ELEMENTS OF THE INFINITESIMAL CALCULUS WITH NUMEROUS EXAMPLES AND APPLICATIONS TO ANALYSIS AND GEOMETRY BY JAMES G. CLARK, A. M. Professor in William Jewell College WILSON, HINKLE & CO. 137 WALNUT STREET 28 BOND STREET CINCINNATI NEW YORK COPYRIGHT 1875 BY WILSON, HINKLE & CO. ELECTROTYPED AT ECLECTIC PRESS FRANKLIN TYPE FOUNDRY WILSON, HINKLE & CO. CINCINNATI CINCINNATI PREFACE, THE Infinitesimal Calculus is generally considered to be the most difficult branch of pure mathematics to which the attention of the student is directed. It is certainly the most powerful instrument of investigation known to the mathematician, and its philosophy is as profound as its methods are far-reaching and comprehensive. But we believe that its difficulties, in so far as they are not purely algebraic, are due quite as much to the manner in which its first principles are usually exhibited, as to any inherent obscurity in the subject itself. In the preparation of the following treatise, the attempt has been made to remove all grounds for that feeling of uncertainty which often possesses the student at the very outset, and from which he rarely finds it possible afterward to extricate himself. With this end in view, considerable space has been devoted to an exposition of the doctrine of limits, which has been made the basis of both the Differential and the Integral Calculus. Many demonstrations might have been abridged, and apparently simplified, by the adoption of the ordinary method of infinitely small quantities; but this would have been, in the opinion of the writer, at the expense of a sound philosophy, for which, in a work intended primarily for educational purposes, the advantage of mere brevity could offer no compensation. (ijii iv PREFACE. The work is founded mainly upon the excellent philosophical treatise of M. Duhamel, and a large number of the examples have been derived'from the works of Hall, Walton, and Todhunter; but many other volumes, whose titles it is needless to mention here, have been consulted, and the writer would hereby acknowledge his indebtedness to all the treatises (American and foreign), relating to the subject, which it has been his privilege to read. He would also tender his obligations to Prof. Schuyler, of Baldwin University, for some valuable criticisms by which he has endeavored to profit; and he is sure that all of his readers will unite with him in this sincere expression of thanks to the publishers for the faultless style in which they have prepared the work for presentation to the public. LIBERTY, MISSOURI, | JAMES G. CLARK. May 24th, 1875. P. S.-Should any who may use the work as a text-book find it too extensive for their purposes, it may be conveniently abridged by omitting the following chapters: Differential Calculus-VI, IX, X, XI, XIX, XX; Integral Calculus —IX.... XVI, inclusive. C ONTENTS. THE DIFFERENTIAL CALCULUS. CHAPTER I. PAGE The Method of Limits. 9 Fundamental Propositions........ 11 Magnitudes Considered as Limits...... 13 Increments and Derivatives........ 18 Geometrical Illstrations........ 21 CHAPTER II. Functions - classified......... 23 Derivatives and Differentials........ 26 Differential Coefficient........ 28 Differentiation in general.29 CHAPTER III. Differentiation of Functions of one Variable.... 34 CHAPTER IV. Successive Differentiation........ 59 Notation........ 60 Leibnitz's Theorem........ 62 Cauchy's Theorem........ 63 CHAPTER V. Taylor's Formula.... 68 Maclaurin's Formula...72 CHAPTER VI. Convergence of Series.. 81 CHAPTER VII. Estimation of the Values of Functions which assume IndeterminIate Forms.. 86 (v) vi CONTENTS -DIFFERENTIAL CALCUL US. CHAPTER VIII. PAGE Maxima and Minima........ 96 Rules for determining the same. 97 CHAPTER IX. Differentiation of Functions of two or more Variables.. 109 Successive Differentiation of Functions of two or more Variables 110 Differentiation of Implicit Functions. 113 CHAPTER X. Extension of Taylor's Formula to two Variables. 121 Extension of Maclaurin's Formula to two Variables 123 Lagrange's Theorem.. 124 CHAPTER XI. Maxima and Minima of Functions of two Variables.. 129 CHAPTER XII. Change of Independent Variable....... 138 Elimination of Constants and Functions..... 146 CHAPTER XIII. Tangents and Normals to Plane Curves —Linear Coordinates. 150 The same when referred to Polar Coordinates.... 157 CHAPTER XIV. Asymptotes —General Method of Determining. 160 Special Method for Homogeneous Equations. 166 Circular Asymptotes. 171 CHAPTER XV Differentials of Arcs and Plane Areas.. 172 Angle of Contact... 176 Concavity and Convexity...... 178 CHAPTER XVI. Curvature.. 180 Radius of Curvature. 181 Contact of Curves........ 183 Evolutes and Involutes......... 186 CONTENTS-INTEGRAL CALCUL US. vii CHAPTER XVII. PAGE Singular Points of Curves. 195 Tracing of Curves from their Equation..... 206 CHAPTER XVIII. Envelopes of Plane Curves.. 208 CHAPTER XIX. Tangents and Normals to Curves of Double Curvature. 214 Tangents and Norn-als to Curved Surfaces... 218 Differential Equations of Curved Surfaces.. 220 Envelopes of Curved Surfaces.. 226 CHAPTER XX. Curvature of Surfaces. 231 THE INTEGRAL CALCULUS. CHAPTER I. First Principles.... 237 CHAPTER II. Direct or Simple Integration.... 241 Integration by Substitution....... 248 Integration by Parts......... 249 CHAPTER III. Integration of Rational Fractions.. 259 CHAPTER IV. Integration of Irrational Functions.... 276 Binomial Differentials.. 282 Formulas of Reduction for Algebraic Functions.. 284 CHAPTER V. Formulas for Logarithmic Functions. 296 Formulas for Trigonometric Functions. 301 viii CONTENTS- INTEGRAL CALCUL US. CHAPTER VI. PAGE Approximate Integration.. 321 Definite Integrals.325 Differentiation and Integration under the sign... 330 CHAPTER VII. Rectification of Curves.337 Quadrature of Plane Areas.338 CHAPTER VIII. Quadrature of Surfaces of Revolution..... 349 Volume of Surfaces of Revolution..... 350 Quadrature of Surfaces in general.. 351 Volume of Surfaces in general... 353 CHAPTER IX. Integration of Functions of two or more Variables... 365 CHAPTER X. General Theory of Differential Equations. 371 CHAPTER XI. Differential Equations of the First Order and Degree.. 376 Factors of Integration.. 378 CHAPTER XII. Integration by Separation of the Variables..... 383 CHAPTER XIII. Equations of the First Order and Higher Degrees. 398 CHAPTER XIV. Singular Solutions of Equations of the First Order. 405 CHAPTER XV. Equations of the Second and Higher Orders. 415 Integration by Series........ 424 Linear Equations.....427 CHAPTER XVI. Integration of Simultaneous Equations..... 434 THE DIFFERENTIAL CALCULUS. CHAPTER I. THE METHOD OF LIMITS. SEOTION. —Definitions and Fundamental Propositions. 1. QUANTITY, when made the subject of mathematical investigation, is to be considered under one of two aspects; viz., as constant or as variable. A constant quantity is one whose value remains fixed after having once been assigned; while a variable quantity is one whose value is either continually changing, or may be supposed capable of such change. Thus, the distance between two fixed points, measured on the straight line joining them, is constant; the distance from the center of an ellipse to its circumference is variable. 2. Whatever may be the law according to which a quantity varies, it will usually be found that there is some one value toward which the variable may be made to approach indefinitely, without ever reaching it; and it will sometimes occur that there are two such values within which the variable is confined. Thus, in the series 1,,,,. the terms themselves are evidently converging toward zero, while the sum of the terms approaches two indefinitely. (9) 10 THE DIFFERENTIAL CALCULUS. The distance between two fixed points being unity, the distance from either of them to a variable intermediate point must vary between zero and unity. This distance can never be either zero or unity; for in either case the third point would coincide with one of the fixed points, and thereby violate the hypothesis that it is between them. 3. The extreme value toward which the value of a variable may be made to approach, or converge, indefinitely, is called the limit of the variable. The inferior limit of a variable is the value toward which it converges in decreasing. The superior limit of a variable is the value toward which it converges in increasing. It is evident that zero and infinity are the inferior and superior limits of f -ii,e numbers. 4. An infinitesimal is a variable which has zero for its limit. For example, the difference between the value of any variable and its limit is an infinitesimal; since, as the variable tends toward its limit, this difference becomes less and less indefinitely, without ever reaching zero. It is evident from this definition that however small may be any given value of an infinitesimal, it will, in tending toward its limit, zero, become smaller still, without ever being actually reduced to zero. We may, therefore, say that an infinitesimal is a variable which may assume a value less than that of any assignable finite magnitude, however small; and when we attribute such a value to an infinitesimal, it becomes what is usually called an infinitely small quantity. The theory of the Calculus does not, in general, require us actually to attribute infinitely small values to these variables, and therefore we have preferred to use the term infinitesimal as descriptive of the class rather than the term infinitely small, which is of more limited application. 5. Proposition.-If two variables are equal in every stage of their variation, and each tends toward a limit, their limits are equal. THE METHOD OF LIMITS. 11'For, since the variables are equal, they may be substituted each for the other. Performing this substitution, we shall have the first tending toward the limit of the second, and vice versa; and since it is obviously impossible for a variable to tend toward two different limits at the same time, it follows that the two limits must be equal. Corollary 1. —It is evident that the limits of the two members of an equation which involve variables are equal. Corollary 2.-A constant quantity may be called its own limit, and hence if one member of an equation is constant, the value of that constant is the limit of the other member. 6. Proposition.-The limit to the algebraic sum of any number of variables is equal to the algebraic sum of their limits. Let x, y, z... be the variables, and designate their limits by lima. (x), lim. (y), etc. Let a, l3, y, etc., be the differences between these variables and their respective limits. Then we shall have x-lim. (x) +- a. y-lim. (y)- +. z lim. (z) +- y. x+y+z=lim. (x)+lim. (y)+lim. (z).. +a~a++y... (a). Now, the limits of a,, y.. being zero [Art. 4], the limit to their sum is evidently zero. The limit of the second member of (a) is, therefoie, lim. (x) + lim. (y) + lim. (z); and since the limits of the two members are the same, we shall have lim. (x +- y +- z) -= lim. (x) +- lim. (y) + lim. (z). 7. Proposition.-The limit to the product of two or more iariables is equal to the product of their limits. Employing the same symbols as before, we have lim. xyzz lim. (x). lim. (y). lim. (z) + terms involving a, 1, y as factors. 12 THE DIFFEREN1TIAL CALCUL US. The limit to each of these last terms being evidently zero, we have, by taking the limit of each side of the equation, lim. (xyz) lim. (x). lim. (y). lim. (z). Corollary 1. —If the product of several variables is constant, the product of their limits is also constant, and equal to that of the variables themselves. Let xyz = c. Then lim. (xyz) = lim. (c) -e. Corollary 2. —If two variables are reciprocal, their limits are also reciprocal. Let xy- 1, or x-1. Y Then lim. (xy) lim. (x). lim. (y) -lim. (1) =1. limn (x) 1 lmlir. (y) 8. Proposition. — The limit to the quotient of two variables is equla to the quotient of their limits. Silnce =x ( ), we have Y (/ Put 1 =z; then yz =1, and lim. (y). limn. () =1..linm. (z)- lir. () - lim. (y)' and by substitution in (1), Hill. (~ x lim. (x). lim (' -lim. (x). 1 (y him. (x) Y1 \ y himl(y lim. (y)' 9. Proposition. —The limit to the nth power of a variable is equal to the n"h power of'its limit. 1st. When n is an integer, we have, by Art. 7, lirn. (x) lim. (xx... lim. (x). lirn. (x).. l im. (x) = linm. (x) ". 2d. Let nt be any fraction, as P; whence x= xq. q MAGNITUDES AS LIMITS. 13 Put x~ y; whence x= y, and xq yP. Then we have lim. (x)- lim. (yq)-= lim. (y) q. Consequently, I li. (x) i-lim. (y), and Ilim. (x) lim. (y)} li. Iyp)Pl m. xv) SECTION II.-Magnitudes Considered as Limits. 10. Since any quantity may be considered as the limit to a variable, and since any number of variables may have the same limit, it is evident that we may regard any given quantity as the limit to a variable of simpler form than itself. Whenever, therefore, we wish to determine a relation between two quantities, we may often facilitate the operation by considering them as the limits of simpler variables, then finding the relation between these new variables, and finally, by means of the propositions established in the preceding section, passing to the limits, the relation between which will be the required relation between the given quantities. 11. Magnitudes may be considered as the limits of variables from several different points of view. 1st. We may consider any number as the limit to the sum of the terms of a converging series of finite quantities. EXAMPLE.-Unity is the limit to the sum of the series A, 4, i, etc. 2d. A quantity may be considered as the limit to the sum of an infinite series of infinitesimals. EXAMPLE. —The length of a given straight line is equal to the sum of the lengths of the parts into which it may be divided. The greater the number of parts, the less is the length of each. If, then, the number of parts is variable, the length of each part is evidently, a variable which has zero for its limit, while the number of parts has infinity for its limit. The length of the line is evidently equal to the limit to the sum of its infinitesimal parts. 3d. We may regard any quantity as the limit to the ratio of two variables. 14 THE DIFFERENTIAL CALCUL US. Ex.-k is the limit to the ratio of the two variables I + i~+ I + - +.- ~', and 1 + + t + I + 1 6 + 12. Proposition.- The limit to the sum of an indefinite number of positive infinitesimals is not changed when we replace them by others whose ratios to them have unityfor their limit. Let a, b, c,... 1 be infinitesimals whose sum tends toward a fixed limit as their number is increased. Let a', b', c',... 1' be other infinitesimals, such that each of the ratios a,, b, etc., shall have unity for its limit. Now, it is a principle of Algebra, that if we have a series of fractions with positive denominators, the ratio of the sum of the numerators to the sum of the denominators will be intermediate in value between the greatest and least of the fractions. Hence, a +- b + c +.. 1I a'+b'+'+ +... I' is comprised between the greatest and least of the fractions ab c etc. If, therefore, the limit to each of these fractions is unity, it follows that lim. a b c... =1; or a'+b'+c'+ 1; or lim. (a+ b +c+... /) lim. (aC'+b' + C.. I'). Corollary 1.-If either of these limits be equal to a constant, the other will be equal to the same constant. Corollary 2.-If the limit to the ratio of the corresponding elements of the two sums be 1, the limit to the ratio of the two sums themselves will evidently be 1; and this principle enables us, under certain circumstances, to reduce the investigation of the limits of semis to that of the limits of ratios, which, as we shall see in the sequel, is a much simpler process tlmn the former. MAGNITUDES AS LIMITS. 15 13. Proposition.- The limit to the ratio of two variables is not changed when we replace them by others whose ratios to them have unityfor their linit. Let a and b be two variables. Let a' and b' be two others, a b such that lrn. a 1, and lim. b= 1; whence a b Now, since a' b' a we have lim. a =lim. X limrn b X lim. a b b'.-Xim? But lim. b1, and lim. a 1. rim. a lim. a' b Corollary.-This proposition is evidently true, according to Art. 12, if for the word ratio we substitute sum. 14. Proposition.- When the limit to the ratio of two variables is unity, their difference is an infinitesimal, and conversely. Let a and a' be two variables and S their difference. Then a' — a- S, and dividing by a', 1 a,+ a Hence, if the limit to a is unity, the limit to must be a'a zero, and 3 must be an infinitesimal. Conversely, if the limit to a is zero, or if a is all infinitesimal, the limit to a must be unity. NOTE.-In the preceding demonstration a and a' are assumed to be variables whose limits are different from zero. Corollary 1. —The limit to the ratio of two variables is not changed when we replace them by others from which, they differ by infinitesimals. 16 THE DIFFERENTIAL CALCUL US. This follows directly from the preceding propositions, but its great importance warrants us in giving the following independent demonstration, for which we are indebted to Professor Schuyler. Let a and b be two variables, the limit to whose ratio is r; and let x and y be two infinitesimals. Then lim. - -r, and lim. b 1 b a r Also lim. ab _1. b a+x b+y *lim. a X +x ~ lim. bX lim.a-x-1; a b + y a b-a-y -Xlin a+ lim. a+x; r b+y b+y'. a lim. a+ y Corollary 2.-The limit to the sum of any number of variables is not changed when we replace them by others from which they differ by infinitesimals. This is evident, since the infinitesimals disappear in taking the limits. 15. Of Different Orders of Infinitesimals.-W-e have seen that the limit to the ratio of two variables may be any finite number or zero, and it may also be itself a variable, either infinitesimal or otherwise. Thus, let R designate the radius vector of an ellipse, let i be an infinitesimal increment to the radius, and let A be the semi-major axis. Then lim. R+i_ R A A and since R is itself variable, while A is constant, 1 is a variable quantity. The statement made above is true whether the variables have finite limits or whether they are infinitesimals, and this circumstance gives rise to infinitesimals of different orders. MAGNITUDES AS LIMITS. 17 Let x be any variable whose limit is different from zero, and let a, b, c, d, etc., be infinitesimals; let the limit to the ratio of a and x be zero; then we shall call a an infinitesimal of the first order; and ally infinitesimal whose ratio to a has a finite limit, will also be an infinitesimal of the first order. It is evident that an infinitesimal of the first order is simply an i6ifilzitesimal part of a variable whose limit is finite. Similarly, an infinitesimal of the second order is an iifinitesimCal part of on~e of the first order; one of the third order is an infinitesimal part of one of the second order; and so on. 15'. Proposition.-The product of two infinitesimals of the first order is of the second order. Let x be an infinitesimal of the second order, and let y, z be two of the first order; then we may evidently, in accordance with the preceding definitions, write -x z; whence x — yz. Y 15". Proposition.-To determine algebraic expressions for the irfinlitesimals of different orders. Let a, a,, a2, a3... a. be infinitesimals arranged in order, a being the one with which the others are to be compared. Designate lim a, by k, a finite quantity. Then we shall have a a, k_ in which 8 is an infinitesimal which disappears at a the limit. Whence al, a (k q-,) which will be the general expression for an infinitesimal of the first order. Also, — a (k + a), or a32a" (I+ 3) (2); 3 a2 (k+- ) or a: a' (k -a -) (3); a and, generally, D. C.-2. 18 THE DIFFERENTIAL CALCUL US. a,= al (k-e,) (n) for the infinitesimal of the nth order. Corollary to (15), (15'), (15").-The corollaries to Prop. 14 are true of infinitesimals as well as of finite variables. For, let a and b be two infinitesimals, and let x and y be infinitesimals of higher order than a and b. Then, word for word, as in Cor. 1., Prop. 14, we may prove that lim.- lim. a b b-b+y' Corollary 2, Prop. 14, may be extended to the case of infinitesimals as follows: Since lim. a 1, we have lim. (a) lim. (a x); a+x and since lim. b 1, we have lim. b = lim. (b + y). b+y lim. (a + b) - lim. (a) + lim. (b) - lim. (a + x) +-lim. (b+y)=lim. (a+-x+bd —y). Therefore, in general, the limit to the stem or ratio of two infinitesimals is not changed whent they are replaced by others from which they differ by ibfinitesimals of higher order than themselves. SECTION II,-.-Increments and Derivatives, 16. When two variables are in any way dependent upon each other, any change attributed to either of them will usually effect a change in the value of the other. Such changes are called Increments. 17. Let y be any function of x, designated by F(x), and let h be an infinitesimal increment of the first order assigned to x, converting y into y + k. Then we shall have y= F(x); y + k - F(x + h); k F(x + h) F(); k F(x + )- F(x). h h INCREMENTS AND DERIVATIVES. 19 and lim. k lim. { F(x +- h)- F(x) rim. klh The limit to the ratio of k and h is called the derivative of y with respect to x, and is designated by F'(x). We therefore have k __ Now, since [Art. 4] the difference between a variable and its limit is an infinitesimal, we shall have (calling this difference d) k - F'(x) +, and k= h F'(x) + 6A. Again, since a and h are infinitesimals of the first order, the product [Art. 15'] is an infinitesimal of the second order, and we nmay therefore substitute h F'(x) for k whenever the latter appears as one of the terms of a ratio or series whose limits we wish to find. 18. The derivative of a function will enable us to determine the manner in which the function varies between any assumed values. Thus we have seen that k = h F' (x) + l- = h (F' (x) + 6). Now, if for any given value of x, F'(x) is not zero, 6 being an infinitesimal will, as it tends toward its limit zero, necessarily become less than F'(x), and the sign of the factor F'(x) + 6 will depend upon that of F'(x). Therefore, supposing h to be positive, k will, when 6 has reached a value less than F'(x), have the same sign as F'(x). Consequently, F(x) is increasing for all values of x which render F'(x) positive; and F(x) is decreasing for all values of x which render F'(x) negative. 20 THE DIFFERENTIAL CALCUL US. The reverse will evidently be true if we consider h as negative. Corollary. —it is obvious that, of two functions of x, that changes the more rapidly whose derivative for any given value of x is the greater. 19. Proposition.-Let y = F(x) be a function of x, subject to the following conditions: 1st. That the increment k of y produced by the increment h of x shall tend toward zero as h tends toward zero. 2d. That, if in passing from the value x, to X, the value of x varies continually in the same sense, then y will also vary continually in the same sense in passing from yo to Y. Then we say that for every value of x between xo and X, the derivative F'(x), or lim., has a determinate value which is, in general, finite. For, if we divide the interval X —xo into n equal parts, and designate by k,, k2, etc., the corresponding increments of y, we shall have Y —yo-k+lk2+... k.,; X-xo-nh; and therefore, Y- yo k+k. --... k,,_ -, + ~ ~ AX- x nh n That is, the constant ratio Y- yo is the arithmetical mean x — x0 of the ratios,', etc. h' h Let n be indefinitely increased. The terms of each of these ratios will then tend to zero, and the limit to each ratio will be a value of lim. k, or of the derivative F'(x). Y-Yo will then be the mean of the derivatives of F(x) X- xo taken for all possible values of x between Xo and X; and since the value of this mean, in general, is finite, it follows that the derivatives themselves must be finite. GEOMETRICAL ILL USTRAT7IONS. 21 This does not, however, preclude the possibility of there being, within the prescribed limits xo and X, particular values of x for which the value of the derivative may be either zero or infinity. 20. If for all values of x between xo and X the value of the derivative be zero, it is clear that all the values of y between yo and Y will be equal to each other, and therefore independent of x. For, in that case, we shall have Y-Yo - 0; therefore, Y- y0 O, and Y= o; X- X0 and the same will be true if for Y we substitute any intermediate value of y. GEOMETRICAL ILLUSTRATION d. 21.-lst. Let it be required to draw a tangent to the curve gee at the point c, and let y -F(x) be the equation of that curve, referred to the rectangu- Y lar axes oX, oY. Designate the co-ordinates of c by x and y, and those of e by x+h, y-+k. Draw the secant line ce, and the tangent ed. If, now, we suppose X the point e to move toward c, the line ce will tend to coincide with cd, the position of which will evidently be the limiting position of ce, which the latter is always approaching, but which it can never reach without ceasing to be a secant, and thereby violating the original hypothesis. In this movement of the point e the lines df and ef evidently tend toward zero, as also does cf; moreover, df and ef, as they approach c, tend toward equality with each other; df, ef, and cf are, therefore, infinitesimals, and the limit to the ratio of df and ef is unity. The position of the tangent cd will be determined if we 22 THE DIFFERENTIAL CALCUL US. can find the tangent of the angle def which it makes with the axis of x. The value of this tangent is (if we consider the trigonometrical radius as unity) qf; and it might seem impossible to determine this ratio without first finding the value of df, which evidently depends on the position of the tangent line itself, the very thing for which we are searching. But, since the limit to the ratio of df and ef is unity, we are [Art. 14, Cor. 1] at liberty to substitute ef for df, and the determination of the ratio df is thus reduced to that of the limit to the ratio ef We therefore have tang dcf - df_ -lim. ef lim. k F'(x). Hence, the tangent of the angle which the tangent line to a curve makes with the axis of abscissas-the axes being rectangular-is equal to the derivative of the ordinate of the point of tangency. 2d. Let it be required to find the area of a right-angled triangle, ABD. Designate the base BD by b, and the altitude AD by a. Divide the altitude into any A Fig.2 number n of equal parts, ED, FE, etc., and complete the rectG angles gD, oF, etc. Now, each F n.. rectangle, as gD, will differ from li t7 the corresponding trapezoid BE DB_ ~q -B by an elementary triangle, Bgq, and the sum of all the rectan. gles will differ from the given triangle by the sum of thes6 elementary triangles. But, if n be increased indefinitely, the area of each of these triangles will decrease, and the limit to each triangle is evidently zero. The given triangle is obviously equal to the limit to the suml of the rectangles and the sum of the DEFINITIONS. 23 triangles. But the limit to the sum of the triangles is zero, since the limit to each is zero. Hence, the given triangle is the limit to the sum of the rectangles. Now, let Bq - go - ern, etc. h; also, let ED = FE, etc. a Then we shall have rectangle qE=(b —)a-; oF (b-2h)a; mG-(b —3h)a, etc. Hence, ABD - lim. { (b-h) + (b- 2h) +... (b - ) a } lim. {nba_-ah (I + 2 } n)} - lim. {ab-ah. n +? — lim.' abh ah } 2ao2 2But, whatever may be the number of rectangles, we have always I-u b, and the limit of h is zero. Hence, since the limit to the sum is equal to the sum of the limits, ABD - lim. (ab) - 2 lim. ab _ ab i 21 2' CHAPTER II. FUNDAMENTAL PRINCIPLES OF THE DIFFERENTIAL CALCULUS. SECTION I. —Definitions. 22. When any relation exists between two or more variables by means of which their values may be determined, they are said to be functions of each other. 24 lTHIIlE DIFFERENYTIAL CALCUL US. In such cases it is usual, as a matter of convenience, to select one or more of the variables in terms of which to express the others. The variables so selected are called independent, and the others dependent variables, or simply functions. 23. Functions are either explicit or implicit. An explicit function is one whose value may be deterimined by performing the operations indicated. An implicit function is one whose evaluation requires the solution of an equation. In the equation y a + bx + cx2, y is an explicit function of X. In the equation ax2 + bxy + cy2 +d.r + ey + f O, y is an implicit function of x. 24. Functions are either algebraic or transcendental An algebraic function is one whose relation to the independent variable is, or may be, expressed in a finite number of algebraic terms. The examples in Art. 23 are algebraic functions. Every function which can not be so expressed is a transcendental function. y —log x; y-ax; y:sin x; y-sin-' x, are examples of transcendental functions. 25. When we wish to indicate, in a general way, that y is an explicit function of x, we make use of some such expression as y —-'(x); y -f(x); y -,(x), the characters F, f, e designating different functions. These expressions are read simply, y equal to the F fiuction of x, etc. WYhen we wish to indicate that y is an implicit function of x, we write F(x. y) -- 0; f(x. y) -- 0; (x. y) - O, nand reacd, the F ftncctionl of x atld y eqtacl to zero, etc. DEFINITIONS. 25 26. When a function depends for its value directly upon one or more independent variables, it is called a simple function. Thus, y = 4x + 3z is a simple function of x and z. When several functions are accumulated upon each other, the first is said to be a function of functions. Thus, if y=- F(z); z =f(u); u= — (x), y is a function of functions. When a variable is a function of several variables, each of which is a function of the same variable, the first is said to be a compound function of the last. Thus, if y -F(z. u. t); z -f(x); u — (x); t — (x), y is a compound function of x. If one variable is a function of another, the second is an inverse function of the first. 27. A variable is continuous when, in passing from one value to another, it passes successively through all intermediate values. WVhen this condition is not fulfilled, the variable is discontinuous. A function is continuous when, in making the variable on which it depends vary continuously, it is constantly real, and also varies continuously. A function may be continuous for all values of the independent variable within certain limits, and discontinuous beyond those limits. 28. We have seen in the first section of Chapter I., that when two variables are always equal their limits are also equal, and that in order to find the limit to any algebraic combination of variables, it is sufficient to replace each variable by its limit. Hence, if F(x. y. z...)=f(x. y. z..), we shall also have lim. F(x. y. z...)lim. f(x. y. z...). I). C.-3 26 THE DIFFERENTIAL CALCUL US. If, now, a. b. c... are the limits of x. y. z..., then lim. F(x. y. z...)= F(a. b. c...), and lim. f(x. y. z...) =f(a. b. c...). Hence, finally, F(a. b. c...)f(a.b. c...). If, therefore; we wish to discover the relation which exists between several variables, it is sufficient to consider them as the limits to other variables which may be of simpler form or of a character more readily dealt with; to determine the relation between these new variables; and, finally, in the algebraic expression of this relation, to replace them by their limits. 29. It has been demonstrated in Art. 19, that the ratio of the two infinitesimal increments of a function and its variable has, in general, a determinate finite limit, which we have called the derivative of the function with respect to the variable. It is the primary object of the Differential Calculus to investigate and establish methods of determining the derivatives of fmunctions under all possible forms and combinations. SECTION II.-Derivatives and Differentials. 30. The infinitesimal increment of a variable or function is called its difference, and is designated by the symbol J. Thus, Jx is the difference of x. The derivative of a function with respect to a given variable is designated by the same symbol which denotes the function, affected with an accent. Thus, if we designate a function by y -F(x), then F'(x) is the derivative of y with respect to x; -and since the derivative of a function is the limit to the ratio of the infinitesimal increments of the func DERIVATIVES AND DIFFERENTIALS. 27 tion and the independent variable, we shall have, in accordance with our notation, F'(x) =lim. y (1), in which By and ax are infinitesimals. V31. If, now, we represent by a a certain quantity which tends toward zero at the same time with Ax, we shall have, in accordance with the theory of limits, y zF'(x) ~- a, or By = Ax. F'(x) - aAx, Ax in which adx is an infinitesimal of higher order than Dx; and therefore [Art. 17], whenever By enters into the terms of a ratio or series whose limits we wish to find, we mnay replace it by ax. F'(x). Consequently, in all such cases, and in such cases only, we may write, as rigorously exact, the equation Ay = dx. F'(x) (2), or F(x) (3). 32. Equations (2) and (3) being rigorously exact only when they are limiting equations, it is found expedient in practice to substitute for Ay and ax the symbols dy and dx, which are used only in limiting equations or expressions, and represent quantities whose ratio is equal to the limit to the latio of By and Ax. Instead, then, of the above equations, we may write the following: dy F'(x) (4), and dy- F'(x). dx (5), which are rigorously exact: The expressions dy and dx are called the differentials of y 28 THE DIFFERENTIAL CALCULUS. and x; and dY, being the quantity by which it is necessary to multiply dx in order to obtain dy, is called the differential co-efficient of y with respect to x. It represents symbolically the derivative of y with respect to x, together with the algebraic process by which it is obtained. A comparison of equations (2) and (5) will show that dy and dx may be considered as infinitesimals whose ratios to Jy and dx, respectively, have the limit unity, and which may be of simpler form than these last, or vice versa; and, if x be taken as the independent variable, dx may evidently be considered as equal to dx. 33. If we have in the same investigation several functions (y, z, u, etc.) of the same variable x, and represent by zy, dz, etc., their differences with respect to JAx and by dy, dz, etc., their corresponding differentials, we shall have Jy Jy. zx A az dr. dz Ax z J Az and, passing to the limits, t3y zy lim. Ay lim. - 1z, or, Jr lim. X dy dy x dz dz j-~ Hence, the derivative of one variable with respect to another is equal to the ratio of their derivatives taken with respect to any third variable of which they are both functions. DIFFERENTIATION IN GENERAL. 29 SECTION III.-Differentiation in General. 34. The operation of finding the derivative and differential of a function is called differentiation. All functions, whatever may be their form, are either simple functions or made up of simple functions, and we proceed to show how the differentiation of any function may be made to depend upon that of the simple functions of which it is composed. [1] DIFFERENTIATION OF SIMPLE FUNCTIONS. Let y- F(x) be any simple function of x. Then we shall have, as has previously been shown, zy - F(x + x) — F(x); Jy F(x + x) -F(x) Jx aJx lim. dy lim. F(x + x) - F( x). Jx ax' r, d F(x), and dy - F'(x). dx. Hence, to differentiate a simple function of a variable, give to the variable an increment, and find the corresponding increment of the function. The limit to the ratio of these increments will be the derivative of the function. Place this derivative equal to the differential co-efficient, multiply by the differential of the variable, and the result will be the differential of thefunction. Scholium.-If we suppose the independent variable to vary uniformly, so that Jx is the same for all values of x, dx may be taken as the rate of variation of x, and dy, being the infinitesimal change in y due to the corresponding change in x, may be considered as the rate of variation of y. 30 THE DIFFERENTIAL CALCUL US. [2] DIFFERENTIATION OF A FUNCTION OF FUNCTIONS. Let u be a function of x, determined by the series of operations u-=F(z); z =f(y); y =(x); and let it be required to find the derivative of u with respect to x. Let Ju, Az, zy be the differences of u, z, y. Then we shall have au _ au Az Ay and, passing to the limits, du da dz dy F'(z). f(y).'(x) d' d' -' dye d' Hence, the derivative of a function u of x, through several intermediate functions, is equal to the product of the derivatives of the whole series of functions. [3] DIFFERENTIATION OF INVERSE FUNCTIONS. Let y= F(x), and x=-f(y). Since these two equations are the same under different forms, they will give the same increments for the variables. We shall therefore have F'(x) lim. y_ 1 1 1 ax lim. f'(y) f'(F(x)) The equation F'(x) er (F(x)) expresses the method of performing this differentiation. DIFFERENTIATION IN GENERAL. 31 [4] DIFFERENTIATION OF COMPOUND FUNCTIONS. In order to determine the derivative of a compound function we shall first find an expression for the infinitesimal increment of a function of several variables. Let y-= F(u. v), and let zy, Ju, Jv be their increments. Then we shall have y-.F(u + DJu, v + Jv) - F(u. v), which can be written y=-F(u+-Au, v)-F(u. v)+F(u+ Ju, v+-v)-F(u+ Ju, v), or ~y= ~F(-,+, Av)-F(u. v) ~(u ~ +F(U+, v+Av)-F(u4 — u, v)va' In this equation the co-efficient of Ju differs by an infinitesimal from the derivative of F(u. v) with respect to u, v being regarded as constant; and in accordance with the theory of limits this co-efficient may be replaced by that derivative, dy de' In like manner, the co-efficient of Av differs by an infinitesimal from the derivative of F(u + Ju, v) with respect to v, iu being regarded as constant; and this latter derivative differs by an infinitesimal from the derivative of F(u. v) with respect to v. Hence we may substitute dy for the dedv rivative of F(ue + u, v), and this for the co-efficient of dv. We therefore have 4y- dy d u + dy. Av. du dv It must be observed, however, that since quantities can be substituted for others from which they differ by infinitesi 32 THE DIFFERENTIAL CALCUL US. mals only in cases where limits are to be taken, we must, in the general case, add to the above value of zy an infinitesimal a, which shall disappear when we pass to the limit. We thus have Jy_ dy. u + dy v + a, an equation which is rigorously correct, and is independent of any relation between u and v. If now we have y-F(u. v. z..), where iu, v, z. are functions of x, we shall have +y- Y u dy v + zY Jz +..+ a. Jdu'dv dz Dividing by Dx, and passing to the limits, we obtain dy dy du dy dv dy dz dx du dx dv dx dz d ( or, multiplying by dx, dy —dy du + dy. dv + dy dz... (B). du dv dz These two equations, (A) and (B), are formulas for obtaining the derivative and differential of a compound function of a variable, and are much more readily remembered than the rules which might be derived from them. NOTE.-In these equations dy dy, etc., are the partial du' dv derivatives of y with respect to each of the variables, and dy in the first member of (A) is the total derivative of y dx with respect to the variable x, on which all the others depend. If any one of the variables, as u, should be equal to x, then we would have DIFFERENTIATION IN GENERAL. 33 dy du dy dx dy du' -- d' d vx d- d x' a partial derivative of the same form with the total derivative in the first member. In like manner, dy in the first member of (B) is the total differential of y, while dy in the second member is a partial differential. Particular attention is necessary in using these expressions to prevent the confu.sion of total with partial differentials and derivatives. The difficulty is sometimes obviated by inclosing the total derivatives or differentials in brackets. [5] DIFFERENTIATION OF IMPLICIT FUNCTIONS, (a). Let F(x. y)- 0 be an equation expressing that y is an implicit function of x. Taking the derivatives according to formula (A) in the last case, we have dF(x. y) + dF(x. y) dy_0. dx dy dx Whence dF(x. y) dF(x. y) dy dx dx X- dF(x. y)' and dy - dF(x. y) d dy dy These expressions may be somewhat abbreviated by substituting u for F(x. y). (b). If we have m —1 equations with m variables, the differentials of these variables may be determined in terms of the variables and the differential of any one of them. Let F(x. y. z..) =; f(x. y. z..) —0; (x. y. z...) 0, etc., be the equations. 34 THE DIFFERENTIAL CALCULUS. Differentiating with respect to x, y, z..., and representing the first members of the equations by F, f, q, etc., we have dF dF dF dF dx + dy+ - dz0, df dx df dy + d dzO, dx dy dz d, ax + dy d dz = O, d~.dydy From these mn-1 equations the values of dy, dz, etc., can be determined in terms of x, y, z... and dx. CHAPTER III. DIFFERENTIATION OF FUNCTIONS OF ONE VARIABLE. 35. Problem 1.-To differentiate y=- x ~+ a. WVe have y+ Jy - Jyx - zx a; whence Jy - Jx, and 4y_1 ax -dy- lim. -y 11; and dy — dx. Therefore, if a constant be connected with a variable by the sign + or -, it disappears in differentiation. NOTE. —In this and the succeeding problems it is to be understood that y is a function of x. DIFFERENTIATION IN GENERAL. 35 Problem 2. —To differentiate y -ax. We have y --:Jy —a(x+ Jx) — ax - aJx, whence Jy - aJx, and z=y a. dy-_ lim. y - a, and dy - adx. dx- J Therefore, if the variable be multiplied by a constant, the differential contains that constant as a factor. Problem 3.-To differentiate any power of a variable. Let y -= x. Then we shall have y+Jy-(x+ jx)m=-x"+mxm-l jx+m(m1) xm-2 (Jx)2+etc. Whence, by subtraction, ly m= xm-_l x + m(m -1) xn-2 (Jx)2 + etc.; and, dividing by Jx, Jy mxml +m(n- 1) X"-2 Jx + etc. jx 1. 2 lim. * Jy dy mxem, and dy —mx dx. Therefore, the derivative of any power of a variable is found by diminishing the exponent by unity and multiplying the result by the original exponent; and the differential is found by multiplying this derivative by the differential of the variable. NoTE.-Since the binomial formula is true for all values of m, this rule for differentiating is correct for all powers. 36 THE DIFFERENTIAL CALCUL US. Problem 4.-To differentiate y_ =/x. We have x — y; whence dx- = 2ydy (by Prob. 3). dx dx. 2y 2- = / Therefore, to differentiate a radical of the second degree, differentiate the variable under the radical sign and divide the result by twice the radical. Problem 5.-To differentiate y-=as + bz + cu + kv + etc., where y, s, z, etc., are functions of x. Let the increment Jx be given to x; then y, s, z, etc., will be converted into y + Jy, s + As, z + Dz, etc., and we shall have y+ Aya(s+ s) +-b(z+ Jz) +c(u+ Du) +k(v+ Jv) +etc. Jy _ ads + bAz + cDeu -+ kAv + etc., Ay as z Deu Av x a+ + bx-+ c+ - k - etc., zix dzx lx Jx zx and l y. dy zs. lz. u dv urn. a a. lim. +b. lim. a+-c. lint +-. lim. +etc., Az ix zix''x' x dy ds + bdz cdu kdv etc. ordx a dx dx dx- d etc. dy =ads + bdz + cdu + kdv + etc. Therefore, the differential of the algebraic sum of any number of functions of the same variable is found by taking the algebraic sum -of their differentials. Problem 6.-To differentiate the product of any number of functions of the same variable. DIFFERENTIATION IN GENERAL. 37 (a). Let us take y - uz. Then we shall have y + ~Jy= (ut + Ju). (z + Jz) uz + ~UJZ + ZJU + Ju. Jz, y + uZz + zJu + Ju. Az. zy = zy + zzu + Ju. Zz; and -y i z zl u Jz. zx Jx dz ~ JS Passing to the limits, and observing that, since dz enters as a factor into the last term, the limit of this term is zero, we have dy dz du d- - + z-, and dx dx dx dy - udz + zdu. (b). Let the function be y uzv. Designating the product of u and z by s, we have y - sv, and, as in the previous case, dy - sdv + vds. (1). But, since s - uz, we have ds = udz + zdu. by substitution in (1), dy uzdv + uvdz + vzdu. The same process can evidently be extended to any number of functions, and we shall have the following general rule: To differentiate the product of any number of functions of the same variable, multiply the differential of each function by the product of all the other functions, and take the algebraic sum of the products so obtained. 38 THE DIFFERENTIAL CALCULUS. Corollary.-Dividing the equation dy = uvdz + uzdv + vzdu by y = uvz, we obtain dy dz dv du y Z V it Wherefore, if the diferential of each function be divided by the function itself, the sum of thle quotients will be equal to the differential of the product of the function.s divided by the product. Problem 7.-To differentiate the quotient of two functions of a variable. Let the expression be y -. Then we shall have y+ q- Jd; whence Zy - + -Ju u vJu — utv (1). v + Jv v v,+ vJv Dividing by Jx, and passing to the limits, observing that the limit to the denominator in the second member of (1) is v2, we have duz dv dy - v -u and dy vd-t- udv dx v Therefore, to differentiate a fraction, multiply the denominator by the differential of the numerator, the n~umerator by the differential of the denominator, subtract the second product from the first, and divide by the square of the denominator. Corollary 1.-If the denominator be constant, then the second term in the differential vanishes, and we have vdu du dy= v -, DIFFERENTIATION IN GENERAL. 39 Corollary 2.-If the numerator be constant, then the first term in the differential vanishes, and we have dy - udv V2 Problem 8.-To differentiate y= log (x). We have y + y y log (x + Ax) log {(1 + ) -og x+ iog l + ) y=. log(l+ -)M( - 2. etc.) x x 2x2' dx M -,2 etc. ); Ylx x 2x' passing to the limits, dy( )_ Mndx Therefore, to differentiate the logarithm of a variable, divide the differential of the variable by the variable and multiply by the modululs of the system. Corollary.-If the logarithm be taken in the Naperian system, the modulus is unity, and we have dy - dx We shall in all cases, unless it is otherwise stated, understand our logarithms to be referred to the Naperian system. Problem 9. —To differentiate y_ ax Passing to logarithms, wve have log y=.x log, a. 40 THE DIFFERENTIAL CALCUL US. Therefore, by differentiation, dy dx log a; dy=y dx loga; or Y da= -a dx log a. Corollary.-If a is equal to e, the Naperian base, then log a - log, e - 1, and therefore, d exZ-e dx. Problem 10.-To differentiate y -sim x. In order to solve this problem it is necessary to demonstrate in the first place that the limit to the ratio of the sine and tangent is unity. For this purpose it is sufficient to observe that, since tangc — i we have s- cos, Cos tang and therefore, lim.( sin )lim. (cos) -. Fig. 3 X /~ We may also notice that, since the length of the arc of a circle is intermediate in value between the sine / and tangent, we shall have s' l are _ lim.( sin t l _ =l I arc im/ sin lim~ t lim. ( arc ) lim. ) 1. tang tang - sin arc Now, resuming our equation y =sin x, and giving to x an increment, zx, we shall have y + by + sin (x- + x) =-sin x. cos'x - + cos x. sin /x; Jy = sin x. cos Jx- + cos x. sin Jx - sin x; DIFFERENTIATION IN GENERAL. 41 ay sill x sin Ax sin x Ax Ax Cx A - x and lim. AY-lim.( I six) lim. (cos Ax) +-l sill AX x ) + lim.'Sill Dx. l im'(cosx)- im. (sin X But lim. (cos Ax) 1, and lim.( sin x ) 1 lim. A im. ( sinx )+lim (Cos x) lim( sin or dy_ cos x, and dy = cos x dx-=d sin x. dx Problem 11. —To differentiate y = cos x. We have y - cos x = sin (90~ - x).. (by Prob. 10), dy = cos (90~ -x) d(900 - x) = sin x d(- x) =- sin x dx=d cos x. Problem 12.-To differentiate y = tang x. We have sill x Cos X.*. (by Prob. 7), dy- cos x d sin x-sin x d cos x c_ Cos2 dx= 1- dx-sic x dx- -d CX dx dx- =sec2 x dx = d tang x. COS'D X COS. X D. C.-4 42 THE DIFFERENTIAL CALCUL US. Problem 13. — To differentiate y = cot x. We have y = tang(900 - x). dy sec2(90~ -x) d(90~-x) — cosec2 x dx - d cot x. Problem 14.-To differentiate y -=sec x. We have y sec x -- \ cos2 x - Cos x c os x = tang x. sec x dx d sec x. Problem 15. —To differentiate y cosee x. We have y = cosec x = sec(900 - x). dy = d sec(90- x) = tang(900 - x). sec(900 - x) d(90~0 - x) _- cot x. cosec x dx - d cosec x. "Problem 16.-To differentiate y - versin x. We have y = versin x - 1 - cos x. dy = d(l - cos x) =-sin x dx =d versin x. Problem 17. —To differentiate y = coversin x. We have y = coversin x = versin(900 - x).. dy=d versin(900- x) =-sin(900 - x) d(90~-.x) - cos x dx - d coversin x. DIFFERENTIATION IN GENERAL. 43 Problem 18. —To differentiate y - sin-' x. We have x sin y. dx = cos y dy= 1/1- sin' y dy-= V1 — xl dy; whence, dy - dx - d sin-l x. Problem 19. —To differentiate y = cos-' x. We have x = cos y. dx - sin y dy= — /l —cos2 y dy — 1/1 - x dy; whence, dy - - dx = d cos'l x. V/I - -' Problem 20.-To differentiate y = tang-' x. We have x - tang y. dx-=sec2 y dy- (1 + tang' y) dy= (1 + x2) dy; dx whence, dy - 1 + d tang-' x. Problem 21.-To differentiate y = cot-' x. We have x = cot y dx — cosec y y d -(1 + cot2 y) dy - (1 + x2) ly; dx - - cot —1 y. whence, dy = -- 1 d cot-' x. Problem 22. —To differentiate y -see-' x. We have x = sec y. 44 THE DIFFERENTIAL CALCUL US. dx = tang y. sec y dy- I/sece y- 1. sec y dy =Vx1/2 —1 dy; whence, dy dx d sec-' x. xl/x — 1 Problem 23.-To diferentiate y -cosec-l x. We have x - cosec y. dx= -cot y. cosec y y = i/cosec" y-1. cosec y dy — xX2 — l y; whence, dy- dx - d cosec-' x. Problem 24.-To differentiate y -versin-' x. We have x = versin y... dx= sin y dy= 1/2 versin y - (versin y)2 dy = 1/2x2- x dy; whence, dy _dx -- d versin-1 x. i/h2x -x2 Problem 25.-To differentiate y = coversin-l x. We have x = coversin y.'. dx =- cos y dy = -1/2 coversin y - (coversin y)2 dy _- 1/2x- x. dy; whence, dy - dx d coversin'- x. 1/2x- x2 EXAMPLES. 45 36. The preceding problems embrace the methods of differentiating all the known simple functions of a single variable. Problems 1 to 7, inclusive, embrace all the algebraic functions; 8 and 9 are the logarithmic, and its inverse, the exponential, functions; 10 to 25 are the direct and inverse circular or trigonometric functions, in each of which the radius is taken as unity. The methods of differentiating all simple functions being known, the differentiation of functions of functions, compound functions, and implicit functions, can be readily effected by means of the formulas already provided for such cases. 37. EXAMPLES. 1. Differentiate y - 3x + 7. According to Problem 1, the term 7 will disappear in differentiation, and by Problem 2, we shall have dy 3, or dy - 3dx. dx 2. Differentiate y - x5. According to Problem 3, we have dy 5.. dy 5x4dx. dx 3. Differentiate y= 312x. We have, by Problems 2 and 4, 3d(2x) 6dx 3dx 2dy 2x 2/2x - V2x 4. Differentiate y = 5x4 + 7x - 4x2 + 6x -5. We have, by Problem 5, dy - 5d(x4) + 7d(x3) - 4d(x2) + 6dx; 46 THE DIFFERENTIAL CALCUL US. differentiating each term, dy = 5(4xdx) + 7(3x2dx) - 4(2xdx) + 6dx - 20x3dx + 21x2dx - 8xdx + 6dx. 5. Differentiate y (4x5)(3x'). We have, by Problem 6, dy = 4.5d(3x2) + 3x2d(4x) - 4x5(6xdx) + 3x2(20x'dx) - 24x6dx + 60x6dx = 84x6dx. 6. Differentiate y = ax' (bx2 + cx + h). We have dy = 3ax2dx(bx2 + cx + h) + ax3(2bxdx + cdx) - 3abx4dx + 3acx'dx + 3ahx2dx + 2abx4dx + acx3dx - 5abx4dx + 4acx3dx + 3ahx2dx. The same result would be obtained by performing the indicated multiplication and then differentiating. 7. Differentiate y = 48x9. By Problem 3, we have, dy = 432x8dx. Again, resolving 48x9 into three factors, we have y =2x2. 4xV. 6x4. Differentiating this by Problem 6 (b), we have dy = 24x7. 4xdx + 12x6. 12x2dx + 8x5. 24xedx - (96x8 + 144x8 + 192x8)dx = 432x8dx. 3x2 8. Differentiate y - EXAMPLES. 47 We have, by Problem 7, dy- 6xd(3x2) - 3x2d(6x) 36x2dx - 18x2dx dx. 36x2 36x2 9. Differentiate y. Since a is constant, we have 5x4dx dy =10. Differentiate y —= We have, by Problem 7, Cor. 2, 5 — ax4dx -- 5adx X10 11. Differentiate y _ log(3x4). We have, by Problem 8, dy d(34) _ 12xdx _ 4dx 3X4 3x4 x 12. Differentiate y- a3x. We have, by Problem 9, dy - a3X log a d(3x) — 3ad3 log a dx. 13. Differentiate y a=. We have dy ae log a d(x2) - 2xax log a dx. 14. Differentiate y - eax. We have dy = eaxd(ax) = aeaxdx. 48 THE DIFFERENTIAL CALCUL US. 15. Differentiate y sin (x2). We have, by Problem 10, dy = cos (x')d(x2) _ 2x cos (x)'dx. 16. Differentiate y cos (5x). We have, by Problem 11, dy — cos (5x)d(5x) =-5 cos (5x) dx. 17. Differentiate y -- tang (ax). We have, by Problem 12, dy = sec2 (ax)d(ax) = a sec2 (ax) dx. 18. Differentiate y = sec (ax). We have, by Problem 14, dy = tang (ax). sec (ax)d(ax) = a tang (ax). sec (ax)cx.'19. Differentiate y- versin (ax). We have, by Problem 16, dy _ sin (ax)d(ax) = a sin (ax)dx. 20. Differentiate y -a sin - x. a We have, by Problem 18, dx a a GCad dy = -— 2 2/ *S 221. Differentiate y =a cos-l. We have, by Problem 19, dx -a- - dy a adxr d a'2 V-' EXAMPLES. 49 22. Differentiate y a tang-' x. a We have, by Problem 20, dx a a a2dx 1adx 23. Differentiate y - a cot-' x. a We have, by Problem 21, dx a — a - a2dx dy -- aX2 + a2 24. Differentiate y-a sec-'. We have, by Problem 22, dx a a2dx dy x - xa / __X _ a 25. Differentiate y= a cosec- We have, by Problem 23, dx - a- a2dx dy = x2 /v — _a2 a~~- 1 26. Differentiate y-a versin-1 x. We have, by Problem 24, dx a a adx D y. - - ~ x 11/2ax -XD. C. —5 50 THE DIFFERENTIAL CALCULUS. 27. )Dif'erentiate y a coversin-' x. a We have, by Problem 25, dx a — ~dy - a -adx d - I z V2ax-_X 37'. The preceding examples have been given simply for the purpose of illustration. The following are intended as examples for practice; they are given without regard to order, and it will be noticed that some of them are quite complicated. 1. Differentiate y - (a + bxm"). Assume a + bx - z... y z', dy nzn-' dz, and dz mbxm-l dx. by substitution, dy- bmn(a + bx)n-lx-l. dx. We may also differentiate this example by the rule for powers and obtain the same result. 2. Differentiate y — x(1 + x2)(1 + x)2. We have dy — (1 + x) (1 + x)2dx+x(1 +x)22xdx+x(1-x2)2(1-+x)dx —? 3. Differentiate y = i/ax + bx2. We have a+ 2bx dx. 2y - 2Vax + bx2 4. Differentiate y 5x_ (X2 + 3)3 We have 20 3)303(2 ++ 3)32 2x..5x4 d2 - (x- +- 3)dx EXAMPLES. 51 5. Differentiate y = log (a + bx + cx2 + hx7). Ans. dy_ b + 2ex + 3hx2 dx a +- bx - cx2 q- hx' o 6. Differentiate y - +1/1 + x= -- x + (1 + x2)y}. We have -- dy=~ Sx+ (1 +x2l)I dJx+ (1 + x2)}t x- 1 + - 2) 2 + (1 +- x2) 2xjdx -g,-1+ X" d 1/11+-x ~x +_ 1/ i + + 7. Differentiate y -= i/x. /1+- 1 -= x (x+ +- I)'. We have dy =- ~x- (xm + 1) dx + Axy(xm + 1)-%. ~x-dx (x -+ 1)%dx dx 4x3' 6x4(x% + 1)Y 51/x+3 dx. 12!~l~f/(T/x + 1)2 8. Differentiate y=- x x i/ i - xt, o 9. Differentiate y 1/ 1 1/-x 1/ +x -1 —x Multiplying both terms of the second member by the numerator, we have (/i + x + 1/_x)2 1 + t/1-1 2 Y 2x x dy? dx 52 THE DIFFERENTIAL CALCUL US. 0 10. Differentiate y -\/x + V/x -- 6 etc., continued indefinitely. Squaring this equation, we obtain' y2X+ y, or y2-y =x. dx 2y dy - dydx, anddy - 2y 1.1. Differentiate y =log (x + /1 + x'). We have d _- dx + + x2}/ __ 1+x(1 + x2)- dx- dx x+l +x2 x+V l +x2 1~x2 12. Differentiate y - x(a2 + x2)1/a2 - xL. Taking the logarithm of each side of this equation, we have log y - log x + log(a2 + x2) +- log(a2 -x). dy dx d(a2 + x2) d(a- x2) y x a2+x 2 a2- x'2 from which the value of dy may be easily found. This method of passing to logarithms may often be resorted to with advantage. 1-+x 0 13. Differentiate y 1 +x2 We have dy (1 -+2_ ) dx- (1 )2dx =1 )2x 1 — 2x dx. (1+ + xe) (1 q- x)2 14. Differentiate + xb (a + x)2(b + x)" We shall find dy - 2(b - x) + 3(a +- x)dx. (a + x)3(b -+- x)' EXAMPLES. 53 15. Differentiate y - (a + x)m(b + x)n. We have log y= —m log(a + x) + n log(b +- x). dy m d(a + x) n d(b %x);'' y a+x bb+x * dy (a x)m(b+ x) n + } dx. za + vx 16. Differentiate y - /Ans. dy - - a(v x -- a v) dx. 2W/xl/a + x(v/a + ll/x)2 17. Differentiate y 1/i + x V1-x Passing to logarithms, we have log y = — log(1 + x) - log(1 - x). dx. dy- ( x)/ _ 18. Differentiate y x + 1/- x2 } We have dy = n{x +- 1/1 -x2 }n-ld{x + V/1-X2 } =? 19. Differentiate y = log 1/1 +x x. We have y- log+{lt +-x —x} - log{1/l + x2 + x}. *'. dy-? 20. Differentiate y = ab+x. We have dy ab- ab: log a dx. 54 THE DIFFERENTIAL CALCUL US. o 21. Differentiate y xx. We have. log y x log x. dy_ ldg x. dx + dx. dy =xx(log x + 1) dx. o 22. Differentiate y=x'x, the notation signifying the x' power of x. We have log y - xX log x. dy-x xxx logx(logx+-1)+ 1}dx 23. Differentiate y -= e. Ans. dy- n ez x'n-' dx. 24. Differentiate y - ezx. Ans. dy = exx xx(log x + 1) dx. 25. Differentiate y x". 26. Differentiate y =log(log x). We have dyd log x dx log x x log x 27. Differentiate y (log x)n. Ans. dy =n(log x)'"- dx X 28. Differentiate y e'o~g 29. Differentiate y = elog Va2+x:2. ex -e e-Z 30. Differentiate y e + e, Ans. dy e- dx. (Cx + e7X) 2 EXAMPLES. 55 31. Differentiate y- x 1' 32. Differentiate y - eol~g. 33. Differentiate y- x5 log3 x + 2x3 log4 x + 4x e2z. 34. Differentiate y _el_ + x — X 2 2x Ans. dy e- 2-2 dx. (1 - x)/1 —X 35. Differentiate y 4 sin5 x. We have dy =20 sin' x d sin x- 20 sin4 x. cos x dx. - 10 sin3 x. 2 sill x. cos x dx - 10 sin3 x. sin 2x dx. 36. Differentiate y sin nx. 37. Differentiate y = tang" x. 38. Differentiate y = sill 3x. cos 2x. We have dy sin 3x d cos 2x + cos 2x d sin 3x - 2 sin 3x. sin 2x dx + 3 cos 3x. cos 2x dx = (cos 3x. cos 2x + 2 cos 5x) dx. 39. Differentiate y =log sill x; y =log cos x; y log tang x; y =log cot x; y = log sec x; y = log cosec x; y =log versin x; y- log coversin x. 40. Differentiate y = sinx x. 41. Differentiate y = sin(sin x). We have dy- cos(sin x) d sin(x) = cos(sin x). cos x dx. 42. Differentiate y =(cos x)sin. We have log y = sill. log cos x. 56 THE DIFFERENTIAL CALCUL US. ~= dY7cosxdx. logcos x + sinx dcos x y cos X cos x and dy = (cos x)sin -1 cos2 x. log cos x —sin2 x} dx. 43. Differentiate y = log a + b tang x 44. Differentiate y - sin-' x 1/1 +X2 We have dii 11+X2 dy_ -? 45. Differentiate y sin-'(sin x). We have d sin x cos x dx dy- _ dx. 1/1- sinz x cos x 46. Differentiate y log i/sin x + log r/cos x. 47. Differentiate y-log cos x + V- 1 sin x}. Ans. d _ 48. Differentiate =y eea(a sin x - cos x) a2 + 1 We have dy _ Sa s a eax(a sin x- cos x) + aea cos x + ela sin x$ eax Sill x dx. 49. Differentiate y = cos' acos- + b }' b cos x + a EXAMPLES. 57 38. In the next article we present some examples to be solved by the methods for functions of functions and compound functions. For convenience we repeat the formulas for these cases. 1. If ut F(y) and y -f(x); du du dy dv- dy'd-x' 2. If u- F(y. z...), y —f(x), z -p(x).. du dlu dy + du dz dx dy' d x dz- dx 39. Ex. 1.-Given u-aY and y= b; find du-. We have du aY log a; dy- b log b. dy dx du u aY bx log a. log b. Udx~d 2. Given u log y, y- log x; find d-. We have du _1. dy 1 dy y' d x du 1 dx xlog x 3. Given u=- l/z, z x + (x + a)"; find du 2ax — x2 du 4. Given u = tang-' y, y 2a-+ x2; find du 5. Given u =sin-' y, y e" cos x; find dudx 6. Given u- ey, y tang-' x; find du 7. Given ut-xy, ypxx; find dt 58 THE DIFFERENTIAL CALCUL US. 8. Given u- sec-' y, y — s-; find du 9. Given u =sin-' (y- z), y = 3x, z - 4x3; find du We have du 1 du -1 dy /1 — ( — )2 d 1/1 (y —z)2 dy 3; dz 12x2. dx dx du 3 -12x2'dx V1 (3x 4x3)2 10. Given u -yv, y - e, v - xx; find dU dx' 11. Given u- Y, y e-e-, z - ex+e-; find du z ax-,d' 12. Given u tang-'y, y - x+ 1 x2; find dl. Ans. du V1 - X dx 2V1 _X,(1 + xi/ iX=2) fndu 13. Given u=tang-'y, y-= —osx; find d Ans. d - dx 2 14. Given u -log y, y v/1 +x+- -x; find du dix15. Given y2 — 2axy+x2 —b2 -0; find dy Assume u - y2 - 2axy + x2 - b2 = 0. Then we have u - F(x. y) =0, and (Art. 34 [4]) du du du dY (1) dTx + dx' d SUCCESSIVE DIFFERENTIATION. 59 Nwdu2 duNow, d 2x - 2ay; 2y - 2ax. by substitution in (1) and reduction, dy ay - x dx y-ax' It will be noticed that equation (1) in this example is the same as the formula for implicit functions in Art. 34. 16. Given u -- y -+ 3axy + x - 0; to find dy. dx' Employing equation (1) as above, we have d 3x2 + 3ay; d 3yu + 3ax; dy x2 x+ ay'x y' + ax' 17. Given u =y4 + 4y'x + 6y2X2 + 4yx3 + X4 —O to find dy CHAPTER IV. SUCCESSIVE DIFFERENTIATION. 40. The derivative of a function of x will, in general, be itself a function of x, and will therefore have its derivative and differential, which may, in their turn, be differentiated; and so on. The process of obtaining the derivatives and differentials of derivatives and differentials is called successive differentiation. In determining the successive derivatives of a function, we shall suppose the differential of the independent variable 60 THE DIFFERENTIAL CALCUL US. to maintain a constant value, as we are evidently at liberty to do, and this will introduce great simplicity into the operation. 41. Notation.-Let y F(x), and dy — F'(x). dx By differentiation we obtain d{ d} - dx designating by F"(x) the derivative of F'(x). Differentiating again, we have -_ F"'(x); dx and so on. The notation used in the first members of the above equations being inconvenient, it is in practice replaced by the following: d(dy) is denoted by d2y; dfd(dy)j by d3y; dx. dx by dx'; dx. dx. dx by dx3; and so on. The above equations will then become y-=F(x); dyF'(x); dY=F"(x); y'"(x); F (x), in which F'(x), F/,(x).. Fn(x) are the 1st, 2d.. nth derivatives of F(x) or y; dy, d2y... dny are the 1st, 2d... nth differentials of F(x) or y; dx, dx2... dxl are the lst, 2d... nth powers of the differential of x; dy/ dy d'yae th ~ye dy'7/... d_/y are the dst, 2d.. nth differential co-eficients of F(x) d' d- d, ith respect to x. or y, wvith respect to x. EXAMPLES. 61 The student must be careful not to confound dny with dy'r. 42. EXAMPLES. d Y _ n xn-l d2 n(-1) x-2 dSY -it (nl)(n-2) x-3; d —Ynm(n —l) (n-2)... 3. 2. 1. 2 y log x;dy l d2y _ 1 y _ 2 dy_ 2. y —iogx;;~ x; dx2 — x-'~; dXy *dx x' dxL x't dx3 F dx 3. y sin x; dy - X d - — sin x; dY COS; dx dxx d _ sin x; etc. dx4 4. y = ax; dy ax log a; diy - aZ log2 a;.ya; ~ oga dX2 d - ax log3 a; etc. (VI 5. y=-; d e; d e; d -; etc. 6. y = u v, in which u and v are functions of x. We have already found dy dv du dax d +d Differentiating both sides of this equation with respect to x, and observing that dv duare functions of x, we have d2y d2v du dv d21t dv'du d_ + 2du dv +du dxI dx dx (p5x' 62 THE DIFFERENTIAL CALCUL US. Differentiating again, we have d3y d3v du d2v +du d2v 2c2u dv dL d'I d. r d X- dx- dx dv d2ut cld3 d:v d du d-v 3dv d2u+ C(3 -X3-l - 3- +3dx c2 +. Similarly we shall find d4y d4v du d'v +d2t d2v 4dv d'3i d4ui dX4 dX4 7XdX-, dp-. + 6 - + 4dv' + dx4 udx -dx d dX' dx2 xdx d..x A simple inspection of the foregoing results shows that the co-efficients follow the same law as in the Binomial formula, and it may readily be shown that this will always be the case. The resulting formula for dy is known as Leibnitz' Theorem. 43. We give here two examples in the successive differentiation of implicit functions of a single variable, reserving until a subsequent chapter the notation and formulas adopted for such cases. 1. u-=x3+3axy+y3- O. We have already found [Ex. 16, Art. 39,] dy x - +ay dx y2 + ax Designating dy by it', we shall have, since u' is a function of x directly and also indirectly through y, d2y d_ d' _ di' dy + clt' dx.2 d dy- dx dx Now, dut' _ 2x(y +- ax) + a(x -+ ay) d.x (y - + ax)2 CA UCHY'S THEOREM. 63 and du' - a(y2 + ax) + 2y(x' + ay) dy (Y2 + ax)2 by substitution in (1) and reduction, d2y _ 2a3xy dx2 (y-2 +a)3' 2. u=y -- 3y+x=O. We shall find dy 1 dx 3(1 -y2) Placing this equal to u', and observing that u' is a function of x through y, we have du' d2y _ du' dy (1). d,:x dx2 dy' dx du' 2y Now, by substitution in (1),-y).'. by substitution in (1), d2y 2y dx2 9(1- - y):' We may also obtain the value of dy as follows: Expanding the value of dY by actual division, we have dxdx dy (1 + y2 _ y4 _ etc.) dXY2 - (2y 4y etc.) (2y + 6y + etc.) 44. Problem. — To find an expression for the ratio of any finite increments of two functions of the same variable. 64 THE DIFFERENTIAL CALCUL US. Let F(x), f(x) be two functions of x, and let x,, X be two arbitrary values of x, such that X=x o+h. It is required to find an expression for the ratio F(X) - F(xo) or F(xo + h) - F(xo) f(X) f(Xo) f(xo + h)-f(xo) Let us suppose that the derivative f'(x) is positive for all values of x from xo to X. Also, let A and B be greater than the greatest and less than the least values of the ratio F'(x) between x, and X. f' () Then we shall have F'(x)B; f'(x) f'(x) and therefore F'(x) < Af'(x); F'(x) > Bf'(x). Now, F'(x), Af'(x), Bf'(x), are the derivatives of F(x), Af(x), Bf(x). Consequently [Art. 18, Cor.], F(x) increases less rapidly than Af(x), but more rapidly than Bf(x). Therefore, F(X) - F(xo) < A If(X) — f(x) }, and > B f(X) -f(x) }.. F(X)- F(xo) < A, and > B. j(X) -f(xo) If, then, the ratio (x) be continuous between the values f'(x) xo and X, which will be the case if F'(x), f'(x) are continuous, there must be between x, and X some value of x which will render this ratio exactly equal to F(X) - F(xo) Callf(X) -f(xo) CA UCHY'S THEOREM. 65 ing this value of x, xo+ oh (0 being less than unity), we have F(X) - F(xO) _ F(xo + Oh) f(X) f'(xo) f - Oh) () If we had taken f'(x) negative, the inequalities in the above demonstration would have been reversed, and we would have been led in the same manner to the equation (1). This equation is therefore general, provided f'(x) retains the same sign for all values of x between xo and xo + h. 45. Let us suppose that for xo we have F(xo) =0, f(Xo) -0; then, designating oh by h', we shall have F(xo + h) F'(xo -- t') f(xo + h) f'(xo + h') If at the same time F'(xo)- 0, f'(xo) O, then we shall have in the same way F'(xo + h') _ F"(xo + h") f'(xo + h') f"(xo + h") h" being less than h', and F"(x) being continuous; hence, F(xo + h) _ F"(xo + h") f(xo + h) f"(xo h") Continuing thus, we shall find that if F(xo )=-O, F'(o )=O.. F-l(Xo)_=O,f(xo)=O, f(xO)=O..fn-1 (xO)0O, and the ratios F(x) F'(x) Fn-1() are continuous, f(x) f'(x) fW-i(X) then F(xo + h) - Fn(xo + oh) (2) f(xo +'h) fn(xo + oh) D. C.-6 66 THE DIFFERENTIAL CALCUL US. in which 0 represents some positive quantity less than unity. Corollary 1. —If all the foregoing conditions be fulfilled except F(xo) - O, then F(xo + h) - F(xo) _ Fn(xo + h) (3) f(xo + h) f"(Xo +- oh) Corollary 2.-If x - 0, then F(h) F'"(h) f(h) fn((Oht) or, writing x for h, F(x) F(X) (4), f(x) fP(Ox) an equation which leads to the following Theorem.-If two continuous functions of x, together with their first n — 1 derivatives, are zero for x-O 0 and if the first n derivatives of one of them have the same sign for all values of x between xO and X; then the ratio of the functions will be equal to that of their nth derivatives, in both of which the value of x is some value between x, and X. 46. The conditions which have been imposed upon f(x) in the foregoing equations will evidently be fulfilled if we take f(X) - (X - Xo)"; for, we shall have f(o) - 0; f'(xo);... fn-'(Xo) - 0; fn(xo) 1. 2. 3..., (n - 1) n; f(xo + h) =f(h) = 1'. Hence, equation (3) becomes F(xo + h) -- F(xo) 1. 2 F-(xo + Oh) (5); CA UCHY'S THEOREJl. 67 and if F(xo) 0, then F(xo + h) F(o 2... h) (6). If, also, o — 0, then F(h) -I F (o7) or F(x) 2 F"(x) (7). 1. 2... n If F(xo) be not zero for xo- 0, then F(x)- F(xo) 1. 2 n F(Ox) (8). 47. Corollary 1.-If F(x) tends toward zero at the same time with x, then F(x)_ F"( ox)_ - _ o x"' 1.2...x For, we must in such case have F(O) = 0, F'(O)- O.0 F2-1(0)- 0; otherwise, F(x) would be infinite for x -0. The conditions attached to equation (7) are consequently satisfied, and hence.(x) F"(2x) Corollary 2. —If in (5) we make n- 1, and replace xo by x, we shall have F(x + h)- F(x) -h F'(x - + h) (9). From this we infer that if the derivative of an expression taken with respect to x is zero for every value of x, the value of the expression is independent of x. 68 THE DIFFERENTIAL CALCUL US. For, let F(x) be such an expression; then we shall have h F'(x + oh) = O, and F(x + h) — F(x) 0, or F(x + h) --- F(x). Therefore F(x) has the same value for every value of x, which can only be the case when it is independent of x. Also, if two functions of x have the same derivative with respect to x, they differfrom each other by a constant. For, the derivative of their difference being equal to the difference of their derivatives, and this being zero, their difference must be independent of x, and therefore constant. APPLICATIONS OF THE DIFFERENTIAL CALCULUS TO ANALYSIS. CHAPTER V. TAYLOR'S AND MACLAURIN'S FORMULAS. 48. The preceding chapters contain the most important principles of the Differential Calculus, so far as it relates to functions of a single variable. We propose now to exhibit some applications of the theory, beginning with certain useful formulas for the development of functions into series. 49. Taylor's Formula has for its object the development of F(x + h) in terms of the ascending powers of h. This formula may be demonstrated as follows. From equation (9), Art. 47, we have F(x + h) - F(x) - h F'(x + oh), the only condition of which equation is that F(x) and its TA YLOR'S FORMULA. 69 derivative F'(x) are continuous between the limits x and x+h. This equation may be written F(x + h) - F(x) -- F'(x) + R1, or F(x + h)- F(x) — h F'(x) RI; R1 being a function of h which vanishes when h 0. Now, the first derivative of R1 is evidently zero when h 0, and its second derivative is F"(x + h). We therefore have [Art. 46] RI -- F"(x + oh), and consequently, F(x + h) - F(x) h F'(x) -- - F"'(x + oh), 1. 2 which may be written F(x-+-h)- F(x)-h F'(x) F"(x) + R2, or F(x + h) - F(x) — h F'(x) --' F"() = R2. R2 is a function of h which, together with its first and second derivatives, vanishes when h-0, as may be readily seen, and its third derivative is F"'(x + h). Hence [Art. 46], 2 1. 2.3 h); and F(x + h)- F(x) - h F(x)- h F() x h), 1. 2 1. 2. 3 70 TIIE DIFFERENTIAL CALCUL US. nx hh F,(x + Oh). or F(x + h) = F(x) + h F(x) + L. 2 F"(x) + F h). Continuing indefinitely in this manner, we shall find F(x+h) F(x)+ F'(x)+ 2 1. " (x)+1.2. 3 + F- (x)+ h _ F-(x+ ) (1). +l. 2... 0, —1) ~'~-~~) 1. 2... n1 If now the last term of this equation tends toward zero as n increases, then F(x + h) is the limit to the sum of the terms of the series F(x), hF'(x), h F"(x).., and we may write h2 ha F(x+ h) =F(zx)+ h F'(x)+!-'2 F/(x)+-. 2 3 F"/(x)-... etc. (2). If we designate F(x) by y, F(x + h) by y', and substitute for the derivatives F'(x), F"(x), etc., the differential coefficients which are respectively equal to them, we shall have F(x + h) y + h dy h2 d2y ha d3yd dx- 1.2' 2 - etc. (3). dX Equations (2) and (3) are, both, forms of Taylor's formula. NOTE.-It must be borne in mind that this formula depends upon the conditions that F(x) and all of its derivatives are continuous between the limits x and x -+h, and that 1. 2 n F(x + h) tends toward zero as n increases. If the last condition is not satisfied, then formula (1) gives the development of F(x + h). 50. The term 12 Fn(x + oh) tends toward zero as 1. () is finite for all values of. nFor, n increases whenever F n(x) is finite for all values of n. For, TAYLOR'S FORMULA. 71 under this supposition, whatever may be the value of h, as soon as n passes this value, the co-efficient 1. 2 n will be multiplied (in finding the next succeeding terms) by the factors h +' h - 2' etc., which form a continually decreasing series; and the limit to the product of these factors, and therefore to the term containing that product, will obviously be zero. Therefore, whenever F (x) and all of its derivatives are finite and continuous, Taylor's formula gives the exact development of F (x +- h). 51. Formula (1) of Art. 49 imposes no condition upon the derivatives which are of an order superior to the ntl". These may be discontinuous for certain values of x between the extremes x and x + h, and the formula be still exact; and thus the development by this formula may, in a given case, be exact up to a certain term, and inexact beyond that term. For example let us take F(x) -f(x) + (x -- xo)m+q q(x). The derivatives of this expression for the particular value xo of x will be finite as far as Fm(xo) if those off(x) and p(x) are so, but beyond this term they will be infinite. The development will then be exact only up to the term preceding that which contains Fm(xo), and in order to render it complete it will be necessary to add a term corresponding to the last term of formula (1), which depends upon the m"'t derivative. Taylor's formula will, therefore, in general terms, give the exact development of F(x + h) only up to the first derivative which becomes infinite for any value of x between the limits x and x + h. By means of formula (1) we may determine the limits of the error committed in stopping the development at ainy 72 THE DIFFERENTIAL CALCUL US. given term of Taylor's formula. In fact, if we take the first n terms, the exact quantity which it is necessary to add in order to obtain the value of F(x + h) is hn F"(x + oh). 1. 2...n If, then, we designate by A and B the smallest and greatest values of Fn(x) between x and x + h, the error committed in taking the first n terms will be between Ah, Bh, A- and Bh' 1.2...n 1.2... n' 52. Maclaurin's Formula. Maclauriz's formula has for its object the development of a function of a single variable in terms of the ascending powers of that variable. If in formula (1), Art. 49, we make x = 0, and then substitute x for h, we shall have F(x)=F(O)+ F'(O)+2 + + Fn(Ox) (4). If, now, as n increases, the last term tends toward zero, F(x) is the limit to the sum of the terms F(O), x F'(O), etc., and we may write F(x) = F(O) + x F'(0) + - 2 F"(O) +... etc. (5). If F(x) be designated by y, and we indicate, by inclosing them in brackets, the values assumed by F(x), F'(x), etc., when x- =0, then (5) may be written F(x) -y=[y]+x[F'(x)] + 12[F"(x)]+... etc. (6); F~lc)=y= yl $2CF'~~ -1. 2 MACLA URIN'S FORMULA. 73 or, replacing the derivatives by the differential co-efficients, F() y= [y] + Xd2[1 + etc. (7). Equations (5), (6), and (7) are different, but equivalent, forms of Maclaurin's formula. This formula is attended with limitations similar to that of Taylor. It ceases to give the exact development of a function, whenever that function or any of its derivatives become infinite for finite values of x; but it will be observed that while Taylor's formula may fail for but a single value of x between x and x + h, if Maclaurin's formula fails for one value of x it fails for all other values. 53. EXAMPLES. 1. Develop y = sin x. Forming the successive derivatives, we have F'(x)=cos x; F"(x)_ —sin x; F'"l(x) — cos x; Fiv(x)=sin x, etc. Making x - 0, these become F(O) - sin 0 = 0; F'(0) cos 0 - 1; F"(O) = —sin 0=0; F"'(O) — cos 0 - 1; Fi(O) = sin O - 0; etc. Therefore, by substitution in Maclaurin's formula, which is applicable here since none of the derivatives are infinite, we have y7= sin x=-x- X+ XI' -+etc. y —sin- 1. 2. 3 1. 2. 3.4. 5 1. 2. 3. 4. 5. 6. 7+ 2. Develop y = cos x. The derivatives are F'(x) - sin x; F"(x) = - cos x; F"'(x) = sin x; Fiv(x) — cos x; etc. D. C.-7 74 THE DIFFERENTIAL CALCULUS. Making x 0, we have F(0) = cos 0 = 1; F'(0) - sin 0 - 0; F"(0) - - cos 0 - 1; F"'(0) - sin 0 = 0; FiV(0) =-cos 0 1; etc. Therefore, by substitution in Maclaurin's formula, X2 x4 x6 y = cos x =- 1 — x' etc. 1.2 1.2. 3.4 1.2. 3. 4.5. 6etc. 3. Develop y = ax. We have F(x) - a; F'(x) - aZ log a; F"(x) - a" log2 a; F"'"(x) = a log3 a; etc..., making x — 0, F(0) - a -- 1; F'(O) -- log a; F"(O) - log2 a; F"'(0) -log3 a; etc. Therefore, by Maclaurin's formula, x2 xg log3ae y = aZ- 1 + x log a + 1 log2 + a 3- etc. Corollary 1.-If a = e, then log a = log e- 1, and * e + $-+1 + 12 3 + etc. Corollary 2.-If x - 1, then ex = e, and I 1. e-1+1+ 1 + 13 etc., a formula for the base of the Naperian system of logarithms. EXAMPLES. 75 4. Euler's Formulas.-If in the series ex - 1x + 1 2 3 etc., we put x- zi-, and x — -zl/-1, we shall have = + z2 Zi/-1+ _ etc.) 1 1 1. 2 1. 2. 3 12. 3. 4 etc _ _-X Z2 _ _ _ _ _ _ _ _ _ _ _ etc. 1- --- 1. 2 3 1. 2. 3 + 12. 3. 4' Adding and subtracting these two equations, we have ezv +e-z'-l= 2(1-1 t2 + 2 cos z; 1. 2 1. 2. 3. 4 e-e- -2 z —.- etc. e~z -ejz==21/- (z 1. 2. 3 + 1. 2. 3. 4. 5' - 21/V 1 sin z. eZ —i - e2 —- eZV-1 + e-z i-1 sin z- - Cos zs* nz=- 2i/-1 2 and by division, sin z ez/ - e-zf-1 - = tang z-1 Cos z ez + e-z 1:~ These are Euler's formulas for the sine, cosine, and tangent of an arc in terms of imaginary exponentials. 5. Demoivre's Formula. —Resuming the equations of the last example, e i z1. / — 2 1. + 234 1- 1.-2 1.2.-3 1.2. 3. etc., 76 THE, DIFFERENTIAL CALCUL US. e-zI ~ ~ z _ -1 etc., 1 ---- 1.2. 12. 4 we notice that, by Exs. 1 and 2, the second members of these equations are respectively equal to cos z -- i —1 sin z and cos z —/ — 1 sin z..-. substituting mnx for z, we have emx I-1 - cos mx ~- / —1 sin 7nx. But e+taX r~-1 {e.+Vx}lXi} - {cos x 4- I/- 1 sin x}m, as above. Hence, cos nx V 1/-1 sin mx cos x -- 1/ — sin x}. This formula serves to convert powers of sines and cosines into series whose terms involve sines and cosines of multiple angles. 6. Develop y =(x + h)". We have?F(x +- h) - (x - h)n; F(x) = x"n; F'(x) - nx-l; F"(x) n 2n(n - 1)xn-2; etc. Therefore, by Taylor's formula, F(x + h) (x - + h)= x' + nx'-, h + - _1. -2 h2 + etc., the well known binomial formula. 7. Develop F(x +- h) - log(x + h). We have F(x+- h) - log(x+h- ); F(x) =log x; F'(x); EXAMPLES. 77 I,, 1. 2 1. 2.3 F"(x)- x2; F"(x)- x_2; F'y(x)= - 2- 3; etc. Therefore, by Taylor's formula, F(x + h) — log(x + h) -- log x +- h+ 3 4 etc. x 2x2 3 4X24 Corollary 1. —If x = 1, then log x =0, and we have log(1 - h) = h — h h3 h4 etc. 2 +3 4+ Corollary 2.-From the last equation, we have, by writing u for h, log(l +-u) u- 2+ t 4+ etc. - I - log(1 + u-u) -+ -2 - + etc. lo(11.. 2' l log(+u)-log(1+ )1 og +u =logu = (u — u-) - 1 (u2_ -2) q-+ A(ut3 -_u-3) - (u4- -tt4) + etc. Let u —exV-/; whence log u - xV/ —T, and xl/V - (e1'-1 -- -) -I(e2/2 V- ge-2x4-1) + ~(e3x V1_ e-3x -i>) - etc. ezV — e-x' -1 e2xV -=i ee-2xVT —1 rv/-1 2 V- 1 e3x Y — __ e-3xl —, etc. i/-i 78 TIlE DIFFERENTIAL CALCUL US. by Example 4, s2 - si n 2x + sin 3x - I sin 4x, etc.; and, by differentiation and reduction, 2 cos x - cos 2x - cos 3x- cos 4x + etc. 8. By Examples 1, 2, and 4, we have de,-"/ cos x + /- 1 sin x, e-xVu-u -cos x- 1/- sin x. by di e-isio_ cos x - / — 1 sin x.. by division, _1 ezcsx+1/1sinx e-cos x — / V1 sin x 1 + 17/-1 tang x 1 — — 1 tang x. passing to logarithms, 2xV/-1-log(1+1/ —1 tang x)-log(1 —/'-1 tang x) (1 But, log(1 + u) - log(1 — U) 2(u + -+ +- etc., as may be seen by substituting ~ u for h in Ex. 7, Cor. and subtracting the results. log(l + i/- 1 tang x) - log(1 - i/- 1 tang x) 211/- 1 tang x + t/- 1 tang x}3 +- V — 1 tang x - +etc.. EXAMPLES. 79 by substitution in (1), 2 x / — 2 1/- 1 tang xI +q I l/-1 tang x}3 +HV"-1 tang x]5 etc.}, or, x tang x- A tang3 x +. tang5 x etc. 9. Develop F(x + h) - sin (x - h). We have F(x + h) = sin (x + h); F(x) - sin x; F'(x) - cos x; F"(x) = - sin x; F"'(x)- cos x; Fiv (x) =sin x; etc. Therefore, by Taylor's formula, F(x + h) - sin(x + h) h h2 h3 _sin x + 1 cos x sin x + etc. =sinx 1 1. 1 23. 4 — etc. +cosxh 1- 2 3 +etc.} = sin x cos h + cos x sin h. 10. Develop F(x- +h) -=cos(x + h). 11. Develop y — sin-lx. We have dy _ (1 2)dx x- i 1-x dy 2 1. 3 1. 3. 5 - + j + — 2 x4 + -3 x6 etc.; d2y 1. 2. x- 1.13.4 x 1. 3. 5. 6, i 9 f- 2- - 9 + etc.; dX. 2..2. 2 ~.2. 3. T: 80 THE DIFFERENTIAL CALCUL US. d3y 1. 32. 4 2+1. 3. 5. 6 d~ y- 1 3 2.4 13526x'+ etc.; 1/.- I 2. 22q1 2.3. 2x+ et d4y 1. 2. 32. 4 1. 3. 4. 52. 6 dx —1.2. 22 -x —- x3+ etc.; d5y 12. 2.232. 4 1.32. 4. 52. 6 d.1-l. 2.22 - 1.2. 3.23 x2+etc.; Now, making x- =O, we have, (Y) ~; ( dy d2y d ( d4)y; (d)5 y12 32; etc. by Maclaurin's formula, 12. XI 1 2.3 2. x5 y - sin- -x 1 2 3 1 3. 3. 4. x + etc. 12. Develop y-=tang-'(x). We have dy 1 — 1 2 dxl-x2 - + Cfl x2 x4; e x etc.; dx- 2 x + 4x3- 6x+ 8x7-etc.; dx2 dys- 2 3. 4- 5. 6x4+ 7. 8 6- etc.; dy _ x)-; dX2y )-; - d xY) 2; etc. dx) (d2 dy CONVERGENCE OF SERIES. 81 by Maclaurin's Formula, e X- X7 y - tang-l x -- x - 5- -7+etc. -t 7 ~ etc. Corollary. —If in this series we make x -tang 450~- 1, we have y arc of 45~ -- + -- — etc. - r; or -- 4(1 a —+ 5-{ - ~etc.) This is a slowly converging series for the ratio of the circumference of a circle to its diameter. Scholium. —In the last two examples we have expanded the values of dy, a method which is applicable whenever dx such expansion will give rise to converging series. When this is not the case, the method ceases to give exact results; for if the series is not converging, we evidently have no right to say that it represents exactly the value of the expression with which it is connected by the sign of equality. In all the preceding examples all of the successive derivatives have proved to be finite, and therefore the Formulas of Taylor and Maclaurin have been applicable. Had this not been the case, it would have been necessary to make use of formula (1) instead of (2) or (3), and (4) instead of (5), (6), or (7). CHAPTER VI. THE CONVERGENCE OF SERIES. 54. The formulas developed in the preceding chapter give exact results only when they give rise to converging series: it is consequently important for us to consider the laws of the convergence of series in general. 82 THE DIFFERENTIAL CALCUL US. Let u0, u,, u2.... un be a series, of which u, is the general term; and let Sn= UO+ U1+ Us+ * Until be the sum of the first n terms. If, as n increases, SS, tends continually toward a finite limit, the series is converging. If this is not the case, the series is diverging. Suppose S to be the limit toward which S,, approaches. Then, when n — o, the sums S,, Sn,, Sl+2, etc., will differ by infinitesimals from S and from each other. Now S,+1 —S —u: S,+2 — S,,atn+Un+: etc.; and, since these differences are infinitesimals, it follows that when n oo, we must have u-O-0; uA,,+-t,+1,=-0; etc. And, conversely, whenever u,, and all the succeeding terms are zero for n — o, the series tends toward a finite limit, and is therefore converging. 55. These principles furnish us with two simple rules for determining, in many cases, the convergence of series. RULE 1.-If we represent by U,, the numerical value of u,, and designate by L the value of the limit toward which f/U. approaches as n increases, then the series will be converging if L < 1, and diverging if L > 1. For, let L < 1. If we designate by p a number less than 1 and greater than L, so that L

1 the series is diverging. RULE 2.-If, as n increases, un decreases, and the ratio Un +I Un converges toward a limit 1, the series will be converging when 1 < 1, and diverging when 1> 1. Let e be a quantity less than the difference between 1 and 1, so that the two quantities, 1 — e and 1 + e, shall be, at the same time with 1, less or greater than 1. Supposing n to increase continually, UJl, will finally be comprised between 1 — e and 1 + e, and the terms of the series, i,,, u+tn, U,,+, etc., will be comprised between the corresponding terms of the two series, U,,, u,( — ), U,(1 —e)2, etc., u,, ui-(l + e), t(1 +- e)', etc. Now these series are both converging if 1 < 1, and diverging if i > 1. Hence, in the first case, the given series, u,, Un,+1, etc., will be converging, and therefore u,, u,, u2, etc., will also be converging, while in the second case it will be diverging. Corollary.-The two limits L and 1 are the same. For, designating any number by m, the ratios Umn+1 Um+2 Umn-2 Um+3 Um Um' Um+.' VUn.2' Um-n-l' 84 THE DIFFERENTIAL CALCUL US. and consequently their geometrical mean,ur, will differ from 1 by a quantity e, whose limit is zero. Consequently, 1 1 i/mUn= i Un(i+} (l e); Ur } Passing to the limits, by making n Ao, we have liml U,/ _L-lim(l 4 e)I Umj =-1. 56. We will now apply these rules to Taylor's and Maclaurin's formulas. Let en be the value of the expression 2. (x), and let O be the limit of Hji_ or 2_+~'. The two formulas will be converging when the numerical value of x is less than T, and diverging in all other cases. For, if we designate the general term of these formulas by un —, (x. ), we shall have, in the first case, lim (u,,) Ox < 1; lim ( +') x < 1; and in the second case, lim (u,~)T= OPx> 1; lim( U"+ )=Px > 1. Therefore, since in the first case the limit is less than 1, the series is converging; while in the second case, the limit being greater than 1, the series is diverging. 57. EXAMPLES. 1. Let F(x) — eZ. Is the development of ex a converging series? EXAMPLES. 85 We have F(O) 7= 1; Fn(0)-; th 2..;1 -'- 1; (P=li — i; m -- nl + 1P P Every numerical value of x is less than 1, and therefore the development of ex is converging for every finite value of x. 2. Let F(x) = cos x. In this case (P — 1. 1 ). The value of this expression is evidently less than 1, and therefore the development of cos x is a converging series. 3. Let F(x) - sin x. The development is a converging series. 4. Let F(x)- log(l + x). We have, by differentiation, F,"(O) (- 1)"-'I1. 2. 3.. - 1.; 1 ci,,+ 1 + 1 I 1. Therefore the development of log(1 + x) is not converging for any value of x greater than unity. 5. Let F(x + hl) = log(x + h). Is the development of this function converging? 58. Although the formula of Maclaurin gives, in general, the expression for the development of F(x) whenever it is converging, yet this is not always the case. If, for example, we take F(x) =e-2+ e- x. 86 THE DIFFERENTIAL CALCUL US. the development of this function, by Maclaurin's formula, is converging; but instead of being the development of the entire function, it is merely that of the first term, e2. Nevertheless, whenever the functions represented by F(x), F(x-~h), call be developed, by any known process, into converging series, arranged according to the ascending powers of x or h, the resulting series will be identical with those given by Maclaurin's and Taylor's formulas: for two converging series, arranged according to the ascending powers of the same quantity, and whose sums are equal, must, by the theory of indeterminate co-efficients, be equal term for term. We may remark that it is an open question whether a function which can not be developed by Maclaurin's formula is essentially incapable of development by any process whatever. CHAPTER VII. E STIMATION OF THE VALUES OF FUNCTIONS WHICH ASSUME INDETERMINATE FORMS FOR CERTAIN VALUES OF THE VARIABLES. 59. WVhen the relation between two functions of the same variable is that of a quotient, product, or exrponential, a particular value of the variable may render one or both of the functions equal to zero or infinity, and thus reduce the given expression to one of the forms,, —, o X 0, 0X, etc. These are usually called indeterminate forms, because, being the symbols of operations which it is impossible to effect, they offer, in themselves, no clue by means of which their values may be determined. They are, properly speaking, the limits toward which the forms of the expressions from which they originate converge as the value of the variable INDETER2MINATE FORMS. 87 tends toward some particular value, say x0; and the limits toward which the values of the given expressions tend, as x tends toward its value xo, are evidently the real values symbolized by the expressions when they assume the peculiar forms above enumerated. The Calculus affords a simple method of finding these real values in most cases. 60. Functions of the form. Let F(x), f(x) be two functions of x, such that F(xo) = 0, f(xo)- =0 Then we shall have F(x.o) o0 f(xo) O We have' already seen [Art. 45] that if F(x), f(x) and their first n -- 1 derivatives reduce to zero for x -x, then F(xo+ h) F"(xo+ Oh) f(xo+ hl) f?(Xo+ Oh) Hence, lim F(xo + h) lim F'(x +; or, f(xo + h) f (Xo+ Oh); F(xo) F"(xo) f(Xo) f"(Xo) Therefore, The value of a function which reduces to the form -0 for a particular value xo of x is equal to the ratio of the 0 first derivatives of the numerator and denominator which do not reduce to 0 or so for the value Xo of x. Corollary.-If either of the derivatives should be 0 or oo while the other is finite, the real value of the function will then be 0 or so. 88 THE DIFFERENTIAL CALCUL US. Scholium. —If all the derivatives should reduce to 0 or 0, this method ceases to be applicable. In such cases the proper plan is to substitute x0 + -h for x in each of the functions, to find by some algebraic process the resulting value of the fraction, and, finally, to pass to the limit by making h =0. EXAMPLES. 1. Let X31 _ for x 1. We have F(x) - x4- 1, f(x) x- 1.. F'(x) - 4x = 4 and f'(Z) 3- x2 a3. F(x) F'(x) 4 f (x) -f'(cx) 3 2. F(x) _ee- x 0 for x - 0. f(x) sill - 0 Ans. F(x) F'(x) 2 2 for x 0. f(x) f() - 2,f or x- O. 3. F(x) sin x 0 for x-0. A(x)- X2 - Ans. F(x) F'(x) cos x 1 f (x) f' (x) 2x 0F(x) logx 0 for x —. f(x) x —1 0' Anls. F(x) F'(x) 1 for x 1. f(x)- f'(xj- xFx a — b 0 O 5. for x — 0. Ans. gf(x) log b EXAMPLES. 89 F(x) x4 —5x+ 9x- 7x+2 0forx 6.6 12x2-10x+3O for x=1. f (x) -- X' — 6x/~ 12x ~- 10X q- 3 - We have F'(x) 4x3 — 15x2+ 18x - 7 0 f'(x) - 4x — 18x0 0 24x —10 - 0' F"(x) 12x2- 30x + 18 0 O f"(x) - 12x2 — 36x + 24 0' F"'(x) 24x - 30 1 for f"'(x) 24x- 36 -- orx. * F(x), for x- 1. f(x) 7. F(x)_tang xsin x 0 for xz0. f(x) sin x - 0 Ans. ~, for x -0. 8. F(x) / x — a _ 0 for x a. f(X) - a2x__ - In this example all the derivatives become infinite, for x- a. We therefore place x a + h, and after reducing the resulting expression as much as possible, put h = 0. We thus have, 1"x — a / h h- -1 1/ 2_a -- j/'2ah + h -- h(2a +h) - (2a + h) — - for x -- a, and h - 0. F(x) 1/'x- /a+Vx —a 0 9Vx2__a2 -- O for x -a. Placing x a + h, since the derivatives are all infinite, we have, D. C. — 90 THE DIFFERENTIAL CALCUL US. (a+h) — a+ hm hL3-+'a-'h, etc. 1 + etc. (2ah + h2) -- h(2a + th) -(2a + h)t I2 I A) -(-2-adsfor x =a, and h —0. -- (2a + h)- (2a)' 10. F(x) ex- 1-log(l x) _O for x- 0. Instead of differentiating as usual, we shall evaluate this expression by substituting for ex and log(1 + x) their expanded values. These are e=-lx 1- + 2.. 3+ etc. X2 XX3 log(1 + x) x -2- +3 etc.'f(x)1 -6' etc., -1 when x -0. f(x) 6 61. Functions of the form bIf F(x)_ for any particular value of x, then f(x) - F(x) and -1 will each be zero for the same value of x, and we f(x) may solve this case by applying the preceding method to 1 1 the functions (x)' (x We have, therefore, F(xo) oo f(xro) 0 f (Xo) -- 1 0 F(xo) Hence, by differentiation, f'(xo) F(xo) f (xo) 2 fF() 2 f () f (xo) F() Fl(x~)' F'(xo) - F(xoo) EXAMPLES. 91 Whence, by reduction, F(xo) F'(x0) __ F-(xo). f (xo) - f'(xo).- f (Xo) and the method in this case is therefore the same as for the 0 form -. It is to be observed, however, that if F(x) is infinite for a finite value of x, so also are all of its derivatives; and the above method will not be of practical value for finite values of x, unless we can detect in the terms of the derived functions a common factor, which; being canceled, will leave a result whose value can be determined. EXAMPLES. 1. F(x) xz_, for x =o. f(x) ex o Differentiating n times, x disappears from the numerator, and we have, F(x) _ (1 - 1)(i- 2).. 3. 21 _ 0, for x -s- C. f(x) e) 2. o F(x) log(x)- for =. f(x) () We have, F(x) F'(x) x — X> - f'(5> -- -x — 0, for x -0. f (x) - f'() -., 62. Functions of the form Xc X O. Let F(x), f(x), be two functions of x, such that for x-=x we shall have F(xo) - x, f (xo) 0. 92 THE DIFFERENTIAL CALCUL US. Then we shall have, identically, F (xo) X f (xo)- f (' F(xo) which reduces this case to the first. Therefore, by differentiation, F'(xo) If the second number of this equation should not prove to be determinable, we must continue to differentiate as before, and it is obvious that the method will have the same limitations as in the preceding case. 63. Functions of the forms 00, 0oy, 0o, etc. Let u- F(x)f(x), in which either or both of the functions F(x), f(x), may reduce to 0 or oo, for the value x0 of x. Passing to logarithms, we have, log u -f (x) log F(x) - 0 X oo. We may, therefore, find the value of log u, or of f(x) log F(x) by one of the preceding methods, and we shall have, finally, o eIog u EXAMPLES. I.. F(x)'f(x) — x- 00, for x - 0. Passing to logarithms, we have, log fx-x log x- - O X o, for x O0. EXAMPLES. 93 log x c_ log x — x log x- (lg) - O, as in Ex. 2, Art. 61. * x-lo~gx — e~- 1, for x - 0. 2. F(x)f(x) (1 -+ x)=- 1%, for x -O. We have 1 10 log(1 - x) = 1 log(1 -- x) =-O for x =0. x 0 by differentiation, 1og(1 + X)P_ +. 1, for x =0. (1 +XX )- e, for x_-0. 3. F(x)f() x)- c~, for x =o. We have log x- log x-, for x o. X cO by differentiation, 1 1 1 log x~-=_ 1 0, and x;- 1, for x- oo. 64. General Examples on this Chapter. 1 (2ax-X4)2 (ax)_ - 0 when x- a. Ans. 16a - ( 9 94 THE DIFFERENTIAL CALCUL US. X2. — r Xm+ 0i 2. _ when x 1. Ans. 2 1- X'P This expression may be put under the form l+-2' 1 —x' and since the latter factor alone reduces to -~, we may find its real value when x- 1, and multiply the result by the first factor, whose value is evidently-. 1 — Sill X+Cos O sr 3. winx -cos hen x -- Ans. 1. sinx + cosx-1 0 2 2X2 —a') sin x a Sl2e 0 4 4 o- a 02 when x -a. Ans.X COS In this example sin x reduces to 1, and we may proceed 2a as in Ex. 2. cos'-(1 — x) 0 5. 2x —, - - when x - 0. Ans. 1. 6. tang rX —7X 0 when x 0. ~ 2xa tang,xX - Assuming x =0 + h, and expanding tang rX = tang rh, by Maclaurin's formula, we have, tang h - h = + -- 3 - + etc. tang 7rh -7rh 1 7_ 2x2 tang rh -2h- 2h(7rh +- ~(h)3+ etc.) r+:-' 1h2+- etc. —, x O,'r h —- when h 0, or x - 0. 2h(rh q - ~(:-h)'+L etc.) - 6 iNDETERMINATE FORMS. 95 7. (1 + 1) -o when x- 0. Ans. 1. x 8. (/tang x_ 1' when x - O. Ans. 1. -9.V -1 when x=0. 1 — 1/ /1 when x - 0 10. (sin x)tang x 1~ when x = 11. log tang 2x- 0_ when x - 0. log tang x cc 12. ( ) o when x =0. 13. x _ 1 OD-co when x -1. x-1 log x This may be put under the form x log x- (x- O when 1) 0 when x 1Ans. 14. (x) —1 - 1 when x =1. Ans. 15. (Axm+Bxm-'+ Mx +YN) =co when x -. Ans. 1. 16. (cos ax)"cO e2lx -1 when x 0. Ans. -. 22, 17. Show that when x — o, ax.,, c_ o01 0, according as m > or < n; a and b being both greater than unity. 96 THE DIFFERENTIAL CALCUL US. CHAPTER VIII. MAXIMA AND MINIMA. 65. When for a particular value xo of x the corresponding value F(xo) of F(x) is greater than the values F(xo+ h) and F(x,- h), in which h is an infinitesimal, the value F(xo) is said to be a maximum value of F(x). If F(xo) be less than F(xo -h) and F(xo- h), then F(xo) is said to be a minimum value of F(x). It results from these definitions that, for a maximum, F(x, + h) — F(xo) must be negative, whatever may be the sign of h; and that, in order to a minimum, F(x, +h) — F(xo) must be positive, whatever may be the sign of h. [1] It is evident that in order that these conditions may be fulfilled, F(x) must be a decreasing function on one side of F(xo), and an increasing function on the other side of F(x,). Now, we have already seen [Art. 18] that when F(x) is decreasing, F'(x) is negative; and when F(x) is increasing, F'(x) is positive. Hence, in order that F(x) may be a maximum or minimum, F'(x) must, in passing through the particular value F'(xo), change from a state of increase to one of decrease, or vice versa: and, as a quantity can change its sign only in passing through 0 or oo, we have, as a necessary condition for a maximum or minimum value of F(x), F'(Xo) - 0, or F'(x0) = o. [2] Now, it has already been shown [Art. 46] that F(Xo + h) - F(xo) = Fn(xo + Oh), an equation which may be written F(xo + h) -F(xo) 1 F"(xo) +Rnh} -. 2.. MAXIMA AND MINIMIA. 97 in which n denotes the order of the first derivative which is finite for the particular value xo of x, and Rh is an infinitesimal. The sign of the second member of this equation will evidently depend upon that of its first term 1. 2 (x). Hence, If n be an odd number, this term and, therefore, the first member of the equation will change sign with h, and F(xo) can not be either a maximum or minimum. If in be an even number, the second member and, therefore, the first member also will not change sign with h: it will be positive if F"(x0) is greater than zero, and negative if F"(xo) is less than zero. But F(xo) is a maximum if F(x + h) - F(xo) < 0, and a minimum if F(xo + h) - F(xo) > 0. Therefore, F(xo) is a maximum if F"(xo) < 0; and F(xo) is a minimum if F"(Xo) > 0. 66. The preceding investigation leads to the following Rule for determining maxima and miniima values of explicit functions of a single variable. 1st. Form the successive derivatives of the function. 2d. Place the first derivative equal to zero or infinity, and solve the resulting equation. The values of x so found are the only values for which the function is to be examined. 3d. Substitute these values of x in the remaining derivatives. If the first derivative whose resulting value is finite be of an even order, and less than zero, the corresponding value of F(x) is a maximum; but if it be greater than zero, the corresponding value of F(x) is a minimum. If the first finite derivative be of an odd order, there is neither maximum nor minimum. D. C.-9 98 THE DIFFERENTIAL CALCUL US. 67. If ALL of the derivatives be either zero or infinity, the preceding method ceases to apply. In such cases the question may be determined in the following manner: Substitute xo + h for x in the function. Then, if the term which contains the lowest power of h changes sign with h, there is evidently neither a maximum nor minimum value of F(x) for the value x, of x. But if this term does not change sign with h, then the value of F(xo) will be a maximum when this term is negative, and a minimum when it is positive: since, in the first case, F(xo) will be greater than the immediately preceding and following values of F(x), and in the second case it will be less than those values. It is possible, and sometimes preferable, to use this method even in cases where the general rule is applicable. 68. If it be required to determine maxima and minima values of implicit functions of a single variable, we may, according to the rules established for these functions, determine the successive derivatives, and then proceed as above. Thus, if we have, du it=F(x,y) =O, then dx - du (1). dy But for a maximum or minimum dy 0, and, therefore, dx du du du O, or - oo. Either of these equations, together with the given equation, F(x,y) = O, will enable us to determine values of x and y for which the function is to be examined, and we may then apply the usual tests to the successive derivatives formed from equation (1). 69. In applying the preceding methods, we may be greatly facilitated by bearing in mind the following simple and selfevident principles: MAXIMA AND MINIMA. 99 1st. If F(x) is a maximum or minimum, a F(x) is also a maximum or minimum. Therefore, the given function may be multiplied or divided by any constantfactor without affecting the character of the result. 2d. If F(x) is a maximum or minimum, F(x) -+ a will be a maximum or minimum; but a — F(x) will be a minimum when F(x) is a maximum, and a maximum when F(x) is a minimum. 3d. If F(x) is a maximum or minimum, will be a minimum or maximum. 4th. If F(x) is a maximum or minimum and positive, SF(x)In will also be a maximum or minimum, n being positive. But if F(x) be negative, IF(x)}2, will be a maximum when F(x) is a minimum, and a minimum when F(x) is a maximum. 5th. If F(x) is a maximum or minimum, log F(x)} will be the same. 6th. It is not admissible to assume x equal to infinity in the search for maxima and minima, for in that case x can not have a succeeding value. 7th. The basis of the whole theory is that F(x) must be continuous, at least in the neighborhood of the particular values to be examined. 70. EXAMPLES. 1. Divide a number a into two parts whose product shall be a maximum. Let x be one of the parts, and a- x the other. Then, F (x) - x(a- x) ax- x2. F'(x)= a-2=x O... x- a F"(x) -- 2. 100 THE DIFFERENTIAL CALCUL US. The first derivative which does not reduce to zero being of an even order, it follows that the value 2a of x, which reduces the first derivative to zero, will render the value of F(x) a maximum. This value is -. 2. Find the value of x which shall render F (x) = X2 - 8x + 5, a maximum or minimum. We have, F'(x)=-2x-8 =0..'. x=4. F"(x) = + 2. The second derivative being positive, x - 4 renders F(x) a minimum. The minimum value of F(x) is - 11. 3. Find the values of x which shall render F (x) 2x3- 9ax2 12a2x - 4a3, a maximum or minimum. We have, F' (x) 6x2- 18ax - 12a2 =0... x-a and x 2a. F"(x) = 12x- 18a - 6a, for x a, - + 6a, for x _ 2a. Therefore x = a renders F(x) a maximum, and x -2a renders it a minimum. These values are, respectively, a3 and 0. 4. Find whether F(x) = b + c(x - a)3 has maximum or minimum values We have, 7 4 F'(x) 3 C (x -a) 0O..'. x-a. 28 (x- a)- O for a, F"(x) = - (x - a) 0, for x. a, EXAMPLES. 101 and all succeeding derivatives are infinite. We infer that the given function has neither maximum nor minimum values. 5. Determine the maximum and minimum values of Fl (x) = b + c(x-a).2 We have, F' (x -2 -c (x2 a)-*, for x-=a. -FWc(x)) 3 2 4 F"(x) — 9 (x — a) —-, for x =a, and all the succeeding derivatives are infinite. Let x -a +- h. Then, F(x) =-b + chi. As this does not change sign with h, and the term containing h is positive, we infer that x = a renders F(x) - b a minimum. If c be negative, F(x) b will be a maximum. 6. Find the maximum and minimum values of F (x)-mn sin(x - a) cos x. We have, F' (x)=m cos(2x-a) 0.. -. x - a +- F"(x) = - 2m sin(2x - a) a 7r 2m, for x - +, a 2-+ 2nm, for x 2 7 X E 4, 102 THE DIFFERENTIAL CALCUL US. Therefore, the first value of x renders F(x) a maximum, and the second renders F(x) a minimum. These values of F(x) are, respectively, m (1- sin a) and - 2 (1-+ sin a). 2 2 7. Find the maximum and minimum values of y in the expression F(x,y) -=y- 3ax + xd =0. We find, dy a2-_ * a. dx- y2 d2y 2a ~d~x~2 for x + a. 2a -,+ for x = -a..'. x= + a renders y a maximum, and x=- a renders y a minimum. 8. Find the greatest and least ordinates of the curve a2y - ax2 + x3 0. We find, dy 2ax- 3x2 2a dx 2. x O or Cdx2 a -- 2 2a' 2 2a -, for x 2 EXAMPLES. 103 x = 0 renders y 0, a minimum; and x - renders 4a y- 2 a maximum. 9. Find the number of equal parts into which a number a must be divided in order that their product may be a maximum. Let x = number of parts, and a the value of each part. x Th~en a a a a Then a.. a a __ the required product. Assume F(x) a ), or by taking the logarithm, F (x.) =-X log( - )x log a- x log x. F'(x) log a- log x -1 -- 0; log a logx=log( a ) 1. 1 F"(x)- Hence, log( x 1, or - -e, and x a x X e The maximum product is ee. 10. Find the value of x which renders F(x),-(2+ a maximum. We have, F'(x) - (a2-2) — 0... x O or aThe value x Va 4 renders =F(x) - a maximum. 104 THE DIFFERENTIAL CALCUL US. NoTE.-Whenever we are certain, from the nature of the problem, that there must be a maximum or minimum, we may neglect the second derivative, unless the value of x, which reduces the first to zero or infinity, should also reduce the second to the same values. This can often be determined by simple inspection. 11. Find the greatest cylinder which can be inscribed in a given right cone with a circular base. In Fig. 4, let AC b, VC — a, Fig. 4. DF = x. v Then we have, VC: VQ -AC: DQ AC x VQ _ b(a -) x). VO a G B and the volume of the cylinder is 7r(a - x)2b'x a2 Therefore, omitting constant factors, F (x) = x(a -- X)2. F'(x) a — 4ax + 3x2 = 0... x = a, or a We see at once that the second value of x is the one required. This value substituted in (1) will give the volume of the maximum cylinder. 12. The content of a right cone being given, find its form when its surface is a maximum. Let be the given content, x the radius of the base, and y the altitude of the cone. EXAMPLES. 105 Then we shall have, 3X2- a3 whence x2y a3, and y- a3 3 3' The area of the base will be,x2, and the convex surface,xl/x$-I y2. Hence, the entire surface = 7x2 + 7X V/x2 + y2 =,X2 + -1/X6 + a6. x.'. omitting the constant factor r, F(x)- = + - (x + ) X We find, x —= and y- 2a. These two elements determine the form of the cone. 13. Find the point on the straight line, joining two lights of unequal intensity, which receives the least amount of illumination from them both. Let a and b denote the intensities of the lights at a unit's distance, designate by c the distance between the two lights, and let x be the distance of the required point from the first light. Then, since the intensity of light varies inversely as the square of the distance, that of the first light at the distance x will be a and that of the second will be (b - at the distance e-x. Hence, (C -- X)2 a b F(x) - - - + (x) 106.THE DIFFERENTIAL CALCUL US. 2a 2b 2 bx- 2a(e - x)' _ x3(C -x -0. X3 a X 3a cla * -(C=X)3 b c-x- b; $ -, 14. Find the greatest rectangle which can be cut from a given trapezoid. 15. Find the greatest cylinder which can be cut from a given frustum of a cone. 16. Divide a into two such parts that the product of the mth power of the first by the nth power of the second shall be a maximum. We have, F (x) =x"(a- x). F'(x) =x"-l(a - x)n- ma - (n + n)x= O.?na X -O, xa, x=- + m+n F"(x) Ila -- (n +n )x] -'(-D - (in + n)x-(a - ) n-'; and by substituting the values of x in this expression, we find that x 0 renders F(x) a minimum if in be even, x - a renders F(x) a minimum if n be even, and x = — a n j+- n renders F(x) a maximum without reference to the characters of m and n. 17. From two points, A and Fig. 5 B, draw lines, AP, BP, to a A given line, CD, so that their sum may be a minimum. Take the given line as the c r D EXAMPLES. 107 axis of abscissas, draw AC and BD perpendicular to CD, and take C as the origin. Let CP-x, CA - a, CD=b, BD = d, PD b- x. Then we shall have F (x) = /I+ + l/ 2d+ (b —x)2. X b-x F'(x) = /a'z - (bx - 0; V ( aA 2+ 29 1/d2- (b - =x)2 x b-x **' x* x Vd-2 - /d b- ( x) The first member of this equation is the cosine of APC, and the second member is the cosine of BPD. Whence, APC = BPD, which defines the position of P. 18. Given the length of the arc of a circle; find the angle which it must subtend at the center in order that the corresponding segment may be a maximum. Let a denote half the length of the arc, and let x be the radius; then a is half the angle of the segment and x we shall find, Area of segment = F(x) _c a- X2 sin a a x X C a a ad a.' "(x) — a {2x sin cos + x cos - sin - x x dx x + X, s, a a +x2 sin —cos -- 0. by reduction, cos — a cos — x sin x 0, or Cos — 0. X X X X 108 THE DIFFERENTIAL CALCUL US. a 7r 2and 2 and -2 -, or the required angle is 180~. x 2 x 19. Show that of all circular sectors of the same perimeter the one which has the greatest area is that whose arc is double the radius. 20. Find the sides of the greatest rectangle which can be circumscribed about a given rectangle whose sides are a and b. Ans. Each side a - b 21. Find the sides of the maximum rectangle which can be inscribed in a given ellipse. 22. Find the least ellipse which can be circumscribed about a given parallelogram. Let 2a and 2b be the sides of the parallelogram, and designate by a their included angle. Then the ellipse will have a pair of conjugate diameters parallel to the sides of the parallelogram, and designating them by 2A', 2Bf, we shall have, A.'2 a2+-B'2 b —A1 2B'2 (1). Also, designating the semi-axes of the ellipse by A and B, we have A'B' sin a_ AB, and the area 7tAB - =rA'B' sin a (2). Taking A' as the independent variable, we have F(A') -;rA'B' sin a = A'B'. Substituting in this equation the value of B' derived from (1), and differentiating, we shall find that Area of ellipse: area of parallelogram: 2. 23. Find the length of the longest line that can be drawn from the focus of a hyperbola perpendicular to a tangent. FUNCTIONS OF TWO OR MORE VARIABLES. 109 NOTE.- The formula for the length of a perpendicular -ex hich x is the abscissa of the point of tangency. 24. Two focal chords are drawn in an ellipse at right angles to each other: find their position with reference to the major axis when their sum is a maximum. 25. Two vessels are sailing at right angles to each other, with velocities a and b per hour. Show that if at any given instant their distances from the point of intersection of their courses arie p and q, respectively, then their least distance from each other is aq - bp CHAPTER IX. DIFFERENTIATION OF FUNCTIONS OF TWO OR MORE INDEPENDENT VARIABLES. 71. It has been shown in Art. 34 that if we have t F(x,y), in which equation x and y may be either mutually dependent or entirely independent of each other, then we shall also have du du dzu — dx -x dy + a (1). In this equation Dx and Jy are the arbitrary but infinitesimal increments of x and y; a is an infinitesimal with du dii reference to Ax and Jy; dx; dy,, are the partial derivatives of u with respect to x and y; and Au is the total increment produced in iu by the increments assigned to x and y. 110 THE DIFFERENTIAL CALCULUS. Now we have seen [Art. 15, et seq.] that whenever infinitesimals occur in the terms of a series or ratio whose limit is to be taken, they may be replaced by others which differ from them by infinitesimals of higher order; and we have also seen [Art. 32] that the differentials of variables may be considered as differing from their infinitesimal differences by infinitesimals. Hence, if we take the limits of the quantities in equation (1), we may write dUdx + du dy for the second member, and du for the first member, thus giving due dx -+ t dy (2) Now as Ju is the total increment produced in n by a change in x and y, so we may call du the total differential of it, while the terms du, du dy, are the partial differentials of u with respect to x and y. We have, therefore, the following Rule.-The total differential of an explicit function of two (or more) variables is equal to the sum of its partial differentials. 72. Problem.-To find the successive differentials of u- = F(x,y). We have adu = dx + dy dy (1). du du Differentiating this equation, and observing that du du d' dyf' are themselves functions of both x and y, we have FUNCTIONS OF TWO OR MORE VARIABLES. 111 d {Ud dx d{ dUdx { du }dy d u- dx+ dy+ dx dx dy dx dd d {u dy + dy dy. d2u d2du d2u d2u Or, -dxp= d+ dx dy+ d d dy2 Or, d2 d 2 d dd dx dy dydx - ddy2 (2). This expression may be simplified by observing that d2u d2u dxdy - dydx for, designating by Dx the increment which a function receives in virtue of the increment Jx to x, we shall have jdu u J+ xz} du ud{u} Dividing by Jx, which may be regarded as constant, X dy dy u} Ax zx dy and passing to the limit, d du} dr du (dy{ } {Tx} or d2u d2u = or, or- - dx dy dxdy dydx Hence, equation (2) may be written d2u d2u (12 dy 2. d2u - dX2 - 2 dxdy dx dy + dy2 dx2 dy2 c~l 112 THE DIFFERENTIAL CALCULUS. In the same manner the higher differentials may be obtained. NOTE.- In the successive differentiation of functions of several independent variables, the following notation is used: d"+t m p+ etc. UL dxe' dy- dze etc. indicating that n + m + p + etc. differentiations have been performed, n with respect to x, m with respect to y, etc., and it is evident from the preceding demonstration that it is immaterial in what order the operation is performed, provided the entire series of differentiations is effected. 73. Differentiating equation (2) we shall find d d3uu J~U dd dy 3. d3u d + d 3 d3 dx2 dy + 3 x +d dyd_ dd4u d4 d4 d d d4u == dx4 + 4 dx3 dy + 6 dX dye + 4 dxu dyd dx + %dy4. It will be observed that the co-efficients and indices in the different terms of these differentials are the same as those in the corresponding powers of a binomial, and this law will hold good in all cases. Hence we may write the following symbolic formula: in the development of which by the binomial theorem the in the development of which by the binomial theorem the exponents are to be placed over the d in du, and over the x and y in dx and dy. FUNCTIONS OF TWO OR MORE VARIABLES. 113 This formula has no significance when taken in its literal sense, but it is extremely valuable as an aid to the memory. 74. Problem.-To differentiate u = F(v,z), in which v and z are functions of both x and y. Since u is a function of x and y, we htve du du du -dx + d dy. dx dy But since v and z are functions of x and y, we have du du dv du dz dx- dx + dx; dx dv ddx dz dx dud du dv du dz ~-dy = - dy dy; dy Y dvdy dz dy dv dv dz dz dy. v-x y dz dy Hence, by substitution and reduction, du du d dv + dZ dz. A second differentiation will give dd d2u d2u dz+ du d u d d2d d -- dv2 + 2 dv ddz +- dZ2+ d+ d2v + d2. The general solution of this problem will be as follows: Differentiate as though v and z were the only variables, and substitute for dv, dz, d2v, d2z, etc., their values taken from the equations which connect v and z with x and y. 75. Implicit functions. -Before showing how the differentiation of implicit functions of two or more variables is effected, we wish to establish the formula for the second derivative of an implicit function of a single variable. D. C.-10 114 THE DIFFERENTIAL CALCUL US. Let u =-F(x,y) = 0 be such a function. Then, as we have already found [Art. 34], du dy du (1). dy dx dx Designating this expression by v, and differentiating, we have dvdy dv y +3;=0 (2). t dv d2t dy d2u dv du dy du d2y +d2u dy dy d + dxdy' dx dxdyr dy d dx2 by substitution in (2), d2u 2 d2u dy du2tldy\2 dud2y 0 (3). dX2 dxdydx dy?\dx +j dy2 Placing in (3) the value of dy derived from (1), we have, after transposition and reduction, d2u du d2u du du d2u du \2 \ dy d2y dy )-2 dxdy dxdy+$ dxf\T This equation gives the value of d2Y in terms of known quantities, but it is so complicated that in practice it is generally better to follow the method than to apply the formula itself. The third and higher derivatives may be obtained in a similar manner, but their forms are quite. unmanageable. 75'. Problem. —To differentiate u - f(x, y, z) = O, in which z is an implicit function of x and y. IMPLICIT FUNCTIONS. 115 For the first derivative we shall have at once du d du a dz + 0 (2); du dz du + =o (2); du du du ddx + dy + dz=O (3) Equations (1) and (2) will give the partial derivatives of z with respect to x and y, while (3) will give the value of the total differential of z. For the second derivatives we shall find from (1) and (2), as in the last article, d2u d+ 2 dz dz u Idz \2 du d2z d -2dx dzJ- d- dx' J d d dz dx2 (4) d2u+ d2u dz d2u(dz du d2z (5);~'+2dyd —- 0 (5). dy dyd d zdzdy dy dz dy2 Differentiating (3), d2u d2 2 d2u d2 4 d2u dx2d - dy +- dz +-2 + dxdy 2 d dxdz dx2 dy- dz dxdy dx dz +2 d2UdYdz+duz O (6); and, differentiating (1) with respect to (y), d2u dz d d2u dz d2u dz dz du d2z dxdyyx + uc (7). dxdy dx dz dy dzz dy dx L y d+ zxdy From equations (4), (5), and (7) we can obtain the values of d2z d 2 and d 2 in terms of known quantities, while dx2 dyv' dx dy from (6) we find the value of d2z. 116 THE DIFFERENTIAL CALCUL US. The expressions for the higher derivatives and differentials are too complicated to be presented here; and, fortunately, they are seldom required in practice. Corollary. —Let u = F(x, y, z) be a homogeneous function of the nt1' degree; that is, such a function that if we multiply each variable by t, the result will be the same as if we had multiplied the function by tV. Then F(tx, ty, tz) - tn F(x, y, z). Differentiating with respect to t, we have dF dtx dF dty dF dtz - nt_ F(x, y, z). dtx dt -dty dt +dtz dt Observing that dtx dty dtz dt = X' dt Ye dt-'Zo and making t- 1, we have xdF dF + dF nF(x, y, z); whence it follows that if we multiply the partial derivatives of a homogeneous function of the nth degree by their respective variables, the sum of the products so obtained will be n times the function. Scholium. — We have confined our investigations in this chapter to functions of two variables, but the student may readily see that the method will be the same for any number of variables whatever; and the fact is that by far the most numerous and important class of problems to which the Calculus is applied involve functions of only one or two independent variables. EXAMPLES. 117 76. EXAMPLES. 1. ut X3y2+ Y3 X2. du d_ du 2 3x2 + 2xy3; du 3y2 x2+ 2x3 y; dx dy du = 3x2 y2 dx + 2xy3 dx + 3y2 x2 dy + 23 ydy. d2u d2u, d- 6xy2 + 2; d 2 y + 2x3; d2u d2u dx dy dy dx d2u = 6xy2 dx2 + 6x2 ydxdy + 2y3 dx2 + 6xy2 dxdy + 6x2 ydxdy + 2x3 dyZ + 6xy2 dxdy + 6x2 ydy2 = (6xy2 + 2y7) dx2 + 12 (x2 y + xy2)dxdy + (6x2 y + 23) dy2. 2. Zu==v2 z; v= /x2+ y2; z x2y2. We have du z dzu V2; dv xdx - ydy dz = 2x2 ydy + 2y2 xdx. du du.'. du —dvd+ z dz 2vz { xdx + } + 2v2 (x2ydy + y2xdx) =? What is the value of d2u? 3. u - sin v cos z; v =2x+ 3y, z -4x 5y. 118 THE DIFFERENTIAL CALCUL US. We have du du dv - cos v coS Z; - in v sin z; dv=2dx + 3dy; dz=4dx+ 5dy. du ="cos v cos z (2dx +- 3dy) — sin v sin z (4dx + 5dy) = (2 cos v cos z-4 sin v sin z) dx + (3 cos v cos z -5 sin v sin z) dy. Find the value of d2u. 4. u -4 +- 2a2 y - ay3- 0. In this example y is an implicit function of x. We have u 2ax2- 3ay2; du 4x3+ 4axy; dy dx d2u d2u d2u d y-i 6ay; dx 12x2 + 4ay; ddy4ax. dy 43 + 4axy. * dx 2ax2- 3ay2' d2y tyL (12x2+ 4ay)(2ax2- 3ay2)2 dY -_ 8ax (4x3+ 4axy) (2ax2- 3ay2) — (2ax2 — 3ay2)' q+ (4x3 + 4axy) (- 6ay) -? 5. u = X3 3axy + y3- O. Find dy d2y d2y 2a3xy Find and -?'Ans. y dx dX2 d 2 (y - 2(a? — Yx)3 EXAMPLES. 119 6. u = ax3+ x3y - ay' 0. dy d2y Find dy and d-2 and show that dy 1 when x=O, y 0. dx dx 2 1 dx 7. u - y3 + 3x + 2xy - z2 - 0, in which z is an implicit function of x and y. We have du du du dtz-6x + 2y; dy = 3y2 + 2x; ~- =- 2z; d2u du2 d2u d_ -_ - 6; 6y; dz2 —2 d'u d2t d'u -d 2; -0; 0. dxdy dxdz 0 dydz - These values substituted in the formulas for implicit functions give dz 3x-+ y dz 3y2 + 2x dx z' dy 2z (6x + 2y)dx + (3y2 + 2x)dy - 2zdz 0, from which the value of dz may be found. Also, 6-2 3 2z d 6y2{3.~ 2x z - 0; 6ydy2 -- 6dx2 + 4dxdy - 2d2 - 2zd2z -0; from which equations d2z d2 and d2z, can be found. d~,~," 120 THE DIFFERENTIAL CALCUL US. X2 y2 22 8. Uz-j -y+2 —-1=0o. Find d dz d2z d, and d2z. 9. u= ax2 +- by2 +- c2 + 2exy + 2fxz-t- +2hyz. We hatve du du 2(ax + ey +fz); du du -- 2(by +- ex + Ihz); du -z = 2(cz - fx + hy). *'. by substitution in the formula for homogeneous functions, xdu + du + zdu = 2x(ax + ey +fz) + 2y(by + ex + hz) + 2z(ez +fx + hy) = 2(ax2+ by2+ cz2 + 2exy + 2fxz + 2hyz) - 2u. DEVELOPMENT OF FUNCTIONS. 121 CHAPTER X. DEVELOPMENT OF FUNCTIONS OF TWO INDEPENDENT VARIABLES. 77. Let u = F(x,y), and let it be proposed to develop Uzl F(x + h, y + k) into a series. The required development may evidently be obtained by expanding F(x + ht, y + kt) with reference to t as a new variable, and then making t equal to unity in the result. Differentiating this expression with reference to t, we have dF dFh dF h+ +-k dt cdx dy d2F d2Fh d22dF hkF+ dFk2, dt2 dx'm dx-dy dy2 d-F d F hn+ n dn-dy h"k +... d+ F -dt - dx dxe,'dy If in these expressions we make t = O, they will become the partial derivatives of F(x,y), and by substitution in Maclaurin's formula, we have D. C.-11. 122 THE DIFFERENTIAL CALCUL US. F(x + ht, y + kt) =( F(x, ) + h + dy ) ( d21t d- F d 22 hk+ d 1t2 Y\ dx2 dxdyhkdy y 12 + 1 2..n d h+ + dy"F' in the last term of which x + oht, y + Okt, must be substituted for x and y, in accordance with the limitations of Maclaurin's formula. Now, making t 1 in this equation, we have F(x + h, y + k) ~)) ~dF dF d2F h2 d2Fh - F(x,y) )- dd-h+-d k+ ddF 112 - dd hk \ Y dx dy d~ k 1.2~ dx dy d2F k2 + dyf 1.2 + dn-1F hn-I d1lF h"-2k + dxe" — 1. 2.. (n-1) + dx"-2dy 1. 2.. (n-2) + dy-' 1. 2... (n —1) + {d-"-hn+ -''.'. - -. 1.2.. n (1), x and y being replaced in the last term by x + Oh, y + Ok. The application of this formula is restricted to those cases in which the first n derivatives are continuous between x and x + h, y and y- +k. If the terms contained in the } tend toward zero as n increases, then formula (1) will give the exact development of F(x + h, y + k), and may be considered as the TA YLOR'S FORMULA EXTENDED. 123 extension of Taylor's formula to functions of two independent variables. 78. If in (1) we make x 0, y 0, and replace h and k by x and y, we shall have dF x d2F x2 F(x,y) - F(0,0) + dx 1- dx 1.2 dF y d2F y2 dy 1 dy2 1.2+ + ddy +.... (2) d F $. - dxn 1.2..n+ d:F "y t dye 1.2..n' x and y in the } being replaced by Ox, Oy. If the terms in the J tend toward zero as n increases, then formula (2) will give the exact development of F(x,y), and may be considered as the extension of Maclaurin's formula to functions of two independent variables. It is to be understood that in this formula the values of the various derivatives are to be found under the supposition that x = 0 and y = 0 after differentiating. 79. EXAMPLE. Develop u = e sin y. We have du sin d'u du etc. dx - e sn y dx; etc. 124 I'HE DIFFERENTIAL CALCULUS. du d2u d e" cos Y; sin y; y dy dy2 d~u d4u d -- e= cos y; d e4 sin y, etc. making x = 0 and y = 0, dF dF (u) - F(0,0) -- 0; dF 0, - I, etc., etc. *. by substitution in (2), x2Y Y etc. u- e"sin y — y-xy-_1.2 1. 2.3' 80. Problem. - Given u — F(y), and y -z + xf(y) in which x and z are independent variables; to develop u in terms of x. Since u is a function of x, we have, by Maclaurin's formula, du\ x d2(u \ X2 u - F(y) -= (u) +( d 1 -- 2 1- + et.; and it remains to determine (u),(du ) etc. 1st. Since y -z + xf(y), we have y z when x =0; hence, (u) (F(y)) F(z). 2d. Designating f (y) by Y, we have y z + x Y. dy dY dy dy dY dy dx dy dx dz dy dy Eliminating x from these two equations, we have dy - dy d —x dz LAGRANGE'S THEOREM. 125 and consequently, dF(y) dy dF(y) - y dF(y) dy yd Fy dy dx dx dy dz dz a relation which is independent of the character of F(y). If in this equation we make F(y)- Yn, it will become dYn dYn — ydx dz Now we have d y dF(y) } dF(y) d Y + y, dF(y) y dF(y) d Y + yn dF(y) dz dx dz dx dz dz dz dx dF(y) d Y- + Yn d2F(y)_ d yn dF(y) dx dz dz dx dz dx d yn+idF(y)} dz + dz. making n equal to 1, 2, 3, etc., in succession, d f ydF(y) d f y2dF(y) d 2dF(y) ) dx V dz J z d z dx d z f d2 { y dF(y) };etc. dz{ dz If now we differentiate successively the equation dF(y) y dF(y) dx dz we shall have dF(y) d y dF(y) - d J y2dF(y.} dx' -x dz dz; 126 THE DIFFERENTIAL CALCULUS. d3F(y) d2 y2dF(y) d2 f y3dF(y) d"F(y) d"-l yn dFy) t dxn dz"-L dz' Finally, making x 0, whence (y) = z, and f(y) or Y=f(z), and substituting the above expressions in the development of u or F(y) by Maclaurin's formula, we have =F (y) = F(z) +x { (z) dz } x2 d (f(Z,\ 2dF(z)l + 1.2 dz (f(z) dz xe d2 3 dF'z) 1.2. 3 () d f ) dz} + etc., which formula is known as Lagrange's Theorem. dFz Corollary. — If F(y) = y, then F(z) = z, and dz 1; whence z + (f( )) + x2 d(f(Z))2 + d2(f (z)) + etc., y -z +xffz) ) +..2 dz + 1.2. 3 dz2 + etc., a formula for the development of y =- z + xf(y). These two formulas are of frequent application in the sciences of Mechanics and Physical Astronomy. 81. EXAMPLES. b 1 1. Given y —ay b -O, or y- b + a y3 to expand y in terms of a LAGRANGE'S THEOREM. 127 We have 1 b x a-; z- -; f(y)-y3; f(z)=z3; d (f (z)) d(z6)- 6i 6(b; d' (f(z)) ab etc. dz -dz 6z5 — 6 a dz2;etc Hence, by substitution in Lagrange's theorem (Cor.), Y ()+(-3 - )( 12 2etc. We may in a similar manner expand the unknown quantity in any algebraic equation, and thus approximate to the roots of the equation. 2. y = b + cay. Expand y in terms of c. Here z b; x -c; f(y) =aY; f(z) ab. d(f(z))2 a_,f(z3 2 dz =fz 2 log a. a2b; d(f(z = 32 log2 a. a3; etc.. yb+ca=-b + cab+ 2 log a. 1- a 32 log2aC a3b; etc. +-32logsa. i 23 Corollary 1. —If a-e and b=- 1, then y2 33 e tc y =+ ce =1+ce + 2e2' 3 2 e 3 etc. 1. 2~31.2.3ec 128 THE DIFFERENTIAL CALCUL US. Corollary 2. - If c 1, whence y 1 + ey, or y-= log (y - 1), then e2 es y = log(y — 1)-= 1 +e+- 2. 1-e2 + 3. etc. 1.2.23 3. y=a + x log y. Expand y in terms of x. We have z=a; x=x; f(y) = log(y); f(z) log(a); d(f(z))2 2 log(a). d2(f(z))3 3 loga(2 loga); etc dz a dz2 -- a 2 log a x2 y-a - x log y a x log a -l a 1.2 + 3 (2-log a) 1 3 etc. a2 1. - etc. 4. Develop yz - +e sinl y. We have - e; f(y) - sin(y); f(z) = sin z; (f(z))2= sin2 z; etc. ~ d(f(z)) 2 2 sin z cos z —sin 2z; dz d2(f(Z))3 6 sin z cos2 z -3 sin3 z dz2 -3 sin zf2 cos2 z - sin2 zj= 3 sin z Icos2 z + cos 2z} 9. 3 - -si 3z — sin z; etc. 4 4. y- =z+ e sin y=z+ e sin z — 1-2 sin2z 9 si 3z-sin etc. 1. 2. 3 4 4 t L~Cj MAXIMA AND MINIMA. 129 5. Givenu = sin y; y -- z+e sin y. Develop u. We have F(y)=siny; F(z) — sinz; f(z) =sinz; xe. dF(z) 1. f(z) dz sin z cos z =- sin 2z; d f 2dF(z) d (sin2 co z) = sin 3 dz (f(z)s -d- 4 sin 3z- -4 sin z; by substitution in Lagrange's theorem, u =sin y -= sinz + -e sin 2z + 2 {4 sin 3z- 4 Sin z + etc. CHAPTER XI. MAXIMA AND MINIMA OF FUNCTIONS OF TWO INDEPENDENT VARIABLES. 82. If F(x, y) be a function of two independent variables, such that for the particular values x, yo of x and y, F(xo, yo) is greater than F(x, + h, yo + k), then F(x,, Yo) is said to be a maximum of F(x, y); and if F(x,, yo) be less than F(xo + h, Yo + k), F(xo, yo) is said to be a minimum of F(x, y). 130 THE DIFFERENTIAL CALCUL US. In order to determine the tests for these two cases let us designate h by adx; k by ady; F(x + h, y -+k) by f(a); and F(x, y) by f(O). Then F(x + adx, y + ady) - F(x, y) _=f(a) -f(O). Now it is evident that in order that F(x, y) may be a maximum, we must have F(x + adx, y + ady) — F(x, y) < 0; and that F(x, y) may be a minimum, F(x + adx, y + ady) - F(x, y) > O. That is, for a maximum, f(a) -f(O) < 0; and for a minimum, f(a) -f(O) > 0. These conditions require that for a maximum of F(x,y), f(O) must be a maximum of f(a); and for a minimum of F(x, y), f(O) must be a minimum of f(a). Now, in order that f(O) may be either a maximum or minimum, we must have, as in the case of functions of one variable, f'(O) - 0 or o; and, supposing f(a) and its derivatives to be continuous, the first of its derivatives which does not reduce to zero must be of an even order, negative for a maximum, and positive for a minimum. If we have f(a) - F(x + adx, y + ady), then, by differentiating with respect to a, we have dF dF f'(a) d(x + adx) d(y + ady) dy MAXIMA AND MINIMA. 131 Making a - O, this reduces to dF dF f'(O) = dx dy l du du dx dx + dy, which is the value of du. Observing, then, that if u = F(x, y) we shall have f(O) = u; f'(0) = du, etc., we obtain the following Rule.-Form the successive differentials of u. Then, 1st. For either a maximum or minimum, du du du du du —dx +d- dy =0; whence - 0, d 0; du du _ or dx + dyoo. d"' u d" u d"u 2d. dn u= d" + n dx-dy +.. d dy" dx" dx'-1 dy du" must be of an even degree and negative for a maximum, or positive for a minimum. On account of the complicated forms of the higher differentials the application of this method is almost entirely restricted to the first and second differentials. Let us suppose, then, that the second differential does not reduce to zero for those values of x and y which render du either zero or infinity. 132 THE DIFFERENTIAL CALCULUS. Then we shall have, for a maximum, d2u d2u du dy2 0; du — dx 2 dx dy 2 < 0; d2 d xdy dy 2 and for a minimum, d2ue d2u d2u dh&u dX2 - 2 -y dxdy q d.2 > 0. ddX 2 dxdy dx dydy2>0. Now, in order that this expression may have constantly the same sign, whatever may be the value of its middle term, we must have, in accordance with the theory of quadratics in Algebra, d2u d\2 dt2u dxcdy 0, 0; d2u d2u d2u d2u dx2 > ~ dy2 dxdy dx dy We may remark that the sign of d2u will depend on the signs of d2 d2u not only when( ddy d2u d2' but signs of \Idxdy <&2 dy' also when the relation between the two terms of this expression is that of equality. The consideration of this case would involve an additional test for the complete determination of maxima and minima, but the resulting expression is rather complicated, and the case seldom occurs in practice. 83. EXAMPLES. 1. Find the values of x and y which render u = x3 + y3 3axy a maximum or minimum. We have du XX d - 32 — 3ay -- O,. y a du 3y2- 3ax O,.'. x- - dy a.x O or a, and y O or a. d2u d2u d2u Also, 6x -; d -; -;dxdy 3a; c~~x ~ d1d 2 dxdy 134 THE DIFFERENTIAL CALCUL US. dxdy d dy2 - -+ 9a2 forx, y 0, = - 27a' for x _ a, y = a. x = a, y = a, render the value of u a minimum. 2. u- -x2 + xy + y2+' + X I We have du a3 du a3 dx 2x +y- = 0; -=2y +x — - 0. From these two equations we derive x = y, and therefore a d2_ 2a3 - d2u + 2a3 Also, d-u 2 2a-; d -; 2a dXl x -- x d y d y when x -y a we have d d2u d2u 2 d2u d2u d2 d2 dx2 +' dy2 +' \dxdyj n. x ASYMPTOTES. 167 Representing xY by s and dividing by x", we have F(s) + -I mf(s) +. =.; whence, making x- =o, F(s) = 0, or F(Y ) and the value of a is evidently to be found among the real roots of this equation. To find the value of b corresponding to any real value of a, put y a — ax + t, whence Y - a + t, and the equation x x of the curve becomes xmF( a + )+ -Xf+ a.... +0. But since F(a) F(Y ) 0, we have [Art. 46] \ + x x ); whence -l'tF' a - x - x"f a. -- 0, or tF'( a+ x X) + m f(+ xt 0. If, now, in this equation we suppose x =o, which renders t = b, and if F'(a), f(a), have finite values, then we shall have 1st, for n < m-1, lim(t)= b =0; 2d, for n m — 1, lim(t) b - f(a). w-lt/a 168 THE DIFFERENTIAL CALCUL US. 3d, for n> m -1, lim(t) - b oc, and there is no asymptote. in the first case the equation of the asymptote is y - ax; in the second case the equation of the asymptote is f(a) y= axIn the first case, in which m < rn - 1, the equations y ax, F(a) 0, give F xi) = 0, and consequently xmF( Y - O, which is also the equation of asymptotes: so that whenever it < n - 1, or nm > n + 1, the function of the degree m placed equal to zero gives us the equation of the asymptotes. EXAMPLES. 1. The curve Ay2 + Bxy + Cx2 + Dy + Ex + F + 0. We have x2 A{ +B-+C} +x {DY +E}+F0. X2 ~'. ( )-Y F(a) Aa2+ Ba + C= O, and -B + 1/B2- 4AC a 2A f (x -)-(a)Da -f(a) - DaE; E x F'(a) 2Aa - B; EXAMPLES. 169 and the equation of the asymptote is Da + E y — ax- 2Aa + B The values of a are imaginary if B2- 4 AC< 0, and the value of b becomes infinite if B2 - 4AC - 0. The asymptote is real only wheh B2 - 4AC > 0. This example is the solution of the problem of asymptotes as applied to the conic sections. 2. The folium of Descartes, y'3 x3 - 3exy = 0. We have X{ +t} $x{ 3c -O=0. Hence, F(a) a' + 1 = 0, and a — 1 f(a) _ 3ca - c F' (a) 32 - a the equation of the asymptote is y - x - c. 3. The curve y3_ X -cx2 O0. We have x3 { - 1 } - X, _ =0. Hence, F(a) = a'- 1 = 0, and a = 1; fa c c F' (a) - 3 a -'3.'. the equation of the asymptote is y-x+ 3. D. C.-15. 170 THE DIFFERENTIAL CALCULUS. 4. The curve y'(Ay + Bx) = A2y2 + B2x2. Placing this under the form 3A.Y + B Y2 - x2 {A2I B2}=0, we find for the equation of the asymptote B Y - AB x+2A. 5. The curve y4- x4 + 2cx2y = O. Placing this under the form z4{-4 -1 }+ {2c} Y 0, we have for the equation of the asymptote y-rx —2~ 6. The curve Ay3- Bx3 + Oxy = —- 0. The equation of the asymptote is 3 B C Y -- x \lA - 3eA'B 103. If the equation of a curve be given in te polar coordinates, it is evident that the curve will rectilinear asymptote whenever for an infinite value radius-vector the perpendicular on the tangent is fin In order, then, to determine whether there is an tote to a polar curve, we must find from the equa the curve all those values of 0 which render r infini by substitution in the expression for the perpendicul whether the length of this line is finite. CIRCULAR ASYMPTOTES. 171 EXAMPLES. 1. The hyperbolic spiral, rO - a. We have dr a r -o when 0 O; also dO 90 r2 r4292 a2. perp - pr dr) Vir2 o4 -a2 a a when 0 — 0. Hence, this curve has an asymptote parallel to the axis, and at a distance a from it. 2. The lituus r2 a2. We have dr a r -o when 0- 0; also dr do 20 I/f r2 2r2o0 2a2 perp O a 1/40(r20) + r2 r when r — o=. Hence, the axis is an asymptote to this curve. 104. Circular Asymptotes. -When the polar equation of a curve is of such a form that when 9o- =o the value of r is finite and equal to a, the curve will make an infinite number of convolutions about the pole before reaching the circumference of a circle whose center is the pole, and whose radius is a. This circle is then an asymptote to the curve; it will be exterior to the curve if every finite value of r is less than a, and if r > a the curve will be exterior to the circle. 172 THE DIFFERENTIAL CALCULUS. As an example, let us take the curve aO2 r 02 1 This may be put under the form a 1 and if 0o - oo we have r - a. Every finite value of 0 will render r > a, and therefore the curve is exterior to the circle. We shall also find two rectilinear asymptotes to this curve. CHAPTER XV. DIFFERENTIALS OF THE ARC, AREA, AND INCLINATION OF A PLANE CURVE. CONVEXITY AND CONCAVITY. 105. Before discussing the subjects of this chapter it is necessary to demonstrate that the limit to the ratio of an are and its chord is unity. Fig. 9 Let y _F(x) be the equation d of the curve abc, and let x, y, be the coordinates of the point b. Take another point, c, so near b the former that the chord be, and all chords drawn from b to points between b and c, shall lie on the same side of their corresponding arcs. Then we may assume that the arc be is contained between the chord be and the tangent bd, and the proposition ARCS OF PLANE CURVES. 173 above stated will be proved if we can show that the limit to the ratio of the chord and tangent is unity. Now, we have bd2- be2 de2 - ce2; and the limit to the ratio of de and ce being evidently unity, the limit to their difference is [Art. 14] zero. lim (bd2- bc2) 0; lim( b) 1; li b — 1, and lim bd 1. The limit to the ratio of the chord and tangent being unity, and the are being always between these two, it follows that the limit to the ratio of the chord and arc is unity, and, therefore, the chord may be substituted for the are in any ratio or series whose limit is to be found. 106. Problem.-To find an expression for the differential of an arc of a plane curve. Designating the chord be (Fig. 9) by ich, be by Jx, ce by Jy, we have Zch -- /(x)2 + (ay)2; whence - L d Y) Passing to the limits, and observing that the limit to Jch - A are ch is the same as that to -ar, we have ax ax dx X dx', or d arc -- 1/dx + d/2 (1). 174 THE DIFFERENTIAL CALCUL US. Corollary. —If the curve be referred to polar coordinates, we have x= r cos 0, y r sin 0. dx - cos Odr -r sin OdO; dy = sin Odr + r cos ode. Substituting these values in (1), and reducing, we have d arc = /r2do +- dr2' (2). 107. Problem.-To find an expressionl for the differential of the area of a plane curve, referred to rectangular axes. Let abg be an area included Fig. 10 between the arc ab, the ordinate Y bg, and the abscissa ag. It is required to find the limit* to the area bgfc corresponding to the increment gf of the abscissa. Completing the rectangles bf and cg, the area bgfe is comprised be- 0 x tween these two rectangles, and a the limit to their ratio is evidently equal to the limit to the ratio of either of them to the area bgfc. Now, since these two rectangles have the same base, the limit to their ratio is equal to the limit to the ratio of their altitudes, and this limit is evidently unity. Hence, the limit to the ratio of either of the rectangles to bgfe is unity. Now, gfeb = yx.!.In this, and in several succeeding propositions, we mean by the term limit the quantity which can replace a variable in a ratio or series whose limit we wish to find. As the word will also be used frequently in its ordinary sense, the student must carefully distinguish between the two meanings, when they occur in the same demonstration. AREAS OF PLANE CUR VES. 175 b bgfc _J area nd rn fc I area - and limg.b-li m yfeb- yJx feb yJx d area ydx d area - ydx (3). 108. Problem. -To find an expression for the differential of an area when referred to polar coordinates. The problem to be solved is to find the limit to the area Oab in- rig. cluded between the two radii Oa, Ob, and the arc ab. Let 0 be the origin, OX the fixed axis, let x, y, be the rectangular co6rdinates of a, and x- + x, y + y, those of b., Then we shall have Oab - D area - dcba + aOd - Obe: a area deba aOd Obe lim im - + lim d lJ ADx dJ D x' dcba yD x aOd xy xyL Now, lim y li m lim - - lim Obe lim 2 { (x+ Jx)(y+,Jy) 1 xy xdy 1 Ax 2 Jx -2 dx+ d y' lim area d area lir Ax dx 1 xy 1 xy Mcy =Y+ 2 dx 2 dx 2dx 2 y -1 y- xdY}; and -2 { - 3~ 1'76 THE DIFFERENTIAL CALCUL US. d area - (ydx -xdy). Finally, substituting for x and y their values, r cos 0, r sin 0, and differentiating, we have d area = 1 r2do (4). 109. Problem. - To find an expression for the differential of the angle which a tangent line to a curve makes with the axis of abscissas. Designating this angle by q, we have dy tang q - =x Hence, by differentiation, sec2 cpd — d2 dx; d2 dy2Y,X dx dx_ dxY dx * d sec2 c - 1 tang2 +( (5d This value of dq is called the angle of contact of the curve and tangent. It may be considered as the limit of Ap, which is the angle made by two lines which are tangent to the curve at points whose abscissas differ from each other by J x; and in all cases where t is the independent variable, di, may be taken as equal to At. 110. Problem.-To find the angle of contact when the curve is referred to polar coordinates. ANGLE OF CONTACT. 177 This angle is the limit to the angle fcg or ecd made by two tangents fe and gd. Fig. 12 Now, we have ecd = cdx - cex;; cdx - Obd + bOx; cex =Oae + aOx. c. dx -ex Obd -Oae + bOx- aOxObd- Oae -+ bOa. Again, Obd - Oae = J Oae, and bOa = JO..'. ecd- =ztJp JOae + JO,.J. Oae. do lim — lim J + lim, or dp dOae do do + -, and d- dOdae, + do (a). ddo Now, Oae = [Art. 95], and tang do-r dr 2 d2r * sec~2 tdt, - r \\ do; and (dr ddr -- d2r idar e\ d2r a d I +,do - i d o o- d o. dr 2 ^do J do do=d -' do- d- d (~L)2{1+r2(+)d} 178 THE DIFFERENTIAL CALCUL US. Substituting this value of dOae in (a), we have dr2 r d2rd d= (d do (6). 111. Concavity and Convexity. —A curve is said to be concave to a given straight line, at any of its points, when the two parts of the curve immediately adjacent to such point are contained between the straight line and the tangent at that point. It is convex to the straight line when the tangent lies between the curve and the straight line. Let cabk be a curve concave to the axis of x, and let gad, hbe, be two tangents drawn at Fig. 13 gh points a and b, infinitesimally near k each other. It is evident that the angle hex is less then gdx, and that, in passing from a to b, q is a decreasi ng function. If, then, x be the abscissa of a the sign of d dx must be wegative [Art. 18], and, therefore, d2y d, dx2 -- -a negative quantity. The denominator of this fraction being essentially positive, the sign of the fraction will be the same as that of its, numerator. Hence, for a point where the curve is concave to the axis of x d y must be negative; and in the same way we may show dx2 that for a point where the curve is convex to the axis of x, dxy dt be positive. must be positive. CONCA VITY AND CONVEXITY. 179 It is obvious that these conditions will be reversed when the point under consideration is below the axis of x; and combining the two cases, we see that the general condition for concavity is that y and dy shall have opposite signs, dx2 and for convexity the condition is that they shall have the same sign. If the curve be referred to polar coordinates, it is easy to see that if a perpendicular p be drawn from the pole to the tangent, then, in any portion of the curve which is concave to the axis, p will increase as r increases; while for a convex portion, p will decrease as r increases. Hence, for concavity, dp must be a positive quantity; and for convexity, d must be a negative quantity. The analytical expression for dp may be readily deduced from the equation already given in Art. 95; I2+ (dr 2 We have dr 2 r+ d02 dp 2r- + ( )-r d ) _do ( dr dr 2 +( d (o - r'~ + 2r r2 r do{tr r+(+ do 180 THE DIFFERENTIAL CALCULUS. It follows from this result, that if r and r+ (dr be positive, the sign of dp will be the same as that of d o and dr do this will evidently depend upon the sign of the numerator of dp in formula (6) of this chapter. In applying this formula, it is necessary to consider 0 as always positive. CHAPTER XVI. CURVATURE AND CONTACT OF CURVES; EVOLUTES AND INVOLUTES. 112. Curvature.-The curvature of any are of a plane curve is the external angle contained between the tangents drawn through its extremities. If in every curve this angle varied directly as the length of the arc, as is the case in the circle, we might obtain the curvature of a unit of arc by dividing that of the whole arc by the length of the arc. But in every case except the circle, because of the variation of curvature, the quotient so obtained will be only the mean or average curvature of a unit. If, now, beginning at any point of a curve, we take an arc of arbitrary but infinitesimal length, its mean curvature will evidently vary as the arc tends toward zero, and it will tend toward a certain limit, which is in general determinable, and is called the curvature of the are at the given point. Let Js and ZJ be the arc and the angle between the tangents drawn through its extremities. Then Jd will be the mean curvature of the unit of the arc, and Jp dq lim ds is ds CUR VATURE. 181 will be the expression for the curvature at the point at which the are begins. 113. The curvature of a circle being uniform, we shall have for the curvature of any point, lim -- J _ as deS' or, since ds = RJ, R being the radius, a_ -Rt RI - curvature of a circle at any point. is - Riq' A Now, whatever be the curva- Fig. Ia ture at any point of a plane Y curve, there is obviously a circle a which has the same curvature, and this circle can be placed tan- / \ gent to the curve at that point, with its radius coinciding in di- o _\ x rection with the normal to the curve at that point. This circle is called the circle of curvature of that point of the curve; its center is the center of curvature, and its radius is the radius of curvature of the given point. 114. Problem. -To find an expression for the radius of curvature of any point of a plane curve. We have Curvature of curve = curvatiure of circle, or dep 1 ds ds -R; whence R-. Substituting in this the-values of ds and dqc, viz: 182 THE DIFFEREN'IAL CALCUL US. d'Y dx dx~ ds = 1/dx2 + dy2, and dc = dx it becomes 1+( dy \2 3 R dx (7); d2y and substituting the values of ds and dq in terms of polar coordinates, viz: x r cos 0, y = r sin 0, we have (.1+ dr -3 j= +dr 2 ( dd2ri (8). 115. Problem. — To determine the radius of curvature in terms of a new variable, t. Applying the formulas for changing the independent variable [Art. 85], we have, at once, R= — t )+( dt or dx d2y dy d2x dt dt& dt dt d- d2y - dyd2x (9). Corollary 1. —If the new variable be s, we have dX2+ dy2 dx d2x + y d y 0. ds' -ds- 6ds ds' ds ds — CONTACT OF CUR VES. 183 d2x dy d2y ds ds2 -- ds ds2 d' and substituting in the above value of R, we have dx 1 _~Y_ R-d d2y (10). ds' ds2 Corollary 2.-If the curve be referred to polar coordinates, and we make p the independent variable, we have at once, by comparing formula (8) with the value of dp in Art. 111, R dr(d dp 116. Contact of Curves. —Let Y F(x), y —f(x) be the equations of two curves referred to the same axes. Giving to x an infinitesimal increment, h, and expanding by Taylor's formula, we have Y —=Y + (qh- - d- y 1 -- etc (+ 1).: y+( d- ) 1 dx 2 1- +etc. (2). If, now, in the equations of these two curves we have for any given value of x, Y=-y, the two curves will evidently have a common point. If, also, _ d Y dy they will have a common tangent; and if, at the same time, d2 Y dSy, they will, after passing the common point, diverge from each other less than if d2 Y> d2y dxe < dxY 184 THE DIFFERENTIAL CALCUL US. Two curves, whose first derivatives only are the same, for the same values of y and x, are said to have contact of the first order; and, generally, if the first in derivatives are the same, the curves have contact of the nth order. 117. The order of contact which one curve, which is given by means of its general equation, may have with another entirely given, depends on the number of arbitrary constants in the equation of the first curve. Contact of the first order requires two conditions, viz: dY dy Y=y, and d d dx - dx hence, the equation of the first curve must contain two arbitrary constants by means of which these conditions may be determined. Contact of the second order requires three conditions, viz: dY dy d2Y dy. i d Y' dx d' dx" dx' hence, the equation of the first curve must contain three constants, and so on. The general equation of the straight line contains but two arbitrary constants; hence, in general, the straight line can not have contact of a higher order than the first with any plane curve. The general equation of the circle has but three arbitrary constants; hence, in general, the highest order of contact possible between a circle and another curve is the second. We say in general, because for certain particular points in a curve the order of contact between it and the circle may be of a higher degree than the second. The circle which has contact of the second order with another curve at a given point is called the osculatory circle of that point. OSCIULATORY CIRCLE. 1185 118. Problem. —To determine the radius of the osculatory circle of any point of a plane curve. The formula for the radius of curvature being R -1- ( y 1 32 dXy we shall have the required radius by substituting in this formula the values of dy d2y, taken from the equation of d, dX'x the circle. But since the circle has contact of the second order with the given curve, the values of dY, (dI, are the same for the circle and for the curve. Hence, the radius of the required circle may also be found by taking the values of dy dYy from the equation of the curve. The d' dx' two results being identical, it follows that the radius of the osculatory circle at any given point of a curve is equal to the radius of curvature of the curve at that point; and, consequently, the osculatory circle is the circle of curvature, and its center is the center of curvature. Corollary.-If we differentiate the value of R with respect to x, and place the first derivative equal to zero, we shall have one of the conditions for a maximum or minimum value of R. Performing this operation we have 3 ( dy - d3Y +( dy } o (1). If, now, we take as the equation of the osculatory circle (X- a)2 + (y- b)2 - R2, I. C. —16. 186 THE DIFFERENTIAL CALCUL US. and differentiate this three times, we obtain, after reduction, 3 ( d2y dy I+ (dy ) o (2), an equation of the same form with (1). Now, since dy d Y, taken from the equation of the circle, are equal to the same quantities taken from the equation of the curve, it follows that d3y must be the same in both equations. But this, in connection with the foregoing, is the condition that the two curves shall have contact of the third order. Hence it follows that at those points for which the value of the radius of curvature is either a maximum or mininmum, the circle and curve have contact of the third order. 119. Evolutes. —The curve which is the locus of the centers of all the osculatory circles of a given curve, is called the evolute of that curve, and the given curve is called the involute of its evolute. 120. Problem. - To find the equation of the evolute of a plane curve. Let y- F(x) be the equation of the curve, and let (x - a)2+ (y_-b)2= R2 be the equation of the osculatory circle. Differentiating this equation, we obtain dy (x -- a) + (y- b) d- ~o 1+ dy(dx + y b) d2y - in which the derivatives are the same as those taken from the equation of the curve. EVOL UTES. 187 If we combine these equations with that of the curve so as to eliminate x and y, the resulting equation will be a relation between a and b, the coordinates of the center of the osculatory circle, and it will therefore be the required equation of the evolute. 121. If we differentiate the equation (x- a) + (y —b) dy _-O dxwith reference to x, considering all the quantities as variable, we shall have 1+( dy 2da ( yyb) d' db dy x dx-(yd - -b)dx dx X But+ ( 2+ (y-b) d2 -0; da _ db dy dy da d -dx dx dx db Now db is the tangent of the angle which the tangent da line to the evolute makes with the axis of x, and dy is the xd tangent of the angle which the tangent line to the curve makes with the axis of x. The above equation indicates that these two lines are perpendicular to each other, and since a line passing through Fig. 15 the point of tangency, perpendicular to the tangent, is the normal, it follows that the tangent to the evolute is the normal to the involute. 122. Resumning the equation (x -- a)2 + ( - b)= R2) 188 THE DIFFERENTIAL CALCUL US. and differentiating with respect to a, we have d( ddy db+(y-b) db da da da But (x —a) d+ b) d(y-b); -- 0. da () y-b)da d a) + - dx 0. db dR -a dda Now db _ dx (Art. 121) b da - - dy x-a by substitution in the last equation, (x- a) {1-( da ) -} - R da (1). Again, (x - a)2+ (y- b)2 (x- a)2{ +( Yxa) -(x a)2 { + ( db } - R2; da db \2 _ (Xda) } 2 R (2). From (1) and (2) we have 1{ + ( 2; dR- ldb + da2 =d (are of evolute). Now we have seen (Art. 47, Cor. 2) that when two variables have the same derivative with respect to allother EVOLUTES AND INVOLUTES. 189 variable, they differ from each other by a constant. Therefore, designating the arc of the evolute by S, we have R- = S+ c. If 1', R", S', S", be two radii of curvature and the corresponding arcs of the evolute, we shall have R' = S' + c, and R"- S" +- c.. R' R" — S'- S"; or the difference between two radii of curvature is equal to the are of the evolute intercepted between them. 123. From the foregoing properties of the evolute it is easy to see that if a string be wound round the evolute of a curve, and then be unwound, keeping the part unwound always tangent to the curve, the end of the string will describe the involute, and every point in the string will describe a curve similar to the involute. 124. Problem. — To find the equation of the involute to a curve. If we eliminate a and b between the following equations: F(a, b)- 0, the equation of the evolute; da dy dy db d; (x -a) + (y b) dy x O there will result a relation between x and y, which will be the equation of the involute. This will generally be a differential equation. 125. EXAMPLES. 1. Find the radius of curvature of all the conic sections. The equation of these curves is y =_ mx + nx2. 190 THE DIFFERENTIAL CALCUL US. dy _ m+ 2nx. d2y _ m2 m2 x 2(mx+nx2)) dx 4(rnx+nx1jd2)y Now since {1~ 2 3 (normal)l -{ 1 +- ( dwy) } (no~mM)[Art. 93], we have R (normal)3 (normal)3 (normal)3 3 d2y 1 2 (semni-parameter)2 4dx 2. The radius of curvature of each of the conic sections. 3. The logarithmic curve y- a". We have dy lo g Y d2y 1 dy _ y dx -m odxa m dx m2 (-( dy 3)' ___4 __. __ 2 + y2m d2y my dx2 4. The cycloid. We have already found [Art. 94, Ex. 3] dy _ 2r-x d2y _r dx' x dx x V12rx -x2 1+2r-x2 R - - x- 2 /2r (2r- x) x V1 2rx - x- twice the normal. EXAMPLES. 191 5. The logarithmic spiral r a. We have dr r d2r r d - = aO log a -- dO' -.R _ r+( dr?I r2 r~-t- 2 +\- rdt? {r2+Lr} {:2 }2 m -- normal. 6. The hyperbolic spiral rO = a. We have dr a r2 d2r 2a 2r3 dO 80 a dO~ 00 a0 a 7. The catenary, Y -2- e+q-e }'Ans. R=- C 8. The evolute of the parabola y2 = 2px. We have dy_ p. d2y p2 dx y dX2 y 192 THE DIFFERENTIAL CALCUL US. These values, substituted in the differential equations of the circle, give y2 a-p x a —a -— 2x- p; whence x= --- p 3 2 4 y-b= -+ y; whence y= —bp2; y2b3p3. These values, substituted in the equation Fig.16 of the parabola, give 2 2 b_ 2_ (a p), or 3p1A 27p the equation of the evolute, which is the semi-cubical parabola. 9. The evolute of the ellipse A2y2 +- B2x2 A2B2. We have dy B2x d2y B4 dx A2y; dx - A2y3 Substituting these values in the differential equations of the circle, we obtain x(A4y2+ B4x2) x(A4B2-A2B2X2 + B4x2) -a A'B 2 -A4 B2 8 2 A4a and x A3 a X 2- and xB EXAMPLES. 193 b y(AiyA+B42) - y(A2B4 -A2B2y2Y+ A4y2) Y — A2B4 A2B4 8 2 B4b 2 _B3b_ y3__A2 B2' and y2 (A2 B2) These values of x2 and y2, substituted in the equation of the ellipse, give, for that of the evolute, A3 a7 + B2 b3 - (A2 - B2). 10. The evolute of the cycloid. The equation of the cycloid is y r versin-L r + 1/2rx — $. r dy |_ 2r -x d2y r *'dx- - x; dx2- xV/2rx x2 Substituting these values in the Fig. 17 differential equations of the circle, we obtain x-a+(y- b) J; 2r x0; A 2r (y —b)r 0. x x/ 2rx x: y - b-=2 V/ 2rx -, y b + 2 / 2rx - x2; x-a+2(2r-x)-O, x-4r-a. These values of x and y, substituted in the equation of the cycloid, give, for the equation of its evolute, b r versin- 4r-a / 2r (4r- a) - (4r- a)2. D. C.-17. 194 THE DIFFERENTIAL CALCUL US. This is the equation of a cycloid equal to the given curve, as will appear by transferring the origin from C to the point A, whose coordinates are a - 2r, b - r versin-1 2. We thus have r r n-2 b-r versin- 2- }- 1/2ra - a, or r b - r versin7- a 1/2ra- r2 which is the equation of the cycloidal arc AE. 11. The evolute of the logarithmic spiral. In finding the evolute of a curve whose equation is referred to polar coordinates, it is generally best to transform the equation to r and p as coordinates. For this purpose, let S be the pole, P a point on the given curve, PO the P radius of curvature of the point P, PY the tangent at P, SP the radius-vector of P, SY the perpendicular on the tan- s gent PY. Then will 0 be a point on the evolute, PO the tangent to the evolute at 0, SO the radius-vector of 0, and SN the perpendicular on PO. o Let SP-r, SY-p, SO=-r,, SN —p,, PO=R. Then we shall have SO -= SP2 + P02 — 2PO X PN, or r2 r2 R2- + -2Rp (1), since PN= SY=p; also p, - V/r2 p (2), [curve, p- =f(r) (3), the equation of the given and R = r (4) [Art. 115, Cor. 2]. dp SING ULAR POINTS. 195 From the equation of the logarithmic spiral r = a0, we find p- nr, n being a function of the modulus; dr r d -p -; -r,~~~~ ~r2 p -l V/r2 p2 p r / l -n2; r 2- r2+R2- 2Rp -- r2 r2 2 n- (1 —n 2)- p2 the equation of the evolute is Pi nr,, which is the equation of a logarithmic spiral similar to the given curve. CHAPTER XVII. SINGULAR POINTS OF CURVES. 126. A singular point of a curve is one which possesses some peculiarity which distinguishes it from other points of the curve. Such points are the points of greatest and least curvature; the points where the tangent is parallel or perpendicular to either of the axes; multiple points, or points through which several branches of the curve pass; points of inflection, or points where the curvature changes from convexity to concavity; cusps, or points where two branches of the curve, which are tangent to each other, terminate; conjugate points, or points whose coordinates satisfy the equation of the curve, while the points themselves are entirely detached from the curve; stop points, or points at which a branch of the curve suddenly terminates; salient 196 THE DIFFERENTIAL CALCUL US. points, or points at which two branches terminate without being tangent. The Calculus affords a very simple means of detecting and determining the positions of such points; and when they are found, the curve may be readily traced through them. 127. Problem. —To find the points at which a given curve is parallel or perpendicular to the axis of x. Let u =-F(x, y) = 0 be the equation of the curve. du du dy du du Then du =- dx +- ~dy _ O, and * -- @dy dx dy dx dx * dy If the curve is parallel to the axis of x at the point x', y', then 2-= 0; and if it is perpendicular to the axis of x at the point x', y', then dy. The values of x and y, which satisfy the two equations d- 0, and dY- =o are the coordinates of the required points. 128. Problem. —To find the multiple points of a curve. Let u - F(x, y) = 0 be the equation of the curve. Then du+ du dy _ 0 (a). dx dy dx- Fig. 19 Now, since a multiple point is a point through which several branches of the curve pass, each branch will have a tangent line at that point, and, consequently, dy must have several values. 0 X dx Hence, since du and du remain fixed, equation (a) can dx dy not be satisfied for more than one value of dy unless du dx dx - 0, and d- 0. dy - MULTIPLE POINTS. 197 We must therefore have, for a multiple point, the three equations du du u= F(,y)=O (1); d:0 (2); d: 0 (3) and the coordinates of multiple points are to be found among those real values of x and y which satisfy these three equations. du du Again, since - O, and d- 0, the expression for 0 dy will assume the indeterminate form —, and its values may be found by the rules established for such cases; or they may be found by differentiating equation (1) several times, and placing equal to zero the first differential equation so found in which the values of dy are determinate. dx EXAMPLES. 1. Find the multiple points of the curve u =4 - 2ay' - 3a2y'- 2ax2' + a-4 0. We have du d= 4x3- 4a2x-0. ~. x- O, x = a; du d -- 6ay'- 6a2y - O,.'. y =O, y —a. The values of x and y which satisfy the equation of the curve are x-O, y=-a; xz-a, y=O; x —a, y=O; and these are the coordinates of the only points to be examined. 198 THE DIFFERE.NTIAL CALCULUS. Now, by differentiating the equation of the curve twice, we have d2+2 d2'u dy ud2U ) ~; and, forming the partial second derivatives from the equation of the curve, and substituting them in this- last equation, we obtain 122 — 4a2 — (12ay + 6a)(d )= 0; whence' dx 6ay+ 3a w 3yO, 2 - when x = O, y - a.. (d)_ -4 when x =+a, y= 0, -= _4 when x- -a, y=O, =- J- when x-O, y — a. The three points are therefore all multiple points, and through each pass two branches of the curve. 2. The curve u =x4+ 2ax2y-ay3=0.. (1). We have du d — 4x 4axy=0'.. x= 0, =y-=0; d -2ax2- 3ay2 O0,.~ x-O, y O; dy 2e-3 EXAMPLES. 199 and these are the only values of x and y which satisfy all three equations. Differentiating the equation of the curve u F(x, y) O0 three times, we have d3u du dy du y d3u dy 33 [\_ d + 3 d -dy d- + 3 ddyd \d/ djdy' 0.(2); and from the equation of the curve, d3u d3u d3u d3u d -24x; x2 4a; dy2dx; 6a. Substituting in (2), and making x — 0, we have dy dy) dx dx dY o, and dy 2 d dThe origin, x = O, y = O, is therefore a triple point. NOTE. —In this example we have differentiated three times, because the values of dy obtained from both the dx first and second differential equations are indeterminate for the particular values of x and y. 3. The curve u = $4 + xy2- 6ax2y + a2y2- 0 We have du 43+ 2xy2- 12axyO *, * x=, y0 du 2x2y-6aX2+2a2y=0,.. x=0, y=-O. 200 THE DIFFERENTIAL CALCUL US. Forming the second derivatives, and substituting in the equation d2u d 2u dy d2u dy\ 0 we have dy2 dy The origin is therefore a double point, and the two branches at that point have a common tangent, the axis of x. 4. The lemniscata, u - (x2 + y) -- + a2yx a22 O. We find the origin to be a double point, and dy - 1. dx - 5. The folium of Descartes, u —- y x3 - 3cxy =0. We find dy O0 or oo for x 0, and y=O. Therefore the origin is a double point, and the two axes are tangent to the curve at that point. 129. Problem.-To find the conjugate points of a curve. It follows from the definition of a conjugate point that if x, y be the coordinates of such a point, the increment Jy, corresponding to an increment ax, will be imaginary; and this being the case, the limit to the ratio of Jy and Jx will usually be imaginary, and, consequently, will admit of at least two values. We may therefore, in general, determine the coordinates of a conjugate point in the same manner as those of a CONJUGATE POINTS. 201 multiple point; and if the values of dy prove to be imaginary for any real values of x and y, the corresponding points are certainly conjugate. But even though Jy should be imaginary, dy may be real, and in such cases we may determine whether a point is conjugate by substituting x + h for x in the equation of the curve. If the value of y, corresponding to x ~+ h, be imaginary, the point is conjugate. EXAMPLES. 1. The curve ay- x + bx2 = 0. We have dy 3x2- 2bx 0 i do 2ay - lf $-~= and yz- 0; d-x 2ay -- 0 Y sFig. 20 6x -2b b 6x-2a dy -. Therefore, dy dy dy\' j b and d7Y j_ bx dx a dx 01 - X The origin is therefore a conjugate point. 2. The curve (c2y - x)2 = (x - a)5 (x - b)6. The value of dy in this equation is real for the values dx of x and y, which evidently satisfy the equation, viz: 3 b3 x=-a, y = — and x=-b, y- -. 202 THE DIFFERENTIAL CALCUL US. If, however, we substitute b ~+ h for x in the given equation, the resulting values of y are imaginary. The point x = b, y =2 is therefore a conjugate point. 130. Problem. — To find the points of inflection of a curve. The direction of curvature being Fig.21 determined by the sign of d'y it follows that, in passing through a point of inflection, d2y must change its sign, and, therefore, for a point of inflection we must have ox d2y -0 (1), or d (2). But, as a quantity does not necessarily change its sign in passing through zero or infinity, it follows that the value of dy2 may be zero or infinity without the corresponding point being a point of inflection. We must, therefore, after finding the values of x and y which satisfy (1) or (2), substitute x - h for x in the value of dLy dX2 If these substitutions cause d2Y to change sign, the point under consideration is a point of inflection. EXAMPLES. 1. The curve a2y x3. We have dy 3x dx2" _ 6a= 0 when x = 0. O + h renders positive, and 0 — h renders it negative. Therefore the point x = 0, y = 0, is a point of inflection. CUSPS. 203 2. The curve y = 3x + 18x2 — 2x3. We have dy3 36x - 6x2: dy -36 12x - 0, when x -3. dxf dX2 If x -- 3 + h, d2y is negative; and if x 3- h, d2y is positive. dx Hence, the point, x = 3, y - 117, is a point of inflection. 131. Problem.-To determine the cusps of a curve. Since a cusp is a point at which two branches of a curve terminate, it follows that it is a multiple point. But since the two branches are tangent to each other, they possess at that point a common tangent line; and, therefore, at that pointdy must have two real and equal values. dx Y Y y 0 X 0 X Fig. 22 Again, since the branches terminate at the cusp, the ordinates to the curve will be real on one side of the cusp, 204 THE DIFFERENTIAL CALCUL US. and imaginary on the other side. Therefore, if x is the abscissa of a cusp point, and x + h renders y real, x - h will render y imaginary. Again, if the two branches of the curve lie on the same side of their common tangent, their curvatures will be of the same character; and, therefore, the values of d~ for the dX2 two branches will have the same sign. Such a cusp is called a ramnphoid cusp. If the two branches lie on opposite sides of the tangent, their curvatures will have opposite characters, and the two values of dY will have contrary signs. Cusps of this chardx2 acter are called ceratoid cusps. NOTE. — In the particular case in which we find the common tangent to the two branches to be perpendicular to the axis of x, it is generally best to consider y as the independent variable, and find the values of d2x. dy? EXAMPLES. 1. lb - y2 - 3X3 - 0. We have du du- 9e2 —0,.'. x-0; dx du d 2y -0,.. y = O. dy d~u d~u d2h d= u 118x; ddy- 0; dy — 2. dx + dxdy dy2.~ 18x + 2(dy) —0, and dy -— = 0when x-=0. dx -r - CUSPS. 205 x 0 - h renders y imaginary, and d~ changes sign dx2 with y. Hence, the origin is a cusp, the axis of abscissas is the common tangent, and the two branches lie on opposite sides of this tangent. 2. y = b +cLX + (- a)2. We have dY 2ex + 5 (x - a)-; dy 2e- (x a)2. The values x = a, y = b + ca2, satisfy the equation of the curve; and, in consequence of the fractional exponent in the second term of dy, this derivative has two equal values, 2ca, for x = a. If x > a, y is real; and if x < a, y is imaginary. If x = a + h, dY _2c + - h2, which has two values, both positive when h is an infinitesimal. Therefore the point x = a, y = b + ca2, is a cusp; the two branches are both convex to the axis of x, and they are situated on the same side of their common tangent. 3. The curve y4 axy2 + x4 - 0. We shall find a ceratoid cusp at the origin. 132. The remaining singular points, such as stop points and salient points are but seldom met with. According to the definition of a stop point, its test will be similar to that for a cusp, except that dy and dLy will each have but a dx dx single value. The test for a salient point is the same as that for a cusp, except that dy has two unequal values. dx 206 THE DIFFERENTIAL CALCUL US. 133. Tracing of Curves. —If we have given the equation of a curve, the form of the curve may be traced by determining the character and position of all of its singular points, the number and directions of its asymptotes, and as many of its ordinary points as may be convenient. Having located these points, the curve may be drawn through them by hand with a considerable degree of accuracy. It will frequently occur that, by a simple inspection of the equation of the curve, we can determine many of the properties of the curve itself, and it is advisable to do this whenever it is practicable. We append a few examples. Fig. 23 1. Trace the curve whose equation is?= 2~-' 22 XZ - a2 X, Ust When x =0, O y = 0..'. the curve passes through the origin. 2d. When x=-, y= ~; when x= o, y = — o. the curve extends to infinity in four directions. 3d. When x ~ a, y = 0. the curve cuts the axis of x at two points. 4th. Since y -oo when x = -, the line whose equation s an asymptote to the curve. is x — is an asymptote to the curve. TRACING OF CUR VES. 207 5th. If we expand the value of y3, and extract the cube root of the result, we shall have Y — 2 {x+ 6 + terms involving negative powers of x. the line whose equation is iYo-2 6 y is a second asymptote. 6th. Forming the first derivative, we have dy 6x4- 4ax3 - 2a2x' + 2a3x a _=x — ( - 2 -- when x- -O, - a; 3 (2x — a) (X4- a2x23 O0 when x < 0 and > - a. Hence, at the points corresponding to these values of x the tangents are respectively perpendicular and parallel to the axis of x. Fi.2 7th. We shall find points of inflection at the points whose abscissas are x -a, x - - a, x > O and < a and a cusp at xx the origin. 2. The curve X4I _yX2 + ay O. y' We shall find dy -0 and - 1, when x —0, y -0. dx ~. the origin is a triple point. We shall find no asymptotes and no other singular points. 208 THE DIFFERENTIAL CALCULUS. 3. The folium of Descartes, y3 - 3cxy + x3 0. 4. The curve y2 a2+ -x a 2 t- x 5. The curve y=2a|2a-X 6. The curve 2 + 1 7. The curve (y2.q X2)2 = a2(x2 Y8. The curve r = a (1- cos 0). This curve can be traced by assigning different values to 0, and determining the resulting values of r. In the same manner may be traced all curves given by their polar equations. CHAPTER XVIII. ENVELOPES OF CURVES. 134. If in the equation of any plane curve it= F(x, y, a), we assign to the arbitrary constant a a series of different values, the resulting equations will be the equations of a series of curves different from each other in form and position, but all belonging to the same class. If we suppose a to change by infinitesimal amounts, the curves of the series will differ in position by infinitesimal amounts; and, as a general rule, any two adjacent curves of the series will intersect. Let us suppose that the points of intersection of every two adjacent curves are connected by straight lines. There will thus be formed a polygon, the lengths of whose sides will depend upon the actual ENVELOPES. 209 value of the increment assigned to a. Now, since this increment is an infinitesimal, it is obvious that the smaller it becomes, the more will the polygon tend toward coincidence with a certain determinate curve; and as the limit to the increment is zero, the limit to the polygon is this curve. This curve is called the envelope of the series, and it may be defined to be the locus of the limiting points of intersection of the consecutive curves of the series. 135. It is clear that when all the consecutive curves of the series intersect each other two and two, the sides of their polygon of intersection will approach toward coincidence with their common tangents as the points of intersection approach each other, and that the limiting curve, or envelope, will be tangent to all the curves of the series. But, as the consecutive curves do not necessarily intersect, the envelope is not necessarily tangent to all of the curves. 136. Problem. - To determine the equation of the envelope of a given series of curves. Let F(x, y, a) -- 0, and F(x, y, a + h) - O, be the equations of two adjacent curves of the series. The values of x and y which satisfy these two equations are evidently the coordinates of the point of intersection of the two curves, and they will converge toward the values of the coordinates of the limiting point of intersection as h tends toward zero. If, therefore, we combine the equations so as to eliminate a, and then pass to the limit by making h equal to zero, the resulting equation will be the equation of the locus of the limiting points of intersection, or of the envelope. Now, if we subtract the first equation from the second, we shall have a new equation, which may be employed instead of the second without affecting the result. rD. C.-1S. 210 THE DIFFERENTIAL CALCUL US. We have, therefore, F(x, y, a + h) -- F(x, y, a) = 0; or, dividing by h and passing to the limit, d [F(x, y, a) I _ 0. da Hence, to find the equation of the envelope it is sufficient to eliminate a between the two equations F(x, y, a) = 0, and d (x a) 0. EXAMPLES. 1. The envelope of a series of parabolas whose vertices and axes are the same. We have y2 2px, or u - y2 2px - 0. du. du 2x -0. dp These two equations give us for the equation of the envelope x - 0. The envelope is therefore the axis of y, as might have been inferred. 2. The envelope of a series of circles whose centers are all on the same straight line, the axis of x, and whose radii are proportional to the distances of their centers from the origin. Let u -(x — a) + y2 r —- 0 be the equation of one of the circles. Also, let r ca, or r2 c2a2. EXAMPLES. 211 Then, u (x- a)2 + y2 c2a __ 0; du — 2 (x -a) -2ac =O. da * a_-, and, by substitution in the equation of the circle, we have, as the equation of the envelope, ~ cx -A- ex Y 1 c' This is the equation of two straight lines passing through the origin, and making equal angles with the axis of x. These two lines constitute the envelope of the series of circles, as may be readily seen. 3. The envelope of a series of concentric ellipses, the sum of whose axes is constant, the axes of all the curves of the series being coincident in direction. The equation of one of the curves is x2 y2 a' + b' = 1, and we have the condition a + b - c. Differentiating these equations with respect to a, we have x2 Y2 3da q- db - O; da + db O0. db X2 y2.db -1, and C ~ k cc' d~ ~ a: 212 THE DIFFERENTIAL CALCUL US. X2 y2 X2 x2 2 a b - ab c whence, 3 x x; c _, _ 2(); and ( ) Y, the equation of the envelope. 4. A given straight line slides between two rectangular axes; find the equation of the curve to which it is always tangent. Let c be the length of the line, a and b its intercepts on the axes. Then the equation of the line will be x + b- 1, a b and we shall also have the condition a2 + b2 6C2 Differentiating with respect to a, we have x y d 0b db db a x ay y 22__c__a22 3V1x7 a2 b 3 ~ 2 a ba -O b-a x; aa -'a2 + 2 2 b cy a-X3+y3 X.3+. y EXAMPLES. 213 and by substitution in the equation of the given line, and reduction, we have, as the equation of the envelope, X y 5. The envelope of a series of equal circles whose centers all lie on the circumference of a fixed circle. Let X,2 + y2- r,2 = 0 (1), be the equation of the fixed circle; and (x --,)2 + (y _ y)2 - r2 = 0 (2), be the equation of one of the movable circles. Differentiating with respect to x,, we obtain X, + dye- 0, or- d, E — y/,-x ( 3) -2(- (x x) 2 (y - ) day' (4) Combining (1), (2), (3), and (4), so as to eliminate x, and y,, we have X2 + 2 = (r -I- r)2. This is the equation of two concentric circles which constitute the envelope. 6. The envelope of the consecutive normals to an ellipse. Let x', y' be a point on the ellipse. Then y y Y' i (x-x') (1) is the equation of the normal line; and the given point being on the curve, we have a2 y'2 - b2 x'2 - a2 b2 (2). 214 THE DIFFERENTIAL CALCUL US. Differentiating (1) and (2) with respect to x', and eliminating y' between the resulting equation and (2), we find 8 2 t2 a8 X3 2 (a2 — b-2) Differentiating (1) and (2) with respect to y', and eliminating x', we obtain 8 2,2 __b y3 (a2 — b2)3 Substituting these values of x'2 and y'2 in (2), we have, for the equation of the envelope, 0 23_ + bF =y- (a2- b)3, which we have previously found to be the equation to the evolute of the ellipse. The result established in this example for the ellipse is true for all curves whatever, as might be readily proved. CHAPTER XIX. TANGENTS AND NORMALS TO CURVES OF DOUBLE CURVATURE AND SURFACES. DIFFERENTIAL EQUATIONS OF SURFACES. 137. A curve of double curvature is one all of whose points are not in the same plane. It may be determined by the equations of its projections on any two of three coordinate planes, or by the equations of the surfaces of which it is the intersection. CURVES OF DOUBLE CURVATURE. 215 The tangent line-to such a curve, at a given point, is the limit to the secant line drawn through that point and another at an infinitesimal distance from it. The length of the curve, or of any portion of it, is the limit to the length of the broken line or polygon inscribed in it. Let x, x + ax; y, y +- y; z, z+ dzz; be the coordinates of two points on the curve, and let as be the length of the chord joining them, Jx, etc., being infinitesimals. Then, supposing the coordinate planes to be rectangular, we shall have Js -= zix" + Ay' + Az'; and passing to the limit, observing that the limit to the ratio of the chord and arc is unity, we have d are = l/dxr + dy + dz (1). 138. Problem. —To find the equation of a tangent line to a curve of double curvature. The equations of the secant line passing through the points whose coordinates are x', x'+- Ax', y', y'+jy', z', z' — Az', are x ix' - y' (z and since the tan'gent is the limit to the secant, we have, by passing to the limit, dx' dy' x- x' —, (Z-z' ); Y-y'= dz' (z_-Z) (2), the equations of the tangent line. Corollary 1.-If the curve be given by the equations of two intersecting surfaces, F(x, y,z) 0y,f(x,y, z) = O, 216 THE DIFFERENTIAL CALCULUS. we shall have dF dx' dF dy' dF dx' dz' dy' dz' dz' df dx' tdfdy' + df 0 dx' dz dy' dz' dz' - dx' dy' Finding the values of 4dy,, and substituting in (2), we have dld dF dF (x - x')- + (-y - ) dd + (Z- -) - o (3); (x- x) d~(Y _Y') + (z-z) df -o dx' Y'> d,, $ (~ dz' - another form for the -equations of the tangent. Corollary 2.-The secant passing through the two points will make, with the three axes, angles whose cosines are, respectively, ax Ay Az as' as' as Hence the tangent will make with the same axes angles whose cosines are dx dy dz ds' ds' ds Representing these angles by a, P, y, we have dx dx 1 cos a cds = dx2+ ay+ zd,.1 ~ dy )2I -+d d 2 _4_d z' NORMAL PLANE. 217 dy dy _ dy dx co s ds l/dx2+dy +dz2 1+ (dy )2+ (dz dx ji d+ ()2 (dz )2 dz dz dz dx COS y - l/dx2 + dy2 +dz( dy ) dx /1 j dX\2+dy 2 +(z ) ( ) 139. A normal plane is a plane' perpendicular to the tangent line at the point of tangency. The equation of a plane passing through a point x', y', z', being of the form A (x - x') + B (y - y) + C (z - z') =, A B or (x ) + C (y y') + (Z -')= 0; if the plane is perpendicular to the tangent, we must have A __ dx' B dy' Hence, C dz C d' dx' dy' (x - x') d' + (Y — y) d + (z - z') = 0, or (4), (x — x') dx' + (y -- y') dy' + (z -z') dz' - 0 is the equation of the normal plane passing through the point x', y', z'. D. C.-19. 218 THE DIFFERENTIAL CALCUL US. Every line in the normal plane, which passes through the point of intersection of the plane and curve, is called a normal line. Every plane which passes through the tangent line is called a tangent plane. It is evident that tangent planes will, in general, intersect the curve. 140. Tangents and Normals to Surfaces.-A tangent plane to a surface at any given point is the locus of all the tangent lines to the surface which can be drawn through that point. WVe shall determine the equation of this locus, and at the same time show that it is a plane. Let it F(x, y, z) = O, or z- =f(x, y), be the equation of a surface. If a curve be traced through any point x', y', of this surface, it will, in general, be a curve of double curvature; and a tangent line to the curve, at the point x', y', will also be tangent to the surface. The differential equations of this tangent line will be dx' dy' (x - x',), ( -- Z'); (Y - _/) d'- (Z- z) Differentiating the second form of the equation of the surface, we have 1( adz \zdx/ + d \ dy' \d - dz' dy dz' in which the ( )s denote the partial derivatives of z, with respect to x' and y'. If we eliminate d,, dy,, from the last three equations, we have, as the equation of the required locus, dz' dz' (z —'))- (x,-')d + (y-i- d- (5); (x-x') -i~d TANGENTS AND NORMALS TO SURFACES. 219 and since this equation is of the first degree between the variables x, y, z, it follows that it is the equation of a plane. Corollary.-If we take the first form of the equation of the surface u = F(x, y, z), we have dF dF dz' dx' dz' dx'- 0; dF dF dz' dy' + dz' dy' Finding the values of dz' di from these equations, 1,' j, and substituting in (5), we obtain __ dF dF (x x) dF + (y-) dF-+ (z -z) dF = (6); another form for the equation of the tangent plane. Comparing this with equations (3), we learn that the tangent line to a curve of double curvature, at a given point of that curve, is the intersection of the two planes which are tangent at that point to the surfaces of which the curve is the intersection. 141. A normal line to a curved surface is a line perpendicular to a tangent plane at the point of tangency. Its equations are, therefore, (x - x') (Z- Z'); dz' @-I')- dy' (Z Z')... (7); dy' or (x - x') dE (Z - ) dF dF - Zf) d;.. (8dF (Y Y') (z dyf~~(8) 220 THE DIFFERENTIAL CALCULUS. Corollary. —If we designate by o', 0", 0"', the angles which a normal line makes with the axes, or those which the tangent plane makes with the three coordinate planes, we shall have [Art. 138, Cor.] dz' d~' cos 0' 1 + (dZ' )2 ( d' )' dz' 1+ d+' ( 2+ ) dz' )2 142. normal plane is a plane perpendicular to a tan142. A normal plane is a plane perpendicular to a tangent line at the point of tangency. The equations of the tangent line beinig =dx' cdy' ~(X $') - dx' (Z-Z'); (y — y') (Z - ), it follows, at once, that the equation of the normal plane will be dx' dy' (x- x') d + (y - y') dy +- (Z-z) - (9). 143. Differential Equations of Surfaces.- If we resolve the two equations of any curved line with reference to any one of the constants which enter into them, -we shall have two new equations of the form u - c, u, — c,, in which ut and u, are functions of x, y, z, and c and c, are constants. EQUATIONS OF SURFACES. 221 If, in these last equations, we attribute to c and e, a series of arbitrary values in succession, the line represented by these equations will change its position, and perhaps its form, without describing any determinate surface. But if we impose some fixed relation upon c and c,, as c, - (c), the two equations will represent a line whose form and position will be determinate for each particular value of c; and if we eliminate c between these equations, the resulting equation, -= X (u), will obviously be the equation of the locus of the entire series of lines, which will necessarily be a surface generated by the motion of the line u = c, u, = c,, according to the laws prescribed by the relation c,- p (c). The function t may be eliminated by the methods already established for the elimination of functions, and the resulting equation will usually be a differential equation. Applications. - 1st. To determine the general differential equation of all cylindrical surfaces. A cylindrical surface is one which may be generated by the motion of a right line whose consecutive positions are parallel to each other, and which rests constantly upon a given curve. Fig. 25 Let x tz + a, y = sz + b, be the equations of the moving straight line. Then u x —tz a, u,y - sz = b. the equation of the surface described by the line is y -- sz = P (x - tz), in which s and t are constant by the conditions of the problem. To eliminate?, differentiate with respect to x and y. 222 THE DIFFERENTIAL CALCUL US. We thus obtain dz dt d(x-tz) dx d(x- tz) d'v dz dp d(x- tz). 1 -- y d (x - t) dy whence, by division and reduction, t d+ s d 1 (10), the required equation. dz dz In applying this equation the values of dz, -, must dy x be taken from the equation of the fixed curve upon which the straight line rests. 2d. The general differential equation of all conical surfaces. A conical surface is one which may be generated by the motion of a right line which passes through a fixed point and rests upon a given curve. Let a, b, c, be the coordinates of the ig.26 fixed point. Then x-a =t(z —c); y -b s(z- c) will be the equations of the moving line, in which t and s are variable. We shall have x —a y-b U -= t; u y - 8. z — Z -- C y b (x-a) EQUATIONS OF SURFACES. 223 Differentiating with respect to x and y, we have y-b dz d, d( zx — (z-c)2 dx d dx d -az c (z- d z c 1 y-b dz d d z ---- z -c (z -C)2 dy d x - dy \ z-c(- ddz x -- a dz vhence, by division and reduction,.z c (x- a) dz-+ (11), ad dy he required equation, in which the values of dz, d, are o be taken from the equations of the given curve. 3d. The general differential equation of all surfaces of revoution. The characteristic property of this class of surfaces is hat every section made by a plane perpendicular to the xis is a circle whose center is in the axis; and we may uppose the surface to be generated by the motion of a rariable circle whose consecutive positions are parallel to ach other, and whose center remains constantly upon he axis. 224 THE DIFFERENTIAL CALCUL US. {ow, since every section of a sphere made by a plane is a circle, we may take the equations of a sphere and a plane as the two equations of the circle which generates the given surface. Let x tz + a, y sz + b, be the equations of the axis; then the equation of a plane perpendicular to that line will be z + tx + sy = c. If we take the center of the sphere at the point where the axis of the surface cuts the plane of XY, its equation will be (x- a)2 + (y - b)2 + z2 r2 and we shall have, for the equation of the surface, (x - a)2 + (y b)2 + z2- (z + tx + sy). In order to eliminate the unknown function q, we might proceed as in the last two cases; but, for the sake of simplicity as well as of variety, we shall adopt another method. It is obvious, from the nature of a surface of revolution, that the normal line to any point of the surface intersects the axis. The equations of the normal to a curved surface are dz' dz' x x+d,(z z) =O; Y-Y+dz(z- z') - 0; and in order that this line may intersect the axis, whose equations are x tz +a, y=sz+b, we must have the relation x' - a -+ dz' db +z''-ad+zr'td y'-b+zdy' dz' - dz' EQ UATIONS OF SURFACES. 225 each member of which equation is the value of z for the point of intersection. Reducing this equation, and omitting the accents, we have, for the required equation of all surfaces of revolution, (x - a - tz) dz - (y - b - sz) d + (x —a)s —(y- b)t =O (12), in which the values of dz dz must be taken from the dy dx equation of the curve which generates the surface. 144. If the algebraic or finite equation of a surface be given, its differential may be obtained under a more symmetrical form than those given in the preceding article. Let u = F(x, y, z) = 0 be the equation of the surface. Then dz d d dd u du Then dx dx dz' dy dy dz Substituting these values in the preceding equations, we have, for the equation of cylindrical surfaces, du du du t + s~ +~ o (13); for the equation of conical surfaces, (x a) + (y-b) du + (z c) du o (14); for the equation of surfaces of revolution, du du (x - a - tz) - (y - b - sz) d du -- f(x — a)s - (y- b)tl 0 (15). — dz 0) 226 THE DIFFERENTIAL CALCUL US. 145. Problem. —To find the equation of the cylinder or cone which incloses a given curved surface. dz dz du du du If we form the derivatives,, or d, -, from d, d, or dy dz the equation of the given surface, and substitute them in the differential equation of the cylinder or cone, the resulting equation will evidently be that of a surface which contains the curve of contact of the given surface with the cylinder or cone: and if this equation be combined with that of the given surface, the resulting equation will be the equation of the curve of contact itself. Then, having this equation and the axis of the cylinder or the vertex of the cone, as the case may be, the equation of the cylindrical or conical surface may be found as in the last articles. 146. Envelopes of Surfaces. — If in the equation of a surface, F(x, y, z, a) - 0, the constant a be made to vary, we shall have the equations of a series of surfaces, all belonging to the same class, but differing in form and position. Any two adjacent surfaces will usually intersect; and a polyhedron may be constructed through the points of intersection. As the increment assigned to a tends toward zero, this polyhedron will tend toward a certain determinate surface, which is called the envelope of the series. Precisely as was done in the case of envelopes of plane curves may we show that the equation of the envelope of a series of curved surfaces may be found by eliminating a between the equations F(x, y, z, a) = 0, and d F(x, y, a) da 147. EXAMPLES. 1. The equations of the tangent line and normal plane to the helix, or screw. EXAMPLES. 227 The equations of this curve are z z y=a sin - x a cos —, mna ma in which a denotes the radius of the base of the cylinder on which the curve is formed, and m is the tangent of the inclination of the curve to the plane of the base. We have, by differentiation, dx 1. z dy 1 z - - sin m — - Cos - dz m a dz m ma Hence, by substitution in the differential equations of the tangent and normal to a curve of double curvature, x - -- sin Z z z'}, Mn ma y- y'i- cos z z- z, m ma are the equations of the tangent line; m(z -z') = (x - x') sin (Y - y') cos m is the equation of the normal plane. 2. The equation of a tangent plane to the ellipsoid. We have X2 y 22 u —, - - - -— 1 o; du 2x. du 2y. du 2z dr x- a-' dy- b2' d 2- c' Substituting these expressions in the general equation of a tangent plane, we have 228 THE DIFFERENTIAL CALCULUS. 2x' 2y',2z' (x i+y _ _ y,_) + 22' ((~ _, cz_ z') - O, or xx' Y Y z' a- + +b2+c 3. Find the equations of the tangent line and normal plane to the curved line formed by the intersection of an ellipsoid and sphere, referred to the same center and axes. The equations of the two surfaces are z2 y2.z2 F(x, y, z). - b2 - 1 0, the ellipsoid; f(x, y, z) = x2 + y2 + z2 -r2 0 the sphere. We have, from these two equations, dx a2 b2- 2 \ z dy b2 c2 -a2 \ z dz c2 a x-b 2x' dz- c- a b2 y Substituting these values in the equations of the tangent line and normal plane to a curve of double curvature, we have, for the equations of the tangent line, x - x'\ Y zy-'l z \. a b' ~ b~ Ce a' a- -b and for the equation of the normal plane, q - b2f -a) Y - y' a2(b2 c2) xx +b2(c2 a2) x Y + c2(a2- b2) = 0. 4. The equations of the tangent line to the curve formed by the intersection of a sphere with a right cylinder, the radius of the base of which is one-half that of the sphere, and whose surface passes through the center of the sphere. EXAMPLES. 229 Taking r as the radius of the base of the cylinder, the center of the sphere as the origin, the axis of z parallel to the axis of the cylinder, and the axis of x, the line passing through the center of the cylinder, we have, for the equations of the curve, x2 + y2 z 4r2; y2 + X2 = 2rx. We shall find, for the equations of the tangent, y(y- y') - (r - x) (x - x'); z(z - z') = r(x - x'). 5. Find the equations of the curve of contact of a sphere and cone whose vertex is on the axis of y, and at a distance 2r from the origin. The equation of the sphere is u = x2 + y2 + z'_ r2= 0. du du du: d —=-2x; -2y; d - 2z. dx dz The coordinates of the vertex of the cone are a =O; b =-2r; c -O. Substituting these values in the equation of all conical surfaces, we obtain, x.2x (y- 2r)2y+z.2z- 0, or x+y2+- 2 - 2ry = 0. This is the equation of a surface which contains the curve' of contact; and combining it with the equation of the given sphere, we obtain, for the equations of that curve, ra 3r2 Y 2, and x2 + z2 - These are the equations of a circle perpendicular to the axis of y, and at a distance - from the origin. 230 THE DIFFERENTIAL CALCULUS. 6. The envelope of a series of ellipsoids, the sum of whose axes is constant. The equation of the ellipsoid is 2 2 Z2 a3 + 1 +,' and the given condition is a + b + c = constant == k. Differentiating with respect to a and b, we have x2 2z de y2 z de 0; -- 3 da -0; b' cd dbde de 0;da 1 + db 0. x+ yZ z2 2 2 2 y2 z2 x2 a3 b" c" oa + b + c - k a3 3 $ y3y_ y z z * a k' b3 -k' c k' and, by substitution in the equation of the ellipsoid, we obtain for that of the envelope, X )+ +( Y )X+( 2 Z _ 7. The envelope of a series of planes +- + =1, a b c in which dbe - constant n m3. m3 Ans. zyz 27 CURVATURE OF SURFACES. 231 CHAPTER XX. THE CURVATURE OF SURFACES. 148. The curvature of a surface at any point is determined by finding that of the various plane sections passing through that point. Among all these sections, those which are made by normal planes, and which are therefore called normal sections, are especially important. Problem. — To determine the radius of curvature of a given point of a normal section of a curved surface. Let z =f(x, y) be the equation of the surface; and, for the sake of simplicity, let the plane of XY be taken as the tangent plane. The normal plane will then contain the axis of Z. Fig. 27 Let ZOa be the normal Z plane, and designate by m the tangent of the angle which this plane makes b with the plane of XZ. A c circle, situated in the nor- x mal plane, and tangent at the origin to the normal YX section bOc, will have, for its equations, y = mx, X2+ y'+ - 2Rz = 0; whence x2(1 + m2) + z- 2Rz — 0 (a). If this circle be the osculatory circle of the section, the values of must be the same for the circle and the dX2section. section. 232 THE DIFFERENTIAL CALCUL US. Now, equation (a), differentiated twice, gives X1+m2+(z__R)d2+ (ld z )2 dz But at the origin we have z ~ - 0; whence 1 + m — R d -o. ~ R - (1> ). dx x Differentiating the equation of the surface, and observing that dy - m, we have, dx dz dz dy dz dz dz dx dy d+ do m dy + dx; d&z dz.2 d2z d2z dxP dxe2 dxdym + dym, in which the values of x, y, z must be placed equal to zero, after differentiation, in order to refer the derivatives to the origin. Designating d by r, dz by ds, by t, and substitutdby dx dy dy2 ing in (1), we have, for the radius of curvature, R= 1 + 2 (2). r +- 2sin + tm2 149. The greatest and least values of the radii of curvature at any point are called the principal radii of that point; and the corresponding sections are called the principal sections. The positions of these sections may be ascertained by determining the values of m which will render R a maximum or minimum. CURVATURE OF SURFACES. 233 For this purpose, differentiating (2) with respect to m, and placing the result equal to zero, we have dR _ 2sm2+ 2m(r- t) - 2 0 dm - (r + 2smt + tmn)2 sm +- m(r - t)- s -- O (3), and by solving this equation,,t = r) + V'4s' + (t - r)2 2s in (t -- r) -1V/4s2+ (t _ r)2 2s These are the two values of m which determine the positions of the principal sections; and since their product is equal to — 1, whence m' m" + 1 = 0, it follows that the planes of the two principal sections of every point of a curved surface are at right angles to each other. 150. If we take the planes of these sections as the coordinate planes of XZ and YZ, the expression for R may be much simplified. For, since in that case the two values of m become 0 or tang 0~, and wo or tang 90~, we must have s - 0, and therefore the value of R becomes R- 1 n'- (4). r + ti. If we designate by p and p' the maximum and minimum radii, we shall have 1, 1 p -and p'= - D.r t D. C. —2 0. 234 THE DIFFERENTIAL'ALCUL US. For, putting (4) under the form 1 R, we have for n = -oo R = —; and from (4), when mt0, It r Designating by q the angle whose tangent is n, substituting the values of p and p' in (4), and reducing, we have -1 — cos2 + - sin2 q. R p P Now, designating by v the curvature of any section, and by v', v", the curvatures of the principal sections, we have 1 I I, 1 v p p and, therefore, v v' cos2 q -t v" sin2... (5), a relation between the curvatures of the principal sections of a point and that of any other normal section through the same point. 151. We deduce from (5) the following important conseq uences: 1st. If we pass through the normal two planes which are equally inclined to that of one of the principal sections, the curvatures of the two plane sections will be equal to each other. For the angles which they make with the plane of the principal section, being designated by q' and — p', we shall have, by substituting these two values for qp in (5), equal values for v. CURVATURE OF SURFACES. 235 2d. The sum of the curvatures of any two normal sections which are equally inclined to the principal sections, is constant and equal to the sum of the curvatures of the principal sections. For, designating the curvatures by v,, v., we have v1 v' COS2 ~ + v"I sin2', V2- v' sin' 2 + v" cos2',.'. v, -2 v' +- v"- a constant. 3d. If q- 45~ we have cos qz- sin = 11/ 2.. Hence, in this case, v' + v".-.2 Therefore, the curvature of each of the sections whose planes bisect the angles between the planes of the principal sections is the arithmetical mean of the curvatures of those sections; and, therefore, any two planes which are equally inclined to one of these nmedial planes, give sections the sum of whose curvatures is v'+ v", since these planes make equal angles with those of the principal sections. 152. Problem. —To find tihe radius of curvature of an oblique section. Fig. 28 Let XZ be the normal plane, AOB the normal section, aOb the oblique section made by the plane Z' Ox'. A Let OD - x'; DC- z'; a = angle between XZ and - X Y'Z'; CE- z; FE - y; OF -x. 236 T'HE DIFFERENTIAL CALCULUS. d2z d2z Then, at the point 0, we shall have - 0, ddy 0, and therefore the value of R may be written d s dz since 1 + m2-1 + dx2 - ( - dx' If we designate the radius of curvature of the oblique section by R', and its arc by s', we shall have i ds' \2 d2z' dx'2 d2z d2z' But z = z' cos a, and d cos a. Also at the point 0, ds' dx' ds dx' dx dX'. -R cos a. Therefore the radius of an oblique section = radius of normal section X cosine of their included angle = projection of the radius of the normal section upon the plane of the oblique section. THE INTEGRAL CALCULUS. CHAPTER I. FIRST PRINCIPLES AND DEFINITIONS. 1. The fundamental conception upon which the Integral Calculus is based, is that any magnitude may be considered as the limit to the sum of an infinite series of infinitesimals; and the grand problem of which this Calculus affords the solution may be thus stated: "Given an infinite series of infinitesimals, to find the limit to its sum." 2. Let y = F(x) be any function of x, continuous from y, to Y, corresponding to the values xo and X of x. Let the interval X- x0 be divided into a number of parts, each equal to Jx, so that we shall have X X+ - x + x + x +.... + ax. Then, for every increment assumed by x, y will acquire a corresponding increment, which may be represented by Jy, and we shall have Y yo + Jy +, y + Jy +... + Jy, in which the different values of zy will, in general, be unequal to each other. Now it has been shown in the Differential Calculus [Art. 31], that,y - F'(x)Jx + aJx, in which a.Jx is an infinitesimal of the second order. (237) 238 THE INTEGRAL CALCULUS. Hence, Y- y0+ F'(x)Jx + azxI + IF'(x)zJx+ aJxI +- F'(x) ax + aAJx + etc.; or, F(X) — F(x,) + F'(x)Jx + aaJx} + lF'(x)ax + aJx) + etc. (1). Observing that lim F(X) = F(X), lim F(xo) - F(xo), and that in any series, the limit to whose sum is to be taken, we can replace F'(x)zJx + adx by F'(x)dx, we shall have, by taking the limit of (1), F(X) - F(xo) + lim zIF'(x)dx (a); designating by z the sum of the terms F'(x)zx + aJx, etc. This equation may be abbreviated into F (X) - F(xo) +j F'(x)dx, Xo whence F(X) - F(xo) - F'(x)dx (2), in which the second member denotes the limit to the sum of the series F'(x)dx + F'(x)dx + etc., and is read the definite integral of F'(x)dx between the limits x0 and X; and in finding the successive terms of the series, F'(x) is to be taken for every possible value of x, from x0 to X inclusive. Now F'(x)dx is the differential of F(x), and it follows from the form of equation (2) that the definite integral of F'(x)dx, between the limits x, and X, is to be found by determining the function F(x) whose differential is F'(x)dx, and taking the difference between its two values F(xo) and F(X). 3. If, instead of determining the definite integral, we find simply the quantity F(x), of which F'(x)dx is the differential, we shall have the general, or indefinite, inte GENERAL PRINCIPLES. 239 gral of F'(x)dx, which includes the definite integral as a particular case. This integral is denoted thus: Yfr(x)dx = F(x) and the first member is read, the integral of F'(x)dx. When an indefinite integral has been thus found, another more general will be determined by adding to the former an arbitrary constant C, since F'(x)dx is the differential not only of F(x) but also of F(x) + C. We thus have, for the most general integral of F'(x)dx, F'(x)dx = F(x) + C (3). NOTE. — In the foregoing expressions the symbol f is merely an elongated S, and it is used only to represent the limit to a sum. It will be seen that it bears exactly the same relation to the symbol 2 that d in the Differential Calculus bears to a. 4. Having found the expression for the general or indefinite integral, the value of C may be found, provided we can ascertain the value of the integral for any particular value of x. By the very nature of an integral, the value x0 of x reduces fF'(x)dx to zero, since the integral may be supposed to originate at that particular value of x. Hence, we have o =- F'(xo)dx = F(o) + C; 240 THE INTEGRAL CALCULUS. and by substitution in (3), fF'(x)dx = F(x) - F(xo) (4). A comparison of (2) and (4) teaches us that the general integral is equivalent to the definite integral taken between the limits x, and x, the latter of which is variable; and the definite integral between x, and X may be found by substituting X for x in the value of the indefinite integral given by equation (4). 5. If in equation (4) we substitute X1 and X2 for x, we shall have fF'(X2)dx -F(X2) — F(o) F'(X,)dx F(X) — F(o); and by subtraction, F'(X2)dx — F'(XI)dx - F(X2) - F(X,). Now F(X2) - F(X1) being the difference between the two definite integrals taken between the limits x,, X1, and x,, X2, is evidently equal to the definite integral taken between the limits X1 and X2; and, in accordance with the adopted notation for definite integrals, we may write X2 f F'(x)dx = F(X2) - F(X1) (5). We also have, in accordance with the same notation, fF'(X2)dx fX F'(x)dx, and Xo F'(Xi)dx f F'(x)dx..xo DIRECT INTEGRATION. 241 X2 X2 XI Hencef F' (x)dx -d F'(x)dx (6). X1 Xo xo We may, therefore, find the definite integral between two given limits X1 and X2 by obtaining its values between x,, X1 and xO, X2, and subtracting the results. Or, since the constant C will disappear by subtraction, we may substitute X1 and X2 for x in the value of the general integral, and take the difference between the results. INTEGRATION OF FUNOTIONS OF ONE VARIABLE. CHAPTER II. DIRECT INTEGRATION, IN GENERAL. 6. The process of finding the indefinite integral, being the same as that of finding the function of which the given function, F'(x)dx, is the differential, is obviously the reverse of differentiation; and the general problem to be solved may be restated thus: "Given any differential function, to determine the function of which it is the differential." The process by which this operation is effected is called integration. 7. Every proposition in the Differential Calculus will evidently have its correlative proposition in the Integral Calculus; and it is by reversing the operations of the former that the primary methods and rules of the latter are established. We shall now present these methods in the shape of formulas, which are to be remembered. D. C.-21. 242 THE INTEGRAL CALCULUS. FORMULAS FOR DIRECT OR SIMPLE INTEGRATION. 1. Since d(ax) - adx;.. adx d(ax) ax+C= adx. Hence, a constant factor can be moved from one side to the other of the integral sign. 2. Since du +- dv dz -d(u v - z);.. du+dv+dz =fd(u+v+z)-=u+v+z+C, =du fdv ~iz. Hence, the integral of the sum of any number of differentials is equal to the sum of their integrals. 3. Since d(uv) = udv + vdu; u. d+ vdu = d(uv) =uv + C. 4. Since d( ) -vd u; vd J udv =fd() + U C. 5. Since d(x"+l) -(m q+ 1)xmdx, whence m dx - d(x + ). m+1' m-1 From this formula we obtainx the following useful From this formula we obtain the following useful DIRECT INTEGRATION. 243 Rule. —To integrate a function which is equal to the product of a power of the variable by the differential of the variable, increase the exponent of the variable by unity, and divide the result by the exponent thus increased. NOTE.. —It will be observed that the rule fails when m = - 1. x 6. Since d log x _;'dx. *. log x + C. From this formula we derive the following Rule.-The integral of a fractional function whose numerator is the differential of the denominator, is equal to the logarithm of the denominator. 7. Since d(ax) ax log a dx; a. log a dx= a+ C, and faxdx oa + C. log a If a = e, the Naperian base, we have log a = log e I1; e. dx —=e+. 8. Since d sin x — cos x dx; cos xdx = sin x + C. 9. Since d cos x - -- sin x dx; -s. in x dx -- cos x + C. 244 THE INTEGRAL CALCULUS. 10. Since d tang x = sec2 x dx;. sec2 dx = tang x + C. 11. Since d cot x - - cosec2 x dx;.. -cosec x dx = cot x + C. 12. Since d sec x- tang x sec x dx; J tang x sec x dx = sec x + C. 13. Since d cosec x- -- cot x cosec x dx; *. -cot x cosec x dx = cosec x + C. 14. Since d versin x - sin x dx; sin x dx =versin x + C. 15. Since d coversin x = - cos x dx;.. -cos x dx coversin x + C. 16. Since d sin-'x - 2;f vJ* iV = sin-l x + C. dd 17. Since d cos-'x d; j dx =- cos-' x + C. 18. Since d tang-' x 1 x2; J -- =- tangl x + C. DIRECT INTEGRATION. 245 dx 19. Since d cot-' x - 1 - x' *. +X -- cot-l x + C. dx 20. Since d sec- x= d x /x'- 1 *. x_ - = sec- x + C. dx 21. Since d cosec-1 x - _; xV/x- 1; _ f — 1 -- cosec- x + C. dx 22. Since d versin- x - d 1/2x -2 J/2 — = versin-1 x + C. 23. Since d coversin-l x - 1/2x- X2 dx. V2 - coversin-1 x + C. These twenty-three cases embrace all the known forms which can be directly integrated; and before any given function can be integrated, it must be brought, by algebraic processes, to correspond with some one of these known forms. It is in this algebraic reduction that the chief difficulty of the Integral Calculus consists; and it may be said, in fact, that the whole body of this Calculus is made up of artifices by means of which the simplification of complicated or unusual expressions may be effected. 246 THE INTEGRAL CALCULUS. NOTE. — We have seen that when n = - 1, the fifth of the above formulas fails. For we then have fx dx =f-1dx _ whereas we know, from the sixth formula, that f-d -- log x + C. This is, however, only an apparent failure; for if we write, as we are at liberty to do, Cm~= Xin+Y xm+-l am+l fxmdx -- x+ C x+ a+- C' Mm+1 +0, we shall have, when m -- 1, X-' dx - 0 The value of this 0 may be found, as for indeterminate expressions, by differentiating the numerator and denominator of l with respect to m, and putting m= 1 in the result. We thus obtain rxm dx + -am- + C'= log x - log a +C' - log x + C", when mn - 1. 8. It will be observed that each of the above forms comes under the general form fF'(x)dx, and it is evident that in this expression x may represent any variable, and therefore any function of a variable. DIRECT INTEGRATION. 247 Wherefore, substituting t(x) for (x), we have fF't (x) d (x) =d (x) - F (x) + C. The expression under the integral sign is composed of two parts, viz: (a), a function of a function or variable; (b), the differential of that function or variable; and whenever the given expression can be decomposed into two sifch parts, its integration may be effected at once by referring it to one of the fundamental forms. The following examples will serve for illustration: 1. f(a + bx + cx)' (b + 2cx)dx +(a + bx + cx') d(a + bx + ex') (a + bx + CX2)3 C 3 +6. 2. b+ 2cx - d d(a + bx + cx2) Ja+ bx +cx' a + b + ex' _ log (a + bx + cx') + C. 3. fe2 dx =f-j 2xd(2x) 2 fe d(2x) 2 e2 + 4. 2 sin x cos xdx = 2 sinx d sin x sin' x + C. 248 THE INTEGRAL CALCULUS. 5. f(a + bx2)3 xdx = 2f 2b (a + bx2)3 2bxdx -b f(a + bx2)3 d(a + bx2) 1 (a+ bX2)4 2b 4 + X 9. Integration by Substitution.-When the expression F'(x)dx can not be integrated directly, the operation may sometimes be effected by substituting for x some function of a new variable z. Thus, if x -,(z), then dx _,'(z)dz, and F'(x) = F' (z). F'(x)dx -f' (z) }'((z)dz. The relation between x and z being arbitrary, we may often choose it in such a manner as to render the integration comparatively easy; the only condition to be observed being that the definite integral with respect to z, taken between any two limits zo and Z, shall be the same as that taken with respect to x between the corresponding limits x, and X. We shall then have the following equation: X Z s F'(x)dx - a Ff'(z) t'(z) dz. Xo Zo APPLICATION. X; F(x + a)dx. Put x + a = z; then dx - dz, and X Z X+ a F(x + a) dx F(z) dz J F(z) dz x o ZXo -+ a X+ a x+ laF(x + a)d(x + a). x=o - a INTEGRATION BY PARTS. 249 10. Integration by Parts. -When the expression to be integrated is of the form udv, the integration may often be effected in the following manner: Since d(uv) = udv + vdu, we have uv =fudv + fvdu; udv d uv S- vdu. This is called the formula for integration by parts. It reduces the integration of udv to that of vdu, which may be simpler. EXAMPLE. Slog x dx. Put log x = u, dx = dv;.'. du = v- x, and ogxdx=xlogx fx = x log x - - C. 11. Use of Constant Factors.-In effecting the reduction of a complicated expression it may often be advisable to introduce a constant factor. This may always be done provided that we take care to divide by this factor also; and since any constant factor within the integral sign may be placed before it as a coefficient without affecting the value of the integral, it follows that we may multiply the expression to be integrated by any constant factor, provided we place the reciprocal of that factor before the sign of integration as a coefficient. There is an example of this operation in Ex. 5, Art. 8. As another example let us take the expression f(1 + 3x4) x3 dx. 250 THE INTEGRAL CALCUL US. We have f(1 + 3x')5x3 dx = 1 (1 + 3x4)5 12x dx - (1 + 3x4) d(1 + 3x4), 12 1 (1+ 3X4)6 12 6 12. GENERAL EXAMPLES IN INTEGRATION. 1. Integrate (7x5 + 6ax2 + 6)dx. We have (7x5 + 6ax2 + 6)dx f7x5dx + f6ax2dx +f6dx 7 6 = 6 + 3 ca- + 6x-+ C. 2. Integrate (ax" + b)m x"- dx. We have f(axn + b)" x'-l dx = f(axn + b)m an~- L dx an b)md(ax + n b) = d(axx" b) N + + C. ~~an - V ~~an m + 1 3. Integrate (2ax + x2)3 (a + x)dx. We have f(2ax + x2)3 (a + x)dx (2ax + x2)3 (2a + 2x)dx -- 2ax + x2)3d(2ax + x2) - (2ax + x- + C. EXAMPLES. 251 x2 dx 4. Integrate 1 -x 1 b+ X+ We have f e - l + x 31 d(+ +-) 3 log(1 - 3) + C. 5. Integrate dxx b + ex We have a _c - b +x- _C log( + ex) C, + ex -- b- ex C- log(b +- cx)7 + C. 7. Integrate + -X + x 1dx. 7. Integrate 8a 3x- - b h We have r 7xdx 7 r-6xdx _ 7 aJ -8a 3x -- 6J 8a-3x - - 6 log(8a - 3x2) +C, - log(8a - 3x2) 6 + log c = log - 7, (8a - 3x2) 6 by making C = log c, which is evidently permissible, since log c is a constant. 252 THE INTEGRAL CALCULUS. dx 9. Integrate 1 ( + (x + a)2 (x + b)2 Multiplying numerator and denominator by (x + a) - (x - b)2, we have 1 21 f =(a+ bX2) 2dX - _l(a 2+b)2 (- 2ax -dx),2 a-b - + b 1 ak -b 3 ( a+a) -- (3 +b)2 dx 11. Integrate (a bx2)2 We have (e have b) 2d1 2 (aX-2+ b) —2 (- 21x3 dx), and by this transformation we have made the quantity in the second parenthesis the exact differential of that in the first. Integrating this last form, we have (azx-2+ b) 2 + c = - + c. a a V a + bx: The method of transformation illustrated in this example may often be employed to advantage. 11. Integrate 3 (1 - X2)2 Ans. - + C, _/ X2 EXAMPLES. 253 12. Integrate adx - (ax + bx2)2 We have f f(ax + b2 axdx (axl+b (2 axdx, f(ax-1+ b) ax-2 dx (ax-~ b) - (a x-2 dX) 1 2x - 2(ax-'+ b)-2 - x + C. /ax + bx2 13. Integrate adx x i/bx +A CPx We have f= aa(bx + c2x2) 2x-'dx= 7-f(bx-~+c2) 2 bx-2 dx. Ans. _ 2a bx + C2 X2 i+ C Ans. - X x 4 14. Integrate (3ax2 + 4bx3W) (2ax + 4bx2) dx. In order to integrate a differential function directly, it is necessary, as we have seen, that one of its factors shall be the differential of the expression of which the other factor is a function. It generally happens that this is not the case, but the second factor can often be brought to the proper form by introducing an unknown constant factor into the expression, and determining the value of this constant by some one of the ordinary algebraic methods. In the present example, let us suppose that the multiplication of the second parenthesis by A will make it the differential of the first. Then we shall have Sf= -A f- 3ax2+ 4bx3)3 (2Aax + 4Abx2)dx. 254 THE INTEGRAL CALCULUS. We must now determine A by the condition d(3ax2 + 4bx) - 2Aax dx + 4Abx2 dx. But d(3ax2 + 4bx3) = (6ax + 12bx2)dx;. 6ax + 12bx2 = 2Aax + 4Abx2; and since this relation is independent of x, we must have 6a - 2Aa, 12b - 4Ab, from each of which equations we have A = 3. Hence we have, finally,!- ( 3ax2 + 4bx3)> (6ax + 12bx2)dx 3 (3aX2 + 4bx3) +C. If the two values of A had not been the same, we would have inferred that there was no constant factor which would reduce the given expression to the required form. 15. j(4ax+ 5bx2+ 6x)' (ax+ 2bx + 3)dx. Let A be the required factor. Then f f(4ae+ 5bx2+ 6x)4 (Aax2+ 2Abx + 3A)dx. We must have Aax2+ 2Abx + 3A - 12ax2+ 1Obx + 6.. Aa -12a, or A 12; 2Ab —lOb, or A =5; 3A - 6, or A - 2. These values of A being inconsistent with each other, we infer that the given expression can not be integrated in this form. EXAMPLES. 255 16. Integrate tang 2x sec 2x dx. We have f- f tang 2x sec 2x d2x - sec 2x + C. 17. Integrate sec2 V/2x x- dx. We have f -=sec2 /2x (1/2-di2x) =-/ 2 tang Vx + C. 18. Integrate tang x dx. " fsin x dx fd cos x We have f -ncs x d - JdCOS —- log cos x 1 -log - log sec x +- C. Cos x dx 19. Integrate a( cos x) We have 1 2 ec2 2d x = tang x + C. __ 2 x 1 f,' 2 a cos 2 20. Integrate sin3 3x cos 3x dx. We have f=-1 jf4 sin3 3x cos 3x d3x - 12 4 sin3 3x d(sin 3x) = 1 sin4 3x + C. 21. Integrate X + dx. We have X + —dx dx - 1/1 -- 2 + sin1x + C. 256 THE INTEGRAL CALCULUS. ~xdx 22. Integrate 1 Ans. tang-' e + C. X~dx 23. Integrate 1 x" Dividing the numerator by the denominator, we have fx- 4dx -x2 dx + dx- +dx X5 X3 -- 3 -- x- tallg x +- C. 24. Integrate - 6x dx 1/4 -9X4 We have 3x2 ff' I 3 xdx 2 - X J9J j r =, +C.= sin 1 3- C. 1 — 225. Integrate esec o sec2 x sin x dx. We have = esec x see x tang x dx -ese + C. 26. Integrate dx dVx bdx a We have bx b=__ x b sin- + C -b sina - C. -- ~ ~]1-( b EXAMPLES. 257 27. Integrate a2 dx a2+ 62 x2 bx We have - tang bx-+ C. aJb ( bX )2 1 a a bx 29. Integrate dx l/ 2xa _-bxZ We have 252 1~F dx 2b x 2= b: b versin-l + C. dx 30. Integrate x The denominator is equal to + x+ 1 ) d$ 4 dx f J3f(x+) f) {l42~}2 r d {, 2x _+ 1 2 ( (2x+1. - 2x+ 1 -22+ C D. C.-222. 258 THE INTEGRAL CALCULUS. 1/ ~ dx 31. Integrate 2 We have'Jr) 1- 2f dx0 -l- sin +C. 2 -25 1-2 v / ~1 —2x' - 3 v Z dd s 32. Integrate 2 4dx We have fJ= 6 =,2-(4x-~: — (4x)z =6 versin-1 (4x3) + C. _________ 1 33. Integrate V/2 — - Ans. I C -3 34. Integrate 1 d We have f= 2 x 2-(-X3 12dX 4 x -2 - 1 ) d 2 - an 1 X-3 1 - C. NOTE. -It will be observed that the last four examples are really the same under different forms. 35. Integrate bxdx as/- a 6.-bX RATIONAL FRACTIONS. 259 xn-' dx 36. Integrate /a - b2 x 37. Integrate dx x / 4x4- 3 38. Integrate- 4dx x V4x- xi 39. Integrate 3 2ax - X2}2 13. The foregoing examples will afford a sufficient illustration of direct integration. We pass now to the consideration of those cases in which the integration can not be directly effected, or in which it imay be more readily effected by the methods of substitution, or decomposition into parts, or by development into series. We shall find it convenient to classify functions as rational or irrational algebraic functions, and logarithmic, exponential, and trigonometric functions. CHAPTER III. INTEGRATION OF RATIONAL FRACTIONS. 14. Every rational differential function of a variable may be considered as composed of two parts, one of which is an entire fiuction, and the other is a rational fraction in which the degree of x in the numerator is less than in the denominator. The entire function can always be integrated immediately by the processes already explained. As for the fractional part, the method will be to decompose it into a series of partial fractions, each of which can be integrated separately. 260 THE INTEGRAL CALCULUS. The integration of rational fractions presents three cases: 1st. When the factors into which the denominator can be divided are real and unequal. 2d. When the factors of the denominator are real and equal. 3d. When the factors of the denominator are imaginary. 15. Case 1. —Factors of the denominator real and unequal. Let -F(x)dx be a rational fraction whose integral is to f (x) be determined, and let a, b, c,.... 1, be the values of x derived from the equation f(x) = 0; then will x - a, x -b,...~.x - 1, be the factors of f(x). Assume F() A-.A B C L f(x)- x- a - ~ x — (1), in which A, B, C, etc., are unknown constants whose values may be found by the method of indeterminate coefficients. We have then F (x) A dx B+ dx L dx dx — a ~ — x- 1 f(x) x-a x-b x-l each term of which may be easily integrated. It is obvious that if F(x).be rational, the above assumption is legitimate, since, from equation (1), we have F(x) A(x —b)(x —c).. (x-l)+B(x-a)(x-c).. (x —)+.. etc. fx) - f(x) and since this equation is entirely independent of the value of x, the values of A, B, C, etc., may be found by placing the coefficients of the like powers of x in the two numerators equal to each other. RATIONAL FRACTIONS. 261 16. From equation (1) we have F(x) Af (x) + Bf (x) + f(x) + etc.... (2). x —a x — b x — c If in (2) we make x — a, all the terms in the second member will reduce to zero, except the first, and we shall hlave F(a) - Af(a) x- a But when x - a, f(x) — f(a) - O, and x - a = 0; whence f(a) 0 fa x - a 0 =f'(, according to the methods established for finding the values 0 of functions which reduce to the form 0 Therefore we have F(a) -- Af'(a), and A- F(a) In the same manner we have F(b) - Bf'(b), and B -F(b) f'(b) F(c) F(c) - Cf'(c), and C -= (); and so on. This method of finding the values of A, B, C, etc., will usually be preferable to that of indeterminate coifficients. EXAMPLES. a dx 1. Integrate a dx x2+ bx a A B Assume + 2+ bx x + X- 262 THE INTEGRAL CALCULUS. Then a - (A +- B)x+ Ab x2 bx - x+ bx.. a =(A+B)x+Ab; A A+-B-=0, and B= -A; a a Ab- a, and A=-, whence B= a b b Substituting these values of A and B, we have a x a dx a a dx a x2+C-: b b- x b x-b -- log x -log(x + b)} b log x b{ x + 6, adx 2. Integrate -+ x by the second method. We have F(x) - a, f(x) -- X2 + bx; whence x - 0, or x ---- b.' A- F(a)_ a F(b) a A,(a) b-and B- (b) b *'. as in the preceding example, j b log x - log(x + b) + C log { x+ b17. Case II. —The factors of the denominator real and equal. RATIONAL FRACTIONS. 263 Let F(J) dx be a rational fraction whose denominator f(x) contains mn factors equal to x - a, and assume f(x) - (x - a)m,(x). If we take F(x) A + B f(x) (x - a) n (x - a)~-l L P + L _ + (1) x -- a,(X) the values of A, B,.. P, may be determined as by the first method in the preceding case, and the original fraction will be thus decomposed into a series of partial fractions, each of which is integrable by the ordinary methods. The values of A, B, C, etc., may, however, be determined more elegantly in the following manner: We have, from (1), F(x) A (x) + B(x -a)~(x) +... * P(x-a)m (2). Making x = a in this equation, we have F(a) =A p (a),.. A = F (a),' (a) Differentiating (2), and making x - a, we have F'(a) = At'(a) + BE(a), from which B may be found. Differentiating (2) m - 1 times in succession, and after each differentiation making x - a in the result, we introduce at each step a new constant whose value may be obtained at once in terms of known quantities. 264 T'HE INTEGRAL CALCULUS. The general equations from which the constants are to be found are the following: F(a) =A T(a), F'(a) -A'(a) +- B T(a), F"(a) - A q" (a) + 2Bp'(a) + 2. 1 CT(a), F"'(a) A q"'(a) + 3 B3p"(a) + 3. 2. CT'(a) + 3. 2. 1. D+(a), (3) Fm-l(a) = A m-1 (a) + (r - 1) B tm-2(a) + (n- I) (m - 2) CG,-3(a) -- *..v +- (m -1)(m -2)(m-3).. 3.2.1LT(a) The constants corresponding to the multiple factor x — a having been determined, we may operate in the same manner upon the fraction P, an; so on, until the original fraction has been decomposed into all the parts of which it is susceptible. NoTE.-The mode of decomposition adopted in this case is the only admissible one; for if we had pursued the method employed in Case I, the number of independent equations obtained would have been less than the number of constants, A, B, etc., to be determined. EXAMPLES. 1. Integrate — 4x- dx. We have x3. - 6x2+ 9x F(x) = x2 — 4x + 3; f(x) — x 6x2+ 9x = (x- 3)2 x. x2 — 4x + 3 A B C Assuming -- 6+ 9- (x-3) x X1 —6x2+ 9x (x-3Y2 x-3 + x RATIONAL FRACTIONS. 265 reducing to a common denominator, and placing the numerators equal to each other, we have X2- 4x + 3 =Ax + Bx2 - 3Bx + Cx2 — 6Cx + 9C..B. B+C —; A -3B- 6C — 4; 9C=3. 1 2 C= -; B-= -; AO0. 2 1 - 3 log (x - 3) + 3 log x + log c. = log ex (x - 3)2 i. 2. The same by the second method. We have F(x) = x- 4x- +3; f(x) (x -- 3)2x. a=3; t (x)-=x; F(a)= 0; +(a) 3.'' A= F(a); B F(a) - A(a) 2 A (a) (a) 3 Substituting these values of A and B in formula (2), and reducing, we have P- F(x) _ 2 1 * f(x) - 3(x-) and, as in the previous case, f=-34fj + - 3x- = logex(x - 3)2f3 18. Case III. -The factors of the denominator imaginary. When the denominator of the proposed fraction contains imaginary factors, the method of decomposition may be entirely similar to the foregoing. I). C. —23. 266 THE INTEGRAL CALCULUS. Since, however, for every factor of the form x - a - b I/-1, there must be another factor of the form x a + be/'-1 (otherwise the product of the factors would not be real), it follows that for every partial fraction of the form M X1/ —i x - a - b V/there will also be one of the form M + N1/- 1 x - a + b / — If we add these two fractions together, the imaginary terms will disappear, and we shall have their sum, 21M(x -a) + 2bN (x -a)2+ b2 which may be written Ax +B (x -_ a)2 + b2' and this will be the form of the partial fraction corresponding to the two imaginary factors x - a F b 1/- 1, or to the single equivalent quadratic factor (x - a)2 + b2. To integrate a)+ b dx, assume (x -a)2 + b2 2~62 (d-a)2+ dx. RATIONAL FRACTIONS. 267 The integral of the first term is evidently equal to 1 A log s (x - a)' + bY2 and that of the second term is easily seen to be equal to Aa+ B x- a b tang-1 b 18'. If the denominator should contain n equal quadratic factors, then we should have, by Case II, F(x) Ax + B Cx + D f(x) -- (x - a)2+ b2'"n + (x-a)+ b2 Lx + M (x_ a) b' and the general term to be integrated in such cases is Ax +B dx. -(x a)2-+- +b2n To integrate this expression, assume } A (x x a)'+bP+ dx d. J - } — a)' +b23l J }(x a) +b23 We easily find the integral of the first term to be A 2(n'- 1) (x - a)2 + b' n-'' and we have now to find that of the second term. Let x - a bz, and the second term may be written Aa+B dz b2n-l (1 + z)n; so that the operation is reduced to finding the integral of this last expression. 268 THE INTEGRAL CALCULUS. Now we have 1 + z2 — z2 1+ -z z2 (1 - z (1 + Z2)) (1 2)- " (1 - + z2)" 1 z2 - (1 + z2)n"- (1 -+ z2)n r dz r dz r z2dz J (1 + z)n -J (1 + z2)'- - 1 J (1+ z) The last term of this equation may be integrated by the formula for integration by parts, viz: udv = v -fvdu. zdz Let z - u and - dv. (1 + z)' Then v -- 2(n ) (1 +)- and duz - dz.. __lzdz _z 2( - ) I +- ( f)"- ( +z _r rd dz + 2 (n-1)J Z(1 2)21-l; rnd d z.J'(1 + Z2) -2(n -1)(1 + z2)'7l + { 2 (t - 1) J}( +z2)n-i We have thus reduced the integration of the original expression to that of (1, in which the exponent of the denominator is one less than the original exponent. EXAMPLES. 269 The continued application of.the formula for integration by parts, reducing the exponent of the denominator by unity, at' each step, will finally bring us to the form f z2 tang z tang-' x - a C; and thus the original fraction is completely integrated. 19. EXAMPLES OF RATIONAL FRACTIONS. 1. Integrate J 6x; + 1ix-6 dx. Placing the denominator equal to zero, we find the values of x to be 1, 2, and 3, and therefore the factors of the denominator are x-1, x-2, x-3. Assume x — 7x+1- F(x) A B C x-6x2+ 11x — 6 = f(x) - 1 - -2B + x2 -3 Then A F (1) _ 1 — 7 1 5 f'(1) - 3(1)2 12(1) + 11 2; B= F(2) 9 f'(2) C F(3) __ 11 f'(3) 2 * 1 5 dx + 9 j 2 1 x 5 11 -- -52- log(x —1) 9 log(x-2) - y log(x — 3) +C log c (x - 2)19 m //,, 1)5 (.,- 3)1_ 270 THE INTEGRAL CALCULUS. 2. Integrate (x 3) (x - 1) dx. (x +- 3) (x + 1)2 We have F(x) = 2x - 5; f(x) -= (x + 1)2(x + 3);.'. by Case II, a —— 1; n- 2; p(x)= x + 3. Assume 2x -- 5 A B P (X+-3)(x + 1)2 (x+1) + + + + 3 (). Then F (a) A. (a),. A —; 2' i1 F'(a) =-Aq'(a) + B (a).. B If we suppose a - 3, and $(x) = (x + 1)2, we shall find A = - 4, and this will evidently be the value of P in (1). 7rdx +11 r dx 7 1 1 1 x -~ - 1 7 2 +- 1 + log log C. 5x — 2 3. Integrate x:~ 6x_4 8x dx. The factors of the denominator are x, x + 2, x + 4. 5x- 2 A B C X3+ 6x2_ 8x x +x 2 x 4 Ans. f=- log {(x+-2)1 } 4 x (x 4]n ~ EXAMPLES. 271 xdx 4. Integrate (x ~ 2) (x +- 3)2 Ans. fs —- +3 log{ (x+23) X2q- 3x - 2 5. Integrate 2+ 3x +1 dx. x3-+ X -x-2x An log (X1)5 Ans~. - 3 log x 1/x2+ 2x x2 dx 6. Integrate (x + 2)2 (x - 4)2 We have F(x) - x2; f(x) = (x + 2)2 (x + 4)2; a —2; n-2; q (x)=(x+4)2. 5x+ 12 (x+4) —. j- x2+ 6x +- 8x + log + 42 2x dx 7. Integrate 1) (x This will be integrated by Case III, since the real factors of the denominator are quadratic. 2x Ax +- B Cx + D Assuming -(xq+ 1) (x2+ 3)- x2+- 1 (x2 + 3) we find A = 1, B - O, C — 1, D 0. *' f x 4 xd1 m x3 =- log c X2+ 1 x28. Integrate dx 8. Integrate (X2+ 1) (X2+ 4) 272 THE INTEGRAL CALCULUS. WVe shall findf= 3 4 - 1J x + 2 r _2-) 1 dx 3J( x) 3 l+ x { 2 tang-' - tang x} + C. 9. Integrate x dx (x + 1) (x + 2) (x2+ 1) Assuming _ x A B Cx+D ( +1)(x +2)(x'2+1)= X+1+ x+2+ 2+1 1 dx 2 dx we find — 2$+1+ 5 2 1 xdx 3 rdx + l-o-J x2+ 1 lo x2+ 2 - log(x + 1) + 2 log(x + 2) +123 + 20 log($+ 1)+ 10 tang-1 x + C. x dx 10. Integrate (x + 1)(x + 2)(X2+ 3) 1(x + 2)8 V tanl x An.f - log (x+l)/x 3 3 3tang- } +C. dx 11. Integrate x3 1 Assume 1 A Bx+C Assume x3- I 1 + -1 — x 1+ x-+ x + 1 EXAMPLES. 273 1 1 2.'. A -; B_ —-- C=- 3; and 1 3 _ dx 1_ x-2 dx. 3x1 3 Jx x The Integral of the first term is - log(x - 1). To integrate the second term assume ~31 3 i 3 x + Z dx - dz, and x + 2 - z + z + dzz _+2_ ___ 1 zdz 2dz 3 2+ 3 3 2 + 3 2+ 3 1f 2zdz 1 d3 3+ -3 1 ++ 4Z2 1 log 3 1 2z =z 4- } 1o — 3 tang — z + C. - 6 4. the whole integral 1 log(x-1)- 6 log 1 tan (g2x +- 1. ) 274 THE INTEGRAL CALCULUS. dx 12. Integrate 1 — dx The factors of the denominator are 1+ x, 1-x x + x2, 1-x + x2. assuming 1 i~ A B Cx~+- D Ex + F — 2 l+ x + 1 —x + +1X- x - 1- xx we find A B 6 C -; 6 6 6 1 1 D — -F; E —-- * =~S~t~ TJ- ~-I~ t_ L.+ ++-S+x dx dx - x-2 d - 6J 1 + X 1 log(i- 1 + + 2, dx + 1- 3 dx.r 2x1 1 3dx ZZJdx+ x + 12J 13+ 11. 1~ X. J2 12s1+x~ - log6 1-X + + log { +x + x} 2 __ f 2 r -1 13 3d 2V- + 1 -x+ 2 2 2 d + / 21/3j+ )2 1 4( EXAMPLES. 275 = log 12 +2x+ x2) (1 + x- + XI) (1g - 2x + x2) (1 - x + x2) + 2x1/ 3 t ang) ) ( 2 1tang )} + C x4 dx 13. Integrate 2 - 3x' AnC. S= x3 2x 2 2- tang-l 3 + Ans.j -W -qW g 2 14. Integrate 1 X - x 2 dx. I+- x + x2[ xP Ans. -- 2 log + -tagx }+C. 15. Integrate 4+ dx x 6x + 1) + 9 3 ns. 6(xl+ 1) - 2x +3 1 x —1 + tang' + (. 181/2tang- /2 16. Integrate 1-X dx Ans. - log( xx i 2 ag1 x 1/2 + +u- 1/ tang-' 1 + C. 2_1/2_ X2 276 THE INTEGRAL CALCULUS. CHAPTER IV. INTEGRATION OF IRRATIONAL FUNCTIONS. 20. The number of irrational functions whose integrals have been determined in finite terms, is comparatively small, and in all cases the object has been effected by transforming the given irrational function into an equivalent rational function of another variable. The case most frequently presented may be expressed, in general terms, as follows: Fix, 1/a - bx -+ x dx; and the object is to rationalize the radical part of this expression. Case I. -(a.) If the term containing x2 has the positive sign, assume V/a +- bxx z +x (1). Z2 a Then, a + bx -- z-+ 2zx; x -2z dx -= - 2 (z2- b z + a) (b - 2) dz /a+ b +x2- Z2-bz + a /a q- bx qX- 2 x z — These values, substituted in the original expression, will reduce it to a rational function of z, of the same character with F. We will then integrate with respect to z, and in the result substitute for z its value derived from (1), viz: z — /a + bx + x2- x; IRRATIONAL FUNCTIONS. 277 which operation will evidently give us the integral in terms of x. (b/.) If a is not negative, assume Va + bx + x=- /a + xz; whence b + x = 2z v a + xz2; x _ 2z- " b; z2 v a-bz + dza dx -2( dz) dx=2~ (1 Z2)2 r/ a + bx + x2 z/ab~ + a The substitution of these values in the original function will render it rational in terms of z. (c.) If the expression a + bx + x2 can be decomposed into two binomial factors, x - a, x - i3, assume V a + bx + x2= (x - a)z; whence a + bx + X2-= (x- a) (x — i) - (x- a)2z2; dx ( 2a) 2; (- ab; d (12z2 ) dz; (-/ a a)z bz + -- _ Z These values, substituted in the original function, will render it rational in terms of z. Case II. —If x2 be negative, we may make use of the second or third of the preceding transformations. The first can not be used in this case, since it will not rationalize the given radical,- nor can the second be used if a be negative. 278 THE INTEGRAL CALCULUS. 21. If an irrational function contains two radicals of the forms I/a + x, v/b + x, it may be easily rationalized thus: Let j/a + x = z; whence x=- z- a, dx=2zdz; 1/b+x 1/b+ z2 — a. This reduces the two radicals to a single one involving z, to which the methods of the last article may be applied. 22. When the only irrational quantities contained in the given expression are of the forms If(x) -, If(x) q, if(x) s etc., they may be rationalized by assuming f(x) = znqs; whence if(x)}= zm8; If(x)l - z ps; etc.; and the given expression is thus rendered rational in terms of z. 23. APPLICATIONS. 1. Integrate dx la + bx + x2 Assume V/a +- bx + x = z - x. z2 a Then a + bx = z2- 2xz; x-b 2z; dx = 2(z -x)dz dx dx _ 2dz b+2z' z x V a bx- x b +2z 2dz dz b ) fo f =f-nb -log{ + Z log - + x +1/a -+ bx +- x } + C. IRRATIONAL FUNCTIONS. 279 If b — 0 we have r-dx -log x +V/a + x + C; a ~ x2 and if a=- 1, then dx -- logIx + l /l +XI +~ C. These two last integrals are of very frequent occurrence. 2. Integrate bx / a + bx x2 Assume V/a — bx - x —- V/a+ xz. Then b - x = 2z /a+ xz2; dx _ dx 2dz V/a + xz /a + bx-x2 1 + Z2' 2J d1 2 tang-z — 2 tang - { + } + dx 3. Integrate a + bx-x2 under the supposition that the denominator contains two real binomial factors. Assume V/a + bx - x2 - (x - a) z, and X2 — bx - a = (x - a) (x - ). Then -x = (x -a)z2; dx(1 + Z2) - 2(x -- a)z dz; dx 2dz (x - a) z + z2 280 THE INTEGRAL CALCULUS. f* ==.-_ 2f 2 = >- 2tang- z - 2 tang-' f - + C; Z -- a and, substituting for P —x, x — a, their values found by solving the equation x2 —bx -- a =O, we have - b 2 - -- 2 tangIb - 2x + vb2+4a +C also~ a~~bx-X2x- b -- Vb-] 4a. If b -= 0, we have -i /a 2 -- 2 tang-' 1 a + C;x and if a = 1, then dx - 2 tang-' C. 4. Integrate ax + x-bxdx. V a ~ bx -6x dX. Assume aX + 3 a(x- ) + VIa + bx - x a + bx - x2 a + bx — x' Then we shall have - bd a d + +~dx J — x — + bx -- x 2 The integral of the first term is — a i/ a+ bx-x, and that'of the second term is given in the previous examples. IRRATIONAL FUNCTIONS. 281 dx 5. Integrate x bx I dz Assume x - Then dx =- z, and dx dz xV/a+b+ x z a+ z 1 dz Vaz" + bz -- 1 the integral of which can be obtained as in the previous examples. dx 6. Integrate x2Va + bx___ x2 1 dz Assuming x --- we have dx - dz z dz r zdz b -- 12 J az2+ bz 1 / b b _1 az+ 2- 2 a 1 az2+ bz __+_ _ 1 az + - d+ - ~12 dz dz 1a az + bz 1 2 -2a- V -z z 1 2 b dz ta last term aa zh- bz -1- a the last term of which is integrable as in the preceding examples. 1). C.-24. 282 THE INTEGRAL CALCULUS. BINOMIAL DIFFERENTIALS. 24. The remaining irrational forms which admit of integration in finite terms belong, for the most part, to the class of binomial differentials, which may be written xm(a + bxn)' dx. In this expression m and n are either whole numbers, or they may be made integrals by the methods of Art. 22, so that the only necessary fractional exponent is p. [1]. To integrate this expression, assume (a + bx'")P - (); then x { z }dx lb {Z b > adz (2); m {z a }~ n(3) ~-=( - -- ab Multiplying (1), (2), and (3), together, we have 1 z-a am+l xm(a + bxn)Pdx = b Z b dz (4). If m +1 be a whole number, the only irrational quantity in the second member of equation (4) is zp, and this may be r rationalized by Art. 22. Thus if p- - we may assume q z- tq, which is the same thing as taking a + bx' - t; and by pursuing the method above illustrated, we may easily rationalize and integrate the given expression in terms of t. [2]. The given expression may be written x4 (a 4- bx') P dx - x'm + (ax-' + b) 1 dl.. BINOMIAL DIFFERENTIALS. 283 If we put ax-"+ b = z, then, according to the preceding case, the expression can be integrated whenever m q+ unp + _- m - 1- + p is a whole number. - n II Hence there are two cases in which a binomial differential can be rationalized. 1st. When the exponent of the variable without the parenthesis, increased by unity, is exactly divisible by the exponent of the variable within the parenthesis. 2d. When the fraction formed in the preceding case, increased by the exponent of the parenthesis, is a whole number. These two rules are called the conditions of integrability of a binomial differential, and when either of them is fulfilled, the integration may be effected according to the general method above indicated. We give an example under each case by way of illustration. 1. Integrate x3(1 + x2)2 dx. Here, - + 1 4 -2, a whole number, and therefore 71 2 this comes under the first case. Let 1 + x'- z2 Then X2- z2- 1Z X (Z2 - 1)2, x dx - (Z2- 1) dz (1); also (1 + x2) - z (2). Multiplying (1) and (2) together, we have x (1 +- x2) dx ( — 1) Z2 dZ - zd Z -Z z2 dz. 284 THE INTEGRAL CALCULUS. * 1 1 1 1 1- (15 + X2)2 (1 + X2)2+ C. 2. Integrate x-4 (1 + X2) 2dx. Here, + -4 +1 1 -2, a whole ne~ 2 2 — number, and therefore the second method is applicable. Put x-4(1 + X2)-2 - -5(X —2+ 1)- 2; and let X-2+ 1 - Z2. Then -3 dx -- z dz (1), x —2 z2 1 (2); also (x-2 + 1)- (3). Multiplying (1), (2), and (3), together, we have 1 x-5 (X-2+ 1)-2 dx - (Z2 1) dz. z3 Z2 - /s \ a2 +3x" - ) 3C. FORMULAS OF REDUCTION. 25. Whenever a binomial differential can be rationalized, it may be integrated as in the last article; but usually the methods there set forth give rise to such complicated results that it is, as a general rule, more elegant to integrate by the process of systematic reduction, the formulas for which we propose now to establish. FORMULAS OF REDUCTION. 285 26. Problem. -- To integrate xm(a + bx")P dx by reducing the exponent m of the variable without the parenthesis. Let xm(a + bx") P dx = udv -uv -fvdu; and assume Z - x-+lfl; dv x"-l (a -- bx") p dx. Then du = (m - n + 1) xm-." dx; v —nb (p +-1) (a + bx") P+I. xm (a + bx?) P dx xm-n+l(a + bxn) P''l -nn( b])+- dx (1). nb (p ~ 1) ~) nh(p+ This formula diminishes the exponent m to n — in, but increases p, which is generally objectionable. The last term of (1) must therefore be converted into an expression in which the exponent p shall not be increased. Now we have (a + bx"l)p+l- (a + bx') (a+ bx"') P= a(a+ bx'") p+ bx"(a+bx")P..'. xm-n(a + bx") +1-~ axm-L(a + bxn")p + bx'.(a + bx')'l Substituting in (1), we have fxm (a + bxz-) dx Xmn-ltl(a + bxe)P+1 a(rn - n-1) C a b.)d m( -nlxm(a~~ be")v dx, from which we obtain, by transposition and reduction, fxm (a + bx") dx xm-41'l(a — + bx)P+- a(m -n - +1)fxm (a+ b)p dx -~- b~ip- ii n+1i ) [A]. 286 THE INTEGRAL CALCULUS. This formula reduces the exponent of xm to m - n. It may be used whenever m is positive and greater than n, and its continued application will finally reduce the expression to be integrated to a form which can be more readily manipulated than the original. 27. Problem.- To obtain a formula for increasing m;-to be used when m is negative. We have from [A], by clearing of fractions, transposition and reduction, fxm-t (a + bxe)P dx xm-n+l (a + bxL)P+ — b(np + m + 1) xm(a + bx9)P dx a(m -n +) 1)(2). If in this formula we write - m for m - n, and - m + n for m, we shall have x7m (a t+ bxt P dx_ x-m+(a + bXn)P+l b(m-np-n- 1)_ x~-m++(a + bxl)P dx [B]. a a(m - 1) 28. Formulas [A] and [B] are to be used whenever the value of p is the same as that of the fractional exponent in some one of the fundamental forms; but as in a given example p may be any fraction whatever, it is necessary to establish formulas for changing the value of p also. 29. Problem.- To integrate xm (a + bxn)P dx by diminishing the value of p. Let (a + bx) = u; x dx = dv. Xm-l Then du -- nbp(a +- bx') P-1 x"-' dx; v m& 1; FORMULAS OF REDUCTION. 287 and substituting in the formula for integration by parts, we have fxm (a + bxe) dx ~M+1 (a + bx-) nbp xm-l(a +bxn)- mn~bpjfxm+(a + bx)"-' dx (3). Now, applying [A] to the last term of this equation, substituting m + n for m, and p - 1 for p, we have Jm+" (a + bxn) P- dx x"ml (a + bx)r- a (m + 1)xm (a + bxn) P- dx b(np+m+ l) and by substitution in (3) we have xm (a + bx) P dx X+l (a + bxn) p + anpifxn (a + bxL)p-l dx n p + P + 1[C]. 30. Problem. —To obtain a formula for increasing p;-to be used when p is negative. We have from [C], by transposition and reduction, Sxm(a + bxm) -1 dx -m(a + bx")+ (na + tp + 1)xm(a + bxn) dx (4). If in (4) we write — p for p 1, or — p 4-1 for p, we shall have the required formula, fxm (a + bx>)-P dx xm+I(a+bxn)-P+'-(mnnp+n+ 1 )fx(a+ bxn)-pldx a'n(p-l) [D]. 288 THE INTEGRAL CALCULUS. 31. EXAMPLES. Xm dx dx. 1. Integrate 1 = -xM Applying formula [A], making m = in, a 1, b -1, 1 It:2, p —2 2 we have _dx_ -- xm-/V1 — x n- 1 xm-2 dx x 1/-r-x1X + - m J1 i1- (a). If m be even, the continued application of this formula will reduce the integral to the form f1'I x2= -. sin-l x + C; and if m be odd, the final expression to be integrated will be -= 1 x2+C. 1st. When m is even. Put m- 0, 2, 4, 6, etc., in succession. Then, from (a), we shall have d —x =sinl- x + C. __xx1- 2 ~ d r X - xl/lX2 1 _ dX J}/1 US 2 +212/1 x"'' d 4dX -X3 1/1- x+ 3 r x'2dX T/1 —x 4 I+ /1 — x' e ~dx - X5IV1- XS 5rX4dX fJ1/1i-2 4 + X fi X 6 ~1x _ EXAMPLES. 289 Hence, by reduction, -2 x/1- X+ 2 sin-I x + C; 1 3 r x~dx X3 1 3 + - - sin- x + C; J { 1 xI+ 1 5 13 5} V 1/1_~r —6 4 2 4 6 sm-x;X2 1+ 1 3- m-1r x 1 n-3 m- 1 -5 +m-4 -n-2 mm1 3 5 m —1''.+2 4 6" -in {1 3 5 m —3 m-1 } n-Ix C If we take the value of this integral between the limits x =0 and x =1 [Art. 5], then, since for x 0, f-0, and for x =- 1, sin7- x = I-, we shall have r xdx_ 1 3 5 m 3 m- 1 / 1 — X- 2 4 6" m- 2 m 2 D. C.-25. 290 THE INTEGRAL CALCULUS. 2d. If m be an odd number, we shall have f l__ -- 1 — x2~+ c; x I/__ 61 142/x dx ~ 1/1 —~~ — 3 1-x'; r. ~dx 4 +- 4 r x3dx J/ 2 —-- 3-/ +'5'- -/'-J 1i-; rsdx el I6/1 x' 6 r x~dx J/1 - -X 7 + 7J V1-x; Hence, by reduction, dJ 1/1 lx _ — x + C;,/V'l-' —---— ~ —-- -3 - 2+ I 3C; + I 1 V 1 x2+CC; + 3 5 J; {dx 1 6 1 4 6 -TFi 4+ 6 x2 1 2 4 6 }1 X2+C; ~g~~~ ~~~~ ~~ g~ ~ ~- gg ~~ ~ -;4 52 EXAMPLES. 291 r xmdx _ 1 + 1 n1 m-3 m m — 2 mn 1 mn-3 m-xM-5 - - m V /1 —x -+ C. + m m4 W — 2 mt 1 2 4 n\+ + 1 3 5 m If the integral be taken between the limits x-O and x- 1, we shall have xmdx 1 2 4 6 m-1 i/1/ -x 1 3 5 7. m 3d. If, finally, m be taken equal to infinity, the values of the definite integrals in the last two cases will be equal to each other, and we shall have 1 3 5 7 1 2 4 6 8 2 2 4 68 i 3 5 7 - 2. 2.4.4.6.6.8.8... 2 -1 3.3.. 5. 5. 7. 7. 9' which is Wallis's theorem for finding the ratio of the circumference of a circle to its diameter. 2. Integrate d 2 2 xm' = x-~(1 -- x) 2 dx. This example will be integrated by formula [B]. We have f- m -1 )-1 Proceeding as in the last example, we shall reduce the 1 integration to that of (1 - X2)2 dx sinl- x + C, if m be even, and to that of x-1(1 — X2) dx, if mn be odd. 292 THE INTEGRAL CALCULUS. To obtain the integral of this expression, assume 1/1 - x = 1 + xz; whence x -2z - xz2; dx dx 2 dz dx dz 1 +xz - -'l' 1 +z' x /l —x'2 X-L (1- x2) 2 dx - log z + C =log / + x3 dx 1 3. Integrate I' 2 ax - x2) 2dX_ x (2 a-X) 2 dx. Formula [A] applied to the last form of this expression gives r x dxm- I/2ax -x /2 ax - x2 xn a (2 -Idx + a-(2n- i1 ))f In / 1/ 2 ax x' Making m 1, 2, 3, 4, 5, etc., in succession, we have - = 1/ 2 ax -x a f xdx _- /2axx. +afV2 dx l/2 ax - x' 2 ax - x' x2dx 3x. 3 r xdx 22;" 2 2 1,2ax ax x + J r dx = s 2a 12a-2+ 3aJ12ax-xx i EXAMPLES. 293 Jf1/2axtx =- - 3 f 2 2ax-x + az fxJ dx = -- -2ax x2+ - affd 5 3 a dx + 4 3a2aj/2 ax x2 I,2x_~ —~-_ -, x+~ aj{/~axd 1x4 x 9 x3 97 x2 2 975 x a =-'-5- 4a- J-5 4'3a - -a-3 753 a4)l/2 ax — x 9 7 5 7 5 3dx + 5 4 3 2 _X1/2ax — x2 x- 2m1 x-2 2m- 1 -3 xm-2 2 + a+- a- a9 7 5 3 2 ax _ x 2m 1 2m3 2m 1 2-5 3 I versin~ x m m-1 m-2'2 I + 3aM versin-+-+ C. rn m m1 m 2 2 1 a 294 THE INTEGRAL CALCULUS. If we take the integral between the limits x O and 2a x _ 2a, we shall have versin-' 7r, and therefore, a f2a xindx am. 3... 2m — 5. 2m- 3. 2in 1 /2 ax-x 2 1. 2.... m- 2. -- -1. rn 4. Integrate (1- 2x2)2 dx. Let m =O, a=l, b= —2, n-2, p- Then we shall have, by applying formula [C], x 3(1 —2x') —~ 4f( (1 — 2x 2)2+ I-( - 2x2)2 dx. Applying formula [C] to the last term, we have 4 (1- 2x) 4 ( x - x 2x)2+ 2' ('3 4, 3 ( 21 x2)2 dx - (1 2x2)~2 x (1 - x2t~z~f(1 —i2j<) 3 1 3 ~dxr1 + 8 x (1 2x2)- -,] 22 8 3 - 4 (1 — 22)2- 8 x (1 - 2x2)2 3 3 + 8 12sin-xl2-2+C. 5. Integrate dx x-l (a + bx) 2 dXo x (a + bx)2 3 Let m=- l, n=-l, - --- o EXAMPLES. 295 Then we shall have, from formula [D], f_ ( Ib2 1+ a —fx-'(a+ bx) 2dx. a (a + bx)2 To integrate the last term, assume a + bx -- 2; 2z dz b _ then dx 2; x- -_; (a +bx) 2 _ Z f. *. xJ - d(a + b ~log { 1 d z- 1/} aa { r1 (/a+bx' — l/a x (a + bx) a (a + bx) a- 1oa g { g a a+ bx +lj/ + C. dx 6. Integrate 5 x (1 + 2x)T Ans. 2 + 3x 4 + log { 1/1 + 2x- 1 (1+2) 2 12 +2x+1J 7. Integrate xi xdx 1 + x2 Ans. 2 V S1 ) {log( 1/+1 -2X) - tang_- 2-/X } + C. The last two examples may be integrated without the use of the reduction formulas. 296 THE INTEGRAL CALCULUS. CHAPTER V. LOGARITHMIC, EXPONENTIAL, AND CIRCULAR FUNCTIONS. 32. There have been no general methods yet established for integrating, in finite terms, all possible forms in which these functions may be presented; but there are many cases in which the integration can be effected, either by reducing the given transcendental expressions to equivalent algebraic forms, or by reducing them to other expressions of the same kind, but of simpler form. 33. A large number of expressions coming under the general form P? dx, in which z is transcendental, may be integrated in the following manner: Let fP dQx _ Q; fd -; d- S; etc. Then the formula for integration by parts will give zn dx Qz= n 9- QZ-1 dx dx dx _ R- dx (jQZ'_i dz dx — -- (- d d Rz-2 dx _ —(n -- 2) s z-3 dzx. dx. Hence, by substitution, Pz, dx - QzL - neRz + n - (n -1) Szn-2 - it ( t - 1 (it - 2) Tz"-3 _ + [EE] FORMULAS OF REDUCTION. 297 APPLICATIONS OF FORMULA [E]. 1st. Let P- 1, and z- log x. x dx Then Q jdx-x; R=Jx X d-x; S-=x, etc. f log x dx -- x log" x - n log"-' x + in ( — l) logn-2 x.. +n(n — 1)...3.2. 1}+C [F]. 2d. Let P - 1, and z - sin-' x. Then we shall find Q-x; R —- 1 -x2; S —x; etc.. J(sin- x)" dx - x (sin-' x)I+ n 1/ - x -x2(sin- x)"-' - n (n — 1) (x) (sin-l x)nt-n (?t —1)(- 2)1/1-x2 (sin -lx)"-+... [G]. 3d. Let P = xm-', and z = log x. Then Q = Xm-l dx - I-xm; R- -4-f x'- dx - Xm, etc.. fxm - log" x dx m log -- -- logX1 x n (n logn2 X1 =.. +n()(n ( —2).+ [H 1. m. n(i1t -)(it -2). 3. 2. 1 \ 4_-, 298 T'HE INTEGRAL CALCUL US. 4th. If in the first formula [F] we put log x- z, whence x C e, dx - e dz, we shall have ezdz d- ezzn- nz+ n (n - 1) z-2... +t n(n —1)(n -2).. 3. 2. 1 + C [I]. 5th. Making the same substitution in [H], we have xm-l dx - e(m-l)z eZ dz = emz dz..'. Zemz dz e- { z n + n(n-1)Z mt?in m - n(n-l) __ 3.2.+ C [K]. 6th. In formula [G] we have z - sin-' x; whence x - sin z, dx -- cos z dz, and 1/x1 - x - cos z. Substituting these values in [G], we have Jz Cos z dz - sin zzn- n (n 1) z-2 } + cos z { nz —-. (n — 1)(n —2)z3+... + C [L]. 7th. If in formula [F] we make x = a, whence dx - a- log a dz, log x - z log a, we shall have, after dividing through by logn+1 a, azzitz az - { 2 nz- n(n- 1) z-2 fa —d-y~ log a log' a. -1) (...3. 2.1 }+C [3.]. logThis formula" embraces [I] as a particular case. This formula embraces [I)] as a particular case. FORMULAS OF REDUCTION. 299 34. When n is negative, the above method no longer gives converging results, and the integral must then be determined by other means. The following examples will serve for illustration: 1st. dx og1 st x dx Put xm+ -u, A- dv; then du = (m + 1)x dx, and v (n1)o (n —1) logn-' x Substituting in the formula for integration by parts, we have xm dx nm +1?n+1 ( xmdx (n -1) log', x - 1 lor x"md+ x nmfl - n+~1 r xmdx log-_1 x (n — 2) logn -x ~ n -2 log2 - x; xm dx xm+l n+1' xm dx J log"-2 x - - (n -3) logn-x + -3J logn x; cc; i ~~ CC cc c cc 99 c cc cc r(xmdx xm+l f m+ 1 1 J logn x n1-1 og-' n-2 log-2x (m — 1)2 1 (n- 2) (n - 3) log-3x... (mn - 1)n- 2 1 (n- 2) (n- 3)... 3.2. 1 log x (m + 1)1- r ax (n 1)(n-2)(n- 33)...32.1J logx N]. 300 THE INTEGRAL CALCULUS. 2d. fa dx If in [N] we put x = a, whence dx a" log a du, and log x -u log a, we shall have r a(mfl)u d a(m+l)u f 1 J log"-a - n -1 log- a (n 2)u"-` log"-2 a } (m + 1)n- ( du + (n )(t -2)(?- 3) 3... 2. i. - [']. Making m-n 0, and writing x for it, this formula becomes, after multiplying through by log"-' a, J aX dx a x log a J (it - 1)x'* -2 x2 log a x"-2 log"10 a - (n — 2) (n - 3) + (nq - 2)(- - 3)... 3. 2. 1 +109og'a ax [0o]. ( 1)(i 2)...3.2. 1 x Finally, making a - e, whence log a - log e 1, we have S dx 1e' 1+~ x J ~,, -- -- (n - 1) xt-1 1+n_2 X2 Xn-2 - (n - 2)(n — 3) + 2 + ( — 2)(-3)... 3. 2. 1} + ~ 1 ex dx o] (n - 1) (m -2)... 3.2. 5 x The values of the last terms of the last four formulas can not be found except by approximation. FORMULAS OF REDUCTION. 301 35. Trigonometric or Circular Forms. — The principal form to be integrated is sin"m x cosy x dx, and it presents four cases for consideration: 1st, to reduce Vn; 2d, to reduce n; 3d, to increase rn; 4th, to increase n. Case I.-To reduce m. Let sinm-l x = u, cos" x sin x dx = dv; then cos"+' x du =- (- 1) sinm-2 X cos X dx, and v - + - Substituting these values in the formula for integration by parts, we obtain rj~. s'.~ sine- 1x cosn+l x in os x dx sin x cos d — - + m-1 f inm2 x cos"l+2 x dx (1). Now, Jsinm-2 x cos+2 x dx — sinm-2 cos x cos2 x dx -sinm-2 x cose x (1- sin2 x) dx =Jsinm-2 x cos x dx — sinm x cos x dx. Therefore, by substitution in (1), and reduction, sin — l X cosl xxc sinm x cos+ x dx- - + X -lnf m2sin2 xcos x dx [P]. Case II.-To reduce n. If in [P] we write 90 — x for x, we shall have 302 THE INTEGRAL CAL&CULUS. fsinr(9o — x) cosn(90 — x) dx sin- (90 — x) cos"+' (90 — x) m +- n + m+-lnfsin-2 (90go X) coS" (90~ — x) dx' or, observing that sin (90~ — x) cos x, cos (90 — x) = sin x, and changing m to n, we have sinm x cos" x dx - sinrs x cos n-L + 1 fsinm x cos"-2 x dx [Q]. The continued application of these two formulas will reduce the exponents m and n as near as possible to zero, and if they are both whole numbers, the final term to be integrated will be presented under one of the following forms, viz: d, fcos x dx, sin x dx, sin x cos x dx. The only exception is when m + n — 0, which will receive attention hereafter. Case III.-To increase m. If in formula [P] we put - m for mn, we shall have fsin-, cosn x dx - sin..1 x cos'+1 x — (m + 1)fsinmn2 x cos, x dx -z +~ n. in-m-2 x cos" x dx - sin-c"l' x cosnl x + (m - n)Jsin-m cos" x dx m + 1 FORMULAS OF RED UCTION. 303 If in this equation we write m- 2 for m, it becomes sin-r x cos x dx - sin-m+l x cos+1 x + (m -n- 2)sin-+2 x cosn x dx m - I[R] This formula is not applicable when m = 1, but in that case we may, by diminishing n, reduce the final integral to Ccosxdx r dx n, or sin x.sin x Case IV.-To increase n. If in formula [Q] we write - n for n, we shall have sinm x cos-n x dx sinm+l x cos'-1L x -(n + ) sinm x cosn-2 x dx m -- n sinm x cos —2x dx sinm+l x cos7-1 x - (m - n) inm x cos-n x dx If in this equation we write n - 2 for n, it becomes fsinm x cos- x dx sinm+l X COS-n+l x -- (n — n + 2fsinm x cos-+2 x dx n - [IS]. This formula is not applicable when n 1, but in that case we may, by diminishing m, reduce the final term to be integrated to rsin x dxr dx Cos x cos X 304 THE INTEGRAL CALCULUS. 36. Problem.-To integrate sin-m x cos- x dx. If in formula [P] we make m and n negative, we have fsin- x cos-, x dx sin-"zx cos-n+ x + (n + sin —-' x cos- x x dx la3 + qz fsin-m-2x cos-" x dx — sin-m-ix cos~l+'+ (mn + n)j sin x-'cos-xdx m + 1 If in this equation we write in - 2 for in, it becomes fsin-"n x cos-z x dx - siln-+' x cos-"+' x + (n + n — 2)sin-m+2 x cos-" x dx m — 1 [T]. In precisely the same manner we obtain from [Q] fsin-m x cos-n x dx sin-m+l x cos-nf+l + (n + n - 2) sin-mx c os-n+2 x dx nW-i [U]. By the continued application of these two formulas we may reduce the exponents rn and n as near to zero as possible, and the final expressions to be integrated will be of the forms r dx r dx r dx Jsinx J cosx sin x cos 37. The integration, under all possible conditions, of the form fsinm x cosn x dx, FORMULAS OF REDUCTION. 305 in which m and n are whole numbers, may be made to depend, as we have seen, upon the forms fdx, cos x dx, fsi. x dx, sin x cos x dx, f(sinx cosx dx dX r dX dx, C -. x,, fd,n c Jcosx -.sinx sin x cos x sin x os x Cin X dx, cos- dx J sfosx x J sinu x each of which expressions we shall now integrate. fdx x- + C; cos xdx sin x + C; fsin xdx - cos x + C; fsin x cos x d -= in x d (sin x) sin2x + C; f sil x dx d cos x cos x dx 1 — -- cs — -- log cos x = log sec x + C; cos x dx = log sin x +C; dx sin x J sin x cos x O dx Cos2 x fsec2 x dx d tangx log tang x +C; sin x - tang tang x cos x dr xsin xdx dx dx _ 2 sinx x X X jx 2 sin x cos sin 2 cos 2 - log tang -2 + C; d dx d(900- x) cosx J sin(900x ) sin(90- Jx) - log tang(90 -- 2 ) D. C.- 26. 306 T'HE INTEGRAL CALCULUS. sin x To integrate dx. If in the formula COSm X fsinm x cose x dx - sinm-l x cosn' x -+ (m - 1) sinm-2 x cos'+2 x dx It + 1 we write - m for n, we shall have inm - C xf__mXsinx dx fsinmn x cosm- x dx - =dx tang-mx dx cangm-s x M tng- jtang -xdx [V]. The application of this formula will rfAiuce the form to be integrated to dx orftang x dx - six dx, both of which are known. If in formula [V] we write (90~- x) for x, we have cosm x dx _ ot x dx sinm x tangm-1 (90~- x) _tang_2 (900 - x) dx _ o tm-i X -rfcoth-2 X dx [ W]. This formula reduces the final term to be integrated to fdx or cot x dx- dx sin x both of which are known. FORMULAS OF REDUCTION. 307 NOTE.-If, in the expression sinm x cosn x dx, the exponents m and n be fractional, the integration -nmay 1 ef ected by transforming the given expression into an equivalent algebraic form, and then applying some one or more of the formulas applicable to such cases. 3 5 For example, let m —, n -—; then the expression to be integrated is 3 5 sin 2 x cos3 x d(1). Assume sin x u. Then cos x-(1 - u2); du 1 dx 1 -(1- 2) 2 du; and by substitution in (1) we have. 5 F 5 fsin2 x cos x CLx -f u(1 - ut2)6 (1- u2)- du -f 2 ( —') 3 dit. Finding the integral of this in terms of it, and then replacing it by its value sin x, we shall have the required integral in terms of sin x. 38. In all of the preceding formulas the integration has been effected in terms of the powers of the trigonometrical functions, sine, cosine, etc. The reduction of these formulas to numerical results, in practical operations, renders necessary an excessive amount of arithmetical computation. This labor may be obviated, in general, by converting the powers of sines, etc., into the sines and cosines of multiple arcs before the integration is performed. For this purpose we employ the three well-known trigonometrical formulas [Ray's Surveying, Art. 96]: 308 THE INTEGRAL CALCULUS. 1 1 sin a sin b- 2 cos (a -b) --- cos (a - b) (1), __ 1 1 sin a cos b- 2 sin (a - b) + [ sin (a + b) (2), 1 1 cos a cos b cos (a — b) +- cos (a + b) (3). If, for example, it be required to integrate sin3 x cos' x dx, we have sin3 x cos2 x dx - sin x (sin x cos x)' dx -sin x( + sin 2x 3)dx [by (2)] 2 -- sinx( os4x)dx [by (1)] sinx sin x cos 4x dx - sin x dx + sin 3x dx -— 16 sin 5x dx [by (2)1 f*fJ sin x dx + sin 3x dx sin 5x dx 1 1 1 - 8 cos x 48 cos 3x + -80 cos 5x + C. 39. We shall close this chapter by integrating severa: expressions of frequent occurrence. 1st. feax cos bx dx; 2d. feax sin bx dx. FORMULAS OF REDUCTION. 309 Putting eax dx = dv, and u - cos bx, sin bx, respectively, we have, from the formula for integration by parts, feoas~ds= 1. b a ax cos bx dx - I eax cos bx + a- eax sin bx dx, a a From these two equations we obtain by elimination, a cos bx - b sin bx a feax cos bx dx - I, a sin bx dx a sin bx - b cos bx ea esixx a2+ b2 e [Y]. 3d. f dx 3d fa+b cos" x +e sin2 x We have dx r cos2 x r sec xdx + a sec2 x - b + c tang2 x -+ b + c tangs x fa-Kbd tang x a j d tang x - a + b + (a +- c) tang2 x a + b a + tang2 x 1~~~l~~a- -] ba+ +b a b tangx -t- - ------ tang-1 tang x + C. /(a +- b) (a + c) a btang b tang x C. 310 THE INTEGRAL CALCULUS. 4th. faX x when a > b. a+ b' cos x 1 1 Since cos x - cos- x - sin -1 x, we have dx 1 2 1 — 2 2 a + b cos2 x - b sin2 x 2 2 V 2a — b2 tang- a b tang 2 } C 5th. da when a 1, then we may write — ~ z~ -x' x6etc. 1z 1 1 1.*.! tang-l 1x -1 +...etc.+C. tngx 3x - 5+ 7x7' x oo renders the first member equal to 2, and all the terms in the second member, except C, equal to zero. Therefore C- 2' and tan g- 3 5 - etc.. 2 t 3x 5X + 7x. This formula is applicable only to arcs between 45~ and 90~. DEFINITE INTEGRALS. 325 DEFINITE INTEGRALS. 42. It has already been shown, in the first chapter, that in order to determine the value of a definite integral, when taken between two given limits, it is sufficient to substitute in the indefinite or general integral the two limiting values of the variable, and take the difference between the results. The formula for this operation is fx F'(x) dx = F(x2) - F(xO), or f2 F' (x) dx f 2 F' (x) dx X —f F'(x) dx; x, being the value of x at which the integral is supposed to originate. This formula ceases to be applicable when the function is not continuous between the limits F(x,) and F(x2), or when any intermediate value of the function is equal to infinity. In such cases we may divide the interval x. —x, into such parts that the function may be continuous throughout each part, then integrate between the new limits, and take the sum of the results. Thus, if a be a value of x between xl and x2, such that F(a) o, we may integrate between the limits xl and a, then between a and x2, and take the sum of the integrals; for, by the very nature of integration, we must have F'(x) dx = F'(x) dx + 2 F'(x) dx. Corollary. —It is evident that we must also have f XI 326 THE INTEGRAL CALCUL US. NOTE. -The student will observe that the term limit is not employed in the preceding remarks in its technical sense, as the value toward which a variable may be made to converge indefinitely without ever reaching it; but in its ordinary sense, as a definite value of the variable itself. EXAMPLES. 1. flxm dx. J 0 We have f dx x - <1 C. Substituting the two values of x, I and 0, and subtracting the results, we have fr1 dx 1 m+l- Om+l 1 mn +-1 m q- 1 2. dx We have d Jx ~ e' + C. rCdx 1 1 Ans. - --- O. rf dx 3.ojO 1$-Ans. tang-1 oo - tang-10 0 - 1 1 5. e-az sin bz dz. From formula [Y], Art. 39, we have e-az sinbzdz - a sin bz + b cos bz ea, e7 sin bz d - ~2+ b A2 e EXAMPLES. 327 Substituting in this formula the two values of z, so and 0, and subtracting the results, we have e-az sin bz dz a b J 0 a2 + b2 6. e-az cos bz dz. Applying formula [X], Art. 39, we have W a 0 — 2 — b2 7.b dx Ans. log b -- loga= log a ) 8. -- b dx. Ans. log (- b) -log (- a) -log( 7- 9. J'b dx - a x In this example the limits are of different signs, and the function is not continuous from b to -a, being equal to - o for x -- 0. Taking the integral between the limits, b, 0, 0, - a, we have rb dx rb dx ro dx -a x JO x +-a x = log (b) - log (0) +- log (0) - log (- a) = an indeterminate expression. In this and similar cases we may proceed as follows: Integrate first between the limits b and em, and then between the limits - en and - a, in which m and?n are arbitrary constants, and e is an infinitesimal. 328 THE INTEGRAL CALCULUS. We will thus obtain b dx -log (b) — log (e) — log (m); em x — en dox= log (e) + log (n)- log (a). — (a x ~ e*+f a log _ + logm) The sum of these two integrals is independent of e; it is dependent on the values of n and m, and is therefore indeterminate. But if we pass to the limit by making e equal to zero, we shall still have Jbb+JB resl_,J bb 0 b dc fJ emf f aa oi a -a x log (a )+ log(2). If, finally, we suppose n equal to m, whence log ( ) log 1 0, we shall have, for this particular relation between n and m, b dx log( b -a- a' which is called the principal value of this definite integral. 10. Find, by substitution in the proper formulas, the values of the following definite integrals: fX" ea- dx. Ans. 1. 2. 3... n. n x" e-axd dx A-s. l c/0 anl-l~~~ EXAMPLES. 329 7r 2sinnx dx. Ans. 1. 35 (n1) 2r when n is even; f 2. 4.6...n 2 2. 4. 6... (n —) when n is odd. 1.3.5...n Cosn x dx. Ans. the same as in the last example. 11.0cos (x cos o) do. Developing cos (x cos o) into a series, we have X2 Cos2 0 cos (x cos O) =1- 1.2 X COS4 X( COS2 n + 1.2.3.4 + 1.2.3... 2m + etc. Substituting this series in the given expression, it becomes f-f " do - I1jgf x2 cos2 Od + i f 7x4 cos4 odo +.... + 1 2 3 2mf xcos2m odo + etc. (1). Now we have, by Ex. 10, 7r 2 cosflxdx 1. 3. 5 d.. 1n - when n is even.',o 2. 4. 6... it 2 Cos2 1. 3. 5... 2mn-1.'0 csoo 2.4. 6... 2m 71 X2 COS2 a d- x1 2 f0 Jx2cos20dO - x2;..-28. i). C.-2S. 330 THE INTEGRAL CALCULUS. f 02x4 cos4 Odo = 2 4 4;, frxoco s6OdO= 1.3.5 x65; 0o x6 COS6 do 52. 4. 6 Substituting these values in (1), it becomes cos (x cos a) do,- 1- 22 x4 X6 e +22. 42 22. 42 62 etc. Since cost o do sin" 0 do, it follows that the above result is also the value of O cos (x sin 0) do. 12. Find the value of f sin 0 sin (x sin 0) do. 43. Differentiation and Integration under the Signs. The integral of a given differential function being the limit to the sum of all the infinitesimal elements of which it is composed, we may differentiate each of these elements with respect to a new variable, and take the limit to the sum of the new set of differentials so obtained. This will evidently be the same as differentiating, with respect to the new variable, the quantity under the integral sign; or, since the sum of any number of differentials is equal to the differential of the sum of the functions, and since the limit to the sum is equal to the sum of the limits, it follows that this operation will be the same as differentiating the itself. All the processes, therefore, for differentiating ordinary functions will apply, without change, to integral functions. DIFFERENTIATION OF INTEGRALS. 331 Problem.-To differentiate with respect to x, u =Z F(x. z) dz, in which zo and Z are functions of x. We have -z: ZF(x,z) dz =f F(x, Z) dz- F(x, zo)dz; du du dzo du dZ du dx - dzO dx dZ dx dx Now, dz -F(x, z); duz F(X, Z); dufZ dF(x,z) dz dx z==,o dx d Z dZ.Fx F(x, z) dz —F(x, Z)I dx - F(x,zo)j dz+ C Z d F(x, z) dz. dx 7 d.x Corollary 1.-If zo and Z are not functions of x, then dz~ 0 O and dZ 0 dx dx and we shall have dx fZF(x, z) dz Z dF( dz an expression which shows that in such cases the order of performing the differentiation and integration is immaterial. Corollary 2.-We have at once, if u =fF(x, z)dz + C; du -dF(x d) dC d F dz + -dx dx.dx dx 332 THE INTEGRAL CALCULUS. 44. If, instead of differentiating, we wish to integrate with respect to x the expression TZF(x, z) dz, in which z is independent of x, it will be sufficient to observe that i f dx F(x,z)dz andfz dzf (., z)dx have the same derivative with respect to x, and as they are both zero when x- xo, they are equal to each other. [This is, of course, under the supposition that neither of the expressions becomes infinite or indeterminate within the given limits.] 45. By means of differentiation and integration under the integral sign we may often find the values of definite integrals with more facility than by any other process. We shall give here a few examples by way of illustration. 1. We have already found C o 1 e-ax dx - - 0 O a If we differentiate this expression nx times with reference to a, we shall have -c~.1 2. 3... (it —1) x-1 e-ax d - 2 3 ( ) (1). Making a 1 in this equation, we have f xL"-' e-dx 1.2.3...3 (-) (2). Designating this integral by r(n), we have r(n) - 1.2.3... (1n- 1), and f e-ax X —1 dx - - e-x x7-1 dx -- ) (3). aU a, DEFINITE INTEGRALS. 333 If we put e-x- = y, we shall have for x-=O, y 1l; for x oo, y O, and x log(, dx dy Y /Y.. e- x-l dx = log } dy J log dy- r(n) (4). If we put xn = z, then dz e-C - e-z, x'-1' dx -, and 1 00()e 1 O C0o 1 ( e-x x?"-1dx = - 6Z itdz -ra) (5)7 If in (5) we put n-2 we shall have e-x ex- dx - 2 e dz r( - ) The second member of this equation having no obvious numerical significance, we shall determine the real value of this integral by another process. Assume K d 0 e-2 dz. Then, since the value of the integral is independent of the variable, we shall also have K O e-Y2 dy; and by multiplication, K2 -f e-?2 dyf0 2 e- dz. 334 THE INTEGRAL CALCULUS. This may be written (since y and z are independent of each other) K2 - e(Y2f+z2) dy dz. Now, let y = tz, whence dy = z dt. Then we shall have K2 0- 2 2 2 e 0 -4 -— f; K 2 =J0 J' e(t 2z2+z2)z dt dz =J dt f e-z2(l+t2) dz rCd r e-Z2 (l1+t2) 2z (1 + t2) dz 2(1 + t2) If in the last term we make z2(1+ t2) = u, it will become r i 6-Z2(1+t2) 2z (1+ t2) dz _ O e- du, o 2(1+ t2) 2(1+t2) 1 e-g du 2 (1+ t2) f e 2 (1]+ t) [See Ex. 2, Art. 42]. K~ f K2 dt r1 dt 2(1+t2) 2 0 l+t2 2 (tang-' - tang- 0) 4. K e - 2 d2 2 (6); and therefore r( )-2 V. DEFINITE INTEGRALS. 335 By means of (6) we can find the value of tu =f Xe(X2+ -) dx. Differentiating this expression with respect to a, we have du n dx 2a2 da Jo — x If we put x -a-, we shall find, since the value of the integral is independent of the variable, du - 2u. da du u:-. _ — 2 da; whence log 1=-2a, and U C -- ce -2a. To determine the value of c, put a - 0; then t6 =c- c e-2 dx i- by (6). ~O 2 f ie(x0 a2 )dX -/ e-2a (7)If we integrate 0 e-dx —- between the limits a b and a = c, we shall have f7 ~ x e-b - dx e log ) (8). oo 45'. The integral 0 e-x - -dx is called the Eulerian integral of the second species. It has a remarkable relation to the integral 0 x(b-") (1- x)"-1 dx, which is called the Eulerian integral of the first species, the names being derived from that of the mathematician who first studied them. 336 THE INTEGRAL CALCUL US. To develop this relation. If in the expression Xo(b-1) (1- x)n-1 dx, we put x 1 - we shall have f ox$(b-~) (1- x)"- dx f- dz since z - o when x -0, and fj~ +~f - Now, applying formula [A] to this last integral, we have, as may be readily seen, H0 z'-1 dz 1.2.3... (-1) (1 + Z)'L+b (n+- -n)(...)(...)(n+b-2)(n+b —1) 1. 2. 3... (n — )... 1. 2. 3... (b - ) 1. 2. 3... (b -1)b... (n+b -1) But 1.2.3... (n -1) r (n); 1.2.3... (b- 1)=r(b); 1. 2. 3... (n+ b- - ) - r(n + b). Hence, z ( +1 -- r (in) r (b) o (i+z)?1+b r(n +b) or by substitution f1 Xb-1 (1-x)n-1 dx -{ f 7x-lx e- dx } { f 71 e dx f xn+b1- e —X dx RECTIFICATION. QUADRATURE. 337 If in this equation we make b = 2, and n = 1, we shall have (xf7xx {f — e-xx xe dx} f x2 e-x dx But x dx, and e- dx - 1; by substitution and reduction, 0 X C- dx = 2f x e- dx. GEOMETRICAL APPLICATIONS OF THE INTEGRAL CALCULUS. CHAPTER VII. RECTIFICATION OF CURVES. QUADRATURE OF PLANE AREAS. 46. Problem.-To determine a formula for the length of an are of a plane curve; i. e., to rectify it. Let y — F(x) be the equation to a plane curve, and designate by s the length of the arc. Then, since [Diff. Cal., Art. 106] ds- Vdx + dy2, we shall have s 8- fds =f dx2 + dy2 dy2 Cd~ / dx2 dx 1 + dy.1+ d2 D. C.-29. 338 THE INTEGRAL CALCULUS. The value of this integral, taken between the limits x,, x2, or yi, y2, will evidently be the length of the arc between the points whose coordinates are xl yl and x2 y2. Designating this arc by S, we have S X2 2d~x4+ y (1), or,S — 2 dy 1 e+ d (2). Corollary.-If the curve be referred to polar coordinates, we shall have [Diff. Cal., Art. 106] r r J do" #S _Jdr 1 + r2 (3), or X a ~'-dr 2 ~SfI do ~'r02+ dr (4). 47. Problem. —To determine a formula for the area included between an are, its two extreme ordinates, and the axis of abscissas. Fig. 29. Designating the length of the arc by A, and by xl x2 the abscissas of its two extreme points, we have b [Diff. Cal., Art. 107] d A - ydx. A =X2 ydx (5). In applying this formula the value of y must be expressed in terms of x from the equation of the curve. RECTIFICATION. QUADRATURE. 339 Corollary. —If the curve be referred to polar coordinates, then we shall have [Diff. Cal., Art. 108] 1 Fig. 30. dA 2 r2dO. A 1 f2r2do (6), will be the formula for the area included between two radii-vectores and their corresponding are. 48. Problem.-To determine a formula for the length of an arc of a curve of double curvature. We have [Diff. Cal., Art. 137] ds = =/dx-+ dy2+ dz' (a) -dX 1+ dy + dz s x2dx1 ) dy 2( dz )2 (7), or:dx dx 2 f2dy + ( 1 + )+(y) (8), or y _r sin 0 sin; z - r cos 0, 340 THE INTEGRAL CALCULUS. in which 0 is the angle between Fig. 31. the radius-vector and the axis z of z; and p is the angle between the projection of the radius on the plane X Y and the axis of x. 0 lX Differentiating these equations x with respect to 0, and substituting the values of dx, dy, and dz in (a), we obtain d 2 do Jr2+( dr)+ r2 sin ( di, 2 *S 2 r( ) + r2 sin2( d) (10) EXAMPLES. 49. 1. The length of the parabola. The equation of the parabola is y2 -= 2px; whence dx Y Ydy Integrating according to formula (C), we have, by taking the integral between the limits y, and y2, S 1 (/p2+ y) + Plog+ Y 2(s/"y y2+Vp y2 -2p (pY Vp2 + y V ) p P log a Y2j Vp + IY+ } =2p (y2 1/? + — yj —yl r/p+ yl L3') + 2 logy++ VP-, y ~ P 2 Y 1+ ]/ P -+ -Y l"~~~~~~V~ E/ 7- C1 EXAMPLES. 341 If y, = 0, then we shall have for that portion of the curve included between the vertex and the point whose ordinate is y, I' Y 1/p+y2+ 1 g log I Y ~ V/P Y. 2P 2 P 2. The length of the ellipse. The equation of the ellipse is a~ y2 + b2 xs2 - a2 b2; whence dy ) (b 2x { a2(1_-e2)x2 dx ) ay) a2y (1- e2)2 x2 (1- e2) X2 and (1-e2) (a2 — X2) a2 -xI' 1 _ dY. + (1-e2)x2 _a- e2$2 * S X/ ((a2 _e2 X2) dx. j x a2 — xa To obtain this integral, we have, by expansion, 12_2f 1 e2x2 1 1 e4x4 1 1 3 e6x6 ( e)- 2 a 2 4 a' 2 4 6 a5 etc. 2 e22 1 r dx 1 e2 r X2dX Hence f( 2 a2_ ex2 )~dX =af-dx 2 aJ I/2- x2 1 1 e4 x4dx -2 4 Ta3/JlVa2x2 1 1 3 e6f r xdx -'2 4 6 aaJ x __,,2 etc. - -Z ~-a-D V d-z ) 342 THE INTEGRAL CALCULUS. Putting x- az, we shall have rS= f rdz2 1 2 fzi2dz4 r 7 z4dz J 1/1-z 2 _ z 2 e, 1 3 6zdz ) 2 e 6 etc. each term of which can be obtained by the formula for SJ iiiX [Art. 31, Ex. 1]. To find the length of the elliptic quadrant, we must integrate between the limits x = 0 and x -a, or z 0 O and z — 1. But (1 zm'dz 1 3 5 m —1 ir JoV1/1-zx 2 4 6 2 rl i'dz 1 3 5'" _./1 —z~ r Zdz 1 3 x Jo-1/1 -z- 2 42 7i zSdz 1 d rl dz.0 1 — 1z 2 2; 1/ 1- z 2 We have, therefore, A Quadrant- { — e 1 32 5 22 42 - e6 etc.} and the entire circumference of the ellipse 2, wa{1~+ 3 1 3 5 EXAMPLES. 343 3. The length of the cycloid. We have previously found for the cycloid [Diff. Cal., Art. 94] dy x ~r —x dy 2r - x l+dx a —Tx * + d -- and S= 2X dx -2 IT - XTo determine the length of one-half the cycloid, we must integrate between the limits x, - O0 and x2- 2r. We thus obtain semi-cycloidal are - 4r, and, therefore, entire cycloid - 8r- 4(2r) = four times the diameter of the generating circle. 4. The length of the spiral of Archimedes. We have r aO; dr — ado; 1 +( r2 do - r. dr / a.. SL fr I/a2+ r2 dr. This integral may be obtained from that of the length of the parabola by changing y into r and p into a. 5. The length of the logarithmic spiral. We have dr r=aO; dr aO log ado; d-, aO log a. SX do ~0 )dr 02 d r (I i+ log2a) 2 102 2 1 1I = ( aO do i/1 + log2 a -- y- r/1+ log2 a I aO2 - ae, 1 a01 og a in which is the modulus of the system of logar —},ithms. in which m is the modulus of the system of logarithms. 344 THE INTEGRAL CALCULUS. 6. The length of the catenary. The equation of the catenary being y =- (ec+e C), we shall find ds y dy S fy2 ydy 2y vC y _ _/ yLIf the arc begin at the point where y = c, we shall have S= 1/ y2-c'. 7. The length of the helix. The equations of the helix are z z y a sin -; x acos ma VWa dy 1 z dx 1. z. = Cos - - sin dz m ma dz m ma Substituting these values in formula (9), Art. 48, we have _x, z.1 z S _ z2 dz ] 1+ sin 2 + 1 COS2 I1m ma m ma = z2 dz 1+ 1 I/ J+ m' m if the arc begin at the point where z- 0. 8. The length of the loxodrome, or rhumb-line, described by a vessel whose direction makes a constant angle with the meridian. EXAMPLES. 345 The equations of this curve are l/x2 + y2 -e + e- a tag - 2r, X2 -- y 12 + Z2 r2. Changing to polar coordinates, as in Art. 48, Cor., we have S J~2do/ 1+a+ 01 a 1/+1 — a2 -aVl+ (02- 01), the values of 0 being the co-latitudes of the two terminal points. 9. The area of the parabola. We have y2 = 2px, whence y = 1/ 2px. A rf 12 px dx 2,. _ 2 1/ 2p l X22 3 X2 Y1Xis difference of rectangles Fig. 32. 2 described on the abscissas 3 and ordinates of the two b extreme points. If x=- 0, yl i0, then a d x 2 A= - xy; or the area of the portion Oabd is equal to two-thirds of the rectangle whose sides are Od and bd. 346 THE IYNTEGRAL CALCULUS. 10. The area of the ellipse. We have y = b/ a2- x2. a A -(a2 X)2 dx. a x, Integrating by parts, we obtain f (a-2 x_ 2 dx = x a — x2 j x / a2 — x2 + a2 sin-l — (a2 x2)2dx a a b 12 f(a2-x2) dx { - 2xl/ a-2 sinl- + C; and A - a 2 x1aX 2 a2sin-l2 ---- a2 2_n1 2 l/a x- a2sinl x Integrating between the limits x 0 and x a, we have for the area of a quadrant, ab 7rab A -- sin-' 1 -- 2 4 and for the area of the entire ellipse, A =- ab - V;ra2.,.b' - mean of circles described on the two axes. 11. The area of a cycloid. We have A - ydx - xy - xdy. Now, dy= 2r-x dx. X EXAMPLES. 347 Hence,fx dy = x 2r=x dx =v 2rx-x dx =-f /r' - (x - r)" dx -_ 2 (x-r x -r 1/ 2rx - x2 + - sin- x- + Ce 2 2 r [by the last example]. XA=2y2-xy1- x22 12rx2-x-. 2 2 r ~x1 r V2;2 r2 sinl x — r 2 +r + 2 1/2rxl- xi+ - sin1 -r Taking x, = 0, yl 0, x2 2rq,y2 7rr, we find for onehalf the area of the cycloid 37-r2 A-2, and for the whole area, A - 37rr2 — three times the area of the generating circle. 12. The area of the catenary. We have A =ydx = c e 4- e- c dx. = 2(e-X e- ) =cy2 — _ 2. This is the expression for the area included between the axis of x, the curve, and the two ordinates, y, = c, Y2 = y. 13. The area included between the arc and two focal radii of the parabola. Fig. 33. The polar equation of the parabola is r +c os 1 a v x 1+ os 2 cos2 - o 348 THE INTEGRAL CALCUL US. A —A rXdo ______d 0d (1+ tang2 ) 2 1 2 tang C os4 0 cos 2 0 - 2 1 p ~= 4'sec22 d 2q- p tang2 o dsec odbetween the limits 01 and 02. If 0, 0, then we have for the area VFb, A p2 { tang - + ta t -ang- } 14. The area of the spiral of Archimedes. We have r - aO.... A - 2 rdo a2~02dO. 1 a2 o3 A- -2- r do 2 a do - 6 aI o — o2 If 0 -- 0, then 10 1 r3 r3 A -d2 (3 —a2 _ 6 a 6 a 15. The length and area of the lemniscata, r2 a2 cos 20. 16. The length and area of the ellipse, the pole being at the focus. 17. The length and area of the hyperbolic spiral, rO _ a. 18. The length of the semi-cubical parabola, ay2 = x3. 19. The area of the cissoid, y2 (2a - x) - x3, 20. The area of the curve, r _- a sin 30. QUADRATURE OF SURFACES. 349 CHAPTER VIII. QUADRATURE OF CURVED SURFACES. CUBATURE OF SOLIDS. 50. Problem.- To determine a general formula for the area of a surface of revolution. Fig. 34. Let abcde be the generating curve, and let the axis of X be taken as the axis of revolution. If the curve be revolved about this a k- x axis, it will describe a surface of revolution, and each side of the polygon formed by the chords ab, be, etc., will describe the surface of a conical frustum. It is plain that the surface of revolution will be the limit to the sum of the surfaces described by the chords, as these are indefinitely increased in number. Designating by x and y the coordinates of a point b, by x + dx, y + dy those of c, and by Jch the chord be, we shall have Surface described by be- 2- 2yr { -- } Jch. Hence, { The whole surface 2r 2y + ayech; and described by the polygon - 1 2 and The surface of revolution limit of surface described by polygon - lim 2r {2y- 2y } deh 2rf yds, in which s represents the arc of the generating curve. 350 THE INTEGRAL CALCULUS. Designating the area by A, and substituting for ds its value Vdx2 + dy2, we have A -- 2 fy Vdxt+ dy (1), the integration to be made between the given limits. 51. Problem.-To determine a general formula for the volume of a solid of revolution. It is evident that the volume in question is the limit to the sum of the conical frustums of which it is composed. Now we have (see last figure) Frustum described by bckh 1- xd zJXy2+ (yHj y)2+y(y+jy)} = -1 Jx S 3y2 3yJy+( (jy)2.. Entire Solid = lim z 3 -,.x 3 3y2 y y (Dy)2} Sy2 dx; or designating the solid by V, we have v, y2 dx (2). Corollary.-In any solid of revolution the section perpendicular to the axis is a circle. If, in any solid which hlis an axis, the section perpendicular to this axis be a curve whose equation may be written y =f(x), then it is easily seen that if the area of this section be designated by F(x), the volume will be given by the formula V =fF(x) dx (3), the integration to be made between given limits. QUADRATURE OF SURFACES. 351 52. Problem.-To find the area of any curved surface given by its equation. Let F(x, y, z) - 0 be the equation of the surface referred to three rectangular coordinate axes, and let f(x, y) =- 0 be the equation of its intersection with the plane of XY. If we suppose the surface to be intersected by two series of planes parallel to XZ and YZ, respectively, it will be divided up Fig. 35. Z III! /h into a series of curvilinear quadrilaterals. Connecting the angular points of these quadrilaterals by straight lines, we shall have inscribed within the curved surface a polyhedral surface, each face of which is a rectilinear quadrilateral; and the given surface will evidently be the limit to the sum of all these rectilinear quadrilaterals which are inscribed within it. If, also, the distances between the intersecting planes are Jx and Jy, respectively, then the area of the projection of each of the quadrilaterals upon the plane XY will be the product of Jx and Jy. Again, the area of the projection of a surface upon a given plane being equal to the area of the surface itself 352 THE INTEGRAL CALCULUS. multiplied by the cosine of the angle between its plane and the plane of projection, it follows that axJy = — a cos a, or za ZIX/JY COS a in which Da designates the area of the quadrilateral, and a is the angle between its plane and that of XY. Now, passing to the limits, we shall have limJa da; imxiy dx dy; lim cos a - cos t - cosine of angle between the plane of XY and the tangent plane to the surface at the point x, y, z 1 = [Diff. Cal., Art. 141]. 4 ( 2 (dz'2 Hence, da dxdy 1+( d )2+ (dz )2 The area of the surface will be found by integrating this expression twice, once between the limits xI, x2, and once between the limits y,i- f(x,), and y-2 f(x2), derived from the equation f(x, y) _ 0. The integration with respect to x, in effecting which we may regard y and dy as constant, will give us the limit to the sum of the quadrilaterals contained between two planes parallel to XZ, and the second integration will give the entire surface. We therefore have AzfJo2 f2 dx dy 1~+(d- ( + (4). CUBATURE OF VOLUMES. 353 53. Problem.-To determine the volume of any solid bounded by curved surfaces. If we suppose the body to be intersected by three systems of planes, respectively parallel to the coordinate planes, it will thus be divided into a series of rectangular parallelopipeds and of oblique polyhedral figures, one face of each of the latter being a curvilinear element of the surface of Fig. 36. iZ / A i/ /2,///\s\1 the given body. If the number of intersecting planes be indefinitely increased, the limit to the sum of these oblique figures will evidently be zero, and the limit to the sum of the parallelopipeds will be the volume of the body itself. Designating the distances between the intersecting planes by ax, Jy, dz, and by Jv the volume of a parallelopiped, we shall have Jv _= Jx Jy Zz, and, passing to the limit, dv =- x dy dz, or V=- d =fdx dy dz. D). C.- 30. 354 THE INTEGRAL CALCULUS. If this expression be integrated with respect to x, supposing y and z to remain constant, we shall obtain the volume of a section of the body bounded by four planes, two of them parallel to XZ, and two parallel to YX. Then, integrating with respect to y, we shall have the sum of all the sections contained between the two last mentioned planes; and, finally, integrating with respect to z, we shall have the entire volume. We may therefore write r fx2 fY2 fZ2 dxydz (5). in which the integration is to be performed in the order indicated above. 54. Problem. — To determine the volume of a body when referred to polar coordinates. Fig. 37. " -\X Let the body be intersected by a series of planes passing through OZ, then by conical surfaces described by the revolution of Oc, Od, etc.,'and finally by a system of spheres having their colnmon center at 0. We shall thus have the CUBATURE OF VOLUMES. 355 body divided up into a series of elementary parts, such as abedefgh, and the volume of the body will be the limit to the sum of these parts. Designating the coordinates of g by r = Og, 0O gOA, — AOX, we shall have gd- Jr; gf =-rJO; gh = r cos olp; and volume abc -h - r'2r cos OJJodrp; or, passing to the limits, and taking the sum, V =jr2 dr cos 0 do d. It is easily seen, as in the last proposition, that this expression must be integrated' three times, and we shall have, finally, V 2 f r, r2 dr cos O do dp (6). r/ 1 1 NOTE. -This formula is based on the assumption that abc - h is a parallelopiped. This is not rigidly correct, but the difference between abc h and the parallelopiped is evidently an infinitesimal which disappears in taking the limit to the sum. NOTE. —In double and triple integration between limits it is necessary to pay special attention to the order of integration, if the limits of the different variables be dependent upon each other. It is obvious that a change in the order of integration may necessitate a change in the limits between which the integral is to be taken. 55. EXAMPLES. 1. The surface generated by the revolution of a semiellipse about its major axis. We have y b 1/a i x2; 2/dx2+ dy 2_ =d.x- dy y a — tZa -- X,: 356 I'HE INTEGRAL CALCULUS. A -2, b'x/ a- e'x' dx aJ xx ba X2 e2X2 d ex e x, a a 27 ba eX2 \ e X + 1 sill-' en2 x- b, 12 ex exl 1 — 2 S- n -. e a a 2 a Taking the integral between the limits- a and + a, we have, for the entire surface of the ellipsoid, A 2.! ba l e/ V1-e2 + sin-' e -- 2ra2 (1 6e2) A+ eV sin-' } sin'4 e If e = 0, the ellipsoid becomes a sphere, sin e, and A -4ra2. 2. The surface generated by the revolution of a cycloid about its base. Here we have d ax_ x dx x and since in the formula for the surface of revolution y is the distance from a point on the curve to the axis, we may write y - 2r - x. Hence, A - 2 f2 (2r - x) — 2r dx 4r (x2- 2x) - 2r 1/2r - 2 3 3 - ( X2 2+ XI2 EXAMPLES. 357 Taking twice the integral between the limits 0 and 2r, we have, for the entire surface, 64 3. The surface generated by the revolution of an ellipse about its minor axis. Changing the formula to 2rJx /1 + ( d )'dy, since the axis of y is now the axis of revolution, we have A -27a2 2 l-e log+e 1 for the entire surface of the ellipsoid. 4. The surface generated by the revolution of a cycloid about its axis. We have, as in example 2, " A+-25! y T (dx ) Ix A 27f2r xJy N2r dx, 4ry 1/2rx - 4r 1/2rf I/2r - x dx. Taking the integral between the limits x 0 and x _ 2r, we have, for the entire surface, 3 5. Find the area of the surface formed by the intersection of two equal circular cylinders at right angles to each other. 358 THE INTEGRAL CALCULUS. Let the origin be at the point Fig. 38. of intersection of the axes, and T let the axes of the two cylinders be taken as the axes of Y and Z. Then the equations of the cylinders will be x2 + z2 - a2, and a b x2 + y2 a2. dz x dz Hence, and - 0.' dx~z dy A-TS dy d 1+( d- dz \ a d( ddx This must be integrated, first between the limits y = 0 and y - /a - x9, and then between the limits x = 0 and x - a. We thus obtain A - aoa dx -- a2, which is the expression for one-eighth the entire surface. 6. The volume generated by the revolution of an ellipse about its major axis. We have at once, f7(X2 y ju' a, _ (a2 — x2) d 4 ab __ 2 -- 3ab — 7b2. 2a (between the limits x = - a and + a) 2=- circumscribing cylinder. EXAMPLES. 359 7. The volume of the paraboloid of revolution. Ans. 1 circumscribing cylinder. 8. The volume of the solid formed by revolving a cycloid Fig. 39. about its base. We have dy 2r- x dx'- X the origin being at the vertex. It will facilitate the integration to transfer the origin to the middle point of the base, and reverse the axes. For this purpose we have y = 2r-x, x=2r —y. Hence, dy d (2r - x) __ y,and dx- d(2r —y) 2r- y dx =- dy Y' 2 — y.V — YS y2 ~ _Y dy- _r,Y2 y- (2r —y)-~ dy. Integrating by formula (A), we obtain V= r(2r - y)2 + ry+ y - ~rr3e versin-1 Y + C. 2 r Taking twice this integral between the limits y = 0 and y = — 2r, we obtain for the entire volume, V-= 5;2 r3 - 2rr X -,r2 - circumscribing cylinder. 360 THE INTEGRAL CALCUL US. 9. The volume of the solid described by the revolution of a cycloid about it axis. Fig. 40. This integral can not be conveniently obtained by the use of rectangular co- A B ordinates; we shall therefore transform to polar coordinates, the pole being at E C, the center of the generating circle, when the latter is in the positionf indicated in the figure. Let A be a point on the curve, VH- x, AH-l y, DC - r, and DCH - 0. Then we shall have VH- x VC - CH= r — r cos 0; AH=-l y AD + DHl — VD + DH- rO + r sin 0. V.. V=y dx r3 (0 + sin 0)2d (1- cos 0). Integrating this expression between the limits 0 - 0 and - 7r, we have for the entire volume, V-= r3{3 -3 210. The volume of the ellipsoid with three unequal axes. Every plane section perpendicular to the axis of x being an ellipse, we may determine the volume by the application of the formula F(x)dx. The equation of the ellipsoid is x2 y2 z2 a + 2+, EXAMPLES. 361 and the semi-axes of an elliptic section at the distance x C- b from the origin are V/al -- V/a2- x2. The area of a a this section is F(x) a2 (r2 x2). Hence, vZ 2f 2F(x) dx _ 2 2 2)_dx 4.ba f2 a x (aS- x a) d " bca (between the limits x - -a and x + a) - circumscribing cylinder. 11. The volume of the paraboloid with elliptic section. The equation to this paraboloid is bz2 + ay2 - 2abx. Finding the area of the elliptic section, as in the preceding case, and integrating between the limits x- 0 and x x, we have V —2 circumscribing cylinder. 12. The surface of the tri-rectangular spherical triangle. The equation of the surface is'r2 -_ y2 + z2 r2. Hence, /+l dQ-)+(-d) d and A ffdydx J ( ) (dydx r A=fs~dz~l~i fd,, dydy( rx 362 THE INTEGRAL CALCULUS. This expression must be integrated once between the limits y-O and y - r2_- x2, and once between the limits x - O0 and x = r. We thus obtain, since dy y__ wn _-= -- sin-1 - when y- 2= Vr2_x2_ y2 — in/r 2r__ x-2 2 r 2 A — 2- r dx -- r2. 13. The volume of the tri-rectangular spherical sector. We have V= f dxdy dz, in which the limits of z are 0 and l r2 — x- y2; those of y are 0 and /r2 — x2; those of x are 0 and r. v. - dx dy r x - y2 Now /r-x2- y2 dy- y x y 1 X2 y + -2 (r'-x2) sin-' Y 1,, (r- x2) whenl y 1/r2 —. T V= J -xdx - I 6r3 whenx- r. 14. The volume of the hemisphere, referred to polar coordinates. We have V=fffr2dr cos doa d, and the limits of r are 0 and r; 1 those of O are-0 and —; those of q are 0 and 2zr. EXAMPLES. 363 Integrating between these limits, we have V=ie rSJJ~cos 0 do dt r3 sin 7 2 3.r-3 2- r, 2. - 3 55'. When the formula V = ffdx dy dz leads to the integration of complicated algebraic expressions, we may frequently simplify the operation by integrating with respect to one of the variables, and then transferring the other two to polar coordinates. For this purpose let the given expression be integrated with respect to z. Then we shall have V- Jfz dx dy. The formulas for transformation being x = r cos 0, y r sin 0, we shall have dx - -dr yd (1), dy- Y dr + xdo (2). r Now, in effecting the double integration, when we wish to determine x, we suppose y to be constant, and therefore dy- O0; and if we wish to determine y, we suppose.x to be constant, and therefore dx - 0. Adopting the latter hypothesis, we have 0 x - dr - ydo, r dy = dr +.xdO; whence, by eliminating do, ydy - rdr (a). 364 THE INTEGRAL CALCULUS. From this equation we infer that when dy - O, dr is also equal to zero. We have then, from (1), dx = - ydO (b). Multiplying (a) and (b) together, we have dx dy = - rdr do, and, therefore, neglecting the - sign, which merely indicates the direction in which 0 is reckoned from the axis of X, the formula for integration becomes V -- z do r dr. EXAMPLE I.-A sphere is intersected by a right circular cylinder whose axis passes through the center of the sphere: find the intercepted volume. Taking the origin at the center of the sphere, and the axis of the cylinder as the axis of z, we have for the equations of the Fig. 41. surfaces, a x2 + y2 + z2 a2; x+ - y 2- b;2 a being the radius of the sphere, a and b being the radius of the cylinder. Now it is easily seen from the e figure that the limits of z are 0 and 1/a2 r2; those of o are 0 and 2~., and those of r are 0 and b. We therefore have V f-fz do r dr -fr a l/a — r2 do r dr =27,f/a2 — r2r dr FUNCTIONS OF TWO OR MIORE VARIABLES. 365 This is the expression for that portion of the volume which lies above the plane of XY. The entire volume is twice this, or 47; 3 fy - (a' -- b')'. EXAMPLE 2.-A sphere is intersected by a cylinder, the radius of whose base is half that of the sphere, and whose axis bisects the radius of the sphere at right angles: find the intercepted volume. The equations of the surfaces are x2 + y2 + z2= a; x2 y2 ax. The limits of r are 0 and a cos 0; those of o are 0 and - 7,. Ans. - entire volume { } 4 3' INTEGRATIONI OF FUNCTIONS OF TWO OR MORE VARIABLES. CHAPTER IX. DIFFERENTIAL EXPRESSIONS. 56. Hitherto our attention has been directed exclusively to the integration of functions of a single variable, comprehended under the general form F(x)dx. We propose now to examine those differential expressions which involve two or more variables, and to investigate the general methods of integrating them, i. e., of determining the functions of which they are the total differentials. 366 THE INTEGRAL CALCULUS. Let us consider the expression Mdx + Ndy (1), in which M- (x, y), N= (x, y). We observe, in the first place, that (1) is not necessarily the exact differential of any function of x and y. For, if it were so, designating this function by iu, we should have dul du M= -, Ndx dy' and sincedx dy dy [Diff. Cal., Art. 72], dMl diV we must also have dM d- (2) dy dx If, then, M and N do not satisfy the indentity represented by (2), there is no function of x and y of which (1) is the exact differential. We shall see, however, that, if this condition is fulfilled by M and N, the function in question does exist, and also how it may be determined. For this purpose, we remark, that since M is the partial derivative of the function with respect to x, the function itself must be included in the general, or indefinite, integral of Mdx with respect to x, y being regarded as a constant. It is, then, included in the expression x Mdx + v, Xo in which v is an unknown function of y. It remains to determine the function in such a manner that its derivative with respect to y shall be N. Now this derivative is CxdM' dv x -dM dx -- -, or its equivalent S do -Cy FUNCTIONS OF TWO OR MORE VARIABLES. 367 Rf dN dv No dx d y- +; or, in fine, dv (X, y) - (xo, y)+ dy Therefore, 4 (x, y) -- (xo, y) +- - N =- - (x, y), and dv - - (Xo, ) + d- = 0; or, by transposing an integrating with respect to y, v = —i (xo, y) dy + C. We have thus demonstrated that when equation (2) is satisfied there exists a function of x and y of which (1) is the exact differential, and the foregoing analysis shows that the integration may be exhibited in the following equation: l(Mdx + Ndy) f (x, y) dx + (xo, y) dy + C (3). 57. Let us take, in the next place, a function of three variables, Mdx + Ndy + Pdz (4); in which M=- (x,y, z); N - (x, y,z); P (x, y, z). If (4) is the exact differential of any function of x, y, and z, then, as in the preceding case, we must have dll dN dM dP dN dP dy - dx dz - dx dz - dy These conditions being fulfilled, the function in question is necessarily included in the indefinite integral of Mdx with respect to x, and it will therefore be given by the formula IX lcMdx +- v in which v is an unknown function of y and z. 368 TH'ItE INTEGRAL CALCULUS. The partial derivatives of the function with reference to y and z are, respectively, N and P. Therefore, rx dJl d vAd d v dNd dv J x0 dy' + dy J x dx' + dy -N —4(xo, y, z) + P~ I -dx+ — dx +-c x0 dz_ dz - xo dx dz -P — (xo, yoz)+ dz. dyd and therefore, by the last case v - r + (xo, y, z) dy + x (x, yoz) dz. The function in question, then, exists necessarily when the conditions (5) are satisfied, and is given by the formula f(Mdx + Ndy + Pdz) af q (x, y, z) dx +fy "~ (xo y, z) dy + (xo, yo, z) dz + C (6). - Yo o 58. When a given differential expression satisfies equations (2) or (5), which are called the conditions of integrability, it is said to be an exact differential; and its integral may be found by the application of formulas (3) or (6). It is obvious that in these formulas we may take o- = 0, yo= 0, and zo — 0, and this observation leads to the following simple Rule for integrating exact differentials.-Integrate Mdx with respect to x; then integrate all the terms in dy which do not contain x; and lastly, integrate all the terms in dz which do not contain x or y. The sunl of the results will be the required integral. EXAMPLES. 369 59. It is to be understood, of course, that the variables are entirely independent of each other, and when this is the case, no expression which does not satisfy the condition (2) or (5) can be completely integrated in finite terms. If there is a known relation between the variables, it may render possible an integration which otherwise could not be effected. 60. EXAMPLES. 1. Integrate dx - + adx + 2by dy. We have 1I__ dM dN M —+ a 1f+x —; N 2by; -- O d The expression is therefore an exact differential, and its integral is fSf a+ / 1K a )dx +2by dy ax + log x + 1/1+ x2 ] + by-2+ C. 2. Integrate (a 2y +x) x+ (b3- a2x)dy. We have M cca2y + x; N-b3' + a2x; dM 2 dN dy dx af y + x3) dx +b dy-a2xy + 4+ b3y + C. 3. Integrate dx-_y: + y yVx_ +y~. 1. n1 We have 31 — 1 N - - x } dI 2_ + NI/ x -2 dl dN dy - dx 370 THE INTEGRAL CALCULUS. fof x X + /+floy _0oC ({X~ 1/Xjlog{ y + C - log {c (x + VI Yq x Y)j. 4. Integrate (3xy' — x2) dx - (3x2y -- 6y- l)dy. Ans. 32 x" 2y- y + C. 5. Integrate (6x2y -x3) dx + (2a2x - 3by3) dy. Ans. No Integral. 2dx 2xdy 6. Integrate 2x We shall find dM dN dy dx 2dx _- 2 log{C(x+/x y2)} 7. Integrate ydx + dy xydz a -- a — z (a - z)2 dM 1 dN dM y dP Ve have dy a-z dx dz (ac — z) dx; dN x dP dz (a z)2 dy Hence the expression is integrable, and we have. aa- z~ DIFFERENTIAL EQUATIONS. 371 8. Integrate a dx- b dy + by- axdz. z zd This expression satisfies the conditions of integrability, and we have f ax - by z x dx + ydy -]- z dz z dx xdz d 9. Integrate x/ vx y - z -. x z z dz. Ans. V - - t1ang-' x 2 1 C. Ans. l X +Z' + talg- - + _ z+C CHAPTER X. DIFFERENTIAL EQUATIONS. 61. The integration of a differential equation which involves two or more variables, consists in the determination of a finite equation of which the given equation is a consequence. Up to the present time this operation has been effected in but a comparatively small number of cases, and we shall, after establishing certain general principles relating to the subject, confine ourselves to the examination of some of the simplest forms in which the integration has been accomplished. 62. Integration by Maclaurin's Formula.-Let us take the general equation F(x, y,d dy.. dmy -0 (1), which is said to be of the m"& order, and of the degree indicated by the highest exponent with which any derivative is affected. 372 THE INTEGRAL CALCULUS. From this equation we shall obtain, by solution, the value ofda in terms of _ y.'"d dm-l Y; and, by successive differentiation, the values of the derivatives superior to the m}ith will be given in terms of the same quantities. The integral of (1) will be a finite equation between y and x, which may be written y =f(x); and by expanding this function of x by Maclaurin's formula, we shall have ~Y- +( dx ~ -'(dy )d 1 2'+ "dxrM~-) 1. 2. 3... (n -) + etc. (2). If, in this general development, we replace all the derivatives of superior order to Im - 1 by their values taken from equation (1) and its derivatives after making x - 0, the resulting value of y will evidently satisfy (1), and the general value of y will be embraced in all the particular values which can be derived from this development. If, then, all possible values of y could be developed by this formula, equation (2) would give us the complete solution of the problem of integration; but as Maclaurin's formula will not give us the development of every function of x, we can obtain in this manner only those values of y for which none of the derivatives become infinite or indeterminate for the particular value zero of x. 63. Integration by Taylor's Formula. —If we suppose yo -f(xo) and y =f(xo + h), we shall have h - x - o, and therefore Taylor's formula may be written dy (x - xo)2 42y Y - yo + * (x3. — _1o) dV+ e1.2 dX (Xy - Xo)m-~ dm-lY +.." + 1.2.3...(m-1) 1m, etc. (3). 1.~~~~~~~d 2....(i- )d' DIFFERENTIAL EQ UA TIONS. 373 Substituting in this equation the values of the derivatives superior to the mtlt, found by placing x xo in (1) and its derivatives, we shall obtain a value of y which will evidently satisfy (1), and from which (1) might, reciprocally, be obtained by differentiation. As is the case with the development by Maclaurin's formula, so, in this development, equation (3) would give us the complete solution of the problem of integration provided none of the derivatives became infinite or indeterminate for the value x0 of x; and this arbitrary value of x might always be so selected that all the derivatives would remain finite if they depended only on xo, but as they also depend on the corresponding value yo of y, it may occur that a certain function y — f(x), while satisfying (1), will render certain derivatives infinite or indeterminate whatever may be the value of xo. It follows from the above that neither (2) nor (3) can give all the functions -which satisfy (1). Within the limits in which they are sufficiently convergent they will serve to give approximate values of the function of which (1) is the differential equation. Equation (3) is called the general integral of (1), and (2) is only a particular integral corresponding to the case in which x - 0. 64. Equation (3) satisfies the given differential equation, whatever may be the values of the m quantities dy d'-l y Yo, dx dxm-' and since these all disappear in performing the m differentiations which will lead from (3) to (1), it follows that they are arbitrary constants. Therefore, the general integral of a differential equation of the mth order contains m arbitrary constants, which are no other than the values of y and its first m- 1 derivatives for the particular value xo of x. 374 THE INTEGRAL CALCULUS. Conversely, every finite equations between x and y, which includes m arbitrary constants and satisfies (1), is the general integral of (1). For, if we develop the value of y deduced from such an equation, the first m coefficients of the development will include x0 and the m constants; these coefficients may have any value, whatever be the value of x0; they may then be regarded as entirely arbitrary; and -as the succeeding coefficients depend oil them according to the law indicated by equation (1), the entire development can not differ from that given by equation (3). Whence the truth of the proposition is apparent. 65. When in the general integral of a differential equation we give particular values to one or more of the arbitrary constants, the result is called a particular integral. When we can satisfy a differential equation by a finite equation which is not included in the general integral, this finite equation is called a singular solution, or singular integral. We shall devote a subsequent chapter to the consideration of singular solutions. 66. We have said that every differential equation is a consequence of its finite, or primitive, equation; but it is easy to see that it is not always a necessary, or direct, consequence of this equation. For, since the primitive contains wm arbitrary constants, if we differentiate it n times, we shall have n + 1 equations from which n1 of the constants may be eliminated, and the resulting equation may be very different from the given one; and by varying the number and method of eliminations, we may obtain as many new differential equations as we please, each of which will necessarily satisfy the primitive, while none of them can be deduced directly from it. If, however, we eliminate by any number of different processes the same constants, the results of all the eliminations will be ideztical with each other. DIFFERENTIAL EQ UATIONS. 375 For, let us suppose that by varying the method we arrive at the two following equations of the order m: dx —,y, d F d dxly, dx' d- X. dxm l) dx~ ---- f (,#y do' dxm-y' These functions, F and f, being the mth derivatives of y, are necessarily equal to each other, whatever may be the dvalue of x. Let -, *o be the values of y, nluao~l. Let( dy dm- y dy. dxm-1 for the value xo of x. Then, whatever be the value of xo, the two expressions F xo, yo dx... dx are necessarily equal to each other for all values of yo, (dY), etc., and the uneliminated constants; consequently, all of these expressions must enter in the same manner into the two functions F and f, thereby rendering them identical. Whence it follows that in whatever way we eliminate the same constants we shall in all cases arrive at identical results. 67. The last proposition leads to the following important consequences, viz: That, every differential equation of the order m can be deduced from m equations of the order m — 1, each of which contains one arbitrary constant; from m(1 2) equations of the order m - 2, each of which contains two arbitrary constants; and so on. For let y - F(x) be a finite equation, and let us eliminate one constant between itself and its first derivative; then eliminate one constant between the new equation so 376 THE INTEGRAL CALCULUS. obtained and its first derivative; and so on until m constants have disappeared. The final equation thus obtained will be the differential equation of the mth order, and, from what precedes, this and each of the other equations will be idenltical with those which we would have obtained by any other process of eliminating the same constants. But the order of elimination determines the forems of the resulting differential equations, and the number of these equations of any particular order, as n, will evidently be equal to the number of combinations of m quantities taken nl in a set. Whence the proposition as above stated. 68. It follows from the foregoing analysis that we may integrate an equation of the mYt order by determining, if possible, its m first integrals-the first integral of a differential equation of the order m being a differential of the order m-1. We shall thus obtain m equations containing x, y.. d-, each of which contains one arbitrary constant, and by eliminating the derivatives we shall have a finite equation involving x, y, and m constants, which will therefore be the general integral of the given differential equation. CHAPTER XI. INTEGRATION OF DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND DEGREE. 69. The most general form of these equations is Qdy + Pdx O, or Q dy + -- 0, ill which P and Q are functions of x and y. It may always be solved by the general method of development by' Taylor's or Maclaurin's formula. In some cases the resulting development can be summed in finite terms, but usually the summation is not possible. EQUATIONS OF FIIRST ORDER AND DEGREE. 377 EXAMPLE. dy. ay + bx3 -0. By differentiation we obtain d + a dx + 3bx2 - O, d3y + a + 6bx O0, d41-y + a!ddY + 6b 0, dx4 de d4+my d3+my 0 dx4+mr a d3+m - Making x = 0, we have ( dx) ay () dX2 a( dx y; d3 )=- a yo; ( ) yo — 6b; (d5y -- yo Y 6ab (d+ dad3+mx Hence, by substitution in Maclaurin's formula, a2x2 fa3 x3 t y yo J + 1. 2 1. 2. 3 etc. a 1.2.3.-4 1.2.3.4. 5 + 1.2.3.4. 5. 6 etc; D. C.-32. 378 THE INTEGRAL CALCULUS. or, by substitution [Diff. Cal., Art. 53] of the value of e-ax, Yo e-ax 6b - a2 a e_~X1Y ~- aax -; -- a - - a 1.2 1.2'.3 and, observing that y, - 6b is constant, 6b a2x2 aex'i 1-ax+y= cea~x + —{ - i1. 2 1. 2.3 70. If Q be a function of y only, and P a function of x only, or if by any transformation the given equation can be converted into an equivalent one, each of whose terms is a function of but one of the variables, the integral may be found by integrating, by known processes, each of the terms separately, and taking the sum of the results. This method, which is much resorted to, is called the separation of the variables, and we shall devote the next chapter to its consideration. 71. Of factors by which the first member of an equation may be rendered an exact differential.-If the first meinber of the equation Qdy + Pdx -O, were an exact differential, it would be necessary and sufficient that the integral should be a constant in order to satisfy the equation. The condition of integrability of Qdy + Pdx is, as we have seen, dQ dP If this condition be not fulfilled, the dx dy expression is not an exact differential; nevertheless, there always exists, as we shall now demonstrate, a factor by the introduction of which into the given expression, the latter will be transformed into an exact differential. For this purpose, let us observe that the integral of the given equation contains an arbitrary constant c; that the given equation results from the elimination of this constant between the integral and its first derivative; and that, however the elimination be effected, the result obtained will give the value of d- identical with that derived from the solution of the given equation Qdy + Pdx — 0. FACTORS OF INTEGRATIOX. 379 This granted, let the primitive be represented by dq, dxd p (x, y) - c; whence T dx+ dy -0. The value of dy derived from this equation being idendx tical with that derived from Qdy + Pdx = 0, we must have dP dx P d — Q dy Substituting the value of P derived from this equation in Qdy + Pdx O0, we obtain Q dpy + = dx 0; dy and therefore, if we multiply the given equation by 1 dy it will reduce to dy + +p dx x 0, dy dx each term of which is integrable. 72. Having demonstrated the existence of a factor which will render the given equation integrable, we shall in the next place show how this factor may be determined. Designating it by v, we must have, necessarily, dvQ dvP dv dP dQ, or Q- _vs dP _ & d P dx dy d or Q Pdy V dy d- W from which equation v may be obtained by integration. If v should be a function of both the variables, this equation would be more difficult to integrate than the original one, and we should gain nothing by attempting to discover the factor. 380 THE INTEGRAL CALCULUS. If v be a function of but one of the variables, say x, its value may be easily found. In this case, since d- 0, we must have dv I dy dQ o Q~- vdy- -jx-,or 1 dv_ 1.dP dQ 7dxQX dy d x f' and it is obvious that if the second member of this equation be independent of y, v may be found by direct integration. Designating 1 dP d1 - by (x), we have - dv, (x) dx; whence log v fap(x) dx, and v=egi e Jxio Substituting this value of v in the given equation Qdy + Pdx - o, and integrating according to the methods in Chapter, IX, we have, for the complete integral of the given equation, PX p 4o(x) dxdf1 ) Y Qdy = C ~Jo Xoo We do not multiply the term Qdy by v, because we make use of only those terms of Qdy which do not contain x. 73. After one factor has been discovered, it is easy to obtain others. For since v renders the expression v(Qdy +Pdx) an exact differential of some function, as u, of x and y, it is evident that we shall obtain by multiplying by t(u), v, (lt) (Qdy + Pdx) =, (u) du;' and, the second member of this equation being integrable, the. first member is also integrable. EXAMPLES. 381 74. If the form to be integrated be dy + Pdx 0, which is called the linear equation of the first order, the factor of integration is readily obtained. In this case, since Q - 1 and dQ - O, we shall have dP dP independent of y, and therefore P = Xy + X, in which X and X1 are functions of x. The equation will then be of the form dy + ( Xy + XI) dx = 0. The factor v becomes v- edx, and we shall have eXdXdy + XyefX'dx + Xlef xx d -0; whence we obtain by integration with reference to y as the first variable, yef Xdx+ Xiefxddx - C, or Y e d{ c-f x, ef Xdx}. 75. EXAMPLES. 1. Integrate ydx - x dy = 0. We have dP dy dQ dx e ha dy -- dx1. cry dx dx d 382 THE INTEGRAL CALCULUS. 1 dv 1 2 dv 2dx v dx x — ( —1) x; v - x; logv=-2logx=logy; V —. Multiplying the given equation by this value of v and integrating, we have ydx- xdy y C. e' y dx -- x dy ____ y__ __ c..'. y _ cx.. xs x 2. Integrate dx + (a dx + 2 by dy) 1/1 - Ix= O. Here P-1 +a /Vl x2-; Q 2by 1l/ 1 +'. dP 0 dQ 2bxy dy dx 1/1+x2; 1 f dP dQ x Q [ dy dx - 1- +2' dv xdx (1 + X2) V +.x1x J fJ i+a/l + z- dx +-f2 by dyax + log (x + 1/1 + X2) + by2 _ c; and x - 1/1 i+- x e-(ax+by2) — Ce-(ay2). 3. Integrate xdy — y dx-=xdx + ydy. Ans. tang-' Y - log /x - + y2 + C. x SEPARATION OF VARIABLES. 383 4. Integrate dy + y d + dx = 0. This is a linear equation, and we have y edx{ C-X } =-e { C-f4 e 4 dx }. The last term in this result may be found by development. CHAPTER XII. INTEGRATION BY SEPARATION OF THE VARIABLES. 76. This method consists, as has already been stated, in the transformation of the given equation into an equivalent equation, each of whose terms is a function of a single variable. When this transformation is possible, the integral of the given equation can be found by integrating each term of the new equation, and taking the sum of the results. We shall give several cases to which this method is applicable. 77. Case I.-Let the form be Xdy + Ydx = 0. Dividing by XY, we have dy dx y + -, in which the variables are separated. 384 THE INTEGRAL CALCULUS. EXAMPLE. x2 dy + y2 dx 0. We have dy dx 0; y x x +-t- y cxy. 78. Case II.-Let the form be XYdy + X Y, dx = 0. Dividing by XY,, we have XI dx - 0, dy +, in which the variables are separated. EXAMPLE. 3xy4dy + (xe2y2 + y2) dx - 0, o y2 (X2 +- 1)dx+ 3xy4 dy - 0. Dividing by y2x, we have x + 1 dx+- 3y2dy= 0. x by integration, 2 +logx x+y'=c, or X2 - Ce- (X2 + 2y3) 79. Case III.-Let the form be Mdx + Ndy - 0O SEPARATION OF VARIABLES. 385 in which M and N are homogeneous functions of x and y, of the mth order; i. e., functions such that if we multiply each of the variables by any quantity k, the result will be identical with that obtained by multiplying the functions themselves by ktm. Let y - ux; whence dy = udx + xdu. The functions M and N will be equal to xm multiplied by functions of u, and the equation, when divided by xm, may be written F(u) dx + f(u) u dx + x du} = O, or iF(u) + uf((u) } dx + xf(u) du - 0, or dx f(u) du x F(u) + uf(u) in which the variables are separated. Designating the integral of the second member by +(u), we have log x cqu; whence x = eCe(u) - Ce(u), in which t is a known function. EXAMPLES. 1. (ax + by) dx (max + ny) dy. We have y = ux; dy = udx + x du; whence, by substitution, du a + bu du a + (b -m) u -nu u~xa, q- - )or x - d orx -- d). C.- 33.+nu D. C.-33. 386 THE INTEGRAL CALCULUS. dx ((m + nu) du and log x - a (m + nu) du log x a +(b- m) u - nu2 -a (b- m) u - nu' The integration of the second member may be readily effected by the methods.for the integration of rational fractions. 2. (ax + by + m) dx =(px + qy +- ) dy. Assuming x =x' + a, y = y' + P; aa + bi + m - O; pa + qP -l- n = O; we have a- b t-m; /qr ma-bp; aq-bp' aq - bp dx=dx'; dy= dy'; and therefore, by substitution and reduction, (ax' + by') dx' —- (px' + qy') dy' which is homogeneous and is integrable as in the last example. If aq- bp 0, the foregoing transformation can not be effected. We then have, however, q- = and the given a equation becomes (ax + by) (a dx -p dy) = a (n dy - n dx). Taking ax + by = z, whence dy - dz - a dx e will have, by eliminating y, adx (an +- pz) dz am - n mb -i (b + p)z' in which the variables are separated. SEPARATION OF VARIABLES. 387 80. Case IV.-Let the form be the linear equation dy + Xydx + X1 dx = 0. Putting y = uv, in which u and v are arbitrary functions of x, we have u dv + v du + Xuv dx + X1 dx =0. Since u and v are arbitrary, we may assume du + Xu dx 0; whence du — Xdx, and u __e: - Xdx u dv +X dx 0; whence e J dv+X,dx 0, and v= xi e Xdxdx + C. by substitution, y =-uv Cae C- t e -or de 81. Case V.-Let the form be dy + Xy dx = Xi y"+l dx, which is known as Bernoulli's equation. 1 _ 1 Assuming u =-, whence y u t, and 1 1 dy -- n du, we have du - nXu dx + nX1 dx - 0, a linear equation, whose integral is, by Case IV, 1 e cefXdx} 388 THE INTEGRAL CALCULUS. 82. EXAMPLES, AND GEOMETRICAL APPLICATIONS. 1. Integrate xdx + ydy- my dx. Assuming y = xz, we have dx zdz x + 1-m+z =0 (1), or _________1 mdz __ d log x + l _mz +z) dz - + -mz + _z — by integration and reduction, log x 1- my + y x + e Y WI m x 2 4 + log ___ log 1/ x2 - mxy + y' +V24 log f 2y mx - x m/ 2-4 } -C 1/ l m — 4 l 2y — mx + x /m2 _4 This solution is real when m > 2. It is indeterminat when m- 2, but in that case (1) is readily integrable. I m < 2, the factors of 1 - mz + z2 are imaginary, and th integration must be effected by the method established fe such cases. 2. Integrate x2ydx + y~dx = 3xy2dy. Assuming y = vx, we find dx 3vdv x 1 - 2v2 EXAMPLES. 389 3 logv 1o a log 3 1- -2v - 4 lo - 2v2 } and x 1 3 x-{- --, or (x2- 2y2)3 = Cx'. 1 2 —2 3. Integrate x dx + y dy = x dy - ydx. AssuIning y = vx, we have xdx + vx2 dv + v2 xdx X2 dv. dx 1 -v * X- + V2' and by integration and reduction, log V/x +- y' = tangl - + C. 4. Integrate xa ydx - ydy - x3 dy. Let y vx; then we shall have, by substitution and reduction, dv dx dv V x V4 ~ x o -3 log y log- - eV. Integrate x y ce xy d- y d dy. 5. Integrate x'dy - x ydx + y'dx - xy2dy O 0. Assuming y = vx, we find x (1 - ) dv = 0 (a). dv - O and v- c; whence y = ex. The reduced equation (a) may also be solved by making x O0 and 1- v2 -0, or v = —- 1; whence y = c X O, 390 THE INTEGRAL CALCULUS. or y -- x. These values of y are evidently particular integrals corresponding to the values c X- and c ~ 1. 6. Integrate xy dy - y2 dx = (x + y)2 e- dx. _ xex Ans. x- ce-+.Y 7. Integrate dy + b2y2 dx = a2 xdx. The integral of this, which is known as Riccati's equation, has been found only for certain special values of m. (a). If m- 0, we have at once dy + b2 y2dx = a2 dx; whence dx -= dy _a2 b2 y2 * 2b logc a by, or a- b" 6y" 2ab a - by a + by C e2abx a - by (b). If m -- 2, let y v-., Then we shall have dv b2d dx Substituting ux for v, and reducing, we have dx du x a2u2 + u - b2 an equation which is immediately integrable by the rules for rational fractions. 1? (c). Assuming y - bx + v, we obtain, by differentiation and reduction, dx dv +- bv2 2 = a2 x-+ dx. x2 RICCATI'S EQU ATION. 391 If m - 4, this equation becomes xr c~x dv - b2 2 dx a2 dxo dx dv x2 a _ b2 v'' from which we easily find, by integration, ab + x - b2 x2 y 2ah ab- x + by - e (d). If in the equation dx 2 dv -+ b v2 -_ a- xm+2 dx, 1 we assume v - -, we shall have u dt if in this equation we make xm+2dx —=, and put a2 =P 2(m + 3), b2'= a(m+3), n~4 it reduces to du +- u2 c2 dt: a2 tn dt. This equation, being of the same form as the original, is necessarily integrable when n-=- 4 or m 8 (e). If n is not equal to - 4, we may, by means of the processes employed in cases (c) and (d), reduce this last equation to a similar equation, du' + -jq2 iu2 dt' = a12 t' -' dt', 8 which can be integrated when n' -4, whence n - 12 and m 12 5 392 T2IIE INTEGRAL CALCULUS. Similarly, if n' be not equal to - 4, we may convert this last equation into an equivalent one du" + p"ft2 u"2 d" - a"f2 tn"'" d'", which is integrable when n" - 4, whence n-, 12 16 in - -, and m= 7; and so on. The given equation is integrable, therefore, whenever the value of mn is a term of the following series: 8 12 16 O, 8, etc.; I 5 7 all of which are included in the general form 4r 2r-1 in which r may be any positive whole number. (f). If in the original equation we make y- -, it becomes du + a2 U2X dx - b2 d; dv and if we make x'7 dx aI 2 (m -- 1); m +m __ we shall have du + P2 Uz dv = as vn dv, an equation entirely similar to the original, and therefore 4r integrable when n 2r1; whence m 4r and 4r m -- 2r- 1 2r + 1 EXAMPLES. 393 Combining the last two cases we say, then, that Riccati's - 4r equation is integrable whenever n = 2r ~l and by pursuing the methods indicated in the foregoing discussion, we may reduce any particular equation to an equivalent one in which the exponent mn 4, or - 2, or 0, which will be directly integrable by one of the first three cases. 8. dy + y2dx- x 3dx (1). We have m 4= =2 (1) Hence we must put, 1 according to (f), y -; whence du + u'x -dx dx (2). Now, making x 3 dx _; _3 +. + 4 ~2 1 3 2a2 =;-4 - n; we have 4 4 3+ 3+1 du - 3u dv --- 3v- dv. Since in this equation the exponent of v is - 4, we 1 z must, according to (c), put u -= 3 -+ - -; whence, by substitution, 3dv dz V z-1 and by integration, 6 {z 1 6 - - _ log z - 1 or +1 -= ce. 394 T'HE INTEGRAL CALCULUS. Substituting in this last equation the values of z and v, we have ce_6x 3v2u +v- 3 3x- 3y- + x- 3 3vzu v — 3 23- y-~ - x- ~ -t 3 ce~ -3v2u4 v+3 3x- a'+ x 3 + 3 3 + yx3 (1- 3x ) 3 + yx (1+ 3x3) or, finally, Ce 3 yx (1+ 3x) 3 + yx3 (1- 3x ) which is the complete integral of the given equation. 9. dy - y2 dx - 2x - dx (1). 1 v Let Y- + Then (1) becomes dv -v2 =2x- +2dx (2). Let v-= Then (2) becomes dzeu $- 2u" 2 dx Let - 2u2dX dt whence t= (3. Let x dx dt; whence t x Then (3) becomes du + 6t2 dt - 3t-4dt (4). Let tt fI+1 s GEOMETRICAL APPLICATIONS. 395 Then (4) becomes dt - 3 s (5), by integrating which we obtain c - 1 = 3- tang-' s 1/2; whence /2 — tang {(3 1/ )C tang{( 3 1/2 ) +C} (6). But, from the above substitutions, we have 6t2u - t 6t2 - tv 6t2 _ t (x2y + X) 6 = 6v 6 (x'y + x) 2 7 4 6X3- (x3y + x) 6 (x2 y + x) ____6(x3Y-+x) — -tang{3<- +2C} 3 1/ 2 x (xy + 1) X3 10. Find the curve in which the subtangent at any point is equal to the sum of its coordinates. We have, without regard to the sign of the subtangent, dx d= x -+ y; ydx-xdy-ydy. Making x = yv, we have, by differentiation and reduction, dy - dv. Y. log - v, and Y - e; whence the equation of the required curve is y -= ce Y. 396 THE INTEGRAL CALCULUS. 11. The curve whose subtangent is constant. dx Here y d -a, or ydx ady. dy d/ dy - dx and log( x.'. y ce, the equation of the logarithmic curve. 12. The curve whose subnormal is constant. Here Y d - a, or y dy = a dx, from which we have at once y2'- 2ax + C - the parabola. 13. The curve which cuts at a constant angle all lines whose equation is F (x, y, a) 0,.(1) in which a may have all possible values. If we designate by m the tangent of the given angle, by x' y' the coordinates of a point on the required curve, and by a the Fig. 42. angle which the tangent line to Y a the given curve makes with the axis of abscissas, we shall have Now the tangent of a is the value of dy derived from the equation (1) for the particular point of intersection x' y'. dF ~~~dyf'dx Hence, tang a dF dy GEOMETRICAL APPLICATIONS. 397 and by substitution in (2), m { dy} d F dy' dF 0 (3). mldy' dx' dx' dy'- d' dx'= If we eliminate a between this equation and F(x', y', a) -0, the resulting equation will be a relation between x' and y', and will therefore be the equation of the required curve. This curve is called a Trajectory. 14. The curve which cuts at an angle of 450 all straight lines drawn through a given point. Let the given point be the origin. Then we shall have F(x, y, a) = y - ax = 0. dF dF dy'1; dx — a; m tang 45~ = 1. Therefore, by substitution in (3), xdx + ydy - xdy + ydx = 0, a homogeneous equation, from which we obtain, by integration, log +YI/ X - tang-l Y.C x which is the equation of the required curve. Transferring to polar coordinates by means of the equations y = r sin 0, x - r cos 0, whence tang 0 =- Y, and X2 + y2 =_ r, we have r log - 0, or r - cee - the logarithmic spiral. 398 THE INTEGRAL CALCUL US. 15. The curve which cuts at right angles all parabolas with coincident vertices and axes. Here F (x, y, a) - y2 2px = 0; dF dF Y2 dy 2y; -- 2p x; — -oo. The value of m being infinite, the coefficient of m in (3) must be zero. 2y + - dy O, or 2xdx + ydy- 0; from which we obtain, by integration, 1 2 d x2 + y2 C2- an ellipse. CHAPTER XIII. INTEGRATION OF EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREES. 83. The number of general forms of these equations which have been integrated is comparatively limited, and we shall confine ourselves to the examination of a few of the more important cases. 84. Case I.-When an equation can be solved with reference to d each resulting value of -k will give us a new equation of the form Pdx + Qdy = O, which can be integrated by the methods already established. Let t (x, y, c) = O,, (x, y, c') =0, O. (x, y, c") = 0..., be the various integrals obtained in this manner. Then the equation (x, y,c) (x, y,c') I (x, y, c")... = o EQUATIONS OF THE FIRST ORDER. 399 will evidently contain each of these particular integrals, and will therefore be the complete integral of the given equation; and since each of the factors is separately equal to zero, we may take the same value for the constant in all of them. EXAMPLES. 1 dy } 2 dx We have d-y = + a, whence y ax + c and y=- ax + c; or (y - ax - c) (y + ax- c) - 0, which is the complete integral of the given equation. 2. ( ) 6( d) )2+ 11( d ) 16 —0. We have dy 1, 2, and 3; whence, dx y=-x+ c; y=2x+ c; y=-3x+ c; and the complete integral is (y - x - c) (y - 2x - c) (y - 3x - c) - 0. 3 3a2( ay )2_ dy i d3 ) ( Putting this under the form dy + b { dy 1 } { dy d + 21 2' ~ wd r fdx Va _ix o nrdx Va _2' x~ -O we easily obtain the integrals. The complete integral is (y- c)3- bX (sin-L )~ 2)a 400 THE INTEGRAL CALCULUS. 85. If we have, generally, F( ) = 0, and can find a, a root of the similar finite equation F(z) - O, we may evidently satisfy the given equation by taking a = dy whence y = ax + c, and a = c. We may then write x as the complete integral of the given equation F Y( Y-) = 0. EXAMPLE. To find the curve which has the property s ax + by. ds dx' wence -Ve have - a+ b d; whence ax'dx (dx )= -dx the required equation is 1+( Y 2 ) a + b x This equation may be resolved into two factors of the first degree, each of which, placed equal to zero, is the equation of a straight line. 86. Case II.-Let the form be the homogeneous equation (dy + F +f Y 0 \x dxQ ) +. +f()=0. If we put y - u, whence y ux, and dy u+ x du x dx~ X the equation becomes U + r +F(u1) u~+ x dx + +f(uc) -- 0. EQUATIONS OF THE FIRST ORDER. 401 Solving this equation, if possible, we shall have m values du of u + x -, each of which may be put under the form du dx du u + x d- =Q(u), or x, (i) — u in which the variables are separated. EXAMPLE. Y-$ X X X +( d) } - Assuming Y = u, we have x dy - + ddy 1 2 $; -= 1 +.2 + 2ux _$ + y du 1 +- u2 + 2ux = 0; dx 2udu x Z~~~1 X2 log log (1 + ) =log _ = log x x2 -~ - yand ex = x2 +Y 2. 87. Case III. —When the given equation can not be resolved with respect to dy it may be integrated provided we can solve it with respect to one of the variables. D. C.-34. 402 THE INTEGRAL CALCULUS. Let the resolved equation be of the form y F(xF( +( Y ) dRepresenting d by p, and differentiating, we obtain pdx = F(p) dx + xF'(p) dp + f'(p) dp, or dx + F)x F'(p)dp= f'(p)dp This is a linear equation [Art. 80], and will give, by integration, a relation between x and p. Eliminating p between this and the given equation, we shall have a finite relation between x and y. EXAMPLES. 1. y= (+p)x +2. We have F(p) - 1 +p; f(p) -p"; F(p) -p — 1; F'(p) dp = dp; f'(p) dp = -2pdp. dx + xdp - 2pdp, the integral of which is x ejfc 22 e pdp - = e- + 2(l- p). Substituting in this the value of p derived from the given equation, we have x 1 Ce 2 2 44y - 4 + X2 + 2 = 0. 2. y=- 2px + 1/1 + p. EQUATIONS OF THE FIRSTr ORDER. 403 We have F(p) 2p; f(p) =-1+p2; F(p)- p =p; F'(p)dp- =2dp; f'(p)dp-= (1 +p) 2 pdp. _ dp - dx + 25dp+ / ldp___ 0, the integral of which is r2dp r 2dp x -e J { C e 1 + p} lg {C log Pp dp P~ {-fJ/4ji 1Jl _ } 111 88. Case IV.-Let the resolved equation be y =-px +-f(p), which is known as Clairault's form. Differentiating with respect to p, we have 0 = px +f'(p)p dp (a). This equation gives dp 0; whence p c, and therefore - ex + f (c) The given equation may therefore be integrated by replacing p by an arbitrary constant. 404 THE INTEGRAL CALCULUS. Equation (a) also gives us x + f'(p) O. If we eliminate p between this result and the given equation, we shall obtain an integral without an arbitrary constant, and therefore not included in the general integral; i. e., it will be what we have called a singular solution. EXAMPLES. 1. y-=px + n i/ 1+ p2. Here f (p) = /1 + p2. the general integral is y = cx + n 1/1 + c2. 2. y = px-ap (1 + p2)-2. The general integral is y c ex - ac(l + C2)2. 3. ay (Y + (2x-b) d-x- o. Assuming y2 u, whence 2ydy = du, we have a du )+- (4x - 2b) du - 4u 0. du b du a du\2 \r 3- 4 - Wx-.. u x dx -2x x +' which is of Clairault's form. The general integral is be ac2 U = y2= C -- 2 4+ 4. -axy(~)d ~ (bx2- ay2- ab) = bxy. 4. rtxy dv dx SING ULAR SOLUTIONS. 405 Assuming x2 = v, and y2 = u, this equation reduces dx dx d b-a d-x which is of Clairault's form. Therefore the general integral is y2- z2 abc z -- CX __ b +- a] CHAPTER XIV. SINGULAR SOLUTIONS OF EQUATIONS OF THE FIRST ORDER. 89. Let F(x, y, a) = 0 (1) be the general integral of a differential equation of the first order; a being the constant which by elimination between (1) and its first derivative dF dF dy _ 0 (2) dx dy dx has led to the given differential equation. We remark that in every equation of the form (1), we may replace a by ~, a function of x and y. For, assuming F(x, y,,) = F(x, y, a), we may determine p in such a manner as to render this an identical equation; and consequently F(x, y, ) c O (3), may be considered as an integral of the given equation. 406 THE INTEGRAL CALCULUS. This integral, not being included in (1), is what we have called a singular integral, and it remains now to determine P. The given equation is, as we have seen, the result of the elimination of a between (1) and its first derivative, and since a may be replaced by I, it follows that the given equation is also the result of the elimination of q between F(x, y, ) - 0 and its first derivative. Now, the first derivative of (3) is dF dF dy dF d 0 (4) dx dy dx dep dx and in order that the elimination of p may produce the same result as that of a between (1) and (2), we must evidently have dF dcp dq' dx_ dF _o -0 (5). This equation may be satisfied in two ways. 1st. If we put 0- 0, we obtain p = c, and the result of substituting this value of p in (3) will be the general integral. dF 2d. If we put dF- 0, and substitute the resulting value of p in (3), the equation thus obtained will be a singular solution if it is not included in (1). This result will be indentical with that obtained by eliminating a between the dF equations F(x, y, a)- 0 and -0; and we have therefore the following Rule for obtaining Singular Solutions. —Find the general integral of the given equation, which will necessarily involve one arbitrary constant. Differentiate the integral with respect to this constant, and elininate the constant between the integral and its derivative. SING ULAR SOL UTIONS. 407 The equation thus obtained will be a singular solution, provided it is not itself included in the general integral. 90. It is evident that (5) may be satisfied by placing dF dF= -co and hence we may also find a singular solution by eliminating the constant between the equations dF F(x, y, a) O=0, and d F o. But in all cases it is necessary to observe whether either or both of these methods will not reduce (5) to the form 0 If this should be the case, there would be no singular 0 solution unless the real value of ~ should be zero. 91. It will be observed that the rule above given is identical with that for finding the envelope of a series of curves [Diff. Cal., Art. 136]. Whence it follows that the curve, of which the singular solution is the equation, is the envelope of the series of curves given by causing the constant in the general integral to vary continuously. 92. The foregoing consideration will lead to the following simple method of determining, in many cases, the singular solution without first obtaining the general integral. Let the differential equation be written f(x, y, p) = 0 (6), dy in which p — Now, since (6) is the differential equation of both the general and the singular solutions, and since the envelope is, in general, tangent to each of the curves of which the general integral is the equation, it follows that for any values of x and y which satisfy both the general and the singular solutions, (6) ought to give two equal values of p or dsy one of 408 THE INTEGRAL CALCULUS. which is the tangent of the angle which the tangent line to the envelope makes with the axis of x, and the other is the tangent of the angle which the tangent line to the given curve makes with the same axis. If, therefore, 7 (x, y, p) = 0 (7) be an equation which expresses the condition that (6) shall give two equal values of p for the same values of x and y, the singular solution ought to satisfy both (6) and (7), and therefore the equation resulting from the elimination of p between them. If, then, we effect this elimination, and the resulting equation satisfies (6), it will be the singular solution. 93. If in (6) the function f be of such a form that it can have but one value for a given value of p, we may readily find equation (7). In fact, it is shown in algebra that when an equation has two equal roots, there is a factor common to itself and its first derived polynomial (which is just what we have called in the Calculus the first derivative); consequently, in the case under consideration, there must exist a value of p common to the two equations (6) and df 0. Hence, in all cases where p is not involved in radical terms, we may find the singular solution by eliminating p between the two equations f(x, y, p) O0 and -df _ 0, the dp latter being the form assumed by (7) in this case. 94. EXAMPLES. 1. Y = xp + a /1-+ p. This being of Clairault's form, the general integral is y = cx ~ a / 1 — c2._ dy _ +2,e x - ac(1 -2)-0. EXAMPLES. 409 Eliminating c between these two equations, we have X2 + y2 = a2, which is the singular solution. 2. y - y-=xP —I/l+p2 The general integral is y- cx- ac (1 + c2)* dy + 2)- 32 de =x-a(l+e)) +ae2(l+C2) 2 0. Eliminating c, we have for the singular solution, 2 2 2 x2 + y- ag. 3. f — y - (y -- x) p+ (a -- )p2 = 0. We have - - 2y — 2(a-x)p=-O. Eliminating p, we have for the singular solution, (X + y)2- 4ay = 0. 4. f = ayp2 + (2x- b)p - y 0. f_ 2ayp + (2x- b) 0; dp2 and the singular solution is 4ay2 + (2x -- b)2 0. 5. f p2 X -py + m- 0. Ans. y2 4mx = 0. 6. f =2 - 2xyp + (1 + X2)p2 — 1 - 0. Ans. y2 -1 + x2. 7. f= (px - y) (px- 2y) + x3= 0. Ans. y2 - 4X3 = 0. D. C.-35s. 410 THE INTEGRAL CALCULUS. 95. MISCELLANEOUS EXAMPLES IN EQUATIONS OF THE FIRST ORDER. 1. xdy — ydx -= dx V/x+ y2. Let x = vy. Then dx = vdy + ydv, and / x2 + Y = y / + v'. Hence vy y - vy dy -y2 dv (vdy + ydv) y 1/1 + v, and dy Adv dv Y vV1l+v2 v 1 V -l v —o 1og - - log /l+v l -log v log + * Y _ Y, or I/x2 + y2=e y, and'C -Vx2 +y y X --- 2 + 2cy. 2. ydy = (xdy + ydx) / 1 + y2. We have _ydy + ydx + xdy. We have1 + +y2 =yx +- c; or X v1+ -y2 __ Y Y MISCELLANEOUS E XAMPLES. 411 3. y2 dy = 3xydx - x dy. Let x = vy. Then, by substitution and reduction, we have dy _ 3vdv y - 1-2v2 log Y — 3 log(1- 2v'), and 3 C _ y2 (1- 2v2) - (y2_ 2x2)T 4 2 Y2 -c3 y3 + 2X2. 4. xdy - ydx __ cado. We have xdy-ydx - d tangad = d (aO). d tang-l Y = d(aO), and tang-1 Y - aO + c; or Y -=tang (aO + c), and y = x tang (aO + e). 5. xdx + ady = b /d&2 +dy2. We have x + a -d -b d. x2 -- 2ax d + ad) b2+ b2 (d 412 THE INTEGRAL CALCULUS. Solving this equation with reference to dy we have dy ax b 2(b-a2) -- - X - + a2 _ b2 The integral of the last term is xV _a2-b2 + a2 b2 + ab 2dy$- + -~ -+- log (x + /x2+ a2 __ bi). ax bxV 2 /x + a e b a Y 2(b2 a2)~ 2(b2- a) a2 b2 + a 2- b+ 2 — log(x + V/x2+a2 a2) +. b-? log (x - 1/x - + a' — b 2) C. 2xdy - ydx 6. xdy- ydx — x2+ y2 1/ x2 + y2 Substituting vy for x, we have by differentiation and reduction, ydv - vdy /1 +. * * - fdv 1~/ + v2 1 - v1/1 + vI + - log (v +1/1 + v )+C; or y2 2 y x | + + 2 log y + 1 +. x2 21 x 1/yX +- yl MISCELLANEOUS EXAMPLES. 413 dv du * i/Av- +v C + VA2+Bt + C = Assume B 2aA; C - Ab; v x — a; u = y - a. Then we have by substitution and reduction, dx dy __ rx __+ b - I/y2 - b - a.. log Ix +- /x-+ b-a} d- +log y + I/y"+- b — a2' log c, and [x+l/x +b-aC;y+ ly +b — a2}_; or {V ++ B + V+ {u + 2B +y+ J+U+ }C} C. + u~-Xu+- =e. 8. dx + xdx - du + udx Assume u = yv. Then we shall have (1 + x3) dx = ydv + vdy + yvcd. The quantities y and v being arbitrary, we may assume vdy + yv dx = O; whence dy + y dx = O, and dy _-dx, or log - x, and y = cex. Y C Also (1 + ) dx -- y dv; whence dv (1 + x3) dx (1 + x3) dx y ce -- e dx + d eX3 dx. c C, 414 THE INTEGRAL CALCULUS. 1 1e * v --- eq- adx c c - e + c -ex3- 3e6X2 + 6xe -6e- + c - e X3 _- 3x2 + 6x- 5; + c C. Xe. (x3 -3x2 + 6x - 5 ) + c ex3- 3x2 + 6x -5 + 7' 9. ( (x2 + xy + y2) ( dxy + xy (x2 + xy + y2) dy_ 3 y3 0. This may be resolved into the three equations dy dy dx dY dz xy 0; dyy2 0; the integrals of which are y -- + c; y- e2 +-c'; y=- C. The general integral is EQUATIONS OF THE SECOND ORDER. 415 CHAPTER XV. EQUATIONS OF THE SECOND AND HIGHER ORDERS. 96. There is no general method of integrating equations of an order superior to the first. The usual mode of procedure is to make such transformations upon the given equation as will reduce the order of the derivatives contained therein. We shall present a few instances in which this method may be successfully adopted, beginning with equations of the second order, the general form of which is F[x, y, -3x, -d )= 97. Case I. —Let the form be ( d~2y )== 0y Resolving this equation with reference to dy, we have d2y Multiplying by dx, and integrating, we have dY dx = dy X dx = x2 + C1. de- dx Multiplying again by dx, and integrating, we have YfX2cxfd +C x= x3 + 6 x+ aC. EXAMPLE. d2- az'. We have d2Y dx = ax4 dx. dX2 416 THE INTEGRAL CALCUL US. dy1_ I 1 *dx -5 ax -+ C5; dy=- 5 ax5dx + CIdx; 1 Y - 56ax + C x + C,. 98. Case II.-Let the form be F(X, dy' d_'y )=0. F x,A -, If in this equation we put dy p whence d d we shall have F x(, p, ~-~ an equation of the first order with respect to p. Solving this equation, if possible, we shall obtain the value of p or dy which may then be integrated by the dx methods heretofore established. EXAMPLE. { ~(dy a} 2 d2y { dxc( - 2x dx2 Assuming dy - p, we have 3 a2 dp whence (1- +-) p2) 2- dx dp 2x dx a~-3 — a a2 EQUATIOMNS OF THE SECONYD ORDER. 417 From this equation we obtain - dy x2 + ab dx la4 (x 2 + ab)22 (X2 + ab) dx dy,and fa4 — (x2 + ab)2 2 _ (x2 -+ ab) dx a - (x -- ab)2 2 which is the equation of the elastic curve. 99. Case III.-Let the form be -' d- 2y )-0. We have at once F(p, d)= 0 an equation of the first order. EXAMPLE. d__ a b ( dY ) We have ~dp WVe have d- a + bp2; whence dx adp,and a + bp 1 b x- V'~ tang- a — p e, or P a tang(xv'ab cl/ab). Also, d -y pdy pdx= pdp p; dy - pdx a. bp 2 418 THE INTEGRAL CALCULUS. y — 1b log la + bpl2 + c'; and y -= -- log la + a tang2 (x - c) V1/ b + c'. 100. Case IV.-Let the form be / d2y )-0. F y, -d2 J In this case we may, if we choose, change the independent variable, which process will convert the equation into one of the form dydx d\x WY.dy dy2 ) ~X the integral of which may be found as in Case II; or we may proceed as follows: dy Assuming -- p, we have d2y dp dp dy dp dxr dx - dy dx P dy Hence, by substitution in the given equation, IF(ympy O.d which is of the first order. EXAMPLE. dxY + a2y = 0. Assuming dx - p, this equation becomes p + a2y = 0; whence, by integration, pdy EQUATIONS OF THE SECOND ORDER. 419 P2 a2 y C 2 22 dye- 2 a 2,or. d.2 2c a2 y, and dx2= 2c - ay, or dy dx - dx / 2c - a2 y*X x J~iZ _dy 1 sin/L ay__ c'.'. 22 / —smay + C. sv~c- ay a Vc 101. Case V.-If the form be F(y, dry d2yx )-o the methods of the last case will still be applicable. EXAMPLE. dy d Y i{( dy ) a (d2y ) This reduces to which is of Clairault's form. The integral is therefore p = cy + n V 1 + a2 cs. dy c + + 11 I/ + a2 C2, and 1og + 2 + x - log { cy + n V1 + a2c c'. 0 420 THE INTEGRAL CALCULUS. 102. Case VI.-Let the given equation FIX y dy d2y) F x, y,~ dX' ) — 0 be a homogeneous function of each of the variables. Assuming dx- yt, we shall have d2 y dt dx2= Y x + yt, and dividing by y-, the equation will become one of the first order between x and t. It will however be difficult, in most cases, to integrate the resulting equation, and therefore the following method is generally to be preferred. dy _ y dp v Let y ux; dy dr d x Then we shall have dy = pdx = udx + xdu, and dx d X p-u But, since dp, we have vdx; or dx dp dx x x v. du _dp and v du dp (p -). p —u v Substituting in this last equation the value of v, taken from the given equation, we shall have an equation of the first order, from which we can fiud p in terms of u. Then, dx du by means of the equation - d we can find x in x p- ut terms of u or -Y x EXAMPLE. d 2 (y dy 2 EQUATIONS OF THE HIGHER ORDERS. 421 Making the substitutions indicated above, we have v = (u- p)2, and therefore from the equation vdu -- dp (p - u) we have dp (p- u) du. This being a linear equation, its integral is p -u + 1 + ce~u. dx du Substituting in the equation dk -----, we have x p -u dx du e-" du x 1 + -eu c + -e-u log - - log (c + e-U); whence C- CX e-u- or by substituting the value of u, y _ —-x log{ x This case applies not only to those equations which are essentially homogeneous, but also to those which can be made so by supposing x and y to be of the dimension unity, dydY to be of the dimendxto be of the dimension, and sion - 1. The above example is of the latter character. 103. Equations of the Higher Orders.-By the methods illustrated above, the order of any equation may be diminished by unity, and the successive application of these' methods will finally reduce the equation to one of the first, order. Usually, however, the resulting equation will not be integrable. We shall present two cases in which the integration is successful. Case I. —Let the form be F -dd yK, dL Y), — 422 THE INTEGRAL CALCULUS. d'y- _ we s d'y _dp and If we assume dx'y - p, we shall have - d and the given equation becomes F (P dp )=o which is of the first order. EXAMPLE. d3y __ d2y dd Y- a -dY' We have dp —ap; whence =dp adx, and ax p log _ - ax; d2 Y cea% Now put dy _u. Then d y - du and therefore dx- dx- dx du - ce dx - eax d (ax). dy = eazdx + c'dx; and a y -- e' x - ". 104. Case II. -Let the form be Mdn-2. dn. \ EQUATIONS OF THE HIGHER ORDERS. 423 Assuming dxy we have d- dp, and dxx-y - we- dx deY deP Therefore the equation becomes F (p ddp ) o which is of the second order. EXAMPLE. d4y - d2y dx4 dx2 We have ddP - dx' P. Multiplying this equation by d dx, we have dp d2 d dp d p dp p dx, -or dx x ldx dx ddx)- pdp Now we have ydx, ydx2 d +Spd C/p2 +6+ and y=fd dx Jf Vp2 + e + C2 dp =fdp + c2 dx - =p + c,2X + e, (2). 424 THE INTEGRAL CALCULUS. If p be eliminated between (1) and (2), we shall have a finite relation between x and y. From (1) we have p + -/p2 + e = eex; whence p -- 2c ex 2 e 2z e'Y- 2c., eex i cl ex c 3x c x + c3 105. Integration by Series. — It has already been shown that the general integral of an equation may be found by developing according to Taylor's formula. We shall now give a few examples in which the same result is attained by the method of indeterminate coifficients. 2 dy d2y=o' 1. ny +- + dx Assume y = al xa + a2x f + a3x + a, 4x + etc., in which a, f, y, etc., are ascending powers of x. Then we shall have ny - na, x +- na2 x -+ na3 XY+- na4 x8 + etc.; 2 dy _ 2al ax-2 + 2a2xJS-2 + 2a37xv-2 + 2a4 xS-2 + etc.; d2y Y? al a (afi1) (afi-2 - 1) xP-2 + as (Y- 1) xv-2 + a4 ( - 1) x8-2 + etc. 2 dy d2y.'. ny +-{ 2 d 2Y - ala(a+1)x~-2+ nalxa + a2 f (i + 1) x0-2 - na2 x + a37(yt + 1)x-2 + na3 xv + a, (8 + 1)x8-2 + na4xs + etc. = 0. INTEGRATION BY SERIES. 425 This being an identical equation, the coefficients of the various powers of x must be separately equal to zero. The lowest power of x being a - 2, we must have a (a + 1) = 0, whence a= 0, or a -- 1. 1st. Let a = - 1. Now since a, is, by supposition, different from zero, the term nza, xx can not be zero, and its coefficient nal must therefore be reduced with the coefficients of the next terms. We must consequently have a. - 2 or a > - 2. If a > p - 2, then we must have the coefficient of the third term, f (i- + 1)- O0, which can only be satisfied by making f = 0, since A is greater than a. Hence we have a ——; — 0; y -2-a, or y- 1; a- 2_- A, or $- 2; etc.; and placing the coefficients of the different powers respectively equal to zero, we have a3 7 (y- +1)- +na, 0; a, (- 1) + na, = 0; etc.; al n_ al n2 whence a,, a,-n2 whence a 1.2 a- 1.2. 3. 4 a2 __ a2 n2 a4 - 1.2.3' a6- 1. 2.3.4.5' etc. Therefore, by substitution in the value of y, we have __ a, i x~x nx n2 X3 1.2 1.2.4 etc 4 + a2( 1.2.3 + 1. 2.3.4.5 etc. a co s x 1/n — a2 sin x 1/'2. - n1/x If we take a - i - 2, we shall find y- - cos x X1/, which is a particular integral. D. C.- 36. 426 THE IN27CEGRAL CALCULUS. 2d. Let a - 0. In this case we call not have a > - 2, for this would render the coefficient of x$-2, or 13 (13 + 1), equal to zero,-an impossible condition, since iP> a. We therefore have P - 2 _ a, from which we deduce - 2; - 2 =-, etc.; and we shall find I?xI2 _2x4 a sin x l Y -- a, ~ 1ZX" 32 etc. - y -a~xo 1 2. 3 + 1.2.3.4et x V- n a particular integral. 2. dXy + 1 dy + dx" x dx Assuming y = al xa + a2 x + a3 xe + a, x8 + etc., we have 1 dy al axQ-2 + a2 xs-92 + a3lyXY-2 + a.4 Jx-2, etc. x dx dY _ a a a( — 1)xa-2 + a,(i-1)Xz-2 + a3y ( — l)x-2, etc. The sums of the second members are identically equal to zero, and placing the co6fficient of x,-2 equal to zero, we have a +a(a - 1) =-, or a — 0. If we put — 2 < a, we shall have, by equating to zero the co6fficient of x-2, P = 0, which is impossible since 3 > a. Nor can we put p -2 > a, for this would render a, = 0. Therefore we must have 2 -2 = a, and similarly y 2 13; a - 2 - y, etc., whence we have a =0; f=2; r=4; = 6; etc. We shall find a, a, -a, a2- 2a,; aa 2".4"' a= 22.-4.6', etc.; whence X2 X2 X6 apriuainerlsicitiv lv,b etc.o a palt{c1 22 22e 42 2". 4. i ec62. a particular integral, since it involves but one constant. LINEAR EQUATIONS. 427 106. Integration of Linear Equations. -The general form of these equations is dxS + A dx + B dY+ ~' + T + UyV (1), dxtm dx'n1 dem2 dx in which y and its derivatives are of the first degree, and A, B,... T, U, V are functions of x. This equation can not be integrated in its general form, but it possesses several remarkable properties which we proceed to notice. 1st. If the last term be wanting, so that &4 y d - ly dyx Ad ~m_-' + + Uy= 0 (2); the sum of the m particular integrals will also be an integral of the equation. For if y', y", y"'... be the particular integrals, we shall have cl-7 +.+ TY + Uy' = 0; dn.~ Td U y; ~ +..d T + Uy — 0; cc it cc cs cc i c Adding these equations we shall have the same result as if we had substituted y' + y" + y"' + etc. for y in (2). Therefore, if y', y", etc. are the m particular integrals, their sum y' + i' + y"' + etc. will also be an integral. Moreover, the sum of any number of particular integrals will be a particular integral; and since the product of any particular integral by a constant factor will evidently be a particular integral, it follows that c' y' + c" y" +- + cm ym, 428 THE INTEGRAL CALCULUS. in which c', c". c. are arbitrary constants, will be the general integral; so that we shall have y - c' y " + y... y + + mM. 2d. Let the last equation be the general integral of (2). Then it is obvious that the general integral of (1) may be obtained from this by replacing the arbitrary constants by functions of x. We shall now show how these functions of x may be determined, and as the method of demonstration would be the same for equations of every order, we shall, for the sake of simplicity, suppose the equation to be of the third order. Differentiating the equation y = c' y' + c" y" + c"' y"', we have dy = cy' + c" dy" + c"' dy"' + y' de' -' y" do" + y"' do"'. Assuming, for the first condition by which e', c", c"' are to be determined, the equation if de' +- y" de" + yi" de"' = O, we have dy = c' dy' + e" dy" + o"' dy"' (a). Differentiating again, we have ey = dc'2y'+ e" d2y"+ e" d2ay"+' + dy " do"+ dy"'dc"'. Assuming, for the second condition between c', c", c"', dy'dc' + dy" de"- + dy" de"' - O, we have d2 y c'd2y' + c" d2y" ~ "dl c"2y"' (b). Differentiating again, we have d3y -c' dB y' + c" d3y" + c'" d3y'" + d2y' do' + d2y" dc" + d2y"' do"' (c). LINEAR EQUATIONS. 429 Assuming, for the third condition between c', e", c"', d2yf d' + d'- y" dy" + dy"'- d de"' — Vdx, we shall have the three equations of condition: y' dc' + y" dc" + y"' de"' = 0, dy' do' + dy" dc" + dy"' deo"' 0, d2y'd' + d y" dy " + d2 y"' de"' = Vdx3. From these three equations the values of de', de", de"', and consequently those of c', c", c"', can be determined in functions of x by elimination and integration; and if the values of dy, d2y, d3y in equations (a), (b), and (e) be substituted in the equation d + d2y + dy dX3 de - - B- Uy d V the resulting equation will be identical. Whence it follows, in general, that if we can find y = C'y' + t"y" +... + Cmym, the general integral of d-y + + Uy = 0, that of dmy +Uy=V dx"' -Uy- V can be found by replacing c', c", etc., by functions of x, determined as above. 107. Linear equations of the particular form dl y A dm-ly B d.~m-y U +dx~ ax -- b dxml + b) d- (ax+b)2 dxm2 (ax+b)m may be readily integrated by assuming y (ax -- b)d. 430 THE INTEGRAL CALCUL US. For, differentiating this expression and substituting in the given equation, we have a (a 1)... - m(a- +1) am + Aa (a - 1)... (a - m+ 2) am-l +... + U=- O. Let a', a", a... e be the values of a derived from this equation. Then the particular integrals will be y'- (ax + b)a'; y" -= (ax + b)a-";... ym__ (ax + b)am; and the general integral is y= c' (ax+ b)a' + c" (ax + b)a" +.. + c (ax + b)a. 108. Linear equations in which the coefficients are all constants, are also readily integrable. Let us take the equation dy A - + * + T- + Uy V (1). dxm cAdx,-' dx If we assume y - y + -V, the last term will disappear, and we may therefore suppose the second member to be zero. Let y_ eax. Then, by differentiation and substitution, we shall have am + Aam-l +... Ta + U- 0 (2). If, now, we can find the?n roots of this equation, a, a", a"'... a, it is evident that we shall have for the m particular integrals the following values: y- _ ea z; y, - e=,r;.. ym = em;; and the general integral will be y - C ea'x + c" eafx +... cm eam (3). LINEAR EQUATIONS. 431 109. If the equation (2) has imaginary roots, the corresponding terms of the general integral will be imaginary also, but they may easily be rendered real. Let the imaginary roots be of the form a I-/ 1/- 1. Then the corresponding terms of the general integral will be of the forms Ae(a+l/-lI)x + Be(a-PV~i)x; or eax(Aevi1 x- + Be-Pi-1 x); or exA!cos Ax + 1/-1 sin fix] + eaxBlcos ix - -V1l sin ix. A and B being arbitrary, we may determine them by the conditions A +B=M; (A - B) 1/-1 N; whence we have for the values of the imaginary terms MeaX cos /2x + Neax sin 3x, and similar forms may be found for any two conjugate imaginary terms in the general integral. 110. If (2) has equal roots, the value of y in (3) will not be the general integral, since it will not then contain m constants. We may, however, find the general integral in this case as follows: Suppose that the equation has two equal roots. If we change the coefficients by infinitesimal amounts, (2) will no longer have equal roots, and (3) will be the general integral. The limit to this general integral will evidently be the general integral of the given equation. Let a' and a" be the two equal roots. If for a" we write a' + a, we shall evidently have for the general integral of the new equation y c' ea' X + c" e(a'+8)x + etc. =C'+ eaea'x ( 1 - x + 1 + etc. + etc. 432 THE INTEGRAL CALCULUS. Making c' + c" = A, and c" S = B, we have y - ea'A+Bx +~{ 1.2 x2ea'+ ~ ~ }- etc.; and, passing to the limits, y - ea'x A + Bx] + c"' ea"'X +... etc., which is the general integral of the given equation. Similarly, if there were three equal roots, we should find y - ea'x A + Bx + CX2? + civeaivZ ". ~ etc.; and so on for any number of equal roots. 111. EXAMPLES OF LINEAR EQUATIONS. 1. d2U + b2U = 0. do2 du dsao u d2 eao. Let u - eao. Then - aeae, and d2 a2eao. a2 eaoe b2 eae - O, and a2 + b2 = 0. a=. -- b 1/ —1, and U - C eb9 —1 + C" e —bV-i - (c' + c") cos bo + (' - e") V1/ 1 sin bo -- M cos bo +- N sin bo. 2. do2 +a2 e + b2- 0O. Putting b2 a2 k, and u + k,, we have d2 o dO +- a2 - - O, which is integrable by the last example. EXAMPLES OF LINEAR EQUATIONS. 433 3. d'y 6 d2y 3. d y 6d Y + 11 d- 6y 0. dx d' d 6y - 0. Putting y = eax, we have a3- 6a2 +11a - 6 - 0, the roots of which equation are 1, 2, 3. Therefore the general integral is y =- ex + C'" e2x + C- e3x. 4. + d d y + 16y 0. ~4. ddx' ~ 8dx Putting y = eax, we have a2 + 8a + 16 = 0. This equation has two roots equal to - 4. The general integral is, therefore, y e A +- Bx. d' s ds 5. dt.-+ 2k ds + fs= 0. Putting s e"t, we have nm2 + 2km + f O, the roots of which are - k - /-1 i/f — k'. Therefore the general integral is 8s e —kt c eatV-l - c"+ ee-at —1- e-kt [M cos at + Nsin ate, in which a = i/f- kV. 6. d2y 5 d + 6y x. dx- dx 6. The solution of the equation d~Y dy d2 5 d- 6y 0, is y =c el3 + c" ex. D. C.-37. 434 THE INT'EGRAL CALCULUS. Differentiating, we have dy de' do" dyx 3c e3x+ 2c e2x+ e3x + e2d dx do' dc" 2dx+ dx de,=' e-3xxdx, * c'-C e 31 x -3x d"- e-2Xx dx,... c" C"+ 1 1 e-2x dy -- 9e+' e1 e" e -X 3e3x de+ C'ex~+ Cte2~ + 4 (x + 6de) 7. d3y 3 dY~ + 3 dy Y;. d' de" dx Ans. y-ex A+Bx + Cx'. CHAPTER XVI. INTEGRATION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS. 112. It is a problem of frequent occurrence in the applications of the Calculus, to determine a finite equation between two or more variables from the data furnished by a number of differential equations between the same variables. number of differential equations between the same variables. SIMULTANEOUS EQUATIONS. 435 We shall in the present chapter offer a few examples in which this operation can be effected. 113. Let there be the two equations, dxt - 0 ay 0; d -j-bx=O. Differentiating the first equation with reference to t, we have d2x dy dt- + a dt - and the elimination of dy between this and the second equation gives d2x 7- -abx =- 0. dt2 The integral of this equation is x - c' emt + c" e-m (1), in which m- v/ab. Differentiating (1), and substituting the value of dx in the first equation, we have y = c" Ik e-mt -- c' emt (2). The elimination of mt between (1) and (2) will give a finite equation between x and y. 114. Let the two equations be dt + ax + by =; dt a'x + b'y 0. -~-~ax by-O;-:4-~ x — y-0 436 1'HE INTEGRAL CALCUL US. If we multiply the second by 0 and add the result to the first, we shall have d(x+y) + (a + a'o) x + (b + b) y O. Putting x -+ Oy - v; a + a' O - a; b + b', this equation becomes dv - av - 0; whence v - ceat. Now determining 0 from the two conditions a+a'O — a; b+b'o —aO, let its two resulting values be denoted by 0' and 0", and let the corresponding values of a be a' and a". Then we shall have x+ 0 y-ct'ea't, X + Ot" y = Ct ealt, from which the relation between x and y can be found by the elimination of t. 115. Let the equations be dt + ax+ -byT (1), dy dty + a'x + b'y T' (2), in which T and T' are functions of t. Multiplying (2) by 0, and adding the result to (1), we have d (x + y) +-(a + a'O)x +- (b + b'O) y T-+ T'O (3). elf SIMULTANEOUS EQUATIONS. 437 Assuming a -+ a' - a; b - b' - aO; x + Oy v, (3) becomes dv - - t av - T + T'O. This is a linear equation of the first order, and its integral is v - et {c + (T + T'o) e-tdt ~ If we determine the values of a and o as in the last example, we shall have the two equations x+'y-eG't{c i-+fT~ T'2')e-a'tdt} x + O"y ea,'t c +f (T + T'")e-rat dt, from which the values of x and y may be determined in terms of t. dx EXAMPLE. d 4- 4 +3y t dt + 2x + 5Y= e. We have d(x + oy) + (4 + 2o)x + (3 + 50)y = t + Oe. dt Now, assuming 4+ 20 -a; 3 + 50 = - ao = 40 + 202, we have 2082- 3; whence'= - 1, o"= - 4-2=-a'= 2, and 4+3 —a"- 7. x y=e e-2tc +(tet)e2tdt} x 2- e y - {c- ~ (t + 2 et) de dt. These equations are readily integrable. 438 THE INTEGRAL CALCULUS. 116. Let us now take the three equations d + axz + by + czT (1), -dt + a'x + b'y + c'z = T' (2), dz - + a"x + b"y + c"z = T" (3), in which T, T', T" are functions of t. Multiply (2) by 0, and (3) by 0l, and add the results to (1). We thus obtain d (x + oy + 0, z) d(+ + 1z)+ (a+a' + a"O1)x+ (b+b' + b"0)y + (C + do + c" O,)z = T+ T'O + T"01. Putting x + By + Olz - v; a + a'O + a"1 -- - a; b b'+O +b'' - a; c -c'tO+c"0o - - aOl, this equation becomes - dv a — T T' -+ T" 1,, which is linear, and may be integrated as in the preceding examples. Now, from the three equations of condition between 0 and 0o, there will result two cubic equations from which we can find three values each for o and O1. Calling these values 0', o", 0"', 0', 01", 0,s"', and designating the corresponding values of v by v', v", v"', we shall have the three equations x + O'y + 0,'z - v', X +- 0f y + 0u" z - V", x + O"'y + Oi"'z = —v'", from which x, y and z can be determined in terms of t. SIMULTANEOUS EQUATIONS. 439 117. Let the given equations be the following of the second order: dt - + ax + by + c =O (1), dt2 dtL + a'x+ by+e'- (2). Multiplying (2) by o and adding to (1), the resulting equation may be written d2t~-5, -O a ~ a'C — d"O dt2 { + 0 + -a- c+d + (a + a o) { x +b - bro'O )+ Em +0 aa'O (Y a - a' o. e - e'o0 b +- b'o Making x +y+ - + -v - a O; a4- a'0 - a + a' = -- n2, this equation becomes d2v 0 dt v the integral of which is v - Cent + C' e-". Designating by o', 0" the values of o derived from the equation b b'e0, we shall have a +- a'O x + Coy + C+ e Ce' + C/e-nC" a + a'' - of C + C'o e C + C'e/" EXAMPLE. dt2 d2y + x-8y + 50. dt2 440 THE INTEGRAL CALCULUS. We have; a' -3; 1; b — 4; b' - - 8; c-+ 3; c'-5; and we shall find o' —— 4; O" -- 1; n'=1/7; n" 2; c + c'O' 17 c + c'" 1 a + a'O' 7' a + a'O" 2 1 Hence, x -4y + a-t) 7+ C'e-0T7, x - y+ -1 = Ce2t+ CI'e-t, from which we may find an algebraic equation between x and y by the elimination of t. 118. As a last example let us take the two equations d2X mx dt2 + r- (1), dt2 (2), in which r - /x2 + y2. If we multiply the first equation by y, the second by x, and subtract, we shall have dt2 - y dt2 0, the integral of which is dy dx dt Y dt =c (3). Transposing the first term of (1), and multiplying by (3), we have d2 m f x2 dy dx d ( y c dto - = - dt XY dt m r the integral of which is -G dt m Y +a (4). SIMULTANEOUS EQUATIONS. 441 Similarly, we shall obtain from (2) the equation c dt- z + b (5). Multiplying (4) and (5) by y and x, respectively, and adding the results, we find dxx } (x+ y2) xddt- Y -dt + m) + -ay + bx (6), or mr + ay + bx = C2 (7), or mn2(x2 + y2) = c2 (ay + bx)}2 (8). This is the partial solution of the problem "to find the motion of a particle when attracted to a fixed point by a force which varies inversely as the square of the distance;" and equation (8), being of the second degree, shows that the path of such a particle will be a CONIC SECTION. White's Graded School Arithmetics. COMIPLETE IN THREE BOOqKS. I. PRIMARY. II. INTERMEDIATE. III. COMPLETE. THE GREAT WANT in Arithmetic is a practical combination of the analytic and inductive methods of teaching-a complete and philosophic union of Mental and Written Arithmetic. The preparation of the Graded School Series of Arithmetics was undertaken to meet this urgent need, and it is now submitted to American teachers in the hope that it may be found a practical solution of one serious difficulty which has so long prevented the adoption of more rational courses of study in graded schools. PROMINENT FEATURES. 1. The Series combines Mental and Written Arithmetic in a practical and philosophical manner. 2. It faithfully embodies the inductive method of teaching. 3. The successive books are adapted, both in matter antd method, to the grades of pupils for wehich they are respectively designed. 4. The problems are sufficiently numerocs, varied, and progressive to afford the requisite drill ansd practice. 5. 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