ANALYTIC GEOMETRY AND CALCULUS BY FREDERICK S. WOODS ANI» FREDERICK II. BAILEY I'Kol ■ |.;.>m.i:s ol" MATllKMVI'lis IN TIIK MASSACHUSETTS l\-llli ll. 01 I I I M Noi.ucv GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO COPYRIGHT, 1017, BY FREDERICK S. WOODS AND FREDERICK H. BAILEY ALL KIUHTS RESERVED 217.2 GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. PREFACE The present work is a revision and abridgment of the authors' "Course in Mathematics for Students of Engineering and Applied Science." The condensation of a two-volume work into a single volume has been made possible partly by the omission of some topics, but more especially by a rearrangement of subject matter and new methods of treatment. Among the subjects omitted are determinants, much of the general theory of equations, the general equation of the conic sections, polars and diameters related to conies, center of curv- ature, evolutes, certain special methods of integration, complex numbers, and some types of differential equations. All these subjects, while interesting and important, can well be postponed to a later course, especially as their inclusion in the present course would mean the crowding out, or less thorough han- dling, of subjects which are more immediately important. The rearrangement of material is seen especially in the bring- ing together into the first part of the book of all methods for the graphical representation of functions of one variable, both algebraic and transcendental. This has the effect of devoting the first part of the book to analytic geometry of two dimen- sions, the analytic geometry of three dimensions being treated later when it is required for the study of functions of two variables. The transition to the calculus is made early through the discussion of slope and area (Chapter IX ), the student being thus introduced in the first year of his course to the concepts of a derivative and a definite integral as the limit of a sum. The new methods of handling the subject matter will be recognized by the teacher in places too numerous to specify here. The articles on empirical equations, the remainder in Taylor's series, and approximate integration are new. zT^nm /*. « iv PKEFACE It is believed that this book can be completely studied by an average college class in a two years' course of 180 exer- cises. Teachers who wish a slower pace, however, may omit the last chapter on differential equations, or substitute it for some of the work on multiple integrals. The book contains 2000 problems for the student, many of which are new. It is, of course, not expected that any student will solve all of them, but the supply is ample enough to allow the assignment of different problems for home work and classroom exercises, and to allow different assignments in successive years. F. S. WOODS F. H. BAILEY COXTEXTS Chapter I — CARTESIAN COORDINATES Article Page 1. Direction of a straight line 1 2. Projection 2 3. Number scale 3 4. Coordinate axes 4 5. Distance 5 6. Slope G 7. Point of division 8 S. Variable and function 9 9. Functional notation 12 Problems 18 Chapter II— GRAPHS <>F ALGEBRAIC FUNCTIONS 10. Equation and graph 20 11. Intercepts 21 12. Symmetry and impossible values 28 13. Infinite values 25 11. Intersection of graphs 29 15. Real roots of an equation 3 I Problems 36 Chapter III — CHANGE OF COORDINATE AXES 16. Introduction 39 17-18. Change of origin 39 19. Change of direction of axes 42 20. Oblique coordinates 43 21. Change from rectangular to oblique axes 44 22. Degree of the transformed equation 45 Problems 45 vi CONTENTS Chapter IV — GRAPHS OF TRANSCENDENTAL FUNCTIONS Article Page 23. Definition 49 24. Trigonometric functions 49 25. Inverse trigonometric functions 52 26. Exponential and logarithmic functions 52 27. The number e 53 Problems ' 55 Chapter V — THE STRAIGHT LINE 28. The pointrslope equation 57 29. The slope-intercept equation 57 30. The two-point equation 58 31. The general equation of the first degree 58 32. Angles 59 33. Distance of a point from a straight line 63 Problems 64 Chapter VI — CERTAIN CURVES 34. Locus problems 69 35-37. The circle 70 38-40. The ellipse 74 41-43. The hyperbola 77 44-45. The parabola 80 46. The conic 83 47. The witch 84 48. The cissoid 85 49. The strophoid 86 50. Use of the equation of a curve 88 51. Empirical equations 89 Problems 92 Chapter VII — PARAMETRIC REPRESENTATION 52. Definition 106 53. The circle 108 54. The ellipse 108 55. The cycloid 109 56. The trochoid 110 CONTENTS vii Article Page 57. The epicycloid Ill 58. The hypocycloid 112 59. The involute of the circle 112 Problems 113 Chapter VIII — POLAR COORDINATES 60. Coordinate system 118 61. The spirals 120 62. The straight line 121 63. The circle 122 64. The limacon 123 65. Relation between rectangidar and polar coordinates . . . 124 66. The conic, the focus being the pole 125 67. Examples 126 Problems 127 Chapter IX— SLOPES AND AREAS 68. Limits 130 69. Theorems on limits L32 70. Slope of a curve 134 71. Increment 135 72. Continuity L36 7:!. Derivative L36 71. Differentiation of a polynomial L37 75. Sign of tin' derivative L38 76. Tangent line 140 77. The differential 141 78. Area under a curve 1 13 79. Differential of area 1 l17 154. Straight line determined by two points 317 155. Straight line passing through a known point in a given direction 318 156. Determination of the direction cosines of a straighl line . 319 157. Distance of a point from a plane 320 158. Problems on the plane and the straight line 321 x CONTENTS Article Page 159. Space curves 322 160. Direction of space curve and element of arc 324 161. Tangent lint' and normal plane 326 Problems 32G Chapter XV — PARTIAL DIFFERENTIATION 162. Partial derivatives 335 163. Higher partial derivatives 338 164. Increment and differential of a function of two variables . 339 165. Extension to three or more variables 342 166. Directional derivative of a function of two variables . . 343 167. Total derivative of z with respect to a; 344 168. The tangent plane 345 169. Maxima and minima 348 170. Exact differentials 349 171. Line integrals 353 172. Differentiation of composite functions . 357 Problems ' 361 Chapter XVI — MULTIPLE INTEGRALS 173. Double integral with constant limits 369 174. Double integral with variable limits 371 175. Computation of a double integral 373 176. Double integral in polar coordinates 374 177. Area bounded by a plane curve 376 178. Moment of inertia of a plane area 377 179. Center of gravity of plane areas 379 180. Area of any surface 381 181. Triple integrals 385 182. Change of coordinates 388 183. Volume 389 184. Moment of inertia of a solid 390 185. Center of gravity of a solid 392 186. Attraction 393 Problems 394 Chapter XVII — INFINITE SERIES 187. Convergence 405 188. The comparison test for convergence 406 189. The ratio test for convergence 407 CONTEXTS xi Article Page 190. Absolute convergence 409 191. The power series 410 192. Maclaurin's and Taylor's series 412 193. The remainder in Taylor's series 416 194. Relations between the exponential and the trigonometric functions 418 195. Approximate integration 419 196. The theorem of the mean 422 197. The indeterminate form J} 123 198. Other indeterminate forms \-j:> 199. Fourier's series 127 Problems 1.">1 Chapteb XVIII — DIFFERENTIAL EQUATIONS 200. Definitions |:',s 201. The equation Mdx + Ndy = 0, when the variables can be separated Ill 202-203. The homogeneous equation Mdx + Ndy=0 142 204. The linear equation of the first order . . .... Ill 205. Bernoulli's, equation llii 206. The exact equation Mdx + Ndy = 117 207. The integrating factor lis 'Jos. Certain equations of the second order 149 209. The linear equation with constant coefficients 154 210. The linear equation of tin' first order with constant coefficients \~>~< 211. The linear equation of the second order with constant coefficients t56 212. The general linear equation with constant coefficients . 160 21:5. Solution by undetermined coefficients L62 214. Systems of linear differential equations with constant coefficients 165 215. Solution by series 166 Problems 471 Answers 481 Index .514 ANALYTIC GEOMETRY AND CALCULUS CHAPTER I CARTESIAN COORDINATES 1. Direction of a straight line. Consider any straight- line connecting two points A and B. In elementary geometry, only the position and the length of the line are considered, and consequently it is immaterial whether the line be called AB or BA; but in the work to follow, it is often important to consider the direction of the line as well. Accordingly, if the direction of the line is considered as from , , , A to B it is called AB, but if A BO Fn; 1 the direction is considered from B to A it is called BA. It will be seen later that the distinc- tion between A3 and HA is the same as that between -\-a and — a in algel >ra. Consider now two segments AB and BC on the same straight line, the point B being the end of the first segment and the beginning of the second. The segment AC is called the sum of AB and BC and is expressed by the equation AB + BC=AC. (1) This is clearly true if the points are in the position of fig. 1, but it is equally true when the points are in the position of fig. 2. Here the line BC, being opposite in direction to AB, cancels part of a c b it, leaving AC. FlG - 2 If, in the last figure, the point C is moved toward A, the sum AC becomes smaller, until finally, when C coincides with A, we have jb + BA = 0, or BA = -AB. (2) l 2 CARTESIAN COORDINATES If the point C is at the left of A, as in fig. 3, we still have AB + BC = AC, where AC=~ CA by (2). It is evident that this addition is analogous to algebraic addition, and that this sum may be an arithmetical difference. From (1) we may obtain by transposition a formula for sub- C A b traction ; namely, Fig. 3 BC = AC-AB. (3) This is universally true, since (I) is universally true. 2. Projection. Let AB and MN (figs. 4, 5) be any two straight lines in the same plane, the positive directions of which are respectively AB and MN. From A and B draw straight lines perpendicular to MN, intersecting it at points A' and B' respec- tively. Then A'B 1 is the projection of AB on MN, and is positive if it has the direction MN (fig. 4), and negative if it has the direction NM (fig. 5). Denote the angle between MN and AB by , and draw AC parallel to MN. Then in both cases, by trigonometry, AC = AB cos . But AC = A'B', and therefore A'B' =AB cos (f>. Hence, to find the projection of one straight line upon a second, multijrty the length of 'the first by the cosine of the angle between the positive directions of the two lines. The projection of a broken line upon a straight line is defined as the algebraic sum of the projections of its segments. NUMBER SCALE 3 Let ABODE (fig. 6) be a broken line, MX a straight line in the same plane, and AE the straight line joining the ends of the broken line. Draw AA', BB', CO', DD', and EE' perpendicular to MN; then A'B', B'C, CD', D'E', tm&A'E' are the respective projections on JAY of AB, BC, CD, DE, and JA. But, by § 1, A'B' + B' C' + ( "D 1 + D'E' = A ' /•/'. Hence the projection of a broken line upon a straight line is equal to the projection of the straight line joining the extremities of the broken line. 3. Number scale. On any straight line assume a fixed point as the zero point, or origin, and lay off positive numbers in one direction and negative numbers in the other. If the line is horizontal, as in fig. 7, it is usual, hut not necessary, to lay off the positive numbers to the right of and the negative numbers to the left. The numbers which we can thus lay off are of two kinds: the rational num- „ hers, including the integers and the _I _ '. _\ _\ — J — \ — £ — ^-^ — common fractions; and the irrational F , _ numbers, which cannot be expressed exactly as integers or common fractions, but which may be so expressed approximately to any required degree of accuracy. The rational and the irrational numbers together form the class of real numbers. Then any point M on the scale represents a real number, namely, the number which measures the distance of M from 0: positive if M is to the right of 0, and negative if M is to the left of 0. Conversely, any real number is represented by one and only one real point on the stale. Imaginary, or complex, numbers, which are of the form a+hs/— 1, cannot be laid off on the number scale. The result of § 1 is particularly important when applied to segments of the number scale. For if x is any number corre- sponding to the point M, we may always place x = 031, since both CARTESIAN COORDINATES x and OM are positive when M is at the right of 0, and both x and OM are negative when M is at the left of 0. Now let M and .!/., (fig. 8) be any two points, and let x x = OM and % — OM. Then o M x M„ OM = x 2 On the other hand, MM = OJf -10 12 3 Fig. 8 03/ It is clear that the segment M X M 2 is positive when M 2 is at the right of M x , and negative when i!f 2 is at the left of M x . Hence the length and the sign of any segment of the number scale is found by subtracting the value of the x corresponding to the beginning of the segment from the value of the x corresponding to the end of the segment. 4. Coordinate axes. Let OX and OY be two number scales at right angles to each other, with their zero points coincident at 0, as in fig. 9. y Let P be any point in the plane, - and through P draw straight lines perpendicular to OX and OY re- spectively, intersecting them at M . and N. If now, as in § 3, we place x = OM and y = OX, it is clear that to any point P there corresponds one and only one pair of numbers x and y, and to any pair of numbers corresponds one and only one point P. If a point P is given, x and y may be found by drawing the two perpendiculars MP and XP, as above, or by drawing only one perpendicular as MP. Then 0M=x, MP=0X=y. On the other hand, if x and y are given, the point P may be. located by finding the points M and X corresponding to the num- bers x and y on the two number scales and drawing perpendiculars to OX and Y respectively through M and X. These perpen- diculars intersect at the required point P. Or, as is often more Fig. 9 DISTANCE convenient, a point 21 corresponding to x may be located on its number scale, and a perpendicular to OX may be drawn through 31, and on tins perpendicular the value of y laid off. In fig. 9, for example, 31 (corresponding to x) may be found on the scale OX, and on the perpendicular to OX at M, MI' may lie laid off equal to y. When the point is located in either of these ways it is said to be plotted. It is evident that plotting is most conveniently performed when the paper is ruled in squares, as in fig. 9. These numbers ./■ and // are called respectively the abscissa and the ordinate of the point, and together they are called its coordi- nates. It is to be noted that the abscissa and the ordinate, as defined, are respectively equal to the distances from <> V and OX to the point, the direction as well as the magnitude of the distances being taken into account. Instead of designating a point by writing x = a and y = — b, it is customary to write /'( ), the abscissa always being written first in the parenthesis and separated from the ordinate by a comma. OX and OY are called the axes of coordinates, but are often referred to as the axes of x and y respectively. 5. Distance. Let I^0\, y^ and /_!(./„, //..) be two points, and at first assume that /;/.' is parallel to <>ne of the coordinate axes, as <>X (tig. Hi). Then //, = //,. Now .!/, .)/.,, the projection of /',/'., on OX, is evidently equal to %P y 15ut "j/J/,= .r,- ^ (§ 3). nence P^x-x^ (1) In like manner, if ./;,= x , I\P 3 is parallel to OF, and ^ == y g _y i . (2) p, Pn ; Jtf, M, ' Fig. 10 If x 2 =f= x x and // o ^ y x , I\r, is not parallel to either axis, bet the points be situated as in tig. 11, and through ]'. and /.! draw straight lines parallel respectively to OX and )'. They will meet at a point 7/, the coordinates of which are readily seen to be (./.,, y^. By (1) and (2), J[U = x — x , in: .",• CARTESIAN COORDINATES But in the right triangle P l RP 2 , whence, by substitution, we have tt-V(* t -*O a +(y t -tf. (3) It is to be noted that there is an ambiguity of algebraic sign, on account of the radical sign. But since P X P 2 is parallel to neither coordinate axis, the only two directions in the plane the positive directions of which have been chosen, we are at liberty to choose either direction of I[P 2 as the positive direction, the other becoming the negative. It is also to be noted that formulas (1) and (2) are particular cases of the more general formula (3). Ex. Find the coordinates of a point equally distant from the three points P 1 (l,2),P^-i;-2),«ndP i (2,-5). Let P(x, y) be the required point. Then P X P = I\_P and P„P = P 3 P. But P X P = V(x -iy 2 + (y- 2) 2 , P 2 P = V(x + l) 2 + (y + 2) s P S P = V(x - 2f + (// + 5)2. Therefore V(z - l) 2 + (y - 2) 2 = V(x + l)' 2 + (y + 2) 2 , V(x + l) 2 + (// + 2) 2 = V(ar - 2) 2 + (y + 5) 2 ; whence, by solution, x = | and y = — f . Therefore the required point is (I, - I)- 6. Slope. Let %(x x , y^ and P 2 (x 2 , y 2 ) (figs. 12, 13) be two points upon a straight line. If we imagine that a point moves along the line from P^ to P 2 , the change in x caused by this motion is measured in magnitude and sign by x 2 — x v and the change in y is measured by y % — y x . We define the slope of the straight line as the ratio of the change in y to the change in x as a point moves along the line, and shall denote it by the letter m. We have then, by definition, V» — Vi m = ^—^l. (1) SLOPE A geometric interpretation of the slope is readily given. For if we draw through 1[ a line parallel to OX, and tlirough P 2 a line parallel to OY, and call R the point in which these two lines intersect, then x 2 — x x = J^E and y 2 — y l = RP 2 ; and hence It is clear from the figures that the value of m is independent of the two points J[ and I 2 and depends only on the given line. We may therefore choose 1[ and J', (as in figs. 12 and 13) so that 1{R is positive. There are then two essentially different Fig. 12 Fig. 13 cases, according as the line runs up or down toward the right hand. In the former case; ///_; and m are positive (tig. 12); in the latter case 1U\ and m are negative (fig. 13). We may state this as follows: The slope of a straight line is positive when an increase in x causes an increase in y, and is negative when an turn-use in x causes a decrease in y. When the line is parallel to OX, y = y , and consequently m = 0. If the line is parallel to )', ./•.,= x^, and therefore m = oo (§ 13). Ex. Find a point distant 5 units from the point (1, —2) and situated so that the slope of the straight line joining it to (1, — 2) is g. Let P(x, y) be the required point. Then (x - l) 2 + 0/ + 2) 2 = 25, y±l = i x-l 3 Solving these two equations, we find two points, (i, 2) and (— 2, — 6). 8 CART ES LAN ( !0< > RD1NATES 7. Point of division. Let P(x, if) be a point on the straight line determined by ll(*\, y^) an( l P(. X 2> #»)» s0 situated that There are three cases to consider, according to the position of the point P. If P is between the points 7/ and J£(fig. 14), the Jf, if Jf g Fig. 15 segments i^P and P 1 P 2 have the same direction, and PfP < 7^ ; accordingly Z is a positive number less than unity. If P is beyond P 2 from if (fig. 15), P X P and P X P 2 still have the same direction, but P 1 P > P X P, ; therefore / is a positive number greater than unity. Finally, if P is beyond P 1 from B, (fig. 16), P X P and J[P 2 have opposite directions, and I is a negative number, its nu- merical value ranging all the way from to 00. In the first case P is called a point of internal division, and in the last two cases it is called a point of ex- ternal division. In all three figures draw P X M V PM, and P 2 M 2 perpendicular to OX. In each figure 0M= OM x + M x M\ and since P X P =l(P x P^), M v M=l(M i M t f) by geometry. Therefore 0M= 0M 1 + I (^ JQ ; whence, by substitution, x — x y + I (x 2 — xf). (1) By drawing lines perpendicular to 07we can prove, in the same way, y = y + / („ _ „ ). (2) M o M, M, Fig. 16 VARIABLE AXD FUNCTION 9 In particular, if formulas become x bisects the line 7//', / = }, and these V\ + .'i-i • r i + X -2 9. II Ex. 1. Find the coordinates of a point § of the distance from J\(2, 3) to ]'.,(•■), -:!). If the required point is P(x, y), x = 2 + |(3 -2) = 2§, y = 3 + f(-3-3)=|. Ex. 2. Prove analytically that the straight line dividing two sides of a triangle in the same ratio is parallel to the third side. Let one side of the triangle coincide with OX, one vertex being at 0. Then tin; vertices of the triangle air 0(0, 0), A(x v ••), B(x 2 , //,) (fig. 17). Let CD divide the aides OB and AB so that 0C = l(OB) and AD = I {All). If the coordinates of C arc denoted by (z 8 , //..) and those of 1> by (..-.,, //.,), then, by the above formulas, x 3 = lx v x i = x 1 + /(■':, -x x ) Since ?/., = i/ i , CI) is parallel to 0A. X Fig. 17 8. Variable and function. A quantity which remains un- changed throughout a given problem or discussion is called a constant. A quantity which changes its value in the course of a problem or discussion is called a variable. If two quantities are so related that, when the value of one is given the value of the other is determined, the second quantity is called a function of the first. When the two quantities are variables, the first is called the independent variable^ and the function is sometimes called the dependent variable. As a matter of fact, when two related quantities occur in a problem, it is usually a matter of choice which is called the independent variable and which the function. Thus, the area of a circle and its radius are two related quantities, such that if one is given, the other is deter- mined. We can say that the area is a function of the radius, and likewise that the radius is a function of the area. 10 CARTESIAN COORDINATES The relation between the independent variable and the function can be graphically represented by the use of rectangular coordi- nates. For if we represent the independent variable by x and the corresponding value of the function by ?/, x and y will determine a point in the plane, and a number of such points will outline a curve indicating the correspondence of values of variable and function. This curve is called the graph of the function. Ex. 1. An important use of the graph of a function is in statistical work. The following table gives the price of standard steel rails per ton in ten successive years : 1895 $24.33 1900 $32.29 1896 28.00 1901 27.33 1897 18.75 1902 28.00 1898 17.62 1903 28.00 1899 28.12 1904 28.00 If we plot the years as abscissas (calling 1895 the first year, 1896 the second year, etc.) and plot the price of rails as ordinates, making one unit of ordinates correspond to ten dollars, we shall locate the points P v P 2 , . . ., P 10 in fig. 18. In order to study the valuation in price, we join these points in succession by straight lines. The resulting broken line serves merely to guide the eye from point to Tig. 18 point, and no point of it except the vertices has any other meaning. It is to be noted that there is no law connecting the price of rails with the year. Also the nature of the function is such that it is defined only for isolated values of x. Ex. 2. As a second example we take the law that the postage on each ounce or frac- tion of an ounce of first-class mail matter is two cents. The postage is then a known function of the weight. Denoting each ounce of weight by one unit of x, and each two cents of postage by one unit of ?/, we have a series of straight lines (fig. 19) parallel to the axis of x, representing corresponding values of weight and postage. Here the function is defined by United States law for all positive values of x, but it cannot be expressed in $kQ $30 Sjd $10 T n i; n s < /• 1\ V. 1 i: > O '9 '9 5 'I 7 'i S >9 J '0 ) '0 1 '0 8 '0 ; '0 v " ] fir 1 r lo z. 2 ->z. 3 oz. h oz. * VARIABLE AND FUNCTION 11 elementary mathematical symbols. A peculiarity of the graph is the series of breaks. The lines are not connected, but all points of each line repre- sent corresponding values of x and y. Ex. 3. As a third example, differing in type from each of the preceding, let us take the following. While it is known that there is some physical law connecting the pressure of saturated steam with its temperature, so that to every temperature there is some corresponding pressure, this law has not yet been formulated mathematically. Nevertheless, knowing some correspond- ing values of temperature and pressure, we can construct a curve that is of con- siderable value. In the table below, the temperatures are in degrees centigrade and the pressures are in millimeters of mercury. Temperature Pressure 100 7(50 105 906 110 1074.7 115 1208.7 120 1400.5 125 1743.3 130 2029.8 135 2853. 7 140 2717.9 145 3126.1 150 3581.9 Y [_ i 1 1 t _j 1 . 1 f: _ 7 _ / r -1 1 r / / r 1- L A t - 7 : o i~- 1.0" ^ Let 100 represent the zero point of temperature, and let each unit of x repre- sent 5 degrees of temperature; also let each unit of y represent loo millimeters of pressure of mercury, and locate the points representing the corresponding values of temperature and pressure given Fig. 20 in the above table. Through the points thus located draw a smooth curve (fig. 20); that is, one which has no sudden changes of direction. While only the eleven points located are exact, all other points are approximately accurate, and the curve may be used for approximate computation as follows : Assume any temperature, and, laying it off as an abscissa, measure the corresponding ordinate of the curve. While not exact, it will nevertheless give an approximate value of the corresponding pressure. Similarly, a pressure may be assumed 12 CARTESIAN COORDINATES and the corresponding temperature determined. It may be added that the more closely together the tabulated values are taken, the better the approximation from the curve; but the curve can never be exact at all points. Ex. 4. As a final example, we will take the law of Boyle and Mariotte for perfect gases; namely, at a constant temperature the volume of a definite quantity of gas is inversely proportional to its pressure. It follows that if we represent the pressure by x and the corresponding volume by //, then )/ — -, where k is a constant x and x and y are positive variables. A curve (fig. 21) in the first quad- rant, the coordinates of every point of which satisfy this equation, repre- sents the comparative changes in pressure and volume, showing that as the pressure increases by a cer- tain amount the volume is decreased more or less, according to the amount of .pressure previously exerted. This example differs from the 1 (receding in that the law of the function is fully known and can be expressed in a mathematical formula. Consequently we may find as many points on the curve as we please, and may therefore construct the curve to any required degree of accuracy. 9. Functional notation. When y is a function of x it is cus- tomary to express this by the notation y =/0> Then the particular value of the function obtained by giving x a definite value a is written /(«). For example, if Fig. 21 f(p) 3.c 2 + l, then /(2)=2 3 +3.2 2 +l=21, /(0)=0 8 +3.0 2 +l = l, / ( _3) = (-3)»+3(-3) 9 + l = l, /0)=a 8 +3a 2 +l. PROBLEMS 13 If more than one function occurs in a problem, one may be expressed as /(a;), another as F(x), another as (.r), and so on. It is also often convenient in practice to represent different functions by the symbols / x («), f. 2 (*), / 3 (X)> ete - Similarly, a function of two or more variables may be ex- pressed by the symbol f(x, y). Then /(a, 6) represents the result of placing x—a and y—b in the function. Thus, if f(x, y) = a- 2 +3.///-4// J , /(a, J)=a a +3a*-46 a . PROBLEMS 1. Find the perimeter of the triangle the vertices of which are (3, 4), (- 2, 4), (2, 2). 2. Find the perimeter of the quadrilateral the vertices of which are (2,1), (-2, 8), (-6, 5), (-2, -2). 3. Prove that the triangle the vertices of which arc (— 3, — 2), (1, 4), (— 5, 0) is isosceles. 4. Prove that the triangle the vertices of which are (— 1, 1) ; (1, 3), (- V3, 2 + V3) is equilateral. 5. Prove that the quadrilateral of which the vertices are (1, 3), (3, 6), (0, 5), (- 2, 2) is a parallelogram. 6. Prove that the triangle (1, 2), (3, 4), (- 1, 4) is a right triangle. 7. Prove that the triangle the vertices of which are (2, 3), (_2, 5), (-1, -3) is a right, triangle. 8. Prove that (8, 0), (0, - 6), (7, - 7), (1, 1) are points of a circle the center of which is (4, —3). What is its radius? 9. Find a point equidistant from (0, 0), (1, 0), and (0, 2). 10. Find a point equidistant from the points (—4, 3), (4, 2), and (1,-1)- 11. Find a point equidistant from the points (1, 3), (0, 6), and (- -*, 1). 12. Find the center of a circle passing through the points (0, 0), (4, 2), and (6, 4). 13. Find a point on the axis of x which is equidistant from (0, 5) and (4, 2). 14 CARTESIAN COORDINATES 14. Find the points which are 5 units distant from (1, 3) and 4 units distant from the axis of y. 15. A point is equally distant from the points (3, 5) and (— 2, 4), and its distance from OY is twice its distance from OX. Find its coordinates. 16. Find the slopes of the straight lines determined by the fol- lowing pairs of points : (1, 2), (— 3, 1) ; (3, — 1), (— 5, — 1) ; (2, 3), (2, - 5). 17. Find the lengths and the slopes of the sides of a triangle the vertices of which are (3, 5), (- 3, 2), (5, 2). 18. A straight line is drawn through the point (5, 0), having the slope of the straight line determined by the points (— 1, 2) and (4, — 2). Where will the first straight line intersect the axis OY? 19. One straight line with slope — \ passes through (2, 0), and a second straight line with slope 1 passes through (— 2, 0). Where do these two lines intersect ? 20. The center of a circle with radius 5 is at the point (1, 1). Find the ends of the diameter of which the slope is J. 21. Two straight lines are drawn from (2, 3) to the axis OX. If their slopes are respectively ^ and — 2, prove that they are the sides of a right triangle the hypotenuse of which is on OX. 22. Find the coordinates of a point P on the straight line deter- mined by P (- 2, 3) and P (4, 6), where ^ = -• 1 1 2 O 23. A point of the straight line joining the points (3, — 1) and (5, — 5) divides it into segments which are in the ratio 2 : 5. What are its coordinates ? 24. On the straight line determined by the points ^(4, 6) and P 2 (— 2, — 5) find the point three fourths of the distance from P x to P 2 . 25. Find the points of trisection of the line joining the points Pj(-3, -7) and P 3 (10, 2). 26. The middle point of a certain line is (— 1, 2), and one end is the point (2, 5). Find the coordinates of the other end. 27. To what point must the line drawn from (1, —1) to (4, 5) be extended in the same direction, that its length may be trebled ? P, P L P x (x v y^ and P. 2 ( : '\,, //.J, such that —— = - > prove that PKOBLEMS 15 28. One end of a line is at (2, — 3), and a point one fifth of the distance to the other end is (1, — 2). Find the coordinates of the other end. 29. Find the lengths of the medians of the triangle (3, 4), (-1,1), (0,-3). 30. The vertices of a triangle are ^4(0, 0), B{— 2, 5), and C(4, 3). Show that the slope of the straight line joining the middle points of Ali and BC is the same as the slope of AC. 31. Find the slopes of the straight lines drawn from the origin to the points of trisection of the straight line joining (— 2, 4) and (4, 7). 32. Given the three points A(— 2, 4), P(4, 2), and C(7, 1) upon a straight line. Find a fourth point D, such that '- — = — '- — -• b 1 DC BC 33. If P(x, y) is a point on the straight line determined by P, P L y^), such that —— = j > prov _ l y r., + l.jr x _ /,//,_, + / g//l 34. Given four points J\, /'.,, P p I\. Find the point halfway between P 1 and P 2 , then the point one third of the distance from this point to P g , and finally the point one fourth of the distance from this point to P 4 . Show that the order in which the points are taken does not affect the result. 35. Prove analytically that the lines joining the middle points of the opposite sides of a quadrilateral bisect each other. 36. Prove analytically that in any right triangle the straight line drawn from the vertex of the right angle to the middle point of the hypotenuse is equal to half the hypotenuse. 37. Prove analytically that the straight line drawn between two sides of a triangle so as to cut off the same proportional parts, meas- ured from their common vertex, is the same proportional part of the third side. 38. OABC is a trapezoid of which the parallel sides OA and CB are perpendicular to OC. D is the middle point of AB. Prove analytically that OD = CD. 16 CARTESIAN COORDINATES 39. Prove analytically that the diagonals of a parallelogram bisect each other. 40. Prove analytically that if in any triangle a median is drawn from the vertex to the base, the sum of the squares of the other two sides is equal to twice the square of half the base plus twice the square of the median. 41. Prove analytically that the line joining the middle points of the nonparallel sides of a trapezoid is one half the sum of the parallel sides. 42. Prove analytically that if two medians of a triangle are equal, the triangle is isosceles. 43. Show that the sum of the squares on the four sides of any plane quadrilateral is equal to the sum of the squares on the diagonals together with four times the square on the line joining the middle points of the diagonals. 44. The following table gives to the nearest million the num- ber of tons of pig iron produced in the United States for the years indicated. Eepresent the table by a graph. 1007 25,000,000 1011 23,000,000 1908 10,000,000 1912 29,000,000 1909 25,000,000 1913 31,000,000 1910 27,000,000 1914 23,000,000 45. The following table gives to the nearest thousand the number of immigrants into the United States during the year 1914. Exhibit this table in the form of a graph. January 45,000 July 00,000 February 47,000 August 38,000 March 93,000 September 29,000 April 120,000 October 30,000 May 108,000 November 20,000 June 72,000 December 21,000 PROBLEMS IT 46. The average yearly precipitation at the different meteoro- logical stations of one of the states for the years indicated was as follows. Represent the table by a graph. 1905 0.74 in. 1910 0.79 in. 1000 2.09 in. 1911 0.30 in. 1907 1.5(5 in. 1912 0.98 in. 1908 1.20 in. 1913 1.06 in. 1909 3.02 in. 1914 0.41 in. 47. The daily maximum temperatures at a certain town in the United States for the first ten days of A.ugust, L915, were respec- tively 100°, 99°, 93°, 89°, 92°, 94°, 96°, 95°, '.Mi , 97°. Construct a graph showing the variation in temperature. (Place 90° at the zero point of the temperature scale.) 48. At a certain place the readings of the height of the barometer taken at noon and midnight for a week were, in order, as follows: 29.4, 29.6, 29.8, 30.1, 30.4* 30.8, 30.9, 30.4, 29.8, 29.3, 29.4, 29.7, 29.6, 29.8. Construct a graph showing the changes in atmospheric pressure. 49. On a certain Swiss railroad, the stations with their distance in miles from the first station and their elevation in feet above sea level are as follows. Represent graphically the profile of the railroad. Station A B C I) E F 11 / J K Distance 8 mi. 9.5 mi. 12mi. 13mi. l.",ini. I8rai. 20 mi. 22.5 mi. 25 mi. 28mi. Elevation 1437' 1440' 1580' 1620' 1555' L558 1665' 2305' 216C 3295 I960' 50. In a test on the tensile strength of a steel rod, originally 10 in. long, subjected to a varying load, the following readings were made, the applied load being expressed in pounds per square inch of cross section and the elongation being measured in inches. Illustrate the test graphically. Load 1000 5000 10,000 20,000 30,000 (50,000 65,000 70,000 Elongation .0015 .0031 .0070 .0108 .0212 .0230 .0248 18 CARTESIAN COORDINATES 51. A varying load, expressed in pounds per square inch of cross section, was applied to the end of a concrete block originally 5 in. tall, and the corresponding compression was measured in inches, with the results expressed in the following table. Illustrate the test graphically. Load 100 200 300 400 500 600 700 800 900 1000 Compression .0004 .0010 .0021 .0035 .0054 .0078 .0107 .0130 .0178 52. A body is thrown vertically upward with a velocity of 100 ft. per second. If v is the velocity at the time t, v = 100 — gt. Assuming g = 32, construct the graph showing the relation between v and t. 53. The space s through which a body falls from rest in a time t is given by the formula s = \ gt 2 . Assuming g = 32, construct the graph showing the relation between s and t. 54. A body is thrown up from the earth's surface with an initial velocity of 100 ft. per second. If s is the space traversed in the time t, s = 100 1 — \ gt 2 . Assuming g = 32, construct the graph showing the relation between s and t. 55. Make a graph showing the relation between the side and the area of a square. 56. Make a graph showing the relation between the radius and the area of a circle. -p 57. Ohm's law for an electric current is C = — > where C is the current, E the electromotive force, and R the resistance. Assuming E to be constant, plot the curve showing the relation between the resistance and the current. 58. Two particles of mass m x and m 2 at a distance d from each other attract each other with a force F given by the equation F = — ^-^- • Assuming m x = 5 and m 2 = 20, construct the graph showing the relation between F and d. 59. If f(x) = .t 4 - 4 x 2 + 6 x - 1, find /(3), /(0), /(- 2). 60. If f{x) = x s - 3 x 2 + 1, show that/(2) + 2/(0) =/(l). 61. If f(x) = x 3 - 3 x 2 + 5 x - 6, find /(a), /(- a), f(a + h). 62 - If •^) = ^f^' find/(3),/(0),/(-l). PROBLEMS 19 63. If /(a-) = jA ~ a ( 7" + ' , prove that/(- x) =/(*). a. -j- A 64. If /(a:) = x b + 5 a; 3 - 9 x, prove that /(—«) = - /(a;). 65. If /(a-) = x 2 + 2 aa: - a 2 , prove that /(«) +/(- a) = 0. 66. If f(x) = (x - ^) (x 2 - |) , prove that /(a) -/(£)■ 67. If /(a-) = |rff > prove that /(«.) . /(- a) = 1. 68. If/Car) = * 4 +g*'+5g + l > prove that /(a-) =/(i)- 69 . If fjx) = x z + a* and f % (x) = 2 ax, prove that /, (a)- «/,(«) = 0. 70. If f x (x) = Va; 2 - 4 and f 2 (x) = Vz 2 + 4, prove that /.(•+D+/.(— D-*» 71. H/.(a:)=-\/- + \ 2 a=d /,(*)= \~ - "\F> P'«™ that [/ 1 w] 2 -[/,(-'-)] 2 =[/ 1 (")] 2 - 72. If /(a-) = 2y + 1 , prove that /[/(.r)] = a-. 73. If /(a-, y)= x 2 + // - 5, find /(0, 0), /(l, 0),/(0, 1),/(1, 2) 74. If/(.r, y)=^iy, prove that /(a, &) = -/(&, a). x i) 75. If /^aj, y) = x + ij and f,(x, y) = x - y, prove that 76. If f^ V)=^ + l and f % {x, y) = ^ + £ , prove that ' ./jr. //) = [/;<,-, y)] 2 - 2. 77. If /^aj, y) = a; + 3 y and / 2 (a;, y) = 3 a: + 9 //, prove that xA(x,y)+uf>(*,y)=[fi(*>y)T' CHAPTER II GRAPHS OF ALGEBRAIC FUNCTIONS 10. Equation and graph. If f(x) is any function, and we place *=/(*). we may, as already noted, construct a curve which is the graph of the function. The relation between this curve and the equation y =f(x) is such that all points the coordinates of which satisfy the equation lie on the curve; and conversely, if a point lies on the curve, its coordinates satisfy the equation. The curve is said to be represented by the equation, and the equation is called the equation of the curve. The curve is also called the locus of the equation. Its use is twofold : on the one hand, we may study a function by means of the appearance and the properties of the curve ; and on the other hand, we may study the geometric properties of a curve by means of its equation. Both methods will be illustrated in the following pages. Similarly, any equation in x and y expressed by represents a curve which is the locus of the equation. To construct this curve we have to find enough points whose coordinates satisfy the equation to outline the curve. This may be done by assuming at pleasure values of x, substituting these values in the equation, and solving for the corresponding values of y. Before this computation is carried out, however, it is wise to endeavor to obtain some idea of the shape of the curve. The computation is then made more systematic, or in some cases the curve may often be sketched free-hand with sufficient accuracy. 20 INTERCEPTS 21 The following plan of work is accordingly suggested: 1. Find the points in which the curve intercepts the coordi- nate axes. 2. Find if the curve has symmetry with respect to either of the coordinate axes or to any other line. 3. Fhid if any values of one variable are impossible, since they make the other variable imaginary. 4. Find the values of one variable which make the other infinite. Each of the above suggestions is illustrated in one of the following articles : 11. Intercepts. The curve will have a point on the axis of x when y = and will have a point on the axis of y when x = 0. Hence we may find the intercepts on one of the coordinate axes by placing the other coordinate equal to zero and solving the resulting equation. Ex. 1. y = .5 (./• + 2) (.-• + ..-,) (.,• - 2). If y = 0, x= — 2 or —.5 or 2, and there are three points of the curve mi the axis of x. If x — 0, y=—\, and there is one intercept on the axis of //. If x<— 2, all three factors are negative; therefore y < 0, and the corresponding part of the curve lies below the axis of x. If — 2 < x <— .5, the first factor is positive and the other two are negative; therefore y > 0, and the corresponding part, of the curve lies above the axis of x. If — .5 < x < 2, the first two factors are positive and the third is negative; therefore y<0, and the corresponding part of the curve lies below the axis of x. Finally, if x > 2, all the factors are positive; therefore //><», and the corresponding part of the curve lies above the axis of x. Assuming values of X and finding the corresponding values of y, we plot the curve as represented in fig. 22. Ex. 2. y = .5 (x + 2.5) (x - l) 2 . If y = 0, x = — 2.5 or 1, and there are two points of the curve on the axis of x. If x = 0, y = 1.25, and there is one intercept on the axis of ?/. Fig. 22 22 GRAPHS OF ALGEBRAIC FUNCTIONS If x < — 2.5, the first factor is negative and the second factor is positive ; therefore y < 0, and the corresponding part of the curve lies below the axis of x. If — 2.5 < x < 1, both factors are positive ; therefore y > 0, and the corresponding part of the curve lies above the axis of x. Finally, if x > 1, we have the same result as when — 2.5 < x < 1, and the curve does not cross the axis of x at the point x = 1 but is tangent to it. Assuming values of x and finding the corre- sponding values of y, we plot the curve as repre- sented in fig. 23. Since it will be shown in § 31 that an equation of the first degree in x and ?/, Ax + By+ C=Q, Fig. 23 always represents a straight line, and since a straight line is determined by two points, it is generally sufficient in plotting an equation of the first degree to find the intercepts on the two axes and draw a straight line through the two points thus determined. The only exception is when the straight line passes tlrrough the origin, in which case some point of the straight line other than the origin must be found by trial. Ex. 3. Plot the line 3a; — 5y + 12 = 0. Placing y = 0, we find x = — 4. Placing x = 0, we find y = 2f. We lay off OL = - 4, OK = 2|, and draw a straight line through L and K (fig. 24). Fig. 24 Fig. 25 Ex. 4. Plot the line 3 x — 5 y = 0. Here, if x = 0, y - 0. If we place = 1, we find y = f . The line is drawn through (0, 0) and (1, f) (fig. 25). SYMMETRY 23 12. Symmetry and impossible values. A curve is symmetri- cal with respect to the axis of x when to each value of x in its equation correspond two values of y, equal in magni- tude and opposite in sign. This occurs in the simplest manner when y is equal to plus or minus the square root of a func- tion of x. Any value of x which makes the function under the radical sign positive gives two points of the curve equi- distant from the .r-axis. Values of x which make the function under the radical sign negative make y imaginary and give no points of the curve. These values of x we call impossible values. Similar remarks hold for symmetry with respect to the axis of y. How symmetry with respect to other lines may sometimes be determined is shown by Ex. 5. Ex.1. y=± V(x + 2) (x - 1) (x - 5). If x=— 2, 1, or 5, y = 0, and the graph intersects the axis of X at three points. The lines x=— 2, x — 1, and x = 5 divide the plane (fig. 26) into four sections. If x < — 2, all three factors of the product are negative; hence the radi- cal is imaginary and there can be DO part of the graph in the correspond- ing section of the plane. If — 2, the first two factors are positive and the third is negative; hence the radical is imaginary and there can be no part of the graph in the corresponding' section of the plane. Finally, if x > 5, all three factors are positive ; hence the radical is real and there is a part of the graph in the corre- sponding section of the plane. Therefore the graph consists of two separate parts and is seen (fig. 2G) to consist of a closed loop and a branch of infinite length. 24 GRAPHS OF ALGEBRAIC FUNCTIONS Ex. 2. y = ± V(jc + 2) (a; - l) 8 . This will be written as i/ = ±(x-l)Vx + 2. The line x = — 2 divides the plane (fig. 27) into two sections. Proceeding as in the previous example, we find the radical to be real if x > — 2 and imaginary if x < — 2. Therefore there is a part of the graph to the right of the line x = — 2, but there can be no part of the graph to the left of that line unless x can have a value that makes the coefficient of the radical zero; and this coefficient is zero only when x equals unity. Hence all of the graph lies to the right of the line x = — 2, as shown in fig. 27. To every value of x correspond two values of y which are in general distinct but become equal when x = 1. Hence the curve crosses itself when x = 1. Comparing this example with Ex. 1, we see that by changing the factor x — 5 to x — 1 we have joined the infinite branch and the loop, making a continuous curve crossing itself at the point (1, 0). Fig. 27 Ex. 3, y = ±V(x + 2) 2 (x- l) = ±(x + 2)Vx-l. The line x = 1 divides the plane (fig. 28) into two sections. If x > 1, the radical is real and there is a part of the graph in the corresponding sec- tion of the plane. If x < 1, the radical is imaginary and there will be no points of the graph excel >t Ior sucn values of x as make the coefficient of the radical zero. There is but one such value, — 2, and therefore there is but one point of the graph, (— 2, 0), to the left of the line x = 1. The graph con- sists, then, of the isolated point A and the infinite branch (fig. 28). Comparing this example also with Ex. 1, we see that by changing the factor x — 5 to x + 2 we have reduced the loop to a single point, leaving the infinite branch as such. INFINITE VALUES 25 Ex.4. y = ±V-(x + 4:)(x + 2) 2 (x-fy = ± (x + 2) V-(a; + 4)(x-4). The lines x = — 4 and x = 4 divide the plane (fig. 29) into three sections. If — 4 < x < 4, the radical is real and there is a part of the graph in the corresponding portion of the plane. If x<— 4 or x > 4, the radical is imagi- nary; and since in the corresponding sections there is no value of x which "makes x + 2 zero, there can be no part of the graph in those sections. It is represented in fig. 29. Ex. 5. 2 x" + f + 3 x - 4 y - 5 = 0. Solving fur y, we have 2 ±y/~. y 3 x + 9, Fig. 29 or, after the expression under the radical sign has been factored, y = 2±V-2(*-|)(* + 8> The lines x = — 3 and x = •?, divide the plane (fig. 30) into three sections, and proceeding as before, we find that the curve is entirely in the middle section (that is, when — 3 -~ 1 i i Fig. 30 13. Infinite values. If the expression defining a function con- tains fractions, the function is not defined for a value of x which makes the denominator of any fraction zero. But if x = a is a value which makes the denominator zero, but not the numerator, and if x is allowed to approach a as a limit, the value of the func- tion increases indefinitely and is said to become infinite. The graph of a function then runs up or down indefinitely, approach- ing the line x — a indefinitely near but never reaching it. 20 GRAPHS OF ALGEBRAIC FUNCTIONS This is expressed concisely by the formula Other important formulas involving infinity GO X G = 00, which may be explained in a similar manner. For example, to c c obtain the meaning of — , we may write - and then allow x to oo x increase indefinitely. It is obvious that the quotient decreases in numerical value and may be made as small as we please by tak- ing x large enough. This is the meaning of the formula — = 0. 00 It is evident that y is real for all values of x ; also, if x < 2, y is nega- y is tive, and if x > 2, y is positive. Moreover, as x increases toward negative and becomes indefinitely great ; while as x decreases toward 2, y is positive and becomes indefinitely great. We can accordingly assign all values to x except 2. The curve is represented in fig. 31. It is seen that the nearer to 2 the value assigned to x, the nearer the corresponding point of the curve to the line x = 2. In fact, we can make this distance as small as we please by choosing an appropriate value for x. At the same time the point recedes indefi- nitely from OX along the curve. Now, token a straight line has such a position with respect to a curve that as the two are indefinitely prolonged the distance between them approaches zero as a limit, the straight line is called an asymptote of the curve. It follows from the above definition that the line x = 2 and also the line y = are asymptotes of this curve. In this example it is to be noted that the asymptote x = 2 is determined by the value of x which makes the function infinite. It is clear that all equations of the type 1 represent curves of the same general shape as that plotted in fig. 31. INFINITE VALUES 27 Ex. 2. + 2 If x 2 or if x is infinite; hence these two values may not be assigned to x, all other values, however, being possible. The curve is represented in fig. 32. By a discussion similar to that of Ex. 1 it may be proved that the lines x = — 2 and x — 2, which correspond to the values of x which make the function infinite, and also the line y = 0, are asymptotes of the curve. This curve is a special case of that represented by 1 " c> II II H « V. "1 1 Fig. 32 and it is not difficult to see how the curve r< -presented by X — (l J— h X — C will look for any number of terms. Ex. 3. (*-2) s All values of x may be assumed except 2. The curve is represented in fig. :!:'>. It is evident that the* lines x — 2 and ij = are asymptotes. This curve is a special case of that repre- sented by 1 y (x-af which is itself a special case of (x _ a)2 (x _ 6)a Fig. 33 28 GKAPHS OF ALGEBltAIC FUNCTIONS Ex. 4. f 1 a- 3' We solve for y, forming' the equation y = ± ■ The line x = (fig. 34) divides the plane into two sections, and it is evident that there can be no part of the curve in that section for which x < 3. Moreover, this line x = 3 is an asymptote, as in the preceding examples. The curve, which is a special case of that represented by 1 x — a is represented in fig. 34. It is to be noted that the axis of x also is an asymptote. f Ex. 5. y = a; 2 + 1 To plot this curve we write the equation in the equivalent fori: , = , + 1. (1) It is evident that all values except may be assigned to x, that value being excluded as it makes y infinite. Let us also draw the line y = x, (2) a straight line passing through the origin and bisecting the first and the third quadrants. Comparing equations (1) and (2), we see that if any value x x is assigned to x, the corresponding ordinates of (1) and (2) a?i + and and that Moreover, the numerical are respectively they differ by — • value of this difference decreases as greater numerical values are assigned to x v and it can be made less than any assigned quantity however small by taking x 1 Fig. 35 sufficiently great. It follows that the line y = x is an asymptote of the curve. It is also evident that the line x = 0, determined by the value of x which makes the function infinite, is an asymptote. The curve is represented in fig. 35. INTERSECTION 29 14. Intersection of graphs. Let /„<>,£> =0 (1) and /„(«,y)=0 (2) be the equations of two curves. It is evident that any point common to the two curves will have coordinates satisfying both (1) and (2), and that, conversely, any values of x and y which satisfy both (1) and (2) are coordinates of a point common to the two curves. Hence, to find the points of intersection of two curves, solve their equations simultaneous///. The simplest case which can occur is that where each equation is of the first degree and hence (§ 31) represents a straight line. In general there is a single solution, which locates the single point of intersection of the two straight lines. If no solution can be found, it is evident that the lines are parallel. Other important cases are the two following: Case I. f(x, y) = and f n (x, y) = 0. Let / 1 (^^)= ' (1) /„O>30=0, (2) be a linear equation and an equation of the nth degree, where n > 1. The degree of a curve is defined as equal to the degree of its equation. Accordingly this problem is to find the points of intersection of a straight line and a curve of the rath degree, and the method of solution is as follows: Solve (1) for either x or y and substitute the result in (2). If, for example, we solve (1) for ?/, the result of substituting this value in (2) will in general be an equation of the rath degree in x, the real roots of which are the abscissas of the required points of intersection. The ordinates of the points of intersection are now found by substituting in succession in (1) the values of x which have been found. If two roots x 1 and x 2 of the equation in x are equal, the cor- responding ordinates are equal and the two points coincide. 30 GRAPHS OF ALGEBRAIC FUNCTIONS We may regard this case as a limiting case when the position of the curves is changed so as to make x x and x 2 approach each other ; that is, so as to make the points of intersection of the straight line and the curve approach each other along the curve. Accordingly the straight line represented by equation (1) is, by definition, tangent to the curve represented by equa- tion (2). In general the tangent line simply touches the curve, without cutting it, as in the case of the circle. Ex. 1. Find the jioints of intersection of 3a; -2y- -4 = (1) and x" — iy = 0. (2) Solving (1) for y and sub- stituting the result in (2), we have x 2 — G x + 8 = 0, the roots of which are 2 and 4. Substi- tuting these values of x in (1), we find the corresponding values of y to be 1 and 4. Therefore the points of inter- section are (2, 1) and (4, 4) (fig. 36). Fig. Ex. 2. Find the points of intersection of x - 4 y - 9 = (1) and x 2 - 4 y = 0. (2) Solving (1) for y and sub- stituting the result in (2), we have x 2 - G x + 9 = 0. The roots of this ecpiation are equal, each being 3. Hence the straight line is tangent to the curve. Substituting 3 for x in (1), we find y = Z ; hence the point of tangency is (3, |) (fig. 37). Fig. 37 INTERSECTION 31 Ex. 3. Find the points of intersection of 3x-2v/-5 = and x 2 — 4 y = 0. (1) (2) Proceeding as in the two previous examples, we obtain x 2 - 6 x + 10 = 0, the roots of which are 3 ±V— 1. Hence the straight line does not intersect the curve (fig. 38). The corresponding values of y are 2 ± % V^l. Ex. 4. Find the points of intersection of y = 2x (1) and y 2 = x(x-Z) 2 . (2) Substituting the value of y from (1) in (2), we have ar[(x-3) a -4z] = 0, or x[x 2 -10x + !)] = 0. Its roots are 0, 1, and 9. The corresponding values <>f y are found from (1) to be 0, 2, and 18. Therefore the points of intersection are (0, 0), (1, 2), and (9, 18) (fig. 39). Ex. 5. Find the points of intersection of Fig. 40 y = 3 x + 2 (1) and y = x s . (2) Substituting in (2), we have X s - 3 x - 2 = 0, or (.r-2)(x + l) 2 = 0. Its roots are 2, —1, — 1. The corresponding values of y, found from (1), are 8, — 1, — 1. Therefore the points of intersection are (2, 8) and (— 1, — 1), the latter being a point of tangency (fig. 10). Fig. 39 32 GRAPHS OF ALGEBRAIC" FUNCTIONS 12). (1) (2) Ex. 6. Find the points of intersection of 2x + y-4 = and if- — x (a? After substitution we have x 3 — 4 x 2 + 4 x — 16 = 0, or (x — 4) (x 2 + 4) = 0, the roots of which are 4 and ± 2 V— 1. The corresponding values of y, found from (1), are - 4 and 4 =F 4 V^. The only real solution of equations (1) and (2) being x — 4 and y = — 4, the straight line and the curve intersect in the single point (4, - 4) (fig. 41). Case II. f m (x, y) = and f n (x, z/) = 0. Let / m O, #) = (1) be an equation of the with degree, and /„£*, y) = (2) be an equation of the nth. degree, where m and n are both greater than unity. The method is the same as in the preceding case ; that is, the elimination of either x or y, the solution of the resulting equation, and the determination of the corresponding values of the unknown quantity eliminated. The equation resulting from the elimination is in general of degree 7nn, and the number of simultaneous solutions of the original equations is mn. If all these solutions are real and distinct, the corre- sponding curves intersect at mn points. If, however, any of these solutions are imaginary, or are alike if real, the correspond- ing curves will intersect at a number of points less than mn. Hence two curves of degrees m and n respectively can intersect at mn points and no more. Tig. 41 and Ex. 7. Find the points of intersection of = = 0. y x 2 + y" _ o (1) (2) Subtracting (1) from (2), we eliminate y, thereby obtaining the equa- tion x 2 + 2 x — 8 = 0, the roots of which are — 4 and 2. Substituting 2 INTERSECTION 33 anr l _ 4 i n either (1) or (2), we find the corresponding values of y to be ± 2 and ± 2 V— 2. The real solutions of the equations are accordingly x = 2, y = ± 2, and the corresponding curves intersect at the points (2, 2) and (2, - 2) (fig. 42). From the figure it is also evident that the value — 4 for x must make y imaginary, as both curves lie entirely to the right of the line x = — 4. Ex. 8. Find the points of intersec- tion of and 3 y 3 a: (1) y* - 6 x = u. (2) Substituting in (2) the value of y from (1), we haw x 4 - 27j; = 0. This equation may be written s(z-3)(x* + 8x + 9)' = 0, the roots of which are 0, 3, and — V-3 Fig. 42 Substituting these values of x in (1), we find the corresponding values of y to be 0, :'>. and 3T3V- :$ Therefore the real solutions of the ationa are x = 0, y = and x= 3, y= '■'>. If we had substituted the values of x in (2), we should have at first seemed to find an additional real solution, y =— 3 when a: = :>. But — 3 for y makes x imaginary in (1 ), as no part of (1) is below the axis of x. Geometrically, the line x = -\ intersects the curves (1) and (2) in a common point and also intersects (2) in another point, Therefore the only real solutions of these equations are the ones noted above, and the correspond- ing curves intersect at the two points (0, 0) and (3, 3) (fig. 43). We see, moreover, that any results found must he tested by substitution in both of the original equations. The remaining two solutions of these equations, found by letting - 3 ± 3 V^3 x = > are imaginary. 34 GRAPHS OF ALGEBRAIC FUNCTIONS Ex. 9. Find the points of intersection of 2 x 2 + 3 y* = 35 (1) and xy = 6. (2) Since these equations are homogeneous quadratic equations, we place y = mx (3) and substitute for y in both (1) and (2). The results are 2 a; 2 + 3 ra 2 x 2 = 35 and mx 2 = 6, whence x 2 - 35 (4) 2 + 3 m 2 and x 2 = — • m (5) Therefore 35 6 «\\ 2 + 3 m 2 m v ' from which we find m = % or f . If m= i) then, from (5), x=±2; and from (3) the corresponding values of y are ± 3. If m = §, in like manner we find x = ± \ ^Q an d y = ± 1 V6. Therefore the curves intersect at the four symmetrically situated points (2, 3), (- 2, - 3), (3 Vo", 2 Vo), (- 1 Vti, - § Vo) (fig. 44). Ex. 10. Find the points of intersection of 2y 2 = x-2 (1) and x 2 - 4 y 2 = 4. (2) Eliminating ?/, we have x 2 - 2 x = 0, the roots of which are and 2. When x = we find fro m e ither (1) or (2) y — ± V— 1, and when x = 2 either (1) or (2) reduces to y 2 = 0, whence y = 0. Therefore these two curves are tangent at the point (2, 0) (fig. 45). 15. Real roots of an equation. It is evident that the real roots of the equation f(x) = determine points on the axis of x at which the curve y =f(x) crosses or touches that axis. Moreover, if x x and x„ (x x < x 2 ~) are two values of x such that X Fig. it REAL ROOTS 35 f(x^) and /(# 2 ) are of opposite algebraic sign, the graph is on one side of the axis when x = x^ and on the other side when x = x 2 . Therefore it must have crossed the axis an odd number of times between the points x = x x and x = x 2 . Of course it may have touched the axis at any number of intermediate points. Now if f(x) has a factor of the form (x— a~) k , the curve y=f(x) crosses the axis of x at the point x = a when k is odd, and touches the axis of x when k is even. In each case the equa- tion f(x) = is said to have k equal roots, x = a. Since then a point of crossing corresponds to an odd number of equal roots of an equation and a point of touching corresponds to an even number of equal roots, it follows that the equation f(x) = has an odd number of real roots between x x and # 2 , if /(',) and f(x^) have opposite signs. The above gives a ready means of locating the real roots of an equation in the form/(.r)= 0, for we have only to find two values of #, as x x and x 2 , for which f(x) has different signs. We then know that the equation has an odd number of real roots between these values, and the nearer together x and x , the more nearly do we know the values of the intermediate roots. In locat- ing the roots in this manner it is not necessary to construct the corresponding graph, though it may be helpful. Ex. Find a real root of the equation i 8 + 2/- 17 = 0, accurate to two decimal places. Denoting x s + 2 x — 17 by /"(•>') al "^ assigning successive integral values to x, we nnd/(2)= — 5 and/(3) = lG. Hence then' is a real root of»the equation between 2 and 3. We now assign values to x between 2 and 3, at intervals of one tenth, as 2.1, 2.2, 2.3, etc., and we begin with the values nearer 2, since; /(2) is nearer zero than is /(3). Proceeding in this way we find/(2.3) =— .233 and/(2.4) = 1.624 ; hence the root is between 2.3 and 2.1. Now, assigning values to x between 'J.:; and 2.4 at intervals of one hun- dredth, we find /(2.31)=— .054 and /(2.32) = .127 ; hence the root is between 2.31 and 2.32. To determine the last decimal place accurately, we let x = 2.315 and find /(2.315) = .037. Hence the root is between 2.31 and 2.315 and is 2.31, accurate to two decimal places. If /(2.315) had been negative, we should have known the root to be between 2.315 and 2.32 and to be 2.32, accurate to two decimal places. 36 GRAPHS OF ALGEBRAIC FUNCTIONS PROBLEMS Plot the graphs of the following equations: 1. 3* + 4// -7=0. 18. //=(.r + l)(.r-4)(.x-3) 2 . 2. 2x-5y+G = 0. 19. i/=(x-l)(x + 3)(x* + 2). 3. a-4-7y = 0. 20. y=(x-l) 2 (2x 2 4-6a: + 5). 4. 4:r — 3 = 0. 21. y 2 = (x - 2) (a; 2 - 9). 5. 3 y + 8 = 0. 22. y 2 = {x + 3)(6 a - a: 2 - 8). 6. y = 4a 2 +4a-3. 23. 9 y 2 = (x + 2) (2 x - If. 7. y = 4 sr - 2 x + 3. 24. 4 y 2 = a; 3 + 4 a: 2 . 8. y = 6 - x - x 2 . 25.9 // = (a 2 - 1) (4 x 2 - 25). 9. 7/ = - 3 x 2 + 4 x. 26. y 2 = (1 - a 2 ) (4 a; 2 - 25). 10. y = (a4-2)(.r-l)(.r-3). 27. 4 y 2 = 9 x i - a 6 . 11. y = ( ft - 2 - 1) (2.x + 9). 28. y 2 = (2a + 3)(4a 2 -9). 12. y = a 3 + 4 a- 2 . 29. y 2 = (x - 2) 2 (3 - 2 a;). 13. y = (x-3)(2x + lf. 30. y 2 =(2 + a-a 2 )fa + 2) 2 . 14. y = a; 8 -8a; a +15aj. 31. f = x 2 (x - 5) a (2 x - 3). 15. y = 2 ft 3 +3 a 2 -14 «. 32. 4y 2 =(a--l) 2 (4a 2 -4a-3). 16. y = a 8 - a 2 - 4 a; + 4. 33. f = a (as + 2) 2 (x + 3) 2 . 17. y = x s - a 2 x. 34. y 2 = (2 a; - 3) (a 2 + 1). 35. y 2 = (x - 1) (2 x - l) 2 (.r 2 -f 3 x + 3). 36. a- 2 4- y 2 — 4 a; 4- 6 y + 9 = 0. 37. K 2_4y_|_4y2 = 0. 48. (y - a-) 2 = 16 - a- 2 . 38. ^-y 2 - 2a + 4 y- 4 = 0. ' 49. (x + y) a = y 2 (y + 1). 39. 9a 2 + 36y 2 -96y + 28 = 0. 50. a 2 -4ay -5y 2 + 9 y 4 = 0. 40. a 3 + 3 a; 2 - y 2 - a; - 3 = 0. 41. y 3 = a(a 2 -9). 42. y 3 = a 2 (a 4- 3). 43. (y + l) 3 =(;*4-l)(a 2 -9). 44. a- 2 —if {% + y) = 0. 45. x 2 — if 4- if 4- 2 y = 0. 46. (y + 3) 2 = x (x - 2) 2 . 47. (y-2) a = (x-2) 2 (a;-5). 55. y 2 (a 2 + a 2 ) = a 2 (a 2 51. Q + ^ = 1 52. (DM!)- 53. a;* 4- y * = a* 54 ■(1) +(!)-• PROBLEMS 37 56. aY + b 2 x 4 = a 2 b 2 x 2 . m . _ , 1 J 67. 4y = 2» + - — 57. 16aY=Px 3 (a 2 —2ax). 2x 58. xy = 16. 68. y = x + — • 59. xy = — 16. 3 61. fr + iy.-iy TO. ,-«> + §. ^^F+V'^W 63. jf = -j x 2 — 5 a; — 6 x-1 72. xy 2 = 4a 2 (2a-x). * 3 J 2 a — x 2 _ X ~ X ' * - (x +l)(x + 3)' ?4 _ x 2 f„+x) 65. x 2 y 2 + 25 = 9/ : BW. x(x+l) 75. y 9 (a 5 a + a 9 ) = a«a ■' a - 1 76. I/O-' + a*) = a-(u - x). Find the points of intersection of the following pairs of loci : 77. 3 x _ j, _ 2 = 0, 5 x - 3 // + 2 = 0. 78. 6 x - 24 y +19 = 0, 12'x + 3y + 4 = 0. 79. 2 x - y- 2 = 0, X s + y 2 = 25. 80. 2 x - 3 y + 9 = 0, ./•- + y* + 2x + 4 y -8 = 0. 81. 4 x + 5 y - 20 = 0, x- + y 2 - 2x - 3 = 0. 82. 2y + 3x-5= 0, X s - 2x - 2y + 4 = 0. 83. 3 x - 2 y + G = 0, if + 4 y + x + 7 = 0. 84. x - 4 y + 1 = 0, 4 y 2 + 4 y - 4 x + 5 = 0. 85. 3 x + 2 y - 7 = 0, 5 x 2 + 4 y* = 21. 86. 7x - 2 y + 4 = 0, 21 x 9 - 4 //- - 12 = 0. 87. x - 2 y = 0, xV + 30 = 25 y 2 . 88. 2x-y-l=0, 4y 2 =(x + 2)(2x-l) 2 . 89. x + 2 y - 2 = 0, y + x 2 // = 1. 90. x 2 + y 2 = 25, 16 x 2 + 27 y 2 = 576. 91. X s + y 2 = 12, x 2 - 8 y + 8 = 0. 4xy=l, 2x 2 + 2y 2 = l. 92 38 GRAPHS OF ALGEBRAIC FUNCTIONS 93. 32// 2 -9:z 8 = 0, %f-§x = 0. 94. 2y* = 3 -x, y 2 = T7^ — 95. x 2 - f = 0, x 2 + f - 4 y - 4 = 0. 96. 7^ = 25-5^^-2 = ^. 97. jf-2^, 16(v / -2) = ( a; -l) 3 . Find the real roots, accurate to two decimal places, of the following equations : 98. x s + 2x- 6 = 0. 101. x* - 4x 3 4- 4 = 0. 99. a; 3 + x + 11 = 0. 102. x* - 3 a; a + 6 x - 11 = 0. 100. a; 4 - 11 x + 5 = 0. 103. x 3 + 3 ar 4- 4 a; + 7 = 0. CHAPTER III CHANGE OF COORDINATE AXES 16. Introduction. So far we have dealt with the coordinates of any point in the plane on the supposition that the axes of coor- dinates are fixed, and therefore to a given point corresponds one, and only one, pair of coordinates, and, conversely, to any pair of coordinates corresponds one, and only one, point. But it is some- times advantageous to change the position of the axes, that is, to make a transformation of coordinates, as it is called. In such a case we need to know the relations between the coordinates of a point with respect to one set of axes and the coordinates of the same point with respect to a second set of axes. The equations expressing these relations are called formulas of transformation. It must be borne in mind that a trans- formation of coordinates never alters the position of the point in the plane, the coordinates alone being changed because of the new standards of reference adopted. 17. Change of origin. In this case a new origin is chosen, but the new axes are respectively parallel to the original axes. Let OX and Y (fig. 40) be the original axes, and O'X' and O'Y' the new axes intersecting at 0', the coordinates of 0' with respect to the original axes being x o and y Q . Let P be any point in the plane, its coordinates being x and y with respect to OX and F, and x' and y' with respect to O'X' and 0' Y'. Draw PMM' parallel to OY, intersecting OX and O'X' at M and M' respectively. 39 1 r 3 r p t \M 0' x M' Fig. 46 40 CHANGE OF COORDINATE AXES Then OM = x, MP = y, 0'M' = x', M'P = y', NO'=x , ON=y . But OM= NM' = NO' + O'M ', and MP = MM' + M'P = ON+ M'P. Therefore x = x + x', y = y o + y', which are the required formulas of transformation. Ex. 1. The coordinates of a certain point are (3, — 2). What will be the coordinates of this same point with respect to a new set of axes parallel respectively to the first set and intersecting at (1, — 1) with respect to OX and OY1 Here x = 1, y — — 1, x = 3, and y = — 2. Therefore 3 = 1 + x' and — 2 = — 1 + y', whence x' — 2 and y' = — 1. Ex. 2. Transform the equation y 1 — 2 y — 3 x — 5 = to a new set of axes parallel respectively to the original axes and intersecting at the point (-2, 1). The formulas of transformation are x = — 2 + x', y = 1 + y'. Therefore the equation becomes (1 + y'f - 2 (1 + 30 - 3 (- 2 + O - 5 = 0, or y' 2 - 3 x' = 0. As no point of the curve has been moved in the plane by this transformation, the curve has been changed in no way whatever. Its equation is different because it is referred to new axes. After the work of transformation has been completed the primes may be dropped. Accordingly the equation of this example may be written y 2 — 3 x — 0, or y 2 = 3 x, the new axes being now the only ones considered. 18. One important use of transformation of coordinates is the simplification of the equation of a curve. In Ex. 2 of the last article, for example, the new equation y 1 = 3 x is simpler than the original equation. It is obvious, however, that the position of the new origin is of fundamental im- portance in thus simplifying the equation, and we shall now solve examples illustrating methods of determining the new origin to advantage. CHANGE OF ORIGIN 41 Ex. 1. Transform the equation y 2 - 4 y - x 3 - 3 x 2 — 3 x + 3 = to new axes parallel respectively to the original axes, so choosing the origin that there shall be no terms of the first degree in x and y in the new equation. The formulas of transformation are x = x + x' and y = y + y', where suitable values of x and y are to be determined. The equa- tion becomes (i/o + y'f - 4 0/o + y') - (*o + x'Y - 3 (* + *y - 3 (*o + «0 + 3 = °> or, after expanding and collecting like terms, y* + (2^-4)/- x' 3 -(3x + 3)x' 2 -(3x 2 + C/ + 3)x' + G/o 2 ~ 4 Z/o - *o 8 ~ 3 x o - 3 X + 3) = 0. By the conditions of the problem we are to choose x and ?/„ so that 2# -4 = 0, 3x^ + 0x0+3 = 0, two equations from which we find x = — 1 and y = 2. Therefore (— 1, 2) should be chosen as the new origin, and the new equation is y'- — x' 3 = 0, or y" — x 3 after the primes are dropped. Ex. 2. Transform the equation 10 x 2 + 25 if + 04 x - 150 y - 111 = to new axes parallel respectively to the original axes, so choosing tin- origin that there shall be no terms of the first degree in x and y in the new equation. We may solve this example by the method used in solving Ex. 1, but since the equation is of the second degree, the following method is very , desirable. Rewriting, we have 16 (x 2 + 4 x) + 25 O 2 - 6 y) = 111 ; whence 16 (x 2 + 4 x + 4) + 25 (y* - 6 y + 9) = 400, or 10(x + 2)'-+25(?/-3) 2 =400. Placing now x = — 2 + x / , y = 3 + y', we have as our new equation 16 x' 2 + 25 y' 2 = 400, the new origin of coordinates being at the point (— 2, 3) with respect to the original axes. 42 CHANGE OF COORDINATE AXES 19. Change of direction of axes. Case I. Rotation of axes. Let OX and OY (fig. 47) be the original axes, and OX' and OY' be the new axes, making Z $ with OX and OY respectively. Then Z XOY' = 90° + and ZYOX'= 90°- (f>. Let P be any point in the plane, its coordinates being x and y with respect to OX and OY, and a/ and y with respect to OX' and OY'. Then, by construction, OM=x y OX=y, OM' = x', and M'P = y'. Draw OP. The projection of OP on OX is OJf, and the projection of the broken line OM'P on OX is OM' cos <£ + Jf 'P cos (90° + <£), or OJf' cos - M'P sin . Therefore M = OM' cos (/> - M'P sin 0, by § 2. In like manner the projection of OP on OY is (9 A 7 ", and the projection of the broken line OM'P on OY is OJf'cos (90°- <£) + ilPPcos<£. Therefore OA 7 " = OM' sin <£ + M'P cos (/>, by § 2. Replacing OJf, 0A", OJf', M'P by their values, we have x = x' cos (f> — y' sin 0, y = x 1 sin + y' cos . Ex. 1. Transform the equation xy = 5 to new axes having the same origin as the original axes and making an angle of 45° with them. , , Here = 45°, and the formulas of transformation are x = V V2 Substituting and simplifying, we have as the new equation x 2 — y 1 = 10. Ex. 2. Transform the equation 34 x 1 + 41 y 2 — 24 xy = 100 to new axes with the same origin as the original axes, so choosing the angle <£ that the new equation shall have no term in xy. V2 OBLIQUE COORDINATES 43 The formulas of transformation are x = x' cos — y' sin <£, y = x' sin + y cos , where is to be determined. Substituting in the equation and collecting like terms, we have (34 cos a <£ + 41 sin 2 - 24 sin <£ cos (f>) x 2 + (34 sin 2 + 41 cos 2 + 24 sin <£ cos <£) y» + (24 sin 2 <£ + 14 sin cos- 24 cos 2 <£) xy = 100. By the conditions of the problem we are to choose so that 24 sin 2 + 14 sin<£ cos — 24 cos 2 = 0. One value of satisfying this equation is tan -1 1. Accordingly we sub- stitute sin <£ = i?- and cos = $, and the equation reduces to x 1 + 2 y 1 = 4. Case II. Interchange of axes. If the axes of x and y are simply interchanged, their directions are changed, and hence such ii trans- formation is of the type under consideration in this article. Hie formulas for such a transformation are evidently x — y',y = x'. Case III. Rotation and interchange of axes. Finally, if the axes are rotated through an angle and then interchanged, the formulas, being merely a combination of the two already found, are x = y' cos <£ — x' sin , y = y' sin + x' cos . A special case of some importance occurs when = 270°. We have then x = x', y =— y'. Cases II and III, it should be added, occur much less frequently than Case I. If both the origin and the direction of the axes are to be changed, the processes may evidently be performed successively, preferably in this order : (1) change of origin ; (2) change of direction. 20. Oblique coordinates. Up to the present time we have always constructed the coordinate axes at right angles to each other. This is not necessary, however, and in some problems, indeed, it is of advantage to make the axes intersect at some other angle. Accordingly, in fig. 48, let OX and OY intersect at some angle to other than 90°. 44 CHANGE OF COORDINATE AXES We now define x for any point in the plane as the distance from OF to the point, measured parallel to OX, and y as the distance from OX to the point, measured parallel to OY. The algebraic signs are determined accord- ing to the rules adopted in § 4. It is immediately evident that the rectangular coordinates are but a special case of this new type of coordinates, called oblique coordinates, since the new definitions of x and y include those pre- viously given. In fact, the term Carte- sian, or rectilinear, coordinates includes both the rectangular and the oblique. Oblique coordinates are usually less convenient than rectangu- lar ones and are very little used in this book. If necessary, the formulas obtained by using rectangular coordinates can be trans- formed into similar ones in oblique coordinates by the formulas of the following article. When no angle is specified the angle between the axes is understood to be a right angle. 21. Change from rectangular to oblique axes. Let OX and OY (fig. 49) be the original axes, at right angles to each other, and OX' and OY' the new Y axes, making angles (f> and (f>' respectively with OX. Then co = $ - . Let P be any point in the plane, its rectangular coordinates being x and y, and its oblique coordi- nates being x' and y'. Draw PM parallel to OY, PM' parallel to OY', M'N parallel to OY, and RM'N' parallel to OX. Then ZBM'P='. But 031= ON + NM = ON + M'N' = 031' cos + 31' P cos ', 3IP = 3IN'+ N'P = N3f + N'P = 031' sin (f> + 3f'P sin '. Therefore x — x' cos <$> + y' cos ', y = x' sin + y' sin '. p / ^-x' '-^v -R ^V\ ! \ M N Fig. 49 PROBLEMS 45 Ex. Transform the equation — - — "— = 1 to the lines y = ± - x as axes. «- b- a Here let = tan -1 ( — '-), and <£' = tan -1 -- The formulas of trans- _.. , \ a] a formation become 0'+ y'), y = ,- — - (- x ' + /)• Vo 2 + b- Va 2 + //- a 2 + b- Substituting and simplifying, we have as the new equation xy = — i 4 Unless b = a, the axes are oblique and w = 2 tan -1 -- a 22. Degree of the transformed equation. In reviewing this chapter we see that the expressions for the original coordinates in terms of the new are all of the first degree. Hence the result of any transformation cannot be of higher degree than the original equation. On the other hand, the result cannot be of lower degree than the original equation ; for it is evident that if any equation is transformed to new axes and then back to the original axes, it must resume its original form exactly. Hence, it' the degree had been lowered by the first transformation, it must be increased to its original value by the second transformation. But this is impossible, as we have just noted. It follows that the degree of an equation is unchanged by any single transformation of coordinates or by any number of successive transformations. In particular, the proposition that any equation of the first degree represents a straight line is true for oblique, as for rectangular, coordinates. PROBLEMS 1. What are the new coordinates of the points (3, 4), (— 3, 6), and (4, — 7) if the origin is transferred to the point (2, — 3), the new axes being parallel to the old ? 2. Transform the equation a- 2 + 9// 2 — 4.z+18?/ + 8 = 0to new axes parallel to the old axes and meeting at the point (2, — 1). 3. Transform the equation 2x 2 + 2?/ 2 — 2x -{- 2 >/ — 7=0 to new axes parallel to the old axes and meeting at the point (^, — £). 4. Transform the equation a- 2 — if + 2x — 3 = to new axes parallel to the old axes and meeting at the point (— 1, 0). 46 CHANGE OF COOBDINATE AXES 5. Transform the equation f - 3 if + 3 x 2 + 3y/-fl2*+ll=0 to new axes parallel to the old axes and meeting at the point (— 2, 1). 6. Transform the equation f — 4 a* — 6// + 5 = to new axes parallel to the old axes, so choosing the new origin that the new equation shall contain only terms in if and x. 7. Transform the equation x 2 + 2x + 4?/ — 3 = to new axes parallel to the old, so choosing the new origin that the new equation shall contain only terms in x 2 and y. 8. Transform the equation 2 x 2 — 4 // + 12 x + 16 y — 7 = to new axes parallel to the old, so choosing the origin that there shall be no terms of the first degree in the new equation. 9. Transform the equation 4a; 2 + 9 y 2 — Ax +12 y + 4 = to new axes parallel to the old, so choosing the origin that there shall be no terms of the first degree in the new equation. 10. Transform the equation xy — 3 y + 2 x — 12 = to new axes parallel to the old, so choosing the origin that there shall be no terms of the first degree in the new equation. 11. Transform the equation 6xy — 10.x + 3 y — 19 = to new axes parallel to the old, so choosing the origin that there shall be no terms of the first degree in the new equation. 12. Show that any equation of the form xy + ax -4- by -f- c = can always be reduced to the form xy = k by choosing new axes parallel to the old, and determine the value of k. 13. Show that the equation y 2 -\- ay + hx -f- c = (b ^= 0) can always be reduced to the form if + bx = by choosing new axes parallel to the given ones. 14 . Show that the equation ax 2 + by 2 -f- ex -\- dy -\-e = 0(a=f=0,b=£0) can always be put in the form ax 2 + by 2 = k by choosing new axes parallel to the old, and determine the value of k. 15. What are the coordinates of the points (0, 2), (2, 0), (2, - 2) if the axes are rotated through an angle of 60°? 16. What are the coordinates of the points (1, 2), (2, 2), (2, -1) if the axes are rotated through an angle of 45°? 17. What are the coordinates of the points (1, 2), (—1, —2), (1, — 2) if the axes are rotated through an acute angle tan -1 !? 18. Transform the equation 2x 2 -f 2 y 2 — 3xy — 7= to a new set of axes by rotating the original axes through an angle of 45°. PROBLEMS 47 19. Transform the equation 4.r 2 + 2 ~v3xy + 2 f — 5 = to a new set of axes by rotating the original axes through a positive angle of 30°. 20. Transform the equation 4 .r 2 — 12 xy + 9 f — 14 = to a new set of axes making a positive angle tan -1 f with the original set. 21. Transform the equation 5 .r 2 -f 4 xy + 8 f — 36 = to a new set of axes by rotating the original axes through a positive angle tan -1 (— i). 22. Transform the equation 4 x 2, + 15 xy — 4 if — 34 = to a new set of axes making a positive angle tan -1 | with the original axes. 23. Show that the equation x* + f = ./■- -f- 2 I ./// — y 2 — 150 = to one which has no xy-term, by rotating the axes through the proper angle. 27. Transform the equal ion 1 6 x* - 2 I xy + 9 //" - 30 x - 40 y = to one which has no scy-tenn, by rotal ing the axes through the proper angle. 28. Transform the equation 11 x* -f 2 1 .#•// + 34 f -100. r - 50// — 100 = to one which has no xy-tevm., by rotating the axes through the proper angle. 29. Transform the _ equation 1 1 .r- - (') V:'» xy + 5 f + (22 — 12 V3)x -(20 + G V3)y +3-12 V3 = to a new set of axes making an angle of G0° with the original axes and intersecting at the point (—1, 2) with respect to the original axes. 30. Transform the equation 4 a- 2 + 25 f = 100 from rectangular axes to oblique axes with the same origin and making the angles tan -1 J and tan _1 (— f) respectively with OX. 31. Transform the equation 9 x 2 — 4 f = 36 from rectangular axes to oblique axes with the same origin and making the angles tan -1 | and tan -1 (— |) respectively with OX. 48 CHANGE OF COORDINATE AXES 32. Transform the equation 9 x 2 — 4 if = 36 from rectangular axes to oblique axes with the same origin and making the angles tan _1 | and tan -1 3 respectively with OX. 33. Prove that the formulas for changing from a set of rectangular axes to a set of oblique axes having the same origin and the same axis of x are x = x' + y' cos u>, y = y' sin w, where The function is therefore p-1000 — . increases without limit. 7 = e 1000 , and when * e 1000 discontinuous for x = 0. The line y = 1 is an asymptote (fig. 63), for as x increases without limit, being positive or negative, - approaches 0, and y approaches 1. 1 + (i* As x approaches zero positively, y ap- proaches zero. As x approaches zero nega- tively, y approaches 10. As x increases indefinitely, y approaches 5. The curve (fig. 6-i) is discontinuous when x = 0. PROBLEMS Plot the graphs of the following equations Fig. Pig. 64 1. y = £ sin 2x. 2. 2/ = 3sin^- 4. y = sin( x 5. y = 2s'm3(x — --)• 3. y=s8m[x + 6. y x-1 50 GRAPHS OF TRANSCENDENTAL FUNCTIONS 7. y = isin(2aj + 3). / *■> 30. y I. y = cos 3x. ). y = 3 cos -j- 31. y 17. y = sin -— — - sin 7ra: >(<*-!) 30. ?/ = sec (a; 8. y = cos3 i- 31. y = sm * — 10. y = 2 cos 3 (a; + 2). 32 - 2/ = cos^x + 2). li. y = 2 cos (2*-l). 33> y^^-i^Z". 12. y= versa;. * "+" x 13. y-2 + BnSft 34 ' y = ta ^>+ 1 )- 14. y = 2 — icosa;. 35. y = tan" 1 -— — -• 15. y = sin x -f sin 3 a;. 36. y = e x ~ x . 16. y = \ sin a* — \ sin 2a\ 37. y = xe~ x . irx 1 . 38. y = a^e - *. i 39. y = ice*. 41. y = e 1 -*. l + x 19. y = 1 + cos x — £ cos 3 x. 40. y = xe x 20. y = sin 2 a;. 21. y = sin a; 2 . 1 42. y = e 1 ^*. 22. y = a; sin-« a; 43. y = \{f — e ~ x ). 23. y = a; 2 sin-. 44. y = £(e* + e"*). 24. y = etna;. 45. y 25. y = i tan 2a. 26. y = 2tan-« 47. y = e~ Sx sin 2jb. 46. y = e - * eosar. 2 27. y = 4tan^p. 48. y = log ~ 28. y = seca\ 49. y = log sin x. 29. y = csca;. 50. y = log tana;. CHAPTER V THE STRAIGHT LINE 28. The point-slope equation. If the slope of a straight line and a point on the line are known, the equation of the line is readily found. Let LK (fig. 65) be any straight line, I^(x v y^) a known point on it, and m its slope. TakeP(x, y~), any point on the line. Then, by § 6, lzh= m . x — x If m is not infinite, we may clear of fractions and obtain y-y x = m(x-xj. Fig. 65 This is an equation which obviously satisfied by the coordi- nates of any point on LK and by those of no other point. Hence it is the equation of LK. If the line is parallel to OX, m — 0, and the equation of the line is , ON y=y x - (2) If the line is parallel to Y, m = and drop the perpendicular J//', we have, in the two cases represented by figs. 66 and 67, tan (/> MP LM MP But — — is equal to the slope of the LM line, by (2), § 6. Therefore tan d> = m. If the straight line is parallel to OY, (/> = 90° and tan = oo. If the line is parallel to OX, no angle is formed, but since m = 0, we may say tan = ; whence <£ = 0° or 180°. 2. Parallel lines. If two lines are parallel, they make equal angles with OX, and hence their slopes are equal. It follows that two equations which differ only in the absolute term, such as Fig. 67 and Ax + By + C x = Ax+By+C =0, represent two parallel lines. It is to be noticed that these two equations have no common solution (§14). Ex. 1. Find the equation of a straight line passing through (— 2, 3) and parallel to3x-5?/ + 6 = 0. First method. The slope of the given line is %. Therefore the required y — 3 = § (a; + 2), or 3ar-5y + 21 = 0. Sea/ml method. We know that the required equation is of the form 3x-5y + C = 0, where C is unknown. Since the line passes through (— 2, 3), 3 (-2) -5 (3) + C=0, whence C = 21. Therefore the required equation is 3 x — 5 y + 21 = 0. ANGLES 61 3. Perpendicular lines. Let AB and CD (fig. 68) be two lines intersecting at right angles. Through P draw PR parallel to OX, and let RPD = 1 and RPB = 2 . Then tan 1 = m y and tan 2 = m 2 , where m x and ra 2 are the slopes of the lines. But, by hypothesis, ^ = ^+90°; whence tan <£ 2 = — cot 4> l which is the same tan <£ x That is, two straight lines are perpendicular when the slope of one is minus the reciprocal of the slope of the other. This theorem may be otherwise expressed by saying that two lines are perpendicular when the prod- uct of their slopes is minus unity. It follows that two straight lines whose equations are of the type Ax+By + C x = and Bx - Ay + C\=0 are perpendicular. Ex. 2. Find a straight line through (•">, 3) perpendicular to 7x+0y+l = 0. First method. The slope of the given line is — l- Therefore the slope of the required line is £. Therefore the required line is y - 3 = 2 (a; - 5), or 9 x - 7 y - 24 = 0. Second method. We know that the equation of the required line is of the form Qx — 7y + C = 0. Substituting (5, 3), we find C =— 24. Hence the required line is 9 x — 7 y — 24 = 0. Ex. 3. Find the equation of the perpendicular bisector of the line join- ing (0, 5) and (5, — 11). The point midway between the given points is (§, — 3), by § 7. The slope of the line joining the given points is — VS by § 6. Hence the required line passes through (§, — 3), with the slope f^. Its equation is y + 3 = & (x - |), or 10 x - 32 y - 121 = 0. Fm G2 THE STRAIGHT LINE 4. Angle between two lines. Let AB and CD (fig. 69) inter- sect at the point P, making the angle BPD, which we shall call /3. Draw the line PR parallel to OX, and place BPB = ^ andJSPjr> = £, Then = ^-^; tan (^ hence tan /3 = tan (., _tan ?V 1 + But tan 1 = m l and tan c/> 2 = m 2 , where ?n 2 is the slope of CD and w 1 is the slope of AB'. Therefore tan ft 1 + m^ If <£> 2 is always taken greater than c^, tan /3 will be positive or negative according as /? is acute or obtuse. Ex. 4. Find the acute angle between the two lines 2a:-3?/ + 5 = and x + 2y+2 = 0. Since the second line makes the larger angle with OX, we jilace m % =- |, m x = $. Then, by substituting in the formula, tan/3= ~ *~J" =- ? Here /3 is an obtuse angle, and the supplementary acute angle is tan -1 J. Ex. 5. Find the equation of a straight line through the point (— 2, 0), making an angle tan -1 § with the line 3x + 4?/ + G = 0. Here tan/3 is given as §, and one of the slopes m z or m x is known to be — ^. Since it is unknown which of the slopes is — ^, the problem has two solutions : (1) Place m 2 =— |. Then, by substituting in the formula, 2 - | - to. . 17 - = - 1. whence m. = . 3 1-flBj * ( The equation of the required line is then y - = - V- (x + 2), 17a: + 6y + 34 = 0. DISTANCE FROM A STRAIGHT LINE 63 (2) Place Then whence The equation of the required line is then or x + 18 y + 2 = 0. 33. Distance of a point from a straight line. Let LK (fig. 70) be a given straight line with the equation 0, C: Ax+By and let P l (x l , y^) be a given point length of the perpendicular J^E drawn from 1\ to LK. Draw the ordinate ML\ and let it intersect the line LK in the point Q. Then the abscissa of Q is a?, and its ordinate may be denoted by y 2 . Since Q is on the line LK. we have It is required to find the F Ax x +By % +C=0, Fig. 70 whence Then OT-Jfi-Jf, J./-, + />'//, + (? It is clear that this expression is a positive quantity when (x lt y^) lies above the line LK and is a negative quantity when (x^ y^) lies below ZA". It is also evident from the triangle T[QB, and from a like triangle in other cases, that the length of P^ is A numerically equal to QP X cos cj>. But tan £ = — — i and hence B B We have, then, cos '. 19. A straight line making a zero intercept on OF makes an angle of 45° with OX. Find its equation. Y^ 20. A straight line making a zero angle with OX cuts OF at a point 3 units from the origin. Find its equation. 21. Find the equation of the straight line through (2, — 3) parallel to the line 2 x 4- y = 7. 22. Find the equation of the straight line through (— |, — 2) parallel to the line 3x — 2^4-2 = 0. 23. Find the equation of the straight line passing through (— 1, — 1) parallel to the straight line determined by (— 2, G) and (2, 1). 24. In the triangle A (—2, -1), 5(3, 1), C(— 1, 4) a straight line is drawn bisecting the adjacent sides AB and BC. Prove by computation that it is parallel to AC and half as long. 25. Find the equation of the straight line passing through the point of intersection of x — 3y + 2 = and 5x -\- 6y — 4 = and parallel to 4x + ?/4-7=0. 26. Find the equation of the straight line parallel to the line a-f-3y — 5 = and bisecting the straight line joining (— 2, — 3) and (5, 5). M THE STRAIGHT LINE 27. Find the equation of the straight line through the origin perpendicular to the line 3 a- + 4 «/ — 1=0. 28. Find the equation of the straight line through (2, — 3) per- pendicular to the line 7 x — 4 y + 3 = 0. 29. Find the equation of the perpendicular bisector of the straight line joining the points (— 5, —1) and (— 3, 4). 30. A straight line is perpendicular to the line joining the points (_ 4 } 0) and (4, — 1) at a point one third of the distance from the first point to the second. What is its equation ? 31. Find the equation of the straight line perpendicular to 2x — 3 y -f- 7 = and bisecting that portion of it which is included between the coordinate axes. 32. Find the equation of the straight line through the point of intersection of 6x — 2 ?/ + 8 = and 4a;- 6y -\- 3 = and per- pendicular to 5x + 2y + 6 = 0. 33. Find the equation of the perpendicular bisector of the base of an isosceles triangle having its vertices at the points (4, 3), (— 1, — 2), and (3, - 4). 34. Find the acute angle between the lines x — y -\- 4 = and 3x-y + 6 = 0. 35. Find the acute angle between the lines 2x — y + 8 = and 2x + 5y-4 = 0. 36. Find the acute angle between the lines x + y — 5 = and 4x + y-8 = 0. 37. Find the acute angle between the line 3x — 2 y + 6 = and the line joining (4, — 5) and (— 3, 2). 38. Find the acute angle between the straight lines drawn from the origin to the points of trisection of that part of the line 2x -\- 3y — 12 = which is included between the coordinate axes. 39. Show that x — y + 3 = bisects one of the angles between the lines 4x-3v/+ll=0 and 3 x — 4 y '+ 10 = 0. 40. Find the vertices and the angles of the triangle formed by the lines 3a: + 5?/^14 = 0, 9 a; — ?/ + 22 = 0, and x — y — 2 = 41. Find the equations of the straight lines through the point (— 3, 0) making an angle tan _1 i with the line 3x — 5y + 9 = 0. 42. Find the equations of the straight lines through (4, —3) making an angle of 45° with the line 3 x + 4 y = 0. PKOBLEMS 67 43. Find the equations of the straight lines through the point (— 1, — 1) making an angle tan -1 \ with the line 3x-\-2y — 6 = 0. 44. Find the equations of the straight lines through the point (2, 1) making an angle tan -1 2 with the line 2x — y + 4 = 0. 45. Find the equations of the straight lines through the point (3, 1) making an angle tan -1 3 with the line x + 3y — 3 = 0. 46. Find the distance of (2, 1) from the line y = 3x -\-7. 47. Find the distance of (2, — f) from the line x -{- 2 y — 4 = 0. 48. Find the distance of the point (b, — a) from the line bx -f- ay = ah. 49. The equations of the sides of a triangle are respectively 3 x + 5 y ■ — 16 = 0, x — y = 0, and 3cc + y + 4 = 0. Find the dis- tance of each vertex from the opposite side. 50. The base of a triangle is the straight line joining the points (— 3, 1) and (5, — 1). How far is the third vertex (6, 5) from the base ? 51. The vertex of a triangle is the point (5, 3), and the base is the straight line joining (— 2, 2) and (3, — 4). Find the lengths of the base and the altitude. 52. Find the equations of the medians of the triangle formed by the lines 2x - 3y 4- 11 = 0, 3* + y - 11 = 0, and x + Ay = 0. 53. Find the foot of the perpendicular drawn from the point (- 1, 2) to the line 3x-5y-21 = 0. 54. Find the distance between the two parallel lines 2x+3y— 8=0 and 2x + 3y- 10 = 0. 55. Find the distance between the two parallel lines 3x—5y+l = Q and3x-5y-7 = 0. 56. A triangle has the vertices (2, 4), (3, —1), and (— 5, 3). Find the distance from the vertex (2, 4) to the point of intersection of the median lines. 57. A straight line is drawn through (2, — 3) perpendicular to the line 3x — 4 ?/ 4- 6 = 0. How near does it pass to the point (6, 8) ? 58. Determine the value of m so that the line y = mx 4- 3 shall pass through the point of intersection of the lines y = 2x + l and y = x 4- 5. 59. A straight line passes through the point (— \, 4), and its nearest distance to the origin is 2 units. What is its slope ? 08 THE STKAIGHT LINE 60. One diagonal of a parallelogram joins the points (3, — 1) and (—3, — 3). One end of the other diagonal is (2, 3). Find its equation and its length. 61. Perpendiculars are let fall from the point (9, 5) upon the sides of the triangle the vertices of which are at the points (8, 8), (0, 8), and (4, 0). Show that the feet of the three perpendiculars lie on a straight line. 62. Find a point on the line 2x -\-3y — 6 = equidistant from the points (4, 4) and (6, 1). 63. Find a point on the line 5x — 3?/ + 15 = the distance of which from the axis of x equals § its distance from the axis of y. 64. A point is equally distant from (3, 2) and (— 3, 4), and the slope of the straight line joining it to the origin is $. Where is the point ? 65. A point is 8 units distant from the origin, and the slope of the straight line joining it to the origin is — \. What are its coordinates ? 66. A point is 5 units distant from the point (1, — 2), and the slope of the line joining it to (0, — 8) is \. Find the point. 67. Find the points on the straight line determined by (1, 1) and (— 2, — 3) which are 15 units distant from either of the given points. 68. Prove analytically that the locus of points equally distant from two points is the perpendicular bisector of the straight line joining them. 69. Prove analytically that the medians of a triangle meet in a point. 70. Prove analytically that the perpendiculars from the vertices of a triangle to the opposite sides meet in a point. 71. Prove analytically that the straight lines joining the middle points of the adjacent sides of any quadrilateral form a parallelogram. 72. Prove analytically that the perpendicular bisectors of the sides of a triangle meet in a point. 73. Prove analytically that the perpendiculars from any two ver- tices of a triangle to the median from the third vertex are equal. 74. Prove analytically that the straight lines drawn from a vertex of a parallelogram to the middle points of the opposite sides trisect a diagonal. CHAPTER VI CERTAIN CURVES 34. Locus problems. A curve is often defined as the locus of a point which has a certain geometric property. It is then usually possible to obtain the equation of the curve by expressing this property by means of an equation involving the coordinates of any point of the locus. This is illustrated in the following examples : Ex. 1. Find the locus of a point at a distance 3 from the straight line ix+By — 6 = 0. Let (x, y) be any point of the locus. By § 33, the distance of (a?, ?/) from the given straight line is ± '— Hence, by the conditiun.s of the problem, 4 * + 3 // - _ •"> which reduces to 4x + 3y — 21 = 0, or 1 .,+:'.// + = 0. These are the equations of two straight lines parallel to the given line. Ex. 2. Find the locus of a point at ;i distance 9 from the point (- 5, - 3). Let (x, j/) be any point of the locus. Its distance from (—5, — •"> ) is, by § 5, V(x + 5) a + (// + '■'>)';. Hence, by the conditions of the problem, V(x+ ."')- + (// + :'.)- = n, which reduces to x 2 + if + K) x + (i // — 47 = 0. This is the equation of the required locus. The curve may be plotted from the equation or may be drawn with compasses, as it is obviously a circle. In the following articles we shall employ the methods just illustrated, to obtain the equations of certain important curves. An equation thus obtained may be used both for plotting the curve and for examining its properties. 70 CEKTAIN CURVES 35. The circle. A circle is the locus of a point at a constant distance from a fixed point. The fixed point is the center of the circle, and the constant distance is the radius. Let (7i, &) be the center C (fig. 71), and let r be the radius of the circle. Then if P (x, y~) is a point on the circle, x and y must satisfy the equation • (x-hy+(y-ky=r\ (1) by §5. Conversely, if x and y satisfy the equation (1), the point (x, y~) is at a distance r from (A, k) and therefore lies on the circle. Therefore (1) is the equation of the circle. Equation (1) expanded gives x 2 +y 2 -2 hx - 2 Icy + Jr + P and if this is multiplied by any quantity A, it becomes Ax 2 + Ay' 2 + 2 Gx +2Fy + C=0, (2) where tf+k 2 -r 2 = Ex. The equation of a circle with the center (J, — ^) and the radius § is which reduces to 12 x 2 + 12 y" — 12 x + 8 y — 1 = 0. 36. Conversely, the equation Ax 2 +Ay 2 -\- 2 Gx + 2Fy + C= 0, where A =#= 0, represents a circle if it represents any curve at all. To prove this we will follow the method of Ex. 2, § 18, and write the equation in the form <"+SM'+3'- F 2 -AC THE CIRCLE 71 There are then three possible cases : 1. G 2 + F 2 -AC>0. The equation is then of the type (1), § 35, where h = , k = , r = — — - , and therefore represents a circle with the center ( -, ) and the radius r-? 5 \ A A G 2 +F 2 -AC \ A 2. G 2 +F 2 -AC=0. The equation is then which can be satisfied by real values of x and y only when G F x — and y = Hence the equation represents the point (' F\ — , J. This may be called a circle of zero radius, regarding it as the limit of a circle as the radius approaches zero. 3. G 2 + F 2 — A C < 0. The equation can then be satisfied by no real values of x and y, since the sum of two positive quantities cannot be negative. Hence the equation represents no curve. Ex. 1. The equation 2 x 1 + -2 >/- + •-' x - '2 y - ."> = may be written (,■+ \f + {!l- lf = Z, and represents a circle with the center (— .1. \ ) ;tn«l the radius ^ :!. Tins circle can now be drawn with compasses, the methods of Chapter II not being required. Ex. 2. The equation x* + //- — 2 x + 1 y + 5 = may be written (ar-l) s +(y + 2) 2 = 0, and is satisfied only by the point (1, — 2). Ex. 3. The equation .c 2 + y* — 2 .*• + 4 // + 7 = may be written (,:-l) 2 + (// + 2) 2 =-2, and represents no curve. 72 CERTAIN CURVES 37. To find ' equation of a circle which will satisfy given conditions, it necessary and sufficient to determine the three quantities h, k, r, or the ratios of the four quantities A, G, F, C. Each condition imposed upon the circle leads usually to an equa- tion involving, these quantities. In order to determine the three quantities J.- .. necessary and in general sufficient to have three equations. Henc in general, three conditions are necessary and sufficient U. ;iine a circle. It is l tant to enumerate all possible conditions which may be ..^ id upon a circle, but the following three may be menti< 1. ?£ftt the condition be imposed upon the circle to pass through the known point (x, y^). Then (x x , y^) must satisfy the equation of the circle ; therefore h, k, and r must satisfy the condition Q Xl ~hy+(y-ky = r\ 2. Let the condition be imposed upon the circle to be tangent to the known straight line Ax + By + C = 0. Then the distance fro.n the center of the circle to this line must equal the radius; therefore, by § 33, h, k, and r must satisfy the condition. Ah+Bk + C =±r. y/A 2 +B 2 The sign will be ambiguous unless from other conditions of the problem it is known on which side of the line the center lies. 3. Let it be required that the center of the circle should lie on the line Ax + By + G = 0. Then h and k must satisfy the condition A7i + Bk+G=0. Ex. 1. Find the equation of the circle through the three points (2, — 2), (7, 3), and (G, 0). The quantities h, k, and r must satisfy the three conditions l-i '0 2 + (-2- - ^') 2 = r 2 , (7- -]0 2 + ro- - l:f = r\ THE CIRCLE 73 Solving these, we have ft = 2, k = 3, and r = 5. ' efore the required equation is (x - 2)2 + ( y - 3)2 = 25, or x 2 + # 2 — 4 x — 6 y — 12 = 0. Ex. 2. Find the equation of the circle which passes 4l — >ugh the points (2, — 3) and (— 4, — 1) and has its center on the line 3 j n x — 18 = 0. The quantities ft, k, and r must satisfy the condi i'>ns (2 - 7,) 2 + (- 3 - kf = r 2 , (_.l_/,)2 + (_l_ J.)2 = ,.2 ; 3 A- + h - 18 = 0. Solving these equations, we find ft = |, /.■ = 'J, /•'- = J -| "'. herefore the required equation is (a? _§ )8 + (y _J 3 L ) a = xj Aj or z 2 + // 2 - 3 x — 11 y - 40 = 0. Ex. 3. Find the equation of a circle which is tangent to the lines 17 x + y - 85 = and 13 a; + 11 y + 50 = 0, and has its center on the line 88 x + 70 y + 15 = 0. The quantities ft, k, and r must satisfy the conditions 17* + * -35 = ± r, V290 13 ft + 11 jfc + 50 ±r, a 290 88 ft + 70 A- + 15 = 0. These equations have the two solutions ,_., , , = V200. ,. . _ . ,., 3V290 and « = 0, K = — 'tj , r = — • 20 Hence each of the two circles 3 x- + 3 //- + 5 x - 5 y - 20 = and 40 x 2 + 40 if - 400 x + 520 y + 2429 = satisfies the conditions of the problem. 74 CERTAIN CURVES 38. The ellipse. An ellipse is the locus of a point the sum of the distances of which from two fixed points is constant. The two fixed points are called the fori. Let them be denoted by F and F' (fig. 72), and let the axis of x be taken through them, and the origin halfway between them. Then if F is any point on the ellipse and 2 a represents the constant sum of its distances from the foci, we have F'P+FP=2a. (1) From the triangle F'PF it follows that F'F<2a. Hence there is a point A on the axis of x and to the right of F which satisfies the definition. We have, then, F'A+FA = 2a, or (F' + OA) + (OA- OF) =2a; whence Let us now place 0A = a. OF OA = e > diere e ) r+f = cc(l-e% (5) or ~+ * . =1. (6) a- a"(l— e-) v y Since e < 1, the denominator of the second fraction is positive, and we place a; 2 ?/ 2 thus obtaining — + %- = 1. (8) We have now shown that any pqint which satisfies ( 1 ) has coordinates which satisfy (8). We may show, conversely, that any point whose coordinates satisfy (8) is such as to satisfy (1). Let us assume (8) as given. We can then obtain (G) and (">), and (5) may be put in each of the two forms x 2 + 2 aex + a 2 e 2 + y 2 = a 2 + 2 aex + t '-'.<-, x 2 — 2 aex + ore 2 + y 2 = a 9 — - cu x + t '-'. i -, the square roots of which are respectively F'P = ± (« + ex), FP = ± (o - ex). These lead to one of the four following equations: /■"/■ + FP = 2a, F'P- FP= 2a, -F'P+ FP= 2 a, -F'P- FP= 2a. Of these, the last one is impossible, since the sum of two negative num- bers cannot be positive; and the second and third are impossible, since the difference between FP and F'P must be less than F'F, which is less than 2 a. Hence any point which satisfies (8) satisfies (1), and therefore (8) is the equation of the ellipse. 39. Placing y = in (8), § 38, we find x = ±a. Placing x = 0, we find y = ±b. Hence the ellipse intersects OX in two points, A (a, 0) and A'(— a, 0), and intersects OF in two points, B(Q, b~) and B'(0, — 6). The points A and A' are called the vertices of the 76 CERTAIN CUEVES ellipse. The line AA', which is equal to 2 a, is called the major axis of the ellipse, and the line BB', which is equal to 2 b, is called the minor axis. Solving (8) first for y and then for x, we have = + - Var- an d a a /m 2 These equations show that the ellipse is symmetrical with respect to both OX and OY, that x can have no value numeri- cally greater than a, and that y can have no value numerically greater than b. If we construct the rectangle KLMN (fig. 73), which has for a center and sides equal to 2 a and 2 b respectively, the ellipse will lie entirely within it; and if the curve is constructed in one quadrant, it can be found by symmetry in all quadrants. The form of the curve is shown Fig. 73 in figs. 72 and 73. 40. Any equation of the form (8), § 38, in which a > b, repre- sents an ellipse with the foci on OX. For if we place, as in § 38, b 2 = a 2 (l— e 2 ), we find, for the eccentricity of the ellipse, Va^b 2 M B L A' v^ J A X 1 V B' K and may fix F and F\ which in § 38 were arbitrary in position, by the relation OF = — OF' = ae. The foci may be found graphically by placing the point of a com- pass on B and describing an arc with the radius a. This arc will intersect A A' in the foci ; for since OB=b and OF= Va' 2 — b 2 , BF= a. It may be noted that the nearer the foci are taken together, the smaller is e and the more nearly b = a. Hence a circle may be considered as an ellipse with coincident foci and equal axes. THE HYPERBOLA 77 Similarly, an equation of the form (8), § 38, in which b > a, re presen ts an ellipse in which the foci he on OY at a distance Vb 2 — a' 2 from 0. In this case BB' = 2 b is the major axis and A A' = 2 a is the minor axis. Finally, any equation of the form a 2 represents an ellipse with its center at the point (li, Tc) and its axes parallel to OX and OY respectively; for if the axes arc shifted to a new origin at (//, /r) by the formulas of £17, this equation assumes the form (8), § 38. Ex. 1. Show that 4 x 2 + if + 4 x - 12 y - 1 = is the equation of an ellipse, and find its center, semiaxes, and eccentricity. Following the method of Ex. 2, § 18, we may write the equation in thef ° rm Hx+ W + 6(y-iy = 8, 3 Hence this curve is an ellipse with its center at (— \, 1) and its major /- 4 Vij 1 and minor axes equal respectively to 2 V 2 and — - — Its eccentricity is — —• '■'' vy Ex. 2. Find the equation of an ellipse with the eccentricity J and its foci at the points (— 1, 4), (7, 1). Since the center is halfway between the foci, tin- center is the point (3, 4). The major axis of the ellipse is parallel t<> OX, Bince it contains the foci. Since each focus is at a distance ae from the center, ae = 4. But e = J, therefore a = 12. Then, from (7), § 38, b- = a- (1 - e 2 ) = 128. The equation of the ellipse is therefore Qr-3) 2 Q/-4) 2 144 128 which reduces to 8 x 2 + 9 y" — 48 x — 72 y — 936 = 0. 41. The hyperbola. An hyperbola is the locus of a point the difference of the distances of which from two fixed ptoints is constant. 78 CERTAIN CUEVES The two fixed points are called the foci. Let them be F and F' (fig. 74), and let FF' be taken as the axis of x, the origin being halfway between F and F'. Then if P is any point on the hyperbola and 2 a is the constant difference of its distances from F and F'> we have either F'P FP- -FP=2a (1) F'P = 2 a. (2) Fig. 74 Since in the triangle F'PF the difference of the two sides FP and F'P is less than F'F, it follows that F'F >2a. There is therefore at least one point A between and F which satisfies the definition. Then F'A-AF=2a, or (F'0 + OA)-(OF-OA-)=2a; whence OA = a. We may therefore place . — = e. where e>l. OA The quantity e is called the eccentricity of the hyperbola. Then the points F and F' are (± ae, 0), and equations (1) and (2) become and ^/(x + aey+if - y/(x - aef +tf=2a ^/(x-aey+y 1 - yf(x + ae)' 2 + f =2 a. (3) (4) By transposing one of the radicals to the right-hand side of these • equations, squaring, and reducing, we obtain from either (3) or (4) (l-_ e s)s3+/ = a*(l_ e 3), d z + d 2 (l-e~) (5) (<0 THE HYPERBOLA 79 But since e>l, , 0), D is (— p, 0), and the equation of ES is x = — p. Draw from P a line parallel to OX, intersecting ES in X. If F is on the right of ES, P must also lie on the right of ES, and, by the definition, SY V p ^^ a D \F FP = NP. Fig. 76 If, on the other hand, F is on the left of ES, P is also on the left of ES, and In either case But, by § 5, and hence which reduces to FP=PX=-M'. FP*=NP\ FP' 2 = (x-pY-hf, NP=x+p-, ( x -py + f = (x+ P y, y 2 =4ju: (1) Any point on the parabola then satisfies equation (1). Conversely, it is easy to show that if a point satisfies (1), it lies so that FP=±XP, and hence lies on the parabola. Equation (1) shows that the curve is Symmetrical with respect to OX, that x must have the same sign as p, and that // increases as x increases numerically. The position of the curve is as shown in fig. 76 when p is positive. When p is negative, F lies at the left of 0, and the curve extends toward the negative end of the axis of x. Similarly, the equation x 2 = 4 py represents a parabola for which the focus lies on the axis of y, and which extends toward the positive or the negative end of the axis of y according as p is positive or negative. In all cases is called the vertex of the parabola, and the line determined by O and F is called its axis. 82 CERTAIN CURVES A more general equation of the parabola is evidently (y — ky= kp(x — h) or Qx — hy = 4 p (y — &), the vertex in either case being at the point (A, F). The work of locating the parabola in the plane is illustrated in the following example. Ex. Show that y 2 + y — 3 x + 1 = is a parabola, and locate it in the plane. The equation may be written ?y 2 + y = 3 x - 1, or if + y + \ = 3 x — 1 + \, which reduces to (// + \)" = 3 (.r — \). Hence the vertex is at the point (\, — J); the equation of the axis is y + 1- = 0, or 2 y + 1 = ; the focus is at the point (I + 4, — I), or (1, — J-) ; and the equation of the directrix is x — \ = — J, or 2 x + 1 = 0. 45. If IKx^ y^) and i^(# 2 , y.^) are two points on the parabola y=4^(fig. 77), then y" = 4 JFi« y 2 2 = 4^ 2 ; (1) whence yl yl which may be wi itten (?y.y (2*/ 2 ) 2 (2) Fig. 77 if both numerator and denominator of the left-hand fraction are multiplied by 4. From the symmetry of the parabola, 2 y x = Q^ and Q 2 P 2 = 2 ?/ 2 ; OM , (2) becomes and since x x = OM l and x, Q^i OM 1 Q,P, OIL (3) That is, the squares of any two chords of a parabola tvhich are perpendicndar to the axis of the parabola are to each other as their distances from the vertex of the parabola. THE CONIC The figure bounded by the parabola and a chord perpendic- ular to the axis of the parabola, as Q l OP l (fig. 77), is called a parabolic segment. The chord is called the base of the segment, the vertex of the parabola is called the vertex of the segment, and the distance from the vertex to the base is called the altitude of the segment. 46. The conic. A conic is the locus of a point the distance of which from a fixed point is in a constant ratio to its distance from a fixed straight line. The fixed point is called the focus, the fixed line the directrix, and the constant ratio the eccentricity. We shall take tin; directrix as the axis of y (fig. 78), and a line through the focus F as the axis of x, and shall call the focus (<■, 0), where c represents OF and is positive or negative according as F lies to the right or the left of 0. Let P be any point on the conic; connect P and /\ and draw I'X per- pendicular to Y. Then, by definition, FP = ±e.NP, (1) J 2f r / / ,( \ & \ \ \ \ \ \ \ \ \ \ \ Fig. 7k according as/' is on the right or the left of OY. In FP 2 =e t .NP i . both But FP 2 =(x-cY+g any point on the conic 2 by £ 5, and NP = x. Therefore for 1. 47. The witch. Let OB A (fig. 80) be a circle, OA a diameter, and LK the tangent to y the circle at A. From draw any line intersect- ing the circle at B and LK at C. From B draw a line parallel to LK and from C a line perpendic- ular to LLC, and call the intersection of these two lines P. The locus of P is a curve called the witch. Fig. 80 THE CISSOID 85 To obtain its equation we will take the origin at and the line OA as the axis of y. We will call the length of the diameter of the circle 2 a. Then, by continuing CP until it meets OX at M and calling (x, y) the coordinates of P, we have 31 = x, MP = y, OA = MC = 2 a. In the triangle 03IC, MP MC OB OC OB . PC ~oc 2 ~' (1) Draw AB. Then OBA is a right angle, and consequently OB .00 = OA 2 ; also C* = 03f+ MC*. Therefore that is, and iinally, MP MC JL 2a OA OW+MC 4 a 2 x- + 4 ) = 2asin 2 0. MP v whence Therefore Now But sin = Substituting in (5), we have whence 2 a This equation is satisfied by the coordinates of any point upon the cissoid. It may be written OP y/a? + tf 2 ay 2 x 2 + f x s = ±x 4n? x From this it appears that the curve is symmetrical with respect to OX, that no value of x can be greater than 2 a or less than 0, and that the line x = 2 a is an asymptote. 49. The strophoid. Let LK and ES (fig. 82) be two straight lines intersecting at right angles at 0, and let A be a fixed point on LK. Through A draw any straight line intersecting ES in D, and lay off on AD in either direction a distance DP equal to OD. locus of P is a curve called the strophoid. (1) (2) (3) (4) (5) (6) (7) (8) The THE STHOPHOID 87 To find its equation take LK as the axis of x and RS as the axis of ?/, and let OA = a. By the definition, the point P may fall in any one of the four quadrants. If we take the positive direction on AD as measured from A towards D, we have OD = PD when P is in the first quadrant, OD=-PD when P is in the second quadrant, when P is in the third quadrant, and - OD = I'D when P is in the fourth quadrant. These four equations are equivalent to the single equation o!r = ■FiA (1) Draw PM parallel to OT and denote the ai igle J/J/' by 0. Then 0D = : a tan (2) and PD = ./Sec 0. (3) Substitutii ig in (1), we have a 2 tan 2 = a? Bee* 6. (4) But tan = MP y (5) MA a- X Substituting in (4), we have which may be reduced to y = ± x-y\— — • (7) This shows that the curve is symmetrical with respect to OX, that no value of x can be less than — a or greater than + a, and that x = — a is an asymptote. 88 CERTAIN CURVES 50. Use of the equation of a curve. The use of the equation of a curve in solving geometrical problems is illustrated in the following problems : Ex. 1. Prove that in the ellipse the squares of the ordinates of any two points are to each other as the products of the segments of the major axis made by the feet of these ordinates. We are to prove that (fig. S3) MJ\ 2 _ A'M X ■ M X A Mjn~ A'M 2 . M^A Let P x be (x v y x ) and let P 2 be (x 2 , y 2 ). Then ■,,2 whence But y x = M 1 P V a + x x = A'O + OM 1 = A'M V a-x 1 = OA - OM x = M X A, y 2 — M 2 P 2 , a + x 2 = A'M 2 , a — x 2 = M 2 A. Hence the proposition is proved. Ex. 2. If M l P l is the ordinate of a point P t of the parabola y 2 = <±px, and a straight line drawn through the middle point of M 1 P 1 parallel to the axis of x cuts the curve at Q, prove that the inter- cept of the line MM on the axis of y equals %M 1 P V Let P x (fig. 84) be (x v y x ). Then y? * x. = — — , from the equation of the ip parabola. By construction, the ordinate of Q F y^^ 7^ I f. Vi Since Q is on the parabola, its is found by placing y = ^ in if- = 4 px. Then Q is Fig. 84 \l6p 2/' and \Wp H/' M is (x v 0), which is the same as (-p-> ). Hence the equation of MM is, by § 30, ^P I 8^ + 3^-2^ = 0. The intercept of this line on OFis § y x = ^M l P v which was to be proved. EMPIRICAL EQUATIONS 89 51. Empirical equations. We have met in § 8 examples of related quantities for which pairs of corresponding values have been found by experiment, but for which the functional relation connecting the quantities is not known. In such a case it is often desirable to find an equation which will represent this relation, at least approximately. The method, in general, is to plot the points as in § 8 and then fit a curve to them. At best this work is approximate, the result depending largely on the judgment of the worker, and in complicated cases it demands methods too advanced for this book. We shall discuss a few simple exam- ples, to illustrate merely the fundamental principles involved. The simplest case is that in which the plotted points appear to lie on a straight line, or nearly so. If the two related quan- tities are x and y, the relation between them is expressed by the equation y=mx + b, (1) where m and b are to be determined to fit the. data. In practice the points are plotted, and it appears that a straight line may be so drawn that the points either lie on it or are close to it and about evenly distributed on both sides of it. The straight line having been drawn, its equation may be found by means of two points on it, which may be either two of the original data or any two points of the graph. This method is illustrated in Ex. 1. Closely connected with this case are two others. Suppose the relation between the two quantities .'■ ami y is known or assumed to be of the form „ . ON V = "•' (2) or y = a ^ x i (&) where a, b, and n are to be determined to fit the given numerical values of x and y. By taking the logarithms of both sides of these equations, we have respectively log y = n log x + log a (4) and log y = (log b~) x + log a ; (5) or, if we place log y = y', log x = x', log a = b', log b = m, y' = nx'+b', (G) y' = mx + V. (7) <)0 CERTAIN CURVES We may now plot the points (V, y'~) or (x, y'~) and determine the straight line on which they lie approximately. The equations (0) and (7) having thus been found, the return to equations (2) and (3) is easy. This method is illustrated in Ex. 2. When the use of a straight line either directly or by aid of logarithms fails, the attempt may be made to fit a parabola y = a + bx + ex 2 (9) to the points of the plot. Since three points are sufficient to de- termine the constants of the equation, the parabola may be made to pass through any three of the plotted points. This parabola may then be tested to see if it passes reasonably near to the other points. This method is illustrated in Ex. 3. Other curves with equations of the form y = a + bx + cx 2 + dx*-\ + lx n may also be used. In this case the number of points through which the curve may be exactly drawn is equal to the number of arbitrary coefficients. In all these cases it is often convenient to use different scales for x and y, the proper allowance being made in the calculations. This is illustrated in Ex. 2. Ex. 1. Corresponding values of two related quantities x and y are given by the following table : X 1 2 4 G 10 V 1.8 2.2 2.9- 3.9 6.1 Find the empirical equation connecting them. We plot the points (x, y) and draw the straight line as shown in fig. 85. The straight line is seen to pass through the points (0, 1) and (2, 2). Its equation is therefore y = .5 x + 1, which is the required equation. ] T L, Si 1 Fig. 85 EMPIKICAL EQUATIONS 91 Ex. 2. Corresponding values of pressure and volume taken from an indicator card of an air compressor are as follows : p 18 21 20.5 33.5 44 62 V .635 .556 .475 .397 .321 .243 Find the relation between them in the form pv n — c. Writing the assumed relation in the form p = cv~* and taking the logarithms of both sides of the equation, we have logp = - nlogw + loge, or y = — nx + b, where y = logp, x = log v, and b = log c. The corresponding values of x and y arc X = 1 1 tg V -.1972 - .2549 - .3233 - .41112 -.4935 -.6144 y = logp 1.2553 1.3222 1.4232 1.5250 1.0435 j 1.7924 We assume on the ar-axis a scale twice as large as that on the //-axi the points (.r, ?/), and draw the straight line as shown in fig. 86 construction should be made on large-scale plotting paper. The line is seen to pass through the points (—.05, 1.075) and (—.46, 1.6). Its equation is therefore :s: x \ '.BO-Jtf-JiD-SS-aO-.SB-.SO-lS—lO- 0B o Fiu. 86 y = — 1.28 x+ 1.01. Hence n = 1.28, log c = 1.01, c = 10.2, and the required re- lation between p and V is pv 1 - 2 * - 10.2. Ex. 3. Corresponding values of two related cpuantities x and y are given by the following table : X 1 2 3 4 5 y 1.37 .08 .41 .54 1.05 Find the empirical equation connecting them. 92 CERTAIN CURVES If we plot the points we assume is in fig. 87, they suggest a parabola. Accordingly w = a + hx + ex 2 and determine a, h, and c, so that the curve will pass through the first, third, and last points. The equations for a, b, and c are 1.37 = a + b + c, .11 = a + 3 b + 9 r, 1.05 = a + 5 h + 25 c ; w H'UI'I « = 2 .45, & = - 1.28, and c = .2. The required equation is therefore r l O l 2.45 - 1.28 x + .2 x 2 Fig. 8< If we substitute for x in this equation the values 2 and 4, we find the corresponding values of y to he .69 and .53. This shows that the curve passes reasonably near to the points of the plot which were not used in computing the coefficients. PROBLEMS 1. Find the equations of the locus of a point the distance of which from the axis of x equals five times its distance from the axis of y. 2. Find the equations of the locus of a point the distance of which from the axis of a; is 3 more than twice its distance from the axis of y. 3. Find the equation of the locus of a point the distances of which from (2, — 1) and (— 3, 2) are equal. 4. Find the equations of the locus of a point equally distant from the lines 3 ,r — 5 y — 15 = and 5x- 3y + l = 0. 5. Find the equations of the bisectors of the angles between the lines 9ir + 2?/ — 3 = and 1 x — 6 y + 2 = 0. 6. Find the equations of the bisectors of the angles between the lines 3.x + 4y— 7=0 and 12 x — 5 y + 1 = 0. 7. A point moves so that its distance from the axis of y equals its distance from the point (5, 0). Find the equation of its locus. 8. Find the equation of the locus of a point the distance of which from the axis of x is one half its distance from (0, 2). 9. A point moves so that the square of its distance from the point (0, 3) equals the cube of its distance from the axis of y. Find the equation of its locus. PROBLEMS 93 10. Find the equation of the locus of a point the distance of which from the line x = 3 is equal to its distance from (4, — 2) . 11. Find the equation of the locus of a point which moves so that the slope of the straight line joining it to (a, a) is one greater than the slope of the straight line joining it to the origin. 12. A point moves so that its distance from the origin is always equal to the slope of the straight line joining it to the origin. Find the equation of its locus. 13. Find the equation of the locus of a point the distance of which from the line 3 a? + 4 y — 6 = is twice its distance from (2, 1). 14. Find the equation of the circle having the center (3, — 5) and the radius 4. 15. Find the equation of the circle having the center i — | } g | and the radius 2. 16. Find the points at which the axis of x intersects the circle having as diameter the straight line joining (1, 2) and (— 3, — 4). 17. Find the equation of the circle having as diameter that part of the line Sx — 4// + 12 = which is included between the coor- dinate axes. 18. Find the equation of the circle having as diameter the common chord of the two circles x* + y*+±x — 4y — 2 = and x- + f - 2 x + 2 y - 14 = 0. 19. Find the ('([nations of the circles of radius a winch are tangent to the axis of y at the origin. 20. Find the center and the radius of the circle ./■- + //- + 26 x + 1 6 // - 42 = 0. 21. Find the center and the radius of the circle 2 x* + 2 y 2 -r- f> .r + 3 y -10 = 0. 22. Find the equation of a straight line passing through the cen- ter of the circle x' 2 + if — 4.r -f- 2y — 5 = and perpendicular to the line x — 2 // + 1 = 0. How near the origin does the line pass '.' 23. Prove that two circles are concentric if their equations differ only in the absolute term. 94 CERTAIN CURVES 24. Show that the circles x 2 + if + 2 Gx + 2Fy + C = and a ,2 +Z/' 2 + 2r;'.r + 2F'y + C = are tangent to each other if V(G-6") 2 +(F- J P') 2 = Vg 2 + J" 2 - C ± Vx — 5 y 4- 15 = 0. What is its equation ? 32. A circle of radius 5 passes through the points (4, — 2) and (5, — 3). What is its equation ? 33. The center of a circle which passes through the points (— 2, 4) and (— 1, 3) is on the line 2x — 3y + 2 = 0. What is its equation ? 34. A circle which is tangent to OX passes through (— 1, 2) and (6, 9). What is its equation ? 35. The center of a circle which is tangent to the two parallel lines x — 2 = and x — 6 = is on the line y = 3 x — 6. What is its equation ? 36. The center of a circle is on the line 2x + y 4- 3 = 0. The circle passes through the point (3, 1) and is tangent to the line 4a: — 3y — 14 = 0. What is its equation ? 37. The center of a circle is on the line x 4- 2y — 10 = and the circle is tangent to the two lines 2x — 3y+9 = and 3x — 2y + 1 = 0. What is its equation ? 38. Given the ellipse 9 x 2 + 25 y 2 = 225, find its semiaxes, eccentricity, and foci. • PROBLEMS 95 39. Given the ellipse 3 a- 2 + 4 y 2 = 2, find its semiaxes, eccen- tricity, and foci. 40. Find the vertices, eccentricity, and foci of the ellipse 4 x 2 + 2 if = 1. 41. Find the center, vertices, eccentricity, and foci of the ellipse 4 a- 2 + 9 if + 16 x -18 y -11 = 0. 42. Find the center, vertices, eccentricity, and foci of the ellipse 16 x 2 + 9 if - 16 x + 6 y - 139 = 0. 43. Find the equation of th» ellipse when the origin is at the left-hand vertex and the major axis lies along OX. 44. Find the equation of the ellipse when the origin is taken at the lower extremity of the minor axis and the minor axis lies along OY. x 2 f 45. Determine the semiaxes a and b in the ellipse -j -f- j- = 1 so that it shall pass through (2, 3) and (—1, — 4). 46. Find the equation of an ellipse if its axes are 8 and 4, its center is at (2, — 3), and its major axis is parallel to OX. 47. Find the equation of an ellipse if its axes are § and £, its center is at (1, —1), and its major axis is parallel to OY. 48. If the vertices of an ellipse are (± 6, 0) and its foci are (±4, 0), find its equation. 49. Find the equation of an ellipse when the vertices are (± 4, 0) and one focus is (2, 0). 50. Find the equation of an ellipse when the vertices are (0, 2) and (0, — 4) and one focus is at the origin. 51. Find the equation of the ellipse the foci of which are (± 4, 0) and the major axis of which is 10. 52. Find the equation of the ellipse the foci of which are (0, ± 3) and the major axis of which is 12. 53. Find the equation of an ellipse when its center is at the origin, one focus is at the point (— 4, 0), and the minor axis is equal to 6. 54. Find the equation of the ellipse the foci of which are (1, ± 2) and the major axis of which is 6. 55. Find the equation of an ellipse the eccentricity of which is § and the foci of which are (0, ± 5). 96 CERTAIN CURVES 56. The center of an ellipse is at the origin and its major axis lies on OX. If its major axis is 6 and its eccentricity is i, find its .equation. 57. The center of an ellipse is at (— 2, 3), and its major axis is parallel to OF and 8 units in length. Its eccentricity is ^. Find its equation. 58. The center of an ellipse is at (1, 2), its eccentricity is \, and the length of its major axis, which is parallel to OY, is 8. What is the equation of the ellipse ? 59. Find the equation of an ellipse the eccentricity of which is ^ and the ordinate at the focus is 4, the center being at the origin and the major axis lying on OX. 60. Find the eccentricity and the equation of an ellipse if the foci lie halfway between the center and the vertices, the center being at the origin and the major axis lying on OX. 61. Find the equation and the eccentricity of an ellipse if the ordinate at the focus is one third the minor axis, the center being at the origin and the major axis lying on OX. 62. Find the eccentricity of an ellipse if the straight line connect- ing the positive ends of the axes is parallel to the straight line joining the center to the upper end of the ordinate at the left-hand focus. 63. Given the hyperbola — — £- = 1. find its eccentricity, foci, and asymptotes. 64. Given the hyperbola 4 a- 2 — 9 if — 36, find its eccentricity, foci, and asymptotes. 65. Find the center, eccentricity, foci, and asymptotes of the hyperbola 9 r 2 - 4 f - 36 x - 24 y - 30 = 0. 66. Find the center, eccentricity, foci, and asymptotes of the hyperbola 2 x 2 - 3 f + 4 x + 12 y + 4 = 0. 67. Find the equation of an hyperbola if its transverse axis is V3, its conjugate axis Vf, its center at (1, — 2), and its transverse axis parallel to OX. 68. Find the equation of an hyperbola if its transverse axis is 5, its conjugate axis 3, its center (— 2, 3), and its transverse axis parallel to OY. 69. Find the equation of the hyperbola when the origin is at the left-hand vertex, the transverse axis lying on OX. PROBLEMS 97 70. Find the equation of an hyperbola if the foci are (± 4, 0) and the transverse axis is 6. 71. Find the equation of an hyperbola if the foci are (0, ± 3) and the transverse axis is 4. 72. An hyperbola has its center at (1, 2) and its transverse axis is parallel to OX. If its eccentricity is § and its transverse axis is 5, find its equation. 73. Find the equation of an hyperbola when the vertices are (7, 1) and (— 1, 1) and the eccentricity is |. 74. Find the equation of an hyperbola the vertices of which are halfway between the center and the foci, the center being at and the transverse axis lying on OX. 76. Find the equation of the hyperbola which has the lines y = ± | x for its asymptotes and the points (±2, 0) for its foci. 76. Find the equation of the hyperbola which has the asymptotes y = ± | x and passes through the point (2, 1). 77. Find the equation of an equilateral hyperbola which passes through (3, — 1) and has its axes on the coordinate axes. 78. Show that the eccentricity of an equilateral hyperbola is equal to the ratio of a diagonal of a square to its side. 79. If the vertices of an hyperbola lie two thirds of the distances from the center to the foci, find the angles between the transverse axis and the asymptotes. 80. Express the angle between the asymptotes in terms of the eccentricity of the hyperbola. 81. An ellipse and an hyperbola have the vertices of each at the x 2 ir foci of the other. If the equation of the ellipse is — + ~ = 1, find that of the hyperbola. Find the equations of the directrices of the two curves. .'- y 2 82. Show that -3 — + -h — — , = 1, where k is an arbitrary quantity, represents an ellipse confocal to -5 + '— = 1 when k 2 b 2 but < a 2 , a 2 being greater than U 1 . 83. Find the vertex, axis, focus, and directrix of the parabola lf + ±y-G X + T=0. 98 CERTAIN CURVES 84. Find the vertex, axis, focus, and directrix of the parabola 4 .r 2 + 4 x 4- 3 y - 2 = 0. 85. Determine p so that the parabola y- = A])x shall pass through the point (— 2, 4). 86. The vertex of a parabola is at the point (2, 3), and the parab- ola passes through the origin of coordinates. Eind its equation, its axis being parallel to OX. 87. The vertex of a parabola is at the point (— 1^, 2), and the parabola passes through the point (— 1, — 1). Find its equation, its axis being parallel to OY. 88. Find the equation of the parabola when the origin is at the focus and the axis of the parabola lies on OX. 89. Find the equation of the parabola when the axis of the curve and its directrix are taken as the axes of x and y respectively. 90. The vertex of a parabola is (3, 2) and its focus is (5, 2). Find its equation. 91. The vertex of a parabola is (— 1, 2) and its focus is (— 1, 0). Find its equation. 92. Find the equation of the parabola of which the focus is (2, — 1) and the directrix is the line y — 4 = 0. 93. The vertex of a parabola is at the point (— 2, — 5) and its directrix is the line x — 3 = 0. Find its equation. 94. The vertex of a parabola is at (5, — 2) and its directrix is the line ?/ 4- 4 = 0. Find its equation. 95. The focus of a parabola is at the point (4, — 1) and its direc- trix is the line y — x = 0. Construct the curve from its definition and derive its equation. What is the equation of its axis ? 96. The altitude of a parabolic segment is 8 ft. and the length of its base is 14 ft. A straight line drawn across the segment perpen- dicular to its axis is 7 ft. long. How far is it from the vertex of the segment ? 97 . An arch in the form of a parabolic curve, the axis being verti- cal, is 40 ft. across the bottom, and the highest point is 12 ft. above the horizontal. What is the length of a beam placed horizontally across the arch 3 ft. from the top ? PROBLEMS 99 98. The cable of a suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and 300 ft. long, is sup- ported by vertical wires attached to the cable, the longest wire being 90 ft. and the shortest being 20 ft. Find the length of a supporting wire attached to the roadway 50 ft. from the middle. 99. Any section of a given parabolic mirror made by a plane pass- ing through the axis of the mirror is a parabolic segment of which the altitude is 8 in. and the length of the base is 12 in. Find the perimeter of the section of the mirror made by a plane perpendicular to its axis and 6 in. from its vertex. 100. Given the ellipse 4 x 2 -+- 9 y 2 = 36, find its foci and directrices. 101. Given the ellipse 5 x 2 + 3 f — 1, find its foci and directrices. 102. Given the hyperbola 5 x 2 — 10 y 2 = 50, find its foci and directrices. 103. Find the equation of an ellipse when the foci are (± 3, 0) and the directrices are x = ±7. 104. Find the center, vertices, foci, and directrices of the ellipse 9x 2 + 25 f + 30 x + 40 y - 184 = 0. 105. Find the center, vertices, foci, and directrices of the hyper- bola 5 x 2 - 4 // + 10 x + 16 y - 31 = 0. 106. Find the equation of a circle through the vertex and the ends of the double ordinate at the focus of the parabola f = Apx. 107. Find the equation of the circle through the vertex, the focus, and the upper end of the ordinate at the focus of the parab- ola f- 8. v = 0. 108. Find the equation of a circle which passes through the vertex and the focus of the parabola f = 8 x and has its center on the line x — y + 2 = 0. 109. Find the equation of the locus of a point which moves so that the slope of the straight line joining it to the focus of the parabola x 2 = 8 y is three times the eccentricity of the ellipse 16 a: 2 +9/ -144 = 0. 110. Find the equation of the cissoid when the origin is at the center of the circle used in its definition, the direction of the axes being as in § 48. 111. Find the equation of the cissoid when its asymptote is the axis of y and its axis is the axis of x. 100 CERTAIN CURVES 112. Find the equation of the strophoid when the asymptote is the axis of y, the axis of x being as in § 49. 113. Find the equation of the strophoid when the origin is at A (fig. 82), the axes being parallel to those of § 49. 114. Show that the lines y = ± x intersect the strophoid at the origin only, and find the equation of the curve referred to these lines as axes. 115. Find the equation of the witch when LK (fig. 80) is the axis of x and OA the axis of y. 116. Find the equation of the witch when the origin is taken at the center of the circle used in constructing it, the axes being par- allel to those of § 47. 117. Show that the locus of a point which moves so that the sum of its distances from two fixed straight lines is constant is a straight line. 118. Find the equations of the locus of a point equally distant from two fixed straight lines. 119. A point moves so that its distances from two fixed points are in a constant ratio k. Show that the locus is a circle except when k = 1. 120. A point moves so that the sum of the squares of its dis- tances from the sides of an equilateral triangle is constant. Show that the locus is a circle and find its center. 121. A f>oint moves so that the square of its distance from the base of an isosceles triangle is equal to the product of its distances from the other two sides. Show that the locus is a circle and an hyperbola which pass through the vertices of the two base angles. 122. A point moves so that the sum of the squares of its dis- tances from the four sides of a square is constant. Find its locus. 123. A point moves so that the sum of the squares of its dis- tances from any number of fixed points is constant. Find its locus. 124. Find the locus of a point the square of the distance of which from a fixed point is proportional to its distance from a fixed straight line. 125. Find the locus of a point such that the lengths of the tan- gents from it to two concentric circles are inversely as the radii of the circles. PROBLEMS 101 126. A point moves so that the length of the tangent from it to a fixed circle is equal to its distance from a fixed point. Find its locus. 127. Find the locus of a point the tangents from which to two fixed circles are of equal length. 128. Straight lines are drawn through the points (— a, 0) and (a, 0) so that the difference of the angles they make with the axis of x is tan -1 -- Find the locus of their point of intersection. 129. The slope of a straight line passing through (a, 0) is twice the slope of a straight line passing through (— a, 0). Find the locus of the point of intersection of these lines. 130. A point moves so that the product of the slopes of the straight lines joining it to A (— a, 0) and /; (a, 0) is constant. Prove that the locus is an ellipse or an hyperbola. 131. If, in the triangle ABC, tan .1 tan \ 11 = 1' and Ml is fixed, show that the locus of C is a parabola with its vertex at A and its focus at B. 132. Given the base 2b of a triangle and the sum s of the tan- gents of the angles at the base. Find the locus of the vertex. 133. Find the locus of the center of a circle which is tangent to a fixed circle and a fixed straight line. 134. Prove that the locus of the center of a circle which passes through a fixed point and is tangent to a fixed straight line is a parabola. 135. A point moves so that its shortest distance from a fixed circle is equal to its distance from a fixed diameter of that circle. Find its locus. 136. If a straight line is drawn from the origin to any point Q of the line y = a, and if a point P is taken on this line such that its ordinate is equal to the abscissa of Q, find the locus of P. 137. AOB and COD are two straight lines which bisect each other at right angles. Find the locus of a point P such that PA ■ PB = PC • PD. 138. AB and CD are perpendicular diameters of a circle and M is any point on the circle. Through M, AM and BM are drawn. AM intersects CD in N, and from 7Y a straight line is drawn parallel to AB, meeting BM in P. Find the locus of P. 102 CERTAIN" CURVES 139. Given a fixed straight line All and a fixed point Q. From any point R in AB a perpendicular to AB is drawn, equal in length to RQ. Find the locus of the end of this perpendicular. 140. is a fixed point and AB is a fixed straight line. A straight line is drawn from 0, meeting AB at Q, and in OQ a point P is taken so that OP -OQ= k 2 . Find the locus of P. 141. Let OA be the diameter of a fixed circle. From B, any point on the circle, draw a straight line perpendicular to OA, meeting it in D. Prolong the line DB to P, so that OD:DB= OA : DP. Find the locus of P. 142. A perpendicular is drawn from the focus of an hyperbola to an asymptote. Show that its foot is at distances a and b from the center and the focus respectively. 143. Two straight lines are drawn through the vertex of a parab- ola at right angles to each other and meeting the curve at P and Q. Show that the line PQ cuts the axis of the parabola in a fixed point. 144. In the parabola if = A,px an equilateral triangle is so inscribed that one vertex is at the origin. What is the length of one of its sides ? 145. Prove that in the ellipse half of the minor axis is a mean proportional between A F and FA '. 146. Show that in an equilateral hyperbola the distance of a point from the center is a mean proportional between the focal distances of the point. 147. If from any point P of an hyperbola PK is drawn parallel to the transverse axis, cutting the asymptotes in Q and R, prove PQ • PR = « 2 . If PK is drawn parallel to the conjugate axis, prove PQ . PR = _ 6 2 . 148. Prove that the product of the distances of any point of the hyperbola from the asymptotes is constant. 149. Prove that in the hyperbola the squares of the ordinates of any two points are to each other as the products of the segments of the transverse axis made by the feet of these ordinates. 150. Straight lines are drawn through a point of an ellipse from the two ends of the minor axis. Show that the product of their intercepts on OX is constant. PROBLEMS 103 151. P 1 is any point of the parabola y 2 = 4^.r, and P X Q, which is perpendicular to OP v intersects the axis of the parabola in Q. Prove that the projection of P X Q on the axis of the parabola is always 4^>. 152. Show that the focal distance of any point on the hyperbola, is equal to the length of the straight line drawn through the point parallel to an asymptote to meet the corresponding directrix. 153. Show that the following points lie approximately on a straight line, and find its equation : X 4 9 13 20 22 25 30 ?/ 2.1 4.6 7 12 12.0 14.5 18.2 154. For a galvanometer the deflection />, measured in millimeters on a proper scale, and the current /. measured in microamperes, are determined in a series of readings as follows : J) 29.1 48.2 72.7 92.0 IIS. II 140.0 i <;.-..< i 199.0 I 0.0493 0.0821 0.123 0.164 0.197 0.234 0.274 0.328 Find an empirical law connecting 1> and /. 155. For a copper-nickel thermocouple the relation between the temperature t in degrees and the thermoelectric power in microvolts is given by the following table : t 50 ' 100 ! 150 200 p 24 2.". 26 26.9 27.6 Find an empirical law connecting t and p 156. The safe loads in thousands of pounds for beams of the same cross-section but of various lengths in feet are found as follows : Length 10 11 12 13 14 15 Load 123.6 121.5 111.8 107.2 101.3 90.4 Find the empirical equation connecting the data. 104 CERTAIN CURVES 157. The relation between the pressure p and the volume v of a gas is found experimentally as follows : Pressure 20 23.5 31 42 59 78 Volume 0.619 0.540 0.442 0.358 0.277 0.219 Find an empirical equation connecting /> and v in the form pv n = c. 158. The deflection a of a loaded beam with a constant load is found for various lengths I as follows : I 1000 900 800 700 600 a 7.14 5.22 3.64 2.42 1.50 Find an empirical equation connecting a and I in the form a = JcF. 159. The relation between the length I (in mm.) and the time t (in seconds) of a swinging pendulum is found as follows : I 63.4 80.5 90.4 101.3 107.3 140.6 t 0.806 0.892 0.960 1.010 1.038 1.198 Find an empirical equation connecting I and t in the form t = Id' 1 . 160. For a dynamometer the relation between the deflection 6, as follows : J7T 400 L 6 40 86 120 160 201 240 280 320 362 I 0.147 0.215 0.252 0.293 0.329 0.360 0.390 0.417 0.442 Find an empirical equation connecting / and $ in the form I = k$". 161. In a chemical experiment the relation between the concen- tration y of undissociated hydrochloric acid is connected with the concentration x of hydrogen ions as shown in the table : X 1.68 1.22 0.784 0.426 0.092 0.047 0.0096 0.0049 0.00098 y 1.32 0.676 0.216 0.074 0.0085 0.00315 0.00036 0.00014 0.000018 Find an empirical law connecting the two quantities in the form y = kx n PROBLEMS 105 162. Show that the values of x and y as given in the following table are connected by a relation of the form y = ca x , and find c and a. X 8 10 12 ' 14 16 18 20 y 3.2 4.6 7.3 9.8 15.2 24.6 36.4 163. In a certain chemical reaction the concentration c of sodium acetate produced at the end of the stated number of minutes t is as follows : t 1 2 3 4 c 0.00837 0.0070 0.00586 J 0.00402 0.00410 Assuming that the law is of the form <■ — ab* } find the equation connecting the concentration with the time. 164. The molal heat capacity at constant temperature is for water vapor at various temperatures as follows: Temp. 10 100 500 700 1000 Cap. 8.8 8.0 8.4 8.6 9.1 Determine the law in the form C = a + bt + rf ~- 165. Assuming Boyle's'law, pv = c, determine <■ graphically from the following pairs of observed values: V 39.02 42.17 45.80 18.52 51.89 60.47 65.97 V 40.37 38.32 85.32 33.29 31.22 26.86 24.53 166. The distance p of an object from a lens and the distance p 1 of its image are found by experiment as follows : p 320 240 180 140 120 100 80 60 p' 21.35 21.80 22.50 23.20 23.80 24.60 26.20 29.00 here f is the focal length of the lens, Assuming the law - -\ : = — P P ./ compute / graphically by plotting the reciprocals of p and p' CHAPTER VII PARAMETRIC REPRESENTATION 52. Definition. Consider the two equations (i) where ^(0 and f 2 (f) are two functions of an independent variable t. If we assign to t any value in (1), we determine x and y and may plot a point with these coordinates. In this way a value of t determines a point in the plane. So other values of t determine other points, which together deter- mine a curve. The two equations (1) then represent the curve. The vari- able t is called a parameter, and the equations (1) are called the parametric representation of the curve. It is sometimes easy to eliminate t from the equations (1) and obtain thus a Cartesian equation of the curve, but this y elimination is not essen- tial and is not always desirable. Ex. 1. x = t' 2 , y Giving t in the values — 3, - 1, 2, 3, we find sponding points = t. succession 2, -1, 0, the corre- (9, -3), Fig. 88 (4, -2), (1, -1), (0, 0), (1, 1), (4, 2), (9, 3). These points, if plotted, may be connected by the curve of fig. 88, and as many inter- mediate points as desired may be found. In this case we may easily eliminate t from the equations and obtain x = y 2 . The curve is a parabola. 106 PARAMETRIC REPRESENTATION 107 Ex. 2. x = t 3 + 2 fi, y |, -1, -i, 0, i, 1, we •), (1, 0), (|, |), (0, 0), Giving t in succession the values — 2, - find as corresponding points (0, — (J), {%, — (J, - t)> (3, 0). These points give the curve shown in fig. 89. If more details as to the shape of the loop are wanted, more values of t must be assumed intermediate to those we have used. Elimination of / in this example is possible hut hardly desirable. Ex. 3. ./■ = . The elimination of t gives the equation £j + yt = n t. The curve is called the four- etaped hypocycloid (§ 58). As the examples show, the param- eter t is in general simply an inde- pendent variable to which values are assigned at pleasure. In problems of mechanics, however, the pa- rameter frequently represents time. In this ease the curve of equa- tions (1) represents the path of a moving point, the position of the point at any instant being given by the equations. Any of the above examples may be interpreted in this way. Other illustrations will be found in the examples of §§53 and 54. In some cases, also, it is possible to give a geometric interpretation to the param- eter t. This is illustrated by the curves which follow, where in each case the parameter is a certain angle. 108 I .» A 1 ; A M ETRIC REPRESENTATION 53. The circle. Let P(x, y) (fig. 91) be any point on a circle with its center at the origin and its radius equal to a. Let be the angle made by OP and OX. Then, from the definition of the sine and cosine, x = a cos cj>, y = a sin c/>, are the parametric equations of the circle with as the arbitrary parameter. Ex. A particle moves in a circle at a con- stant rate k. Then, if s represents the arc traversed in the time /, s — kt and , y = b sin . The elimination of cf> from these equa- tions gives I- +[ i 1, showing that Fig. 92 the locus of P is an ellipse. 4> is called the eccentric angle of a point on the ellipse, and the circle x 2 +y 2 = a 2 is called the auxiliary circle. Ex. A particle Q moves at a constant rate along the auxiliary circle of an ellipse ; required the motion of its accompanying point P. As in § 53, = —■ Hence the equations of the path are kt y bsm It THE CYCLOID 109 55. The cycloid. If a circle rolls upon a straight line, each point of the circumference describes a curve called a cycloid. Let a circle of radius a roll upon the axis of x, and let C (fig. 93) be its center at any time of its motion, N its point of contact with OX, and P the point on its circumference which Fig. 93 describes the cycloid. Take as the origin of coordinates, 0, the point found by rolling the circle to the left until P meets OX. Then ON= arc I'X. Draw MP and CX, each perpendicular to OX, PR parallel to OX, and connect C and P. Let angle X('P= . Then x = O M = OX - MN = arc XI'- PR = a — a sin (f>. ij = MI> = XC-RC = a — a cos (f>. Hence the parametric representation of the cycloid is x= a($ — sin ), y = a (1 — cos <£). By eliminating <£, the equation of the cycloid may be written _ „-i a — y _V2 ay - y\ but this is less convenient than the parametric representation. At each point where the cycloid meets OX a sharp vertex called a cusp is formed. The distance between two consecutive cusps is evidently 2 ira. 110 PARAMETRIC I IK PRESENTATION 56. The trochoid. When a circle rolls upon a straight line, any point upon a radius, or upon a radius produced, describes a curve called a trochoid. Let the circle roll upon the axis of x, and let C (figs. 9-4 and 95) be its center at any time, N its point of contact with the axis of x, P(x, y) the point which describes the trochoid, and K the point in which the line CP meets the circle. Take as the origin the point found by rolling the circle toward the left until A' is on the axis of x. Then ON = arc NX. Draw PM and CN perpendicular to OX, and through P a line parallel to OX, meeting CN, or CN produced, in R. Let the radius of the circle be a, CP be h, and angle NCP be , ami denote the angle OCP by 0, the angle KOC by cf>. Thru arc KX= /-, an- NP = ad ; whence /«/> = ad. We now have x = 031= OL+L .V = OC cos KOC-CP cos SPC = (a + li) cos — a cos (<£ + 0) a + b a y = MP = LC-EC = 0C sin KOC - OP sin SPC = (a + b) sin <£ — a sin ($ + 0) = (a + b) sin $ — a sin <£>. = (a + A) cos ^-aci >s • 112 PARAMETRIC R EP RES EKT ATION The curve consists of a number of congruent arches, the first of which corresponds to values of 6 between and 2 7r, that is, to Similarly, the Mi arch corre- 2(k 1) air , 2 kair TT and — : — • Hence values of <£ between and — — spon ds to values of between b o the curve is a closed curve when, and only when, for some value of k, — — is a multiple of 2 it. \ia and b are incommensurable, b this is impossible, but if - = *1 , where — is a rational fraction in its lowest terms, the smallest value of k — q. The curve then con- sists of q arches and winds p times around the fixed circle. 58. The hypocycloid. When a circle rolls upon the inside of a fixed circle, each point of the rolling circle describes a curve called the hypocycloid. If the axes and the notation are as in the previous article, the equa- tions of the hypocycloid are x = (b — a.) cos (f> + a cos — y = (b — a) sin — a sin ■ + cos 3 $) = 4 a cos 3 <£ = b cos 3 <£, y — a (3 sin c/> — sin 3 (/>) = 4 a sin 3 (£ = b sin 3 <£. This is the four-cusped hypocycloid of Ex. 3, § 52. 59. The involute of the circle. If a string, kept taut, is unwound from the circumference of a circle, its end describes a curve called the involute of the circle. Let (fig. 98) be the center of the circle, a its radius, and A the point at which PROBLEMS 113 the end of the string is on the circle. Take as the origin of coordinates and OA as the axis of x. Let P(x, y) be a point on the involute, PK the line drawn from P tangent to the circle at K, and <£ the angle XOK. Then PK represents a portion of the unwinding string, and hence KP — arc AK=atf>. Now it is clear that for all positions of the point K, OK makes an angle — -^ with OY. Hence the projection of OK on OX is always OK cos = a cos , and its projection on OY is F " • '- ,s (7j-\ jj- (f) — — \ = a sin . Also KP always makes an angle (f> — — with OX and an angle ir — with (> Y. Hence the projection of KP on OX is KP cos I — — ) = atf> sin <£, and its projection on C> F is jKP cos (7r — <£) = — « cos (/>. The projection of OP on OX is #, and on 01' is y. Hence, by the law of projections, § 2, x = a cos <$> + d(f> sin (f>, y = a sin (f>— acf) cos w = -• * 2 lX £ 3. £C = ± Vl + 9 1 1 .'/ ± Vi + 9 e 4. x = f, y a 2 + * 2 5 . x = - o > 1 1 = l + t- J 6. x = 2 a shr, y 7 . cr = e* sin £, ?/ = t(l+l*) 2 a sin 3 <£ cos <£ e* cos £. x = a + a sin cf>, ij = r< ■ — a- cos e£. 3 a a Q iC — —£- cos 4> — -x cos 3 <£, 3 a . y = — sin <£ a' = 2 a cos — a cos 2 <£, y = 2 a sin <£ 114 PARAMETRIC REPRESENTATION - sin 3 <£. 10. a* = 2 a cos <£ — a cos 2 , y = 2 a sin — a sin 2 <£. 11. A projectile moves so that the coordinates of its position at any time t are given by the equations x = Q0t,*y= 80£ — 16Z 2 . Plot its path. 12. Find the parametric equations of the parabola y 2 = 4tpx when the parameter is the slope of a straight line through the vertex. x 2 ?/ 2 13. Find the parametric equations of the ellipse — + — =1 when the parameter is the slope of a straight line through the center. 14. Find the parametric equations of the cissoid when the param- eter is the slope of a straight line through the origin, the axes of coordinates being as in fig. 81. 15. Find the parametric equations of the cissoid when the param- eter is the angle A OP (fig. 81). 16. Find the parametric equations of the strophoid when the parameter is the angle MAP (fig. 82). 17. When a circle rolls upon the outside of a fixed circle, a point on the radius of the rolling circle at a distance h from its center describes a curve called an epitrochoid. Find its equations. 18. When a circle rolls upon the inside of a fixed circle, a point on the radius of the rolling circle at a distance h from its center describes a curve called an hypotrochoid. Find its equations. 19. If a circle rolls on the inside of a fixed circle of twice its radius, what is the form of the curve generated by a point of the circumference of the rolling circle ? 20. AB is a given straight line perpendicular to OX at the point C, where OC — a. Through any straight line is drawn, meeting AB at D. On OX a point M is taken, to the left of C, so that CM = CD. Finally, through M a straight line is drawn perpendicular to OX, intersecting OD at P. Find the parametric equations of the locus of P, using the angle XOD as the parameter. Find also the Cartesian equation, name the curve, and sketch the graph. PROBLEMS 115 21. A fixed circle of radius a with its center at intersects OX at A. The straight line BC is tangent to the circle at .4. Through any straight line is drawn, intersecting the circle at D and intersecting BC at E. Through D a straight line is drawn parallel to OY, and through E a straight line is drawn parallel to OX. These lines inter- sect at P. Find the parametric equations of the locus of P in terms of the angle XOD as parameter. Find also the Cartesian equation and sketch the curve. 22. A circle of radius a has its center at 0, the origin of coordinates. The tangent to the circle at any point A meets OX at .1/. Through M a straight line is drawn parallel to <>Y, and through .1 a straight line is drawn parallel to OX. These lines intersect at P. Find the par- ametric equations of the locus of P, using the angle MOA as the parameter. Find also the Cartesian equation and sketch the curve. 23. A circle of radius a has its center at the origin of coordinates 0. Through O any straight line is drawn, intersecting the circle at A. The tangent to the circle at .1 intersects OFat /;. Through B a straight line is drawn parallel to 0X } meeting OA produced at /'. Find the parametric equations of the locus of P in terms of the angle XOA as parameter. Find also the Cartesian equation. 24. Let OA be the diameter of a fixed circle and LK the tangent at A. From O draw any straight line intersecting the circle at B and LK at C, and let /' be the middle point of BC. Find the para- metric equations of the locus of /', using the angle AOP as the parameter, OA as the axis of y, and as the origin. Find also the Cartesian equation. 25. A circle of radius a has its center at the origin of coordinates 0, and the straight line -I B is tangent to the circle at A(a, 0). From any straight line is drawn, meeting All at /•; and the circle at D. On OE, OP is taken equal to DE. Find the parametric equations of the locus of P in terms of the angle AOP as parameter. 26. The straight line All is perpendicular to OX at A (a, 0). From a straight line is drawn to any point C of All. The straight line drawn from C perpendicular to OC meets OX at M. The perpendic- ular to OX at .1/ meets' OC produced at P. Find the parametric equations of the locus of P in terms of the angle XOC as parameter. Find also the Cartesian equation. 116 PARAMETRIC REPRESENTATION 27. OBCD is a rectangle with OB = a and BC = c. Any line is drawn through C, meeting OB in E, and the triangle EPO is con- structed so that the angles CEP and EPO are right angles. Find the parametric equations of the locus of P, using the angle DOP as I he parameter, OB as the axis of x, and as the origin. Find also the Cartesian equation of the locus. 28. A fixed circle has as diameter the straight line joining the origin and the point A (0, 2a). Any point B of the circle is connected with A and 0, and BM is drawn perpendicular to OX, meeting OX at M. On MB, MP is laid off equal to BA. Find the parametric equations of the locus of P in terms of the angle XOB as parameter. Find also the Cartesian equation. 29. Let AB be a given straight line, a given point a units from AB, and k a given constant. On any straight line through 0, meet- ing AB in M, take-P so that OM • MP= k 2 . Find the parametric equations of the locus of P, using as the origin, the perpendicular from to AB as the axis of x, and the angle between OX and OP as the parameter. Also find the Cartesian equation. 30. ABC is a given right triangle of which the sides AB and BC about the right angle at B are always equal to a and b respectively. The triangle moves in the plane XOY so that A is always on OF and B is always on OX. P is the middle point of the hypotenuse AC. Find the parametric equations of the locus of P, using the angle XBC as the parameter. 31. Let be the center of a circle with radius a, A a fixed point on the circle, and B a moving point on the circle. If the tangent at B meets the tangent at A in C, and P is the middle point of BC, find the equations of the locus of P in parametric form, using the angle A OB as the arbitrary parameter, OA as the axis of x, and as the origin. 32. A fixed circle has as diameter the straight line joining the origin of coordinates and the point A (2 a, 0), and LK is tangent to the circle at A. From O any straight line is drawn, meeting the circle at D and the tangent LK at E. On OE a point P is so taken that PD = DE in both length and direction. Find the parametric equa- tions of the locus of P in terms of the angle AOE.&s parameter. Find also the Cartesian equation. PROBLEMS 117 33. A and B are two points on the axis of y at distances — a and -f- a respectively from the origin. AH is any straight line through A, meeting the axis of x at H. BK is the perpendicular from B on All, meeting it at K. Through K a straight line is drawn parallel to the axis of x, and through H a straight line is drawn parallel to the axis of y. These lines meet in P. Find the parametric equa- tions of the locus of P, using the angle BAK as the parameter. Also find the Cartesian equation. 34. Q is the point on the auxiliary circle of the ellipse corresponding to the point P of the ellipse. The straight line through P parallel to OQ meets OX at L and OY at M. Prove PL = b, and PM= a. 35. If a projectile starts with an initial velocity v in an initial direction which makes an angle a with the axis of x, taken horizontal, its position at any time t is given by the parametric equations x = vt cos a, y = vt sin a — \ yP. Pind the Cartesian equation of the path of the projectile and its nature and position. 36. From the equations of problem 35 determine when and where the projectile strikes a point on the axis of x. 37. From the equations of problem 35 determine when, and for what value of x, the projectile passes through a point which is at a distance h below the horizontal. 38. From the equations of problem 35, what elevation must be given to a gun that the projectile may pass through a poinl b units distant from the muzzle of the gun and lying in the horizontal line passing through the muzzle ? 39. From the equations of problem 35, what elevation must be given to a gun to obtain a maximum range on a horizontal line passing through the muzzle ? 40. A gun stands on a cliff // units above the water. From the equations of problem 35, what elevation must be given to the gun that the projectile may strike a point in the water b units from the base of the cliff ? AC CHAPTER VIII POLAR COORDINATES 60. Coordinate system. So far we have determined the posi- tion of a point in the plane by two distances, x and y. We may, however, use a distance and a direction, as follows: Let (fig. 99), called the origin, or pole, be a fixed point, and let OM, called the initial line, be a fixed line. Take P any point in the plane and draw OP. Denote OP by r and the angle MOP by 0. Then r and 6 are called the polar coordinates of the point P (r, #), and when given will completely determine P. For example, the point (2, 15°) is plotted by laying off the angle MOP — lb° and measuring OP = 2. OP, or r, is called the radius vector, and 9 the vectorial angle, of P. These quantities may be either positive or negative. A negative value of 6 is laid off in the direction of the motion of the hands of a clock, a positive angle in the opposite direction. After the angle 6 has been constructed, positive values of r are measured from along the terminal line of 6, and negative values of r from along the backward extension of the terminal line. It follows that the same point may have more than one pair of coordinates. Thus (2, 195°), (2, -165°), (-2, 15°), and (— 2, — 345°) refer to the same point. In practice it is usually convenient to restrict 6 to positive values. Plotting in polar coordinates is facilitated by using paper ruled as in figs. 100 and 101. The angle 6 is determined from the num- bers at the ends of the straight lines, and the value of r is counted off on the concentric circles, either towards or away from the number which indicates 6, according as r is positive or negative. When an equation is given in polar coordinates, the corre- sponding curve may be plotted by giving to 6 convenient POLAR coordinates 119 values, computing the corresponding values of r, plotting the resulting points, and drawing a curve through them. Ex. 1. r = a cos 6. a is a constant which may be given any con- venient value. We may then find from a table oi natural cosines the value of r which corresponds to any value of 6. By plottiug the points corre- sponding to values of 6 from 0° to 90°, we obtain the arc ABCO (fig. 100). Values of 6 from 90° to 180° give the arc ODEA. Values of from 180° to 270° give again the arc ABCO, and those from 270° to 360° give the arc ODEA. Values of 6 greater than 360 'can clearly give no points not already found. The curve is a circle (§ 63). Ex. 2. r = a sin 3 0. As 6 increases from 0° to 30°, r increases from to a; as increases from 30° to 60 , r decreases from a to ; the point (?•, 6) traces out the Loop 0A0 (fig. 101). As 6 increases from G0° to 90°, r is negative and decreases from to — a ; as 6 increases from 90° to 120°, r increases from —a to 0; the point (r, 6) traces out the loop OB 0. As 8 increases from 120° to 180°, the point (r, 6) traces out the loop OCO. Larger values of $ give points already found, since sin 3 (180° + '(9) = - sin 3 ft sin 3 (G0° + 6) = — sin 3 6. The three loops are congruent, because This curve is called a rose of three leaves. 120 POLAR COORDINATES Ex. 3. r 2 = 2 rt 2 cos 2 0. Solving for r, we have ±aV2cos2 0. Hence, corresponding to any values of which make cos 26 positive, there will be two values of r numerically equal and opposite in sign and two corresponding points of the curve symmetrically situated with respect to the pole. If values are assigned to 6 which make cos 2 9 negative, the cor- responding values of r will be imaginary and there will be no points on the curve. Accordingly, as 6 increases from 0° to 45°, r decreases nu- merically from a to 0, and the portions of the curve in the first and the third quadrant are con- structed ; as increases from 45° to 135°, cos 2 6 is negative, and there is no portion of the curve between the lines 6 = 45° and 6 = 135°; finally, as 6 increases from 135° to 180°, r increases numerically from to a, and the portions of the curve in the second and the fourth quadrant are constructed. The curve is now complete, as we should only repeat the curve already found if we assigned further values to 6; it is called the lemniscate (fig. 102). 61. The spirals. Polar coordinates are particularly well adapted to represent certain curves called spirals, of which the more important follow: Ex. 1. The spiral of Archimedes, Fig. 102 In plotting, 6 is usually considered in circular measure. When 9 = 0, r — 0, and as 6 increases, r increases, so that the curve winds infinitely often around the origin while receding from it (fig. 103). In the figure the heavy line represents the portion of the spiral corresponding to positive values of 6, and the dotted line the portion corresponding to negative values of 6. Ex. 2. The hyperbolic spiral, Fig. 103 As 6 increases indefinitely, r approaches zero. Hence the spiral winds infinitely often around the origin, continually approaching it but never THE STRAIGHT LIXE 121 reaching it (fig. 104). As 6 approaches zero, r increases without limit. If P is a point on the spiral and NP is the perpendicular to the initial line. NP 6 Fig. 104 Hence, as 6 approaches zero as a limit, NP approaches a (§ 95). Therefore the curve comes con- stantly nearer to, but never reaches, the line LK, parallel to OM at a distance a units from it. This line is therefore an asymptote. In the figure the dotted portion of the curve corresponds to negative values of 6. Ex. 3. The logarithmic spiral, r = &*. When 6 = 0, r = 1. As 9 increases, r increases, and the curve winds around the origin at increasing distances from it (fig. 10")). "When 6 is negative and increasing numerically without limit, r approaches zero. Hence the curve winds infinitely often around the origin, continu- ally approaching it. The dotted line in the figure corresponds to negative values of 6. A property of this spiral is that it cuts the radius vectors at a constant angle. The student may prove this after reading § 103. We shall now give examples of the derivation of the polar equation of a curve from the definition of the curve. 62. The straight line. Let LK (fig. 106) be a straight line perpendicular to OD. Let the angle MOD be denoted by a, and let OD = p ; then p is the normal distance of LK from the pole. Let P(r, 6} be any point of LK. Then, by trigonometry, Fig. 105 OP cos D OP r cos (0 — a) OD, (1) which is the equation of the straight line. 122 POLAR COORDINATES I f a = and p — a, we have the special equation r cos 6 = a, or r = a sec 6. (2) If the straight line passes through the origin, p = 0. The equation of the line then becomes cos (6 — a) = 0, or simply 6 — — + a, which is of the form 6 = c. (3) 63. The circle. Let C(b, a) be the center and a the radius of a circle (fig. 107). Let P(r, 0) be any point of the circle, and draw the straight lines OC, OP, and CP. By trigonometry, we have OP 2 + OC 2 - 2 OP • OC cos POC=CP\ Noting that cos POC = cos (0 — a), OP = r, OC = h, and CP = «, and substituting in the equation, we have the result r 2 -2rb cos (6 - a) + 6 2 = a 2 (1) as the polar equation of the circle. When the origin is at the center of the circle, b = and (1) becomes simply Fig. 107 When the origin is on the circle, b = a and (1) becomes r-2«cos(6>-«)=0; which may be written r = a Q cos 6 + a y sin 0, (3) where a and a x are the intercepts on the lines 6=0 and 6 = — respectively. When the origin is on the circle and the initial line is a diameter, (3) becomes ? . _ a CQg ^ ^ When the origin is on the circle and the initial line is tangent to the circle, (3) becomes . n ^ rx r = a sin 0. (5) THE LIMAgON 123 64. The limaQon. TJirough any fixed point (fig. 108) on the circumference of a fixed circle draw any line cutting the circle again at D, and lay off on this line a constant length measured from D in either direction. The locus of the points P and Q thus found is a curve called the limagon. Take as the pole, and the diameter OA as the initial line, of a system of polar coordinates, and call the diame- ter of the circle a and the constant length b. Then it is clear that the entire locus can be found by caus- ing OD to revolve through an angle of 3»>0° and laying off Dl'=b, always in the direction of the terminal line of AOD. Let P be (>•, 0), where 6 = AOD. Then r = OD+DP when 6 is in the first or the fourth quadrant, and r=— (>I) + DP when 6 is in the* second or the third quadrant. But it appears from the figure that OD = OA cos 6 when 6 is in the first or the fourth quadrant, and that OD = — OA cos 6 when is in the second or the third quadrant. Hence, for any point on the limac.on, r = a cos 6 + b. -,„,'(-;,) -ct>» : '(-af) Fn;. 109 In studying the shape of the curve there are three cases to be distinguished : 1. b>a. r is always positive; the curve appears as in fig. 108. 2. b < a. r is positive when cos 6 > •> negative when h 1 ^ c,osd< , and zero when cos = The curve appears as a a in fig. 109. 3. b = a. The equation now becomes Q r — a (cos 6 + 1) = 2 a cos 2 - • 124 POLAR COORDINATES Here r is positive, except that when = 180° r is zero. The curve appears as in fig. 110 and is called the cardioid. The cardioid is an epicycloid for which the radius of the fixed circle equals that of the rolling circle. The proof of this is left to the student. 65. Relation between rectangular and polar coordinates. Let the pole and the initial line OM of a sys- tem of polar coordinates be at the same time the origin and the axis of re of a system of rectangular coordinates. Let P (fig. HI) be FlG 110 any point of the plane, (x, y) its rectangular coordinates, and (r, 0) its polar coordinates. Then, by the definition of the trigonometric functions, cos sin 6 (1) Whence follows, on the one hand x = r cos 0, y — r sin 6 ; and, on the other hand, r =Vx 2 +y 2 , si? i/} — Fig. Ill COS0 (2) Vx 2 +y 2 V;r' 2 + u By means of (1) a transformation can be made from rectangular to polar coordinates, and by means of (2) from polar to rectangular coordinates. Ex. 1. The equation of the cissoid (§ 48) is x 3 y = 7T-. — z • Substituting from (1) and making simple reductions, we have the polar equation . r = ^ — THE CONIC 125 Ex. 2. The polar equation of the' lemniscate (Ex. 3, § 60) is 7- 2 =2a 2 cos2 0. Placing cos 26 = cos 2 6— sin 2 and substituting from (2), we have the rectangular equation (x 2 + r) 2 =2« 2 (2- 2 -r)- 66. The conic, the focus being the pole. From § 46, the equa- tion of a conic when the axis of x is an axis of the conic and the axis of y. is a directrix is We may transfer to new axes having the focus of the conic as the origin and the axis of the conic as the axis of x by placing x = c + x', y = y\ thus obtaining x' ' 2 + y' a = f -( ./•' + rf. If we now take a system of polar coordinates having the focus as the pole and the axis of the conic as the initial line, we have x r =rco8 0, y , = ram0. The equation then becomes r 9 =e a (rcos^ + c) 9 , which is equivalent to the two equations ce ce r — » r = • 1 — e cos 6 1 + >■ cos 6 Either of these equations alone will give the entire conic. To see this, place 6 = d l in the second equation, obtaining — ce r = 1 1 + e cos X Now place 6 = it + 6 X in the first equation, obtaining r = — r { . The points (r^ 6^) and (— r^ tt + # x ) are the same. Hence any point which can be found from the second equation can be found from the first. I hereiore r = 1 — f cos 6 is the required polar equation. 120 POLAR COORDINATES 67. Examples. Polar coordinates may be used with great advantage in the solution of problems involving a number of straight lines radiating from a given point, the given point then being taken as the pole of the system of coordinates. This use is illustrated in the following examples: Ex. 1. Prove that if a secant is drawn through the focus of a conic, the sum of the reciprocals of the segments made by the focus is constant. Let P 1 P 2 (fig. 112) be any secant through the focus F, and let FP 1 = r v FP 2 = r 2 , and the angle MFP 1 = 0. Then the polar coordi- nates of P 1 are (r v 6) and those of P 2 are (r 2 , 6 + tt). From the polar equation of the conic, we have r. — £! , Fig. 112 Hence I + I 1 — e cos (6 + 7r) 1 + e cos I 2 Ex. 2. Find the locus of the middle of a circle all of which pass through a Take any circle with the center C (fig. 113), and let be any point in the plane. If is taken for the pole, and OC for the initial line, of a system of polar coordinates, the equation of the circle is r 2 - 2 rb cos d + i 2 - a 2 = 0. (1) Let P X P 2 be any chord through O, and let OP 1 = r v OP 2 = r 2 . Then 1\ and r 2 are the two roots of equation (1) which corre- spond to the same value of 0. Hence ?•, + r„ = 2 b cos 6. points of a system of chords fixed point. If Q is the middle point of P X P 2 and we now place OQ — r, we have ri + b cos 0. But this is the polar equation of a circle through the points and C. Plot the following curves 1. r = a sin 2 0. 2. r = a cos 3 0. 3. r = a sin-' Z 4. r = a cos 5. r 6. r 8 = a 2 sin 0. 7. j- 2 = «°- sin 3 0. 8. r 2 = a 2 sin 4 0. 9. >•= «(l + sin0). 10. r= a (2 + sin0). 11. r= « (1 + cos 2 0), 12. r= a (1 — cos 2 0). 13. /■ = a(l+ cos 3 0). 14. r = a (2 + cos 2 0). 15. r = a(l + 2sin0). 16. r = a(l + 2 cos 2 0). 17. >• = «(! + 2 cos 3 0). PROBLEMS 18. o , • 30 r = 2 + sin — • 19. r 2 =« 2 (l-COS0). 20. r = « 2 (l + 2cos20) 21. r = a tan 0. 22. /• = a tan 2 0. 23. , ;• = a tan - • 24. r = a sec 2 0. 25. /■ = a sec • 127 26. ?• = a sec 2 -- 27. r = a(l + scc0). 28. r = a(l + 2 sec 0). 29. r= «(2 + sec0). 30. /• cos = a cos 2 0. 31. r = 32. r = 1 33. >• = - COS0 sill 2 0. Plot each pair of the following curves in one diagram and find their points of intersection : 34. rcos 35. r cos 36. r cos (d — ~) =a V2, r = 2 a cos 0. 37. r* = a a sin0, r*=a 3 sin20. 38. r = a(l + sin 2 0), r* = 4 a 2 sin 2 0. 39. >- 2 = a 2 sin 0, z- 2 = a 2 sin 3 0. 128 POLAR COORDINATES 40. is a fixed point and LK a fixed straight line. If any straight line through intersects LK in Q, and a point P is taken on this line so that OP ■ OQ = A- 2 , find the locus of P. 41. A straight line OA of constant length a revolves about 0. From A a perpendicular is drawn to a fixed straight line OM, inter- secting it in B. From B a perpendicular is drawn to OA, intersecting it in P. Find the locus of P, OM being taken as the initial line. 42. is a fixed point of a circle of radius a, and OM is a fixed straight line passing through the center of the circle. A straight line is drawn from O to any point P x of the circle, and from P 1 a straight line is drawn perpendicular to OM, meeting OM at Q. From Q a straight line is drawn perpendicular to OP v meeting OP 1 at P. Find the equation of the locus of P, taking O as the origin of coor- dinates and OM as the initial line. 43. MN is a straight line perpendicular to the initial line at a distance a from 0. From a straight line is drawn to any point B of MN. From B a straight line is drawn perpendicular to OB, inter- secting the initial line at C. From C a straight line is drawn per- pendicular to BC, intersecting MN at D. Finally, from D a straight line is drawn perpendicular to- CD, intersecting OB at P. Find the locus of P. Transform the following equations to polar coordinates : 44. xy = 7. 46. x i + x 2 f — ah? = 0. 45. x 2 + f - 8 ax - Say = 0. 47. (.« 2 + iff = d 2 (x 2 - if). 48. Find the polar equation of the strophoid when the pole is and the initial line is OA (fig. 82). Transform the following equations to rectangular coordinates : 49. r cos (o- 7 ^) + r cos (d + ^) = 12. 50. r = « sin 0. 51. r = a tan 6. 52. Find the Cartesian equation of the rose of four petals r = a sin 20. 53. Find the Cartesian equation of the cardioid r = a(l— cos 0). 54. Find the Cartesian equation of the limacon r = a cos 6 + b. PROBLEMS 129 55. In a parabola prove that the length of a focal chord which makes an angle of 30° with the axis of the curve is four times the focal chord perpendicular to the axis. 56. A comet is moving in a parabolic orbit around the sun at the focus of the parabola. When the comet is 100,000,000 miles from the sun the radius vector makes an angle of 60° with the axis of the orbit. What is the equation of the comet's orbit ? How near does it come to the sun ? 57. A comet moving in a parabolic orbit around the sun is observed at two points of its path, its focal distances being 5 and 15 million miles, and the angle between them being 90°. How near does it come to the sun ? 58. If a straight line drawn through the focus of an hyperbola parallel to an asymptote meets the curve at P, prove that FP is one fourth the chord through the focus perpendicular to the transverse axis. 59. The focal radii of a parabola are extended beyond the curve until their lengths are doubled. Find the locus of their extremities. 60. If 1\ and P 2 are the points of intersection of a straight line drawn from any point to a circle, prove that <>/\ • OP a is constant. 61. If Pj and P 2 are the points of intersection of a straight line from any point O to a fixed circle, and Q is a point on the same L' OP ■ OP straight line such that (>(} = — — L ^> find the locus of (). oi\ + <>P, 62. Secant lines of a circle are drawn from the same point on the circle, and on each secant a point is taken outside the circle at a distance equal to the portion of the secant included in the circle. Find the locus of these points. 63. From a point O a straight line is drawn intersecting a fixed circle at P, and on this line a point Q is taken so that OP • OQ = k 2 . Find the locus of Q. 64. Find the locus of the middle points of the focal chords of a conic. 65. Find the locus of the middle points of the focal radii of a conic. 66. If P 1 FP 2 and Q X FQ 2 are two perpendicular focal chords of a conic, prove that — h — is constant. ' * P X F • FP 2 T Qf • FQ 2 CHAPTER IX SLOPES AND AREAS 68. Limits. A variable is said to approach a constant as a limit, when, under the laiv which governs the change of value of the variable, the difference between the variable and the constant becomes and remains less than any quantity which can be named, no matter how small. If the variable is independent, it may be made to approach a limit by assigning to it arbitrarily a succession of values follow- ing some known law. Thus, if x is given in succession the values 13 7 2 n -l *x=2' x *=r x *=r •••' x ^-¥~' and so on indefinitely, it approaches 1 as a limit. For we may make x differ from 1 by as little as we please by taking n sufficiently great ; and for all i i 7 « larger values of n the differ- 1 + 1 'f i- ence between x and 1 is still smaller. This may be made evident graphically by marking off on a number scale the successive values of x (fig. 114), when it will be seen that Fig. 115 the difference between x and 1 soon becomes and remains too minute to be represented. Similarly, if we assign to x the succession of values 1111 , 1V- 1 1 2 2 3 3 4 4 5 ' " v ' n+1 x approaches as a limit (fig. 115). 130 LIMITS 131 If the variable is not independent, but is a function of x, the values which it assumes as it approaches a limit depend upon the values arbitrarily assigned to x. For example, let y =f(x), and let x be given a set of values, approaching a limit a. Let the corresponding values of y be y* y* y 8 > y* •••» y.» •••• Then, if there exists a number A such that the difference between y aifd A becomes and remains less than any assigned quantity, y is said to approach A as a limit as x approaches a in the manner indicated. This may be seen graphically in fig. 116, where the values of x approaching a are seen on the axis of abscissas, and the values of y approach- ing A are seen on the axis of ordinates. The curve of the function is continually nearer to the line y = A. In the most common cases the limit of the function depends only upon the limit a of the independent variable and not upon the particular succession approaching a. This is clearly of values that the case if the x assumes in graph of the function is as drawn in fig. 116. Ex. 1. Consider the function x- + 3 x - 4 y= x-l ' and let x approach 1 by passing through the succession of values x = 1.1, !F = 1.01, x = 1.001, a: = 1.0001, •••. Then y takes in succession the values y = 5.1, y = 5.01, y = 5.001, y - 5.0001. It appears as if y were approaching the limit 5. To verify this we place X = 1 + h, where It is not zero. By substituting and dividing by h, we find 132 SLOPES AND AREAS y = 5 + h. From this it appears that y can be made as near 5 as we please by taking h sufficiently small, and that for smaller values of h, y is still nearer 5. Hence 5 is the limit of y as x approaches 1. More- over, it appears that this limit is independent of the succession of values which x assumes in approaching 1. Ex. 2. Consider y = = as x approaches zero. 1 - Vl - x Give x in succession the values .1, .01, .001, .0001, • • •. Then y takes the values 1.9487, 1.9950, 1.9995, 1.9999, • • •, suggesting the limit 2. In fact, by multiplying both terms of — = by 1 + Vl — x, we find 1 - VI - x y = 1 + Vl— x for all values of x except zero. Hence it appears that y approaches 2 as a; approaches 0. We shall use the symbol = to mean "approaches as a limit." Then the expressions Lima- = a and x = a have the same significance. The expression ~Limf(x') = A x = a is read " The limit of f(x), as x approaches «, is Ay 69. Theorems on limits. In operations with limits the follow- ing propositions are of importance : J 1. The limit of the sum of a finite number of variables is equal to the sum of the limits of the variables. We will prove the theorem for three variables ; the proof is easily extended to any number of variables. Let X, F, and Z be three variables, such that Lim X=A, Lim Y=B, him Z=C. From the definition of limit (§68) we may write X = A+a, Y = B+b, Z=C+c, where a, b, and c are three quantities each of which becomes and remains numerically less than any assigned quantity as the variables approach their limits. Adding, we have A'+ F+ Z = A + B + C+ a + b + c. LIMITS 133 Now if e is any assigned quantity, however small, we may make a, b, and c each numerically less than - , so that a + b + c is o numerically less than e. Then the difference between X+Y+ Z and A+B + C becomes and remains less than e; that is, Lim (A' + Y+ Z)=A+B + C = Lim X+ Lim Y+ Lim Z. V 2. The limit of the product of a finite number of variables is equal to the product of the limits of the variables. Consider first two variables A' and F, such that Lim X= A ami Lim Y= B. As before, we have A= A + a and Y=B + b. Hence XY=AB+bA+aB+ab. Now we may make a and b so small that bA, aB, and ab are each less than -, where e is any assigned quantity, no matter how small. Hence Lim XY= AB = (Lim A') (Lim F). Consider now three variables A, F, Z. Place XY = U. Then, as just proved, LimPZ=(LimP)(LimZ) . that is, LimA]'Z = (LimA)')(LiinZ) = (LiinA)(Lim)')(LimZ). Similarly, the theorem may be proved for any finite number of variables. v / 3. The limit of a constant multiplied by u variable is equal to the constant multiplied by the limit if the variable. The proof is left for the student. 4. The limit of the quotient if two variables is equal t*> the quotient of the limits of the variables, provided the limit of the divisor is not zero. Let A' and Y be two variables, such that Lim .V= J and Lim Y==B. Then, as before, X=A+a, Y=B + b. X A+a t A' A A + a A aB — bA IIence y=i^i,' mi y-jriiTi-^^M- 134 SLOPES AND APEAS Now the fraction on the right of this equation may be made less than any assigned quantity by taking a and b sufficiently small. Hence Lim LimX Lim Y ' The proof assumes that B is not zero. 70. Slope of a curve. By means of the conception of a limit we may extend the definition of " slope," given in § 6 for a straight line, so that it may be applied to any curve. Let P x and P 2 be any two points upon a curve (fig. 117). If i? and P 2 are connected by a straight line, the slope of this line is y.-yi If P 2 and P x are close enough to- Fig. 117 gether, the straight line P,P 2 will differ only a little from the arc of the curve, and its slope may be taken as approximately the slope of the curve at the point P v Now this approximation is closer, the nearer the point jP is to P v Hence we are led naturally to the following definition: The slope of a curve at a point P x (x^ y^) is the limit approached by the fraction y*-Vx ivhere x n and y are the coordinates of a second point P 2 on the curve and where the limit is taken as P^ moves toward i? along the curve. Ex. 1. Consider the curve y = x 2 and the point (5, 25) upon it, and let «i = 5. !lx = 25 - We take in succession various values for x 2 and y„, corresponding to points on the curve which are nearer and nearer to (x v y^), and arrange our results in a table as follows : x a ?/ 2 x 2 — x x y., - v x y-2 - V\ x 2 — x x G 30 1 n 11 5.1 20.01 .1 1.01 10.1 5.01 25.1001 .01 .1001 10.01 5.001 25.010001 .001 .010001 10.001 INCREMENT 135 The arithmetical work suggests the limit 10. To verify this, place ar 2 = 5 + h. Then y 2 = 25 + 10 h + h 2 . Consequently ' 2 _ ' * = 10 + A, and y« — iii 2 l as x„ approaches x,, h approaches and " _ approaches 10. Hence the x i x \ slope of the curve y — x' 2 at the point (5, 25) is 10. Ex. 2. Find the slope of the curve y = - at the point (3, ^). "We have here x x — 3, y x = \. We place x„ — 3 + A, »/o = 3 + A rrv 7. ~ '' 1 'A. — '/l 1 1 hen x„ — x, = «, ?/., — ?/, = > aud — -■ 9 + 3A x 8 — z 2 9 + 3* As P 2 approaches 1\ along the curve, h approaches 0, and the limit of - is — ;t> hence the slope of the curve at the point (:5, .',) is — 1. In a similar manner we may find the slope of any curve the equation of which is not too complicated ; but when the equa- tion is complicated, there is need of a more powerful method y — y for finding the limit of — • This method is furnished by x a - x x the operation known as differentiation, the first principles of which are explained in the following articles. 71. Increment. When a variable changes its value, the quan- tity which is added to its first value to obtain its last value is called its increment. Thus, if x changes from 5 to 5|-, its incre- ment is i. If it changes from 5 to 4;j, the increment is — \. So, in general, if x changes from r x to .>,, the increment is * x 2 — Xj. It is customary to denote an increment by the symbol A (Greek delta), so that A.r = x 2 — x^ and x„ = x x + A#. If y is a function of x, any increment added to x will cause a corresponding increment of y. Thus, let y =f(x) and let x change from x 1 to x n . Then y changes from y x to y 2 , where y=f(xr) and y 2 =/(,•) =/(.r l + A*). Hence Ay = y„- y x =f(x l + A.r) -f(x^. 136 SLOPES AND AKEAS 72. Continuity. A function y is called a continuous function of a variable x ivhen the increment of y approaches zero as the increment of x approaches zero. It is clear that a continuous function cannot change its value by a sudden jump, since we can make the change in the function as small as we please by taking the increment of x sufficiently small. As a consequence of this, if a continuous function has a value A when x= a, and a value B when x = b, it will assume any value C, lying between A and B, for at V least one value of x between a and h (fig. 118). V In particular, if f(a) is positive and f(V) is negative, f(x) = for at FlG n8 least one value of x between a and b. When Ax and Ay approach zero together, it usually happens that — approaches a limit. In this case y is said to have a derivative, defined in the next article. 73. Derivative. When y is a continuous function of x, the deriva- tive of y with respect to x is the limit of the ratio of the increment of y to the increment of x, as the increment of x approaches zero. The derivative is expressed by the symbol — ; or, if y is expressed by f(x), the derivative may be expressed by f'(x). Thus, if y =f(x), dv j., , N T . Ay T . f(x + Ax~) —f(%) -f —f(x\ = Lim -£ = Lim 1 ^ — J - — J -±^-. dx ijioAx ax-=o Ax The process of finding the derivative is called differentiation, and we are said to differentiate y with respect to x. The process involves, according to the definition, the following four steps : 1. The assumption of an increment of x. 2. The computation of the corresponding increment of y. 3. The division of the increment of y by the increment of x. 4. The determination of the limit approached by this quotient as the increment of x approaches zero. DIFFERENTIATION OF A POLYNOMIAL 137 Ex. 1. Find the derivative of y when y = x 3 . (1) Assume Ax = Ji. (2) Compute Ay = (x + /<) 3 - x 3 = 3 x 2 h + 3 xJr + h s . (3) Find ^ = 3 x 2 + 3 xh + h 2 . ,1, (4) The limit is evidently 3 x 2 . Hence —■ = 3 x 2 . Ex. 2. Find the derivative of - • 1 x (1) Place y = - and assume Ax = //. (2) Compute A y = ^-i = -^A_. (3) Find ^ = __1— . Ax ' «■ + ■** ! rf (4) The limit is cleariy = i and therefore '-— = -. x" ax j- It appears that the operations of finding the derivative oif(x) are exactly those which are used in finding the slope of the curve y=f(x). Hence the derivative is a fund inn which gives I In' slope of the curve at each point of it. 74. Differentiation of a polynomial. The obtaining of a deriv- ative by carrying out the operations of the last article is too tedious for practical use. It is more convenient to use the definition to obtain general formulas which may be used for certain classes of functions. In this article we shall derive all formulas necessary to differentiate a polynomial, 1. — ^- — -=waof -1 , where n is a positive integer and a any constant. Let y = ax". (1) Assume Ax = h. (2) Then Ay = a (x + h) n — ax n -•( ,-ij + "(" 1 > a »-'/ t '+ ...+/<' (3) U«.^. + »£gJi - .-. J + ... + i (4) Taking the limit, we have -^ = nax" ~\ 138 SLOPES AND AKEAS d (ax) = a, where a is a constant. This is a special case of the preceding formula, n being here equal to 1. The student may prove it directly. dc 3. — = 0, where c is a constant. d.v Since c is a constant, Ac is always 0, no matter what the value of x. Hence Ac Ax dc 0, and consequently the limit — = 0. 4. The derivative of a polynomial is found by adding the derivatives of the terms in order. This is a special case of a more general theorem (3, § 82). The proof of the special case before us may be easily given by the student or may be assumed temporarily. Ex. Find the derivative of f(x) = G x 5 - 3 x i + 5 x s - 7x 2 + 8 x - 2. Applying formulas 1, 2, or 3 to each term in order, we have f {x) = 30 x- 4 - 12 x 3 + 15 x 2 - 14 x + 8. 75. Sign of the derivative. A function of x is called an increasing function when an increase in x causes an increase in the function. A function of x is called a decreasing function when an increase in x causes a decrease in the function. The graph of a function runs up toward the right hand when the function is increasing and runs down toward the right hand when the func- tion is decreasing. Thus x~— x — 6 (fig. 119) is decreasing when x < \ and increasing when x>\. The sign of the derivative enables us to determine whether a function is increasing or decreasing in accordance with the follow- ing theorem : When the derivative of a function is posi- tive, the function is increasing; when the derivative is negative, the function is decreasing. SIGN OF THE DEKIYATIVE 139 To prove this, consider y =f(x), and let us suppose that — is positive. Then, since -~ is the limit of — , it follows that — - dx dx Ax Ax is positive for sufficiently small values of Ax; that is, if Ax is assumed positive, Ay is also positive, and the function is increasing. Similarly, if ~ is negative, Ay and Ax have oppo- site signs for sufficiently small values of Ax, and the function is decreasing by definition. Ex. 1. If y = x 1 — x — 6, — = 2x — 1, which is negative when x < k dx and positive when x>\. Hence the function is decreasing when x< \ and increasing when x> \, as is shown in fig. 119. Ex. 2. If y= l(x 3 -3o; 2 -9x + 27) : |x»-|r-|=|(x+l)(x-3). Now — is positive when x< — 1, negative when -KK3, and positive dx when x > 3. Hence the function is increasing when x< — 1, decreasing when a: is between —1 and 3, and increas- ing when ./•>:; (fig. 120). It remains to examine the cases in which -~ = 0. Referring to the two ex- dx amples just given, we see that in each the values of x which make the deriv- ative zero separate those for which the function is increasing from those for which the function is decreasing. The points on the graph which correspond to these zero values of the derivative can be described as turning points. Likewise, whenever f'(x) is a continuous function of x, the values of x for which it is positive are separated from those for which it is negative by values of x for which it is zero (§ 72). Now in most cases which occur in elementary work, f'(x) is a continuous function. Hence we may say, The values of x for which a function changes from an increas- ing to a decreasing function are, in general, values of x which make the derivative equal to zero. Fig. 120 140 SLOPES A^'J) AREAS The converse proposition is, however, not always true. A value of x for which the derivative is zero is not necessarily a value of x for which the function changes from increasing to decreasing or from decreasing to increasing. For consider l(a; 3 -9o; 2 -h27^-19). Its derivative is x % — 6 x + 9 = (x — 3) 2 , which is always positive. The function is therefore always increasing. When x = 3 the derivative is zero, and the corresponding shape of the graph is shown in fig. 121. 76. Tangent line. A tangent to a curve is the straight line approached as a limit by a secant line as two points of intersection of the secant and the curve are made to approach coincidence. Let P x and P 2 be two points on a curve. Then if a secant is drawn through P x and P 2 of a curve (fig. 122) and the point P 2 is made to move along the curve toward i^, which is kept fixed in position, the secant will turn on P x as a pivot and will approach as a limit the tangent P X T. The point P x is called the point of contact of the tangent. From the definition it follows that the slope of the tangent is the same as the slope of the curve at the point of contact ; for the slope of the tangent is evidently the limit of the slope of the secant, and this limit is the slope of the curve, by § 70. The equation of the tangent is readily written by means of § 28 when the point of contact is known. Let (x^ y^) be Fig. 122 denote the value of -~ when dx the point of contact, and let x = x t and y — y x . Then (x x , y^) is a point on the tangent and -~\ is its slope. Therefore its equation is ;).(• (1) THE DIFFERENTIAL 141 Ex. 1. Find the equation of the tangent to the curve y = x* at the point (x v y x ) on it. Using formula (1), we have y - > h = :j % l ( x - ar x ). But since (x v y x ) is on the curve, we have y 1 = a;*. Therefore the equa- tion can be written ,, ., „ 3 y = :> xf x — 2 arf. T Ex. 2. Find the equation of the tangent to y = x* + 3 x at the point the abscissa of which is 2. dy dx x + :i. Ifz^then^lOandg)^?. ^ m Therefore the equation is y - lo = 7(. # -2), or y-7x — 4. If ?T (fig. 123) is a tangent line and the angle it makes with OX, its slope equals tan (/>, by ^ -'52. Hence ,7„ tan (/> = l.r 77. The differential. Lot the function /(./) be represented by the curve y=f(.r), and let /' and A'T = (tan RPT) PR =f'(x) & x - Fig. 124 The quantity f'(^x)Lx is called the differential of y and is represented by the symbol dy. Accordingly dy=f(x)b*\ (1) 142 SLOPES AND AREAS This definition is true for all forms of the function/ (x) and is accordingly true when y=f(x) = x. In this case/'(V) = l, and formula (1) gives dx = Ax. (2) Substituting from (2) into (1), we have the final form dy=f(x)dx. (3) To sum this up : The differential of the independent variable is equal to the increment of the variable ; the differential of the function is equal to the differential of the independent variable multiplied by the derivative of the function. It is important to notice the difference between Ay and dy. The figure shows that, in general, they are not equal, but that they become more nearly equal as Ax approaches zero. Without using the figure, we may proceed thus: Since Lim ~f- =f'(x\ Az = AX where Lim e = ; and hence Aa; = Ay =f'(x) Ax + eAx = dy + eAx. Ex. 1. Let y = x 3 . We may increase x by an increment Ax equal to dx. Then Ay = (x + dx) 3 - x z = 3 xHx + 3 x (dxf + (dxf. On the" other hand, by definition, dy = 3 x 2 dx. It appears that A.y and dy differ by the expression 3 x (dx) 2 + (dx) 3 , which is very small compared with dx. Ex. 2. If a volume v of a perfect gas at a constant temperature is under the pressure p, then v = - , where k is a constant. Now let the pressure be P increased by an amount A/> = dp. The actual change in the volume of the gas is then the increment a , _ ^ _ & — kdp k f ty I 1 The differential of v is, however, p + dp p p(p + dp) p 2 [ 1 + 'Jp kdp AREA 143 It is to be emphasized that dx and dy are finite quantities, subject to all the laws governing such quantities, and are not to be thought of as exceedingly minute. Consequently both sides of (3) may be divided by dx, with the result /'(*) = _dy dx Then That is, the derivative is the quotient of two differentials. This explains the notation already chosen for the derivative. So, in general, the limit of the quotient of two increments is equal to the quotient of the corresponding differentials. For let y=f(x) and z = $(x). Ay=f'(x)Ax + e^Ax, Az = <\>'(x) Ax + e^Ax, dy = f'(x)dx, dz = '(x)dx t Az 0'OO + e,' T . Ay T . /'(*)+«, f'(x) dy Lnn -r 1 = Lim ., v J — l = J m . y ' = -# . and Whence Az '(x) + € 2 $(x) dz 78. Area under a curve. Let LK (fig. 125) be a curve with equation y =/(*), and let OE = a and OB = b. It is required to find the area bounded by the curve LK, the axis of x, and the ordinates at E and B. For convenience, we as- sume in the first place that a < b and that f(x) is positive for all values of x between a and />. We will divide the line EB into n equal parts by placing Ax = J/, .V, M 3 M t M 6 M a J/ 7 J/ a Fig. 125 and laying off the lengths EM^ = M X M^ X B = Ax. (In fig. 125, n = 9.) 144 SLOPES AND AREAS Let OM t = Xj, OM 2 =x 2 , • • •, OM n _ 1 = x n _ v Draw the ordinates ED=f(a), Mfi~f(xd, M 2 B 2 =f(x,), . . ., ^.jP,,.^/^.,), and BC. Draw also the lines DR^ P X B^ B,B 3 , • • •, JjJ_j.fi,,, parallel to OX Then /(a) Ax = the area of the rectangle EDB M , f(x t }Ax = the area of the rectangle M^B^M^ /(x„) Ax = the area of the rectangle M 2 B,B 3 M 3J /(* B _i)A* = the area of the rectangle M n _ x P n _ x B n B. The sum /(a) Ax +/(s 1 ) Ax +/(.r 2 ) Ax + . . . +/(x, t _ 1 ) Ax (1) is then the sum of the areas of these rectangles and equal to the area of the polygon EDB X I^B. 2 • • • B n _ 1 B l _ 1 B n B. It is evident" that the limit of this sum as n is indefinitely increased is the area bounded by ED, EB, BC, and the arc EC. The sum (1) is expressed concisely by the notation 'S'/C^Ax, i = where 2 (sigma), the Greek form of the letter S, stands for the word " sum," and the whole expression indicates that the sum is to be taken of all terms obtained from /(x f ) Ax by giving to i in succession the values 0, 1, 2, 3, • • •, n — 1, where x Q — a. The limit of this sum is expressed by the symbol r. f(x) dx, where | is a modified form of S. /(x) dx = Lim V/(x.) Ax = the area EBCD. » = <*» t = It is evident that the result is not vitiated if ED or B C is of length zero. AREA 145 Ex. Required to find the area bounded by the curve y = — , the axis of x, and the ordinates x = 2 and x — 3 (fig. 126). (1) We may divide the axis of x between x = 2 and x = 3 in 10 parts, 3-2 placing Ax = ,„ = .1. 10 We make then the following calculation Ax ^ a = 2, /(a)A* = .08 x x = 2.1, /(xJAx = .0882 x 2 = 2.2, /(x 2 )Ax = .0968 x 3 = 2.3, * 4 = 2.4, /(x 3 )Ax = .1058 /(x 4 )Ax = .1152 x 5 = 2.5, /(x 5 )Ax = .1250 * 6 = 2.(5, /(x 6 )Ax = .1352 a- =27 x 7 t, /(.-•- )Ax = . 1458 •\ x 8 = 2.8, /(* 8 )Az = .1568 x fl = 2.9, /(z 9 )A* = .1682 1.2170 The first approximation to the area is therefore 1.217, which example, the value of the sum (1) for n = 10. (2) As a better approxi- mation the Btudent may compute the sum for n = 20 3 — 2 .05. The forth ind Ax 20 result is 1.2118. (3) If we take 3-2 100 calculation is very tedious. The result, however, is 1.26167. These successive and Ax = .01, 100 the Fig. 126 determinations appear to be approaching a limit. By subsequent methods it will be shown that this limit is 1/j. It is obvious that the direct calculation of the sum (1) is very tedious, if not practically impossible, if the number of terms is very large. Some other method must be found to determine the limit of the sum as n increases indefinitely. This method is fur- nished by the discussion in the following sections. 146 SLOPES AND AREAS 79. Differential of area. Let any one of the rectangles of fig. 125 be redrawn in fig. 127 and relettered, for convenience, MNRP. Draw also QS and complete the rectangle MNQS. Let A denote the variable area EMPD. Then MN=Ax, RQ = Ay, MNQP=AA, MNRP = MP . MN= yAx, MNQS = NQ. MN ~ 0/ + %) Aft But, from the figure, that is, whence MNRP < MNQP < MNQS ; yAx < A A < (y -\- Ay) Ax, AA i\X AA dA Now as Ax approaches zero as a limit, — - approaches Ax AA dx y is unchanged, and y + Ay approaches y. Hence , which lies between y and y + A?/, also approaches y ; that is, £-»-/<-> a) In the differential notation we have dA = f(x)dx. (2) To find the area it is therefore necessary first to find a func- tion whose derivative is f(x) and whose differential is f(x) dx. 80. The integral of a polynomial. The process by which a function is found from its derivative or its differential is called integration, and the result of the process is called the integral of the derivative. Integration is expressed by the symbol I • thus, ff(x)dx = F(x), (1) THE DEFINITE INTEGRAL 147 where F(x) is a function of which the derivative is f(x). The process may be carried out in the simpler cases by reversing the rules for differentiation. Thus, I 2 xdx = x 2 + c, I 3 x\lx = x 3 + c, by the formulas of § 74. In these results c may be any constant whatever, since — = 0. In fact, any derivative has an infinite number of integrals dif- fering by a constant. The most general form of formula (1) is / f(x)dx = F(x)+C, (2) where F(x) is any particular function whose derivative is f(x) and C is any arbitrary constant, called the constant of integration.' To integrate a polynomial we need to know that its integral is the sum of the integrals of its terms and that the integral of each term is found either by the formula ./' ax" dx = + c n+1 or by the formula I adx = ax + c. These are simply the formulas of § 74 reversed. Ex. f (,:•+ :,,-+ 7x + B)dx = ^ + •'.;' + - •''" + 3a: + C. J \ o - 81. The definite integral. Return now to the problem of area. From §79, dA=f(x)dx, whence, by use of § 80, A = F(x) + C. (I) This is the area of the figure EMPJ) (fig. 127), in which the line MP can be drawn anywhere between ED and BC. But it' the line MP coincides with ED, .1=0 and x= a. Substituting these values in (1), we have o = F(a^ + C whence C = — F(a). 148 SLOPES AND AREAS Formula (1) now becomes A = F(xy-F(a). The area A becomes the area EBCD when x = b. Then area EBCD = F(K) - F(a). This gives us our desired method of evaluating the limit of the sum (1), § 78, and may be expressed by the formula i: f(x)dx = F(V)-F(a). (2) The limit of the sum (1), § 78, which is denoted by / f(x)dx, U a is called a definite integral, and the numbers a and b are called the lower limit and the upper limit* respectively of the definite integral. This result gives the following rule for evaluating a definite integral : To find the value of I f(x) dx, evaluate J f(x) dx, substitute x = b and x = a successively, and subtract the latter result from the former. It is to be noticed that in evaluating j f(x)dx the constant of integration is to be omitted, since — F(ji) is that constant. However, if the constant is added, it disappears in the sub- traction, since [^(7>) + C] - \F(a) + C] = FQ>) - F(a). In practice it is convenient to express F(li) — F(a) by the symbol [FCx)~\ b a , so that X f(x)dx=[F(z~)f a . Ex. The example of § 78 may now be completely solved. The required areais r*x 2 , Tx 3 13 27 8 19 , . I 5^nTol = T5-15 = 15 =1 -- * The student should notice that the word "limit" is here used in a sense quite different from that in which it is used when a variable is said to approach a limit (§ G8). THE DEFINITE INTEGRAL 149 In the foregoing discussion we have assumed that f(x) is always positive and that a < b. These restrictions may be removed as follows: If f(x) is negative for all values of x between a and b, where a < b, the graphical representation is as hi fig. 128. Here f(a) Ax = — the area of the rectangle EMJl^I), f(x^) Ax = — the area of the rectangle JM^M^BJ^, etc., so that f f(x) dx = - the area EBCD. In case /(^) is sometimes positive and sometimes negative, we have a combination of the foregoing results, as follows: I Ifa b, Ax is negative, since Ax = The only change necessary in the above statement, however, is in the algebraic signs, the areas above the axis of x being now negative and those below positive. It is usual to arrange the work so that Ax shall be positive. It is obvious, however, that C a f(x~)dx = - ff(x)dx. Jb J a Also, from the areas involved, Cf(x) dx = f / dx +f f(x) dx. 150 SLOPES AND AKEAS PROBLEMS Find approximately, by a numerical calculation, the slope of each of the following curves at the point given : 1. y = x 2 at (2, 4). 2. y = x 2 at (3, 9). 3. y = v* at (1, 1). 4. y = x s at (2, 8). Find from the definition, without the use of formulas, the deriva- tives of the following expressions : 7. 4a; 3 . 9. x 5 — x. 2^ 8. 5 x 2 + 7x-2. io. -. X * x 2 12. V.* , Find by the formulas the derivatives of each of the following polynomials : 13. 4.r 3 -3x 2 + 2x-l. 15. x s + 7 x' -6x 3 + 7x-3. 14. x 4 + 7 x 2 - x + 3. 16. ^: 6 -|ic 5 + |a; 4 + 2 2 -7 x. 17. Prove that the derivative of ax s + bx 2 + ex + e is the sum of the derivatives of its terms. 18. By expanding and differentiating show that the derivative of (4* + 3) 3 is 12(4x + 3) 2 . 19. By expanding and differentiating show that the derivative of (x + a)' 1 is n(x + a)" -1 . Find the values of x for which the following expressions are respectively increasing and decreasing, and draw their graphs : 20. x 2 + 6 x — 4. . 23. x 4 — 2x 2 + l. 21. x 3 -3x 2 + 7. 24. 2 x- 3 - 15;c 2 + 36 a; -270. 22. x 4 + 4x-6. 25. x 3 -3x 2 -9x + 27. 26. If a stone is thrown up from the surface of the earth with a velocity of 100 ft. per second, the distance traversed in t seconds is given by the equation s = 100 1 — 16 1 2 . Find when the stone moves up and when down. PROBLEMS 151 27. A particle is moving in a straight line in such a manner that its distance x from a fixed point A of the straight line, at any time t, is given by the equation x = t 3 — 9 t' 2 4- 24 t 4- 100. When will the particle be approaching A ? 28. A piece of wire of length 20 in. is bent into a rectangle one side of which is x. When will an increase in x cause an increase in the area of the rectangle and when will it cause a decrease ? 29. In a given isosceles triangle of base 20 and altitude 10 a rec- tangle of base x is inscribed. Find the effect upon the area of the rectangle caused by increasing x. 30. A right circular cylinder with altitude 2x is inscribed in a sphere of radius a. Find when an increase in the altitude of the cylinder will cause an increase in its volume and when it will cause a decrease. 31. A right circular cone of altitude x is inscribed in a sphere of radius a. Find when an increase in the altitude of the cone will cause an increase in its volume ami when it will cause a decrease. 32. On the line 3a: + y = G a point /' is taken and the sum s of the squares of its distances from (5, 1 ) and (7, 3) computed. Find the effect on s caused by moving P on the line. Find the turning points of the following curves and draw the curves : 33. y = 2 ;>■» - 9 x 8 . 35. y = \ x 4 - 2 X s + ], 34. y = 2x* + 3x* - 1 2x - 18. 36. y = x* - 2 a* + 4. 37 . Find the equation of the tangent to the curve y — 4 x 2 + 4 a- — 3 at the point the abscissa of which is — 1. 38. Find the equation of the tangent to the curve y = x 3 + 4 x 1 at the point the abscissa of which is — 3. 39. Show that the equation of the tangent to the curve y = ax 2 4- 2 bx A- c at the point (x v // t ) is y = 2 (ax i 4- 1>) x — ax? + c. 40. Show that the equation of the tangent to the curve y = x 3 + ax + b at the point (x v y^ is y = (3 x* 4- a)x — 2 x 3 4- b. 41. Find the area of the triangle included between the coordinate axes and the tangent to the curve y = x 3 at the point (3, 27). 152 SLOPES AND AREAS 42. Determine the point of intersection of the tangents to the curve y = x 3 — 5 x 4- 7 at the points the abscissas of which are — 2 and 3 respectively. 43. Determine the point of intersection of the tangents to the curve y = x 3 — 3x4-7 at the points the abscissas of which are 2 and respectively. 44. Find the angle between the tangents to the curve y = x 2 — 4 x 4- 1 at the points the abscissas of which are 1 and 3 respectively. 45. Find the angle between the tangents to the curve y = x 3 — 3 x 2 4- 4 x — 12 at the points the abscissas of which are — 1 and 1 respectively. 46. Find the equations of the tangents to the curve y = x 3 4- x 2 that have the slope 8. 47 . Find the equations of the tangents to the curve 2 x 3 4- 4 x 2 — x — y = that have the slope \. 48. Find the points on the curve y = 3 x 3 — 4 x 2 at which it makes an angle of 45° with OX. 49. Find the points on the curve y = x 3 — x 2 4- 2x 4- 3 at which the tangents are parallel to the line y = 3 x — 7. 50. How many tangents has the curve y = x 3 — 2x 2 + x — 2 which are parallel to the line 7 x — 4?/4-28 = 0? Find their equations. 51. Find approximately the area bounded by the straight line y ='2x4-3, the ordinates x = 1 and x = 2, and the axis of x, by considering the area as the sum of rectangles the bases of which are .2 in the first approximation and .1 in the second approximation. Also find the area exactly by elementary geometry. 52. Find approximately the area between the axis of x and the portion of the curve y = x — x 2 which is above the axis of x, by considering the area as the sum of rectangles the bases of which are .2 in the first approximation and .1 in the second approximation. 53. Find approximately the area bounded by the curve y = -> the ordinates x = 2 and x = 3, and the axis of x, by considering the area as the sum of rectangles the bases of which are .2 in the first approximation and .1 in the second approximation. PEOBLEMS 153 54. Find the area bounded by the curve y = Va-, the ordinates x = 1 and x = 4, and the axis of x, by considering the area as the sum of rectangles the bases of which are .5 in the first approxima- tion and .2 in the second approximation. 55. Find by integration the area described in Ex. 51. 56. Find by integration the area described in Ex. 52. 57. Find the area bounded by the curve y = x 3 — 2ar + Sx — 1, the ordinates x = 2 and x = 4, and the axis of x. 58. Find the area bounded by the axis of x and the portion of the curve y = 9 — x 2 above the axis of x. 59. Find the area between the axis of x and that part of the curve y = 10 — 11 x — 6 x 2 which is above the axis of x. 60. Find the area between the axis of x and that part of the curve y = x s — 3 x 2 — 9 x + 27 which is above the axis of x. 61. Find the area bounded by the axis of x and the portion of the curve y = x a + 3x 2 — 4 below the axis of x. 62. Find each of the two areas bounded by the curve y = 150 a; — 25 x 2 — x s and the axis of x. 63. Find the area bounded by the axis of x, the curve y = 2x 3 + 3 a; 2 -f 2, and the ordinates through the turning points of the curve. 64. Prove that the area of a parabolic segment is two thirds of the product of its base and altitude. 65. Find the area between the parabola y = \ x 2 and the straight line Sx-2y-4: = 0. 66. Find the area of the crescent-shaped figure between the curves y = x 2 + 5 and y = 2 x 1 + 1. CHAPTER X DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 82. Theorems on derivatives. In order to extend the process of differentiation to functions other than polynomials, we shall need the following theorems: 1. The derivative of a function plus a constant is equal to the derivative of the function. Let u be a function of x which can be differentiated, let c be a constant, and place .. _ ., ._■_ „ y — u -f- c. Then if x is increased by an increment Ax, u is increased by an increment Au, and c is unchanged. Hence the value of y becomes u + A^ + c. Whence Ay = (u + Ah + e) — (u + e) = Au. Therefore — = — i Ax Ax and, taking the limit of each side of this equation, we have dy __ du dx dx Ex. 1. 7/=:4a,- 3 + 3. 2. The derivative of a constant times a function is equal to the constant times the derivative of the function. .Let u be a function of x which can be differentiated, let c be a constant, and place „, __ „. y — ta. Give x an increment Ax, and let Au and Ay be the corre- sponding increments of u and y. Then Ay = c (ii + A?f ) — cu — c Au. THEOREMS ON DERIVATIVES 155 TT Ay Au Hence — z- = p— — , Ax Ax and, by theorem 3, § 69, T . Ay T . Au Lmi — — = c Lim — • Aa; Ax Therefore -~- = c — - > aa; aa; by the definition of a derivative. Ex. 2. # = 5(x 3 + 3x 2 + l). l ll = 5 — (x 3 + 3 x 2 + 1) = 5 (3 x 2 + 6 x) = 15 (x 2 + 2 x). dx dx 3. The derivative of the sum of a finite number of functions is equal to the sum of the derivatives of the functions. Let u, v, and w be three functions of x which can be differen- tiated, and let _ u + v + UK Give x an increment Ax, and let the corresponding increments of u, v, w, and y be Au, Av, Aw, and A//. Then Ay = (u + Am + v + Av + w + Aw) — (?* + v + w) = Au + At' +Aw; , A?/ Am At' Aw whence -f- = — + — + -— • Aa; Aa; Aa; Aa; Now let Aa; approach zero. By theorem 1, § 69, T . Ay T . An, T . Av T . Aw Lim — ^ = Lim ■ — - + Lim \- Lim ; Aa- Aa- Aa- Aa; that is, by the definition of a derivative, dy _ du ,dv dw dx dx dx dx The proof is evidently applicable to any finite number of functions. Ex. 3. y = x 4 - 3 x 3 + 2 x 2 - 7 x. ^ = 4 x 3 - 9 x 2 + 4 x - 7. dx 156 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 4. Tlie derivative of the product of a finite number of functions is equal to the sum of the products obtained by multiplying the derivative of each factor by all the other factors. Let u and v be two functions of x which can be differentiated, and let y _ wu# Give x an increment Ax, and let the corresponding increments of u, v, and y be Aw, Av, and Ay. Then Ay = (u + Aii) (v + Av) — uv = uAv + v Au + Au • Av Av Av Au , Aw . and -rf- = it — + ^ t~ + -t— • Av. Ax Ax Ax Ax If, now, Ax approaches zero, we have, by § 69, T . Ay T . Av , T . Am , , . Am T . . Lim — - = u Lim - — \-v Lim \- Lim — - • Lim Av. Ax Ax Ax Ax But L im Av = 0, and therefore dy dv , du Again, let y = uvw. Regarding uv as one function and applying the result already obtained, we have dy dw d (uv) -JL—UV—--\-W ■ J dx dx dx dw , dv du] = uv — — f- w \u — + v — dx L dx dx] dw , C?U . $W — UV—- + UW— + VW—' dx dx dx The proof is clearly applicable to any finite number of factors, Ex. 4. y = (3 x - 5) (a: 2 + 1) x s . = (3 x - 5)(x 2 + 1)(3 x 2 ) + (3 x - 5)x 3 (2x) + (x 2 + l)x 3 (3) = (18 x s - 25 x 2 + 12 x - 15)x 2 . THEOREMS ON DERIVATIVES 157 5. The derivative of a fraction is equal to the denominator times the derivative of the numerator minus the numerator times the deriva- tive of the denominator, all divided by the square of the denominator. Let y = - , where u and v are two functions of x which can be v differentiated. Let Ax, Am, Av, and Ay be as usual. Then u + Au u vAu — uAv ±y v + Av v v 2 +vAv Ah Av v ~k u IT Ail Ax Ax and — - = — ; Ax v~+v Av Now let Ax approach zero. By § 69, T . Au T . Av v Lim u Lim — _ . Ay A.r Ax Lim — - = ; ■ Ax v~+ v Lim Ay du dv dy dx dx whence T 2 — 1 Ex.5. y = - a —± x* + 1 dx V dy _ (.r 2 + l)(2x)-(j- 2 -l)2j- _ 4:r dx (x 2 + l) s (.<•- + l) 2 ' 6. If y is a function of x, then x is a function of y, and the " derivative of x with respect to y is the reciprocal of the derivative of y with respect to x. Let Ax and Ay be corresponding increments of x and y. Then Ax _ 1 Ay~ Aji Ax whence Lim — — = ^ Lim^ Ax dx 1 that is, -— = —-. dy dy dx 158 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 7. If y is a function of u and u is a function of x, then y is a function of x, and the derivative of y with respect to x is equal to the derivative of y with respect to u times the derivative of u ivith respect to x. An increment Ax determines an increment Au, and this in turn determines an increment Ay. Then evidently Ay _ Ay Au Ax Au Ax ' whence Lim — - = Lim — — • Lim — - ; Ax Au Ax dy _ dy du dx du dx 1 2 + 3 x 2 2 4 + 6 a: 2 that is, Ex. 6. y = m 2 +3h + 1, where u = The same result is obtained by substituting in the expression for y the value of u in terms of x and then differentiating. This result has an important application to the differential. For suppose we have y =/(>)j u = ^ ^ (1) By substitution, we obtain y=/[ (2) and the formula proved above gives us **(*)=/'00'.f(*> (3) By use of § 77 we obtain from (1) dy =f'(u) du, du = '(V) dx, (4) and from (2) we have dy = F'(x)dx. (5) It is important to know that the two values of dy in (4) and (5) agree. In fact, by means of (3) and the second part of (4), (5) becomes dy=f(ii)y(i^dx=f(,i)du. Hence it is not necessary, in applying § 77 to find a differential, to ask whether x is an independent variable or not. DERIVATIVE OF U N 159 83. Derivative of u n . If u is any function of x which can be differentiated and n is any real constant, then d(u n ) m , die ax dx To prove this formula we shall distinguish four cases: 1. When n is a positive integer. d (w n ) d (w n ) du dx du dx i du (by 7, § 82) = ™-V (Byl, §74) 2. When w. is a positive rational fraction. Let n = — i where p and q are positive integers, and place y = u\ By raising both sides of this equation to the qt\\ power, we have y q = u p . Here we have two functions of x which are equal for all values of x. If we give x an increment A.r, we have A(y) = AO*)> A(yO _ Ap/ p ) Ax Ax md therefore iSfl^Jfl; dx dx whence qy q ~ l -^ = pu 1 " 1 — * 1J dx r dx since p and q are positive integers. Substituting the value of y and dividing, we have dx q dx Hence, in this case also, d(u n ~) __ B _ 1 du ^ 160 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 3. When n is a negative rational number. Let n = — m, where m is a positive number, and place , 1 y = U m = —' u m d(u m ) Then ft- p dx u lm „ i du mu m — dx u 2m (by 5, § 82) (by 1 and 2) -m-ldu = — mu - — dx Hence, in this case also, d C M ") . i du dx dx 4. When n is an irrational number. The formula is true in this case also, but the proof will not be given. It appears that 1, § 74, is true for all real values of n. Ex. 1. y = (x 3 + 4 x 2 - 5 x + 7) 3 . ^ = 3 (x 3 + 4 x 2 - 5 x + 7) 2 — (x 3 + 4 x 2 - 5 x + 7) dx dx = 3 (3 x 2 + 8 x - 5)(x 3 + 4 x 2 - 5 x + 7) 2 . 3/— 1 2 Ex. 2. w = V x 2 + — = xs + z- 8 . x d rfv 2 _i _ . -^ = -x 3 _ 3x~ 4 rfx 3 2 3 3^x * 4 ' Ex. 3. r/ = (x + l)Vx 2 + l. ^ = (x + l)^ 2 + 1 )%(x 2 + l)i^ + 1 ) dx dx dx = (x + 1)Q| (x 2 + I) - * • 2 x] + (x 2 + 1)* = x(x ± l 1 + (x2 + 1) i (x 2 + 1)4 2 x 2 + x + 1 VxHTL FORMULAS 161 dy = \t x \~* d I x \ dx~3\x 3 + V dx\x 3 + l) _ 1 l x 3 + 1 \* 1 - 2 x 3 3\ X / (x 3 + l) 2 l-2x 3 3 xf (x 3 + 1)* 84. Formulas. The formulas proved in the previous articles are a) (2) (3) *£»*_,,*+„*», (4 ) d(u + c) du dx dx d (cii) dx du d(u + v) dx du dv dx dx dx dx dx , /u\ du dv \vj dx dx dx v' 2. dx __L dy~dy dx dy _ dy du dx du dx dy dy du dx dx du Formula (9) is a combination of (7) and (8). (5) te =nu £' (6) (7) (8) 00 162 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 85. Higher derivatives. If y =f(x), then -j- is in general a function of x and may be differentiated with respect to x. The result is called the second derivative of y with respect to x and is indicated by the symbol -t~(-^}' which is commonly abbreviated into ——• dx 1 Similarly, the derivative of the second derivative is called the third derivative, and so on. The successive derivatives are commonly indicated by the following notation : y =f(x), the original function ; -r~ =f(x), the first derivative ; ax -z-(-rM = -r4 =f"(z), the second derivative; dx\dx/ dx 2 v y A-(pL\ = fl=f»( x \ the third derivative; dx \azr/ dx d n y -7-7 =/ (B) (^)» the nt\\ derivative. It is noted in § 9 that /(a) denotes the value of /(re) when x = a. Similarly, /'(a), f"(a), f' n (a) are used to denote the values of /'(#), f"(x), f"'(x) respectively when x = a. It is to be emphasized that the differentiation is to be carried out before the substitution of the value of x. Ex - lf /(*)=^h4' find /''(°)- + 1 -*+Q. + l roo 2z 3 -6a; 2 -6x + 2 (x* + l) 3 Therefore /"(0) = 2. 86. Differentiation of implicit functions. Consider any equation ^ the form /(*, = 0, dx that is, 2x + 2y^ = 0; dx , dii x whence — = dx y The derivative may also be found by Bolving the equation for y. Then ± v 5 - x 1 , f/y — x x ^ Vs - x 2 y Ex. 2. Given f- ■ xy -r 1 = 0. Then d(y z ) f _ y°+x 2 Ex. 4. If y z — xy — 1 = 0, we have found - dx 3 y 2 — x dy d(Sy 2 -x) Then d"y ^-i)t-'-i dx*~ Qiy*-xy (3y*-xy (Sy* x) V y( 6 / - V J J Sf-x »\3y*-x - 1 ) $if-xf (3 »»-*)■ 87. Tangents and normals. It has been shown in § 76 that the tangent to a curve y =/■ - 3), or a? + 2y + 3 = 0, and the equation of the normal is y + 3 = 2 (./• - 3), or 2 z — y - 9 = 0. Ex. 2. Prove that the norma] to a pa- rabola at any point makes equal angles with the axis of the parabola and the focal radius drawn to the point. Lei /'(('p //i) '"' an .v point of the parabola y* = ipx (fig. 129), and let F(p, 0) be the focus. Then 11\ is the focal radius of /',. and let /',.Y be the normal to the parabola. To prove ZI'\I\ — Z/'/yY. By differentiation we have 2 y— = 1 />, whence the slope of the tan- ° p ' Ij v gent at P l is — and the Blope pf the normal is — t — • It follows that !l\ - J' tan F2ITP X 8ji Slope of FP. = -^— ; •'i - P therefore 1 _^ L /_^ I _\ - Ill (*1 + P) 2p if we replace yf by 4_pz 1 , since y x 2 = ipx v P x being a point of the parabola. Since tan FP X JY = tan FNP V the angles are equal. 1GG DIFFEEENTIATION OF ALGEBRAIC FUNCTIONS The angle of intersection of two curves is the angle between their respective tangents at the point of intersection. The method of finding the angle of intersection is illustrated in the following example: Ex. 3. Find the angle of intersection of the circle x 1 + y 2 = 8 and the parabola x 2 = 2y. The points of intersection are P t (2, 2) and P 2 (- 2, 2) (fig. 130), and from the symmetry of the diagram it is evident that the angles of intersection at P x and P 2 are the same. Differentiating the equation of the circle, dy dy we have 2 x + 2 y — = 0, whence — = , dx ax y and differentiating the equation of the parab- dy Fig. 130 1 and the slope ola, we find -2. = x. dx Hence at P 1 the slope of the tangent to the circle of the tangent to the parabola is 2. Accordingly, if /3 denotes the required angle of intersection, — 1 — ° tan ft = o =3, or /3=tan-!3. 88. Sign of the second derivative. Since the second derivative is the derivative of the first derivative, the sign of — ^ shows dy dx 1 whether — is an increasing or a decreasing function. dx d 2 v The significance of -~ f or the graph y =/(*) is obtained from the fact that -~- is equal to the slope ; hence — ~ is the deriva- G/'JO (X'X d 2 v tive of the slope. Therefore, by § 75, if — ^ is positive, the slope is increasing ; if — ^ is negative, the slope is decreasing. We may have, accord- ingly, the following four cases: cW 1. f>0, dx Fig. 131 Since both the ordinate and the slope are increasing, the graph runs up toward the right with increasing slope (fig. 131). SIGN OF THE SECOND DEEIVATIVE r Id" 2. ^>0, g<0. dx dor The graph runs up toward the right with decreasing slope (fig. 132). 3. ^<0, dx dx 2 >0. 132 The graph runs down toward the right. The slope, which is negative, is increasing algebraically and hence is decreasing numerically (iig. 133). 4. ^<0, dx dx* <0. \p08l- The graph runs down toward the right, and the slope is decreasing algebraically (fig. 134). The consideration of these cases leads d 2 y to the following conclusion : If — '\ tive, the graph is concave upward ; if ^-™ is negative, the graph is concave downward. If a curve changes from concavity in one direction to con- cavity in the other direction at any point, that point is called a d 2 y point of infection. It follows that at such a point \, changes sign, either by becoming zero or by becoming infinite. These two cases are illustrated in the following examples : Ex. 1. Examine the curve y = -^(x 8 — G x 2 ) for points of inflection. dy_l 2 _ dx ~ I X " *' cPy dx 2 x-1 Fig. 135 The curve (fig. 135) is concave downward when x<2, is concave upward when x>2, and accordingly then' is a point of inflection when x — 2. The ordinate of this point is — 1£. 108 DIFKEPvKNTIATION OF ALGEBRAIC FUNCTIONS- Ex. 2. Examine the curve y = (x — 2)* for points of inflection. dy _ 1 dhj _ 2 da: 3 (.,-2)1 It is evident that dx* 9 (a: -2)1 co if x = 2, and that d*y no finite value of x makes — 2. = 0. If a: < 2, ^ > ; and if a: > 2, f —£ < 0. Hence the point dx* dx" l Fig. 136 for which x — 2 is a point of inflection, since on the left of that point the curve is concave upward and on the right of that point it is concave downward (fig. 130). The ordinate of this point is 0. 89. Maxima and minima. lif(x) changes from an increasing function to a decreasing function (§ 75) when x increases through Fig. 137 the value a and /(«) is finite, f(a) is called a maximum value of /(.>) (figs. 137, 138); and if f(x) changes from a decreasing O x=a Fig. 139 O x=a Fig. 140 function to an increasing function when x increases through the value a and f(a) is finite, /(«) is called a minimum value oif(x) (figs. 139, 140). Since the derivative of an increasing function is positive and the derivative of a decreasing function is negative, it follows MAXIMA AND MINIMA 169 that the derivative of the function must change sign at either a maximum or a minimum value and hence must become either zero or infinity. Accordingly we have two cases : I. If — = or oo ivhen x = a, and ~ > when x a,f(a) is a maximum value of y =f(x). II. If -~ = or cc iv hen x = a, and ~~ < when x < a, and , dx dx all -~ > wAew x > a, f(a) is a minimum value of y = f(x). If, however, — - changes sign by becoming infinite and at the same time y becomes infinite (fig. 33), the function is discon- tinuous and there is no corresponding maximum or minimum value. In order to apply the above tests it is necessary to factor- 1 -, as shown in the following examples: Ex. 1. Find the maximum and the minimum value of /(./) = x 5 - 5 x* + 5 x 3 + 10 x- - 20 x + 5. We find f'(x) = 5 x* - 20 x 3 + 15 x- + 20 x - 21 > = 5(x 2 -l)(x 2 -4x + 4) = 5(x + l)(x-l)(x-2)*. The roots of /'(x) = are — 1, 1, and 2. As x passes through — 1, /"(./) changes from + to — . Hence x = —1 gives /'(./) a maximum value, namely - 1. As x passes through + 1, /"(.') changes fnnii — to +■ Hence x = + l gives f(x) a minimum value, namely — 4. As x passes through 2, ./"(') does not change sign. Hence x = 2 gives f(x) neither a maximum nor a minimum value. Ex. 2. A rectangular box is to be formed by cutting a square from each corner of a rectangular piece of cardboard and bending the resulting figure. The dimensions of the piece of cardboard being 20 by 30 in., required the largest box which can be made. Let x be the side of the square cut out. Then if the cardboard is bent along the dotted lines of fig. 141, the dimensions of the box are 30 — 2x, 20 — 2 x, x. Let y be the volume of the box. Then y = x(20-2x)(30-2x) = 600 x - 100 x 2 + 4 x 3 . ^ = 600 - 200 x + 12 x 2 . dx 170 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS Equating this to zero, we have 3 a,- 2 - 50 a + 150 = 0. 25 x = — ±5 3 Vf -.8.9or 12.7 Hence dy _ , ZQ\S~ TO dx 12(j:-3.9)(k-12.7). dy changes from + to — as x X X 30-2X Cm, ll ©1 aa: Fig. 141 is through 3.9. Hence x = 3.9 gives the maximum value 105(3+ for the capacity of the box. x = 12.7 gives a minimum value of y, but this has no meaning in the problem, for which x must lie between and 10. Ex. 3. y = V(x-l){x-2f = (x- 1)* (x - 2)1 dy_ 3x-4 dx 3 V(x-lf(x-2) — — when x = f , and changes from + to — as x passes through J. il.r when x = a, f(a) is a minimum value of y =f(x). Fig. 142 MAXIMA AND MINIMA 171 It is evident that these tests can be used to advantage when it may be difficult or impossible to factor ~-i and that they fail if —4 also becomes zero. dzr Ex. 4. Light travels from a point A in one medium to a point B in another, the two media being separated by a plane surface. If the velocity in the first medium is i\ and in the second v 2 , recpiired the path in order that the time of propagation from .1 to B shall be a minimum. It is evident that the path must lie in the plane through A and B perpendicular to the plane separating the two media, and that the path will be a straight line in each medium. We have, then, fig. 143, where MX represents the intersection of the plane of the motion and the plane separating the two media, and ACB rep- resents the path. Let MA = a, KB = b, MN= c, and MC = x. Then AC= Va a + x* and CB = V(c — x) 2 + b' z . The time of propagation from .1 to B is therefore M X Fig. 143 whence and dx dx 2 D l-V is the angle made by A C with the normal at C and is called the angle of incidence, and if/ is the angle made by CB with the normal at C and is called the angle of refraction. Hence the time of propagation is a minimum when the sine of the angle of incidence is to the sine of the angle of refraction as the velocity of the light in the first medium is to the velocity in the second medium. This is, in fact, the law according to which light is refracted. In practical problems the question as to whether a value of x for which the derivative is zero corresponds to a maxi- mum or a minimum can often be determined by the nature of the problem. In Ex. 2 above, it is evident that there must be a maximum volume of the box and that there can be no minimum value. Accordingly, when we have found — = if x — 3.9 or 12.7, since 12.7 is unreasonable in our dx problem we conclude, without further discussion, that x = 3.9 corresponds to the maximum volume. 90. Limit of ratio of arc to chord. The student is familiar with the determination of the length of the circumference of a circle as the limit of the length of the perimeter of an inscribed regular polygon. So, in general, if the length of an arc of any curve is required, a broken line connecting the ends of the arc is constructed by drawing a series of chords to the curve as in fig. 144. f ^ 144 Then the length of the curve is denned as the limit of the sum of the lengths of these chords as each approaches zero and as their number therefore in- creases without limit. The manner in which this limit is obtained is a question of the integral calculus and will not be taken up here. We may use the definition, however, to find the limit of the ratio of the length of an arc of any curve to the length of its chord as the length of the arc approaches zero as a limit ; that is, as the ends of the arc approach each other along the curve. LIMIT OF RATIO OF ARC TO CHORD 173 Accordingly, let P x and P 2 (fig. 145) be any two points of a curve, P X P 2 the chord joining them, and P X T and RT the tangents to the curve at those points re- spectively. We assume that the arc PP 2 lies entirely on one side of the chord P 1 P and is concave toward the chord. These condi- tions can in general be met by taking the points P x and P 2 near enough together. Then it follows from the definition that l P 1 T+TP 2 > arc PJ>, > P X P 2 ; whence fiI±S>«55 >L P X P 2 P X P 2 If TR is the perpendicular from T to PJ^ and if the angles P„P X T and PJ\T are denoted by a and ft respectively, then J[ T— l[R sec a, and TP = EP 2 sec ft = ( /;/.] - PJi) sec ft. Therefore P/P+ TP 2 = 2>B sec a + (/?/* - ijfi) sec ft = iJ/J sec ft -f /;/.' (sec a - sec ft), , P X T + TJ^ i?i^ sec ft + iffi (sec a - sec ft) = sec ft + -^— - (sec a — sec ft). Now, as 7,* and P 2 approach each other along the curve, a and ft both approach zero as a limit, whence sec a and sec ft approach PR unity as a limit; and since — — is always less than unity, it fol- P T + TP lows that the limit of — — is unity. arc PP. Hence — ^- J — - lies between unity and a quantity approaching 1 . arc PR unity as a limit, and therefore the limit of — - is unity ; that is, the limit of the ratio of an arc to its chord as the arc approaches zero as a limit is unity. 174 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 91. The differentials dx } dy, ds. On any given curve let the distance from some fixed initial point measured along the curve to any point P be denoted by s, where s is positive if P lies in one direction from the initial point and negative if P lies in the oppo- site direction. The choice of the positive direction is purely arbi- trary. We shall take as the posi- tive direction of the tangent that which shows the positive direction of the curve and shall denote the angle between the positive direc- tion of OX and the positive direction of the tangent by <£. Now for a fixed curve and a fixed initial point the position of a point P is determined if s is given. Hence x and y, the coordinates of P, are functions of s which in general are con- tinuous and may be differentiated. We will now show that dx ds cos<£, dy . , ■—- = sin , ds dy ds sin cf>. 1 and (1) RATE OF CHANGE 175 If the notation of differentials is used, equations (1) become dx = ds • cos (f>, dy = ds • sin ; whence, by squaring and adding, we obtain the important equation , „ ., ds =dx +dy. (2) This relation between the differentials of x, y, and s is often represented by the triangle of fig. 147. This figure is convenient as a device for memorizing formula (1), but it should be borne in mind that EQ is not rigorously equal to dy (§ 77), nor is PQ rigorously equal to ds. In fact, RQ = Ay and I'( i >=As, but if this triangle is regarded as a plane right triangle, we recall immediately the values of sin <£, cos , and tan

■), a change of A./- units in ./■ causes a change of Ay units in //, and the quotient ' gives the Ax ratio of these changes. If this ratio is equal to m, Ay = mAx\ that is, the change in y is on times the change in x. Hence, if m were independent of A./-, a change of one unit in u- would cause a change of m units in //, and — * would consequently measure the change in y per unit of change in ./•. But m does depend in general upon A.r, and hence does not give an unam- biguous measure of the relative changes in x and y. To obtain such a measure, it is convenient to take the limit of -— as Ax Ax approaches zero and to call this limit the rate of change of y with respect to x. We have then -y- = rate of change of y with respect to x. Ex. 1. Coefficient of expansion. Let a substance of volume v be at a temperature t. If the temperature is increased by A^, the pressure remain- ing constant, the volume is increased by Ac. The change per unit of vol- ume is then — , and the ratio of this change per unit of volume to the 176 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS change in the temperature is - — • The limit of this ratio is called the coefficient of expansion ; that is, the coefficient of expansion equals In other words, the coefficient of expansion is the rate of change of a unit of volume with respect to the temperature. Ex. 2. Elasticity. Let a substance of volume v be under a pressure p. If the pressure is increased by A/>, the volume is increased by — Ay. The change in volume per unit of volume is then The ratio of this change per unit of volume to the change in the pressure is — , and v A/> the limit of this is called the compressibility ; that is, the compressibility is the rate of change of a unit of volume with respect to the pressure. The reciprocal of the compressibility is called the elasticity, which is therefore equal to — v — • dv In many cases it is convenient to take time t as the inde- pendent variable. Then — - and — measure the rates at which 1 dt dt x and y respectively are changing with the time. If both x and y can be expressed in terms of t, these rates may be found by differentiating ; but if y is expressed in terms of x and x is expressed in terms of t, -~- may be found by the formula dy _ dy dx dt dx dt which is a special case of (8), § 84. Ex. 3. A stone thrown into still water causes a series of concentric ripples. If the radius of the outer ripple is increasing at the rate of 5 ft. a second, how fast is the area of the disturbed water increasing when the outer ripple has a radius of 12 ft.? Let x be the radius of the outer ripple and A the area of the disturbed water. Then A = ttx 2 . dA n dx and = lirx — • dt dt By hypothesis, — = 5. dA Therefore — — = 10 ttx ; dt and when x = 12, dA = 120 7r, the required rate. tit RECTILINEAR MOTION 177 This problem may also be solved by expressing J. directly in terms of t. By the conditions of the problem, x = 5 t, and therefore A = 25tt1 2 ; whence dt When x = 12, t = 2i and — dt = 120 7r, as before. 93. Rectilinear motion. An important application of the con- cept of a derivative is found in the definition of the velocity of a moving body. We shall confine ourselves in this article to rectilinear motion, that is, to motion which takes place in a straight line. Let a body move along a straight line AL (fig. 148), and let its distance from a fixed point A, at any time t, be denoted by 8. Then, if the body is at the point P at the time t, AP=s. * ? $ L The velocity of the body is then defined FlG - 148 as the rate of change of the distance 8 with respect to the time t. More in detail, if t is increased by the increment At, let s As be increased by the amount As=PQ. Then — is the average velocity of the body dining the period At. Since this average velocity depends in general upon the magnitude of At, we take As the limit of — as At approaches zero, and call this limit the At velocity of the body at the point P. Hence, if v denotes the velocity, T . As ds v = Lim -— = —-• At dt If the velocity is constant and equal to Jc, the motion is said to be uniform, and s = kt. We note that if v > 0, an increase of time corresponds to an increase of s, while if v < 0, an increase of time causes a decrease of s. Consequently, the velocity is positive when the body moves in the direction in which s is measured, and negative if it moves in the opposite direction. 178 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS Ex. 1. If a body is thrown up from the earth with an initial velocity of 100 ft. per second, the space traversed, measured upward, is given by the equation .v = 100 t - 16 t\ Then ds 100 - 32 t. When t < '.) J., v>0; and when £ >3 i, v <0. Hence the body rises for 3^ sec. and then falls. The highest point reached is 100 (3 \) - 10 (3 1) 2 = 156 \. Ex. 2. A man standing on a wharf 20 ft. above the water pulls in a rope attached to a boat at the uniform rate of 3 ft. per second. Required the velocity with which the boat approaches the wharf. Let A (fig. 149) be the position of the man and C that of the boat. Let AB = We wish t h = o li = 20, AC = s, and BC Now therefore x = Vs 2 - 400 ; dx s df dt V* 2 - 400 (/ But, by hypothesis, s is decreasing at the rate of 3 ft. per second ; there- fore — = — 3, and the required expression for the velocity of the boat is (Ac _ - 3 .9 dt Vs 2 - 400 To express this in terms of the time, we need to know the value of s when t = 0. Suppose this to be s ; then and s = s t dx_ dt ~ 3t 3 s n + t Vsj - 400 - o s7t + y p When the motion of the body moving in a straight line is not uniform, the velocity at the end of an interval of time is not the same as at the beginning. Then, if v + Av denotes the velocity of the body at Q (fig. 148), — is the average change of velocity per unit of time during the period A*. The limit of this ratio is called the acceleration; that is, the acceleration of a body moving RECTILINEAR MOTION 179 in a straight line is the rate of change of the velocity with respect to the time. Hence, if a denotes the acceleration, _dv_ r o = L > and the original equation may be written S = s n + vj + 1 at\ 180 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 94. Motion in a curve. When a body moves in a curve, the discussion of velocity, acceleration, and force becomes more com- plicated as the directions as well as the magnitudes of these quan- tities need to be considered. We shall not discuss acceleration and force, but will notice that the definition for the magnitude of the velocity, or the speed, is the same as before, namely, ds — *• where s is distance measured on the curved path, and that the direction of the velocity is that of the tangent to the curve. Also as the body moves along a curved path through a distance PQ = As (fig. 150), x changes by an amount PR = Ax and y changes by an amount BQ = Ay. We have then Ax ds Lim — = — = v = velocity of the body in its path, T . Ax Lim—— = At dx dt ' = v x = component of velocity parallel to OX, Lim -^ = At dy ~dt~ = v y = component of velocity parallel to OY. Otherwise expressed, v represents the velocity velocity of the projection of P upon OX, and v y of the projection of P on OY. Now, by (8), § 84, and by § 91, of P, v x the the velocity dx dx x dt ds ds di = V cos <£, (1) and dy dy y dt ds = v sin ■ -. " - x , L , . . / 3 a 6 a strophoid y = ± x-v — — — at the point I — — > — 44. Find the point at which the tangent to the curve y = x 8 at (1, 1) intersects the curve again. 45. Find the equations of the tangent and the normal to the ellipse 3x 2 + 5 y 2 = 32 at a point the abscissa of which is equal to its ordinate. 46. Find a point at which the tangent to the curve xy — 5x 2 — 4 = has the slope 1. 47. Find the length of the portion of the normal to the parabola y 2 = 8 x at (2, 4) included between the axis and the directrix of the parabola. Find the equation of the tangent to each of the following curves at the point (a? , y^: 48. y* = x b . 50. xl + y$ = ai. 49. Vx + Vy = Va. 51. X s -f y 3 - 3 ax// = 0. PROBLEMS 183 52. Prove that the equation of the tangent to the parabola y 2 = kpx at the point (x , y^ is y 1 y = 2p(x + a x ). 53. If the slope of a tangent to the parabola y 2 = Apx is in, prove • • P that its equation is y = mx -\ 54. Prove that the equation of the tangent to the ellipse -5 + jt 2 = 1 at the point (x v y^ is -~ + ^r = 1, and that the equa- a 2 //'- tion of the tangent to the hyperbola — — '-; =1 at the point (x v yS) a 2 b 1 55. Prove that the equations of the tangents with slope m to the X 2 I/ 2 ./'" if ellipse -^ + 7^ = 1 and the hyperbola — 2 — •— = 1 are respectively y = mx ± V (/-//<- + //' J and y = mx ± ~v/ 2 = 45, y 2 = 12a. 62. a- 2 4- y" - 12 x + 16 = 0, v/ 2 a" 4-3 63. if = a 3 , y 2 = -^ 65. SB* = 8 if - 4 ay, y 5 -a '" y a 2 4-4« 2 X* 64. a 2 # = 4, y = • 66. a// = if. ,f ^ + 4 ;/ ' * 2a -a 67. a 2 -3,W, If-y^- 68. y 2 = 6(a-3), 4^=(a-3) 2 (a-l). 69. Prove that the parabolas y* = 4 ax + 4 a 2 and y 2 = — 4 &a + 4 b 2 are confocal and intersect at right angles. 72. Find a point on the ellipse — 2 + yj = 1 sucli that the tangent 184 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 70. Show that for an ellipse the segments of the normal between the point of the curve at which the normal is drawn and the axes are in the ratio a 2 : V 2 . x 2 t/ 2 71. Find the coordinates of a point on the ellipse ~a + T2=l such that the tangent there is parallel to the line joining the positive extremities of the major and the minor axes. "*" b 2 there is equally inclined to the two axes. 73. Prove that the portion of a tangent to an hyperbola included by the asymptotes is bisected by the point of tangency. 74. If any number of hyperbolas have the same transverse axis, show that tangents to the hyperbolas at points having the same abscissa all pass through the same point on the transverse axis. 75. If a tangent to an hyperbola is intersected by the tangents at the vertices in the points Q and R, show that the circle described on QR as a diameter passes through the foci. 76. Prove that the ordinate of the point of intersection of two tangents to a parabola is the arithmetical mean between the ordi- nates of the points of contact of the tangents. 77. If, on any parabola, P, Q, and R are three points the ordinates of which are in geometrical progression, show that the tangents at P and R meet on the ordinate of Q. 78. Show that the tangents at the extremities of the chord of a parabola, which is perpendicular to the axis of the parabola at the focus, are perpendicular to each other. 79. Prove that the tangents described in Ex. 78 intersect on the directrix of the parabola. 80. Prove analytically that if the normals at all points of an ellipse pass through the center, the ellipse is a circle. 81. Prove that any tangent to the parabola if = Apx will meet the directrix and the straight line drawn through the focus, per- pendicular to the axis of the parabola, in two points equidistant from the focus. 82. Find in terms of x x and p the length of the perpendicular from the focus of the parabola if = ±px to the tangent at any point (x v y x ). PROBLEMS 185 83. If from two given points on the axis of a parabola which are equidistant from the focus perpendiculars are let fall on any tangent, prove that the difference of their squares is constant. 84. Show that the product of the perpendiculars from the foci of an ellipse upon any tangent equals the square of half the minor axis. 85. Find the equation and the length of the perpendicular from the center of the ellipse —^ + j^ = 1 to any tangent. 86. If two concentric equilateral hyperbolas are described, the axes of one being the asymptotes of the other, show that they intersect at right angles. 87. Prove that an ellipse and an hyperbola with the same foci cut each other at right angles. 88. Prove that the normal to an ellipse at any point bisects tin- angle between the focal radii drawn to the point. 89. Prove that the normal to an hyperbola at any point makes equal angles with the focal radii drawn to the point. Determine the values of x for which the following curves are (1) concave upward ; (2) concave downward : 90. a = 4 x s - 6 x 8 + 3. 91. y = ** - 12 x°- + 2. Find the points of inflection of the following curves : 92. y = 2a- 3 + 9a- 2 -2x-5. 96 y= 1 + *_. 93. y = 3a; 4 -4x»-6!B» + 4. 94. y = (x + 6 a)(x — ay. J ^ J Find the turning points and the points of inflection of each of the following curves and then draw the curve : 99. y = (x — 2) 2 (x + 2). 102. y = x* - 4 X s + 16. 100. y = x 3 - 3« 2 - 9x - 5. 103. if = x(x 2 — 4). 101. y = x {x — l) 3 . 186 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 104. It is required to fence off a rectangular piece of ground to contain a given area, one side to be bounded by a wall already constructed. If the length of the side parallel to the wall is x, will an increase in x cause an increase or a decrease in the total amount of fencing ? 105. The hypotenuse of a right triangle is given. If one of the sides is x, find the effect on the area caused by increasing x. 106. The stiffness of a rectangular beam varies as the product of the breadth and the cube of the depth. From a circular cylindrical log of radius a inches, a beam of breadth 2 a; is cut. Find the effect on the stiffness caused by increasing x. 107. A right cone is generated by revolving an isosceles triangle of constant perimeter about its altitude. If x is the length of one of the equal sides of the triangle, will an increase in x cause an increase or a decrease in the volume of the cone ? 108. A gardener has a certain length of wire fencing with which to fence three sides of a rectangular plot of land, the fourth side be- ing made by a wall already constructed. Required the dimensions of the plot which contains the maximum area. 109. A rectangular plot of land to contain 216 sq. rd. is to be inclosed by a fence and divided mto two equal parts by a fence parallel to one of the sides. What^a*^ be the dimensions of the rectangle that the least amount of fencing may be required ? 110. A gardener is to lay out a flower bed in the form of a sector of a circle. If he has 20 ft. of wire with which to inclose it, what radius will he take for the circle to have his garden as large as possible ? V. 111. An open tank with a square^ase and vertical sides is to have a capacity of 4000 cu. ft. Find the dimensions so that the cost of lining it with lead may be a minimum. 112. A rectangular box with a square base and open at the top is to be made out of a given amount of material. If no allowance is made for the thickness of the material or for waste in construction, what are the dimensions of the largest box that can be made ? 113. Find a point on the line y = x such that the sum of the squares of its distances from the points (— a, 0), (a, 0), and (0, b) shall be a minimum. PROBLEMS 187 114. A piece of wire 12 ft. in length is cut into six portions, two of one length and four of another. Each of the two former portions is bent into the form of a square, and the corners of the two squares are fastened together by the remaining portions of wire, so that the completed figure is a rectangular parallelepiped. Find the lengths into which the wire must be divided so as to produce a figure of maximum volume. 115. The strength of a rectangular beam varies as the product of its breadth and the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a circular cylindri- cal log of radius a inches. 116. What are the dimensions of the rectangular beam of great- est volume that can be cut from a log a feet in diameter and b feet long, assuming the log to be a circular cylinder ? 117. A log in the form of a frustum of a cone is 20 ft. long, the diameters of the bases being 2ft. and 1 ft. A beam with a square cross section is cut from it so that the axis of tin- beam coincides with the axis of the log. Find the beam of greatest volume that can be so cut. 118. Find a point on the axis of x such that the sum of its dis- tances from the two points (1, 2) and (4, 3) is a minimum. 119. Find the point on the circle jc 9 4- if = BAB' > BCB', whence t>a>p. Dividing through by r, we have tap r r r that is, tan li > h > sin h. Dividing by sin h, we have 1 h cos h sin h , . . _ sin h i or, by inverting, cos h < — - — < 1. h TRIGONOMETRIC FUNCTIONS 193 Now as h approaches zero, cos h approaches 1. Hence 1 h which lies between cos h and 1, must also approach 1 ; that is, T . sin h ^ Lim — = — = 1. 96. Differentiation of trigonometric functions. The formulas for the differentiation of trigonometric functions are as follows, where u represents any function of x which can be differentiated : d . du — sinw = cosw — -> (1) dx dx (2) (3) (4) (5) d (hi — esc u = — esc u ctn u ■ • (6 ) dx dx 1. By f 8), § 84, — sin u = — sin u • — • J w 3 dx du -/./■ To find ■ — ■ sin u, we place y = sin u. (hi Then if u receives an increment An, y receives an increment Ay, wliere a • , , a n o / , A "\ • A " Ay = sin (?t + At< ) — sin u = 2 cos I u -f — - 1 sin — > the last reduction being made by the trigonometric formula . , a + b . a — b sm a — sm 6=2 cos — - — sin — - — • Then we have . An . Au sm -— sin Ay / Au^ ° d du —- COS u = dx — sm u — - * dx d L o du — tan u = sec w — - » dx d t o rfw — ctn ?* = dx i — CSC U — - i i d du — sec u = dx seew tan u dx = 2 COS w + A** V 2 / Ait ■(-+t)-t 194 TRANSCENDENTAL FUNCTIONS Let Aw approach zero. By 2, § C>9, sin — — Lim — - = Lim cos ( u + -77- ) Lim Aw But Lim Aw \ 2 / Aw by _dy Aw cZw ■ Am \ Lim cos ( w + — 1 = cos w, . Aw sin — and Lim = 1. (By § 95) d . Hence — sm w = cos w, du , c? . c?w and — - sm w = cos w — - ■ dx dx 2. To find — cos w, we write dx cos w = sin 1 — — w 7T Then — cos u = — sin ( — — w ) £e \2 / dx dx i-)=S-) (by(1)) /tt \ (7w —"Hi 7*)* rfw = — sin w — - • dx 3. To find — - tan w, we write dx sin u tan w = cos w TRIGONOMETRIC FUNCTIONS 195 „, d d sin u Then — tan u = dx dx cos 11 d . d cos u — sin u — sin u — cos u cosV X Cb7(5),§84) (cos w + snrit) — COS" W (by (1) and (2)) = sec u — > dx 4. To find — - ctn ?/, we write dx GOSU ctn m = — sin w ™, (7 , c? cosw I hen — ctn u = — — : dx dx sm u d d . sm u — cos u — cos ii ■ — sm u (by (5), §84) (by (1) and (2)) — shr u — co&*u du sin*-// dx ., dx dx whence dy _ 1 du dx cscy ctn y dx But cse y = u, and ctn y = Vw 2 — 1 when y is in the first or the third quadrant, and ctn y = — vw 2 — 1 when y is in the second or the fourth quadrant. If the quadrant in which an angle lies is not material in a problem, it will be assumed to be in the first quadrant. This applies particularly to formal exercises in differentiation. Ex. 1. y — sin -1 Vl — .r 2 , where y is an acute angle. ^ = . I ±a-x*)t= — -L=- dx Vl - (1 - x 2 ) dx VI - x 2 This result may also be obtained by placing sin -1 V 1 — x 2 — EXPONENTIAL AND LOGARITHMIC FUNCTIONS 199 Ex. 2. y = sec- 1 Vi x 2 + 1 x + 2. dx V-i x- + 4x + 2 \ (4 r + 4x- + 2)-l 4 x + 2 1 ( I , - + -1 .,- + 2) (-2X + 1) 2 z 2 + 2 x + 1 i 98. Limit of (1 + h) h . In obtaining the formulas for the differ- entiation of the exponential and the logarithmic functions it is i necessary to know the limit of (1 + // /' as // approaches zero, the rigorous derivation of which requires methods which are too advanced for this book. We must content ourselves, therefore, with indicating somewhat roughly the general nature of the proof. We begin by expanding (1 4- h )'' by the binomial theorem and making certain simple transformations; thus, i, KH,. KH(M (i+ h) h =i + ~ h+ — p— /r+ k '< :! + • • • _ 1 1 (1-7Q (t-7>)(1 -2/0 l + [| + [I + "" + * where R represents the sum of all terms involving h, Jr. K\ etc. Now it may be shown by advanced methods that as It approaches zero R also approaches zero, and at the same time approaches e (§ 27). Hence i Lim (1 + JiY = e. 99. Differentiation of exponential and logarithmic functions. The formulas for the differentiation of the exponential and the logarithmic functions are as follows, where, as usual, u represents 200 TRANSCENDENTAL FUNCTIONS any function which can be differentiated with respect to x, log means the Napierian logarithm, and a is any constant: d , log„£ du ^^ l0g oW = _6a_ (1) dx u dx d , 1 du a -—log u = -—-, (2) dx u dx d U Ul dU Q — a =a loera — ? (6) dx & dx W d v _ u du . dx dx The proofs of these formulas are as follows : 1 .B y( 8),§84,|log„« = Alog„«.|. To find — log a M, place y = log a M. Then if u is given an increment Aw, y receives an increment Am, where A , , , A N , A# = log a (w + Am) — log M Am . /. A?/,\ A " u \ u the last transformation being made by the formula p log M = log l/ p . Then Am m \ m / — may be taken as h of § 98. Hence Therefore and Ay = Am u ° \ u 3hes zero, the fractioi Limfl A« = \ AuY" + — =e. M / dy^ du - losr„ e u &a dy dx log a e £?M M cfa: EXPONENTIAL AND LOGARITHMIC FUNCTIONS 201 2. If y = log u, the base a of the previous formula is e ; and since log e e = l, we have dy dx _ldu u dx 3. If y- = a H , we have \ogy-. = log a" = u log a. Hence, by formula (2), 1 dy y dx , du whence dy dx „, du = « u loga— -• dx 4. Uy = e ", the pr evious dy dx formula becomes dx Ex. 1. y = log(x- - 1 x + 5). dy 2 x - I dx x 1 — 4 x + 5 Ex. 2. y = e~ x2 - l Il = -2xe-^ dx Ex. 3. y — e-^ cos far. — = cos bx — (e~ ar ) + e - "* — (cos bx) = — oe" ^ cos bx — be~ ox sin hr dx rfx ' dx = — e~ "- r (« cos for + & sin bx). 100. Sometimes the work of differentiating a function is simplified by first taking the logarithm of the function. Ex. l. Let Vir?- l-x 2 Then io gy = log^^— - = Uog(l-.r 2 )- ilog(l+.T 2 ). 1 dy _ X a; .'/ dx 1 - x" 1 + X 2 -2x (1- -aP)(l + x>y dy -2xy dx (1- (1- -x 2 )(l + -2 a: -x 2 )(l + ■'■') .^)Vi + x 2 X 2 -2x 202 TRANSCENDENTAL FUNCTIONS Bence and (l+,: 2 )Vl This method is especially useful for functions of the form ?/'', where u and v are both functions of x. Such functions occur rarely in practice and cannot be differentiated by any of the formulas so far given. By taking the logarithm of the function, however, a form is obtained which may be differentiated. Ex. 2. Let y = x eiux . Then log y = log(x 8in *) = sin x ■ log x. Therefore - -f- = (sin x) - + cos x ■ log x, II dx x and — = x Biax ~ 1 • sin x + x Biax cos x • log x. dx 101. Applications. The applications of differentiation discussed in the previous chapter are evidently applicable to problems in- volving transcendental functions. Ex. 1. Find the turning points and the points of inflection of the curve (Ex. 5, § 24) . , t • o v ' s ' y = smi + | sin 2 x. We find — = cos x + cos 2 x = 2 cos 2 .c + cos x — 1 ." dx = (2 cos x — 1) (cos x + 1). Equating — to zero, we have cos x = — or cos x = — 1. dx 2 If cos ar = — , x = — V 2 »ir or — — + 2 «7r. As x passes through either of these values, — changes sign, and hence these values correspond to turning dx o _ points of the curve. In fact, x = — + 2 mr gives maximum values of y — - Vs, and x — — ~ + 2?nr gives minimum values of y — — '- V3. APPLICATIONS 203 If cosx dy x = it ± 2 nw. As x passes through these values, — does dx not change sign. Hence these values do not correspond to turning points of the curve. To examine for points of inflection, we find dx 2 — sin x — i sin — sin x (4 cos x + 1). Then velocity acceleration bk cos bt, -- - Irk sin bt = b% This is zero when x = + 2nirOTTr + 2 rnr, or when x = cos _1 (— \). As x passes through any of these values, — { changes sign. These values dx 2 correspond, therefore, to j'oints of inflection. Ex. 2. A particle of mass m moves in a straight line so that .s- = /sin ///, where t = time, s = space, and b and k are constants. (h_ dt ~ df- force = F= ma - — mb-s. Let be the position of the particle when I = 0, and let <~>.\ = k and OB — — k. Then it appears from the for las for .-.and v thai the particle oscillates forward and backward hetween Band I. It describes the distance (>. I in the time — and moves from A' to .1 and bach to /<' in the time - — 2 b b The formula F=— ndi-s shows that the parti.de is acted on l>v a force directed toward and proportional to the distance of the particle from 0. The motion of the particle is called simple harmonic motion. Ex. 3. A wall is to be braced by means of a beam which must pass over a lower wall b units high and standing n units in front of the first wall. Required the shortest beam which can lie used. Let A B = 1 (fig. 152) be the beam, and let (' lie the top of the lower wall. Draw the line CD parallel to OB, and let EBC = d. Then l = BC+ CA = ECcsc6 + nr^cO = b esc 6 + a sec 0. - = — b csc 6 ctn + a sec $ tan i a sin 3 — bcos z Placing we have dl dd a sin 3 sin 2 # cos-0 = 0, to find the minimum, = b cos 3 0, whence tan 8 = — • 204 TRANSCENDENTAL FUNCTIONS When 6 has a smaller value than this, a siu 3 6 < b cos 3 9 ; and when has a larger value, a sin 3 9 > b cos 3 6. Hence I is a minimum when tan# = — . a? Then / = Acac0 + «sec0 _ b \ai + bi a V n* + 6$ 6t a^ = (at + 6f)t. 102. The derivatives in parametric representation. When a curve is defined by the equations we have, by (9), § 84, !^ = £. dx dx (1) *, If it ; is required to find — ^, we may proceed as follows: c? 2 y c? (dy\ dt\dx) dx 2 dx\dx) dx (2) Ex. di For the cycloid x = a( — sin d>), y = a(l— cos <£), flty a sin <4 <7 _ r _ v = ctn • dx dx a (1 — cos ) 2 dx 2 d\ 2) dx cosec** — 2 2 a(l — cosd>) 1 4 a sin 4 ^ 2 DIRECTION OF A CURVE Formula (2) may be expanded as follows: Therefore ±(dy\ dt\dx) (dy\ d 2 y dx d 2 x dy dx \dt dt 2 dt df dt l_x it d 2 y d 2 y dt 2 dx d 2 x ~dt ~~ dt 2 dy dt dx 2 (IT 205 (3) 103. Direction of a curve in polar coordinates. The direction of a curve expressed in polar coordinates is usually determined by means of the angle between the tangent and the radius vector. Let P (r, 0) (fig. 153) be any point on the curve, PT the tangent at P, and -v/r the angle made by PT and the radius vector OP. Give d an increment Ad = POQ, expressed in circular measure, thus fixing a second point Q(r + A>; 6 + A6) of the curve. To determine Ar describe an arc of a circle with center at and radius OP, intersecting OQ at B. Then and OQ=r+Ar BQ=Ar. Draw also the chord PQ and the straight line PS perpendicular to OQ and meeting it in S. Then SP=r sin Ad, OS =r cos Ad, SQ=OQ-OS= r + Ar-rcosAd Ad = Ar + 2 r sin 2 — • 206 TK A NSC UN DENTAL FUNCTIONS As A0 approaches zero the chord PQ approaches the limiting position PT and the angle 11 QP approaches i/r. But in the triangle SPQ, SP rsmA6 tan SQP = — = SQ A , „ . ,A6 Ar + 2 r sin 2 — sin Ad r Ad . Ad Ar . A0 Sm T + r sm A6> ' 2 A(9 2 Hence, taking the limit, we have tan^ = ^-. (1) dd If it is desired to find the angle MNP = , it may he done by the evident relation , , ' A ,~ N = yfr + 0. (2) 104. Derivatives with respect to the arc in polar coordinates. In the triangle PQS (fig. 153), SP siii SQP = chord PQ SP arc PQ mcPQ chord PQ rsinA0 axcPQ As chord PQ sin Ad A6 arcP l and „ are the values of for the points If and i^ respec- tively, then ., — <^) j is the total change of direction of the curve between J^ and i?. If 3 -0 1 = A$, expressed in circular meas- A<£ unit of the arc Fig. 155 ure, the ratio — - is the average change of direction per linear Regarding as a function of 8 and 208 TRANSCENDENTAL FUNCTIONS taking the limit of have -j-i which as Acb as As approaches zero as a limit, we is called the curvature of the curve at the and the circle is a curve of constant point ij". Hence the curvature of a curve is the rate of change of the direction of the curve with respect to the length of the arc (§92). If ~- is constant, the curva- ds ture is constant or uniform ; otherwise the curvature is va- riable. Applying this defini- tion to the circle of fig. 156, of which the center is C and the radius is a, we have A. Therefore Ad> 1 rT deb 1 —-£ = -. Hence -?- = - As a ds a curvature equal to the reciprocal of its radius. 106. Radius of curvature. The reciprocal of the curvature is called the radius of curvature and will be denoted by p. Through every point of a curve we may pass a circle with its radius equal to p, which shall have the same tangent as the curve at the point and shall lie on the same side of the tangent. Since the curvature of a circle is uniform and equal to the reciprocal of its radius, the curvatures of the curve and the circle are the same, and the circle shows the curvature of the curve in a manner similar to that in which the tangent shows the direction of the curve. The circle is called the circle of curvature. From the definition of curvature it follows that . _ ds P ~dj>' If the equation of the curve is in rectangular coordinates, ds_ by (9), §84, p = ±. dx RADIUS OF CURVATURE 209 To transform this expression further, we note that ds — dx + dy ; whence, dividing by dx and taking the square root, we have Since ^ = tan-M^j, (by § 91) dry d<\> dx 2 dx 1+ /^y Substituting, we have \dx) HS! d 2 y d.r 1 In the above expression for p there is an apparent ambiguity of sign, on account of the radical sign. If only the numerical value of p is required, a negative sign may be disregarded. Ex. Find the radius of curvature of the ellipse — + f- = 1. a- U l Here dy h-.r and cPy _ b* dx? ay Therefore _ OV+ b*x 2 )% 9 " a*b* Another formula for p, that is, , MS 1 ] <>>r may be found by defining q> as the angle between Y and the tangent and interchanging x and y in the above derivation. 210 TRANSCENDENTAL EUNCTIONS 107. Radius of curvature in parametric representation. If x and y are expressed in terms of any parameter t, the radius of curvature may be found as follows: But and whence ds ds dt d4> = d$ dt (By (9), § 84) ds dt~ W(iH!)' dy dy dt (by (2), § 91) = = tan — = tan — ; dx dx "dt (dx\ d 2 y (dy\ d 2 x d(f> \ dt) dt 2 \dt)dt 2 dt HI . \dt! . \dt) dx d 2 y dy d 2 x dt dt 2 dt dt 2 dxV (dyV dt \dt Therefore, by substitution, 2-if dx d 2 y dy d 2 x It ' ~df ~ ~di ' df Ex. Find the radius of curvature of the cycloid x = a — a sin<£, II = a — a cos . Here the parameter t of the general formula is replaced by cj>. Therefore RADIUS OF CURVATURE dx 211 tty a — a cos , d 2 x . , = a sin g> : a sin (p, dhi , — = «cos<^. Hence, by substitution, p — [q 2 (1 — cos ~ d$ dd (By (9), §84) From § 104, i-^+S) and, from § 103, 4> = yfr + 6. Then tf<9 j/ _l log Va a + ^-« 40. y = log Vx 2 + 4 x 4- 3. X 1 2 x — 3 • 47. // = log sin x. 41 - •'/ = 12 log 2x4-3 ' 48. y = log(sec 3x4- tan 3 x). 42. y J-log Ss-A i^rfn- 2v3 3x4-V3 49. // = log — 43. y = log . x 1 1 + sin 2 2 tan x 4- 1 V3 - 4 x 4- x 2 44. y = log(3 x 4- V9x 2 + 2). 5 °' -' 7 = log tan x 4- 2 45. y = \ log (x :! 4- Vx° — « G ). 51. // = log ctn x — esc 2 x. 52. y = logVar , + 4 + ^-j-j. 53. y = 3 Vx 7 ^ 4 + log (a 3 + Vx 4 -" 4 f. 1 , 1 - cos 2 54- ^=8 10g l + cos2x 4sin 2 2x 55. y = x [(log ox) 2 - 2 log aa + 2]. 56. y = log(x 2 + Vx 4 — l) - sec- 1 x 2 II I TRANSCENDENTAL FUNCTIONS 57. y 58. y 59. y 2 x tan- 1 2 x — log Vl + 4 x\ | log (2 .r 2 + 1) + V2 tan" 1 x V2. 60. ?/ = x sec a; tan -1 a« log Vl + where t is the time and s the distance from a point 0. Find when the particle is moving forward and when backward. Find also the greatest distance which the particle reaches from 0, and the force which acts upon it. 124. A motion of a particle in a straight line is expressed by the equation s = 5 — 2cos a £ Express the velocity and the acceleration at any point in terms of s. 125. Two paxticles are moving in the same straight line, and their distances from the fixed point on the line at any time t arc respectively x = a cos ht and as' = a COS ( ht -f- .,)' A and a being constants. Find the greatest distance between them. 126. If s = ae** + be~ kt , show that the particle is acted on by a repulsive force which is proportional to the distance from the point from which s is measured. 127. If a particle moves so that s = e -i«*(a sin ht + b cos ///), find expressions for the velocity and the acceleration. Hence show that the particle is acted on by two forces, one proportional to the distance from the origin and the other proportional to the velocity. Describe the motion of the particle. 128. A revolving light in a lighthouse \ mi. offshore makes one revolution a minute. If the line of the shore is a straight line, how fast is the ray of light moving along the shore when it passes the point of the shore nearest to the lighthouse ? 129. A, the center of one circle, is on a second circle with center at B. A moving straight line, AMN, intersecting the two circles at M and N respectively, has constant angular velocity about A. Prove that BN has constant angular velocity about B. 218 TRANSCENDENTAL FUNCTIONS 130. BC is a rod a feet long, connected with a piston rod at C, and at B with a crank , I /;, h feet long, revolving about A. Find C's velocity in terms of AB's angular velocity. 131. A body moves in a plane so that x = acos t-\-b,y = asmt -f- c, where t denotes thne and a, b, and c are constants. Find the path of the body, and show that its velocity is constant. 132. The parametric equations of the path of a moving point are, in terms of the time t, x = a cos Id, y = b sin ht, where a, b, and 'k are constants and a > b. Prove that the path is an ellipse. Find the velocity of the point in its path. Find when the velocity is a maximum and when a minimum. 133. A particle moves so that x = 2 cos t — cos 2 1, y = 2 sin t — IT sin 2 t, where t is the time. Find its velocity in its path when t — — • 134. If a wheel rolls with constant angular velocity on a straight line, required the velocity of any point on its circumference ; also of any point on one of the spokes. 135. Prove that a point on the rim of the wheel of problem 134 is moving parallel to the straight line on which the wheel rolls, with a velocity proportional to its distance from OX. 136. Show that the highest point of a wheel rolling with constant velocity on a road moves twice as fast as each of the two points in the run whose distance from the ground is half the radius of the wheel. 137. If a wheel rolls with constant angular velocity on the cir- cumference of a fixed wheel, find the velocity of any point on its circumference and on its spoke. 138. If a string is unwound from a circle with constant angular velocity, find the velocity of the end in the path described. 139. A man walks along the diameter, 200 ft. in length, of a semi- circular courtyard at a uniform rate of 5 ft. per second. How fast will his shadow move along the wall when the rays of the sun are at right angles to the diameter ? 140. How fast is the shadow in the preceding problem moving if the sun's rays make an angle a with the diameter ? 141. A man walks across the diameter of a circular courtyard at a uniform rate. A lamp, at one extremity of the diameter perpen- dicular to the one on which he walks, throws his shadow on the wall. Required the velocity of the shadow along the wall. PROBLEMS 219 142. A ladder b feet long leans against a side of a house. Its foot is drawn away in the horizontal direction at the rate of a feet per second. Find the path described by the center of the ladder and the velocity of the center in its path. 143. Find ■—■ and ■— for the curve x = a (cos 4- sin ), y = a (sin cf> — cos ). 144. Find — and -^ for the curve x = a cos 3 <£, y = a sin 3 <£. 145. Find -,' and , '., for the curve x = e* sin t, >/ = e* cos t. 146. Prove that the logarithmic spiral r = e a6 cuts all radius vectors at a constant angle. 147. Prove that the angle between the normal and the radius vector to any point of the lemniscate is twice the angle made by the radius vector and the initial line. 148. Prove that the angle between the cardioid r = a(l — cos 0) and a radius vector is always half the angle between the radius vector and the initial line. 149. I£j> is the perpendicular distance of a tangent from the pole, prove that p ^hW 150. If a straight line drawn through the pole perpendicular to a radius vector OP meets the tangenl in .1 and the normal in B, show that OA = r 1 —- and OB = — • dr ad 151. Show that for any curve in polar coordinates the maximum and the minimum values of ;• occur in general when the radius vector is perpendicular to the tangent. 152. Sketch the curve r = 2 4- sin 3 6, and find the angle at winch it meets the circle r = 2. 153. Sketch the curve 7- 2 = a 2 sin-> and determine the angle at which it intersects the initial line. 154. Sketch the curves i~ = a' 2 sin 2 6 and r 2 = « 2 cos2 0, and show that they intersect at right angles. •2'20 TRANSCENDENTAL FUNCTIONS 155. If a particle traverses the cardioid /• = a(l — cos 6) so that 6 makes uniformly two revolutions a second, find the rate at which /• changes, and the velocity of the particle in its path : (1) when = | ■ (2) when 6 = w. 156. Find the velocity of a point moving in a limacpn r = a cos 6 + b when changes uniformly. f) 157. When a point moves along the curve r = 4 sin 3 - at a uniform o rate of 2 units per second, find the rates at which 6 and r are changing : (1) when 6 — — ; (2) when 6 = ir. 158. Find the radius of curvature of the curve ar + y^ = a*. 159. Find the radius of curvature of the catenary y = -(e a + e a ). 160. Show that the catenary y = -(e a + e a ) and the parabola 1 ' y = a + — x 1 have the same slope and the same curvature at their common point. 161. Find the radius of curvature of the curve if = £ 7 (x — 2) 3 at the point for which x — 3. 162. Find the radius of curvature of the cycloid i « — x /T. — ; 5 y = a cos -1 ±-\2ax — x 2 at a point for which x = — • 163. Find the radius of curvature of the curve y = e~ 2x sin 3x at the origin. 164. Find the least radius of curvature of the curve y = logx. 165. Find the points of greatest and of least curvature of the sine curve y = sin x. 166. Show that the curvature of the parabola y = ax 2 + hx + c is a maximum at the vertex. 167. Show that the product of the radii of curvature of the curve y = ae a at the two points for which x = ± a is a 2 (e + e~ l ) z . PEOBLEMS 221 168. Find the radius of curvature of the four-cusped hypocycloid x = a cos 3 , y = a sin 3 <£. 169. By use of the parametric equations of the ellipse find the points where the radius of curvature is a maximum or a minimum, and the values of these radii. 170. Find the radius of curvature of r = a (2 cos — 1). 171. Find the radius of curvature of the lemniscate i a = 2 a a cos 2 0. 172. Find the greatest and the least values of the radius of curvature of the curve r = a sin 3 - • 173. If the angle between the straight line drawn from the origin perpendicular to any tangent to a curve and the radius vector to the point of contact of the tangent is either a maximum or a mini- r* mum, prove that p = — , where p is the length of the perpendicular. CHAPTER XII INTEGRATION 109. Introduction. In § 80 the process of integration was defined as the determination of a function when its derivative or its differential is known, and was denoted by the symbol I ; that is, if „, N 7 ,„, N J f(x)dx = dF(x), then ff(x) dx = F(x). (1) The expression f(x) dx is said to be under the sign of inte- gration, f(x) is called the integrand, and F(x) is called the integral of f(x)dx; sometimes F(x) is called the indefinite integral, to distinguish it from the definite integral defined in § 81. The determination of the indefinite integral is important in a wide range of problems, and for that reason we shall now deduce formulas of integration. We ought to note first, however, that a more general form of (1) is p jf(x)dx = F(x)+C, (2) where C is the constant of integration (§80). In each of the formulas we shall derive, C will be omitted, since it is inde- pendent of the form of the integrand, but it must be added in all the indefinite integrals determined by means of them. 110. Fundamental formulas. The two formulas fcdu = cfdu (1) and j(du + dv + dw -\ ) = fdu + fdv + fdw -J (2) are of fundamental importance, one or both of them being used 222 INTEGRAL OF U" 223 in the course of almost every integration. Stated in words they are as follows: (1) A constant factor may be changed from one side of the sign of integration to the other. (2) The integral of the sum of a finite number of functions is the sum of the integrals of the separate functions. To prove (1), we note that since cdu = d(cu), it follows that I cdu = I d(cu) = cu = <" for all values of n except n = — 1. In the case n = — 1, the expression under the sign of inte- gration in (1) becomes — , which is recognized as c?(log?<). Therefore / - — = logw. (2) 224 INTEGRATION In applying these formulas the problem is to choose for u some function of x which will bring the given integral, if pos- sible, under one of the formulas. The form of the integrand often suggests the function of x which should be chosen for u. Ex. 1. Find the value of f(ax*+ bx +-+ ^)dx. Applying (2), § 110, and then (1), § 110, we have flax* + bx + -+ ^Sdx = fax 2 dx+fbxdx +f-/lx + f~ 2 d x = afx*dx + bjxdx + c f— + efx-*dx. The first, the second, and the fourth of these integrals may be evaluated by formula (1) and the third by formula (2), where u = x, the results being 11 e respectively - ax 3 , - bx 2 , , and c log x. Therefore f(ax 2 + bx + - + ~)dx = - ax s + -bx* + c\ogx--+ C. J \ x x-j 3 2 x •s/ Ex. 2. Find the value of f(.r 2 + 2)xdx. If the factors of the integrand- are multiplied together, we have f(x 2 + 2) xdx =f(x 8 + 2 x) dx, which may be evaluated by the same method as that used in Ex. 1, the result being \ x* + x 2 + C. Or we may let x 2 + 2 = u, whence 2 xdx = du, so that xdx = \ du. Hence f(x 2 + 2)xdx =j\udu = ljudu 2 2 = i (a* + 2) 2 + C. Instead of actually writing out the integral in terms of u, we may note that xdx = ^d (x 2 + 2) and proceed as follows : f (x 2 + 2) x dx = f (x 2 + 2) 1 d (x 2 + 2) .= }jf(x 2 +2)d(x 2 +2) = \(x 2 +2y+C. Comparing the two values of the integral found by the two methods of integration, we see that they differ only by the constant unity, which may be made a part of the constant of integration. INTEGRAL OF U N 225 Ex. 3. Find the value of C (ax 2 + 2 bx) s (ax + b)dx. Let ax- + 2 bx = u. Then (2 ax + 2 b) dx = du, so that (ax + b) dx = I du. Hence f (ax 2 + 2 bxf (ax + b) dx = f \ u 3 du ^ = \ (ax 1 + 2 bxy + C. Or the last part of the work may be arranged as follows : f (ax* + 2 bx) z (ax + b) dx = f (ax 2 + 2 bx)* \ d (ax* + 2 6a:) = I f(ax 2 + 2 bxf d (ax 2 + 2 6x) = I (ax 2 + 2 bx)* + C. Ex.4. Find the value of f * C* + V'** . J ax 2 + 2 bx As in Ex*. 3, let ax 2 + 2 bx = u. Then (2 ax + 2 b)dx = du, so that (ax + b) dx = I du. Hence /4 (ax + b) dx _ r 2 )-<"' dx. Let t" J + b= u. Then >'■' ad.r = du. Hence /<- •+/,)-,"■ ■dx= r+i* j ■ a du /sec 2 (ax + />)ox _ r 1 c/m tan (ax + b) + c J a u 1 rdu a J u -logu + C - log [tan (ax + b) + c] + C, /• sec 2 (ax + V)dx _ rl d [tan (ax + b) + c] «/ tan (ax + 6) + c J a tan (ax + b) + c _ 1 /• '■>) + C. Ex. 2. Find the value of f sec (e a3? + b) tan (e** + b) e^xdx. Let e 03 ? + b = u. Then e ax2 2 axdx = du. Therefore J^sec (e** 2 + 6) tan (e^ + 6) e a *\r dx = -L J sec (c«* + &) tan (c a ' 3 + b) d ( e**' + h) = J_ sec ((•«•' 2 + &)+ C. 228 INTEGRATION It is often possible to integrate a trigonometric expression by means of formulas (1) and (2) of § 111. This may happen when the integrand can be expressed in terms of one of the elementary trigonometric functions, the expression being multiplied by the differential of that function. For instance, the expression to be integrated may consist of a function of sin x multiplied by cos xdx, or of a function of cosrr multiplied by (— sin xdx), etc. Ex. 3. Find the value of / Vsina: cos 3 xdx. Since d (sin x) = cos xdx, we will separate out the factor cos xdx and express the rest of the integrand in terms of sin x. Thus Vsina; cos z xdx = vsina;(l— sin 2 x) (cos xdx). Now place sin x = u, and we have I Vsin x cos s xdx = J u? (1 — u") du = I («2 _ u ^)du = $ «t - f u% + C = ^ T sinlx (7- 3 sin 2 x) + C. Ex. 4. Find the value of f sec 6 2 xdx. Since d(tan 2 x) = 2 sec 2 2 xdx, we separate out the factor sec 2 2 xdx and try to express the rest of the integrand in terms of tan 2 x. Thus sec 6 2 xdx = sec 4 2 a: (sec 2 2 xdx) = (1 + tan 2 2 x) 2 (sec 2 2xdx) = (1 + 2 tan 2 2 x + tan 4 2 x) (sec 2 2 xdx). Now place tan 2 x = u, and we have f sec 6 2 xdx = £ f (1 + 2 m 2 + u 4 ) du = Jr tan 2 x + ^tan 3 2 x + T Vtan 5 2 x + C. Ex. 5. Find the value of I tan 5 xdx. Placing tan 5 x = tan 3 a; tan 2 x = tan 3 a; (sec 2 x — 1), we have / tan 5 xdx = f tan 3 x sec 2 x dx — f tan s xdx = ^tan 4 x — | tan s xdx. TRIGONOMETRIC FUNCTIONS 229 Again, placing tan 3 x = tanx (sec 2 x — 1), we have J tan 3 xdx — f tanx sec 2 xdx — J tanxtfx = i tan 2 x + log cos x + C. Hence, by substitution, / tan 5 x dx = ^ tan 4 a: — i tan 2 a; — log cos x + C. When the above method fails, the integral can often be brought under one or more of the fundamental formulas by a trigonometric transformation. Ex. 6. Find the value of j con- xdx. Since cos 2 x = J (1 + cos 2 x), we have i cos' 2 xdx = \ f (1 + cos 2 x) dx = ■• f lLr + lfcoa2xd(2z) = l.r+ \ sin •_'.; + C. Ex. 7. Find the value of / Bin'z C0B*xdx. Placing sin 2 x cos 4 x = (sin x cos ar) a cos 8 v, we have sin^x cos 4 x = | sin 2 2 x(l + cos 2 x). Therefore J sin 2 x cos 4 xr/x = \ j sin-2xdx + l I sin 2 2 x cos2xtfx. Using the method of Ex. 6, we have | sin- 2 x dx — \ f (1 — cos 4 x) ilx = h x ~ \ s i n 1 •''• Writing sin 2 2 x cos 2 x dx = sin 2 2 x (cos 2 x dx) and placing sin 2 x = u, we have / sin 2 2 x cos 2 xdx = £ j u 9 <£« = J sin 3 2 x. Combining these results, we have, finally, f sin 2 x cos 4 x 0). Making this substitution and evident reductions, we have as our required formulas du , u _, N = sin 1 - , (1) / Va 2 - u 2 a f du 1 _ 1 u . . tan l -, (2) a -f- w a a /; ^ 1 , ?* , ON sec -. (o) Vm 2 - a. 2 « « Only one of the possible values has been given for each integral, as that single value is sufficient for all work. Referring to 1, § 97, we see that sin -1 - must be taken in the first or the fourth quadrant ; if, however, it is necessary to INVERSE TRIGONOMETRIC FUNCTIONS 231 have sin -1 - in the second or the third quadrant, the minus a sign must be prefixed. In like manner, in (3), sec -1 - must be taken in the first or the third quadrant or else its sign must be changed. J y/9 - 4 , Ex. 1. Find the value x 2 - -sin -1 h C. Letting 2x = u, we have du = 2 dx, and f dx = 1 r ^(^Q J V !) - 4 x- 2 J A 9 _ (O x) 2 - /rfa; — • z V 3 a: 2 — 4 If we let V3a: = u, then ^/u = v3 da?, and we may write f dx r fl(VHx) _ = 1 scc -i^ a? ^a;V3^-4 J V3*- V(V3*) 2 - * 2 2 — • \ 1 ,■ - x* Since \'l.c-r = v / l - (.<• - 'J)-, , have r-j^= = r rf * = r ^*- 2 > J VI x - x- ^1- (,■ - 2)- J \ 1 - (./■ - 2) 2 x — 2 = sin" 1 " t) " + C. /ill' — 2 ./•- + 3 x + 5 To avoid fractious and radicals, we place dx 8 -A-- n 1 )- + 31 Therefore f dx = r 4 dx = 2 r rf(^ + 3) ^2x 2 +3a- + 5 J (4a: + 3) a +31 "J (4 a: + 3) 2 + 31 = — — =tan _1 — + C. V31 V31 The methods used in Exs. 3 and 4 are often of value in dealing with functions involving ax 2 + hx + c. 232 INTEGRATION Ex. 5. Find the value of f (?+*)& . Separating the integrand into two fractions, that is, 5 + 1 x 4 5 + 4 x 4 and using (2), § 110, we have / (x 3 + x) dx _ r x 8 dx r xdx 5 + 4x 4 ~J 5 + 4x 4 J 5 + 4x 4 ' ■d 4. r xZ(Ix i r i6x 3 f/x i . ,_ , . . x But i^M^ = I f iJ.^M^ = W l0g(5 + 4a:) ' and f *^ =1 f **«* = J- tan -i^. J5 + 4x 4 4 J 5 + (2*3)3 4 V5 V5 Therefore /■ (x 3 + x) r/x 1 . „ , . 4N , 1 , , 2 x 2 , _ I H 7^7- = 77: 1o 8' (5 + 4 x 4 ) + —tan- 1 — - + C. J 5 + 4x 4 10 bV w J 4 Vo V5 Ex. 6. f^Y^~2 = P* 11 "" 1 *]^? = tan_1 ^ - tan-i (- 1). There is here a certain ambiguity, since tan _1 V3 and tan -1 ( — 1) have each an infinite number of values. If, however, we remember that the graph of tan _1 x is composed of an infinite number of distinct parts, or branches, the ambiguity is removed by taking the values of tan -1 v3 and tan _1 (— 1) from = tan -1 b — .. 1 + x 2 tan -1 a and select any value of tan -1 o, then if b = a, tan -1 J must be taken equal to tan -1 a, since the value of the integral is then zero. As b varies from equality with a to its final value, tan _1 i will vary from tan -1 a to the nearest value of tan- 1 ^. The simplest way to choose the proper values of tan _1 & and tan -1 a is to take them both between and — • Then we have / V3 dx _ir I 7r\ _ 7 tt -l 1 + x 2 ~ 3 \ 4/ 12 ' Ex. 7. J 2 — = sin -1 - = sin -1 \ — sin -1 0. Va 2 -x 2 L «Jo The ambiguity in the values of sin -1 \ and sin -1 is removed by notic- ing that sin -1 - must lie in the fourth or the first quadrant and that the a two values must be so chosen that one comes out of the other by continuous change. The simplest way to accomplish this is to take both sin -1 \ and sin -1 between and — • 2 2 Then /»2 dx Jo Va 2 - LOGARITHMIC FUNCTIONS 233 114. Closely resembling formulas (1) and (2) of the last article in the form of the integrand are the following formulas : f- r ^-; = l0g(m-V^T7») > (1) J V u'- + a- f-^L= = \o g (u+V^^), (2) J Vtt 3 — a A , C du 1 , a — a 1 . a — u . . and I — - = -— log or-— log — • (3) Jir-a- 2 a a u + a 2 a s a + u w To derive (1) we place u = a tan . Then du = a sec 2 <£c/$, and W 2 + a' 2 = a sec . Therefore / , = I sec &J& V w 2 +a 2 J = leg (sec and proceed as in the derivation of (1). Formula (3) is derived by means of the fact that the fraction — may be separated into two fractions, the denominators of which are respectively u — a and u + a ; that is, 1 1/1 1 a\u — a u + a I ri 1 1 \ du hi r du Then fL-** r/_L__u J W — a' 2 a J \u — a u + a) _!_{ P du C du 2 a \ J u - a J u + a = 2^ l0g (M _ ^ ~ l0g (M + ^ ^ 1 . u — a = —— log 2a b u+a 234 INTEGRATION The second form of (2) is derived by noting that r du r - du , . I = / = log(« — «•). The two results differ only by a constant, for a + u u-\- a and hence log = log(— 1) + log , a + u u + a and log(— 1) is a constant complex quantity which can be ex- pressed in terms of V— 1. dx Ex. 1. Find the value of f - V 3 x 2 + 4 x To avoid fractions we multiply both numerator and denominator by V3. dx _ V3 dx -y/Srlx Then V3 x 2 + 4 x V9 a; 2 + 12 x V(3 a; + 2) 2 - 4 Letting 3 x + 2 = w, we have c/w = 3 tfx, and r dx _ 1 -J r 3 r/x ■' V« a; 2 + 4 x -n/3^ V(3 x + 2) 2 - 4 = ^-log (3 x + 2 + V(3 x + 2) 2 - 4 ) + C V3 = -^=log (3 x + 2 + V9x 2 + 12x) + C. V3 /<7x 2 x 2 + x - 15 Multiplying the numerator and the denominator by 8, we have - r dx _ 2 r 4 dx J 2 x 2 + x - 15 "J (4 x + 1)'~ - (ll) 2 = 1_ (4x + l)-ll 11 & (4x + l) + ll + L - 1 2 x — 5 1 2 a; — 5 1 This may be reduced to —-log + C, or —-log - — — - log 2 + C, 11 2x + b 11 x + 3 11 and the term — ^ log 2, being independent of x, may be omitted, as it will only affect the value of the constant of integration. Ex. 3. Find the value of ( -± '- J 2 x 2 + x - 15 If 2 x 2 4- x - 15 = u, du = (4 x + 1) dx. Now 3 x + 4 may be written as J (4 x + 1) + ^-. LOGARITHMIC FUNCTIONS 235 Therefore f (3* + 4)/$ x 2 + 12 x), by Ex.1. Hence the complete 3 V3 integral is 11 £ V3 x 2 + 4 x 4- — ^plog (3x42 + V'.»x 2 + 12x) 4- C. y 3 V 3 Ex.5. Find the value of jsecxdx. /r dx r coQxdx secxdx = I = I J cosx J cosx t/ cos'x J sm- x - 1 2 & 1 + sin x 1 . 1 + sin x -log- : — 4- C 2 1 — sin x 1, (1 + sinx) 2 ~ 2 IOg l-sin 2 x +C ilog(l± 2 °\ c< sinx\ 2 _ 2 \ cos x = log (sec x + tan x) 4- C. 236 INTEGRATION (17) J Vu~ — a /du 1 , u — a 1 . a — u .+ ^ -j = = — log — ■ — or — log , (18) m 2 — a 2 2a ° w + a 2a ° a + u fe"du = e u , (19) Ca H du = - a". (2(T) J log a v 238 INTEGRATION 117. Integration by substitution. In order to evaluate a given integral it is necessary to reduce it to one of the foregoing standard forms. A very important method by which this may be done is that of the substitution of a new variable. In fact, the work thus far has been of this nature, in that by inspection we have taken some function of x as u, In many cases where the substitution is not so obvious as in the previous examples, it is still possible by the proper choice of a new variable to reduce the integral to a known form. The choice of the new variable depends largely upon the skill and the experience of the worker, and no rules can be given to cover all cases. We shall, however, suggest a few substi- tutions which it is desirable to try in the cases denned. I. Integrand involving fractional 'powers of a + bx. The substi- tution of a power of z for a + bx will rationalize the expression. x-ilx (1 + 2 x)i Here we let 1 + 2 x = z 3 ; then x — h (~ 3 —.1) and dx = % z 2 dz. Ex. 1. Find the value of I Therefore f — — — = - f (z' - 2 z 4 + z)dz J (1+ 2x)i SJ = 3 io- 2 (5~ 6 -16z 3 + 20) + whence sin z = — === > so that, by substitution, « Vx 3 + a 2 ./' dr (x 2 + rt'-)2 a 2 Va; 2 + a 2 If we try to find the value of f Vx 2 + a 2 dx by the substitution x = a tan z, we meet the integral a 2 C sec 3 zdz, which is not readily found. Accordingly for a better method see Ex. 6, § 119. V. Integrand involving Vx 2 — a 2 . Let x = a Ex. 7. Find the value of ix 3 Vx 2 — trdx. Let x = a sec z. Then dx = a sec z tan z r/z, and Vx 2 — u 2 = a tan z. Therefore J x 3 Vx 2 — a 2 dx = a 5 ( tan 2 z sec 4 zdz = a 5 I (tan 2 z + tan 4 z) sec 2 zdz = rt 5 (i tan 3 z + l tan 5 z) + C. x Va- 2 — a 2 But sec z = - , whence tan z = > so that, by substitution, we have a a fx 3 Vx 2 -a 2 dx = T V V(.r 2 -« 2 ) 3 (2 o 2 + 3r) + C. We might have written this integral in the form C x 2 V x 2 — a 2 (x dx) and let z 2 = x 2 — a 2 . VI. Integrand of the form , „ , = • Let S J J (Ax + B)Vax* + bx + c Ax + B = — z C dx Ex. 8. Find the value of I : + l 1 Then a; J (2x 2\z + iy y$x 2 dx = + 8a; 1 2 z' + 3 Let 2 a and V5 x 2 + 83 : + 3 f.v* + 6; : + 5. SUBSTITUTION 241 Therefore /• dx ___ r dz r dz J (2 x + 1) Vo a. 3 + 8 x + 3 J Vz 2 + t> z + 5 J V(z + 3) 2 - 4 = - logO + 3 + VY 2 + tiz + 5) + C. But 2 = > and hence 2x + l log (z + 3 + Vz 2 +6z + 5) = - log 6 x + I + 2 V.j -r- + 8 , t - + 3 2 x + 1 Or + 1 log * + - log 2. 3 a: + 2 +V5a; 2 + 8x + 3 Therefore / /* = leg 2J+1 + C, J (2 x- + 1) V5 x- 2 + 8 x + 3 3 z + 2 + \ 5 x- + 8 x + 3 log 2 having been made a part of the constant of integration. 118. The evaluation of the definite integral I f(x)dx may *£«' be performed in two ways, if the value of the indefinite integral is found by substitution. One method is to find the indefinite integral as in the pre- vious article and then substitute the limits. Ex . 1. Find fVa 2 — x« («) dz = <» = F(x), where F(x) is obtained by replacing z in (2) by its value in terms of .r. Then But F(h)-F(a)= C f(x)dx, and *W-^(«o)= f^OOcb. Hence f /(.r) cfo = f " '<£ (.2) efe. Applying this method to the example just solved, we have by Ex. 5, § 117, I v« s — x' 2 dx = r/ 2 I cos 2 ~<7~ = $a?(z + £ sin 2 z) + C, where a; = a sin z. When x = 0, z = 0, and when x — a, z — —, so that z as a; varies from to a. Therefore ( Va' 2 — x 2 dx — a 2 f' 2 cos 2 zdz Jo Jo -[*"(■ + t-«')I _ 7Trt 2 In making the substitution care should be taken that to each value of x between a and b corresponds one and only one value of z between z Q and z % , and conversely. Failure to do this may lead to error. Ex. 2. Consider j 2 cos <£ dcf>, which by direct integration is equal to =F dx Let us place cos (f> = x, whence = cos -1 a; and dd> = — where Vl - x 2 the sign depends upon the quadrant in which is found. We cannot, INTEGRATION BY PARTS 243 therefore, make this substitution in J 2 cos <£ d, since lies in two differ- ent quadrants ; but we may write ~ 2 f 2 cos d = f cos d + J * cos d, and in the first of the integrals on the right-hand side of this equation place = , and in the second d> = cos -1 x, dd> = — • Then V1-- VI -x- f^cos * a* = r-^= - r°_^^ = o r 1 _^ = = 2 . J_ Z ^ J Vl - X 2 J ! Vl - X 2 J Vl - ., - 2 119. Integration by parts. Another method of importance in the reduction of a given integral to a known type is thai of integration hy parte, the formula for which is derived from the formula for the differential of a product, c?(w) = udv + vdu. From this formula we derive directly that uv I udv + I vdu, which is usually written in the form / udv = uv — I vdu. In the use of this formula the aim is evidently to make the original integration depend upon the evaluation of a simpler integral. Ex. 1. Find the value of f xe*dx. If we let x = u and tFdx = ' dx = .rr>- - & + ( ' = ( X _l )fi x + C. It is evident that in selecting the expression for dr it is desirable, if possible, to choose an expression that is easily integrated. 244 INTEGRATION Ex. 2. Find the value of J sin -1 x^x. dx Here we may let sin -1 a: = u and dx — dv, whence du = ' and v = x. Substituting in our formula, we have VI — a; /C x c sm~ 1 xdx = x sin _1 x — I — = xdx VI - x 2 = a: sin- 1 .*: +Vl- x 2 + C, the last integral being evaluated by (1), § 116. Ex. 3. Find the value of / xcosPxdx. Since cos 2 x = i(l + cos 2 x), we have /l r x 2 1 r x cos 2 xdx = - J (x + x cos 2x)dx = — + - J x cos 2 xdx. Letting x = u and cos 2 xdx = dv, we have du = dx and v — 1 sin 2 x. Therefore / x cos 2 xdx — — sin 2 x — - j sin 2 xdx = - sin 2 x + - cos 2x + C. 2 4 /x 2 1 /x 1 \ x cos 2 xf/x = 1 — ( - sin 2 x + - cos 2 x)+ C 4 2 \2 4 / = \ (2 x 2 + 2 x sin 2 x + cos 2 x) + C. Sometimes an integral may be evaluated by successive inte- gration by parts. Ex. 4. Find the value of / x 2 e x dx. Here we will let x 2 = u and e x dx = dv. Then du = 2 x dx and v = e x . Therefore fx 2 e x dx = x 2 e x — 2 Cxe x dx. The integral / xe x dx may be evaluated by integration by parts (see Ex. 1), so that finally Cx 2 e x dx = x 2 e x - 2 (x - 1) e x + C = e x (x 2 - 2 x + 2) + C. Ex. 5. Find the value of J e 000 sin bx dx. Letting sin bx = u and e°~ x dx — dv, we have /e ax sin bxdx = - e ax sin bx — - I e ax cos &x rZx. a a J INTEGRATION BY PARTS 245 In the integral I e"* cos bxdx we let cos bx = u and e ax dx = dv, and have /e ax cos bxdx — — e"* cos 6x -\ — | e aj " sin bx dx. Substituting this value above, we have /e™ sin bxdx = -e ax sin bx (- e"* cos bx + - I e ai sin bxdx). a a \a a J / Now bringing to the left-hand member of the equation all the terms containing the integral, we have (1 + — ) I e ax sin bx dx = - e** sin bx e" cos bx, a- J J a «'- whence /• . , e ax (a sin bx — h C< is bx ) e^ sin bx dx = * : ' a 2 + U 1 x dx J Ex. 6. Find the value of f Vx 2 + a?dx. Placing Vx 2 + a 2 = u and dx = dr, whence du = — = = and v = x, we have A ■'" + "" fVx* + a*dx = x Vx* + a* - f xVx . (1 ) J J Vx 2 + « 2 Since x 2 = (x 2 + a 2 ) — a 2 , the second Integra] of ( 1 ) may be written as r (x 2 + a 2 )dx _ „ r dx J Vx 2 + a 2 J Vx 2 + «'-' which equals / vr + crdx — a* j — • J J a ./- + a 2 Evaluating this last integral and substituting in (1), we have f Vx 2 + u 2 dx = x Vx 2 + a 2 - f Vx 2 + crdx + a 2 log (x + Vx 2 + a 2 ), whence fVx 2 + irdx = }, [x Vx 2 + a 2 + a 2 log (x + Vx 2 + a 2 )]. 120. If the value of the indefinite integral if(ai)dx is found by integration by parts, the value of the definite integral X/(V) dx may be found by substituting the limits a and b, in the usual manner, in the indefinite integral. 246 INTEGRATION Ex. Find the value of C 2 x 2 sin x dx. To find the value of the indefinite integral, let x 2 = u and sinxdx — dv. Then j x 2 sin x dx = — x 2 cos x + 2 I x cosxdx. In / x cos a; (for, let x = w and cos xdx = dv. Then I x cos a; Jx = x sin x — / sin x dx = x sin x + cos x. Finally, we have j x 2 sin xdx = — x 2 cos x + 2x sin a; + 2 cos a; + C. Hence f 2 x 2 sin x c/x = — x 2 cos x + 2 x sin x + 2 cos x = tt-2. The better method, however, is as follows : 6 lif(x)dx is denoted by udv, the definite integral j f(x)dx Xb J a udv, where it is understood that a and b are the values of the independent variable. Then I udv = [uv~\ — / vdu. To prove this, note that it follows at once from the equation /*>b s%b />6 s*b [uv~\ = j d(u,v*) = j (udv + vdu) = j udv + I wc?m. Applying this method to the problem just solved, we have C 2" x 2 sin x rfx = I — x 2 cos x + 2 C 2 x cos x rfx = 2 r 2 x cos x dx Jo 2 sin x rfx o PARTIAL FRACTIONS 247 121. Integration by partial fractions. A rational fraction is a fraction in which both the numerator and the denominator are polynomials- If the degree of the numerator is equal to, or greater than, the degree of the denominator, we may, by actual division, separate the fraction into an integral expression and a fraction in which the degree of the numerator is less than the degree of the denominator. For example, by actual division, 2* 5 -.r 4 + .r 3 +3./--3i;.r + :)(; n _ , r » + 3^_4 J . + 20 "•" ..4 1/; ' { ) x A - 10 z*-16 It is evident, then, that we need to study the integration of only those fractions in which the degree of the numerator is less than the degree of the denominator. If the denominator of such a fraction is of the first degree or the second degree, the integration may be performed by formulas (2), (14), (18), §116, as in Ex.3, § 114. If the denominator is of higher degree than the second, we can separate the fraction into partial fraction* the sum of which will equal the given fraction. For example, a 3 + 3 J .3-4a:4-20 _ 1 1 x-1 . r '_16 ~x-2 x+2 + a?+4' as the reader can easily verify. The three fractious on the right-hand side of (2) are the partial fractions of the fraction on the left-hand side of (2). It is to be noted that their denominators are the rational factors of the denominator of the fraction of the left-hand side of (2). Substituting in (1), we have 2 :r 5 _ . r -» + a *+ ;j ,r- 30 x + 30 a; 4 - 10 248 INTEGRATION r 2 x 5 - x 4 + x a + 3 x 2 - 30 x + 36 , Hence J - 4 — 3— ax dx ^_ :r+log(r _2 ) _i og(:r+ 2) + ilog(^+4)-itan- 1 ^ , . (.r-2)V* 2 +4 1, _ x x = x 2 -x + \og^ -^ 2 tM 2' The separation of a fraction into partial fractions, as in (2), is evidently a great aid in integration. We shall illustrate this process in the following examples : Ex. 1. hnd the value of I — dx. J (x + 3) (x 2 — 4) The factors of the denominator are x + 3, x — 2, and x + 2. We assume x 2 + 11 x + 14 _ ^ £ C a (x + 3) (x 2 - 4) x + 3 x - 2 x + 2 ' ; where A , B, and C are constants to be determined. Clearing (1) of fractions by multiplying by (x + 3) (x 2 — 4), we have x 2 +llx + 14 = .4(x-2)(x + 2)+i3(x+3)(x + 2) + C , (x + 3)(x-2), (2) or x 2 +11 x + 14 = (A+B + C)z a + (55 + C)x + (-4.4 +6 B - 6 C). (3) Since .4, B, and C are to be determined so that the right-hand member of (3) shall be identical with the left-hand member, the coefficients of like powers of x on the two sides of the equation must be equal. Therefore, equating the coefficients of like powers of x in (3), we obtain the equations . , d , ^_i 5B+ C = ll, - 4 A + 6 B - 6 C = 14, whence we find ,4 = - 2, B = 2, C = l. Substituting these values in (1), we have x 2 + 11 x + 14 2 2 , 1 (x + 3)(x 2 -4) x + 3 x-2 x + 2 . r x 2 + llx+14 , r2dx , r 2dx , r dx and i ( 3 ; + 3)(x 2 -4) ^ = "Jxn + i.^2 + JxT2 = -2 log(x + 3) + 2 log(x - 2) + log(x + 2) + C = (X + 2H.-2) 2 c> b (x + 3) 2 PARTIAL FRACTIONS 249 Ex.2. Find the value of f 4 ^ + x + l dx. J x 3 — 1 The real factors of x 3 — 1 are x — 1 and x 2 + x + 1. Hence we assume 4 x 2 + x + 1 = .4 Bx + C x 3 - 1 a; - 1 x 2 + x + 1 " Clearing of fractions, we have 4 x 2 + x + 1 = A (x 2 + x + 1) + (Bx + C) (x - 1) = (.4 + B) x 2 + (.4 - B + C)x + (-4 - C). (2) Equating coefficients of like powers of x in (2), we obtain the equations A + B = 4, A - B + C = 1, A-C = l, whence A - 2, B = 2, C = 1. „ 4x 2 + x + 4 2 2x + l lience = + /'' r-i/r ; (x + 2) 2 (x-2 B 8 - 1 x - 1. X 2 + X + 1 and f 4 * 2 + * + 1 ,/x=f^l + f(2* + l)rf* J x 3 - 1 J x - 1 J X" + X + 1 = 2 log (x - 1) + log (x 2 + x + 1) + C = log [(x - 1) 2 (X 2 + X + 1)] + C. •2 x*dx _ Here we assume (x + 2) 2 (x-2) (x + 2) 2 x + 2 x-2 w Clearing of fractions, we have 2 x 2 = .4 (x - 2) + B (x 2 - 4) + C(x + 2) 2 = (B + C)x* + (A + \C)x + (-2A -4B+4C). (2) Equating the coefficients of like powers of x in (2), we obtain the equations B + C — 2 A + 4 C = 0, - 2 /I - 4 B + 4 C = 0, whence y4 =- 2, 5 = f, C = .1. 2x 2 = 2 g & ^ (x + 2) 2 (x - 2) (x + 2) 2 x + 2 x - 2 ' J (x + 2) 2 (x - 2) J (x + 2) 2 J x + 2 J x - 2 = x~T^ + I l0S ( '' + 2) + ^ l0g <* " 2) + ° — =— + log V(x + 2) 3 (x - 2) + C. x + J 250 INTEGRATION Ex.4. Find the value of f ^ X + + ( * ~ ^ dx. Now (x + 1) (x 8 + 1) = (x + l) a (ar 2 — x + 1), and we assume 3x 3 + 3x-6 J . /; | Cx+ I> (x + 1) (a; 8 + 1) (x + If x + 1 x 2 - x + 1 Clearing (1) of fractions, we have 3 x 3 + 3 x - 6 = A (,i- 2 - x + 1) + B (x 8 + 1) + (C x + D) (x + l) 2 = (B+ C)x s + (A + 2 C + Z>)x 2 + (- .1 + C+ 2D)x + (A + B + D). (2) Equating coefficients of like powers of x in (2), we obtain the equations B + C = 3, A + 2C + D = 0, -A+C + 2D = 3, A + B + D = - 6, whence .4 = - 4, 5 = 0, C = 3, 7) = - 2. Substituting these values in (1), we have 3 x s + 3 x - 6 - 4 3 x - 2 (x + 1) (x 3 + 1) (x + 4) 2 x 2 - x + 1 f 3.*3 + 3x- one fraction of the form — 5 is assumed. ax -f- ox + c 5. The numerators assumed are determined and the integration of the partial fractions is completed. If (ax + by in 3 is replaced by (ax + 6) B , and the correspond- ing n fractions are assumed to be A + * +...+-*-, (ax + of 1 (ax -\- by 1 ax + b and if ax 2 + bx + c in 4 is replaced by (ax 2 + bx + c) n , the cor- responding n fractions assumed being Ax + B Cx + D Px + Q (ax 2 + bx + c) n (ax 2 + bx + c) n _1 " + OX + CJ than unity. Since d (ax 2 + bx + (?) = (2 ax + i) dx, we may, as in Ex. 3, §114, let Ax + B = -£-(2ax + b') + B- — , and obtain the equation C Ax+B dx=J ± rd(ax*+bx+c) / _ Ab\ C dx _ J (ax 2 +bx+cj 3 -2aJ (-iyr[( w 2 +« 2 )»- 1 v J J (w 2 +a 2 ) B_1 J This is a special case of (4), § 122. 252 INTEGRATION 122. Reduction formulas. The methods of integration derived in this chapter are sufficient for the solution of most of the problems which occur in practice. If the reader should meet any integrals which cannot be evaluated by these methods, he should refer to a table of integrals, in which the integrals have been either completely evaluated or expressed in terms of simpler integrals. Some of this latter type of integrals, known as reduc- tion formulas, have been tabulated below for convenience. = * -fr r zrrr I % n (a + bx n ydx, (1) (np + m+V)b (np + m+l)bj J v J ^far(a + bx»y-idx, (2) Cx m (a + bx n ydx _ x m + 1 (a + bx n ] np + m + 1 np + ni C x m (a + bx n y p dx ^Xa + tey«_(np + n + m+V)b f^ + . (fl+tey ^ (3) (m+V)a (m + l)a J K T J ' ^ J fx m (a + bx' l ) p dx = _ary+bz»y~ y + n+m+l C^ a+ ^y+i dx . (4) n(jt?+l)a n(p+l)a J K J K J j sin m a; cos" xdx /' sin" 1 # cos"x^» /sin m ^ cos n_2 a: cfe, (5) + w 4- 2 /* — / sin m :zcos" + 2 2;efo, (6) n+1 J K J x m 71 + 1 j sin m :z cos n xdx -\ — j sin m ~ 2 xcos n xdx, (7) m + n J sin m_1 :r cos" +1 f< m + n Sin"" a; cos" xdx sin m+1 a; cos" +1 x m m + 1 m + n + 2 r . ; — I S11 l xcos n xdx. (8) PROBLEMS 253 These formulas do not always hold. For example, (1) and (2) fail if np + m + 1 = 0, (3) fails if m + 1=0, etc. In these cases, however, it is not necessary to use these formulas, as the integration may be performed by elementary methods. There are also integrals which cannot be expressed in terms of the elementary functions. For example, / — J V(l-z*)(l-*V) cannot be so expressed; and, in fact, this integral defines a function of x of an entirely new kind. PROBLEMS Find the values of the following integrals : Jl. f (4 x 3 + 3r + 4x-3) dx. 12. f- x + e _" x dx. 2. f(a* - x* + i - i)«fa 13. f(2 + Sx)*dx. J \ '* . J x " J Vl+3a;+a: 8 7. / V2 + e^tfe. 18. / ^t— jr-ete. J J <- c — cos 2x Pe**dx C dx 'Je** + 2' 19 J (x -a) [log (x -«)]«' 9. / — r- 20. / -^<£c. J xlogx 2 J X 10 -/(i + ^.-/ n-fp + f)*.*.- n. r 7 i±^ 5 esc x + ctn x J •' + J J 1 — COS X J • » •' + - /ci.s.i -f/x 83 ^ r x — 2 1 + COS X J Vl — X 2 2^' 7 84 - I -7== d». sec aa J y,/ _ ,,■ /. f sinxc/x Vl-cosxr7x. 85. j 1+cos 2 : J Vl+cos2x J V4x 70 il.r IS ■ 159. /%,-(.,. + Va- 2 + « 2 )^. 174. f^f^ 160. | a- sin 3 a; (fa. 175. f — — fsee->2xdx. 176. f^±< Ixsec-^xdx. 177. f * + 4 , + „".,.., / ^— cZa;. 202. I -t-= r-' J 1— x J sura cos^x 191. f *+^El dx. 203. f-'",- J aj-Vl-a; J sin 8 x f da f dx r* 204. -r-= j cos°a; ^y sura cos x I 7^ — T 205. i - Ja(a 8 + 2)* J« cus 5 ^; CHAPTER XIII APPLICATIONS OF INTEGRATION 123. Element of a definite integral. In §§78 and 81, by means of the area under a curve, we have defined the definite integral by the equation " /(*) dx = Lim 5)/(a0 Ax, (1) £ and have shown that this limit may be evaluated by the formula h f(x)dx==F(h)-F(a), (2) I. where dF(x)=f(x)dx. Since any function f(x) may be graphically represented by the curve y —f(x), formulas (1) and (2) are perfectly general. We shall proceed to give certain applications. The general method of handling any one of the various problems proposed is to analyze it into the limit of the sum of an infinite number of terms of the form f(x) dx. The expression f(x) dx, as well as the concrete object it represents, is called the element of the sum. 124. In finding the element of integration, it is often not possible to express the terms of the sum (1), § 123, exactly as f(Xi) Ax, the more exact expression being [/(#<) + ej Ax, where the quantities e t are not fully determined but are known to approach zero as a limit as Ax approaches zero. It is consequently of the highest importance to show that Lim V e.Ax = 0, so that Lim 2 [/OO + eJAx=Umy / f(x i -)Ax= Cf(x)dx. ELEMENT OF A DEFINITE INTEGRAL 261 For that purpose, let 7 be a positive quantity which is equal to the largest numerical value of any e. in the sum. Then - 7 ^ e. = 7 and — 27A.T = Se.Ar = ^7A.r. But 27AX = 7lA.r = 7 (b — a) and Lim ~2yAx = since 7 approaches zero as A.r approaches zero. Hence Lim "Ze.Ax = 0. Hence the quantities e t which may appear in expressing the sum do not affect the value of the integral and may be omitted. Quantities such as Ax and e-, which approach zero as a limit, are called infinitesimal*. Terms such as f(x) A.r, which are formed by multiplying Ax by a finite quantity, not zero, arc called infini- tesimals of the same order as Ax. Quantities such as e Ax, which are the products of two infinitesimals approaching zero together, are called infinitesimals of higher order than either infinitesimal. The theorem above proved may be restated in the follow- ing way : In forming the element of integration infinitesimals of higher order than f(x)Ax may be disregarded. Ex. Consider the area under a curve (§78). We have obtained it, by means of rectangles, as i = „-i Lim V /(x,)Ax. (1) Suppose that in place of the rectangles we used the trapezoids formed by drawing the chords DP V l\l'.,, etc. (fig. 125). The area of one Buch trapezoid is -, . A , , A A 1 f(Xi)Ax + \ A//Ar. But I Ay is a quantity which approaches zero as a limit when Ax approaches zero, and may be denoted bye,-- Hence, it' we used the trape- zoids, we should have for the required area Lim"j? [/(*,)+ «*] Ax. (2) " ~ ■ « = We see then directly that in this example i'=n-l i = n-l Lim J) [/(*«) + */] Ar = Lim J f(x t ) Ax. 262 APPLICATIONS OF INTEGRATION 125. Area of a plane curve in Cartesian coordinates. This problem was used to obtain the definition of a definite inte- gral, with the result that the area bounded by the axis of .r, the straight lines x = a and x— b (a< 6), and a of the curve y =f(x~) which lies above the axis given by the definite integral f. ydx. portion of x is (i) It has also been noted that either of the boundary lines x= a or x = b may be replaced by a point in which the curve cuts OX. Here the element of integration ydx represents the area of a rectangle with the base dx and the altitude y. Similarly, the area bounded by the axis of y, the straight lines y = c and y — d(c < d), and a portion of the curve x =f(jf) lying to the right of the axis of y is given by the integral / d xdy, (2) where the element xdy repre- sents a rectangle with base x and altitude dy. Areas bounded in other ways than these are found by express- ing the required area as the sum or the difference of areas of the above type, or by writing a new form of the element as illustrated in Ex. 2. Ex. 1. Find the area of the ellipse — + '(- = 1. « 2 Ir It is evident from the symmetry of the curve (fig. 157) that one fourth of the required area is bounded by the axis of y, the axis of x, and the curve. Hence, if A is the total area of the ellipse, 4 = 4 2 b f y dx = 4 C - Vo 2 - x 2 Jo Jo a l H n — 2 , 2 • 1*1 — \x V a* — x* + a- sin - 1 - a L «J dx TTClb. AEEA 263 Ex. 2. Find the area bounded by the axis of x, the parabola y- = 4 px, and the straight line y + 2x — ip = (fig. 15S). The straight line and the parabola intersect at the point C ( p, '2 p), and the straight line intersects OX at B (2p, 0). The figure shows that the required area is the sum of two areas OCD and CBD. Hence, if A is the required area, r 1 px dx + f<>. The area may also be found by con- sidering it as the limit of the sum of such rectangles as are shown in fig. 159. The height of each of these rectangles is Ay, and its length is x„ — x v where x„ is taken from the equation of the straight line and x 1 from that of the parabola. The values of y range from y = at the base of the figure to y — 2p at the point C. Hence -[■»-?-i£]M* In the above examples we have replaced y in J ydx by its value f(x) taken from the equation of the curve. More generally, if the equation of the curve is in the parametric form, we replace both x and y by their values in terms of the independent parameter. This is a substitution of a new variable, as explained in § 118, and the limits must be correspondingly changed. Ex. 3. Let the equations of the ellipse be x = a cos <£, y = b sin . Then the area A of Ex. 1 may be computed as follows: J. = 4 | ydx— —A j ah sin 2 t! = 4 ah j ~ sin 2 <£ tlcf> = trab. Fig. 159 264 APPLICATIONS OF INTEGRATION 126. Infinite limits or integrand. If the curve extends in- definitely to the right hand, as in figs. 160-1G2, it is possible to consider the area bounded by the curve, the axis of as, and a fixed ordinate x = a, the figure being unbounded at the right hand. Such an area is expressed by the integral f(x) dx = Lim F(b) - F(a), which may be written concisely as jf f(x) dx = F(oo ) - F(d). There is no certainty that this area is either finite or deter- minate. Where it is so, the area bounded on the right by a movable ordinate approaches a defi- 7 nite limit as the ordinate recedes indefinitely from the origin. [2vTT = ^. (Fig. 160) Ex. 1 Va Ex. 2 Ex. I rs-Hr- 1 - I sin xdx = [— cos x^° — indeterminate. (Fig. 162) (Fig. 161) Similarly, the area may be unbounded at the left hand, and the lower limit or lM^ y/M^ both limits of the definite integral may be infinite. In like manner let f(x) become infinite at the upper limit, and the curve y =f(x) approach x = b as an asymptote. Then the area bounded by the curve, the axis of x, an ordinate x = a, and an ordinate near the asymptote x = b may approach a definite value as the latter ordinate approaches the asymptote. Such an area may be expressed by the integral Lim f f(x) dx = Lim F(b - A) - F(a), or, more concis 3el y> f /O) dx = F(b) - F(a). Ex. 4 J" a dx T . ,xl a = sin- 1 - o Va 2 — x 2 L "-1° MEAN VALUE .. (Fig. 163) 265 a - Fig. 1G4 Failure to do this (Fig. 164) Similarly, f(x) may become infi- nite at the lower limit or at both limits. If it becomes infinite for any value c between the limits, the integral should be separated into two integrals having c for the upper and the lower limit respectively may lead to error. X +l dx — ■ i ./■'- Since — becomes infinite when x = (fig. 165), w x' 1 separate the integral into two, thus: J-i x- J -\ x- Jo .i- Had we carelessly applied the incorrect formula xrs-t-as we should have been Led to the absurd result - 2. FlG - 165 127. The mean value of a function. In fig. 166 let the curve DPC be the graph of the function /(a;). Then r f(x) dx = area ADPCB, Let m = AX and M=AH be respectively the smallest and the largest value assumed by f(x) in the interval AB. Construct the rectangle A BKH with the base AB and the alti- tude AM— M. Its area is AB • AH=(b - a)M. Construct also the rectangle ABLN with the base AB and the altitude AN— m. Its area is AB • AN= (6 — a)m. 260 APPLICATIONS OF INTEGRATION Now it is evident that the area A BCD is greater than the area ABLN and less than the area ABKH. That is, (b - a) m < f f(x) dx < (b - a) M* Consequently I f(x) dx = (b — a) /x, U a where fx, is some quantity greater than m and less than M, and is represented on fig. 166 by AS. But since f(x) is a continuous function, there is at least one value £ between a and b such that /(|) = fr anf l therefore f. f(x)dx = (b-d)f(g). (1) Graphically, this says that the area ABCD is equal to a rec- tangle ABTS whose base is AB and whose altitude AS lies between AN and AH. From (1) we have /(D=^J^/(*)^ (2) where f lies between a and b. The value l r h a?we of /(#) in the interval from a to b. This is, in fact, an extension of the ordinary meaning of the average, or mean, value of n measurements. For let y , y v y. 2 , • • ■, y n _ x correspond to n values of x, which divide the interval from a to b into n equal parts, each equal to Ax. Then the average of these n values of y is yo + ^+^4- • •• + ?/„_! n This fraction is equal to Oo + ffi+yaH +y n -,)Ax = y^x+y.Ax-hy^Ax-l + y n _,Ax nAx b — a *A slight modification is here necessary if f(x) = fc, a constant. Then f(x) dx = (b — a) k. AREA 267 As n is indefinitely increased, this expression approaches as a 1 r h 1 r b limit / ydx = I f(x) dx. Hence the mean value t>- a Ja h - a Ja of a function may be considered as the average of an " infinite number " of values of the function, taken at equal distances between a and b. Ex. 1. Find the mean velocity of a body falling from rest during the time t v The velocity is gt, where g is the acceleration due to gravity. Hence the 1 rh mean velocity is I gtdt — \gt v This is half the final velocity. Ex. 2. Find the mean velocity of a body falling from rest through a distance .s' r The velocity is V'2 gs. Hence the mean velocity is — f o h V2glds = %V2^ r This is two thirds the final velocity. 128. Area of a plane curve in polar coordinates. Let (fig. 167) be the pole, OM the initial line <>f a system of polar coordinates (r, 0), OA and OU two fixed radius vectors for which 0= a and B = fi respectively, and All any curve for which the equation is r =/(#). Required the area AOH. The required area may be divided into n smaller areas by dividing the angle AOB—^ — a into n equal parts, 6 — a each of which equals = A#, n and drawing the lines 0P X , OP^ 0P 3 , • • -, OP n _ v where AOP 1 = %OP i =P 2 OP 3 = • • •=P n _ 1 05=A^ (In the figure w = 8.) The required area is the sum of the areas of these elementary areas for all values of n. The areas of these small figures may be found approximately by describing from as a center the circular arcs AL\, 2£B a , P 2 P^ • • •, ^_i-K„« Let OA = r , OP, = r v OP = r,, • • • , 0P_ X = r n _ v 268 APPLICATIONS OF INTEGEATION Then, by geometry, the area of the sector A 0/>\ = } 2 r'^AO, the area of the sector J^OB 2 = ^ r'^AO, the area of the sector B_ 1 OB n = -* rl_ x A6. The sum of these areas, namely is an approximation to the required area, and the limit of this sum as n is indefinitely increased is the required area. Hence the area A OB -*/V* The above result is unchanged if the point A coincides with 0, but in that case OA must be tangent to the curve. So also B may coincide with 0. Ex. Find the area of one loop of the curve r = a sin 3 6 (fig. 101, § 60). As the loop is contained between the two tangents $ = and $ = — > the required area A is given by the equation 1 Jo 2 Jo cos .6 8 a ira 2 "2 d6= U\ 129. Volume of a solid with parallel bases. Fig. 168 repre- sents a solid with parallel bases. The straight line Off is drawn perpendicular to the bases, cutting the lower base at A, where h = a, and the upper base at B, where h = b. Let the line AB be divided into n parts each equal to =Ah, and let n planes be passed through each point of division parallel to the bases of the solid. Let A o be the area of the lower base of the solid, A 1 the area of the first section parallel to the base, A 2 the area of the second section, and so on, A n _ x being the area of the section next below the upper base. Then A Q Ah represents the volume of a cylinder with base equal to A o and altitude equal to Ah, A^Ah represents the volume of a cylinder VOLT. ME 269 standing on the next section as a base and extending to the section next above, and so forth. It is clear that A Ah + A^h + J 2 A7i + ... + A n _ x Mi = j?A t Ah is an approximation to the volume of the solid, and that the limit of this sum as n indefinitely increases is the volume of the solid. That is, the required volume /' is V= C Adh. To find the value of this integral it is necessary to express .1 in terms of h, or both A and // in terms of some other independent variable. This is a problem of geometry which must be solved for each solid. It is clear that the previous discussion is valid if the upper base reduces to a point, i.e. if the solid simply touches a plane parallel to its base. Similarly, both bases may reduce to points. 108 Ex. 1. Two ellipses with equal major axes arc placed with their equal •pendicular. A variable ellipse move axes coinciding and their planes p so that the ends of its axes are on the two given ellipses, the plane of the moving ellipse being per- pendicular to those of the given ellipses. Required the volume of the solid generated. Let the given ellipses be ABA'W (fig. 169) with semiaxes OA =a and OB = b, ami ACA'C with semiaxes OA =a and OC=C, and let the common axis be OX. Let NMN'M' be one position of the moving ellipse with the center P where OP — .r. Then ii A is the area of NMN'M', n „ „ ,„ A = ir • PM ■ PN. (By Ex. 1, § 125) 169 270 APPLICATIONS OF INTEGRATION But from the ellipse ABA'B' - n + ^f- = 1 a 2 lr and from the ellipse ACA'C - + J —^- = 1 be PM -PN = -(a 2 - x 2 ). a 2 Therefore Consequently the required volume is — — (a 2 — x 2 ) dx = - Trdbc. ■ a a- o The solid is called an ellipsoid (§ 143, Ex. 5). Ex. 2. The axes of two equal right circular cylinders intersect at right angles. Required the volume common to the cylinders. Let OA and OB (fig. 170) be the axes of the cylinders, OY their common perpendicular at their point of intersection 0, and a the radius of the base of each cylinder. Then the figure represents one eighth of the required volume V. A plane passed perpendicular to OF at a dis- tance ON = y from intersects the solid in a square, of which one side is NP = -VOP 2 - ON 2 = Va 2 - y 2 . Therefore \ V= FnP* dy = f* (a 2 - y 8 ) dy = § a 8 Jo t/0 and F=V l « 3 - 130. Volume of a solid of revolution. A solid of revolution is a solid generated by the revolution of a plane figure about an axis in its plane. In such a solid a section made by a plane perpen- dicular to the axis is a circle, or is bounded by two or more concentric circles. Therefore the method of the previous article can usually be applied to find the volume of the solid. No new formulas are necessary. The following examples illustrate the method. Ex. 1. Find the volume of the solid generated by revolving about OX the figure bounded by the parabola y 2 — \px, the axis of x, and the line x = a. VOLUME 271 The area to be revolved is shaded in fig. 171. Let P(x, y) be a point on the parabola. Then any section of the solid through P perpendicular to OX is a circle with radius MP = y. Hence in the formula of § 129 we have A = try 2 and dh — dx. Hence the required volume V is ■fjirfdx. But from the equation of the pa- rabola y- — 4 par. Therefore '•= 4 WT dx ipmr Ex. 2. Find the volume generated by revolving around the line x the figure described in Ex. 1. If P (fig. 172) is a point on the curve, a section of the required solid through P and perpendicular to All is a circle with radius PN = u — x. Hence in the general formula of § 129 A = ir (n — x)- and dh — dy. When x = a, y = 2-Vpa. Hence the volume V is given by V = f*^~ a Tr(a - xfdy = Trf 2y/Pl '(« 2 - 2 ax + But from the equation of the parabola x = - — Hence Jo \ 2/> li)ji-/ Id Ex. 3. Find the volume of the ring solid generated by revolving a circle of radius a about an axis in its plane b units from the center (ft > a). Take the axis of revolution as OY (fig. 173) and a line through the center as OX. Then the equation of the circle is (x - ft) 2 + y* = a' 2 . A line parallel to OX meets the circle !->A>a Fm. 172 16 4 J in two points, A where x = x 1 = ft — Va 2 — y 2 and B where x = x 2 = ft + Vo 2 — y 2 . A sec- tion of the required solid taken through AB perpendicular to OY is bounded by two concentric circles with radii x t and x 2 respectively. Hence in § 129 A = TTxf — 7TJ' 2 , and dh = dy. The summation extends from the point L where y —— a to the point A' where y = + a. Hence, for the volume V, T = ^ J_ (''-" _ ' ri ") dy = 4 irh i_ , x _ ll ~ dy -a-h. APPLICATIONS OP INTEGRATION 131. Length of a plane curve. To find the length of any curve AB (fig. 174), assume n — 1 points, //, 7f, • • •, %_ v be- tween A and /.' and connect each pair of consecutive points by a straight line. The length of AB is then defined as the limit of the sum of the lengths of the n chords AP X , I^B,, P,P Z , • • ., P l _ 1 B as n is increased without limit and the length of each chord approaches zero as a limit. By means of this definition we have already shown (§§ 91 and 104) that Fig. 174 d8 = Vda?+dtf in Cartesian coordinates, and in polar coordinates. Hence we have and ds = ^dr+r\W> 8=fV,h*+d!f 8= fy/d)*+i*dff\ CO (2) (3) (4) To evaluate either (3) or (4) we must express one of the variables involved in terms of the other, or both in terms of a third. The limits of integration may then be determined. It may be noticed that (4) can be obtained from (3). For we have /. • a x — r cos 6, y = r sm 0. Then dx = cos 6 dr — r sin 6 dQ, dy = sin 6 dr + r cos 6 dd, and dx 2 + dy 1 = dr* + r\W\ Ex. 1. Find the length of the parabola if = ^ P x from the vertex to the point (h, I). From the equation of the parabola we find 2 ydy = ipdx. Hence formula (3) becomes either t/U \ ±p- 1 p J a LENGTH 273 Either integral leads to the result * = — Vi 2 + 4p a + p log !—■ ip 2p Ex. 2. Find the length of the epicycloid from cusp to cusp. The equations of the epicycloid are (§ 57) x = (a + 1>) cos — a cos , y = (" + '0 sin — a sin — (j>. Ileuce Then dx = \- (a + 6) sin <£ + (« + &) sin ^— ■ <^>1 ,/<£, rf y = ["(„ + 5) cos <£-(« + &) cos ^-±-^lrf<£. /,• = (a + 6) y-J - 2 /sin sin i±i <£ + c< is <£ c< >s 5!±* ^ ,/<£ = (« + 6 ) \/ 2 - 2 cos - = — (a + &). 132. The work of the previous article may be brought into connection with §124 as follows: kT7%y Since y_(^ + c^^_ N ^A,/ A, V (/./" + (/// >HBr | 1+ /%y ,. T . V(A.»-)'-+(A//) a .. N VA./7 A./- 1 then Lim v 7 ^=^- = Lnn — - Lim — == 1. +m v ( /.r + J/z- and V(A.r) 2 + (A t y) 2 = Vj.r + ,ty 2 + e V^ + df- By §124 the term eV,lr + thf will not affect the limit of 2V(A,-) 2 + (A^. ^ 7 Ps 7p 4 1 8 Fig. 175 274 APPLICATIONS OF INTEGRATION 133. Area of a surface of revolution. A surface of revolution is a surface generated by the revolution of a plane curve around an axis in its plane (§ 130). Let the curve AB (fig. 175) revolve about OH as an axis. To find the area of the surface generated, assume n — 1 points, //, P v P 6 , • • ., i£_u between A and B and connect each pair of consecutive points by a straight x 6 line. These lines are omitted in the figure -^ since they are so nearly coincident with the iV 4 arcs. The surface generated by AB is then n 2 defined as the limit of the sum of the areas ^' No of the surfaces generated by the n chords AP V B^P,, i£j£, • • ., P n _ x B as n increases without limit and the length of each chord approaches zero as a limit. Each chord generates the lateral surface of a frustum of a right circular cone, the area of which may be found by elementary geometry. Draw the lines AJSf , I{N V B^N^ • • • perpendicular to OH, and P lace N A = r D , N X P X = r v K 2 P 2 = r if . • -, A^J^ = r a . Then the frustum of the cone generated by PP t + x has for the radius of the upper base 2f i+1 J^ +v for the radius of the lower base N t P, and for its slant height I*P i+v Its lateral area is therefore equal to (N.P+N. P ^ Therefore the lateral area of the frustum of the cone equals This is an infinitesimal which differs from 2 wr$s by an infinitesimal of higher order, and therefore the area generated by AB is the limit of the sum of an infinite number of these terms. Hence, if we represent the required area by S, we have r S=2tt rds. WORK 275 To evaluate the integral it is necessary to express r and ds in terms of the same variable and supply the limits of integration. Ex. Find the area of the surface of revolution described in Ex. 1, § 130. Here r = y and ds = Vdx 2 + dy 2 , where x and y satisfy the equation y" = ipx. Consequently we may place r = 2 Vpx, and, as in Ex. 1, § 131, ih _ /* + P dj . Then S = 4 tt VJ, f" Vx + p dx = §wVp[(a + p)* -/<-]• 134. Work. By definition, the work done in moving a body against a constant force is equal to the force multiplied by the distance through which the body is moved. Suppose qow that a body is moved along OX (fig. 17G) from A (x — a) to B (x = /<) against a force which is not - , . , t , . « , y constant but a function of a; A 5S * M * * * U and expressed by/(:r). Let the Fl °" 176 line AB be divided into n equal intervals, each equal to Ax, by the points M^ M % , M z , . . •, M n _ x . (In fig. 176, n = 7.) Then the work done in moving the body f rom A to M i would be f (a*) Ax if the force were constantly equal to /(«) through- out the interval AM. Consequently, if the interval is small, /(rt)Ax is approximately equal to the work done between A and M x . Similarly, the work done between M l and M a is approxi- mately equal to/(^) A.r, that between JA, and 3f g approximately equal to/(:r,)A:r, and so on. Hence the work done between A and B is approximately equal to f(a)Lx+f(xdLx+f(x^*x+. ..+f(z n _JAz. The larger the value of n, the better is this approximation. Hence we have, if W represents the work done between A and /.', \Y - Lim£ /(.•/) A.r = Cf(x)dx. 135. Pressure. Consider a plane surface of area A immersed in a liquid at a uniform depth of h units below the surface. The submerged surface supports a column of liquid of volume hA, the weight of which is whA, where w (a constant for a given liquid) is the weight of a unit volume of the liquid. 270 APPLICATIONS OP INTEGRATION This weight is the total pressure on the immersed surface. The pressure per unit of area is then wh, which is defined as the pressure at a point h units below the surface. By the laws of hydrostatics this pressure is exerted equally in all directions. We may accordingly determine, in the following manner, the pressure on plane surfaces which are perpendicular to the surface of the liquid: Let BRQ (fig. 177) be a plane surface so immersed that its plane is perpendicular to the surface of the liquid and inter- sects that surface in the line TS. Divide BRQ into strips by drawing lines parallel to TS. Let the depth of a line of the first strip be h Q , that of the second strip ^^M^^ be h^ that of the third strip be 7;.,, / =\ and so on. Call the area of the first strip (A^) , that of the second / \ p strip (A.i)^ that of the third strip FlG 177 (A.4) 2 , and so on. Then the pres- sure on the first strip is approximately wh (AA) , that on the second strip is approximately u'h^AA)^ that on the third strip is approximately w\(AA)^ etc. Therefore the total pressure on BRQ is approximately w [K(AA) + hiAA)^ . • . + h B _ 1 (^AAX-t'] = u>X ^(A^),.. This approximation is better the greater the number of strips, since we have taken the whole strip as lying at the level of the same line. Therefore the total pressure P is the limit of the above sum as n = oo ; that is, "="'/ lid A. To evaluate the integral it is necessary to express h and A in terms of the same variable and supply the limits. In finding dA the strips may be taken as rectangles, as in finding the area. Ex. A parabolic segment with base 2 h and altitude a is submerged so that its base is in the surface of the liquid and its axis vertical. Let I?QC (fig. 178) be the parabolic segment and let CB be drawn through the vertex of the segment perpendicular to TS. According to CENTER OF PRESSURE 277 the data RQ = 2b, CB = a. Draw a horizontal strip LXA\L V with its bottom line cutting CB at .1/. Let CM = x ; then the depth h of the line LN is a — x and the breadth MM X of the strip is dx. R B Q * dx. Consequently dA = (LN)dx. But, from § 45, LN 2 CM. RQ 2 CB ' whence a and therefore a 2 Therefore, since x = at ( ', ; given by P = u — (a X) Fig. 178 = a at B, the total pressure P .,-ilx = — irhir. 15 136. Center of pressure. From mechanics we take the fol- lowing principles: 1. The resultant of a set of parallel forces is equal to the sum of the forces. 2. The moment of a force about a line al righl angles to the line of action of the force is defined as the product of the force and the shortest distance between the two lines. 3. The moment about a line of the resultant of a number of forces is equal to the sum of the moments of the forces. Now in the pressure problem of § 135, the pressure on each one of the elementary strips is a force approximately equal to whAA acting at right angles to the area. By the second principle stated above, the moment of this force about TS is h(tchAA), and the limit of the sum of the moments of all the forces is fh(whdA) = w fir J A By the first principle stated above, the resultant of the pres- sures on all the rectangles is the total pressure P. If this acts at a distance h below the surface of the liquid, we have, by the third principle, _ hP = w JcJA, from which h can be found. AC 278 APPLICATIONS OF INTEGRATION The point at which P acts is called the center of pressure. The formula above gives the depth of the center of pressure- Ex. Find the depth of the center of pressure of the parabolic segment of the example in § 135. From the discussion just given, 2 b . v „ i , 32mj6o 8 — (a — x) 2 x^c?a Ph F But P = T 8 5 wba? (Ex., § 135). Therefore h = \ a. By symmetry the center of pressure lies in CB, and is therefore fully fixed. 137. Center of gravity. Consider n particles of masses m x , m 2 , m s , ••-, m n , placed at the points P[(x v ^), i£0 2 , # 2 ), 50«v y.). •-. ^fe y.) (%• 179 ) re - spectively. The weights of these particles form a system of parallel forces equal to m-yff, m 2 g, m 3 g, • . ., m n g, where g is the acceleration due to gravity. The principles of mechanics stated in § 136 are therefore applicable. The resultant of these forces is the total weight W of the n particles, where i—n W = m x g -f m„g + m z g -f- . . . + m n g = 9^ m>i- This resultant acts in a line which is determined by the con- dition that the moment of W about any line through is equal to the sum of the moments of the n weights. Suppose first the figure placed so that gravity acts parallel to OY, and that the line of action of W cuts OX in a point the abscissa of which is x. Then the moment of W about a line through perpendicular to the plane XOY is gx^m p and the moment of one of the n weights is gmfr. Hence gx V m i = g V m^. Similarly, if gravity acts parallel to OX, the line of action of the resultant cuts OF in a point the ordinate of which is y, where _^ ^ CENTER OF GRAVITY 279 These two lines of action intersect in the point G, the coordi- nates of which are __< ^ * = %— ' 3/ = %— C 1 ) Z, m i Zf ^ Furthermore, if gravity acts hi the XO Y plane, but not paral- lel to either OX or Y, the line of action of its resultant always passes through G. This may be shown by resolving the weight of each particle into two components parallel to OX and OY respec- tively, finding the resultant of each set of components in the manner just shown, and then combining these two resultants. If gravity acts in a direction not in the XO V plane, it may still be shown that its resultant acts through G, but the proof requires a knowledge of space geometry not yet given in this course. The point G is called the center of gravity of the » 'particles. If it is desired to find the center of gravity of a physical body, the solution of the problem is as follows: The body in question is divided into n elementary portions such that the weight of each may be considered as concentrated at a point within it. If m is the total mass of the body, the mass of each element may be represented by Am. Then if (>. //, ) arc the coordinates of the point at which the mass of the ith element is concentrated, the center of gravity of the body is given by the equations V.r.Am V if Am * = Lim^- — , y=Um^r — ; Z/Am 2j Am I xdm | ydm whence * = ~~r — "' ^ = ~7 ' ® I dm I dm To evaluate, the integrals must be expressed in terms of a single variable and the limits supplied. It is to be noticed that it is not necessary, nor indeed always possible, to determine x 9 y { exactly, since, by § 124, Lim V (x t -f e.) Am = Lim Va;.Am, »=« « = i »< = * ,- = i if e, approaches zero as Am approaches zero. 280 APPLICATIONS OF INTEGRATION Ex. 1. Find the center of gravity of a quarter circumference of the circle at 2 + ?/ 2 = a 2 , which lies in the first quadrant. Let the quarter circumference be divided into elements of arc ds (fig. 180) ; then, if p is the amount of mass per unit length, dm = p ds. The mass of each element may be considered concentrated at a point (x, y) of the curve. Hence Jpxds . fpjjds y fpds f f ds If p is assumed constant, it may be removed from under the integral signs and canceled. The denominator of each fraction is then equal to s, a quarter circumference. To compute the numerators, we have, from the equation of the curve, ds = Vdx 2 + df = - dx y dy, where s is assumed as measured from A so that dx is positive and dy negative. Therefore and Hence f xds = — f a dy — a 2 , | ij ds = f a dx = « 2 , / ds = — , a quarter circumference. _ _ 2 a x = y = Ex. 2. Find the center of gravity of a quarter circumference of a circle when the amount of matter in a unit of length is proportional to the length of the arc measured from one extremity. As in Ex. 1, dm = pds, but here p = ks, k being a constant. Then dm = ksds. The integration is best performed by use of the parametric equations of the circle (§ 53). Then /sxds ( 2 a 3 (j> cos , Jo (* t — o)a f»* rh* d

J-y z )dy Fig. 183 J pirx 2 dy j ('r-if)dy 3 (a + c) 2 4 ' 2a + c ' Ex. 6. Find the center of gravity of the surface of the sjmerical seg- ment of Ex. 5. Divide the surface into elementary bands as in § 133. Then dm = 2 irpx ds, where p, the amount of mass per unit area, is assumed constant. This mass may be considered concentrated at (0, y). Hence, using the notation and the figure of Ex. 5, we have ds = — - , and therefore x _ r^ dS fj' d> -> a + c C xds J> ATTRACTION 283 138. Attraction. Two particles of matter of masses m 1 and m 2 respectively, separated by a distance r, attract each other with a force equal to k l , 2 > where k is a constant which de- r pends upon the units of force, distance, and mass. We shall assume that the units are so chosen that £=1. Consider now n particles of masses m^ m 2 , m g , • • •, m n lying in a plane at the points P v P, be required to find their attraction upon a particle of unit mass situated at a point A in their plane. Let the distances AP V AP, • • • , AP^ be denoted by r x , r 2 , • • •, r n . The attractions of the indi- vidual particles are P H (rig. 184). Let it Fig. 184 but these attractions cannot be added directly, since they are not parallel forces. To find their resultant we will resolve each into components along two perpendicular axes AX and AS respectively. If we denote the angle XAP t by #,, we have as the sum of the components along AX, X=^cos0 1 +^cos0 2 + r'r x r': + -\ cos e n r: and for the sum of the components along AY, Wl- sin H — - sin 0„ + 4- _? sin n . The resultant attraction is then R=Vx*+ Y 2 and acts in a direction which makes tan" J with AX. 281 APPLICATIONS OF INTEGRATION Let it now be required to find, the attraction of a material body of mass m upon a particle of unit mass situated at a point A. Let tlie body be divided into n elements, the mass of each of which may be represented by Am, and let Jf be a point at which the mass of one element may be considered as concen- trated. Then the attraction of this element on the particle at A is — - > where r t = I? A, and its component in the direction AX is — - cos 6 { , where 6 i is the angle XAI?. The whole body, there- fore, exerts upon the particle at A an attraction whose com- ponent in the direction AX is Ajos0 Lim 2^H A ™ dm. Similarly, the component in the direction AY is Ex. Find the attraction of a uniform wire of length I and mass m on a particle of unit mass situated in a straight line perpendicular to the wire at one end, and at a distance a from it. Let the wire OL (fig. 185) be placed in the axis of y with one end at the origin, and let the par- / tide of unit mass be at A on the // axis of x where A O = a. Divide OL //','' into n parts, 0M V M X M V M 2 M 8 , ■■■, ///'/'' M n _ i L, each equal to - = Ay. Then, if p is the mass per unit of length of the wire, the mass of each element is Am = pAy. We shall consider the mass of Fig. 185 -X each element as concen- trated at its first point, and shall in this way obtain an approximate expression for the attraction due to the element, this approximation being the better, the smaller A// is made. The attraction of the element M { M i + 1 on A is then approximately pAy _ pAy AM? « 2 + Vl where y. = 0M ( . PROBLEMS 285 The component of this attraction in the direction OX is pAy r\A-h/r p"A// ^ ' cos O A M. = — " - — - , H " + ^ («* + *,')* and the component in the direction OY is f** sin QAM. = P yAy ■ Then, if A' is the total component of the attraction parallel to OX, and Y the total component parallel to OY, we have X = Limy 1 P r,A// = pa f l r + f) 1 To evaluate the integrals for A' and )', place y = atan#. Then, if a = tan -1 - = OA /-, A' = - I co% Odd = - sina= .sin a, a Jo a id Y=~ f Bin 0i/6--(l- cos a) = -' (1 - cos a), since /p = ?n. If R is the magnitude of the resultant attraction and /i tin- angle which its line of action makes with OX, B = VP+P= sin - ). 25. Find the area inclosed by the four-cusped lrypocycloid x = a cos 3 0, y = a sin 3 0. PROBLEMS 287 26. Find the entire area bounded by the curve x = acosO, y — b sin 3 0. 27. Find the mean value of the lengths of the perpendiculars from a diameter of a semicircle to the circumference, assuming the perpendiculars to be drawn at equal distances on the diameter. 28. Find the mean length of the perpendiculars drawn from the circumference of a semicircle of radius a to its diameter, assuming that the points taken are equidistant on the circumference. 29. Find the mean value of the ordinates of the curve y = sin x between x = and x = ir, assuming that the points taken are equidistant on the axis of x. 30. A number n is divided into two parts in all possible ways. Find the mean value of their product. 31. If the initial velocity of a projectile is v , and the angle of elevation varies from to — > find the mean value of the range, using the result of problem 36, Chap. VII. 32. In a sphere of radius r a series of right circular cones is inscribed, the bases of which are perpendicular to a given diameter at equidistant points. Find the mean volume of these cones. 33. A particle describes a simple harmonic motion defined by the equation 8 = a sin kt. Show that the mean kinetic energy I - 1 during a complete vibration is half the maximum kinetic energy if the average is taken with respect to the time. 34. In the motion defined in problem 33, what will be the ratio of the mean kinetic energy during a complete vibration to the maximum kinetic energy, if the average is taken with respect to the space traversed ? 35. Find the area described in the first revolution by the radius vector of the spiral of Archimedes r = a 6. 36. Show that the area bounded by the hyperbolic spiral rO = a and two radius vectors is proportional to the difference of the lengths of the radius vectors. 37 . Find the total area of the lemniscate r 2 = 2 a 2 cos 2 6. 38. Find the area of a loop of the curve r = a sinnfl. 39. Find the area of a loop of the curve r 2 = a 2 sin nd. 288 APPLICATIONS OF INTEGRATION 40. Find the area swept over by the radius vector of the curve 7j- r = a tan 6 as changes from to — • 41. Find the total area of the cardioid r = a(l + cos 6). 42. Find the area of the limacon r = 2 cos + 3. 43. Find the area of the curved strip of the plane which has two portions of the initial line for two boundaries and the arc of the spiral r = ad between 6 = 2 ir and 6 = 6 ir for the other boundary. 44. Find the area of the loop of the curve r 2 = a 2 cos 2 6 cos 6 which is bisected by the initial line. 45. Find the area of a loop of the curve r 2 sin 6 = a 2 cos 2 6. 46. Find the area of the kite-shaped figure bounded by an arc of a parabola and two straight lines from the focus making the angles ± « with the axis of the parabola. 47. Find the area bounded by the curves r = a cos 3 6 and r = a. 4 48. Find the area inclosed by the curves r = and 4 J 1 - cos e 1 ~ i + cos e ' 49. Find the area cut off one loop of the lemniscate r 2 = 2 a 2 cos 2 6 by the circle r = a. 50. Find the area of the segment of the cardioid r = a(l + cos 9) cut off by a straight line perpendicular to the initial line at a distance | a from the vertex. 51. Find the area of the loop of the curve (x 2 + y 2 ) 3 = ka 2 x 2 y 2 . (Transform to polar coordinates.) 52. Find the total area of the curve (a; 2 + iff = 4 a?x* + 4 «y. (Transform to polar coordinates.) 53. Find the area of the loop of the Folium of Descartes, x z + if — 3 axy = 0, by the use of polar coordinates. x 2 if 54. On the double ordinate of the ellipse — : + 7^ = 1 as base an isosceles triangle is constructed with its altitude equal to the dis- tance of the ordinate from the center of the ellipse and its plane perpendicular to the plane of the ellipse. Find the volume gener- ated as the triangle moves along the axis of the ellipse from vertex to vertex. PROBLEMS 289 55. Find the volume cut from a right circular cylinder of radius a by a plane through the center of the base making an angle 6 with the plane of the base. 56. Two parabolas have a common vertex and a common axis but lie in perpendicular planes. An ellipse moves with its center on the common axis, its plane perpendicular to the axis, and its vertices on the parabolas. Find the volume generated when the ellipse has moved to a distance h from the common vertex of the parabolas. v 57. An equilateral triangle moves so that one side has one end in OY and the other end in the circle x 2 + f = a 2 , the plane of the rectangle being perpendicular to OY. Required the volume of the solid generated. 58. In a sphere of radius a find the volume of a segment of one base and altitude h. 59. Find the volume of the solid generated by revolving about OY the plane surface bounded by OY and the hypocycloid x l + y* = «». 60. Find the volume of the solid formed by revolving about the line x — 3 the figure bounded by the parabola f = 8.r and the line x = 2. 61. Find the volume of the solid formed by revolving about the line y = —a the figure bounded by the curve y = sinx } the lines 77- x — and x = — > and the line y = — a. 62. A right circular cone with vertical angle 2a has its vertex at the center of a sphere of radius a. Find the volume of the portion of the sphere intercepted by the cone. 63. A variable equilateral triangle moves with its plane perpen- dicular to the axis of I/ and the ends of its base respectively on the parts of the curves y 2 =16u.r and f = 4 ax above the axis of x. Find the volume generated by the triangle as it moves a distance a from the origin. 64. Find the volume of the solid formed by revolving about OX the plane figure bounded by the cissoid f — > the line x = a, and the axis of x. 65. A right circular cylinder of radius a is intersected by two planes, the first of which is perpendicular to the axis of the cylinder, and the second of which makes an angle 6 with the first. Find the 290 APPLICATIONS OF INTEGRATION volume of the portion of the cylinder included between these two planes if their line of intersection is tangent to the circle cut from the cylinder by the first plane. 66. On the double ordinate of the four-cusped hypocycloid x* + y* = Q>* as base an isosceles triangle is constructed with its altitude equal to the ordinate and its plane perpendicular to the plane of the hypocycloid. Find the volume generated by the triangle as it moves from x = — a to x — a. 67. Find the volume of the solid formed by revolving about OY 8 a). 73. The cap of a stone post is a solid of which every horizontal cross section is a square. The corners of all the squares lie in a spherical surface of radius 8 in. with its center 4 in. above the plane of the base. Find the volume of the cap. 74. Find the volume of the solid formed by revolving about the line x = — 2 the plane figure bounded by that line, the parabola y 2 = &x, and the lines y = ± 2. PROBLEMS 291 75. Find the volume of the solid formed by revolving about the line x = 2 the plane figure bounded by the curve if = 4 (2 — x) and the axis of y. 76. A variable circle moves so that one point is always on OY, its x 2 if center is always on the ellipse — 2 + 'j^ = 1, and its plane is always perpendicular to OY. Required the volume of the solid generated. 77. Find the volume of the solid generated by revolving about x 3 the asymptote of the cissoid y 2 = the plane area bounded by the curve and the asymptote. 78. Find the volume of the solid formed by revolving about OX the plane figure bounded by OX and an arch of the cycloid x — o ((f) — sin ), y = a (1 — cos ). 79. Find the volume of the solid generated by revolving the cardioid /• = a(l + cos 0) about the initial line. 80. A cylinder passes through two great circles of a sphere which are at right angles to each other. Find the common volume. 81. Find the length of the semicubical parabola //- = (x — 2) 3 from its point of intersection with the axis of x to the point (6, 8). 82. Find the length of the catenary y = „ I ' > " + e " ) from x = to x = It. 83. Find the total length of the four-cusped hypocycloid X s + y J = « ¥ - 84. Show that the length of the logarithmic spiral r = c" e bel ween any two points is proportional to the difference of the radius vectors of the points. Q 85. Find the complete length of the curve r = «sin 3 -- 86. Find the length of the curve y = a log — y : from the a 1 — x 2 . . a origin to * = - • a 87. Find the length from cusp to cusp of the cycloid x = a ( — sin <£), y = a (1 — cos ). 292 APPLICATIONS OF INTEGRATION 88. From equidistant points on an arch of the cycloid x = a ( — sin ), y = a(l — cos ), perpendiculars are drawn to the base of the arch. What is their average length ? 89. From a spool of thread 2 in. in diameter three turns are unwound. If the thread is held constantly tight, what is the length of the path described by its end ? e x -f- 1 90. Find the length of the curve y = log from x = 1 to x = 2. e ~ 1 91. Find the mean distance of all points on the circumference of a circle of radius a from a given point on the circumference. 92. Find the length of the spiral of Archimedes, r = a$, from the pole to the end of the first revolution. 93. Find the length of the curve 8a 3 ^/ = x 4 + 6«¥ from the origin to the point x = 2 a. 94. The parametric equations of a curve are x = 50 (1 - cos 9) + 50 (2 - 0) sin 9, y = 50 sin 9 + 50 (2 - 9) cos 9. Find the length of the curve between the points 9 = and = 2. 95. Find the length of the cardioid r = a (1 + cos 6). 96. Find the mean length of the radius vectors drawn from the pole to equidistant points of the cardioid r — -(1 + cos 9). 9 97. Find the length of the curve r = a cos 5 - from the pole to o the point in which the curve intersects the initial line. 98. Find the length of the tractrix (§ 200) V = T ? log , fl -Va 2 - x 2 from x = h to x = a. 99. Find the area of a zone of height h on a sphere of radius a. 100. Find the area of the surface formed by revolving about OX the hypocycloid x = a cos 3 0, y = a sin 3 9. 101. Find the area of the surface formed by revolving about the line x = a the portion of the hypocycloid x = a cos 3 0, y = a sin 3 0, which is at the right of OY. PROBLEMS 293 102. Find the area of the surface formed by revolving about the a - _- tangent at its lowest point the portion of the catenary i/ — -(e"-\-e «) between x = — h and x = h. 103. Find the area of the surface formed by revolving about the initial line the cardioid r = a (1 + cos 6). 104. Find the area of the surface formed by revolving an arch of the cycloid x = a(cf> — sin ), y = a(l — cos ) about the tangent at its highest point. 105. Find the area of the surface formed by revolving about OY a + vV — ./•- the traetrix (§ 200) // = 7^ log- — Va? — x*. 2 a — V"- — ./•' 106. Find the area of the surface formed by revolving the lem- oiscate r 3 = 2 a 2 cos 2 6 about the initial line. 107. Find the area of the surface formed by revolving the lem- niscate y 3 = 2 a 2 cos 2 6 about the line 9 = 90°. 108. A positive charge m of electricity is fixed at 0. The repul- sion on a unit charge at a distance x from (> is -■ Find the work done in bringing a unit charge from infinity to a distance a from O. 109. Assuming that the force required to stretch a wire from the length a to the length a -f-as is proportional to > and that a force of lib. stretches a wire of 36 in. in length to a length .01 in. greater, find the work done in stretching the wire from •">•'> in. to 39 in. 110. A body moves in a straighl line according to the formula x = cf, where x is the distance traversed in the time t II' the re- sistance of the air is proportional to the square of the velocity, find the work done against the resistance of the air as the body moves from x = to x = a. ill. Assuming that below the surface of the earth the force of the earth's attraction varies directly as the distance from the earth's center, find the work done in moving a weight of w pounds from a point a miles below the surface of the earth to the surface. 112. Assuming that above the surface of the earth the force of the earth's attraction varies inversely as the square of the distance 294 APPLICATIONS OF INTEGRATION from the earth's center, find the work done in moving a weight of m pounds from the surface of the earth to a distance a miles above the surface. 113. A wire carrying an electric current of magnitude C is bent into a circle of radius a. The force exerted by the current upon a unit magnetic pole at a distance x from the center of the circle in a straight line perpendicular to the plane of the circle is known to be -• Find the work done in bringing a unit magnetic pole from infinity to the center of the circle along the straight line just mentioned. 114. A spherical bag of radius 5 in. contains gas at a pressure equal to 15 lb. per square inch. Assuming that the pressure is in- versely proportional to the volume occupied by the gas, find the work required to compress the bag into a sphere of radius 4 in. 115. A piston is free to slide in a cylinder of cross section S. The force acting on the piston is equal to pS, where p is the pres- sure of the gas in the cylinder, and a pressure of 7.7 lb. per square inch corresponds to a volume of 2.5 cu. in. Find the work done as the volume of the cylinder changes from 2.5 cu. in. to 5 cu. in., (1) assuming pv = k, (2) assuming pv 1A — k. 116. Find the total pressure on a vertical rectangle with base 8 and altitude 12, submerged so that its upper edge is parallel to the surface of the liquid at a distance 5 from it. 117. Find the depth of the center of pressure of the rectangle in the previous problem. 118. Find the total pressure on a triangle of base 10 and altitude 4, submerged so that the base is horizontal, the altitude vertical, and the vertex in the surface of the liquid. 119. Show that the center of pressure of the triangle of the pre- vious problem lies in the median three fourths of the distance from the vertex to the base. 120. Find the total pressure on a triangle with base 8 and altitude 6, submerged so that the base is horizontal, the altitude vertical, and the vertex, which is above the base, at a distance 3 from the surface of the liquid. PROBLEMS 295 121. Find the depth of the center of pressure of the triangle of the previous problem. 122. The centerboard of a yacht is in the form of a trapezoid in which the two parallel sides are 3 and 5 ft. respectively in length, and the side perpendicular to these two is 4 ft. in length. Assuming that the last-named side is parallel to the surface of the water at a depth of 1 ft., and that the parallel sides are vertical, rind the pressure on the board.* 123. Find the moment of the force which tends to turn the center- board of the previous problem about the line of intersection of the plane of the board with the surface of the water. 124. A dam is in the form of a regular trapezoid with its two horizontal sides 400 and 100 ft. respectively, the longer side being at the top and the height 20 ft. Assuming that the water is level with the top of the dam, find the total pressure. 125. Find the moment of the force which tends to overturn the dam of the previous problem by turning it on its base line. 126. Find the total pressure on a semiellipse submerged with one axis in the surface of the liquid and the other vertical. 127. Find the depth of the center of pressure of the ellipse of the previous problem. 128. The gasoline tank of an automobile is in the form of a horizontal cylinder, the ends of which are plane ellipses 20 in. high and 10 in. broad. Assuming n- as the weight of a cubic inch of gasoline, find the pressure on one end when the gasoline is 15 in. deep-. 129. A parabolic segment with base 15 and altitude 3 is sub- merged so that its base is horizontal, its axis vertical, and its vertex in the surface of the liquid. Find the total pressure. 130. Find the depth of the center of pressure of the parabolic segment of the previous problem. 131. A circular water main has a diameter of 5 ft. One end is closed by a bulkhead and the other is connected with a reservoir in which the surface of the water is 20 ft. above the center of the bulkhead. Find the total pressure on the bidkhead. *The weight of a cubic foot of water may be taken as G2 1 lb. = J., ton. 296 APPLICATIONS OF INTE( J RATION 132. A pond of 10 ft depth is crossed by a roadway with vertical sides. A culvert, whose cross section is in the form of a parabolic segment with horizontal base on a level with the bottom of the pond, runs under the road. Assuming that the base of the parabolic seg- ment is 6 ft. and its altitude 4 ft., find the total pressure on the bulkhead which temporarily closes the culvert. 133. Find the pressure on a board whose boundary consists of a straight line and one arch of a sine curve, submerged so that the board is vertical and the straight line is in the surface of the water. 134. Find the center of gravity of the semicircumference of the circle x 2 + if = a 2 which is above the axis of x. 135. Find the center of gravity of the arc of the four-cusped hypocycloid x* + y s = a 3 which is above the axis of x. 136. Find the center of gravity of a parabolic segment. 137. Find the center of gravity of the area of a quadrant of an ellipse. 138. Find the center of gravity of a triangle. 139. Find the center of gravity of the area bounded by the semicubical parabola ay* = x s and any double ordinate. 140. Find the center of gravity of the area bounded by the pa- rabola x 2 -\- At/ — 16 = and the axis of x. 141. Find the center of gravity of half a spherical solid of con- stant density. 142. Find the center of gravity of the solid formed by revolving aj a v/ 2 about Y the surface bounded by the hyperbola — — 'j- 2 — 1 and the lines y = and y = b. 143. Find the center of gravity of a hemisphere. 144. Find the center of gravity of the surface of a right circular cone. 145. Find the center of gravity of the area bounded by the curve y = sin x and the axis of x between x = and x = it. 146. Find the center of gravity of the area between the axes of coordinates and the parabola x 1 + y* — a*. PROBLEMS 297 147. Find the center of gravity of a uniform wire in the form of the catenary y = - (e a -f e~«) from x = to x = a. 148. Find the center of gravity of the solid formed by revolving about OX the surface bounded by the parabola f = &jpx, the axis of x, and the line x — a. 149. Find the center of gravity of the plane area bounded by the two parabolas f = 20x and x 2 = 20 y. 150. Find the center of gravity of the plane area bounded by the parabola f = 4x, the axis of //, and the line y = 4. 151. Find the center of gravity of the solid formed by revolving about OY the plane figure bounded by the parabola f=Aj,x, the axis of y, and the line y = /.'. 152. Find the center of gravity of the surface of a hemisphere when the density of each point in the surface varies as its perpen- dicular distance from the circular base of the hemisphere. 153. Find the center of gravity of that part of the plane surface bounded by the four-cusped hypocycloid se = acos 8 0, y = a sin 8 0, which is in the first quadrant. 154. Find the center of gravity of the plane area bounded by the ellipse -5 + T7, = 1, the circle x 2 + y* = a 2 , and the n.xis of y. 155. Find the center of gravity of the plane area common to the parabola x 2 — 8 y = and the circle aP+y 2 — 128 = 0. 156. Find the center of gravity of the plane surface bounded by the first arch of the cycloid x = a((f> — sin <£), y = a (1— cos <£), and the axis of x. 157. Find the center of gravity of the arc of the cycloid x = a( — sin ), y = a (1 — cos <£), between the first two cusps. 158. Find the center of gravity of the solid formed by rotating about OX the parabolic segment bounded by f = 4a and x = It, if the density at any point of the solid equals -• 298 APPLICATIONS OF INTEGRATION 159. Find the center of gravity of the plane surface bounded by the two circles x a -f if = a 1 , x 1 + // — 2 ax = 0, and the axis of x. 160. Show that the center of gravity of a sector of a circle lies on . a 2 Sm 2 the line bisecting the angle of the sector at a distance - a from 2 the vertex, where a is the angle and a the radius of the sector. 161. Find the center of gravity of the solid generated by revolv- ing about the line x = a the area bounded by that line, the axis of x, and the parabola if = 4=px. 162. Find the center of gravity of the plane area bounded by the two parabolas x 2 — 4tp (y — b) = 0, x 2 — kpy = 0, the axis of y, and the line x = a. 163. Find the center of gravity of the arc of the curve 9 ay 2 — x (x — 3 a)' 2 = between the ordinates x = and x = 3a. 164. The density at any point of a lamina in the form of a para- bolic segment of height 8 ft. and base 6 ft. is directly proportional to its distance from the base. Find the center of gravity. 165. Find the center of gravity of the portion of a spherical surface bounded by two parallel planes at distances h^ and h 2 respectively from the center. 166. Find the center of gravity of the solid formed by revolving about OY the plane area bounded by the parabola x 2 = 4^y and any straight line through the vertex. 167. Find the center of gravity of the surface generated by the revolution about the initial line of one of the loops of the lemniscate r 2 =2« 2 cos2 0. 168. Prove that the total pressure on a plane surface perpendic- ular to the surface of a liquid is equal to the pressure at the center of gravity multiplied by the area of the surface. 169. Prove that the area generated by revolving a plane curve about an axis in its plane is equal to the length of the curve multi- plied by the circumference of the circle described by its center of gravity. 170. Prove that the volume generated by revolving a plane figure about an axis in its plane is equal to the area of the figure multiplied by the circumference of the circle described by its center of gravity. PEOBLEMS 299 171. Find the attraction of a uniform straight wire of length 20 and mass M upon a particle of unit mass situated in the line of direction of the wire at a distance 3 from one end. 172. Find the attraction of a rod of mass M and length /, whose density varies as the distance from one end, on a particle of unit mass in its own line and distant a units from that end. 173. A particle of unit mass is situated at a perpendicular dis- tance 5 from the center of a straight homogeneous wire of mass .1/ and length 12. Find the force of attraction of the wire. 174. Find the attraction due to a straight wire of length 2 / on a particle of unit mass lying on the perpendicular at the middle point of the wire and distant c units from the wire, the density of the wire varying directly as the distance from its middle point. 175. Find the attraction of a homogeneous straight wire of neg- ligible thickness and infinite length on a particle of unit mass at a perpendicular distance c from the central point of the wire. 176. Find the attraction of a uniform wire of mass .1/ bent into 77- an arc of a circle with radius 5 and angle — upon a particle of unit mass at the center of the circle. 177. Find the attraction of a uniform circular ring of radius a and mass M upon a particle of unit mass situated at a distance c from the center of the ring in a straight line perpendicular to the plane of the ring. 178. Find the attraction of a uniform circular disk of radius a and mass M upon a particle of unit mass situated at a perpendicular distance c from the center of the disk. (Divide the disk into con- centric rings and use the result of problem 177.) 179. Find the attraction of a uniform right circular cylinder with mass M, radius of its base a, and length I upon a particle of unit mass situated in the axis of the cylinder produced, at a distance c from one end. (Divide the cylinder into parallel disks and use the result of problem 178.) 180. Find the attraction of a uniform straight wire of length 5 and mass M upon a particle of unit mass situated at a perpendicular distance 12 from the wire and so that lines drawn from the particle 7T to the ends of the wire inclose an angle — • o CHAPTER XIV SPACE GEOMETRY 139. Functions of more than one variable. A quantity z is mid to be a function of two variables, x and y, if the values of z are determined when the values of x and y are given. This rela- tion is expressed by the symbols z —f(x, y), z = F(x, y), etc. Similarly, u is a function of three variables, x, y, and z, if the values of u are determined when the values of x, y, and z are given. This relation is expressed by the symbols u =f(x, y, z), u = F(x, y, z), etc. Ex. 1. If r is the radius of the base of a circular cone, h its altitude, and v its volume, v = ^ 7rr 2 h, and v is a function of the two variables r and h. Ex. 2. If / denotes the centrifugal force of a mass m revolving with a velocity v in a circle of radius >',f= , and f is a function of the three variables m, v, and r. Ex. 3. Let v denote a volume of a perfect gas, t its absolute temperature, and p its pressure. Then — = k, where k is a constant. This equation may be written in three equivalent forms: p = k-, v — k-, t= -pv, by which each of the quantities p, v, and / is explicitly expressed as a function of the other two. A function of a single variable is defined explicitly by the equation y =f(x), and implicitly by the equation F(x, y) = (§ 86). In either case the relation between x and y is repre- sented graphically by a plane curve. Similarly, a function of two variables may be defined explicitly by the equation z =f(x, y~), or implicitly by the equation F(x, y, z) — 0. In either case the graphical representation of the function of two variables is the same, and may be made by introducing the conception of space coordinates. 300 RECTANGULAR COORDINATES 301 140. Rectangular coordinates. To locate a point in space of three dimensions, we may assume three number scales, OX, OY, OZ (rig. 186), mutually perpendicular, and having their zero points coincident at 0. They will determine three planes, XO Y, YOZ, ZOX, each of which is perpendicular to the other two. The planes are called the coordinate planes, and the three lines OX, OY, and OZ are called the axes of x, y, and z respectively, or the coordinate axes, and the point is called the origin of coordinates. Let P be any point in space, and through P pass planes perpendicular respectively to OX, )', and OZ, intersecting them at the points L, M, and N respectively. Then it' we place x — OL, y = 0M, and z = ON, it is evident that to any point there corresponds one, and only one, set of values of x, y, and z ; and that to any set of values of ./-, //, and n\ jj, z there corresponds one, and only one, s'/— L— -{/ point. These values of x, y, and z are called the coordinates of the point, which <)' (i) 302 tiVACK GEOMETRY Then the locus of all points the coordinates of which satisfy (1) is -the graphical representation of the function f(sc, y). To con- struct this locus we may assign values to x and y, as x — x x and y = y l , and compute from (1) the corresponding values of z. There will be, in general, distinct values of z, and if (1) defines an algebraic function, their number will be finite. The corre- sponding points all lie on a line parallel to OZ and intersecting XOY at the point P l (x l , y^), and these points alone of this line are points of the locus, and the portions of the line between them do not belong to the locus. As different values are assigned to x and y, new lines parallel to OZ are drawn on which there are, in general, isolated points of the locus. It follows that the locus has extension in only two dimensions, i.e. has no thickness, and is, accordingly, a surface. Therefore the graphical representation of a function of two variables is a surface* If fQx, y~) is indeterminate for particular values of x and y, the corresponding line parallel to OZ lies entirely on the locus. Since the equations z —f(x, y) and F(x, y, z~) = are equiv- alent, and their graphical representations are the same, it follows that the locus of any single equation in x, y, and z is a surface. There are apparent exceptions to the above theorem if we demand that the surface shall have real existence. Thus, for example, + f + is satisfied by no real values of the coordinates. It is convenient in such cases however, to speak of " imaginary surfaces." Moreover, it may happen that the real coordinates which satisfy the equation give points which lie upon a certain line, or are even isolated points. For example, the equation x 2 + y- = is satisfied in real coordinates only by the points (0, 0, z) which lie upon the axis of z ; while the equation x 2 + if + z 2 = * It is to be noted that this method of graphically representing a function cannot be extended to functions of more than two variables, since we have but three dimensions in space. CYLINDERS 303 is satisfied, as far as real points go, only by (0, 0, 0). In such cases it is still convenient to speak of a surface as represented by the equation, and to consider the part which may be actually constructed as the real part of that surface. The imaginary part is considered as made up of the points corresponding to sets of complex values of x, y, and z which satisfy the equation. 142. Cylinders. If a given equation is of the form F(x, y) = 0, involving only two of the coordinates, it might appear to rep- resent a curve lying in the plane of those coordinates. But if we are dealing with space of three dimensions, such an inter- pretation would be incorrect, in that it amounts to restricting z to the value s = 0, whereas, in fact, the value of z corre- sponding to any simultaneous values of x and y satisfying the equation F(x, y)=0 may be anything whatever. Hence, corresponding to every point of the curve F(x, //) = in the plane XOY, there is an entire straight line, parallel to OZ y on the surface F(.r, y) = 0. Such a surface is a cylinder, its directrix being the plane curve /''(./•, y) = Q in the plane 2 = 0, and its elements being parallel to OZ, the axis of the coordinate not present. For example, .r + //" = )x+ab=0 may be written in the form (x— a) (x— l>) = 0, which represents the two planes x — a = and x — b — 0. Similarly, any equation involving only one coordinate determines values of that coordinate only and the locus is a number of planes. Regarding a plane as a cylinder of which the directrix is a straight line, we may say that any equation not containing all the coordinates represents a cylinder. If the axes are oblique, the elements of the cylinders are not perpendicular to the plane of the directrix. 304 SPACE GEOMETRY 143. Other surfaces. The surface represented by any equa- tion /'(./-, //, z) = may be studied by means of sections made by planes parallel to the coordinate planes. If, for example, we places = <> in the equation of any surface, the resulting equation in x and y is evidently the equation of the plane curve cut from the surface by the plane XOY. Again, if we place 2 = 2 1? where z x is some fixed finite value, the resulting equation in x and y is the equation of the plane curve cut from the surface by a plane parallel to the plane XOY and z l units distant from it, and referred to new axes O'X' and O'Y', which are the intersections of the plane z = z x with the planes XOZ and YOZ respectively; for by placing z = z x instead of 2=0, we have virtually transferred the plane XOY, parallel to itself, through the distance z^ In applying this method it is advisable to find first the three plane sections made by the coordinate planes x = 0, i/ = 0, z = Q. These alone will sometimes give a general idea of the appearance of the surface, but it is usually desirable to study other plane sec- tions on account of the additional information that may be derived. The following surfaces have been chosen for illustration because it is important that the student should be familiar Avith them. Ex. 1. Ax + By + Cz + D = 0. Placing- z = 0, we have (fig. 187) Ax + By + D = 0. (1) Hence the plane XOY cuts this surface in a straight line. Placing y = and then x = 0, we find the sections of this surface made by the planes ZOX and YOZ to be respectively the straight lines Ax + Cz + D=0, (2) and By + Cz + D=0. (3) Placing z = z v we have A x + By + ( 'z 1 + D = 0, (4) which is the equation of a straight line in the plane z — z v The line (1) is parallel to the line (1), since they make the angle tan- 1 / — —J with the parallel lines O'X' and OX and lie in parallel planes. To find the point SURFACES 305 where (4) intersects the plane XOZ, we place y = 0, and the result Ax + Cz 1 + D= shows that this point is a point of the line ("2). This result is true for all values of z r Hence this surface is the locus of a straight line which moves along a fixed straight line always remaining parallel to a given initial position; hence it is a, plane. Since the equation Ax + By + Cz + D= is the most general equa- tion of the first degree in the three coordinates, we have proved that the locus of every lunar equation in rectangular spaa coordinates is a plane. + !>//-, where a > Ex. 2. s &>0. Placing 2 = 0, we have a./ 2 + by* = 0, and hence the XOY plane surface in a point (fig. L88) ,j = (), we have z = il.i-, which is the equation of a with its vertex at ami its axis along OZ. Placing x = <•, we have z = /,n (3) which is also the equation of a pa- rabola with its vertex at and its axis along (>Z. Placing z = z 1 , where z x > 0, we may write the resulting equation in t he form b (1) and As the -1 \ -1 which is the equation of an ellipse with semiaxes \ (I \ h plane recedes from the origin, i.e. as - n increases, it is evident thai the ellipse increases in magnitude. It is also evident thai the ends of the axes of the ellipse lie on the parabolas (•_') and (:">). If we place z = — z v the result may he w litten in the form — y l, and hence there is no part of this surface on the negative side of the plane XOY. The surface is called an elliptic paraboloid, and evidently may l>e gener- ated by moving an ellipse of variable magnitude always parallel to the plane XOY, the ends of its axes always lying respectively on the parabolas 2 = 0, I > 0. Placing z = 0, we obtain the equa- tion ax 2 - by 2 = 0, (1) i.e. two straight lines intersecting at the origin (fig. 193). Placing y = 0, we have (l.r- (2) Fig. 192 the equation of a parabola with its v^i^x at and its axis along the positive direction of OZ. IS x = 0, we have = - by*, (3) the equation of a parabola with its vertex at and its axis along the -negat ive direction of OZ. Placing ./• = ± x v we have Z = (,.r{ - /;//'-, or \f ±(z-ax?), (1) Fig. 193 a parabola with its axis parallel to OZ and its vertex at a distance ax* from the plane XOY. It is evident, moreover, that the surface is sym- metrical with respect to the plane YOZ, and that the vertices of 308 SPACE GEOMETKY these parabolas, as different values are assigned to x v all lie on the parabola z = ax 2 . Hence this surface may be generated by the parabola z = — by 1 moving always parallel to the plane YOZ, its vertex lying on the parabola z = ax". The surface is called the hyperbolic paraboloid. The reason for the name given to this surface becomes more evident if two more sections are made. Placing z = z v where z 1 > 0, we have z x = ax 2 — hy 2 , or f -L x 2 -Lf- = l, (5) an hyperbola with its transverse axis parallel to OX, the ends of the transverse axis lying on the parabola z = ax' 2 . If z = — z v we may write the equation in the form ■v- (6) an hyperbola with its transverse axis parallel to OY, the ends of the transverse axis lying on the parabola z = — by' 1 . Ex. 8. z = kxy, where fc>0. This surface is a special case of the hyper- bolic paraboloid of Ex. 7, in which b=a. The proof of this statement is as follows : If h = a, the equation of the surface of Ex. 7 is z = a(x*-y*). (1) Fig. 194 Revolve the planes XOZ and YOZ about the axis OZ, which is held fixed, through an angle of — 45° into new positions X'OZ and Y'OZ. By § 19, the formulas of transformation are , _ x' + y' _ — x + // Z- , X- , V2 Substituting these values in (1), and simplifying, we have z' = 2 a x'y', which is the equation given above with k — 2a. (2) SURFACES 309 The discussion of the plane sections of the surface (fig. 194) made by the planes parallel to the coordinate planes is left to the student. If b ^ a, we can make a similar transformation by using the formulas of § 21, and the result will be ~' = k/y, only the coordinates will not be rectangular. 144. Surfaces of revolution. If the sections of a surface made by planes parallel to one of the coordinate planes are circles with their centers on the axis of coordinates which is perpendicular to the cutting planes, the surface is a surface of revolution (§ 133) with that coordinate axis as the axis of revolution. This will always occur when the equation of the surface is in the form F{\ f .r + >/% z) = 0, which means that the two coordinates x and y enter only in the combination v.r + //- ; for if we place z = z 1 in this equation to find the corresponding section, and solve the resulting equation for ./•" + //", we have, as a result, the equation of one or more circles, according to the number of roots of the equation in x 1 + y' 1 . Again, if we place x = 0, we have the equation /•'(//, z) = 0, which is the equation of the generating curve in the plane YOZ. Similarly, if we place // = 0, we have /•'(./-, g) = 0, which is the equation of the generating curve in the plain' XOZ. It should be noted that the coordinate which appears uniquely in the equation shows which axis of coordinates is the axis of revolution. Ex. 1. Show that the imparted hvperboloid ', — •- + -= 1 is a surface "'" //_ "'" of revolution. Writing this equation in the form t±* -f-l = Q t a- b- we see that it is a surface of revolution with OY as the axis. c 2 >r Placing z = 0, we have — — 2-=l, an hyperbola, as the generating a- b- curve. The hyperbola was revolved about its conjugate axis. Conversely, if we have any plane curve F(x, z) = in the plane XOZ, the equation of the surface formed by revolving it about OZ as an axis is F(Vx 2 + y 2 , z) = 0, which is formed by simply replacing x in the equation of the curve by Vx 2 + y 1 . 310 SPACE GEOMETRY Ex. 2. Find the equation of the sphere formed by revolving the circle + z 2 = a 2 about OX as an axis. Replacing a by Vy 2 + z 2 , we have as the equation of the sphere, x 2 + y- + z 2 = a 2 . This equation may also be found directly from a figure. Let P x (fig. 195) be any point of the circle, and .let P be any point of the sphere, on the circle described by P v Since P x is a point of the circle, OZ 2 + TP{ = a 2 . (1) But LP^LP=VlM 2 + Mp' 2 - Substituting this value of LP 1 in (1), we have OZ 2 + LM 2 + MP 2 = a 2 , or x 2 + f + z 2 = a 2 , as the equation of the sphere. 145. Projection. The projection of a point on a straight line is defined as the point of intersection of the line and a plane through the point perpendicular to the line. Hence, in fig. 186, L, M, and N are the projections of the point P on the axes of x, y, and z respectively. The projection of one straight line of finite length upon a second straight line is the part of the second line included between the projections of the ends of the first line, its direction being from the projection of the initial point of the first line to the projection of the terminal point of the first line. In fig. 196, for example, the projections of A and B on MN being A' and B' respectively, the projection of AB on MJSF is A'B', and the projection of BA on MN is B'A'. If MN and AB denote the positive directions respectively of these lines, it follows that A'B' is positive when M- Q 4--- F^ A' 1 t B' Fig. 196 PROJECTION 311 it has the same direction as MN y and is negative when it has the opposite direction to MN. In particular, the projection on OX of the straight line i^ drawn from P l (x v y v zj to P 2 (x 2 , y. v z 2 ) is L X L V where OL x = x and 0L 2 = x 2 . But L x L 2 = x 2 — x v by § 3. Hence the projection of P X P 2 on OX is x 2 — x x ; and, similarly, its projections on Y and OZ are respectively y 2 — y t and z 2 — z x . If we define the angle between any two lines in space as the angle between lines parallel to them and drawn from a common point, then the projection of one *trn a second is the product of the length of the first line and the cosine of the angle between the positive directions of the two lines. Then, if $ is the angle between AB and MN (fig. 196), A'B' = ^17? cos 4>. To prove this proposition, draw A'C parallel to .//; and meet- ing the plane ST at C. Then A'C = AB, and A'B' = A'C cos , by § 2, whence the truth of the proposition is evident. Defining the projec- tion of a broken line upon a straight line as the sum of the projec- tions of its segments, we may prove, as in § 2, that the projections on any straight line of a broken line and the straight line joining its ends are the same. We will now show that the projection of a)iy plane area )i/»>n another plane is the product of that area and the cosine of the angle between the planes. Let X'OY (fig. 197) be any plane through OY making an angle with the plane XOY. Let A'B' be any area in X'OY such that any straight line parallel to OX' intersects its boundary in not more than two points, and let AB be its projection on XOY. 312 SPACE GEOMETRY Then (§ 125) area A'B' = C (x[ - x[) dy, (1) the limits of integration being taken so as to include the whole area. In like manner, area AB — I (x 2 — x i )dy, (2) the limits of integration being taken so as to cover the whole area. But the values of y are the same in both planes, since they are measured parallel to the line of intersection of the two planes; and hence the limits in (1) and (2) are the same. Since the x coordinate is measured perpendicular to the line of intersection, x 2 = x' 2 cos <£, x 1 = x[ cos , and (2) becomes area AB = / (./•.( — x[) cos dy = cos j (x' 2 — .Tj') dy = (cos <£) (area A'B'). H>-: 146. Components of a directed straight line. Let B X B 2 (fig. 198) be a straight line, the direction of which is from B x to B,. Through B x and B 2 pass planes parallel respectively to the coordinate planes, thereby forming on BJ! 2 as a diagonal a rectangular paral- lelepiped with its edges parallel to the coordinate axes. The lines B t Q, B X B, and B lt S, considered with respect to both length and direction, are called the components of 7^. It is evident that they are the projec- tions of i^on OX, OY, and OZ respectively. Conversely, the components of a straight line will determine its direction and length, but not its position ; for if the compo- nents are given equal to a, b, and e, we may lay off, from any point Ij, a straight line parallel to OX and equal to a in length, a straight line parallel to OY and equal to b hi length, and a straight line parallel to OZ and equal to c in length. These three lines determine the edges of a rectangular parallelepiped, and hence determine the diagonal drawn from B v That is, if B X Q (fig. 198) is laid off equal to a, B^ equal to b, and B X S equal DISTANCE 313 to c, the rectangular parallelepiped is determined, and hence the diagonal P X P 2 is determined in both length and direction. It is evident that the direction of P l P 2 will not be changed if a, b, and c are multiplied by the same number; in other words, the ratios of the components are the essential elements in fixing the direction of the line. We shall accordingly speak of a straight line as having the direction a: b : c. On the other hand, the length of the line does depend upon the values of a, b, and c, for P X P = Vy^/+i^r+^s' 2 = Va 2 + b 2 + c\ (1 ) 147. Distance between two points. An important application of (1), §110, is in finding the distance between two points 1\{.<\, /j x , zj and /._!(•'".,' //.,, ?.,). Referring to fig. 198, we have, by §U5, a = P 1 Q=x 2 -z l1 b=P 1 R = y % -y v c=P 1 S = z a -z 1 ; whence ^ = V(.r, - .r^ +(^- yj 2 +(*,- z l} 2 . (1) Ex. 1. Find the Length of the straight line joining the points (1, - _\ -1) and (8, — 1, 3). The required length is V(:S - I)- + (- 1 - 2)* + (3 + 1)- =\ 29. Ex. 2. Find a point Vll units distant from each of the three points (1,0,3), (2, -1, 1), (3, 1, •_'). Let P(x, y, c) be the required point. Then (x - l) 2 + (// - 0) a + (z - 3) 2 = 14, (x-2)* + (y + l)* + (z-iy = U, (*-8)» + (y-l)» + (*-2)» = 14. Solving these three equations, we determine the two points (0, 2, 0) and (4, - 2, 4). Ex. 3. Find the equation of a sphere of radius r with its center at Pi(* v yi.*i). If P (x, y, z) is any point of the sphere, (z _ Xi) 2 + (y _ yi) a + (z _ Zi)a = r3 . Conversely, if P (x, y, z) is any point the coordinates of which satisfy this equation, P is at the distance r from P v and hence is a point of the sphere. Therefore this is the required equation of the sphere. 314 SPACE GEOMETRY 148. Direction cosines. If we denote by a, /3, and 7 the angles which a straight line makes with the positive directions of the coordinate axes OX, OY, and OZ respectively, the cosines of these angles, i.e. cos a, cos ft, cosy are called the direction cosines of the line. If the line is drawn through the origin, as in fig. 199, it is evident that the same straight line makes the angles a, /3, 7 or 17 — a, ir — /3, 17 — 7 with the coordinate axes, according to the direction in which the line is drawn. Hence its direction cosines are either cos a, cos/3, cos 7 or — cosa;, — cos/3, —cos 7. Hence the straight line can have but one set of direction cosines after its direction has been chosen. The direction cosines may be deter- mined directly from the components of the line ; for, referring to fig. 198, we see that PQ or cos 7 = a Va 2 + & 2 + c 2 ' c Va 2 + 6 2 + 6" 2 ' 1]E cos/3 = PS -^-1 COS/3 = — , C0S7 = — — , CI) P,P 2 P,P 2 7 JJJJ W b Va 2 + b % + c* (2) Squaring and adding equations (2), we have cos 2 a-f cos 2 /3 + cos 2 7=l; (3) that is, the sum of the squares of the direction cosines of any straight line is always equal to unity. It follows that the direction cosines of any line are not independent quantities. Ex. Since the length of the line of Ex. 1, § 147, is V29, and its respec- tive components are 2, — 3, and 4, it follows that its direction cosines are 2 3 4 V29' V29' V29 ANGLE 315 149. Angle between two straight lines. Given two straight lines having the respective directions a x : b x : c x and a % : b 2 : c n . If they are drawn from a common point P (x, y, z) (fig. 200), let the segment of the first line extend to P x and the segment of the second line extend to P 2 , so that the coordinates of P x are x 4- dji y + b x , and z + e t , and the coordinates of P 2 are x + a 2 , y + ?>„, and z + <- n . It follows that the components of P X P 2 are a n — a ± , b 2 — b^ and c n — Cj. Then if 6 is the angle between these two lines, we have, by trigonometry, ■ PP 2 + PP?-T\p; _ cos 6 = — *- i- 2 -. rn 2PP,.PP 2 K } But Pli 2 =a; + b: + e:, P' _, Fig. 200 Pi^O* + &» + <& whence, by substitution in (1) and simplification, P.nsfl= «,«, + ¥, + *!', (2) V«f + ft»+«fVo£ + tf + <£ If cosa x , eos/3^ COS7J are the direction cosines of /'/;. and cos« 2 , cos/3 , C0S7 2 are the direction cosines of J'J.U formula (2) may be written, by (2), § 148, cos 6 = cos « 1 cos a 2 + cos /S x cos /3 o + <•< is 7 t c< H3 7.,. (3) If the lines are perpendicular to each other, cos — 0, and (2) and (3) reduce respectively to a i a 2+ I, A+ , \ ( '-Z =0 ( 4 ) and cos a i cos « 2 + cos fi x cos /? 2 + cos 7 1 cos 7, = 0. (5) If the lines are parallel to each other, cos a x = cos «„, cos /3 X = cos /3 2 , cos 7 X = cos 7„, whence it is easily shown that ?i = h = ?i. (6) a., b., 2 = 0. (2) STRAIGHT LINE 317 The angle between these planes is the same as the angle between their respective normals, the directions of which are respectively the directions A x : B 1 : C\ and A.,: />'.,: C g . Hence, if 6 is the angle between the two planes, by (2), § 149, , A x A« + B,B„+C,C n cos 6 = 1 - 1 ; 1 = • Va? + a; + c* V a* + J5 a a + cy 2 The conditions for perpendicularity and parallelism of the planes are respectively i A », Ci 153. Equations of a straight line. In space of three dimen- sions a single equation in general represents a surface; hence, in general, a curve cannot he represented by a single equation. A curve may, however, be regarded as the line of intersection of two surfaces. Then the coordinates of every point of the curve satisfy the equations of the surfaces simultaneously : and, con- versely, any point the coordinates of which satisfy the equations of the surfaces simultaneously is in their curve of intersection. Hence, in general, the locus of two simultaneous equations in x, //, and z is a curve. In particular, the locus of the two simultaneous linear equations \x+B x y + C' 1 2+/> 1 =0, A 2 x + B„y + C^+J) 2 =0, is a straight line, since it is the line of intersection of the two planes respectively represented by the two equations. We will now find the equations of the straight line determined by two points, and the equations of the straight line passing through a known point in a given direction. 154. Straight line determined by two points. Let the given points be B,(x v y v z x ) and B,(x 2 , y. 2 , z 2 ). Then the direction of 1{B is # 2 — .r, : //., — y l : z.,— z v Let P(x, y, z) be any point of the line. Then the direction of P^P is x — x\ : y — y x : z — z x . 318 SPACE GEOMETRY Since I[P and I(P 1 are parts of the same straight line, and hence parallel, it follows that x ~ x \ _ V ~ V\ _ z ~ z \ Here are but two independent equations in x, y, and z. This result proves the converse of the statement above, that two linear equations always represent a straight line ; for we have any straight line represented by two linear equations. It is to be noted that, if in the formation of these fractions any denominator is zero, the corresponding component is zero, and the line is perpendicular to the corresponding axis. Ex. Find the equations of the straight line determined by the points (1, 5, - 1) and (2, - 3, - 1). x — 1 _ y — 5 _ z + 1 2^T~ -3- 1 + 1 Hence the two equations of the line are z + 1 = 0, since the line is par- allel to the XOY plane and passes through a point for which z = — 1, and 8 x + y — 13 = 0, formed by equating the first two fractions. 155. Straight line passing through a known point in a given direction. If the direction of the line is given as a : b : c, the equations of the line are evidently x — x, y — y, z — z. ^ . (!) a b c for in the formula of § 154 we may place If the direction of the line is given hi / terms of its direction cosines, the derivation y Fig. 201 of the equations is as follows : Let P 1 (x v y v 2j) (fig. 201) be a known point of the line, and let I, m, and n be its direction cosines. Let P(x, y, z) be any point of the line. On P^P as a diagonal construct a parallele- piped as in § 146. Then if we denote P^P by r, we have P x Q=lr, %B= mr, %S= nr. But P x Q=x—x vl PR—y — y v I£S=z — e 1 , whence x— x,= lr. y — y= mr, z — 2 = nr. STRAIGHT LINE 319 Eliminating r from these last three equations, we have x - *i _ y - Vx _ z - z i , ^ 2 ) l m n which are but two independent linear equations. 156. Determination of the direction cosines of a straight line. If the equations of the straight line are hi any one of the forms of §§ 154 and 155, the determination of the direction cosines is very easy, for the denominators of the fractions in those formu- las are either the direction cosines of the line or else give the components for the line, from which the direction cosines are quickly computed. If the equations of the straight line, however, are in any other form, as „ . _ ^ . ./,./• + /; i/ / + r, + /, i = o, (1) J,r + n ji + cj 4 D a = 0, (2) let its direction cosines be /, m, and n. Since* the line lies in both planes (1) and (2), it is perpendicular to the normal to each. Therefore, by (4), § 149, A J 4 B x m 4- C x n = 0, AJ, 4 B t m 4 C a n = ; also l* + m* + n*=l. (§148) Here are three equations from which the values of I, m, and n may be found. Ex. Find the direction cosines of the straight line 2z+3y+«— 4 = 0, 4 x + >/- z + 7=0. The three equations for I, m, and n are 2 Z + 3 m 4 n = 0, 4 I + m — n = 0, /- + m* + n" = 1, the solutions of which are z = _2_ ; 3_ ? 5 V38' V38' V3s' ,_ 2 3 5 V38 V38 V38 320 SPACE GEOMETRY Since cos (180° — ) = — COS = 0. (1) From P. (fig. 202) draw the required perpendicular P X N and also a line parallel to the axis of z and let it cut the plane in R. Then for the point R, x = Xf y = y x , and z is determined from the equa- tion of the plane as -Ax 1 -By 1 -D Hence RP 1 _ Ax 1 +By 1 + Cz 1 +I) _ C But P 1 N=RP 1 cos 7 where x ) + fc a K x + B Jf +''/+ ",) = o. (3) Equation (3) is the equation of a plane, since it is a linear equa- tion, and furthermore it passes through the given straight line, since the coordinates of every point of that line satisfy (3) by virtue of (1) and (2). Hence (3) is the required plane, and it may be made to satisfy another condition by determining the values of k^ and /-., appropriately. Ex. 1. Find the equation of the plane determined by the point (0, 1. | and the line 4 x + 3 y + 2z - 4 = 0, 2x — 11 y - I : - 12 = 0. The equation of the required plane may be written / M (4x + 3 y + 2 z - 4) + /,-,( 2 x - 11 y -4a- 12) = 0. (1 ) Since (0, 1, 0) is a point of this plane, its coordinates satisfy (1), and hence fcj + 23 / , = 0, or l\ = - 23 I:,. Substituting this value of /c' x in (1), and reducing, we have as the required e( l uatiuu > 9x + 8y + 5z-8 = 0. 322 SPACE GEOMETRY Ex. 2. Find the equation of the plane passing through the line 4 x + 3 y + 2 z — 4 = 0, 2 .r — 11 y — 4 s — 12 = 0, and perpendicular to the plane 2 x + y-2z + l = 0. The equation of the required plane may be written ^(4 x + 3y + 2z-i) + k 2 (2 x - 11 y - 4 z - 12) = 0, (1) or (4 k x + 2 fc 2 ) x- + (3 ^ - 11 £ a ) y + (2 h x - 4 k 2 )z + (- 4 ^ - 12 fr a ) = 0. Since this plane is to be perpendicular to the plane 2 x + y — 2z + l = 0, 2 (4 fcj + 2 fc 2 ) + 1 (3 k, - 11 ifc 2 ) - 2 (2 ^ - 4 k 2 ) = 0, whence k„= — 7 k v Substituting this value of k 2 in (1), and reducing, we have as the required equation, x — 8y — 32 — 8 = 0. 2. Plane determined by three points. If the equations of the straight line determined by two of the points are derived, we may then pass a plane through that line and the third point, as in Ex. 1. The result is evidently the required plane. Ex. 3. Find the equation of the plane determined by the three points (1, 1, 1), (-1, 1, 2), and (2, - 3, -1). The equations of the straight line determined by the first two points are x - 1 y - 1 z-1 -1-1 1-1 2-1 which reduce toy — 1 = 0, x + 2z — 3 = 0. The equation of the required plane is now written in the form k 1 (y-l) + k 2 (x + 2z- 3)^=0. Substituting (2, — 3,-1) in this equation, we have - 4 k t - 3 fcg = 0, or & 2 = — | k v Substituting this value of k 2 in the equation of the plane, and simplify- ing, we have as our required equation, 4a:-3# + 8z-9 = 0. 159. Space curves. We saw in § 153 that, in general, the locus of two simultaneous equations in x, y, and z is a curve — CURVES 323 the curve of intersection of the surfaces represented by the equations taken independently. Let / 1 ( a ;,y, 2 )=0, /,(*y f «)=0, (1) be the two equations of a space curve. If we assign a value to one of the coordinates in equations (1), as x for example, there are two equations from which to determine the corresponding values of y and z, in general a determinate problem. But if values are assigned to two of the coordinates, as x and y, there are two equations from which to determine a single unknown, z, a problem generally impossible. Hence there is only one independent variable in the equations of a curve. In general, we may make x the independent variable and place the equations in the form y = cj> 1 (x~), z = 4> 2 (x), (2) by solving the original equations (1) of the curve for y and z in terms of x. The new surfaces, y = 1 (z), z = (f>j.r), deter- mining the curve, are cylinders ($ 142), with elements parallel to OZ and Y respectively. The equation y = <\> x (x) interpreted in the plane XOY is the equation of the projection (§145) of the curve on that plane. Similarly, the equation z = i (x) i interpreted in the plane ZOX, is the equation of the projection of the curve on that plane. Hence, to find the projection of the curve (1) on the XOY plane we eliminate z from the two equations. Similarly, to find the projection on the XOZ plane we eliminate y, and to find the projection on the YOZ plane we eliminate x. Finally, the three equations *=/x(9. y-/.(0. a =/»(0 ( 3 ) are parametric equations of a curve. They may generally be put in the form y = 1 (_x), 3 = <£.,(#), by eliminating t from the first and second equations, and from the first and third equations. 324 SPACE G EOMETK Y Ex. The space curve called the helix is the path of a point which mows around the surface of a right circular cylinder with a constant angular velocity and at the same time moves parallel to the axis of the cylinder with a constant linear velocity. Let the radius of the cylinder (fig. 203) be a, and let its axis coincide with OZ. Let the constant angular velocity be w and the constant linear velocity be v. Then if 6 denotes the angle through which the plane ZOP has swung from its initial position ZOX, the coordinates of any point P (x, y, z) of the helix are given by the equations I>nt 6 = oit, and accordingly we may have as the parametric equations of the helix, x = a cos (at, y = a sin wt, t being the variable parameter. n Or, since t = —■> we may regard 6 as the CO variable parameter, and the equations are x = a cos 6, y = a sin 6, z = led, where k is the constant — • Fig. 203 160. Direction of space curve and element of arc. Let P(.r, y, z) be any point of a curve, and Q (x + Ax, y + Ay, z + Az) be any second point of the curve. Then the direction cosines of the chord PQ are A.r Ay Az ' Vaz 2 + Ay 2 + A/ y/Ax 2 + Ay 2 + Az' 2 y/Ax* + Ay 2 + Az' 2 As the point Q approaches the point P along the curve, Ax, Ay, and Az each approach zero as a limit, and the direction cosines of the chord PQ approach the direction cosines of the tangent to the curve at P as limits. To determine these limits, denote by s the distance of the point P from some fixed point CURVES 325 of the curve, s being measured along the curve. Then the arc PQ = As; and As = as PQ approaches the tangent. Ax Ax As Now Vax 2 +a/+az 2 As VZ7+a/Ta7 2 ' Ax dx whence Lim Asi0 VAl- 2 +Ay + A2 2 ds for Lim — ^==^===^ = 1. WAx'+A/ + Az 2 Proceeding in the same way with the other two ratios, we , dx dy dz ,, ,. ,. . . , . nave — , -f-, — as the direction cosines of the curve at any as as as J point, since the directions of the tangent and the curve at any point are the same. whence ds = Vdx' 2 + dy 1 + dz' 2 , (1 ) a formula for the differential of the arc of any space curve. It also follows from (1) that we may speak of the direction of the curve as the direction dx : dy : dz. Ex. 1. Find the direction of the helix x = (i cos 6, y = a sin 6, z = kd, at the point for which = <>. Here dx = — a sin dO, dy — a cos 6 dO, d: = k d$. Therefore, at the point for which $ = 0, the direction is the direction : a d6 : k dd, and the direction a k cosines are 0, Va- + k- Vo s + k* Ex. 2. Find the length of an arc of the helix corresponding to an increase of 2 ir in 6. Using the values of dx, dy, and dz found in Ex. 1, we have ds = Va 2 + P dd ; whence s = / Va 2 + k 2 d$ e, + 2tt 2 7rV« 2 + k-. 326 SPACE GEOMETRY 161. Tangent line and normal plane. If J^(x v y v s x ) is the point of tangeney, the equations of the tangent line are, by (1), § 155, x — x, y — il z — 2, * = 2 — -n = 1, f-jN dx x dy x dz l ^ J where dx x , dy v dz x are the respective values of dx, dy, dz at the point P v The plane perpendicular to a tangent line at the point of tangeney is called the normal plane to the curve. By § 151, the equation of the normal plane to the curve at ij* is dx x (x - aQ + dy^y - yj + dz^z - sQ = 0. (2) Ex. Find the equations of the tangent line and the equation of the normal plane to the helix x = a cos 9, y = a sin 6, z= kd at the point for which 6 = 0. Here x x — a, y x = 0, z x = 0, and dx 1 = 0, dij = add, d~ 1 = kdd. Hence the equations of the tangent line are x — a y — z — ~ ud8 ~ kdd ' which reduce to x — a = 0, ky — az = The equation of the normal plane is 0(x - a) + ad$(y - 0) + kdd(z - 0) = 0, which reduces to ay + kz = 0. PROBLEMS 1. Describe the surface if — 4// — 2x — 0. 2. Describe the surface y(z — 1) = 1. 3. Write down the equation of a right circular cylinder of radius a. 4. Show that the surface ax + by = cs 2 is a cylinder, and describe its directrix and generatrix. 5. Describe the surface x 2 + y 2 + z 2 — 6 z = 0. 6. Describe the surface xyz = a 3 . 7. Describe the surface 9x 2 + 4 « 2 = 12 # - 24. 8. Describe the locus of the equation x — (z + 2) 2 = 0. PROBLEMS 327 9. Show that the surface (ax 4- byf = cz is a cylinder, and describe its directrix and generatrix. 10. Describe the surface 36 x 2 —12x + < dif + ±z 1 = 0. 11. All sections of a given right cylinder made by planes parallel to the plane XOZ are ellipses of which the longest chord is 10 in. and the shortest is 8 in. What is the equation of the cylinder '.' 12. Describe the surface x* 4- if + z 3 = '/ f . 13. Show that the surface z = a — vV J 4- //- is a cone of revolul ion, and find its vertex and axis. 14. Find the equation of a prolate spheroid, i.e. the surface gen- erated by revolving an ellipse about its major axis. 15. Find the equation of an oblate spheroid, i.e. the surface gen- erated by revolving an ellipse about its minor axis. 16. Describe the locus of the equation x a 4- -1 xy 4- 4 if — 4 .-.- = 0. 8 a 9 17. Describe the surface z = —, 5 : — ;• ■'- + //- 4- 4 it- is. Find the equation of the cone of revolution formed by revolving the line z = 2x about OX us an axis. 19. Describe the locus of the equation -.>■'- — 3x — 2 = 0. 20. Find the equation of a parabolic cylinder the elements of which are parallel to OX and the directrix of which is in the plane YOZ. //- - A 21. Describe the surface '— + ■— + —.=1. n- Ir r 4 22. Describe the surface if —(2a — //)(.-.'- + x 2 )— 0. 23. Describe the surface x' 2 — f — 2 ./■ + 4 y — 0. 24. Find the equation of the cone of revolution formed by revolv- ing the line 3// = 2x +1 about the line y =1 in the plane A'OT as an axis. What are the coordinates of the vertex of the cone ? 25. Show that the surface ./•'■ + 2 f - 3 z 2 + 2 x - 12 y + 1 2 z + 7 = is a cone with its vertex at the point (—1, 3, 2). What are its cross sections made by plunes parallel to the plane XOY? 26. Describe the surface (x — a)x 2 + (•'• + «) if + -~") = 0. 27 . Find the equation of the ring surface formed by revolving the ellipse 4-^=1 (a > b) about <>Y as an axis. 328 SPACE GEOMETRY 28. Describe the surface 4 s 2 = y 2 (9 — a; 2 ). 29. If P(x, y, z) is situated on the straight line drawn from P&v I/v -x) to I\(x 2 , y i} z 2 ) so that P 1 P = k(P x P£, prove that x = x 1 + k(x 2 -x 1 ), y = y 1 + A;(y 3 -y ] ), z = z 1 + k (« a - « x ). i //i + & *i + *a\ ig the middle point of 2 2 2 the straight line joining P 1 (a; 1 , y^ zj and P 2 (x 2 , y 2 , z 2 ). 31. Find the equation of the sphere constructed on the straight line joining (3, — 1, 3) and (5, 3, 5) as a diameter. 32. Find a point of the plane x+3y+z=0 equally distant from the three points (1, 1, 1), (0,12, 1), (2, 1, 2). 33. Find the points distant 5 from the points (—2, — 2, 1), (3, - 2, 6), (3, 3, 1). 34. Find the point of the plane x + 2y + 3p — 6 = equally distant from the points where the plane is pierced by the three coordinate axes. 35. Find the equation of the sphere passing through the points (- 1, 1, - 5), (- 2, 4, 3), (- 5, 0, - 2), (7, 1, - 1). 36. A point moves so that its distances from two fixed points are in the ratio k. Prove that its locus is a sphere or a plane according as k =£ 1 or k = 1. 37. Prove that the locus of points from which tangents of equal length can be drawn to two given spheres is a plane perpendicular to their line of centers. 38. A straight line makes the same angle with the three coordi- nate axes. What is that angle ? 39. Prove that a straight line can make angles 60°, 45°, 60° respectively with the coordinate axes. 40. Find the direction cosines of the straight line determined by the points (1, 3, 5), (2, - 1, 4). 41. A straight line makes an angle of 30° with OX and equal angles with O Y and OZ. What is its direction ? 42. Find the angle between the two straight lines joining the origin to the points (1, 2, 1) and (3, — 1, 3). PROBLEMS 329 43. Prove that the three points (5, 3, - 2), (4, 1, - 1), (2, - 3, 1) lie on one straight line. 44. Through the point (1, — 3, 1) of the straight line having the direction 1:2:3a straight line is drawn to the point (4, 2, 0). Find the angle between the two lines. 45. Prove that P^-l, 2, 1), P 2 (2, 3, 5), and P,(4, 5, 3) are the vertices of a right triangle. 46. Find the equation of a plane passing through the point (— 2, 3, — 4) parallel to the plane x — 3y -{- 7 z — 11 = 0. 47. Find the equation of a plane passing through the point (5, — 2, 7) equally inclined to the three coordinate axes. 48. Find the equation of a plane perpendicular to the straight line joining the points (1, 3, 5) and (4, 3, 2) at its middle point. 49. Find the equation of a plane passing through the point (1, 1, 2) perpendicular to the straight line determined by the points (1, - 1, 1) and (3, 1, 3). 50. What is the angle between the planes 2x-\- y — 72 + 11 = 0, 5a5-2y + 5«-12 = 0? 51. Find the angle between the planes 3x + 2 y — 4 = 0, 2y -\-3z + 13 = 0. 52. Find the equations of the straight line determined by the points (6, 2, - 1) and (3, 4, - 4). 53. What are the equations of the straight line determined by the points (2, 3, 5) and (1, -1,5)? 54. Find the equations of a straight line passing through the point (0, 3, 5) perpendicular to the plane x + 3y + 5z — 9 = 0. 55. A straight line is drawn through the point (4, 6, — 2) parallel to the straight line drawn from the origin to the point (1, — 5, 3). What are its equations ? 56. A straight line making angles 60°, 45°, and 60° respectively with the axes of x, y, and z passes through the point (2, — 2, 2). What are its equations ? 57. A straight line passes through the point (2, — 5, 2) parallel to OY. What are its equations ? 58. Find the direction cosines of the line 4cc — 3y — 4 = 0, 12a; _ 3s -15 = 0. 330 SPACE GEOMETRY 59. Find the direction cosines of the line 3x + y — 1 z — 6 = 0, 2a;-3y + 4»-7=0. 60. Find the equations of the straight line passing through (1, 3, — 5) parallel to the line y = 3 a- — 14, 7 x — 2z = 17. 61. Prove that the three planes x — 2 y + 1 = 0, 7 // — z — 4 = 0, 7x- 22 + 6=0 are the lateral faces of a triangular prism. 62. Find the angle between the line 3x — 2 y — 4 = 0, y + 3z + 5 = and the plane 3 x + y - 2 z + 31 = 0. 63. Find the distance of the plane 2* + 3s +11= from the origin. 64. Find the locus of points distant 3 from the plane x + y + z + 3 = 0. 65. Find the locus of points equally distant from the planes x + 2 y + 3 z + 4 = 0, x - 2 y + 3 z - 5 = 0. 66. Find a point on the line 3x — 2 y — 11=0, 2x — y— z — 5 = equally distant from the points (0, 1, 1) and (1, 2, 1). 67. Find the equation of the plane passing through the point (2, — 3, — 2) perpendicular to the line 2 x + y — 5z — 7=0, y + 2 z -4 = 0. 68. Find the equation of a plane four units distant from the origin and perpendicular to the straight line through the origin and (1, - 5, 6). 69. A straight line is drawn from the origin to the plane 2x + y + 2 z — 5 = 0. It makes equal angles with the three coordinate axes. Find its length. 70. Find the coordinates of a point on the straight line determined by (- 1, 0, 1) and (1, 2, 3) and 3 units distant from (2, - 1, 1). 71. Find the foot of the perpendicular drawn from (3, — 2, 0) to the plane 2x + y — 4«+17=0. 72. Find the length of the projection of the straight line joining the points (1, 2, 1) and (2, —1, 2) upon the straight line determined by the points (2, 1, 3) and (4, 4, 6). 73. Find the equation of the plane determined by the three points (1, 3, - 2), (0, 2, - 10), and (- 2, 4, - 6). ' 74. Find the direction of the normal to the plane determined by the three points (1, 2, 3), (-1, - 2, - 3), (4, - 2, 4). PROBLEMS 331 75. Find the point of intersection of the lines fs + 2y-3 = 1 f3y + 5* + 15 = 01 \x + y- 2*- 9 = 0/ and l3z-2*-15 = 0/ 76. Prove that the lines \2x-2>/-z + 3 = 0J U.,-2.~-3 = 0j intersect at right angles. 77. Prove that the two lines fx + 2y-z + 7 = \ Ux-7y + 8s + 19 = 01 \2x-y + 2z + U = 0} an U-3y + 3s + 4 = J are coincident. 78. Prove that the two lines r3a._2.y-7=01 (x - z = 1 \2 // - 3 « + 7 = J 13 .r - 4 // + 3 c - 8 = J can determine a plane, and derive its equation. 79. Prove that the two lines f2,+3 // + 4 = 01 f* + 2y + * + 2 = 1 \2y + « + 3 = J ,U l2x-y + 2 2 -9 = 0/ cannot determine ;i plane. 80. Prove that the two lines Cx - 2 // - 10 = 01 f 7a; - 3 --11=0 1 \4 y - z + 17 = J aD< 17a; + 14 y - « + 43 = Oj can determine a plane, and derive its equation. 81. Find the equation of a plane passing through the line ./•+// + 3-- — 7 = 0, 3x + 2y — a = and perpendicular to the plane 2 x + // - 2 .? + 11 = 0. 82. Find the equation of a plane passing through the points (—2, 3, —2), (2, — 1, 2) perpendicular to the straight line deter- mined by the points (0, 0, 0), (1, 2, 1). 83. Find the equation of the plane determined by the point (1, 5, — 2) and the straight line passing through the point (6, — 2, 4) equally inclined to the coordinate axes. 84. Find the equation of the plane passing through the points (0, 3, 2), (2, - 3, 4) perpendicular to the plane Gx + 3y-2z + 3 = Q. 332 SPACE GEOMETRY 85. Find the equation of a plane determined by the point (2, 3, 2) and the straight line passing through (1, — 1, 1) in the direction 1:2:3. 86. Find a point on the line 5x + 3y — 1=0, 3y — 5z — 11 = equally distant from the planes 3 x + 3y — 2 = 0, 4 x + y + z + 4 = 0. 87. Find the equations of the projection of the line x -f- y + z -2 = 0, a- + 2?/ + s-2 = upon the plane 3x + y + 3z — 1 = 0. 88. Find the length of the projection of the straight line joining the points (2, 3, 4), (0, —3, 1) upon the straight line ' ='— — = '" i} ■ 89. Prove that the plane 5 x -f- 3 y — 4 z — 35 = is tangent to the sphere (x + l) 2 + {y - 2) 2 + (« - 4) 2 = 50. 90. Find the center of the circle cut from the sphere x 2 + y 2 + z 2 = 49 by the plane 4 cc + 6 y + 12 * - 49 = 0. 91. Find the equation of a plane passing through the line x + 3y + 3z + l = 0, y + 2z + l = Q and parallel to the line 2x + y-z = 0,3x + 2z-7 = 0. 92. Find the center of a sphere of radius 7, passing through the points (2, 4, — 4) and (3, —1,-4) and tangent to the plane 3 x — 6 y + 2z + 51=0. 93. What kind of line is represented by the equations x 2 + z? _ 4 y = 0, y-2 = 0? 94. What kind of line is represented by the equations x 2 — 9y -36 = 0, x + 5 = 0? 95. What is the projection of the curve y 2 + z 2 — 6 x = 0, z 2 = £y on the plane XOY? 96. What is the projection of the curve x 2 + y 2 = a 2 , y 2 + ,- 2 = a 2 on the plane XOZ ? 97. Find the projection of the curve x 2 + 3 y 2 — z 2 = 0, a? 2 + y 2 — 2 a; = on the plane XOZ. 98. Find the projection of the curve x 2 + 2 y 2 — z 2 = 1, 2 x 2 — y 2 = 8«on the plane YOZ. 99. Show that the curve x 2 + y 2 = a 2 , y = z is an ellipse. (Rotate the axes about OX through 45°.) 100. Find the projections of the skew cubic x = t, y = t 2 , z = £ s on the coordinate planes. PROBLEMS 333 101. Prove that the projections of the helix x = a cos 6, y = a sin 0, z = kd on the planes XOZ and YOZ are sine curves, the width of each arch of which is kir. 102. What is the projection of the curve x = e', y = e~\ z = t V5 on the plane XOV? 103. Turn the plane XOZ about OZ as an axis through an angle of 45°, and show that the projection of the curve x = e', y — e~ \ z = t V2 on the new XOZ plane is a catenary. 104. Show that the curve x = tr, y = 2t, z = t is a plane section of a parabolic cylinder. 105. Prove that the skew quartic x = t, y = t s , z = t* is the inter- section of an hyperbolic paraboloid and a cylinder of which the directrix is the cubical parabola y = x s . 106. The vertical angle of a cone of revolution is 90°, its vertex is at O, and its axis coincides with OZ. A point, starting from the vertex, moves in a spiral path along the surface of the cone so that the measure of the distance it has traveled parallel to the axis of the cone is equal to the circular measure of the angle through which it has revolved about the axis of the cone. Prove that the equations of its path, called the conical helix, are .;• = t cos t, y = t sin t, z = t. 107. Show that the helix makes a constant angle with the elements of the cylinder on which it is drawn. 108. Find the angle between the conical helix x=tcos t, y=t8in t } z = t and the axis of the cone, for the point t = 2. 109. Show that the angle between the conical helix x = t cos t, y = t sin t, z = t and the element of the cone is tan -1 — -=• J V2 110. At what angle does the curve x = a (1 — cos 6), y = a sin 6, z = <>/)_ Um f(^y + ^~)-f(^y) (2) cy ^ y =o Ay Graphically, if 2 =/(./-, y) is represented by a surface, the rela- tion between z and x when y is held constant is represented by the curve of intersection of the surface and the plane y = const., and — is the slope of this curve. Also, the relation between z ex and y when ./• is constant is represented by the curve of inter- section of the surface and a plane x = const., and — is the slope of this curve. Thus, in fig. 204, if PQSR represents a portion of the surface z =f(x, y), PQ is the curve y = const., and PI! is the curve x = const. Let P be the point (x, y, z), and LK=PK'=Ax, LM= PM' = Ay. 336 PARTIAL DIFFERENTIATION Then LP=f(x,y), KQ=f(x + Az, y), MR=f(z,y + Ay-), K'Q=f(x + Ax,y)-f(x,y), M'Ji=f(x, y + Ay) -f(x, y), and dz t • K 'Q 1 * C — - = Lim = slope ot PQ, dx PK' dz T . M'B . , „ — - = Lim : = slope of PR. dy PM' l z p_____ — —^7 R M ^^lS^==^ N' /^\ / A I h T Fig. 204 Ex. 1. Consider a perfect gas obeying the law ct We may change the temperature while keeping the pressure unchanged. If At and Av are corresponding increments of t and v, then c(t + A/) ct cAt Av — — '- = P P P and dv _ c dt ~ p Or we may change the pressure while keeping the temperature un- changed. If Ap and At" are corresponding increments of p and v, then A..- ct ct = ctAp p p< and *P p + Ap _ct_ -2- .PAP So, in general, if we have a function of any number of variables f(x, y, — • •, g), we may have a partial derivative with respect to each of the variables. These derivatives are expressed by the sym- df of df bols dx dy ••> -T- » or sometimes hyf x (x, y, 0»/»0*y» •■•■»*)» DEKIVATIVES 337 •■;f g (z, y, • • •, z). To compute these derivatives, we have to apply the formulas for the derivative of a function of one variable, regarding as constant all the variables except the one with respect to which we differentiate. Ex. 2. /= x 3 — 3 x 2 y + y 3 , Ex. 3. /= sin (a 2 + y 2 ), V _ dx 3x 2 - 6xy, c.r 2 x cos (x a + .'/-•) of _ -3x 2 + 3y 2 Ex. 4 • / = lo g Vx 2 y cos (x 2 + /r) 2 + !f + z-, sf = ex . ./• x- + .'/- + z 2 ' .'/ T + to In (1) — means the limit of the ratio of the increment of x to an increment of r when is constant. Graphically (fig- 205), OP = r is increased by dx PR PQ = Ar, and PR = Ax is thus determined. Then — = Lim — — = cos 6. cr PQ 338 PA RTIAL DIFFERED TIATION Also — in (2) means the limit of the ratio of the increment of r to that of ./■ when y is constant. Graphically (fig. 206), OM = £ is increased by MN = PQ = Ax, and JIQ = A/- is thus determined. Then — = Lim — — = cos 6. dx dr Sx c6 dX PQ It happens here that — = — But — in (1) and — in (2) are neither equal . . Br dx dd V J ex K J L nor reciprocal. 7 Q 1 R \° J / - p Q dy rt 2 Jyi INCREMENT AND DIFFERENTIAL 339 Similarly, the third partial derivatives of f(x. number, namely, , y) are four in urn, dx\cx-J of cx s ' a /ay\ dyW)~ = g / gy \ a 2 /an = da?\dy) . JZ_, drdy d /dj\ dx\cy'y _g/gyv dy\dxdy) df\dx) ay cxcy' 1 a ; ' + ''/* So, in general, ^ ai S signifies the result of differentiating f (■>'■< >/) p times with respect to x and q times with respect to ?/, the order of differentiating being immaterial. The extension to any number of variable's is obvious. 164. Increment and differential of a function of two variables. Consider z=f(x, ?/), and let x and y be given any increments Ax and Ay. Then z takes an increment A.?, where A2 =/(.,• + Ax, n + Ay) -/(./■. // ). (1) In fig. 204, A7>* = 2 ' =/< ./• + A/-, v/ + Ay ) and J!FS=z' -z = Az. (2) If :r and y are independent variables. A./' and Ay are also independent. Thus the position of S in fig. 201 depends upon the choice of LK and LM, which can be taken at pleasure. The function z is called a continuous function of x and y if A.z approaches zero as a limit when A./- and Ay approach zero as a limit in any manner whatever. Tims, in fig. 201, if 2 is a continuous function of x and y, the point S will approach the point P as LK and LM approach zero, no matter what curve the point N traces on the plane XOY or the point S on the surface. We shall assume that z and its derivatives are continuous functions. 340 PARTIAL DIFFERENTIATION The expression for Az may be modified as follows : The line N'S may be separated into two portions by drawing from Q a line parallel to K'N meeting NS in N". Then N'S = N'N" + N"S= K' Q + N" S. (3) The line K'Q is connected with the slope of PQ by the relation K , ~ „ Lim - — r = slope of PQ = ~, PA' ex the limit being taken as PK' = Ax approaches zero. K'Q dz Hence — - = he,, PK' dx * where e x approaches zero as Ax approaches zero, so that a "M! + %> a *- (4) Also the line N" S is connected with the slope of QS by the relation N ,, „ Lim — ~— = slope of QS, the limit being taken as QN" = Ay approaches zero. But as Ax = 0, the curve QS approaches the curve PR. Hence we are justified in saying N" S dz Lim = slope of PR = — , QN" F dy the limit being taken as both Ax and Ay approach zero. N"S dz Hence -= |-e„, QN" dy 2 where e 2 approaches zero as Ax and Ay approach zero, so that **«-(!+■,,) ah (5) since QN" = PM' = Ay. Substituting from (4) and (5) in (3) and then in (2), we have j, o Az = ■£ Ax + ^ Ay + e x Ax + e^Ay. (6) DIFFERENTIAL 341 In a manner analogous to the procedure in the case of a function of one variable (§ 77), we separate from the incre- ment the terms e J Az + e„Ay, call the remaining terms the total differential of the function, and denote them by dz. The differentials of the independent variables are taken equal to the increments, as in § 77. Thus, we have by definition, when z is a function of two independent variables x and y, , dz , dz , ,„^ dz = — dx + — dy. (7) dx Cy K y In (7) dx and dy may be given any values whatever. If, in particular, we place either one equal to. zero, we have the partial differentials, indicated by d x z and d y z. Thus d x z = — dx, d z = — dy. ex " dy A partial differential expresses approximately the change in the function caused by a change in one of the independent variables ; the total differential expresses approximately the change in the function caused by changes in all the inde- pendent variables. It appears from (7) that the total differen- tial is the sum of the partial differentials. Ex. The period of a simple pendulum with small oscillations is •Wj- T U . \ TV -I J y.o Let / = 100 cm. with a possible error of }, mm. in measuring and T = 2 sec. with a possible error of T ^ ff sec. in measuring. Then dl = ± J G Also d g = ±£dl.-%£dT, and we obtain the largest possible error in g by taking dl and dT of oppo- 10.36. site signs, say J/ = 5(j» dl i [TJTJ- Then dg - The ratio of error is ■s* 7T 2 = : 1.05 7T dg = .'11-0 I " dT _ T " .0005 + .01 .0105 = 1.05%. 342 PARTIAL DIFFERENTIATION 165. Extension to three or more variables. The results of the previous article may be extended to the cases of three or more independent variables by reasoning which is essentially that just employed, without the geometric interpretation, which is now impossible. For example, consider u=f{x, y, z~). (1) Let x, y, z be given increments Ax, Ay, Az, and let u' =f(x -{-Ax, y + Ay, z + Az). Then Am = u' — u. For convenience, introduce new functions u x =f(z + Ax, y + Ay, z), u 2 =f(. x + A ^' y> g > (2) Then Au — U 1 — Mj + Mj — U 2 + U 2 — U. Now S T . u — u du , u„ — u du , Lim \ = , whence \ = + e, ax= Aa; dx Ax ox x . w. — u du„ , u, — w„ dw,, , Lim -±- — - = — - , whence -±- — - = — - -+- e , a^o Ay dy Ay dy T . w' — u, du, , m' — u, du, Lim — - — i = — i , whence — - — - = -^ + e , Az = A« 02 Az cs that w — u = — As + eAx, 2 eb * w 1 -M„ = - J Ay + eAy, 1 - dy "'-"■= & A2+e ^- But — and — approach — and — as Ax and Ay approach dy dz dy dz , du du ^ du, du , TT , ON zero so that — *- = — + e ± and — ± = — + e_. Hence (2) may , tJJ dy dy * dz dz be written ■* ^ Aw = — Ax + — Ay + — Az + e,Ax + e^Ay + e^Az, (3) dx dy dz where 4 = e 2 + e 4 an( ^ , e s = e s + e 5- DIRECTIONAL DERIVATIVE 343 Then du is the part of this expression which does not contain e v e' 2 , or e' 3 , with the increments of the independent variables replaced as usual by their differentials. That is, du = — ate 4 du -\ dz. (4) OX C I i ' cz The extension to more variables is obvious. 166. Directional derivative of a function of two variables. The result of § 164 may be used to find the slope of PS (fig. 204), which is a curve cut out of the surface z —f(z, y) by any plane through LP. Draw the lines FN 1 and LN as shown in the figure, and let r , r „, rt . ft LN = PX' = Ar, where r is the distance measured from some point on the line LN produced. Denote by 6 the angle KLX= h" /'X'. which is equal to the angle made by the ['lane of PS with the plane ZOX. Then __ . _ „ . T.k A >• I M A a 6, LK Ax n LM Ay = cos 0, = — - = sin LX A/- LN A/- and the slope of PS= Lim ' - = Lim -— = — 1 the limit being PN A/- dr taken as S approaches P along PS. From (6), § 164, A* cz Ax cz Ay Ax A// — = 1 - + e he.— - A/- fa; A/- cy Ar J Ar " Ar = — cos 9 + — sin 4- e cos # 4- e sin 0. ax- e y Taking the limit, we have 'lz = c -l cos # + ££ sin # = s i ope of P & ar ex £*/ Now — measures the rate of change of z in the direction dz dz LK, — the rate of change in the direction LM. and — the rate C ' J . . dV . dz of change in the general direction LN. The derivative — is called the directional derivative in the direction of /\ 344 PARTIAL DIFFERENTIATION" Ex. The temperature u at any point of a plate is given by the formula u — — - ■ Find at the point (2, 3) the rate of change of temperature in the direction making an angle of 30° with OX. w.i.™, 8u - 2x du - 2 n dx (x* + fy > ty (x 2 + ff d at the point in question du dx~ 4 169' du 6 169' Hence the required i rate of change is du 4 = cos 30° 169 - -5- sin 30° 169 2V3 + 3 - .038. 169 167. Total derivative of z with respect to x. In fig. 204, let the point S approach the point P along any curve whatever on the surface, and not along the curve PS, as in § 166. Then the point N describes a curve on the plane XO Y, the equation of which may be taken as y = $ (:r), and | = f(*) (D is the slope of the curve described by A 7 ". During this motion of the point S, z is a function of x ; since it is in general a function of x and y, and y is a function of x. Hence z has a derivative with respect to x and dz T . Az — = Lim ax Ax Dividing the expression for Az as given in (6), § 164, by Ax, and taking the limit, we have dz = fa + dzdy^ dx dx dy dx This is the total derivative of z with respect to x. TANGENT PLANE 345 This result (2) has an important application when the curve along which S moves is on the plane XO Y. For then z = 0, a constant, and, from (2), f + TT = ' (3) cx cydx where -j- is the slope of this curve, as shown in (1). We express this result in the following theorem: 1. The value of -j- may he found from the equation /(*,£) =o by the formula ¥. + d l^l = 0. Cx Cy ax Again, let z be defined as an implicit function of x and y by the equation F ^ ^ ^ _ 0> If we hold y constant temporarily, the case reduces to the one dz discussed in theorem 1, with z in place of y and — in place 7 OX of -j— Similarly, if we hold x temporarily constant, we get theorem 1 with change of letters. Hence : 2. The values of— <■ (1) and the tangent line to PR in the plane x — x 1 is 346 P APT I AL 1) I F F E R EN T I A TION Both of these lines lie in, and hence determine, the plane of which the equation is for this equation reduces to (1) when y = y and reduces to (2) when x = % x . This plane is called the tangent plane to the surface at the point We shall prove that the plane (3) contains all tangent lines to the surface z=f(x, if) which pass through P. The line through the two points P and S has the direction Ax : Ay : As. Its equations are therefore x - x x = y- y x Ax Ay Az A- Ax (x-x\ (4) (5) As the point S approaches the point P, the line (5) ap- proaches as a limit a tangent line at P, and the equations of this tangent are 'dz dz dy ex cy dx/i l (6) An easy combination of these equations gives (3) as an equation satisfied by any tangent line. Hence the theorem is proved. If the equation of the surface is given in the form F(x, y, Z )=0, (?) TANGENT PLANE 347 the equation of the tangent plane may be found without solving for z. For, from theorem 2, § 107, d_F dz _ dx dx~~d£ ~dz~ cF dy _ dF cz Substituting these values in (3) and making a few simple changes, we have as the equation of the tangent plane, (fKH^M + (f),H)- - The straight line perpendicular to the tangent plane at the point of eontaet is the normal to the surface. Its equations are (9) (10) x — X y - //, z - *, (-) _1 y - Hx z - z , \exl UyA \d*h Ex. 1. Find the tangent plane and the normal line t<> the paraboloid - = a.r' 1 + by 2 . dz dz Here — = 2 ax and =2 hi/. Hence the tangent plane is dx dij 2 ax, (x - x, ) + 2 by, (// - y, ) - (z - z x ) = 0, or 2 ax,x + 2 by x y - 2 axf - 2 by? -z + z,= 0. But since 2 ax, + 2 by, = 2 ~ r this may be written 2 «x r r + 2 6y^ -z — z 1 = 0. x ~ x i _ V ~ V\ _ g- 8 ! . The normal is —i — -* 2 ox x 2 by, 348 PARTIAL DIFFERENTIATION Ex. 2. Find the tangent plane and the normal line to the ellipsoid TT dF 2 x dF 2 y dF Here dx a' 2 dy 0' z dz c 2 Hence the tangent plane is o r o w o - a 2 i 2 c 2 since -L + ^L 4. II = l, a 2 ft 2 c 2 The normal line is _L = 2 1! = I 11 . £1 2/i £1 a 2 b" c 2 If a curve is denned as the intersection of two surfaces by the equations f(x,y,z)=0, F(x,y,z)=0, its tangent line is evidently the intersection of the two tangent planes to these surfaces. The equations of the tangent line are therefore two equations of the form (8). The direction cosines of the tangent line can be found by the method of § 156. The normal plane may be found by the method of § 151. 169. Maxima and minima. In order that the function f(x, y) shall have a maximum or a minimum value for x = x % , y = y x , it is necessary, but not sufficient, that the tangent plane to the surface z =f(x, y~) at the point (.^, y x , z^) should be parallel to the plane XOY. This occurs when (y)= 0, (f-)=°- These are therefore necessary conditions for a maximum or a mini- mum, and in case the existence of a maximum or a minimum is known from the nature of the problem, it may be located by solving these equations. Ex. It is required to consflrnct ont of a given amount of material a cistern in the form of a rectangular parallelepiped open at the top. Required the dimensions in order that the capacity may be a maximum, if no allowance is made for thickness of the material or waste in construction. EXACT DIFFERENTIALS 349 Let x, y, z be the length, the breadth, and the height respectively. Then the superficial area is xy + 2 xz + 2 yz, which may be placed equal to the given amount of material, a. If v is the capacity of the cistern, axy — x 2 t/ 2 V = XlIZ = - — • 2(x + y) rr\ &° _ ( a ~ - x ll ~ x2 ) V 2 ? v _ ( a ~ 2 xy — y") x 2 d~x~ 2(x + y) 2 JJj~ 2(x + y) 2 For the maximum these must be zero, and since it is not admissible to have x = 0, y = 0, we have to solve the equations a — 2 xy — x 2 = 0, a — 2 xy - y 2 = 0, 1 fa ,/. In § 164, (1) was called a total differential ; it will now be called an exact differ- ential, to emphasize the fact that it may be exactly obtained by differentiation. 350 PARTIAL DIFFERENTIATION Now expressions of the form (2) arise in practice by other methods than by differentiation, or they may be written down at pleasure. For example, we may write arbitrarily the two following expressions :' (4 x» - 2 ,-/) dx + (4 f - 2 xy) dy, (3) (a?+xy)dx + y*dy. (4) It is important, therefore, to know whether an expression of the form (2) is always exact ; that is, whether it is always possible to find z=f(x, y) so that (2) is equivalent to (1). In discussing this question we note first that if (2) is equivalent to (1), we must have — = M, — = N, (5) ox dy d' 2 z dM dN whence t-^- = -— = -—. oxcy dy ex Hence, if Mdx + Ndy is an exact differential, it is necessary that d_M_d_N dy ~ dx ' ^ ' From this it appears that (4) is not an exact differential, since — = x and — = 0. On the other hand, (3) may possibly dM . . dN , be exact, since — — = — 4 xy and — — = — 4 xy. dy dx Let us now assume that the condition (6) is met, and try to find z. We may integrate the first equation of (5) consider- ing y as a constant. The constant of integration then possibly contains y and must be expressed as a function of y. Then ■/ Mdx + (y}. (7) Substituting this in the second equation of f5), we have yCMdx + V/ 3 x 2 whence /'(#) = if — —• But /'(?/) should be a function of ?/ alone, and the last equation is absurd. Equations (1) and (2) are therefore false. LINE INTEGRALS 353 171. Line integrals. The expression Mdx+Ndy occurs in certain problems involving the limit of a sum as follows : Let C (fig. 207) be any curve in the plane XOY connecting the two points L and K, and let M and N be two functions of x and y which are one-valued and continuous for all points on C. Let C be divided into n segments by the points P v i£, P z , - • -, P n _ v and let Ax be the pro- jection of one of these segments on OX and Ay its projection on OY. That is, Ax = x i+1 — x„ Ay = y i + 1 — y n where the values of Ax and Ay arc not necessarily the same for all values of i. Let the value of M for each of the n points L, P v P,, • • -, ° p IG 2 o7 P l _ 1 be multiplied by the correspond- ing value of Ax, and the value of N for the same point by the corresponding value of Ay, and let the sum i=n-l %[M(x„ y { )Ax+N(z { , yJAy] be formed. The limit of this sum as n increases without limit and Ax and Ay approach zero as a limit is denoted by / J(C) ( Mdx+Ndy), '(C) and is called a line integral along the curve C. The point K may coincide with the point L, thus making C a closed curve. Ex. 1. Work. Let us assume that at every point of the plane there acts a force which varies from point to point in magnitude and direction. We wish to find the work done on a particle moving from L to A' along the curve C. Let C be divided into segments, each of which is denoted by As and one of which is represented in fig. 208 by PQ. Let F be the force acting at P, PR the direction in which it acts, PT the tangent to C at P, and 6 the angle PPT. Then the component of F in the direction PT is F cos 6, and the work done on a particle moving from P to Q is F cos 6 As, except for infinitesimals of higher order. The work W done in moving the particle along C is, therefore, TF = Lim V Fcoa 6As= f Fcos Ods. 354 1 'A RTIAL 1 ) 1 F FE I ', EN T I ATION Now letabe the angle between PR and OX, and 4> the angle between P7 7 and ".V. Then = -sin a. Therefore X(7'' cos <£ eo.s a + 7' 7 sin sin a) ds. But /•' cos a is the component of force parallel to OX and is usually denoted by A'. Also F sin a is the component of force parallel to ()Y ami is usually denoted by )'. Moreover, COSffids = dx and fiin j 1 Jl I J Fig. 209 3 LINE INTEGRALS 855 Similarly, if the line NQ a intersects C in Q x and Q. 2 , where iVQj = x, and X'l, — ./;.„ we have l= X>-X> ./(C) the last integral being- taken also in the direction opposite to the motion of the hands of a clock. By adding the two values of .1, we have 2.1 = f (-ydx + xdy). ■ , If we apply this to find the area of an ellipse, we may take .;• = a cos , y = h sin (§ 54). Then . I = \ C^abdtft = irab. If the equation of the curve C is known, the line integral may be reduced to a definite integral in one variable. In general, the value of the line integral depends upon the curve C and not merely on the position of the points L and A". This is illustrated in Ex. 4. If, however, Mdx + Ndy is an exact differential rfz, we shall have ^(Mdx + Ndy)=pdz = z^ where z and z are the values of z at the points A and K. This result is, in general, independent of the curve <\ though special consideration may be necessary if z may take more than one value at A or A". The integral of an exact differential taken around a closed path is, in general, zero ; while the line integrals of other dif- ferentials around a closed path are not zero. (ydx - xdy). (0, 0) Let us first integrate along a straight line connecting and P x (fig. 210). The equation of the line is y = — r, and therefore along this line y dx — x dy = 0, x i and hence the value of the integral is zero. 356 PARTIAL D I F F EE EN TI ATION Next, let us integrate along a parabola connecting O and P v the equation V? of which is y 2 — — x. Along this parabola C **- «-5fer^*-i ■'u'/i- Next, let us integrate along a path consisting of the two straight lines OM x and M X P V Along OM v y = and dy = ; and along M 1 P V x — x l and dx = 0. Hence the line integral reduces to .— f l x 1 dy = —x 1 y v J o Finally, let us integrate along a path consist- ing of the straight lines ON 1 and N 1 P V Along ON v x = and dx = ; and along N 1 P V y — y x and e/y = 0. Therefore the line integral reduces t0 f Xl y 1 dx = x 1 y v "0 Ex. 5. Here r vd — ydx + xdy Fig. 210 Therefore I, (x , y ) ydx + xdy x 2 + y 2 ydx + xdy (1 ( tan- !■" c z + y 1 r>dd=e x -< If the curve C does not pass around O, Q x will be the angle shown in the figure (fig. 211). If, however, C is drawn around the origin, the final value of 6 is 2 ir + 6 V and the value of the integral is 2ir + 6 1 — O . The value of this integral around a closed curve is zero if the curve does not inclose the origin, and is 2 it if the curve winds around the origin once in the posi- tive direction. Ex. 6. Work. If X and Y are compo- nents of force in a field of force, and — — = — - > then the work done in moving a particle between two points is inde- pendent of the path along which it is moved, and the work done on a particle moving around a closed curve is zero. Also there exists a function <£, called a force function, the derivatives of which with respect to x and y give the components of force parallel to the axes COMPOSITE FUNCTIONS 357 of x and y. Such a force as this is called a conservative force. Examples are the force of gravity and forces which are a function of the distance from a fixed point and directed along straight lines passing through that point. If the components of force X and Y in a field of force are such that — 9* — i then the work done on a particle moving between two points cy dx depends upon the path of the particle, the work done on a particle moving around a closed path is not zero, and there exists no force function. Such a force is called a nonconservative force. Ex. 7. Heat. If a substance is brought, by a series of changes of tem- perature, pressure, and volume, from an initial condition back to the same condition, the amount of heat acquired or lost by the substance is the mechanical equivalentof the work done, and is not in general zero. Hence the Line integral Q = C(M(>, //). T>1 Of df C II ., 'II . . Then ti = t,^ = >( " , , r - < 2 > The proof of this formula is like that of (1), the only difference being that T . Af cf T . Am du Lun-^-=f-, Lim— = — • Ax ox Ax ex 358 PARTIAL DIFFERENTIATION Ex. 1. If 2 = sin-, y dx Place u = - • y Then by (2), dz x d /x\ 1 x — = cos 1 - ) = - cos - • dx y dx \y/ y y Similarly, dz x e (x\ x x — = cos 1 - 1 = cos - • d v y < y w y y Ex. 2. If z = f(x" + y' 2 ), show that x — - y— = 0. dy ■ ax Place u = x- + y' 2 . Then f=f(u ) ^ = 2xf(u )> ex dx dz du — =f(u) — = 2yf(u); dy J V } dy JJ V h whence x — ' — y — = 0. By J dx 3. Consider /(w, v), where ?£ = 0(V), v = -\jr(x}. From (6), § 164, with a change of letters, /}/* ?* f Af = — An + — Av + e, Au + e, Av. du dv If we divide by Ax, and take the limit, we have df_cfdudfdv dx du dx dv dx Ex. 3. Let z — tan - , where u = ;r 2 , r = loo;. I roni (2), — = sec 2 - — - = - sec 2 - , du \ v} du \vi v o dz I ., u\ 8 ln\ u .u — = ( sec 2 -) — (-)= sec 2 - dc \ vj do \vl v'" v Hence, from (3), — = - (sec 2 -) 2x — — (sec 2 -) - dx v \ vj v' 2 \ v/ x 2 x loo- X _ x I .,2 \ sec 2 (log xf \log xl COMPOSITE FUNCTIONS 359 4. Consider _f(u, v), where u = (x, y), v = ty(x, y~). Then %JL*L + %.* cx at cx cv ex \ (4) cf cf cu cf cv by cu dy dv dy The proof is like that of (3). c - d- Ex. 4. If z =f(x - y, y - x), prove — + __ = 0. Place x — y — «, y — x = v. Then z = /'(«, v), and cz _ cf du if dv _ cf _ cf cx in (i- dv dx du dv dz _ df du dfdv __ df df £# 8u dy cr c;i cu c r By addition the required result is obtained. Ex. 5. Let it be required to change ! and -■ ■ from rectangular coor- dx By dinates (x, //) to polar coordinates (r, #), where ./• = rcosO, y — /-sin#, and /is a function of x and y. From Ex.5. §16 whence cr dx = cos 9, cr dy = sin 0, cO dx = sin e d0 dy _ cos r df cr ci- CO8 0- df d$ sin8 r df = cf cr >ii + df COS* r 5. If, in (4), we multiply the equations (4) by dx and dy, add the results, and apply the definition of § 104, we have df^&du + tfdv. cu cv This result is easily generalized. Hence the form of the differ- ential df is the same whether the variables used are independent or not. 6. Higher derivatives of a composite function may be found by successive applications of the foregoing formulas. 360 PARTIAL DIFFERENTIATION The extension of all the foregoing relations to cases involving more variables is obvious. 02 y 32 y Ex. 6. Required to express — - H in polar coordinates, where V is a function of x and y. dx 2 dy 2 „ v . dV dV ,. aFsin0 t rom JirX. 5, — = — cos a - , dx dr dd r BV dV . a , dVcoad dy dr dd r Then, by (4), d 2 V dVdV a BVain'Oldr , dVdV Q aFsin0~la0 Vd 2 V a d 2 V sin0 , a F sin 01 a - ^V cos & — TT^Ti ^ ~K 7T cos " \_dr 2 drdd r dd r 2 J , ra 2 F Q e 2 Fsin0 dV . a aF cos 01/ sin0\ + -k cos xr — sin x LdrdO dd 2 r or dd r ]\ r J &V 9A a 2 F sin cos ? 2 Fsin 2 , dV sin 2 o 5Fsin0cosi = ^ co ^- 2 ^0 — r— + w— + -d7-T- + 2 Te^>— Similarly, d 2 V d 2 V . 2/1 , n c 2 Fsin0cos0 , a 2 Fcos 2 , aFcos 2 o aFsin0cosi dy* dr 2 drcd r dd* r 2 dr r dd r 2 drcd r d 2 V dx 2 a 2 f _ i dy* d 2 v , l a 2 F , l aF Hence — - H = — - + — — r- H 3/- 2 r 2 80 2 r a/- Ex. 7. If z = f, (x + at) + L (x - at), show that — a = a 2 — . T i. TU 3,i 1 S " Sl ' 1 ^W Let x + at = u, x — at = v. 1 hen — = 1, — = a, — = 1, — = — a, dx dt dx dt , , /ox a* ay: a u , rf/ 2 to .//; , a/ 8 and, by (2), — = 4 1 [- _Zl _ = ^1 + ^2 , ax «u 3x aw ox a« do Bz_df k a_« ^to_ a (]f 1 _ a (]f 1 dt du dt dv dt du do Differentiating these equations a second time, we have d 2 z = d 2 f x du d% do = d% ( d% ^ dx 2 du 2 dx dv 2 dx du 2 do 2 a< 2 " tf« 2 a< " rfw 2 dt U du 2 ° dv 2 ' By inspection the required result is obtained. PROBLEMS 361 PROBLEMS Find ;r- and — > when 1. .-- ^- 4. s = log (>/ + Vy 2 - -a?). X u 5. s = sin — ! — • x — y 6. « = e"i + log|- Va^ + y 2 2. s = tan -1 -- x .x — y 3. «== sin" 1 z « a; 7. If z = sin (a- 2 — 2 a*?/ + ?/), prove -r- + t^ 1 = 0. V 9 x 9 » * OX ClJ 8. From 2 a- 2 - 3 if + G .77/ + 2 .~ 2 = 0, prove a- r- + y j- = ». - • V & 0« « 9. If z = e x sin - > prove a- ^- + y ^- = 0. x L ex 9 dy 10. If z = y 2 + tan O j ), prove x 2 ~ + y -^= 2 y 2 . 11. If s 2 = log (a-?/) + sin -1 - > prove sa; ^ + ~y 5- = 1. v *' x ex ' ry 12. If z = -5 = e (l2 + ^, prove y ^ - x f 1 = 0. x 1 + //' 0a- dy 13. If z = V?/ 2 — a; 2 sin -1 - > prove jb ^- + y s~ — «■ 14. Show that the sum of the partial derivatives of u = (x — y) (y — z)(z — x) is zero. 15. Ifa- = « + r,y = ;-;,find(^and(|) ; 16. If x = e r sin 0, y = e r cos $, find ( — ) and f - -) • 17. If x = e" sec r, y — e u tan v, find (5-) and ( — ) • 18. If a; = re", y = re' 6 , find f ^-J and l-^J • 19. If « = (a- 2 + if) tan- 1 ^, find « 362 PARTIAL DIFFERENTIATION l :; .v 20. Ii' z = e y sin (x — ?/), find 21. If z = log(x a + y 2 ), prove ^4 + Tl = 0- 22. If ;■ = tan (y 4- aa) + (// — a#)^j prove ~ = a? — ~ 2 - Verify ^— r- = ;— tt- 1 when : 23. » = a^ + 2ye*. 24. "^ 25. g = log (a + Vy 2 + x*). V./' 2 + v 2 26. g = =! -^-- + 2/ 27. Calculate the numerical difference between Ag and rfg, when g = cb 8 + y 3 - 3 afy, a; = 2, y = 3, A* = cZz = .01, and Ay = / 2 , find the rate of change of potential at any point: (1) in a direction toward the origin ; (2) in a direction at right angles to the direction toward the origin. 35. If the electric potential I' at any point of the plane is given by the formula V= log — > find'the rate of change of B V(x + a)" + y* potential at the point (0, a) in the direction of the axis of //. and at the point (a, a) in the direction toward the point (— a, 0). 36. On the surface « = 2tan -1 — , find the slope of the curve through the point (1, 1. t ) whose plane makes an angle of 30° with the plane XOZ. 37. On the paraboloid z = 2 .r 1 -p :> //-, what plane section perpen- dicular to the plane XOY &xid through the point (2, 1, 11) will cut out a curve with the slope zero al that point'.' 38. In what direction from the point (.'•,. //,) is the directional derivative of the function z = hey a maximum, and what is the value of that maximum derivative? 39. Find a general expression for the directional derivative of the function u = e~ ■" sin x + - e~ 3 v sin 3 x at the point ( — j ). Find also the maximum value of the directional derivative. 40. If s = 4 .r 2 + 2 //" and y = -, find ^- 43. 364 PARTIAL DIFFERENTIATION v 41. If a point moves on the surface * = k tan -1 - so that its X when a" 3 + y a + ,-s 2 — log v/,? 2 = c. ex OlJ d- dz 46. Find y and Y' when ^ 3 + ^ + y ^ z + ^ = °" 47. Find ^ and y , when (x 2 + y 2 + * 2 ) 3 = 27 xyz. 48. Find the equations of the tangent plane and the normal line to the ellipsoid x 2 + 3 y 2 + 2 z 2 = 9 at the point (2, 1, 1). 49. Find the equations of the tangent plane and the normal line to the surface xy 4- yz + «* =1 at the point (1, 0, 1). 50. Find the equations of the tangent plane and the normal line to the surface z = (ax + by)' 2 at the point (x v y v z x ). 51. Find the tangent plane to the cone x 2 + y 2 — z 2 = and prove that it passes through the vertex and contains an element of the cone. 52. Show that the sum of the squares of the intercepts on the coordinate axes of any tangent plane to the surface x* + y* 4- z 1 = a 5 is constant. 53. Show that any tangent plane to the surface z = Jcxy cuts the surface in two straight lines. 54. Find the equations of the tangent line and the normal plane to the curve xyz = 1, y' 2 = x at the point (1, 1, 1). 55. Find the equations of the tangent line and the normal plane to the curve x = sin z, y = cos z at the point (l, 0, —)• 56. Find the equations of the tangent line and the normal plane to the curve of intersection of the cylinders x 2 + y 2 = 25, v/ 2 +s 2 = 25 at the point (4, 3, — 4). PROBLEMS 365 57. Find the equations of the tangent line and the normal plane to the curve of intersection of the ellipsoid 24.r 2 + 16//--|-3.? 2 = 288 and the plane 2a; + 8y-f5« = 0atthe point (2, — 3, 4). ?/ 58. Find the angle at which the helix x 2 + if = a 2 , z = k tan -1 '- intersects the sphere x 2 + if + z 2 = r(r > a). 59. Find the angle at which the curve f — z 2 = a 2 , x = l>(y + z) intersects the surface x 2 + 2 zy — c 2 . 60. Find the minimum value of the function z = £x 2 — 3xy + 9y 2 + 5x + 15y + 16. 61. An open rectangular cistern is to be constructed to hold 1000 cu. ft. Required the dimensions that the cost of lining should be a mininium. 62. Divide the number a into three parts such that their product shall be the greatest possible. 63. Find a point in a plane quadrilateral such that the sum of the squares of its distances from the four vertices is a minimum. 64. Find the volume of the greatest rectangular parallelepiped inscribed in an ellipsoid. 65. Find by calculus the point in the plane 2 ./• -f ■"> // — ('. .-. -(-5 = which is nearest the origin. 66. Find the points on the surface 2r J 4- 1 f — .-;'- — (/>./■ + 5 y -(- 18 = which art' nearesl the origin. 67. Find the highest point on the curve of intersection of the hyperboloid aj 9 + //" —*-'" = 1 and the plane as + y + 2« = 0. 68. Find the volume of the greatest rectangular parallelepiped which can be inscribed in a right elliptic cone with altitude h and semiaxes of the base a and b, assuming that two edges of the parallelepiped are parallel to the axes of the base of the cone. 69. Through a given point (1, 1, 2) a plane is passed which with the coordinate planes forms a tetrahedron of minimum volume. Find the equation of the plane. 70. Find the point inside a plane triangle from which the sum of the squares of the perpendiculars to the three sides is a minimum. (Express the answer in terms of K, the area of the triangle ; a, b, c, the lengths of the three sides ; and x, y, z, the three perpendiculars on the sides.) 366 PARTIAL DIFFERENTIATION Prove that the following differentials are exact and find their integrals : 7 1 . (5 x* - 3 afy + 2 x,f) dx + ( 2 xh, - s 8 + 5 ,f) dy. 72. (y + -\dx+(x + -\dy. 73 a l + 2 ? + a V 2^±1 ay «y * , r 2 + y3 Q J' 3 , 75. !i,fe _/£*-? + ^W y \y" y/ l 77. (cos 2 2C — // sin x) da; + cos a- dy. 78. e 2 sin (x + y) da; + e' 1 [sin (a; + y) — y cos (a: + y)~\dy. (0,0) (1) along a straight line, (2) along a parabola with its axis on OX. Jr«a.o) [(.r 2 + y" 2 )^.r + xdy\ (0,1) (1) along a straight line, (2) along a circle with its center at 0. r &, i) 81. Find the value of j [y*dx + (xy + y")dy], «7(0, 0) (1) along a parabola with its axis on O.Y, (2) along a broken line consisting of a portion of the ,r-axis and a perpendicular to it. 82. Find the value of / I ," " „ + ? ' f'7 xdx r; (1) along the curve x = t, y = t 2 , (2) along a broken line consisting of a portion of the a--axis and a perpendicular to it. 83. Find the value of J/">(5, (3,0) PROBLEMS 30' (5, 4) _ y ( J X _|_ x dy Vx 2 - f (1) along the curve x = 3 sec 0, y = 3 tan 0, (2) along a broken line consisting of a portion of the cc-axis and a perpendicular to it. 84. Find, by the method of Ex. 3, § 171, the area of the four- cusped hypocycloid x = a cos 8 , y = a sin 8 <£. 85. Find, by the method of Ex. 3, § 171, the area between one (arch of a hypocycloid (§ 58) and the fixed circle. 86. Find, by the method of "Ex. .">, ^ 1 7 1 . the area between one arch of an epicycloid (§ 57) and the fixed circle. 87. If u =f(x, y) and y = F(x), find dx* dx*\t // ' dxi 1 1 dxdy < '/ i x W' T „ , „ '/"'/ cx~ \dy/ dxdy dxdy dv 88. If f(x, y) = 0, prove —7' - [cy /x\ t - d~ 89. If z =/(-), prove x ' +yi~ = 0. 90. If f(lx -f my + nz, x 2 -+- y' 1 + » 9 )= <>, prove I ly - mx) + (ny - mz) ^ + (A- - nx | ^ = 0. 91.If« = ^ + 2/(i + lo gy ), P rove| = 2 y -^|. ?/\ , ,M -i d ' z , o„.. ^* , ■•''-'■- 93. If c = x K + ^r •' , ] >rove x 2 -T-T, + 2 xy -=-=r + f — = 0. \xj r \x/ and F is any function of x and ?/, 0a V d*V 2 d 2 V prove a a -^-5 — -5-5 = a- 5— «- • 98. If .x = e" cos v, y = e" sin ?;, and F is any function of x and y, d 2 V find r — — in terms of the derivatives of V with respect to x and y. dado 99. If x = e u cos v, y = e" sin -v, and F is any function of x and y, d 2 V d 2 V ./PV , d 2 F\ P rove ^ + ^ = e ~"'W + W e« _i_ g-o ^0 e -e 100. If x = 1 — — > y = ) n~ — > and F is any function of d 2 V fflV d 2 V 1 d 2 V ldV x and y, prove ■=-= — -z-= = -^-r g ^ i ^— • • n l dx 1 dy 1 Ci- 2 r 2 d6 2 r dr 101. If x = e v sec u, y = e v tan it, and <£ is any function of x and y, / d 2 4> d\ (d 2 4> d 2 \ I 2 a a 2 <*> 102. If x -t y = 2 e e cos , x — y = 2 ie 8 sin <£, and F is any func- ' a 2 r a 2 F , a 2 F tion of x and y, prove _ +_ « 4ay ^ 103. If x=f(u, v) and y=(u, v) are two functions which satisfy the equations ^- = -r 21 > ^- = — -^-> and V is any function of d 2 V d 2 V (c 2 V 8 2 V\V/df\ 2 (d/Vl x and y, prove ^ + -^ = (^ + -^)[{^) +{ Fu ) [ CHAPTER XVI MULTIPLE INTEGRALS 173. Double integral with constant limits. By definition, f(x) dx = Lira 2) /OO Ax, and f(x) dx is the element of the integral. In the problems of Chapter XIII it has been possible to form the element f(x)dx immediately by elementary theorems of geometry and mechanics. There are problems, however, in which it is advantageous to determine the element itself as a definite integral. For example, let us find the volume bounded by the planes z = 0, .r = a, x = b(a n,j ' /,ir - (4) Integrals (3) and (4) are called double definite integrals, t la- limits in this case being constants. As any function f(x, y ) may be represented graphically by the surface 2 =/(./■, //), we are led to the general definition of the double definite integral, J j /('•, y)dydx = C J ./(,-. y)dxdy, as equal to the limit, as m and n are both increased indefi- nitely, of the double sum X ' %/(&, //,)A.rA//, (5) where Ar, A//, and (x c //, ) have the meanings already defined. The integral is called the double integral of f(x, // ) over the area bounded by the lines x = a, x = b, y = <; y = d. This definition is independent of the graphical interpretation, and therefore any problem which leads to the limit of a sum (5) involves a double integral. 174. Double integral with variable limits. We may now extend the idea of a double integral as follows: In- stead of taking the integral over a rectangle, as in § 173, we may take it over an area bounded by any closed curve (fig. 213) such that a straight line parallel to either OX or OT intersects it in not more than two points. Drawing straight lines parallel to OY and straight lines parallel to OX, we form rectangles of area AxAy, some of which are entirely within 372 MULTIPLE INTEGRALS the area bounded by the curve and others of which are only partly within that area. Then %% f& y-) Ax Ay, (1) where the summation includes all the rectangles which are wholly or partly within the curve, represents approximately the volume bounded by the plane XOY, the surface z =f(x, y), and the cyl- inder standing on the curve as a base, since it is the sum of the volumes of prisms, as in § 173. Now, letting the number of these prisms increase indefinitely, while Az=0 and Ay = 0, it is evident that (1) approaches a definite limit, the volume described above. If we sum up first with respect to y, we add together terms of (1) corresponding to a fixed value of x, such as x t . Then if MB is the line x = :r., the result is a sum corresponding to the strip ABCD, and the limits of y for this strip are the values of y corresponding to x = x i in the equation of the curve. That is, if for x = x { , the two values of y are MA=f 1 (x i s ) and MB =/ 2 (x t .), the limits of y are f x (x^) and f % (x^). As different integral values are given to i, we have a series of terms corre- sponding to strips of the type ABCD, which, when the final summation is made with respect to x, must cover the area bounded by the curve. Hence, if the least value of x for the curve is the constant a and the greatest value is the constant b, the limit of (1) appears in the form * X f(x,y)dxdy, (2) 7i (x) where the subscript i is no longer needed. On the other hand, if the first summation is made with respect to x, the result is a series of terms each of which corre- sponds to a strip of the type A'B'C'D', and the limits of x are of the form ^(y) and „(j/)> found by solving the equation of the curve for x in terms of y. Finally, if the least value of y for the curve is the constant c and the greatest value is the constant d, the limit of (1) appears in the form n«2(Z/) f(x, y)dydx. (3) .At/) J a J,\(. DOUBLE INTEGRALS 373 While the limits of integration in (2) and (3) are different, it is evident from the graphical representation that the integrals are equivalent. So far in this chapter f(x, y) has been assumed positive for all the values of x and y considered, i.e. the surface z =f(x, y) was entirely on the positive side of the plane XOY. If, however, f(x, y) is negative for all the values of x and y considered, the reasoning is exactly as in the first case, but the value of the integral is negative. Finally, if f(x, y) is sometimes positive and sometimes negative, the result is an algebraic sum, as in § 81. 175. Computation of a double integral. The method of comput- ing a double integral is evident from the meaning of the notation. Ex. 1. Find the value of f I xydxdy. Jo Jo As this integral is written, it is equivalent to j ( j xydyjdx, the integral in parentheses being computed first, on the hypothesis that y alone varies. Ex. 2. Find the value of the integral j j xydxdy over the first quadrant of the circle x- + y 2 — a 2 . If we sum up first with respect to y, we find ;i series of terms corresponding to strips of the type ABCD (fig. 214), and the limits of y are the ordinate's of the points like A and B. The ordinate of .1 is evidently 0, and from the equation of the circle the ordinate of B is Va* — x' 2 , where OA = x. Finally, to cover the quadrant of the circle the limits of x are and a. Hence the required integral is i .i i) Fig. 214 Jo Jo J o |_ 2 Jo ~2L 2 4 Jo 374 MULTIPLE INTEGRALS 176. Double integral in polar coordinates. Let us assume that we have a function f(r, 0) expressed in polar coordinates and an area bounded by a curve (rig. 215) which is also expressed in polar coordinates. As in § 1-11, we may graphically express the function by placing z =f(r, 6), where the values of z correspond- ing to assigned values of r and 6 are laid off on perpendiculars to the plane of r and 6 at the points determined by the given values of r and 6. It follows that the graphical representation of f(r, 0) is a surface. Let us now try to find the volume bounded by this surface, the plane of r and 6, and the cylinder standing on the curve as a base. Proceeding in a manner analogous to that in § 174, we divide the area into elements, such as ABCD (fig. 215), by drawing radius vectors at distances Ad apart, and concentric circles the radii of which increase by Ar. The area of ABCD is the dif- ference of the areas of the sectors OBC and OAD. Hence, if OA = r, area ABCD=\ (r4-Ar) 2 A0-ir 2 A6> = (r-fe) Ar A0, where e=J-Ar. Then the volume of the element standing on the base ABCD is f(r, 0)rArA<9+/(r, 0)eArA0, and the total volume is the limit of the double sum 5J5jr/<>, 0)rArA0+/(r, 0)eArA0], or, what is the same thing, the limit of the double sum X X f( r > ^ r Ar Ae = fff( r > e ^> r dr de - 0-) If the summation in (1) is made first with respect to r, the result is a series of terms corresponding to strips such as A 1 B l C\D iy and the limits of r are functions of 6 found from the equation of the boundary curve. The summation with respect to 6 will then add all these terms, and the limits of 6 taken so as to cover the entire area will be constants, i.e. the least and the greatest value of 6 on the boundary curve. DOUBLE INTEGKALS 375 If, on the other hand, the summation is made first with respect to 0, the result is a series of terms corresponding to strips such as A 2 B 2 C\ 2 D 2 , and the limits of 6 are functions of r found from the equation of the boundary curve. The summation with respect to r will then add all these terms, and the limits of r will be the least and the greatest value of r on the boundary curve. Now f(r, 6) may be any function, and (1), which is inde- pendent of the graphical representation, is called the double definite integral over the area considered. Furthermore, the area of ABCD has been denoted in (1) by rdrd0, i.e. by the product of AB and AD, for AB = dr and AD = rdO. Ex. Find the integral of /•- over the circle r = 2acos0. If we sum up first with respecl to r, the limits are and '2 a cos 6, found from the equation of the boundary curve, and the result is a series of terms corresponding to sectors of the type A OB (fig. 216). To sum up these terms so as to cover the circle, the limits of 6 are and • The result is ,£ "**-/. , ,[3. "* = f\4:a*COB*dd$ Fig. 21G We might have solved this problem as follows: Since the Initial line is a diameter of the circle and the values of r' 1 at corresponding points of the two semicircles are the same, it is evideni that the required integral is twice the integral taken over the semicircle in the first quadrant. By this method the result is J-. ,, jiiai o Jo la i f 2 co8 i 0dd Such use of symmetry as was made in the second solution above is so often of advantage that the student should always note when there is symmetry, and arrange his work accordingly. 376 MULTIPLE INTEGRALS 177. Area bounded by a plane curve. Let us in (2), § 174, denote ,f\{x) by y x and f 2 (x) by y 2 , and omit f(x, y). The result is ph r „ I I 'dxdy, (1) J a J V\ which is evidently the area bounded by the curve in fig. 213. But nh nyi nb yjdx, (2) where (j/ 2 — y^)dx is the area of the rectangle ABCD. In the same way we may transform (3) of § 174 into I (x^-x^dy, (3) which will represent the same area that is represented by (2), (.r 2 — x^) dy being the area of the rectangle A'B' C'D'. It is evident that, if the area bounded by a plane curve expressed in rectangular coordinates is found by double integration, the result of the first integration is an integral of the type given in § 125. Ex. Find the area inclosed by the curve (y - a; -3)2 = 4-^ (fig. 217). Solving the equation of the curve for y in terms of x, we have y x + 3 ± V4 Accordingly we let y 1 = x + 3 — V-i — x 2 and y 2 = x + 3 + V4 — x 2 , whence y„ — y 1 = 2 V4 — x 2 , and take for the element of area a rectangle such as ABCD. Its area is 2 V4 — x 2 dx. Since the curve is bounded by the lines x = — 2 and x = 2, — 2 and 2 are the limits of integration. Hence the area = ( 2 v4 — x 2 dx = 4 ir. In like manner the area bounded by any curve in polar coordinates may be expressed by the double integral If' drd9, (4) MOMENT OF INERTIA 377 the element of area being that bounded by two radius vectors the angles of which differ by A0, and by the arcs of two circles the radii of which differ by A>\ We may transform (4) into forms similar to (2) or (3), or we may make the double integration, substituting both sets of lhnits in each problem. If the first integration of (4) is with respect to r, the result before the substitution of the limits is A rdO, which is exactly the expression used in computation by a single integration. 178. Moment of inertia of a plane area. The moment of inertia of a particle about an axis is the product of its mass and the square of its distance from the axis. The moment of inertia of a number of particles about the same axis is the sum of the moments of inertia of the particles about that axis. From this definition we may derive the moment of inertia of a lamina of uniform thick- ness k, and of density p, about any axis as follows : Divide the surface of the lamina into elements of area dA. Then the mass of any element of the lamina is pkdA. Let B i be the distance of any point of the /th element Prom the axis. We may then take as the moment of inertia of the /th element E'rpkdA, the exact expression evidently being ( R? + e^pkdA (§124). If the lamina is divided into n elements, V llfpk dA is an approximate expression for the moment of inertia of the lamina. Then, if / represents the moment of inertia of the lamina, i=n 1= Um V RfpkdA = jA'-p/cdA, (1) "=x 1 = 1 J where the integration is to include the whole lamina. If in (1) we let k=l and /o = l, the resulting equation is ■/- dA, (2) where I is called the moment of inertia of the plane area which is covered by the integration. When dA in (1) or (2) is replaced by either dxdy or rdrdO, the double sign of integration must be used. 378 MULTIPLE INTEGRALS Ex. 1. Find the moment of inertia, about an axis perpendicular to the plane at the origin., of the plane area (fig. 218) bounded by the parabola y- = -1 ((.c, the line y = 2 a, and the axis OY. We divide the area into elements by straight lines parallel to OX and OY, Then (1. 1 = dx dy, and R 2 = x 2 + //-, whence the expression for the moment of inertia of any element is (x 2 + y 2 )dxdy. If the integration is made first with respect to x, the limits of that if integration are and — , since the operation is the summing of elements 4 a of moment of inertia due to the elementary rectangles in any strip corre- sponding to a fixed value of y ; the limit is found from the axis of y, if and the limit — is found from the equation 4 a of the parabola. Finally, the limits of y must be taken so as to include all the strips parallel to OX, and hence must be and 2 a. Therefore Jo Jo Jo \U)2 a s 4 a J 178 105' Y ::::z / / u -L - f J Fig. 218 Ex. 2. Find the moment of inertia about OTof the plane area bounded by the parabola y' 2 = 4 ax, the line y = 2 a, and the axis Y. Since the above area is the same as that of Ex. 1, the limits of integration will be the same as there determined, but the integrand will be changed in that x 2 + y 2 is replaced by x' 2 , for R = x. Hence f o °f o *«x*dydx m*C* dJf o Ex. 3. Find the moment of inertia, about an axis perpendicular to the plane at O, of the plane area (tig. 219) bounded by one loop of the curve r = a sin 2 6. We shall take the loop in the first quadrant, since the moments of inertia of all the loops about the chosen axis are the same by symmetry. CENTER OF GRAVITY 379 We divide the area into elements of area by drawing concentric circles and radius vectors. Then dA = rdnlQ (§ 176), and R- = /■-, whence the element of moment of inertia is r^drdd. If the first integration is made with respect to r, the result is the moment of inertia of a strip bounded by two successive radius vectors and a circular arc ; hence the limits for r are and a sin 2 $. Since the values of 6 for the loop of the curve vary from to — > it is evident that those values are the limits for $ in the final integration. Therefore / = f - f" sl " r*d9dr = I a* C *8in< 2 6<16 Pig. 219 179. Center of gravity of plane areas. If the center of gravity of any physical body can be expressed by two coordinates x and y, we proved in § lo7 that / xdm I >i dm P'" ./' lm where x and y are the coordinates of the point at which the element of mass r regarded as concentrated. We may now place d in = pil.nl '//, or dm = prdrd0 1 where p is the mass per unit area, in which ease the above integrals become double integrals. Ex. 1. Find the center oi gravity of the segment »>t the ellipse — + 1- = 1 rt 2 or cut off by the chord through the positive ends of the axes of the curve. This is Ex. 4, § lo7, and the student should compare the two solutions. The equation of the chord is bx + ay = ab. Dividing the area into elements dxdy (fig. 220), we have dm = pdxdy, but we may omit p since it is constant. Hence, to determine x and y, we .have to compute the two integrals CCxdxdy and CCydxdy over the area A CUD, and also find that area. 380 MULTIPLE INTEGRALS The area is the area of a quadrant of the ellipse less the area of the triangle formed by the coordinate axes and the chord, and accordingly is \(Trab)- \ah= \ ah {ir -2). For the integrals the limits of integration with respect to y are Vi = ab — bx and b /— y l being found from the equation of the chord, and y 2 being found from the equation of the ellipse. The limits for x are evidently and a. f"P' JO J„b- x dx dy= fV^xVa 2 J \fl f fa ° ydxdy = — f°(- b°-x" + dPz)dx Jo Jab-bx CpJo b 2 a. Therefore 2 6 3(*-2) " 8(»-2) Ex. 2. Find the center of gravity of the area bounded by the two circles r = a cos 0, r = b cos 6. (J> > a) It is evident from the symmetry of the area (fig. 221) that y = 0. As p is constant, the denominator of x is the difference of the areas of the two circles, and is equal to ^ = l "(W-**). Since x — r cos 6, and the ele- ment of area is rdrdO, the numer- )r of x be comes Fig 221 fl f hC ° Se r 2 " 77 J a cos cos 0d6dr = K^ 3 - a 3 ) f 2 cos Odd = J»(8" -a 3 ). Therefore x = b 2 + ab + a 2 2 (b + a) AREA 381 180. Area of any surface. Let C (fig. 222) be any closed curve on the surface f(x, y, z) = 0. Let its projection on the plane XOY be C. We shall assume that the given surface is such that the perpendicular to the plane XOY at any point within the curve C meets the surface in but a single point. In the plane XOY draw straight lines parallel to OX and OY, forming rectangles of area AxAy, which lie wholly or partly in the area bounded by C. Through these lines pass planes paral- lel to OZ. These planes will intersect the surface in curves, which intersect in points the z projections of which on the plane XOY are the vertices of the rectangles ; for example, M is the projection of P. At every such point as P draw the tangent plane to the sur- face. From each tangent plane there will be cut a parallelo- gram* by the planes drawn parallel to <>Z. We shall now define the area of the surface /(•<■, Ih *0=0, bounded by the curve C, as the limit of the sum of the areas of these parallelo- grams cut from the tangent planes, as their number is made ti> increase indefinitely, at the same time that Ax==0 and Ai/ = 0. It may be proved that the limit is independent of the manner in which the tangent planes are drawn, or of the way hi which the small areas are made to approach zero. If AA denotes the area of one of these parallelograms in a tangent plane, and 7 denotes the angle which the normal to the tangent plane makes with OZ, then (§ 145) A.r A^ = AA cos 7, (1) * This parallelogram is not drawn in the figure, since it coincides so nearly with the surface element. 8S2 M U LTIPLE IN TEGlt ALS since the projection of A.'i on the plane XOY is A.rAv/. The direction cosines of the normal are, by (9), § 168, proportional dz dz to dx c// 1 ; hence cos 7 : 1 n^ar and hence and a -'=^ + ( IJ + (SJ a ^ < 2 > x—xx^RSFM^ & According to the definition, to find A we must take the limit of (3) as A.i-=0 and A t // = 0; that is, ■114 -'s) + @)** O) where the integration must be extended over the area in the plane XOY bounded by the curve C. Ex. 1. Find the area of an octant of a sjihere of radius a. If the center of the sphere is taken as the origin of coordinates (fig. 223), the equation of the sphere is x 2 + y 2 + z 2 = a 2 , and the projection of the required area on the plane XOY is the area in the first quadrant bounded by the circle y 2 + y 2 = a 2 and the axes OX and OY. From (1), —=--, v J dx z dy z' ^FWFW- Therefore A = I = -ira I dx = - Jo Jo yVr -./;-- u 2 2 Jo 2 AREA 383 Ex. 2. The center of a sphere of radius 2 a is on the surface of a right circular cylinder of radius a. Find the area of the part of the cylinder intercepted by the sphere. Let the equation of the sphere be X s + f + z* = 1 a\ (1) the center being at the origin (fig. 224), and let the equation of the cylin- der be f + ~ 2 - 2 ay = 0, (2) the elements of the cylin- der being parallel to OX. To find the projection of the required area on the plane XOY it is necessary to find the projection on that plane of the line of intersection of (1) and (2). Hence (§ lo'.t) we must eliminate : from (1) and (•_'). The result is 1 a- = 0. (3) From (2), 0, - ^(iHtr=^§ Jfl! = and dA adydx \ •_' a~y - f itive side of t!u Since the area on the positive side of the plane XOY is symmetrica] ■with respect to the plane YOZ, it is twice the area on the positive side of the latter plane. Hence we may find this latter area and multiply by 2. If the first integration is with reaped to x, the lower limit is evidently and the upper limit, found from (:'>), is V4 a 2 — 2 ay. For the final integration with respect to y the limits are <) and 2a, the latter being found from (3). adydx Therefore .1 J n Jo Vi* 8 the limits of integration all being determined from the equation r — a cos 2 6. This integral we multiply by 4 to obtain the required area. Jo Jo Va 2 - >•-' = 4rr r ¥ (l-sin2 0W0 Jo Therefore If the required area is projected on the plane YOZ, we have (5) '-J5M -'l) + (i" where the integration extends over the projection of the area on the plane YOZ; and if the required area is projected on the plane XOZ, we have fU ^B+ay** (6) where the integration extends over the projection of the area on the plane XOZ. TRIPLE INTEGRALS 385 181. Triple integrals. 1. Rectangular coordinates. Let any volume (fig. 226) be divided into rectangular parallelepipeds of volume AxAyAz by planes parallel respectively to the coor- dinate planes, some of the parallelepipeds extending outside the volume in a manner similar to that in which the rectangles in § 174 extend outside the area. Let (a;,., y p z^) be a point of intersection of any three of these planes and form the sum as in § 174. Then the limit of this sum as n, m, and p increase indefinitely, while Ax = 0, Ay = 0, Az = 0, so as to include all points of the volume, is called the triple integral of ,/'(./-, y, z) throughout the volume. It is denoted by the symbol Iff /'(./■. y, z)dxdydz, (1) the limits remaining to be substi tuted. If the summation is made first with respect to z, x and y remaining constant, the result is to extend the integration through- out a column of cross section Ax Ay ; if next ./• remains constant and y varies, the integration is extended so as to combine the columns into slices; and, finally, as x varies, the slices are combined so as to complete the integration throughout the volume. The volume of the parallelepiped with edges d.r, dy, dz is the element of volume c/F, and hence Fig. tk\ dV= dxdydz. (2) 2. Cylindrical coordinates. If the x and the y of the rectan- gular coordinates are replaced by polar coordinates r and 6 in the plane XOY, and the z coordinate is retained with its original 386 MULTIPLE INTEGRALS Fig. significance, the new coordinates r, 0, and z are called cylindri- cal coordinates. The formulas connecting the two systems of coordinates are evidently x = r cos 0, y — r sin 0, z — z. Turning to fig. 227, we see that z = z 1 determines a plane parallel to the plane XOY, that = Q x determines a plane MONP, passing through OZ and making an angle 1 with the plane XOZ, and that r — r x determines a right circular cylinder with radius r 1 and OZ as its axis. These three surfaces intersect at the point P. The element of volume in cylindrical coordinates (fig. 228) is the volume bounded by two cylin- ders of radii r and r + Ar, two planes corresponding to z and z + Az, and two planes corresponding to and + A0. It is accordingly a cylinder with its altitude equal to Az and the area of its base approximately equal to rA0Ar (§ 176). Hence, in cylin- drical coordinates, dV=rdrdddz, (3) / Fig. 228 and the triple definite integral in cylindrical coordinates is Iff f(r, 0, z)rdrd0dz, (4) the limits remaining to be substituted. 3. Polar coordinates. In fig. 229 the cylindrical coordinates of P are OM= r, MP = z, and ZLOM= 0. If instead of placing OM = r we place OP = r, and denote the angle NOP by (/>, we shall have r, , and as the polar coordi- nates of P. Then, since 0N= OP cos TRIPLE INTEGRALS 387 and 0M= OB sin , the following equations evidently express the connection between the rectangular and the polar coordi- nates of P: z = r cos (f>, x = r sin cos 6, y = r sin (f> sin 6. The polar coordinates of a point determine three surfaces which intersect at the point. For 6 = 6^ determines a plane (fig. 230) through OZ, making the angle with the plane XOZ\ 4> = (f> l determines a cone of revo- lution, the axis and the vertical angle of which are respectively OZ and 2^; and r = r^ drier- mines a sphere with its center at and radius r . The element of volume in polar coordinates (fig. 231) is the volume hounded by two spheres of radii r and r + Ar, two conical surfaces corresponding to (f> and (f> + A$, and two planes corresponding to 6 and 6 + A$. The volume of the spherical pyramid O-ABCD is equal to the area of its base ABCD multiplied by one third of its altitude r. To find the area of ABCD we note first that the area of the zone formed by completing the arcs AD and BC is equal to its altitude, rcos$ — r cos ((f) + A<£), multiplied by 2irr. Also the area of ABCD is to the area of the zone as the angle Ad is tO 2 7T. Hence area ABCD and vol O-ABCD ■■ Similarly, vol 0~EFGII= l(r+ A/-) 3 A0 [cos - cos ( + A<£)]. rA6 [r cos (f) — r cos ((f) + A$)] I r 3 Ad [cos - cos ((f) + A(£)]. 388 MULTIPLE INTEGRALS Therefore vol ABCDEFGH = J [(/• + Ar) 3 - r 3 ] A0 [cos -eos(<£+ A<£)]. i [(r + A;-) 3 - >•'] = ' /2 Ar + rAr* + I AT- 3 = (r 2 + e 1 )Ar. cos (/> — cos ((£ + A<£) = — A cos c£ = (sin <£ + e 2 ) A<£. (§77) Hence vol ABCDEFGH = (f sin + e 8 ) A>- A0 A<£, which differs from r 2 sin ArA0A by an infinitesimal of a higher order. Accordingly we let dV = r 2 sin drd /(r, (/>, 0) r 2 sin dr d<$> dd, (6) the limits remaining to be substituted. It is to be noted that dV is equal to the product of the three dimensions AB, AD, and A£J, which are respectively rd, r sin dO, and dr. 182. Change of coordinates. When a double integral is given in the form jjf(x, y)dxdy, where the limits are to be substi- tuted so as to cover a given area, it may be easier to determine the value of the integral if the rectangular coordinates are replaced by polar coordinates. Then f(x, y) becomes f(r cos 6, r sin 0), i.e. a function of r and 6. As the other factor, dxdy, indicates the element of area, we may replace dxdy by rdrdd. These two elements of area are not equivalent, but the two integrals are nevertheless equivalent, provided the limits of integration in each system of coordinates are taken so as to cover the same area. In like manner the three triple integrals HP Iff Iff f(x, y, z)dxdydz, f(r cos 6, r sin 6, z)rdrdddz, f(r sin cos 0, r sin $ sin 0, r cos (/>) r 2 sin drdd6 VOLUME 389 are equivalent when the integration is taken over the same volume in all three and the limits are so taken in each as to include the total volume to be considered. 183. Volume. In § 181 we found expressions for the element of volume in rectangular, in cylindrical, and in polar coordinates. The volume of a solid bounded by any .surfaces will be the limit of the sum of these elements as their number increases indefi- nitely while their magnitudes approach the limit zero. It will accordingly be expressed as a triple integral. Find the volume bounded by the ellipsoid /,- = 1. From symmetry (fig. 232) it is evident that the required volume is eight times the volume in the first octant b6unded by the Burface and the coordinate planes. In summing up the rectangu- lar parallelepipeds dxdydz to form a prism with edges paral- lel to OZ, the limits for z are and V. x 2 y* the latter being found from the equation of the ellipsoid. Summing up next with respect to y, to obtain the volume of a slice, we have as the lower limit of y, and 6 a/I V Fi - as the upper limit. This latte 1, found by letting ■ VI limit is determined : = in the equation by solving the equation — 4- '— of the ellipsoid; for it is in the plane z = that the ellipsoid has the greatest extension in the direction <>)', corresponding to any value of x. Finally, the limits for x are evidently and o. Therefore Jo t/0 Jo ^dxdydz dxdy 390 MULTIPLE INTEGEALS It is to be noted that the first integration, when rectangular coordinates are used, leads to an integral of the form //< (Zo-z^dxdy, where z 2 and z y are found from the equations of the boundary surfaces. It follows that many volumes may be found as easily by double as by triple integration. In particular, if z x = 0, the volume is the one graphically repre- senting the double integral (§ 174). Ex. 2. Find the volume bounded by the surface z = ae-^ + v 2 ) and the plane z = 0. To determine this volume it will be advantageous to use cylindrical coordinates. Then the equation of the surface becomes z = ae~ r ', and the element of volume is (§ 181) rdrdOdz. Integrating first with respect to z, we have as the limits of integration and ae~ r2 . If we integrate next with respect to r, the limits are and go, for in the plane z = 0, r = cc, and as ~ increases the value of r decreases toward zero as a limit. For the final integration with respect to 6 the limits are and 2tt. Therefore V= f" K f°°f ae ' rdOdrdz Jo Jo Jo = a f 2n f"re- r \l9dr Jo Jo In the same way that the computation of the volume in Ex. 2 has been simplified by the use of cylindrical coordinates, the computation of a volume may be simplified by a change to polar coordinates ; and the student should always keep in mind the possible advantage of such a change. 184. Moment of inertia of a solid. Following the method of § 178 we divide the volume of the solid into n elements Ay and multiply each element by its density p. Then, if R t is the distance of any point of the ith. element from the axis about which the moment of inertia is to be taken, we may take It'fpAv MOMENT OF INERTIA 391 as the moment of inertia of that element. If / denotes the moment of inertia of the solid, Vi?. 2 pAv is an approximate expression for /. Finally, if we let n = oo and at the same time let each element of volume approach zero as a limit, we have 1= Limit 2} Hfpkv = J Kydv, where B, p, and civ are to be expressed in terms of the same variables and the proper limits of integration substituted. In particular, if civ is replaced by any one of the three elements of volume determined in § 181, the integral becomes a triple integral. Ex. Find the moment of inertia of a sphere of radius a about a diameter if the density varies directly as the square of the distance from the diameter about which the moment of inertia is to be taken. We shall take the center of the sphere as the origin of coordinates, and the diameter about which the moment is to be taken as the axis of r. The problem will then be most easily solved by using cylindrical coordinates. The equation of the sphere will be r 2 + « 2 = a 2 , and dv = rdrd6dz ) R = r, and p = kr' 2 , so that we have to find the value of the triple integral *///•*»**■ Since the solid is symmetrical with respecl to the plane 2 = 0, we shall take and \ f = a. f ^ f f°r cos <£ • r 2 sin (fydOd^dr Then ~ = Jo Jo J\ f' n f" f V sin tf> (Id ddr Jo Jo Jo The denominator is the volume of a spherical cone the base of which is a zone of one base with altitude a (1 — cos ex) ; therefore its volume equals § ira 3 (1 — cos a). f f f >' 3 cos $ sin dr = \ a* C C cos <£ s'mcf}d6dcf> = 1 a 4 (l-cos 2 «) f 2n d9 Jo = 1 :7ra 4 (l — cos 2 a:). Therefore z = j} (1 + cos a) a. ATTRACTION 186. Attraction. The formula 'cos 6 dm P ir (§ 138) for the component of attraction in the direction OX is entirely general. Similar formulas for the components in the direc- tions OY and OZ may be deduced. The application of these formulas requires us to express B, cos 6, and dm in terms of the same variables, and to substi- tute limits of integration so as to include the whole of the attracting mass. In general, the integral, after the substitution of the variables, will be a double or triple integral. Ex. Find the attraction due to a homogeneous circular cylinder of density p, of height h, and radius of cross section a, on a particle in the line of the axis of the cylinder at a distance l> units from one end of the cylinder. Take the particle at the origin of coordinates (fig. 233), and the axis of the cylinder as ()Z. Using cylindrical coordinates, we have dm = prdrdddz and R = Vz~ + r 2 . From tlie symmetry of the figure the Fig. 233 ■sultant components of attrac- tion in the directions OX and O Y are zero, and cos 6 = resultant component in the direction OZ. 1 represent X2 7T r% a r\b i JO Jb Vr 2 for the Therefore, letting A represent the component in the direction OZ, + h we have dddrdz, (c 2 + r 2 )2 where the limits of integration are evident from fig. 233. A = p f*' ["I r - r ) dOdr Jo Jo VVi 2 + r 2 V(h + h) 2 + /•-/ = P f 2w (k + V6 2 + a* - V(/> + ?>y 2 + « 2 ) ./■ + 4 U 2 . 31. Find the moment of inertia about OF of the area of the loop of the curve y 2 = x 2 (2 — x). 32. Find the moment of inertia of the area of the cardioid r — a (1 + cos 6) about an axis perpendicular to its plane at the pole. 33. Find the moment of inertia of the area of a circle of radius a about an axis perpendicular to the plane of the circle at any point on its circumference. 34. Find the moment of inertia about its base of the area of a parabolic segment of height h and base 2 a. 396 MULTIPLE INTEGRALS 35. Find the moment of inertia about OX of the area bounded on the left by an arc of the curve y 2 = ax -4- a s and on the right by an arc of the curve x" + if = a' 1 . 36. Find the moment of inertia of the area of one loop of the lemniscate r* = 2 a 2 cos 2 6 about an axis in its plane perpendicular to the initial line at the pole. 37. Find the moment of inertia about the initial line as an axis of the area of the cardioid r = a (cos 6 -\- 1) above the initial line. 38. Find the moment of inertia of the area bounded by a semi- circle of radius a and the corresponding diameter, about the tangent parallel to the diameter. 39. Find the moment of inertia of the area of a loop of the curve v = a cos 2 6 about the axis of the loop as an axis. 40. Find the moment of inertia of the area of the circle r = a which is not included in the curve r = a sin 2 6 about an axis perpendicular to its plane at the pole. 41. Determine the center of gravity of the half of a parabolic segment of altitude 9 in. and of base 12 in. formed by drawing a straight line from the vertex of the segment to the middle point of its base. 42. Find the center of gravity of a lamina in the form of a parabolic segment of altitude 7 in. and of base 28 in. if the density at any point of the lamina is directly proportional to its distance from the axis of the lamina. 43. Find the center of gravity of the area of a loop of the curve a y = «V - x 6 . 44. Find the center of gravity of the area bounded by the parabola x* -4- ?/ 2 = a* an d the circle x 2 + if = a 2 . 45. Find the center of gravity of the area bounded by the cardioid r = a (cos 6 + 1). 46. Find the center of gravity of the area bounded by the parabola x 2 = 4 aii and the witch y = -= : — - • J J x 1 + 4 a 2 47. A plate is in the form of a sector of a circle of radius a, the angle of the sector being 2 a. If the thickness varies directly as the distance from the center, find its center of gravity. PROBLEMS 397 48. Find .the center of gravity of the area in the first quadrant bounded by the curve x$ + //'• = "•'■ and the line x 4- y = a. 49. The density at any point of a lamina in the form of a loop of the curve r == a cos 2 8 is directly proportional to its distance from the point of the loop. Determine its center of gravity. 50. Find the center of gravity of the area bounded by the limacon r = 2 cos $ 4- 3. 51. Find the center of gravity of the area bounded by the curve v = a sin - as $ changes from to 2 it. 52. Find the center of gravity of the area bounded by the cissoid x 3 f = and its asymptote. 53. Find the center of gravity of the area cut off from the lemnis- V3 cate r 2 = 2 a 2 cos 2 6 by the straight line rcos0= -jr- a. ■'■'- 'r 54. From a homogeneous elliptic plate, —, + ',., — F is cut a circular plate of radius /•(/• < -J with center at f-j OJ. Find the center of gravity of the part left. 55. Find the area of the surface cut from the] >aral >< >loid //"+ v- = 4 cos cos d6d(f> dr. %J0 i/0 t/asin0 Xtt /-» 2 tt /»a cos / I r sin 3 dd0 dr. Jo Jo PEOBLEMS 399 75. Find the volume bounded by the surface x^ + y^ + z^ = a? and the coordinate planes. 76. Find the volume of a cylindrical column bounded by the surfaces y = x 2 , x = if, z = 0, z = 12 + y — x 2 . 77. Find the volume bounded by the plane z = and the cylinders x 2 + y 2 = a 2 , f = a 2 — az. 78. Find the volume bounded by the surfaces r 2 = bz, z = 0, r — a cos 0. 79. Find the volume bounded by the sphere x 2 -\- f + z 2 = 5 and the paraboloid x 2 + if — 4 z. 80. Find the volume bounded by the cylinder z 2 = x -\- y and the planes x = 0, y = 0, z = 4. 81. Find the volume of the paraboloid y 2 + z 2 = 2 x cut off by the plane y = x — 1. 82. Find the volume bounded by a sphere of radios " and a right circular cone, the axis of the cone coinciding with a diameter of the sphere, the vertex being at an end of the diameter and the vertical angle of the cone being 90°. 83. Find the total volume bounded by the surface (x 2 ■+■ y 2 + z 2 ) 3 = 27 a s xyz. (Change to polar coordinates.) 84. Find the volume bounded by the plane XOY, the cylinder .'"" + if" — 2aa! = 0, and the right circular cone having its vertex at <>, its axis coincident with UZ, and its vertical angle equal to 90°. x 2 i/ 2 z* 85. Find the total volume bounded by the surface — 2 + 'j^ + — = 1. 86. Find the volume bounded below by the paraboloid x 2 + y 2 = az and above by the sphere x 2 + if + z 2 — 2 az = 0. 87 . Find the volume bounded by the surface b 2 z 2 = if (a 2 — x 2 ) and the planes y — 0, y = b. 88. Find the volume cut from a sphere of radius a, by a right circular cylinder of radius -> one element of the cylinder passing through the center of the sphere. 89. Find the total volume bounded by the surface (x 2 + y 2 + z 2 ) 2 = axyz. 400 MULTIPLE INTEGRALS 90. Kind the volume in the first octant bounded by the surfaces * = (a + !/f, x 1 + if = « a - 91. Find the volume of the sphere j' 2 + // 2 + z 2 = a? included in a cylinder with elements parallel to OZ, and having for its directrix in the plane XOY one loop of the curve r 2 = a 2 cos 2 0. 92. Find the volume bounded by the surfaces az = xy, x + y + z = a, z = 0. 93. Find the total volume which is bounded by the surface x* + //■'> + ~i = J. 94. Find the total volume which is bounded by the surface r 2 -f- z" = 2 ar cos 2 6. 95. Find the moment of inertia about its axis of a hollow right circular cylinder of mass M, the inner radius and the outer radius of which are respectively r x and v % : (1) if the cylinder is homoge- neous ; (2) if the density of any particle is proportional to its distance from the axis of the cylinder. 96. A solid is bounded by the plane z = 0, the cone z = r (cylindri- cal coordinates), and the cylinder having its elements parallel to OZ and its directrix one loop of the lemniscate r = 2 a 2 cos 2 6 in the plane XOY. Find its moment of inertia about OZ if the density at any point varies directly as its distance from OZ. 97. Find the moment of inertia of a homogeneous right circular cone of density p, of which the height is h and the radius of the base is a, about an axis perpendicular to the axis of the cone at its vertex. 98. A ring is cut from a homogeneous spherical shell of density p, the inner radius and the outer radius of which are respectively 4 ft. and 5 ft., by two parallel planes on the same side of the center of the shell and distant 1 ft. and 3 ft. respectively from the center. Find the moment of inertia of this ring about its axis. 99. A mass M is in the form of a right circular cone of altitude h and with a vertical angle of 120°. Find its moment of inertia about its axis if the density of any particle is proportional to its distance from the base of the cone. 100. The radius of the upper base and the radius of the lower base of the frustum of a homogeneous right circular cone are respec- tively &, its axis. PEOBLEMS 401 101. The density of any point of a solid sphere of mass M and radius a is directly proportional to its distance from a diametral plane. Find its moment of inertia about the diameter perpendicular to the above diametral plane. 102. Given a right circular cylinder of mass M, height //, and radius a, the density of any particle of which is /.• times its distance from the lower base. Find the moment of inertia of this cylinder about a diameter of its lower base. 103. Find the moment of inertia about (>Z of that portion of the surface of the hemisphere z = V«" — x'~ — f which lies within the cylinder x" + if = ax. 104. A homogeneous solid of density p is in the form of a hemi- spherical shell, the inner radius and the outer radius of which are respectively i\ and /;,. Find its moment of inertia about any diam- eter of the base of the shell. 105. A homogeneous anchor ring of mass .1/ is bounded by the surface generated by revolving a circle of radius a about an axis in its plane, distant b(b > a) from its center. Find the moment of inertia of this anchor ring about its axis. 106. The density at any point of the hemisphere z = "\Ar — x' 2 — >f is k times its distance from the base of the hemisphere. Find the moment of inertia about OZ of the portion of the hemisphere in- cluded in the cylinder x' 2 + //" = OX. 107. Through a homogeneous spherical shell of density p, of which the inner radius and the outer radius are respectively a^ and "„ a circular hole of radius b(b the vertex of which is on o the surface of the sphere, and the axis of which coincides with a diameter of the sphere. Find its center of gravity if the density at any point is k times its distance from the axis of the cone. 123. Find the attraction of a hemisphere of radius a on a particle of unit mass at the center of its base if the density at any point of the hemisphere varies directly as its distance from the base. 124. A homogeneous solid of density p is bounded by the plane z = 3 and the surface z % == x 1 + if. Find the attraction of this solid on a particle of unit mass at the origin of coordinates. 125. A portion of a right circular cylinder of radius a), the center of which coincides with the center of the base of the cylinder. Find the attraction of this portion of the cylinder on a particle of unit mass at the middle point of its' base 126. A portion of a right circular cylinder of radius a is bounded by a spherical surface of radius b(b>a), the center of winch coin- cides with the center of the base of the cylinder. Find the attraction of this portion of the cylinder on a particle of unit mass at the middle point of its base, the density of any particle of the cylinder being proportional to its distance from the axis of the cylinder. 127. Show that the attraction of a segment of one base, cut from a homogeneous sphere of radius a, on a particle of unit mass at its vertex is 2ir/ip(l — - ^l_), where p is the density of the sphere and h is the height of the segment. 128. A ring is cut from a homogeneous spherical shell of density p, the inner radius and the outer radius of which are respectively 4 ft. and 5 ft., by two parallel planes on the same side of the center of the shell and distant 1 ft. and 3 ft. respectively from the center. Find the attraction of this ring on a particle of unit mass at the center of the shell. 129. The density of a hemisphere of mass ^^ and radius a varies directly as the distance from the base. Find its attraction on a particle of unit mass in the straight line perpendicular to the base 404 MULTIPLE INTEGRALS at its center, and at a distance a from the base in the direction away from the hemisphere. 130. A solid of mass M is bounded by a right circular cone of vertical angle 90° and a spherical surface of radius 2 ft., the center of the spherical surface being at the vertex of the cone. If the density of any particle of the above solid is directly proportional to its distance from the vertex of the cone, find the attraction of the solid on a particle of unit mass at the vertex of the cone. 131. The vertex of a right circular cone of vertical angle 2 a is at the center of a homogeneous spherical shell, the inner radius and the outer radius of which are respectively a x and « 2 . Eind the attrac- tion of the portion of the shell outside the cone on a particle of unit mass at the center of the shell, in terms of the attracting mass. 132. The density at any point of a given solid of mass M in the form of a hollow right circular cylinder is directly proportional to its distance from the axis of the cylinder. If the height of the cylinder is 2 ft., and its inner radius and outer radius are respec- tively 1 ft. and 2 ft., find its attraction on a particle of unit mass situated on its axis 2 ft. below the base. CHAPTER XVII INFINITE SERIES 187. Convergence. The expression « l + a 1 + a I "+a 4 + a B + ---, (1) where the number of the terms is unlimited, is called an infinite semes. An infinite series is said to converge, or to be convergent, when tin- sum of the first n terms ajyproaches a limit as n increases without limit. Thus, referring to (1), we may place s„ =a 1 +a 2 , Then, if Lim8 n = A, the series is said to converge to the limit ./. The quantity A is frequently called the sum of the series, although, strict ly speaking, it is the limit of the sum of the first n terms. The convergence of (1) may be seen graphically by plotting * r *.,, *.., • • •, 8 n on the number scale, as in § 3. A series which is not convergent is called divergent. This may happen in two ways : either the sum of the first n terms increases without limit as n increases without limit ; or s n may fail to approach a limit, but without becoming indefinitely great. Ex. 1. Consider the geometric series a + ar + ar" + \ + k = h HHI + ?>H1+Hi = l. and in this way the sum of the first n terms of the series may be seen to be greater than any multiple of ^ for a sufficiently large n. Hence the harmonic series diverges. 188. The comparison test for convergence. If each term of a given series of positive numbers is less than, or equal to, the corre- sponding term of a known convergent series, the given series converges. If each term of a given series is greater than, or equal to, the corresponding term of a known divergent series of positive numbers, the given series diverges. Let « 1 +« 2 + rt 3+ fl, 4 + ••• C 1 ) be a given series in which each term is a positive number, and let K+h+h+\+--- (2) be a known convergent series such that a k ^ b k . Then, if s n is the sum of the first n terms of (1), s' n the sum of the first n terms of (2), and B the limit of s' n , it follows that CONVEEGENCE 407 since all terms of (1), and therefore of (2), are positive. Now as n increases, s n increases but always remains less than B. Hence s H approaches a limit, which is either less than, or equal to, B. The first part of the theorem is now proved ; the second part is too obvious to need formal proof. In applying this test it is not necessary to begin with the first term of either series, but with any convenient term. The terms before those with which comparison begins form a poly- nomial, the value of which is of course finite, and the remaining terms form the infinite series the convergence of which is to be determined. + -, + - + -+•■• + Ex. 1. Consider 1 2 s :P i* (n-1)" Each term after the third is less than the corresponding term of the con- vergent geometric .scries i + i + ^ + i + i + ... + 5 l i + .... Therefore the first series converges. Ex. 2. Consider % V2 V3 Vi V5 VTi Each term after the first is greater than the corresponding term of the divergent harmonic scries Therefore the first series diverges. 189. The ratio test for convergence. If in a series of positive numbers the ratio of the (>i+l)s. term to the nth term approaches a limit L as n increases without limit, then, if L<\, the series converges ; if L>1, the series diverges ; if L — 1, the series may either diverge or converge. Let a, + a a + a 3 -\ + a n + a n+1 H (1) be a series of positive numbers, and let Limi- i±i =X. We have three cases to consider. " =: ° " 408 INFINITE SERIES 1. L<1. Take r any number such that i1. Since - JL±1 approaches I as a limit as n increases without limit, this ratio eventually becomes and remains greater than unity. Suppose this happens for the wth and all subsequent terms. Then ^ a m+\ ~^ a m1 «m + 2> a m+l>V Each term of the series (2) is greater than the corresponding term of the divergent series a » + «» + «« + ffl J • ( 4 ) Hence (2) and therefore (1) diverges. 3. L =1. Neither of the preceding arguments is valid, and examples show that in this case the series may either converge or diverge. In applying this test, the student will usually find - R±1 hi the form of a fraction involving n. To find the limit of this CONVERGENCE 409 fraction as n increases without limit, it is often possible to divide numerator and denominator by some power of n, so as to be able to apply the theorem (§13) that Lim- = 0, or some other known theorem of limits. Ex. 1. Consider 2 3 4 5 n 3»- The nth term is — — - and the (n + l)st term is (n + l)st term to the nth term is , and 3ft j T . n + 1 T . n 1 Lim = Lim = - n =» 3fl „=„ 3 3 + 1 The ratio of the Therefore the given series converges. Ex. 2. Consider 1 + 3 8 4 4 + r + The nth term is — ■ and the (n + l)st term is (» + l)'- the (n + l)st term to the nth term is (n + !)» + ' n + 1 - = £+*)" (n + L)»" \ n / Lto(!i±l)-=Lim(l + l) n = » \ ft / n = ao \ n/ Therefore the given series diverges. The ratio of , and (§98) 190. Absolute convergence. The absolute value of a real num- ber is its arithmetical value independent of its algebraic sign. Thus the absolute value of both + 2 and — 2 is 2. The abso- lute value of a quantity a is often indicated by \a\. It is evi- dent that the absolute value of the sum of n quantities is less than, or equal to, the sum of the absolute values of the quantities. A series converges when the absolute values of its terms form a convergent series, and is said to converge absolutely. Let a i + a 2 + a z + a i~\ •be a given series, and (i) (2) 410 INFINITE SERIES the series formed by replacing each term of (1) by its absolute value. We assume that (2) converges, and wish to show the convergence of (1). Form the auxiliary series («i + KI) + («k + N) + («. + |«,|) + (a 4 + |aJ)+.... (3) The terms of (3) are either zero or twice the corresponding terms of (2). For a k = — \a k \ when a k is negative, and a k = \a k \ when a k is positive. Now, by hypothesis, (2) converges, and hence the series 2 W+2|«.l+2|« 1 |+2|aJ+... (4) converges. But each term of (3) is either equal to or less than the corresponding term of (4), and hence (3) converges by the comparison test. Now let s n be the sum of the first n terms of (1), s' n the sum of the first n terms of (2), and s'J the. sum of the first n terms of (3). Then ' s n = s'J-s' n , and, since s" and s' n approach limits, s n also approaches a limit. Hence the series (1) converges. We shall consider in this chapter only absolute convergence. Hence the tests of §§ 188, 189 may be applied, since in testhig for absolute convergence all terms are considered positive. 191. The power series. A power series is defined by % + v ; + v ;2 + a z x * ^ •■ a n x " -\ — > where « , a , a , a 8 , ■ • ■ are numbers not involving x. We shall prove the following theorem : If a 'power series con- verges for x = x v it converges absolutely for any value of x such that \x\<\x x \. For convenience, let \x\ = X, \a n \ = A n , jzj = A' a . By hyjDothe- sis the series a + a x x x + a„x\ + a z x\ -\ h a n x[ l H (1) converges, and we wish to show that A + A X X+ AX' + A Z X* + • • • + A n X' 1 + • • • (2) converges if X1. The region of convergence extends on the number scale between — 1 and + 1. 412 INFINITE SERIES Ex. 2. Consider 1 + Z + £. + ± + . . . + JL_ - + . . .. 1 [2 13 | ». — 1 — -j the (n + l)st term is — , and their ratio is -• |n — 1 [n ra Liin - = for any finite value of x. Hence the series converges for any value of x and its region of convergence covers the entire number scale. Ex. 3. Consider 1 + x + (2 x 2 + [3 X s + • • • + Ira-l a;"- 1 + • • •. The nth term is \n — l a;" -1 , the (n + l)st term is \nx n , and their ratio is nx. This ratio increases without limit for all values of x except x = 0. Therefore the series converges for no value of x except x — 0. A power series defines a function of x for values of x within the region of convergence, and we may write f(x) =%+ a x x + a^x" + a 3 x 3 -\ + a„of + • • • , (4) it being understood that the value of f(x) is the limit of the sum of the series on the right of the equation. The power series has the important property, not possessed by all kinds of series, of behaving very similarly to a polynomial. When a function is expressed as a power series it may be integrated or differentiated by integrating or differentiating the series term by term. The new series will be valid for the same values of the variable for which the original series is valid. If the method is applied to a definite integral, the limits must be values for which the series is valid. Similarly, if two functions are each expressed by a power series, their sum, difference, product, or quotient is the sum, the dif- ference, the product, or the quotient of the series. For proofs of these theorems the student is referred to ad- vanced treatises. 192. Maclaurin's and Taylor's series. We have noted that any convergent power series may define a function. Conversely, it may be shown (see § 193) that any function which is con- tinuous and has continuous derivatives may be expressed as a power series. When a function is so expressed it is possible to «o=/(°)' «x=/(0). V=|/'(°)> « 3 =|/"'(°)' • • "> ^=^/ (n) (0). MACLAUEIX'S AND TAYLOE'S SEEIES 413 express the coefficients of the series in terms of the f miction and its derivatives. For let f(x)= %+ a r v + ajr+ a 3 x 3 + «/+•••+ a„af + By differentiating we have f(x)=a 1 + 2 a 2 x + 3 «,** + 4 a 4 x 3 H + na^' 1 -] , f"(x)= 2 « 2 + 3 • 2 rt 8 .r + 4 • 3 «/+ • • • + n(n-l)«yf- f + • • •, f"(x) = S. 2a 8 +4 • 3 • 2a 4 «+ • • ■ +»(n-l)(n-2)a.af-»+ ■ • ■, / ( %r) = [>( W -1)(h- 2) • • • 3 • 2]a n + ■ • -. Placing £=0 in each of these equations, we find =/(0), a=f(0-),a = Consequently we have /(.O=/(0) + /X0), + -^.. + ^,- + ... + -^. + ..,(i) This is called Maclaurin , 8 series. Again, if in the right-hand side of f(x) = a Q + a t x + ./.,/•- + ay + . . . + ,/„.>•" + . . . we place x = a 4- •>•', and arrange according to powers of a/, we have f(x) = ?> o + 6/ + 6 a a/ a + ft/ 8 + • • • + ?>,/» + . . ., or, by replacing ./•' by its value a; — a, /(.r) = 6 + ^(s - a)+ \(x - a) a + 6,(a: - a ) :, + . . • + K(x- a)*+ By differentiating this equation successively, and placing x=a in the results, we readily find ».=/oo. Woo. & a =,4/'oo, ».-[l/"(«).-. j -=^/ (b) («)- Hence /w-««)+c»-«)f«+^/'(«)+^Vw+ • • • +-*^^.r00+---- (2) This is Taylors series. 414 INFINITE SERIES Another convenient form of (2) is obtained by placing x — a = h, whence x = a + h. We have then /(« + h) =/(«) + hf(a) +~f\a) +^f"(a) + . ■ ■ If If Maclaurin's series (1) enables its to expand a function into a series in terms of ascending powers of x when we know the value of the function and its derivatives for x = 0. By means of the series the function may be computed for values of x for which the series converges. Practically the computation is con- venient for small values of x. Taylor's series (2) enables us to expand a function in terms of powers of x — a when the value of the function and its derivatives are known for x = a. The function is said to be expanded in the neighborhood of x = a, and the series can be used to compute the value of the function for values of x which are near a. Ex. 1. Expand e x into a power series and compute its value when x = \. Since f(x) = e*, f (x) = e*, f" (x) = eF, etc., /(0) = 1, f (0) = 1, f" (0) = 1, etc. Hence, by Maclaurin's. series, x .r 2 x 3 x i e * = 1 + l + j. + ii + j4 + -"- This converges for all values of x. If we place x = J, we have e* = 1 + i + is + t> a«~ 3 x s + • ■ -. MACLAURIN'S AND TAYLOE'S SERIES 415 This is the binomial theorem. If n is a positive integer, the expansion is a polynomial of n + 1 terms, since f (n + 1) (x) and all higher derivatives are equal to 0. But if n is a negative integer or a fraction, the series converges when x is numerically less than a. Ex. 3. Find the value of sin 61°. Let f(x) = sinx, then f'(x) = cosx, f (x) = — sinx, etc., provided x is expressed in circular measure. 61° expressed in circular measure is ^Vo 7r = — + -^— . Since sin - and cos - are known to be respectively - v3 3 180 3 3 * J 2 and — , it will be convenient to use Taylor's series with a = - ■ Formula 2 J 3 (it , ,\ . I,, 7T h- . IT ft 8 7T - + h = sin - + h cos - — — sin - — — cos - + • • • ^3 / 3 3 [2 3 [3 a ivi + J»-^-ip + Placing /* = — — and computing, we have ^(ll*) sin 61° = .S74G. The expansion of a function may sometimes be obtained by- special devices more conveniently than by direct use of the for- mula (1) or (2): This is illustrated by the following examples : Ex. 4. Required to expand sin _1 x. We have sin-ix= f X = f X (l-x*)-ldx Jo VT^~7 2 J ° Io\ 1 + h 2 + hi xi + hT^ x6 + --)' lx (1,yEx - 2) 1 x* 1-3 x 5 1-3-5 x 7 ~ X + 2'32-4'52-4-6'7 „ . m i sm _1 x Ex. 5. To expand Vl - x 2 x 3 3 x 5 5 x 7 ByEx.4, sin -i x = x + | + i£. + ^ + ..- i r 2 3 x 4 5 x 6 by Ex.2, ( l_, T l = l + |. + ^_ + !_+... Hence, by multiplication, sin- 1 j 2x 3 Sx 5 16 x 7 Vf^~c*~ 3 15 35 416 INFINITE SERIES 193. The remainder in Taylor's series. Let us write /(*) =/(«) + O - «■)/' 00 + ^w^-f" 00 + • • • and attempt to determine R. For that purpose place In the right-hand member of equation (1), with P in the form (2), replace a everywhere, except in P, by z, and call F(z) the difference between f(x) and this new expression. That is, let F(z) be defined by the equation f(?) =f(p) -m - o - *)/' oo - (J ~^r oo - ■ ■ • 5~ y ( * } — khT p ' (3) where x is considered constant. Differentiate (3) with respect to z, still holding z constant. All the terms obtained cancel, except the last two, and we have f> <» = - ^f^> +i) (z) + (*^y P = ^£): [P _ /(re+1)(g)]- (4) IZ: Now when 2 — 2*, P(2)=0, as is at once apparent from (3). Also when z — a, F(^z) = 0, as appears from (3) with the aid of (1). Hence F(z) must have a maximum or a minimum for some (unknown) value of z between z = a and z — x. That is, F'Q) = where f lies between a and x.* * The theorem that if F(z) = Ofor z = a and z = b, then F'(z) = for some value of z between z — a and z = b is called Rolle's Theorem. It is geometrically evident on drawing a graph. Of course F(z) and F'(z) must be continuous and hence finite. REMAINDER IX TAYLOR'S SERIES 417 From (4), it follows that p=/ (b+1) (£>, and hence, from (2), S = ^fff^(?)- (5) This is the remainder in Taylor s Theorem. It measures the difference between the value of the function /(./) and the sum of the first n + 1 terms in (1). It is evident that if R approaches zero as n is indefinitely increased, the Taylor's series converges and represents the function. We have, then, in this case, a proof of the possi- bility of a series expansion for the function, which was assumed in § 192. Generally also it will be sufficient to test the convergence of the series by one of the methods of §§ 188 and 189. For usually if the series converges, it properly represents the func- tion. Examples can be given in which this is not true, but the student will certainly not meet them in practice. The remainder may be said to measure the error made in cal- culating the value oif(x) by means of n + 1 terms of a Taylor's or Maelaurin's series. It is therefore often important to know Bometihing of the magnitude of R. Now H can usually nut be found exactly, since £ is unknown, but it can sometimes be seen that B cannot exceed some known value, and this is enough for practice. This is illustrated in the examples. Ex. 1. What error is made by calculating < 8 by 5 terms of Maelaurin's series? (See Ex.1, § 102.) When/(x)= e? r ,/< B+1 >(a:) = <■. Hence, in Maelaurin's series for e*, n + ] where £ lies between and x. In the present example n = 4 and x = 3. Therefore 11 =^e^ = — — & 5 29160 418 INFINITE SERIES where £ lies between and J. Since the largest value of £ gives the largest value of e$, we may write -ft < 23 i so e3 < ^^ f (1) x 3 x 5 x 1 sin * = ;c __ + ___ + ... > (2) cos, = l-| + |-^+.., (3) where the laws governing the terms are evident. It is possible to show that in each case R approaches zero as the number of the terms increases without limit, no matter what the value of x. Hence the series converge and represent the functions for all real values of x. The series (1) may be used to define the meaning of e* when x is a pure imaginary quantity and the definitions of § 26 no APPROXIMATE INTEGRATION 419 longer have a meaning. We write as usual i = V— 1 and replace x in (1) by ix. We obtain - i + T + "]2~ + "]3 _ + "]^ + •"• Then, since i* = — l, i 3 = — i, &*=+l, etc., fa A * 2 * 4 \, ■/ * 3 , * 5 \ •-f-i+g— H*;ii + s— ■> But the two series here involved are equal to cos a; and sin a; respectively by (3) and (2). Hence we have e ix = cos x + i sin x. (4) Similarly, e~ ix = cos # — i sin #, (5) and, from (4) and (5), (6) (7) The results (4)-(7) are of great importance in some appli- cations, notably to the simplification of certain results in the solution of differential equations. It may be proved from (1) that t? l e Xt = e Xl+x -. Then e ~ iy = (fe- ilJ = e x (cos y - i sil i y). (9) 195. Approximate integration. When it is not possible, or convenient, to evaluate the integral sin a; = >■'■ — e 2i cos # e* + e~ ix I. f(x)dx (1) exactly, the function f(x) may be expanded into a power series and the integral computed to any required degree of accuracy. This procedure leads to the following three rules: 1. TJie prismoidal formula. Let us take the first four terms of Taylor's series for f(x) in the neighborhood of x = «, writ- ing them hi the form f(x) =A + B (x -a)+C (x - a) 2 + D (x - a) 9 . (2) 420 INFINITE SEEIES Substituting this in (1), we have I f(x)dx = A(b-a}+lBQ)-ay + ±C(b-ay + \D(l>-ay = ^[6^+3jB(5-a)+2C(6-a) 2 +fZ>(6-a) 8 ]. (3) Now, from (2), /(«) = 4 /(J) = .4 + /? (J _ a) + C(J - ay + D(b- a) 3 , and / fe±±\ = l + }5(i-«)+j C(J - a) 2 + | Z>(5 - a) 3 ; from which it appears that equation (3) can be written in the f0 ™ jj^J^t[ m + if (^) +m ]. (4) This is the prismoidal formula. If the integral (1) is interpreted as an area, the result (4) may be expressed as follows : The area bounded by the axis of x, two ordinates, and a curve may be found approximately by multi- plying one sixth of the distance between the ordinates by the sum of the first, the last, and four times the middle ordinate. If the .integral (1) arises in finding the volume V of a solid with parallel bases, then formula (4) becomes F=|(2? + 4Jf+5), (5) where h is the altitude of the solid, B the area of the lower base, b the area of the upper base, and M the area of the section midway between the bases. Of course the prismoidal formula gives an exact result when f(x) can be exactly represented in the form (2), where any of the coefficients may be zero. The most important and frequent cases in which (5) is exact are those in which f(x) is a quadratic polynomial in x. In this way the student may show that the formula applies to frustra of pyramids, prisms, wedges, cones, cylinders, spheres, or solids of revolution in which the gener- ating curve is a portion of a conic with one axis parallel to the axis of revolution, and also to the complete solids just named. APPROXIMATE INTEGRATION 421 The formula takes its name, however, from its applicability to the solid called the prismoid, which we define as a solid hav- ing for its two ends dissimilar plane polygons with the same number of sides and the corresponding sides parallel, and for its lateral faces trapezoids. Furthermore, the formula is applicable to a more general solid two of whose faces are plane polygons lying in parallel planes and whose lateral faces are triangles with their vertices in the vertices of these polygons. Finally, if the number of sides of the polygons of the last defined solid is allowed to increase without limit, the solid goes over into a solid whose bases are plane curves in parallel planes and whose curved surface is generated by a straight line which touches each of the base curves. To such a solid the formula also applies. The formula is extensively used by engineers hi computing earthworks. 2. Simp8orC% ndr. When f(x) is not exactly expressed by (2), the prismoidal formula will in general give better results the nearer b is to a. Hence we may obtain greater accuracy by dividing the interval 1> — , (1) or, placing x = a +• h, /(a + A)=/(a) + A/'(f), (2) where £ is between a and a +• h. INDETERMINATE FORMS 423 This result either in the form (1) or the form (2) is called the theorem of the mean, and has a very simple graphical interpre- tation. For let LK (fig. 234) be the graph of y =/(»), and let OA = a,OB=a + h. Then AB = h, f(a) = AD, /(« + li) = BE, and f(a + h) -f(a) h the slope of the chord BE. If now £ is any value of 2", /'(£) is the slope of the tangent at the corresponding point of LK. Ilenee (2) asserts that there is some point between I) and E for which the tangent is parallel to the chord BE. This is evidently true if f(x) and f(x) are continuous. Formula (1) may be used to prove the proposition which we have previously used without proof ; namely, If the derivative of a function is always zero, the function is a constant. For let /'(./) be always zero and let a be any value of x. Then, by (l),f(x) —f(a) = 0. That is, the function is a constant. From this it follows that t/r<> functions which have tlw same derivative differ by a constant. For if f'(x) = Xx), then d (x) + C. 197. The indeterminate form -. Consider the fraction and let a be a number such that f(a) (1) 0. If and 4>((a) = 0. Therefore for h*Q As a: is made to approach a, 7i approaches zero, and | x and £ 2 approach a. Hence Lim .zw = m. (2) INDETERMINATE FOEMS 425 If, however, f'(a) = and $'(a) = 0, the right-hand side of (2) becomes - • In this case we take more terms of Taylor's series and have -,n whence Lim ^ v J = ' „ v y > unless /"(a) = and <£"(«)= 0. In the latter case we take still more terms of Taylor's series, with a similar result. Accordingly we have the rule : To find the value of a fraction which takes the form - when x= a, replace the numerator and the denominator each by its deriv- ative and substitute x = a. If the new fraction is also - > repeat the process. Ex. 3. To find the limit approached by -when x = 0. sin x By the rule, Lim ^ ~ e ~* = |"fl±iZ!l =- = 2. x =o sinx L cosx J x =o 1 gSC — O cog x "f* € — x Ex. 4. To find the limit approached by when x = 0. xsmx If we apply the rule once, we have T . e x - 2 cos x + e~ x [e x + 2 sin x — e~ x ~\ Lim = = -. x =o xsinx L sinx + xcosx J x =o We therefore apply the rule again, thus : T . e* — 2cosx + e~ x fe* + 2 cosx + e~ x l 4 _ Lim ; = — ; = - = 2. x =o xsmx L 2 cosx — x smx 1^=0 2 198. Other indeterminate forms. If /(«) = oo and <£(V)=ao, the fraction , '' takes the meaningless form — when x = a. The value of the fraction is then defined as the limit approached 426 INFINITE SERIES by the fraction as x approaches a as a limit. It may be proved that the. rule for finding the value of a fraction which becomes - holds also for a fraction which becomes — . . oo The proof of this statement involves mathematical reasoning which is too advanced for this book and will not be given. Ex. 1. To find the limit approached by ° (n > 0) as x becomes infinite. 1 By the rule, Lim * X = Lim = Lim = 0. .t=o) x n x=oa nz" -1 x=oo nx n There are other indeterminate forms indicated by the symbols • 00, oo - 00, 0°, oo°, 1", The form • oo arises when, in a product f(x) • ^>(^)» we have f(a)= and <£(V) = oo. The form oo — oo arises when, in f(x) — (V), we have f(a) = oo, (a) = oo. - These forms are handled by expressing f(x) ■ (#) or f(x) — (f>(x), as the case may be, in the form of a fraction which becomes - or ^ when x = a. The rule of § 197 may then be applied. Ex. 2. x 3 e~^. x s When x = co this becomes co • 0. We have, however, x 3 e _x2 = — j> which becomes — when x = co. Then co x 3 3 x 2 3 x 3 Lim — = Lim = Lim = Lim = 0. x=«>e l! i=»2i^ x=o,2e^ x =«,±xe 3? In the same manner Lim x n e~ ^ = for any value of n. Ex. 3. sec x — tan x. When x = - this is co — co. We have, however, 1 — sin x sec x — tan x — , cos a; which becomes - when x = — • Then 2 Lim (sec x — tan x) = Lim ■ = Lim . „. n COSX . n 2 2 FOURIER'S SERIES 427 The forms 0°, oo°, 1" may arise for the function [ZOO]**" 9 when » = «• If we place u = [/(a;) ]*<*>, we have log u = (x) log/(a;). If Lim (:r) log/(.r) can be obtained by the previous methods, x = o the limit approached by u can be found. i Ex. 4. (1 - jy. When x = this becomes 1". Place then log u = — 2-i '- ■ x now Lim ioga-*) = r^ri =_i. x = X LI - xJ J - = o Hence log u approaches the limit — 1 and u approaches the limit - • e 199. Fourier's series. A series of the form -^ + « x cos ./• + a„ cos 2 x + • • • + a„ cos wa; -+- • • • + 6, sin & + & a sin 2x+ h 6„ sin »./• H , (1) where the coefficients a Q , a x , • • •, 6 t , ?> 2 , • • • do not involve a-, is called a Fourier's series. Every term of (1) has the period * 2 7r, and hence (1) has that period. Accordingly any function defined for all values of x by a Fourier's series of form (1) must have the period 2tt. But even if a function does not have the period 2 ir, it is possible to find a Fourier's series which will represent the function for all values of x between — ir and ir, provided that in the interval — ir to ir the function is single- valued, finite, and continuous except for finite discontinuities,! *f(x) is called a periodic function, with period k, if f(x + k) =f(x). t If x x is any value of x, such that f(x l — e) and f(x x + e) have different limits as e approaches the limit zero, then f(x) is said to have a finite discon- tinuity for the value x = x v Graphically, the curve y = /(x) approaches two distinct points on the ordinate x = x v one point being approached as x increases toward x v and the other being approached as x decreases toward x v 428 INFINITE 8EEIES and provided there is not an infinite number of maxima or minima in the neighborhood of any point. We will now try to determine the formulas for the coeffi- cients of a Fourier's series, which, for all values of x between — it and 7r, shall represent a given function, f(x), which satisfies the above conditions. Let f(x) = -£ + a 1 cos x + a 2 cos 2 x + • • • + a n cos nx + • • • u + b x sin x + £„ sin 2 x + • • • + b n sin nx + • • • • (2) To determine a Q , multiply (2) by dx, and integrate from — it to 7r, term by term. The result is I f(x)dx = a Tr, whence a = — \ f(x) dx, (3) since all the terms on the right-hand side of the equation, except the one involving « , vanish. To determine the coefficient of the general cosine term, as a n , multiply (2) by cos nxdx, and integrate from — -rr to tt, term by term. Since for all integral values of m and n L £ sin mx cos nx dx = 0, cos mx cos nx dx = 0, (m =£ n) and / cos 2 nx dx = tt, all the terms on the right-hand side of the equation, except the one involving a n , vanish, and the result is £ f(x) cos nx dx = a n 7Tj T i r n whence a n = — I f(x) cos nx dx. (4) It is to be noted that (4) reduces to (3) when n = 0. FOURIER'S SERIES 429 In like manner, to determine b n , multiply (2) by sin nxdx, and integrate from — ir to 7r, term by term. The result is ^JO sin nx dx. (5) For a proof of the validity of the above method of deriving the formulas (3), (4), and (5), the reader is referred to advanced treatises. Ex. 1. Expand x in a Fourier's series, the development to hold for all values of x between — tt and tt. By (3), by (4), and by (5), 1 r* i = — I xdx = 0, TT J - n 1 /*"■ i„ = - I x cos nxdx = 0, f> n = — / x sin 7r«/-7r nxdx 2 - cos nir. Hence only the sine terms appear in the series for x, the values of the coefficients being determined by giving n in the expression for h n the values 1, 2, :!,••• in succession. Therefore b l = 2, b a = — |, b s = §, • • •, and sin :? x '(2ir,0) /(37T,0) Tlic graph of the function x is the infinite straight line passing through the origin and bisecting the angles of the first and the third quadrant. The limit curve of the series coincides with this line for all values of x between — it and it, but not for x = — it and x = tt; for every term oi the scries vanishes when x — — tt or x = tt, and therefore the graph of the series has the points (± tt, 0) as isolated points (fig. 235). By taking x x as any value of x between — tt and tt, and giving k the values 1, 2, 3, • • • in succession, we can represent all values of x by x x ± 2 kir. But the series has the period 2 tt, and accordingly has the same value for x l ±2 kir as for a;,. Hence the limit curve is a series of repetitions of the part between x = — tt and x — tt, and the isolated points (±2 k-rr, 0). It should be noted that the function defined by the series has finite discon- tinuities, while the function from which the series is derived is continuous. 235 430 INFINITE SERIES It is not necessary that/(a:) should be denned by the same law throughout the interval from — tt to it. In this case the integrals defining the coefficients break up into two or more integrals, as shown in the following examples : Ex. 2. Find the Fourier's series for f(x) for all values of x between — tt and it, where f(x) = x + tt if — rr < x < 0, and /'(./•) =ir — xii. 0 (i) y = | + 2mnx, (2) ^4 /7 eight terms of the form — > and 2* terms of the form 8' (k = 1, 2, 3, • • •), converges when a > 1. (2 , )a 2. By comparison with the series in problem 1 or with the harmonic series (Ex. 2, § 187) prove that the series !+*+£+£+ + A + 71° converges when « > 1, and diverges when a ^ 1. 432 INFINITE SERIES By comparison with a geometric or a harmonic series establish the convergence or the divergence of the following series : 3. 1 + 1 + 1 + 1 + . . . + — L- + • • •. [2 [3 [4 |w + 1 2 2 2 2 s 2" _1 3 3-5 3 ■ 5 ■ i 3 ■ 5 • i • • • (2n + 1) 3 4 5 6 to + 2 5 ' 2 + 3T2 + 4T3 + 5T4 + "*" + (n + l)n + "" By comparison with the series of problem 2 establish the con- vergence of the following series : e.i + | + i + ...+_L_ + .... 1 • 3 ' 5 • 7 ' 9 • 11 ' (4 n - 3) (4 n - 1) 8 1 I 1 I 1 I * L2-3- 42- 3- 4. 63- 4. 5. 6 + rc(» + l)(n + 2)(n + 3) + ""' 9 -i + -5 + i^ + --- + ^TT + '-- By the ratio test establish the convergence or the divergence of the following series : 1 1 10 * 2- l + 2 8 - 3 + 2 5 - 5 H 2 a — 1 (2w-l) e 5 5 a 5 3 5"" 1 11. 1 + - + - + - + ... + =■+.... 1 12 [3 | w. — 1 2 2 2 2 3 2" / 12 -i— 2 + 2T-3 + 3^ + --- + ^lf + ---' 13 ' 5 + 3T5 2 + 475 3+ '" ''(n+l)^"*""' 12 3 n "•S + S + 3 + -+S + -- PROBLEMS 433 -i+f+l+f +•••+£+•••• . 3 3 2 3 3 3" \* 1 , 1 3 , 1 3 2 , 1 3"- 1 16 ' 2 + 5-2> + 5>-3> + -'- + 5^-^r + --- Find the region of convergence of each of the following series x s X 5 £C 2n_1 17. a; + - + -+... + + . . .. 3 5 2 n — 1 a; ar 2 * 3 ./" 18 -2- 2 + S + (P + "- + (^+--- ic X 2 X s .r" 19 - ^r-r> + Tr-r + r-7i + --- + 1-2 ' 3- 4 ' 5- 6 ' ' (2n-l)2n ' 1 X X 2 x n ~ l 20. - + - 2 + - 3 +. .• + — + .... 03 1 CC 8 , 1 X 5 , , . .. , 1 X 2n_1 21 -3-3-3- 3 + 5-3^ + --- + (- 1 )"- 1 2^i-3^ + --- nn . x 2 , 1 • 3 . 1 . 3 • 5 . , 22 - 1 --i + TT2 xi -rYT3 x * + --- 1.3.5-2.-3 ^ ' 1.2-3...n-l + ' Find the following expansions and verify the given region of ^ convergence : 2 3 . sinx = x -- + --...+ ( -i r - + L - '— ' (— GO 0 ' and '*'< (» + ixi + *r' whe " x< °' 1 4- x 54. Show that, in the expansion of log- ■ (problem 26), 2x n + 2 X 2\x n + 2 \ \ R \<7 — T^i ^TTi when x >°> and \ R \<7 — , ox/1 i n» + 2 1 ' (n 4- 2)(1 — x) n + 2 ' ' (n 4- 2) (1 + .*•)" + - when x < 0, where n is the exponent of x in the last term retained in the expansion. 55. By integrating the expansion of -, to obtain the expan. 1+J " \x" + 2 \ sion of tan _1 a\ show that for the latter expansion \R\ < -; > where 1 ' n + 2 n is the exponent of x in the last term retained in the expansion. 56. Show that, in the expansion of (1 + x) k , | n + 1 aml i^i < A i^i 1 ( i+4^ i^ 1 i wtog<0 ' if n - 1: + 1 > 0. 57. From the result of problem 53 estimate the error made in computing log 1.2 from three terms of the series. Bow many terms of the series are sufficient to compute log 1.2 accurately to 6 decimal places ? 58. From the result of problem 53 how many terms of the expansion of log (1 + x) are sufficient to compute log .9 to 5 decimal places ? 59. From the result of problem 54 how many terms of the expan- 1 4- i- sion of log are required to compute log § to 4 decimal places ? 60. Using the result of problem 55, find how many terms of the expansion of tan -1 a; are sufficient to compute tan -1 1 to four decimal places. Also estimate the error made in computing tan -1 ^ from 5 terms of the series. 430 INFINITE SERIES 61. From the result of problem 56 find how many terms of the binomial series are sufficient to compute Vl02 to four decimal places. 62. Compute J , approximately, (1) by the prismoidal formula, (2) by Simpson's ride, taking Ax = 1, (3) by the trapezoidal rule, taking Ax = 1. r z dx 7, (i + ^ 2 ) 2 ' Compute I — — ■ — jrj, approximately, (1) by the prismoidal formula, (2) by Simpson's rule, taking Ax = i, (3) by the trapezoidal rule, taking Ax = i. 64. Compute J log cos x dx, approximately , (1) by the prismoidal formula, (2) by Simpson's rule, taking Ax = — > _ (3) by the trapezoidal rule, taking Ax = — • Find the limit approached by each of the following functions as the variable approaches its given value : _ _ 2 COS 2 X — 1 . 7T 65. ,X: 7r 6 sin 3 x 66. sin x — x . 72. )j = 0. x — tan x sin 2 x cot 5 x a 2x _ px 67. - > x = 0. 2x (*-!)' 68.^ ^,x^f 2 sin x - 1 6 e *_ e -*_2x 69. : >x = 0. x — sin x log sin - 70. r-^x = 7r. (x - tt) 2 73. cotx 74. 1-logx 2 . A x > x = U. ^ 75. log (X — 7r) _2 i , X = 7T. taaf 76. logx > x = oo (n > 0). x n 77. (7r — 2 x) tan x, x = — - PROBLEMS 437 86. {f + £C)« x = oo. 87. a' 1 ^, x=l. 88. (cosa-) C8CJr , x==0. 89. (1+ sin xy, x = 0. 90. a- J , x = 0. Expand each of the following functions into a Fourier's series for values of x between — w and ir : 91. x\ 92. e"*. 93. /(aj), where /(x)=—9rif — 7T X — 7T. a.' — 7r tan a; _1 "-.a-ii a; — 1 log a; 82. 83. a..™* xi0i 84. (sina;) ta,,J ", a- = 0. CHAPTER XVIII DIFFERENTIAL EQUATIONS 200. Definitions. A differential equation is an equation which contains derivatives. Such an equation can be changed into one which contains differentials, and hence its name, but this change is usually not desirable unless the equation contains the first derivative only. A differential equation containing x, y, and derivatives of y with respect to x, is said to be solved or integrated when a relation between x and y, but not containing the derivatives, has been found which, if substituted in the differential equa- tion, reduces it to an identity. The manner in which differential equations can occur in practice and methods for their integration are illustrated in I ii n a the following examples : Ex. 1. Required the curve the slope of which at auy point is twice the abscissa of the point. By hypothesis, — = 2 x. ax Therefore y = x % + C. (1) Any curve whose equation can be derived from (1) by giving C a definite value satisfies the condition of the problem (fig. 238). If it is required that the curve should pass through the point (2, 3), we have, from (1), 3 = 4 + C; whence C=-l, and therefore the equation of the curve is » = *»-!. Fig. 238 But if it is required that the curve should pass through (—3, 10), we have, from (1), 10 = 9 + C; whence C = l, and the equation is y = x 2 + 1. 438 EXAMPLES 439 Ex. 2. Required a curve such that the length of the tangent from any point to its intersection with OF is constant. Let P(x, y) (fig. 239) be any point on the required curve. Then the equation of the tangent at P is Y-y = ^(X-x), dx where (A', F) are the variable coordinates of a moving point of the tangent, (x, y) the constant coordinates of a fixed point on the tangent (the point of tangency), and — is derived from the, as yet unknown, equation of the curve'. The coordinates of II, where the tangent intersects OY, are then X Y dx and the length of PR is -\\x- + •'-(—)"■ he co Representing by a the constant length of the tangent, we have which is the differentia] equation is clearly (1) if tl>. 239 -*/ y/cp - ./■- : V a « -*-+" leg dx + C a ± V a- — . + C.(2) The arbitrary constant C Bhowa thai there is an infinite number of curves which satisfy the conditions of the problem. Assuming a fixed value for C, we see from (1) and (2) that the curve is sym- metrical with respect to OY, that x- cannot be greater than a 2 , that — = and y = C dx when x = a, and that — becomes infinite dx as x approaches zero. From these facts and the defining property the curve is easily sketched, as shown in fi"\ 240. The curve is called the tractrix. 440 DIFFERENTIAL EQUATIONS Ex. 3. A uniform cable is suspended from two fixed points. Required the curve in which it hangs. Let A (fig. 241) be the lowest point, and P any point on the required curve, and let PT be the tangent at P. Since the cable is in equilibrium, we may consider the portion AP as a rigid body acted on by three forces, — the. tension t at P acting along PT, the tension h at A acting horizontally, and the weight of AP acting vertically. Since the cable is uniform, the weight of AP is ps, where a- is the length of AP and p the weight of the cable per unit, of length. Equating the horizontal components of these forces, we have , , , and equating the vertical components, we have t sin = ps. From these two equations we have tan _ dy h ' dx O Fig. 241 where - = a, a constant. P This equation contains three variables, x, y, and s, but by differentiating with respect to x we have (§ 91) (1) cPy _ ds dx 2 dx ■■^m the differential equation of the required path. To solve (1), place — = p. Then dx dp a— - dx (1) becomes = Vl + v \ or log(j; dp _ Vi + P n - dx . a whence + Vl+^ 2 ) = :- + C. (2) Since A is the lowest point of the curve, we know that when x = 0, p = 0. Hence, iu (2), C = 0, and we have whence, since p p + Vl + p 2 = e a , p = h\e a -e «*/; e ") + C. y = lv VARIABLES SEPARABLE 441 The value of C depends upon the position of OX, since y = a + C when x = 0. We can, if we wish, so take OX that OA = a. Then C" = 0, and we have, finally, the equation of the catenary (fig. 61, § 27). The order of a differential equation is equal to that of the derivative of the highest order in it. The simplest differential equation is that of the first order and of the first degree in the derivative, the general form of which is or Mdx+2Tdy = Q, (1) where M and N are functions of % and //, <>r constants. In the following articles we shall consider some cases in which this equation can be readily solved. 201. The equation Mdx + Ndy = when the variables can be separated. If the equation (1), § 200, is in the form /^)e7x-+/ 2 Cy) and sin- 1 ?/ = i//. Equation (1) is then + \p = c, whence sin(<£ + if/) = sine; that is, sin cos \p + cos <£ sin ^ = k, where k is a constant. But sin = x, sin \p = y, cos = Vl - x 2 , cos ^ = Vl — ?/ 2 ; hence we have xVl-y 2 + y Vl-x* = k. (2) In (1) and (2) we have not two solutions, but two forms of the same solution, of the differential equation. It is, in fact, an important theorem that the differential equation M dx + Ndy = has only one solution involv- ing an arbitrary constant. The student must be prepared, however, to meet different forms of the same solution. Ex. 3 • (1 -* 2 )f + ^ = ax. This is readily written as (l-x *)dy + x (y - a) \dx = --0, or dy x i = 0. (1) If /j(«) + tj/j(v) =£ 0, this can be written * ./iOO + ^O) where the variables are now separated and the equation may be solved as in § 201. If /jOO + w/jOO = 0, (1) becomes dv = ; whence w = 6' and y = ex. Ex. (x- 2 — if) dx + "2 .r//'/// = 0. Place y = vx. There results (1- v*)dx + 2v(xdv + vdx) = 0, die 2vdv _ Integrating, we have logx + log(l + r 2 ) = c ; whence x(l + r 2 ) = c, ■or ./•"- + f — ex. 203. The equation (a x x + b t y + cj rfz + (a 2 x + b„y + c 2 ) dy = (I) is not homogeneous, but it can usually be made so, as follows : Place x = x' + h, y = y' + k. (2) Equation (1) becomes (ttjx'4- /; 1 /+ OjA + ^Z; + c x ) (/x-'+ (« 2 x + \,/+ a 2 A + bjc + c 2 ) r// = 0. (3) 444 DIFFERENTIAL EQUATIONS If, now, we can determine h and k so that a 1 h + b 1 k+c 1 = 0^ > (4) a 2 A+6 2 fc+c 2 = 0j (3) becomes (o 1 .x- / + b^f) dx + (" 2 r ' + h.'/) "V/ ~ ®> which is homogeneous and can be solved as in § 202. Now (4) cannot be solved if a-J) 2 — •, y) as in § 170, and set it equal to a constant. Ex. (4 a* + 10 xy* - 3 >/*) dx + (15 x- ;/ - - 12 xy» + 5 y 4 ) dy = 0. Here = 30 xtf — 12 y 3 = ^— , and the equation is therefore exact. dy dx Hence its solution is /'(./•, //) = c, where c?/ ox it may be proved that there exists an infinite number of func- tions of x and y such that if (1) is multiplied by any one of them it is made an exact equation. Such a function is called an integrating factor. No general method is known for finding integrating factors, though the factors are known for certain cases, and lists can be found hi treatises on differential equations. Sometimes an integrating factor can be found by inspection. In endeavoring to do this the student should keep in mind certain common differentials, such as d(uv) = vdu + udv, ?' du — u dv , , ;■ u vdu — udv d tan -1 - = — , V U + V , , u v du — u dv d log - = , V uv d (u 2 + v 2 ) = 2 (u du + v dvy CERTAIN EQUATIONS OF SECOND ORDER 449 Ex. {x 2 - f) dx + 2 xydij = 0. We may write this equation in the form x 2 dx — y 2 dx + xdty*) = 0. The last two terms of the left-hand member of the equation form the numerator of dl — J. Consequently we multiply the equation by — , and have dx + **<*>- f& ^i x- the solution of which is x + — = c, x or x 2 + y 2 = ex. It is to be noted that it is not necessary to use the method of § 206 to solve the equation, for when the integrating factor is found by inspection, the solution is at once evident. 208. Certain equations of the second order. There are certain equations of the second order, occurring frequently in practice, which are readily integrated. These arc of the four types: We proceed to discuss these four types in order: 1 d *y tr~\ By direct integration, 2; -//»*+* y = I lf(x) dx 2 + c x x + c % . (Ty This method is equally applicable to the equation — — =/(#). 450 DIFFERENTIAL EQUATIONS Ex. 1. Differential equations of this type appear in the theory of the bending of beams. Each of the forces which act on the beam, such as the loads and the reactions at the supports, has a moment about any cross sec- tion of the beam equal to the product of the force and the distance of its point of application from the section. The sum of these moments for all forces on one side of a given section is called the bending moment at the section. On the other hand, it is shown in the theory of beams ■pi r that the bending moment is equal to — — , where E, the modulus of elas- ticity of the material of the beam, and I, the moment of inertia of the cross section about a horizontal line through its center, are constants, and 72 is the radius of curvature of the curve into which the beam is bent. Now, by 8 10G, 3 * dfy 1 dx* R HS)j where the axis of x is horizontal. But in most cases arising in practice -j- is very small, and if we expand — by the binomial theorem, thus : 1 = W 1 _3/W I R dx 2 [_ 2\dx) J C B we may neglect all terms except the first without sensible error. Hence the bending moment is taken to be El — - This expression equated to dx' 2 the bending moment as defined above gives the differential equation of the shape of the beam. y We will apply this to find the shape of a beam uniformly loaded A and supported at its ends. 4 7j Let / be the distance between the ^ r TP 242 supports, and to the load per foot-run. Take the origin of coordinates at the lowest point of the beam, which, by symmetry, is at its middle point. Take a plane section C (fig. 242) at a distance x from and consider the forces at the right of C. These are the load on CB and the reaction of the support at B. The load on CB is w I- — x\, acting at the center of gravity of CB, which is at the distance 1 (I \ 2 2 ~ X W \2~ X ) of — - — from C. Hence the moment of the load is ~ j which is taken negative, since the load acts downward. The support B supports CERTAIN EQUATIONS OF SECOND ORDER 451 half the load, equal to — • The moment of this reaction about C is therefore — ( - — x). Hence we have "* r~>, wi (i \ w(i \" w/r~ ,\ The general solution of this equation is EI y =1 %{-8~~ii) +CiX+c *' But in the case of the beam, since, when x = 0, both y and — are 0, we have c x = 0, c 2 = 0. dx Hence the required equation is to /Px* x*\ EIy = 2[-8- 12> 2 ^y -Ax dlJ \ The essential thing here is that the equation contains — and and the equation , dx 'l.r- dx becomes -j-=f(x, />), which is a differential equation of the first order in which p and x are the variables. If we can find p from this equation, we can then find y from -^-=p. This method has been exemplified in Ex. 3, § 200. '•SM*!> The essential thing here is that the equation contains -~- and —^i but does not contain x. As before, we place -£-=p, but now write —4 = ,=,,—?>,' so that the equation becomes , dxr dx dy dx ay p — =f(y, jo), which is a differential equation of the first order in which p and y are the variables. If we can find p from tins equation, we can find y from -y- = p. 452 DIFFERENTIAL EQUATIONS Ex. 2. Find the curve for which the radius of curvature at any point is equal to the length of the portion of the normal between the point and the axis of x. . r- /dy\ a ~] % The length of the radius of curvature is ± — (§ 106). The equation of the normal is (§ 87) — - dx dx2 This intersects OX at the point [x + w — , ). The length of the normal I — 7w V dx ' is therefore y -v 1 + ( — ) • The conditions of the problem are satisfied by either of the differential d 2 y dx* 21 f + + &• y\l+ 2 ■ C 1 ) b = — 2 L . Replacing p by — , we have l = dx. p y dx Vfir^ Transforming this equation to - = dx and integrating, we have 4 i- 1 whence Z/ = 7T\ e e i + e Cl / (2) CERTAIN" EQUATIONS OF SECOND ORDER 453 This is the equation of a catenary with its vertex at the point (c 2 , Cj). If we place — - = p, and — - — p — in (2), we have dx dx- ay whence — = —± — '- • U 1 + 1>~ The solution of this equation is y Vl whence p = Replacing p by — » we have ax J^U = dx. Integrating, we have — v c j 2 _ y- = j: — ^ f or (x — c„) 2 + // 2 = c 2 . This is the equation of a circle with its center on OX. If we multiply both sides of this equation by 2 ' ' s x — 5 sin x = 2 cos a; — 3 sin ./•. Also, the solution of (1) or (2) is expressed by the equation * = 7, + „ i /r- + 1 .. + „,_,/, + „/ ' < 3 > where the expression on the right hand of this equation is not to be considered as a fraction but simply as a symbol to ex- press the solution of (2). Thus, if (2) is the very simple equation Dy =f(x), then (3) becomes y=±f(x)=jf(x)dx. (4) In this case — means integration with respect to x. What the more complicated symbol (3) may mean, we are now to study. 210. The linear equation of the first order with constant coefficients. The linear equation of the first order with con- stant coefficients is or, symbolically, (D—a)y=f(x). (1) 456 DIFFERENTIAL EQUATIONS The solution of this equation is given in § 204. Hence y = —L- f(x) = ce°* + e-je—fix) dx. (2) The solution (2) consists of two parts. The first part, ce ax , con- tains an arbitrary constant, does not contain f(x), and, if taken alone, is not a solution of (1) unless f(x) is zero. The second part, e ax I e~ ax f(x)dx, contains /(:r), and, taken alone, is a solu- tion of (1), since (1) is satisfied by (2) when c has any value, including 0. Hence e ~/W(s)<*-»»-£(*-*)-.(*-%■ = (^+JP*> + *)* (1) where p = — (a + &), g = aft. This result, obtained by considering the real meaning of the operators, is the same as if the operators D — a and D — b had been multiplied together, regarding D as an algebraic quantity. Similarly, we find (i)-6)(2)-a)y = [2) 3 -(a+&)D+a5]y=(i)-a)(D-S)y. That is, the order in which the two operators D — a and D — b are used does not affect the result. Moreover, if (lP+pD + q)y is given, it is possible to find a and b so that (1) is satisfied. In fact, we have simply to factor D 2 + pD + q, considering D as an algebraic quantity. This gives a method of solving the Linear equation of the second order with constant coefficients. For such an equation has the form r > , ay 'iii j., x or, what is the same thing, (D*+pD + q)y=f(x), (2) where p and q are constants and /(a?) is a function of x which may reduce to a constant or be zero. Equation (2) may be written whence, by (2), § 210, (D - V)y = jf—f(?) = ^""+ f*.Jr-f(x)to. 458 DIFFERENTIAL EQUATIONS Again applying (2), § 210, we have = c ^* + $* f e - bx (c'e ax + e ar Ce- a: 'f(xyx\dx. (3) There are now two cases to be distinguished: I. If a =£ b, (3) becomes y ^c/ x + Cl e^+e ix CU a -^ x Ce-^f(^d^dx. (4) II. If a = 5, (3) becomes y = 2 + - 1) = er* f(x - Y)<*dx = x-2. Therefore the general solution is .'/ = Oi + V)''- '+■'•- 2. Ex. 3. Consider the motion of a particle of unit mass acted on by an attracting force directed toward a center and proportional to the distance of the particle from the center, the motion being resisted by a force pro- portional to the velocity of the particle. 460 DIFFERENTIAL EQUATIONS If we take s as the distance of the particle from the center of force, the ds attracting force is — ks and the resisting force is — h — > where k and h are positive constants. Hence the equation of motion is or (D* + hD + fc)s = 0. (1) The factors of the operator in (1) are We have therefore to consider three cases : I. h 2 - 4 k<0. The solution of (1) is then s = e - ( C , cos 1 + C 9 sin t } , -i . ( V4k-h* t or s = ae l sin I t — pj. The graph of s has the general shape of that shown in fig. 62, § 27. The particle makes an infinite number of oscillations with decreasing ampli- tudes, which approach zero as a limit as t becomes infinite. II. A 2 - 4 k> 0. The solution of (1) is then -(-- ^ A2 - 4i A. (h . V/i3-4t > hi 4*. The particle makes no oscillations, but approaches rest as t becomes infinite. III. h 2 - 4 k = 0. The solution of (1) is _h s = (c 1 + c 2 e "• The particle approaches rest as t becomes infinite. 212. The general linear equation with constant coefficients. The methods of solving a linear equation of the second order with constant coefficients are readily extended to an equation of the ?ith order with constant coefficients. Such an equation is or, symbolically written, (B" + a^"- 1 + • • • + a n _^D + a n ~)y =f(x). (2) Ex. 1 dx LINEAR EQUATIONS 463 dx 2 dx J This may be written (D + 3) (D - 2) y = e 4 '. The complementary function is rp<-'" + c 2 e -8 *. To find the particular integral, we place . _ , 4 .. and substitute in the equation. We obtain li.k- 4 ' =e 4 '. To satisfy the equation, we must have 11.1 = 1, whence .1 = ^\. Therefore the particular integral is 1 — 1 4 ' I and the general solution is y - Ci ^x + c 2 e-"+ V,' 4> - Ex.2. ^ + ^ = sin2x. = ,',,, A = — ._,',,. Therefore the particular integral is I = — J sin 2 .c + { \, cos 2 x, and the general solution is H = <\ + V + c 8 e-* - -A sm 2 x + to c os 2 x. Ex.3. $K + & = *«-. oar aa; Substituting y = e x z, we have This may be written (D + 1) (£> + 2) z = x' 1 . (2) fz + :i ll f + 2z = x>. (1) 4D4 DIFFERENTIAL EQUATIONS The complementary function is therefore c x e~ x + c 2 e~ 2x . To find the particular integral, place I = Ax 2 + Bx + C, and substitute in (1). We obtain 2 Ax" + (6 .1 + 2 S) .*• + (2A + BB + 2 C) = a,- 2 . Therefore 2 .4 = 1, G .1 + 2 B = 0, 2 ,1 + 35 + 2 C = 0, whence A = h, B = — i}, and C = J. Hence the particular integral is 7"_l~2_3~i7 and the general solution of (1) is z = Cl e- X + c 2 e~ 2x + ^x 2 - |x + \, whence y = c\ + c 2 e- x + e*(^ x 2 — § a; + |). This may be written (D - 1) (D + 2) (Z> - 2)y = e 2 *. Since 2) — 2 is a factor of P (Z>), we place I = Axe 2x , and substitute in the equation. We obtain 4Ae 2oc = e 2x , whence A = \ and I = \ xe 2x . The general solution is y = Cl e x + c 2 e- 2c + c 3 e 2x + ^a;e 2a; . ~ - d?y , Ex. 5. — - + y — sin x. dx 2 This may be written (D 2 + l)y = sin a;. By III, we write I = Ax sin x + i?x cos x, and substitute in the equation. There results — 2B sin x + 2A cos a; = sin x. 1 a: Therefore B = , A — 0, and I — — - cos x. The general solution is -2 ?/ = Cjc" + c 2 e~ ix — - cos x = C 1 cos x + C 2 sin a; — - cos x. LINEAR EQUATIONS 465 Ex.6, fir 2|& + £ = **- + «-. tlx 6 ax- ax This may be written D(D - l) 2 y = xe 2x + e 3x . The complementary function is c 1 + (c 2 + c 3 x) e x . The particular integral I is the sum of I 1 and I 2 , where I x corresponds to the term . + 2) z = x. Placing 2 = ylx + .6, we find /l = ^, ZJ = — §. Therefore A = I (- * - 5) e 2 *. To find I 2 , we substitute y = Ae Zx in the equation (D 8 — 22^+ /))// = e 8 *. We find 7 2 = T V e 3 *. Hence I =\{2x- 5)e a * + ,V e 3x - The general solution of the equation is V = c, + ('•., + c 8 z) <■•■ + J (2 z - 5)< - ' + ,',, < Ba >. 214. Systems of linear differential equations with constant coefficients. The operators of the previous articles may be em- ployed in solving a system of two or more linear differential equations with constant coefficients, when the equations involve only one independent variable and a number of dependent vari- ables equal to the number of the equations. The method by which this may be done can best be explained by an example. ,-. dx (Jii + 8) x + (D 2 -77) + 12) y = - 2 e 2 ', (4) since (D — 4)e 2 ' = — 2 e 2 '. By subtracting (4) from (3) we have (2 D - 5)x = 2 e 5 « + 2e 2 ', (5) the solution of which is a - = c x e** + § e 5 ' — 2 e 2 '. (6) Similarly, by operating on (1) with (D — 2) and on (2) with D — 1, and subtracting the result of the first operation from that of the second, we have (2D-5)y = -3e 5t + e 2 ', " (7) the solution of which is y = c 2 e^' — § e u — e 2( . (8) The constants in (6) and (8) are, however, not independent, for the values of x and y given in (6) and (8), if substituted in (1) and (2), must reduce the latter equations to identities. Making these substitutions, we have l( c i — ««)«*' + e 5t = e 5t , whence it is evident that c„ = c v Therefore, replacing c x by c, we have x = ce% 1 + 2e 5 '-2e 2 ', y = ce^ — 3. e 5« — g'2^ as the solutions of the given equations. 215. Solution by series. The solution of a differential equa- tion can usually be expanded into a series. This is, in fact, an important and powerful method of investigating the function defined by the equation. We shall limit ourselves, however, to showing by examples how the series may be obtained. The method consists in assuming a series of the form y = a o x m + ajc m + 1 + a 2 x m + ' 2 + • . • , where m and the coefficients a Q , a % , a 2 , • • • are undetermined. This series is then substituted in the differential equation, and m and the coefficients are so determined that the equation is identically satisfied. SOLUTION BY SERIES 467 Ex.l. ,g + ( ,-8)2-2, = a We assume a series of the form given above, and write the expression for each term of the differential equation, placing like powers of x under each other. We have then x -3| =m(m-l)a x m - 1 + (m+l)ma 1 x m + ■■■+ (m + r+l)(m + r)a r+1 x m + r + • ■ • , ax '"",/"+ ••• + (/// + 0","''" , + ' dx —3 mil,,.!-'"- L — 8(m+l)0|:* i -3(m+r+l)a r+1 :C»+' -2y = -2a x'» — 2i Adding these results, we have an expression which must be identically equal to zero, since the assumed series satisfies the differentia] equation. Equating to zero the coefficient of z* -1 , we have m(m- l)a o =0. (1) Equating to zero the coefficient of x m , we have (m +1)0/1-3)0, + (m-2)a = 0. (-2) Finally, equating to zero the coefficient of x m * r , we have tlie more general relation (m + r + l)(m + r-3)a r+] + (m + r- 2)^ = 0. (3) We shall gain nothing by placing a = in equation (1), since a x ra is assumed as the first term of the series. Hence to satisfy (1) we must have either A , m = or /// = 4. Taking the first of these possibilities, namely m = 0, we have, from (2), and from (3), a_ +1 = — a.. (i) (r + l)(r-3) l ; This last formula ( 1 ) enables us to compute any coefficient, a r . ,. when we know the previous one, n r . Thus we find ) Returning now to the second of the two possibilities for the value of m, we take m = 4. Then (2) becomes 5 a, + 2 a = 0, /• + 2 and ('■]) becomes a r . , = ^,.. ("6") (r + 5)(r + l) 468 DIFFERENTIAL EQUATIONS Computing from this the coefficients of the first four terms of the series, we have the solution * = 4'-f*' + 6V-6^7r' + -)- (7 > We have now in (5) and (7) two independent solutions of the differential ('(liiation. A more general solution is y = q.'/i + coj 2 > and this may be shown to be the most general solution. (I V uV Ex. 2. Lcgendre's equation. (1 — a; 2 ) — - — 2 x — + n (n + 1) y = 0. , il.r- dx • Assuming the general form of the series, we have — ~ = m(m— l)fl .r"'~ 2 + (m + l,)ma 1 2r m— J + (m+ 2)(m + l~)a 2 x m + •••, - X *d^= —m(m-l)a x»> , -2x C -^= -2ma.jr m - ■■•, dx n (n + \)y= n (n + 1) a x m + • • • . Equating to zero the coefficients of x m ~' 2 , x m_1 , and x' n , we have m (m - 1) a = 0, (1) (m + l)ma 1 = 0, (2) (m + 2) (m + l)o, - (m - n) (m + n + 1) a = 0. (3) To find a general law for the coefficients, we will find the term contain- ing x m+r ~ 2 in each of the above expansions, this term being chosen because it contains a r in the first expansion. We have d 2 >i -\- ■■■ + (m + r) O + r - l)a r x m + r - 2 + ■■-, - x*j¥ = (m + r - 2)(m + r - 3)a,._ 2 x m + r -* , - 2 x <] j- = 2 (m + r- 2)a,._ 2 x m + r -* , n(n +l)y = ••• + n(n + l)a,._. 2 x m + r - 2 + •••. The coefficient of x m + r -' 2 equated to zero gives (m + r) (m + r - 1) a r - (m -n + r-2) (m + n + r — V)a r _ 2 = 0. (4) »(» +1) « 2 = g «o5 w „ x (n-r + 2)(n + r- and from (4), « r = — ~ , — ~, -1) By means of (0) we determine the solution SOLUTION BY SERIES 469 We may satisfy (1) either by placing m = or by placing m = 1. We shall take m — 0. Then from (2), a x is arbitrary ; from (3), (5) (6) >„(■ (»-l)(» + 2) , 3 , (»-l)(»-3)(« + 2)(n + 4) \ + M |8 [5 _ / Since a and Oj are arbitrary, we have in (7) the general solution of the differential equation. In fact, the student will find that if he takes the value m = l from (1), he will obtain again the second series in (7). Particular interest attaches to the cases in which one of the series in (7) reduces to a polynomial. This evidently happens to the first series when n is an even integer, and to the second series when n is an odd integer. By giving to a or a, such numerical values in each case that the polynomial is equal to unity when x is equal to unity, we obtain from the series in (7) the polynomials >4 1\ = x, D 3 , 1 p,=l*-\*. 4 4-2 4-2 4 ■J. • 2 „ • 7 , „ 7 • 5 , , 5 •3 . 2 each of which satisfies a Legendre's differential equation in which n has the value indicated by the suffix of P. These polynomials are called Legendre's coefficients. AC 470 DIFFERENTIAL EQUATIONS Ex. 3. BesseVs equation. x l — - + x — + (x~ — n?)y = 0. ax* dx Assuming the series for y in the usual form, we have x 2 ^ = m(m - l)a x<» + (m +l)ma 1 x m + i + (m + 2) (m + l)r,,x"' + 2 + as ^ = wfl x m + ( m + 1 ) a x x m + 1 + (m + 2) « 2 x» + 2 + • • • , — n 2 y = — n 2 a x m — rfa 1 x m+1 — n' 2 a 2 x m + ' 2 — • • •, x 2 y — a x m + 2 + Equating to zero the coefficient of each of the first three powers of x, we have (w«-n«)a = 0, (1) [( B , + l)»-„rja 1 .= 0, (2) [(m + 2) 2 -ra 2 ]a 2 + a = 0. (3) To obtain the general law for the coefficients, we have ,d*y dx 2 dy c 2 — | = • • • + (m + r) (m + r - 1) a r x m + r + •••+(?» + r) a r x m + '" + • • •, — n 2 y = ... — ri 2 a r x m + r — • • • , z 2 ?/ = • • • + n,._ 2 i' m + r + • • •. Equating to zero the coefficient of x m + r , we have [(m + r) 2 - n 2 ] a r + a r _. 2 = 0. (4) Equation (1) may be satisfied by m = ± n. We will take first m = n. Then from (2), (3), and (4) we have Oj = 0, a 2(2n + 2) r(2n + r) By use of these results we obtain the series / x 2 x 4 \ • A = **■ I 1 ~ 2(2» + 2) + 2.4.(2» + 2)(2» + 4) ""'")' (5) Similarly, by placing wi = — n, we obtain the series y2 = a ^"( 1 + 2(2n-2) + 2.4.(2n-2)(2»-4) + ---)- (G) PROBLEMS 471 If, now, n is any number except an integer or zero, each of the series (5) and (6) converges and the two series are distinct from each other. Hence in this case the general solution of the differential equation is V = CiVi + '■■j.v- If n = the two series (5) and (6) are identical. If n is a positive integer, series (6) is meaningless, since some of the coefficients become infinite. If n is a negative integer, series (">) is meaningless, since some of the coefficients become infinite. Hence, if n is zero or an integer, we have in (5) and (6) only one particular solution of the differential equa- tion, and another particular solution must 1><- found before the general solution is known. The manner in which this may be done cannot, however, be taken up here. The series (5) and (6) with special values assigned to a define new- transcendental functions of x, called Bessel's Junctions. They are important in many applications to mathematical physics. PROBLEMS Solve the following equations : 1. x(l-y)dx + y(l-x)dy = 0. 2. sec 2 yd x -|- gob 2 xdy = 0. 3. fdx - (x 2 4- 2 xy)dy = 0. 4. (x Vj/ 2 + f — y*)dx + xydy = 0. V V 5. [(.r — ;/),■' -j- x"]dx + xc' ill/ = 0. 6. (2 X + 3 y - 4) dx + < 3 x + y + 1) dy = 0. 7. (./• + i, - .-,),/,• + (x 4- y - 3)dy = 0. 8. (y — x* — l)dx + xdy=0. 10. dx + (x — y)dy = 0. 9. xdy — (y + x*e? x )dx = 0. 11. (a: + 1 fydx +(x + l fdy = dx. 12. (y + xy*)dx-dy = 0. 13. (1 + -rfdy - [(1 + x)y 4- x'f^lx = 0. 14. (2 x 4- !,<■■" ) dx 4- (cos y + xe nr ) dy = 0. . 1 3 x 2 + ^ 2 + 2 xf - 2 £\dx 4 (3 f +4+2 x*y 4- 2 J W = 0. 1 // |\ . I\ y - \ 472 DIFFERENTIAL EQUATIONS 18. xdx + ydy = (x 2 + y 2 )dx. 19. xdy — ydx == V.r — y*dx. 20. ye v dx — (xe •" + y*)di/ = 0. 21.3 sin (a- + y) dx + [3 sin (x -f y) — 2y cos (x + y)~] dy = 0. 22. x GOS 2 ydx — esc xdij = 0. 23. c/ic + (x tan // — sec y)dy = 0. 24. (,// - ■Vx 2 + y 2 )dx - xdy = 0. 25. (a- 2 - ?/ 8 ) tfa; + 3 xi/dy = 0. 26. (2x — 5y + 5) <2as -f (4 a; — y + 1) dy = 0. ■c-- ■ (2y ■ ?/ WW * 1N i^ = 0. j+ri^+yp+f yp- - .r 2 + 1) dx + (a- 2 - 1) dy = 0. 29. x 2 Vl - v/Va- - if Vl - a;V// = 0. 30. 2 a^da; - (4 x 2 + y 2 ) dy = 0. 31. xdx + ydy + (a- 2 + y*)%(ydx — a-rfy) = 0. 32. e* (a- 2 + // -f 2 a;) da: -f- 2 ye*dy = 0. 33. (3 xy + 2 e^ajcfa; - dy = 0. 34. (2 — xy)ydx + (2 + xy)xdy = 0. 35. (5 y 2 - 6 xy) dx + (6 a: 2 - 8 xy + y 2 ) r/y = 0. 36. x(l + a' 2 )f/y - [(1 + a- 2 )// + a 7 / 2 ]W^-3-» 65. Solve -yr, = - Vy, under the hypothesis that, when x =— 1 ,/ = and ^ = 0. 66. Solve -—^ — 2 if + G //, under the hypothesis that, when x = — j 7/ = and -^ = 3. 67. Solve tfy y-^ = tan 3 // -4- tan //, under the hypothesis that, when dy x = — > y = and -y 1 = 1. '/' 2 >/ 68. Solve -=-j = tan // + tan 3 //, under the hypothesis that, when n "" j ^ 1 a = 0, ii — — and — = 1. 474 DIFFERENTIAL EQUATIONS Solve the following equations : 69. ^--^-0,, = 6-7x-3* 2 . 1 1. 1- dx 70 . pL + &-a t -84*2* «.£ + •* + «,-*■. = 20cos3x. dar dx dx- dx 79. 4--4-4^ — 3 ?/ = 2 sin x + cos 2 x. dx 2 dx 80. V?-2^ + 3'/ = e-- r cosx. 82. t4 + 9y = 2 sin 3 «• dx 2 dx ^ dx 2 J d 2 « „ dy 84. — 4 — 7 + 1 + 10 // = xe 2 * + sin x. dx dx 85. ^{ + 2^ + 4 y= 2x + 3 e 2 ''. dx- dx 86. — ^-2-^ — Zy = xr" + cos 2 x. dx 2 dx J 87. p{ - 4 ^ + 8 y = 4 x 2 - 15 cos 3 x. dx - dx 88.^ + 2/ = x3 + x. 89. 'p{ + 3'p{-p-3i, = 3x 2 +10 sin 3*. dx s dx- dx 90 . ^ 3 -2^-4^ + 8y = x w 2 " 94. -^- + 3-^ + 3-^ + y = 5e* sin dx 8 dx- dx ■hi 96 - ^> + -7? + « = - 3 + -- r/ 4 y d*V 97. ti + 534-36v= 20 e 3 * cos 3 x, d*y dSi 98. -2 + 2 -4 = 2 x + 25 e f sin 2 x. r/./' 4 dx 8 dSf , d*y dx* dx 4 100. dt' dy dt o, d*y dP d r -2 — dt + X = 6' 2 '. 101. dx dt' hy = e ', x — dy Tt = e -'. dx dy - + J + 4y = 81n 3*. d*X ,//, d*X ,/// ^ + * ' *""' dt 3 ' ',// ^ dt 102 ' 7^ + Si = shl '. 10 6. ^ + 2^ + ^ = 2, dx d*y 7i + M' = '- st dx dy ,*-«.-,+* 107. £-„ V= 0, ft-.,-.. + l g + *-a 476 DIFFERENTIAL EQUATIONS Solve the following equations by means of series : 108 - *% +(•'■- **>%-»» = a io9.(.-^g + «4 + e,-ft no. rfg + w*+c— «)ir = a 111. (1 + a 2 )-^ + £c ~ - ny = 0. 7 oar ofcc 113. & + 3^ + ^ = 0. tfar da: 114. Prove that any curve the slope of which at any point is proportional to the abscissa of the point is a parabola. 115. Find a curve passing through (0, — 2) and such that its slope at any point is equal to three more than the ordinate of the point. 116. Find the curve the slope of which at any point is propor- tional to the square of the ordinate of the point and which passes through (1, 1). 117. Find the. curve in which the slope of the tangent at any point is n times the slope of the straight line joining the point to the origin. 118. Find in polar coordinates the equation of a curve such that the tangent of the angle between the radius vector and the curve is equal to minus the reciprocal of the radius vector. 119. Find in polar coordinates the equation of a curve such that the tangent of the angle between the radius vector and the curve is equal to the square of the radius vector. 120. Find in polar coordinates the curve in which the angle be- tween the radius vector and the tangent is n times the vectorial angle. 121. A point moves in a plane curve such that the tangent to the curve at any point and the straight line from the same point to the origin of coordinates make complementary angles with the axis of x. What is the equation of the curve ? PROBLEMS 477 122. Show that if the normal to a curve always passes through a fixed point the curve is a circle. 123. Find the curve in which the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact. 124. Find the curve in which the perpendicular upon the tangent from the foot of the ordinate of the point of contact is a constant a. 125. Find the curve in which the length of the portion of the normal between the curve and the axis of x is proportional to the square of the ordinate. 126. Derive the equation of a curve such that the sum of the ordi- nate at any point on it and the distance from the point to the axis of x, measured along the tangent, is always equal to a constanl how long will it take to Va traverse half the distance to the center ? 147. A body moves through a distance . ' 27. (10, 17). Page 15 28. (- 3, 2). 29. § V5, ', \/26, J, Vl49. 31. 00, 3. 32. (16, - 2). Page 18 59. 02, - 1, - 13. 62. ! \ 16, 0, jV§. Page 19 73. - 5, - -1. - 1. 0. Page 37 CHAPTER II 77. (2, 4). 83. (-4.-3). (-3'. -1§). 88. (.'„ 0), (2, 3). 78. (- I, §). 84. (5, 1!). 89. (0,1). (1, .'.,. 79. (3, 4), (- is, -4i). 85. (2, },), (1, 2). 90. (± 3, ±4). 80. (-3,1). 86. (-1. -l.\). 91. (±2V2,2). 82. (- I ± } 2 VZ, Y T |V5). 87. (± 4, ± 2)", (± 3, ±1},). 92. (± 1, ± },). Page 38 93. (0, 0), (2, ± 1<). 96. (±1, •■;-;). 99. - 2.07. 102. 2.41. 94. (1, ± 1). 97. (3, 2£), (- 1, 1 I). 100. .40. 2.05. 103. - 2.52. 95. (±2V2, 2). 98.1.40." 101.1.12,3.93. Page 45 CHAPTER III 1. (1,7), (- 5,9), (2, -4). 2. x 2 + 9*/ 2 = 5. 3. x 2 + y* = 4. 4. x*-y*=4. Page 46 5. y 3 + 3x 2 = 0. 7. jc- + 4y = 0. 9. 4s 2 + 9y 2 = l. 6. ?y 2 = 4 j. 8. 2 x 2 - 4 ;/ 2 = 9. 10. xy = 6. 11. Sxy = 7. 15. (V3, l), (l, -V3), (l - V3, - 1- V§). 12. o& - c 16. (|V2, iV2), (2V2, 0), (1V2, -|V2). 14 £l + * _ e 17- (2, 1), ('- 2, - 1), (- f, - 21). " 4a 46 18. x 2 + 7y 2 = 14. 417.2 481 4S: ANSWEJJS Page 19. 20. 21. 22. Page 32. Page 14. 15. 16. Page 27. 28. 29. 30. Page 43. 44. 45. 49. 50. Page 64. Page 3. 4. 5. 6. Page 10. 13. 14. 15. 16. 17. 47 5 x- + y 2 = 5. 24. 2 xy ± 40 = 13 y 2 - 14 = 0. 25. 3x 2 + 2y a 4 x 2 + 9 y" = 36. 26. 3 x 2 - 2 y 2 x*-y* = 4. 27. y 2 = 2x. 48 24x 2 -15y 2 = 200. 65 6x + a.v — a* = 0. 21. Ux + 4y -18 = 0. 22. 2Sx-Uy + 26=0. 23. 66 4x-3y = 0. 4x + 7?/ + 13 = 0. 4x + 10 2/ + 1 = 0. 24 x - 21y + 109 = 0. 40. (- 2, 4), tan- 41. 2x— 9y + 6 42. 7 x + y - 25 67 8x + 2/ + 9 = 0, 4x+72/+ll=0. y — 1 = 0, 4x 4- 3y- 11 = 0. x-3==0, 4x-3?/-9 = 0. §VlO^ jfV34, 4V2. ffVl7. 0. : 6. = 30. 10x 2 4- 5 ?/ 2 - 22x + 4 2/ - 20 = 0. 205 x 2 4- 520?/ 2 = 4264. 4 xy + 13 = 0. CHAPTER V 1 = 2x + y- Gx-4y 5 x + 4 •(/ 25. 12 x + 3 ?/ -2 26. 2x + 6y — 9 = f 24 77 102/4 2/ — 5 -35 = 1 = 0. = 0. 31. 36 x 32. 4x- 33. 2x- 34. tan-11. fi; (-3, -5), tan- 1 = 0, 7x- 6y + 21 = : 0, x - 7 y - 25 = 0. (3, 1), 35. tan~ 112. 36. tan- 1 3_ 37. tan~ 15. 38. tan tan- - J 4. 1 9 51. Voi, ffVeT. 52. 5x- 2y- 0, 4x+ 5?/— 11 x — 1 y + 11 = 53. (2, - 3). 54. T 2 ^Vl3. 55. T * 7 V34. 56. 2V2. 57. 9±. 68 5x_2?y-4 (H, *>■ (- 5, - 3£), = ; 2 V29. (" 21, If). 92 5x — Sy + 4 = 0. x+v/+8 = 0, 4x-4 2/-7=0. 2x+ 8;//- 5 = 0, 16x-4j/-l 21x- 77 y + 96 = 0, 99 x + 27 y 93 y 2 + \y CHAPTER VI 0. 86 = 0. 65. (±3 2Vl7, x T 8_VT7). 66. (4, -6), (2|, -61). 67. (-8, -11), (10, 13); (- 11, - 15), (7, 9). 7. 2/ 2 -10x + 25 = 0. 8. x 2 - 3 y"- - 4 y + 4 = 0. 9. x 2 + (y - 3) 2 = ± x 3 . I). 2x + ll 91 x 2 - 24 xy + 84 y 2 - 364 x - X 2 + y 2 _ Q X + 1Q?y + 18 = 0. 5 x 2 + 5 2/ 2 + 8 x - 6 ?/ - 15 = (-1 + 2V3, 0). x 2 + 2/ 2 + 4x-32/ = 0. 11. x z = ay. 12. x 4 + x 2 2/ 2 152 y + 464 = 0. x 2 + y 2 + 2 x - 2 2/ - 6 x 2 + 2/ 2 ± 2ax = 0. 2 x + 2/ - 3 = ; § V5. ANSWEES 483 27. 37. 29. x 2 + y 1 + 3 x - 4 y = 0. 30. x- + y- — 4 x + 4 y + 4 = 0, x- + j/ 2 - 20 x + 20 y + 100 = 0. 5 + 240x— 240y + 225 = 0. 2 y + 40 = 0. 34 // + 81 = 0. Kt Page 94 25. x 2 + y 2 -2x-2y = 0. 26. 2x 2 + 2y 2 + 3^-2 = 0. 5 x 2 + 5 y 2 ± 9 y - 80 = 0. 7x 2 + 7 // 2 - 19x + 11 y - 6 = 0. 4 x 2 + i y 2 - 60 x - GOy + 225 = 0, 64 x 2 + 64 x 2 + y 2 - 2x4- 12 j/ + 12 = 0, x 2 + y 2 -16x x 2 + y 2 + 26x + 16y-32 = 0. x 2 + y 2 — 6x - 10 y + 9 = 0, x 2 + y 2 + 18 x - x 2 + 2/ 2 -8x- 12^/4- 48 = 0. x 2 + ?/ 2 + 4x - 2 // - 20 = 0, x 2 + if + 24 x - 42 „ 13 x 2 + 13 y 2 - 156x - 52 y + 295 = 0, 13 x 2 4- 13 r - 52 x - 104 y + 259 = 0. 5,3; |; (±4,0). 95 iV6, ^Vf; I; (±i.V6,0). (0, ±^V2); iV2; (0, ± fc). (-2,1); (-5,1), (1,1); iV5; (-2±V5, l). (*.-*>; G.-41MM*); 1x7; <-.- ' ; | ^7). & 2 X 2 + a 2 y 2 _ 2 „,,,' x _ _ 6 2 x 2 + ahj 2 - 2 a 2 by = 0. i a/385. 1 VlG5. Page 39. 40. 41. 44. 45. Page 56. 57. 58. 59. 60. 61. Page 70. 71. 72. 73. 74. 75. 46. x 2 + 4 y 2 - 4x + 24y + 24 = 0. 47. 36 x 1 + 25 if - 72 x + 60y + 60 = 0. 48. Sx 2 + 9// 2 = 180. 49. :;./- + 4// 2 = 48. 50. <..,•-' + 8 // 2 + 16.y- 04 = 0. 51. Bx 2 + 25,/ 2 = 225. 52. 4x 2 + 3,/ 2 =108. 53. 9x 2 + 25 y 2 = 225. 54. 1 3 <- + 9 if - 26 x — 104 = 0. 55. :;,/- = 4 a 2 ; V\/5. 65. (2, - 3) ; Wl3 ; (2 ± Vl3, - 3)_; 3x - 2 y - 12 = 0, 3x + 2 y = 0. 66. (- 1, 2) ; \ a 10 ; (- 1, 2 ± ^VlO) ; y - 2 = ± J V6 (x 4- 1). 67. 4 x 2 - 20 .v 2 - 8 x - 80 // - 79 = 0. 68. 100 x 2 - 36y 2 + 400 x + 216 y + 301 = 0. 69. &*x 2 - , y — (a + b) sin — A sin 0. 18. x = (b — a) cos + h cos 0, y = (b — a) sin — h sin 0. 19. Straight line. 20. x = a (1 — tan ), y = a tan (1 — tan 0) ; x 2 — a (x — y). Page 115 21. x — a cos 0, ?/ = a tan ; x 2 (a 2 + ?/ 2 ) = a 4 . 22. x = a sec — 3 ; decreasing if x < — 3. 21. Increasing if x < or x > 2 ; decreasing if < x < 2. 22. Increasing if x >— 1 ; decreasing if x <— 1. 23. Increasing if — ll; decreasing if x < — 1 or < x <1. 24. Increasing if x < 2 or x > 3 ; decreasing if 2 < x < 3. 25. Increasing if x < — 1 or x > 3 ; decreasing if — 1 < x < 3. 26. Upward if t < 8£ ; downward if t > 3£. Page 151 2 l'l 5. „. ._ „ ' . ... _ ' 29. Increase if x < 10 ; decrease if x > 10. „ ' , J ' .,' /n ,' 60. (— /, — rf^), (U, i), 0. = 0. 30. Increase if x < ; V3 4a 31. Increase if x < — ; 3 decrease if decrease if x> x> a 4a 3 ' (2, - 8|). 36. (H, 2 T V). 37. 4x + ?/ + 7 = 38. 3x-y + 18 = 32. s increases if x > § ; decreases if x < §. 41. 54. Page 152 42. (4§, 55f). 43. m, 3). 44. tan-i|. 48. 49. 50. (1, _!),(- 1,_^). (1, 5), (- J, 2/ 7 ). 7z-4y-17 = 0, 45. tan- 1 f. 189 x- 108 ?/ -209 = 0. 46. 8x-2/+ 12 = 0, 51. 5.8; 5.9; 6. 216x-27?y-176 = :0. 52. .160; .165. 47. x - 2 ?/ + 9 = 0, 53. .423; .414. 27x-54?/-7 = 0. I ANSWERS 487 Page 153 54. 4.411; 4.566. 57. 88|. 60. 108. 63. 2.5. 55. 6. 58. 36. 61. 6|. 65. i 56. |. 59. 3Hff. 62. 90,000; 677 T V. 66. lOf. Page 181 CHAPTER X 1. 6x 2 + 18x + 7. 5 + 1) H 4x(x + l) 2. 0(x + 2)(2x 2 + 8j V(4x 3 + 6x2-5)2 3. 2a . 2x 3 + x-l (x - a) 2 Vx 4 + x 2 -2x 16x " (x- + 4)2 (x 3 + 8) 2 2x2(6 -x) (4-x) 2 n 6(x + 2) (x 2 + 4x + l) 2 6 38(x + l) 18. 4X . ' (3x 2 + 6x + 5) 2 V (X s + l) 4 j2 + 4j. + 1 7. -• 19. 3(6x-8)(8x- l)(x-l)». (X 2 + X + I) 2 „ 4/ i 1 1 3 ^\ X3 X* X* 9.(8x>-2)(l + |). Vx + 1 20. :i (_' i J - l)(4x» - 10x- + 2x + 1). 2 x 8 — x + 1 22. 10. _w 4 _!\_-j_/i + !y 24. V Vx 2 + 1 x + 2 (x 2 + 1)§ (5x 4 -7x 8 +6x 2 +3x-3)(x 2 -2x + 3)* (x 8 + \)\ 1/ 1 1 12. 18x(x + l)(2x 3 + Sx 8 + (ij- lS. 12x 2 (x 3 -l) 3 . 25. 1- 8\a 1 + x x 1-x \ x- - 1 Page 182 31. 32. 33. 34. 35. 36. 37. 38. 39. „. » . X + A 'I 1 + X- 3x 3 (1 - X) A X 2 - 1 27. (x + l)vx 2 -l og 2x 2 + «2 2x a 2 Va 2 + x 2 2(x + 2/) 2x + 3y 2 2 y 4 — 4 x 3 y — x* y*-8xy* + x* 2x + y x + 2y Vx+~y + \ 7 x - y vx + j/ — Vx — y x 4 4a 5 x 3 ~y*'' "**" a 2 Va 2 + x 2 «"' « 2 x2 2xi " (<>'-- x 2 )f « 2 30. - 2x + 2/ ;0. x + 2j/ x 2 Va 2 + x 2 488 ANSWERS 2ax 2a(4a-3y) 45. 3a: + 5y ± 16 = ' 3 y 2 - 2 ay ' (3 y - 2 a) 8 ' 5x-3?/±4=0. 41 1-y 2 . 2(?/ 2 -l)(4x + y) 46. (± 1, ± 9). 2xy-l (2X2/- 42. x — ly + 5 a = j Ix + y ■ - 15 ft = 0. 43. 31x + £ I y 4. 9 a = ; 8x-31 1/ + 42 ft = 0. 44. (-2, - 8). Page s 183 56. tan-i3, tan-ia. 57. tan-i2. 58. ■t . — , tan- 2 *ft- 59. — , tan- 2 17. Page 184 47. 8V2. 48. 2y x y= 5x£x — Sxf. 49. x x ~ ^x + y y ~^y = o^. 50. x^x + y{~^y = ($• 51. x : 2 x + i/fy — ftZ/jX — ax x y = ftx^. 60. -, tan- 1 2. 2 61. tan- 1 3. 62. 0. 63. 0, tan- 1 T 2 j 64. tan- 1 .\ . 65. 0, tan- 1 1. 66. tan- 1 3. 67. tan- 1 ?-. 68. tan- 1 V2, ,5V6 tan-i 24 71.(±?V2,±|V2). t2.(±-£=,±— £=)• 82. Vp(pT^). \ 2 2 / V Vft 2 + 6 2 Vft 2 + 6 2 / Page 185 ft 2 fr 2 85. a-y x x — b^x^y = ; ■y/b 4 x{ + a t y{ 90. Upward if x > J; downward if a; < J. 91. Upward if x < — V2 or x > V2 ; downward if — V2 < x < V2. 92. (- U, 111). 99 , (2> o), (- I, 913); (§, 4|f). 93. (1, - 3), (- I, 3 i 4 ) . 100. (- 1, 0), (3, - 32); (1, - 16). 94. (0, Gftl). 101. (\, - ^); (1, 0), (h, - T V). 95. (± — V3, % a). 102 - (3. ~ n ); (°. 16), (2, 0).' V 3 2 ; / 2 2v / 2~\ 96. (0, 0). 103. ± — , T —=■ ; (0, 0), (± 2, 0). 97. ( ; 0). V V3 V3 / 2 '4 y Page 186 104. Increase if x >V2ft ; decrease if x < V2 ft (ft = given area). 105. Increase if x < -— ; decrease if x > — - - (ft. = hypotenuse). V2 V2 106. Increase if x < - ; decrease if x > - • 2 2 107. Increase if x < ^p ; decrease if x > ^p (p — perimeter). ANSWERS 4Si» 108. Length is twice breadth. 109. 12 rd., 18 rd. 110. 5 ft. Page 187 114. 4 portions 1 ft. long ; 2 portions 4 ft. long. 115. Breadth = —V§, 3 depth =—V6. 116. Breadth = depth. 117. 13' r ft. long. 118. (21, 0). 111. Side of base = 20 ft., depth = 10 ft. 112. Depth = one half side of base. 113. (*,*). \6 6/ (l^H- 120. Height = twice radius of base. 121. J-. V2 122. Altitude = -V2; P 4 base = - (p = perimeter). 4 Page 188 123. Altitude = * radius of sphere. 124 ^5*. V ^ sf ' sl 125. 200 cu. in.; 2547 cu. in. 127. Height of rectangle = radius of semicircle ; semicircle of radius — • IT Page 189 132. a Ian miles on laud miles in water. Vn 2 - lm Vn 2 - m? 133. l^i hr. 134. 7hr. 135. Area of ellipse = - area of rectangle. 128. -L in. V3 129. "VS. 3 130. 8 mi. from point on bank near- est to A, 131. He travels 8.', mi. on land. 136. Altitude = \ \ 2 radius of semi- circle. 137. Altitude = § altitude of seg- ment. 138. \ 150 mi. per hour. 139. Velocity in still water —mi. per hour. 140. 144 w cu. ft. per hour. 141. (1,3), (5, -5). Page 190 142. 2 V? ft. per second. 143. 4 7r times distance from vertex. 144. .1 in. per second. 145. .02 in. per second. 146. .00 cm. per second. 147. 34.9 sq. in. per second. 148. Forward if t < 1 or t > 5 ; backward if 1 < t < 5. 149. d max. when t = .85 ; backward if 2 < t < 4. 150. - v (y = distance of top of lad- der from ground, x = dis- tance of bottom of ladder from wall). 490 ANSWEIIS Page 191 151. 6 COS0 (0 is the angle between the wire on which the bead slides and the straight line drawn from the bead to the fixed point). 152. 281 J ft. per minute. 153. 150 ft. per second. 154. y 8 2V(t 2 + l) 4 + 4t 2 x* + 4 ' " («2 + l) 2 155. When x = |. 156. 20 ft. per second ; 10 V5 ft. per second ; (100, 20). 157. .22 ft. per se cond. 158. Ellipse; 2 J 144 ~ 2 f - Page 1. 2. 3. 4. CHAPTER XI 212 2sin 3 2xcos2x. 11. xctn(x 2 + a 2 ) esc 2 (x 2 + a 2 ) . 3 sin 4 3 x cos 3 3 x. , _ x . x . _ , 12. ctn 6 -. sm 3 ax cos 3 ax. 3 sin 2 2 x. cos 2 (l-2x). cost 3 x sin 3 3 x. 14 . sec | * Un 5 5 . sin 3 2 x 5 5 15. esc bx (esc bx — ctn bx) . 2 Vl-4x 2 1 13. sec 5 - tan - . 2 2 cos 3 2x 15 8. sin 3 (2x + 1). 16 9. sec 2 x(l + tanx) 2 . 0. tan 4 -. 17 19. V4x-x 2 2 (x + 2) V2x X V3-6x 2 - 1 X 4 V 3 x — x 2 Vx-x 2 Page 213 99 3 °3 V8-6x-9x 2 2a °4 x 2 + a 2 1 01 > x 2 — 4 x + 5 1 ^fi (x + l)Vx 2 + 2x 1 27. Va 2 - x 2 2a 2 x x 4 + a 4 2 x 2 + I ' 1 33. 35. xV9x 2 -l 2 (x + 2) Vx 2 " X 2 - + 4x 1 (x 2 + 1) Vx 4 + x 2 1 + 1 (x 2 + a;)V4x 2 + 4x — 1 4a 2 x 2Va .r* + a* 36. sin- 1 Vl- 37. 2xtan~ 1 - 38. 39. 40. 41. 42. 43. 44. 45. 2x 2 -4x + x + 2 x 2 + 4x + 3 4x 2 -9 1 3x 2 -l 2-x 3 — 4x + x 2 3 V9x 2 + 2 X 2 Vx 6 ANSWERS 491 46. X 50. 6 . 4 + 5 sin 2 x 53. 6x 3 + 4x x Va 2 + x 2 Vx 4 - a 4 47. ctnx. 51. 2csc2x(ctn2x-l). 54. esc 3 2 x. 48. 3 sec 3 x. 49. — sec - • 2 52. * 3 • (x 2 + 4) 2 55. 56. (log ax) 2 . 2 /x 2 - 1 X \x 2 + l" Page 214. 57. 2 tan- 1 2 x. 65. ia c ° 82 *- sin2x • loga. 74. 4 „ 3x + 2 66. (e* + er*)*. s/ e -2.r + e -2* 58. l + 2x 2 67. (ae) b + "- 75. 0. 59. tan- 'ax. 68. x 2 e ar . 76. x 2 - a 2 - — - — - log (x + a + Vx 2 + 2 a (x + a) 3 -<*) 82. V2e- ei*. I- ( eI - e2T ) lo g( 2 - V2 e* — e 2 * 83. .vx'(l + logx). 84. 2/x4~ + logx + (log.r)-l. 85. ^g+logx). 86. = (cos Vx • log tan Vx + sec Vx). 2Vx x 3 tan x 88. -~— 12 x tan- » - + a log (x 2 + a 2 )l . x 2 + a 2 L a J Page 215 89 y(l-x 2 -y 2 ) 92 xy\ogy-y 2 ' x(l + x 2 + 2/ 2 )' xylogx-x 2 90- ~V Q o y(ysecxy-Iogy) y 2 — x 93. -: -■ 9L 2sin(x + 2y)-cos(x + 2y) _ X{1 ~ y SeC Xy) 2cos(x + 2y) — sin(x + 2y)' 94. — e'J~ x ; e 2 »~ x . 492 ANSWERS 95< s + y . 2(x 2 +y 2 ) ' x-y' (x - y) s % 2z + y 10(x 2 + s/ 2 ) ' x - 2 y ' (x — 2 yf tan x tan 2 y sec 2 x + tan 2 x sec 2 y 104. tan-i3, tan-i^. 97. tan y ' tan 3 y 105. ('• i)- 106. (* 1 V2' 1 107. e- a- 108. 6- .), 4. 98. y2 ; tltlM. ' x(y — x) x (y - x) 3 102. tan-!2V2. 103. tan-if. Page 216 110. Turning points when x = ± -1 V2 ; points of inflection when x = or ± 1 V6. 111. Turning points when x = or 2 ; points of inflection when x = 2 ± V2~. 112. Turning point when x = 3 ; points of inflection when x = or 3 ±VS. 113. Turning points when x = k-jr or (2 k + 1)- ; points of inflection when x = (2 fc + 1) - • V3 — 1 114. Turning points when x = cos -1 ; 2 2 7t points of inflection when x = ktr or 2 for ± 3 115. Turning point when x — ae ; .point of inflection when x = ae%. 116. 12 ft. 1 119. At an angle tan- 1 A' 120. lain. 117. 70°. 118, V2' ■ with the ground. 121. 5V5ft. Page 217 122. 24 sq. ft. ; 14.71 sq. ft. per second. 12 4. 2 V(5 - «) (a - 3) ; 4(4 - s). 123. Greatest distance =13 ; 125. a . force = — (7 — 2s). 128. -rr miles per minute. Page 218 130 (b sin 6 -\ — ^— ) times angular velocity of AB, where 0= CAB. " V Va 2 — 6 2 sin 2 BJ 131. Circle. 132. kva 2 sin 2 &£ + b 2 cos 2 kt; maximum at end of minor axis; minimum at end of major axis. 133. 2V2. 134. 2 «w sin - ; w Va 2 — 2 ah cos + h 2 , where w is the constant angular 2 velocity. , __ 2 aw (a + &) . 6 (a + b) w /-= — — . 137. — ■ — - sin - — Va 2 — 2 ah cos 9 + h 2 . b 2 b ANSWERS 493 138. a0w, where a is radius of circle, ad is distance through which point of string in contact with circle has moved along circle, and w is the constant angular velocity. 139. — ft. per second, where x is distance of man from center V10000 - x 2 of diameter. 140. ft. per second, where x is distance of man from Vl0000-x 2 sin 2 a center of diameter. 2 a 2 - — times man's rate, where x is distance of man from center pf a 2 + x : diameter. Page 219 142. Circle; feet per 2Vb 2 -aH* second, where at is distance of foot of ladder from side of house. 143. tan ;^. a* Page 220 155. 4 7ra, 4 ira V2 ; 0, 8 ira. 156. w Va 2 + 2 ab cos 8 + b 2 , where w = rate of 6. 157. 2, V3; }, 1. 165. Greatest when x = (2 k + 1) - ; Least when X = few. Page 221 168. S a sin* cos*. 169. t- * 170. a ( 5 - 4 ^^) f , 171 2a 2 . b a 9 — 6 cos 3 r 144. — tan d> ; — sec 4 esc*. 145 ens/ _ sin/ cos£ + sini' 2 e*(cos£ + sint) 3 152. tan- 1 |. 153. 0,*. 2 158. S(axy)i. 162. 2 a. 159, .'/- 163. ,; \ Hi. 161. a 164. J-V2. CHAPTEE XII Page 253 22. 2 Va 2 - x 2 . Page 254 23. x - log (e* + 1) . 30. Bin* 3 x (i sin 3 x - , 2 T sin 2 3 x 24. log[log(x +Vx 2 -a a )]. + T V sin*3x). 26. Jsin*(2x + 1). 3lJcos35_2cos?. 27. l - (tan ax + sec ax). ^ ^ 3+ 2 .sin 2 4x ^ 28. - ^(ctnx + 1) 3 . 6Vshil^ ' 29. ^(3cos 5 2x-5cos 3 2x). 33. 2sinx - log(secx + tanx). 494 ANSWERS 34. § sin 8 x. 46. - 4 eta - - - ctn 3 - - - ctn« - . 1 % 4. 3 4 5 4 35. -(tan ax — ctn ax). ._ . . . , „,. , , . ., , . „ a v ' 46. log sin (x + 2) + \ ctn 2 (x + 2) 36. tan 2 X + sec 2 x - X. - \ ctn 4 (x + 2) . 37. - 5 ctn 3 x — ^ esc 3 x — x. 47 1 „2x/_ „ „2x\ x\ * 10 ctn»^(5 + 3cta2^y n 3 ax tan ax a ) v csc ax + sin ax) . a , „ 9 3 x( „x _\ 50. -x sec- sec 2 - — 7). x 7\ 3\ 3 / 38. 2 (tan- + ctn . \ ' 48. — tan 3 ax tan ax + x. 39. £sin(2x - 1) - ^sin 3 (2x-l) 3a a + fa sin 5 (2 x - 1) . 49 _ 1 fMB r/j , + s}n ^ _ / x 40. 3 sin — cos 41. - tan 2 - + 3 log cos 2 3 3 51. \x- 1 L S in(6x + 2). 42. J tan (3 x + 2) + \ tan 3 (3 x + 2) . 52. 1 x — T ^ sin (4 - 6 x). 43. ?sec 3 ^. 53. x+lcos4x. 9 _2 54. J g x- J^sinl2x. 44. glj Vtan 2 x (7 tan 2 x + 3 tan 3 2 x) . Page 255 55. I x + J, sin x + ^ sin 2 x. 72 1 s[n _ l 3x _ 56. fe x - ^4 sin 4 x - J s sin 3 2 x. 3 5 57. - § V(cos 2 x + l) 3 . 73. ^tan-i 22 58. log (sec x + tanx). 6 V3 1 ,2x 59. cosx + sinx. 74. g"~~ cft irsin(a-6)x sin(a + ft)x ~j j ,2x + l 60> ^ — T TTT — " 75. - tan- 2\_ a-b a + J ""e 3 lTsin(a + 6): 2 L a + ft y»(a-6)» 1. 76> _2_tan-i^±l a- ft J V7 V7 62. i-xsin6 — |cos4x. 77 _ ^ n -i x ~ 2 , 63. ^tCosOx— Jr cos4x — icos2x. V6 1 . , 4 x — 5 64. I(tan2x + sec2x). 78. -—sin- 1 — v 2 65. — log (1+ cosx). 2x— 9 x 79. sin- 1 — — — -■ 66. — 2 ctn x. 9 80. sec-^x + l). 67. x + ctn x — esc x. 7 x j 81. logVx 2 +9 + -tan-!-. 68. '-x + (8 sin 2 ax + sin 4 ax). o * 82. ^log(3x 2 + 2)--^tan 2 V 2 cos - • 2 83_ _Vl — x 2 — 2sin- 1 x. Vo V& 70. log (sec x + tan x) . g4 in _ 1 » _ Va 2 - x 2 . V2 a V2 , 85. - tan- 1 (cosx). 71. _(cos 3 2x-3cos2x). ^ ^ log ( 2 x + Vi^T^) • ANSWERS 495 Page 256 87. £log(3x + V9x 2 -2). 12 °3x + V2~ 89. Jlog(4x + 3 + 2V4x 2 + Vx). 90. I log ?LZ*. 6 ° x 91. 1 , 2x + 3-V6 — log =• V 5 2 x + 3 + VS 1 ~ log (3 x - 1 + V9x 2 -6x + 9) -log^+- 7 . 10 °x-3 94. — log(2x + 2+V4x 2 + 8x-14). V2 95. ^log(x 2 + 6x + 12)-GV3tan-i^i^. 2 V3 96.Ilog(x a + x- 6)+ llog^|. 97.1log(2x 2 + 5x + l) + ^log 4X + 5 - V !!. 2 34 4j . + 5+ Vl7 98. ilog(6x 2 + 7x-3) + 3 log 3 -^i. 2 2 6 2x + 3 99. ] log(2x a + 6x+ 9) - * tan-i 2x + 3 . 100. log (3x8 + 2x + 8) -IlJtan-i 3g + 1 1 12 2V2 101. 3sin-i^i_V3 + 2x-x8. 102. -V8-4x — 4x* + 7sin- 103. V4x-x 2 + uin-i — 1 104. tan-i(x + tana). 105. sec a tan- 1 (r- sec a + tan a). 106. sin- 1 (e* cos « + sin or). 107. log (x + Vx a - 1 ) + sec- 1 x. 1 108. -log(x a +Vx*-a*)- 9 110. tan- 1 4tan- + 8 ■\ 7 111. — !— tim- 4V2 tan x vT -2V2 \ 2 tan 2x — 3 + 2V2 3 tan - - 1 109. -log ! 3 tan - + 1 2 Page 257 pft + cx a b + ex 120. - c(l + log a) 121. 21og(e* + l)-x. 122. llog (e ^-2)_l tan- + 2 + V5 113. — log - Vs tan- + 2-V5 2 123. log(e* — e~ x ). 124. l(x 2 + 2x + 6)j\_2T^3. 125. x + 2 + 4 Vx + 2 + 4 log (Vx + 2 - l) 126. ^ (x 2 - 2 a 2 ) Vx 2 + a 2 . 127. J (x 3 - 9) v x 3 + 3. 496 ANSWEES V4 141. 21og(2x + V4x 2 -9) V4x 2 -9 129. • 27 V(9 - x 2 ) 8 130. 1 ^(3x 2 -8)(4 + x 2 )f. 142. 143. ix 2 - 18 6V3 + 2x 2 x 2 + 18 132. Vx 2 - 25 - 5 ! ,__ V9x 2 -4 9 ,3x 133. sec- 1 8x 2 16 2 134. ! > , g (4x-l)(3x + l)3. x 3_9 x 2_ 81x _ 8 i 135. + 271og(x+! 2(x + 3) 136. ^_i-log(3 + 4x 3 ). 12 16 GV ' 137. -x(2x 2 -a 2 )Vo 2 ^x 2 +— sin-i-. 8 v ' 8 a x 138. Vx 2 + 9 144. - x (5 a 2 - 2 x 2 ) V« 2 - x 2 8 ' 3 a 4 . , x + sin- 1 -- 8 a 145. ^(7x-5)(x + l)i T 2 4- 2 , T 146. 5-±j£ Vx^^l-Ssec- 1 -- x 2 2 Vx 2 - 1 1 4 Vx 2 + 4 1 . Vx 2 + 4-2 -log— 140. V3 - x 2 + V3 log V3 - V3 - x 2 147. _^liV2-x 2 . 6x 3 148. I (x 4 + 4) Vx 4 + 1 1 Vx 4 +1-1 + - log 4 Vx 4 + 1 + 1 149. J} (2 x 2 - 1) Vl + 4x 2 . 150. —tan- 1 - 2 a a 2 (a 2 + x 2 ) age 258 151. 3x 2 3(5- -10 -x 2 )f 152. sin -1 2x (X- -3 2)Ve 153 ^log 2 ° 2x-l 2x + Vl6x 2 -12x + 3 154. 1 . -sm 4 2 3- 2x 155. 1 10, x + 2 V2 ' V2 x 2 + 4 x + 4 156. x (log ax — 1). 161. xsec- 1 2x-Uog(2x + V4x 2 -l). 162. - sec- 1 3 x - — V9x 2 -1. 2 18 163. 5 1 7 e 3r (9x 2 -6x+ 2). 164. l[(2x 2 -l)sin2x+2xcos2x]. 165. x[(logx) 2 - 21ogx + 2]. 166. sin 6 x cos 6 x. 4 12 72 167. Je 2x (2sinx — cosx). 168. T L e? (3 sin 3 x + cos 3 x) . 169. 1 [xVx 2 - 1 - log(x + Vx 2 - l)]. 170. \ [sec x tan x + log (sec x + tan x)]. 157. — r(m + l)logx-l] (m + l) 2 " V 158. x tan- 1 ax log (1 4- a 2 x 2 ) . 171 g_ ■ \ os x " ~~ *. 2a "2 & x 159. x log (x + Vx 2 + a 2 ) - Vx 2 + a 2 . ^ g ^ + ^ x-2 160. £ (sin 3 x - 3 x cos 3 x). & x 2 (x 4- 2) 3 ANSWERS 497 173. lo£j x . 1 * i x — , + - tan- 1 - Vx2 + 4 2 2 174. x — tan-!x + 1( (3 + 1)8 3 Vx 2 + 1 175. log- x a 4x + 7 Page 259 181. log V <* +!>(»- 3)' , x-2 182. log(x-2)(x 2 + 2x + 5)§ .,35 + 1 + - tan- 2 2 183. log x — 4 x— 4 184. 1 f - 8g + log/-£-Y. x(x-l) °\x-l/ 185. log Vx^T - + ^lo, V - c3 + 2 -^ 194. 3Vj 3 +2 (JY2 ~~ \ x-+2 + \ 2 1 + sin - log Bin 5/3 + sin 9 -Y , . x 3 2 1 2/ 1 — sin - x ' 195. ±log 1+COs4x cos4x 16 1 — cos4x 8sin 2 4x 196. 1 [a a x 2 + « 2 -a 2 log(x + Vx 2 + u 2 )]. x 197. 198. v 1 + x» vT+^ 199. kx(2x 3 + a 2 )Vx 9 ~+ -/-' a* 8 ^1 + S log(x + Vx 2 + a 2 ). 203. }x(x 2 + 10)Vx 2 + 4 + 01og(x+Vx 2 + 4). 3sin 2 x-2 3, 7T~- 5~ + ^l°g(secx + tanx). 2 sin x cos 2 x 2 ; COS X 1 , -2sh^ + 2 l0g(C8CS - Ctna! >" + ^log V5 2Vl-x + l + V5 192. 2cos 2 x 4 °l + sinx 204. log (sec x + tan x) — esc x. 205 sinx(2 + 3cos 2 x) 8cos 4 x + I log (sec x + tanx). 498 ANSWERS CHAPTER XIII Page 285 h _ h 1. 2. 2. 2. 3. Ja 2 . 4. a 2 (e°-e <*). 5. 4 Tra 2 . Page 286 7. 7r-2 1og2. 8. j}(e* + 2). 15. 4a 2 . 9- f. 16. 8. 10. |(3 r- 2) a 2 . 17i aWs. a- + a- h—Vw- 20. J(4-7r)u' 2 . 21. A(tt-2)« 2 . 22. Tra 2 . 23. a|V2. 24. 3 Tra 2 . 11. 13J. 12. |a&. 13. 2irab. 18. A (6 -a) 15 ■ 19. ^aft. °V ? — a 25. gTTO 2 . Page 287 26. fmz&. 27. iTTrt. „„ 2 a 28. 7T 29. .6366. 30. Jn 2 . 2u 2 31. — 1. Try 32. 34. 35. 37. §Trr 3 . §• + 7r 3 a 2 . 2 a 2 . 38. 4n 39. *. n Page 288 40. i(4-7r) a*. 41. g Tra 2 . 42. 11 TT. 43. le^a 2 . 44. a -V2. 3 45. a 2 [V2- ■log ;(V2 + i)]. 46. - (3 tan— + tan 3 -) c 2 , where 2c is the length of the chord through the focus perpendicular to the axis of the parabola. 47. J to 2 . 49. j(3V3 - 7r)a 2 . 51. ^Tra 2 . 53. fa 2 . 48. 21i. 50. ^(STr + 9Vs)a 2 . 52. 2 tt (a 2 + b 2 ) . 54. § 6a 2 . 61. — (2a 2 +l) + 2wa. 4 62. §7ra 3 (l- cos a). M 9a 3 V3 63. 5120 64. § Tra 3 (3 log 2 -2). 65. Tra 3 tan 6. Page 289 55. |a 8 tan0. 57. | a 3 V§. 56. 2Tr/i 2 Vp 1 p 2 , where p 1 and p. 2 are the 58. — (3 a — 3 v respective param- 59. ffoira*. eters of the pa- rabolas. 60. 38|tt. Page 290 66. VW« 3 - 69. 4Tr 2 a 3 . 67. 4 Tra 3 (2 log 2-1). 70. % ha 2 {Sir 68. 2Tra 3 (4-Tr). 71. iha 2 . 72. 2Tr 2 a6d. 73. 1152 cu. in. 74. 22 ^tt. , AN S WEES 499 Page 291 75. *£-irV2. 76. §7ra 2 6. 77. 2n*a*. 78. 5Tr 2 a 3 . 79. § Tra 3 . 80. §(2 + Tr)a 3 , where a is the radius of the sphere. 81. ^(l0Vi0-l). 83. 6a. 86. -(log9-l). 82. |(e*-e~«). 85. |*ra. 87 . 8a . Page 292 88. § o. 89. 177.5 in. 90. log(e 4-e-x). 91. i?. TT 92. ~[27rV47r 2 +l+log(27r + V47r 2 +l)]. 93. 4 1'"' x --• ez. bid 64. x 65. 4(/ + 9 = and 2x + O2 - 1 = 66. (3, -1,2). 73. :;./• + 5// -z- 20 = 0. 74. 7:4: -5. Page 331 75.(3,0,-3). 82. x + 2y + z-2 = 0. 78. x - z = 0. 83. 13x - y -Viz- 32 = 0. 80. 11 x - 2y — bz — 25 = 0. 84. 3 x + 8 y + 21 2 - 66 = 0. 81. 19a: + 10// + 272-70 = 0. Page 332 85. ox _y_z-5 = 0. 92. (0, 1, 2), (^P, Jy^L, - ^J). 86. (- 1, 2, - 1), (2, - 3, - 4). 95. y- + 4 y - x = 0. 87. 3x + y + 32 - 1 = 0, 96. x 2 - z- = 0. x - y + 2 - 2 = 0. 97. 2 ./•'- + z* - Ox = 0. 88. .V 1 .. 98. bi/ 2 -2z 2 + 8z = 2. 90. (1. I, 3). 100. 1/ = x 2 ; 2 = x 3 ; 2 2 = tj s . 91. 13x + 11 y -17z -15 = 0. AC 502 ANSWERS Page 333 102. xy : = 1. 108. COS" -i_L. V6 Page 334 114. 2. 115.^= 2 1 110. cos-i — . 112. e-e-i. 3 111. |( 12 + 5 log 5). 113. 15. — 2 —— = Z — 1; 2x + 2y + z-7 = 0. v — 1 z /- 116. x - 1 = 2 = —7= \ x-y + V2 z = 0. — 1 V 2 117. ^— = 2/ = ^— ; 4x + ?/ + 9z - 39 = 0. 4 9 118. 4x + 2iry - tt 2 = 0, y - z = ; ttx - 2 y - 2z + 2ir =: 0. 119. K_l=^-=l_ = ^ZLl; x + 32/ + 4z-8 = 0. CHAPTER XV Page 361 1 2/ 3 X 3 (X 2 + 2/ 2 )f ' (x 2 + 2/ 2 )i 2 2/ X X 2 + y 2 ' x 2 + i/ 2 3. 2/ 1 x V2 xy — y 2 V2 xy -y- 4. X "VV - ; * 2 (2/+V2/ 2 - X 2 ) 1 V?/-; c 2 5 2/ 2 xy (x - yf x — y X * co° X2/ (x - y) 2 x-y Page 362 \2/ x/ y 2 y 15. 1+2/ (1 + 2/) 2 2/ 2/ x^ + y 2 x 2 + 17. 2/ &-v 2 xVx^" 2y 2 \ly 19. 2 tan- 2 ' 2a " x X 2 + 2/ 2 20. 2 e^sin (x - y). 30. .640 sq. in. ; .650 sq. in. 27. .0004. 31. .49 ft. 28. -.0302; .0002. 32. .0213ft. 29. 11.030 cu. ft. ; 10.996 cu. ft. Page 363 33. .0048. 34. 1 Vx 2 + y 2 ; o. 35. 0; 0. 1-V3 36. 2 Page 364 k 41. y 42. x + y X 43. y Vx 2 - >r X xsin- it ANSWERS 503 37. 4x + 3t/-11 = 0. 38. k v z{ + y 2 in direction tan- 1 -3 . 1 V3 39. _-cos0- — - sin^; 1. 45 Sx * z • z ( 1 - 2 v 2 ) ' 2 - 2 z 2 ' ?/ (2 z 2 - 2) ' 46 2aaJ + 3zV 2yz + 3x 8 y 2 3z 2 + x 2 + y 2 ' 3z 2 + x 2 + 2/ 2 47 0j/z-2x(x 2 + y 2 4-z 2 ) 2 - " 2 z (x a + ?/ 2 + z 2 ) 2 - 9 xy ' 9xz-2j/(x 2 + y 2 + g 2 ) a ^ (y + z)(2x + y + z) . 2z(x 2 + y 2 + z 2 ) 2 -'.»// (x + y)(x + y+2z)' 48. 2x + 3i/+ 2z-9 = 0; _ (z + x)(x + 2y + 2) x-2 _ y-1 _ z-1 (x + 2/)(* + 2/ + 2z)' 2 :: 3 " 2 49. z + 2y + z — 2 = 0; x-l= V = z-l. 50. 2(ax. + by.) (ax + /»/) = z + z,; ^^ = V ~ ?/ * = Lzfl. 1 2a(ax 1 + 6 ! / 1 ) 2 6(ax 1 + 6y 1 ) -1 54. z + y + z — 8 = 0, X — 2 y + 1 = ; 2 x + // - 3 z = 0. 55. x — 1 = 0, 2 // + 2 z — tt = ; 2 y — 2 z + tt = 0. 56. 4x + 3y-26 = 0, 3y-4z-25 = 0; 3x - iy - 3z - 12 = 0. Page 365 57. 4x- Ay + z - 24 = 0, 2x + 8?/ + 5z = 0; 14x + 9y— 20z + 79 = ieVr 2 -a? 5H. sin- 1 rVa? + k* 59. TV 2' 60. 0. 61. 10\/2 , 10^2, 5V2. a a a 62. 3' 3* 3 63. Mean point of the vertices. 64. 8c*c 3V§ 65. (- u _ 1 •" a "\ 4 B> 4:1' 66. ('- 1 5V2\ 67. 68. (-1, 8abh 27 -1,1). 69. 2x + 2y + z — 6 = 70 x = — a- 2aZ ! + 6 2 + c 2 ' y = - 2 67T ! + & 2 + c 2 ' 2cK a- + B 2 + c- 74. -F- + x. 504 ANSWERS Page 366 71. x 5 - x 3 y + x 2 y 2 + y & . 76. Vx 2 + y 2 - y. 72. xy + logxy. 77. \ (x + sin x cosx) + ?/ cosx. 73 . logx - A _ 7g _ eTcos (x + y)t 79. 2i;2f. 80. 1; 1(4 -7T). 81. If; 1^. 75. _e ?- logy. 82 . l + l og 2; £(l+log2). Page 367 83. tan-H; rfn-l|. 8? ^ ^ F , ( ^/ [2 ^ )]2 84. §7ra 2 . 2x 2 T 3x3*/ V ' dy 2 L V ' J 85. —(36- 2a). + &F"(x). b dy im 2 d 2 V 1 dV 86™ (36 + 2a). 95 "+-J1. b dr 2 r dr Page 368 /a 2 F d 2 v\ , . . „. a 2 F h 98. x?/( — )+(x 2 — y 2 ) y hx du 2 dv 2 \dy 2 ax 2 / v " 'dxdy dx dy CHAPTER XVI Page 394 1. 6 -log 2. 7. iTra 2 . 11. 2|. 2. £(2-tt). 8. 2f. 12. 1(15 -8 log 4). 3. ir-2. * I '- 13. «(a + 6)Va6. 4. 1(11 -16 log 2). 10.^(44 + 9^). 14* 5. 1(5 + 6 V3). 30 V ; T5 15. 4 + ^( S in-i|-sin-iJ). n< ^ (gr _ 2) . 16. ( 2 - 7T + 5 sin- 1 — ) a 2 ; V V5/ 18. ^(3V3 + 2tt). /67r-2-5sin-i-Ma 2 - V V5/ 19- Page 395 20. 10 7T. 24. 317. 29. j^-n-a*. 2 25 A 2 - 30 - i|(a'^ + a6 2 )Va6. 21. |(8 + ir). 105 ' 1024 V2 26. -— (e*™-l)(e 2 ™-l). ' 315 22. -(tt + 3-3V3). 10 " 32. ^ira*. 6 27 ' W- 33. §ira*. 23. 22^. 28. {TTtt 4 . 34. r %%ah 3 . 48 * 2 V3 answ: ERS 505 Page 396 35.^(15^32). a 4 a 4 38. — (15 tt 24 v a 4 -32). «•<¥»»■ 12. (V-, 0). 36. -(8 + 3tt). 39. (45tt-128). /o 9 „ v 1920 43. (^,0\. 37. UiraK 40. T %7ra 4 . \15tt / 11 I 18a 18 a \ 46 / 8a(6,+ 8)x V 5(6w-4) / \5(3ir-2)' 5(3 tt- 2)/ «(•?..). .. * ., . .„. 3 a sin a 17 \'n the a v is ui the c ector ' 4 a from the center of the circle. Page 397 48. ( , \105(16-3tt) 304 a \ 53./ a < 37r+2 >,0\ V 6V3 / 105 (10- 3 tt)/ •^■•> 51. ( "■ ,0V V 2(a6-r 2 ) / 50. (ff, 0). 55. ^ttu-. 56. 4ttV5. -■(-35-* 57. 8 a 2 . 58. 2a 2 (7r-2). *(-.). 59. ^(tt-2). 3 V ' 60. 8 a 2 . Page 398 a 2 61. — (20-3tt). 68. ,',; rra 4 . 18 V ' 69. - \((P-Se + 1B). 62. $(20- Sir). 70 . 2 a 2 63. (20-3tt). " 36 48 9 v 71. a(2-V§). 64. 4a 2 [V2-logi Ci+Vi)]. 72. i(4 + 7r)-log(V2 + l). 65. 16a 2 [7r-V2- -log(l + V2)]. ,« «- 66. (71- -2) a 2 . 73. — 40 67. -(7T- 8 V •2). 71. 2 7ra 2 . Page 399 90 76. 4,1,,. 77. § ttu 3 . 80. 81. 82. 102$. Ott" 7ra 3 . 87. \ira?b. O/78 88. — (3tt-4), „„ a 3 89. 360 78.^. 32 6 83. 81. 9 n 3 5 a . 3 2«8 -ij-a . »*(. Vs- -4). 85. 86. § 7ra6c. X7ra 3 . 506 ANSWERS Page 400 91 — (3 ir + 20 — 10 V2) 96 ' § to 4- e-^^-V). 2 c, $. y= A( e A(x- e -A(..-c 2 )). 3x 64. y 4- c 2 = qx 3 65. 1296 2/ = (x + l) 4 - 66. y = Vi tan V§ (x - ^J 7T 67. sin ?/ = x — - • 68. x = log (V2 sin y). 510 ANSWERS Page 474 69. FSBV M+ V -i-+£ + *-L n - ? ' = V+ V^"+*«-- 70. y = c x e* + c 2 e- ■■ 2 72 - ■" = ( c i x + h) e~**+{x- 2) 2 . — -}- (0 sin 2 x + 2 cos 2 x). 73. y = c 2 cos 3 x + c 2 sin 3 x + J e 3 *. 74. ?/ = e 2 (cj cos'— 1- c 2 sin— — ) + x 3 + 3x 2 . 75. .v = (x + c^ t- 3r + c 2 e-6*. 76. y = c 1 + c.,e' Sx + x' 2 . 77. y = c,c-"-- ,: + c 2 e 2x — — (sin 2a + 2 cos2x). 78. y = e~ 2x (Cj cos x + c 2 sin x) + J (3 sin 3 x — cos 3 x). a- 8x 79. y = c x e 2 + c 2 e 2 + ^ (8 cos x — 14 sin x) — ? £ F (8 sin 2 x + 19 cos 2 x) . 80. y — eP° (c, cos x V2 + c„ sin x V2 ) H (5 cos x — 4 sin x). 41 81. y = (c,x + c )e- -2 * + (0 - 20 x + 25 x 2 ). (525 82. y = C-pSinSx + (c 2 Jcos3x. 83. y = qe 2 * + c. 2 e~ x — — (3 sinx + cos x). (x x 2 \ 1 Cj 1 + c 2 e 5 x H (9 sin x + 7 cos x) . 85. y = e-*(c 1 cosxV3+ e 2 sinxV3) + J(2x — 1 + e 2 '). xe x ' 1 86. y = c 1 e 3j ' + c„e -a: (4 sin2x + 7cos2x). 4 65 87. y = e 2x (c 1 cos2x + c 2 sin2x) + J, (1 + 4x + 4x 2 ) + ^ (3 cos 3 x + 30 sin 3 x). 88. y = c 1 e~ x + e 2 ( c 2 cos + c 3 sin ) — + x + x 3 . 89. y = c-je- 1 " + c 2 e~ x + c 3 e-' ix — J (20 — Ox + 9 x 2 ) + } (cos 3 x — sin 3 x). 90. y = Cl e-*s + e**(c 2 + c 3 x - ^ + ^ + J (1 + x + x 2 ). Page 475 9l . J , = ,(,,_|) +e ,.^ + ?) + ,, c -= J + ^. 92. y = c x & + U 2 - ^J cos 2 x + (c s - |Asin 2 x. 1 e 2x 93. y = Cje x + c 2 e~ * + c 3 e- 2a ' + - (cos 2 x — sin 2 x) + — (12 x — 19). gX 94. y = e~ x (c 1 + c„x + c 3 x 2 ) + — (2 sinx — 11 cosx). 95. y = (c x + c 2 x) cosx + (c 3 + c 4 x) sinx — 4 + x + x 2 . 96. ?/ = e 2x (< i + C 2 X + t) + e_2x ( c s + C 4 X )- ANSWEES 511 97. y = c 1 e 2x + c 2 e~ 2x + c s cos 3 x + c A sin 3 x — - — (sin 3 x + 3 cos 3 x). 98. y = e, + c„x + c 3 cos xVi + c 4 sinx V2H (13 sin 2 x + 16 cos 2 x). 6 17 99. y = c 1 + c 2 x + c 3 x 2 + c i e~ ix + T2 -^(cos4x — sin4x). 100. x = (Cjt + c 2 ) e< + e 2 ', -(2r-2)g] \ III "7 + C2X ( 1 + !Lzl x2+ (»- 1 H--^ x 4 + ("-i)(n- 7 Q)(n-25) x6 + (n - 1) („ - 9) (n - 25) ■ ■ • [n - (2 r - l) 2 ] ^ r 1 2 r + 1 ^ + ...). 512 ANSWERS 112. y = cJ\ +~ + „ f r n + • \ 2-5 2-4.5-9 [2 - 4 - 6 • • • 2 r] [5 - 9 - 13 - - • (4 r + 1)] + c £i( 1 + 2-3 2- 4-3-7 [2-4-G...2r][3-7-ll.--(4r-l)] 113. w = Cl (l- — + — + + (- l) 1 - 22r|2r_+l '\ 2 2 |2 2*14 T 4 r 115. x = log(y + 3). 118. r{0 + c) = l. 116. k (x — 1) y — y + 1 = 0. where 119. r 2 = 2(0 + c). A; is the constant ratio. 120. r n = c sin nd. 117. y = ex". 121. x' 2 - y 2 = c. Page 477 123. 124. x' 2 + y' 2 = ex. a I+f _^±£. y = -(e a +e «). 129. y = ce k , where x = a is the fixed ordinate, and & is the constant ratio. 125. y = ^( ek(J - c) + e - kCr - c) h 130. y = -(e A " + e * ), where fc where k is the constant ratio. 2 is the constant ratio. 126. . Va 2 — 2 ay — a x + c = 4- a log — 131. 2/ = c ± 1 Vfc 2 x — 4 x 2 k 2 . . 2 Vx . ± —sin- 1 , where* is 4 A; Va 2 — 2 ay + a ± 2 Va 2 - 2 ay. 127. r = ce *■ the constant ratio. 128. y = ax 2 + 6. 132. s = 25 (2) s . Page 478 134. p = 14.7e -,00004A , 135. $1218. 136. 67 niin 137. EIy = /lx 2 — wl \2 x 3 \ 138. EIy = w/l 2 x 2 2\~2~~ -?* -) 12/ 139. EIy = w/lx 2 X 2\T~"e :)• 133. c = .01 e- -183* 140. (x - r x ) 2 + (y - c 2 ) 2 = c 2 141. Cl2/ 2 -^-(x + c 2 ) 2 = l. 142. V3^ ANSWERS 513 Page 479 -..* a ittt i_ , . ., 149. About 7 mi. per second. 145. - V3A:, where A; is the constant „ ,- — - * 2 150. V2gh. ratio. 151. 20 sec. 146. J (4- V2) Vol 152. 234| ft. ; 4.3 sec. (g = 32.) 147. ■ 153. J. » 1X / mg , . V mg + fcoj v i + Page 480 154. 1.8 sec. (£/ = 32.) 155. s = c, cosht + c n sin /i( 4 — cosfcf, where h- is the constant ratio K 1 — fc a s = Cj cos Af + c 2 sin ht H sin kt, if h = k. It t^P-ih* t^/p-th- 156. s = e~ 2 (qe 2 + c 2 e a ) , a(/t 2 - k 2 ) coskt + akl sin kt .. , „ , + (tf-*y + (tt)' ' lf * >2 * ; -£/ «V4/i 2 - J 2 . fV4A 2 -/ 2 \ 2 I c, cos |-c„sin — I \ x 2 2 2 / a (A 2 - fc 2 )cosfe<4- akl sin W + (A 2 -Jfc 2 ) 2 +(tt) 2 '**<2A; Vl 2 ' (A* -*»)* + (2 M:) 3 (/(- and / arc the constant ratios.) 157. h = fc, I very small. s = e INDEX (The numbers refer to the pages) Abscissa, 5 Acceleration, 178 Angle, 59, 166, 205, 315, 316 eccentric, 108 vectorial, 118 Arc, differential of, 174, 206, 324 limit of ratio to chord, 172 Archimedes, spiral of, 120 Area, of any surface, 381 as double integral, 376 as line integral, 354 of plane curve, 143, 262, 267 of surface of revolution, 274 Asymptote, 26, 79 Attraction, 283, 393 Axis of conic, 76, 80, 81 Bemouilli's equation, 446 Bessel's equation, 470 Bessel's functions, 471 Bisection of straight line, 9 Cardioid, 124 Catenary, 54, 441 Center of gravity, 278, 379, 392 Center of pressure, 277 Change of coordinates, 39, 124, 386, 387, 388 Circle, 70, 108, 122 auxiliary, 108 involute of, 112 Cissoid, 85, 124 Comparison test, 406 Components, of a straight line, 312 of velocity, 180 Concavity of plane curve, 167 Cone, 306 Conic, 83, 125 Constant of integration, 147, 222 Continuity, 136, 339 Convergence, 405, 409 Coordinates, Cartesian, 1, 4, 301 cylindrical, 385 oblique, 43, 301 polar, 118, 386 Curvature, 207 Curves, in space, 322 intersection of, 29 Cusp, 109 Cycloid, 109 Cylinder, 303 Degree of plane curve, 29 Derivative, 136 directional, 343 higher, 162, 338 in parametric representation, 204 partial, 335 sign of, 138, 166 total, 344 Differential, 141, 339, 359 of arc, 174, 206, 324 of area, 146 exact, 349 total, 341 Differentiation, 136 of algebraic functions, 154 of composite functions, 357 of implicit functions, 162 partial, 335 of polynomial, 137 successive, 162, 338 of transcendental functions, 192 Direction in polar coordinates, 205 ',11 INDEX U5 Direction cosines, 314, 319, 324 Directrix of conic, 80, 83 Discontinuity, finite, 427 Distance, of point from plane, 320 of point from straight line, 63 between two points, 5, 313 Divergence, 405 Division of straight line, 8 e, the number, 53 Eccentricity of conic, 74, 78, 83 Element of definite integral. 2H0 Ellipse, 74, 108 area of, 262 Ellipsoid, 307 volume of, 270 Epicycloid, 111 Epi trochoid, 114 Equations, differential, 438 empirical, 89 parametric, 106 roots of, 34 Factor, integrating, 448 Focus of coiii,-, 74, 78, 80, 83 Force, 179 Forms, indeterminate, 423. 12"> Fourier's series, 427 Fractions, partial, 247 Function, 9, 300 Bessel's, 471 complementary, 456, 458, 461 composite, 357 implicit, 163, 300 notation, 12 periodic, 427 transcendental, 49 Graph, 10, 20, 49, 301 Helix, 324 Hyperbola, 77 Hyperboloid, 306, 307 Hypocycloid, 112 four-cusped, 107 Hypotrochoid, 114 Increment, 135, 339 Infinitesimal, 261 Infinity, 25 Inflection, point of, 167 Integral, 146, 222 constant of, 147, 222 definite, 147, 260 line, 353 multiple. 369 particular, 456, 458, 461 triple, 385 Integrand, 222, 264 Integration, 222 approximate, 41H collected formulas of, 236 by partial fraction.-. 2 17 by paiis. 21:; of a polynomial. 1 Pi by reduction formulas, 252 by substitution, 238 Intercepts, 21 Intersections, 2!» Involute of circle, 1 12 Legendre's coefficients, 169 Legendre's equation, 168 Lemniscate, 120, L25 Length of a curve, 272. 325 Limacon, 123 Limit, 130 of ratio of arc to chord, 172 Of ^\ 1,2 h of (i+ ft)*, iog theorems on, 132 Limits of definite integral, 148, 264 Line, straight, 57, 121, 317 Line integral, 353 Locus, 21). <;'.> Logarithm, Napierian, 54 Maclaurin's series, 412 Maxima and minima, 168, 348 Mean, theorem of, 422 Moment of inertia, 377, 390 516 INDEX Motion, in a curve, 180 rectilinear, 177 simple harmonic, 203 Normal, to curve, 104, 326 to plane, 310 to surface, 347 Number scale, 3 Operator, 455 Order of differential equation, 441 Ordinate, 5 Origin, 3, 118 Parabola, 80 segment of, 83 Paraboloid, 305, 308 Parameter, 106 Parts, integration by, 243 Plane, 305, 316 Point, of division, 8 turning, 139 Pole, 118 Polynomial, derivative of, 137 integral of, 146 Pressure, 275 center of, 277 Prismoid, 421 Prismoidal formula, 419 Projectile, path of, 181 Projection, 2, 310, 323 Radius of curvature, 208 Radius vector, 118 Rate of change, 175 Ratio test, 407 Reduction formulas, 252 Region of convergence, 411 Remainder in Taylor's series, 416 Revolution, surface of, 270, 274, 309 Rolle's theorem, 416 Roots of an equation, 34 Rose of three leaves, 119 Segment, parabolic, 83 Series, 405 Fourier's, 427 geometric, 405 harmonic, 406 Maclaurin's, 412 power, 410 Taylor's, 412 Simpson's rule, 421 Slope, 6, 134 Space geometry, 300 Sphere, 310, 313 Spirals, 120 Strophoid, 86 Substitution, integration by, 238 Surfaces, 304 of revolution, 270, 274, 309 Symmetry, 23 Tangent, to plane curve, 30, 140, 164 to space curve, 326 to surface, 345 Taylors series, 412 Tractrix, 439 Transformation of coordinates, 39, 124, 386, 387, 388 Trapezoidal rule, 421 Trochoid, 110 Value, absolute, 409 infinite, 25 mean, 265 Variable, 9 Vector, radius, 118 Velocity, 177, 180 Vertex of conic, 75, 79, 81 Volume, of any solid, 389 element of, 385, 386, 387 of solid with parallel bases, 268 of solid of revolution, 270 Witch, 84 Work, 275, 353, 356 *f RETURN CIRCULATION DEPARTMENT TO— * 202 Main Library LOAN PERIOD 1 HOME USE 2 3 4 5 6 ALL BOOKS AAAY BE RECALLED AFTER 7 DAYS 1 month loans may be renewed by calling 642-3405 6-month loans may be recharged by bringing books to Circulation Desk Renewals and recharges may be made 4 days prior to due date DUE AS STAMPED BELOW 8EC. 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