a — + + Ax Ax Ax Ax c . T . .. ["AzTI dv T . .. fAz/l da Since Limit - — = — , Limit — = - — -, jA^=0 dx \_Ax\ Limit [Az*] ax=0 = o, we may apply I and II, § 8, and obtain ( 6 ) _AxJ Iax=0 dx d , \ dv , dn — {iiv)=u — + v~ dx dx dx To find — (uvw), consider nvw as made up of the two dx factors uv and w ; then, by (6), d , \ dw . d(uv) — (uv ■ w) = uv b w — — % dx dx dx or by (6) again, ( 7 ) dw , dv , du = uv — — b wu —— + wv — — dx dx dx III. To differentiate a quotient. Since a r- = u + An v Au — u Av v + Av v v 2 + v Av ’ Au Av v u - — A fu\_ Ax Ax A x\vj v 2 + vAv Then since Limit [_v 2 + vAv~\^ x = S) = v 2 , we may apply I, II, III, § 8, and have du _ dv (8) d fu\ _ dx dx dx \vj v 2 DIFFERENTIATION 35 From (6), (7), and (8) we have the rules : I. The derivative of an algebraic sum of any number of vari- ables is equal to the same algebraic sum of the derivatives of the variables. II. The derivative of a product of any number of variables is equal to the sum of all the products formed by multiplying the derivative of each variable by ail the remaining variables. ITT . The derivative of a quotient equals the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. To these we may add the following: IV. The derivative of a constant is zero. V. The derivative of a constant times a variable equals the constant times the derivative of the variable. Rule V comes from (6) and IV, if we place u equal to a constant. EXAMPLES 1 . Workout — (1 + x 2 ) (1 — 2x 2 ). dx Rule II is first applied, and we get — (1 + X 2 )(l -2r 2 ) = (l + X 2 ) — (I - 2X 2 ) + (l - 2X 2 ) — (I + X 2 ). dx dx dx By Rule I, £ 0 - »**) = £ (0 ~ £(***>> d_ dx (i + x 2 ) Since by V, — (2 x 2 ) = 2 — x 2 , and from (4) § 15, dx dx = 2r, we have finally, d_ dx (1 -f x 2 )(l — 2X 2 ) = (1 +x 2 ) • — 4 x+ (1 — 2 X 2 ) -2X= -2r(l+ 4X 2 ). 36 DIFFERENTIATION 2. Work out d ! sin x \ dx\\og e x) Rule III we use first and find d / sinr\ dx Vlog e x) log, x — sin x — sin x — log e x dx dx 0°geX) 2 By (4) § 1 5j d • d , i — sin x — cos x, — lo£, x = - • dx ’ dx 6 * / sin .r \ ;r cos .r log, x — sin x dx \ loge x) x (log e x) 2 EXERCISE 5 Prove the following differentiations : 1 d \ -j d fsinx\ _xcosx-smx dx^ 1 -^- 1 - 2 *- 3 - dx[-r )- — ^ — 2 d ( x \ - I — x 2 4 d / 1 + .r 2 \ _ 4 J To simplify, multiply numerator and denominator by (I + x 2 ) 2 . DIFFERENTIATION 45 Then, since (i + x 2 ) 0 — i, we have, reducing, d / i - dxy VY i + * + x J (i + x 2 y 2 . Find dx ' i cos .r + cos x For convenience, set / = log e "^j — — cos Since log, then by I and V, 4 — COS XI,. . I . , . = - log ( I — cos x) — log ( I + COS X), + cosr 2 sV ' 2 45 v ' S = i - c°s*)-i£j°g.(i+cos*). Applying IX, we have — (i — cosx) — (i + cosr) dy i dx I dx dx ~~ 2 i — cos .r 21+ cos .r , and by XII, dy i [ sin x | sin x \ _ sin x _ i dx 2 \ i — cos x' i + cos x) i — cos ' 2 x sin x ... A lo ge J 1 T _ cg i£ = _I_. dx ' x + cos .r sin x 3 . Find ^ arc tan Setting the function equal to y. we have, by XIX, die 9 — e e + e~ e dy dQ\ 2 / 2 , dd~ (e*-e-o\ 2 ~ 4 + e 2 e -2+e- 2 o^ Y } 1 + [ 2 / i 2 _ ^ + 46 DIFFERENTIATION EXERCISE 7 Prove the following differentiations : i d , o , , \ r ~ 5 7 x i — 2 jr 2 — I 1. — (;r z + i) Vx a — x = . dx 2 (x 3 — _r)£ 2 . A-Jj + 1 ’ r * l dx' 1 ! - x (j _ jr) Vi - jr2 3 . At. x dx\ \J i . i)=7t x 2 l (l-a: 2 )2 4 / 3X 3 + 2 \_ 2 ^ \ar(j.- 3 + i)t / jr 2 (ar 3 +i)l 5. — (i — 2 x + 3 ar 2 — 4_r 3 )(i + a;) 2 = — 20 x 3 (l + x ). dx 6. ~(i — 3 .r 2 + 6 ^ 4 )(i + x 2 ) 3 = 60 .r 5 (i + jr 2 ) 2 . dx 7 - -^(5 i2 + 2i )= 2(4- + i)5 l2+2z l°gc5- ax 8 . — x n a x = x n ~ l a x (n 4- x\o g e a ). dx x ( 1 - T) 2 ’ 9 d L rxio§&x + j (I _ x y\ = Jog « jr Li— or J(i — 10 . _ 2£? + 6* 6 \ ajr3 ^. V a a 2 a 3 ) 11 . — logeO* + e-*) = e * ~ e —. dx 6 v 7 e* + e~* 12. log e (i + Vx)) = l ——.. dx 2(1 + Vx) 13. A tan 2 5 0 = io tan 5 d sec 2 5 d. dd 14. 4; sin 3 & cos $ = sin 2 $ (3 cos2 $ — sin 2 0) . dd 15. A log sec 6 = tan 6. dd DIFFERENTIATION 47 16. (tan 2 6 — log sec 2 6) — 2 tan 3 9. da 17. sin n6 sin n 6 = n sin ” -1 6 sin {n + I )0. da 18. — arc sin ( 3^-4 X s ) = 3 ^ VI to d x a 19. — arc sec — = — . ^ & xVx 2 — a ! 2 20. — arc esc dx 2 ;r 2 — 1 21 . a — ■ arc sin dx 1 — x 2 1 + x 2 2 I + X 2 ' 22 . — arc tan dx x + a 1 — ax 1 1 + x 2 ' 23. 24. 25. d e? — e~ x — 2 — arc cos — = — dx e x + e~ x e? + e~ d I 2 — 1 — arc sec \ = - — - dx ' 1 T x 2 \J 1 ~ ( arc cot - + log e \l - — - ) = dx \ x 'x + a / 2 ax 2 x 4 — a 4 ' 19. Differentiation of Implicit Functions. If an analytic relation is given between two variables not solved for either variable in terms of the other, then either variable is said to be an iviplicit function of the other. For example, in x 1 — y 1 + 9 = o either x or y is an im- plicit function of the other variable. In such a case either variable may be chosen for the independent variable, and if we can solve explicitly for the other (as in the above example for y, giving y = Vx 2 — 9), then we can differentiate as before. But it is generally better not to solve the equation, but to differentiate the given relation as it stands. 48 DIFFERENTIATION Thus, to find ~ from ax x 2 — 3 xy + 2 y 2 Then dy\ d y 2x- 3 [y + x^j+4y± = o, dx and d y - 2X ~ 2>y dx 3 x-4y' To justify this process is beyond the limits of this text- book. One thing is to be noted, namely, that only those values of the variables which satisfy the original relation can be substituted in the derivative. exercise 8 dy Find -d- from the following equations : 1 . y 2 — 2 xy = a 2 . Ans. dy y dx y — x x 2 V 2 2. — j + — = I. a 2 b 2 Ans. dy dx b 2 x 3 . ax 2 + 2 bxy + cy 2 + 2 fx + 2 ay + h = o. Ans. y = _?£jj l±f. dx bx + cy + g 4 . x s + y 3 — 3 axy = o. 5 . x$ +yi = a%. . dy x- — ay Ans. -y- = 7, dx y- — ax a dy yi Am - s=— r X* 6. Given r = a ( i — cos 6 ) ; show = a sin 6 - dv 7 . Given r 2 = a 2 cos 2 6 ; show dr Jd a- sin 2 i DIFFERENTIATION 49 20. Derivatives of Higher Orders. Since the derivative of a function of a variable x with respect to x is also in general a function of jr, we may differentiate the derivative itself, that is, carry out the operation, d_ dx This double operation is indicated by the more compact notation, d 2 r, x and this new function is called the second derivative. the same way, d d 2 r, s d 3 ,, \ In is the third derivative, and in general, d n dx“ fix) is the nth. derivative of f{x), that is, the result of differen- tiating fix) n times. The following notation is also used, -£f(x) =f'(x), j*-J{x)=f\x), -, £- n f(x)=f^(x). The operation of finding the successive derivatives is called successive differentiation. EXERCISE 9 1 . Given f(x) = 3 x 4 — 4 x 2 6 a* — 1, then fix) = 12 — S x + 6 ; f"(x) = 36 jr 2 — 8, etc. el. calc. — 4 50 DIFFERENTIATION 2. Given 3 . Given 4 . Given 5 . Given 6. Given f(x)=e ax \ prove f (n \x) = a n e az . f(x) = log e (i - x) ; prove f M (x) = y — x 3 log e x\ prove d*y _ 6 dx* x (— i) n | 7 z — i 0 *)* , . r , d 3 y 2 cos x y — l°ge sin x; find ^ = y = e 2x ( x 2 — 3 + 3) ; find d z y dx 3 7 . Given -b + ys — z* or b 2 x 2 + a 2 y 2 = aW, to find d 2 y b 2 From Ex. 2, Exercise 8, ^ = _ dx bbc_ a y d-y dx 2 dx 2 a *y^x)~ Vx^ay d 2 y dx 2 a 2 b 2 y — b 2 a-x‘^- a*y 2 ’ dy then substituting for and reducing, d 2 y _ b 2 (a'y+b 2 x 2 ) _ b* a 2 y s d 2 y _ 4 fi 2 dx 2 8. From y 2 = 4 fix, find ^4 = = - S 9 . From y 2 — 2 xy = a 2 , prove dx d 2 y _ a 2 dx 2 — (y — x) 3 xy CHAPTER III APPLICATIONS 21. Tangent and Normal. For all applications of the Calculus to Geometry the fact established in § 14 is of fundamental importance, viz. Theorem. The value of the derivative of y with respect to x found from the equation of a curve in rectangular coor- dinates gives the slope of the tangent at any point on that curve , or — = slope of tangent. dx If we wish the slope at any particular point (x',y'), we have to substitute x' and y' respectively for x and y in the general expression for — • Let (~-\ be the value of , dx \dx) — after this substitution, then we have from Analytic dx Geometry, Equation of the tangent at (x 1 , y') is (1 7 ) - v '= (d £) [x - x ' ) - y-y Since the normal is perpendicular to the tangent, and dy\ from ( 1 1 ), [-£) = 7^7, we find Eqtiation of the normal at (pc’ , y') is (18) 51 y-y’ = - (^j (x - x'). 52 APPLICATIONS EXERCISE 10 1 . Find equations of tangent and normal to the parabola / 2 = 4 x + 1 at the point whose ordinate is 3. Substituting 3 for/, we find x = 2, hence (pc' , /') is (2, 3). Differen- .. .. dy 2 . /^/Y 2 t,atm& »=7 ■■[■£) =- 3 - Ans. tangent, 7 . x — 3/ + 5 = 0; normal, 3 jr + 2/ — 12 =0. 2. Find equation of tangent to the ellipse b^x 1 + a 2 / 2 = aW at (x', /'). ^u'br-)- a^y'/ = a 2 ^ 2 . 3 . Show (Fig. 12) that the subtangent M Y T X =— y'( — \ , and the , , \ ' \dy 1 subnormal M X N X = /' \-d- \ • 4. Prove that the subnormal in the parabola / 2 = 4 px has the con- stant length 2 p. 22 . Sign of the Derivative. An important question is the following : Is the function increasing or decreasing as the variable passes through a given value a ? The phrase “ passing through a ” is understood to mean that the series of values assumed by the variable is an increasing sequence including a, i.e. on the graph of the function we proceed from left to right. In Fig. 12, as we APPLICATIONS 53 pass through P x the ordinates are decreasing, while at P 2 the ordinates are increasing, and since the ordinates repre- of the tangent, we have the result : The function f{x) is increasing or decreasing as x passes through a according as f(a ) is greater or less than zero. At P 3 and P i (Fig. 12) the tangent is parallel to XX 1 , and therefore f'(x) vanishes at these points. For such values of x, therefore, the rule just given does not enable us to answer the question proposed. If, now, for any value of x, say x=a, the second deriva- tive l ~t,, or f"(x), is positive, then as ,r passes through a, dx- the first derivative f\x), or tan 7, must be an increasing function of x, i.e. 7 must be increasing as x passes through a ; and therefore as we pass along the curve from left to right, the tangent is rotating counter-clockwise, and the curve is accordingly concave upward (as at (a), Fig. 13). On the contrary, if f"(d) < o, the reasoning shows the tangent to be rotating clockivise as we pass along the graph through x — a, and hence the curve is concave downward ((f), Fig. 13). Fig. 13 54 APPLICATIONS The result is : A curve is concave upward or downward as x passes through a according as the value of the second derivative r , , -jx 2 for x = a is greater or less than zero, dfy As before, if = o, the rule just given does not enable us to decide. If = o for x = a and changes sign as *- passes through a, then at x = a we have a point of inflec- tion (Aj and P b , Fig. 12). exercise 11 1. Show that the following functions are either always increasing or always decreasing, and draw the graphs in each case : (a) tan*; (b) (c) log*; (d) 2. Show that y = sin* has a point of inflection at each intersection with XX 1 . 3. Determine the points of inflection of y = (* — a) s + b. Atis. ( a , b). 23. Maxima and Minima. A function fix) is said to be a maximum for x = a when f(a) is the greatest value of fix') as *- passes through a. A function fix) is said to be a minimum for x = a when fia) is the least value of fix) as ** passes through a. In other words, a maximum value is greater than any other in the immediate vicinity, and similarly for a mini- mum value. It is not to be inferred that a maximum value is the greatest of all values of the function ; on the con- trary, a function may have several maxima. APPLICATIONS 55 Graphically, at a maximum we have a highest point (. P l and P 3 , Fig. 14), at a minimum a lowest point \ p 2 and Pfi Fig. 14 Since, by definition, if f(a) is a maximum, f(x) must be an increasing function for x < a and a decreasing function for x> a, we have (§ 22): Theorem. If f(a ) is a maximum value of fix), then the first derivative fix) must change sign from positive to negative as x passes through a. By similar reasoning for a minimum, we find a change in sign from negative to positive must occur in fix). In either case, therefore, /'(hr) must change sign. If we now assume that fix) is continuotis for x — a, we see that f(a) = o ; that is, the tangent at a highest or lowest point must be horizontal and P 2 in Fig. 14). If, on the con- trary, fix) is not continuous for x = a, then the change in sign occurs by passage through co ; i.e. the tangent becomes parallel to YY' , as at P 3 and P 4 . This case is, however, of minor importance, and is omitted from further con- sideration. Furthermore, if f\a) o indicates a lowest point (P 2 ). 56 APPLICATIONS We have therefore the following Rule for determination of Maximum and Minimum values of a function fix). Find the first derivative fix), and get the roots of the equation f'( x) = 0. First Test. If f'(x) changes sign as x passes through any root a of the equation f'(x) = 0, then f(a) is a maxi- mum or minimum value according as the change is from + to — , or from — to + . Second Test. Find the second derivative f'(x) ; then, if a is any root of f'(x) = 0. f(a) is a maximum if f" (a) < 0. and a minimum if f" (a) > 0. If, however, f'(a) = 0, we must use the first test. examples 1 . Examine the function x 3 — 3 x 3 — 9 x + 5 for maxima and minima. Placing f{x)=x 3 — 3 x 2 — gx+ 5, then f'(x)=3x 2 —6x—g, and the roots of 3 x 3 — 6x— 9 = 0 are x= 3 and — 1. Now f"(x) = 6x—6, and f" (3) = 12, /"(— 1) = — 12, hence by the Rule, Second Test, f(2>) — — 22 is a minimum value, and y( — 1) = 10 is a maximum value of the function. The student should draw the graph. 2 . Examine the function ^ f or maxima and minima. (x+ i) 3 Here Differentiating and reducing, we find r, v (X — i ) 2 fix) - !—■ J y ’ (x+ I ) 3 f (x) (^ — 1 ) 6 r ~ 0- J K 1 ix+ iy APPLICATIONS 57 The roots of /' (x) = o are therefore x= I, x= 5. We now apply the First Test, since it is unwise to form the second derivative. Taking account of the signs only, we have When*x< 1, f'(x) = — £ — ^ — — When x> 1, /'(x) = — ^ — - = + Hence f{x) is a minimum when x = 1 . When x < 5, /'(x) = - ^ = + 1 , . . y + I Hence /' (x) is a , Trl _ n , \ ( + )( + ) f maximum when x= c. When x>5, f\x)~- ’ = - j ^ /(i) = o is a minimum, and /(5) = ^ a maximum value of the function. EXERCISE 12 1 . Examine the following functions for maxima and minima : (*) x 2 — 3 x+ 5. Minimum value 1 1 . Ans. (*) X Max. value min. value — \. Ans. I +x 2 (0 6 x+ 3 x' 1 — 4 x 3 . Max. value 5, min. value — J. Ans. (d) x 3 — 3 x 2 + 6x. No max. or min. values. Ans. 0) <2X 2 + 2 bx + c. If a > 0, min. value — — — , if a < 0, a then — — — is a maximum. Ans. a (/") iov/8 x — x 2 . Max. value 40. Ans. This function is a maximum or minimum according as 8 x — x- is a maximum or minimum, hence f a constant factor or a radical sign may be dropped in investigations of this sort. * We consider values of x differing only very slightly from the number on the right of the inequality sign. t If u is any polynomial in x containing 710 multiple factors, we may show that fu is a maximum or minimum only when u is a maximum or minimum. For if 58 APPLICATIONS 2. Divide the number a into two such parts that their product shall be a maximum. Hint. If ur is one part, then a — x is the other, and the function to be examined is x(a — x) or ax — x~. Equal parts. Ans. 3. Divide the number a into two such parts that the product of the in th power of one and the «th power of the other shall be a maximum. In the ratio m : n. Ans. 24. The subject of Maxima and Minima is one of the most important in the applications of the Calculus to Ge- ometry, Mechanics, etc. It is often necessary to derive the expression for the function to be investigated, and in testing this, attention should be paid to the remark in Example i (/) of the preceding exercise. EXERCISE 13 1 . A box with a square base and open top is to be constructed to contain 108 cubic inches. What must be its dimensions to require the least material ?* Base 6 inches square, height 3 inches. Ans. 2. The strength of a rectangular beam varies as the product of the breadth b and the square of the depth d. What are the dimensions of the strongest beam that can be cut from a log whose cross section is a circle a inches in diameter ? f Breadth is % a inches. Ans. H — so that fix') f(x) = Va, fix) = — - — — , and f"(x) = — - — 1 vanishes only if — = o, and then f" ( x ) has the same sign as dx dx 2 108 * HINT. Let JV be the side of the base, y the height, then x^y ~ 108, i.e. y - x- and since the material is x 2 + 4 xy, we find by substituting for y the function *2 + 432 . t Hint. The strength therefore equals W 2 multiplied by some constant, which may be dropped by the remark of § 24. But d 2 = a 2 — ^ 2 ; hence the function is 1 5 (a 2 — b 2 ) , b being the variable. APPLICATIONS 59 3 . Find the dimensions of the stiffest beam that can be cut from the same log as in 2, given that the stiffness varies as the product of the breadth and the cube of the depth. Breadth \ a inches. Ans. 4 . The equation of the path of a projectile (see Fig. 16) is ,-2 y = x tan a. gx< 2 VS COS" « where a is the angle of elevation and v 0 the initial velocity. Find the greatest height v 0 z sin 2 u Ans. 5 . Find the dimensions of the rectangle of greatest area that can be inscribed in the ellipse b 2 x 2 -\-a 2 y' 2 — a 2 b 2 . Ans. Sides are aV 2 and bV2. 6. Find the altitude of the right cylinder of greatest volume inscribed in a sphere of radius r. Altitude = IT. Ans. v 3 6o APPLICATIONS 7 . Assuming that the brightness of the illumination of a surface varies directly as the sine of the angle under which the light strikes the surface and inversely as the square of the distance from the source of light, find the height of a light placed directly over the center of a circle of radius a when the illumination of the circumference is greatest. From Fig. 17 , the bright- ness at P is given by k sin 0 kx X \d — - Fig. 17 d 2 d 3 V ( a 2 + x 2 ) Hence the brightness is a maximum when •2\3 ( (a 2 + ,(a 2 + * 2 : x' 2 (d 2 + x' 2 ) 3 is a maximum. .*•= . Ans. V2 25 . Expansion of Functions. By actual division — I “f" X -f- X -f- • • • -f- X n -j” 09 ) V ' I — X where n is some positive integer. In this simple way we may find for the function 1 an equivalent polynomial all of whose coefficients save that of x n+l are constants. By transposition (19) becomes (20) I — a: — ( I + x + x 1 + A" 3 + ••• + x n ) = I — x „7!+l Now let x be some number numerically less than 1, say x=.5, and suppose we wish the value of correct within one one-hundredth, i.e. correct to two decimal places. Let us then determine for what values of n the term 1 1 —x equality x n+l when x=.$ is less than .01, i.e. solve the in- 1 1 - -5 • 5" +1 <.oi. We find n>6. APPLICATIONS 6l Furthermore, if ;r is numerically less than .5, 1 — x and x n+1 are less than for .r = .5, so that taking n — 7 (J.e. >6), ■ — ~ x 8 < ,oi for every value of a- not numeri- cally greater than .5. And we now see from (20) that the function ^ 1 may be replaced by the polynomial I + x + x 2 + x 3 + x* + x 5 + x 6 + x~ for all values of numerically equal to or less than .5 if results correct only to hundredths’ place are desired. Precisely the same reasoning holds for any value of x numerically less than unity, since for any such value jtr n+1 can be made as small as we please by taking n sufficiently great. But this reasoning does not hold for any value equal to or exceeding 1 numerically. We may then state this theorem : For any value of x numerically less than unity , the func- tion — - — may be represented with any desired degree of accuracy by a sufficiently great number of terms of the polynomial, I + x + x 2 + x 3 + •••. The Differential Calculus enables us to obtain a similar theorem for many other functions, as will now be explained. In all practical computations results correct to a certain number of decimal places are sought, and since the process in question replaces a function perhaps difficult to calcu- late by a polynomial with constant coefficients, it is there- fore of great practical importance in simplifying such computations. 62 APPLICATIONS 26 . Theorem of the Mean. If f(x) and f(x) are con- tinuous as x varies from a to b, then there is at least one value of x, say x v between a and b, such that (21) f(b)-f(a) b — a =f(* 1 ). In Fig. 11, f(b)—f(a)=CB, b — a = AC, A 6 )-/(a) b — a = slope of AB, and at each of the points P x and P 2 the tangent is parallel to AB, and hence (21) is true if x x is the abscissa of P x or P 2 .* Multiplying (21) out gives (22) /(b) =f(a) + (b - a)f'(x 1 ), where it must be remembered a >x x > b. A more general theorem than (21) is enunciated as follows : If f(x) and the (;/ + 1) successive derivatives f'(x), f"(x), f (v+ 1 ) (x) are continuous when x varies from a * This proof of the theorem of the mean is not mathematically rigorous, but merely illuminates the significance of (21). The student should draw other figures, and especially such that the necessary conditions of the continuity of f[x) and f'{x) fail. APPLICATIONS 63 to b, then there is at least one value of ;r, say x v between a and b such that (23) /W=/W + ( -^/ 0 ) + ^|^/"W + ( -^j /"(?)+ -+ ( -^ r / < '>(«) (* - a) \»+ 1 n + 1 The proof of (23) is beyond the scope of this book.* The student should, however, carefully note the law by which the expression on the right is constructed. Putting for b in (23) the variable -r, we get Taylor s Theorem , (24) f(x) = f (a) + - --- - /*(«) + ( -| 2 - )2 /^) + ••• (x — aY +x . n/ . + — / 1 (-^1), where a(o) +jy/"(o)+ ... +jf / ( ">(o). By “ expansion of a function ” is meant the forming of this polynomial. Of course 71 is indefinite, and must be taken great enough to give the desired degree of accuracy. It is of greatest theoretical importance to determine for what values of x the polynomial represents the function when n is taken indefinitely great. This consists in exam- ining for what values of x Limit V \n + 1 for this term is the difference between the function and the polynomial. EXERCISE 14 1. Expand sin x. Since f(x) = sin x, and for x = o, /(o) = sin o = o ; then f( x ) = cos x, and for x = o, /'( o) = 1 ; f" (r) = — sin x, f (o) = o ; f"(x) = — COS X, f"(o) = — I i f n (x) = sin x, ffo) = o ; etc. etc. x 3 x 5 x~ x 9 Hence sin x= x — , 1- , . 1- . etc. [3 LL IZ. 12. 2. Show that the expansion of cos a- is X 2 X 4 X* X s COSX= I b . ,-w + 1-5- - etc - h Li II 11. * Namely, for all values of x such that the 11 remainder” 2-n+l T+T /["+« (^i) is less than the limit of error. This question is often difficult to settle. APPLICATIONS 65 3 . Expand e x . Since f(x) = e x , and all its derivatives are likewise e z , while e° we obtain x *2 *s ^4 ^5 e T = 1 H 1- 1 h . h , hi h • • • • ■ ll ll [4 [5_ Putting x = 1, we find 1 1 1 1 e — I H h, h, h, h 1 ll ll 1 4 h the expression given in § 10. The expansions of sin x, cos x, and e x are remarkable in that they hold for every value of x, positive and negative. 4 . Prove the following expansions l -2 - v -3 ( v y2 -y-O y-4 yO \ T - T + y- T + y--) (£) (1 + x) m = i + mx + m ( m ~ 1 „ 2 II. 13 ** + W(W - l)(w-2) ^ + (£) is the binomial formula. These expansions hold only for values of x numerically less than 1 . Taylor's Theorem (24) differs from (25) in that we are to consider values of the variable x near some given num- ber a , since (24) is a polynomial in (x — a) in the same sense that (25) is a polynomial in x. It is evident that no greater difficulty arises in the application of (24) to a given function than has been already pointed out. 27 . Differentials. From (23) we are able to find an ex- pansion for the increment of a function in powers of the increment of the variable as follows : Write b = x + Ax, a — x, ,\ b — a = Ax, and (23) be- comes, after transposing fix), (26) fix + Ax) -fix) = Ax f{x) + ^p-/' \x) + • • •, or (27) A fix) = fix) Ax +f"ix) Now, if we suppose Ax to diminish toward zero, the first term fix) Ax of the right-hand member will ultimately el . calc . — 5 66 APPLICATIONS greatly exceed the sum of the remaining terms, since these contain higher powers of Aw. For this reason fix) Aw is called the principal part of the increment of fix). Also, when we wish to emphasize the fact that the variable Aw is to approach zero as a limit, we write dx, called differen- tial x, instead of Aw, and the principal part of the incre- ment f(x)dx we call the differential of the function; that is, (28) d fix) = fix) dx. The following definitions are fundamental: A differential (or infinitesimal) is a variable whose limit is zero. The differential of the independent variable is an incre- ment of that variable whose limit is zero. The differential of the dependent variable is the princi- pal part of the increment of that variable, and equals the product of the derivative and the differential of the inde- pendent variable (28). From (28), we see that if y is a function of w, then (29) dy = ^dx. EXERCISE 15 1. Prove by (28) and (29) the following differentials : (a) d (3 w 2 ) = £>xdx. dx if d log e x = — • dx (e) d Vi — x = — ■ 2 V 1 — x (f) d sin 2 x = 2 cos 2 xdx. (f) de z — e T dx. sec 2 (d) dx m = mx m ~^dx. d tan ( ..'j — pi (h) If y = wlog e w, then dy =(1 + log e w)^w. 2. If / = uv , then dy = ( u — - + v \ dx = u — dx + v — dx, or dy = udv + v du. \ dx dx / dx dx dx. APPLICATIONS 67 3 . Show that ! 0 - v du — u dv v 2 4 . State the rules I-V for differentiation in terms of differentials instead of derivatives. 28. We may write (27) after replacing A,r by dx, ( 30 ) A f{x) =f(x)dx + dx 2 + pj-' dx +~ ^j. Now, since by (28) f{x)dx is the differential of the function, (30) shows that A fix) and df{x) differ by a term containing the factor dx 2 . Such a quantity is called a differential of the second order ; in general, any quantity containing as a factor the product of tzuo differentials is thus designated. The increment of a function differs from its differential by a differential of the second order. EXAMPLES 1. Differential of a product uv. Let u — AB, v = AC, then uv = area ABCD. If du = BB’, dv = CC, then A {uv) = area AB' CD’ — area ABCD = area CDC' E + area BB'DE' + area DE' ED' = u dv + v du + du • dv. Now du ■ dv is a differential of the second order, .-. principal part of A (uv) is udv + vdu; i.e. d{iiv') = udv + v du. (Cf. Ex. 2, § 27.) 68 APPLICATIONS 2 . Differential of an area. Consider the area aAPM bounded by any curve, the axis XX' and the ordinates aA, MP, and call this area u. Then if MN = dx, A u = axtaaAQN— area aAPAI= zrta MPQN. \u = y dx + area PSQ. But area PSQ < dx- dy. .-. PSQ = k dxdy, where k is some number < i. Hence area PSQ is a differential of the second order, and du=ydx. The differential of the area bounded by any curve , the axis XX', and two ordinates is the product of the ordinate of the curve and the differ- ential of the abscissa. 3. Differential of the volume of a solid of revolution. Let the solid be generated by revolving a curve APQ around XX', and denote the volume APA'P by v. If dx= MX, then \v = volume AQA'Q' — volume APA'P', or Sv = volume of the cylin- der PSP'S' + volume gener- ated by the curvilinear A PSQ. Now the volume of the cylin- der PSP'S' = iry'-dx, since y — PM = radius of base and dx = altitude. The volume generated by the curvilinear A PSQ < volume generated by the rectangle PRSQ, and this last volume = ttNQ~ • MN — ttMP ' • MN — tt (2 y dy -f dyfdx. Fig. 2i APPLICATIONS 69 We see therefore that Av = 7ry 2 dx + a differential of the second order, i.e. dv = iry-dx. The differential of the volume of a solid of revolution generated by revolving any curve around the axis XX' equals n tunes the product of the square of the ordinate and the differential of the abscissa. 4 . Show that the differential of the area 71 bounded by a curve AP and two radii vectores OA and OP is given by du — \ r 2 dO, where (r, 6) are the polar coordinates of P. CHAPTER IV INTEGRATION 29. Indefinite Integral. Integration consists in finding a function of which a given differential expression, such as x dx, sin x dx, — , etc., is the differential. The function n thus found is called the integral of the given differential expression, and the operation is indicated by prefixing the integral sign Thus, since J'x dx = 1 j-2 . O --V i d (|^ x 2 ) = x dx, J* dx = x, ^ sin x dx = — cos x, etc. In general, § fix) dx means to find a function F( x) such that dF(x) = f{x)dx, i.e. Ax)= f* F(x) - Constant of Integration. Since d( ^ x 2 + C ) also equals a' dx, no matter what the constant C is, we have ^ x dx = jX 2 + C, where C is any constant whatever, called the constant of :foi 7° integration. We see, therefore, that a given differential INTEGRATION 71 expression may have infinitely many integrals, found by giving to the constant of integration different values. Thus J' f{x) dx = F{x) + C, and since C is unknown and indefinite , F(x ) + C is called the indefinite integral of f{x)dx. Of course, the same differential expression has an in- definite number of distinct integrals, but what has just been said shows that the difference of any two of these must be a constant. 30. Rules for Integration. From Rule V in differentia- tion, if v is any function of ;v, and k a constant, then dL( KV \ = k—, i.e. dinv) = k dv. dx dx Integrating, we have, since if two differential expressions are equal so are their integrals equal, j " k dv = j* d{KV), or, since J*d(/cv) = KV, kv =J^/cdv. But *f dv = KV. (30 .’. J' k dv — kJ ' dv. XXIII. A constant factor may be written either before or after the integral sign. The chief application of XXIII is to be found in cases like the following : 72 INTEGRATION To work out J* xdx . If we multiply xdx by 2, we have an exact differential, since d (x 2 ) = 2 xdx, but by XXIII, J*2 xdx — x 2 ; 2 xdx — 2 xdx, -i From (31) we may also write xdx= — 2 (32) ff(x)dx=~ j % Kf(x)dx Integral of a Sum of Differential Expressions. If u and v are functions of x, then d(u + v) = — (11 4- v) dx = du + dv. dx J*(du + dv) =j*d(u + v) = u + v = j' du +J'dv. This result gives Rule XXIV. The integral of any algebraic sum of differential ex- pressions equals the same algebraic sum of the integrals of these expressions taken separately. That is, eg., j (x + 3) d x= (xdx + 3 dx) = xdx + j3dx=%x 2 +3x + C. 31. From any result in differentiation may always be derived an integration formula, and we now proceed to obtain some of the simpler ones, making use of § 18. INTEGRATION 73 Since by VIII, d(v m+1 ) = (in + i )v m dv, then, integrating, m + i)v m dv = ( in + i (XXIII) ( 33 ) From IX, v m+l m + i dlog e v = dv 5 V ( 34 ) In the same way we might go through with each formula in § 1 8. It will suffice for our purpose to tabulate a few of the results : XXV. f v m civ = v>n 1 + C ( m d - J m + 1 XXVI. f^ = loge v + C. J V XXVII. ^a v clv = ° v +C. loge a XXVIII. y sin v dv = — cos v + C. XXIX. ^ cos v dv = sin v + C. XXX. f dv — arc sin v + C. ' d a- — v 2 a XXXI. f ,^- = 1 arctan^ + C. J a% + v% a a 74 INTEGRATION 1 . Find EXAMPLES dx S' V i — x This is the same thing as J* (i — x)~ 7 dx, which resembles XXV. For put i — x — v, then — dx = dv, or dx = — dv. J*(i — x) 2 dx = ^v 7 — dv = — ^v -dv. by XXV and by substituting again, T v 2 dv = — + C, J i f — — x ■ — — 2 Vi — x + C. J V i — x 2 . Work out Si 3 ax dx ¥7 2 Taking out constant factor 3 a (XXIII), this becomes x dx C x dx 3 > a \ tt J c- — b-x and this resembles XXVI. dv For put c 2 — b 2 x 2 = v,.-. — 2 b 2 xdx = dv, or xdx — 2 b- _ dv r xdx r id 1 3 a C dv 3 a , ^ J c 2 — b 2 x 2 j v 2 b- J v 2 b' 2 C ?> axdx = _ 3_^ log (cl _ b 2 x 2 ) + C. J c 2 - b 2 x 2 2 b 2 J 3 . Find S dx 9 + 4 X 2 ' This resembles XXXI, if a = 3, 2 x = v. Then 2 dx = dv, and since the given integral by (32) is the same as 1 C z dx or I r dv_ t 2 J 3 2 + (2 x ) 2 2 J a 2 + v 2 we find by XXXI, f — — — = - arc tan — + C. J 9 + \x 2 6 3 INTEGRATION 75 By studying the above examples the student will see that integration depends upon comparison of the given integral with certain standard forms. To be able to tell quickly what form the given integral resembles is abso- lutely essential. Tables of standard forms* have been constructed con- taining all integrals occurring in ordinary work. EXERCISE 16 1 . Prove the following integrations : {a) J {ax + bx 2 ) dx=\ax 2 + \ bx 8 + C. (Use XXIV.) sin xdx = log c ( i - cos x) + C. (*> /“ (c) J y/a 2 - ar 2 xdx _ _ i( a 2 _ x iy + c. (Use XXV, v = a 2 - x 2 ). W iO»3X+C. (^) y £ -x dx — — e-* + C. <£) S (/> y dx xdx Va 2 +x dx '• Va 2 + x 2 f C. V i — 4 . 1 arc sin (2 x) + C. {h) f— h- = -logTi -ar) + C. 2 J i — ar (0 y ^ xdx — §xi + C ; + C . J J X° 2 X* C sin x (j) i tan xdx = - log c cos a- + C. (Put tan ar = and use J ° x cos ar XXVI.) (k) fan* xdx =\x- f sin2ar + C. (Put sin 2 ar = |(i - cos 2 ar).) 2. Special Devices in Integration. (a) By partial fractions , when we have to integrate a rational frac- tion times dx, and this fraction can be replaced by partial fractions. E.g . , B. O. Peirce’s A Short Table of Integrals, Ginn & Co., 1899. 76 For example, INTEGRATION dx JV i A B a 2 — x 2 ~ a — x a + x 2 a Putting and clearing of fractions, i = x{A — B) + a(A + B). .-. A — B = o, a(A + B) = i, or A = B = =i>G^D +c (b) By change of variable. Find J" V a 2 — x 2 dx. Substitute x=azosQ\ dx = — a sin 6 dd, y/a 2 — x 2 — fa 2 — a 2 cos 2 6 = a sin 0 , and J* y/a 2 — x 2 dx = — a 2 j sin 2 8 dQ = — ^-0 + — sin 20 + C by Ex. i (£). Now 0 = arc cos sin 2 0 = 2 sin 8 cos 6 = 2y[i ••• i’ V ^ 7 x' 2 dx = arc cos - + - xV a 2 — x 2 . 2 a 2 3 . Prove J; = INTEGRATION 77 Let now a equal the initial sum of money ; that is, the sum started with, so that P = a when t = o ; substituting these in this equation, we have log e a=C, so that (i) becomes log ,, P = — t + log e a, or, transposing, I Of t P - log e a = — t, or log e (-) = — ioo \a J ioo P = ae m . Ans. 5. Find the relation between s (space) and t (time) for uniformly accelerated rectilinear motion. d v d'l) Since the acceleration — is constant, say f, we have — = f. dt dt Multiplying by dt , dv = f dt , and integrating, v —ft + C. To determine C, let the initial velocity be v 0 , i.e. v = v 0 when t — o, or v 0 — o+C. v =ft + v 0 . Since v — ft + v Fib) -Fib 5 )=fix 6 )Ax. Adding the six equations (39) and (40), we find (41 ) Fib) - F{a) =fix x )Ax +f(x 2 )Ax + /( x s )Ax +fi x i)& x +fi x 5 )& x F fix^jA.r. But fix-^Ax = area of the rectangle aPP x b v fix 2 ) Ax = area of the rectangle b-J) x P 2 b 2 , etc., so that the sum on the right equals the area aPP x p x P 2 p 2 P z p 3 P,p,P h p h Qb ; t.e. (42) Fi b ) —F{a) = area between the broken line FF iPi-PnQ and XX', EL. CALC. — 6 82 INTEGRATION and this is true independently of the number of parts into which ab is divided. Hence for any number n of equal parts (43) Fib) —F(a) = f{x^)\x +f(x 2 )kx 4 +f{x n ) Ax, ( 44 ) and A.r = b — a Equations (43) and (44) hold when n increases without limit, and then A;r becomes dx (§ 27), i.e. a variable whose limit is zero. F(b)— F(a)= L™* (f(x l )dx+f(x 2 )dx + ... +f(x n )dx), or, by (38), (45) f a f(x)dx = (f(x 1 )dx+f(x 2 )dx + ... 4 -f{x n )dx). And now we see very clearly why I f(x)dx gives the d area under th£ curve, for as n increases, the broken line PPxPxP^p^ ••• p b Q approaches the curve itself, and the sum fix-P)dx + -.. f(x n )dx always represents the area under this broken line. Integrating between limits is accordingly spoken of as “summing up”; the integration sign j is historically a distorted S, the first letter of swn. But let the student not forget that the definite integral is not a sum, but the limit of a sum , the number of terms increasing without limit , and each term itself diminishing toward zero. The problem of finding the area is then to be thought of thus : Divide the interval on xx' into an}' number of equal parts, and at a point within each division erect an ordinate to the curve ; construct the rectangles on the divisions as bases, with the corresponding ordinate as INTEGRATION 83 altitude. Then finding the area consists in summing up these rectangles and taking the limit of this sum as the number of divisions increases without limit. As an example of the great number of problems in Physics and other branches of Mathematics which involve in their solution definite inte- grals, consider the following : To determine the amount of attraction exerted by a thin, straight, homogeneous rod of uniform thickness and of length l upon a material point P of mass in , situated in the line of direction of the rod. dx - Fig. 25 Imagine the rod (see Fig. 25) divided up into equal infinitesimal portions (elements) of length dx. If M = mass of rod, then -jdx — mass of any element. The law of attraction being Newton’s Law, i.e. attraction = product of masses h- square of distance, then M idx attraction of element dx on P = ■ (x+a) 2 and the total attraction is the sum of these from x = o to x — l. M Force = f -Jo mdx 1 l Mm n dx (x+a) 2 l Jo (x+a) 2 or integrating, Force = ( 1 f d \ = — . Answer. 5 l \ 1+ a a) a(a + l) CHAPTER V PARTIAL DERIVATIVES 35. Functions of More than One Variable. In the pre- ceding chapters we have been concerned with functions of one variable ; i.e. the variable function depended for its value upon the value of a single variable. Such functions do not by any means suffice for the applications of the Calculus. In fact, the student is already familiar with many examples of a variable whose value depends upon those assigned to two or more distinct variables. Thus the area of a rectangle is a function of two variables, viz. the two sides ; the volume of a gas depends upon both the pressure and the temperature ; the volume of a parallelo- piped depends upon the three edges, etc. Notation. If the value of a variable u depends upon two variables, x and y, and can be computed when values are assumed for x and y, then we write precisely as in § 3, (46) u =f(x, y). Similarly for a function of three variables, u = cf)(x, y, 2), etc. 36. Partial Differentiation. As in § 12 the important question arising here is how to determine the manner of variation of the function when the variables change in value. But we have greater latitude here than in § 12. For in (46) we can ask ourselves, 84 PARTIAL DERIVATIVES 85 first , how does n vary when ;r alone varies and y remains constant ? or second , how does it change when remains constant and 1 y varies ? or third, in what manner does u vary when both and y change independently of each other ? Thus let u- xy, x and y being respectively the base and altitude of a rectangle ; if y remains constant (say y = b), 11 gives the area of all rectangles of a certain altitude b ; and if x = a constant, say a , then u represents the area of all rectangles with common base a. But if x and y both vary independently, then we are to consider all possible rectangles. Now the first and second cases do not differ in the least from § 12, for we really have in the first, u a function of x alone, and in the second, u a function of y alone. We can therefore form, first, the increment quotient (§ 13) when x alone varies, and this is ( 47 ) An _ f{x + Ax, y) — f(x, y) _ Ax Ax second , the increment quotient when y alone varies, which is r.st Au - f(' r > y + Ar) -f(x, y) (4§) Ay Ay For example, in the area of rectangle already used, u — xy, A u (x + Ax)v — xy ' A u . -t— = t — = y, and --- reduces to x. Ar A.r J A y Finally, we can, as in § 14, find the limits of the func- tions in the right-hand members of (47) and (48), in (47) when Ax approaches zero, in (48) when Ay approaches 86 PARTIAL DERIVATIVES zero. The results are called the partial derivatives of u or f[x, y) with respect to .v and y respectively, and this step of passing to the limit we indicate on the left by replacing the A’s by round d’s, so that ( 49 ) du T . . — — Limit ox ffx + Ax,y)-f(x,y) \ Ax Ai=o’ ( 50 ) du T . — = Limit dy 7(l y + A v) ~/(l j>A Ay Jav= o ’ The partial derivatives — , — are then to be calculated dx dy by the rules of Chapter II, the independent variables being respectively ;r and y. EXERCISE 18 1 . Find the partial derivatives of : 0 * = iog,(r). (2) rc = arc tan - (3) u = xv. A us. du _ 1 dx X Ans. du _ dx X 2 A ns. du _ dx yx y- du _ ' dy~ y_ . 2 +y 2 ' i • §11 dy y du dy x 2 +y 2 Partial Differe 7 itials : By § 27, (29), the differential of u, when x alone varies is — dx, and when y alone varies equals — dy ; these are dx " dy called the partial differentials of u. C r\ — dx = partial differential of u, when x alone varies ; I dx r) — dy = partial differential of u, when y alone varies. dy PARTIAL DERIVATIVES s; 37 . Total Differentiation. We have yet to discuss the third case of § 36, viz. required the change in u when .rand / vary independently. If Ar, A/, and Az< are the increments of these variables, then from (46) we have (52) A u = /(r + Ar, / + Ay) -f(x, y). By adding and subtracting f(x, y + Ay) in the right-hand member, (52) becomes ( 53 ) Aa=/(r+Ar, y + Ay) -/(r, y + Ay) +f(x, y + Ay) -f(x, y). Consider now the last two terms, f(x,y + Ay) — f(x, y). This is the increment of u or f (x, y) when / alone varies. Hence, by (27), § 27, (54) f (+7 + A y) ~ = Ay + terms in higher powers of Ay. In the same way the first two terms of (53) give us, if we set u' =/(r, y + Ay), ( 55 ) / 0 + Ar,/ + Ay) -f(x,y + Ay) But also = dff Ax+ terms involving Ar 2 , etc. dx it! = f{x,y + Ay) = /(r,/) + |"A/ + terms in A/ 2 , etc., = — + terms in Ay, dx dx by (26), § 27. Differentiating with respect to r, we find ( 56 ) since H —f(X, /). Consequently, from (56), (55), and (54)' (S 3 ) becomes / c 7 \ Au = — Ar + — Ay + terms of higher degree in Ar, Ay. > dx dy Now letting Ar and Ay approach zero, i.e. become the infinitesimals dx and dy , then, as in § 27, calling the principal part of Au the total differential of u, we have ( 58 ) du = Yx clx + dy dy ' 88 PARTIAL DERIVATIVES From (51) and (58), then, we have the theorem: The total differential of a function of several variables equals the sum of the partial differentials. Example. In § 28, Example 1, was demonstrated the result d ( xy ) = xdy + y dx, which agrees with (58). EXERCISE 19 Find the total differentials of the following : (a) u = log,(£). A ns. du = / y\ xdy — y dx {b) u = arc tan 1 - 1 . Ans. du = ^ + a — ( c ) u = x y . Ans. du = xi'~ 1 (y dx + xlog e xdy). 38 . Total Derivative. We may in (57) assume that and y are not independent, but are functions one of the other, say y a function of x. Then u becomes also a func- tion of ;r alone, and we may therefore form the total derivative — - dx Dividing (57) by Ax and taking the limit for A.r = o, and .•. Ay = o, we have the result ( 59 ) du _ du _ f du \ dy dx dx \dyj dx a very important formula. Suppose in the illustration of the rectangle, § 36, we wish the deriva- tive with respect to the base x of the area u of all rectangles whose altitude / is double the base. Then du du dy u = xy, y = 2 x, ~=y, 3- = *, f L = 2 > dx dy dx du dx = y + 2x=4x. and (59) gives PARTIAL DERIVATIVES 89 Or, we may substitute for / before differentiation; i.e. u = x • 2x = 2x 2 , .-. — = \x, as before. dx Equation (59) is especially important as affording a proof of the method given in § 19. For in the example of that article, set u = ar 2 _ 3 xy + 2y -2 _ 3 . i.e. (60) u = o, and du dx o, or du + (3u\^ = 0 . dx \dy / dx du ' dy _ dx _ 2 x—t, y _2x— 3/ dx du~ —3 a- +4/ 2>x - 4/’ dy the same answer as before. This formula (60) is very useful. For further study of the Calculus the student is referred to : G. A. Gibson, An Elementary Treatise on the Calculus. London, 1901. Young and Linebarger, The Eleme 7 its of the Differential and Integral Calculus. New York, 1900. McMahon and Snyder, Elements of the Differe 7 itial Calculus. New York, 1898. Murray. A71 Eleme 7 itary Course m the I/itegral Calculus-. New York, 1898. CHAPTER VI ADDITIONAL EXAMPLES IN CONNECTION WITH THE GIVEN EXERCISES EXERCISE 1. PAGE 10 6 . Given f(pc) — x 3 — 6 x 2 + 5 ; then by § 3, /( 2) = (2)* - 6(2)' 2 + 5 = 8 — 24 + 5 = — n. Similarly, /(- 3) = (~ 3) 3 - 6( - 3) 2 + 5 = - 27 - 54 + 3 = - 76. Also f(x — t) = (x — i) 3 — 6 (x — i)' 2 4- 5, which reduces to = x 3 — 9 x 2 + 15 .r — 2. 7 . Given = 3 x 3 — 2 x 2 + 6 x, prove A-(o)=o; /■(!)= 7; F (- 2)= - 44; F(i -y) = 7 - 11 y + 7 y 2 - 37 s - 8. Given % _i_ i (x) = , prove the following: x — i (~ i)— ° ; <^>(i+^) = 2 + X 9. In Example 7, prove y± y - EXERCISE 2. PAGE 22 X \ / 6. Draw the graphs of 5 1 ; 1 Vi—x 2 ; csc.r; cot2t 7. Prove the following : Limit resell =00; Limit P A ll =00: Limit + 1 1 L Lx-iJ, 90 ADDITIONAL EXAMPLES 91 EXERCISE 3. PAGE 24 It is often convenient to place the variable y equal to the given func- tion of x. Then when x changes to x + h, y is replaced by y + A/, that is, A y is the increment of the function. The following examples illustrate this, and in them the increment of x is denoted by A.r instead of h, as heretofore. 9 . Given y = x " 1 — 3^+6; find A y. Substituting x + tS.x and y + Ay for x and y respectively, we have y + Ay = (x y A.r)' 2 — 3(x+ Ax ) + 6 = x 2 y 2 x- Ax + (Ax ) 2 — 3 x — 3 Ax + 6. Subtracting from this the original equation, we find Ay = 2 x • A.r + (A.r) 2 — 3 Ax, or Ay = (2 x — 3 + Aar) A.r. Ans. 10. Given y = x 3 ; find Ay. Ay = (3 x 2 + 3 • A.r + (A.r) 2 ) Ax. Ans. 11. Given y = 1 ^ ' x ; find A_y. x * x 2 — i + x • A x \ . Ay = ! Ax. Ans. x 2 + x ■ Ax 12. Given y = -^r ; find Ay. y/x Ay = — — Ax. Ans. (x + Ax) Vx + xy/x + Ax EXERCISE 4. PAGES 32-33 As above, if y is set equal to the given function of x to be differ- entiated, the General Rule of Differentiation (page 29) may be stated as follows in Four Steps : 1st step. 2d step. 3d step. 4th step. Substitute sc + Ax and y + Ay for x and y, respectively. Subtract the original equation from the new equation, and reduce. Divide both sides of this equation by Ax. Find the limit of this result when Ax approaches the limit zero, replacing || by 92 ADDITIONAL EXAMPLES 9 . Differentiate j = 3 j 2 + 2J + 5. Carrying out the Four Steps : i°. Substituting x + Ax and / + A/ for x and/, / + A/ = 3(x + Ax) 2 + 2 (x + Ax) + 5, or / + Ay = 3 x 2 -f 6 x - Ax + 3 (Ax) 2 + 2 x + 2 Ax + 5. 2 0 . Subtracting the original equation and factoring, we have A/ = (6 x + 3 Ax + 2) Ax. 3 0 . Dividing by Ax, Ay = 6 x + 3 Ax + 2. Ax . i\y dy 4°. Letting Ax approach the limit zero and replacing by 4^ = 6 x + 2. Ans. dx X 10. Differentiate / = 3 - y 1 + x 2 Working out the Four Steps gives : x+ Ax reducing, 3 °- y + A/ = A/ = A/=- 1 + (x Ax) ■ x + Ax x 1 + (x + Ax) 2 1 + x' 2 1 — x 2 — x ■ Ax (1 + x 2 )(i + (x + Ax) 2 ) Ay 1 — x 2 — x • Ax Ax - (1 + x 2 )(i +(x+ Ax) 2 ) dy _ 1 — 3 Ax. Ans. * ' dx ( 1 + x 2 ) 2 11 . Differentiate y = }x 3 -x 2 - 3X+ 5. ^ = x 2 — 2 x — 3. Ans. dx J 12 . Find the slope of the tangent to the curve/ = £x® - x 2 - 3 x+ 5 at the points where x = o, x=3, x= — 1, x = 1. From Example 1 1 and page 27, the general expression for the slope at any point is given by the value of i.e. slope = x 2 — 2 x — 3. ADDITIONAL EXAMPLES 93 Hence, substituting, slope at x — o is — 3 ; slope at x = 3 is 9 — 6 — 3=0; slope at x— — 1 is 1 + 2 — 3= 0; slope at x= 1 is 1 — 2 — 3 = — 4. 13. Find the general expression for the slope of the tangent to each of the following curves : (a) y = x i — 8 x 2 + 16. 4 x 3 — 16 x. 2 x 8 — 2 x 2 V) y = (0 y = 4 + x 2 8 a 3 (4 + x 2 ) 2 ' — 1 6 a 3 x Atis. A ns. A ns. x 2 + 4 a 2 (x 2 + 4 a' 2 ) 2 ' 14. Determine the coordinates of all points on each of the curves of Example 13 at which the tangent is parallel to the axis of x. Hint. Since at these points the slope of the tangent is zero, place the general expression for the slope equal to zero, and solve for ar. {a) (o, 16), (±2,0); (b) (2, i), (-2, (c) (o, 2 a). Ans. EXERCISE 8. PAGE 48 dy x — y 8. Given x 2 — 2 xy = a 2 ; prove ~ = — — • 9 . Given ( x — a) 2 + (y — b) 2 = r 2 \ prove ^ = — ^ ^ 10. Given x 2 + y 2 = a 2 ; prove dy_ dx dr 11. Given r 2 sin 2 6 = a 2 ; prove — - = — r cot 2 0 . do 12. Given the curve/ 3 = ; show that the axis of / is the tangent at the origin. EXERCISE 9. PAGE 49 If the variable / is set equal to the function of x to be differentiated, then the first differentiation gives the value of -]-■ The second de- & dx rivative, obtained by differentiation from this, is represented, as in d 2 y § 20, by (read, “the second derivative of / with respect to x"). 94 ADDITIONAL EXAMPLES Then further successive differentiation will give the third, fourth, fifth d^y d A y d h y derivative, and so on, which are symbolically ^4, etc. d 4 y 10. Given y — e~ z cos x ; prove — — 4/. d‘ 2 y 2 X 11. Given / = arc tan .r ; prove = — dx 2 ( 1 + x ' 1 ) 2 1 d 2 y 1 — x ^ 12. Given / = log e (1 + x 2 ) 2 ; prove — d i y . dx 2 (1 + x : ) 2 13. Given y = xcos x; prove —4 = x cos •*■ + 4 sin ar. . d 2 y 14. Given / = sin 2 2 ;r ; prove = 8 cos 4 x. d 2 y cl 2 15. Given x 2 + 2 xy — a 1 ; prove ^ d^y 16. Given y = tan 2 x + 8 log cos x + 3 x 2 ; prove -j~ — 6 tan 4 x. r . * 3 d 4 y [4 17. Given / = — ; ^ = (7^' d 3 y 4 a 3 18. Given / = (x 2 + a 2 ) arc tan - ; prove ^ ^ 2 - d^y — 24. x 19. Given y 2 + y = x 2 ; prove -dr, = — — 7 7 r dx 3 (i + 2/) 5 „ d 2 y a(m 2 — 1) 20. Given y 2 — 2 m xy + x 2 — a = o ; prove -yy. = -A ~ y ' ’ r dx 1 (/ — mx ) 3 ;r d 2 y 2 x(x 2 -3) 21. Given / = s ; prove -yy. = 7 1 + x 2 ' F dx 2 (1 + a.- 2 ) 3 EXERCISE 10. PAGE 52 5. Find equations of tangents and normals to the curve y 2 — 2 x 2 — x a at x = 1 . Alls. Tangents are / = 4 x + ^ ; / = — 1- x — Normals are / = — 2 x + 3 ; y = 2 x — 3 . 6 . Prove that subtangent and subnormal to the ellipse b 2 x 2 + a 2 y 2 oP> x ' 2 b^x 1 = a 2 b 2 at (x r , /') are — -p - — and — respectively. (Use formulae of Example 3.) ADDITIONAL EXAMPLES 95 7 . Find equations of tangents and normals to 2 x' 2 + 3/2=35 at / — 3. Ans. Tangents are 4.x± 9 / = 35. Normals are 9 a- T 4/ = 6. 8 . Find equations of tangent and normal to / = x 3 at (x', /'). Ans. Tangent is 3 x' 2 x — y — 2 y' = o. Normal is x + 3 x" 2 y — x'(3 x ' 4 + 1) = o. X 9 . Find equations of tangents and normals to y = — , at = 1 . Ans. Tangents are y = ± l- Normals are x = ± 1. 10 . Prove that the equation of the tangent to x 3 — 3 axy + y 3 = 0 at {x', /') may be written x" 2 x — ax’y — ay' x +y' 2 y = o. EXERCISE 11. PAGE 54 4 . Prove that the curve/ = and x = ± aVy x a 2 + :r 2 has points of inflection at x= o 5 . Examine the curve a 2 y = i x 3 — ax 2 +2 a 3 for points of inflection. Ans. ( a , | a) . 6. Examine y = x + 36 x 2 — 2 x 3 — x* for points of inflection. Ans. At x = 2, x = — 3. EXERCISE 12. PAGE 57 Referring to Fig. 14, page 55, we see that the conditions for maximum and minimum values of a function may be stated as follows, the graph of the function being supposed drawn : Maximum : tangent horizontal and graph below the tangent (e.g. at P^) ; Minimum : tangent horizontal and graph above the tangent (^.at Pg). If we set y equal to the given function of x, then the preceding may be stated thus : Maximum: and ^^<0: ax dx 1 Minimum : / - = o and > o. dx dx 2 96 ADDITIONAL EXAMPLES This is identical with the Second Test, page 56, and the process is summarized in the following rule : 1st step. Set y equal to the given function of x and work out „ dy , d-y expressions for ^ and 2d step. Set the expression found for equal to zero, and solve A (lOO for x. 3d step. Substitute each value of x thus found in the expression If for ^ 5 . If this result is negative, the corresponding value of y is a maximum ; if positive, the correspond- ing value of y is a minimum. To illustrate, solve the problem : To find the maximum and minimum values 2 X Following the rule, we have -f- 4 i°. Assume/ = then, differentiating and reducing, dy _ x 2 — 4 . dx ~ 2 x' 1 ’ . d 2 y _ 4 dx 2 x 8 Setting the expression dx or 3 0 . If x — 2, then x 2 - 4 = 0, x = ± 2. d ~y _ 4 _ r . dx 2 2 3 2 ' if ur= —2, then Therefore, since and d 2 y _ 4 _ 1 dx 2 — 2 3 2 x = 2 gives y — 2, x = — 2 gives y — — 2, then 2 is a minimum value, and — 2 a maximum value of the given function. ADDITIONAL EXAMPLES 97 4. Find maximum and minimum ordinates in each of the following curves, and draw the loci : (a) _y = 2 x 3 — 2 x 2 — 12 x + 4. Ans. Max. value n, min. value - 16. (£) y = 2 X 3 — 21 X 2 + 36 x — 20. Ans. Max. value — 3, min. value — 128. 6 + * N 1 H II Vsi Ans. Max. value 10, min. values ± 9. (d) y = (x — l) 3 (x — 2)“. Ans. Max. value min. value 0. H' ° H II Ans. Max. value -• e ( f ) y = sin 2 x — x. Ans. Max. values when x = (n + £)tt; Min. values when x = (n — -J)7r, n being any integer. , N sin x ■’' = ,+ tan * Ans. Max. value \ V2. (/i) y = sin x( I + cos x). Ans. Max. value |V3, min. value — 5V3. (0 y = x* - 4. Ans. Min. value — 4. (j) y = x s - 8. Ans. No max. nor min. values. CO* 1 II Ans. Max. value 4. EXERCISE 13. PAGE 58 8. Find the altitude of the cone of maximum volume which can be inscribed in a sphere of radius r. Ans. f r. 9. Show that the radius of the base of the cylinder of greatest lateral surface which can be inscribed in a sphere of radius r is \r V2. 10 . Find the shortest distance from the point (2, 1) to the parabola y 2 = 4x. Ans. V2. 11. Determine the area of the greatest rectangle which can be in- scribed in a given triangle whose base is 2 b and altitude a. Ans. \ab. 12. If the shape of a window is a rectangle surmounted by a semi- circle, and if the perimeter is given, show that the maximum light is admitted when the height and breadth of the window are equal. 13. Determine the altitude of the minimum cone which can be circumscribed about a given sphere of radius r. Atis. 4 r. el. calc. — 7 98 ADDITIONAL EXAMPLES EXERCISE 14. PAGE 64 Prove the following expansions : 5 . e sinx = I + x + z x 2 — % x 4 — T \x 5 + — . jj-3 ^*5 6. arc tan .r = .r — 1 !-•••. 3 5 7 x 2 5 ar 4 7 . sec X= I + , (- -r . — I . ip [4 X 2 X 4 X 6 8. log sec x = 1 1 b •••• & 2 12 45 2 9 . e z sec x = i + x + x 2 + b — . 3 10 . sin ( i -b 2 x) = sin i +2 cos i • x + 2 sin I • x 2 + •••. 11 . log (I -IB? 1 ) = log 2 + l X + | X 2 - •••• 12. Vi + ^x + 12 x 2 = I + 2 x + 4X 2 + ■■■. EXERCISE 16. PAGE 75 Prove the following integrations : 6. ^ cos 2 xdx = \ x + \s\n 2 x + c. (Put cos 2 .r = |(i -b cos 2 7 ‘ J|f$= log( * + * 8 ) 2 + c - 8. (* '* ^ X - — x — $ x 2 + $x s — log (x — i) + c. J x - b i Hint. Divide numerator by denominator before integrating. 9 ‘ f(* ax 2 + 4bx 3 )*(2 ax + 4 bx 2 )dx — 1(3 ax 2 + 4 Ar 3 )7 + c 10 11 12 • f ■ f ■f bar* dx = -a 21 + c. 2 log a ■ j* cos (mx + n ) dx= ~ sin (tnx + n)+ c. xdx V 4 — x‘ dx s~- m — \ arc sin x 2 + c. 2 = arc tan ( x ~ r ) + c - 13 . 2 x + x 2 Hint. Write denominator in the form 1 + (x— i) 2 . ADDITIONAL EXAMPLES 99 14 ‘ I dx V3 -4x- 4 x 2 j arc sin - . 2X + 1 + c. 15 • S x 2 + x _ 2 dX== l0g ^ ~ l ^ X + 2 ^ 8 + C ' 16 'I x 2 dx (x + 2)' 2 (;tr + i) X + 2 + log(jr+ 1) + c. EXERCISE 17. PAGE 79 8. Find the area bounded by y= 9 — x' 2 and the axis XX'. Ans. 36. 9. Find the area bounded by y — to ;r = 8. x 1 + x 2 and XX' from the origin Ans. \og e v'65 = 2.087. 10. Find the area bounded by y 2 = 9 x and y = 3 x. Ans. Hint. The area required is the difference of the areas bounded by the individual curves, the axis XX', and the ordinates at their points of intersection. 11. Find the areas bounded by the following curves. (a) y = x 2 — 4 and XX'. Ans. iof. (6) y = 8 + 2 x 2 -x* and XX'. Ans. 29xf- (r) y 2 = g x and x — 4. Ans. 32- (d) xy =4 and x + y = 5. Ans. 6.1 14. 12. Find the volumes of the solids of revolution generated by revolv- ing around XX' the following areas : ( a ) Of Example 8. Ans. 2595 *-• (b) Of Example 11 (a). Ans. 34fs 7 T- (c) Of Example 11 (c). Ans. 72 IT. ( d ) Of Example 10. Ans. fir. Hint. This is given as the difference of the volumes of the two solids generated by the two areas mentioned in the previous Hint. ELEMENTS OF DESCRIPTIVE GEOMETRY By CHARLES E. FERRIS, Professor of Mechanical Engineering, University of Tennessee $1.25 T HE leading engineers and draughtsmen, as investigation shows, do nearly all their work in the third quadrant or angle. 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