GIFT OF W. G. 3. Shonhan ELEMENTARY ANALYSIS BY PEECEY F. SMITH, Ph.D. PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY AND WILLIAM ANTHONY GEANVILLE, Ph.D. INSTRUCTOR IN MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON Entered at Stationers' Hall Copyright, 1910, by Percey F. Smith and William Anthony Granville all bights reserved 510.9 e. GINN AND COMPANY • PRO- PRIETORS ' boston • U.S.A. PREFACE The text of this volume is, to a considerable extent, identical with portions of corresponding chapters in Smith and Gale's "Elements of Analytic Geometry" and Gran- ville's "Elements of the Differential and Integral Calcu- lus." The new material is contained in the chapters on Curve Plotting (Chapter V) and Functions and Graphs (Chapter VI). At the same time, the parts which have appeared in previous books of the series have been thor- oughly revised and, to a considerable extent, rewritten, to the end that the aim of the authors might be accomplished, — namely, to prepare a simple and direct exposition of those portions of mathematics beyond Trigonometry which are of importance to students of natural science. In this connection attention may be called to the intentional avoidance of anticipating difficulties, — a feature which is not common in textbooks. To particularize, processes which are natural are introduced without explanation, and exact definition is not given until the student is familiar by practice with the matter in hand. Again, in the deriva- tion of certain formulas in the Differential Calculus the evaluation of particular limits is not undertaken until the student sees that this work must be done before the prob- lem can be solved. V 401141 vi ELEMENTARY ANALYSIS In many instances, when deemed wise, a general discus- sion is introduced by concrete examples. This feature, so common in school texts, is strangely absent from books in- tended for use in colleges and technical schools. Interest in the subject is usually aroused in this way, and it is the hope of the authors that this stimulus may not be lacking when the volume is» studied. New Haven, Connecticut CONTENTS CHAPTER I FORMULAS FOR REFERENCE SECTION PAGE 1 . Formulas and theorems from geometry, algebra, and trigonometry 1 2. Three-place table of common logarithms of numbers .... 4 3. Logarithms of trigonometric functions 5 4. Natural values of trigonometric functions 5 5. Special angles. Natural values 6 6. Rules for signs 7 7. Greek alphabet 7 CHAPTER II CARTESIAN COORDINATES 8. Directed line 8 9. Cartesian coordinates 9 10. Rectangular coordinates 10 11. Lengths 13 12. Inclination and slope 16 13. Areas 21 CHAPTER III CURVE AND EQUATION 14. Locus of a point satisfying a given condition 28 15. Equation of the locus of a point satisfying a given condition . . 28 16. First fundamental problem 30 17. Locus of an equation 35. 18. Second fundamental problem 36 19. Third fundamental problem. Discussion of an equation ... 41 20. Symmetry 45 21. Further discussion 46 22. Directions for discussing an eqviation 47 23. Points of intersection 50 vii viii ELEMENTARY ANALYSIS CHAPTER IV STRAIGHT LINE AND CIRCLE SECTION PAGE 24. The degree of the equation of any straight line '. . 54 25. Locus of any equation of the first degree 55 26. Plotting straight lines 56 27. Point-slope equation 58 28. Two-point equation 59 29. The angle which a line makes with a second line 63 30. Equation of the circle 68 31. Circles determined by three conditions 70 Locus problems 74 CHAPTER V CURVE PLOTTING 32. Asymptotes 77 33. Natural logarithms ' 79 Exponential and logarithmic curves 80 Compound interest curve 82 34. Sine curves 83 35. Addition of ordinates 88 CHAPTER VI FUNCTIONS AND GRAPHS 36. Functions 91 37. Notation of functions 100 CHAPTER VII DIFFERENTIATION 38. Tangent at a point on the graph 102 39. Differentiation 103 40. Derivative of a function 105 CHAPTER VIII FORMULAS FOR DIFFERENTIATION 41. Theorems on limits Ill 42. Fundamental formulas 113 CONTENTS ix SECTION PAGE 43. Differentiation of a constant 114 44. Differentiation of a variable with respect to itself 114 45. Differentiation of a sum 115 46. Differentiation of the product of a constant and a function . . 115 47. Differentiation of the product of two functions 116 48. Differentiation of a power of a function 117 49. Differentiation of a quotient 117 50. Differentiation of a function of a function . 123 51. Differentiation of a logarithm 124 52. Differentiation of an exponential function 127 53. Proof of the power rule 129 54. Differentiation of sin u 132 55. Differentiation of cos v 135 56. Differentiation of tan .u : 135 57. Proofs of XIII-XV 136 58. Inverse circular functions 139 59. Differentiation of sin-iv and tan- 1 u 140 60. Implicit functions 143 CHAPTER IX SLOPE, TANGENT, AND NORMAL 61. Tangent and normal 146 CHAPTER X MAXIMA AND MINIMA 62. Rule for the determination of maximum and minimum values . 151 63. Derivatives of higher orders 156 64. Geometrical significance of the second derivative 158 65. Second test for maxima and minima 161 m. Points of inflection , 162 CHAPTER XI RATES 67. Velocity and acceleration . ^. 167 Time rates • • 168 ELEMENTAEY AI^ALYSIS CHAPTER XII DIFFERENTIALS SECTION PAGE 68. Definition of differential 175 69. Formulas for finding differentials 176 70. Infinitesimals 179 CHAPTER XIII INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 71. Integration 180 72. Constant of integration. Indefinite integral 182 73. Rules for integrating standard elementary forms 183 74. Trigonometric and other substitutions 194 CHAPTER XIV CONSTANT OF INTEGRATION 75. Determination of the constant of integration by means of in- itial conditions 196 Compound interest law 197 CHAPTER XV THE DEFINITE INTEGRAL 76. Differential of an area 201 77. The definite integral 202 78. Geometric representation of an integral . . 204 79. Calculation of a definite integral 204 CHAPTER XVI INTEGRATION A PROCESS OF SUMMATION 80. Introduction 207 81. The fundamental theorem of the integral calculus 207 82. Areas of plane curves 211 83. Volumes of solids of revolution 216 84. Miscellaneous applications of the integral calculus 219 ELEMENTARY Al^ALYSIS CHAPTER I FORMULAS FOR REFERENCE I. Occasion-55^ill arise in later chapters to make use of the following formulas and theorems proved in geometry, algebra, and trigonometry. 1. Circumference of circle = 2 7rr.* 3. Volume of prism = Ba. 2. Area of circle = irr^. 4. Volume of pyramid = J Ba. 5. Volume of right circular cylinder = wr'^a- 6. Lateral surface of right circular cylinder = 2 irra. 7. Total surface of right circular cylinder = 2 7r7'(r 4- a). 8. Volume of right circular cone = ] irr^a. 9. Lateral surface of right circular cone =7rrs. 10. Total surface of right circular cone = 7rr(r -f- s) . II. Volume of sphere := | irr^. 12. Surface of sphere = 4 Trr^. 13. In a geometrical series, r — 1 r — 1 a = first term, r = common ratio, I = nth term, s =: sum of n terms. 14. loga?) = loga +log&. * 16. loga^ = wloga. - \o^ a. *In formulas 1-12, r denotes radius, a altitude, B area of base, and .9 slant height. 1 15. I0g^ = l0grt- b - log b. 17. log Va =-\oga. n 18. log 1=0. 19. logaa = 1. 20. logi V.-'«--7>rv^^ V \*i'0-v-«iafc-*. ELEMENTARY ANALYSIS Functicns of an aixjU In a ri^p triangle. In any right triangle of whose acute angles is A^ the functions of A are defined as follows ; _ opposite side 21. sin A — -— , hypotenuse cos^ = ^dj^cent side hypotenuse ^^ ^ ^ opposite^ide adjacent side ' CSC A = ^^yPQtenuse ^ opposite side' sec A = ^^yPQ^enuse ^ adjacent side ' . A _ adjacent side opposite side From the above the theorem is easily derived : 22. In a right triangle a side is equal to the prod- uct of the hypotenuse and the sine of the angle op- posite to that side, or of the hypotenuse and the cosine of the angle adjacent to that side. Angles in general. In Trigonometry an angle XOA is considered as generated by the line ^ OA rotating from an initial position OX. The angle is positive when OA rotates from OX counter-clock- loise^ and negative when the direction of rotation of OA is clockivise. The fixed line OX is called the initial line, the line OA the terminal line. Measurement. of angles. There are two important methods of measur- ing angular magnitude, that is, there are two unit angles. Degree measure. The unit angle is ^i^ of a complete revolution, and is called a degree. Circular measure. The unit angle is an angle whose subtending arc i^ equal to the radius of that arc, and is called a radian. The fundamental relation between the unit angles is given by the equation 23. 180 degrees = ir radians (ir^: 3.14159 •••)• Or also, by solving this, 24. 1 degree = -^ = .0174 • • • radians. ^ 180 owa* 25. 1 radian : 180 : 57.29 ... degrees. FORMULAS FOR REFERENCE 3 These equations enable us to change from one measurement to another. In the higher mathematics circular measure is always used, and will be adopted in this book. The generating line is conceived of as rotating around through as many revolutions as we choose. Hence the important result : Any real number is the circular measure of some angle, and conversely, any angle is measured by a real number. 26. cotx=: ;secic=: ;cscic= tan X cos x sin x 27. tana; = ?i^; cotx = 55?^. cos X sin X 28. sin2 X + cos2 x = 1 ; 1 + tan^x = sec% ; 1 + cot^ic = csc^ x. 29. sin (— x) = — sin x] esc (— x) = — esc x ; cos ( — x) = cos a:; ; sec (— a:) = sec x ; tan ( — x) = — tan x ; cot ( — x) = — cot x. 30. sin (tt — x) = sin x ; sin (tt + x) = — sin x ; cos (tt — x) = — cos X ; cos (jr -{-x) = — cos x ; tan (tt — x) = — tan x ; tan (tt + x) = tan x. 31. sin r^-— X j = cosx; sin( ^+ X ) = cosx; ^ — X J =: sin X ; cos (^+xj = — sinx; t;i ! 1 ( ^ — X 1 = cot X ; tan f '^ + x ) = — cot x. 1.2 y ' V2 y 2 TT — x) = sin(— x) = — sin x, etc. • >: + ?/) = sin x cos ?/ + cos x sin ?/. X — y)— sin x cos ?/ — cos x sin ,'/. X + ?/) = cos X cos ?/ — sin x sin i' , X— y)= cos X cos y + sin x sin y. ' , . + y)=: t^^^ + tany ^ 38. tan (x - y) ^ tan x - tan y _ 1 - tan X tan ?/ 1 + tan x tan ?/ 39. i !; : c =: 2 sin x cos x ; cos 2 x = cos'^ x — sin^ x ; tan 2 x = _ 2 tan x ELEMENTARY ANALYSIS 40, ' X , /l — cos X X , ll -\- COS X . X , /l — COS X + COSX 2 >( 2 2 >( 2 41. sin2 X = I ~ I COS 2 x ; cos'^ x = ^ + I cos 2 x. 42. sin A - sin ^ = 2 cos ^ (^ + ^) sin i (A - B), 43. cos ^ - cos jB =— 2 sin ^ (A -{- B) sin | (^ - ^)- 44. Theorem. Law of cosines. In any triangle the square of a side equals the sum of the squares of the two other sides diminished by twice the product of those sides by the cosine of their included angle ; that is, a2 = &2 ^_ c2 _ 2 &c cos A. 45. Theorem. Area of a triangle. The area of any triangle equals one half the product of two sides by the sine of their included angle ; that is, area = | a6 sin C = J 6c sin A = ^ ca sin B, 2. Three-place table of common logarithms of numbers. N 1 2 3 4 5 6 7 « 9 1 000 041 079 114 146 176 204 230 255 279 2 301 322 342 362 380 398 415 431 447 462 3 477 491 505 518 532 544 556 568 58" • -' 4 602 613 623 634 643 653 663 672 6^ 5 099 708 716 724 732 740 748 756 703 771 6 778 785 792 799 806 813 820 826 832 8ro 7 845 851 857 863 869 875 881 886 S-- 8 903 908 914 919 924 929 934 939 9-t + 9 10 11 954 959 964 968 973 978 982 987 037 000 004 009 013 017 021 025 029 041 045 049 053 057 061 064 068 072 076 12 079 083 08^5 093 097 100 104 107 ill 13 114 117 121 127 130 134 137 140 143 14 146 149 152 155 158 161 164 167 170 173 15 176 179 182 185 188 190 193 196 190 201 16 204 207 210 212 215 218 220 223 22o 228 17 230 233 236 238 241 243 246 248 250 253 18 255 258 260 262 265 267 270 272 274 276 19 279 281 283 286 288 290 292 294 297 299 FORMULAS FOR REFERENCE 3. Logarithms of trigonometric functions. Angle in Radians Angle in Degrees log sin log cos log tan log cot .000 0° 0.000 90° 1.571 .017 1° 8.242 9.999 8.242 1.758, 89° 1.553 .035 2° 8.543 9.999 8.543 1.457 88° 1.536 .052 3^^ 8.719 9.999 8.719 1.281 87° 1.518 .070 40 8.844 9.999 8.845 1.155 86° 1.501 .087 5° 8.940 9.998 8.942 1.058 85° 1.484 .174 10" 9.240 9.993 9.246 0.754 80° 1.396 .262 15° 9.413 9.985 9.428 0.572 75° 1.309 .349 20° 9.534 9.973 9.561 0.439 70° 1.222 .436 25° 9.026 9.957 9.609 0.331 65° 1.134 .524 30° 9.699 9.938 9.761 0.239 60° 1.047 .611 35° 9.759 9.913 9.845 0.165 55° 0.960 .698 40° 9.808 9.884 9.924 0.080 50° 0.873 .785 45° 9.850 9.850 0.000 0.000 45° 0.785 log cos log sin log cot log tan Angle in Degrees Angle in Radians 4. Natural values of trigonometric functions. Angle in Radians Angle in Degrees sin COS tan cot .000 0° .000 1.000 .000 QO 90° 1.571 .017 1° .017 .999 .017 57.29 89° 1.553 .035 2° .035 .999 .035 28.64 88° 1.536 .052 30 .052 .999 .052 19.08 sr 1.518 .070 40 .070 .998 .070 14.30 86° 1.501 .087 5° .087 .996 .088 11.43 85° 1.484 .174 10° .174 .985 .176 5.67 80° 1.396 .262 15° .259 .966 .268 3.73 75° 1.309 .349 20° .342 .940 .364 2.75 70° 1.222 .436 25° .423 .906 .466 2.14 65° 1.134 .524 30° .500 .866 .577 1.73 60° 1.047 .611 35° .574 .819 .700 1.43 55° .960 .698 40° .643 .766 .839 1.19 50° .873 .785 45° .707 .707 1.000 1.00 45° .785 cos sin cot tan Angle in Degrees Angle in Radians 6 ELEMENTARY ANALYSIS 5. Special angles. Natural values. Angle in Radians Angle in Degrees sin COS tan cot sec CSC 0° 1 GO 1 GO IT 2 90^ 1 GO GO 1 IT 180^ - 1 00 - 1 GO Sir 2 270^^ -1 GO 00 - 1 27r 360° 1 00 1 00 Angle in Radians Angle in Degrees sin cos tan cot sec CSC 0° 1 - ; GO 1 GO TT 6 30^=^ 1 2 2 3 ' V.^., ■- 3 2 IT 4 45'^ V2 2 V2 - 2 1 1 V2 V2 TT 3 60° V3 2 1 2 V3 V3 3 2 2V3 3 TT 2 90° 1 00 GO 1 The student sliould understand that the numerical values of the functions given in the second of the above tables are to be used when formal exactness only is required ; that is, when it is sufficient to leave the results in radical form. Explicit numerical values are given in Art. 4. The same remarks apply to the values of the angle in radians. FORMULAS FOR REFERENCE 6. Rules for signs. ^ __ -V - Quadrant sin cos tan cot sec CSC First + + + + + + Second .... + - - - + Third .... - - + + - - Fourth .... - + - - + - 7. Greek alphabet. ETTERS Names Letters Names Letters Names Aa Alpha I I Iota PP Rho B^ Beta K/c Kappa S (7 S Sigma Ft Gamma A \ Lambda T T Tau A5 Delta M/x Mu Tu Upsilon E e Epsilon N V Nil * Phi zr Zeta s^ Xi Xx Chi H^ Eta Oo micron Theta Htt Pi 12 w Omega CHAPTER II CARTESIAN COORDINATES 8. Directed line. Let X'X be an indefinite straight line, and let a point 0, which we shall call the origin, be chosen upon it. Let a unit of length be adopted, and assume that lengths measured from to the right are x>ositive, and to the left negative. -5-4-3-2-1 0-hl-+2+3-l-4+5 unit X' O ^ X 1 Then any real number, if taken as the measure of the length of a line OP, will determine a point P on the line. Con- versely, to each point P on the line will correspond a real number ; namely, the measure of the length OP, with a posi- tive or negative sign according as P is to the right or left of the origin. The direction established upon X^X by passing from the origin to the points corresponding to the positive numbers is called the positive direction on the line. A directed line is a ~B 2 X "b^ straight line upon which an origin, a unit of length, and a positive direction have been assumed. An arrowhead is usually placed upon a directed line to indi- cate the positive direction. If A and B are any two points of a directed line such that OA = a, OB = b, then the length of the segment AB is always given by b — a-, that is, the length of AB is the difference of the numbers cor- 8 CARTESIAN COORDINATES 9 responding to B and A. This statement is evidently equiva- lent to the following definition : For all positions of two p)oints A and B on a directed line, the length AB is given by (1) AB^OB-OA, where O is the origin, (I) Cll) an) (IV) +3 -h6 -4 0+3 -3 H-5 -6 -2 O 4. M B A A B B A Illustrations. In Fig. ^ ^- I. ^li?= 01^-0^^=0-3= +3; ^^=0^-05=3-6 = -3; II. AB=0B-0A=-4.-?,= -l', 75^=0^-05=3-(-4) = +7; III. ^5=0i>^-0.1=+5-(-3) = +8; ^^l=r0.4-OB=-3-5 = -8; IV. AB=OB-OA=-Q-{-2) = -4:- BA^0A-0B=-2-{-&)^+^, The following properties of lengths on a directed line are obvious : (2) AB =-BA. (3) AB is positive if the direction from A to B agrees with the positive direction on the line, and negative if in the con- trary direction. The phrase " distance between two points " should not be used if these points lie upon a directed line. Instead, we speak of the length AB^ re- membering that the lengths AB and BA are not equal, but that AB = — BA. 9. Cartesian^ coordinates. Let X'X and Y'Y be two directed lines intersecting at 0, and let P be any point in their plane. Draw lines through P parallel to X'X and Y' Y respectively. Then, if OM=a, ON=h, * So called after Eene Descartes, ISOB-IOHO, who first introduced the idea of coordinates into the study of Geometry. 10 ELEMENTARY ANALYSIS the numbers a, b are called the Cartesian coordinates of P, a the abscissa and b the ordinate. The directed lines X'X and F' Y are called the axes of coordinates, X'X the axis of abscissas, Y'Y the axis of ordinates, and their intersection the origin. The coordinates a, b of P are written (a, b), and the symbol P(a, b) is to be read: "The point P, whose coordi- nates are a and b^ Any point P in the plane determines two numbers, the coordinates of P. Conversely, given two real numbers a' and 5', then a point P' in the plane may always be constructed whose coordinates are (a', V). For lay off OM^ = a', 0N^= b', and draw lines parallel to the axes through JP and N\ These lines intersect at P\a\ b'). Hence Every point determines a pair of real numbers^ and conversely, a pair of real numbers determines a point. The imaginary numbers of Algebra have no place in this representation, and for this reason elementary Analytic Geome- try is concerned only with the real numbers of Algebra. 10. Rectangular coordinates. A rectangular system of coordinates is determined when the axes X^X and F' Y are perpendicular to each other. This is the usual case, and will be assumed unless otherwise stated. The work of plotting points in a rectangular system is much * simplified by the use of coordinate ov plotting paper, constructed by ruling off the plane into equal squares, the sides being parallel to the axes. In the figure several points are plotted, the unit of length CARTESIAN COORDINATES 11 > Y (6, 7) {-4 6) X' qo, 0) X {-0 -4) {0, -4) I 1 being assumed equal to one division on each axis. The method is simply this: Count off from along X'X a number of divisions equal to the given abscissa, and then from the point so determined a number of divisions equal to the given ordinate, observing the Rule for signs : Abscissas are positive or negative according as they are laid off to the right or left of the origin, Ordinates are positive or nega- tive according as they are laid off above or beloiv the axis of x. Eectangular axes divide the plane into four portions called quadrants; these are numbered as in the figure, in vs^hich the proper signs of the coordinates are also indicated. rt Second First (f v-f) X Fourth X' Thxrd (-,-) Y' As distinguished from rectangular coordinates, the term oblique coordinates is employed when the axes are not perpen- dicular, as in the figure, p. 10. The rule of signs given above applies to this case also. In the following problems assume rectangular coordinates. 12 ELEMENTARY ANALYSIS PROBLEMS 1. Plot accurately the points (3, 2), (3, - 2), (-4, 3), (6, 0), (-5,0), (0,4). 2. What are the coordinates of the origin ? Ans. (0, 0). 3. In what quadrants do the following points lie if a and b are positive numbers: (—a, b)? (—a, —b)? {b; — a)? (a\ b)? 4. To what quadrants is a point limited if its abscissa is positive ? negative ? its ordinate is positive ? negative ? 5. Plot the triangle whose vertices are (2, —1), (—2, 5), (-8,-4). 6. Plot the triangle whose vertices are ( — 2, 0), (5^3—2, 5), (-2,10). 7. Plot the quadrilateral whose vertices are (0, — 2), (4, 2), (0,6), (-4,2). 8. If a point moves parallel to the axis of x, which of its coordinates remains constant? if parallel to the axis of y? 9. Can a point move when its abscissa is zero ? Where ? Can it move when its ordinate is zero ? Where ? Can it move if both abscissa and ordinate are zero ? Where will it be ? p 10. Where may a point be found if its abscissa is 2 ? if its ordinate is — 3? r. 11. Where do all those points lie whose abscissas and ordi- iiates are equal ? 12. Two sides of a rectangle of lengths a and b coincide with the axes of x and y respectively. What are the coordinate^ of the vertices of the rectangle if it lies in the first quadrant ? in the second quadrant ? in the third quadrant ? in the fourth quadrant ? 13. Construct the quadrilateral whose vertices are (—3, 6), (_3^ 0), (3, 0), (3, 6). What kind of a quadrilateral is it? 14. Show that {x, y) and {x, '~y) are symmetrical with re- spect to X'X; (Xj y) and {— x, y) with respect to F'F; and {x, y) and { — x^ ~ V) with respect to the origin. CARTESIAN COORDINATES 13 15. A line joining two points is bisected at the origin. If the coordinates of one end are {a, — b), what will be the coor- dinates of the other end ? 16. Consider the bisectors of the angles between the coordi- nate axes. What is the relation between the abscissa and ordinate of any point of the bisector in the first and third quadrants ? second and fourth quadrants ? 17. A square whose side is 2 a has its center at the origin. What will be the coordinates of its vertices if the sides are parallel to the axes ? if the diagonals coincide with the axes ? Arts, (a, a), (c(, — a)^ (— a, — a), (— a, a) ; (a V2, 0), (- a V2, 0), (0, aV2), (0, - aV2). 18. An equilateral triangle whose side is a has its base on the axis of x and the opposite vertex above X'X. What are the vertices of the triangle if the center of the base is at the origin ? if the lower left-hand vertex is at the origin ? 11. Lengths. Consider any two given points Pli^iy Vl), ^2(^2, 2/2). ' Then in the figure 03f^ = x\, 03L = We may now easily prove the im- portant Theorem. The length I of the line j^t-q joining ttvo points Pi{x^, 2/i)? ^2(^^27 2/2) ^^ ^J given by the formula (I) l = V{xi^'jc2y + {yi-y2y- Proof Draw lines through P^ and Po parallel to the axes to form the right triangle P^iSPu Mnt^-^^. 14 ELEMENTARY ANALYSIS Then and hence P1P2 = '^jSP2^ + P^S' Q.E.D. (-5, The same method is used in deriving (I) for any positions of Pi and P2 ; namely, we construct a right triangle by draw- ing lines parallel to the axes through P^ and Pg- The hori- zontal side of this triangle is equal to the difference of the abscissas of P^ and P2, while the vertical side is equal to the difference of the ordinates. The required length is then the square root of the sum of the squares of these sides, which gives (I). A number of different figures should be drawn to make the method clear. EXAMPLE Find the length of the line join- ing the points (1, 3) and(— 5, 5). Solution. Call (1, 3) Pj, and (-5,5)P,. Then Xi = 1, ?/i = 3, and x2 = — 5,y2 = 5i and substituting in (I), we have I = V(l + 5/+(3-5)2 = V40 = 2VIO. It should be noticed that we are simply finding the hypote- nuse of a right triangle whose sides are 6 and 2. Remark, The fact that formula (I) is true for all positions of the points Pi and P2 is of fundamental importance. The application of this formula to any given problem is therefore simply a matter of direct substitution. In deriving such gen- eral formulas, it is most convenient to draw the figure so that the points lie in the first quadrant, or, in general, so that all the quantities assumed as known shall he positive. (H X CARTESIAN C0ORMN ATES U / 15 PROBLEMS 1. Find the projections on the axes and the length of the lines joining the following points. (a) (-4, -4) and (1,3). Ans. Projections 5, 7 ; length = V74. (b) (- V2, V3) and (V3, V2). Ans. Projections V3 + V2, V2 - V3 ; length = VlO. (0 (0, 0) and g, -^y Ans. Projections -, -V3; length = a. (d) (a -i-hj G + a) and (c + ay 6 + c). Ans. Projections c — b, b — a-, length = V(6 — c)^+ (a — b^). 2. Find the projections of the sides of the following tri- angles upon the axes : (^) (0, 6), (1, 2), (3, - 5). (b) (1,0),(~1, -5),(-l, -8). (c) (a, b), (b, c), (c, (^). (d) (a, - b), (b, - c), (c, - d). (e) (0,7y), (-^, -y)^^-x,0). 3. Find the lengths of the sides of the triangles in problem 2. 4. Work out formula (I) : (a) if 0^1 = 0*2; (6) if?/i=2/2. 5. Find the lengths of the sides of the triangle whose ver- tices are (4, 3), (2, -2), (-3, 5). 6. Show that the points (1, 4), (4, 1), (5, 5) are the vertices of an isosceles triangle. 7. Show that the points (2, 2), (-2, -2), (2V3, -2V3) are the vertices of an equilateral triangle. 8'. Show that (3, 0), (6, 4), (—1, 3) are the vertices of a right triangle. What is its area ? 16 ELEMENTARY ANALYSIS 9. Prove that (- 4, - 2), (2, 0), (8, 6), (2, 4) are the ver- tices of a parallelogram. Also find the lengths of the diagonals. 10. Show that (11, 2), (6, -10), (-6, -5), (-1, 7) are the vertices of a square. Find its area. 11. Show that the points (1, 3), (2, V6), (2, — V6) are equi- distant from the origin ; that is, show that they lie on a circle with its center at the origin and its radius VlO. 12. Show that the diagonals of any rectangle are equal. 13. Find the perimeter of the triangle whose vertices are (a, 6), (-a, b), (-, c + a), (c -\- a, b -{- c). Ans. G~b 5. Find the slopes of the sides of the triangle whose vertices are (1, 1), (—1, —1), (V3, — V3). Ans. 1, 1±^, Lz:^. 1_V3 1+V3 ^ 6. Prove by means of slopes that (— 4, — 2), (2, 0), (8, 6), (2, 4) are the vertices of a parallelogram. 7. Prove by means of slopes that (3, 0), (6, 4), (—1, 3) are the vertices of a right triangle. 8. Prove by means of slopes that (0, - 2), (4, 2), (0, 6), (—4, 2) are the vertices of a rectangle, and hence, by (I), of a square. 9. Prove by means of their slopes that the diagonals of the square in problem 8 are perpendicular. 20 ELEMENTARY ANALYSIS 10. Prove by means of slopes that (10, 0), (5, 5), (5, — 5), (— 5, 5) are the vertices of a trapezoid. 11. Show that the line joining (a, h) and (c, — d) is parallel to the line joining (—a, — h) and (— c, d). 12. Show that the line joining the origin to (a, h) is perpen- dicular to the line joining the origin to {—h, a). 13. What is the inclination of a line parallel to Y'Y? per- pendicular to Y'Y? 14. AYhat is the slope of a line parallel to Y' Y? perpen- dicular to Y'Y? 15. What is the inclination of the line joining (2, 2) and 4 16. W^hat is the inclination of the line joining (—2, 0) and 4 17. What is the inclination of the line joining (3, 0) and 3 / 18. What is the inclination of the line joining (3, 0) and 3 19. What is the inclination of the line joining (0, — 4) and 6 20. What is the inclination of the line joining (0, 0) and (-V3, 1)? . ^ Stt 6 21. Prove by means of slopes that (2, 3), (1, —3), (3, 9) lie on the same straight line. 22. Prove that the points (a, & + c), (h, c + ct), and (c, a + b) lie on the same straight line. 23. Prove that (1, 5) is on the line joining the points (0, 2) and (2, 8) and is equidistant from them. CARTESIAN COOKDINATES 21 24. Prove that the line joining (3, — 2) and (5, 1) is perpen- dicular to the line joining (10, 0) and (13, — 2). 25. Find the coordinates of the middle point of the line joining (4, —6) and (—2, —4). Ans. (1, —5). 26. Find the coordinates of the middle point of the line joining (a + b, c + d) and (a — b, d — c), Ans. (a, d). 27. Find the middle points of the sides of the triangle whose vertices are (2, 3), (4, —5), and (—3, —6); also find the lengths of the medians. 28. Prove that the middle point of the hypotenuse of a right triangle is equidistant from the three vertices. 29. Show that the diagonals of the parallelogram whose vertices are (1, 2), (—5, —3), (7, —6), (1, —11) bisect each other. 30. Prove that the diagonals of any parallelogram mutually bisect each other. 31. Show that the lines joining the middle points of the opposite sides of the quadrilateral whose vortiOes are (6, 8V (- 4, 0), (- 2, - 6), (4, - 4) bisect each other. 32. In the quadrilateral of problem 31 &how by means of slopes that the lines joining the middle points of the adjacent sides form a parallelogram. 33. Show that in the trapezoid whose vertices are (—8, 0), (— 4, —4), (— 4, 4), and (4, — 4) the length of th© line joining the middle points of the non-parallel sides is equal to one half the sum of the lengths of the parallel sides. Also prove that it is parallel to the parallel sides. 13. Areas. In this section the problem of determining the area of any polygon, the coordinates of whose vertices are given, will be solved. We begin with Theorem. The area of a tria.yigle whose vertices are the oriyin, -f*i(^i7 2/i)? ^^^ Afe 2/2) **s given by the formula 22 (IV) Y ELEMENTARY ANALYSIS Area of triangle OPiPz = I {pciyo — oc^Vi) - Po(x,,y,) Proof, In the figure let M, JC a = Z XOP^, /3 = ZX0F,, , e = ZP,OP,, ^^y-^' By 45^4^ C.P /I . i ^- 1^-- ^ (2) Area A OP1P2 = 1 OP^ - OP, sin (9 = iOP,.OAsin(^^a) [by(l)] (3) = i OPi . OP2 (sin ^ cos a - cos /? sin a). (by 34, p. 3) But in the figure OP/ sin^ ^^^=-/L-cos^ = OP2 _M,P,_ y. OP, OP, 0M^_ x^ sina= — - — i = -!^J— , cos a OP^ OP/ OP, OP, Substituting in (3) and reducing, we obtain Area A OP^P, = i {x,y, — x^y,). EXAMPLE Q.E.D. Eind the area of the triangle whose vertices are the origin, (-2, 4), and (-5, -1). Solution. Denote ( — 2, 4) by Pi, ( -^ 5, - 1) by P2. Then x, = — 2, 2/1 = 4, x2 = — 5, y ,■= — !. Substituting in (IV), Area = 1 [_ 2 • - 1 - (- 5) • 4] = 11. Then Area = 11 \init squares. If, however, the formula (IV) is applied by denoting (- 2, 4) by P2, and (- 5, - 1) by Pi, the result will be - 11. r-2 A) Yk \ / \ i / \ / \ (1,1) J \ 5 -^ L ^ ^^ X (-5 -1) CARTESIAN COORDINATES 23 The two figures are as follows : The cases of positive and neg- ative area are distinguished by Theorem. Passing around the perimeter in the order of the ver- tices 0, Fi, P2, if the area is on the left, as in Fig, 1, then (IV) gives a posi- tive result; if the area is on the right, as in Fig, 2, then (IV) gives a nega- tive result. Proof In the formula (4) Area A OP^P^ = i OP^ - OP2 sin the angle is measured fi^om OP^ to OP2 within the triangle, P^ p^ Hence is positive when the area lies .V^^~~^^^^^II_Pi ^^^""""-^-^2 to tlie left in passing around the ^^^X^ \ Y^v-'""''^ perimeter 0, P^, P^, as in Fig. 1, since Q O ^ is then measured counter-clockwise (1) ^^^ (p. 2). But in Fig. 2, is. measureJi clockwise. Hence 6 is negative and sin in (4) is alsp negative. S Q.E.D. Formula (IV) is easily applied to any polygon by regarding its area as made up of triangles with the origin as a't common vertex. Consider any triangle. ^ Theorem. The area of a triangle whose vertices are P^ {x^, 2/1); P2{^2, 2/2), A (^3, 2/3) i8 given by (V) Area A P1P2P3 = i(^i2/2 - a?22/i + i^22/3 - 003y2 + ocsiji - Xiy^). This formula gives a positive or negative result according as the area lies to the left or right in pass- ing around the perimeter in the order Pi P2 P3. Proof Two cases must be distin- guished according as the origin is within or without the triangle. 24 . ELEMENTARY ANALYSIS Fig. 1, origin within the triangle. By inspection, (5) Area A P,P,P, = A OP,P, + A OP,P, + A OP,P^, since these areas all have the same sign. , Fig. 2, origin imthout the triangle. By inspection, (6) Area A P^P.P^ = A OP^Po + A OP2P3 + A OPs^i, since OP1P2, OP3P1 have the same sign, but OP^A the opposite sign, the algebraic sum giving the desired area. By (IV), A OP,P, = i (x,y, - x.jj,\ A OP2P3 = \ {x.y^ - x^y^), A OP3P1 = ^ {x^y^ - x,y^. Substituting in (5) and (6), we have (V). Also in (5) the area is positive, in (6) negative. q.e.d. An easy way to apply (V) is given by the following Rule for finding the area of a triangle. x^ t/i First step. Write down the vertices in two columns, X2 2/2 abscissas in one, ordinates in the other, repeating the x^ y^ coordinates of the first vertex. x^ y^ Second step. Multiply each abscissa by the ordinate of the next row, and add results. This gives x^y^ + ^^'2^/3 + ^zVi- Third step. Multiply each ordinate by the abscissa of the next row, and add results. This gives yiX2 + ^2^3 + 2/3^1- Fourth step. Subtract the residt of the third step fro7n that of the second step, and divide by 2. This gives the required area, namely, formula (V). It is easy to show in the same manner that the rule applies to any polygon, if the following caution be observed in the first step : Write down the coordinates of the vertices in an order agreeing with that estal)lished by passing continuously around the perime- ter, and repeat the coordinates of the first vortex. CARTESIAN COORDINATES 25 EXAMPLE Find the area of the quadrilateral whose vertices are (1, 6), (-3, -4), (2, -2), (-1,3). Solution. 1 -1 -3 2 1 6 3 -4 -2 6 Plotting, we have the figure from which we choose the order of the ver- tices as indicated by the ar- rows. Following the rule : First step. Write down the vertices in order. Second step. Multiply each abscissa by the ordinate of the next row, e^lj^ add. This gives lx3 + (-lX -4)4-(-3x -2)-f2x6 = 25. Third step. Multiply each ordinate by the abscissa of the next row, and add. This gives 6 X -1+3 X -3 + (-4x2) + (-2x1) = -25. Fourth step. Subtract the result of the third step from the result of the second step, and divide by 2. 25 4. 25 .*. Area = -^- = 25 unit squares. Ans. The result has the positive sign, since the area is on the left. PROBLEMS 1. Find the area of the triangle whose vertices are (2, 3), (1, 5), (-1, -2). Ans. -V-. 2. Find the area of the triangle whose vertices are (2, 3), (4, -5) (-3, -6). Ans. 29. 3. Find the area of the triangle whose vertices are (8, 3), (-2,3), (4, -5). Ans. 40. 4. Find the area of the triangle whose vertices are (a, 0), (- a, 0), (0, h). . Ans. ab. 26 ELEMENTARY ANALYSIS 5. Find the area of the triangle whose vertices are (0, 0), (P^h Vi), (^2, 2/2)- Ans, ^^^2 - ^2^1 . 6. Find the area of the triangle whose vertices are (a, 1), (0,6),(c,l). Ans. (°^-<^K^-l) . 7. Find the area of the triangle whose vertices are (a, &), (h, a), (c, — c). Ans. \ (or — W). 8. Find the area of the triangle whose vertices are (3, 0), (0, 3 V3), (6, 3 V3). Ans, 9 V3. 9. Prove that the area of the triangle whose vertices are the points (2, 3), (5, 4), ( -^*^ lj is zero, and hence that these points all lie on the same straight line. 10. Prove that the area of the triangle whose vertices are the points (a, h + c), (b, c + a), (c, a-{-b) is zero, and hence that these points all lie on the same straight line. 11. Prove that the area of the triangle whose vertices are the points (a, G-]-a), (— c, 0), (— a, c — a) is zero, and hence that these points all lie on the same straight line. 12. Find the area of the quadrilateral whose vertices are (-2, 3), (-3, -4), (5, -1), (2, 2). Ans. 31. 13. Find the area of the pentagon whose vertices are (1, 2), (3, - 1), (6, - 2), (2, 5), (4, 4). , Ans. 18. 14. Find the area of the parallelogram whose vertices are (10, 5), (-2, 5), (-5, -3), (7, -3). Ans. 96. 15. Find the area of the quadrilateral whose vertices are (0, 0), (5, 0), (9, 11), (0, 3). Ans. 41. ' 16. Find the area of the quadrilateral whose vertices are (7, 0), (11, 9), (0, 5), (0, 0). Ans. 59. 17. Show that the area of the triangle whose vertices are (4, 6), (2, —4), (—4, 2) is four times the area of the triangle formed by joining the middle points of the sides. CARTESIAN COORDINATES * 27 18. Show that the lines drawn from the vertices (3, — 8), (__ 4^ 6), (7, 0) to the medial point of the triangle divide it into three triangles of equal area. 19. Given the quadrilateral whose vertices are (0, 0), (6, 8), (10, — 2), (4, — 4) ; show that the area of the quadrilateral formed by joining the middle points of its adjacent sides is equal to one half the area of the given quadrilateral. CHAPTER 111 CURVE AND EQUATION 14. Locus of a point satisfying a given condition. The curve ^ (or group of curves) passing through all points which satisfy a given condition, and through no other points, is called the locus of the point satisfying that condition. For example, in Plane Geometry, the following results are proved : The perpendicular bisector of the line joining two fixed points is the locus of all points equidistant from these ]3oints. The bisectors of the adjac^t angles formed by two lines are* the locus of all points equidistant from these lines. To solve any locus problem involves two things : 1. To draw the locus by constructing a sufficient number of points satisfying the given conditio^ and therefore lying on the locus. 2. To discuss the nature of the locus, that is, to determine properties of the curve. Analytic Geometry is peculiarly adapted to the solution of i^pth parts of a locus problem. , 15. Equation of the locus of a point satisfying a given condition. Let us take up the locus problem, making use of coordinates. We imagine the point P(x, y) moving in such a manner that the given condition is fulfilled. Then the given condition will lead to an equation involving the variables x and ?/. The following example illustrates this. * The word "curve" will hereafter signify anij continuous line, straight or curved. 28 CURVE AND EQUATION 29 EXAMPLE The point P(x, y) moves so that it is always equidistant from j^(_ 2, 0) and B{—3, 8). Find the equation of the locus. Solution. Let P{x, y) be any point on the locus. Then by the given condition (1) FA = FB. But, by formula (I), p. 13, FA = -\/{x + 2y -h (2/ - 0)2, FB = V(x-{-3y-\-{y- Substituting in (1), (2) ^(x + 2y-{-(y-0y sy =^(x-^3y+(y-sy. Squaring and reducing, (3) 2x-16y + 69 = 0, In the equation (3), x and y are variables representing the coordinates of any point on the locus; that is, of any point on the perpendicular bisector of the line AB. This equation is called the equation of the locus ; that is, it is the equation of the perpendicular bisector CF. It has two important and characteristic properties : 1. The coordinates of any point on the locus may be sub- stituted for X and y in the equation (3), and the result will be true. For let Pi (xi, y^ be any point on the locus. Then F^A = F^B, by definition. Hence, by formula (I), p. 13, (4) VK + 2y + y,' = V(^i + 3)2 + (2/1^=^ or, squaring and reducing, (5) 2a;i-16?/i + 69 = 0. But this eqaiation is obtained by substituting x^ and .Vi foi" ^ and y, respectively, in (3). Therefore x^ and y^ satisfy (3). 30 ELEMENTARY ANALYSIS 2. Conversely, every point whose coordinates satisfy (3) will lie upon the locus. For if Pi (xi, ?/i) is a point whose coordinates satisfy (3), then (5) is true, and hence also (4) holds. q.e.d. In particular, the coordinates of the middle point of A and Bj namely, a; = — 2^, ?/ = 4 (III, p. 19), satisfy (3), since 2(_2-i-)-16x 4 + 69 = 0. This discussion leads to the definition : The equation of the locus of a point satisfying a given condi- tion is an equation in the variables x and y representing coor- dinates such that (1) the coordinates of every point on the locus will satisfy the equation ; and (2) conversely, every point whose coordinates satisfy the equation will lie upon the locus. This definition shows that the equation of the locus must be tested in two ways after derivation, as illustrated in the exam- ple of this section. The student should supply this test in the examples, p. 31, and problems, p. 32. Erom the above definition follows at once the Corollary. A point lies upon a curve when and only when its coordinates satisfy the equation of the curve. 16. First fundamental problem. To find the equation of a curve which is defined as the locus of a point satisfying a given condition. The following rule will suffice for the solution of this prob- lem in many cases : Rule. First step. Assume that P{x, y) is any point satisfying the given condition and is therefore on the curve. Second step. Write down the given condition. Third step. Express the given condition in coordinates, and simplify the result. The final equation, containing x, y, and the given constants of the problem, will be the required equation. CURVE AND EQUATION 31 EXAMPLES 1. Find the equation of the straight line passing through Sir Pj (4^ — 1) and having an inclination of Solution. First step. Assume P(xj y) any point on the line. Second step. The given condition, since the inclination a is — , may be written (1) Slope of PiP = tan a = - 1. Tliird step. From (II), p. IT, (2) Slopeof PiP= tan « = -^^^-^:i^2 = -^^^ - [By substituting {x, y) for (a;,, y^, and (4, — 1) for (a;^, y^."] y{ ^ ^: \ r '\ V i \ it. J \ \ a,- ■1) — — i-.S .'. from (1), x — 4z 1, or (3) x-{-y — 3 = 0. Ans, 2. Find the equation of a straight line parallel to the axis of y and at a distance of 6 units to the rierht. f^ M Solution. First step. Assume that ^-^ P(x, y) is any point on the line, and draw NP perpendicular to 1^ Second step. The given condition ►^ may be written (4) NP = 6. Third step. Since NP= OM==x, (4) becomes ^^ (5) x = 6. Ans. 3. Find the equation of the locus of a point whose distance from (—1, 2) is always equal to 4. 32 ELEMENTARY ANALYSIS Solution. First step. Assume that P{x, y) is any point on the locus. Second step. Denoting ( — 1^ 2) ) by C, the given condition is (6) Pe=:4. TJiird step. By formula (I), p. ^. 13, PG = -V(x-\-iy-i-(y-2f. Substituting in (5), -V(x+iy + (j/-2y=4.. Squaring and reducing, (7) x^ + y^ + 2x-^y-ll==0. This is the required equation, namely, the equation of the circle whose center is (—1, 2) and radius equal 4. '\t^. PROBLEMS 1. Find the equation of a line parallel to OF and. (a) at a distance of 4 units to the right. (b) at a distance of 7 units to the left. (c) at a distance of 2 units to the right of (3, 2). (d) at a distance of 5 units to the left of (2, — 2). >l • * ^ 2. Find the equation of a line parallel to OX and (a) at a distance of 3 units above OX. (b) at a distance of 6 units below OX. (c) at a distance of 7 units above (—2,-3). (d) at a distance of 5 units below ^4, — 2). 3. What is the equation of XX' ? of YY'? %^' ^ ' 4. Find the equation of a line parallel to the line x = 4z and 3 units to the right of it. Eight units to the left of it. 5. Find the equation of a line parallel to the line y = — 2 and 4 units below it. Five units above it. CUKVE AND EQUATION 33 6. What is the equation of the locus of a point which moves always at a distance of 2 units from the axis of x ? from the axis of ?/ ? from the line ic = — 5 ? from the line 2/ = 4 ? 7. What is the equation of the locus of a point which moves so as to be equidistant from the lines a; = 5 and a; = 9 ? equi- distant from ^ = 3 and ?/ = — 7 ? 8. What are the equations of the sides of the rectangle whose vertices are (5, 2), (5, 5), (-2, 2), (-2, 5)? In problems 9 and 10, P^ is a given point on the required line, m is the slope of the line, and a its inclination. 9. What is the equation of a line if (a) Fi is (0, 3) and m = - 3 ? ^ns. Z x-\-y -?>=.^, ^(6) Pi is (- 4, - 2) and m = | ? Ans. x-3y-2 = 0. (c) Pi is (-2, 3) and m=^? ^ns. -s/2x-2yf6 J _ +2V2 = 0. ((^) Pi is (0, 5) and 711 = ^? Ans. V3 a;-2 ?/ + 10 = 0. / "^ (e) Piis (0, 0) andm = -|? ^ns. 2a^4-3?/ = 0. (/) Pi is (a, 5) and m = ? ^tis. ?/ = &. (9') Pi is (— a, Z>) and 771 = 00 ? ^dliis. x = — a. 10. What is the equation of a line if ^ (a) Pi is (2, 3) and a = 45°? Ans. x-y-{-l = 0. ^ (b) Pi is (- 1, 2) and r^e = 45° ? ylns. a; - 2/ + 3 = 0. (c) Pi is (— a, — 6) and a = 45° ? Ans. x — y = b — a. (d) Pi is (5, 2) and r^ = 60° ? ^ris. ■\/3x-y + 2 -5V3 = 0. (e) Pi is (0, - 7) and a - 60° ? ^ns. VSx-y~7 = 0. (/) Pi is (- 4, 5) and a = 0°? Ans. y = 5. (g) Pi is (2, - 3) and a = 90° ? A71S. x = 2. Qi) Piis(3,-3V3)anda:=rl20°? Ans. ^3x^y = 0, (i) Pi is (0, 3) and a - 150° ? Ans. VS x-i-S y -9 =0. (j) Pi is (a, b) and a = 135° ? ^ns. a; + 2/ = a + 6. 34 ELEMENTARY ANALYSIS 11. Find the equation of the circle with (a) center at (3, 2) and radius = 4. Arts, x^ -\- y^ — 6 X — 4:y — 3 = 0, (b) center at (12, —5) and r=13. Aris. x^ + y'--24.x + 10y = 0, (c) center at (0, 0) and radius = r, Ans. x^-\-y^ = rl (d) center at (0, 0) and r = 5. Ans. x^-\-y^ = 25. (e) center at (3 a, 4 a) and r = 5 a. Ans. x^ -\- y^ — 2 a{3 X + 4: y)= 0. (/) center at (6 + c, b — c) and r = c. .4ns. a;- + 2/' - 2(^ + c)a: -2(b-c)y + 2b^ + c^ = 0. 12. Find the equation of a circle whose center is (5, — 4) and whose circumference passes through the point (—2, 3). 13. Find the equation of a circle having the line joining (3, — 5) and (— 2, 2) as a diameter. 14. Find the equation of a circle touching each axis at a dis- tance 6 units from the origin. 15. Find the equation of a circle whose center is the middle point of the line joining (—6, 8) to the origin and whose cir- cumference passes through the point (2, 3) . 16. A point moves so that its distances from the two fixed points (2, —3) and (—1, 4) are equal. Find the equation of the locus. Ans. 3x — 7y-\-2 = 0. 17. Find the equation of the perpendicular bisector of the line joining . (a) (2,1), (-3/ -3). Ans. 10 a; +8^ + 13 = 0. (b) (3, 1), (2, 4). Ans. x-3y-{-5 = 0. (c) (-1^ _1)^ (3, 7). Ans. x + 2y-7=:0. (d) (0, 4), (3, 0). Ans. 6x-Sy + 7 = 0, (e) (xi, ?/i), (X2, 2/2). Ans. 2 (x^ — 0^2) ^ + 2 (^1 - 2/2)2/ + ^2^ — ^i + ^2' — Vi^ = 0. 18. Show that in problem 17 the coordinates of the middle point of the line joining the given points satisfy the equation of the perpendicular bisector. I CURVE AND EQUATION 35 9. Find the equations of the perpendicular bisectors of the sides of the triangle (4, 8), (10, 0), (6, 2). Show that they meet in the point (11, 7). 17. Locus of an equation. The preceding sections have illustrated the fact that a locus problem in Analytic Geometry leads at once to an equation in the variables x and y. This equation having been found or being given, the complete solu- tion of the locus problem requires two things, as already noted in the first section (p. 28) of this chapter, namely, 1. To draw the locus by plotting a sufficient number of points whose coordinates satisfy the given equation, and through which the locus therefore passes. 2. To discuss the nature of the locus, that is, to determine properties of the curve. These two problems are respectively called : 1. Plotting the locus of an equation (second fundamental problem). 2. Discussing an equation (third fundamental problem). For the present, then,^ we concentrate our attention upon some given equation in the variables x and y (one or both) and start out with the definition : The locus of an equation in two variables representing co- ordinates is the curve or group of curves passing through all points whose coordinates satisfy that equation,^* and through such points only. * An equation in the variables x and y is not necessarily satisfied by the coordinates of any points. For coordinates are real numbers, and the form of the equation may be such that it is satisfied by no real values of x and y. For example, the equation iB2 + ?/2 + l = is of this sort, since, when x and ?/ are real numbers, cc^ and ?/2 are necessarily positive (or zero), and consequently x^ + y'^-\-l is always a positive number greater than or equal to 1, and therefore not equal to zero. Such an equation therefore has no locus. The expression " the locus of the equation is imagi- nary " is also used. An equation may be satisfied by the coordinates of 2i finite number of points 36 ELEMENTARY ANALYSIS From this definition the truth of the following theorem is at once apparent : Theorem I. If the form of the given equation be changed in any way {for example^ by transposition, by multiplication by a constant, etc.), the locus is entirely unaffected. We now take up in order the solution of the second and third fundamental problems. 18. Second fundamental problem. Rule to plot the locus of a given equation. First step. Solve the given equation for one of the variables in terms of the other.* Second step. By this formula compute the values of the vari- able for tvhich the equation has been solved by assuming real values for the other variable. Third step. Plot the points corresponding to the values so determined, t Fourth step. If the points are 7iumerous enough to suggest the general shape of the locus, draw a smooth curve through the points. Since there is no limit to the number of points which may be computed in this way, it is evident that the locus may be drawn as accurately as may be desired by simply plotting a sufficiently large number of points. Several examples will now be worked out. The arrangement of the work should be carefully noted. only. For example, q;2 + ?/2 = o is satisfied by a; = 0, ?/ = 0, but by no other real values. In this case the group of points, one or more, whose coordinates satisfy the equation, is called the locus of the equation. * The form of the given equation will often be such that solving for one variable is simpler than solving for the other. Alvmys choose the simpler solutio7i. t Remember that real values only may be used as coordinates. CURVE AND EQUATION 87 \YJk ^. Y \y / Y ^ y^ / Y (0.2) A r r (-3 0) X if EXAMPLES 1. Draw the locus of the equation Solution. First step. Solving for ?/, ?/ = I a; + 2. Second step. Assume values for x and compute y, arranging results in the form : Thus, if a; = l,2/ = | -1 + 2 = 21, aj = 2,2/ = |-2 + 2 = 3i etc. Tliird step. Plot the points found. Foiirth step. Draw a smooth curve through these points. 2. Plot the locus of the equation y = x^ — 2 X — 3. Solution. First step: The equation as given is solved for y. Second step. Computing y by assuming values of Xj we find the table of values below : X y X y 2 2 1 2! - 1 ^ 2 H -2 i 3 4 - 3 4 4* -4 -! etc. etc. etc. etc. X 2/ X 2/ o — o -3 1 -4 - 1 2 -3 - 2 5 3 -3 12 4 5 -4 21 5 12 etc. etc. 6 21 etc. etc. 38 ELEMENTARY ANALYSIS Third step. Plot the points. Fourth step. Draw a smooth curve through these points. This gives the curve of the figure. ■^ 3. Plot the locus of the equation x^ + y^ + 6x-16 = 0. Solution. First step. Solving for y, y=± Vl6 — Qx — x^, Second step. Compute y by assuming values of x. X y X 2/ ±4 ±4 . 1 ±3 - 1 ± 4.0 2 -2 ± 4.9 3 imag. -3 ±5 4 u -4 ± 4.9 5 u - 5 ±4.6 6 u -6 ±4 7 ii. — 7 ±3 -8 -9 imag. yJ^l^OY example, ii x=% y = ± Vl6 — 6 — 1 = ± 3 ; ifx = 3,y=±VW- an imaginary number ; 18-9= ±V-11, if aj = - 1, ?/ = ± Vl6 + 6 - 1 = ± 4.6, etc. TJiird step. Plot the corresponding points. Fourth step. Draw a smooth curve through these points. The student will doubtless remark that the locus of example 1, p. 37, appears to be a straight line, and also that the locus of example 3 (above) appears to be a circle. This is, in fact, the case. But the proof must be reserved for later sections. CURVE AND EQUATION 39 PROBLEMS. 1. Plot the locus of each, of the following equations. (a) x-i- 2 y — 0. (m) y = x^ — x. - x^ (b) x-^2 y = 3. ,(n) y = x' (c) 3x-y+ 5 = 0. (o) x^ + ?r = 4. (d) y = 4.x\ {p) x''-\-\f = ^. \e) a;2 + 4?/ = 0. (q) x'-{-y' = 25. (/) y = x'-3, (r) x' + f-^9x=0. (g) x' + 4.y-5 = 0. (s) x'-{-y'-{-4.y = 0. (h) y^x' + x-i-l. (t) x'' + y^-6x-16 = 0. (i) x = y^^2y-3. (u) x" ^y'' - Q> y -1(5 = 0, (/) 4,x = y^, ' . (y) 4.y = x'^ — S. \t) y=x^-i, CjygjQ^ 2. Show that the loilowing equations have no locus (footnote p. 35). (a) x^ + y'-{-l = 0. (e) (x + lf + f- -\- 4. = 0, (b) 2x'-t3f = r^S. (/) x' + f + 2x + 2y + S=0. (c) a;2 + 4 = 0. (g) 4.x' + 7f-i-S x-^5 = 0. (d) x^-\-y' + S = 0, (Ji) y^ + 2x' + 4. = 0. (i) 9x' + Ay'-]-lSx-\-Sy + 15 = 0. Hint. Write each equation in the form of a sum of squares, and reason as in the footnote on p. 35. The following problems illustrate the Theorem. If an equation can be put in the form of a product of variable factoids equal to zero, the locus is found by setting each factor equal to zero and plotting each equation separately. 3. Draw the locus of 4 a;^ — 9 ^^ = 0. Solution. Factoring, (1) C2x-3y){2x-\-3y) = 0, Then, by the theorem, the locus consists of the straight lines (2) 2x~5y = 0, (3) 2x + 2,y = 0. 40 ELEMENTAEY ANALYSIS Proof. 1. The coordinates of any point {xi, y^ which satisfy (1) ivill satisfy either (2) or (3) . For if (xi^ 2/i) satisfies (1), (4) {2x,~3y,){2x, + Sy,)=0. This product can vanisli only when one of the factors is zero. Hence either 2x,-3y, = 0, and therefore (o^j, 2/1) satisfies (2) ; or 20^1+32/1 = 0, and therefore (x^, 2/1) satisfies (3). 2. A point (xi, 2/1) on either of the lines defined by (2) and (3) will also lie on the locus of (1). For if (x^, 2/1) is on the line 2x — 3y = 0y then (Corollary, p. 30) (5) 2o^.,-3y, = 0. Hence the product (2 iCi — 3 yi)(2 Xi-\-3 y^) also vanishes, since by (5) the first factor is zero, and therefore (xi, y^ satis- fies (1). Therefore every point on the locus of (1) is also on the locus of (2) and (3), and conversely. This proves the theorem for this example. q.e.d. 4. Show that the locus of each of the following equations is a pair of straight lines, and plot the lines. {a) x'-f^O. (f) f-5xy + 6y = 0. (h) 9i»2-2/2 = 0. (^) xy~2x^--3x = {), (c) x' = <^y\ (h) xy-2x = 0. (c^) x^-4:X-5 = 0. (i) xy = 0. (e) y^-6y^7. (j ) 3 x^ -\- o'jj — 2 y^ -i- 6 X — 4: y = 0. (k) x'^ — y^ + x-}-y = 0, (m) x^ -2xy+y^-{-6x-6 y=0. (I) x^-xy~h^x-5y = 0. (n) x^ -iy^ -^ 5x -^10 y = 0, (0) i«2 + 4 i»^/ + 4 / + 5 a; + 10 ?/ + 6 = 0. CURVE AND EQUATION 41 (q) x^*— 4, xy — 5 y^ -]- 2 X — 10 y = 0, (r) Sa^— 2xy — y^-{-5x — 5y = 0. (s) x^-3xy-4.y'- = 0. (t) x' + 2xy-j-y' + x-\-y = 0. . (li) of — 3 xy = 0. (v) y^ + 4: xy = 0. 5. Show that the locus of Ax^ + Bx + C = is a pair of parallel lines, a single line, or that there is no locus accord- ing as A = B^— 4 AG is positive, zero, or negative. 6. Show that the locus of Aoiy^ + Bxy + Gy^ = is a pair of intersecting lines, a single line, or a point according as A = B^ — 4, AG is positive, zero, or negative. 19. Third fundamental problem. Discussion of an equa- tion. The method explained of solving the second funda- mental problem gives no knowledge of the required curve except that it passes through all the points whose coordinates are determined as satisfying the given equation. Joining these points gives a curve more or less like the exact locus. Serious errors may be made in this way, however, since tJie nature of the curve between any two successive points plotted is not determined. This objection is somewhat obviated by de- termining before plotting certain properties of the locus by a discussion of the given equation now to be explained. The nature and properties of a locus depend upon the form of its equation, and hence the steps of any discussion must depend upon the particular problem. In every case, however, the following questions should be answered. 1. Is the curve a closed curve, or does it extend-out infinitely far? 2. Is the curve symmetrical with respect to either axis or the origin ? The method of deciding these questions is illustrated in the following examples. 42 eleme:ntary analysis EXAMPLES 1. Plot and discuss the locus of (1) x' + 4:y' = 16. Solution. First step. Solving for x, -y - ^^ / l^ (2) i»=±2V4^. Second step. Assume values of y and compute x. Third step. Plot the points of the table. Fourth step. Draw a smooth" curve through these points. X y X y ±4 ±4 ±3.4 1 ±3.4 -1 ±2.1 li ±2.7 -1| 2 -2 imag. 3 imag. -3 Discussion. 1. Equation (1) shows that neither x nor y can be indefinitely great, since x^ and 4 2/^ are positive for all real values, and their sum must equal 16. Therefore neither x^ nor 4 y'^ can exceed 16. Hence the curve is a closed curve. A second way of proving this is the following : From (2), the ordinate y cannot exceed 2 nor be less than — 2, since the expression 4 — 2/^ beneath the radical must not be negative. (2) also shows that x has values only from — 4 to 4 inclusive. 2. To determine the symmetry with respect to the axes we proceed as follows : The equation (1) contains no odd powers of x oy y\ hence it may be written in any one of the forms (3) {xf + ■4(- 2/)2 = 16, replacing {x, y) by {x, -y) ; (4) (- xy + 4(?/)2 = 16, replacing (x, ?/) by (- x, y) ; (5) (- xf + 4(-?/)2 = 16, replacing {x, y) by (- x, -y). CURVE AND EQUATION 43 The transformation of (1) into (3) corresponds in the figure to replacing each point P(x, y) on the curve by the point Q(Xy —y)' But the points P and Q are symmetrical with respect to XX\ and (1) and (3) have the same locus (Theorem I, p. 36). Hence the locus of (1) is unchanged if each point is changed to a second point symmetrical to the first with respect to XX\ Therefore the locus is symmetrical with respect to the axis of^c^ Similarly, from (4), the locus is symmetrical with respect to me axis of y, and from (5), the locus is symmetri- cal with respect to the origin. The locus is called an ellipse. 2. Plot the locus of (6) 2/'-4^ + 15 = 0. Discuss the equation. Solution. First step. Solve the equation for x, since a square root would have to be extracted if we solved for y. This gives (7) x = \{f + 15). X y 3| 4 ±1 4| ±2 6 ±3 7| ±4 10 ±5 12f dz6 etc. etc. Second step. Assume values for ?/ and compute x. Since ?/^ only appears in the equation, positive and negative values of y give the same value of x. The calculation gives the table. 44 ELEMENTARY ANALYSIS For example, ii y = ± S, then x=: 1(9 + 15) = 6, etc. TJiii^d step. Plot the points of the table. Fourth step. Draw a smooth curve through these points. , Discussion. 1. From (7) it is evident that x increases as y increases. Hence the curve extends out indefinitely far from both axes. 2. Since (6) contains no odd powers of y, the equation may be written in the form by replacing (x, y) by (x, — y). Hence the locus is symmetrical with respect to the axis ofx. The curve is called a parabola. 3. Draw the locus of the equation (8) 4,y = x^. X . y X y 1 i -1 -i n II -n -w 2 2 -2 -2 2| 3|| -^ -m 3 6f -3 -Of H lOil -H - lOil Solution. First step. Solv- ing for y, y- ■■\x\ For example, if Second step. Assume val- ues for X and compute y. Values of x must he taken between the integers in order to give points not -too far apart. ^— ^2"? 7, _ 1 . 1 2 5 _ 1 2 5 _ Q 2 9 (^\o Third step. Plot the points thus found. Fourth step. The points determine the curve of the figure. Discussion. 1. From the given equation (8), x and y in- CURVE AND EQUATION 45 ^ i 1 / f \ -[ 1^^ 1^ ) / 1 / ' J J r X r ) / \ / / f ■X r/ ) ' - — J — ^ crease simultaneously, and therefore the curve extends out in- definitely from both axes. 2. In (8) there are no even powers nor constant term, so that by changing signs the equation may be written in the form K-y) = {-xf, replacing {x, y) by {~x, -y). Hence the locus is symmetrical with respect to the origin. The locus is called a cubical parabola. 20. Symmetry. In the above examples we have assumed the definition : If the points of a curve can be arranged in pairs which are symmetrical with respect to an axis or a point, then the curve itself is said to be symmetrical with re- spect to that axis or point. The method used for testing an equation for symmetry of the locus was as follows : if (x, y) can be replaced by {x, — y) throughout the equation without affecting the locus, then if (a, b) is on the locus, (a, — b) is also on the locus, and the points of the latter occur in pairs symmetrical with respect to XX', etc. Hence Theorem II. If the locus of an equation is unaffected by re- 2olacing y by — y throughout its equatioyi, the locus is symmetrical with respect to the axis of x. If the locus is unaffected by changing x to — x throughout its equ^-^.ion, the locus is symmetrical tvith respect to the axis of y. If the locus is unaffected by changing both x and y to —x and — y throughout its equation, the locus is symmetrical tvith respect to the origin. These theorems may be made to assume a somewhat different form if the equation is algebraic in x and y. The locus of an 46 ELEMENTARY ANALYSIS tlgebraic equation in the variables x and y is called an algebraic curve. Then from Theorem II follows Theorem III. Symmetry of an algebraic curve. If no odd powers of y occur in an equation, the locus is symmetrical with respect to XX' ; if no odd powers of x occur, the locus is sym- metrical with respect to YY\ If every term is of even^ degree, or every term of odd degree, the locus is symmetrical with respect to the origin. 21. Further discussion. In this section we treat of three more questions which enter into the discussion of an equation. Is the origin on the curve ? This question is settled by Theorem IV. The locus of an algebraic equation passes through the origin ivhen there is no constant term in the equation. Proof The coordinates (0, 0) satisfy the equation when there is no constant term. Hence the origin lies on the curve (Corollary, p. 30). q.e.d. What values of x and y are to be excluded ? Since coordinates are real numbers we have the Rule to determine all values of x and y which must he excluded. Solve the equation for x in terms of y, and from this result de- termine, all values of y for ivhich the computed value of x will be imaginary. These values of y must be excluded. Solve the equation for y in terms of x, and from this result de- termine all values of x for which the computed value of y will be im^aginary. These values of x must be excluded. The intercepts of a curve on the axis of x are the abscissas of the points of intersection of the curve and XX. The intercepts of a curve on the axis of y are the ordinates of the points of intersection of the curve and YY^. * The constant term must be regarded as of even (zero) degree. CURVE AND EQUATION 47: Rule to find the intercepts. ? Substitute y = 0, and solve for real values of x. This gives the intercepts on the axis of x. Substitute x=:Oy and solve for real values ofy. TJiis gives the intercepts on the axis of^. The proof of the rule follows at once from the definitions. The rule just given explains how to answer the question : What are the intercepts of the locus ? 22. Directions for discussing an equation. Given an equa- tion, the following questions should be answered in order before plotting the locus. 7s the origin on the locus t %/y ^^^JCy' * (0 x^ + y'^ = 2ry. (d) 2xy = a\ (J) ay'' = ^. (e) t^vl^l, Qc) ahj=:^. ^^ a' h' 3. Draw the locus of the equation y- z=z{x — a)(x — h)(x— c), (a) when a 1. The conic is now called a hyperbola (see p. 48). 6. A point moves so that the sum of its distances from the two fixed points (3^ 0) and (—3, 0) is constant and equal to 10. What is the locus ? Ans. Ellipse l^x^ + 25?/^ = 400. 7. A point moves so that the difference of its distances from the two fixed points (5, 0) and (— 5, 0) is constant and equal , to 8. What is the locus ? Ans. Hyperbola 9 o;^— 167/^=144. 23. Points of intersection. If two curves whose equations are given intersect, the coordinates of each point of intersection must satisfy both equations when substituted in them for the variables (Corollary, p. 30). In Algebra it is shown that all values satisfying two equations in two unknowns may be found by regarding these equations as simultaneous in the unknowns and solving. Hence the Rule to find the points of intersection of two curves whose equa- tions are given. Consider the equations as simultaneous in the coordinates, and solve as in Algebra. Arrange the real solutions in corresponding pairs. These will be the coordinates of all the points of intersection. Notice that only real solutions correspond to common points of the two curves, since coordinates are always real numbers. EXAMPLES 1. Find the points of intersection of (1) • x-ly + 2^ = 0, (2) x' + y''=25. Solution. Solving (1) for x, (3) x=^7y-~25. CURVE AND EQUATION 51 "Fa ^ ^^ \^ ...^ Xi-} H ,3; /^ \ -*" \ \ '■ \ j / \ / N *^ y Substituting in (2), (7 2/ -25/ + 2/2 = 25. Reducing, ,-. y =3 and 4. Substituting in (3) [iio^ in (2)], 0^ = — 4 and +3. Arranging, the points of intersection are (—4, 3) and (o, 4). A7is, In the figure the straight line (1) is the locus of equation (1), and the circle the locus of (2). 2. Eind the points of intersection of the loci of (4) 2a^ + 3if = S5, (5) 3x'-Ay = 0. Solution. Solving (5) for x^, (G) x'=iy. Substituting in (4) and reducing, 9f^ + Sy-105 = 0, /.yz=z3 and — %^-. Substituting in (6) and solving. aj=±2and ±iV-210. Arranging the real values, we find the points of intersection are(+2, 3), (-2, 3). Ans. In the figure the ellipse (4) is the locus of (4), and the pa- rabola (5) the locus of (o). 62 ELEMENTARY ANALYSIS 1,. PROBLEMS Find the points of intersection of the following loci. jx-lly + l = x-{-y — 2 = Ans. a, f). x — y=:5) Ans. (6, 1). ■^y = Sx + 2\ zyx'+f-=4: j' Ans. (0,2), (-f,- y^ = 16x ] x' + f = Al^ xy=^20 J * " Ans. (±5, ±4), f + 4,±5). g y' = 22M] x'=2py] Ans. (0, 0), {2p, 2p). I)- 9 4.x^jry'' = ^\ 4. y — X = 0] Ans. (0, 0), (16, 16). a;2 + 2/' :a" 5. " - " . I ^x + y + a = J ^n.(0,-a),(-^,^). 6. -/=:16 Ans. (±4V2, 4). ^n8. (i 2), (i -2). ^2 + 2/' = 100 10. 9 9 a; ^7is. (8,6), (8, -6). a? A- y- z=i ^ a^ ^ 11- 2 1 h a;^ = 4: ay J ^ns. (2 a, a), (—2 a, a). Find the area of the triangles and polygons whose sides are the loci of the following equations. .12. 3x + y + 4: = 0,3x-5y + 3^ = 0,3x-2y + l=0. Ans. 36. 13. x + 2y = 5.^2x + y = 7,y = x-^l, 14. x-\-y = a, x — 2y = Aa, y — x-{-7a = 0. 15. a; = 0, 2/ = Oj a; = 4, ?/ = — 6. 16. x — y = 0,x + y = OyX — y = a,x + y=b. Ans. f . Ans. 12 a?. Ans. 24. Ans. — 17. y=z3x—^,y^3x + 5,2y = x- 3, 2.?/ = a; + 14. Ans. 5(S. CURVE AND EQUATION 53 18. Find the distance between the points of intersection of the curves 3 a^ — 2 ?/ + 6 = 0, a.- -f ?/- = 9. Ans. || Vi3. 19. Does the locus of y^ = 4tx intersect the locus of 2a^ + 32/ + 2 = 0? Ans. Yes. . 20. For what value of a will the three lines ^x-\-y — 2 = 0^ ax-\-2y — Z = 0, 2 a; — 7/ — 3 = meet in a point ? Ans. a = 5. 21. Find the length of the common chord of x'^-\-y^ = 13 and y^ = 3x-\-3. Ans. 6. 22. If the equations of the sides of a triangle are x-\-7y + 11 = 0, 3x+y-7 = 0, x-3y-{-l=:0, find the length of each of the medians. Aiis. 2 V5, |V2, ^VlTO. yr\- y- ^' CHAPTER IV STRAIGHT LINE AND CIRCLE 24. The degree of the equation of any straight line. It will now be 'shown that any straight line is represented by an equation of the first degree in the variable coordinates x and y. Theorem. The equation of the straight line passing through a point B(Oy b) on the axis of y and having its slope equal to m is (I) y = ntiT' + &. Proof First step. Assume that F(x, y) is any point on the line. Second step. The given condition may be written slope of PB = m. Third step. Since by (II), p. 17, • slope of P^ = ^^, x — [Substituting (ic, y) for (iCi, y{) and (0, h) for (x2, 2/2)] then ^~ = m, or y = mx + h. q.e.d. X In equation (I), m and & may have any values, positive, negative, or zero. Equation (I) will represent any straight line which intP'^ sects the ?/-axis. But the equation of any line parallel t<» ?/-axis has the form x = a constant, since the abscissas of all points on such a line are equal. The two forms, y = mx + b and X = constant, will therefore represent all lines. Each of these equations being of the first degree in x and y, we have 64 STRAIGHT LINE AND CIRCLE 55 Theorem. TJie equation of any draight line is of the first degree in the coordinates x and y. 25. Locus of any equation of the first degree. Tlie ques- tion now arises : Given an equation of the first degree in the coordinates x and y, is the locus a straight line ? Consider, for example, the equation (1) 3i»-2y + 8 = 0. Let us solve this equation for y. This gives (2) y = ix^L Comparing (2) with the formula (I), y = mx + b, we see that (2) is obtained from (I) if we set m = f , 6 = 4. Now in (I) m and b may have any values. The locus of (I) is, for all values of m and 6, a straight line. Hence (2), or (1), is the equation of a straight line through (0, 4) with the slope equal to f . This discussion prepares the way for the general theorem. The equation (3) Ax-hBy-{-C = 0, where A, B, and C are arbitrary constants, is called the general equation of the first degree in x and y because every equation of the first degree may be reduced to that form. Equation (3) represents all straight lines. P^or the equation y = mx + b may be written mx — ?/ + 6 = 0, which is of the form (3) if ^ = m, i? =— 1, C =b ; and the equation x = con- stant may be written x — constant = 0, which is of the form (3) if ^ = 1, ^ = 0, C =— constant. Theorem. The locus of the general equation of the first degree Ax + By+C=^0 is a straight line. Proof Solving (3) for ?/, we obtain (4) 2/ = --^--- ^ ^ -^ B B 56 ELEMENTARY ANALYSIS Comparison with (1) shows that the locus of (4) is the straight line for which G m- B' or If, however, B = 0, the reasoning fails. But if jB = 0, (3) becomes Ax + = 0, C The locus of this equation is a straight line parallel to the Y-axis. Hence in all cases the locus of (3) is a straight line. '\ Q.E.D. Corollary. TJie slope of the line Ax + By-j-C=0 A is m = ; that is, the coefficient of x with its sign changed JB divided by the coefficient of y, 26. Plotting straight lines. If the line does not pass through the origin (constant term not zero, p. 46), find the intercepts (p. 47), mark them off on the axes, and draw the line. If the line passes through the origin, find a second point (p. 37) whose coordinates satisfy the equation. EXAMPLE Plot the locus of3aj — 2/ + ^ = ^- Find the slope. Solution. Letting y = and solving for x, we have .T = — 2= intercept on aj-axis. Letting a? = and solving for y, wei,. - with the o line 0? — 2/ + 6 = 0. Solution. Let m^ be the slope of the re- quired line. Then its equation is by (II), p. 59, (1) ?/ — 5 = mi(a; — 3). The slope of the given line is mg = 1, and since the angle which (1) makes with the given line is - , we have, o . TT mi ^ tan- =— ^ or whence = -(2 + V3). 1-V3 Substituting in (1), we obtain 2/-5 = -(2 + V3)(a^-3), or (2 + V3)i^ + ij - (11 + 3 V3) = 0. In Plane Geometry there would be two solutions of this problem, — the line just obtained and the dotted line of the figure. Why must the latter be excluded here ? In working out the following problems, the student should first draw the figure and mark by an arc the angle desired, remembering that this angle is measured from the second line to the first in the counter-clockwise direction. STRAIGHT LINE AND CIRCLE 67 PROBLEMS 1. Find the angle which the line 3x — y + 2 — makes with 2x-\-y — 2 = 0; also the angle which the second line makes with the first, and show that these angles are siipplemeivf^W 4 4 2. Find the angle which the line (a) 2 X — ^iif + 1 = makes with the line x — 2y + 3 = 0. (b) a; + 2/ + l = ^ i^^^^s with the line i»—?/ + 1 = 0. * ^ (c) Sx — 4:y + 2 = makes with the line x-[-3y — 7 = 0. (d) 6x—3y -{-3 = makes with the line x = 6, *» (e) x — 7y + l = makes with the line x-\-2y — A = 0. In each case plot the lines and mark the angle found by a small arc. Ans. (a) tan-'(-J^); (b) |; (c) tan-H-i/); (d) tan-'(-l); (e) Un-\j%). 3. Find the angles of the-fTiangle whose sides are x + 3y -4 = 0, 3a;-22/ + l = 0, and ^•-?/4-3 = 0. Ans. tan-i(--y-), tan-\i), tan-i(2). Hint. Plot the triangle to see which angles formed by the given lines are the angles of the triangle. 4. Find the exterior angles of the triangle formed by the lines 5x — y-\- 3 = 0, yd-z 2, X- 4: y-\- 3 = 0. Ans. tan-i (5), tan-^(-i), tan~^ (--!/). 5. Find one exterior angle and the two opposite interior angles of the triangle formed by the lines 2a; — 3^— 6 = 0, 3x + 4:y — 12 = 0,x — 3y-{-6 = 0. Verify the results by for- mula 37, p. 3. 6. Find the angles of the triangle formed by 3x-\-2y—4^ = 0, x — 3y-{-(j = 0, and 4:X — 3y — 10 = 0. Verify the results by the formula tan A + tan B + tan C = tan ^ tan B tan C,iiA-{-B-{- (7= 180°. Q8 ELEMENTAKY ANALYSIS 7. Find the lin^passing through the given point and making the given angle with the given line. (a) (2,1), '~,2x-3y + 2=:0. Ans. 5x-y-9 = 0. (b) (l,-3),^,x + 2y + 4:=p0. Ans. 3x + y=:^0. (c) (xi, 2/1), ,y = mx + b. Ans. y-y,= »^ + tan<^ .^_ 1 — m tan ^ (d) («!, Vi), ,Ax+By+G=0. 30. Equation of the circle. Ev^ry circle is determined when its center and radius are known. Theorem. TJie equation of the circle wJiose center is a given point (a, P) and wJiose radius equals r is (V) (oo-ay-\-(y-P)^ = r^ Proof. First step. Assume that P(x, y) is any point on the locus. Second step. If the center (a, /?) be denoted by C, the given condition is Third step. By (I), p. 13, Squaring, we have (V). q.e.d. Corollary. The equation of the circle whose center is the origin (0, 0) and whose radius is r is If (Y) is expanded and transposed, we obtain ( 1 ) x^ + y'' -2 ax -2 ^y + a^ + ^^ - 7^ = 0, STRAIGHT LINE AND CIRCLE 69 # From the form of this equation we observe : Any circle is defined by an equation of the second degree in the variables x and y, in which the terms of the second degree consist of the sum of the squares of x and y. Equation (1) is of the form (2) x'^y' + Dx + Ey + F=0, where (3) D=-2a,E=-2p,2^ndiF=a?^P''-r'. Can we infer, conversely, that the locus of every equation of the form (2) is a circle? By adding \D' + \h to both members, (2) becomes (4) (x + \Df-^r{y + \E)' = \{D'^-E'^-4.F), In (4) we distinguish three cases : If D^ + ^^ — 4 i^ is positive, (4) is in the form (V), and hence the locus of (2) is a circle whose center is (— |-Z), —^E) and whose radius is r = |: V-D^ -{-E^ — 4: F. If D^ + E^ — 4:F=0, the only real values satisfying (4) are x= —^^D, y=—-^E (footnote, p. 35). The locus, therefore, is the single point (— |-Z>, —^E), In this case the locus of (2) is often called a point circle, or a circle whose radius is zero. If i>^ + -E- — 4 i^ is negative, no real values satisfy (4), and hence (2) has no locus. The expression D^-\-E^—4:F is called the discriminant of (2), and is denoted by ®. The result is given by the Theorem. TJie locus of the equation (VI) 00^ i- y^ -\- Doc + Ey + F = 0, whose discriminant is ^ ^^-= ^'^ -\ f? ^ ^i '^ determined as follows : (a) When ® is positive the locus is the circle ivnose center is (— I D, —\ E) and whose radius is r = | Vi^ + ^^ — 4~^ I V®L (b) When © is zero the locus is the point circle (— |-Z>, —^E), (c) When © is negative there is no locus. 70 ELEMENTARY ANALYSIS Corollary. When E = the center of (YI) is on the X-axis, and when D = the center is on the Y-axis. Whenever in what follows it is said that (YI) is the equar tion of a circle it is assumed that © is positive. EXAMPLE Find the locus of the equation a^^ + 2/^ — 4a; + 8?/ — 5 = 0. Solution. The given equation is of the form (YI), where D=-4, E=:S,F=-5, and hence r>K ®=16jr 64+^ = 100 >0. The locus is therefore a circle whose center is the point (2, —4) and whose radius is The equation Ax^ + Bxy + Cy'- + Dx -{-Ey + F= is called the general equation of the second degree in x and y because it contains all possible terms in x and y of the second and lower degrees. This equation can be reduced to the form (YI) when and only when A=0 and B = 0. Hence the locus of an equa- tion of the second degree is a circle only when the coefficients of x^ and y^ are equal and th6 xy-teim is lacking. 31. Circles determined by three conditions. The equation of any circle may be written in either one of the forms (x-ay + (y-(3y=r% or ' x^ + y' + Dx + Ey-\-F=0. Each of these equations contains three arbitrary constants. To determine these constants three equations are necessary, and as any equation between, the constants means that the STRAIGHT LINE AND CIRCLE 71 circle satisfies some geometrical condition, it follows that a circle may be determined to satisfy three conditions. Rule to determine the equation of a circle satisfying three conditions. First step. Let the required equation he (1) {x~af + (jj-lif = T% or (2) x' + f- + Dx -\- Ey + r={}, as may be more coyivenient. Second step. Find three equations between the constaiits a, p, and r \_or D, E, and F^ ivhich express that the circle (1) [or (2)] satisfies the three given conditions. Third step. Solve the equations found in the second step for a, f3, and r [or D, E^ and F']. Fourth step. Substitute the results of the third step in (1) [or (2)]. Tlie result is the required equation. EXAMPLES 1. Find the equation of the circle passing through the three points Pi(0, 1), P,(0, 6), and P^{^, 0). First solution. First step. Let the required equation be (3) x' + y^-{-Dx + Ey-{-F=0. Second step. Since P^, Po, and Pg lie on (3), their coordinates must sat- isfy (3). Hence we have (4) 1 + ^4-P-O, (5) 36 + 6J5; + P=0, and (6) 9 + 3i)+P=0. 72 ELEMENTARY ANALYSIS Tliird step. Solving (4), (5), and (6), we obtain ^=-7, i^=6, i)=-5. Fourth step. Substituting in (3), the required equation is The center is (f , ^) and the radius is f V2 = 3.5. Second solution. A second method which follows the geo- metrical construction for the circumscribed circle is the fol- lowing. Find the equations of the perpendicular bisectors of P1P2 ^^d P1P3. The point of intersection is the center. Then find the radius by the length formula. 2. Find the equation of the circle passing through the points Pi(0, —3) and P2(4, 0) which has its center on the line a; H^r 2 i/ sc 0. First solution. First step. Let the re- i^x quired equation be (7) x' + f + Dx + Ey + F=0. Second step. Since P^ and Pg lie on the locus of (7), we have 9-3J5; + i^=0 16+4i> + P=0. (8) and (9) The center of (7) is ( — ~, D E line, -f+^ 2 E I, and since it lies on the given = 0, . or (10) D^2E = 0. Tliird step. Solving (8), (9), and (10), we obtain i> = --VS^ = |,andP=--V-. STRAIGHT LINE AND CIRCLE 73 Fourth step. Substituting in (7), we obtain the required equation, or 5a^ + 52/'-14.T + 72/-24 = 0. The center is the point (|, — ^-^), and the radius is |- V29. Second solution. A second solution is suggested by Geometry, as follows : Find the equation of the perpendicular bisector of P^ Pg- The point of intersection of this line and the given line is the center of the required circle. The radius is then found by the length formula. PROBLEMS 1. Find the equation of the circle whose center is (a) (0, 1) and whose radius is 3. Ans. x^ + y^— 2y — 8 = 0. (b) ( — 2, 0) and whose radius is 2. Ans. x^ -\-y^-{-4:X = 0. (c) ( —3, 4) and whose radius is 5. Ans. x^ -{- y'^ + 6 x — S y = 0. (c?) (a, 0) and whose radius is a. Ans. xr-\-y^ — 2ax — 0. \ (e) (0, /3) and whose radius is {3. Ans. x'-^ -\-y^ — 2/3x = 0. 'A ^// ,.0. li, fj (/) (0, — ^) and whose radius is ^. - Ayis. x^ -{-y^ + 2 px = 0. /^ tT 2. Find the locus of the following equations. (a)aP + y^~6x-16 = 0. (f) x^ + y'^ - 6 x +4.y -^ 5 = 0. (b)3a^ + 3 y'- 10 x-24.y = 0. (g) (x + If + (y - 2)^ = 0. (c) x' + y' = 0. (Ji'^jx'-{-7y'^-4.x-y = ^. (d)x' + y^-Sx-6y-{-25 = 0. (i) x'+y'-+2ax-^2by+a^-{-b^=:0. (e) af-\-y'^-2x + 2y + 5 = 0. (j) or + y^ + i6x -\-100 = 0. 3. Find the equation of the circle which (a) has the center (2, 3) and passes through (3, — 2) . Ans. x^ + y'- - 4. x^6y -13 = 0. (b) passes through the points (0, 0), (8, 0), (0, — 6). A71S. x^-}-y'--Sx+6y = 0. (c) passes through the points (4, 0), (—2, 5), (0, —3). Ans. 19 x^ + 19 ?/' -f 2 x - 47 y -^312 = 0. 74 ELEMENTARY ANALYSIS (d) passes through the points (3, 5) and X~^y '^) ^^^ ^^s its center on the X-axis. Ans. x^ -{- y^ -\- 4: x — 4:6 = 0. (e) passes through the points (4, 2) and (— 6, —2) and has its center on the F-axis. A7is. x^ -{- y^ -{- 5 y — 30 =0. (/) passes through the points (5, — 3) and (0, 6) and has its center on the line 2x~3y — 6 = 0. Ans. 3i»2 + 32/2_ll4aj_ 642/ +276 = 0. (g) h^ the center (— 1, — 5) and is tangent to the J^axis. Ans. x'4-y^-\.2x + 10y+l=0. (h) passes through (1, 0) and (5, 0) and is tangent to the IF-axis. Ans. x^ + y^ — 6x±2-\/5y-\-5=0. (i) passes through (0/1), (5, 1), (2, -3). Ans. 2x^ + 2y^-10x + y-3 = 0. (j) has the line joining (3, 2) and (— 7, 4) as a diameter. ^ . Ans: x^-{-y^ + 4:X-6y-13 = 0\ (Tc) has the line joining (3, — 4) and (2, — 5) as a diameter. Ans. x^+y^-5x + 9y + 26 = 0. (/) which circumscribes the triangle formed by a;— 6 = 0? X -\-2y = 0, and x~2y = S. Ans. 2x^ + 2y'^-21x + Sy-^60=0. The following problems illustrate cases in which the locus problem is completely solved by analytic methods, since the loci may be easily drawn and their nature determined. LOCUS PROBLEMS 1. Find the equation of the locus of a point whose distances from the axes XX' and YY' are in a constant ratio equal to |. Ans. The straight line 2x —3y = 0. 2. Find the equation of the locus of a point the sum of whose, distances from the axes of coordinates is always equal to 10. Ans. The straight line x-^y — 10 = 0. 3. A point moves so that the difference of the squares of its distances from (3, 0) and (0, — 2) is always equal to 8. Find the equation of the locus, and plot. Ayis. The parallel straight lines 6a; +4?/ +3 = 0,6a;+42/— 13=0. STRAIGHT LINE AND CIRCLE 75 4. A point moves so as to be always equidistant from the axes of coordinates. Find the equation of the locus, and plot. A71S. The perpendicular straight lines x -^-y = 0,x — y =0. 5. A point moves so as to be always equidistant from the straight lines x —4z = and y -\-5= 0, Find the equation of the locus, and plot. Ans. The perpendicular straight lines cc— y— 9 = 0,x-[-y^l=0. 6. Find the equation of the locus of a point the sum of the squares of whose distances from (3, 0) and (—3,0) always equals 68. Plot the locus. Ans, The circle x^-\-y^ = 25. 7. Find the equation of the locus of a point which moves so that it^ distances from (S, 0) and (2, 0) are always in a constant ratio equal to 2. Plot the locus. Ans, The circl^ x^-\-y^=16, 8. A point moves so that the ratio of its distances from (2, 1) and ( — 4, 2) is always equal to 2. Find the equation of the locus, and plot. Ans. The circle 3xr-{-3y^—24:X—4:y=0. In the proofs of the following theorems the choice of the axes of coordinates is left to the student, since no mention is made of either coordinates or equations in the problem. In such cases always choose the axes in the most convenient manner possible. 9. A point moves so that the sum of its distances from two perpendicular lines is constant. Show that the locus is a straight line. Hint. Choosing the axes of coordinates to coincide with the given lines, the equation is x + y = constant. 10. A point moves so that the difference of the squares of its distances from two fixed points is constant. Show that the locus is a straight line. Hint. Draw XX' through the fixed points, and YY' through their middle point. Then the fixed points may be written (a, 0), ( — a, 0), and if the "constant difference" be denoted by k, we find for the locus 4:ax = k or 4:ax =— k. 716 ^^EEEMWTARY ANALYSIS 11. A point moves so that the sum of the squares of its distances from two fixed points is constant. Prove that the locus is a circle. Hint. Choose axes as in problem 10. 12. A point moves so that the ratio of its distances from two fixed points is constant. Determine the nature of the locus. Ans. A circle if the constant ratio is not equal to unity, and a straight line if it is. 13. A point moves so that the square of its distance from a fixed point is proportional to its distance from a fixed line through the fixed point. Show that the locus is a circle. CHAPTER V CURVE PLOTTING 32. Asymptotes. The following problems elucidate diffi- culties arising frequently in drawing the locus of an equation. EXAMPLES 1. Plot the locus of the equation (1) xy-2y-4. = 0. Solution. Solving for y^ (2) y- ^— 2 We observe at once, if x = 2, This is interpreted X y X y -2 -2 1 -4 - 1 -f H -8 -2 -1 If - 16 -4 -1 2 CO -5 -f 2i 16 ^ 8 - 10 -i o 4 etc. etc. 4 2 5 f 6 1 12 0.4 etc. etc. thus : The curve approaches the line x^2 as it passes off to in- finity. The vertical line a; = 2 is called a vertical asymptote. In plotting, it is necessary to assume values of x differing slightly from 2, both less and greater, as in the table. From (2) it appears that y diminishes and approaches zero as X increases indefinitely. The curve therefore extends in- definitely far to the right and left, approaching constantly the axis of x. The axis of x is therefore a horizontal asymptote. If we solve (1) for a; and write the result in the form 4 77 y 78 ELEMENTARY ANALYSIS it is evident that x approaches 2 as 2/ increases indefinitely. Hence the locus extends both upward and downward indefi- nitely far, approaching in each case the line x = 2. This curve is called a liyperhola. n 1 \ \ y \ ^ **" — ^ — — "~^ uO (x)yf'{x), etc. During any investigation the same functional symbol always indicates the same law of dependence of the function upon the variable. In the simpler cases, this law takes the form of a series of analytical operations upon that variable. Hence, in such a case, the same functional symbol will indicate the same operations or series of operations, even though applied to dif- ferent quantities. Thus, if f(x) = x^-9x + U, then f(y^=^y^-9y + U, Also f(a) = a^ — 9 a + 14, /(6 + !) = (& + 1)2 -9(6 + 1)4-14 = 5' -76 +6, J FUNCTIONS AND <>KA?IIS IQl /(0) = 02_9. + 14 = 14,^. ,;,. . ^ /(-l) = (-l)^-9(- (x) =^ log.o x. Find ct>(2), <^(1), <^(5), ^(a-1), (y),(x + l),ct>(^x), 2. Given (x) = e^. Find <^(0), <^(1), <^(-l), (2y), {-x). 3. Given fix) = sin 2 a:. Find /|\ /jY /(- tt), /(- .;), /(7r-a;),/(i7r-^),/(f7r + 5). 4. Given ^ (a;) = cos x. Prove CHAPTER VII DIFFERENTIATION 38. Tangent at a point on the graph. A glance at the figure on p. 94^ that is, the graph of the equation (1) X in which M represents the number of square feet of lumber required to construct an open box with a square base to con- tain 108 cu. ft., makes clear the following facts. When the base is less than 6 ft. square, the material decreases as the size of the base increases. (For the graph is falling to the left of x= 6.) When the base is more than 6 ft. square, the material increases as the base increases. (For the graph is rising to the right of a? = 6.) If we draw the tangent to the graph at any point to the left of a; = 6, the slope is clearly negative, while for any point to the rifh4^f x = 6, the slope is positive. At the lowest point x = 6, the slope is zero. Clearly, then, the facts described are characterized by the slope of the tangent to the graph. We may state in general : If the slope of the graph "^ is positive, then the function in- creases as the variable increases. If the slope of the graph is * The slope of the graph is the same as the slope of the tangent. 102 DIFFEKENTIATION 103 negative, then the function decreases as the variable increases. The slope of the graph is zero at a maximum or minimum. 39. Differentiation. The discussion just given shows plainly that a method for obtaining the slope of the graph will be useful. We turn our attention to this question. Let the figure be the graph of the equation , ^^^^ (1) y=f(x) It is required to find the slope of the tangent at the point P{x, y). First step. Take a second point Q on the curve near P, where coordinates ^ are (x + Ax, y -f A?/). These coordinates satisfy equation (1). Second step. Draw PR parallel to OX. Then BQ = Ay = increment of the function. Also, PR = Aa; = increment of variable. Third step. Draw the secant through P and Q, and call its inclination <^. Then, clearly, tan 6, ?7i > ; if ic = 6, m = 0, results agreeing with the figure. 40. Derivative of a function. The general rule for differ- entiation gives for any function y of x the value of (1) limit A?/ This result is called the derivative of the function with respect to the variable. In words the derivative of a function is the limit of the quotient of the increment of the function by the incre- ment of the variable, when the latter increment varies and ap- proaches the limit zero. It is customary to use as an abbreviation of the expression (1) the symbol -^ (read '^derivative of y with respect to a;''); that is, we place for convenience /ox ^_ hmit Ay The symbol -^ is for the present to be regarded as a whole, dx not as a fraction. In a later section we define dy and dx sepa- rately and also -^ as a fraction. dx 106 ELEMENTARY ANALYSIS Similarly, if 8 is a function t, then ds limit As -i • i • n -ii ^ i ^ — = . ^ ^ — = derivative of s with respect to t. It is useful to write down symholically the results of apply- ing the General Rule tg the general equation (3) y=f{x). We have : First step. y + Ay =f(x + Ax). Second step. Ay = f(x + Ax) — f(x) . Third step. ^ = f(x + Ax)^f(x) , Ax Ax Fourth step. ^ = ?^^\ f(x-\-Ax)-f(x) ^ ^ dx ^^=^ Ax The derivative of f(x) is also a function of x. We indicate this result by f'(x), that is, we use the abbreviation (3) ^,^.^lun\t f(x + Ax)-f(x)^ Summing up, if y=f(x), then (4) -^=f(x)= slope of tangent at (x, y). ctx In words : the value of the derivative of a function equals the slope of the tangent at the corresponding point on the graph. . . EXAMPLES 1. Differentiate 3 x^ + 5. Solution. Applying the successive steps in the General Bule we get, after placing y = Sx' + 5, First step. y + Ay = 3(x+ Ax)^ + 5 = Sx^ + 6x' Ax + 3(Axy + 5. DIFFERENTIATION 107 Second step, y -\- ily = .^ x^ -{- Q> x - ^x -\- 3(Aic)^+ 5 y =3^ +_5 A^/ = 6 aj • Aic + 3(Aa;)^ i Third step, -^ = 6 cc + 3 • Aa;. Aic Fourth step, -^ = 6x. Ans. dx We may also write this ^(Sx'-^5)^6x, \ ^. 1 dx ) ' 2. Differentiate a^ — 2 cc -|- 7. Solution. Place y = x'^ — 2x-\-7. First step, y-^Ay=(x-\- Axf— 2(x + Ax-) + 7 = 0^34.3 a;2 . Ao; + 3 a; . {Axy+(j\xy- 2 a; -2 • Aa;+7. Second step, 2/ + A2/ = o;^ + 3 a;2 . Aa; + 3 a; . (Ax^ + (Aa;)^ - 2 a; - 2 • Ao; + 7 y = QC^ — 2x +7 A2/= 3 cc^ . Aaj + 3 a; . (Aa;)^ + (Axf -2.Aa; Th ird step, ^ = 3 ic^ 3 a; • Ao; + {^xf - 2. Fourth_step. ^ = 3 ic^ _ 2 . ^ns. die Or.-^(aT^-2a;+7) = 3a;2_2. do; 3. Differentiate — • Solution. Place 2/ = ^ * 108 ELEMENTARY ANALYSIS First step. Second step, Tliird step. Fourth step. Ay = (x + Axf c Ay Ax (x+Axy c _ — c ' Aa;(2 x+Ax) x^ x^{x + Axy ^ = -c x\x + Axy 2x 2c Ans. dx x\xy Or A(±\ = =ll. dx \x^J a^ 4. Eind the slopes of the tangents to the parabola y = x^Sit the vertex, and at the point where x = ^. Solution. Differentiating by the General Rule, we get {A) -y- = 2x = slope of tangent line at any point on curve. ax To find slope of tangent at vertex, substitute a; = in {A), giving ^ = 0. dx Therefore the tangent at vertex has the slope zero ; that is, it is parallel to the axis of X and in this case coincides with it. To find slope of tangent at the point P, where x = \, substituting in (J.), giving dx that is, the tangent at P makes an angle of 45° with the ic-axis. The derivative may now be made use of in checking up a plot. 5. Plot the locus of {A) y = a^-12x, find the slope at each point plotted, and check up in the figure. DIFFEREXTIATIOjST 109 Solution. Differentiating by the General Rule, we find (B) ^ = 3 0^2 - 12 = slope at {x, y). CtiC The table gives the results of the calculation, e.g. for x from {A), 2/ = - 11, 1, and from (B\ ^ = > ^' dx X y dy dx X y dy dx -12 -12 1 -11 - 9 - 1 11 - 9 2 -16 -2 16 3 - 9 15 -3 9 16 4 16 36 -4 -16 36 etc. etc. etc. etc. etc. etc. 3 - 12 = - 9 = slope at (1, - 11). The scales used in the figure are those indicated. To con- struct the tangent at any point, proceed as follows : At (0, 0), the slope = — 12. Hence lay off from the origin 1 unit to the left and 12 units up. The tangent at (0, 0) passes through this point (p. 18). Similarly, at (3, —9), slope = 15, hence lay off 1 unit to the right and 15 units up, the connecting line being the tangent. 110 ELEMENTARY ANALYSIS Note that different scales on x and y change the inclination, but the construction for the tangent is clear. Discussion. The graph has a maximum point at (—2, 16) and a minimum at (2, — 16). There are no other horizontal tangents, since -^=z^x^ — 12 = when a; = ± 2, only. (ajX PROBLEMS Use the General Rule in differentiating the following examples. 1. 2/ = 3a;l 4., y = x^. 7. s = t"-2t + 3, Ans. ^ = 6x. Ans. ^^Sx", A71S. ^ = 2t-2. dx dx dt 2. y = x^-3x, 5. r = ae^ 8. y = -' X Ans. ^ = 2x-3. Ans. ^ = 2a^. Ans. ^ = -\- dx dO dx x^ 3. y=ax^+bx+c. 6. p = 2q\ f Ans. ^ = 2ax + b, Ans. ^ = 4:0. Ans. ^ = -i. dx dq dt f 10. Find the slope of the tangent to the curve y = 2i]i^ — 6x + 5, (a) at the point where x = l', (6) at the point where x=0. Ans. (a) ; (b) - 6. 11. (a) Find the slopes of the tangents to the two curves y=: 3x^ — 1 and y = 2x^-{-3 at their points of intersection. (b) At what angle do they intersect ? Ans. (a) ±12, ± 8 ; (b) arc tan -/y. Plot the locus of each of the following, find the slope at each point plotted, and check up. 12. (a) y = x'-l. (/) 2/ = ar^-3 X. (b) y = 3 — x^. (g) y = ^x^~4:X-{-l. (c) y = 2 — 4,x—x^. (h) y = ^x^ — x^ — 3x. (d) y = 4,x-\- x^. (i) y = ^a^ — 3xF-]-5x. (e) y = x^ — 6x, (j) y = x'^—Sx^ -{-10, o CHAPTER VIII FORMULAS FOR DIFFERENTIATION 41. Theorems on limits. In the last step of the process of differentiation the increment of the variable is assumed to '^ vary and approach the limit zero." This statement is made precise by the following definition : A variable is said to approach the limit zero when its numerical value becomes and remains less than any positive number, how- ever small. Again, in differentiation, we start with certain values of x and ?/, and these values re- main fixed in the four steps. The actual variables are clear- ly Aic and A^. The fact is, of course, that the point P(x, y) is fixed, but the point Q moves toward P, and thus Ao; and A?/ both vary. ]^ow in the figure, the position of Q depends only upon the choice of the point N, and the position of N depends upon the value of Aa* only. Hence the slope of the secant is a function of Aic only ; that is, when P is fixed, we have (1) tan a = ^ = a function of Lx = 6 (Aa;). Aaj ^^ ^ The derivative, that is, the slope of the tangent at P(x, y), is the value of limit , / A ^\ 111 112 ELEMENTARY ANALYSIS that is, it is the limiting value of a function of Ace, when the variable ^x approaches the limit zero. ' It is clear that the successive values of <^ (Aa^) are the slopes of the successive secants through P and the successive positions of the point Q. How shall the value of ^^^^ (Aa?) be found ? In all cases thus far the limiting value has been found by substitution directly in the function Aa; = 0. In other words, we have assumed the definition : Tlie limiting value of a function of Ax when Ax varies and approaches the limit zero is the value of the function when Ax equals zero. To make it clear that difficulties arise in applying this definition, consider the two examples : limit Va? -{- Ax — Va? -, limit sin A x Direct substitution leads in each case to the meaningless or indeterminant expression -• In each of these examples, therefore, the definition does not apply, and other methods must be used (see Arts. 51 and 54). In the following pages, repeated application is made of the following theorems, whose proof is here omitted. Given a number of variables whose limits are known; then I. Hie limit of an algebraic sum of any number of variables equals the same algebraic sum of their respective limits, II. The limit of the product of any number of variables equals the product of their respective limits, III. The limit of a quotient of two variables equals the quotient of their respective limits when the limit of the denominator is not zero. FORMULAS FOR DIFFERENTIATION 113 42. Fundamental formulas. For economy of time in differ- entiation, special rules have been devised, which are here set down, the proofs being given in later sections. In each formula, ii, v, w, etc., are assumed to be functions of the same variable x, and a, c, e, and n are constants. I ^ = o. II ^ = 1. III JL(u + v-w)='^ + §L-<^. doc ddo doc ddo IT ^(c^.)=c^- ddc dx^ doc doc doo TI A.(v^)=nv^-^^' dot) doo VI a — (iJC^) = nx^-\ doc Til ^du_^dv d fu\ _ dx^ doc doc \v ) ~ v'^ du doc \cl c VIII ^(iog^v)=Mnedv^ doc V doc Villa #^(lo^v) = -4^ doc V doc IX ^(a^)=anoga^ doc doc IX a _^(^.)^^.^. doc doc X ^(sini;) = cosi;^. doc doc 114 ELEMENTARY ANALYSIS XI :^(cosi;) = -sinv^. doc doc XII ^(tant;) = sec2t;^. doc doc XIII 4- (cot v) = - csc2 1; ^ . XIV ~ (sec v) = sec vUnv^- djc doo XV ^(cscv)=-cscvcoti;^. doD doo XVI -^ (arc sin v) = ^ ^ . XVII #^ (arc tan v) = -J— ^ doo 1 + t;^ c^x 43. Differentiation of a constant. A function that is known to have the same value for every value of the inde- pendent variable is constant, and we may denote it by y = c. As X takes on an increment Ax, the function does not change in value ; that is, Ay = 0, and ^ = 0. Ax But ;i^^^J^V^=0. I .-.^ = 0. dx Tlie derivative of a constant is zero. 44. Differentiation of a variable with respect to itself. Let y =z X. Following the General Rule, p. 104, we have First step, y + Ay = x+ Ax, FORMULAS FOR DIFFERENTIATION 115 Second step. Ay = Ax. TJiird step, —^ = 1. Ax Fourth step, -^ = 1. dx II ...^ = 1. Tlie derivative of a variable with respect to itself is unity, 45. Differentiation of a sum. Let y = u + v — w. By the General Rule : First step. Changing a? to a? + Ax, y ■\- Ay = u-\- Au-{-v-{- Av — w — Aw, Second step. Ay = Au ■+■ Av — Aw. Third step. Ay^Au_^^v_Aw^ Ax Ax Ax Ax rjj .1 . dy du , dv dw Fourth step. -^ = 1 . dx dx dx dx [Applying Theorem I, p. 112.] dot) doc doc doc Similarly for the algebraic sum of any finite number of functions. The derivative of the algebraic sum of a finite number of func- tions is equal to the same algebraic sum of their derivatives. 46, Differentiation of the product of a constant and a function. Let y = cv. First step. Changing a:; to ic + Ax, y -\- /^y =z c (v -{- Av) = CV + cAV. 116 ELEMENTARY ANALYSIS Second step. Ay = c - Av. Third step. J = c — . Ax Ax Fourth step, — ^ = c — . dx dx K [Applying Theorem II, p. 112.] doo doc The derivative of the product of a constant and a function is equal to the product of the constant and the derivative of the function, 47. Differentiation of the product of two functions. Let y = uv. First step. Changing a? to a; + Ax, y-\- Ay = (u-\- Au) {v + Av) = uv -{-u ' Av -\-v ' Au + Au • Av. Second step. Ay = u - Av + v • Au + Au . Av. Third step, ^ = u — + "": f- Au — . Ax Ax Ax Ax Fourth step, -^= u [-v — . dx dx dx Applying Theorem II, p. 112, since when Ax approaches zero as a" limit, Au also approaches zero as a liuiit, and limit I Au — ] = 0. \ Ax/ Y ..d^^^^^^dv^^^ doc doc doc The derivative of the product of two functions is equal to the first function times the derivative of the second, plus the second function times the derivative of the first. To extend V to the product of three functions, proceed thus : FORMULAS FOR DIFFERENTIATION 117 — (uvw) = — (uv • w). where we now regard uv as one function. (V) ,'. — (uv -w)— uv \-w — (uv^ ^ ^ dx^ ^ dx dx^ ^ dw , dv , du = uv \- wu \-wv . dx dx dx The general rule to be read out of thjs result is : The derivative of the product of any number of functions equals the sum of all the products that can be formed by multiplying the derivative of one function by all the remaining functions. 48. Differentiation of a power of a function. In equation (1) of Art. 47, if we assume ^i and iv to be identical with v, we obtain at once dx dx which proves formula VI for n = 3. In a similar manner we may prove VI for any positive integer. For the present we assume VI to hold if n is any constant, proof being reserved for a later section. In VI, putting v = x, we obtain VI a, using II. Power rule. The derivative^ of a function with a constant exponent equals the product of the exponent, the function tvith exponent less one, and the derivative of the function, 49. Differentiation of a quotient. Let 2/=-, '^'^0. V First step. Changing a? to .'i? + Ace, , . u + Au y + Ay = —--—. V -\- Av 118 ELEMENTARY ANALYSIS V • Au — U'Av Second step. Third step. Fourth step. \tl u-\- Au u ay - V + Av V Au Av 7' ?y Ay Ax Ax Ax v(v 4- Av) dy_ du dv V u dx dx v(v-{- Av) dx v^ [Applying Theorems II and III, p. 112.] VII .-. -f {^\ (loo doo The derivative of a fraction is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. When the denominator is constant, set v = c in VII, giving du Vila ^IVl\^^. doc \c I c [Since ^i^-^ = 0.1 L. dx dx J We may also get VII a from IV as follows : du d fu\ 1 du dx dx\cj c dx c The derivative of the quotient of a function by a constant is equal to the derivative of the function divided by the constant. PROBLEMS Differentiate the following : 1. y = x\ FORMULAS FOR DIFFERENTIATION 119 Solution. ^=— (aj3N 3^2 ^^^^^ (by VI a) [^ = 3.] 2. y = ax^ — bx^. Solution. ^ = -^ (ai«* - b^) = - (aa;^) - ~ (ba^) ' (by III) dx dx^ ^ dx^ ^ dx^ ^ ^ ^ ^ = 4 aa^ — 2 6a;. ^ns. (by VI a) 3. 2/ = i«^ + 5. Solution. ^ = A (o^t) _^ 1 (5) (by XII) = I a;^. ^ns. (by VI a and I) 3x^ 7 X , o 7/-0 Solution. ^ = -^ (3 xV)- ^ (7a;-3) + - (Sx^) (by III) da; dx^ ' dx^ ' dx^ ' ^ •' ' = ¥ *^ + 1 «'^ + -T- »'"*• ^ns. (by IV and VI a) 5. y = (x'-3y. Solution. ^ = 5(x2 - 3)* ^_ («2 - 3) (by VI) CtX (IX [ij = x2 — S and n = 4.] = 5(a;2 - 3)^ . 2 a; = 10 x(x^ - Sy. Arts. We might have expanded this function by the Binomial Theorem and then applied III, etc., but the above process is to be preferred. , 6. y z= Va^ — a^. Solution. ^ =^l(a?- x")^ = i (a' _ x^^ — (a' - x') (by VI) dx dx^ ^ ^ ^ dx^ J y J J [z? — a^ — x^ and n = l-] 120 ELEMENTARY ANALYSIS = i(a2 - a^)-i(- 2x)= — — • Ans. 7. 2/=(3a;2 + 2)Vl+5a;l (by V) Soluti(m. ^ = (3 a;^ + 2) - (1 + 5a;^) I + (1 + 6 a;^i - (Sa;^ + 2) da; (^a; da; [m = 3 a;2 + 2 and « = (1 + 5 x^)^ ] = (3a;2+2)i(l +5a;^)-^— (1 + 5 a;2) _^ (1 ^5 a^ylga; (by VI, etc.) = (3 x" + 2)(1 + 5 .T^-2 5 a; + 6<1 + 5 x')^ 5a?(3a)2+2) , ^ / -, , ^ 2 45ar5 + 16a; a^ + ^ ^ns. 8. 2^= — ' Solution. ^ = dx a- — oj- (by YII) ^ 2x(p?-x')-\-x(a? + x'') 1 [Multiplying both numerator and denominator by {aP- — x^)^.] = • Ans. (a^ - xy 9. 2/ = 5a;^ + 3a^'-6. ^^ = 20a;3 + 6aj. 10. y=*dGX^—^dx-\-5e, -^^ = 6cx — Sd. dx 11. y = x^+\ ^ = (a + b)x^+''-\ ctx FORMULAS FOR DIFFERENTIATION 121 12. y = x'^ -\- nx -\- n. -^ = nx'^~^ -{- n. dx 13. f(x)=lxf-'fx^ + 5. f'(x)=2x^-3x. 14. f(x) = (a + b)x^-{-cx-]-(l f(x) = 2{a + b)x + c. 15. — (a + i>^ + c^^) = b +2cx. dx^ ^ 16. — (5 ^"* - 3 ?/ + 6) = 5 m?/"*-i - 3. 17 v = !^. ^ = _Mo. 18. ^ = t'o+/^. 5=/. 19. s = So + Vot + ift'. ^=^Vo+fi. dt 20. /z=l + ^>(9 + c(9-. — = h-h2cO. dO 2\. s = 2f + 3t-{-b. ^ = 4^ + 3. 22. s = a/3 - Z>^- + c. — = 3at^-2bt, dt 23. r = a^l ~ = 2aO, dO 24. r = c^3 + ^6>^ + e(9. — = 3 c^^ _|. 2 d(9 +e. 25. 2/ = 6a;^ + 4:a?^ + 2ici ^ = 21 a?5 + 10 x'* + 3 aji 26. ?/ = V3a.' + A/^ + 1 dy _ 3 ^ ^^ 2V3a^ 3^x^ ^ _ a + bx + CX^ ^hi — n — ^L X dx x^ 122 ELEMENTARY ANALYSIS 1 ■gg x'' — x — x^ + a dy _ 2 x^ + X + 2 x^ — S a 30. 2/ = (2 ar^ + a;2 _ 5)3_ ^ = 6a;(3a! + l)(2a;' + a;2-5)2. 31. f(x) = (a + 6aj2)l y'(^) = ^(^^ _|. ^^f^i, 32. f(x)=(l+4.a:^)(l+2x'). f\x)=4.x(l+3x-^10a^). 33. /(x) = (a + i»)Va-aj. /'(i^) = 2Va — a? 34. /(aJ) = (a+aJ)"»(6+aJ)^ /'(aj)=(a + ^)"*(& + aj)"r-^H — ^1. [_a + a; b + xj 35. 2/ = "- ^=_JL. dy __ a'* + a^o?^ — 4 a?* 37 =_1^ ^^8^V-4j ' *^ 62-aj2' dx (b'-xy 38. 2/ = ^^^=^- ^= — a + x dx (a + xy 39 o.^ ^ ds^Sf+l (1 + 0' ^^ (1 + 0' 41. /(^) = — A^. r(e)= "" Va - bO^ {a - bOf 42. i^(.)=Vr?|- ^'^-)= (i-rvr^ - FORMULAS FOR DIFFERENTIATION 123 *«=(r^)" 43 44. cl>(x) d ny) _ my"" 2a^-l X VI + x^ '1 dx a'y dx ^ X dr _ Va + 3 c<^ d<^ 2V<^ 51. z^ = cc? dv d c 52. p = i2^iiD!. 50. Differentiation of a function of a function. It some- times happens that y, instead of being defined directly as a function of x^ is given as a function of another variable v which is defined as a function of x. In that case y is ^ func- tion of X through v and is called a function of a function, 2v For example, if y- and .l-x\ 124 ELEMENTARY ANALYSIS then 2/ is a function of a function. By eliminating v we may express y directly as a function of x, but in general this is not the best plan when we wish to find -^ • dx Given then y =/('y)? '^ = <)^ (^); and .\y = F(x). To differentiate, chp.nging a; to a? + Ao;, then v and y become V + i^v and y + Ay respectively. The third step will give the values of Ay Av -, Ay — ^, — , and —^^ Av Ax Ax But by direct multiplication Ay Av __ Ay Av Ax Ax Taking the fourth step, we obtain, applying Theorem II, p. 112, after interchanging the members, (^) ^^^ . ^. doc dv doo If y =f(v) ayid V — (x), the derivative of y with respect to x equals the product of the derivative of y with respect to v and the derivative ofv with respect to x, 51. Differentiation of a logarithm. It is required to de- rive a formula for when v is a function of x, and the logarithm is taken in any system whose base is a positive constant a. Let us proceed on the basis of Art. 50, taking at first v for the variable. Let y=log^v. First step. Changing v to 'y + ^^) then y + Ay = log^ (v + Ay). FORMULAS FOR DIFFERENTIATION 125 Second step. Ay = log„ (v + Av) — log„ v. TJdrd step. ^ = log^(v + Av)~log^v ^ Av Av Fourth step. -^ = - ; dv that is, the limiting value of the right-hand member in the third step cannot be found by direct substitution (Art. 41, p. 112). We therefore examine the result of the second step, (2) A2/ = log„(^' + Ay)-log„v, and endeavor to transform the right-hand member. By 15, p. 1, we may write (2) in the form, (3) A.y = log„(^^) = log.(l + ^). Dividing by Av, then (4) f = -llog/l+^' Av Av \ V We observe that the fraction — occurs in the logarithm. V We may introduce this fraction also before the logarithm by multiplying and dividing by v. This gives (5) ^^l.JLlog/l + A!!- Av V Av \ The factor — in front of the logarithm may be written as an Av exponent on the parenthesis (16, p. 1), and hence (6) ^ = -l«g» Av V 1 + ft] Consider now the expression within the square brackets* If we set 2; = — , this will have the form V (7) (l+z){ 126 ELEMENTARY ANALYSIS The fourth step requires the value of limit 2 (8) z = 0(l + z)% since obviously, when Av = 0, also z = — = 0, the value of v V remaining constant in differentiation (Art. 41). Direct sub- stitution of z = in (7) gives, however, the meaningless ex- pression 1*. It then becomes necessary to assume values of z as close to zero as we choose and calculate the value of (7). The limiting value thus obtained, for example, by setting ^ = ^j TTT? T"o7? TO" cm? ^^^-y is the number e (Art. 33, p. 79), the natural base ; that is,^ as a definition^ we set limit 2 (9) e = z = 0(l + zy. The calculation of e to two decimal places is easy. For ex- ample, using a seven-place table of common logarithms, we find the values ^ — ^ TTF TTTO" T170^7 1 (l-{.zy = 2 2.69 2.70 2.71 Eeturning now to (6) and letting Av approach the limit zero, we obtain the required result (10) J = -log.e. dv V Since 'y is a function of x and it is required to differentiate log« V with respect to a?, we must use formula {A), Art. 50, for differentiating a function of a function ; namely, dy _dy ^ dv dx dv dx * The number 7r(= 3.1416), with which the student is familiar in geometry, is defined also as a limit ; namely, in a fixed circle, tt is the limiting value of the ratio of the perimeter of an inscribed regular polygon to the diameter, when the number of sides increases indefinitely. FORMULAS FOR DIFFERENTIATION 127 Substituting the value of -^ from (10) , we get, since y = log« v^ dv doc V ddo When a = e this becomes since log^ e = 1 (19, p. 1). In VIII a, the natural base is omitted in writing down log v ; that is, when no base is indicated it is assumed that natural logarithms are used. Putting a = 10, VIII becomes (11) — logio V = ^^^^0^ dv^Md^^ dx V dx V dx' where M (Art. 33, (6)) is the modulus of the common system. Formula VIII a justifies the introduction of the number e, for in the Calculus the use of natural logarithms renders unneces- sary writing down M, and this results in a great saving of labor. The derivative of the logarithm of a function equals the product of the moduhis of the system, the reciprocal of the function, and the derivative of the function, 52. Differentiation of an exponential function. Let us find the slope at any point (x, y) on the exponential curve (Art. 33) (1) y = e\ Using natural logarithms, then (2) x = \ogy. Differentiating with respect to y, (3) ^ = 1, by Villa, ^ dy y' ^ We wish the value of —' • Now if we had dx dy dx _ ^ 128 ELEMENTARY ANALYSIS differentiated (1) by the General Bide the third step would have given the value of the ratio -^ • Similarly, from (2) we should have found the value of — • But by direct multipli- cation, ^ (4) ^ . ^ = 1 ^ ^ . Ax Ay ' Now take the fourth step. Then we have at once dy dx dx dy or the useful formula ^ ' dy doo When we differentiate (1), we wish the value of -^- From dx (5), and (3), (6) ^ = l^'h = y, , dx dy Referring to (1), we have the formula (7) lie) = e. In other words, the exponential function e"" possesses the remarkable property of being its own derivative. Changing cc to iJ in (7) gives (8) — 6^ = e^ ^ ^ dv . From this result and (A), Art. 50, we derive doc doc The derivative of the natural base with a f auction as exponent equals the whole expression times the derivative of the exponent. To derive IX, make use of the equation FORMULAS FOR DIFFERENTIATION 129 (9) ei^"" = a, which follows at once from the definition of a logarithm (Art. 33, (1)). Hence (10) a'' = e^i^"«. ...Aa^ = Ae-iog« = et'iogaA(^loga)=e^^°s«loga — . dx dx dx dx Hence, substituting from (10), we obtain IX 4^a^ = anoga^^' doo doc 53. Proof of the power rule. We may now complete the proof of VI postponed from Art. 48. Since, as in (9) of the preceding section, then (1) t)'* = e"i«g«. . .. A ^n = A e« log. ^ gn log V A Ui log V) = C^^^^^^ - — • dx dx dx vdx Substituting from (1), this becomes ^ ,n _ ^^^'' ^^ _ n-l ^^ dx V dx dx^ which establishes the formula. EXAMPLES 1. Differentiate y = log Vl — x\ i.(l_a;2)i Solution. ^ = ^ '-— (by VIII a) dx (i_^)i x^-l (1 - xy Ans. 130 ELEMENTARY ANALYSIS 2. Differentiate y — a^\ Solution. ax = 6 oj log a • a^'. Ans, ^ = loga.a^^ — (3a^). dx dx (by IX) 3. Differentiate y = be''+^\ Solution. dx dx , (by IV) dx (by IX a) y = 2 hxe^'+^\ Ans. 4. Differentiate =^oWJ!::- Solution. Simplifying by means of 17 and 15, p. 1, 2/ = i[log(l + a;'0-log(l~a^^)]. d^^l dx 2 ■|(i+.=) |(i-^)- 1+x^ l-\-x^ + l-a;2 2a; (by VIII a, etc.) aj2 1- Ans, PROBLEMS Differentiate the following: 1. 2/ = log(i» + a). 2. 2/ = log(aic+ &). 1+x 3. 2/ = log 4. 2/ = log 1-x l-M" 1-a^* dy _ 1 c^a; x+ a dy_ a dx ax-{-b dy _ 2 dx 1-a^ dy _ 4a; c?a; 1 — a;* FORMULAS FOR DIFFERENTIATION 131 fill 5. 2/ = 6"^ 6. 2/ = e^*+^ 7. y = \og{x^ + x), 8. y = \og{Qi^ — 2x-\-5). 9. 2/ = log,(2a; + i«3). 10. y — x\ogx.: 11. /(a?) = log aj^. 12. f(x)z=\og^x. JO • ^m^. log^x = (logx)3. Use fii-st VI, v=logx, ^=3; and then VIII. dx = ae"". dx ,4g4x+5_ dy _ 2a; + l dx aj^ + a; dy_ 3*2-2 dx ar'-2a; + 5 dy _ dx , 2+3a^ = ^°^»'-2. + x^ dy _ dx = loga5+ 1. f\^)= 8 a; /'(«')= _31og2x 13. /(a;)=log^±^. a — X •^«=„-i^-- 14. f(x) = log (ic + Vl + or). f{x) = ^ ^ . Vl+aJ^ dy _ log a • a^^'e* cZaj ^ = 2a;log6.^>^^ dx ^ = 2log7 '(x + l)7^'+^\ dx ^ = -2x\ogG'e''-^\ dx 132 ELEMENTARY ANALYSIS 19. r = a^, 20. r = a^°s^. 21. s = &''^'\ 22. u = ae'^~\ 23. prrze*i«g«. 24. — [e^(l-.T2)-]^gx(i_2a;-ic2)^ 25. A^^^-l^ 2e^ = a^ log a. dr _ a^^^^loga d(9 = 2 ^e^'+'l du _ _ ae""^ dv 2Vv dp _ dq = e^'^^^(l + logq). 26. — (^V^) = :»e"^(aa; + 2). tf - -'^^i+e ,x 28. y^ -'/-' 29. y = e" -- e-' e 4- e-^ 30. y = = a?"a^ ^. f/a: 1 + e* dx 2^ ^ dij _ 4 -^ = a*a?"-i (n + a; log a). dx 64. Differentiation of sin v. Let us work out d . — sm ^', dv taking, as indicated, v for the variable. Let y = sin v. FORMULAS FOR DIFFERENTIATIOIST 133 First step. Changing 'y to 'y + Av, then y-^Ay = sm(v + Av). Second step. Ay = sin('y + Av) — sin v. If we proceed without transforming the right-hand member, we shall encounter the same difficulty as at the beginning of Art. 51. Applying formula 42, p. 4, by assuming A = v -\- Av, B = V, and /. K^ + J5) = ^ + i Av, ^{A -B) = ^ Av, we obtain . Ay = 2 cos (v-{- ^ Av) sin -J- Av, Third step. -^ = cos (v + i- Av) • '^ — , ^ Av V ^^ / ^Av in which we have written Av beneath the second factor and multiplied numerator and denominator by -|-. If we now let A^' approach the limit zero, and apply III, p. 112, we see that the first factor approaches cos 'y as a limit, but that limit sin ^ Av Av = 1 ^y (1) cannot be found by direct substitution (p. 112). Let us there- fore calculate the value of sin a; y = — , into which (1) transfoi^ms if i» = i Av, for small values of x. Referring to the three-place table of Art. 4, and choosing angles less than 10°, we see that sin x and x (radians) are equal to three decimal places. That is, for small values of x, y equals unity very nearly. We therefore infer that /o\ limit sino^ ^ lU ELEMENTARY ANALYSIS The proof is the following. In the figure, let x = 2iTG AM = dire AM \ the radius J. being taken equal to unity. Then \^ sinx = FM=FM', tan X = TM= TM'. Now M'M< arc M'M< M'T+ TM. . •. 2 sin x<2x <.2 tan x ; whence, dividing through by 2 sin x, ^ 'X' tan X sm X sm x\ cos a? Therefore, taking reciprocals , sm X cos 0? < - <1. X Now let cc approach zero as a limit ; then, since cos = 1, and the value of smi« lies between 1 and cos a?, we must have limit sm x x = = 1. X The result just established enables us to complete the differ- entiation. Fourth step. dy -^ = cos V. dv If 'y is a function of x, then by (A), Art. 50, dy dv dx dx and we thus obtain formula doc doc The statement of the corresponding rules will now be left to the student. FORMULAS FOR DIFFERENTIATION 135 55. Differentiation of cos i;. Let y = cos v. By 31, p. 3, this may be written Differentiating by formula X, dy fir -^ = oos - dx \2 :COS jdx\2 J TT \f dv\ dv = — sm V — dx fsince cos ( - — ?; j = sin r?, by 31, p. 3- XI .-. :^(cosi;)--smt;^ dx dx 56. Differentiation of tan v. Let y = tan v. By 27, p. 3, this may be written sin v y = COS'V Differentiating by formula YII, cos V -r- (sin v) — sin v -zr- (cos v) dy dx^ ^ dx^ ^ dx cos'^v cos^-y dv . „ dv -— + sm^ v ^r- dx dx COS^ V dv; dx cos^'y o dv = sec-'y^^- dx 136 ELEMENTARY ANALYSIS XII /. ^ (tan V) = sec2 v ^ • doc doc 57. Proofs of XIII-XV. By setting (using 26 and 27, p. 3) , cosv 1 1 coti?=-^ — , sec'y = , csci;: sm V cos V sm v and applying VII, X, and XI, we easily prove the three formulas in question. Details are left to the student. PROBLEMS 1. y = tan Vl — x. ^ = sec^ Vr=^ A(i-x)i (by XII) dx dx [i? = Vl — X.] : secVl - X . i(l - ajp(- 1) sec^ Vl — i 2Vl-aj 2. 2/ = cos^a;. This may also be written y = (cos xy. | = 3(cosx)^-|(cosa,) (by VI) [v = COS a: and w = 3.] = 3 cos^ cc ( — sin x) (by XI) = — 3 sin X cos^ a.\ 3. 2/ = sin naj sin"" cc. -^=^irinx~(8mxy+sm-x — (smnx} (by V) [z^ = sin fix and v = sin^x.] = sin nx • n (sin a;)"-^— (sin ic) + sin" x cos 7i.'r— (ncc) CIX (X,X (by VI and X) FORMULAS FOR DIFFERENTIATION 137 = n sin nx • sin*""^ x cos x-\-n sin"" x cos nx = n siii*""^ X (sin ?iaj cos ic + cos nx sin a?) = n sin**"^ a; sin (n + 1) cc. 4:. y = sec aa;. ^- = ^ sec ax tan ao?. dv 5. ?/ = tan(aa^+&). — = asec^(aa; + 6). Ctaj 6. '?/ = sin^a?. -^ = sin 2 a;. •^ da; 7. y = cos^ a:^. ^ z= — 6x cos^ a;^ sin a?^. ^ dx 8. /(?/)= sin 2 2/ cos 2/. f\y)='\LMOs2ycosy. 9. i^(a;) = cot^ 5 a;. F'(x) = 10 cot 5 x cosec- 5 a?. 10. F{0) = t2in0-0, i^'((9) = tan2|9. 11. /(<^) = <^ sin (^ + cos <^. /'(<^) = <^ cos dr 1 1 + r' X VI- ^ dp dq' gtan~^g du 2 2 12. 2/ = a^c sin (sin a?). 4 sin ic 13. ?/ = arctan 3+5 cos aj 14. v=arctan?4-log\/' dv e - e-^ dy _ dx = 1. dy 4 dx 5+3 cos X dy _ 2 ax' dx x^-a"" FORMULAS FOR DIFFERENTIATION 143 15. 2/=log^l±^Y--arctana.. ^ = ^. dy _ X arc sin a; 16. ?/ = Vl — x^ arc sin x—x, J~ ~" , , 17. 2/ = arc sm — c7?/ _ 2 710?""^ ic-" + l ^a; a,^^^+l ^^ d —\dv 18. — arc cos v = — m^ziiz — • dx ^1 _ ^^2 c^a? , ^ d , — 1 c??; 19. — arc cot 'y = — dx 1 4- ^!'^ dx 20. — 860-^?; = dx 'y V'y^ —.\dx 60. Implicit functions. Required the slope of the curve (1) 0^ - 2 2/' = 9. If we solve for y^ we shall obtain ?/ as a function of x^ namely, (2) y^4 fa^-9 Instead of solving for ?y, however, we may in (1) regard y implicitly as a function of x, and differentiate directly. Thus, by III, we have from (1), Remembering that y is & function of x, then fj^^f)='^^/ = ^y% (by VI). Hence (3) gives (4) 2a.-4^^ = 0,or^ = -^. da; dx 2 y 144 ELEMENTARY ANALYSIS To see that this result is the same as would be obtained by differentiating (2), using VI in (2), ,.. dy l/aj2-9V* dfx'-^\ X x . ,^. ^'^ rx=2[-^) 'rx{r2-rrj^_=2y^^^^'^' The general conclusion illustrated by this discussion is the following : When the equation of a curve is given in unsolved form, either coordinate may he considered an implicit function of the other. We may then differentiate directly, and solve for the desired derivative. Thus, to find -^ from dx a?-3xy-^2y^ = 3. Then -f {x') -3^(xy) + 2^ {f) = -f 3. dx dx dx dx ,'.2x-3(y^-x^\-\-4.y^ = 0, (V and VI) y dxj dx c, T . dy _2x ^' dx 6x -4.y' PROBLEMS Differentiate the following : 1. y^ = 4:px. dy _2p ^ dx y 2. a^ + y'' = r\ dy _ X dx y 3. bV + ay = a'b\ dy __ Wx ^ dx a^y 4. f-3y-{-2ax = 0. dy 2 a dx 3(1- f) FORMULAS FOR DIFFERENTIATION 145 5. xl+y^ = a^. ^ = ^Jy. rim* ^ /v» dx 6. a;« + r = ai ^ = -Ji. dx ^x 8. /-2a;«/ + 62 = 0. 'M=jy_. dx y — X dx y^ — ax dx x^ — xylogx Hint. Take the logarithm of both members. 11. p^ = a^cos2^. dp^_ a'sm2$ ^ de p 12. p' cos e = a' sin 3 0. d^^'Sa'cosSO + p'sm 0_ de 2 p cose 13. cos {uv) = cv. dM^ c + usm{^iv) ^ dv V sin (uv) 14. ^ = cos(^4-' 2 2 ^ sq. in. It is easy to see that the rectangle .is now a square. For when x = — -, the altitude = ^25 — x^ = — - (ry w\ = x V2 V2 ^^* ^ Hence this example proves the result : — Of all rectangles which can be inscribed in a given circle, the square has the greatest area. PROBLEMS In the following problems the student will work out the functional relation, and examine this for maxima and minima. 1. Eectangles are inscribed in a circle of radius r. Ex- amine the perimeter P of the rectangles as a function of the breadth x. Arts. Max. for x — r -\/2, 2. Right triangles are constructed on a line of given length li as hypotenuse. Examine (a) the area A^ and (6) the perimeter P as a function of the length x of one leg. Ans, (a) Max. for x = ^ ^ V2. (b) Max. f or aj = i 7i V2. 3. Eight cylinders are inscribed in a sphere of radius r. Examine as functions of the altitude x of the cylinder, (a) volume V of the cylinder, (b) curved surface S. Ans. (a) Max. for x = ^r V3. (b) Max. for x=zr V2. 154 ELEMENTARY ANALYSIS 4. Eight cones are inscribed in a sphere of radius r. Ex- amine as functions of the altitude x of the cone, (a) volume V of the cone, (b) curved surface S, Ans. (a) Max. if a; = | r. (b) Max. if a? = | r. 5. Right cylinders are inscribed in a given right cone. If the height of the cone is h, and the radius of the base r, ex- amine (a) the volume Fof the cylinder, (b) the curved surface S, (c) the entire surface T, as functions of the altitude x of the cylinder. Ans. (a) Max. ii x = ^ li. (b) Max. ii x=^ 1i, 6. Right cones are circumscribed about a sphere of radius r. Examine as a function of the altitude aj of the cylinder, the volume Fof the cone. Ans. Min. if 0^ = 4 r. 7. Right cones are constructed with a given slant height L. Examine as functions of the altitude x of the cone, (a) the volume V of the cone, (b) the curved surface S, (c) the entire surface T. Ans. (a) F=Max. if a7 = iX V3. (b) ISTeither. 8. A conical tent is to be constructed of given volume V. Examine the amount A of canvas required as a function of the radius x of the base. Ans. Min. if x = -|- V2 altitude. 9. A cylindrical tin can is to be constructed of given volume V. Examine the amount A of tin required as a func- tion of the radius x of the can. Ans. Min. if x = altitude. 10. An open box is to be made from a sheet of pasteboard ) 12 in. square, by cutting equal squares from the four corners and bending up the sides. Examine the volume V as [ a function of the side x of the square cut out. Ans. Max. if a? = 3. 11. The strength of a rectangular beam is proportional to the product of the cross section by the square of the depth. Examine the strength /S as a function of the depth x for beams which are cut from a log 12 in. in diameter. A71S. Max. if X = 6 V3. MAXIMA AND MINIMA 155 12. A rectangular stockade is to be built to contain a cer- tain area A. A stone wall already constructed is available for one of the sides. Examine the length L of the wall to be built as a function of the length x of the side of the rectangle parallel to the wall. A71S, Min. if x = twice other side. 13. A tower is 100 ft. high. Examine the angle a sub- tended by the tower at a point on the ground as a function of the distance x from the foot of the tower. Ans. Neither. 14. A tower 50 ft. high is surmounted by a statue 10 ft. high. If an observer's eyes are 5 ft. above the ground, ex- amine the angle a subtended by the statue as a function of the observer's distance x from the tower. Ans. Max. if a; = 10 V30. 15. A line is drawn through a fixed point (a, b). Examine, as a function of the intercept on XX' {=x) of the line, the area A of the triangle formed with the coordinate axes. Ans. Min. when x = 2 a. 16. A ship is 41 mi. due north of a second ship. The first sails south at the rate of 8 mi. an hour, the second east at the rate of 10 mi. an hour. Examine their distance d apart as a function of the time t which has elapsed since they were in the position given. Ans. Min. if t= — 2. 17. Examine the distance e from the point (4, 0) to the points (x, y) on the parabola ?/^ = 4 cc. Ans. Min. if a; = 2. 18. A gutter is to be constructed whose cross section is a broken line made up of three pieces each 4 in. long, the middle piece being horizontal, and the two sides being equally inclined. Examine the area ^ of a cross section of the gutter as a func- tion of the width x of the gutter across the top. Ans. Max. for a? = 8. 19. A Norman window consists of a rectangle surmounted by a semicircle. Given the perimeter P, examine the area A as a function of the width x. Ans. Max. when x = total height. 156 ELEMENTARY ANALYSIS 20. A person in a boat 9 mi. from the nearest point of the beach wishes to reach a place 15 mi. from that point along the shore. He can row at the rate of 4 mi. an hour and walk at the rate of 5 mi. an hour. The time it takes him to reach his destination depends on the place at which he lands. Ex- amine the time. Ans. Min. if he lands 3 mi. from the camp. 21. The illumination of a plane surface by a luminous point varies directly as the cosine of the angle of incidence, and in- versely as the square of the 'distance from the surface. Ex- amine the illumination /of a point on the floor 10 ft. from the wall, as a function of the height a; of a gas burner on the wall. Ans, Max. if aj = 5 V^. 22. The sides of a quadrilateral are given in order and length. When is the area a maximum ? 23. Examine the functions of problems 22-30, p. 99, for maxima and minima. €3. Derivatives of higher orders. Since the derivative of a function of a variable x with respect to x is also in general a function of x^ we may differentiate the derivative itself, that is, carry out the operation, This double operation is indicated by the more compact notation, and this new function is called the second derivative. In the same way, ^ ^2 ^3 f(x) = — f(x) dx dx"^ ^ ^ dx^^ ^ ^ is the third derivative, and in general, dx-^ ^ ^ MAXIMA AND MINIMA 157 is the nth. derivative of f{x), that is, the result of differentiat- ing /(a;) n times. The following notation is also used: 1/(0.) =/'(^), £jix)=r(x), ..., £./(«,)=/(».(«,). The operation of finding the successive derivatives is called successive differentiation. For example, given f(x) = 3 o^^ — 4 o?^ + 6 a; — 1, then /(a;)=12ar^-8a; + 6; /'(a^) = 36a;'^-8, etc. If the independent variable is y, then the second differentia- tion gives the value of d fdy\ _ d'^y dx\dxj ~~ dx^^ the abbreviation indicated being that commonly employed. The symbol — ^ is read, " the second derivative of y with respect (JjX" X." isimiiariy, d (d'y\_d'y dx\dxy~da^' the third derivative. etc. Thus, given y = x^ — 4:X^. Then ^ = Sx^-8x; dx 2 = 6^ 8; d3^ ^ = 6; d3i> fy = 0; etc. dx* J y^j - (1- -xf r\y) = = 16. d*y_ 6 dx* X dhj _ _ 11 (n + l)c 158 ELEMENTARY ANALYSIS PROBLEMS Differentiate the following. r^ 14 3. /(^)=/. 4. y=^oi? log X, G 5. V "= — • 6. ?/=(aj-3)e2^ + 4a;e^ + a;. ^ = 4e^[(a;- 2)e* + a; + 2]. cix 8. /(«) = a*- + &« + c. /"'(a;) = 0. 9. f(x) = log (a; + 1). /' (x) = - -^^ . 10. /(a;) = log if + e-). /" (a;) = - ^/f~g " 11. r=sina^. — - = a^ sin a^ = aV. 12. r = tan^. -^ = 6 sec^ <^ - 4 sec^ <^. d^ dh d^ 14. /(^) = e-' cos ^. /^(f) = — 4 e-*cos ^ = — 4/(f). 15. /(^)=Vse^T^. r(^)=3[/(^)]^-/(^). 13. r = logsin<^. — ^ = 2 cot (^ cosec^ ^. 64. Geometrical significance of the second derivative. It has been observed that the value of the first derivative of a MAXIMA AND MINIMA 159 function determines the slope of the graph. Moreover, we have seen, in Art. 38, that (i) if the derivative is positive^ the function increases as the variable increases ; (ii) if the derivative is negative , the function decreases as the variable increases. If we consider the appearance of a curve in the neighborhood of one of its points P, then clearly two cases are distinguishable : (a) The curve is above the tangent at P, or is concave upward; (h) The curve is below the tangent at P, or is concave down- ward. How are these cases distinguishable analytically ? Consider (a). As we approach P from the left, that is, with increasing values of x, we observe from the fig- ure that the inclination a of the tan- gent increases (a' > a). That is, in (a), the inclination a increases as x increases. Hence by (i), if — is pos- itive at P, the curve is certainly con- cave upward. Consider (b). Here, a decreases as X increases («' X' < 0, the curve is concave downward. Summing up, if (A) — > 0, curve is concave upward; — < 0, concave downward, dx dx N"ow, \if(x) is the derivative of the function /(o:^), then (2) f\^) = tan a, or a = arc tan /' (x). 16Q ELEMENTARY ANALYSIS Differentiating with respect to x, using XVII, ,ox d(._ dx^^^''^ ^ r(x) ^^ dx l-{-f\x) l+f\xy where /" (a?) = second derivative oif{x). Hence (3) becomes da_ f\x) (4) dx i-vr\x) da Since the denominator l+/'^(a?) is always positive, — has the same sign as/"(a;), and conversely. Hence the result : second derivative > 0, curve concave upward ; second derivative < 0, curve concave downward. As an example, consider this question : EXAMPLE Is the curve y = x log x concave up- ward or downward at i» = 1 ? Solution. Differentiating twice, dy dx dry dx^' = logx + 1, 1 ' X d'y Hence when x = l, -~ = 1 = a positive number. (a/X'^ Hence the curve is concave upward at a^ = 1. dy~ Erom the first derivative, dx : log 1 + 1 = 1. Also from the equation of the curve, when x = 1, y = log 1 = 0. Arranged in a table, we have the results set down. The appearance of the curve, in the neigh- borhood of x = l, is now determined, and we may construct a small arc of the curve at that point, as in the figure. X y dy dx dx-' 1 1 + MAXIMA AND MINIMA 161 PROBLEMS 1. Determine the direction of curvature (concave upward or downward) of the following curves at the points indicated. Draw a figure in each case. , ^ . TT 3 TT 5 TT (a) \j=^\iix, ^ = 1^ ~^^ -2' (h) y = 3x-a^, x = l,2, -3. (c) y = e-% x = 0, -1,2. (d) y = xe'"", x = 0, 3. (e) y = e"* cos x, x = 0. (/) y = log{l + x),x = 0,2. (g)y = i±^^ .^ = 1,-1,2. ' x 2. Show that each of the following curves is always concave upward or downward. (a) yz=alogx, (b) y = Ax — x^. (c) y = ae~''. 65. Second test for maxima and minima. At a high point, the graph is concave downward; at a low point, concave upward. The directions on p. 152 may therefore be stated in another form, the first and second steps, however, remain- ing unchanged. Determination of maxima and minima — second method. First step. Same as on p. 152. Second step. Same as on p. 152. Third step. Consider any critical value not excluded by the conditions of the problem, and substitute this in the second deriva- tive of the function. If the residt is negative, the function has a maximum value ; if positive, a minimum value ; if zero, the method of p. 152 must be resorted to. 162 ELEMENTARY ANALYSIS Clearly, this second method should be used only if the second derivative is easily obtained. For example, the method of p. 152 is preferable for the problem solved on p. 152. On the other hand, if we turn to the problem on p. 94, equation (2), X the differentiation is simple, namely, d?M ^^ 864 We may conveniently test the critical value x=:6 ( = for x = 6 \dx d?M~ by substitution in the second derivative. This gives — — dx-- = a positive number. Hence Jf is a minimum when x= 6. In each of the following problems the function is easily differentiated and the second method should be adopted. PROBLEMS 1. A circular sector has a given perimeter. Show that the area is a maximum when the angle of the sector is 2 radians. (Area of a sector equals ^ arc times radius.) 2. A Gothic window consists of a rectangle surmounted by an equilateral triangle. The perimeter is given. What must be the proportions in order to admit as much light as possible ? 3. Apply the second test for maxima and minima to problems 3 (a), 4 (a), 5, 6, 7 (a), 9, 10, 12, 14, 15, and 19 of Art. 62. 66. Points of inflection. The problems on p. 110 illustrate the aid afforded by the use of the first derivative in obtaining MAXIMA AND MINIMA 163 an accurate plot. The second derivative is also of service in this connection. For if we know the sign of the second de- rivative at any point on the curve, we then know if the curve at that point is concave upward or concave downward (p. 159), and we therefore know if the curve at that point lies above or below the tangent. Suppose we have a continuous curve (as in this figure) such that in passing along it from A to (7, the curve changes from concave upwards to concave down- wards. Then a point B must exist such that to the left of B, the curve is concave upwards; to the right of B, the curve is concave downwards. The point B is called a point of inflection. The second derivative (f"(x) or -^ j is positive 2it edich point of the arc AB, and negative at each point of the arc BC. Hence at B the second derivative is zero. (J5) At points of inflectioD, f"(x) or ^^ equals zero. Solving the equation resulting from (B) gives the abscissas of the points of inflection. To determine the direction of curving in the vicinity of a point of inflection, testf"(x) for values of x, first a trifle less and then a trifle greater than the abscissa at that point. Iif"(x) changes sign, we have a point of inflection, and the signs obtained determine if the curve is concave upwards or concave downwards in the neighborhood of each point. The student should observe that near a point where the curve is concave upwards (as at A) the curve lies above the tangent, and at a point where the curve is concave downwards (as at C) the curve lies below the tangent. At a point of in- flection (as at B) the tangent evidently crosses the curve. 164 ELEMENTARY ANALYSIS PROBLEMS Examine the following curves and graphs for horizontal tan- gents, points of inflection, and direction of curving. 1. y = 3x^ — 4:X^ + l. Solution. f{x) = 3x^-4.0^ + 1. Hence f(x) = 12c^-12x' = 12x\x-l). .*. f'(x) = when x = 0, ov x = 1. Differentiating again, f\x) = S6x'-24.x. Using (B), S6x'-24:x = 0. x = i and x = 0. Factoring, f\x) = 36x{x-^). When x<0, f" {x)=+', and when x>0, /' {x)=-. .*. the curve is concave upwards to the left and concave downwards to the right of x = (A in figure). When X < |, /" (x) = —- and when x>^,f" (x) = + . .'. the curve is concave downwards to the left and concave upwards to the right ol x = ^ (B in figure). The curve is evidently concave upwards everywhere to the left of A, concave downwards between A (0, 1) and B (|, ^), and concave upwards everywhere to the right of B. In work of this kind it is well to tabulate the results which aiford a check on the plot. We follow the plot from left to h^ght, and choose for the initial value of a; a value to the left of fT the least root of f'(x) = and f"(x) = 0. The table should in- clude all critical values of x (f'(x) = 0) and also those for which /"(aj) = 0, that is, values determining the points of inflection. Furthermore, intermediate values of x must be included, as in the following table, which the student should carefully study. MAXIMA AND MINIMA 165 X y d'y Eemarks ~1 i 1 1 2 8 1 17 24 -¥ 48 + + + Concave upward f Point of inflection [ Horizontal tangent Concave downward 1 Point of inflection [ Slope of tangent = — ^^- Minimum value of y Concave upward 2. y-- 3. y = 5 — 2x — oi:^. 4. y = 0/^. Ans. Concave upwards every- where. Ans. Concave downwards every- where. Ans. Concave downwards to the left and concave upwards to the right of (0, 0). Ans. Concave downwards to the left and concave upwards to the right of (1, - 2). Ans. Concave downwards to the left and concave upwards to the right of (a, b). Ans. Concave downwards to the left and concave upwards 4a '3 8. x^ — 3 hx^ + a?y = 0. Ans. Point of inflection is 5. ?/ = a^-3a^-9aj + 9. 6. y=za-\-(x— by 7. a-y = ax^ -f 2 a^. ^3 to the right of [ a, 9. y = x^. '•f Ans. Concave upwards every- where. 166 ELEMENTARY ANALYSIS 10. Soc^ — d 0:^ — 27 x + SO. Ans, x = — l, gives max. = 45 ; ives mill. = — 51. x = 3j g 11. 2x^-21x^ + 36x--20. Ans. x = l, x = 6,g: Ans, x = ly g x = 3,g: 13. 2a^-15x^ + 36x + 10. Ans. x = 2, g x.= 3, g 14. a^-9x' + 15x-3. 12.. t-2x^ + 3x-\-l. 3 15. a^-3x^ + 6x-it-10. 16. o(^ — 5x^-\-5x^-{-l. Ans. x = l, g '- ^ = 5, g- Ans. No max. or min. ves max. = — 3 ; ves min. = — 128. ves max. =|-; ves min. = 1. ves max. = 38 ; ves min. = 37. ves max. = 4 ; ves min. = — 28. Ans. x = l, gives max. = 2 ; x = 3, gives min. = — 26; x = Oj gives neither. CHAPTER XI RATES 67. Velocity and , p q acceleration. Con- oj x ' Xx ^ x sider a moving point on the a;-axis. Its distance from the origin (= x) is a function of the time. That is, symbolically, (1) x=f{t). Suppose that P is the position when the time is t seconds, and suppose, further, that the elapsed time from P to Q is A^ seconds. Also let PQ = Ax, Then the quotient Ax (2) — = average velocity of the motion from P to Q. Ax The velocity at P is the limit of the value of — when Q is taken nearer and nearer to Q ; that is (3) f =irJot=-'-"''^*^ Considered without the idea of motion, equation (1) asserts that the variable x changes as t changes. Then we say (4) — = time rate of change of x. Similarly, if y and z are functions of the time, then -^^ and — are the time rates of change of y and z respectively. The point P may move from the point with constant 167 168 ELEMENTARY ANALYSIS speed Vq. Then the distance moved is the product of -the speed (or velocity) and the time. That is, equation (1) now is (5) X = VqL , If, however, the velocity changes, the time rate of change of velocity is called the acceleration^ and equation (4) asserts that (6") acceleration = -— • ^ ^ dt From (3) we may also write, /r7\ ^ j_' f^^ a OS (7) acceleration = — = — - • ^ ^ dt dt^ Similarly, any derivative may be interpreted as a rate of change. For example, ^ = rate of change of y with respect dx to i;);, if 2/ is a function of x. \ The case frequently arises when in the equation both X and y are functions of the time. If the time rate of change in X I = — ) is known, then — can he found. \ dt) ' dt ^ PROBLEMS 1. A particle slides along the curve, y'^ — ^x = 0, so that it moves in the XX^ direction at the constant rate of 3 ft. a second. How fast is it moving in the FF' direction (a) at any point {x, y) ; (h) at the point (4, 4) ? # Solution. Given — = 3, required -^ • dt dt As y and x in the equation y'^ — 4.x = are functions of t, we differentiate the equation with respect to t. This gives (p. 144), m dt RATES 169 Solving for — , (a) ^ = ? ^ = ?ft. a second ; (6) ^ = ^ = ? ft. a second. ^ ^ dt y dt y ' ^ ^ d^ 4 2 2. Find the rate of change of the volume and of the surface of a sphere with respect to the radius. dv Solution. Let y be the volume and x the radius. Required —• cccc Now yz=-7rci^, .-. -^ = 4 7ra^, 3 dx which varies^as the square of the radius. The rate of change of the surface z with respect to the radius x is found from the • ^^ relation z = 4: ttx^ to be — =8 ttx, and it varies as the first dx power of the radius. 3. An express train and a balloon start at points 12 mi. apart. The former runs 50 mi. per hour, and the balloon rises vertically at the rate of 8 mi. per hour. How fast are they separating at the end of 10 min. ? Solution. Let the figure represent the state of affairs at the expiration of any number (=^) of d minutes. That is, y BD is the path of the balloon, AC \s the path of the train, l^ and y = number of miles traveled by balloon, 0.' = number of miles traveled by train, z = distance apart. Evidently, x, y, and z are all functions of the time. Now get the relation between these variables. Evidently, (1) z'' = f+(x + i2y. Differentiate with respect to t. This gives 2z — = 2y^-\-2(x + 12) — ' dt '^ dt ^ ^ dt 170 ELEMENTARY ANALYSIS ^ ^ dt y dt ^ ^ dt \ This result gives the velocity of separation at any time, ' The prc^lem calls for this velocity after 10 min. Hence x = S^ miles, y = U miles, — = 50, ^ =8. / dt \dt We find from (1) the value of 2; = 20.4 mi. Substituting in (2) we find -- = 50.3 mi. Ans. ^ ^ dt This problem shows that in many cases the following rule should be followed. 1. Draw a diagram to represent the state of affairs at any instant {after t seconds). 2. In this diagram mark the variable elements with x, y, z, etc, 3. Find, the equation connecting x, y, z, etc. 4. Differentiate this result with respect to t. 5. Substitute the given data in this result. 4. The radius of a soap bubble is increasing at the rate of 2 in. a second. How fast is the volume increasing, (a) at any time, (5) when the radius is 3 in. ? Ans. (a) 8 vrr^ cu. in. a second, (b) 72 tt cu. in. a second. 5. Solve the problem similar to the preceding, where vol- ume is replaced by surface. 6. At what point of Ex. 1 are the ordinate and the abscissa , increasing at the same rate ? Aiis. (1, 2). \7. Where in the first quadrant does the arc increase twice as fast as the sine ? Ans. 60°. 8. A man 6 ft. tall walks away from a lamp post 10 ft. high at the constant rate of 4 mi. an hour. How fast does the shadow of his head move ? (Use similar triangles to de- termine the relation between the man's distance from the lamp post and the distance of the shadow.) Ans. 10 mi. an hour. / RATES 171 ^^9. Find the rate of change of the area of a square, when the side x is increasing at the rate of k in. a second. A71S. 2 kx sq. in. a second. /^lO. A ladder 25 ft. long rests against a house. A man takes hold of the lower end and moves it aw^ay at the rate of 2 ft. a second. How fast is the top of the ladder descending when the bottom is 7 ft. from the house ? Ans. 7 in. a second. C^yfi. Two particles start together from the origin along y^ — 9x = and of -{- f/ — 34: x = respectively. The former has a speed in the YY' direction of 3 ft. a second, and the latter of 4 ft. a second. Which goes through the point (25, 15) with the greater speed in the XX' direction ? A71S. One on the parabola. 12. Find how fast the man's shadow in problem 8 is lengthening. Ans. 6 mi. an hour. P 13. A circular plate expands by heat so that the radius in- y creases uniformly at the rate of .01 in. a minute. At what rate is the surface increasing when the radius is 2 in. ? Ans. .879 sq. in. a minute. 14. Water runs into a barrel at the rate of 2 cu. ft. a min- ute, but leaks at the bottom at the rate of 1 cu. ft. a minute. Assuming the barrel to be a right cylinder of radius 1 ft. and of height 5 ft., how long will it be before water runs over the top ? Ans. 7 min. 51 sec. /^15. A boy starts flying a kite. If it moves horizontally at the rate of 2 ft. a second, and rises at the rate of 5 ft. a second, how fast is the string being paid out at the end of 3 min. ? Ans. 5.38 ft. a second. ^^0^^. At what point of the curve a??/ -f 25 = v^ill a particle move in the x and y direction at the same rate? Ans. (— 5, 5). /^^JW^A man standing on a dock is drawing in the painter of a boat at the rate of 2 ft. a second. His hands are 6 ft. above the bow of the boat. How fast is the boat moving when it is 8 ft. from the dock ? Ans. f ft. a second. > \ 172 ELEMENTARY ANALYSIS 18. The volume and the radius of a cylindrical boiler are expanding at the rate of 1 cu. ft. and .001 ft. per minute respectively. How fast is the length of the boiler changing when the boiler contains 60 cu. ft. and has a radius of 2 ft. ? Ans. 0.78 ft. a minute. /: 19. An equilateral triangular sheet of rubber is stretched so that it always keeps its shape, but expands at the rate of 3 sq. a Qiinute. How fast is its side increasing when 4 in. long ? Ans. .866 in. a minute. 20. The rays of the sun make an angle of 30° with the hori- zon. A ball is thrown vertically upward to a height of 64 ft. How fast is its shadow traveling along the ground just be- fore the ball hits the ground? (Use s = ^gf and v = gt,) Ans. 110.8 ft. a second. 21. A man is walking over a bridge at the rate of 4 mi. an hour, and a motor boat passes beneath him at the rate of 8 mi. an hour. If the bridge is 20 ft. above the boat, how fast are they separating 3 min. later? Ans. 8.95 mi. an hour. 22. A ship is anchored in 18 ft. of water. The cable passes over a sheave on the bow 6 ft. above the surface of the water. If the cable is taken in at the rate of 1 ft. a second, how fast is the ship moving when there are 30 ft. of cable out ? Ans. -| ft. a second. 23. A boy rows out for a swim against a tide running in at the rate of i mi. an hour horizontally. He dives off and swims parallel to the coast at the rate of 2 mi. an hour. How fast are he and his boat separating at the end of 15 min. ? Ans. 2.06 mi. an hour. 24. Four men standing 5 ft. from a house are hoisting a piano to the third-story window by means of a block and tackle. If the window is 50 ft. up, and the men pull in the rope at the rate of 10 ft. a minute and back away from the building at the rate of 5 ft. a minute, how fast is the piano ris- ing at the end of the first minute ? Ans. 10.98 ft. a minute. KATES 173 25. Find at what points on y = x^ — 4: the rate of increase of y with respect to x is equal to the rate of increase of x with respect to y. Ans. (± |-, — -V). 26. Assuming the volume of wood in a tree is proportional to the cube of its diameter and that the latter increases uni- formly year by year, show that the rate of growth when the diameter is 3 ft. is 36 times as great as when it is 6 in. 27. A rectangular sheet of metal is subjected to a pressure which expands it only lengthwise and at a rate of 2 in. a min- ute. Find how fast the area of the sheet is enlarging, if it is 7 in. wide. Ans. 14 sq. in. a minute. 28. Water flows from a faucet into a hemispherical basin of diameter 14 in. at the rate of 2 cu. in. a second. How fast is the water rising (a) when the water is halfway to the top, (b) just as it runs over ? (The volume of a spherical segment is l7rr^^ + |-7r/i^) A71S. (a) .013 in. a second ; (b) .005 in. a second. 29. Sand is being poured on the ground from the orifice of an elevated pipe, and forms a pile which has always the shape of a right circular cone whose height is equal to the radius of the base. If sand is falling at the rate of 6 cu. ft. per second, how fast is the height of the pile increasing when the height is 5 ft.? Ans. .076 ft. per second. 30. An aeroplane is 528 ft. directly above an automobile, and starts east at the rate of 20 mi. an hour at the same time that the automobile starts east at the rate of 40 mi. an hour. How fast are they separating in 6 min. ? Ans. 19.92 mi. an hour. 31. A ship is 41 mi. due north of a second ship. The first sails south at the rate of 8 mi. an hour ; the second, east, at the rate of 10 mi. an hour. How rapidly are they approaching each other ? How long will they continue to approach ? Ans. For 2 hr. 174 ELEMENTARY ANALYSIS 32. A railroad train is running along a curve in the form of 2/2 = 500 X, tHe axis of the parabola being east and west. If the train is going due east at the rate of 30 mi. an hour, how fast is the shadow moving along the wall of a station which runs north and south, when the train is 500 ft. east of the wall, provided the sun is just rising in the east? Ans. 15 mi. an Jiour. 33. One side of a rectangle inscribed in a circle of radius 5 cm. expands at the rate of 2 cm. a minute. Find (a) how farSt the area of the rectangle is changing when the above side is 8 cm. ; {h) how long the above side is when the area is not changing. (Get the area as a function of the side.) Ans. (a) — 9J cm. a minute; (h) 3.5 cm. a minute. 34. Work out similar problems, for cylinders and cones in- scribed in a fixed sphere, where the radii of their bases vary. 35. A revolving light sending out a bundle of parallel rays is at a distance of |- a mile from the shore and makes 1 revo- lution a minute. Find how fast the light is traveling along the beach when at a distance of 1 mi. from the nearest point of the beach. Ans, 15.7 mi. per minute. 36. A kite is 150 ft. high, and 200 ft. of string are out. If the kite starts drifting horizontally and away from the flyer, at the rate of 4 mi. an hour, how fast is string being paid out ? Ans. 2.64 mi. an hour. 37. A solution is poured into a conical filter of base radius 6 cm. and height 24 cm. at the rate of 2 cu. cm. a second, Avhich filters out at the rate of 1 cu. cm. a second. How fast is the level of the solution rising when (a) one third of the way up, (h) at the top ? Ans. (a) .079 cm. a second, (b) .009 cm. a second. CHAPTER XII DIFFERENTIALS 68. We have already emphasized that the symbol 4 dy dx was to be considered not as an ordinary fraction with dy as numerator and dx as denominator, but as a single symbol de- noting the limit of the quotient Ao? as ^x approaches the limit zero. Problems do occur, however, where it is convenient to give meanings to dx and dy separately. How this may be done is explained in what follows. Definition, li f (x) is the derivative of f(x)j and ^x is an arbitrarily chosen increment of x, then the differential of f{x), denoted by the symbol df(x), is defined by the equation (A) df(x)=f(x)Ax, If now f(x) = X, then /' (x) = 1, and (A) reduces to dx = Ax, Hence, when x is the independent variable, the differential of x(=dx) is identical with Ax. If y=f(x), (A) may be written in the form (B; dy=f(x)dx.* * On account of the position which the derivative fix) here occupies, it is sometimes called the differential coefficient. ' 175 176 ELEMENTARY ANALYSIS Y V 1 dy y -f^dx-'S y -^^ o y M J M'X The differential of a function equals its derivative multiplied by the differential of the independent variable. Let us illustrate what this means geo- metrically. Let /' (x) be the derivative of y =f(x) at P. Take dx = PQ, then dy =f' (x) dx = tan r • PQ = f^.-PQ=QT. Therefore dy, or df(x)j is the increment (= QT) of the ordi- nate of the tangent corresponding to dx. This gives the following interpretation of the derivative as a fraction : If an arbitrarily chosen increment of the independent variable xfor a point P{x, y) on the curve y =f(x) be denoted by dx, then in the derivative dy dx -f (x) = tan T dy denotes the corresponding increment of the ordinate of the tan- gent to the curve at the point P. The student should observe that, while the differential and the increment of the independent variable are always equal {dx = IS.x), the same is not true of the dependent variable. In the figure, A?/ = increment of 2/ = QP\ but dy = differential of y=QT. 69. Formulas for finding differentials. Since the differen- tial of a function is its derivative multiplied by the differential of the independent variable, it follows at once that the formu- las for finding differentials may be obtained from those for finding derivatives (Art. 42) by multiplying by dx. This gives I d(c) = 0. II d (x) = dx. Ill d(io + v — w) =du + dv — dw. IV d (cv) = cdv. V d(uv)= udv + vdu. VI d(y") = nv''~^dv. via d («") = w«"-i dx. VII jfu\ vdu — udv Vila \cj c VIII d(\og^v) = log^e^. IX d (a") = a" log a dv. IX a d (e") = e" dv. X d(sinv)= cosvdv. XI d (cos v)— — sin v dv. XII d (tan v) = sec- v dv, etc. XVI d fare sin «)= — , etc. 177 Vl-v' The term differentiation also includes the operation of find- ing differentials. In finding differentials, the easiest way is to find the deriva- tive as usual, and then multiply the result by dx. :e:xAMPLEs 1. Find the differential of x-{-3 Solution. ^ . fx-\-3\ ^ (x'-\-'S)d{x^3)-{x^-3)d{x' + 3) ^ \x' + 3y (x" + 3)2 ^ {x'-{-3)dx-{x-\-3)2xdx ^ (3 - 6.-^ - x") dx . {x^ + 3f {x' + Sf * ^^' 178 ELEMENTARY ANALYSIS 2. Find dy from Solution. 2Wxdx-2a^ydy=-^. ,'. dy = ---dx. Ayis, o?y 3. Find dp from 9" Solution. 2 pclp = — a^ sin 2 ^ . 2 dO. , a? sin 2 6 -.^ . • . dp = dd, 9 4. Find d [arc sin (3^-4 f)']. Solution. d[arcsin(3^-4f^)]^ ^^ ^~^^'^ = ^^^ . Ans. Vl-(3^-4^/ Vl-^' PROBLEMS Differentiate the following, using differentials. 1. yz=za^ — hx'^+cx + d. dy = (S ax^ — 2 bx + c)dx, 2. y= 2x^—^ x^ + 6aj-H5. dy={5x^ — 2xr^ — 6 a;-^)^^!. 3. 2/ = (a^ — i^)^ c?2/ = — 10 x{ar — a^^^c^aj. 4. ?/ = Vl + aJ^. (^V = — =^=zz dx. ^ Vl+a;2 s- y—-7z — ^* ' ^y= 7, — .dx. {i-^-x'y ^ (i + x''y+'^ 6. ^ = logVl-.?. ^^ = 2(^ir3r)- 7. ?/ = (e^ 4- e-'^y. dy = 2(e2^ - e-^'')dx. 8. y — e^ log ^, d^ = e'l log cj; + - Ic^ic. 9. 8 = ^- "- ~^ . ds=#(^^?^ ^'Vd^. DIFFERENTIALS 179 10. p = tan <^ + sec <^. c?p = i-±^i5_^ dd). 11. r = i tan^(9 + tan (9. dr = seG^edO, 12. f(x) = (log xy, f(x)dx = ^(M^^ . X 13. cl>(t)= ^3. ^'(^C?^: ^^'^^ 70. Infinitesimals. An infinitesimal is a variable whose value decreases numerically and approaches zero as a limit. In Art. 68 it has been shown that the differential and incre- ment of the independent variable are identical. Equation (B) of that section defines the differential of the dependent variable. Clearly, if dx is an infinitesimal, then also dy is an infinitesimal. In the Integral Calculus, we have to do with expressions of the form cj)(x)dx, called ^* differential expressions.^' Then, if dx is an infinitesimal, so also is the product (x)dx. CHAPTER XIII INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY FORMS 71. Integration. The student is already familiar with the mutually inverse operations of addition and subtraction, multi- plication and division, involution and evolution. From the Differential Calculus we have learned how to calculate the derivative f{x) of a given function f{x), an operation indi- cated by or, if we are using differentials, by df(x)=f\x)dx. The problems of the Integral Calculus depend on the inverse operation ; namely, To find a function f(x) vjJiose derivative (A) f(x)={x) is given. Or, since it is customary to use differentials in the Inte- gral Calculus, we may write (B) df(x) =f(x) dx = (x) dx, and state the problem as follows : Having given the differential of a function, to find the function itself The function f(x) thus found is called an integral of the given differential expression, the process of finding it is called inte- 180 INTEGRATION 181 gration, and the operation is indicated by writing the integral sign ^ j in front of the given differential expression. Thus, (0) jf\x)dx=f{x), read, an integral of f'(x)dx equals f{x). The differential dx indicates that x is the variable of integration. For example, (a) If f{x) = x?, then /'(^) dx = 3 Qi?dx, and I 3 x'^dx = ^x^, (h) If fix) = sin X, then /' (x) dx = cos xdXy and I cos xdx = sin x, (c) If f(x) = arc tan x, then /'(a.*) = — — - , and I = arc tan x. J 1+x^ Let us now emphasize what is apparent from the preceding explanations, namely, that Differentiation and integration are inverse operations. Differentiating (C) gives (D) d Cf'(x) dx = fix) dx. Substituting the value oif{x)dx\_=df{x)'] from (B) in (O), we get {E) fdf(x)=f(x). Therefore, considered as symbols of operation, — and I -^ dx dx J * Historically this sign is a distorted »S, the initial letter of the word smn. Instead of defining integration as the inverse of differentiation we may define it as a process of summation, a very important notion which we consider in Chapter XVI. 182 ELEMENTARY ANALYSIS are inverse to each other ; or, if we are using differentials, d and I are inverse to each other. When d is followed by I they annul each other, as in (Z)), but when I is followed by d, as in (E), that will not in gen- eral be the case unless we ignore the constant of integration. The reason for this will appear at once from the definition of the constant of integration given in the next section. 72. Constant of integration. Indefinite integral From the preceding section it follows that since d (ar^) — 3 xHx, we have I 3 :)(?dx = x^ ; since d(x^+2) = ^ x^dx, we have I 3 xHx = a^ + 2 ; since d{oi? — l) = ^ xHx, we have I 3 x^dx = x^ — 7. In fact, since d(x^+ 0)=3 a^dXy where C is any arbitrary constant, we have C3a^dx = a^+a A constant O arising in this way is called a constant of integration.'^ Since we can give C as many values as we please, it follows that if a given differential expression has one integral, it has infinitely many differing only by constants. Hence ]f(x)dx=f(x)+C', s> and since C is unknown and indefinite^ the expression is called the indefinite integral off'(x)dx. * Constant here means that it is independent of the variable of integration. INTEGRATION 183 It is evident that if <^ (x) is a function the derivative of which is f(x), then {x)-\- C, where C is any constant whatever, is likewise a function the derivative of which isf(x). Hence the Theorem. If two functions differ by a constant, they have the same derivative. 73. Rules for integrating standard elementary forms. The Differential Calculus furnished us with a General Rule for Differentiation (p. 104). The Integral Calculus gives us no corresponding general rule that can be readily applied in practice for performing the inverse operation of integration. Each case requires special treatment, and we arrive at the integral of a given differential expression through our previous knowledge of the known results of differentiation. Integration then is essentially a tentative process, and to expedite the work, tables of known integrals are formed called standard forms. To effect any integration we compare the given differential expression with these forms, and if it is found to be identical with one of them, the integral is known. If it is not identical with one of them, we strive to reduce it to one of the standard forms by various methods, many of which employ artifices which can be suggested by practice only. From any result of differentiation may always be derived a formula for integration. The following two rules are useful in reducing differential expressions to standard forms. (a) The integral of any algebraic sum of differential expressions equals the same algebraic sum of the integrals of these expressions taken separately. Proof Differentiating the expression j dit + j d'y — j div, u, v, w being functions of a single variable, we get du + dv — div. (Ill, p. 113) 184 ELEMENTARY ANALYSIS [1] .', I {du + dv — dw)= jdu+ I dv — j div. (b) A constant factor may he written either before or after the integral sign. Proof Differentiating the expression a i dv gives adv. (IV, p. 113) [2] /. j adv = a\ dv. On account of their importance we shall write the above two rules as formulas at the head of the following list of STANDARD ELEMENTARY FORMS [1] ( {du -\-dv - dw) = \du + \dv — (dw. [2] \adv = a \dv. [3] (doc = oc^€. [4] (v-dv=^^^^-C, n^-1 [5] r^-io^v+c J V = log V + log c = log ev. [Placing C — log c] [6] Ca-dv=-^^^C. •/ log a [7] .(e'^dv = e''+ C. [8] i^invdv =- cos v -\- C. [9] Icos V dv = sin v -^ C. [10] t sec^ V dv = tan v -i- C. mXEGRATION 185 [11] (eo^ed^vdv = — cotv -h C. [12] i^ecvtanvdv = ^ecv + C. [13] icsaveotvdv =— cfiev -\- C. [14] Ttan vdv = log sec v + C, [15] ( cot i; dv = log sin v -\- C. [16] r ^^ ^ =:iarctan-+C [17] r ^^ = arc sin - + C. Proo/o/[3]. Since c^(a^+(7) = da^, (II, p. 113) then i dx = X + C. Proof of [4]. Since d fj^\ = v-dv, (YII, p. 113) \7i + ly n + 1 This holds true for all values of n except n = — 1. For, when n = — l, [4] gives Jv-hlv =-J^^ 4- C=^ + C= 00 4- O, which has no meaning. The case when n = — 1 comes under [5]. Proof of [^5']. Since d (log ^ + C) = — , (YIII a, p. 113) we get I ^ — = log V -{- C. J V 186 ELEMENTARY ANALYSIS The results we get from [5] may be put in more compact form if we denote the constant of integration by log c. Thus, =log V + log C = log CO. V Formula [5] states that if the expression under the integral sign is a fraction whose mimerator is the differential of the de- nominator, then the integral is the natural logarithm of the denominator. PROBLEMS For formulas [l]-[5]. Verify the following integrations. x^dx = — h C, by [4], where v-=^x and ri = 6. 2. j aQ(?dx = a\ Mx (by [2]) = ^+C. (by [4]) 4 3. C{2^-bx'-3x + 4.)dx = C2 xhlx - ("5 xHx - Cs xdx + A dx (by [1]) = 2 Coi^dx - 5 Cx'^dx - 3 Cxdx + 4 Cdx (by [2] ) x' 5x' Sx\ . , ^ • = 2-T— 2- + '"+^- ' Note. Although each separate integration requires an arbitrary con- stant, we write down only a single constant denoting their algebraic sum. = 2 a Cx-^ dx - hfx-'dx + 3 c fa;* dx ^^iJ^T) ^2a.'^-b.^ + ^^+C (by [4]) INTEGRATION 187 5. j2ai^-'dx = '^^-\-a 6. C3 7nz'dz = ^^+a 8. C-V2px dx = I x^Jpx + C, 9. C(ai -x^^dx = a^x - 1- a^aj^ + f a^o;* -~+0. Hint. First expand. 10. f(a^-f)Wydy=2y^(^^-^-fl+^-^-t^+a 11. fC Va - VO'd< = ah-2 ati + ^ " ^ + C' 12. f-^,= inxy + c. (nx) " 13. f(a' + b'l^^xdx = ^"-^ + y)' + a. Solution. This may be brought into the form [4]. For let v = a^-\- bV and n = ^ ; then dv = 2 Irxdx. If we now insert the constant factor 2 6^ before ccda;, and its reciprocal before the integral sign (so as not to change the value of the expres- sion), the expression may be integrated, using [4], namely, / ,,,71 + 1 v'^dv = h C, ?i + 1 Thus, j{a' + 5 V)^- xdx = ^,J{o? + h'x'f 2 ft^^xc^a; = 2^-, J(r/ + h'x'fdia' + ?>V) 262* I "^ 3^2 -+- • iVb^e. The student is warned against transferring any function of the variable from one side of the integral sign to the other, since that would change the value of the integral. 188 ELEMENTARY ANALYSIS 14. j Va^ — x'xdx = — ^(a^ — o^)^ + C. 15. J (3 ax' + 4 6.T^)^(2 ax + 4 /^i^^y^^ ^ 1(3 ^^2 ^ 4 ^^s^| _^ ^^ //m«. Use [4], making v = S ax^ -\- 4 bx^ and w = f . 16. fb(6 ax'+S bx')^ (2 ax+A hx')dx= — . (6 ao^^+S hx^)i + C. J 16 Ifiw*. Write this 1 (a^ -|- ys-y^x'^'dz and apply [4]. 18. f— ^^=_2vr^^+c 21- r|^2 = V^" 10S(^' + '^'«'') + ^- This resembles [5]. For let t? = 6^ + eV ; then dv = 'l e'xdx. If we introduce the factor 2 e' after the integral sign, and before it, we have not changed the value of expression, and the numerator is now seen to be the differential of the denomi- nator. Therefore = |^,log(6'^ + 6V) + a (by [5]) 22. J^^=llog(a^2_i)^l^gC'_l(^g^Y^^31. INTEGRATION 189 23. f^^^P^ = logc{:^-3a'x)l J OCT — 3 a^x ^ J 8 a - 6 &«2 s (8 a - 6 6ir)T2 26. r^=:a;-|' + f-log(a; + l) + G •^ iC -f- 1 Zi o Hint, First divide the numerator by the denominator. 27. p^^^dx = x-\og{2x-3f+a 28. f?— =L^ dx = - log (a;** - 7ix) + C. J X" — nx n 29. I ?^^ ^L-^ = -\-'±. -log 9f ^ a J a + hV' nb ^^ ^ Proofs of [6] and [7]. These follow at once from the cor- responding formulas for differentiation, IX and IX a, p. 113. PROBLEMS For formulas [6] and [7]. Verify the following integrations. /' ba'-dx == -^^ + C. 2 log a Solution. Cba^^dx = bCa^^dx. (by [2]) This resembles [6]. Let v = 2x', then dv = 2dx. If we then insert the factor 2 before dx and the factor ^ before the integral sign, we have, b Ca'^dx = - Ca'-2 dx = - Ca'^d(2 x)=- • -^ + C. J 2 J 2J ^ ^ 2 log a (by [6]) 190 ELEMENTARY ANALYSIS 3. C(e'' + a'^) dx=^ fe'^ + -^\ + G. J, X X e^dx = ne"" + C. 5. Ce^"+^^+\x + 2)dx = ^e^"+'^+^ + a n log a m log o */ 1 + log a (e« + e ")da; = a(e« — e «) + (7. 10. ("(3 e^* - l)h^'dt = i (3 e^* - 1)* + C. Proofs of [8]-[13]. These follow at once from the corre- sponding formulas for differentiation, X, etc., p. 113. Proof of [_U-\, J, , _ rsin^^cg^ __ C — sin vdv J cos'y J cos'y / eg (cos ^) cosv = — log COS v+C (by [5]) = log sec V -\-C. [Since — log cos v = — log = — log 1 + log sec v = log sec v. sec V J T> ^ rr-i'— I r 4- ^ Tcos'yd'y ^(^(sin^?) c/ J sin V J smv = log sin v-{- C. (by [5] ) INTEGRATION 191 PROBLEMS For formulas [8]-[17]. Verify the following integrations. 1. I sm 2 axdx = [- (7. c/ 2a Solution. This resembles [8]. For let v = 2 ax, then dv = 2 adx. If we now insert the factor 2 a before dx, and the factor — before the integral sign, we get 2a J sin 2 axdx = — i sin 2 ax -2 adx 2aJ 1 r 1 = — - I sin 2 ax - d (2 ax) = - — — cos 2ax-\- C ^''^ ^"^ (by [8]) cos 2ax p 2a 2. I cos mxdx = — sin mx + C. J m 5 sec^ bxdx = - tan bx + C 4. r/'cos ^ - sin 3 (9"] d0 = 3 sin ^ + ^ cos 3 (9 + (7. 5. I 7sec3«tan3 acZa = |sec3a4- C 6. I A: cos (a + 57/) dy = - sin (a + by) + C. 7. I cosec^ x^ • cc^dT = — 1 cot a^ + (7. 8. I 4 CSC ax cot aoJcfiK = esc ax + (7. */ a ^ C sin a*da? ^ 9. I = log a -f & cos aj (a -f 6 cos a;)^ 192 ELEMENTARY ANALYSIS 10. I e'^''' sin xdx = — e''^'^ + C. 11- r 2/'' , =-J:taii(a-&a^)+a */ cos^ (a — 6^:1:^) 6 cos (log x) — = sin log x + (7. X 13. f^^ = _ncot^ + a ^ sin^^* ^ 14. I ^^ — ^ ^ — = log (x + sm a;)(7. c/ a; + sm 0? 15. r?iB4^ = sec<^ + a J COS- - 9 X? 3 4 _. i o xctx O • 1 /^ 3. I — =: = - arc sm x- + (7. J VI - «* 2 4- r^ = ^arctan^ + C JV9-4a;2 2 3 6. f—i^^arcsin-^+C. „ r 7 ds 7 . /5 , ^ 7. I — = arc sin-\/~s + C. COS ada 1 = - arc a^ + siii^ a a e*dt t^f'^^a Vl - e'' dx \ a arc sin e* + (7. 194 ELEMENTARY ANALYSIS 10. I "^"^ = arc sin (log a^) + O. ^u + &^ 11. a; Vl — log^ X /du Va2-(^ + 6)^ 12. f -^^^ : arc sm + a arc tan — ^ h (7. 74. Trigonometric and other substitutions. A useful de- vice for integration is afforded by simple transformation of the variable. The following examples illustrate this statement. EXAMPLES 1. Work out /: X (XX Hence f^^^^ f Va^ — 01? Solution. Substitute x=^a sin 0. ,'. dx=a cos dOj V«^ — x^ = -y/ct^ — a^ sin^ = a cos 0. a^ sin^ • a cos • dO a cos a2[|(9-i^sin2(9]. = a^ I SI sin2 (9 dO From the figure, ^ = arc sin -, a sin 2 ^ = 2 sin cos = 2 (problem 16, p. 192) x -y/a? — a? r a?dx ^ ^ Vc6^ — X^ — arc sin -x Va^ — x^ + G. Ans, 2 a 2 Integrals inyolving the radical Va'^ — oc'^ may be worked in this way. INTEGRATION 195 2. Work out Solution. Substitute aj = -. y \ da; = ^. Also Va^-f ^ =\/« +-^ = '^ ° 2/' ^ f y ' J_dy y' ^_ r ?/# l.VaV + 1 -/ VaV + 1* Hence T y = - V ^''V' + ^ (by (4)) a^ >^^l+^ +a ^n.. -PROBLEMS Work out : 1. f ^^ ^_V^'-^ + a rputa.=i.' 2. fVo^^^ dx = ^ Va^^^ + ^ a' arc sin - + (7. c/ 2 2 a 3. r ^^ = - 1 arc sin « + C. [Put a; = ^ 5. r-^^^ = -^-arcsin^ + G 6. rVZEZ^^^_VZEZ_arcsm^ + a • J x^ X a CHAPTER XIV CONSTANT OF INTEGRATION 75. Determination of the constant of integration by means of initial conditions. In order to determine the constant of integration, data must be given in addition to the differential expression to be integrated. Let us illustrate by means of examples. EXAMPLES 1. Find a function, whose first derivative is 3x'^ — 2x + 5y and which shall have the value 12 when x = l. Solution. (Sx^~-2x-^o)dx is the differential expression to be integrated. Thus, C(3x^-2x + 5)dx = x^-x^ + 5x+C, where C is the constant of integration. From the conditions of our problem this result must equal 12 when x = l] that is, 12 = l_l+5 + C, or 0=7. Hence x" — x'^ -- 5x -\-7 is the required function. 2. Find the laws governing the motion of a point which moves in a straight line with constant acceleration. Solution. Since the acceleration is constant, say/, we have ^^^ from (6), Art. 6T~] dt 'dt~-'' or dv —fdt. Integrating, (-1) v=ft+C. 196 CONSTANT OF INTEGRATION 197 To determine C, suppose that the initial velocity be Vq ; that ^^^ ^^^ v=zVq when ^ = 0. These values substituted in {A) give ^y^ = + (7, or C= Vq. Hence {A) becomes (B) v=ft + v, '0- ds Since v = — [Art. 67], we get from (B) (Xv ds J,. , dt or ds=ftdt-\-Vodt, Integrating, (G) s = ^ft'-\-Vot+a To determine (7, suppose that the initial space (= distance) be Sq ; that is, let s = Sq when ^ = 0. These values substituted in (C) give S(3 = + + C, or (7 = So. Hence (C) becomes (D) s = ^ft'-\-v,t + So. Formulas (B) and (D) are the required laws. 3. A certain magnitude z varies with the time according to dz the Compound Interest Law, that is, z and — ^ are proportional. Find z as a function of the time t. Solution. By definition, dz — = az, dt ' where a is a constant factor of proportionality. Multiplying both sides by dt and dividing through by z gives — = adt. z 198 ELEMENTARY ANALYSIS Integrating, (1) log z = at^C. To determine C, assume that the value of z when t = \i denoted by % Substituting in (1), C=\ogZo. We may now write (1) in the form (2) log z — log Zq = at, or log — = at. Hence, changing to exponentials (Art. 33), — = e«^^ or ;^ = ZQe""^.. Ans, PROBLEMS Find the function whose first derivative is 1. X — 3, knowing that the function equals 9 when x = 2. Jns. ?^_3a; + 13. 2 2. 3 -\-x— 5 x^, kuQwing that the function equals — 20 whei ^"=^- Ans. 124 + 3^;+-- — 3. (y^ — b^y)dy, knowing that the function equals whei y = ^- Ans. y--^^ + 2b^-4. 4 2 4. sin a + cos cc, knowing that the function equals 2 whei cc = -' Ans. sin a — cos a + 1. Find the equation of a curve such that the slope of th( tangent at any point is 5. Sx-2. Ans. y = '^-2x+a 6. ':)i?-\-^x, the curve passing through the point (0, 3). Ans. 2/ = - + — + 3. CONSTANT OF INTEGRATION 199 7. — , the curve passing through the point (0, 0). y Ans. y^ = 2px, 8. — , the curve passing through the point (a, 0). ay Ans. bV — a^y^ = a^b^. 9. m, the curve making an intercept b on the axis of y, Ans, y = mx + b. Assuming that v = Vo when ^ = 0, find tlie relation between v and t, knowing that the acceleration is 10. Zero. Ans, v = Vo. 11. Constant = A;. Ans. v=Vi) + lct. bf 12. a + bt. A71S. v = Vo + at-] Assuming that s = when ^ = 0, find the relation between s and t^ knowing that the velocity is 13. Constant (= t'o)- Ans. s = Vot. nf 14. m + nt. Ans. s = mt-\ .15. 3 + 2^-3^1 Ans. s = 3t + f-f. 16. The velocity of a body starting from rest is 5f ft. per second after t sec. (a) How far will it be from the point of starting in 3 sec. ? (b) In what time will it pass over a distance of 360 ft. measured from the starting point ? A71S. (a) 45 ft. ; (b) 6 sec. 17. A train starting from a station has after t hr. a speed of f — 21 ^^ + 80 ^ mi. per hour. Find (a) its distance from the station; (b) during what interval the train was moving backwards ; (c) when the train repassed the station ; (d) the distance the train had traveled when it passed the station the last time. Ans. (a) ^t^ — T f -\-4.0f mi. ; (b) from 5th to 16th hr. ; (c) in 8 and 20 hr. ; (d) 46581 mi. 200 ELEMENTARY ANALYSIS 18. The equation giving the strength of the current i for the time t after the source of E.M.F. is removed, is {R and L being constants) dt _m Find I, assuming that J= current when ^ = 0. Arts, i=: le ^. 19. Find the current of discharge i from a condenser of capacity (7 in a circuit of resistance R, assuming the initial current to be Iq, having given the relation di _ dt J'^CR' t C and R being constants. Ans. i = I^e^^, M N CHAPTER XV THE DEFINITE INTEGRAL 76. Differential of an area. Consider the curve AB whose equation is Let CD be a fixed and MP a vari- able ordinate, and let u be the measure of the area CMPD.* When X takes on a sufficiently small increment Aa;, u takes on an increment A?^ (= area JOTQP). Completing the rectangles MNRP and MNQS, we see that areaJf^i^P< area MNQPk area MNQS, oi: JfP. Aa;j is the differential of the area between the curve, the axis of x, and two ordinates. Integrating {A), u = I <^{x)dx. Let j ^{x)dx be worked out and denote the result hjfix) + C, (B) r.u=f{x) + a We may determine (7, as in Chapter XIII, if we know the value of ^i for some value of x. If we agree to reckon the area from the axis of y, i.e. when (C) x = a,u = 2iVQ^ OCDG, and when x=zb, u = area OEFG, etc., it follows that if (D) x=0, then i6 = 0. Substituting (Z>) in (5), 0=/(0) + C,orC=-/(0). THE DEFINITE INTEGRAL 203 Hence from (B) (E) u =f(x) -/(O), giving the area from the axis of y to any ordinate (as MP), To find the area between the ordinates CB and EF^ we observe that (F) area GEFD = area OEFG - area OCDG, But, using (E), area OCDG =f(a) -/(O), area OEFG =f(b) -/(O). Substituting in (F), (G) area CEm = f(fi) - f{a). Theorem. Tlie difference of the values of I ydx for x = a and x = b gives the area hounded by the curve whose ordinate is y, the axis of Xj and the ordinates corresponding to x = a and x = b. This difference is represented by the symbol ydxj or i cl)(x)dx, and is read ^^ the integral from a to & of ydx.^^ The operation is called integration between limits, a being the lower and b the upper hmit. Since {H) always has a definite value, it is called a definite integral. For, if I cf){x)dx = f(x) + C, f{h)+C then Cci,{x)dx = UXx) + O or j cl,(x)dx=f(b)-f(a), the constant of integration having disappeared. /(a) + C 204 i:lementary analysis We may accordingly define the symbol (x), the axis of X, and the ordinates of the curve at x=a, x=b. This definition presupposes that these lines bound an area, i.e. the curve does not rise or fall to infinity, and both a and b are finite, 78. Geometrical representation of an integral. In the last section we represented the definite integral as an area. This does not necessarily mean that every integral is an area, for the physical interpretation of the result depends on the nature of the quantities (j>(x) and x. In the chapter on functions, the variable x .was chosen to represent magnitudes of various kinds — time, length, etc. The corresponding function might be a volume, or any other physical magnitude. The definite integral {H) is then represented in numerical value by the area in the graph of <^(a;), but its actual physical significance might be something quite different. For example, if we turn to equation (6), p. 168, and represent the acceleration as a function of the time by a {t), then, multiplying through by dt and integrating, we obtain the indefinite integral v = | a(t)dt. The difference of two values of this integral, that is, the defi- nite integral, is clearly a change in velocity, and hence is a velocity. Further illustrations occur in later sections. 79. Calculation of a definite integral. The process may be summarized as follows: First step. Find the indefinite integral of the given differential expression. Second step. Substitute in this indefinite integral first the tipper limit and then the lower limit for the variable, and subtract the last result from the first. It is not necessary to bring in the constant of integration,^ since it always disappears in subtracting. THE DEFINITE INTEGKAL EXAMPLES 205 1. Find r x^dx, x^dx= ^ 1 [_o 2. Find | ^inxdx. Solution. I sin xdx = 3. Find r^^,- Solution. I -— : 4 64 1 : 21. ^TIS. — cos a; Jo "1 . x' - arc tan - a a Ans. 1 ± - — arc tan 1 arc tan a a 3^_0 = ^. Ans. 4a 4a PROBLEMS 3. 1. r6x'dx = 3S. 3. P— = 1. 2. I (a-o; — a;^^?^ = T- • 4.1 — =1. Jo ^ '4 Ji X 6. f ^-^ =V3-1. Jo V3-2a; n_tdt_ "Jo aH lo,£r2 '^^ da; V2-3a;=^ 4V3 9. 10. '^ 3 xdx {x) is the derivative oi f{x), then it has been shown in § 77, p. 203, that the value of the definite integral {A) (\(p)dx==f{h)-f{a) gives the area bounded by the curve y = <^(a^), the o^axis, and ordinates erected at a; = a and x—h. Let us now make the following construction in connection with this area. Divide the interval from x = atox = h into any number n of equal subintervals, erect ordinates at the points of division, and complete rectangles by drawing horizontal lines through the '207 208 ELEMENTARY ANALYSIS 6 ^ extremities of the ordinates, as in the figure. It is clear that the sum of the areas of ^ these n rectangles (the shaded area) is an ap- proximate value for the area in question. It is further evident that the limit of the sum of the areas of these rectangles, when their number n is indefinitely increased, will equal the area under the curve. Let us now carry through the following more general con- struction. Divide the in- terval into n subintervals, not necessarily equal, and erect ordinates at the points of division. Choose a point within each sub- interval in any manner, erect ordinates at these points, and through their extremities draw horizon- tal lines to form rectangles as in the figure. Then, as before, the sum of the areas of these n rectangles equals approximately the area under the curve, and the limit of this sum as n in- creases without limit and each subinterval approaches zero as a limit is precisely the area under the curve. These considerations show that the definite integral {A) may be regarded as the limit of a sum. Let us now formulate this result. 1. Denote the lengths of the successive subintervals by Aa^i, Aa?2, Ai»3, ..., ^x^. 2. Denote the abscissas of the points chosen in the subinter- vals by Xi, X2, Xq, •••, x^. INTEGRATION A PROCESS OF SUMMATION 209 Then the ordinates of the curve at these points are (^l), ^(^2), <^(^3), •••, (^n)' 3. The areas of the successive rectangles are obviously 4. The area under the curve is therefore equal to limit n=co (l>(x^)Ax^+ (X2)AX2+ ••• + The discussion gives the equation (B) j\{x)dx = ]^l \{x,)Ax, + {x,)Ax, ct>{x,,)Ax^ + + {x„)Ax^ The equation (B) has been established by making use of the notion of area. The area under discussion is bounded^ that is, it has a closed perimeter consisting of the curve y = <^(ic), the ic-axis, and the lines x = a, x = b. It is therefore tacitly assumed that the function (l>(x) is finite for all values of a; from a to 6 inclusive. That is, the curve y = {x) does not run off to infinity between ic=« and x^h] otherwise expressed, the 210 ELEMENTARY ANALYSIS curve has no " breaks " in it. The student may refer to Art. 32 for examples of curves which have *^ breaks '' in them ; that is, have vertical asymptotes. The property of the function (x) be continuous for the interval x = a to x=:b. Let this interval he divided into n subintervals whose lengths are AXi, Aa?27 '"y ^^n> ^^^ let points be chosen, one in each subinterval, their abscissas being Xi, a?2, •••, x^ respectively. Consider the sum (O) (^ (x{) Ao^i ^cl>(x2)Ax2 + "' + (x^) Ax^. Then the limiting value of this sum ivhen n increases without limit and each subinterval approaches zero as a limit equals the value of the definite integral i <^ (x) dx. The importance in the application of this theorem is this : We are able to calculate by integration any magnitude which is the limit of a sum of the form (C). The difficulty which arises in actual practice is that of de- termining the function <^(a?). No general rule applicable to all cases can be given for overcoming this difficulty. Th^ student should study the examples worked out in the follow- ing pages in order to obtain practice. We may observe that the terms of the sum (C) are of the form {B) cjy (x) Ax, or also <^ (x) dx (since dx — Ax), which may be called the general term, or also an element of the integral. The following directions will be found useful in applying the Fundamental Theorem. INTEGRATION A PROCESS OF SUMMATION 211 First step. Divide the required magnitude into similar parts such that it is clear that the desired result will be found by summing up these parts and passing to the limit. Second step. Choose a suitable variable such that the mag- nitude of each part can be expressed in the form (D). Third stejj. Apply the Fundamental Theorem and integrate. 82. Areas of plane curves. As already explained, the area between a curve, the axis of X, and the ordinates x = a and x = b is given by the formula (^) area j^ y doc, the value of y in terms of x, being substituted from the tion of the curve. Equation {A) is readily memo- rized by observing that the inte- grand ydx represents the area of a rectangle CR of base dx and alti- tude y. The a^Dplication of the Funda- mental Theorem to the calculation of the area bounded by the curve equa- (1) x = 4>{y\ the 2/-axis and abscissas at 2/ = c and y = d, is immediate. First step. Construct the n rectangles as in the figure. The area is clearly the limit of the sum of these rectangles as their number increases indefinitely. Second step. Call any one of the alti- ^x=4>{y) tudes A?/. The base of any one of the rectangles is x. Hence the general ex- pression for the area of the rectangles is X ^y or 214 ELEMENTARY ANALYSIS Another method is to consider the strip PS as an element of the area. If y' is the ordinate corresponding to the witch, and y" to the parabola, the differential expression for the area of the strip FS equals (y' — y")dx. Substituting the values of y' and y^' in terms of x from the given equations, we get area AOOB = 2 x area OCB ■•2a -4 '0 V o;^ + 4 a^ x'' 4taJ \dx :2a2(7r-|). X u "4. Find the area of the ellipse — + ^ = 1. Solution. To find the area of the quadrant OAB, the limits are a? = 0, ic = a ; and h, a Hence, substituting in (^), p. 211, area OAB = ~ f^w'-a^^Ux aJo VhX ^ o 2\l , ct^ y- /~2 I bx . 2 arc sm- 2 a nrdb (problem 2, § 74) Therefore the entire area of the ellipse equals it ah. We may also calculate the integral giving area Q^^a s follows. Let a;=acos <^ (see figure and § 74). Then -y/a^—x^^a sin ^, dx=z — a sin <^ d. Changing the limits : — when a;=0, <^= |^7r; when aj=a, THE UNIVERSITY OF CALIFORNIA LIBRARY