It P' lili liliPiPPIih PI; i^i' iiiiiililiil y iliiipl'i 1 I !i! I'll'" ■■!!|lllill';!'', Ilillilii" ilV ill i!il! ii!!! OF THE UHIVBRStTY OF ^AlFOlfiiJ Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/analyticgeometryOOriggrich ANALYTIC GEOMETRY ■y^^^ THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO ANALYTIC GEOMETRY BY N. C. RIGGS, M.S. ASSOCIATE PROFESSOR OP APPLIED MECHANICS FORMERLY ASSISTANT PROFESSOR OF MATHEMATICS CARNEGIE TECHNICAL SCHOOLS, PITTSBURGH, PA. Neto gork THE MACMILLAN COMPANY 1916 All rights reserved COPTBIGHT, 1910, By the MACMILLAN COMPANY. Set up and electrotyped. Published September, 1910. Reprinted January, March, September, 1911 ; February, July, 1912 ; September, 1913 ; February, July, 1914; August, 1915 ; August, 1916. NorixiooU l$xn% J. S. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. :^ ^ PREFACE In the preparation of this book the author has tried to keep in mind the twofold requirement of a text-book on Analytic Geometry : to bring out clearly the fundamental principles and methods of the subject, and to make it a natural introduction to more advanced work. Since for most students of Analytic Geometry the subject is quite as essential as a preparation for the study of Calculus as it is valuable for its own methods and body of facts, the method and notation of the Calculus have been used in their application to tangents, normals, and maxima and minima in the plane, and to tangent planes and lines in space. The conic sections have not been accorded as much space relatively as in most text-books on the subject, but it is believed that the student's time in the usual brief course can be spent to greater profit in the study of such chapters as those on Trigonometric and Exponential Functions, Parametric Equations, Empirical Equations, Maxima and Minima, and Graphical Solution of Equations, than upon a prolonged course on the conies. Especially is this true for engineering students. The answers to many of the problems have not been given. Where the student can check the answer by graphical means, it is best that he should thus test the correctness of his work, and a complete list of answers tends to take away his incen- tive for doing this. The author is under many obligations to Professors D. F. Campbell, Alexander Pell, C. W. Leigh, and C. I. Palmer, of the Armour Institute of Technology, and to Mr. Paul Dorweiler vi PREFACE of the Carnegie Technical Schools, for valuable criticism and advice, and to Professors Leigh and Palmer for the answers to many of the problems. The imperfections of the book are, however, the author's alone. For the drawing of. most of the figures the author is indebted to Mr. John R. Boyd, and for the remainder to Mr. Edwin 0. Kaul, students in the School of Applied Science, Carnegie Technical Schools. N. C. RIGGS. Carnegie Technical Schools, August, 1910. CONTENTS CHAPTER I Articles 1-18 Graphical Representation op Numbers. Systems op Coordinates PAGE Point and number — Addition and subtraction of segments — Coordi- nates of points in the plane — Cartesian coordinates — Rectangu- lar coordinates — Polar coordinates — Trigonometric functions — Relation between rectangular and polar coordinates . . 1 CHAPTER II Articles 19-38 Projections. Lengths and Slopes of Lines. Areas op Polygons Projections by parallel lines — Orthogonal projection — Projection on coordinate axes — Distance between two points — Angle between two lines — Inclination and slope of a line — Ratio •into which a point divides a line — Angle between two lines of given slopes — Condition for parallel lines, for perpendicular lines — Area of a triangle — Area of a polygon .... 15 CHAPTER III Articles 39-45 Graphical Representation of a Function. Equation op A Locus Function and variable — The graph of a function — Equation of a locus — Locus satisfying given conditions — Intercepts . . 38 viii CONTENTS CHAPTER IV Articles 46-50 Locus OF AN Equation PAGE Examples — Symmetry — Discussion of an equation ... 61 CHAPTER V Articles 51-54 Transformation of Coordinates Translation of axes — Rotation of axes — Applications ... 64 CHAPTER VI Articles 55-71 The Straight Line The equation of first degree — Conditions determining a straight line — The two-point equation — The intercept equation — The point-slope equation — The slope equation — The normal equa- tion — Reduction of Ax + By -^ C = to standard forms — Intersection of lines — Change of sign of Ax + By -\-C — Dis- tance from a point to a line — Equations of the straight line in polar coordinates 70 CHAPTER VII Articles 72-96 Standard Equations of Second Degree The circle — The equation x"^ + y"^ + Dx + Ey + F = — Circle through three points — The parabola — Axis and vertex of the parabola — Parameter of the parabola — The equation y = ax^ -\-bx-\- c — The parabolic arch — The ellipse — Axes, vertices, center of ellipse — The hyperbola — Asymptotes — Axes, cen- ter, vertices of hyperbola — The conjugate hyperbola — The equilateral hyperbola — Reduction of the general equation of second degree in two variables 88 CONTENTS ix CHAPTER VIII Articles 97-107 Trigonometric and Exponential Functions PA6K The sine curve — Periodic functions — The exponential curve — The logarithmic curve — Graphs — Plotting in polar coordinates . 117 CHAPTER IX Articles 108-120 Parametric Equations of Loci Parametric equations of circle and ellipse — Construction of the ellipse — The cycloid — The hypocycloid — The epicycloid — The cardioid — The involute of the circle 129 CHAPTER X Articles 121-124 Intersections of Curves. Slope Equations of Tangents Intersections of curves — Graphical solutions of equations — Slope equations of tangents 142 CHAPTER XI Articles 125-154 Slopes. Tangents and Normals. Derivatives Increments — Slope of a curve at any point — Equation of tangent to a curve — The normal — Derivatives — Formulas of differen- tiation — Geometric meaning of the derivative — Continuity — Limit of ratio of a circular arc to its chord — Radian measure of an angle — Derivatives of the trigonometric functions . . 153 CHAPTER XII Articles 155-163 Maxima and Minima. Derivative Curves Maximum and minimum points of a curve — Determination of maxima and minima — The first derivative curve — Concavity — The second derivative curve 178 X CONTENTS CHAPTER XIII Articles 164-177 The Conic Sections PAGB Construction of conies — Vertices — Classification of conies — Equa- tion of the conic in rectangular coordinates — The parabola — The centric conies — The ellipse — The hyperbola — Axes — ' The equation of the conic in polar coordinates . . . .192 CHAPTER XIV Articles 178-186 Pkopekties of Conics Subtangent and subnormal of the parabola — Property of reflection of the parabola — Focal radii of ellipse and hyperbola — Prop- erty of reflection of ellipse and hyperbola 205 CHAPTER XV Articles 187-195 The General Equation of Second Degree Removal of the terms of first degree — Removal of the term in xy — Locus of the equation 214 CHAPTER XVI Articles 196-201 Empirical Equations Points lying on a straight line — The curve y ■= cx"^ — The curve y = ah' — Some special substitutions — The curve y = a + bx -^ cx^ + "' +kx^ 223 CHAPTER XVII Articles 202-212 COOEDINATES IN SpACB Rectangular coordinates in space — Distance between two points — Polar coordinates — Relation between rectangular and polar coordinates — Relation between the direction cosines of a line — Direction cosines of a line joining two points — Spherical coordinates — Projection of a broken line — Angle between two lines 233 CONTENTS xi CHAPTER XVIII Articles 213-217 Loci and their Equations PAGE Certain lines and planes — Cylinders — Surfaces of revolution — Plane sections of a locus — Locus of an equation in three variables 243 CHAPTER XIX Articles 218-225 The Plane and the Straight Line The normal equation of the plane — The intercept equation of the plane — General equation of first degree in three variables — Distance from a point to a plane — Angle between two planes — Equations of the straight line 249 CHAPTER XX Articles 226-234 The Quadric Surfaces The ellipsoid — Hyperboloid of one sheet — Hyperboloid of two sheets — Elliptic paraboloid — Hyperbolic paraboloid — The cone — The conic sections 257 CHAPTER XXI Articles 235-239 Space Curves The helix — Curve of intersection of two cylinders — Curve of inter- section of cylinder and sphere — General equations of a space curve 268 CHAPTER XXII Articles 240-246 Tangent Lines and Planes Partial derivatives — Tangent plane to a surface — Normal to a sur- face — Tangent line to a space curve 272 ANALYTIC GEOMETRY CHAPTER I GRAPHICAL REPRESENTATION OF NUMBERS. SYSTEMS OF COORDINATES I. POINTS ON A STRAIGHT LINE 1. Point and Number. On a straight line let a fixed point be taken from which to measure distances, and let a definite length be chosen as a unit. If this unit be laid off in succession on the line, beginning at 0, other points of the line are obtained whose distances from are 1, 2, 3, •••, etc. times the unit dis- tance. It is convenient to think of these points as represent- ing the numbers, or of the numbers as representing the points. Thus a point 7 units from may be taken to represent the number 7, and conversely the number 7 may be said to repre- sent the point. Since there are two points of the line at the same distance from P, one to the right, the „ ^ ,. ,. P P, P other to the left, and since ^ there are both positive and -4-3-2-1 1 2 3 4 negative numbers, let it be ^^^" ^' agreed that points to the right of shall represent positive numbers and those to the left of negative numbers. Thus a point 3 units to the right of represents the number 3, and a point 3 units to the left of represents the number — 3. The numbers are also said to represent the points. It can be shown that to every point of the line there corre- sponds a real number, and conversely, to every real number there corresponds a point of the line. The whole system of real numbers may therefore be represented by points on a B 1 2 ANALYTIC GEOMETRY straight line with one number for each point and one point for each number. The point is called the origin. It represents the number zero. 2. Notation. If P is any point of the line and is the ori- gin, the symbol OP is used to denote the number which repre- sents the point P. E.g. if P lies 3 units to the right of 0, then OP is 3 ; while if P lies 3 units to the left of 0, OP is - 3. It is convenient to denote the number which represents a point by a single letter, as x; thus OP=x. Then if P lies to the right of 0, a; is a positive number, and if P lies to the left of 0, cc is a negative number. Different points on the line will sometimes be denoted by P with different subscripts, and the numbers representing these points by x with corresponding subscripts. Thus, in Fig. 1, OP^ = x^ = 2, OP^ = a^a = - 4. 3. Segments of the line. In speaking of any segment of the line, as AB, the first letter named is called the beginning, and the last letter the end, of the segment. Thus A is the beginning, and B is the end, of AB, while B is the beginning, and A is the end, of BA. It is important to represent the value of any segment of the line by a number, and this is done by defining the value of any segment of the line to be the number which would repre- sent the end of the segment if the beginning of the segment were taken as origin. Thus, in Fig. 2, with as origin, P, P3 A P2 B . I I I ^5 = 6, 5^ = - 6,^30 = 5, 0Pi = -6, 0^=-2, P3Pi = -l, P,0 = -2. GRAPHICAL REPRESENTATION OF NUMBERS 3 From the definition of the value of a segment it follows that the value of any segment read from right to left is negative, while the value of any segment read from' left to right is positive. EXERCISE I 1. What numbers represent the points Pi, P2, P3, -4, B, in Fig. 2 ? 2. What are the values of P2P3, P1P2, P3P2, BP3, P3B ? 3. If A be taken as origin, what are the numbers that represent Pi, P2, O, P3, P? 4. If the origin be moved two units to the right, how are the numbers representing different points affected ? How if the origin be moved h units to the right ? to the left ? 4. Change of sign of a segment. Since any segment AB of the line contains the same number of units as BA, but is meas- ured in the opposite direction, it follows that BA = -AB, or BA + AB = 0. 5. Addition of segments. Let A, B, and C be any three points on the line. Then AC = AB + BC. Proof. Three cases arise : (1) B between A and G, (2) A between B and C, (3) be- tween ^ and B. (^) A B C C B (2) (3) Fig. 3. In(l), AO=AB-\-BO', in (2), AC=BC-BA = BC-hAB,hj Art. 4, or AC=:AB-\-BC; in (3), AC = AB-CB = AB-\-BO, A B C > B A c 1^ A C B C A B B C A 4 ANALYTIC GEOMETRY 6. Subtraction of segments. By writing — CB instead of BC in the equation of Art. 5, namely, AC=AB-^BC, that equation becomes AC = AB-CB. The results found in this and the preceding articles lead to the rules for geometric addition and subtraction of numbers that follow. 7. Geometric addition of numbers. Let P^ and P^ be two points on the line represented by the numbers x^ and X2 respec- tively. Then A = x^, OP^ = x^. Three cases arise : (1) both numbers positive ; (2) one number, say x-^, negative, the other positive ; (3) both 0) — I h! h^ 1 numbers negative. rg) Pi O P Pg To represent geometri- p p p Q cally the sum of x^ and x^ lay ^^^ ^ 1 — I •— off from the end of x^ sl ^^^- *• segment, PiP, equal to X2 and measured from Pj in the same direction as x^ is measured from 0. Then 0P=Xi-{-X2. For, in each case, 0P= OP^ + P^P, by Art. 5, = 0P,^0P2 M — 37j ~}~ fl/g. 8. Geometric subtraction of numbers. Consider again the three cases of Art. 7. To represent geometrically the differ- ence Xi — X2 lay off from the end of Xi a segment PiP equal to — X2, i.e. having the same numerical value as X2 but opposite in direction. P Pi P, i 1 1 1 P Pi P2 H 1 1 1 Pi P P2 o GRAPHICAL REPRESENTATION OF NUMBERS 5 Then 0P = Xi-X2. For, in each case, 0P= OP, + F,P= 0P-PP,= 0P^-0P, = x,-X2. Another, and more important, expression of the difference is as follows : x^ - X, = OP, - OP2 = OP, -\-PM = P,0-{-0P,= P2Pi, by Art. 5, or JP1-P2 = a52 — a?i. Hence, the value of any seg- ment of the line is equal to the number that represents the end minus the number that represents the beginning of the segment. This principle will be of frequent use hereafter. Illustration. In Fig. 6, if P„ P.>, P3 are three points on P2 . P4 . Pi Pa - — t 1 1 1 (— — ^! — I 1^ — - 3 - f Fig. 6. ^ ^ the line represented by the numbers 2, — 3, 4 respectively, then PiP2 = -3-2 = -5, P,Pi = 2-(-3) = 5, P3P, = -3-4 = -7, PiP3 = 4-2 = 2, P3Pi = 2-4 = -2. 9. Relative position of points representing numbers. Let x^ and X2 be any two real numbers represented by the points P, and Pa respectively. By Art. 8, -P2A = x,— x^. Now if Xi > 072 then x, — x^ is positive, and conversely. Therefore, if x^^x^, P^P, is positive, and hence Pi lies to the right of Pg ; if a^i < %, AA is negative, and hence Pi lies to the left of Pg, and conversely. Hence, of the two points which represeyit ttvo real numbers the point which represents the greater number lies farther to the right. E.g. in Fig. 6, P,, which represents 2, lies to the right of Pg, which represents —3; P4, which represents —1, lies to the 6 ANALYTIC GEOMETRY right of P^j which represents — 3. This agrees with the state- ment that 2 is greater than — 3, and that — 1 is greater than -3. EXERCISE II 1. Represent geometrically the following pairs of numbers, their sum, the first minus the second, the second minus the first : (a) 3, 2. (6) - 2, 3. (c) .4, - 3. (d) - 5, - 1. 2. In Fig. 5 express the following segments as the difference of the numbers representing the points: P1P2, P2-P1, P2O, OPi, PiO, OP2. vvk^ 3. In Fig. 5 what segments represent Xx — iC2, x^ — Xx^ iCi, X2, — Xi^ 4. In Fig. 6, by means of the principle in Art. 8, find the values of P2P1, P3P1, P4O, P4P3, P3P4, OP3, P3O, P3P2. n. COORDINATES OF POINTS IN THE PLANE 10. Location of a point. To determine the position of a point on a straight line one magnitude is sufficient; namely, the distance of the point, right or left, from a fixed point of the line. The number that represents a point on the line determines the position of the point when the origin is given. In the plane, however, two magnitudes are necessary to determine the position of a point. There are many ways of choosing these magnitudes. Two simple methods, and the only ones used in this book, are to consider the location of the point, (1) with reference to tw^ intersecting straight lines, (2) with reference to a fixed Lnv" and a fixed point. A consideration of these two methods leads to the definitions of (1) Cartesian Coordinates, (2) Polar Coordinates. 11. Cartesian coordinates. Let two intersecting straight lines, OX and OF, be taken as lines of reference and an arbi- Fig. 7. GRAPHICAL REPRESENTATION OF NUMBERS 7 trary length be chosen as a unit. Then to every point P in the plane there can be assigned a pair of real numbers as follows : Through the point P draw lines parallel to OX and Y, meet- ing OX and OY in M and JST respectively. The pair of num- bers which measure JSfP and MP is taken to represent the point P. To every position of P there corresponds one, and only one, pair of such num- bers. In order that to every pair of real numbers there may correspond one, and only one, point, some agreement in re- gard to signs is necessary. To the agreement already made that a segment measured from left to right shall be positive, and one measured from right to left shall be negative, let there be added the agreement that a seg- ment measured upward shall be positive, and a segment meas- ured downward shall be negative. With this agreement in regard to signs there corresponds one, and only one, point in the plane to every pair of real numbers. The lines OX and OY are called the a?-axis and 7/-axis repectively. The segments NP and MP are called respectively the abscissa and ordinate of P, and together are known as the Cartesian coordinates of the point. It should be carefully noted that, from the definition, the abscissa of P is measured from the y-axis to P, and the ordi- nate of P is measured from the x-axis to P. The abscissa and ordinate are most frequently denoted by x and y respectively, though other letters are sometimes used. The point P is denoted by the coordinates inclosed in parentheses and separated by a comma, thus, (x, y) or P(x, y). 8 P2(«2.2/2) ANALYTIC GEOMETRY Y/ To distinguish one point from another, subscripts are often used. Thus in P^i^^^yO Fig. 8, X2 = JSr2P2=03Io = -2, y, = M,P,= 0^2 = 3, x^ = N^Ps = — 3, '4 (2:4 » 2/4) Fig. 8. ys = MsPs = -S, etc. M,M, = M,0+OM, = -x, + x„ M^Mi = M^O + 03fi = -x^-\- xi, etc. EXERCISE III 1. Assume a pair of axes and locate the points (2, 3), (2, —3), (-2, 4), (-5, -6), (0, 2), (4, 0), (-1, 0), (0, -3), (0, 0). 2. In Fig. 8 express as the dif- ference of two abscissas, 3f2Ms, M^Mi, MsM^, MoMi. 3. Express as the difference of two ordinates, N2N3, iViiV4, 4. What segments represent Xi — X2, X3 - Xi, Xi — Xi, X3 - res ? 5. What segments represent y2 - Vu Vi - «/2, ys - 2/2, 2/1 - y* ? 6. Where do all points lie that have the abscissa zero ; that have the ordinate zero ? 7. Where do all points lie that have the abscissa 2 ; that have the abscissa — 3 ; that have the ordinate 2 ; that have the ordinate — 4 ? Fig. 9. GRAPHICAL REPRESENTATION OF NUMBERS 9 8. In Fig. 9 express as the difference of two abscissas, PiB, P^^-, PzQi IB, SN; and as the difference of two ordinates, TPs, PnV, BQ, TS, P2B, PaS. 9. In Fig. 9 let Pi, Pg, Ps have coordinates (2, 3), (4, — 2), and (-3, 2), respectively, and find the values of SN, P2S, BQ, BT, PiN, SPs, P2P, QP3. 12. Segments not parallel to an axis. Segments of lines not parallel to one of the coordinate axes will not have definite signs given to them. They vi^ill generally be considered as positive lengths, but where the two opposite directions along the same straight line are considered, one of them will be counted as opposite in sign to the other. 13. Rectangular coordinates. If the axes in the Cartesian coordinate system are at right angles to each other, the system is called the rectangular system of coordinates. This system possesses the advantage of simplicity, in many problems, over that of oblique axes, and as most of the proper- ties and relations of figures to be studied do not depend upon the system of coordinates used, the rectangular system will be used except where otherwise indicated. 14. Polar coordinates. The position of a point in the plane may be determined by the length of the line joining it to a fixed point, and the angle which this line makes with a fixed direction. In Fig. 10 let be a fixed point and OA a fixed line. Let P be any point in the plane. Then the seg- ment OP and the angle AOF deter- mine the location of P. Fig. 10. The segment OP is called the radius vector, and the angle AOP the vectorial angle of P. Together they are known as the polar coordinates of P. They are usually denoted by r and 6, respectively, and the point indicated by (i-, 6), or P(r, 6). 10 ANALYTIC GEOMETRY The fixed point is called the origin, or pole ; the fixed line OA the initial line, or axis. The line OP is called the terminal line of the angle AOP. With these definitions it is easy to see that any point in the plane may be represented by polar coordinates, both of which are positive, and with the angle less than 360°. In order, however, to represent both positive and negative numbers by points, the following agreement in regard to signs is made: Positive angles will be measured in the counter-clockwise direction from the initial line ; nega- tive angles in the opposite direction. By a negative radius vector will be meant one laid off on the terminal line of the vectorial angle produced back through the pole. 3 27r 3 and 5, would be as indi- Thus, the points (5, Gated in Fig. 11. With the above agreement in regard to signs it follows that to every pair of coordinates there is just one point in the plane, but to every point in the plane there corresponds an indefinite number of pairs of coordinates. J^.gr. thepoint [2, - J may also be represented by [ — 2, (-2,*f)(2,-f) or b. GRAPHICAL REPRESENTATION OF NUMBERS 11 any other pair of coordinates obtained by increasing the angle of any of the above pairs by an integral multiple of 2 tt, the radius vector being unchanged. If d is restricted to being numerically less than 2 tt, the four pairs of values written above are the only ones that represent the given point. Note. The student should remember that the unit of circular measure of an angle is the angle subtended at the center of a circle by an arc equal in length to a radius of the circle. This unit is called the radian. From the definition it follows that tt radians = 180°, where tt =3.14159 •• •. When an angle is represented by a letter or figure without the degree sign (°), it will be understood that the unit of meas- ure is the radian. EXERCISE IV 1. Plot in polar coordinates (2, -30^), f-4, ^V ll, -^), (tt, tt), (tt, 7r°), (3, 2). 2. Plot in rectangular coordinates (—3, 4), (0, —3), (0, 0), (a, 0), (0,.),(.,-2f),(6,|). HI. THE TRIGONOMETRIC FUNCTIONS 15. Definitions of the trigonometric functions. given any angle, P' assume a system of rectangular coordinates and place the vertex of the angle at the origin, with the initial line coinciding with the positive part of the a?- axis ; , , ' Fig. 13. positive angles to be reckoned counter-clockwise and negative angles, clockwise. 12 ANALYTIC GEOMETRY Assume any point P on the terminal line ; let its coordinates be X and y, and its distance from the origin be r, counted always positive. Then, whatever the size of the angle, the following definitions are given : sine of ^ = ordinate/distance = y/r^ cosine of ^ = abscissa/distance = x/r, tangent of A = ordinate/abscissa = y/Xj cotangent of ^ = abscissa/ordinate = x/y, secant oi A = distance/abscissa = r/x, cosecant of ^ = distance/ordinate = r/y. 16. Formulas and tables. A set of the more important formulas connecting the trigonometric functions of angles, and a table of sines, cosines, and tangents are given at the back of the book. 17. The inverse trigonometric functions. The symbol sin*' £c, read " anti-sine a?," is used as equivalent to the words, "an angle whose sine is ic." Thus one value of sin~^ (i) is - , or 30°; another value is — . 6 6 In like manner the symbols cos~^ x, tan~^ x, etc., are used as equivalent to " an angle whose cosine is a;," " an angle whose tangent is ic," etc. EXERCISE V 1. Find by the use of the table the sine, cosine, and tangent of each of the following angles : 20^, 17*^ 20', 185°, 109° 40', 290°, 165^ .2 radian, .72 radian, (-J radian. 2. Given A = sin-i .6, find a value of A in the first quadrant, and one in the second quadrant. 3. Given A = tan-i .4563, find two values of A. 4. Find sin-i(tan25°}, sin(tan-i3.26), sin(sin-i ,35). 5. Show that sin (sin- 1 .5) = .5, and that sin-i(sin30°)= 30°, or 150'', or 390°, etc. GRAPHICAL REPRESENTATION OF NUMBERS 13 18. Relation between rectangular and polar coordinates. Let the origin in the two systems be the same, and let the initial line coincide with the positive part of the pc^, y\ ic-axis. Let P be any point in the plane with rectangular coordinates x and y and polar coordinates r and (Fig. 14), the polar coordi- nates being so chosen that e=Z. XOP and r = OP, where OP is positive. Fig. 14. Then from the definition of sine and cosine. - = cos ^, ^ = sin e, r r or (1) Qc z=r cos 6, y =r sin 6. These equations express x and y in terms of r and 0. From the figure, or from these equations, r and 6 can be expressed in terms of x and y. The resulting equations are r = v^+ y% 9 = tan-^ (^y (2) EXERCISE VI 1. Show how to obtain eqs. (2) of Art. (18) from eqs. (1). 2. Show that if the polar coordinates of F be chosen so that d differs from Z XOP by 180°, and r is the negative of OP, eqs. (1) still hold. 3. Find the polar coordinates of the points whose rectangular coor- dinates are (3, - 7), (4, 3), (- 2, 1), (- 4, - 2). 4. Find the rectangular coordinates of the points whose polar coor- dinates are (2, 30'^), (- 3, 45°), (4, - 60"), (- 2, - 15°). 5. In rectangular coordinates where do all points lie whose abscissas are zero ; whose ordinates are zero ; whose abscissas equal any constant 14 - ANALYTIC GEOMETRY C ; whose abscissas equal their ordinates ; whose abscissas equal the nega- tive of their ordinates ? 6. What is true of the polar coordinates of points which satisfy each of the conditions of example 5 ? 7. In polar coordinates where do all points lie whose vectorial angles are zero ; whose vectorial angles equal 30*^ ; whose vectorial angles equal any constant ,- whose radii vectores equal 5 ; whose radii vectores equal any constant C ? 8. What equation is true of the rectangular coordinates of the points which satisfy each of the conditions in example 7 ? 9. Find the polar coordinates of the point whose rectangular coor- dinates are (3.26, -2.67). 10. Find the rectangular coordinates of the point whose polar coor- dinates are (6.34, 34° 16'). CHAPTER II PROJECTIONS. LENGTHS AND SLOPES OF LINES. AREAS OF POLYGONS L PROJECTIONS 19. Projections by parallel lines. Through the beginning and end of a segment AB let lines parallel to a given direc- tion be drawn to intersect a given line MN in C and D respec- tively. Then CD is called the projection of AB on MJ^, for the given direction. The beginning and end of the projection are to be read in the same order as the beginning and end of the segment. Thus CD is the projection of AB, while DC is the projection of BA. (Fig. 15.) The direction, parallel to which the lines AC and BD are drawn, is called the direction of projection. Evidently, the value of the projection depends upon, (1) the length of the segment, (2) the difference in direction of the segment and the line on which it is projected, and (3) upon the direction of projection. It is evident, also, that the pro- 15 16 ANALYTIC GEOMETRY jections of a given segment on parallel lines are equal, if the direction of projection is the same. 20. Orthogonal projection. If the direction of projection is perpendicular to the line on which the segment is projected, the projection is called orthogonal. B Fig. 16. Thus in Fig. 16 CD is the orthogonal projection of AB on MN. 21. Projection in the direction of one coordinate axis on a line parallel to the other axis. Definitiox. The projection in the direction of the 2/-axis of a segment on a line parallel to the a;-axis will be called the ic-projection of the segment. A similar definition is given for the y-projection of the seg- ment. Consider now the ic-projection of any segment P^Pi. Let the coordinates of Pj and P., be {x'^, y^ and (x^, y^ re- spectively. Three cases may arise : P^Pi may lie wholly to the right of the ly-axis, may cut the ?/-axis, or may lie wholly to the left of the 2/-axis. (Fig. 17.) Let the projection in either case be M^M,, and let the line on which P1P2 is projected meet the ?/-axis at N, Then, in either case, M1M2 = MiN+ NM2 = — Xi-\-x<> = X2 — iCj. PROJECTIONS. LENGTHS AND SLOPES OF LINES 17 Therefore, the x-projection of a segment is equal to the abscissa of the end of the segment minus the abscissa of the beginning. Yi In like manner it can be shown that the y-projection of a seg- ment is equal to the ordinate of the end minus the ordi7iate of the beginning. Example. The a>projection of the segment from Pi(— 1, 3) to P2(3, 2) is 3 — (— 1) = 4, and the ^/-projection is 2 — 3 = — 1. EXERCISE VII 1. Prove that the ^/-projection of a segment is equal to the ordinate of the end of the segment minus the ordinate of the beginning. 2. Find the x- and ^/-projections of the segments from the first to the second of each of the following pairs of points: (2, 3), (—2, 6); (-3, -1), (4, -5); (1, -2), (3,7); («,&), (c, d); (0, 1), (-2, 0); (0,0), (3, -5); (u,v), (s,t). Check the results by drawing the figure in each case. 3. If the axes are at right angles to each other, find the distance from the origin to (3, 7) ; from the origin to (x, y). c 18 ANALYTIC GEOMETRY • 4. If the axes are at right angles to each other, find the distance between (- 5, 3) and (2, - 6). 5. If the axes are rectangular, show that the distance between (iCi, yi) and (X2, y-i) is V(a:i - x^y^ + {yi - y2)\ 6. In rectangular coordinates the point (x, y) moves so as to keep at the distance 5 from the origin. Express this by means of an equation. What is the locus of the point ? 7. What is the a;-projection of a segment parallel to the y-axis ; the t/-projection of a segment parallel to the x-axis ? 8. The vertices of a triangle are A, B, and G. Show that the sum of the projections of AB, BC, and CA on any line is zero, and that the projection of AC = the projection of AB + the projection of BC. 9. Show that the sum of the projections of the sides of any closed polygon taken in order, i. e. so that the beginning of each side is the end of the preceding, on any line is zero. 10. Show that if the sum of the projections of the sides of a polygon taken in order on one straight line is zero, the polygon is not necessarily closed ; but if the sum of the projections taken in order on two non- parallel lines is zero, the polygon is closed. II. LENGTHS AND SLOPES OF SEGMENTS. DIVISION OF SEGMENTS 22. Distance between two points. Numerical examples. Example 1. To find the distance between the two points whose Cartesian coordinates are (2, -4) and (-3, 5), the angle be- tween the axes being G0°. Let (2, -4) be P„ and (-3, 5) be Pa- Through Pi and Pg draw lines parallel, respectively, to the x- and ?/-axes, intersecting in Q. (Fig. 18.) jj,j^ ^g By the law of cosines from trigo- nometry, P^l=QPl+ QPl- 2 QPi- W, cos P^QP^, PROJECTIONS. LENGTHS AND SLOPES OF LINES 19 Here QP^ = 2 - (- 3) = 5, by Art. 21, cosPiQP2 = cos60° = i. .•.P,P2 = V6i = 7.81 ... . Example 2. To find the distance between the points whose polar coordinates are [2, — j and (5, — '^ lPi Fig. 19. Let (2, ^\ be P„ (5, - 1") be P^ and let P^Po = d. (Fig. 19.) By trigonometry, d' = OPI + OPI - 2 OPi . OPa cos Pi OP2 = 4 + 25-2. 2. 5cos^ 6 = 4 + 25 + 20 cos^ 6 = 29 + 17.32 ... . ,'.d = V46.32 = 6.81 nearly. EXERCISE VIII 1. If the angle between the axes is 45°, find the distance between the points (-.3, 5) and (4, 1). 2. If the angle between the axes is 80°, find the distance between (6, 2) and (-3, -4). 3. If the axes are rectangular, find the distance between (a, 6) and (c, d). 20 ANALYTIC GEOMETRY 4. Find the distance between the points whose polar coordinates are (6, 20°) and (4, 2('-)), where 2('-) means 2 radians. 5. In polar coordinates find the distance between ( — 3, - ] and 23. Distance between two points. General formula in rec- tangular coordinates. Let Pi(aaj 2/i) ^'^^ Pgfe V-i) be two points in rectangular coordinates, and let d = PiPg- Through ^2 Pj and Pg draw lines parallel, respec- tively, to the X- and 2/-axes to intersect in M. Fig. 20. Then d = Vp^ii^f ' + j/pf. But PiM= X2 — a^i, JtfPa = 2/2 — 2/i- .-. d= V(:«i- 0^27 + (2/1 -2/2)^ 24. Distance between two points. General formula in polar coordinates. O2. ^2) Fig. 21. Let the two points be (rj, ^1) and (rg, dg)? and let the distance between them be d. (Fig. 21.) PROJECTIONS. LENGTHS AND SLOPES OF LINES 21 There are two cases to consider : according as the difference between the vectorial angles is less than or greater than 180°. In the first case d2 = rlfrl-2 r^r^ cos {0^ - O.), and in the second case d^ = rf-^rl-2 r.r^ cos [360° - (O^ - ^i)]. These reduce to the one form d = Vrf + r| — 2 r^rg cos (^i — O2). EXERCISE IX 1. Find the distance between (—4, 1) and (3, 5), in rectangular coordinates. 2. Find the distance between (3, 2) and (— 4, — 5), in rectangular coordinates. 3. From a certain point three other points, A, J5, and O, are located as follows : A lies 3 mi. N. and 2^ mi. E. from 0, B lies 4 mi. S. and 1^ mi. E. from 0, and C lies 5 mi. W. and 1^ mi. N. from 0. Find the dis- tances between the points A, B, and C, and the distance of each of the points from correct to hundredths of a mile. 4. Find the distance between the points whose polar coordinates are (4, 24°) and (-2, 40°). 5. Find the lengths of the sides of the triangle whose vertices are (5, —2), (—4, 7)j and (7, —3), in rectangular coordinates. 6. Find the lengths of the sides of the triangle whose vertices are (-2, 30°), (4, 25°), and (5, 115°). 25. The angle which one line makes with another. Definition. The angle which one line, L^, makes with another, L2, is the angle, not greater than 180°, measured coun- ter-clockwise from L2 to Li. Thus, in Eig. 22, 6 is the angle which Li Fia. 22. 22 ANALYTIC GEOMETRY makes with L^. The supplement of 6 is the angle which L^ makes with L^. 26. Inclination and slope of a line. The angle which a line makes with the x-axis, or with any line parallel to the a>axis, is called the inclination of the line. This angle is to be measured from the positive direction of the avaxis toward the positive direction of the ?/-axis. In rectangular coordinates, the slope, or gradient, of a line is the ratio of the change of the ordinate to the corresponding change of the abscissa of a point moving along the line. It is counted positive if the ordinate increases as the abscissa in- creases ; negative if the ordinate decreases as the abscissa in- creases. Thus, if, as a point moves along a line, the ordinate increases one unit to an increase of 3 units in the abscissa, the line has a slope of i ; while if the ordinate decreases 1 unit to an in- crease of 3 units in the abscissa, the line has a slope of — \. The inclinations of these lines are, respectively, 6 = tan-^i = 18° 26', and 6' = tan-^ (- i) = 161° 34'. (Fig. 23.) 3 Fig. 23. From the definitions of inclination and slope it follows that slope = tangent of inclination, or, designating the inclination of a line by and its slope by m, m = tan 0. If the axes are not rectangular, the equation, slope = tangent of inclination, is taken as definition of the slope. PROJECTIONS. LENGTHS AND SLOPES OF LINES 23 27. Slope of a line through two points in terms of the rec- tangular coordinates of the points. Let the two points be Pj {x^, y^) and P^ (x.,, y^. Through Pj and P^ draw lines parallel to the coordinate axes to meet in M. (Fig. 24.) Then whether the slope is positive or negative its value is given by slope formula MP, y. V\ PiM HC2 — oci If Pi is the higher point, then slope = ^^^~'^- , which is the same as the above. ^^ ~ ^- Therefore, in rectangular coordinates, the slope of a line through two points is the difference of the ordinates of the points divided by the corresponding difference of the abscissas of the points. 28. Point dividing a line in a given ratio.* Example. To find the point which divides the line from (- 1, 5) to (6, - 4) in the ratio 3 : 2. Let (-1,5) be P^, (6, - 4) be P„ and let the required point be P{x, y). Then, by hypothesis, PiP^3 PP2 2* Through P, P^, and Po, draw lines parallel to the axes as in Fig. 25. Then, from similar triangles, MP ^P.P^S NP2 PP2 2' * In this article and in several following articles the word "line" is fre- quently used in the sense of " segment of a line," where there is no doubt of the meaning. Fig. 25. 24 ANALYTIC GEOMETRY and I.e. and MP^_ _^i^_ 3 NP ~PP2~ ^2' x + l_ 3 6-x ^2' 5-y_ 3 jV + 4 ~2' from which x = 3^, ?/ = — f . Hence the required point is (3^, — |). 29. External division. The point Pis said to divide the line P1P2 externally when it lies on the line produced. (Fig. 26.) The segments into which P divides P1P2 are defined to be PiP and PP2. The first segment is that from the beginning of the line to the point of division, and the second segment is that from the point of division to the end of the line. Since these segments are measured in oppo- site directions, they are opposite in sign. Hence their ratio is negative. The first and second segments must correspond respectively to the first and second terms of the given ratio into which P is to divide PiP2' 30. Example of external division. To find the point which divides the line from (—1, 5) to (4, 7) in the ratio — |. Let (—1, 5) be Pj, (4, 7) be P2, and let the required point be P{x, y). ' Then P^=JL PP. 3 Since P^P must be numerically less than PP^, P must lie nearer to Pi than to Pg, i.e. P must lie on the portion of the line extended through Pj. Fig. 26. PROJECTIONS. LENGTHS AND SLOPES OF LINES 25 Project the segments so as to obtain their a> and i/-projec- tions. (Fig. 27.) Fig. 27. Then M^P P,P PM^ PP^ 2 ~3' and M,P,_P,P_ P^M^ PP^ 2 3* -•• x^l 2 4-a; '3' and 5-.y 2 from which if=-ll, v = 1 Hence the required point is ( — 11, 1). EXERCISE X 1. Find the point which divides the line from (—3, 1) to (6, — 5) in the ratio— |. Ans. (12,-9). 2. Show that the point which bisects the line joining (xu Vi) and 3. Find the ratio in which the line from (2, 0) to (6, 0) is divided by (1, 0) ; by (5, 0) ; by (9, 0). 'A. The point P(2, k) is on the line joining Pi(— 2, 3) and Pa (4, — 7) ; find the ratio into which P divides Pi Pa, and the value of k. 26 ANALYTIC GEOMETRY 31. General formulas for a point dividing a line in a g'ven ratio. Let the line from Pi(it'i, 2/1) to Pal^^a? 2/2) be divided by P(x, y) in the ratio r : 1. There are three cases to consider: (1) P between Pj and P2, (2) P on the line produced through P^ (3) P on the line produced through Pg. In (1) r may have any positive value, in (2) r is negative and numerically less than 1, in (3) r is negative and numerically greater than 1. M, M Project PiP and PP^ on any two lines parallel to the axes. (Fig. 28.) In either of the three cases, ¥iM-P^-r and ^^^-^^^-r or from which - = r, and ^ — ^ = r, ^2 — ^ 2/2-2/ 3C EXERCISE XI 1. Find the point which divides the line from (- 1, 3) to (6, — £) in the ratio 3:2. PROJECTIONS. LENGTHS AND SLOPES OF LINES 27 2. Find the point which divides the line from (3, |) to (- 5, 8) in the ratio — f ; in the ratio — |. 3. Find the external point on the line joining Pi(a, b) and P2(c, d) which is n times as far from Pi as from F2. 4. Find the points which trisect the line joining Piixi, yO and P2(X2, y-z)' 5. The point P divides the line PiPz in the ratio r : 1 ; trace the varia- tion in r as P moves along the line internally from Pi to P2, then on from P2 to 00 , and then, changing to the other side of Pi, comes in from — go to Pi. 32. Angle between two lines of given slopes. Example. Let two lines L^ and L.2 have slopes respectively ; to find the angle which Li makes with ig- Let Li and L2 make angles Oi and O2 respectively with the a>axis, and let the angle which L^ makes with io he . Through the intersection of the lines draw a line parallel to the a;-axis. (Fig. ■ 29.) Then it is seen that (ft = 61 — do- Hence tan = tan (^1 — ^2) tan ^, — tan Oo Fig. 29. But 1 -f- tan ^1 tan O2 tan <9i = - 2, tan O2 = 3. -2-3 . tan <^ 1-2.3 = 1. 33. The angle between two lines. General formula. Let two lines, Li and L2, have slopes mi and mg respectively ; to find the angle which Li makes with L^. Let the angles which Lj and Lo make with the avaxis be Oi and O2 respectively. Then m^ = tan O.1, m^ = tan O.2. 28 ANALYTIC GEOMETRY Let <^ be the angle which Li makes with L^. Through the intersection of L^ and L.2 draw a line parallel to the a>axis. Then (Fig. 30), case (i), case (ii), and in either case Fig. 30. 6i > ^2J tan (j> = tan (^1 — O2) __ tan $1 — tan $2 1 + tan $1 tan 62 _ mi — m2 1 + miTJia 34. Condition for parallel lines, and for perpendicular lines. If the two lines of the preceding article are paral- lel, tan 61 = tan O2, and hence mj = mg. If the two lines are perpendicular, tan <^ = tan 90° = 00 , and hence 1 + mimg = 0. Conversely, if mi = ma, tan = 0, . • . = 0, and therefore the lines are parallel. If l4-mim2 = 0, tan<^ = oo, .'. <^ = 90°, and therefore the lines are perpendicular. Therefore, the condition that two lines of slopes mi and mg be parallel is mi = m2 ; the condition that they be perpen- dicular is 1 + Wjma 0, or mi = < PROJECTIONS. LENGTHS AND SLOPES OF LINES 29 EXERCISE XII For rectangular axes. Draw a figure in each case. 1. Show that the line joining (3, 2) and (— 2, — 13) is perpendicular to the line joining (1, 3) and (4, 2). 2. Show that (— 1, — 2), (3, 2), and (—3, 0) are the vertices of a right triangle. Find the other angles. 3. Where does a line cut the x-axis if it passes through (2, — 3) and is parallel to the line through (—1, 5) and (4, — 2) ? 4. A line is drawn perpendicular to the line through Pi ( — 2, 5) and P2(4, — 3) at its middle point ; find a point P on this perpendicular whose abscissa is 3, and show that P is equidistant from Pi and P^. 5. The vertices of a triangle are (7, 4), (— 2, — 5), and (3, — 10) ; show that the line joining the middle points of two sides is parallel to the third side, and is half as long, by using formulas for slope and distance. 6. Find a fourth point which with the three given in example 5 form the vertices of a parallelogram. 7. Two lines, Li and X2, make tan-i 2 and tan-i — 4 respectively wdth the ic-axis ; find the angle which Ly makes with L^. 8. The vertices of a triangle are Pi(— 1, 5), P2(3, —4), and P3(6, 2) ; find the slopes of the sides and the angle at Pi. 9. Show by their slopes that the line joining (—3, 4) and (6, 1) is parallel to the line joining (7, 2) and (5, |). 10. A line L makes an angle of 45° with the line through (1, 1) and (6, 8) ; find the slope of L and the angle which it makes with the x-axis. 11. Li passes through (4, 5) and (6, — 3). L-2. is perpendicular to Li ; find the slopes of Li and L^. 12. Zi has a slope m. The angle which L^ makes v\dth the a>axis is double the angle which L^ makes with the a;-axis ; what is the slope of L^ ? 13. The slope of one line is 3.728 and of another — .324 ; find the acute angle between them. 14. Find the slope of a line which makes an angle of — 42° with a line of slope .4364. 15. A line passes through (6, —. 3) and has a slope .324 ; find a point on the line with abscissa 1.2. 16. A line cuts the x-axis at (a, 0) and makes tan-i m with the ic-axis ; find where it cuts the ?/-axis. 17. A line passes through (a, 0) and makes tan-^w with a line of slope n ; find its slope, and where it cuts the y-axis. 30 ANALYTIC GEOMETRY III. AREAS OF POLYGONS 35. Area of a triangle in terms of the coordinates of its vertices. Example 1. To find the area of a triangle whose vertices in rectangular coordinates are Pi (—2, 3), P2{4:, —1), and Ps(l,-6). ■ Through the lowest vertex, Pg (Fig. 31), draw a line parallel to the ic-axis, and from the other vertices drop perpendiculars to this line, meeting it Y Pi (-2, 3> -1) in 3/i and M^. Then the area re- quired is equal to area of M^P^PiM^ — area of P.M2P2 — area of MiP^Pi = i 3f,M2(M,P,-^ M2P2) -\P,M2 -iM,P, = 1.6(9 + 5) 5 M2P2 M,P, -i-3 -i-3 = 21. Fig. 31. If P1P2P3 a triangular 9. represents field to a scale of 1 space = n ft., then the area of the field is 21 n^ sq. ft. Example 2. To find the area of the triangle whose vertices in polar coordinates are (3, 60°), (-2, 125°), and (5, 215°). The area required is the sum of the areas of the triangles OP2P,, OPJ\, 0P,P2 (Fig. 32). The area of a triangle is equal to one half the product of two sides and the sine of the included angle. PROJECTIONS. LENGTHS AND SLOPES OF LINES 31 P, (3, 60°) P, (-2, 125°) Pg (5, 215°) Fig. 32. .*. the required area = 10P,' OF, sin P.OPs + iOPs' OP2 sin P3OP2 +10P2- OPisinP^OPi = i . 3 . 5 sin 155° + 1 . 5 • 2 sin90° + i • 2 . 3 sin 115° = i (15 sin 25° 4-10+6 cos 25°) = 10.89. EXERCISE XIII 1. Find the area of the triangle whose vertices in rectangular coordi- nates are (3, - 5), (— 8, 6), and (9, 2). 2. Find the area of the triangle whose vertices in polar coordinates are 3. Find the area of a triangle whose vertices in rectangular coordinates are (0, 0), (xi, yi), and (x2, y^). 4. Find the area of a triangle whose vertices in polar coordinates are (0, 0), (n, ^i), and 0-2, ^2). 5. Find the area of the quadrilateral whose vertices in rectangular coordinates are (-2, 5), (7, 9), (10, -3), and (-6, -9). 36. Area of a triangle. General formula in rectangular coordinates. Let Pi(.Ti, ?/i), P-zipo, 1/2), and P,{x.^, y,) be the ver- tices of a triangle in rectangular coordinates ; to find the area of the triangle. 32 ANALYTIC GEOMETRY Through the lowest vertex (Pg in I'ig- 33) draw a line paral- lel to the a^-axis, and from the other vertices drop perpendicu- lars to this line, meeting it in M^ and M^. Then M,P, area of triangle P1P2P3 = area of trapezoid M^P-^P^M^ + area of triangle P^P^M^ — area of triangle P2M^P^ = ^{M,P, + JW3P3) • M,M, + i P^M, . M,P, - 1 P^M, = K (2/1 - :^2 + 2/3 - 2/2) (^-3 - a^i) + (a^i - ^2) (2/1 - 2/2) -(^•3-a?2)(2/3-2/2)], or, area P^P^P^ = J (a'12/2 + a^gl/s + ^32/1 — ^iVs — ^iVi — ^3^2)- This may be written in the determinant form i»i 2/1 1 1 a;2 2/2 1 ^s 2/3 1 In Fig. 33 the succession of subscripts 1, 2, 3, is obtained by going around the triangle counter-clockwise. If the points had been so lettered that in following the above order it would be necessary to go around the triangle clockwise, the area would have been found to be minus the above expression. This can be seen to be true by exchanging two of the sub- scripts, say 1 and 2, in Fig. 33, and making the same exchange in the formula. The change in the figure changes the order PROJECTIONS. LENGTHS AND SLOPES OF LINES 33 from counter-clockwise to clockwise, and the change in the formula just changes the sign of the whole expression. 37. Area of a triangle. General formula in polar coordi- nates. Let Pi(ri, $i), P^ir^, O^), and Pg^rg, ^3) be the vertices of a triangle in polar coordinates ; to find the area of the triangle. Two cases are to be distinguished, according as the pole lies without or within the triangle. The second case will occur only when the difference between the vectorial angles of two of the vertices is greater than 180°. Fig. 34. In case (1) the area of the triangle P^P^P^ is equal to the area of triangle OP^P-i -\- area of triangle OPzPs — area of triangle OP1P3 = i r^u sin (02 - ^1) + i r^r^ sin (^3 -O^)-^ r^r^ sin (^3 - Oi) = i [ri^o sin (^2 - ^1) rh r^r^ sin (^3 —^^2) 4 ^31-1 sin (^1 - ^3)]. In case (2) the area of triangle OP1P3 must be added to the areas of the other two triangles, instead of subtracted from them, as in case (1) ; but area of OP1P3 is here equal to i ViV^ sin [360° — (^3 — ^1)] which is equal to — ^ ^Vg sin (^3 — ^1). The formula for the area of the triangle sought reduces there- fore to the same as in case (1). Just as in the case of the area in rectangular coordinates, the above formula would give the negative of the area if the sub- 34 ANALYTIC GEOMETRY scripts were so arranged that in following the order 1, 2, S, it would be necessary to go around the triangle clockwise. 38. Area of a polygon. General formula in rectangular coordinates. If the origin be one of the vertices of a triangle whose other vertices are Pi (x^, y^ and P2(x*2, 2/2), the. formula for the area of the triangle given in Art. 36 becomes provided that in going around the tri- angle counter-clockwise the vertices are passed in the order P^, P^, and 0. This area of the triangle OP1P2 may be thought of as generated by a line OP, initially in the position OPi, turning counter-clockwise about to the final position OP2, the point P moving along the line PiP2- With this conception of the area, it must be noted that it is the ab- scissa, x^, of the initial position, Pj, of P which comes first in the formula for the area, ^(xiy2 — .T22/1). If the line OP must turn clockwise from the position OPi to the position OP2, then the expression -J (a^iif/2 — ^^yi) is equal to the negative of the area of the triangle OP1P2. Let ^(«i2/2 — ^22/1) b® den'oted by A. Thus Consider now any polygon whose vertices in rectangular co- ordinates are Pi(xi, y^), P2(x2, y^, ••• Pn^^m 2/n)j the vertices being so lettered that in going around the polygon counter- clockwise the vertices are passed in the order Pj, Po, ••• P„. For definiteness let r? = 6, and let the polygon be as shown in Fig. 37, the origin being outside of the polygon. Let a point P PROJECTIONS. LENGTHS AND SLOPES OF LINES 35 start at Pj, traverse the perimeter of the polygon counter- clockwise, and return to P^. The line OP generates in order the triangles OP^Pi, OP^P^,, ••• OP^P^. Now the area gener- ated by OP which lies without the polygon is generated twice, with OP turning once clockwise, once counter-clockwise; or else is generated four times with OP turning twice clockwise, twice counter-clockwise ; but the area within the polygon is generated once, with OP turning counter- clockwise; or else is generated three times, with OP turning once clockwise, twice counter- clockwise. Therefore if the expression A be formed for each of the triangles OP1P2, OP.Ps, ••• OP^Pi, and their sum taken, all the area generated by OP will be cancelled out except that within the polygon and that area will be counted just once. Therefore the area of the polygon is equal to i (^m - ^'22/1 + a?22/3 - X^V2 + ^sVi - ^42/3 + ^42/5 " ^oVa + ^52/6 " ^sVs 4-a:a2/i-^i2/6)- Fig. 37, A convenient method of arranging the coordinates for the computation of the area is as follows : Write down in succession the abscissas of the vertices taken in order counter-clockwise around the polygon, rejieating the first abscissa at the last; under the ab- scissas write the corresponding ordinates : Xi X2 Xs x^ x^ Xq a?! 2/1 2/2 2/3 2/4 2/5 2/6 2/1 Then multiply each abscissa by the following ordinate and take the sum of the terms obtained; multiply each ordinate by ihe fol- lowing abscissa and take the sum of the terms obtained. The area is half of the first sum minus half of the second. 36 ANALYTIC GEOMETRY EXERCISE XIV 1. The vertices of a polygon taken in order are (6, 1), (9, —4), (3, - 10), (- 3, - 5), (- 6, - 8), (- 12, 0) and (- 4, 6) ; find the area of the polygon. 2. The distances north of a fixed east and west line of four points A^ B, C, D are respectively 32.6 ft., 65.1 ft., 80.3 ft., 51.7 ft., and their dis- tances east of a fixed north and south line are respectively 25.3 ft., 48.2 ft., 94.5 ft., 106 ft.; find the area of the quadrilateral ABCD. * 3. The distances of four points A, B, C, D from a point O are respec- tively 120 ft., 216 ft., 320 ft., and 65 ft., and their directions from are respectively E. 25° N., N. 32° W., S. 74° W., E. 67° S. ; find the area of ABCD. 4. The vertices of a triangle are (3, — 2), (—4, 1), and (—8, ~ 5) ; find (a) the area, (h) the lengths of the sides, (c) the slopes of the sides, (d) the angles. 5. Show (a) by the lengths of the sides, (6) by the slopes of the sides, that the quadrilateral whose vertices are (1, 2), (3, — 2), (— 1, — 3), and (—3, 1) is a parallelogram. Find its area. 6. Show by means of the slopes of the lines that the line joining the middle points of two sides of any triangle is parallel to the third side. Show also that its length is half that of the third side. 7. The vertices of a triangle are Pi, P2, Ps ; find the point which di- vides the line from Pi to the middle point of P2P3 in the ratio 2 : 1. Show that, using either of the vertices in like manner, the same point is obtained, and hence that the three medians of a triangle meet in a point. 8. In the formula for the area of a triangle in rectangular coordinates, substitute the values of the rectangular coordinates in terms of the polar coordinates and obtain the formula for the area of the triangle in terras of polar coordinates. 9. The line joining (a, &) and (c, d) is divided into four equal parts ; find the points of division. 10. Show analytically that the middle points of the sides of any quad- rilateral are the vertices of a parallelogram. 11. Prove that the middle point of the line joining the middle points of two opposite sides of any quadrilateral has an abscissa equal to one fourth the sum of the abscissas of the vertices of the quadrilateral, and find the similar relation for the ordinates. What conclusion can you draw ? PROJECTIONS. LENGTHS AND SLOPES OF LINES 37 12. The point (2, k) is equidistant from (— 5, 7) and (3, 4) ; find k. 13. The point (x, y) is equidistant from (2, — 1) and (7, 4) ; write the equation which x and y must satisfy. What is the locus of (a;, y) ? 14. Express by an equation the condition that the point (a;, y) is dis- tant 5 from (2, 3). What is the locus of the point (a:, y) ? 15. Show that the line joining (4, — 4,) and (— 2, — 1) is perpendicu- lar to the line joining (3, 1) and (I, — 3). 16. Find the angle which the line whose slope is 6,324 nlakes with the line whose slope is — .657. 17. Find the slope of a line which makes an angle of 30° with a line whose slope is 3. 18. The line Lx makes an angle of 40° with the a;-axis, and the line L^ makes an angle whose tangent is 2 with Li ; find the slope of L^. 19. If Li makes tan-i a with the a;-axis, and L^ makes tan-^ h with Lx^ find the slope of L». 20. The angle from Lx clockwise to L^ is tan-i (|), and the angle from L2 counter-clockwise to the x-axis is tan-i(— f) ; find the slope of Li. CHAPTER III GRAPHICAL REPRESENTATION OF A FUNCTION; EQUATION OF A LOCUS 39. Function and variable. One quantity is said to be a function of a second quantity when to every value of the second there corresponds one or more values of the first. Thus in the equation v = gt, which expresses the velocity of a body falling freely in a vacuum in terms of the time, the velocity, v, is a function of the time, t. Again, in the equation pv = a constant, the formula which expresses the relation between the pressure and volume of a gas kept at constant temperature, either of the quantities p or v is a function of the other one. The quantity which may take, or to which may be assigned, arbitrary values is called the independent variable, or often simply the variable, and a function of this variable is often called the dependent variable. According to the above definition of a function any constant may be regarded as a function which takes the same value for all values of the variable. Ii to every value of the variable there is just one value of the function, the function is said to be a single-valued function of the variable. If two, three, or more values of the function exist for every value of the variable, the function is called re- spectively a double-valued, triple-valued, or, in general, a multiple-valued function of the variable. Thus in v=32t, v is a single-valued function of t, and in 2/2 = 4 £c, y is a double- valued function of x. On the other hand, a; is a single-valued function of y, if y be taken as the independent variable. 38 GRAPHICAL REPRESENTATION OF A FUNCTION 39 40. The graph of a function. It is not always possible to express by means of an equation the value of a function in terms of the variable. When, however, there are known several pairs of corresponding values of two quantities, one of which depends upon the" other, a graphical representation of one of the quantities as a function of the other may be made which will exhibit in an instructive way the dependence of one of the quantities upon the other. To illustrate this consider the following examples. Example 1. It was found that when a certain rod of steel was subjected to tension, the values of the extension of the rod in terms of the tension were as shown in the following table, in which T is the number of pounds of tension per square inch of cross-section of the rod and e is the number of units of extension per unit length of the rod, the initial tension being 1000 lb. T 1000 5000 10,000 20,000 30,000 40,000 50,000 51,000 c .0003 .0009 .0019 .0030 .0040 .0053 .0056 T 52,000 54,000 56,000 58,000 60,000 70,000 80,000 £ .0058 .0064 .0075 .0089 .0113 .0272 .0500 Take the values of € as abscissas and the values of T as ordinates and plot the points representing the corresponding values of c and T. Then draw a smooth curve through these points. On the assumption that as the tension changes grad- ually, passing through all values between the first and last values of the tension that are given, the extension also changes gradually, the smooth curve through the plotted points may be taken as a graphical representation of T as a function of c in the sense that the coordinates of any point on the curve are corresponding values of c and T. In general the more points that are determined by known values of the variables the more accurately will the curve represent the function. Of course, too, these points should be somewhat evenly separated. 40 ANALYTIC GEOMETRY Outside the range of values given, no information can be drawn from the curve concerning the values of the function for a given value of the variable. 80,000 60,000 ■-20,000 jm .02 .03 elojigation injncheb Fig. 38. M .05 The curve does not give any information that is not con- tained in the table, but gives the same information in such a way as to bring out relations that are not readily observed from the table. From the curve it is seen that as long as T is less than about 50,000 the extension is proportional to the tension, the points of the curve lying on a straight line approximately, but that when T passes through the value 50,000 the extension increases more and more rapidly as T increases. Also the value of T corresponding to an assumed value of c may be found approximately from the curve by measuring the value of the ordinate of the point of the curve which has the assumed value of c as abscissa. Likewise the value of e corre- sponding to an assumed value of T may be found. Example 2. The following table shows the number B of GRAPHICAL REPRESENTATION OF A FUNCTION 41 beats per minute of a simple pendulum of length L centi- meters for certain values of L : L 10 12 15 20 25 30 40 50 60 70 80 90 100 B 190 172 154 136 120 110 95 85 78 72 67 63 60 200 \ \ \ \ 150 > \ \ a. \ 2 K ^ 100 \ N \ I ■^ ^ >^ ■-J s. 'S p— - 50 25 50 LENGTH IN CM. Fig. 39. 75 Take the values of L as abscissas and the values of B as ordinates and plot the points representing the corresponding values of L and B. The curve drawn through these points shows graphically the manner in which B depends upon L. It also enables one to pick out approximately the value of B for a given value of L within the limits given, or the value of L for a given value of B. 41. Equation of a locus. In each of the two preceding ex- amples a curve was drawn such that the coordinates of all of its points were corresponding values of the function and variable, but no equation was found which expressed the dependence of the function upon the variable. In each of the examples to be next studied some simple locus of points will be considered, and the equation which expresses the dependence of the ordinate of any point of the 42 ANALYTIC GEOMETRY locus Tipon the abscissa of the point will be derived. This equation will be known as the equation of the locus. Definition. The equation of a locus is an equation between the coordinates of any point of the locus. The locus, on the other hand, is called the locus of the equa- tion. 42. Two fundamental problems. The two fundamental prob- lems of Plane Analytic Geometry are : (1) Having given a locus of points determined by certain geometric conditions, to find the equation of that locus. (2) Having given an equation in two variables, to find by a study of the equation the form and properties of the locus which it represents. In this chapter some examples illustrating the methods of finding the equation of a given locus will be considered, and in the next, chapter some methods of obtaining the locus when the equation is given will be studied. 43. Illustrations. Example 1. Consider the locus of a point which moves along the straight line passing through the points Pi (3, -1) and P^ Y Pa (-5. 4) (-5, 4). If any point P(x, y) be taken on this line, the value of the or- dinate clearly de- pends upon the value of the abscissa of the point. That is, 1?/ is a function of x. To find the law, or equation, which expresses the depen- dence of y upon X, draw through P, Pj, and P^ lines parallel to the axes to form the triangles PM^P^ and PiM^P^ as in Fig. 40. P(x, 1/) GRAPHICAL REPRESENTATION OF A FUNCTION 43 Then by similar triangles M.Po I.e. M,P{ (_5)_ 4-(-l) y x-3 -1 which reduces to 5 aj + 8 ?/ = 7. (1) If P{x, y) is a point not on the line through Pj and P^, the triangles PM^P^ and P^M^P^ are not similar, and equation (1) does not hold. Hence equation (1) holds for all points, on the line and for no others. It is therefore the equation of the line. The equation may be solved for y and written The equation is the law of the dependence of y upon x. It may be stated as follows : The ordinate of any point on the straight line jmssing through (3, —1) and (—5, 4) is equal to — I of the abscissa of the point plus |. Equation (1) might also be solved for x, which would ex- press X as a function of y. Example 2. Consider the locus of a point which moves so as to keep always at a distance 6 from the point Pi (3, 2). The locus is a circle with radius 6 and ■ with center at (3, 2). Here again the value of the ordinate of any point on the locus is a function of the ab- scissa of the point. To find the law that expresses the ordinate as a function of the abscissa, consider any point P (x, y) on the circle. The con- dition that Pmust fulfill is that Fig, 41. PiP=6. 44 ANALYTIC GEOMETRY Now P^P= -^{x - 3f + {y- 2)2. 0^^-3)2 4- (2/ -2)2 = 36. (2) Since eq. (2) is true for all points on the circle and for no others, it is the equation of the locus. If the equation be solved for y, the result is 2/ = 2± V36-(a;-3)2. This equation expresses 2/ as a function of x. Since there are two values of y for every value of a;, 2/ is a double-valued function of x. Equation (2) might be solved for x, and x be thus expressed as a function of y. Example 3. A point moves in the plane so as to keep equi- distant from Pi(3, — 2) and Pii— 4, 7) ; to find the equation of the locus. To find the equation of the locus, one must express by means of an equation which contains the coordinates of any point of the locus that geomet- ric condition which is satisfied by all points of the locus and by no others. This property is expressed by the equation P^P=P^P. Expressed in terms of the co- ordinates of the point P, this equation becomes V(a^-3)2 + (2/ + 2)2 = V(x' + 4)2 + (2/-7)l (1) Squaring both members, cancelling, and collecting, there results 7a;-92/-+-26 = 0, (2) which is the desired equation of the locus. For all values of x and y that satisfy (1) also satisfy (2). In retracing the steps from (2) to (1), a double sign is introduced which would give Fig. 42. GRAPHICAL REPRESENTATION OF A FUNCTION 45 P^P = ± P^P. But as P^P and P^P are positive distances, the equation containing the minus sign has no geometric signifi- cance. Equations (1) and (2) therefore are satisfied by pre- cisely the same points. The locus is known from plane geometry to be the straight line which is perpendicular to PiP^ at its middle point. Example 4. A point moves so that the sum of its distances from Pi (4, 0) and PgC— 4, 0) is always equal to 10 ; to find the equation of the locus. Let P (x,y) be any point of the locus. The geometric condition sat- isfied by all points of the locus and by no others is expressed by the equation P2P+PiP=10. Expressed in terms of the coordinates of the point P, this becomes V(a;-4y-f-2/' -f- ^{x-\-4:)\+f = 10. When freed from radicals, this equation becomes 9 ar^-f 25 2/^ = 225. This is the* equation of the locus. It will be shown in Art. 83 that no new points are introduced into the locus by squaring. A point which moves so that the sum of its distances from two fixed points is constant, describes an ellipse. The above locus is therefore an ellipse. Points of the locus may be obtained by describing arcs with Pi and P2 as centers and radii whose sum is 10. The inter- sections of two such arcs are points of the locus. Example 5. A point moves so that the difference of its dis- tances from Pi(5, 0) and P2(— 5, 0) is 8; to find the equation of the locus. Fig. 43. 46 ANALYTIC GEOMETRY Fig. 44. Let P(x, y) be any point of the locus. The geometric condition satisfied by all points of the locus and by no other Y V points is then This equation when expressed in terms of X and y and freed from radicals re- duces to 9a72_16/ = 144, which is the equa- tion of the given locus. It will be shown in Art. 87 that no new points are introduced into the locus by squaring. A point which moves so that the difference of its distances from two fixed points is constant, de- scribes an hyperbola. The above locus is therefore an hyperbola. Points of the locus may be obtained by describing arcs with P^ and P^ a-s centers and radii whose difference is 8. The points of intersection of two such arcs are points of the locus. Example 6. A point moves so that it remains always equidistant from Pj (6, 0) and the 2/-axis ; to find the equa- tion of the locus. Let P{x, y) be any point of the lo- cus. From P draw PM perpendicular to OY. Then the geometric condition to be satisfied by P is [ GRAPHICAL REPRESENTATION OF A FUNCTION 47 expressed by the equation MP=PyP. Expressed in terms of the coordinates of P{Xj y) this is x = ^{x-Qf + f, (1) which on squaring reduces to 2/2 = 12aj-36. (2) This is the equation of the' locus. That no new points were introduced into the locus by squar- ing eq. (1) may be seen as follows : Any values of x and y that satisfy (1) also satisfy (2), but there are values of x and y that satisfy (2) that do not satisfy (1). For in retracing the steps from (2) to (1) a double sign is introduced ; i.e. given eq. (2), there follows x^±^{x-QY + y\ Now it is evident geometrically that no point can be equi- distant from the ^/-axis and (6, 0) and have its abscissa negative. Therefore only the plus sign can be used. Therefore all points whose coordinates satisfy (2) also satisfy (1). No real values of X and y could therefore have been introduced into eq. (1) by squaring. A point which moves so as to keep equidistant from a fixed point and a fixed straight line describes a parabola. The above locus is therefore a parabola. 44. Method of finding the equation of the locus of points which satisfy a given condition. In finding the equation of the locus of points satisfying a given condition, a certain method was followed in the preceding examples. This method will suffice for finding the equation of the locus of points satisfying any condition, if that condition can be ex- pressed by means of an equation. The method may be formu- lated as follows : 48 ANALYTIC GEOMETRY To find the equation of the locus of points which satisfy a given condition, (1) Assume any point P on the locus. (2) Write the equation tvhich expresses the condition that P must satisfy. (3) Express this equation in terms of the coordinates of P and simplify the equation. 45. Intercepts of a locus on the axes. The abscissa of a point where a locus cuts the a>axis is called an jr-intercept of the locus. The ordinate of a point where a locus cuts the 2/-axis is called a /-intercept of the locus. If the equation of the locus is known, the ic-intercepts may be found by letting y equal zero in the equation and solving the resulting equation for x. Likewise the ^/-intercepts may be found by letting x equal zero in the equation and solving the resulting equation for y. EXERCISE XV Derive the equations of the following loci. Find the intercepts of the loci on the axes. Plot the loci. 1. A straight line through (1, 4) and (— 6, 7). 2. A straight line through the origin making an angle of 60*^ with the aj-axis. 3. The X-axis. The y-axis. A parallel to the a;-axis through (5, 2). 4. A straight line through (.3, — 5) with slope 2. 5. A straight line through (a, 0) and (0, 6). 6. A straight line through (0, h) with slope m. 7. A circle with radius 5 and center at (2, — 4). 8. A circle with center at (- 6, 4) and passing through (3, 1). 9. A circle with the ends of a diameter at (5, - 6) and (3, 12). 10. A circle with center at (h, k) and radius r. 11. A circle with center at the origin and radius r. 12. A circle tangent to both axes and radius r. 13. A circle tangent to the ^/-axis at the origin and radius r. GRAPHICAL REPRESENTATION OF A FUNCTION 49 14. The locus of a point which moves so that the sum of its distances from (0, 3) and (0, - 3) is 8. 15. The locus of a point which moves so that the difference of its distances from (0, 3) and (0, — 3) is 4. 16. The locus of a point which moves so as to remain always equi- distant from the point (0, — 4) and the x-axis. 17. The locus of a point which moves so that the sum of its distances from (3, 2) and (- 6, 1) is 12. 18. The locus of a point which moves so that the difference of its dis- tances from (2, 3) and (—5, — 1) is 6. 19. The locus of a point which moves so as to keep equally distant from (—3, 4), and the line parallel to the y-axis through (8, 6). 20. The perpendicular bisector of the line joining (1, 7) and (8, 2). 21. A column of concrete 50 in. long was compressed longitudinally and the following numbers obtained, in which P ■= number of pounds compression per square inch of cross section of the column, and e = num- ber of inches of compression, the initial load being 100 lb. per square inch. P 100 150 200 300 400 600 550 e .0007 .0015 .0034 .0057 .0080 .0093 P 600 600 650 700 800 900 1000 , - ., , e .0108 .0112 .0121 .0139 .0175 .0221 .0275 ^° "°^^ *^ ® Make a graph which shows P as a function of c, and get what informa- tion you can from the curve. 22. A steel rod of diameter .564 in., length 3 in., was subjected to a tensile force. The following measurements were made, in which P = number of pounds tension per square inch of cross section of the rod, X = number of inches extension, the initial load being 1000 lb. per square inch. P 1000 5000 10,000 20,000 30,000 40,000 36,000 37,000 X .0003 .0008 .0018 .0028 .0039 .0058 .0072 P 38,000 39,000 40,000 41,000 42,000 44,000 46,000 50,000 X .0114 .0559 .0596 .0615 .0669 .0800 .0905 .1210 Make a graph which shows P as a function of X. What information do you get from the curve ? E 50 ANALYTIC GEOMETRY 23. The following measurements were taken in an experiment in which an india rubber cord was stretched by hanging a weight to its end. W = weight in kilograms, L = length in centimeters. w .5 1.0 1.5 2.0 2.5 3.0 3.5 L 10 10.1 10.3 10.6 10.9 11.3 11.7 12.2 W 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 L 12.7 ia.3 13.9 14.6 15.3 16.1 16.9 17.9 Make a graph which shows TF as a function of L. 24. In Ex. 23 reduce W to pounds and L to inches, and draw the graph. How does the curve compare with that of Ex. 23? By what choice of scale units could you make the two curves coincide ? CHAPTER IV LOCUS OF AN EQUATION 46. The second fundamental problem. In the preceding chapter some equations of simple loci were obtained from the geometric conditions which the points of the loci satisfied. In this chapter the converse problem of finding the locus when the equation is given will be considered for some simple equa- tions. Illustrations. Example 1. To find the locus of the equation y = 2x-\-l. Any number of points whose coordinates satisfy this equa- tion may be found; for any value may be assigned to x and a corresponding value for y computed from the equation. A few corresponding values so obtained are : X 0, 1, 2, 1, 3, 5, 4, -3, --!/, y 1, 6, 5, 9, -5, -12. Plot the points determined by these pairs of values of x and y. They seem to lie on a straight line. That the locus of the equation is a straight line may be proved as follows : Draw a straight line through two points whose coordinates satisfy the equation, as Pj (0, 1) and P,(2, 5). (Fig. 46.) Take any point F(x, y) on this line and through it draw a line parallel to the a^axis. Erom Pi and P.J drop perpendiculars to this line, meeting it in Jfj and M^. Then from similar triangles, PM^P^ and PM,P,, M^P M,P' x-2 x-0' 51 V ' /p /' T __i 1 X r/ ^4f ii._|m^- 1 Fig. 46. 52 ANALYTIC GEOMETRY which reduces to y = 2x + l. This equation, therefore, holds for every point on the line. Conversely, all points whose coordinates satisfy the equation lie on the line ; for if a point P(x, y) be taken not on the line, the triangles PM^P^ and PM^P^ are not similar, and hence the above equation does not hold. Hence the equation y — 2 x-\-l is satisfied by all points on the straight line through (0, 1) and (2, 5) and by no others. The line is therefore the locus of the equation. Example 2. To find the locus of the equation, This equation may be brought into a form like that of eq. (2) of Art. 43, by completing the squares in the terms containing x and in those containing y as follows, a:^ _ 6 aj 4". 9 + ?/2 + B y + 1 6 = 24 + 9 -f 1 6, or (a;-3)2 + (y + 4)2 = 49. Now the left-hand member of this equation is equal to the square of the distance from (x, y) to (3, —4), and the equation therefore states that this distance is equal to 7. Hence (x, y) must lie on the circumference of a circle with center at (3,— 4) and radius 7. Moreover, the coordinates of any point on this circle satisfy the equation. Hence the circle is the locus of the Fig. 47. eqniation. (Fig. 47.) Example 3. To plot the locus of r 4 X. LOCUS OF AN EQUATION 53 The following pairs of values of x and y are obtained by ar- bitrarily assigning values to x and computing the corresponding values of y. X 0, 1, 1, 2, 3, 4, 5, 6, 10, y 0, ±1, ±2, ±V8, ±Vi2, ±4, ±V20, ±V24, ±V40: From the equation the following facts are readily seen to be true: (1) If X is negative, y is imaginary ; therefore no part of the locus lies to the left of the y-Sixis. (2) Every positive value of x gives two values of y which differ only in sign ; there- fore the points of the locus lie in pairs such that the X-axis bisects at right angles the lines joining the pairs. (3) As X increases, the positive value of y also in- creases, and as x becomes infinite, y also becomes in- finite ; the locus therefore recedes indefinitely from both axes as x increases in- Y ^^-^^ ^^^ ^^ ^^ ' / z X ^ s \ ^ ^:^ : "^^^ '^^^ FiQ. 48. definitely. (4) A small change in x makes a small change in y. The part of the locus which lies in the first quadrant may, therefore, be thought of as generated by a moving point which, starting at the origin, moves along a curve gradually rising as the point moves to the right and passing through the above calculated points. The part of the locus which lies below the a^axis could be obtained from that above the a;-axis by folding the upper part of the plane over upon the lower part, using the a;-axis as an axis of revolution. The locus is therefore approximately the curve of Fig. 48. 54 ANALYTIC GEOMETRY EXERCISE XVI Prove that the locus of each of the equations from 1 to 6 is a straight line. Find the intercepts of the lines on the axes and draw the lines. 1. 3x-4y^6. ^ ^ + y-l 4. y = 7a; + 3. 2. 2x + 5y = 12. '34~' 5. ix-Sy + 9 = 0. Prove that the locus of each of the following equations is a circle, and find the center. and radius. 6. x^-{-y^-4x = 0. 9. x^-\-y^-2ax-2by = r^-a^-b^. 7. x^-\-y2-8x \-2y = S. 10. x:^ + y^ + x -Sy = 1. 8. x^ + y^ = r^. 11. x2 + y^-2ax = 0. Plot the loci of the following equations : 12. 2/2 = 4(x-2). 14. a;2 = 8(y-4). 16. x^=-y. 13. x^ = 6y. 15. y^ = -4x. 17. a:-3 = 2 (y + 1)2. 18. y^ = mx, letting m = j\, 1, 4, 16, 100, — 1, - 100. 19. x^ = my, letting w take different values. 20. x2 + 4y2 = i6. 21. x^-4y^=l6. 47. Symmetry. Before taking up more difficult problems in loci it will be well to discuss briefly the subject of symme- try of a curve with respect to a line 9,nd with respect to a point. Two points are said to be symmetric with respect to a given line when the given line bisects at right angles the line joining the two points. Two points are said to be symmetric with respect to a given point when the given point bisects the line joining the two points. A locus of points is said to be symmetric with respect to a given line when all points of the locus lie in pairs which are symmetric with respect to the given line. The line is then called an axis of symmetry. A locus of points is said to be symmetric with respect to a given point when all points of the locus lie in pairs which are symmetric with respect to the given point. The given point is then called a center of sjrmmetry Illustrations, (a) The points (x, y) and (— x, y) are sym- LOCUS OF AN EQUATION 55 metric with respect to the y-a,xis, the points (x, y) and (x, — y) are symmetric with respect to the aj-axis, and the points {x, y) and (— X, — y) are symmetric with respect to the origin. (p) In 2/^ = 4 X, if (Xj y) is a point of the locus, so also is {x, —y)\ for if the coordinates of either point satisfy the equation, so do the coordinates of the other. The locus is therefore symmetric with respect to the ic-axis. (c) In x^-\-4:y'^ = 16, if {x, y) is a point on the locus, so are {—X, y), (x, —y), and (—a*, —y)', for if the coordinates of the first point satisfy the equation, so do the coordinates of each of the other points. The locus is therefore symmetric with respect to the y-axis, with respect to the avaxis, and with respect to the origin. 48. Tests for symmetry with respect to the coordinate axes and the origin. If an equation is such that it is unchanged by replacing a; by — x, the locus of the equation is symmetric with respect to the ^/-axis. For, whatever value, say x^, be given to x, the resulting equation which determines the cor- responding value, or values, of y will be the same equation as that obtained by substituting — x^ for x. Hence Xi and — ajj give the same values of y. Similarly, if replacing y hy — y leaves the equation un- changed, the locus is symmetric with respect to the avaxis. If replacing a; by — a; and y hj —y leaves the equation un- changed, the locus is symmetric with respect to the origin. In particular, if an equation contains only even powers of x, the locus is symmetric with respect to the y-axis. If it con- tains only even powers of y, the locus is symmetric with respect to the avaxis. If the terms of an equation are all of even degree, or are all of odd degree in x and y, the locus is sym- metric with respect to the origin. (In applying this last test a constant term must be considered as of even degree.) 49. Discussion of an equation. When it is desired to plot the locus of an equation in two variables, it is well to discover 56 ANALYTIC GEOMETRY as many properties and facts concerning the locus as one can by a study of the equation. Some important things to look for are (1) Symmetry. (2) Points where the locus crosses the axes. (3) What values, if any, of one variable make the other imaginary. (4) What finite values, if any, of one variable make the other infinite. (5) How increasing or decreasing one variable will affect the other. (6) What value, if any, does one variable approach when the other variable becomes infinite. 50. Illustrations. Example 1. To plot the locus of ar^-f.4/ = 16. (1) If the equation be solved for x and y, respectively, there results x = ± 2^/^-f (2) and y=± i V16 - af. (3) (1) Equation (1) shows the curve to be symmetric with respect to both coordinate axes and the origin. (2) If y = 0, x=±4:', if x = 0, y = ±2. Hence the curve meets the axes at (4, 0), (-4, 0), (0, 2), and (0, -2). (3) Equation (2) shows that if 2/^>4, x is imaginary. .*. y cannot be greater than 2 nor less than — 2. Likewise, eq. (3) shows that x cannot be greater than 4 nor less than — 4. (4) No finite value of either variable can make the other infinite. (5) From eq. (3) it is clear that as x increases gradually from to 4, taking all values in that interval, the value of y represented by the positive radical steadily decreases from 2 toO. LOCUS OF AN EQUATION 57 (6) Values of x and y are excluded from becoming infinite by (3). The part of the locus that lies in the first quadrant may then be thought of as generated by a point which, starting at (0, 2), moves gradually to the right and downward, until it reaches (4, 0). A few additional points through which the curve passes will then suffice for a fairly accurate drawing of the curve. A few points computed from eq. (3) are X 1 2 3 3.5, y 1.9 1.7 1.3 .96. The curve is therefore approximately as shown in Fig. 49. The curve is an ellipse, as will be shown later. -Ip^^ ^^ ir- / J^ ^l ^ ^^ ^^ y ''^-^ .^---^ (1) Fig. 49. Example 2. To plot the locus of a^-4.y^ = l&. Solving for x and y, respectively, aJ=±2V/T4, (2) 2/=±iVar^-16. (3) (1) Equation (1) shows the curve to be symmetric with re- spect to both coordinate axes and the origin. (2) If a; = 0, 2/ is imaginary; if ?/=0, x= ±4. Hence the locus does not meet the 2/-axis, and meets the a;-axis in (4, 0) and (-4,0). (3) From eq. (2) it is evident that x is real for all real values 58 ANALYTIC GEOMETRY of 2/, and from eq. (3) that y is imaginary for all values of A between — 4 and 4, and is real for all other values of x. (4) No finite values of either variable makes the other in- finite. (5) Considering the value of y corresponding to the positive sign of the radical in eq. (3), and considering positive values of X, it is evident that as x increases y also increases, a small change in x making a small change in y. (6) As X increases indefinitely, y also increases indefinitely. Moreover, as x becomes larger and larger, Va?^ — 16 differs less and less from x. This may be proved as follows : The difference between x and Var^ — 16, i.e. x — ■\/x^ — 16, may be expressed as (a;_Va^_16)(a;+Va^-16) _ Vaj2-16 16 -l-Var^-16 01 ^^ without limit to the value of \ Now, y — \x is easily shown to Now, when x increases indefinitely, this fraction decreases in- definitely and approaches the limiting value 0. Therefore as x increases indefinitely, the value of y approaches nearer and nearer ^x. — 1 2 be the equation of a straight line through the origin and the point (2, 1). Let this line be drawn. (Fig. 50.) The curve will then come nearer and nearer without limit to this line as x becomes infinite. A few points through which the curve passes in the first quadrant are iK 4 5 6 7 10, y 1.5 2.2 2.9 4.6. The part of the locus which lies in the first quadrant may then be thought of as generated by a point which, starting at Fig. 50. LOCUS OF AN EQUATION 59 (4, 0), gradually rises as it moves to the right, passes through the above points, and approaches nearer and nearer to the straight line whose equation is y = ^ x. (Fig. 50.) The complete locus is obtained from the part in the first quadrant by considerations of symmetry. (Fig. 51.) The curve is an hyperbola, as will be proved later. The straight line to which the curve approaches indefinitely near as the point generating the curve recedes indefinitely is called an asymptote of the curve. Example 3. To plot the locus of Fig. 51. y Solving for x, (1) (2) 3?/ + l 2,-2 ■ (1) The locus is not symmetric with respect to either coor- dinate axis or the origin. (2) If 05 = 0, ?/ = — ^ ; if ?/ = 0, « = — i. .*. the curve meets the axes in (0, — \) and {—\,^). (3) No real values of either variable make the other imagi- nary. (4) If X = 3, y is infinite ; \i y = 2, x is infinite. (5) By division, eq. (1) may be written 7 2/ = 2 + (3) From this equation it follows that as x increases from a nu- merically large negative number to 3, y steadily decreases from 60 ANALYTIC GEOMETRY a value a little less than 2 to — oc. As a;, increases through 3, y changes, from — to +, and as x increases from 3, y steadily decreases and approaches the limiting value 2 when x becomes infinite. Y "" p ~ \ \ ' \ \ k. "" ^ "1 ^-^ X \, ^ L \ — _^ L u _ Fig. 52. 4 6 10, 9 ¥ 3. The following points are on the locus : x -5 -3 -2 1 2 ^ 8" 6 5" 3^ 2 The curve may then be sketched as in Fig. 52. The lines a? = 3 and y = 2 are asymptotes of the curve. Example 4. To plot the locus of y = x{x + l){x^2). (1) The locus is not symmetric with respect to either co- ordinate axis or the origin. (2) The locus meets the axes in (0, 0), (— 1, 0), and ( — 2, 0). (3) No real values of either variable make the other imagi- nary. (4) No finite value of either variable makes the other in- finite. (5) Let X take a numerically large negative value ; then y is numerically large, but negative. As x increases from the value LOCUS OF AN EQUATION 61 assigned toward — 2, each of the factors of y remains negative, but decreases in numerical value; y therefore remains nega- tive, but decreases in numerical value until a; = — 2, when y = 0. As X passes through the value —2, the factor x-{-2 changes sign and becomes positive, the other factors of y remaining negative in sign until ic = — 1 ; therefore y is positive for all values of -ic between — 2 and — 1. As a; passes through — 1, 2/ passes through and remains negative for all values of x between — 1 and 0. Asa; increases through 0, y again becomes positive and steadily increases as x increases and becomes infinite when x becomes in- finite. The locus may then be generated by a point which, starting indefi- nitely far to the left and below the origin, steadily rises as it moves to the right until, after cross- ing the X axis at ( — 2, 0), it turns at some value of x between — 2 and — 1, descends to cross the a^-axis at ( — 1,0), turns again at some value of x between — 1 and and ascends to cross the a>axis at (0, 0), and continually thereafter moves to the right and upward, receding indefinitely from both axes. The following points are on the curve ; Y / X Fig. 53. x -8 -6 -4 -3 -2 -f y -336 -120 -24 -6 |. X -1 - + 1 2 6, y ~l 6 24 336. The curve is shown in Pig. 53. Example 5. To plot the locus of f={x + 2)(x-l)(x-S). (1) The locus is symmetric with respect to the a^axis. 62 ANALYTIC GEOMETRY (2) The locus crosses the a^axis at (—2, 0), (1, 0), (3, 0), and the 2/-axis at (0, + v'6) and (0, — VB). (3) If X is less than —2, or is between 1 and 3, y is imaginary. (4) No finite value of either variable makes the other infinite. (5) Since ^^ = when x= —2 and when x=% and is posi- tive for all values of x between — 2 and 1, therefore as x increases from — 2 to 1, the positive value of y must increase from when x= — 2 and then decrease to when x=\* As X increases from 1 to 3, y^ is negative ; y is imaginary. As X increases from 3, y"^ becomes and remains positive and steadily in- creases as X increases. The positive value of ?/) therefore, increases as x increases from 3. (6) When x becomes infinite, y be- comes infinite. The curve then consists of a closed portion between x= —2 and ic = 1, and an infinite branch to the right of a; = 3. The following points are on the curve : a;_2-f-l 013 4 5 6 7 10, y ±2.4 ±2.8 ± 2.5 ±5.5 ±7.5 ±11±14.7±27. The curve is shown in Fig. 54. * At present the student has no means of telling that y does not change from increasing to decreasing and from decreasing to increasing several times as x increases from - 2 to 1 ; nor of telling where a change of this kind takes place. The investigation of such questions will be the subject oJ a later chapter. Y Z ----------t-- - 7 =g==t== X t ^ , -----\--- fv — J- Fig. 54. LOCUS OF AN EQUATION 63 EXERCISE XVII Discuss the following equations and plot the curves : 1. xy = 4:. 2. y = ix^. 3. y-=—x. 4. y = —x^. 5. y^ = xK 6. y=x(x-S). 7. y = -^. 8. ?/=(a;+5)x(x-3). 9. 2/2 ^ 4 a:^ = 4. 10. y^ _ 4 ^^2 _ 4. n. ^2 ^ 4 _ a;2, 12. y2^a;3. 13. y = xK 14. (x + 2) (y + 3) = 1. 15. y = 4x2 + 4. 16. i = ?i^. y a;-2 17. a:V = 4. 18. 2/2=(a;-l)(ic-3)(ic-6). 19. y2=(a;-l)2(x-2). 20. 2/ ^ (x-l)(x-4) 21 y= ^±2 . 22. y=(^+2)(^-5) ^ (a;_l)(x-3) (x+l)(x-3) 23. y = g_ (^ + 3) (x - 2) ^ ^^ »u = a constant. (x + l)Cx-4) 25. i)vi-2 = 6. 26. v = 32«. 27. s=16<2. 28. y- 2 = x-3 29. A light is placed at a distance h ft. above a plane surface. Given that the illumination of the plane at any point varies inversely as the square of the distance from the light, and directly as the cosine of the angle between the incident rays and the perpendicular to the plane ; prove that the illumination at a point in the plane at a distance x from the foot of the perpendicular from the light to the plane is given by I = — , where O is a constant. (X2 + A2)f Plot the curves for A = 20 ft., 30 ft., and 40 ft. CHAPTER V TRANSFORMATION OF COORDINATES 51. Change of axes. The coordinates of a point in the plane depend upon the position of the axes to which the coordinates are referred. A change of axes will change the coordinates. The equa- tions connecting the coordinates of any point in the plane with the coordinates of the same point when referred to another system will next be derived for certain changes of axes. 52. Translation of axes. Assume a set of axes OX and OY and a second set O'X' and O'Y' parallel respectively to the first axes. Let Pj^ 55 0' referred to OX and OF be (A, A:).- Take any point P in the plane and let its coordinates referred to OX and OY he x and y, and referred to O'X' and O'Y' be x' and y'. Then (see Fig. 55), x = NP, x' = N'F, h = NN', y = MP, y' = M'F, k = MM', Kow iVP= ]^N' 4- A^'P, and MP = MM' + M'P, or , a? = 7* + a?'. y = k + y'. This transformation from one set of axes to the other is called " Translation of the axes." 64 Y N Y' N' X M X' ^ A' 0' TRANSFORMATION OF COORDINATES 65 53 Rotation of axes. Let the rectangular axes OX' and OY^ make an angle ^ with OX and Oy respectively. Let any point Phave coordinates (x, y) referred to OX and OY, and {x\ y') referred to OX' and OY'. Let OP=r and AX'OP = cf>'. ThenZXOP=cl>' + e. Then, Fig. 56, Fig. 56. x' = r cos <^', y' = r sin ', x = r cos (<^' + 0)f y = rsm(' + e). Expanding the last two equations, x = r cos <^' cos ^ — r sin ' sin 0, y = r cos (f>' sin ^ + r sin ' cos 0, or x = ie' cos 6 — y' sin 0, y = x' sin 6 + y' cos 9. These equations hold for any point in the plane. They ex- press X and y in terms of x' and y'. To express x' and y' in terms of x and ?^, these equations may be solved for x' and y', or the equations may be derived as follows : InFig. 56, let ZXOP=<^, then x = r cos <^, y = r sin <^, 66 ANALYTIC GEOMETRY a;' = r cos (— 0) = r cos <^ cos $ -\- 7- sin cj> sin 6j y' = r sin (^ — 6) = r sin cos ^ — ?• cos <^ sin 6, ' or x' = x cos ^ + 2/ sin ^, y^ = y cos ^ — a; sin ^. This transformation from one set of axes to the other is known as " Rotation of the axes." 54. Applications. The formulas of translation and rotation of axes may be used to simplify equations, thereby making the construction and classification of the loci easier. Example 1. Consider the equation 12 a;2 - 48 ic + 3 ^2 + 6 2/ = 13. Let the axes be translated to a new origin (Ji, k), the formu- las for which are x = x' + li, y = y' -\-k. Substituting these values in the equation, it becomes 12 ic'2 + 3 2/'2 + (24 7i - 48) oj' + (6 A; -f 6)2/' + 12 7i2 + 3 A:2 _ 48 ^ + 6A;-13 = 0. The quantities h and A; may have any real values assigned to them. If they be so chosen that the terms of first degree in x' and y' drop out of the equation, the equation will be simpli- fied and the locus will be symmetric with respect to the axes O'X' and 0' Y'. To accomplish this it is only necessary to let 24/1-48 = 0, 6 A; + 6 = 0, from which 7i = 2, fc = — 1. The equation then becomes 12 x" + Sy" = 64.. This, then, is the equation of the locus referred to the axes O'X' and OT'. TRANSFORMATION OF COORDINATES 67 The equation is now easily discussed and the locus plotted. Fig. 57 shows the locus and both sets of axes. The student should discuss and plot the locus. Example 2. y'-Sy-\-4.x-h6 0. the new Translate the axes to origin {h, k) by means of x = x'-{-h,y = y'-\-k. The transformed equation is 2/ '2 + 2 % ' + A;2 - 8 / - 8 A; + 4 « ' + 4^ + 6 = 0. Here it is not possible to choose h and k so that the terms of first de- gree in x' and y' will drop out, since the coefficient of x' is 4. They can, however, be so chosen that the term in y' and the constant term will drop out. To ac- complish this it is only necessary to let 2k-S = 0, Fig. 57. Y Y' X' / 0' X ^^^ ' and k'-Sk-\-4:h + 6-. from which A* = 4, h = The equation then = 0, = #• be- comes y"-{-4.x' = = 0. The locus is now easily constructed. (Fig. 58.) Example 3. lU-+24a;?/-f42/2=20. (1) In equations of this form, i.e. equations of second degree in x and y containing a term in the product xyj the term in xy may be made to drop out by a rotation of axes. Fig. 58. 68 ANALYTIC GEOMETRY Let x = x^ cos 6 — y^ sin 0, y = ic' sin ^ + y' cos 6. Substituting in eq. (1), 11 (»' cos B — y' sin ey + 21 {x^ cos O — y' sin ^) (x' sin $ ■\-y' cos ^) + 4 (a;' sin ^ + ?/' cos ^) ^ = 20. (2) Expanding and collecting, ^'-^=20. (3) llcos^^ +24 cos ^ sin ^ +4sin2^ aj'2_22sin^cos^ + 24cos2^ -24sin2^ + 8 sin cos a;y+llsin2^ -24 sin ^ cos (9 + 4cos^^ It is now possible to choose 9 so that the coefficient of x'y^ will become zero ; for it is only necessary to have or or 24 (cos^ - sin2 6) = 14 sin 6 cos By 24cos2(9 = 7sin2^, tan 2 (9 24 (4) whose tangent is -^y*- To satisfy eq. (4), let 2 ^ be the angle in the first quadrant Draw the right triangle with sides 24 and 7 as in Fig. 59. The hypotenuse is then 2^. .-. sin2^ = |i, cos2^ = -/5. Now sin^ l9 = 1 (1 - cos 2 B), cos^ ^ = i (1 + cos 2 ^), and sin B cos ^ = J sin 2 ^. sin2<9=2^^, cos2^ = ||, sin ^ cos ^ = If. Substituting these values in eq. (3) and dividing the resulting equation through by 125, there results 4 a;'2 - 2/'2 = 4. Referred to the new axes the locus is much more easily constructed. The discussion of the equation is very similar to that of example 2, Art. 50. TRANSFORMATION OF COORDINATES Y 69 Fig. 60. The locus and both sets of axes are shown in Fig. 60. The angle through which the axes are turned is tan~^ |. EXERCISE XVIII Simplify the following equations by a translation of axes to remove the terms of first degree where possible, and by a rotation of axes to re- move the terms in xy. Plot the curves and all coordinate axes. 1. x2-6x+4?/2_^8?/ = 5. 2. 4a:2-4?/2 + a;-2?/ = 0. 3. 4a;2 + y2_i2ic + 2y-2=0. 4. ic2 + y-^-4a; + 2y-ll = 0. 5. ?/2-6 2/ + 8 = 4a;. 6. 2a;2-62/2 4.xy-5a; + lly = 3. 7. a;2-j-2a:2/ + «/2-12x + 2 2/ = 3. 8. a;2-a;?/-2«/2_x-4i/-2=0. 9. 3x2 + 2a:y + 3y2 = 8. 10. xy=4^. CHAPTER VI THE STRAIGHT LINE 55. Theorem. Every straight line has an equation of first degree in Cartesian coordinates. Two cases are to be considered : (1) The line parallel to a coordinate axis. If the line is parallel to the a^axis, then all points of the line have equal ordinates. .'. ^ = c, where c is a constant, is true for all points of the line and for no others. It is therefore the equation of the line. Likewise, a line parallel to the ,v-axis has an equation of the form x — c. (2) The line not parallel to an axis. Let the line cut the 2/-axis at JV(0, h). Let P(x, y) be any point of the line. Through P draw PM parallel to the x-axis to meet the 2/-axis in M. Then as P moves along the line, the ratio — will remain unchanged. For if P is any other point of the line, then, by similar tri- angles, PM P'M'' Let this constant ratio be denoted by m. MN Fig. 61. Then PM = m is true for all points on the line, and for no others. 70 THE STRAIGHT LINE 71 From the figure the values of MN and PM are seen to be h — y and — a; respectively. Therefore h y or Fig. 62. This is therefore the equation of the line. It is of first degree in X and y. For a line passing through the origin, the value of h is zero, and the equation becomes y = mx. The relation between the lines y = mx and y=mx + b is shown in Fig. 62. It is important to notice that if the axes are rectangular, the constant ratio m, or — —, is the PM slope of the line. 56. The equation of first degree. Conversely, every equation of first degree in Cartesian coordinates^ with real coefficients, is the equation of a straight line. The general equation of first degree is of the form Ax + By-^C=0. (1) Here again two cases are to be considered : (1) When either ^ or 5 is zero. Suppose A = 0. Then B^O, and the equation may be written C y = B This equation is evidently satisfied by all points on a line parallel to the ic-axis, and by no others 72 ANALYTIC GEOMETRY Likewise, if B = 0, the equation represents a straight line parallel to the ?/-axis. (2) When neither A nor B is zero. Solve the equation for '' A O ^ B B Now this is of the same form as y = mx 4- b, which was found in the preceding article to be the equation of a straight line, and since a straight line can be drawn so that m and b will have any assigned real values, a line can be C A AC drawn so that b = , and m = — — . Th.eny — — —x——, B B B B or Ax -\-By + 0=0, is the equation of this line. Hence Ax -\- By -\- C = is the equation of a straight line. The proofs given in this and the preceding article hold for oblique as well as for rectangular coordinates. It is only in rectangular coordinates, however, that m is the slope of the line. 57. The conditions which determine a straight line. The position of a straight line is determined when there are known either, (1) Two points on the line, (2) A point on the line and the direction of the line, (3) The length and direction of a perpendicular from a fixed point to the line. Considerations of these conditions lead to the following special forms of the equation of the straight line. 58. The two-point equation. Let Pj (.x'j, y^) and Pg (x2, 2/2) be any two points. To find the equation of the straight line through them. THE STRAIGHT LINE 73 Let P{x, y) be any point on the line. Through Pj draw a line parallel to the ^/-axis to meet lines parallel to the ic-axis through P and Pg ill ^ and Jfg respectively. Then by sim- ilar triangles, MP,^MF M^Pi M2P2 wnich is the same as y-vi X — Xi 2/2 — 2/1 a?2 — OCi This equation holds for Fig. 63. every point on the line, and for no others. It is, therefore, the equation of the line. 59. The intercept equation. In the last article let the two given points be (a, 0) and (0, 6). The equation then becomes y — _x — a y ^h x — Clearing of fractions, transposing, and dividing by a6, the equation re- duces to 7:+^=l- a o This is known as the intercept equa- tion of the line, since a and h are the Fig. 64. intercepts of the line on the coordinate axes. In this equation neither a nor b can be zero. 60. The point-slope equation. Let the line pass through -fi(^i) 2/1) and have a slope m. Let P(a', y) be any point on the line. The slope of the line joining Pj and P is ^ ~ ^^ , which by hypothesis is equal to m. X — Xi 74 ANALYTIC GEOMETRY - — ^ = m is an equation which holds for all points on the line and for no others. It is, therefore, the equation of the line. Fig. 65. Clearing of fractions, it may be written y — Vi = m{x — a?i). This equation does not apply to a straight line parallel to the 2/-axis, for which m is infinite. 61. The slope equation. If in the last article the given point is (0, 6), the equation reduces to y = mx -|- 6, which is the slope equation already considered. (Art. 6^.) 62. The normal equation. Let the distance from the origin to the straight line be p, and let the angle which this perpen- dicular makes with the a;-axis be a. The quantity p will be considered positive always. Let H be the foot of the perpendicular from the origin to the line. The coordinates of H are then p cos a and p sin a. The slope of OH is tan a. Therefore the slope of the given line is — cot a. Hence the line passes through {p cos a, p sin a) and has; a slope equal to — cot ol THE STRAIGHT LINE The equation of the line is therefore, by Art. 60, y —p sin a = — cot a(x — p cos a). Y 75 Fig. 66. Replace cot a by -, clear of fractions, and transpose ; the sin a equation then becomes X cos a-\-y sina—p (cos^ a -\- sin^ a) = 0, or 0? COS a 4- y sin a — 2> = 0. 63. Reduction of Ax -{- By -{- C =0 to the slope intercept, and normal forms. The equation Ax -{- By -{- C = may be reduced to the slope, intercept, and normal forms as follows : (a) To reduce Ax -{- By -\- C = to the slope form. Solve the equation for y. There results A C — x , B B' y = which is in the form y = mx -\- b, where m = — — , b C B' b' The method fails when ^ = 0. The equation cannot then be put in the slope form. (b) To reduce Ax -\- By -\- C = to the intercept form. Transpose C to the right-hand side of the equation, and divide 76 ANALYTIC GEOMETRY by C; the resulting equation may be written X . y C A + C' B 1, 7^ which is in the form - 4- " = 1, where a = — — , 5 = — — . a b ' A' B The method fails if either A, B, or C is zero. If (7 = 0, both intercepts are zero. If either A or B is zero, the line is parallel to an axis of coordinates. (c) To reduce Ax-\-By+C=0 to the normal form. X Let X cos a-\-y sin a — p = be the normal equation of the line. The foot of the perpendicular from the origin to the line is then (p cos a, p sin a). The coordinates of this point must satisfy. the equation Ax-\-By-{- (7=0. Ap cos a + Bp sin a + (7 = 0. (1) Also the slope of the perpendicular is the negative reciprocal of the slope of the line ; J) tana = — . [See («) of this article.] (2) .JjL cos a A ± Vl + tan^ a ± VA' + B^ Substituting in (2), sm a — Substituting these values of sin a and cos a in (1), -p = C ± VA^ + B' THE STRAIGHT LINE 77 Substituting these values of sin a, cos a, and p in the normal equation of the line, there results Hence, the equation of a line is reduced to the normal form by dividing the equation of the line through by the square root of the sum of the squares of the coefficients of x and y. TJie sign of the radical should be taken opposite to the sign of C so that p will be positive. 64. Illustration. To reduce 2a; — 4?/ + 7 = to the slope, intercept, and normal forms. (a) Solving for y brings the equation into the slope form in which ^ = i? & = |. (&) Transposing the constant term, 7, and dividing by — 7, brings the equation into the intercept form -i i in which a = — |, & = J. (c) Dividing through by — V2^ + 4^ brings the equation into the normal form — -^ + —^y F = ^> 1.2 '^ in which cos a = , sin « = -^ , p V5 V5 2V5 65. Applications of the formulas. By the use of the formulas derived in this chapter the equations of straight lines which satisfy certain conditions can be easily found. Illustrations. (a) To find the equation of a straight line which passes through (3, — 5) and makes an angle of 30° with the avaxis. 78 ANALYTIC GEOMETRY The slope of the required line is tan 30°, or — -. By sub- V3 stituting iu the equation y — y^ = m{x — x^ there results 2^ + 5 = ^(^-3), which reduces to V3a;-32/ = 15-f 3V3, the required equation. (6) To find the equation of the straight line which passes through (—3, 1) and makes an angle of 60° with the line 4a;-9y = 12. Let the angles which the given line and the required line make with the a>axis be O-^ and 6 respectively, and the slopes of these lines be rrii and m re- spectively. Then m^ = tan 0^, m = tan 6. But mj = |^, and ^ = ^, + 60°. (Fig. 68.) m = tan ^ = tan {O^ + 60°) ^ tan ^1 + tan 60° 1 - tan Oi tan 60° - I+V^ _ 4 + 9V3 . 4V3 9-4V3 ^a Fig. 68. ^ 144 + 97V3 33 Therefore the required equation is 144 4- 97 V3 2/-1 33 (a; 4- 3), or approximately 2/-l = 9.450c + 3). . THE STRAIGHT LINE 79 66. The point of intersection of two straight lines. Let the two lines whose equations are A,x-{-Biy+C, = 0, (1) and A^ + ^22/ + C2 = 0, (2) be denoted by Li and L2 respectively. In eq. (1) x and y may be the coordinates of any point on L^, and in eq. (2) x and y may be the coordinates of any point on L2, and hence x and 2/ in one equation are not the same in general as X and y in the other. If, however, the lines intersect, there is one pair of values of x and y that satisfy both equations; namely, the coordi- nates of the point of intersection. Conversely, if values of x and y can be found which satisfy both equations, they are the coordinates of a point on both lines, i.e. the point of intersection. Therefore, to find the coordinates of the point of intersection of two lines, solve the equations of the lines as simultaneous. What if the lines are parallel ? Illustration. To find the point of intersection of Sx-\-2y = ll and 4:X — 5y = T. Solving the equations as simultaneous, the values of x and y are found to be x = 3, y = l. Therefore the point of inter- section is (3, 1). Let the student plot the lines and check graphically. EXERCISE XIX By substituting in the formulas write the equations of the straight lines which satisfy the following given conditions: 1. Passing through (2, 1) with slope — 2. 2. Passing through (- 3, 7) and (2, - 5). 3. With X- and ^/-intercepts 3 and - 8 respectively k 80 ANALYTIC GEOMETRY 4. With 2/-intercept 6 and slope 2. 5. Passing through the origin with slope — |. 6. With a = SO'' and p = 4. 7. With j9 = 5 and m = — |. 8. Passing through (2, — 5) parallel to 3aj — 2/4-4 = 0. 9. Passing through (0, 0) perpendicular to ax + 6y + c = 0. 10. Passing through (a;i, y{) parallel to y = mx + &. 11. With y-intercept h and perpendicular X,o Ax-^ By ■\- C=0. 12. Passing through (/i, A;) parallel to iccos/S + ysin/3 = q. 13. Passing through (e, /) parallel to Ix + my + n = 0. 14. Passing through the origin and perpendicular to gx -\-fy = c. Reduce, where possible, each of the following equations of straight lines to the intercept, slope, and normal forms, giving the values of a, b, m, a, and p. 15. Sx-4y = 6. 17. 2x-6y = 0. 19. y-25 = 0. 16. y+2x = 4. 18. -x + 2y = 9. 20. Obtain the equation of the straight line which passes through (1, 2) and makes an angle of 60° with the line 2x — 6y = 8. 21. Two lines, Li and X2, intersect in (—3, — 2) ; Li has a y-inter- cept equal to — 6 and makes tan-i | with L2 ; find the equations of the two lines. 22. Find the equation of the straight line of slope — f which passes through the intersection oi2y --x = b and x-~Sy = 1. 23. The vertices of a triangle are (1, 2), (4, — 6), and (—5, — 3) ; find the equations of its sides. 24. Find the equations of the perpendiculars from the vertices upon the opposite sides of the triangle in example 23, and prove that they meet in a common point. 25. Find the equations of the medians of the triangle in exam,ple 23, and prove that they meet in a common point. 26. Find the equation of the line through (h, k) making tan-^m with y = lx-[-b. 27. A line passes through (2, 5) and is distant 3 from the origin ; find its equation. How many solutions ? THE STRAIGHT LINE 81 28. Show that Ax -\- By -\- C = and Ax + By + K= are parallel, and that Ax + By+C = 0, and Bx — Ay + K=0 are perpendicular. 29. What set of lines is obtained by varying b in the equation y = mx + & ? What set of lines by varying m ? 30. Discuss the effect upon the line Ax -\- By + C = of changing each of the constants, keeping the other two unchanged. 31. Find the equation of a straight line which passes through the inter- section of2x— y-\-5 = and a; — 22/4-1=0, and makes an angle of 45° with y = 2x. 32. Prove that ax -\- by -}- c +k(Ix -{- my -f- w) = is the equation of a straight line which passes through the intersection of ax -\- by -h c = and Ix + my 4- n= 0. What is the effect on the line of varying k ? 33. Using the fact expressed in example 32, find the equation of the straight line which passes through (3, —1) and the intersection of 2ic + 4?/— 7 =0 and 7 x — 2 y -f- 13 = 0, by determining the proper value of A:. 34. Find the equation of the straight line which passes through the intersection of x -\-Sy —7 =0 and y — Sx = 2, and makes an angle of 135° with the a;-axis. 35. Find the equation of the straight line which passes through the intersection of 2x ~ 9y = 18 and 7 y + bx = 21, and is parallel to 4x + 6y -S = 0. 36. Find the equation of the straight line perpendicular to Sy = 7 x which passes through the intersection of x -\- 2y = 8 and 4 x = 13 y. 67. Change of sign of AiK -f- Bj/ + C. The expression Ax -\- By + C is j^ositive for all points on one side of the line Ax-\-By-{-C = Oy and is negative for all points on the other side of the line. Proof. I. Srippose\B ^ 0. The line is then not parallel to the ^-axis. Let L be the line whose equation is Ax-\-By+ (7=0. 82 ANALYTIC GEOMETRY For all points on this line ^ B B Let (a?!, 2^1) be any point above the line. Since y^ is greater than the ordinate of the point on the line with abscissa Xi, it follows that or 2/1 4- - a^i + - > 0. ^ B B This is true, then, for any point above the line. There- fore, for all points above the line, Ax^ + By^-\-C>0,iiB>0, and Ax^ + By^ + C<0/\iB<0. In either case the expression has the same sign for all points above the line. If the point is taken below the line, the inequalities are all reversed. Hence the expression has the same sign for all points below the line, but that sign is opposite to the sign of the expression for points above the line. II. Suppose ^ = 0. Then A^O. The expression becomes Ax-\-C, and the equation of the line becomes Ax-^ C = 0. C The line is parallel to the ?/-axis. On this line x = To C the. left of the line aj< , and to the right of the line x> C A Therefore Ax + C has the same sign for all points on one side of the line and has the opposite sign for all points on the opposite side of the line. Hence the theorem is true in all cases. THE STRAIGHT LINE 83 An important special case of this theorem is the following : The sign of the expression Ax -\- By -{- G is the same as, or opposite to, the sign of C according as the point (x, y) and the origin are on the same, or opposite, sides of the line Ax-\-By-{- C=0. This follows at once from the theorem, since the value of the expression Ax -\-By-\- Cis C when the coordinates of the origin are substituted. If (7=0 and A:^0, the student can easily show that the sign of Ax + By is the same as, or opposite to, the sign of A, according as (x, y) lies to the right or left of the line Ax -\-By = 0. 68. Illustration. The expression 3x-\-7 y — S has the value 2 at (1, 1), which is opposite in sign to C, or — 8. Hence (1, 1) and the origin are on opposite sides of the line 3x-\-7y-S = 0. Also the expression 3x-\-7 y — 8 has the value — 1 at (2, ^), which is the same in sign as — 8. Hence (2, I) and the origin are on the same side of the line 3x-{-7y — S = 0. 69. Distance from a point to a line. A numerical example will be first worked through. Let it be required to find the distance from (6, — 3) to the line 3x — 5y = 7. Transform to parallel axes through the given point (6, —3), as a new origin, the equations of transformation for which are x = x'-{-6, y = y'-3. Substituting these values in the equation of the line, it Fig. 71. becomes 3x'-5y' + 26 = 0, which is the equation of the line referred to the new axes. Y Y' ^ <■ X ^ ^ \ // - \ 0' X' - (6, -3) 84 ANALYTIC GEOMETRY The distance from the new origin to the line is given by the formula C P = ±^A'-irB' (Art. 63), which here becomes 26 p = — :::::: = 4.46 nearly. V34 70. General formula for the distance from a point to a line. Let the given point be Pq{xq, y^, and the given line Ax^By-\-C=0. Transform to parallel axes through Pq as a new origin, for . which the formulas of transformation are x==Xq-\-x\ y = yo-{-y'. x[ The equation of the line referred to the new axes is then A(x' + ^o) + B{y' + 2/o) + O = 0, or Ax'-\-By' -{-Axo + Byo-^C=0. In this equation x' and y' are the vari- able coordinates, and the constant Fig. 72. term of the equation is Axq -f ByQ 4- C. Therefore the distance, d, from the new origin to the line is ^ ^ Axq + Bpo + C This distance will be regarded as a positive quantity. The sign of the radical must therefore be taken the same as the sign of Axq + ByQ + O. But Axq -\- By^ -f C has the same sign as C when {x^^ y^) and the origin are on the same side of the line Ax-{-By-\-C=0, and has the opposite sign to C when (xq, 2/o) and the origin are on opposite sides of the line (Art. 67). THE STRAIGHT LINE • 85, Therefore the sign to be taken with the radical is the same as the sign of C when {xq, y^ and the origin are on the same side oi Ax-\-By -\-C=0, and opposite to the sign of C when {xq, y^ and the origin are on opposite sides of the line. If (7=0, the sign to be taken with the radical is the same as or opposite to the sign of A according as {xq, 2/0) lies to the right or left of the line Ax -\-By = 0. EXERCISE XX 1. Find the distance from the point (3, — 6) to the line 7 x — 6y = 1S. 2. Find the distance from the intersection of 2x — 9y = 3 and — by — 4:X = 12 to x—b=6y. 3. The vertices of a triangle are ^(1, 4), B(- 3, - 5), and (7(6, - 4); find the area by finding the lengths of AB and the altitude from C to AB. Check by using the formula for the area of a triangle. (See Art. 36.) 4. The equations of the sides of a triangle are x-\-4y —7 = 0, 3x + y + l=0, and 2y — 6x-\-lS = 0; find the area of the triangle by finding the length of one side and the length of the perpendicular from the opposite vertex to that side. Check by using the formula for the area of a triangle. (See Art. 36.) 5. Find the distance from the intersection of 2 x— 6y =S and 8x +y + 13 = to the line through (- |, 4) and (0, — 3). 6. Find the distance from (9, — 1) to the line through the origin with slope — ^. 7. Find the distance from (xi, yi) to y = mx + 6. 8. Find the distance from (xo, 2/o) to x cos a -\- y sin a = p. 9. Find the equations of the bisectors of the angles formed by the two lines 2x + y — 7 =0 and 4x — Sy — 5 = 0. Show by their slopes that the bisectors are perpendicular to each other. Suggestion. The distances from (xo, yo) to 2 x + y — 7 = and 4iX — Sy — 6 = are, respectively, 2 Xq -f yo - 7 ^^^ 4 Xq - 3 jj/o - 5 ±\/5 ±V25 Now the bisector of an angle is the locus of points equidistant from the sides of the angle. Hence to obtain the equation of the bisector, place 86 ANALYTIC GEOMETRY the above values for the distances equal to each other and remove the subscripts to indicate variable coordinates. The proper signs of the radicals must be chosen, as explained in Art. 70. 10. The three sides of a triangle have the equations 3 ic — 4 y = 7, 6x+12y + 8 =0, and 4x + 3y— 12 = 0; find the equations of the three inner bisectors of the angles, and shov7 by their equations that they meet in a point. 11. Find the equations of the three outer bisectors of the angles of the triangle of example 10, and prove by their equations that two of the outer bisectors and the inner bisector of the remaining angle meet in a common point. 71. Equations of the straight line in polar coordinates. (i) Equation of the straight line through two points. Let ■f'lC*'!) ^i) ^nd ^2 0*2) ^2) t)e any two points in the plane. To find the equation of the straight line passing through them. Let P(r, 0) be a point on the line as shown in Eig. 73. Then area 0PiP2 = area OPiP+area OPP2. I.e. I i\r2 sin (O^ — ^1) = i ^^1 sin {6 — 0^) + ^ rr^, sin {B^ — 6), or rjra sin (^o — ^1) + ^\^ sin {Q — 62) + rr^ sin (d^ — 0) = 0. This equation holds for any position of P on the line be- tween Pi and Pg. If P be so taken that Pg lies between P and Pi, the equation that holds can be obtained from the above equation by interchanging r and ?*2, and and O2. But this interchange only changes the sign of the left member of the equation, and since the right member is zero, the equation it- self is unchanged. If P be so taken that Pi lies between P and Pg, it may be shown in the same way that the same equation holds. Hence the above equation holds for all points on the line, and clearly for no others, and is tlieref ore the equation of the line. Fig. 73. THE STRAIGHT LINE 87 The same equation may be derived at once by equating to zero the area of the triangle whose vertices are P, Pj, and Pg. (See Art. 37.) (ii) Equation of a straight line in terms of the length of the perpendicular from the origin to the line and the angle which this perpendicular makes with the initial line. Let the perpendicular from the origin to the line be of length py and make an angle a with the initial line. Let P(r, ^) be a ^ point on the line. Then (Fig. 74) /^S r cos (a — 6)= p. Since cos (—A) c^:— lLu = cos A, this may be written r(ios{0-a)=p. ^i«-'^4- The student should show that this equation holds for all points on the line. EXERCISE XXI 1. Write the equation of the line through (2, 30'') and (1, 60°). 2. Draw the line whose equation is r cos (0 — 60°) = 5. 3. Draw tlie line whose equation is r sin ^ = 4. 4. Find the intersection of the lines r cos ^ = 8, and r sin ^ = 4. 5. Find the intersection of the lines r cos [^ — sin-i (f)] =2, and r cos [d - cos-i ( j\)] = 4. 6. Derive the equation r cos (0 - a) =phy substituting in a; cos a + ?/ sin a = p, the values x = r cos 6, y = r sin d. 7. Derive the equation of the straight line through two points in polar coordinates by substituting in y — yi _ a^ — a:i y2-yi xi-xx the values a: = r cos ^, y = r sin ft CHAPTER VII STANDARD EQUATIONS OF SECOND DEGREE CIRCLE, PARABOLA, ELLIPSE, HYPERBOLA 72. The circle. The equation of a circle of radius r and center at (Ji, k) is Proof. Denote the center by (7. Let P(x, y) be any point on the circle. The condition that P is on the circle is expressed by the equation (7P = r. In terms of the coordinates of the points it becomes (x-hy^-{y-ky = r', P (^. y) which is therefore the equation of the circle, for it is an equation which is satisfied by all points on the circle and by no others. If the center is at the origin, the equation reduces to 73. The equation ay^ + y^ -\- Doc -{- Ey -[- F = 0. (1 ) An equation of this form, by completing the square in the terms containing x and in those containing y, can be thrown 88 STANDARD EQUATIONS OF SECOND DEGREE 89 into the form of the equation of the preceding article as follows: Add ^J^^-Fto both sides of eq. (1). The result is 4 4 ^ + Dx + ^ + f + Ey + ^ = ^ + ^-F, 4 4 4 4 x+fj+(.+g^ ^-^^-^^ - (2) Now, if D^ -\- E^—4: F is positive, eq. (2) is, by the preceding article, the equation of a circle with center at [ — — , —-^] and radius equal to ^VD^ ^E''-4.F. If D^ + E'^-4.F=0, eq. (2), and hence eq. (1), is satisfied bya;=— — , y= — — only ; for the sum of two terms, neither of which is negative, can vanish only when the terms vanish separately. \iD'^-\-E'^-^F<^, eq. (2), and hence eq. (1), is satisfied by no real values of x and y : for the sum of two quantities, neither of which is negative, cannot equal a negative quantity. Hence the equation . represents (1) a circle, center at (- ^,- ^\ radius = \ VZ^4-^'-4i^, if D2-f J5;--4i^>0, (2) apoint(^-|,-|^,ifi)^4--E^-4i^=0, (3) no locus, if Z)2 _|_ ^ _ 4 2^ < 0. 74. The equation of a circle through three points. The equation of a circle through three given points, not in the 90 ANALYTIC GEOMETRY same straight line, can be found by use of the equation of the preceding article as is illustrated by the following example : Example. To find the equation of a circle through (2, 1), (-1, 3), and- (-3, -4). The equation of the required circle is of the form ^+f+Dx-^Ey+F=0. (1) The coordinates of each of the given points must satisfy this equation. Therefore 5 + 2i)+ E + F=0, 10- D + SE-{-F=0, 25-3D-4.E + F=0. The values of B, E, and F, obtained by solving these equa- tions, are D Y ^ ^^. / \ 1 \ X \ \ / V X / <~- Fig 76. the circle with center and radius as computed ■VS E = l, F=-ii^. Substituting these values in eq. (1) and clearing of fractions, the re- quired equation of the circle is 5(a.-2+ 2/2) -f 13 ic + 7 2/ - 58 = 0. Using the formulas of Art. 73, the center and radius of the circle are found ta be (—1.3, —.7) and r = 3.71. . . . A check on the work is obtained by plotting the points and drawing (Fig. 76.) EXERCISE XXII Find the centers and radii of the circles represented by the following equations. Draw the figures. 1. a;24.y2^25. 2. a;2 + 2/'^-4a; + 6y = 12. 3. a:2 + 2/2 + 8x-6y = 0. 4. 2a;2 f 2 y2 _ 7y + 3a; = 11. 5. (x - 1)2 + (y + 2)2 = 0. 6. (a; - /i)2 + (?/ - A;)2 = 0. 7. m2 + v2 4- M 4- 1> = 0. 8. x2 + y2 _ 4 a; + 6 y + 14 = 0. STANDARD EQUATIONS OF SECOND DEGREE 91 9. x^-{-y^-ix-\-Qy-lS = 0. 10. x'^ -2ax + y^ = 0. 11. x^-2ax-\-y^ -2ay = 0. 12. x'^ -{- y^ - ax - by = 0. Find the equations of circles which fulfill the following conditions : 13. Center at (- 1, 3), radius = 2. 14. Center at (a, 0) , radius = a. 15. Center at the intersection of y + 4ic+l = and 2x—y-\-6 = 0, and passing through (2, — 3). 16. Center at (2, 5) and tangent to3x4-4y = ll. 17. Center on the line y = 2x and passing through (0, 5) and (6, 1). 18. Passing through (0, 2), (- 1, 3), and (5, 0). 19. Circumscribing the triangle whose sides are 5x + Sy — IA = 0, ix — Sy + 6 = 0, and x + 6 ?/ +8 = 0. 20. Inscribed in the triangle whose sides are 5x + l2y — 2 = 0, 4ic + 3y + 5 = 0, and3x-4!/-15 = 0. 21. Tangent internally to the first two sides of the triangle mentioned in example 20, and tangent externally to the other side. 22. Prove that if Pi(xi, yi) is any point without the circle and T is the point of contact of a tangent drawn from Pi to the circle, then, P^2 = (xi - hy + (yi - ky - r^. 23. Show that if the equation of the circle of example 22 is x^ + y^ + Dx + Ey+ F=0, then, KT^ = xi^ + yr^ + Dxi + Eyi + F. 24. Prove that the locus of points from which equal tangents may be drawn to ' a;2 + w2 -f Dix + E^y + Fi = 0, and x'^ + y^ + Di - D2)x + (^1 - E^^y -\-Fi-F2 = 0, or, in case the circles intersect, is that portion of the line not inside of the circles. This line is called the radical axis of the two circles. 25. Show that if two circles intersect, their radical axis passes through their points of intersection. 92 ANALYTIC GEOMETRY 26. Find the equations of the radical axes of the circles of examples 1, 2, and 3, and prove that they meet in a point. 27. Prove that the three radical axes of any three circles taken in pairs meet in a common point. 28. Prove that the radical axis of two circles is perpendicular to their line of centers. 75. The parabola. The parabola is the locus of a point which moves so as to keep equidistant from a fixed point and a fixed straight line. The fixed point is called the focus, the fixed straight line the directrix, of the parabola. To obtain the equation of the parabola, let, at first, the direc- trix be taken as the axis of y and the focus at the point (p, 0). Let P(x, y) be any point on the locus. Join P and F{p, 0), and draw MP parallel to the avaxis to meet the ^/-axis in M, Then the condition that P is the point on the locus is expressed by the equation FP = MP, if MP is positive, and by FP=- MP, if MP is negative. P ix,yy F(p,o) PCx,y) F(p,o) Fig. 77. Evidently MP is positive or negative according as (p, 0) lies to the right or the left of the origin, i.e. according as p is posi- tive or negative. Now FP = -V(x-2)y-{-y^ and MP = x. ■y/{x-pY-\-y^ = a;, if p > 0, (1) and V(a;-p)2-f.2/2 = -x,\ip<0. (2) STANDARD EQUATIONS OF SECOND DEGREE 93 Squaring and transposing, either of these equations becomes f' = 2px-p\ (3)* This equation may be written = 2,(.-|). Let the axes be translated to (-^, j as a new origin. The formulas of transformation are a^ = ^' + |,2/ = y'- The equation of the parabola referred to the new axes is, therefore, y'2 = 2px\ - (4) Dropping, primes, y' = 2poc (5) is therefore the equation of a parabola when the 2/-axis is paral- lel to the directrix through a point halfway between the focus and directrix, the avaxis passes through the focus and is per- pendicular to the directrix, and the focus is at ( ^, ). It is important to note that in eq. (5) the abscissas of points on the parabola vary as the square of the ordinates. 76. The graph of 2/2=2 i>ic. I. p positive. * Equation (3) is not equivalent to both eqs. (1) and (2), but only to (1) if p is positive, and to (2) if p is negative. For on retracing the steps from (3) the eq. \/(x — p)'^ -\- y'^ = ±x\s obtained. Now the + sign can only be used when x is positive, since the radical is counted positive. But if p were negative when x is positive, then \/(x — p)'^ + y'^ would be greater than X. .'. when x is positive, p is positive. Therefore the + sign of x can be taken only when p is positive. Hence when p is positive, eq. (3) is satisfied by precisely the same points as eq. (1). In the same way it can be shown that eq. (3) is equivalent to eq. (2) when p is negative. 94 ANALYTIC GEOMETRY (1) The curve is symmetric with respect to the x-axis. (2) When x = 0,y = 0', when y = 0,x = 0. The curve there- fore meets the axes at (0, 0) only. (3) All negative values of x make y imaginary. The curve, therefore, lies to the right of the ?/-axis. (4) No finite value of either variable makes the other infinite. (5) As X increases, the positive value of y increases, a small change in x making a small change in y. (6) When x becomes infinite, y also becomes infinite. The upper half of the curve may, therefore, be generated by a point which, starting at (0, 0), moves ever to the right and upward, receding indefinitely from both axes. The following points are on the curve : xO 2 2 8 2 p 2p Sp 4tp Sp 50p 200p, y ±^ ±p±pV2 ±2p ±pV6 ±2pV2 ±4.p ±10p ±20p. The curve is shown in Fig. 78 for a certain value of p. F(|,o) p positive Fig. 78. p negative II. p negative. This case differs from that in which p is positive only in making the curve lie to the left of the y-axis instead of the right. The curve is shown in Fig. 78, the values STANDARD EQUATIONS OF SECOND DEGREE 95 of p for the two curves being numerically equal, but opposite in sign. 77. Axis of a parabola. Vertex. The straight line through the focus perpendicular to the directrix is called the axis of the parabola. The point where the parabola crosses its axis is called the vertex. In both cases of Fig. 78 the a>axis is the axis of the parabola, and the vertex is at the origin. 78. Parabola with axis on the ^/-axis and vertex at the origin. If the vertex is at the origin and the focus at (0, ^ j, the equa- tion can evidently be obtained from that of Art. 76 by ex- changing X and y. The equation is therefore The two cases are shown in Fig. 79. p positive. a;2 = 2py. Fig. 79. p negative. 79. The arbitrary constant of the parabola 96 ANALYTIC GEOMETRY Definition. A constant which may have any vahie in an equation is called an arbitrary constant, or a parameter, of the equation. In the equation of the parabola, if = 2})x, since p may have any real value, it is an arbitrary constant of the equation. Corresponding to each value of p there is a definite curve. The curves which correspond to a few different values of p are sketched in Fig. 80. \ \ i Y / ^ ^ ^ \ // /^ \ \ UP A . < \ \ / /^ <■ \ \ / / \ \j/ y P = "10 p = '"' V = X C -^^__ h 1 '\^ T \"^ / \ \ 7~ 1 \ ^. / / \ ^ \_ / 7 f \ ^V / 1 \ y2 _ 2 j3x. Fig. 80. 80. The equations C2/'+/>a?+^2/ + l^=0, C^O, JD^^, Equations of these forms can by a translation of axes be thrown into the forms x^ = 2py and jf^^px respectively. The equations therefore represent parabolas with their axes parallel respectively to the y-axis and the o^axis. STANDARD EQUATIONS OF SECOND DEGREE 97 A numerical example will make this clear. Example. Sx^ -\-2x-\-5y -4. = 0. Complete the square in the terms containing x, and trans- pose the other terms to the other side of the equation : or (x + iy = -%(y-{i). Translate the axes to (— ^, yf) as a new origin by means of the equations The transformed equation is 2/ = 2/' + tI- ''^--f2/'. Y' 0' Y -1 X' /^ \: > / \ Fig. 81. This is the equation of a parabola with axis on the new y-axis, vertex at the origin, and focus at (0, — -j^) referred to the new axes. Referred to the old axes, the vertex is (— \, |-f), the focus is (— 1^, T5^), and the equation of the axis of the parabola is x = -i. 81. The equation y = aoc^ + 6a? + c. This equation represents a parabola with its axis parallel to 98 ANALYTIC GEOMETRY the ?/-axis. For, it may be written \ a 4 ay 4a 4 a \ 2aJ Let a^ + and this equation becomes 2a' 6^ — 4 ac y\ which is the equation of a parabola with axis on the new i^-axis and vertex at the new origin. Hence, referred to the old axes the vertex is at f , — — ^ ), and the axis of the pa- V- 2a' 4a / ^ rabola is parallel to the 2/-axis. The parabola extends upward or downward from the vertex according as a is positive or negative. The sign of 6^ — 4 ac determines whether or not the curve crosses the a;-axis. Let the student show that the conditions are as stated in Fig.' 82. a positive (1) 62_4ac>0. Fig. 82. y = a'3?' -\- h% -^ c. (2) 62_4ac=0 a negative. (3) 62_4«c<0. STANDARD EQUATIONS OF SECOND DEGREE 99 82. The parabolic arch. An arch of height ?i and span 2 1 is in the form of a parabola with vertex at the crown. It is desirable to compute readily the heights of the arch at vary- ing distances from the center of the span. Choose the axes as in Fig. 83, counting y as positive down- FiQ. 83. ward. Since in a parabola, and with this choice of axes, the ordinate varies as the square of the abscissa, therefore This form of the equation enables one to compute readily the heights at varying distances from the center. E.g. the heights at distances from the center of -, -, and 3^ ^. T 15h 3/i , 7h — are respectively -—— — , and —— • EXERCISE XXIII Plot the following parabolas, finding the vertex of each and reducing the equation to the standard form. In each case compute the value of the discriminant, b^ — 4 ac. 1. y = 2x'^-Sx + 5. 2. y=-Sx^ + 4x-l. 3. y = x^-\-4:X + 4. 4:. 5y-2x^+4x-3 =0. 5. 2x^ + Sx + y = 4. 6. Sy'^-2y -\- 4x - S =0. 100 ANALYTIC GEOMETRY 7. Discuss the effect upon the position and form of the curve caused by separately varying the quantities a, 6, and c in the equation y = ax^ -\-bx + c. 8. Discuss the equation x = ay^ + by -\- c. (Compare Art. &1.) 9. A parabolic arch of 60 ft. span is 20 ft. high at the center. Com- pute the heights at intervals of 5 ft. from the center. 10. Through how many arbitrarily assigned points can a parabola with axis parallel to one of the coordinate axes be passed in general ? Name some exceptions. Find the equation of a parabola with axis parallel to the ?/-axis through the three points (1, 0), (3, 2), and (6, 8). Draw the figure. 11. Find the equation of a parabola through the three points of example 10 with axis parallel to the ic-axis. 12. Find the equation of a parabola through the points (— h, yi), (0, 2/2), and (h, yz) with axis parallel to the ?/-axls. 13. Find the equation of a parabola through (1, 0), (3, 2), and (6, 5). 14. Find the equation of a parabola with axis parallel to the ?/-axis passing through (- 20, 0), (0, ^V), and (20, 0). Can a parabola with axis parallel to the aj-axis be passed through these points ? 15. Show that any line parallel to the axis of a parabola cuts the parabola in one and only one point. 83. The ellipse. The ellipse is the locus of a point which moves in the plane so that the sum of its distances from two fixed points in the plane is Y constant. The fixed points are called the foci. To obtain the equation of the ellipse, let the a;-axis be taken through the foci, and the origin midway be- tween the foci. Let the distance between the foci be 2 c. The foci -are then P(:r,i/) Fig. 84. F(c, 0) and F\- c, 0). STANDARD EQUATIONS OF SECOND DEGREE 101 Call the given constant 2 a, where 2 a > 2 c. Let P{x, y) be any point of the ellipse ; then In terms of the coordinates of the points, this becomes ■^(^x - cf + f 4- V (a^ + c)2 + / = 2 a. (1) This is therefore the equation of the ellipse. Equation (1) can be thrown into a more convenient form free from radicals as follows : Transpose the second radical to the right-hand member of the equation and square, a^_2ca;+c2+.y2=4a2-4aV(a;-f-c)2+2/2+a^+2ca;+c2+/. (2) Canceling, transposing, and dividing by 4, cx + a^ = aV(x + cy + yK (3) Squaring, c V -f 2 a^cx + a* = a'x' -f 2 a'cx + a V -f ay. (4) Canceling and collecting terms, (a' -(^)x'-\- aY = a' ip? - c^). (5) Dividing by a^(a^ — c^)j t^^£- = X. (6) All values of x and ?/ that satisfy eq. (1) also satisfy eq. (6), but in obtaining (6) from (1) the operation of squaring was twice performed, and in this process there are introduced values of x and y which satisfy eq. (6) but do not satisfy eq. (1). However, the values so introduced in this case are imaginary, and hence there are no points on the locus of eq. (6) that are not also on the locus of eq. (1). For, start- ing with eq. (6), the steps may be retraced until eq. (4) is reached, where, upon extracting the square root, a double sign is introduced, i.e. cx-\-a?= ± a^ix + cf -{- y^, or —cx = a^ q: aV(x-\-cy + y^ (3') 102 ANALYTIC GEOMETRY Multiply by 4 and add (x -\- cy + y'^ to both sides, a^ + 2caj4-c2 4-?/2-4ca; = 4a2:f 4 aV(a; + c)2 + / + («; + c)2 + /, or {x-cy-{-y' = (2aTV(x-\-cy-j-yy. (2') Extract the square root, ± V{x- cy -^ y' = 2 aT Vix-j- cy + y% or ± V(« - cy + y'- ± V(a; + cy + y^ = 2a. (1') Therefore, if (x, y) is denoted by P, ±PF±PF^ = 2a. Now 2 a is a positive quantity ; hence both negative signs cannot be used. Also FF' = 2c. The difference of PF' and PFis therefore less than 2 c. (Fig. 84.) That difference can- not therefore be equal to 2 a, which is greater than 2 c. Hence the only allowable combination of signs for real values of x and y is given in + PF-]-PF' = 2a. Therefore all real values of x and y that satisfy eq. (6) also satisfy eq. (1). Hence eq. (6) is the equation of the ellipse. Replacing the positive quantity a^ — c^ in eq. (6) by 6*, the equation becomes 84. Graph of ^' + ^' = 1. a^ b^ Solving for y, y = ±- Va^ — x^. a Solving for x, a; = ± - V6^ — y^. (1) The curve is symmetric with respect to both coordinate axes, and the origin. (2) It crosses the avaxis at (a, 0) and (— a, 0) and the y-axis at (0, b) and (0, - b). STANDARD EQUATIONS OF SECOND DEGREE 103 (3) If X is less than — a or greater than a, y is imaginary. If y is less than — 6 or greater than h, x is imaginary. There- fore no portion of the locus lies to the left of x = — a ov to the right of x = a; below y = — b, or above y = b. (4) No finite value of either variable makes the other infinite. (5) In the first quadrant, as x increases from to a, y steadily decreases from b to 0. (6) By (3) neither variable can become infinite. The following points are on the curve, a a 3a 7^ 9a 4 2 T T To ^ X y b .97 b .87 6 Mb The curve is sketched in Fig. 85. .48 6 .44 6 Fig. 85. 85. Axes, vertices, center of the ellipse. Definitions. The chord of the ellipse which passes through the foci is called the major axis of the ellipse; the chord at right angles to the major axis and passing through its center, the minor axis ; their intersection the center, and the ends of the major axis the vertices of the ellipse. Thus in Fig. 85, A' A = 2 a is the major axis, B'B = 2 6 is the minor axis, is the center, A (a, 0) and A' {— a, 0) are the vertices. 104 ANALYTIC GEOMETRY In terms of a and b the foci are F(Va^ — b'-, 0) and since — ^2 G^ — & or c = V<: 6-. 86. The ellipse with major axis on the 2/-axis. In Art. 83 the equation — -j- ^ = 1 was found for the ellipse with center at the origin and with major axis 2 a on the a;-axis. If the major axis 2 a were taken on the ?/-axis, the equation would clearly be obtained from that above by exchanging x and y.. It is therefore, y" ,y? , ^o + 1, where the major axis is 2 a. If, however, the major axis is called 2 b and the minor axis 2 a, the equation becomes the same as that of Art. 83 ; namely, -fi-' This equation therefore represents an ellipse with major axis on the a>axis or the ^/-axis according as a is greater than Y Y a> h Fig. 86. a 7;--->7-' -^— 1 . - - _ ^^xt? - 3^ - . - _ . - ,_ :::fs^:::::::::_::::^?^:.:::::j^: :::::::::::: S-: ::::::::::: ::-zh::: : ±::::h;:::::::i::i??-::: ::::::1:=:::::::::::::?^. -,"- :^^ «;' - _t *» -^ 1 y= 8IN (x+a) Fig. 94. 100. Graph oi y = sin nx^ where n is positive. Let ic' = nx, or x = ~' The equation then becomes y = sin x\ the locus of n which is shown in Fig. 93. Now the substitution of a; = ^ can n be interpreted as shortening the abscissas of all points in the ratio 1 : n without changing the ordinates. If, then, the curve 2/ = sin a; be drawn, the curve y = sin nx can be obtained from it by shortening the abscissas of all points on the curve y = sin x in the ratio 1 : n. This is equivalent to compressing uniformly 120 ANALYTIC GEOMETRY in the direction of the x-axis any portion of the curve y == sin st which begins at the origin into - of its original space, the end n of the curve at the origin to remain at the origin. It is also equivalent to choosing a unit on tlie ic-axis equal to - of the unit on the 2/-axis and then plotting the curve y — sin x. n. If n is less than 1, the contraction of the curve becomes in fact an expansion. Graphs of 1/ = sin 3 a; and of y = sin ( — - ] are shown in rig. 95, together with y = sin x. ^ 2x -y-. gu::i-:=:=;=::::| ■ iTrNTmiTrrKM ml l:l^lZ:t^^}i:V::-:-::: i-i==±^:l=±!: y— SIN X J = sm 3X Fig. 95. i; = siN^ m 101. Graph of y = sin (nx + m). Letting x = x' , the n equation becomes y = sin nx'. Hence translate the axes to the new origin ( , ] and construct the curve y = sin nx. Com- \ n J pare Arts. 99 and 100. Figure 96 shows the locus of ?/ = sin (nx + m) for n = 2, m= — 1. Fig. 96. TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS 121 102. Graph of y =p8m(nx-\-rn). The graph of this equation can be obtained by multiplying each ordinate of y = sin {nx + m) by p. The curve is shown in Fig. 97 for p and 71 both positive. y^p sinC«xh-w) Fig. 97. 103. The exponential curve, y = a'', where a is positive. (Only positive values of y are considered.) (1) The locus is not symmetric with respect to either coordi- nate axis or the origin. (2) It intersects the ?/-axis at (0, 1), but does not meet the a;-axis. (3) For every real value of x there is one real and positive value of y. Only this value is considered. (4) No finite value of x makes y infinite. (5) lia < l,y approaches zero as ar becomes infinite positively. If a > 1, i^ approaches zero as x becomes infinite negatively. (6) If a > 1, 2/ increases always as x increases. If a < 1, y decreases always as x increases. If a = 1, the curve becomes the straight line y = 1. Figure 98 shows a few curves whose equations are of the form y = a'', for certain values of a. Values of y may be computed by logarithms. E.gAi a = e = 2.718 •••* * The quantity 1 + 4-+- +^- \1 ' \1' li^."*"!! \1 by e and is called the natural base of logarithms, in more advanced mathematical work. — = 2.71828 — is denoted It is of much importance 122 then ANALYTIC GEOMETRY Iogio2/ = a5logio2.718 ... = .4343 a;. Y r 1 1 1 / 1 / u o t 11 ij ? \ / \ 1 / \ 1 1 \ , 1 / t 0^ \__ Ur jy ^ _ X Fig, 98. The following points are on the curve y — e*, a; _5 _3 _2 -1 1 2 3 y .007 .05 .14 .37 1 2.7 7.4 20 5, 148. After a few points on the curve have been obtained, other points are easily found by noticing that when x is doubled, y is squared ; when x is tripled, y is cubed ; etc. This follows at once from the law of exponents, a"* = (cu^y. 104. The logarithmic curve, y = \o^a^. This curve is the same as that oi y = a' with x and y interchanged. The curves for y = logio X and y = logg x are shown in Fig. 99. TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS 123 Since log^ x = log^ x log^ a, when the curve y = log^ a; has been constructed for any value of a, the curve y = logf, x can be easily obtained from it by multiplying all the ordinates of the first curve by logj a. E.g. the ordinates of y = log,x are 2.3026 times the correspond- ing ordinates oi y = logi6 a;, since log. 10= ^ 1 logio e .4343 Fig. 99. = 2.3026. 105. Graph of y=e ""% where Y I] i_ r I I 3 t 1 \ V 5 V \ \ ^^^ ""^ Z±I X 1 L e = 2.718 ••• and a is positive. Since e""* = (e-")' — this curve is of the ©■ form Fig. 100. same lorm as y = a% where a^ is less than 1. The curve is therefore as shown in Eig. 100. (Compare Art. 103.) The rapidity with which the curve falls as the tracing point moves from left to right depends upon the value of a. 106. Graph of y = he~'^ sin {nx + m). This graph is easily ob- tained by plotting sep- 124 ANALYTIC GEOMETRY arately the graphs of y = e~"* and y = b sin (nx 4- m) and multi plying together the corresponding ordinates. The form of the curve is shown in Fig. 101. This is an important curve in the theory of alternating currents. y = b e~^^ SIN {nx+m)y where a= —.4, 6=1.5, ii =2, m= ~3. Fig. 101. EXERCISE XXV Plot the following curves. (The letters i, q, t. are variables.) 1. y = cos X. 2. y = tan x. (Divide ordinates of sine curve by those of cosine curve. 3. y = CSC X. (Obtain from sine curve.) 4. y = sec X. 5. y = -^^ • (Examine carefully near x = 0.) sin ic S. y = cos 2 X. 8. y = tan 2 x. 7. 2/ = cos3x. d. y = tan 3 x. TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS 125 10. y = sin (3x— 1). 11. y = cos (x -\- a) . 12. y = cos (nx). 13. y =cos(nx-\-m). 14. Show that the graph ot y = cos x is the same as the graph of y = sin a; moved parallel to the ic-axis the distance - in the negative direction. 15. By what change in position can the graph of y = cot x be made to coincide with the graph of y = tan x ? 16. I = 6e-«'. 17. i = 6(1 — e"«') . (Combine the graphs of i = b and i = &e-«'.) 18. i = bte-"^. 19. gz=6 + c(l + A;Oe-«\ 20. q = asinnt + b sin 3n^ 21. y =x + &mx. 22. 24. y = sin a; + cos x. y = sin2 X. 23. —a)- 26. y = sinii X. 25. ?/ = sin^ X. 28. y = smx-\- sin 2 x. 27. yii = sin X. 30. y^^ = sin X. 29. y = sini*^ X. 32. y = sin a; + sin 3 X + sin 5 x. 31. y = sin" X. 34. q = e-' sin 2 «. 33. ?/ = e-2a:sjnjc^ - 36. y = sin-ix. 35. I = e ""' sin w^. 38. y^sin-ix. cos-ix 37. y = tan~i x. 39. y = sin X -f ^ sin 3 X + ^sin5x + |sin7x. 107. Plotting in polar coordinates. The methods used in plotting a curve in polar coordinates do not differ essentially from those used in plotting curves in rectangular coordinates. The difference comes mainly in the manner of locating the points. The following examples will sufficiently illustrate the methods. Example 1. To plot in polar coordinates the curve whose equation is r = a cos 6. The following pairs of values of r and 6 may be at once written, using approximate values of r : 6 0° 30° 45° 60° 90° 120° 135° 150° 180° r a .S7 a .71a .5 a —.5 a —.71a —.87a —a 210° 225° 240° 270° 300° 315° 330° 360° r —.87a —.71a —.5 a .5 a .71a .87 a a 126 ANALYTIC GEOMETRY An examination of the variation in r as 6 increases from 6 to 360° shows that as 6 increases from to 90°, r decreases from a to ; as ^ increases from 90° to 180°, r is negative and decreases from to— a, the point (r, B) tracing out apart of the curve in the fourth quadrant ; as $ increases from 180° to 270°, r remains negative and increases from — a to 0, the point {r, 6) tracing over again the part of the curve already traced in the first quadrant; as 9 increases from 270° to 360°, r increases from to a, the point p (r, 6) tracing over again the part of the curve already traced in the fourth quadrant. If 6 is allowed to increase beyond 360° or to take negative values, P cos takes on the r = a cos e. same series of values Fig. 102. already obtained, since cos {6 ± 360°) = cos 0, and no new points are obtained. The curve is therefore as represented in Eig. 102. The curve appears to be a circle. That it is so in fact may be proved as follows : Take any point P{r, 0) on the curve ; then r = a cos or cos $ = — Therefore Z OPA must be a a right angle. Therefore the curve is a circle. Example 2. To plot in polar coordinates the curve whose equation is r^ = a^ cos 3 6. For ev^ry value of which makes cos SB positive there are two values of r which differ only in sign, and for every value of B which makes cos 3 B negative the values of r are imaginary. In order then for r to be real, 3 B must be an angle in the first or fourth quadrant. Let the positive value of r be taken for discussion first. TRIGONOMETRIC AND EXPONENTIAL FUNCTIONS 127 The following table shows the changes that take place in r as 6 increases from to 2 tp. e oto;. '!:to '^ 6 2 i-T 3 6 se Oto'^ 2 ^to^ 2 2 ^to2. 2.to^- r a toO imag. Oto a a to 5 TT , ^ 7 TT — to 6 6 6ir .Irr — to — 2 2 — — to — — to e 6 3 i^to§^ 3 2 ie ^to4. 4.to^ 2 r to a a too IItt 6 11 TT 2 2 11 TT imag. to 2 11 to6 7r to a The second column is to be read, as 6 increases from to ^, 3 d increases from to -, and hence r decreases from a to 0, and similarly for the other columns. A few intermediate values of r and 6, computed from a table of natural cosines, are shown for values of ranging from to J- 6 ?*2 = a"^ cos 3 8 Fig. 103. 128 ANALYTIC GEOMETRY 6 5° 10° 15° 20° 25° 30* r .98 a .93 a .84 a .71a .51a Since cos 3 6 takes the same values, either in the same order or in the reverse order, when 6 increases through the other intervals for which r is real as it does when increases from to — , the values of r are the same in those intervals as in the 6 iirst. The curve is shown in Fig. 103. The dotted portion is the part corresponding to the negative values of r. If 6 is allowed to increase beyond 2 tt, or to take negative values, cos 3 d takes the same set of values over again, and the same points of the curve are again obtained. EXERCISE XXVI Plot the following curves in polar coordinates. 1. r = a sin 6. (Prove it is a circle.) 2. r = d. 3. r = a tan 0. 4. r = 2 b, take the center of the ellipse as a center and describe circles of radii a and b. Draw any radius making an angle with the major axis. Through the points where the ra- dius cuts the inner and outer circles draw par- allels respectively to the major and minor axes. Their intersection is a point of the ellipse. Proof. Taking the a;-axis along the major axis of the ellipse the point of intersection P is at once seen to have the coordinates x = a cos 6,y = b sin 6, and is therefore a point of the ellipse from the preceding article. Exercise 1. Construct an ellipse by this method. Fig. 106. 132 ANALYTIC GEOMETRY Exercise 2. Prove that, for the same values of x, the ordinates of the ellipse and circle in Fig. 105 have a constant ratio, -. a Exercises. The sun's rays fall vertically upon a plane; prove that the shadow on this plane of a circular hoop not parallel to the plane is an ellipse. 112. The cycloid. The curve traced by a fixed point on the circumference of a circle as the circle rolls in a plane along a fixed straight line is called the cycloid. The circle is called the generator circle and the point the generating point. To derive the equations of the cycloid : Let the fixed line be taken as icaxis and the point on this line where the gen- erating point touches it as the origin. Take the y-axis per- pendicular to the ic-axis. Let Pix, y) be any position of the generating point, the angle, measured in radians, through which the radius through Phas turned since the generating point left the origin, and a the radius of the circle. Then (Fig. 107) M Fig. 107. x=OM=OH+HM. y = MP=HC+CN. Now 0H= arc HP= aO, HM= - PN= - a sin ^, PARAMETRIC EQUATIONS OF LOCI 133 HC = a, CN= — a cos 6. .'. oc = a^ —a sin 0, 2/ = a — a cos 0. (1) (The student should make sure that these equations hold for either position of the generator circle shown in Fig. 107, and should draw other positions of the generator circle and prove that the same equations hold.) Equations (1) give the values of x and y in terms of a third variable Q. By assigning values to ^, values of x and*?/ may be computed and thus points on the curve located. It is usual to take the two equations (1) as representing the cycloid, but a single equation connecting x and y may be ob- tained as follows : From the second equation, 1 — cos ^ = ^ , or vers ^ = ^ • a a .-. ^=:vers~^^, a and sin B = Vl-cos2^= ^1 - f^^~^X = ^ V2ay-y\ Substituting these values of 6 and sin $ in the first of eqs. (1), there results x = a yers"^ - if V2 ay — y\ (2) 113. Construction of the cycloid. Besides the method of locating points on the cycloid by computing values of x and y from eqs. (1) of Art. 112, the following method may be easily employed : On a straight line lay off a distance OA equal to the circumference of the generating circle. At the middle point B of OA draw a circle equal to the generating circle tangent to OA. Divide OB into a number of equal parts by the points Bi, B,,, -Bg, etc., and the semi-circumference BC into the same number of equal parts by the points (7i, Og, Cg, etc., 134 ANALYTIC GEOMETRY obtained by use of the protractor. Through Ci, Cg, Cg, draw lines parallel to OA. -Pi- C C ^ ■-" O B2 B, B Fig. 108. As the circle rolls back, the point P, now at the top of the cir- cle, generates the cycloid, the point P descending to the level of Ci when the point of tan gen cy moves back to Bi. Hence the point Pj may be obtained by using Ci as a center and BBi as a radius to describe an arc cutting the line through Ci. Similarly with radius equal to BBo and center C2 the point P2 is obtained, etc. Other methods of constructing the cycloLl are employed by draftsmen. Exercise 1. Construct a cycloid by the method explained, dividing the circumference into twelve equal parts. Exercise 2. Construct a cycloid by computing values of X and y by eqs. (1), Art. 112. 114. The hypocycloid. The hypocycloid is the curve traced by a fixed point on the circumference of a circle which rolls internally along the circumference of a fixed circle. To derive the equations of the hypocycloid : Let the radii of the fixed and rolling circles be a and b respectively. Take the center of the fixed circle as origin, and the line through this center and the point of contact of the generating point with the fixed circle as a;-axis. Let P(x, y) be any position of the generating point, d the angle through which the line of centers has rotated, and <^ the angle through which any PARAMETRIC EQUATIONS OF LOCI 135 radius of the generator circle has turned since the generating point left the x-axis. Then (Fig. 109), X = 0M= OH -{-NP=OC cos e+CF cos = (a — b) cos6-\-b cos , y = MP= HC-NC={a- b) sinO-b sin <^. Fig. 109. Now arc PB = arc AB, and therefore b(-{-0) = aO, , a — b 9 = - or 0. .'. 0?= (a — 6)cos0 + b cosi— -0); 115. Construction of the hypocycloid. From the above equations as many values of x and y as desired may be com- puted by assigning arbitrary values to 6. By this means a sufficient number of points may be obtained, through which the curve may be drawn. Another method is as follows : Draw two concentric circles, K and K', with radii a and a — b respectively. Let <^'= ^ -|- <^ ; 136 ANALYTIC GEOMETRY then ad= b'. Compute the value of 6 which makes cf>' = 360° i.e. = - 360°. Let AOB be this angle, constructed by use of a the protractor. Then B is the second point of contact of the generating point with the fixed circle. Divide AOB into any number, n, of equal parts and draw radii to intersect the circle K' at Ci, Co, Cs, etc., and the circle K at Bi, B2, B^, etc. With Ci, C2, •" as centers draw circles of radius b. Fig. 110. The position of the generating point on the first of these circles is obtained by drawing an angle B^C^P^ equal to -th of n 360°; the point on the second circle by drawing an angle B^C^P^ equal to ?ths of 360°, etc. See Fig. 110, where n = 8. 116. The hypocycloid where a = 2 6. Letting a = 2 6 in the equations of Art. 114, there is obtained a? = a cos 0, The latter equation shows that the generating point moves along the a>-axis, and the former that it is at any time in the same vertical as the point of contact of the circles. PARAMETRIC EQUATIONS OF LOCI 137 Hence, if a circle rolls within a Jixed circle of double the diam- eter, every point of the rolling circle moves hack and forth along a diameter of the fixed circle. Moreover, if the circle rolls with uniform angular velocity, every point of it moves with simple harmonic motion.* 117. The four-cusped hypocycloid. The points where the generating point reverses its direction of motion are called cusps. Thus the points of contact of the generating point and the fixed circle are cusps. If a = 4 5 there are four cusps. The curve in this case is of interest because it is possible to eliminate between the equa- tions of Art. 114 and obtain a simple equation connecting x and y. Substituting - for b in eqs. (1), Art. 114, they become a; = — cos^ + |cos3d = |(3cos(9-f cos3^), 2/ = — sin ^- - sin 3 (9 = -(3 sin - sin3 ^). 4 4 4 By trigonometry, cos 3 ^ = 4 cos^ ^ — 3 cos 0, sin3(9 = 3sin^-4sin3d. Substituting these values, there result jc=acos^^, y = asm^O, from which cos^=f-] > a, sm6/ = * When a point moves with uniform velocity along the circumference of a circle the projection of the point on any diameter is said to have simple harmonic motion. 138 ANALYTIC GEOMETRY Squaring, adding, and clearing of fractions, 118. The epicycloid. The epicycloid is the curve traced by a fixed point on a circle which rolls externally on the circum- ference of a fixed circle. Fig. 111. Let the student show from the figure that the equations are ac = (a + 6)cos 6 —ft cos 2/ = (a + 6)siii e - 6 sin b a-\-b, Notice that the equations differ from those of the hypocycloid only in having — b take the place of b. 119. The cardioid. The epicy- cloid for which the rolling and fixed circles are equal is called the car- dioid. Its equations are obtained by letting 6 = a in the equations of the Fig. 112. PARAMETRIC EQUATIONS OF LOCI 139 Fig. 113. preceding article. They then become a? = 2 a cos — a cos 2 9, 2/ = 2 a sin — a sin 2 e. 120. The involute of the circle. If a thread is wound around a circular form and then unwound, kept always stretched, any point of the thread traces a curve called the involute of the circle. To derive its equations: Choose the axes as in Fig. 113. Let a be the radius of the circle, P(x, y) the position of the generating point at any time, and 6 the angle through which the radius to the point of tangency has turned during the unwinding. Then x= 0M= 0N+ NM= 0N+ TP sin 0, y = MP = NT- JST= NT- TPcos 6. Now TP = arc ^ r = aO. .*. a? = a cos + a sin 0, 2/ = a sin e — a cos 0, are the equations of the involute of the circle. EXERCISE XXVII 1. Trove that if a circle of radius a rolls along a straight line, a point on a fixed radius of the circle at a distance h from the center describes a curve whose equations are * jc = a^ — & sin ^, y = a — hco&d. Plot the curve for 6 < a ; for ?) > a. These curves are called trochoids. 2. Devise a method of constructing the cycloid similar to the method of constructing the hypocycloid in Art. 115. 140 ANALYTIC GEOMETRY 3. Carefully construct on coordinate paper a cycloid by the method you have described. By counting the squares between the cycloid and the line on which the circle rolls and the squares in the generating circle, what idea do you get of the area of the cycloid ? 4. By combining the equations of the cardioid (Art. 119) and trans- forming to polar coordinates, show that the polar equation of the cardioid is r = 2 aCl — cos 0), where the pole is the point of contact of the gen- erating point with the fixed circle. Fig. 114. Suggestion. Square and add the equations of Art. 119, move to new origin by letting x = x' + a, y = y' ; substitute x' = r cos^, y' = rsin 6 ; complete the square in the terms in r, and extract the square root. Also derive the polar equation independently from the figure. (Fig. 114.) 5. Taking the origin at the point of the cycloid farthest from the line on which the circle rolls, and the ic-axis parallel to that line show that the equations of the cycloid are X = ad + a sin 0, y z=— a -\- a cos 6, where 6 is measured from the positive direction of the y-axis to the radius of the circle through (a;, y), clockwise rotation being counted positive. 6. Construct a hypocycloid where a = Sb. 7. Devise a method for constructing the epicycloid and apply it to the case where a = 4b. PARAMETRIC EQUATIONS OF LOCI 141 8. Construct the involute of a circle. 9. A circle rolls along a straight line, and a line through the center of the circle turns about a point of the fixed line. Find the equations of the locus of the point of intersection of line and circle, and plot the curve. Ans. x = a (cot 6 + cos ff) , y = a{\ + sin ^) for outer point, X = a(c,otd — cos^), y = a(\ —simd) for the inner point. and Fig. 115. 10. Show that the polar equations of the curves of example 9 are r = a (esc ^4-1) and r = a (esc ^ — 1) respectively. 11. A circle moves with its center always on a straight line, and a second straight line passes through the center of the circle and a fixed point. Find the loci of the pointp of intersection of the second line and the circle. Ans. Using the notation of Fig. 116, X = b tan -{■ asm 6, y = acos 6, for P. X = b tan d — a sin 0, Fig. 116. y =— acos0, for P'. 12. Plot the curves of example 11 for 6 < a, 6 = a, & > a. CHAPTER X INTERSECTIONS OF CURVES. SLOPE EQUATIONS OF TANGENTS 121. Intersections of curves. It has been seen that an equation in two variables can be represented graphically by a curve, every point of which has coordinates which satisfy the equation. Two different equations in the same two variables will then in general represent two different curves. If these curves be plotted on the same diagram they may or may not intersect. The coordinates of the points of intersection, if any, must satisfy both equations, and no other points will have this property. ISTow the values of the variables which satisfy two equations are obtained by solving the two equations as simultaneous. Hence to find the points of intersection of two curves, solve the equa- Y tions of the curves as simultaneous. The real values of the variables so obtained which sat- isfy both equations are the coordinates of the points of intersection of the curves. Example. To find the points of inter- section of the circle ic2 -f 2/2 = 16 and the pa- rabola x^ = Qy. Elimi- nating X from the first Fig. 117. equation by substitut- 142 INTERSECTIONS OF CURVES 143 ing the value of x from the second, there results from which y = 2 or — 8. Substitutmg the first of these values of y in the second equation, there is obtained x= ± 2 V3. The substitution of — 8 in the second equation gives imaginary values of x. Hence the points of intersection are (2v^, 2) and (-2V3, 2), or approximately (3.46, 2) and (- 3.46, 2). On plotting the curves these results are seen to be approxi- mately correct. EXERCISE XXVm Find the points of intersection of the following pairs of curves. Check graphically by plotting the curves and measuring the coordinates of the points of intersection. 1. y? + yp- = 5, y^ =i\x. 2. y = 3 ic + 7, x2 + ?/2 := 9. 3. (a) y = 2 a; + i, y2 = 4 a;. Ans. {\, 1) (Tangent) . (6) y = 2a; + .49, y'^ = ^x. Ans. (.326, 1.141), (.184, .859). (c) y = 2a; 4- -51, y^ = \x. Ans. No intersection. 4. a;2 + 4 ?/2 - 16, a;2 + y = 0. 5. 3 a; - y = 1, 16 a;2 + 9 y^ = 144. 6. a; +2/ = 5, 9 a;2 + 16 2/2 = 144. 7. a;2 -I- ?/2 = 16, a:2 - 2/2 = 9. 8. For what values of6isz/ = 2a; + 6 tangent to a;2 + y2 =; 9 ? 9. For what values of 6 is y = wa; + 6 tangent to a;2 -f- y2 _ ,.2 9 10. For what value of p is y2 = 2 px tangent to?/ = 3a;4-l? 11. Prove that the two segments of any line which cuts xy — O in two points, included between the curve and its asymptotes, are equal. 122. Graphical solution of simultaneous equations. It fre- quently happens that when two equations containing two variables are given it is not possible to eliminate one of the variables, and so obtain an equation with only one variable ; or, if the elimination is possible, the resulting equation may be very difficult or impossible of solution by ordinary methods. 144 ANALYTIC GEOMETRY In such cases, if the coefficients are numerical, an approximate solution may be obtained by carefully plotting the curves and measuring the coordinates of the points of intersection. More accurate solutions may then be obtained by methods illustrated in the following examples. Example 1. To find the intersection of the curves and y = sin X y = 2x + l. (1) (2) Plot the curves carefully on coordinate paper. From the figure the abscissa of the point of intersection is seen to be about — .9. Substitute this value in equations (1) and (2), remembering that .9 radian = .9 of 57°.3 = 51°34', and there results, from (1) y = sin(- 5r34') = - .78, from (2) 2/ = - -8. Y Fig. 118. This shows the assumed value of x to be too small, but very near to the correct value. (Compare Fig. 118.) Try next a = - .88. Then, from (1), y = sin(- 50°25') = - .771, from (2), 2/ = - •'J'^. This shows the assumed value of x to be too large, so n^xt try x=- .89. INTERSECTIONS OF CURVES 145 Then, from (1), y = sin(- 51°)= -.777, from (2), y=- .78. Hence, correct to two significant figures, the solution is x=- .89, y=- .78. Example 2. To solve the equation a^_2a;2 + 4a;-7=0. (1) Let ?/ = a^-2a^ + 4a;-7. ' (2) Then the solution of (1) is the same as the simultaneous solu- tions of (2) and the equation 2^ = 0. (3) Plot the curve of eq. (2). (Figure not shown.) The following are corresponding values of x and y: X 1 2 3 4 -1 -2 -3 y -7 -4 1 14 41 -14 -31 -64 The curve, is seen to cross the a^axis between 1 and 2, at about 1.8. Try this value of x in eq. (2) ; y = 5.832 - 6.48 + 7.2 - 7 = - .448. Hence the value of 1.8 for x is too small. Try next a; = 1.9 ; then y = .239. Hence the value of 1.9 for x is too large. Plot now on an enlarged scale the points representing x and 2/ for a; = 1.8 and 1.9, and join the points by a straight line. Since the interval is small, the curve probably differs but slightly from a straight line in the interval. The line is seen to cross at about .65 of the distance from 1.8 to 1.9. Then 1.865 is probably a close approximation to a root of eq. (1). Substituting this value of x in eq. (2), there results ?/= — .0094. 146 ANALYTIC GEOMETRY The work of computation, arranged according to Horner's method of synthetic division, is as follows : 1-2 4-7 )1.865 ■ 1.865 -.2518 6.9906 - .135 -3.7482 - .0094 By the Remainder Theorem from Algebra, the value of 2/ is the last remainder, — .0094. Since y comes out negative, it shows that in this case the assumed value of x is too small. Try then x = 1.866. 1-2 4-7 )1.866 1.866 - .2500 6.9975 - .134 3.7500 - .0025 Hence y=- .0025. Try next x = 1.867 : 1-2 4-7 )1.867 1.867 -.2483 7.0044 .133 3.7517 .0044 Hence y = .0044. The root therefore lies between 1.866 and 1.867 and is nearer to the former. Hence, correct to four significant figures, a root of eq. (1) is 1.866. Evidently one could by this method obtain a root correct to any desired degree of accuracy. Example 3. To solve the equation <^2 _ sin 2 <^ = 0. (1) This may be treated as in the last example, or it may be more easily solved as follows : Plot separately the curves y = ^' (2) and 2/ = sin 2 <^ (3) on the same diagram. Then a value of <^ at a point of inter- section of the curves of eqs. (2) and (3) is a root of eq. (1). INTERSECTIONS OF CURVES 147 The figure shows that a value of <^ at the intersection is a little less than 1. Try then <^ = .9. Then from (2) y= M and from (3) y = .974 Difference =—.164. Y Fia. 119. Substitute = ^f then from (2) y = 1 and from (3) y = .909 Difference = .091 Plot on an enlarged scale the difference for <^ = .9 and <^ = 1, using <^ as abscissa and difference as ordinate, and connect the points obtained by a straight line. This straight line is seen to cross the axis at about .65 of the distance from .9 to 1. On substituting <^ = .965 there results from (2) y = .931 and from (3) y = .936 Difference =— .005 Let the student show that when x, 2, = m^ + -^ (6) oo^^^py, y = fna^-:i^l^. 2fn 2 (7)xy = c, y = mx± ^yZ—cm. EXERCISE XXX (Use the above formulas in solving these exercises.) 1. Find the equations of tangents to ic^ -\-y^ = \Q which have a slope equal to V3. Check graphically. 2. Find the equations of tangents to 9 a;2 + 16 y"^ = 576 which are parallel toy = x. Check graphically. 3. Find the equation of a tangent to ?/2 = 6 x which is perpendicular to2x— y — 3 = 0. Plot the lines. Where do they intersect ? 4. Write the equation of a tangent to y- = 2px and the equation of a line through the focus perpendicular to the tangent, and prove that they intersect on the ?/-axis. 5. Obtain the slope equation of a tangent to the circle from the equa- tion of the tangent to the ellipse. 6. Find the equations of tangents to y^ — Qx from the exterior point (2, 4). Check graphically. 7. Find the equations of tangents from (7, 1) to x^ + 2/2 _ 25. Check graphically. Ans. 3a; + 4y — 25 = 0, 4a; — 3y — 25 = 0. 8. Find the equations of tangents to 9 x^ _ 25 y'^ = 225 which pass through (— 1, 3). Check graphically. Ans. X — ?/+4 = 0, 3x + 42/— 9 = 0. 9. Find the equations of tangents to 12 x^ + 5 ?/2 = 30 which intersect in (—3, —2). Check graphically. 10. Show by the use of formula (7), Art. 124, that no tangent can be drawn to xy = 8 which has a positive slope. 11. Find the equations of all lines that are tangent to x^ + y2 = 25 and x2 + 4?/2 = 36. Plot. Ans. 11 y =±4\/ll x ± 15V33. 12. Find the equation of a common tangent to y"^ = 2px and x^ = 2py. Check graphically. 13. Find the equation of the common tangent to y^ = 6 x and x^ = 48 y. Check graphically. SLOPE EQUATIONS OF TANGENTS 151 14. Find the equations of tangents to h^x"^ — a^y^ = a%^ that intersect in the origin. Ans. The asymptotes, hx — ay — 0, hx + ay = 0. 15. For what value of m\^y = mx + 8 tangent to y"^ = ix^ Plot. 16. A line is tangent to x^ -\-y'^ = 16 and y'^ = Qx; find its equation. How many solutions ? Plot. 17. Find the equations of lines of slope 2 which are tangent to a:2 4.y2_4^_l_6y_l_5,3 0. Plot. 18. Prove that y — k = m^x — h) ± r\'l -\- m- '\^ tangent to Suggestion. Move the origin to (h, k) ; use formula (1), Art. 124, and then translate the axes to the original position. 19. Find the equations of tangents to x^ + ?/2 — 4 x + 6 1/ — 12 = 0, with slope 2, by using the formula of Ex. 18. 20. Prove that y — k = m{x — h) + -^ is tangent to 2m (y-ky = 2p{x-h)' 21. Find the equation of a tangent to ?/2 — 2?/ — 4x = with a slope equal to 3. 22. Find thp slope equation of a tangent to ^^ ~ ^^^ + ^^ ~ ^') = 1. a^ 62 23. Find the equations of tangents to 4x2 + 9y2-|-8x-36i/ + 4 = with slope equal to — 3. 24. Find the equations of lines with slope equal to 2 which are tangent to aj2 - y2 = 1. 25. Prove that a line with slope numerically less than - cannot be a tangent to 62^2 - a'^y'^ = a%\ 26. Prove that any two tangents to y^ = 2px which are at right angles to each other intersect on the line x = —^, the directrix. 2' 27. Show that any two tangents to the ellipse h'^x'^ + ci^y^ = a%^ which are perpendicular to each other intersect on the circle ^2 -\-y'^ = «2 -\. ^2^ Suggestion. The equations of two tangents to the ellipse which are perpendicular to each other are y = mx-{- y/m'^a^ + 6^, (1) and y = -^- + A/-,+ &'- (2) 152 ANALYTIC GEOMETRY If these equations be regarded as simultaneous, the values of x and g that satisfy thera are the coordinates of the intersection of tlie two tangents. If then m be eliminated between the two equations, an equation will be obtained which is satisfied by the coordinates of the intersection of any two perpendicular tangents. In this case the elimination is easily made as follows ; Eqs. (1) and (2) may be written y — mx— y/m'^cfi + h^t and my -\-x= y/d^ + rri^h^. Then square and add, 28. Prove that the locus of foot of the perpendicular from the focus upon a tangent to the ellipse V^x'^ + a^y'^ = d^lP- is the circle aj^ + y2 _ ^^2^ Check graphically. 29. Show that any two tangents to the hyperbola hH^ - a?-y'^ = a^W' which are perpendicular to each other intersect upon the circle a;2 _|_ ^2 _ Q,2 _ yi^ if a>h, but that there are no perpendicular tangents if a < &. What if a = & ? 30. Prove that the locus of the foot of the perpendicular from the focus upon a tangent to V^x'^ — aV = «^&"^ is the circle x^ + 2/2 = (fi^ Check graphically. 31. Find the equation of the locus of the foot of the perpendicular from the center upon the tangent to W-x^ + a^y^ — a^W-. Ans. (x2 + !/2)2 = a2x2 + 62^,2, 32. By transforming to polar coordinates reduce the equation of Ex. 31 to the form r^ = oP- cos2 + IT- sin2 Q. Construct the curve by use of the circles r = a cos 6^ and r=.hsva.d. (See Fig. 121.) 33. Find the equation of the locus of the foot of the perpendicular from the cen- ter upon a tangent to the equilateral hyperbola x'^ — y^ = a? Ans. (x^ + y-^y = a^Cx^-y^). Yjq 121 ^^- Show that the equa- tion of the locus of Ex. 33 in polar coordinates is r^ = a^ cos 2 d. Plot the curve. This cui-ve is called the lemniscate. CHAPTER XI SLOPES. TANGENTS AND NORMALS. DERIVATIVES 125. Introduction. In this chapter methods will be derived of finding the direction of a curve whose equation is known in rectangular coordinates at any point of the curve ; of finding the equations of tangent and normal to the curve at any point ; and some general methods established which will shorten the work of computing the slopes of curves. These methods will be shown in their application to some numerical cases. 126. Increments. In an equation connecting x and ?/, e.g. 42/ = aj2-2a; + 4, (1) if a value be assigned to x, y takes a value to correspond ; and if X is given a different value, y will in general take a different value. Thus, if x = 0, then 2/ = 1 ; if a? = 1, then y = %\ if x = — 1, then 2/ = J ; if 'ic = 2, then y = l. Any change in x in general brings about a change in y. These changes are most easily seen by referring to the curve which eq. (1) represents. As the point {x, y) traces the curve, both x and y change, and the amount that y changes depends upon the amount that x changes, and also upon the point of the curve from which the change is reck- oned. Thus, if X increases by 1 from the value 1, y increases from J to 1, or 163 154 ANALYTIC GEOMETRY the increase in ?/ is J ; while if x increases by 1 from the value 2, y increases from 1 to J, or the increase in y is f . Again, if x increases from — 3 to —2,y decreases from -L^- to -^-j or it may- be said that the increase in y\^—\. Suppose now that some definite value of x is chosen, and a study made of the changes brought about in y by increasing X by small amounts from this definite value. Let the increase that is given to x be denoted by the symbol Aa?, read " delta oc^ or " increment oo " ; and let the increase brought about in y by this change in ic be denoted by Ay, read "delta 2/," or "incre- ment yP The following table shows values of a;, ?/, Aa;, Ai/, and the ratio — ^, the value 2 being chosen for x from which to reckon Aa; the increments. The values of Ax are arbitrarily assumed. ^y=.y? — 2x-\-^. a? y Aa, Ay Aa? 2 1 3 1.75 1 .75 .75 2.5 1.3125 .5 .3125 .625 2.1 1.0525 .1 .0525 .525 2.01 1.005025 .01 .005025 .5025 2.001- 1.00050025 .001 .00050025 .50025 2 + Ax i+f-? Aa; 2 4 .a.f An examination of this table shows that as the increment in ' X is made smaller and smaller the corresponding increment in y becomes smaller and smaller, and approaches the limiting value zero when Aa; approaches the limiting value zero. The ratio — ^, however, does not approach zero, but approaches the t^x limiting value .5 when Aa; approaches the limiting value 0. SLOPES 155 Fig. 123. 127. Slope of the curve at any point. Look now at the geo- metric meaning of these facts. If P and P' denote the points (2, 1) and (2 -\- ^x,l + A?/) on the curve, then Aa; and A?/ have the values shown in the figure, and the ratio — ^ is the slope of the Aa; ^ secant line through P and P'. As Aa; approaches the limiting value zero, the point P' moves along the curve to the limiting posi- tion P, and the secant line through P and P' turns about P to the limiting position defined to be the tangent to the curve at P. Hence the slope of the tangent line at (2, 1) is .5. Definition. The slope of a tangent to a curve at any point is called the slope of the curve at that point. The method here employed is a general one. By it one can compute the slope of the curve at any point. The table of values need not be computed, as in the preceding article, for this purpose. E.g. to find the slope of the curve at the point where a; = 3 one may proceed as follows : Substitute a; = 3 in the equation ; then 2/ = J- Take a point on the curve near P(3, ^). It may be repre- sented by P' (3 + Aa;, -J + A?/). Since this point is on the curve, its coordinates must satisfy the equation of the curve. .-. 4(J + A2/) = (3 + Aa^)2- 2(3 + Ax)4-4, or y4-4A2/ = ^ + 6Aa!-f-A«^ — ^ — 2Aa;-|-^, or 4 A^/ = 4 Ax -f- Aa; . ^ — 1 I A^. * * Aa; 4 156 ANALYTIC GEOMETRY As Ax approaches the limiting value zero, i.e. as P' moves along the curve to coincide with P, the ratio — ^ approaches Ax the limiting value 1, which is, therefore, the slope of the tan- gent line to the curve at the point (3, J). EXERCISE XXXI 1. Compute the value of Ay when Ax = .01 for x = .5, 1, 10, respec- tively, iny = x^. 2. Compute the slope of the curve 4 y = x^ — 2 x + 4 at the points where x=— 1, x = 0, x = 4. 3. Find the slope of the curve 8 y = x^ + 1 at the points where x = 1, 3, 0,-2. 4. Write the equation of the tangent line to the curve of eq. (1), Art. 126, at the point (3, |). 128. Equation of the tangent to a curve at any point. As a second example let it be required to find the slope of the tan- gent line to the curve 40^ + 2/^=4 (1) at any point (xq, y^) on the curve, and the equa- tion of the tangent line at that point. Let P(xq, 2/o) be any point of the curve and Q (xq 4- Ax, 2/o + Ay) a point of the curve near P. For convenience Ax is taken positive, and then Ay will be positive or nega- tive according as the curve rises or falls to- ward the right from P. In the figure, for either position shown, PM= Ax, MQ = Ay. ^ / Q / r \ M r M \ I / \ X A 1 \ i / Q \ V ^ p M TANGENTS AND NORMALS 157 Then — ^ = slope of the secant line PQ, and hence the limit- ing value of — ^, as Aa; approaches the limiting value zero, is the slope of the tangent line to the curve at {xq, y^. Since (a^o, ?/o) and {xq + Ax, y^ -\- Ay) are points on the curve, the coordinates must satisfy eq. (1). .-. 4«o^ + 2// = 4, (2) and 4(a-o + Ax)2-f-(yo + A2/)2=4. (3) Expanding eq. (3) and subtracting the corresponding mem- bers of eq. (2), there results SxoAx + 4:Kx-\-2 y^Ay + A^^ = 0. (4) Every term of this equation contains either Ay or Aa; as a fac- tor. Take to the right member all the terms containing Aa; and factor the two members of the equation. Then Ay (2 2/0 + A2/) = - Ax (8 ;Bo + 4 Aa;), or A^^_8^_+4A^^ ,^. Aa; 2 2/0 + Ai/ As Q moves along the curve to the limiting position P, both Aa; and Ay approach the limiting value zero, and the right member of eq. (o) approaches the limiting value, 2/0 Hence is the slope of the tangent line to the curve at 2/0 {^0, 2/o). Since the equation of a line of slope m through (a^o, 2/0) is y-y^ = m(x-XQ), therefore the equation of the tangent line to the curve at (a^o, 2/0) is y-yo = — ^(x-xo). (6) 2/0 158 ANALYTIC GEOMETRY This equation may be put into a simpler form as follows : Clear of fractions : y^y - 2/0^ = - 4 aroo; -h 4 x^^, or 4«oaJ + 2/o2/ = 4a;o2 + ?V- The right member of this equation is, by eq. (2), equal to 4. Therefore 4 X(flc -f- ?/o2/ = 4 (7) is the equation of the tangent to 4a;2 + 2/^ = 4, at («o, 2/0). Since in eq. (7) (xq, 2/0) may be any point on the curve, the equation of the tangent line at any particular point may be written by substituting for a^o and 2/0 the coordinates of that point. Thus the tangent at (^, V3), which is a point on the curve, is 2x+V32/ = 4. The student must not fail to recognize the fact that in eq. (7) Xq and 2/0 are the coordinates of a fixed point, the point of tangency, and that x and y are the variable coordinates of any point on the tangent line. 129. The normal. The normal to a curve at any point is the line perpendicular to the tangent at that point. Since its slope is the negative reciprocal of the slope of the tangent, the equation of the normal to the curve of the pre- ceding article at (xq, y^ is 2/-2/o = -r^(a5-a;o)- EXERCISE XXXn Find the equations of tangents and normals to the following curves at the points assigned. Check graphically by plotting the curves and the lines whose equations are found. 1. ?/2 = 4a;at (1, -2). 2. a;2 + 2/2 ^ 25 at (- 3, 4). DERIVATIVES 159 3. 2/2 = a;3 at (x^, yo) ; at (0, 0), (1, 1), (4, 8). 4:. y = mx + & at (Xq, yo). 5. y = x^-\-4:X— bsit the points where the curve crosses the a;-axis. 6. x^-y^= 16 at (5,3). 7. a;y = 8at (2, 4). 8. At what angles does the line y = 3x-\-2 cut the parabola y = x^ + x— 6? (By the angle between two curves is meant the angle between their tangents at the point of intersection.) 9. Find the angles at which x^ 4- y'^ = 25 and ix^ = 9y intersect. 10. Find the point on the curve of example 5 where the slope is zero. 11. Find the point- on the curve y=— x^ — Sx-\-2 where the slope is zero. Find also the point of the curve where the slope is 1. Where 2. 130, Tangent equations for reference. By the method used in Art. 128, the student can show that the following are the equations of the tangents to the given curves at the point (^0, 2/o). Equation of Curve Equation of Tangent y,^ = 2poo. ^ Au. Am Divide by Aa; ; ^=0 Ax Ax As Aa; approaches the limit 0, — ^ and — approach the limits dy dx ~»0 . dy dx Since Xq is any value of x, then dy _ ^1 dxi dx ~ dx or, since y = Cu, c Cu) _ p du Ix dx 138. Derivative of a sum. The derivative of a sum of func- tions with respect to any variable is equal to the sum of the derivatives of the functions with respect to that variable : ^^u + v + w-)=^p + ^ + ^ + -. doc doc doc due Proof. (For two functions.) Let ii and v be two functions of X. Let y = u -{- V. Let X = Xq, then 2/0 = Uq -\- Vq. Let x = Xq -}- Ax, then yQ + Ay = Uq 4- Am + Vo + Av. Subtracting, Ay = Au + Av. or DERIVATIVES Divide by Ax, At, ^ A» ^ A« Ax Ax Ax Let Ax approach the limit ; dy dx du ^r dx dv dx dx dx 165 A similar proof holds for any number of functions. 139. Derivative of a product. The derivative of the product of two functions is equal to the sum of the products of each function times the derivative of the other : d(uv) dx doc doc Proof. Let u and v be any two functions of x. Let y = uv. Let X = Xq, then 2/0 = ^o'^o- Let x = Xq -\- Ax, then yo + Ay = (uo + A?^) (vq + Av). Subtracting, Ay = (uq -f Au) (vq + Av) — Uq Vq = Uq Av + Vq Au -\- Au • Av. Dividing by Aic, J Ax Av , Au , Au J. Uo—- + '^0-7- + -— • Av. Ax Ax Ax Let Ax approach the limit 0; then Au, Av, and Ay each approaches the limit 0, and the limiting values -^, — , and Av . . ^^ ^^ ^^ are, respectively, the derivatives of y, u, and v with respect to x for the value Xq. dy dx = Uq dv dx + V(i . du dx 0, 166 ANALYTIC GEOMETRY or, since Xq is any value of x, and y = uv, d(uv) dv . du dx dx dx In a similar way a formula may be derived for tlie derivative of the product of three or more functions. However, one may make use of the formula just proved to obtain the derivative of the product of more than two functions. Thus, d(uvw) d(vw) , , .du dx dx dx [dw , dv~\ V \-w — dx dxj , du dx dw , dv . du = uv f- uw \-vw • dx dx dx 140. Derivative of a quotient. The derivative of the quo- tient of two functions is equal to the denominator times the derivative of the numerater, minus the numerator times the derivative of the denominator, divided by the square of the de- nominator: ^du_^dv €lQC due d fu\ (Ix dx\vj~ v'^ u Proof. Let y = -i V then, vy = u. Differentiating, using the formula of the preceding article, dy , dv du v-^-{-y — = — - • dx dx dx du dv Solving for ^, f^dc^Jj^, dx dx V Eeplacing y by -, V du dv V u — d fu\ dx dx dx\v) v^ DERIVATIVES 167 141. Derivative of the power of a function. The derivative of the nth power of a function is equal to n times the function to the power w — 1, times the derivative of the function : ^^^^ = lii***-! — , where n is constant. dx^ doc Proof. (1) n a positive integer. Let y = w". Let X = Xq, then y^ = Uq\ Let x = Xq + Ax, then yQ-\-Ay= (uq + Aw)'*. Expanding (uQ-\-Auy by the binomial theorem and sub- tracting, Ay = niiQ^'-^Au -f ^ ^f ~^^ i^o""'^ • ^+ ••• + Aw". 1 • ^ Every term on the right after the first contains Au to a power higher than the first. Set out the factor Au and divide both members by Aaj : Ax [_ 2 -\-Au P-- Ax Now as Aa; approaches the limit 0, so do Au and Ay. The limiting value of the quantity in the parenthesis is therefore nV'. dy *** dx or ± (y^^r^^n-.du^ dx dx (2) n a negative integer. Let n = — my where m is a posi- tive integer. Let y = u'* = u~"' = — • ^ u"^ Differentiate, using the formula for the derivative of a quotient, « ^ _ i dju"") dy _ dx dx dgo ~~ w^** 168 ANALYTIC GEOMETRY But — = 0, by Art. 135, and since m is a positive integer, dx ^L(SPQ . chord PQ. .-. =^ = Go^SPq . ^1^2£dPQ . ^, Aa; arc PQ Ax Therefore, letting Aa; approach the limit 0, _dz^_ /tt _ \ ^ dx \2 Jdx' dx^ dx 149. Derivatives of sine and cosine of an angle not in the first quadrant. The foregoing proofs have assumed the angle to be in the first quadrant. Proofs could as easily be given for the other quadrants, or they may be made to depend upon those above. E.g. to find — ^ for a value of u in the second quadrant. dx Let w=5 + v; then sin w= cos-y, cosw=— sinv, and — = — . 2 ' ' ' dx dx c2(sin u) c?(cos v) . dv ^ ,. , ^mq /. -^^ ^ = ^ -^ = — sm v — , by Art. 148, dx dx dx = 008% — , as before. dx DERIVATIVES 175 Analytic proofs of the above formulas which are independent of the size of the angle are given in text-books on the Calculus. 150. Derivative of the tangent. Let y = tan u. sinw .*. y = cos^^ Differentiating by the formula for a quotient, ^^^ d(sin u) . d(cos u) cos u -^ ^ — sm u -^ ^ ay _ dx dx dx cos^ u du ' / . .du cos u COS u sm u {— sm u) — dx 'dx (cos^u-{- sin^i^) — dx = 2 * ^^ d(tan u) ., du 151. Derivatives of cotangent, secant, cosecant. The student can show that the following formulas hold : dicotu) ^^o du dCsecu) . du ~^— — - = — csc^ u — , -^^ ^ = sec u tan u — , ax dx dx dx c?(csc u) , du -^- — ^ = — CSC w cot w — dx dx 152. Summary. The formulas for the differentiation of the trigonometric functions are here collected and numbered con- secutively with those of Art. 142. VIII ^^^m^ = cost*^. doc doc IX ^^^^^^> = -sini^^. doc doc 176 ANALYTIC GEOMETRY dx doc XT ^(cot^^) ^_csc2^J^. ^ ddc doc XII ^^^^^^ = sect* ton t*^. ^ cix doc XIII ^i^f«^=-csci*cott.^. ■ rfic dx 163. Illustrations. The foregoing formulas, together with those of Art. 142, enable one to find the derivative of any alge- braic expression involving trigonometric functions. The fol- lowing examples will help to make this clear. Example 1. Given y = sin^ 2 a;; to find ^• dx By formula VII, ^ = 3 sin^ 2 x I^^IA. dx dx By VIIIandII,iii5iEM=cos 2c. ^^M dx dx — 2 cos 2 x. .'. -^ = 6 sin^ 2 X cos 2 a;. ^^ Example 2. Given 2; = Vl + 2 tan^ 3 s; to find -77' By VII, ^=i (1+2 tan^ 3 .)-^ ^a+2tan^38) ^ ' dt 2^ ^ dt By IV, I, III, VII, and X, d(l + 2 tan^ 3 s) . , o ^2 o „ o ^« -^ — ! L = 4 tan 3 s • sec^ 6s ' 6 dt dt clz 6 tan 3 s sec^ 3 g (^s ' 'di^-Vl+2 tSiu'Ss'dt' 154. Other derivative formulas. Formulas for the derivatives of the inverse trigonometric functions, sin~^t*, etc.; the logar DERIVATIVES 177 rithmic functions, log„ u ; and the exponential functions, a", w", are derived in text-books on the Calculus. The foregoing formulas will be sufficient for use in showing the application of derivatives to the study of curves, which is given in the next chapter. EXERCISE XXXIV Find the derivative of each of the following functions with respect to its variable : 11. y = sec2 X — csc2 x. 12. y= ^ 1. y = a cos2 rK + 6 sin2 x. 2. y = 4 tan3 2 x. 3. z = sin^ t. 4. s = cos^x — sin^a;. 5. ?/ = cos2(a» + 6). 6. sec^x, CSC2 X 7. y = sec* 3 x. 8. ^ _ tan 2 « 1 + sin « 9. y = sin2a;\/secic. 10. y = tan" mx. Vsin X 13. z = ^ tan3 ^ - tan ^ 4- ^. 14. y = X sina;. 15. y = X ta.nx. 16. y = (sin x)x — l. 17. y = cot 4 X CSC 4 x. 18. z = m cot" qx. 19. ?/ = a;(sina;— cosic). 20. q = a sin" bt. CHAPTER XII MAXIMA AND MINIMA. DERIVATIVE CURVES 155. Maximum and minimum points of a curve. In the discussion that follows the curves are supposed to be such that the ordinate is a single-valued, continuous function of the abscissa. If the curve as a whole is not single valued, it can be divided into portions each of which is single valued. For convenience, such a curve, or portion of a curve, may be thought of as generated by a point moving from left to right. If, as the curve is so traced, the generating point rises to a certain position and then falls, that position is called a maxi- mum point of the curve, and the ordinate at that point is called a maximum ordinate. If the generating point falls to a certain position and C then rises, that position is called a minimum point of the curve, and the ordinate at that point a minimum ordi- nate. Thus A, C, and E are maximum points, and 2/1, 2/3, and 2/5 are maximum ordinates, while B and D are minimum points, and 2/2 and y^ are minimum ordinates of the curve in Fig. 128. According to the above definition a maximum ordinate is not necessarily the greatest ordinate of the curve. The defini- tion requires only that a maximum ordinate shall be greater than the ordinates immediately to the right and left of it. 178 Fig. 128. MAXIMA AND MINIMA 179 At a maximum point the curve is said to change from rising to falling, and at a minimum point to change from falling to rising. 156. Determination of the maximum and minimum points of a curve. The location of the maximum and minimum points of a curve whose equation in rectangular coordinates is known may be determined by use of the derivative. The method employed is a general one, but the solution of the equations is sometimes impossible. In the case of equations with nu- merical coefficients, however, an approximate solution can always be obtained. It was shown in Art. 132 that -^ for any point of the curve is equal to the slope of the tangent to the curve at that point. Then if -^ is positive for a given point of the curve, the tan- da; gent line at that point makes with the ic-axis an angle less than 90°, and hence the curve rises toward the right from that point. If -^ is negative for a given point of the curve, the dx tangent at that point makes with the a>-axis an angle between 90° and 180°, and hence the curve falls toward the right from that point. Of course this rising or falling may continue for a very short distance only. Figure 129 illustrates points of the curve for which -^ is respectively positive, zero, and negative. dx T^^^X" -^ dy. ^ (Jy dy ite>o ii=° dl3, -^ is positive, and hence the curve again rises dx toward the right for all values of a; > 3. As X passes through — 2 from left to right, -^ changes from dx positive to negative, arid hence the point of the curve for which a; ^ — 2 is a maximum point ; i.e. (—2, 5^) is a maximum point. As X passes through 3 from left to right, — changes from dx negative to positive, and hence (3, — 15i) is a minimum point. Figure 131 shows the curve plotted from these considerations and a few additional points through w^hich it passes. The meaning of the dotted curve is explained in the next article. 182 ANALYTIC GEOMETRY 158. The first derivative curve. The faqts of the preceding article are clearly brought out graphically by plotting the curve of eq. (2), using x as abscissa and -^ as ordinate. dx dy ¥m. 131. Eor convenience let -^ be rep- dx ^ resented by z. Then eq. (2) be- comes „ _ Z = Qr — X—K). This curve, being a parabola, is easily plotted. It crosses the a;-axis at — 2 and 3, has its vertex at (|-, — -2^), and its axis parallel to the 2/-axis (Art. 81). The locus is the dotted curve in Fig. 131. The original curve will be re- ferred to as the primitive curve, and the curve just described as the first derivative curve. From the relations established in the preceding article, it follows that for those values of x for which the first derivative curve dy is above the a^axis, that is, 2;, or --^, is positive, the primitive dx curve rises toward the right; for those values of x for which the derivative curve is below the a^-axis, the primitive curve falls toward the right; for a value of x at which the first de- rivative curve crosses the a^-axis from above, in going from left to right, the slope of the primitive curve changes from positive to negative, and hence the primitive curve has a maximum point ; and for a value of x at which the first derivative curve crosses the avaxis from below in going from left to right, the primitive curve has a minimum point. DERIVATIVE CURVES 183 Moreover, the value of the ordinate of the first derivative curve gives one a good idea of the rapidity with which the primitive curve is rising or falling. Thus, if for a certain value of Xj the ordinate of the first derivative curve is positive and numerically large, the primitive curve is rising rapidly toward the right for that value of x ; while if the ordinate of the first derivative curve is negative and numerically small, for a certain value of x, the primitive curve is falling slowly toward the right for that value of x. This is at once evident on remembering that the ordinate of the first derivative curve is equal to the slope of the primitive curve for the same value of x. 159. Concavity. Suppose that for x = Xq the first derivative curve has a positive slope. Let Zq, %, and z^ be the ordinates of points on the derivative curve for the values Xq, Xq — Ax, and Xq + Aa; respectively, and let Ax be chosen small enough so that Zi<. Zq <. Z2* Then, since the values of z are equal to the slopes of the primitive curve for the same values of x, the tangent to the primitive curve must have turned counter-clockwise as x in- creased through Xq from Xq — Aa; to Xq -\- Ax. This *is true whether Zq be positive, negative, or zero. (See Fig. 132.) Exercise 1. In the curve y = a^ ■}- 2 x -\- 4: draw the deriva- tive curve, measure the ordinates at a; = 1^, 2, 2^, and draw the tangents to the primitive curve at points corresponding to the selected values of x. How would the tangent to the prim- itive curve turn as x increases through 2 ? Do the same for aj = -2i., -2, -11. * This is possible since the slope of a curve at any point is the limiting value of the slope of a secant line through that point and a neighboring point of the curve. The secant line, cutting either to the right or left of the given point, can then be brought near enough to the tangent to have a positive slope, since the slope of the tangent is positive. The ordinates of the curve are therefore greater just to the right and less just to the left than the ordinate at the point of tangency. 184 ANALYTIC GEOMETRY a Derivative Curve. b Primitive Curve. Fig. 132. Exercise 2. Prove that if the derivative curve has a nega- tive slope for X = Xq, the tangent to the primitive curve turns clockwise as x increases through Xq. Exercise 3. Illustrate the law stated in exercise 2 by using the curve y — — x^-{-2x — 3. Definitions. If the tangent to a curve turns counter- clockwise as the point of tangency moves to the right through a given point, the curve is said to be concave up at that point ; while if the tangent turns clockwise as the point of tangency moves to the right through a given point, the curve is said to be concave down at that point. A point on the curve where the curve changes from concave up to concave down, or vice versa, is called a point of inflexion. As the point of tangency passes through a point of inflexion, the tangent line changes the direction of rotation. The curve crosses the tangent at a point of inflexion. Slope increasing toward the right. Curve concave up. Slope decreasing toward the right. Curve concave down. Fig. 133. Point of inflexion. DERIVATIVE CURVES 185 The results of this article may be stated as follows: For all values of x for which the first derivative curve is rising toward the right, the primitive curve is concave upward ; for all values of X for which the first derivative curve is falling toward the right, the primitive curve is concave downward ; for a value of X for which the first derivative curve has a maximum or minimum point, the primitive curve has a point of inflexion. 160. The second derivative. The derivative of a function of a variable is itself a function of that variable. This de- rivative may then also be differentiated. Thus, if 2/ = 2ar3 + sin 2a;, ^=6a;2 + 2 cos 2a;, dx and — (^\ = 12 a; - 4 sin 2 a;. ) dx\dx The derivative, ~ -, is called the first derivative of y with re- dx spect to X, and — ( — ) is called the second derivative of y with CLX \CIXJ respect to x. The symbol ^^ is used to denote the second derivative of y with respect to x, thus -^ = ^( ^). ^ ^ ' dx" dx\dxj Similarly, —^ means — — f -^ doif dx\jXx\dx 161. The second derivative curve. The second derivative is related to the first derivative in precisely the same way as the first derivative is related to the primitive function. But it also has an interesting and important relation to the primitive function, now to be explained. 186 ANALYTIC GEOMETRY Suppose the second derivative to be represented by a curve, using X for abscissa and -^ as ordinate. This curve is called dmr the second derivative curve. Then, for all values of x for which the second derivative curve is above the a7-axis, the primitive curve is concave up ; for the ordinate of the second derivative curve is equal to the slope of the first derivative curve for the same value of x, and where the slope of the first derivative curve is positive, the primitive curve is concave up. (Art. 159.) In like manner it is proved that for those values of x for which the second derivative curve is below the i»-axis, the primitive curve is concave down. For a value of x at which the second derivative curve crosses the ic-axis, the first derivative curve has either a maximum or minimum point, and hence the primitive curve has a point of inflexion. (Art. 159.) 162. Summary. The results of the foregoing discussion of this chapter may be summarized as follows : For all values of x for which the first derivative curve is above the a;-axis, the primitive curve rises toward the right ; for all values of x for which the first derivative curve is below the a>axis, the primitive curve falls toward the right; for a value of X at which the first derivative curve crosses the a^axis from above in going from left to right, the primitive curve has a maximum point ; for a value of x at which the first deriva- tive curve crosses the a^axis from below in going from left to right, the primitive curve has a minimum point. For all values of x for which the second derivative curve is above the avaxis, the primitive curve is concave up; for all values of x for which the second derivative curve is below the flj-axis, the primitive curve is concave down ; for a value of x at which the second derivative curve crosses the aj-axis, the primitive curve has a point of inflexion. DERIVATIVE CURVES 187 163. Illustrations. Example 1. Given then and -^ = cos X, dx dx" — sm X. The curves are shown in Fig. 134, and the relations established above are seen to hold. Y y \^ V y IfN >\ 4 y X s ^^^ ;x' y TT Fig. 134. The student should make a careful study of the figure. Example 2. As another illustration, study the curves of Fig. 131. The straight line in the figure represents the equa- tion ^'y ^2x 1 dx^ Example 3. A circular cistern is to be built to have a given capacity ; to find its dimensions in order that the amount of lining required will be a minimum. Let H = depth, D = diameter, and S = area of inner surface. 4 Then S -\-TrDH. But vol. = i)2 + = C, where C is constant. 40 D 188 ANALYTIC GEOMETRY Here S is expressed as a function of the variable D. From the equation it is at once evident that if D is very small the surface is very large, and is again very large when D is very large ; while for intermedi- ate values of D the surface has smaller values. The curve which represents the ■ equation between S and D therefore falls and then rises as D increases from 0, as in Fig. 135. There will therefore be a mini- mum point, which may be found by equating to the value of ^. ■ t dD Equating this expression to 0, and solving for D, i.=^ "8(7 The relation between D and H is most easily obtained by replacing (7 by ^^— — in the expression for — , and equating ^ CHJ the result to 0. Then 2 4Z)^ "^' or B = 2H, EXERCISE XXXV Sketch the following curves, first sketching the first and second deriva- tive curves. Locate maximum and minimum points and points of inflexion. 1. y = a;2-4rc + 5. / ^\ 5 y = cos [x I • 2. y=_x2-rr + 3. "^ \ 6/ 3. 3 2/ = x8- 12«+ 6. 6. 6?/ = 2x3-3a;2_12x-6. 4. y = sin2x. 7. 10 y = 2 ic^ + 9a;2 - 24 a; + 20. DERIVATIVE CURVES 189 8. 20ij = x'^ -^9x'^ + l5x- 20. 13. ?/ = sin x + a. 9. y = x^-Sa'^x + b^. 14. y - l=(x- 2)3. 10.y = xK ^^^ ,3 11. 2/ = a-(a — a:). x - 4 12. y = ax^ + hx + c. IQ. Zy = x^ -^x'^ + Qx - 1. 17. 12 y = 3 x'' - 8 a:^ - 30 ic2 + 72 a; + 24. 18. 8 ?/ = ic^ _ 6 ic2 + 8 a; + 16. 19. In y = ax^ -{■hx + c^ where a t^ 0, show that there is a maximum and a minimum point if h and a are opposite in sign, hut that there is neither maximum nor minimum if a and h are of the same sign, or if 6 = 0. Compare the curves obtained by using the following values of a, h, and c. (1) a = 1, 6 = - 3, c = 2 ; (2) a = 1, 6 = - .03, c = 2 ; (3) a = 1, J) = — .0003, c = 2. If a > 0, and a and c are held fast while h is made to approach the limit from the negative side, what becomes of the maximum and minimum points ? If 6 then becomes positive, how is the tangent at the point of inflexion affected ? 20. It\ y — ay? -\- hx'^ -{■ ox -\- d show that there is a maximum and a minimum point if 62_3Qrc>0, but not otherwise. How does the case where h"^ — 3 ac = differ from that where &2 _ 3 ^c < ? 31. The equation of the path of a projectile, fired at an angle a to the horizontal with an initial velocity F, is y = x tan a ^ • ^ 2 r2 cos2 a Find the maximum height to which the projectile rises. Ans. ^^^ " . 22. Letting B = the range on the horizontal of the projectile described in ex. 21, show that B = ZI^RA^. g Letting a vary, plot the curve which represents JR as a function of a. For what value of a is i? a maximum ? Ans. ^ . 4 23. Prove that the greatest rectangle of a given perimeter is a square. 24. A cylindrical tin can, closed at both ends, is to be made to have a certain capacity. Show that the amount of tin used will be a minimum when the height equals the diameter. 25. Show that the rectangle of greatest area that can be inscribed in a circle is a square. 190 ANALYTIC GEOMETRY 26. Given that the strength of a rectangular beanj of given length varies as the product of the breadth and the square of the depth, find the ratio of depth to breadth of the strongest beam that can be cut from a cylindrical log. Ans. h = V2 • b. 27. Given that the deflection, under a given load, of a rectangular beam of given length, varies inversely as the product of the breadth and the cube of the depth, find the ratio of depth to breadth of the beam of least deflection that can be cut from a cylindrical log. Ans. h = VS • &. Suggestion. Make the reciprocal of the deflection a maximum. 28. A rectangular piece of tin of vyidth b is to be bent up at the sides to form an open trough of rectangular cross section. Find the width of the strip bent up at each side vv^hen the carrying capacity is a maximum. Ans. ^. 4 29. Find the dimensions of the greatest right circular cylinder, the sum of the length and girth of which is 6 ft. Ans. H=2tt., Diam. = - ft. IT 30. Find the dimensions of the greatest rectangular box of square base, the sum of the length and girth of which is 6 ft. Ans. Length = 2 ft, 31. Find the ratio of altitude to radius of base of the conical vessel, of open base, which requires the least amount of material for a given capacity. Ans. Alt. = V2 rad. 32. A point moves along a straight line. At the time t its distance from a fixed point of the line is s : at the time t + At, its distance is As s + As. Then ^ is the average velocity of the point for the time At. As The limiting value of ^ , as At approaches as a limit, is defined to be the velocity, v, at the time t. Hence v = —. dt Given s = 16 «2, find the velocity at any time t. 33. The average acceleration, during an interval of time, of a point moving in a straight line, is the increase in velocity during that time, divided by the length of the interval of time. Make a definition for the acceleration at any instant, and show that the acceleration is dv ^^^ dt dt^ Find the acceleration if s = 16 t"^. DERIVATIVE CURVES 191 34. Plot the curves representing the space, velocity, and acceleration, in terms of the time, if s = 16 t^. 35. Given s = at^ -\- ht + e, where a, 6, and c are constant, show that the velocity in terms of the time is represented by a straight line, and that the acceleration is constant. 36. The formula for the space traversed by a body projected vertically upward, with velocity Vq, is s z=VQt — 16 «2 (s in ft., t in sees.) Find, by differentiation, the velocity and acceleration of a bullet fired upward with initial velocity of 1000 //s. Plot the curves representing space, velocity, and acceleration in terms of the time. How high does the bullet go ? 37. A point moves back and forth along a diameter of a circle of radius a, with simple harmonic motion (Art. 116), making n complete oscilla- tions per unit of time. If s is the abscissa of the point referred to the center, and the point is at the end of the diameter when « = 0, show that s = a cos(2 irnt). Find also the velocity and acceleration at any time, and plot the curves for space, velocity, and acceleration. 38. Since — (x^ + C) is the same as — (x2\ how many primitive dx dx curves are there whose first derivative curve is -^ = 20.? dx Sketch some of the derivative curves. How are they situated with refer- ence 'o each other ? What is the equation of the primitive which passes through (2, 6) ? 39. Find the primitives of which — = cos x is the first derivative dx curve. d^v 40. Find the primitive of which Jl = ^ is the second derivative curve, and which passes through (4, 1) with a slope equal to 3. d^v 41. Show that for the second derivative curve ^ =^ ^' ^ primitive may be obtained which passes through any given point in any given direction. CHAPTER XIII THE CONIC SECTIONS 164. Definition of the conic. A conic section, or simply conic, is the curve of intersection of the surface of a right cir- cular cone and a plane. It can be shown, however, that the following definition is equivalent to the one just given. Definition. A conic is the locus of a point which moves m a plane so that the ratio of its distance from a fixed point in the plane to its distance from a fixed straight line in the plane is constant. This definition will be adopted here. The fixed point is called the focus, the fixed straight line the directrix, and the constant ratio the eccentricity, of the conic. 165. Construction of conies. Let F be the focus, DD' the directrix, and e the eccentricity. Let P be any point on the conic, and M the foot of the perpendicular drawn from P to the directrix. Then, by defini- tion of the conic, FP ^ MP (The lines FP and MP are to be counted as positive, whatever their direction.) This suggests the following method of locating points of the conic : Through F draw a line FB perpendicular to DD\ intersecting DU in B. Through B draw a line BL, making an angle 6 with BF such that 192 Fia. 136. THE CONIC SECTIONS 193 tan = e. Take any point H on BF, and let the perpendicular to BF through H meet BL in K. Then, ^^ = tan ^ = e. With F as a center and a radius equal to HK, describe an arc of a circle cutting HK in P and P', The points P and P' so ob- tained are points on the conic. In this manner, as many points as desired may be obtained, and the conic sketched by drawing a smooth curve through them. Evidently, they lie in pairs which are symmetrical with FB as an axis of symmetry. This line FB is called the axis of the conic. 166. Vertices of a conic. The points of the conic which lie on the line through the focus perpendicular to the directrix are called the vertices of the conic. To obtain these points, draw lines through F inclined 45° and 135° to the line BF. From the points of intersection of these lines with BL drop perpendiculars to BF. The feet of these perpendiculars are the vertices, as the student can easily show. If e = 1, there is only one vertex, but if e ^it 1, there are two vertices. The figures on the following pages show conies constructed f or e = I, e = 1 , and e = f . EXERCISE XXXVI 1. Plot in different figures the conies for e = |, e = 1, e = ^. 2. Plot in the same figure, using the same directrix and focus for all the curves, the conies for e = .9, e = 1, e = 1.1. 3. Assume a unit of distance, and taking the distance from focus to directrix to be 1, 2, .4, 20, respectively, construct the conies for e = 1. 4. Same as example 3 for e = f . 5. Same as example 3 for e = |. 6. Prove that the conic is tangent to the line BL at the intersection of BL and a line through F parallel to the directrix. / \ ^^^ ^^^ / \ z ^ / V / \ 1^ A- / .V / or / ! \ '-^ Jl / _j\ ^ \ / V ^ y ■\ / !j / " / So \ ' V x] ^/ 1 ^ \ x/ ^ / \y i^ ^ \ V \ ."- / \ ' J \^ ^ 194 FiQ. 137 c. 196 196 ANALYTIC GEOMETRY 167. Classification of conies. From the constructions already made, it is evident that the general shape of the conic depends upon the value of e, and that the conies may be divided into three classes, according as e < 1, e = 1, or e > 1. A conic whose eccentricity is less than 1 is an ellipse ; one of eccentricity equal to 1, a parabola; ^ ■ and one of eccentricity greater than 1, u\ pP an hyperbola. (See footnote, Art. 171.) 168. The equation of the conic in rec- tangular coordinates. Let the directrix „ be taken as ?/-axis and the line through "^fTTo) t^^ focus perpendicular to the directrix as the fl?-axis. Let the distance from the directrix to focus be p. Then the coor- dinates of F are {p, 0). Let P{xy y) be any point on the conic, and MP the distance from P to the directrix. Then, from the definition of the conic, ^^^ = e, or FP=eMP. MP ' But FP=:V{x-py + y% and MP=x. . •. (x—py-\-y^ = e^y?^ or (1 - e2)a?2 _ 2 j^a? + 1,2 4.^2 ^ 0. This is, therefore, the equation of any conic when the 2/-axis is the directrix and the a^axis is the line through the focus perpendicular to the directrix. 169. The parabola, e = 1. In the equation just found let e = 1. The conic is then a parabola. The equation reduces to y^ = 2px—p^. This equation of the parabola was obtained in Art. 75, and from the same definition as here used. The equation was dis- cussed in that place. The student should review Arts. 75-78 at this time. THE CONIC SECTIONS 197 170. The centric conies. e=^l. In the equation of Art. 168, divide by the coefficient of a^ and then complete the square in the terms containing x, x^ - ^P X + ^' 4- y' - -P' _ P^ _ -P'^' 1 _ e2 ' (^1 _ g2)2 ' 1 _ e2 (^l_ g2^2 1 _ g2 (J^ _ g2^2' or f p \^ I y^ P'^ Substitute x' = x — --^—, y' = y, 1 — e- which transforms to parallel axes through ( ^ . ) (Art. 52.) The equation then becomes x" + _ p^€^ 1-e^ (1-ey Dividing by the right-hand member brings the equation into the form x'^ ^'2 (1 _ e2)2 1 _ g2 Since this equation contains only even powers of x and y, the curve is symmetric with respect to both coordinate axes, and hence with respect to the origin. The origin may there- fore be called the center of the conic, and the conic called a centric conic. Also, since the conic is symmetric with respect to the center, rotation of the conic in its own plane through 180° about its center will bring the conic back into its original position, hav- ing merely interchanged the points. Let the conic, together with its focus and directrix, be thus rotated. The focus and directrix are brought into new positions which are symmetric with respect to the center. They have remained focus and directrix of the conic, however, and since the new position is 198 ANALYTIC GEOMETRY the same as the old position they must be focus and directrix of the conic in its original position. Therefore every centric conic has two foci and two direc- trices. They are respectively symmetric with respect to the center. 171. The ellipse. eb. * In Art. 83 the ellipse was defined in an altogether different way. The equation of the ellipse derived from that definition and that just derived are, however, the same, which proves that the two definitions are equiva- lent. The property of the ellipse used in Art. 83 as a definition will be shown in a succeeding article to follow from the definition used in thia chapter. A hke remark applies to the hyperbola. THE CONIC SECTIONS Y 199 Fig. 139. The abscissa of the new origin referred to the old in the transformations of Art. 170 is — ^ 1 — e^ I.e. Now a = pe l-e^ A BO = e Also FO = BO-BF=:^ or FO = ae. The relation a\l-e^) = b^ may be written a^e^ = a^-^b^f from which ae= y/aP' - &2. Therefore if the end of the minor axis be taken as a center and an arc described with the semi-major axis as a radius, this arc will cut the major axis in the focus. 173. Summary. In an ellipse whose major axis is 2 a, minor axis 2 6, and eccentricity e, the following relations hold : a-^C* a^ 62. ae = distance from center to focus, — = distance from center to directrix. e 200 ANALYTIC GEOMETRY 174. The hyperbola, e > 1. In eq. (A), Art. 170, the divisor of ^'^ is negative if e > 1. Let then a" = . ft^ e^~l Then both a and h are real. Substituting these values in eq. (A) and dropping primes, it becomes ^ - ^ = 1. This is known as the standard form of the equation of the hyperbola. (See also Art. 87, and the footnote to Art. 171.) 175. Axes of the hyperbola. Letting y = 0, the intercepts on the a^axis are seen to be a and — a. If a? = 0, 2/ is imaginary. Fig. 140. Hence the curve does not cross the y-axis. The length 2 a is called the transverse axis, and 2 b the con- jugate axis. THE CONIC SECTIONS 201 The relation connecting a, b and e is 6^ = a^ (e^ — 1), or a^e^ = a' + b\ This shows that 6 = a according as e = V2. As in the ellipse, the abscissa of the center referred to the d origin, ( since e > 1. old origin, on the directrix, is ^ ^ , , which is here negative, 1 — e'^ 1 — e^ Now a = ., ^ . e^ — 1 .-. OB = -. e Also OF=OB+p e^ — 1 — P^^ ~ e^ - 1 = ae. Since ae = Va^ + 6^, the focus may be obtained by using the center of the conic as a center and the hypotenuse of the right triangle whose sides are a and 6 as a radius and describing an arc to cut the major axis produced. 176. Summary. In an hyperbola of transverse axis 2 a, con- jugate axis 26 and eccentricity e, the following relations hold: ae = distance from center to focus, - = distance from center to directrix. e Compare Art. 173. 202 ANALYTIC GEOMETRY EXERCISE XXXVII 1. Derive the equation of the parabola whose directrix is the line aj = 6, and whose focus is (2, 3). 2. Derive the equation of an ellipse whose directrix is the line ?/ = 4, focus at (0, 2), and center at (0, — 1). 3. Derive the equation of the hyperbola of eccentricity 2, with focus at (0, 4) and the line ic = 2 as directrix. 4. What is the eccentricity of the equilateral hyperbola ? 5. Keeping the major axis unchanged,plot ellipses with eccentricity .1, .5, .9. What limiting position do the foci approach as the eccentricity ap- proaches the limit ? What is the limiting form of the ellipse ? 177. The equation of the conic in polar coordinates. (a) Origin at the focus. Taking the origin at the focus and the initial line perpendicular to the di- rectrix, the polar equation of the conic is easily written. Let P(r, 6) be any point on the conic and MP the length of the perpendicu- lar from P to the directrix. Then, by the definition of the conic, FP=eMP, or r = e(p -f r cos $), from which Fig. 141. ep 1 — c cos 8 If the focus lies to the left of the directrix, then PM=p — r cos $. .'. r = e(p — r cos 6)j from which ^- ep 1 4- c cos 6 Fig. 142. THE CONIC SECTIONS 203 (b) Origin at the center. For the centric conies the equa- tion in rectangular coordinates is the upper sign being for the ellipse, the lower for the hyperbola. Change to polar coordinates by means of x = r cos 6, y = r sin 6. Substituting and clearing of fractions, T^b^ cos^^ ± ?%2 sin2^ = a^b% from which a'b' r2 = 62 cos^^ ± a' sin^^ This equation may be expressed in a somewhat simpler form in terms of the eccentricity and 6. For convenience con- sider separately the equation of the ellipse. It is or since 62cos2^ + a2sin2^ b'^co^'^O + a 2(1 - C0S2^) 62 a' C0S2^, ^._ h'^ l-e^cos^e' e^-(«^-^r Similarly for the hyperbola the equation is 204 ANALYTIC GEOMETRY EXERCISE XXXVIII Determine the nature of the following conies and sketch them: 1. r = 3. r = 5. r^ = 2+3 sin2d 7. r^ = "" 8. r = a sec2.^ 16-20sin2(9 2 9. Show that if the vertex of a parabola is taken as origin and the axis of the parabola as the initial line, the equation in polar coordinates is 2 p cos d r = ^. ^ sni2^ 4 1- - 1 cos e 5 2- - 2 cos 3 2 - C0S2 d 64 2 r = 3 2 + 4 COS ^ 4. r'2 -6 1 - 4 cos20 a ^2 20 CHAPTER XIV PROPERTIES OF CONICS 178. In this chapter a few of the more important properties of the conies are derived. I. PROPERTIES OF THE PARABOLA 179. Subtangent of the parabola. In Art. 130 the equation of the tangent to the parabola 7f==2px at (iCo, y^ was found to be Letting ?/ = in this equation, there results x = — Xq, i.e. 0T= -a-oCFig. 143). .-. TO = Xq. .'. TM==2xo. The line TM is called the subtangent. 180. The subnormal of the parabola. The slope of the nor- mal to the parabola y^ = '2px at the point {xq, y^) is the negative reciprocal of the slope of the tangent at that point; i.e. the slope of the normal is ?h P Fig. 143. The equation of the normal is therefore ^0, 205 206 ANALYTIC GEOMETRY To find where the normal cuts the a;-axis, let y = 0. The result is I.e. in Fig. 143. OiV= o^o +jp. MN=p. The line MN is called the subnormal. Hence, in the parabola the suhnormal is constant and equal to p. 181. Property of reflection of the parabola. In Fig. 143, Art. 179, from the definition of the parabola, z Also TF= TO-\-OF = x, +|. (Art. 179.) FP=TF. Z.FPT = ^FTP=Z.TPH. Let PL be drawn parallel to the axis of the parabola. Then AFPT=ZLPQ. Hence, if the parabola were a reflector, any ray of light from the focus striking the parabola and reflected so as to make the angle of reflection equal to the angle of incidence would be reflected along a parallel to the axis of the parabola. A concave reflecting surface in the form of a surface generated by re- volving a parabola about its axis would therefore reflect all rays from a source at the focus in lines parallel to the axis of the reflector. Definition. The chord of a conic which passes through the focus, per- pendicular to the axis of the conic, is Fig. 144. called the latus rectum. PROPERTIES OF CONICS 207 EXERCISE XXXIX 1. By means of the result found in Art. 179, show how to draw a tan- gent at any point of the parabola. 2. Prove that the tangents at the ends of the latus rectum meet at the intersection of the directrix and the axis of the parabola, and are at right angles to each other. 3. Prove that the distance from the focus of a parabola to a tangent is half the length of the normal from the point of tangency to the axis of the parabola. 4. Prove that any point P of the parabola and the intersections of the axis of the parabola with tangent and normal at P are all equidistant from the focus. 5. Prove that the tangent at any point of a parabola meets the directrix and latus rectum produced at points equally distant from the focus. 6. Show that the normal at one extremity of the latus rectum of a parabola and the tangent at the other extremity are parallel. 7. Show that the directrix of a parabola is tangent to the circle described on any chord through the focus as a diameter. 8. Show that the tangent at the vertex of a parabola is tangent to the circle described on any focal radius as a diameter. 9. Prove that the angle between two tangents to a parabola is equal to one half the angle between the focal chords drawn to the points of contact. 10. Prove that the tangents at the ends of any focal chord of a parabola meet on the directrix. 11. Prove that the length of the latus reetum of the parabola y^ — Ipx is 2 p. 12. Prove that if from a point (a;o, 2/o) two tangents are drawn to the parabola, the equation of the line through the points of tangency is 2/oy =p{x +Xo). 13. By means of the preceding example prove that if tangents are drawn to the parabola from any point on the directrix, the line through the points of tangency passes through the focus. 14. Prove that in the parabola if- = 2 pa;, the ordinate of the middle point of a chord of slope m is — , and hence that the locus of the middle m P points of a system of parallel chords of slope m is the straight line y = — m Draw the figure. 208 ANALYTIC GEOMETRY Definition. The straight line which bisects a system of parallel chords of a parabola is called a diameter of the parabola. 15. Find the equation of the diameter which bisects all chords of slope m in the parabola x^ = 2py. Ans. x = mp. 16. Transform the equation of the parabola y^ = 2 px to the tangents at the extremities of the latus rectum as axes. Suggestion. First, moving to parallel axes through ( — ^, 0), the equation becomes y^ = 2px- p^. Next, rotating the axes through — 45°^ the equation becomes a;2 -2xy-\-y^- 2V2p(x + y) + 2p^ = 0, which becomes a perfect square on the left by the addition of 4 xy. Then extract square root, transpose, extract square root again, and obtain Vx±Vy = ± yp\/2, or Vx ±Vy = ± Va, where a=pV2. 17. Plot the curve x^±y^ = ± a\ "What portions of the curve correspond to the different combination of signs ? II. PROPERTIES OF THE ELLIPSE AND OF THE HYPERBOLA 182. Focal radii of the ellipse. Let P(xo, y^ be any point of the ellipse of semi-axes a and 6, and let r and r' be the radii from the foci F and F^ to P. Through P draw a line parallel to the major axis of the ellipse, meeting the directrices in M and M\ Then from the definition of the ellipse, using the left-hand focus and directrix, M'P or r^ =^eM'P=e[--\-XQ\ = a + exQ. PROPERTIES OF CONICS 209 Similarly, using the right-hand focus and directrix, r = ePM=ef- — XQ\=a — exo. Adding, r-{-r' = 2a. Hence the sum of the focal radii of a point on the ellipse is constant, and equal to the major axis of the ellipse. Fig. 145. 183. Focal radii of the hyperbola. In a manner similar to the above the student can show that in the hyperbola the focal radii are r = exo -\- a and r' = exo — a, and hence r — r' = 2 a. 184. Property of reflection of the ellipse. The focal radii to any point of an ellipse make equal angles with the normal to the ellipse at that point. Proof. In Art. 130, the equation of the tangent to the ellipse at (xq, 2/o) was found to be yvo 1. The slope of the normal at (xq, y^) is therefore J^, and the b-XQ 210 ANALYTIC GEOMETRY equation of the normal is 2/-2/o = ||(«'-^o). In this equation let y = and solve for x, x = Xo f = -— Xq = e'xQ. I.e. in Fig. 145. 0N= e%. F'N= ae + e^XQ = e(a-\- ex^), an d NF = ae — s^Xq = e (a — gxq) . gJ^=^ + ^^o = ^(Art.l82). NF a — exQT Therefore by plane geometry, Z F'PN = Z NPF, which proves the theorem. Hence if the ellipse served as a reflector, a ray of light, or sound, emitted at one focus would be reflected to the other. It is on this principle that whispering galleries are some- times constructed. 185. Property of reflection of the hyperbola. In the hyper- bola the focal radii to any point of the curve make equal angles with the tangent at that point. The proof is left to the student. 186. If a line is drawn to cut the hyperbola in two points, the two segments of the line included between the hyperbola and its asymptotes are equal. Proof. The equations of the hyperbola, its asymptotes, and any line are, respectively, b^x'-ay^a^ (1) b''x'-aY=0, (2) and y=mx-\-c. (3) PROPERTIES OF CONICS 211 Let the points of intersection of line and hyperbola be A(^ij 2/1) ^^^ A(^2) 2/2); and of line and asymptotes be Qi(xi', y^) and Q.lx.J, yJ). Substituting the value of y from eq. (3) in eqs. (1) and (2) respectively, and collecting terms, there results (62 _ a'nv^x' - 2 a^mcx - a'ic' + W) = (4) and (6 — a?m^)x^ — 2 ahncx — oj^c? = 0. (5) The roots of (4) are x^ and x^, and of (5) are x-^ and x^. Now in any quadratic equation the sum of the roots is equal to minus the coefficient of the first power of the variable divided by the coefficient of the second power ; and since the first two terms in eqs. (4) and (5) are the same, therefore Ju\ ~p" Ct/o — *vi -y" iCo • /y» I rp rp ' | rp ' But zLX_2 and ^ "^ ^ are respectively the abscissas of the middle points of P1P2 and Q1Q2. Fig. 146. the middle points of P1P2 and Q1Q2 coincide. Q.E.P. 212 ANALYTIC GEOMETRY EXERCISE XL 1. Prove that the length of the latus rectum of an ellipse or an hyper- bola is . a 2. Prove that the tangents at the extremities of the latus rectum of an ellipse or hyperbola intersect on the directrix. 3. Prove that the line drawn from the focu^ to the intersection of a tangent and the directrix of an ellipse or hyperbola is perpendicular to the line from the focus to the point of tangency. 4. A circle is drawn on the major axis of an ellipse as a diameter. A perpendicular to the major axis meets the ellipse and circle in P and Q respectively. Prove that the tangents drawn at F and Q intersect on the major axis. Hence show how to construct a tangent to an ellipse at a given point. 5. Show that' the distance from the focus to an asymptote of an hyperbola is equal to b. 6. Prove that the product of the perpendiculars from any point of an hyperbola upon the asymptotes is constant, and equal to — • 7. Prove that the product of the perpendiculars from the foci upon a tangent to the ellipse is equal to the square of the semi-minor axis. 8. State and prove a like property of the hyperbola. JW.2 oi2 9. Prove that if tangents are drawn to the ellipse — + f- = 1 from an exterior point (Xq, ?/o), the equation of the line through the points of tan- gency is ^ + ^ = 1 . 10. Prove the statement in example 9 to be true for the hyperbola, with proper changes of sign. 11. Prove that if tangents are drawn to an ellipse or hyperbola from any point on the directrix, the line joining the points of tangency passes through the focus. (Use examples 9 and 10.) 12. Through a fixed point within a given circle, a circle is drawn tan- gent to the given circle ; prove that the locus of its center is an ellipse. Draw the figure. PROPERTIES OF CONICS 213 13. A line y = mx ■}- c cuts the ellipse b^x^ + a^y^ = a^b^ ; prove that if (xi, yi) is the middle point of the chord, then a^mc ,. b'^c b'^-\-a^m-^ b^ + a^m^ 14. From the preceding example, by eliminating c, prove that the locus of the middle points of a system of parallel chords, with slope m, of the ellipse is the straight line &2 y = X. This line is called a diameter of the ellipse. Prove that any line through the center of an ellipse is a diameter. 15. Show that if two lines through the center of the ellipse 6^x2 + a2y2 = ^252 have slopes m and m' such that mm' = , then each line bisects all chords parallel to the other. Draw two such lines. Two such lines are called conjugate diameters. /y2 o<2 16. Prove that in the hyperbola ^ = 1 the equation of the locus of the middle points of a system of parallel chords of slope m is y= — zx. 17. Through the point (xq, yo) on the ellipse b^x^ + a^y^ = a262 a diameter is drawn ; prove that the coordinates of the extremities of its conjugate diameter are a: = ± ^^, y = T — . b a 18. If a' and 6' are the lengths of two conjugate semi-diameters of the ellipse, prove that a'^ -\- b''^ = a^ + b^. (Use example 17.) 19. Prove that the tangent at any point of the ellipse is parallel to the diameter which is conjugate to the diameter through the given point ; and hence that the tangents at the extremities of two conjugate diameters form a parallelogram. 20. Prove that the area of the parallelogram formed by the tangents at the extremities of two conjugate diameters of an ellipse is constant, and is equal to 4 ab. Suggestion. The area in question is 8 times the area of the triangle whose vertices are (0, 0), (xo, yo), and (^ - ^\ . (See example 17.) CHAPTER XV THE GENERAL EQUATION OF SECOND DEGREE IN TWO VARIABLES 187. In the preceding chapters certain equations of second degree in two variables have been studied. It will now be shown that every equation of second degree in two variables with real coefficients is the equation either of one of the conies, a circle, a pair of straight lines, one straight line, a point, or else the equation has no locus. Moreover, the conditions which the coefficients must satisfy in the different cases will be established. 188. The general equation of second degree in x and y is aa?2 + hocy -\^ cij^ + dijc-\- ey + f = 0. (1) Let the origin be moved by a translation of axes to the point (h, k) by means of the formulas x = x' -{-hj y =y' -{-k. Equation (1) then becomes ax'^ + bx'y' + cy" + d'x' + e'y' +/' = 0, (2) where d' = 2 aJi + bk + d, (3) e' = 6/i + 2 cA; -f e, (4) / = ah^ + bhk -f ck^ + dh + ek -\-f. (5) Equation (2) will be simplified if h and k can be so chosen that d' = and e' = 0. Putting d' = and e' = and solving for h and k, J, _2cd—7}e 7. _ 2 ae - bd /n\ 214 THE GENERAL EQUATION OF SECOND DEGREE 215 These values of li and k are definite finite values unless 6^ — 4 ac = 0, in which cases there are no values of h and h that make d' = and e' = 0. Hence there are two cases to consider, I, 6^ — 4 ac ^ 0, and II, h^-4.ac = 0. Case I. h^-^ac^(i, 189. Removal of the terms of first degree. Consider first the case where b^ — ^ac^O. Then if Ti and k have the values shown in eq. (6), d' and e' are both zero, and eq. (2) becomes a«'2 + 6xy + c?/'2 +/ = 0. (7) The value of /' can be obtained by substituting the values of h and k from (6) in (5), but more easily as follows : Multiply eq. (3) by h, eq. (4) by k, and add. The result is dli + e'A: = 2 ah'' + 2 hhk -\-2ck' + dh + ek. To both members of this equation add dh -\- ek -\- 2/. Then d'h + e'k + dh + eA: + 2f=:2{ah^ -f bhk + cF + d/i + ek-\-f)=2f or 2f' = dh-\-ek-\-2f, since d' = e' = 0. Substituting the values of h and k from eq. (6), f _ — (4 acf-]- bde — ae^ — cd^ —fb^ /on The quantity in the parenthesis is of importance in what fol- lows. For convenience let it be denoted by a single letter, H; H= 4 acf + bde - ae^ - cd^ - fh\ Also let I> = h^-^.ac. 190. Removal of the term in ocy. Equation (7) may be re- duced to one lacking the a^y-term by a proper rotation of the axes. Let a;' = oj" cos 6 — ly" sin ^, 2/' = a?" sin -f y" cos 0. Substituting these values in eq. (7), it becomes a!x'''' + b^x'Y + cy" + /' = 0, (9) 216 ANALYTIC GEOMETRY where a' = a cos^ 6 -\-b cos ^ sin ^ + c sin^ 6^ (10) 6' = - 2 a cos ^ sin ^ -f 6 (cos^ - sin^ ^) + 2 c cos ^ sin 0, (11) c' = a sin2 ^ _ 5 sin ^ cos ^ + c cos^ d. (12) Now let be so chosen that b' = 0, i.e. let b (cos^ $ — sin^ ^) = 2 (a — c) cos 6 sin ^, or & cos 2 ^ = (a — c) sin 2 ^, or tan 2 - — ^. (13) tt — c Since the tangent of an angle may have any real value, it is always possible to choose so that b' = 0. With this value of 6j eq. (9) becomes a'i€"' + c'y"'^+f' = 0. (14) 191. Locus of the equation. The nature of the locus of eq. (14) depends upon the signs of a', c', and/', and these signs depend upon the original coefficients of eq. (1). To determine the signs of a' and c' one may proceed as fol- lows: Using the relations 2 sin ^ cos ^ = sin 2 Oy 2 cos^ ^ = 1 4- cos 2 ^, 2 sin2 ^ = 1 - cos 2 ^, eqs. (10) and (12) may be written 2a' = a + c + ?> sin 2^ + (a - c) cos 2^. (15) 2c' = a + c - 5 sin 2^ - (a - c) cos 2^. (16) Adding, a' + c' = a -[- c. (17) Subtracting, a' — c' = 6 sin 2 ^ -f- (a — c) cos 2 0. From equation preceding (13), 6 cos 2 ^ - (a - c) sin 2 ^ = 0. THE GENERAL EQUATION OF SECOND DEGREE 217 Square and add & sin 2 (9 + (a — c) cos 2 ^ = a' — c', and 6 cos 2 ^ — (a — c) sin 2 ^ = 0, and there results 62 + (a _ cf = (a' - c^. Erom (17) (a + cf = (a' + cj. Subtracting, 4 a'd = 4.ac- b\ (18) Since 4.ac — b^ =^0, neither a' nor c' can be zero. Eq. (14) may therefore be written, since from eq. (8),/' = , no f^2 ^ = 1, HH^O, (19) // ' H a'D c'D or a'x"' + c'y"' = 0, itH=0, (20) Two cases must here be considered. (1) Z)<0, ^.e. 4ac-&2>0. Then neither a nor c can be zero, and a and c must be of like signs. It follows also from eq. (18) that a' and c' must be of like signs, and hence of the same sign as a and c, by (17). TT Therefore, if — < the locus of (19) is an ellipse if a' =^ c', and a circle if a' = c'. TT If — > 0, eq. (19) has no locus. Equation (20) is satisfied only by the point x" = 0, y" = 0. (2) 2»0, i.e. 4ac-62 0. (23) THE GENERAL EQUATION OF SECOND DEGREE 219 Then Vaic ± -yjcy =k(xcos 6 + y sin 0). Transform now to axes which make the angle with the axes X and 2/j for which the formulas of transformation are X = x' cos 6 — y' sin 9, y = x' sin 6-{-y' cos 0. Then x cos O + y sin 6 = x', and eq. (21) becomes kV + d'x' + e'y' +/= 0, (24) where d' = d cos + e sin ^, (25) e' = — dsmO + e cos ^. (26) If e' =^ eq. (24) may be written This is of the form yf^-^x"-^x'-l. (27) ^ e' e' e' ^ ^ y = aa^ -j- 5a; -|- c which in Art. 81 was seen to be the equation of a parabola. If e' = eq. (24) becomes kV + d'x' +/= 0. (28) This is a quadratic in x' alone. It is satisfied by ^,_ -d' ±^d"-Ak^f X — • 2 k' Hence eq. (28) is the equation of two parallel lines, one line, ot has no locus according as d'^-4:k'f=0. < 194. Evaluation of e' and of d'^ — 4 F/. The quantities e' and d'^ — 4:kyof the preceding article may be expressed in terms of the coefficients of eq. (1) as follows : From (26) and (22) .'=-c^(±^^)+ff» = l(«VST = ; 6^ = 4 ac, and H becomes H= bde — ae^ — c(P = ± 2 Vac • de — ae^ — cd^ = _(eVaT(^Vc)2, or, from (29), il= -k^e'^. (30) .'. e' vanishes or does not vanish according as H does or does not vanish. Again, from (25) and (22) d' =-(dVa ± eVc), and hence d'^ - 4 A^y = \{dVa± eVcf - 4 hj = i iad"" ± 2 de Vac + e^c - 4/(a + c)^]. But if e' = 0, then eVa = ± ^a/c, from eq. (29). /. d'2 _ 4 ;t2y._ 1 L^2 4_ 2 ^2^ + !^' _ 4/(a + c)2l ^ (g + cf ^ d'-iaf k^ a a-\-c (d'-4:af). Hence the sign of d'^ — 4 k^f is the same as the sign of d^ — 4: af. Therefore if /) = and a is positive, the locus of (1) is a parabola if H=^0, and is two parallel lines, one line, or there is no locus according as d^-4:af= 0,iiH=0. < If a is negative, eq. (1) may be divided by — 1 and then the above conditions hold if each coefficient in H and d^ — A af is changed in sign. This, however, only changes the sign of H and does not affect at all d^ — 4 af. The above conditions hold, therefore, whether a is positive or negative. THE GENERAL EQUATION OF SECOND DEGREE 221 If a = 0, then 6 = 0, and c=^0. Eq. (1) then reduces to c/ + da; + e2/+/=0, (31) or dx= — cif — ey —f. This is a parabola if d ^fc 0. When a = 6 = 0, H becomes — ccZ^, and since c=^0, H van- ishes or does not vanish according as d does or does not vanish. If H=0 eq. (31) becomes c/ + e2/4-/=0, which is satisfied by „ _ — e ± Ve^ — 4 c/ ' ^~ 2~c "' the locus of which is two parallel lines, one line, or there is no locus according as e2_4c/=0. < 195. Summary. The nature of the locus of the general equation of the second degree ax^ 4- bocy + cy^ -^dx-^ey +/= is shown in the following table, in which H= 4 acf+ bde - ae^ - cd^ -fh\ (aH < ellipse, reducing to a circle if 6 = ana a = c, aH > no locus, ^ = a point. jj^^\ il 9^0 hyperbola, I H =0 two intersecting straight lines, H=^ parabola, ' a :^ two parallel lines, one line, or no locus D = H=0 according as d^— 4 a/ = 0, a = two parallel lines, one line, or no locus according as e^ — 4:cf = 0. 222 ANALYTIC GEOMETRY EXERCISE XLI Apply the above test to determine the nature of the loci of the follow- ing equations. 1. x^-2xy-{-Sy^ + 2x-y + 3 = 0. 2. Sx^-4:xy + y^-x + 2y-l = 0. 3. Sx^ + 6xy-2y^-'Sx-{-y = 0. 4. 9x^-6xy + y^-Sx-\-y-2 = 0. 5. x2-4x?/ + 4 2/2 + 2a:-4?/ + l = 0. 6. x2-xy + y2 + 2a; + y + 2 = 0. 7. 3x^-Sxy+3y^ + 6x + Zy+'7 = 0. 8. ^x^-4:xy + y^ + ix-2y + 2 = 0. 9. Ax^ - 12 xy + 9 y^ + x - y + 1 =0. 10. 3 x^ - xy - y^ -\- X -2 y -{- 1 = 0. 11. Show that the locus of 2x^ — 2xy + y^-Sx + y+f=0 is an ellipse, a point, or there is no locus, according as / is less than, equal to, or greater than f . 12. Show that the locus of ax^ + bxy -\-cy^ = is two intersecting lines, one line, or a point, according as 6^ _ 4 ac is greater than, equal to, or less than zero. 13. Show that the locus of icy + dx + ej/ + / = is an hyperbola except when / = de. What is the locus then ? 14. In the equation {Ix -\-my + ny + px + qy-\-r = Q show that 2) = 0, and that H= — {mp — IqY, and hence that the locus of the equation is a parabola except when - = — . What is the locus then ? CHAPTER XVI EMPIRICAL EQUATIONS 196. Statement of the problem. It is sometimes desirable to find an equation of a curve drawn through points determined by pairs of corresponding values of two variable quantities. Frequently these values are found by experiment, and the general law which they satisfy may be known or suspected. The following illustrations will show how, in some of the simpler cases, the law may be tested and the constants of the equation determined. The more difficult problems of this nature can be treated by the use of Fourier's series, a method of wide application, but too diffi,cult to discuss here. 197. Points lying on a straight line. The simplest case that occurs is that where the points whose coordinates are the two measured quantities lie on, or approximately on, a straight line. In this case one has only to select the straight line which seems to best fit the points, and write its equation. The equation of this line is then the equation connecting the variables if the same scale has been used throughout. In plotting the points, however, any convenient scales may be used, and the equation of the line written with any other scale that is desired. The two coordinates in the equation of the line must then be expressed in terms of the two variables be- tween which an equation is sought. The substitution of these values in the equation of the line gives the desired equation. Example. The extension of a certain wire when loaded 223 224 ANALYTIC GEOMETRY was observed to be as shown in the following table, where E is the elongation in inches, and W is the load in pounds. TT 1 2 3 4 5 7 10 12 15 18 E .12 .23 .34 .46 .58 .80 1.16 1.39 1.74 2.09 ^ ^ ..-.".- .1 .11 .11 ..... ."Z'- - ... " <<\\ O . . . . . . . . .;»' f L>rT " ::::::::::: :: i?^: ::: ::: :: ::: ■^ '.".'.".'. '.2' " " " ' j^Pn 'it* ' ' " - - - - ■ -^f S :..::::.:..:::::::::::::: 5 10 15 w. 2 spaces to 1 lb. Fig. 147. 20 On plotting the points whose coordinates are the correspond- ing values of E and W, they are seen to lie approximately on a straight line. The line which seems to best fit the points passes through the origin and the point (38, 22). The equa- 22 tion of this line is therefore i/ = — a?. But in the scale used, ^38 x = 2Wyy = 10E, and hence the equation connecting E and TFislO^ = — TT, or E = .116 W. This equation therefore holds approximately' for the particular wire used and within the limits of the observed values. Exercise. From the following corresponding values of u and V determine the equation connecting them. u 1 1.5 2.3 3.1 3.8 4.2 5.0 5.8 6.5 7.2 8.0 V 5.5 6.4 8.2 9.7 11.0 11.9 13.5 15.0 16.5 18.0 19.5 EMPIRICAL EQUATIONS 225 198. The curve y = Cac\ A number of curves obtained from physical measurement follow the law y = Cic**. If the logarithm of both sides of the equation be taken, there results log y = log C-\-n log x. If now u = log Xj v = log y, b = log C, then v = b-\- nu. This is an equation of first degree in u and v. Hence if u and V be taken as coordinates and the points representing corre- sponding values plotted, these points will lie on a straight line. Conversely, if the points (log x, log y) do lie on a straight line, the equation of the line is of the form v=:nu-\-b, where u = log x, v = log y ; i.e. log 2/ = w log X + log (7, if 6 = log C, or log y = log (Ca?**). y= Cx^. 2.S00 2L000 1.600 1.400 L200 ^^ ^^^ 1.00 L20 Fia. 148. 226 ANALYTIC GEOMETRY The following illustration will show how the constants C and n may be determined when the points lie on a curve of this kind. Illustration. The following represent pressure p and vol-, ume V oi Si gas : V 3 4 5.2 6.0 7.3 8.5 10 p 107.3 71.5 49.5 40.5 30.8 24.9 19.8 Let X = log V, y = log p. Then the values of x and y are X ATT .602 .716 .778 .863 .929 1.000 y 2.031 1.854 1.695 1.607 1.489 1.396 1.297 The points determined by x and y are seen to lie on a straight line, approximately, Fig. 148. The slope of this line is found by measurement to be — |f, or — 1.40. Then, since the line passes through (1, 1.297), its equation is y - 1.297 = - 1.40 (x-1), or 2/ = -1.40 a; + 2.697. But 2.697 = log 497.7. Therefore, since y = logp, x = log v, logp + 1.40 log V = log 497.7, or i)?;i''« = 497.7, which is therefore approximately the formula connecting p and V. The correctness of this formula should be tested by substi- tuting some or all of the values otp and v in the given table. E.g. if V = 5.2 and p = 49.5, then log 'y=. 7160, which multiplied by 1.40 gives log 5.2i'«>= 1.0024 log 49.5 = 1.6946 logpvi«=: 2.6970 pvi*^ = 497.7, which checks the result already found. EMPIRICAL EQUATIONS 227 Where the points do not lie so accurately on the line as in this example, it would be better after obtaining the slope of the line to write pv" = C, and having n, substitute the given values of p and v to find C. Make this computation for each pair of values given, and take the average of the values found for C. Exercise. Find the equation connecting Q and Ji from the following observed values. h .583 .667 .750 .834 .876 .958 Q 7.00 7.60 7.94 8.42 8.68 9.04 199. The curve y = ab^, or y = ae^% where e = 2.71828 .... Certain physical quantities are connected by an equation of the form y = ah' where a and h are constant. If it is thought that two quantities for which several corresponding values are known obey this law, they may be tested, and, if the law is fulfilled, the values of the constants determined as follows: Plot the points whose abscissas are x and whose ordinates are log 2/. If they lie on a straight line, the supposed equation is correct, otherwise not. This follows from the fact that if y=ah% (1) then log 2/ = log a + a; log 6, ' (2) and the converse. Suppose the points («, logiy) lie on a straight line. The slope of this line is then the value of log h (see eq. (2)), from which h may be found. Also the intercept of the line on the axis of ordinates is log a. From this intercept a may then be found. However, it will be more accurate to obtain a from the average of the values of ^ after h is determined from the line. In some cases a is 1, and this will be indicated by the straight line passing through the origin. If it is desired to express y = ah'' in the form y = ae^, one 228 ANALYTIC GEOMETRY has only to let h = e^, for then 6*=(e*)'= = e*''. To determine h^ log6_ log& log e~. 4343* log h = k log e, or k Example. The values of x and y of the following table are thought to be connected by an equation of the form y = ah'. X 2 3.2 4.7 8.5 y 7.086 12.64 26.07 163.0 Form then the following table : X 2 3.2 4.7 8.5 logy .8504 1.1017 1.4161 2.2122 Plot the points (a?, log y). They are seen to lie on a straight line. 10.3 12.6 388.4 1178 10.3 12.6 2.5893 3.0711 A /. ^ y Q^ ^ -^ ^- / ^ ^"^ - y 1 J^ - J^ T / 5 , 10 15 Fig. 149. The slope of this line, computed by using the extreme values of X and log y in the table, is Hence 3.0711 - .8504 ^ 2.2207 12.6-2 10.6 log h = .2095, b = 1.62. .2095. EMPIRICAL EQUATIONS 229 To determine a, a = ^ • log a = logy — x log b = log 2/ - .2095 X. Using X and logy from the table of values, the following values of log a are obtained. .4314, .4313, .4314, .4314, .4314, .4314. The average of these is log a = .4314. .-. a = 2.70. .-.2/ =2.70(1.62'). 200. Some special substitutions. In some other cases, if the law connecting the variables is suspected, the correctness of the supposition may be easily tested by a substitution which will reduce the problem to that of the straight line. For example, if it is thought that the relation is y = a-\-—, of plot the points ( -^j 2/ ) • If these points lie on a straight line, the assumed equation is correct, and the quantities a and b can be found from the graph. In like manner the equation xy = ax-\- by, an hyperbola, may be written y = a-\-by., (1) X or or x = b + a'^, (2) y 1 = - + -, (3) X y and these may be reduced to the straight line form by using u for ^ in (1), u for - in (2), and u and v for = and - in (3). 230 ANALYTIC GEOMETRY ExERcise. Prove that the following points lie on a curve of the form xy = ax + by, and determine a and b. X 1.59 1.96 2.27 3.12 5.00 7.15 16.7 y .885 1.11 1.28 1.85 3.24 5.10 22.0 201. The curve y = a-^bac + cx^ + da^ + • + kx\ When no other equation can be found to fit the given points the equation y =: a -\- bx -^ cx^ -\- dx^ +•••-!- Jcx"" may be assumed, and by substituting the coordinates of the given points enough equations can be obtained for the deter- mination of the constants a, b, c, • • • 1c. The number of terms to assume will depend upon the num- ber and location of the given points. If the curve on which the points lie diverges only slightly and in one direction from a straight line, it will usually be sufficient to assume three terms on the right. This, of course, makes the curve a pa- rabola. But each case must be settled on its merits, and the construction of the curve from the equation which Is found will be the test of the accuracy with which it fits the given points. Example. To find the equation of a curve through the fol- lowing points: X 8 23 39 53 63 y 10 19 27 33 36 These points when plotted are seen to lie on a curve which resembles a portion of a parabola with axis parallel to the 2/-axis. It is worth while then to try y = a-{-bx-\- coi?. Take the two extreme points and the middle point for the de- termination of the coefficients. The equations obtained are 10 = a-|- 86+ 64c, 27 = a-}-39&-f-1521c, 36 = a + 63 6 + 3969 c. EMPIRICAL EQUATIONS 231 Solving these equations, a = 4.63, 6 = .697, c= -.00315. Hence the approximate equation is y = 4.63 + .697 x - .00315 a^. The substitution of the intermediate values of x not used in the computation of a, 6, and c give, for a; = 23, y = l^M, forx = 53, 2/ = 32.72, which are reasonably close to the values of 19 and 33. If greater accuracy is desired, four or five terms may be assumed on the right and then four or five of the given points used to determine the constants. Again, different seta of the given points might be used to determine the constants and average values of the constants so found used. EXERCISE XLH 1. In an experiment to determine the deflection of a beam of varying length the following measurements were made : Length (in.) 12 16 20 24 28 32 36 40 Deflection (in.) .017 .043 .085 .145 .220 .342 .512 .713 Prove that the deflection d and the length L are connected by an equation of the form d = CL», and find the values of m and C. 2. Find an equation connecting x and y to fit the following values : X .6 1.2 1.6 2.2 2.8 3.4 4.3 6.0 y .801 1.70 2.54 3.98 5.58 7.32 10.17 16.22 3. Prove that the following values of x and y satisfy an equation of the form 1 -\- bx and find the values of a and 6. x .5 1.2 2.0 3.4 4.1 5.3 y 1.08 2.41 3.71 5.56 6.33 7.46 232 ANALYTIC GEOMETRY 4. The following numbers are taken from a table : X 1.1 1.4 2.0 2.6 3.4 4.1 6.3 7.8 9.8 y .095 .336 .693 .956 1.224 1.386 1.841 2.054 2.282 Find the equation connecting x and y. Suggestion. Plot the points (log x, y'). 5. Prove that the following values of u and v satisfy an equation of ,the form v = a -\ , and find the values of a and 6: M V 6. Find an equation to fit the following values ot p and y; (Trypv" = C.) V 4.2 4.7 5 5.5 6.2 7 8 9 p 105 92 86 78 68 60 53 46 .5 1.1 1.7 2.3 5.1 6.4 13.6 4.00 2.37 1.84 1.33 1.28 ANALYTIC GEOMETRY OF SPACE CHAPTER XVII COORDINATES IN SPACE 202. Rectangular coordinates in space. As on a straight line one quantity was required to determine the position of a point, and in the plane two quantities, so in space three quanti- ties are necessary. One way of choosing these quantities is the following : Through any point 0, chosen as an origin, draw three mutually perpendicular lines OX, Y, OZ. These lines determine three mutually perpendicular planes XY, XZ, YZ. From any point P in space let perpen- diculars be drawn to the three planes. Then the distances measured from the planes to the point are called the rec- tangular coordinates of the point P. Let distances measured in the direc- tion of OX, OY, and OZ, i.e. to the right, forward, and upward, be counted as positive, and distances in the oppo- site direction, i.e. to the left, backward, and downward, as negative. Then to every set of three real numbers there corresponds a point in space and conversely. The distances SP, QP, and NP (Fig. 150) are called respec- tively the Qc, 2/, and s of the point P, and the point is denoted t>y (x, y, z), or by P(x, y, z). The plane containing OX and 01^ is called the a?2/-plane, and similarly for the others. The three planes containing the axes are known as coordi- nate planes. 233 z Q / / p M X / / N Fig. 150. 234 ANALYTIC GEOMETRY The eight portions of space separated by the coordinate planes are called octants. Two points are said to be symmetric with respect to a plane when the line joining the points is perpendicular to the plane and is bisected by it. EXERCISE XLin 1. Locate the points (1, 3, 2), (-1, 3, 4), (1, - 2, 4), (1, 3, - 2), (2, - 3, - 4), (- 1, - 2, - 3), (- 1, - 2, 3), (- 1, 3, - 2), (0, 1, 2), (2,0,0), (0,0,0). 2. Show that the line OP in Fig. 150 is the diagonal of a rectangular parallelopiped of which the numerical values of x, y, and z are the lengths of the sides. 3. Show that OP = Vx^ + y'^ + z'^- 4. Find the distance from the origin to each of the points, (1, 3, — 2), (3,-1,4), (2,-1, -3). 5. Find the point symmetric to each of the following points with respect to each of the coordinate planes, (2, 3, 4), (—3, — 1, — 2), (3, - 1, 2). 6. Find the point symmetric to each of the following points with respect to the origin, (2, 3, 5), (- 2, 4, 3), (3, - 4, - 1). 7. Prove that (a, &, c) and (—a, — 6, — c) are symmetric with respect to the origin. 8. What is the value of x for any point in the i/sj-plane ? What therefore is the equation of the y^-plane ? What are the equations of the other coordinate planes ? 9. Where do all points lie that have a; = and y = 0? What are the equations of the coordinate axes ? 10. Find the locus of points which satisfy the following sets of conditions : (a) x = y,z = 0. (/) x = 2,y = 3. (b) x = y,z = 2. (g) x^ + y^ = 16, ^ = 0. (d) x = y = z. ^^^ a^-^b^- '''- ' (e) —x = y,y = z. (i) y^ = 4x,z = S. COORDINATES IN SPACE 235 203. Distance between two points in rectangular coordi- nates. Let the points be Pi(xi, y^, z^ and P2(^2j 2^2? ^2)- Through P^ and P^ pass planes parallel respectively to the three coordinate planes. These three planes form a rectan- gular parallelopiped of which P1P2 is the diagonal, and the edges are respectively the differences of the coordinates parallel to the edges. Thus, in Fig. 151, PiiV= x^ — x-^, NM= y^ — 2/1, MP^ = z^— z^. But P^P\ = PiAt' 4- nm" -{-MPI ', d = r,r, = V(p,, - ^,Y + (2/1 - 2/2)' + («i - «2)'. Fia. 151. If the two points are the origin and the point (a;, y^ 2), this formula becomes 204. Point dividing a line in a giveii ratio. If the point (a;, y, z) divides the line from (a^j, 2/1, z^ to (a^g, 2/2? ^2) i^i the ratio r : 1, then ^^^H-r^^ y^ih±nh ^^ ^i + yg2 1 H-r ' 1 -t-r* ' \-\-r The proof is left to the student. f xi + 236 ANALYTIC GEOMETRY EXERCISE XLIV 1. Find the distance between (3, 4, — 2) and (— 6, 1, — 6). 2. Prove that the center of gravity, i.e. the intersection of the medians of tlie triangle whose vertices are (xi, yi, Zi), {X2, 2/2, ^2)* and (0:3, y^, z^), x^ + xs ■ yi + yo-h y3 £i_±_52_±£i\ 3 ' 3 ' 3 j' 3. Show that the lines drawn from the vertices of a tetrahedron to the intersection of the medians of the opposite faces meet in a common point which is I the distance from each vertex to the opposite face. 4. Write the equation which expresses the condition that (x, y, z) shall be equidistant from (0, 0, 0) and (3, 5, 1). What is the locus of (x, y, z) ? 5. Write the condition that (x, y, z) shall remain at the distance 4 from (0, 0, 0) . What is the locus of (x, y, z) ? 6. Find the equation of the surface of a sphere with center at (2, 1, — 3) and radius 5. 205. Polar coordinates. A point in space may be deter- mined, by its distance from the origin and the angles which the line from the origin to the point makes with the rectangular coordi- nate axes. Thus, let OX, OY, OZ,he a set of rectangular axes, and let P be any point in space. Then OP and the angles a, jS, y, between OP and the axes of X, y, and z, respectively, de- termine the position of P. If Fig. 152. ^ ^ , , . , 1 j - j OP=:r, the point may be denoted by (r, a, p, y). The four quantities r, a, fB, y are sometimes called the polar coordinates of P. It is convenient to restrict r, a, ^, y to positive values, and to further restrict the angles to values not greater than 180°. Any point in space may be represented by such values of r, a, ^, y. COORDINATES IN SPACE 237 The angles a, jS, y are called the direction angles of OP, and the cosines of these angles the direction cosines of OP. 206. Relation between rectangular and polar coordinates of a point. From Fig. 153, if the rectangular coordinates of P are x, y, z, then the following rela- tions are seen to hold: 0? = r cos a, 2/ = r cos p, z = rco»y. Since r = Vic^ + 2/^ + ^^ the above equations may be solved for the direction cosines and the following values obtained: a = cc V^ + z^ v^ + 2/'^ z + z^ Fig. 153. cosp = COS 7 = 207. Relation between the direction cosines of a line. Definition. The direction cosines of a given directed line are the direction cosines of a line drawn from the origin in the same direction as the given line. If the three equations of the preceding article, a? = r cos a, y =rG0s/3f z — r cos y, be squared and added, there is obtained x^ -\- y"^ -\- z^ = T^ (cos^ a + cos^ ^ + cos^ y). But ar' + y^ + ^^^r^. .-. cos2 a + cos2 p + cos2 7 = 1. 238 ANALYTIC GEOMETRY Hence, the sum of the squares of the direction cosines of any straight line is 1. 208. Direction cosines of a line joining^ two points. Let ^i(^ij 2/ij ^i) aiid P2(^2> 2/2? ^2) be any two points in space and consider the line as directed from P to Pg- Fig. 154. Let the direction cosines of PjPg ^6 cos a, cos p, cos y. Then cos a _ a?2 —oci ^.-P=^^--v ^2 — gl where d = V (a^ - aja)^ + (2/1 - 2/2)' + (% - 2!2)^- These relations are evident from Fig. 154. 209. Spherical coordinates. Again, take the three mutually- perpendicular axes OX, OY, OZ. Let P be any point in space. Then the position of P is determined by the distance r, or OP, and the angles ^ and <^, where 6 is the angle between OP and the positive OZ, and <^ is the angle be- tween the positive OX and the orthogonal projection of OP upon the a72/-plane. The point is denoted by (r, 0, ). The quantities r, 6, and <^ are called the spherical coordinates of P. Fig. 165. The student can easily show COORDINATES IN SPACE 239 that if P has rectangular coordinates {x, y, z), then the relations between the rectangular and spherical coordinates of the point are ac = r sin 6 cos <|), y = rsmQ sin 90°, CD is negative, i.e. is opposite in direction to 211. Projectioii of a broken line. The projection on any axis of a straight line joining two points is equal to the sum of the projections on the same axis of the sides of any broken line connecting the two points, if the parts of the broken line are directed so that the beginning of each side after the first is at the end of the preceding. This is evident from the definition of projection. COORDINATES IN SPACE Thus in Fig. 157, 241 Fig. 157. ab = ac -\- cd -^ de -\- ef + fb, or proj. AB = proj. AG -\- -pro j. CD + proj. DE + proj. EF + proj. FB. If I, li, I2, "• ?5 are the lengths of AB, AC, CD, ••• FB, respec- tively, and a, a^, a^, ••• a^ are the angles between these lines and MN, then I cos a = Zi cos «! + ?2 cos a^-\- • •• + Z5 cos ag. 212. The angle between two lines in terms of their direction cosines. Let two lines have direction angles «i, ySi, yi, and rtg? Aj 72) respectively, and let be the angle between them. To find the value of Q. Through the origin draw two lines OPj and OP2 having the same directions respectively as the two given lines. Let the coordinates of Pi be {x^, y^, z-^ and let OP^ = i\. On OP. project OP^ and the broken line OM-^MN-\-NP^ (Fig. 158). Since proj. OPi=proj. Oitf + proj. JOT + proj, iVPj, therefore, 7*1 cos d = X]^ cos a2 + 2/1 cos /Sa + «i cos y^ 242 But or ANALYTIC GEOMETRY Xi = riCosa., 2/1 = ^1 COS ySu 2;i = riCosyi. fi cos = riCCa Ui cos 02 + ^1 cos I3i cos p2 4- ^1 cos yi cos yaj cos 6 = cos tti cos a2 + cos pi cos P2 + cos ^i cos 'yg* _/^ y f ^ X ..H x Yi M t w k Fig. 158. (Notice that if one, or more, of the coordinates x^, y^, z^ is negative, e.g. y^, then — ?/i is the length of MN, but 180° — ySj is the angle between MN and OP2 ; hence the middle term is — 2/1 cos (180° ~ ^2)? which is the same as 2/1 cos ft-) EXERCISE XL VI 1. Find the projection of the line from (2, 1, — 3) to (3, — 4, 5) upon each of the coordinate axes. 2. The direction cosines of a line are proportional to 2, 3, and — 4. Find their values. 3. Express in terms of the direction cosines of two lines the condition that the two lines be parallel. The condition that they be perpendicular. 4. Find the angle between two lines whose direction cosines are respectively proportional to 2, — 1, 3 and 1, 3, —2. CHAPTER XVIII LOCI AND THEIR EQUATIONS 213. Certain straight lines and planes. The student has already considered some simple equations of straight lines and planes. For example, a; = a is the equation of a plane parallel to the 2/2-plane. The two equations y — h, z = c, represent a straight line par- allel to the ic-axis, the intersection of the two planes y = b and z = c. The two equations x = y, z = c, represent a straight line, the intersection of the plane z = c and a plane bisecting the dihe- dral angle between the iC2;-plane and the yz--p\sine. 214. Cylinders with elements parallel to a coordinate axis. Consider a circular cylinder with the 2-axis for its axis and with radius r. (Fig. 159.) If any point P be taken on the sur- face of this cylinder, the x and y of the point are the same as the x and y of the projection of the point on the ic?/-plane. But these latter values satisfy the equation of the circle x^-\-y^ = 7^. Hence the coordinates of P satisfy the same equation. The equation of the surface of the cylinder is therefore of -\-y^z= T^, In like manner it may be shown that if a straight line, kept always parallel to the 2;-axis, is moved along any curve in the aji^-plane, a cylindrical surface is 243 z X \ ^ ? X X k w Fig. 159. 244 ANALYTIC GEOMETRY generated which has the same equation as the equation of the curve in the xy-^\sine. Thus the equation y^ = 4:X, interpreted as an equation of a locus in space, is the equation of a cylindrical surface generated by a straight line parallel to the 2;-axis, moving along the curve y^z=4:xin the a^^z-plane. Likewise an equation of the form y = f(z), read "1/ equals/ of 0," i.e. ?/ is a function of z, is the equation of a cylindrical surface generated by moving a line parallel to the ic-axis along the curve y=zf{z) in the 2/2-plane. The student should describe the locus in space of the equa- tion z=f(x). EXERCISE XL VII Describe and sketch the loci in space of the following equations ; 1. ic2 + 02-25. 5. x^ = 2pz. 2. (x - ay +(2/ - &)2 = r2. Q ^^t=l, 3. ic cos a + ?/ sin a =p, «" ^'^ 7 w2 _ ^2 _ 0,2 ^ = 1. 8. ?/ = mz -f- c. 215. Surfaces of revolution. If the equation of a curve in one of the coordinate planes is known, the equation of the sur- face formed by revolving this curve about one of the coordinate axes can be obtained from it. As an illustration, consider the surface formed by revolving about the aj-axis the parabola 2/2 = 4 a;. Let P{x, ?/, z) be any point on this surface. Then (Fig. 160) x=OM,y= MN,z = NP. Since MP= MR, it follows from the equation of the parabola LOCI AND THEIR EQUATIONS 245 that But Mp = MN^'-^Np^f^-^. This is therefore the equation which is true for any point on the surface, and clearly for no other points, and hence is the equation of the surface. The process of obtaining the equation of the surface from that of the curve in the xy-^\di,nQ consists in replacing y by In general, if any curve in the a;i/-plane, F(x, y) = 0, be re- volved about the aj-axis, the equation of the surface formed is F(x,Vy'-bz')=0. EXERCISE XLVin 1. Find the equation of the surface generated by revolving the curve y^ = 4:X about the y-axis. Sketch the figure in one octant. 2. Find the equation of the surface generated by revolving the circle x^ -j-y^ = r^ about the ic-axis ; about the y-&xis. 3. Find the equation of the surface of the spheroid generated by re- volving the ellipse — \-^ = 1 about the y-axis. The spheroid is said to be oblate if a > 6, prolate if a < 6. 4. Find the equation of the surface of a cone generated by revolving the line y = mx about the x-axis. 216. Nature of locus determined by plane sections. It is frequently useful, in trying to determine the nature of a locus, to find the intersection of the locus by a plane. Generally the planes parallel to the coordinate axes, or else containing a coordinate axis, are the simplest ones to use. Example 1. As an illustration, consider the locus of the equation ^+-!^+^=i- (1) a^ 1/ & ^ ' 246 ANALYTIC GEOMETRY If in this equation z be set equal to zero, the resulting equa- tion represents the part of the locus which lies in the plane 2; = 0, i.e. in the ic^Z-pl^-ne. is the equation of the intersection of the locus of eq. (1) and the a72/-plane. This intersection is called the trace of eq. (1) in the cciz-plane. It is an ellipse with semi-axes a and h lying on the axes of x and 2/, and with center at the origin. Likewise the equations of the locus in the xz- and the 2/2;-planes are shown to be respectively the ellipses To find the trace of the locus of eq. (1) in a plane parallel to the 2/2;-plane let x be held constant in eq. (1) and y and z be allowed to vary. Letting x = li. in eq. (1), the resulting equa- tion is 1.2 in which the constant term — is transposed to the right side of the equation. This equation may be written .s>. 2 This is the equation of an ellipse, if li^ < a^, with axes in the planes of xy and xz, the values of the semi-axes being ^,^6Va'-fc^^^^^,^cV«'-fc', LOCI AND THEIR EQUATIONS 247 Hence any section of the locus of eq. (1) by a plane paralle tctiie yz-plsine, and distant less than a from the origin is an ellipse with axes in the planes of xy and xz. As k changes gradually from to a, the semi-axes of the ellipse change gradually from b and c to 0. The locus of eq. (1) may then be thought of as gener- ated by an ellipse of gradually varying di- mensions moving with its axes in the planes of xy and xz. The locus is therefore a surface. Since all sections parallel to three mutually perpendicular planes are ellipses, the figure is called an ellipsoid (Fig. 161.) Example 2. To find the locus of x^-{-2y^ = Az. (2) If z is held constant, z = k, the equation may be written Fig. 161. 4:k 2k = 1, which is the equation of an ellipse if A; > 0, but has no locus if fc < 0. When A; = 0, the equation is satisfied only by the point (0, 0). Therefore a section of the locus of eq. (2) by a plane parallel to the a;?/-plane is an ellipse if the plane is above the fljy-plane, but there are no points below the a^^z-plane which satisfy the equation. If x= 0, eq. (2) reduces to y^ = 2 z, which is the equation of a parabola in the 2/2-plane. If y = 0, eq. (2) reduces to a;^ = 4 z, which is the equation of a parabola in the iC2-plane. The locus of eq. (2) may therefore be -thought of as a surface 248 ANALYTIC GEOMETRY generated by an ellipse, moving in a plane parallel to the xy- plane, its center on the 2;-axis, and so changing in size that the ends of its axes are always Z . on the curves y^ = 2z and x' = 4:z. (Fig. 162.) The figure is called an elliptic paraboloid. 217. Locus of an equation in three variables. In gen- eral an equation in three variables represents a sur- face. For if any one of the variables be held constant, an equation between the other two variables is ob- tained, which in general represents a curve, as was found in the study of loci in two variables. The locus of the equation in three variables is then such that in general its intersections by planes parallel to the coordinate planes are curves. Therefore the locus of the equation is in general a surface. Fig. 162. EXERCISE XLIX Discuss and sketch the loci of the following equations : 1. a;2 + y2 4. ^^2 = ^2. % y'^' = x + z. 3. x + y -\- z = l. 4. a;2 ^ y'2 + 4;^2 ^ 1. 5. ic2 + ?/2 _ ;52 _ 0. 6. a;2 + 4^2 = ^2. 7. a; + y = sin «, CHAPTER XIX THE PLANE AND THE STRAIGHT LINE L THE PLANE 218. The normal equation of the plane. Let p be the length of the perpendicular from the origin upon a plane, and let the direction angles of this perpendicular be a, y8, y. io,o,c) H N^^ vX^P ^ '^'y^^^"^"""'^ M \. X -^ N \/^ ^^^(a,o,o) ,b,o) Fig. 163. Let P (x, y, z) be any point in the plane. Project the line OP and also the broken line OM -\- MN -\- NP upon the perpendicular. (Fig. 163.) These projections are equal. (Art. 211.) .*. a; cos a + ?/ cos /8 + 2 cos y = proj. of OP on OH, (Art. 211) or a? cos a + 2/ cos p + 2! cos y =P- Since this is true for any point in the plane, and for no other points, it is the equation of the plane. 249 250 ANALYTIC GEOMETRY It is known as the normal equation of the plane. 219. The intercept equation of the plane. If the above plane meets the axes in {a, 0, 0), (0, b, 0), and (0, 0, c), then cos a = -^ , cos i8 = ^, cos y = ^ . a be Substitute these values of cos a, cos p, cos y in the equation of the plane and there results a b c 220. The general equation of the first degree in x, y, and z. The general equation of first degree in x, y, and z is Ax + By + Cz + I) = 0. (1) Consider the point Q whose coordinates are the coefficients of X, y, and z ; i.e. the point {A, B, C). (Fig. 164.) Let OQ have direction angles a, fi, y. Then Fig. 164. COS« = A,cOS^=^,COBy = ;^, where OQ = -VA' + B'+0'. Dividing eq. (1) through by ± V^' + -B^ + C% it may be THE PLANE AND THE STRAIGHT LINE 251 written in the form ± -^A^ + B' + C' ± ^A^ + B'+G' ± V^' + ^' + C^ -D ± VA" + B' + C (2) Let the sign of the radical be chosen so that ± V^' + ^'+O^ is positive, and let p = = . Eq. C2) may then be written X cos a' -\- y cos P' -\- z cos y' = i?, in which a', ^', y', are the same as a, fi, y, or are the direction angles of the line from the origin to (— ^, — B, — C), accord- ing as the positive or negative sign of the radical is chosen. In either case the equation is the equation of a plane by Art. 218. Therefore the equation Ax-\- By + Cz + D = 0, in which A, B, C, D, are real quantities, is the equation of a plane. If p is the length of the perpendicular from the origin to the plane, and a, ft, y, are the direction angles of this per- pendicular, then cos a = cos p = ± \/A-' + B' + C-' ± VA^ + B^ + €'^ COS 7 = ■ ^ P = . the same sign of the radical being used throughout, and so chosen that p is positive. 221. Distance from a point to a plane. (The case where the point and the origin are on opposite sides of the plane is the only one discussed here.) 252 ANALYTIC GEOMETRY Let d be the distance from (x^, y^, Zi) to the plane whose equation is X cos a + y cos (3-\-z cos y=p. Through (x^ yi, z^ draw a plane parallel to the given plane. Since the perpendicular from the origin to this plane is in the same •(a;i,?/i,2i) direction as that from the origin to the given plane, the equation of the second plane is X cos a + 2/ cos P-\-z cos y = p\ Since (iCj, 2/1, ^i) is on this plane, «i cos a + 2/1 cos /8 + 2^1 cos y =p'. But d=p' —p. .•. c? = a?i cos a + 2/1 cos p + «! cos y—p. The student should show that if the point and the origin are on the same side of the plane, the above formula gives the negative of the distance from the point to the plane. From the above it follows that the distance from (xu 2/1, ^i) to the plane Ax -\- By -\- Cz -\- D = Fig. 165. is d = Aoci + By I -\-Czi + jy 222. The angle between two planes. Since the angle be- tween two planes is equal to the angle between the normals to the planes, it follows that the angle between the two planes X cos «! + 2/ cos fti-hz cos yi =Pi, and X cos % H- 2/ cos Pi-{-z cos y, = P2 is given by cos 6 = cos «! cos ttg 4- cos /81 cos ^2 + cos yi cos y2, and that the angle between the two planes A,x-\-B,y+C,z + D, = Oy and ^2» + ^22/+C'22 + i>2 = THE' PLANE AND THE STRAIGHT LINE 253 is given by cos 6 = ± A1A2 + B1B2 + C1C2 y/A{' + B{' + Ci^ V^22 + B2' + C22 EXERCISE L Find the lengths and direction cosines of the perpendiculars from the origin upon each of the following planes. Reduce each equation to the normal form. 1. 2x-3i/ + 40 = 6. 2. 3x-6y-2z = 0. 3. Sx-\-4:y=:2. 4. x + y-{-z = l. Find the distance from the following points to the planes : 5. From (3, 1, 2) to 2x-3?/ + 7;s = 2. 6. From (-1, 3, 2) tox + 2y-0 = 5. 7. From (0, 0, 1) to 2 x - y = 4. 8. Find the angle between the two planes of example 1 and example 2. 9. Find the angle between the two planes of example 3 and example 4. II. THE STRAIGHT LINE 223. The equations of a straight line through two points. Let the two given points be Pi(xi, y^, z^) and 1*2(^2} 2/2> ^2)- I^6t P(x, y, z) be any point on the line through Pj and P^. Pro- ject PiP and P1P2 upon the a>axis. Then, by plane geom- etry, M^M ^ PiP M^M^ P1P2' X-Xi ^PiP X2 — Xi Xir2 In like manner it is shown that and ^_^ = AP. Z2-Z^ P1P2 ... a? - xi _ y - yi _ g - jgi " X2-OC1 2/2-2/1 «2 - Zl Fig. 166. (1) 254 ANALYTIC GEOMETRY These equations are therefore the equations of the straight line. 224. The equations of a straight line through a given point and with given direction cosines. In the preceding article if the line makes angles a, y8, y with the axes, and ii d = P1P2, then 3^2 — ^ o V2 — V1 ^2 — ^1 cosa = ^^ -, cos3 = — — ^, cos 7 = -^^ -• d d d Substituting the values of ajg — ajj, 1)2 — Vn 2:2 — 2:1 obtained from these equations in the equation X2-Xy_ 2/2-2/1 ^2-Zi there results, on dividing through by d, cos a cos p cos "Y Hence these are the equations of a straight line through (^1) 2/i> z^ with direction angles a, /3, y. Any equations of the form ag-a^i _ y-V i _ z-zi I m, n are the equations of a straight line through (01^, 2/ij ^^i) with di- rection cosines proportional to /, m, n. For these equations have only to be multiplied by V/^ -\-m^-\-n^ to bring them into the form x — Xx _ y — Vx _ Z—Zx I m n which are the same as eqs. (2), since the denominators in these equations are the direction cosines of a straight line. (Art. 206.) 225. The general equations of a straight line. Since a straight line is the intersection of two planes, the equations THE PLANE AND THE STRAIGHT LINE 255 of two planes may be taken as the general equations of a straight line. Thus and Aa, 8 267 258 ANALYTIC GEOMETRY If a = 6 = c, the equation reduces to that of the sphere a?2 + 2/^ + «^ = a^- 228. The hyperboloid of one sheet. a^ h^ b^. For y = b,ov y = —b, the equation represents two intersect- ing straight lines - + - = 0, and--? = 0. a c a c The hyperboloid of one sheet is sketched in Eig. 170, and a few sections parallel to the ic^-plane are indicated. Sections parallel to the yz-plane are also hyperbolas, and have their major axes parallel to the y- or 2;-axis according as THE QUADRIC SURFACES 259 the distance of the section from the origin is less than oi greater than a. The two sections parallel to the 2/2;-plane at the distance a from the origin are each the pair of straight lines 2 + ? = 0, and 2^-?=0. be bo 229. The hyperboloid of two sheets. a2 ^2 c2 1. Any section of the surface parallel to the a^^z-plane is an hyperbola with major axes par- allel to the a.'-axis, the major axis and conjugate axis both in- creasing as the distance of the cutting plane from the «?/-plane increases, but their ratio re- maining equal to -• b A like remark applies to sec- yig. 170. tions of the surface made by planes parallel to the a;2;-plane, the major axis being parallel to the a^axis and the ratio of the axes being equal to -• c Sections of the surface parallel to the yz--pla,n.G are of the form ^ ,z^ _a^ 1 b- c^ a^ or + 2 = 1. This is the equation of an ellipse if a;^ ->^ ^2^ )^^^ there is no 260 ANALYTIC GEOMETRY locus if a^ < a^. When a; = ± a, the locus is a point (a, 0, 0), or (- a, 0, 0). Since this hyperboloid consists of two separate parts, it is called the hyperboloid of two sheets. Only one part is shown in the figure. The other part is symmetric to the part that is shown with respect to the yz-iglsme. (Fig. 171.) Fig. 171. 230. The elliptic paraboloid. The trace in the fl??/-plane is the origin — -|-^ = 0. The trace in the a;2;-plane is the parabola x^ = — z. The trace in the 2/2;-plane is the parabola y^= —z. Sections of the surface parallel to the a;?/-plane are of the form and are therefore ellipses if h and c are of the same sign, but there is no locus if k and c are of opposite sign. Sections of THE QUADRIC SURFACES the- surface parallel to the a^^i-plane are the parabolas 261 k' with vertices at , and axes parallel to the 2;-axis. a' [o, K f Sections of the surface parallel to the i/^-plane are the parabolas ckn with vertices at [M,f , and axes parallel to the 2;-axis. The locus is sketched in Fig. 172, for c positive. Z i Fig. 172. I 231. The hyperbolic paraboloid. ^ _ ^ — ? The trace in the a^y-plane is the pair of straight lines a b ah The trace in the a^^^-plane is the parabola x^ =i — z. The trace c in the i/^-plane is the parabola y^ = z. Sections parallel to 262 ANALYTIC GEOMETRY the iC2/-plane are the hyperbolas Sections parallel to the xz-iglane are the parabolas s^_z k^ Sections parallel to the 2/0-plane are the parabolas l^_z If W c a'' The locus is sketched in Fig. 173 for c positive. Fig. 173. 232. The cone. a?2 , y^ «2 r.+ 6^ c2 = 0. THE QUADRIC SURFACES 263 Let this surface be cut by the plane y = x tan $. Let »' be the distance from the 2-axis to any point P on the intersection of surface and plane. Then y — x^ sin B, a; = «' cos 0, (Fig. 174), and or a;^^cos^^ a;'^sin^l9 a? "^ h^ 62cos^+ a^sin^^ ,, ^ = 0, - = 0. Fig. 174. This is the equation of the intersection of the plane and sur- face referred to rectangular coordinates in the cutting plane. The equation can be factored into two real factors of first degree in a;' and z, and is therefore the equation of two straight lines. Since ic' = and 2 = reduce both of the factors to zero, the two lines pass through the origin. Hence any plane containing the 2;-axis intersects the surface in two straight lines through the origin. Moreover any plane parallel to the a;?/-plane, 2 = A;, intersects the surface in the curve an ellipse. 264 A.NALYTIC GEOMETRY Hence the locus of ic^ y^ z^ _ -2 + -- = 6^ (? is a cone with vertex at (0, 0, 0), and with the section at the distance c from the a;?/-plane. (Fig. 175.) 233. The right circular cone. In the equation of the preceding article if a = b, the cone becomes a right circular cone. If - be replaced by m, the equa c tion of the right circular cone be- FiG. 175. comes If x = 0, then y^± mz. Therefore the straight lines y = ±mz are the intersections of the cone and the y2;-plane. Hence the quantity m is the tangent of the angle between an element of the cone and its axis, m = tan if>, in Eig. 175. 234. The conic sections. In the equation of the cone a:^ 4- 2/^ — mh^ = 0, let the y- and z-axes be rotated through the angle $ to the new axes OY' and OZ'. The old coordinates y and z of any point in terms of the new coordinates y' and / of the point are then Fig. 176, THE QUADRIC SURFACES 265 given by 2 = 2' cos d — y^ sin 0, y — z^ sin B -\-y^ cos Q. The ^-coordinate does not change. (Fig. 176.) Substituting in the equation of the cone, and collecting terms, the equation of the cone referred to the new axes is Q^ + (sin2 ^ _ ^2 gQs2 0^ ^t2 ^ 2 sin ^ cos ^ (1 + w?) y'z' + (cos^ e-m'' sin2 6) y'^ = 0. If in this equation y' is held constant, the intersection of the cone and a plane parallel to the a;2;'-plane is obtained. Since the X and z' of points in this plane are the same as their pro- jections on the ajg'-plane, the equation of the curve of intersec- tion is of the form x'-{-az'^-hdz'+f=0, where a — sin^ 9 — m^ cos^ 6, c? = 2 sin ^ cos ^ (1 + m})y\ f =(cos^e-7rv' sin' 0)y'^ A discussion of this equation shows that (1) If y' = 0, then both d and / are zero, and the equation becomes x^-\-az'^ = 0. This is the equation of a point if a > 0, i.e. if tan^ 6 > m^; of two intersecting lines if tan- < 7n'; and of one straight line if tan^ = m^. (2) y'^0. (a) If tan^ = m^, the equation is of the form x' + dz'-\-f=0, which is the equation of a parabola. (&) If tan^ ^ m% the equation is of the form x'-\.az"-\-dz'+f=0, 266 ANALYTIC GEOMETRY which is of the type of an ellipse or hyperbola according as a is positive or negative, i.e. according as tan^ 6 is greater than or less than m^. If ^ = 90°, i.e. the cutting plane is perpendicular to the axis of the cone, the equation reduces to which is the equa- tion of a circle. Hence, if a right circular cone is cut by a plane : (1) passing through the vertex, the intersection is a pair of lines, a single line, or a point, according as the angle which the plane makes with the axis of the cone is less than, equal to, or greater than the angle between the axis and an element of the cone ; Fig. 177. (2) ^^^* passing through the ver- tex, the intersection is an hyperbola, a parabola, or an ellipse, according as the angle between the plane and the axis of the THE QUADRIC SURFACES 267 cone is less than, equal to, or greater than the angle between the axis and an element of the cone. In the special case where the plane is perpendicular to the axis of the cone, the intersection is a circle. (See Fig. 177.) EXERCISE Ln Describe and sketch the loci of the following equations : 1. x^ + y^ + ^ z^ = 4:. 2. x2 + y2 _ 4 ^^2 _ 4. 3. z^-\-y'^ = 4:X. 4. a;2-4(y2 + 2;2) =0. 5. x^-\-2z^ = y. 6. z-x^ = y^. y2 «2 3.2 7. ^ — - — s = !• S. pv = Bt, B constant ; p, v, t, variables. 62 c2 a2 9. ^4-?l = by. 10. x^-z^ = 2y. a^ c2 11. (X - 1)2 +(y + 2)2 +(0 + 1)2 =16. 12. x^ + y^ + z^ = ai 13. x^ + y^ + z^ = a^. CHAPTER XXI SPACE CURVES 235. Introduction. In this chapter a few curves in space, which do not lie in a plane, will be considered, and the equations derived. 236. The helix. The helix is a curve traced on the surface of a right circular cylinder by a point which advancies in the direction of the axis of the cylinder at the same time that it rotates around the axis, the amount of ad- vance being proportional to the angle of rotation. To find the equations of the helix, let the axis of % be the axis of the cylinder on which the helix is traced, a the radius of the cylinder, h the amount of advance along the axis to each radian of rotation, and let the a;-axis be chosen to pass through a point of the helix. Then, if Q is the angle of rotation around the axis, the values of a;, 2/, and z of any point on Fig. 178. the curve are x= a cos ^, y = a sin ^, z^hB, (Fig. 178.) 268 SPACE CURVES 269 237. The curve of intersection of two cylinders of unequal radii, with axes intersecting at right angles. Let the axes of the cylinders be the x- and ^/-axes respectively, the radii a and h respectively. (Fig. 179.) The equations of the surfaces are then and f + z' = a\ 0? + Z^=z h\ These equations, regarded as simul- taneous equations, are therefore the equations of the curves of in- tersection. The equations of the curve may be written in the parametric form, / as in the case of the helix, by let- ting z equal some arbitrary function of another variable and then solving the equations for x and y. E.g. if Fig. 179. then and z — a sm 6, y = ± a cos 0, x= ± V^^ — a' sin^ 0. Or z itself may be considered the parameter, and the equations written in the parametric form x= ±-Vb^.- z\ 2/ = ± Va^ — z^^ z = z. 238. The curve of intersection of a sphere and circular cylinder. Let the sphere have its center at the origin and radius a, and let the cylinder have its axis parallel to the x-axis, cutting the 2;-axis at 2; = c, and radius h. 270 ANALYTIC GEOMETRY The equation of the sphere and cylinder are then respectively y? -\- if- ■\- z^ = o?i and / + (2; - cf = h\ These equations, regarded as simultaneous equations, are there- fore the equations of the curve of intersection. The student should sketch the figure. The coordinate z may conveniently be considered the inde- pendent variable and have arbitrary values assigned to it, the corresponding values of x and y being then computed from the equations. Corresponding to one value of z four points are obtained, in general. Exercise. Letting a = 5, 6 = 2, c = 3, find four points on the curve corresponding to ^ = 2. How many points of the curve are there having z=\^ How many having 2 = 5 ? 239. General equations of a space curve. If the equations of two surfaces are known, these equations, regarded as simul- taneous equations, are satisfied by all points common to the two surfaces, and by only those points. The equations of the two surfaces are therefore together the equations of the curve of the intersection of the two surfaces. EXERCISE LIII 1. A screw has 8 threads to the inch. The diameter of the screw is \ inch. What are the equations of the edge of the threads ? 2. A point starts at the base of a right circular cone and traces a curve on the surface, advancing in the direction of the axis of the cone propor- tional to the angle of rotation about the axis. Find the equations of the curve. 3. Similar to example 2, using a hemisphere instead of a cone. 4. Find the polar equation of the projection of the curve of example 2 upon the plane of the base of the cone. Trace the curve of projection. 5. Find the polar equation of the projection of the curve of example 3 upon the plane of the base of the hemisphere. CHAPTER XXII TANGENT LINES AND PLANES 240. Introduction. In the plane a knowledge of derivatives was found to be important in obtaining the equations of tangent lines to curves. In space, also, derivatives play an important part in the deduction of the equations of tangent planes to sur- faces and of tangent lines to curves. But in space a somewhat extended conception of derivatives is necessary, for the number of variables has increased from two to three. 241. Partial Derivatives. Consider an equation which ex- presses 2 as a function of two independent variables, x and y. E.g. z = 2x^^Zxy^ + ^f. (1) If y is regarded as a constant and the derivative of z taken with respect to x, the result is 4.x^Sf. This result is called the partial derivative of z with respect to Xj and is denoted by the symbol — . Thus, ax Similarly, -— = 6xy + 15 y\ dy In eq. (1) let x and y take the values Xq and y^ respectively. Then z takes a corresponding value, Zq. Then z^ = 2x^ + 3 Xoyo^ 4- 5 2/o^ Let X take an increment. Ax. Then z takes a corresponding increment. Let this increment, which is due to the change in X only, be denoted by A^z. Then Zo + A,2 = 2(xq + Axy + S{xq + Ax)yJ^ + 5 y^^ 271 272 ANALYTIC GEOMETRY From the definition of a derivative, it follows that the value of dz dx is the limiting value of -^ as Ax approaches zero. ■ Ax The student can easily check this by computing the value of -^ from the above equations and finding the limiting value. i^X In general, if f(x, y), read "/ of x and 2/," is used to denote any function of x and 2/, then, if ^ =f(^, y), the values of dz and If then dx dz\ dz are defined by and 0. ^0 oy = iimif. /(^o + Ax, yp) -f(xo, yo) dx\x^,y^Ax = Ax x„ y,Ay = Ay u = F(x,y, 2), du is defined by dx xo-yo'^o du ^ i-jjj.^ Fjxp + Ax, 1/0, gp) - F(xq, yo, gp) dx x^, y^, Zf, Ax = Ax and similarly for the other partial derivatives of u. EXERCISE LIV 1. Find the partial derivative of z with respect to x and y for the values x = 2,y = 3,ilz = '6x^- 5xy^ + 2f. 2. In the equation of example 1, letting x = 2, i/ = 3, compute the increment in z due to an increment of . 1 in x. Also the increment in z due to an increment of .1 in y. 3. If u = Sx^ -\-2xyz -\- y^ + 5 x^z + yz^, find the partial derivatives of u with respect to each of the variables x,y,z. 4. In z = 2x^ ^Sxy + y, find the value of — , (1) by differentiating, dx. regarding y as constant ; (2) by giving x an increment, Ax, computing A z AgZ, and finding the limiting value of —^ . Ax TANGENT LINES AND PLANES 273 242. The tangent plane to a surface. Let F{x,y,z) = () be the equation of a surface. ' Let P(iCo5 2/oj ^^o) be any point on this surface, and let the surface be cut by a plane parallel to the 2;-axis and passing through P. (Fig. 180.) The equation of such a plane is y = mx + h. Fig. 180. Let P\xq + \x, yo-\-Ay, Zq + A^:) be any other point on the intersection of the surface and plane. Then yo + Ay = m(.ro + Ao;) + b, and ?/o = 7)iXq + b. .*. Ay = mAxj Ay Ax The equation of the line through P and P' is x-x^ ^ y-y^ ^ z-z^ Ax Ay x-Xq y-yo or or Ay Ax Az ■ Zj-Z^ Az Ax (Art. 223) 274 ANALYTIC GEOMETRY Let P' approach the limiting position P. The line through P and P' approaches the limiting position of tangency at P to the curve of intersection of the plane and surface, and hence of tangency at P to the surface. The equation of this tangent line is therefore 1 711 n Az where n is the limiting value of — ^ as P' approaches P. Ax To find the value of n, let F(x, y, z) be represented by u ; u = F(x,y,z). Since P' and P are both on the surface, therefore F(xq + Ax, ?/o + Ay, Zo 4- Az) = 0, and F {xq, yo, Zq) = 0. ^.^ F(xq + Ax, yo + Ay, Zp + Ag) - F{xq, yp, Zp) _ ^ Ax This equation may be written Fjxp + Ax, j/o 4- Ay, Zp + Az) — F(xq, y^ + Ay, Zq + Az) Ax I F{xp, yp + Ay, Zp + Az) - F(xp, y^, Zp + Az) ^ Ay Ay Ax , F(xp, yp, Zp 4- Az) - Fjxp, yp, Zp) ^ Az ^^^ Az Ace If Ay and Az were held constant, and Aa; allowed to approach zero as a limit, the first term of this equation would approach the limiting value du Since, however, as P' approaches P, Ax, Ay, and Az all approach zero, the limiting value of the first term is du dx TANGENT LINES AND PLANES 275 Likewise the second and third terms approach the limiting values du m and By du dx *0' ^0' *0 *o' 2'o» "o du , du , du \-m 1- w — dx dy dz 0, (2) the values of the partial derivatives being taken at (xq, y^, z^. Equation (2) expresses the value of n in terms of m. If eq. (2) were solved for n, and the value substituted in eqs. (1), the equation of the tangent at P to the curve of intersection of the plane and surface would be obtained. The elimination of m between the eqs. (1) would then result in an equation be- tween the coordinates of points on any tangent line to the sur- face at P, i.e. the equation of the locus of all tangent lines that can be drawn to the surface at P. The elimination of m and n is most easily affected by solving eqs. (1) for m and n and substituting their values in eq. (2). The result is / \ du {x-x^)-- dx + (z-Zq) '^O' ^0' ^0 0. *0' ^0' *0 Since this is an equation of first degree in x, y, and z, it is the equation of a plane. Hence all tangent lines to a surface at a given point lie in a plane. This plane is called the tangent plane to the surface at that point. Since u = F{x, y, z), the symbol — — may be used instead of — dx dx Hence, if F{x,y,z) = 276 ANALYTIC GEOMETRY is the equation of any surface, then + (2/-2/o)-T- + (^-^o)f = *0> 3/0) ^0 is the equation of the tangent plane to the surface at (xq, 2/0, Zq) 243. Illustration. Consider the ellipsoid, Here and H^,y.^)=^,-^ y + ^— 1, h' ' c" dF^2x §F^2y dF^2_z dx a^ ' dy b^ ' dz (? ' and hence the equation of the tangent plane at (a^o, 2/0? ^0) is (^-«^o)^4-(y-2/o)^«H-(^-^o)^^ = 0. Since ^2 -f- 52 -^ c2 1, the equation of the tangent plane becomes a^'^ b^^ & ' 244. The normal to a surface. A line perpendicular to a tangent plane to a surface at the point of tangency is called a normal to the surface at that point. If the equation of the surface is F(x,y,z)=0, the equation of the tangent plane has been found to be dF (X-Xo) dx + (2/-2/o)^- «o' ^'o' ^0 ^y "O- ^0' = 0. "^0' ^0' ^0 The equations of a line perpendicular to this plane and passing through (xq, 2/0, Zq) are therefore X—Xn dF dx 'qi Vq) ^q y-yo w Z — Zq dF dz (Arts. 220, 224.) *0' ''o' "0 TANGENT LINES AND PLANES 277 If the equation of the surface is given in the form then F(x, y,z)=z- f(x, y), and ^ = -^ = -^ M! = _3£ = _^ Ml dx dx dx^ dy dy dy^ dz The equations of the normal then become 1. y-Vo 60 dx "^0' ^0' ''o dz^ dy Z — Zq -1 "O' 2'0' ^0 245. The tangent line to a space curve. Let P(xo, y^, Zq) and P' (a^o -f Ax, 2/0 + ^y, ^0 + ^^) be two points on a curve. The equations of the line through these points are ^ — ^0 y — yo ^ — ^0 Ax Ay Az (Art. 223.) p/^ A2 r /Ay X Fia. 181. If x, y, and 2; are functions of some independent variable, t (compare the equations of the helix, Art. 236), Ax, Ay, and Az will depend upon A^. Let the above equations be multiplied by At. Then x-Xo y-yo z — Zo Ax At Ay At Az At 278 ANALYTIC GEOMETRY As P' approaches coincidence with P, the ratios ^^^ Ai' A«' A« approach the values of the derivatives of x, y, and z respectively at {xq, 2/0, z^. The line through P and P' approaches at the same time the limiting position as tangent to the curve at P. Hence the equations of the tangent to the curve at {xq, y^^ z^ are x-X(, y-yo z-Zq dx dy dz — K^) dt #0 dt t, dt <0 If the equations of the curve are the simultaneous equations two surfaces, f(x,y,z) = 0, {x,y,z = 0, .. dx dii dz the values of — > — > — may be obtained as follows: Since dt dt dt ^ P' and P are on the surface /(^j, y, z)= 0, therefore /(xo -+ A«, ?/o + A?/, Zq + ^z) = 0, and /(i»o, 2/0, 2;o) = 0. . /(a^o + Aar, yo + Ay, ^p + Ag) -/(a?o, yp, gp) ^ q Ai Treating this expression as was done in Art. 242, there results dfdxdfdy §fdz_r. dx dt dy dt dz dt Similarly, d±dx d^dy d^dz^^ ^' dxdt dy dt dzdt ' the values of all the derivatives being taken at the point From these equations there result ^ (% ^ dt _ dt _ dt ^d^ _d_f^d^'~ dfd^_dfd^'~ d^d^_dfd^' dy dz dz dy dz dx dx dz dx dy dy dx TANGENT LINES AND PLANES 279 Multiplying the members of eq. (1) by the corresponding members of this equation, there result as the equations of the tangent line at (xq, y^, z^ to the curve whose equations are f(x,y,z) = 0, and (x, y, z) = 0, « - ^0 y-Po By Bz _BfB_^\ ~ fBf B _ Bf Bct>\ \Bz Bx BxBzJ,^,y^,,^ Z-Zo /BfBcl>_BfB\ \BxBy ByBxJ,^,y^,,^ 246. Illustrations. Example 1. To find the equations of the tangent to the helix at any point. The equations of the helix are x = a cos 0, y = a sin 0, z = be. (Art. 236.) dx . ^ -^ = a cos 6. dO ' — — h de~ Hence the equations of the tangent to the helix at a point where 6 = 6^ are x — a cos Op _ y — a sin Oq _ z — bOp — a sin 6q a cos $q b Example 2. To find the equations of the tangent to the curve of intersection of the cylinders y^-\-z^ = a% aP + z^r^bK 280 ANALYTIC GEOMETRY Let f{x,y,z) = y''-\-z^~a% 1 = 0, g=2„ 1 = 2, ?.-'- ? = «' t = ^^- Therefore the equations of the tangent at {xq, yo, Zq) are x — Xq ^ y — yo ^ z — Zq ^ 2/02^0 ^6^0 — i»o2/o EXERCISE LV Find the equation of the tangent plane to each of the fol- lowing ten surfaces: 1. a;2 + y-2 -\- z'^ = r^. 3. a;2 ,/2 ^2 a2 6-2 c2 ' 5. X_2 y_2_£. a2^&2 c 7. ^+?-!-^! = o. 2. a2 62 ^ c2 4. x2+y2=2pX. 6. a-.2 y2_ a2 62 • 8. i}v = EL 62 9. xyz = c. 10. = 3 a; + 2!/. 11. Prove that the direction cosines of the tangent to the helix are — a sin 00 a cos do 6 vV + 62 Va2 + 62 Va2 + 6^ (Note that the angle between the tangent and the ^-axis is constant.) 12. If the point generating the helix advances in the direction of the axis of the cylinder j^^ of the radius of the cylinder at each revolution, find the angle between the tangent to the helix and an element of the cylinder. 13. Find the equations of the tangent to the helix at the point where d = 30°. Find the equations of the tangent to the curve of intersection of each of the following pairs of surfaces. TANGENT LINES AND PLANES 281 14. 1/2 + ^2 = 1^ x^ + 2 y^ -\- 4: z'^ = 4, Bit a, point where z = \. 15. z + 2 2/2 = 4, x2 + ?/2 _ 2 = 0, at (1, 1, 2). 16. 02 4. 2 2,2 = 4^ a;2 + 2/2 _ ^2 = 0, at (1, 1, V2). 17. Prove that the direction cosines of the normal to F{x, y,z) = at any point (x, y, z) are dF dF dx By -!)- 3 ( a-nc b-nd \ 2. (-29, 27.5), (27, -18). '\l-n' 1-n) . f xi + 2x2 yi + 2 y2 \ f2xi+_X2 2 yi + yg N [ S ' 3 J' V 3 ' 3 j* 287 288 ANSWERS Exercise XII 2. 26° 34', 63° 26'. . ^q _ 6, 99° 25'. 12. ^^ ■ 3. (-^,0). ^-m^ 6. (12, - 1), or (- 6, - 19), 13. 87° 4'. 14. - .3332. or (2, 9). 15. (1,2, - 4.56). 16. (0, - am). 7. 139° 24'. 8. 42° 60'. ^^ m -\- n ^ /Q^ -a(m + w) \ ' 1 — ?w« \ ' 1 — mil ) Exercise XIII 1. 71.5. 2. 22.56. 4. ^ nrg sin (^2 - ^i)- 3. i(xiy2- XiVi). 5. 160. Exercise XIV 1. 185. 2. 1842 sq. ft. 16. 114° 19'. 17. - 4.871. 3. 60305 sq. ft. 18. - 4.186. 4. (d) 41° 3', 38° 27', 100° 30'. ^g a + b ^0. 3.154. 12. 13.5. IZ. x + y = Q. ' 1 - ab 14. a;2 + y2 _ 4 a; _ 6 y = 12. Exercise XV 1. Sx-{-7y = Sl. 12. x^-\-y-2±2rx±2ry-\-r^=0. 4. 2x- 2/- 11 = 0. 14. 16x2 + 7 2/2 = 112. g ^ . y^l 15. 4x2 -5?/2 + 20 = 0. a 6 * 16. x2 + 82/ + 16 = 0. 6. y = mx + b. 17. 63x2 + 143 y'^ -\Sxy + 216 x 7. x2 + i/2_4a; + 8y =5; Inter- — 456 1/ — 1728 = 0. cepts, X = 5 or - 1 ; y = .58 18. 52 x2 - 80 y^ + 224 x?/ - 68 x or -8.58. + 496 !/- 1343 = 0. 10. (X - 70^ -f (2/ - A;)2 = r2. 19. 2/2 + 22 x - 8 y - 39 =0. 11. x^ + y^ = r^ Exercise XVIII 1. x2 + 42,2=l8. . ^ x2=-^y, e = 45°, new 2.64x2-64^2 + 3 = 0. 2 origin (1.77, .93). 3. 4x2 + 2/2:^=12. g Lines x-2 2/=0, and x+2/=:0, 6. Lines x + 2 y = and referred to If axes through 2 X — 3 2/ = 0, referred to ,q _ -j^n II axes through (1, 1). ^ 2 ^2 _^ ^./^ 4^ ^ ^ 450^ ^ = 45°. ANSWERS Exercise XIX 1. 2x + y = 5. 22. 3x + 4y+75=0. 3. 7. 8a;-3.y = 24. 3x + 2y±5\/l3 = 0. ^•^-^=i'-rj^-''> 8. 3x-y = n, 27. - .5642 X + .8257 y = 3, 9. bx— ay = 0. .9780 X + .2088 y = S. 10. y -yi =m(x-x{). 31. 3x4-^+10=0. 11. Bx -Ay + Ab = 0. 33. 107 a; + 134 y - 187 = 0. 20. y-2 = 6Mix-l). 34. _5x + 5 2/= 12. 21. Li,ix + Sy + 18=0 1; 35. 'll8x + 177y = 486. Za, 11x-6y-hS9 = 0. 36. 63 a; + 147 y = 536. 289 Exercise XX 1. 3.84. 2. .383. 5. 1.06. 6. 3.13. w mxi — y\ + b a . . 7. ^' — — • 8. Xq cos a + 2^0 sm a —p. ± Vw2 + 1 Exercise XXI 1. 1.23 r sin - .134 rcos0 = 1. 6. (4.91, 102° 50'). 4. (8.94, 26° 34'). Exercise XXII 1. (0, 0),r = 5. 13. x^ + y^-h2x-6y + 6 = 0. 2. (2, -3),r = 5. 16. x^ + y^ + 2x-6y = 36. 4. (-.75, 1.75), r = 3.02. 16. a;2 _|. 2/2-4x - lOy +20 = 0. 6. (1, -2),r = 0. n. x^ + y^ + 6x-\-12y = 85. 7. (-.5, -.5),r=.707. 18. x^ -{- y^ -llx -17 y -h S0=0. 8. No locus. 19. 27 (x^ + y^)- 66x -\- 16y 9. (2, -3), r=5.10. -250 = 0. 12. («, ^V r = iV^M::P. 20. (x-H)2+(, + |)2=(H)«. V2' 2/' 2 ^ 21. (X -^«-)H (!/ + ¥)'=(¥)'• Exercise XXIII 1. x2 = ^y, F(|, V). ' 9- 19-44 ft, 17.78 ft., 15ft., 3. x2 = y, F(-2, 0). 11.11 ft., 6.11 ft. 4. x2 = f y, F(l, i). 10. 5 y = x2 + X - 2. 6. i/2 = -tx, F(f,^). 12. 2 A2y = (yj ^. yg _ 2 y2)x« + /i(y3 - yi)x + 2 ^2^2- u ANSWERS Exercise XXIV 1. x2=4y, F(3, -2). 12. (V2-l)x2-(V2 + l)y2=10, 2. ?? + l^' = 1,0(2,3). ^=67° 30'. 4 9'^^ 13. 8 x-2 + 28^2 = 13, 3. The lines Sx-y + 'J = 0, c^- 1, 1), ^ = tan-i ^. 3a; + «/ + 5 = 0. 14. The lines 2 x + 7 ?/ = 0, 4. ^^_l!-l of- -3V 7x-2y=0, referred to 1 16 ' V2' / II axes through (-3^, it). 8. ^^y!^l .^tan-12. 15- x^ - t,2 = 2, 0(- 1, - 2), If ^ = 45°. 9. 9%-i>''' = ''°'"- "■.-* = 4(-^)- 10. i/2=4x, = tan-if ; 19. x = 6 and y = a. F(-3,-3). 23. xy-4x-2y + 12=0. 11. x2-y2 = i6, ^ = 45°. Exercise XXVIII 8. ±3V5. 9. ± r Vl + m2. 10. 6. Exercise XXX 1. y = VSx±S. 16. y = :^a; + 3V2, 2. y = x±10. ^ 3. x + 2y + 6 = 0. y = _^2^_3V2. 6. x-2«/+6=0,3x-2y + 2 = 0. 4 12. 2x + 2y+P = 0. 17. y = 2x-7±2VlO. Exercise XXXI 1. .007651,. 030301, 3.003001. 3. f , 3f , 0, f . 2. -1, -hh 4- 4x-4y = 5. Exercise XXXII ' 1. x + y + 1 = 0, x-y = 3. 2. 3x-4y4-26=0,4x+3y=0. 3. Tangents, 2?/oy- 3xo2x + Xo^ = 0, y = 0, Sx-2y=l, Sx-y=A. 6. Tangents, 6x-i/ = 6, 6x + y4-30 = 0. 8. 12° 6', 36° 52'. 9. 73° 41'. 10. (-2, - 9). ANSWERS 291 Exercise XXXIII 2. 2aa;-^. 6. ' 4-1. 4. -^« 8 ^^ -^ 9. (x -f ay-^(x + 6)«-i[rc(n + m) + ma + w6]. 10 ~ ^^ . 14. 8 a; — y = 4. s"+i * 15. x-y+1 =0. 11. 2 anx(ax^ + b)^-\ 16. 4a; - Sj^ + 25 = 0. 12. -2 an- ^^-^^^^^- 21. l(y+yo) = axoa;+^(x+a;o) + c. (x — a)"+i 2 2 13. y = mz+b. ^ i(=c+^„) = «y„,+| (,+,„)+<. Exercise XXXIV 1. (6 - a) sin 2 x. 5. — a sin 2 (ax + 6) . 2. 24tan2 2x(l+tan2 2a-0. g 2sinx - 12 sin 3 a; 3. 4. 1 cos « Vsin t. - 2 sin 2 X. cos^ a; ' cos^ 3 x 8 2(1 + sin t) sec2 2 « - cos t tan 2 t (1 -f sin 0^ 9. .-^l5^(l + 3cos2x). 1*- a^cosx + sinx. 2 cost X 15.x sec^ x -f tan x. 10. wm(tan«-imx + tan«+iwx). ^6. siux + xcosx. 2(sin4x + cos4x) 1^. 4csc4x (1 - 2 csc24x). 11. 12. 13. tan4^. sin^xcosSa; 18. — mng ^*^^"~ ^^ » cosx sin»»+igx 3 ■ 19. (x + 1) sin x+ (x— 1) cosx 20. abn sin «-i 6« cos bt. Exercise XXXVII 1. y^-6y + Sx = 23. 2 ?!+(y±I)!^l 3. 3 x2 - 2/2 « 16 X -f 8 2/ = 0. '6 15 292 ANSWERS Exercise XLI 1. No locus. 2. Hyperbola. 3. Two intersecting lines. 4. Two parallel lines. 5. One line. 6. Ellipse. 7. A point. 8. No locus. 9. Parabola. 10. Hyperbola. Exercise LIII 2. x=~(h- be) cos e, y =^{h- bd) sin 0, z = bd. h h 3. a; = y/a^ - b'^ff^ • cos 6, y = Va^ — 6^^ • sin 0, z = be. A. r = ^{h-b0). 6. r = VcF^^W\ h Exercise LV 1. x^-^-y^y + z^z^r"^. 8. p^p+v^p = B{t + Q. • a2 5-2 "^ c-2 ■ 14 2a;-V6 ^ 2y-V3 ^2g-l 4. a;oic + 2/oy =P(^+^o). * 2V3 V6 --3V2 g ^ , |/oy _ g + gp three other answers. a2 ft2 2c ' ^g x-\^ y-\ ^ z-2 a ^(y^_M = l 3 -1 4 * a2 52 • . INDEX The numbers refer to the pages. Abscissa, 7. Addition of segments, 3. Angle, between two lines, 21, 27, 241. between two planes, 252. Area, of a triangle, 30, 31, 33. of a polygon, 35. Asymptote, 59. of the hyperbola, 108. Axes, of the ellipse, 103, 198. of the hyperbola, 109, 200. .Axis of the parabola, 95. Cardioid, 138, 140. Center, of ellipse, 103. of hyperbola, 109, Change of sign oiAx + By + C, 81. Circle, equation of, 88. through three points, 89. Circular measure of an angle, 172. Concavity, 183. Cone, 262. Conic Sections, 192, 264. classification of, 196. polar equation of, 202. rectangular equation of, 196. Conjugate diameters, 213. Conjugate hyperbola, 109. Continuity of functions, 161. Coordinate planes, 233. Coordinates, Cartesian, 6. rectangular, 9. polar, 9. rectangular, in space, 233. polar, in space, 236. spherical, 238. Cycloid, 132, 140. construction of, 133. Cylinders, equations of, 243. Derivative curves, 182, 185. Derivatives, 159. partial, 271. Diameter, of parabola, 208. of ellipse, 213. conjugate, 213. Differentiation, 161. formulas of, 162, 169, 175. Direction cosines of a line, 237. Directrix, of parabola, 92. of conic, 192. Discontinuity, 161. Distance, between two points, 18, 20, 235, 236. from a point to a line, 83. to a plane, 251. Eccentric angle of ellipse, 130. Eccentricity of a conic, 192. Ellipse, definition of, 45, 100. construction of, 131. Ellipsoid, 247, 257. Elliptic paraboloid, 248, 260. Empirical equations, 223. Epicycloid, 138. Equation of a locus, 41. Exponential function, 123. Focal radii, of ellipse, 208. of hyperbola, 209. Foci of ellipse, 100. Focus of a parabola, 92. Function and variable, 38. General equation of second degree,214. Graph of a function, 39. Graphical solution of equations, 143. Helix, 268. Hyperbola, definition of, 46, 105. equilateral, 92. 293 294 INDEX Hyperbolic paraboloid, 261. Hyperboloid, of one sheet, 258. of two sheets, 259. Hypocycloid, 134. construction of, 135. of four cusps, 137. Inclination of a line, 22. Increments, 153. Intercepts, 48. Intersection, of lines, 79. of curves, 142. Involute of circle, 139. Latus rectum of a conic, 206. Limit of ^— :, 146. sin Locus of an equation, 51. Logarithmic curve, 122. Maxima and minima, 178. Normal, to a curve, 158. to a surface, 276. Ordinate, 7. Parabola, 47, 92, 97. parameter of, 95. Parabolic arch, 99. Parallel lines, condition for, 28. Parametric equations of loci, 129. Periodic functions, 118. Perpendicular lines, condition for, 28. Plane, equations of, 249, 250. Plotting in polar coordinates, 125. Projections, 15, 239. Property of reflection, of parabola, 206. Property of ellipse, 209. of hyperbola, 210. Quadric surfaces, 257. Radical axis of circles, 91. Ratio into which a point divides a line, 23, 26, 235. Rotation of axes, 65. Sine curve, 1 17. " Slope, of a line, 22. of a curve, 155. Space curves, 268. Standard equations of second degree, 88. Straight line, equations of, 70-74, 86, 253. Subnormal of parabola, 205. Subtangent of parabola, 205. Subtraction of segments, 4. Surfaces of revolution, 244. Symmetry, 55. Tangent plane to a surface, 273. Tangents, slope equations of, 148, 149. contact equations of, 156, 159. to space curves, 277. Transformation of coordinates, 64. Translation of axes, 64. Trigonometric functions, 11. Variable, dependent and independent, 38. Vertex of a parabola, 95. Vertices, of ellipse, 103. of hyperbola, 109. of conies, 193. Printed in the United States of America. T HE following pages contain advertisements of a few of the Macmillan books on kindred subjects First Course in Differential and Integral Calculus By WILLIAM F. OSGOOD, Ph.D. Professor of Mathematics in Harvard University Revised Edition. Cloth, xv + 462 pages, $2.00 The treatment of this calculus by Professor Osgood is based on the courses he has given in Harvard College for a number of years. The chief characteristics of the treatment are the close touch between the calculus and those problems of physics, including geometry, to which it owed its origin; and the simpHcity and directness with which the principles of the calculus are set forth. It is important that the formal side of the calculus should be thoroughly taught in a first course, and great stress has been laid on this side. But nowhere do the ideas that underlie the calculus come out more clearly than in its appHcations to curve tracing and the study of curves and surfaces, in definite integrals, with their varied applications to physics and geometry, and in mechanics. For this reason these subjects have been taken up at an early stage and illustrated by many examples not usually found in American text-books. From the beginning the book has been a favorite with the academic classes, and it has now been adopted by some of the best-known and best-thought-of engineering schools in the country. THE MACMtLLAN COMPANY 64-66 Fifth Avenue, New York Applied Mechanics for Engineers A Text-book for Engineering Students BY F. L. HANCOCK Professor of Applied Mechanics, Worcester Polytechnic Institute Illustrated, cloth, i2mo, xii-hjSS pages, $2.00 A new edition with typographical corrections In the preparation of this book the author has had in mind the fact that the student finds much difficulty in seeing the applications of theory to practical problems. For this reason each new prin- ciple developed is followed by a number of applications. In many cases these are illustrated, and they all deal with matters that directly concern the engineer. It is believed that the problems in mechanics should be practical engineering work. The author has endeavored to follow out this idea in writing the present volume. Accordingly, the title "Applied Mechanics for Engineers " has been given to the book. The book is intended as a text-book for engineering students of the Junior year. The subject-matter is such as is usually covered by the work of one semester. In some chapters more material is presented than can be used in this time. With this idea in mind the articles in these chapters have been arranged so that those coming last may be omitted without affecting the continuity of the work. The book contains more problems than can usually be given in any one semester. An appendix giving tables for the use of the student is of importance. These tables include the following : Hyperbolic Functions, Trigonometric Functions, Logarithms of Numbers, Squares, Cubes, etc., and Conversion Tables. THE MACMILLAN COMPANY 64-66 Fifth Avenue, New York COLLEGE ALGEBRA BY SCHUYLER C. DAVISSON, Sc.D. Professor of Mathematics in Indiana University Cloth, i2mo, ig I pages, $1.50 A discussion of those parts of algebra usually treated in the first year's course in college. The author aims that the student shall acquire not merely a comprehension of algebraic processes, but the abiHty to use without difficulty the language of algebra — to express in his own language conclusions ordinarily expressed in symbolic form, and thus gain the ability to generahze easily. A characteristic feature, developed in the course of several years of teaching college freshmen, is the introduction early in the course of the fundamental laws of algebra. When the student once recognizes these foundations and the continuity of the sub- ject is pointed out to him, there will be a higher degree of interest in the facts of algebra accompanying a more intelligent compre- hension of their relations, and, in consequence, they will be more readily retained and more easily applied in later work. PUBLISHED BY THE MACMILLAN COMPANY 64-66 Fifth Avenue, 2iew York Trigonometry By DAVID A. ROTHROCK, Ph.D. Professor of Mathematics in the University of Indiana Cloth, 8vo, xi + 140 pages, $1.40 In this work the author has prepared a text-book which will serve as a basis for courses in plane and spherical trigonometry as ordinarily presented in advanced, secondary, and elementary college courses. The book is not particularly different from a number of other text-books on trigonometry in its plan, but the author has placed special emphasis upon drill work in the trigo- nometrical identities, upon the applications of trigonometry to practical problems, and upon approximate calculations by means of natural functions. For the benefit of those who may wish to pursue advanced courses in mathematics a brief discussion of analytical trigo- nometry is presented in chapter 10. In Part II the elements of spherical trigonometry are developed in so far as to include the ordinary formulae necessary in the solution of right and oblique spherical triangles. Especial attention, too, has been given to the preparation of a really satisfactory set of tables which are included in the back of the book. These tables have been so arranged as to emphasize in the student's mind the advantage of an orderly arrangement of trigonometrical calculations. The book will be found a very satisfactory introductory course in the subject. THE MACMILLAN COMPANY 64-66 Fifth Avenue, New York UNIVERSITY OF CALIFORNIA LIBRARY ^ BERKELEY Return to desk from which borrowed This book is DUE on the last date staxnped below. ^ , ^Oeo="' 4Feb'S0 .^-^^''^ REC'D '-O m'l'^'^'^'i •JAN 2 S ^--^ -^^.''.t'SOr^ ^<^^^ JUL 25 19S9 26Mv.'51ES 1 ^#C•D LD — imir^^"' OCT 1 S iOHS T4^v5^.^ V 201^om'58JP M^ wov «» LD21-100n.-ll,'49(B7146»16)476