'II li! ■mm iiiiii li f iiil li 11 III Hi jiiii^ i'h':--^;.tfi;:'-[M-'f: 'M!i.T;f!.TiLttMU-!l JItliaca, Hem ^otk BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. SAGE 1891 Cornell University Library 3 1924 031 220 795 olin.anx Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031220795 ANALYTIC GEOMETEY BY H. B. PHILLIPS, Ph.D. Assistant Professor of Mathematics in the Massachusetts Institute of Technology. FIRST EDITION TOTAL ISSUE FIVE THOUSAND NEW YOEK JOHN WILEY & SONS, Inc. London: CHAPMAN & HALL, Limited 1916 5 Copyright, 1916, BY H. B. PHILLIPS Stanbope iprcBS F. H. GILSOW COMPANY BOSTON, U.S.A. PREFACE The author of this text beUeves that the differential calculus should be given to the student in college at the earliest possible moment, and that to accomplish this a short course ia analytic geometry is essential. He has, therefore, written this text to supply a course that will equip the student for work in calculus and engi- neering without burdening him with a mass of detail useful only to the student of mathematics for its own sake. The exercises are so numerous and varied that the teacher who desires to spend a longer time on analytic geometry can easily do so; and, indeed, if more than the briefest course is given, the best way to spend the time is in working a large number of varied examples based upon the few fundamental principles which occur constantly in practice. The author is indebted to Professor H. W. Tyler and Professor E. B. Wilson for many suggestions and to Dr. Joseph Lipka for valu- able assistance both in the preparation of the manuscript and the re- vision of the proof. H. B. PHILLIPS Boston, Mass. August, 1915. CONTENTS CHAPTER 1 ALGEBRAIC PRINCIPLES Art. Page 1. Constants and variables 1 2. Equations 3 3. Equations in one variable 5 4. Factors and roots ; 7 5. Approximate solution of equations 9 6. Inequalities 11 7. Simultaneous equations 12 8. Special cases 14 9. Undetermined coefficients 16 10. Functions 18 ■CHAPTER 2 RECTANGULAR COORDINATES 11. Definitions 23 12. Segments 26 13. Projection 27 14. Distance between two points 29 15. Vectors 32 16. Multiple of a vector 35 17. Addition and subtraction of vectors 36 18. Slope of a line 39 19. Graphs 43 20. Equation of a locus 47 21. Point on a locus 49 22. Tangent curves 51 CHAPTER 3 STRAIGHT LINE AND CIRCLE 23. Equation of a straight line 54 24. First degree equation 57 25. The expression Ax + By -\- C 60 V vi Contents Aet. Page 26. Distance from a point to a line 61 27. Equation of a circle 64 28. Circle determined by three conditions 66 CHAPTER 4 SECOND DEGREE EQUATIONS 29. The ellipse 70 30. The ellipse in other positions 71 31. The parabola 74 32. The hyperbola 77 33. The rectangular hyperbola 80 34. The second degree equation 83 35. Locus problems 87 CHAPTER 5 GRAPHS AND EMPIRICAL EQUATIONS 36. Intersections with the coordinate axes 90 37. Real and imaginary coordinates 92 38. Symmetry 93 39. Infinite valjies 95 40. Direction of the curve 96 41. Sine curves 99 42. Periodic functions 102 43. Exponential and logarithmic curves 104 44. Empirical equations 107 CHAPTER 6 POLAR COORDINATES 45. Definitions 113 46. Change of coordinates 116 47. Straight line and circle 117 48. The conic 117 49. Graphing equations 121 50. Intersections of curves 124 51. Locus problems 127 CHAPTER 7 PARAMETRIC REPRESENTATION 52. Definition of parameter 130 53. Locus of parametric equations 132 54. Parametric from coordinate equations 135 55. Locus problems 138 Contents vii CHAPTER 8 TRANSFORMATION OF COORDINATES Abt. Page 56. Translation of the axes 142 57. Rotation of the axes 144 58. Invariants _. 145 59. General equation of the second degree 146 CHAPTER 9 COORDINATES OF A POINT IN SPACE 60. Rectangular coordinates 150 61. Projection 151 62. Distance between two points 154 63. Vectors 155 64. Direction of a line 156 65. The angle between two directed lines 158 66. CyUndrical and spherical coordinates 160 CHAPTER 10 SURFACES 67. Loci , 163 68. Equation of a plane 163 69. Equation of a sphere 167 70. Equation of a cylindrical surface 167 71. Surface of revolution 168 72. Graph of an equation 170 CHAPTER 11 LINES AND CURVES 73. The straight line 176 74. Curves 178 75. Parametric equations 180 ANALYTIC GEOMETRY CHAPTER 1 ALGEBRAIC PRINCIPLES Art. 1. Constants and Variables In analytic geometry much use is made of algebra. Hence a brief review is here given of some algebraic principles and processes used in this book. In a given investigation a quantity is constant if its value is the same throughout that work, and variable if it may have different values. It should be noted that a quantity that is constant in one problem may be variable in another. Thus, ia discussing a particu- lar circle the radius would be constant, but in a problem about a circular disk expanding under heat the radius would be variable. A quantity whose value is to be determined is often called an unknonm. Such a quantity may be either constant or variable. In some cases it is not even known in advance whether it is con- stant or variable. Real Numbers. — The simplest constants are numbers. The proc- ess of counting gives whole numbers. Division and subtraction give fractions and negative numbers. Whole numbers and fractions, whether positive or negative, are called rational numbers. A number, like V2, that can be expressed to any required degree of accuracy, but not exactly, by a fraction, is called irrational. Ra- tional and irrational numbers, whether positive or negative, are called real. The absolute value of a real number is the number without its algebraic sign. The absolute value of « is sometimes written \x\. Thus, I -2] =,|-|-2l = 2. 1 2 Algebeaic Principles Chap. 1 Graphical Representation. — Real numbers can be represented graphically by the points of a straight line. Upon any point of a line mark the number 0. Choose a unit" of length.' On one side of mark positive numbers, on the other negative numbers, making the number at each point equal in absolute value to the distance from to the point. The result is a scale on the line. When the line is horizontal, as in Fig. 1, it is usual, but not necessary, to lay off the positive numbers on the right of 0, the negative numbers on the left. c A B -7 -6 -6 C-4 -3 -2 -1 01 3 30t 66 67 8 Fig. 1. The point A representing the number a divides the scale into two parts. On one side of A, called the positive, lie all numbers greater than a; on the other side, called the negative, he all numbers less than a. At a point B on the positive side of A is located a number b greater than a, at a point C on the negative side of A is a number c less than a. The distance between two points of the scale is equal to the difference of the numbers at those points. This is obvious if the numbers are both positive. Thus AB = OB-OA = h-a. It is still true if one or both are negative. Thus, since c is negative, CO = —c and C£ = CO + OB = -c + 6 = 6 - c. Imaginary Quantities. — The extraction of roots sometimes leads to expressions like V— 1 or a + 6 ^/—l, where a and 6 are real num- bers. These expressions are called imaginary. This means merely that such expressions are not real numbers. It should not be in- ferred that imaginaries cannot be used or that they have no meaning. A quantity may have a meaning in one problem and not in another. For example, in determining the number of workmen needed in a certain undertaking the answer 3f would be absurd since 3f work- men cannot exist. In determining the area of a field the answer Art. 2 Equations 3 — 10 acres would be meaningless since there is no negative area. In determining the ratio of two lengths the answer V— 2 is imagi- nary since the result must be a real number. But in still other problems, notably in work with alternating currents, an interpreta- tion can be given to the process of extracting the square root of a negative number and then such results are entirely real. Art. 2. Equations An equation is the expression of equality between two quantities. An identical equation is one in which the equality is true for all values of the variables. Thus, in {x -yy + 4:xy= {x + yY the two sides are equal whatever values be assigned to x and y. In many equations, however, the equality is true only for certain values of the variables; thus x^ -|- x = 2 is an equation not true for all values of x, but only when x = +1 or —2. Two or more equations are called simultaneous if all are satisfied at the same time. Equations often occur that are not simultaneous. Thus if x'' = 1, then x = l,orx = — 1, but not both simultaneously. A solution of an equation is a set of values of the variables satis- fying the equation. Thus x = 3, ?/ = 4 is one solution of the equation x^ + y"^ = 2b. A solution of a set of simultaneous equa- tions is a set of values of the variables satisfying all of the equations. Eqmvalent Equations. — Sets of equations are called equivalent if they have the same solutions. Thus the pair of simultaneous equations x^ + xy + y^ = 4, x'^ — xy -\- y'^ = 2 is equivalent to x^ + y^ = 3, xy = 1 (obtained by adding and subtracting the original equations) in the sense that any values of x and y, satisfying both equations of one pair, satisfy both equations of the other pair. Similarly, (^ + 2/) {x — 2y) = is equivalent to the two equations x + y = 0, X — 2y = 4 Algebraic Principles Chap. 1 in the sense that if x and y satisfy the equation {x + y) (x — 2y) — 0, then either x + y = Oorx — 2y = Q; and, conversely, if x and y satisfy either of the latter equations, they satisfy the former. The maia problem in handling equations is to replace an equation or set of equations by a simpler or more convenient equivalent set. To solve an equation or set of equations is merely to find a particu- lar equivalent set of equations. Degree of Equation. — The equations of algebra usually have the form of polynomials equated to zero. By a polynomial is meant an expression, such as a;' + x" — 2 or x^y + 3 xy — y-, containing only positive integral powers and products of the variables. The degree of a term like x' or 3 xy is the sum of the exponents of the variables in that term. Thus the degree of x' is three, that of 3 xy is two. The degree of a polynomial is that of the highest term in it. Thus, the polynomials given above are both of the third degree. If a polynomial is equated to zero, or if two polynomials are equated to each other, the degree of the resulting equation is that of the highest term in it. For example, x^ + y^ — x = and xy = 1 are both equations of the second degree. Exercises 1. Determine which of the following equations are identities: (a) x^s" = aj^H-", (6) - + - = 2, (c) 1 + 1 = 5-+^'. ^ ' y X X y xy 2. Expand (x + y)' by the binomial theorem. Is the resulting equation an identity? 3. Show that x = V2 is a solution of the equation x6 + 2x'-a;2 -8a; + 2 =0. 4. Show that a;=— 1, 2/ = 2isa solution of the simultaneous equations x^ + 6xy + y* = 5, x''+y' = 5. 5. Show that the pair of simultaneous equations x' + y^ = 2, x + y = 1 Art. 3 Equations in One Variable 5 is equivalent to the pair x^ - xy -\- y'^ = 2, x+y = 1. 6. Find a set of three equations equivalent to (s^ - 1) (a;3 + 2) = 0. Explain in what sense the three are equivalent to the one. 7. Is x^ — 4 ^2/ + 3 1/^ = equivalent to the pair of simultaneous equations x = y, x = Zyl 8. The symbol V2 is generally used to represent the positive square root of 2. Is a; = v^ equivalent to x^ = 2? 9. Show that Vx + 1 + Vx — 2 = 3 is equivalent to x = 3. 10. The solution of the simultaneous equations X -\-y = i, xy = X can be written X = 1(3 ± VB), 2/ = i (3 =F VI). What do these mean? How many solutions are there? 11. What is the degree of the equation (x + z/)* = 3 xyl 12. If X and y are the variables, what is the degree of ax^ = hxyl Art. 3. Equations in One Variable Quadratic Equations. — The quadratic equation ax^ + 6x + c = can be solved by completing the square. Transposing c, dividing by a and adding 6^/4 c?- to both sides, the equation becomes „ , & ,6^ 6' — 4 oc x^+ -x-]r-T— = a 4 a^ id? Extracting the square root and solving for x, _ 6 -fc V62 _ 4 ac """ 2a If the expression under the radical is positive, the square root can be extracted and real values are obtained for x. If it is negative> no real square root exists and the values of x are imaginary. Solution by Factoring. — Another method for solving quadratic equations is factoring. Thus x2 + 5a;- 6 = 6 Algebraic Principles Chap. 1 is equivalent to (a; - 1) (x + 6) = 0. Since a product can only be zero when one of its factors is zero, the above equation is satisfied only when a; = 1 or a; = —6. If the quadratic cannot be factored by inspection, it can still be factored by completing the square. Thus Sx-' - 2x + 1 = 3 (x^ - ^x + I) = 3[ix - lY + U = 3 (a; - i - I V=2) (x - i + § V^). The solutions of the equation Sa;"— 2a; + l=0 are then a; = i (1 ± ^~2). In this way any equation can be solved if the expression equated to zero can be factored. For example, to solve the equation a;' + x" - 2 = write it in the form a;3 - 1 + x2 - 1 = 0. Since x' — 1 and x^ — 1 both have x — 1 as a factor, the equation is equivalent to (x-1) (x2 + 2x + 2) =0. The solutions are consequently X = 1 and X = —1 ± V— 1. Exercises Solve the following equations: 1. 2x2 + 3x - 2 = 0. 2. x^ + 4 X - 5 = 0. 3. 3 x^ + 5 X + 1 = 0. 4. x^ + X + 1 = 0. Solve by factoring 8. x^ - 3 X - 1 = 0. 9. 2 x^ + X - 2 = 0. 10. x2 - X + 1 = 0. 14. Solve the equation x* + 1 = by reducing it to the form (x'' + 1)2 - 2 x^ = 0. 5. (x2 - 1) (x2 - 2) = 0. 6. (x2 - 4) "■ 7. X^X - 1 ' X + 1 11. x' - 2 x2 - X + 2 = 0, 12. x' - 1 == 0. 13. x« - 1.. Art. 4 Factors and Roots 7 15. Solve the equation a;' + x^ + 4 = by the method of the last example. 16. Factor 4x^ + ixy — y^by completing the square of the first two terms. Art. 4. Factors and Roots It has been shown above that the roots of an equation can be found if the factors of the polynomial equated to zero are known. Conversely, if the roots are known the factors can be found. This is done by the use of the following theorem : If r is a root of a poly- nomial equation in one variable x, then x — r is a factor of the poly- nomial. To prove this, let P = ttx" + bx"-^ + ■ ■ ■ +px + q be a polynomial of the nth degree in which a,b, . . . , p, q are con- stants. If ?• is a root of the equation given by equating this 'poly- nomial to zero, ar" + hr"-^ + • • • + pr + q = 0. Since subtractiag zero from a quantity does not change its value, P = ax"-\- bx"-^ + ■ ■ ■ +px + q— {ar" + br"--^ + • • ■ +pr + q) = a (x" - r") + b ix"-^ - r"-') + ■ ■ ■ + p (x -r). Each term on the right side of this equation is divisible by a; — r. Hence the polynomial, P, has a; — r as a factor, which was to be proved. Number of Roots. — It can be shown that any polynomial equation in one unknown has a root, real or imaginary. Assuming this, it follows that any polynomial of the rath degree in one variable is the product of n first degree factors. In fact, if n is a root of P = 0, then P=ix- n) Q, Q being the quotient obtained by dividing P by a; — ri. Similarly, if r2 is a root of Q = 0, Q= {x- ri) R. Hence P = (a; - ri) (x - r^) R. 8 Algebraic Pkinciples Chap. 1 In the same way R can be factored, etc. Now each time a factor a; — r is divided out the degree of the quotient is one less. After taking out n factors, what is left will be of zero degree, that is, a constant. If a is the constant P = a{x -ri){x -n) . . . (a; — ?■„). Hence P is the product of n first degree factors, a {x — n), {x — Ti), etc. Since a product can only be zero when one of its factors is zero, it follows that the roots of P = are n, r^, . . . , r„. It is thus shown that an equation of the nth degree has n roots. Some of these r's may be equal and so the equation may have less than n distinct roots. Rational Roots. — Though every polynomial equation in one un- known has a root, no very definite method can be given for finding it. If nothing in the particular equation suggests a better method, it is customary to try first to find a whole number or fraction that is a root of the equation. Such roots are found by trial. Some methods that may be useful are shown in the following examples. Example 1. Solve the equation 4 x' + 4 a;^ — a; — 1 =0. Since a; is a factor of all the terms in this equation except the last, — 1, it follows that any integral value of x must be a divisor of —1. The only integral roots possible are then ±1. By trial it is found that a; = — 1 satisfies the equation. Hence a; + 1 is a factor of the polynomial. Factoring, the equation becomes (a;+l) (4x''— 1) = 0. The roots are consequently —1, and ±|. Ex. 2. Solve the equation 27 x^ + 9 a;^ - 12 a; - 4 = 0. Proceeding as in the last example it is found that the equation has no integral root. Suppose a fraction -p/q (reduced to its lowest terms) satisfies the equation. Substituting and multiplying by g', 27 p3 + Qp\- 12pg2 _ 42' = 0. Since all the terms but the last are divisible by p, and p and q have no common factqr, —4 must be divisible by p. For the same reason 27 must be divisible by q. Any fractional root must then be equal to a divisor of 4 divided by a divisor of 27. It is found by trial that Art. 6 Appeoxmate Solution of Equations 9 f is a root. Hence a; — f is a factor. Dividing and factoring the quotient, the equation is found to be 27 (x - f ) (x + f ) (x + i) = 0. The roots are consequently ±f and Exercises 0. Solve the following equations: 1. 2. 3. 4. 5. 6. x' + 4 a:^ -f- 4 X + 3 = 0. x3-2x2-x + 2 = 0. 3 x' - 7 x2 - 8 X + 20 4 x^ - 8 x« - 35 X + 75 = 0. 8x3 - 28x2 +30x - 9 = q. x3 - 4 x= - 2 X + 5 = 0. 7. 4x« + 8x' + 3x'-2x-l=0. 8. 6x<-llx'-37x2+36x + 36 = 0. 9. 3x^-17 x' + 41x2-53x + 30 = 0. 10. 2 X* - 9 x' - 9 x^ + 57 X - 20 = 0. Art. 6. Approximate Solution of Equations If the equation has no whole numbers or fractions as roots, any- real roots can still be found approximately. The method depends on the foUowing theorem: Between two values of x for which a •polynomial has opposite signs must be a value for which it is zero. To show this supppse when x = a the polynomial is positive and when a; = 6 it is negative. Let x, beginning with the value a, gradually change. The value of the polynomial changes gradually. When x reaches 6 the polynomial is negative. There must have been an instant when it ceased to be positive and began to be nega- tive. Now a number can only change gradually from positive to negative by going through zero. There is consequently a value of x between a and 6 for which the polynomial is zero. The theorem can be illustrated by a figure. Let x be the number at the pouit M in a scale OX (Fig. 5), and let the perpendicular MP have a measure equal to the value of the polynomial for that value of X, it being drawn above OX when the value is positive, below when negative. As M moves along OX from A to B, the point P describes a curve. Since the curve is Fig. 5. 10 Algebraic Pkinciples Chap. 1 above at A and below at B, it must cross the axis at some inter- mediate point C. At that point th'e value of the polynomial is zero. Example 1. Find the roots of i' + 3 x^ — 1 = accurate to one decimal place. By substitution the following pairs of values are found: X = -3, -2, -1, 0, +1, a;3 + 3x2- 1 = -1, +3, +1, -1, +3. The polynomial changes sign between x = — 3 and x — —.2, be- tween X = —1 and X = and between x = and x = 1. There is consequently a root of the equation in each of these intervals. To find the root between and 1, make an enlarged table for this region. X = 0, 0.5, 0.6, 1, x' + Sx''- 1 = -1, -.125, +.296, 3. It is thus seen that the root is between 0.5 and 0.6. When x = 0.55 the polynomial is positive. Hence the root hes between 0.5 and 0.55. The value 0.5 is therefore correct to one decimal. In the same way the value —2.9 is found for the root between —2 and —3, and —0.7 for the one between —1 and 0. Since the equation can have only three roots this completes the list. Ex. 2. Solve the equation x' + x — 3 = 0. Since x' + x increases with x it can equal 3 for only one real value of X. To two decimals this root is found to be 1.21. The polynomial then has x — 1.21 as an approximate factor. Dividing by this the quotient is x^ + 1.21 X + 2.46. The solutions obtained by equating this to zero are x= -.6 ±1.4 V^. Exercises Fiiid to one decimal the roots of the following equations: 1. z' - 3 x2 + 1 = 0. 4. a;" - 3 x^ + 3 = 0. 2. a;3 + 3 X - 7 = 0. 5. a^ + x - 1 = 0. 3. x' + a;2 + X - 1 = 0. 6. x* _ 3 j; _ i = q. Art. 6 Inequalities 11 Art. 61 Inequalities An inequality expresses that one quantity is greater than (>) or less than (<) another. Thus, x^ + 1 >x and (x - 1) (x + 2) < are inequalities. The first of these is an identical inequality (true for all values of x), the second is not. As in equations, terms can be shifted (with change of sign) from one side of an inequality to the other and inequalities having the same sign ( > or <) can be added but not subtracted. Both sides of an inequality can be multiplied or divided by a positive quantity, but the sign must be changed (> to < and < to >) when an inequality is multiplied or divided by a negative quantity. The main problem in inequalities is to determine for what values of the variable an inequality holds. How this is done is best shown by an example. Example. Find the values of x for which 5x'^-x-3 x'' (2 - x) This is equivalent to bx'-x-Z >I. 1 >0 x" (2 - x) {x + 1) (x - 1) {x + 3) >0. x" (2 - x) The problem is to determine the values of x for which the expression on the left is positive. Since x' is always positive, the sign of the expression is determined by the signs of the other four factors. The values of x making one of these factors zero are —3, — 1, 1, 2. Mark -++++++4- ►++ — — -1 o Fig. 6. these values on a scale (Fig. 6). If a; < — 3 the thi-ee factors in the numerator are all negative, and (2 — x) is positive. T^ whole expression, having an odd number of negative factors, is negative. 12 Algebraic Principles Chap. 1 If X is between —3 and —1, there are two negative factors, a; + 1 and X — 1, and the whole expression is positive. If x is between — 1 and +1, the only negative factor is {x — 1) and the expression is negative. If x is between 1 and 2, all the factors are positive and the whole expression is positive. If a; > 2 there is one negative factor, (2 — x), and the expression is negative. The expression is positive when x is between —3 and —1, or be- tween 1 and 2. These conditions are expressed by the inequalities — 3 0. 4. x' - 3 x - 1 < 0. 2. x= > X. 1 1 1 f. 3. x=-2x2 + 2x-l<0. x'^x-2'^x + 2 6. Show that x^ — 3x + 3>0is true for all values of x. 7. Find for what values of x, the value of y is real in the equation x'^ +xy + y' = 1. 8. Find the values of x satisfying both the inequalities x2 > X, x2 > 2. Art. 7. Simultaneous Equations Simultaneous equations in more than one unknown are solved by a process called elimination. This is a name applied to any proc- ess by which equations are found equivalent to the given equations but some of which contain fewer unknowns. By a continuation of this process equations may eventually be obtained each containing a single unknown and these can be solved by the methods already given. In other cases it may not be possible to solve the equations completely but they may be reduced to a simpler form. If nothing in the equations indicates a simpler way, there are three general methods that may be useful: (1) Multiply the equations by constants or variables and add or subtract to get rid of an unknown or to obtain a simpler equation. Art. 7 Simultaneous Equations 13 (2) Solve one of the equations for one of the unknowns and sub- stitute this value in each of the other equations. (3) Between one of the equations and each of the others eliminate the same unknown. Proceed with the new equations in the same way until finally (if possible) one of the unknowns is found. Then determine the other unknowns by substituting this value in the previous equations. However the solutions be found they should be checked by sub- stitution in each of the original equations. Example 1. Solve the simultaneous equations x+ y + 2 = 2, 2x- %j + Sz = 9, 3x + 2y- z= -1. Adding the second to the first and twice the second to the third, 3X + 4:Z= 11, 7x + 5z = 17. Subtracting 5 times the first from 4 times the second of these equa- tions, there is found 13 x = 13, whence x = 1. This value substi- tuted in either of the preceding equations gives z = 2. The values of X and 2 substituted in either of the original equations give y =—1. The solution is x = 1, y = —1, z = 2. These values check when substituted in the original equations. Ex. 2. Solve the equations x'' + y^-2x + 4:y = 21, x'^ + y^ -\- x — y = 12. Subtraction gives 5y — 3x = 9. Hence y = f (x -|- 3). This value substituted in the second equation gives 17x2 + 32x- 132 = 0. The roots of this are 2 and — ff. The corresponding values of y are 3 and —-fj. The solutions are x = 2, y = S and x = — fr, y =—-^7. These values check when substituted in the original equations. 14 Algebraic Pkinciples Chap. 1 Exercises Solve the following simultaneous equations : 1. 4:x-5y+ 6 = 0, 6. x^ + y^ + 2x=0, 7x-9y + n=0. 2/ = 3x + 4. 2. x + 2y-z + 3=0, 7. h' +k'' - 8h + 4k + 20 = f, 2x- y -5=0, K' + k? + Qh->r2k-\-lQ = f\ x + 2z -8 = 0. h? + 8h + lQ = r\ 3. x + 2y + z = 0, 8. x'' + iy'' = 5, X — y — z = 1, xy = —1. ■ 2x+ y-z^O. ., = 1 + 1 i. x + 2y + 3z = 3, y^ z x-2y + 3z = l, =1-1^1 x + 4:y + Qz = 6. '^ ~ z'^ x' 5.- + - = l, z = --\ X y X y 1,1^2 10. a;2 + 2/2 + z2 = 6, y z ' X +y + z =2, 11^ 2x -y +3z =Q. z X Art. 8. Special Cases Inconsistent Equations. — Sometimes equations are inconsistent, that is, have no simultaneous solution. This is usually shown by the equations requiring the same expression to have different values. For example, take the equations x+ y+ z = l, 2x + 32/ + 4z = 5, a; + 22/ + 33 = 3. Elimination of x between the first and second and first and third gives i/ + 2z = 3, y + 2z = 2. Any solution of the original equations must satisfy these. Since the expression y +2z cannot equal both 2 and 3, there is no solu- tion. Dependent Equations. — Sometimes the solutions of part of the equations all satisfy the remaining equations. These last give no added information. Such equations are called dependent. Art. 8 Special Cases 15 For example, take the equations^ X + 2/ = 1, x" - 2/2 + a; + 3 2/ = 2. Substituting 1 — x icx y va. the second equation, it becomes 2 = 2. All the solutions of the first equation satisfy the second. The two equations are equivalent to one equation x -\- y = 1. They have an infinite number of simultaneous solutions. Number of Solutions. — In general, definite solutions are expected if the number of equations is equal to the number of unknowns. Thus, two equations usually determine two imknowns, three equa- tions determine three unknowns, etc. This is, however, not always the case. The equations may be inconsistent and have no solution or may be dependent and have an infinite mmiber of solutions. If the equations determine definite solutions, the number of solutions is expected to equal the product of the degrees of the equations. Special circumstances may, however, change this number. It can be shown that unless the number of solutions is infinite it cannot exceed the product of the degrees of the equations. If there are fewer equations than unknowns, the unknowns will not be determined. In this case, if the equations are consistent, there wUl be an infinite number of solutions. If there are more equations than unknowns, usually there wUl be no solution. In particular cases, however, there may be solu- tions. To determine whether there is a solution, solve part of the equations and substitute the values found in the remaining equa- tions. If any of them satisfy all of the equations, there is a solu- tion, otherwise there is none. Homogeneous Equations. — If all the terms of an equation have the same degree the equation is called homogeneous. A set of homo- geneous equations can often be solved for the ratios of the variables when there are not enough equations to determine the exact values. For example, take the homogeneous equations X — y — s = 0, 3x — y — 2g = 0. Solving for x and y x = \z, y=-is. 16 Algebraic Principles Chap. 1 Any value can be assigned to z and the values of % and y can then be determined from these equations. Let % = 2h. The solution is then x='k, y = —k, z = 2k. Since h is arbitrary, x, y, z have any values proportional to 1, — 1, 2. This result can be written x:y:z = 1: —1:2. Exercises Determine whether the following equations have no solutions, definite solutions, or an infinite number of solutions: 1. X + y + z = \, 6. x~y-\-2z = X, 2x -2y + 5z = 7, ' Sx + y - 3 = 5, 2x-3y+4z = 5. Sx + 2y-3z = 2. 2. x + 2y+ z = 0, 7. 3x+y + l = 0, 2x + y - 2 = 5, 2x + 3y-i = 0, 5x + 7y + iz = 2. x + 2y-3=0. 3. x + 2y + l = 0, 8. X - y =0, 3x-y-i = 0, x^ - 2/2 = 1. 2x + 3y + l=0. 9. {X- yy+{y - zy + {z-xy = 1, 4. s + 2/ - z = 4, x« + 2/2 +z^ =2, x + 2y + 3z = 2, (_x+yy + (y+zy + {z + xy = 3. 5x + 8y + 7z = li. 10. w + x+ y+ z = 1, 5. x + y ~ 5 = 0, w + 2x - 3y - 4:z = 2, 3x + 2y -12 = Q, w - x-2y-3z = 3, 2x + y - 6=0. w-bx+2y + 3z = 3. Find values to which the variables in the following equations are proportional: 11. X +y -2z = Q, 13. X +y - z = 0, 3 X - 2/ - 4 z = 0. x2 ^ 2/2 - 5 z2 = 0. 12. X +2y +. z = 0, 14. x'' + y' = 2 z\ 4 2/ + 3 z = 0. y^ = xz. Art. 9. Undetermined Coefficients It is often necessary to reduce a given expression or equation to a required form. This form is indicated by an expression or equa- tion having letters for coefficients and the reduction is made by calculating the values of these coefficient^. In this work frequent use is made of the following theorem: If two polynomials in one variable are equal for all values of the variable, Art. 9. Undetermined Coefficients 17 the coefficients of the same power of the variable in the two polynomials are equal. To show this, suppose, for all values of x, Oo + aiCK + ajx^ + • • • + a„x" = 60 + hx + fcax^ + • • • + bnX"- Then for all values of x (do -bo) + {ai-bi)x+ • ■ • + (a„ - &J x" = 0. If the coefficients ia this equation are not zero, by Art. 4, it cannot have more than n distinct roots. Hence the coefficients must all be zero and oo = 60, ai = 61, etc., which was to be proved. To reduce an expression to a given form, equate the expression to the given form, clear of fractions or radicals, and determine the un- known coefficients by the above theorem. Example 1. To find the coefficients a and 6 such that X _ '^ 1 ^ (x - 1) (s + 3) ~ a; - 1 "^ a; + 3 clear of fractions, getting, x = a{x + Z)+b{x-l) = {a + b)x + 2,a-b. If this equation holds for all values of x, a + 6 = l, 3a-6 = 0. Hence a = j, 6 = f . Conversely, if a and 6 have these values, the above equations are identically satisfied. Therefore X 1 + 3 (x - 1) (x + 3) 4 (x - 1) ' 4 (x + 3) In many cases the expression can be more easily changed to the required form by simple algebraic processes. This is particularly the case with second degree expressions where completing the square may give the required result. Example 2. To reduce the expression l + 4x-2x2 to the form a — b {x — cY, it can be written l-2(x2-2x) =3-2(x-l)2, which is the result required. 18 Algebraic Principles Chap. 1 In reducing equations to a required form it should be noted that multiplying an equation by a number gives an equivalent equation. Thus, X + y = 1 and 2x + 2y = 2 are equivalent. Two equa- tions are then equivalent when corresponding coefficients are pro- portional. Example 3. To find k such that x + 8y — 1 = and 3x — y — 3 + A; (a; + 3 2/ — 1) = are equivalent, write the last equation in the form (3 + k)x + {3k-l)y-{k + 3) -=0. This is equivalent toai + Sj/ — l = Oif corresponding coefficients are proportional, that is, if 3+k 3k-l _ k+3 1 ~ 8 " 1 ■ These equations are satisfied hy k = —5. Exercises Reduce the following expressions to the forms indicated: 1. 2 x2 + 3 a; + 4 = (ax + 6)2 + c. 2. S + 2x-x' = b~(,x- ay. 3. x'' -\- xy + y^ = a (x + myy + 6 (y — mx)'. X + 1 a , b 4- r?::^ — sr = z + : X {x — 2) X X — 2 ax + b , c (x + 3) (x2 + 1) cc" + 1 ' a; + 3 Reduce the following equations to the forms indicated: 6. 3a; — 42/ = 5, y = mx + b. 7. 2x + 3y = 4, ^ + |=1. 8. dx' + 2y'-6x + iy = l, i^^' + l^^l^' = 1. 9. x'-4y^-ix + 8y = 4, (x - ^' _ (y - fc)' ^ ^^ 10. 3 a; - 2/ + 5 = 0, (x + y - 1) + k {x - y + 3) = 0. Art. 10. Functions It is often desirable to state that one quantity is determined by another. For this purpose the word function is used. A quantity Art. 10 Functions 19 y is called a function of x if values of x determine values of y. Thus, if 2/ = 1 — x^, then t/ is a function of x, for a value of x determines a value of y. Similarly, the area of a circle is a function of its radius; for, the length of radius being given, the area of the circle is deter- mined. It is not necessary that a value of the variable determine a single value of the function. It may be that a limited number of values are determined. Thus, y is a, function of x in the equation ' x^ — 2 xy -\- y'^ + X = 1. To each value of x correspond two definite values of y obtained by solving a quadratic equation. If a single value of the function corresponds to each value of the variable, the function is called single valued. If several values of the function correspond to the same value of the variable the function is called many valued. Kinds of Functions. — Any expression containing a variable is a function of that variable, for, a value of the variable being given, a value of the expression is determined. Such a function is called explicit. Thus VxM-1 is an explicit function of x. Similarly, if y = Vx^ + 1, then y is an expUcit function of x. If X and y are connected by an equation not solved for y, then y is called an implicit function of x. For example, 2/ is an implicit function of x in the equation a;2 + 2/' + 2x + 2/ = l. Also X is an imphcit function of y. ExpUcit and implicit do not denote properties of the function but merely of the way it is expressed. An implicit function is rendered expUcit by solving. For example, the above equation is equivalent to 2/ = i (- 1 ± V5-8x-4x2). A rational function is one representable by an algebraic expression 20 Algebraic Pbinciples Chap. 1 containing no fractional powers of variable quantities. For example, a; V5 + 3 is a rational function of x. An irrational function is one represented by an algebraic expres- sion whicli cannot be reduced to a rational form. For example, Vx + 1 is an irrational function of x. A function is called algebraic if it can be expressed explicitly or implicitly by a finite number of algebraic operations (addition, subtraction, multiplication, division, raising to integral powers, and extraction of integral roots). All the functions previously men- tioned are algebraic. Functions that are not algebraic are called transcendental. For example, x ' and 2^ are transcendental functions of x. The terms rational, irrational, algebraic, and transcendental de- note properties of the function itself and do not depend on the way the function is expressed. Notation. — A particular function of x is represented by the nota- tion / (x), which should be read function of x, or/ of x, not/ times x. For example, / (z) = Vx^ + 1, means that / (a;) is the definite function VxM-1. Similarly, y — f {x) means that y is a definite (though perhaps unknown) function of x. The / in the symbol of a function should be considered as repre- senting an operation to be performed on the variable. Thus, if / (x) = Vx^ + 1, /represents the operation of squaring the variable, adding 1, and extracting the square root of the result. If x is re- placed by any other quantity, the same operation is to be performed on that quantity. For example, / (2) is the result of performing the operation / on 2. With the above value of /, /(2) = V2M^ = V5. Similarly, , /(2/ + l) = V(2/ -f- 1)'' + 1 = V2/2-h22/-l-2. If it is necessary to consider several functions in the same dis- cussion, they are distinguished by subscripts or accents or by the Art. 10 Functions 21 use of different letters. Thus, /i (x), f^ (x), f, {x), f (x), f" (x), f" (x) (read /-one of x, /-two of x, /-three of x, /-prime of x, /-second of X, /-third of x), g (x) represent (presumably) dififerent functions of X. Functions of Several Variables. — A quantity u is called a function of several variables if values of u are determined by values of those variables. For example, the volume of a cone is a function of its altitude and the radius of its base; for the volume is determined by the altitude and radius of base. This is indicated by the notation v=f{h,r), which should be read, w is a function of h and r, orvisfoih and r. Similarly, the volume of a rectangular parallelopiped is a function of the lengths of its three edges. If a, b, and c are the lengths of the edges, this is expressed by the equation v=fia, b,c), which should be read, w is a function of a, b, and c, or vis foi a, b, c. Independent and Dependent Variables. — In most problems there occur a nxunber of variable quantities connected by equations. Arbitrary values can be assigned to some of these quantities and the others are then determined. Those taking arbitrary values are called independent variables; those determined are called de- pendent variables. Which are taken as independent and which as dependent variables is usually a matter of convenience. The num- ber of independent variables is, however, fixed by the equations. Example. The radius r, altitude h, volume v, and total surface S of a cyhnder are connected by the equations V = Tr%, S = 2nr'' + 2 irrh. Any two of these four quantities can be taken as independent vari- ables and the other two calculated in terms of them. If, for ex- ample, V and r are taken as the independent variables, h and S have the values ■ht' r 22 Algebraic Pbinciplbs Chap. 1 Exercises 1. If/ (a;) = a;2 - 3 a; + 2, show that/ (1) = / (2) = 0. 2. lifix) =x + i, find/ (a; + 1). Also find/ (a;) + 1. 3. If/ (x) = Vi^^n, find/ (2 a;). Also find 2/ (x). 4. If/(x)=^3,find/(l). Alsofind^-l-^. 6. If lA (a;) = a;< + 2 x^ + 3, show that i/- (-x) = v^ (x). 6. If (^ (x) = X + -, show that [ (x)]' = 4> (x^) + 2. 7. If F (x) = 1^, show that F (a) F (-a) = 1. 8. If/, (x) = 2\ f, (x) = xS find/i [/a (2/)]. Also find /a [/i (t/)]. 9. If / (x, y) = x^ + 2xy - 5, show that/ (1, 2) = 0. 10. If F (x, y) = x^ + xy+ y\ show that F {x, y) = F (y, x). 11. If / (x, 2/) = x^ + 3 x^y + 2/^, show that / (x, wx) = x'f (1, u). 12. If a, b, c are the sides of a right triangle how many of them can be taken as independent variables? 13. Express the radius and area of a sphere in terms of the volume taken as independent variable. 14. Given u = x^ + y^, v = x + y, determine x and y as functions of the independent variables u and v. 15. If X, y, z satisfy the equations x + y + 2 = 6, x-2/ + 2z = 5, 2x -\-y — z = l, show that none of them can be independent variables. 16. The equations x+ y + z = 6, X- ^ + 2z= 5, 2X + 42/+ z = 13 are dependent. Show that any one of the quantities x, y, z can be taken as independent variable. 17. If u, V, X, y are connected by the equations V? + uv — y = 0, uv -\- X ~ y = 0, show that u and x cannot both be independent variables. CHAPTER 2 RECTANGULAR COORDINATES Art. 11. Definitions Scale on a Line. — In Art. 1 it has been shown that real numbers can be attached to the points of a straight line in such a way that the distance between two points is equal to the difference (larger minus smaller) of the numbers located at those points. The line with its associated numbers is called a scale. Proceed- ing along the scale in one direction (to the right in Fig. 11a) the -5-1-3-2-10133 15 Fig. 11a. numbers increase algebraically. Proceeding in the other direction the numbers decrease. The direction in which the numbers in- crease is called positive, that in which they decrease is called negative. Coordinates of a Point. — In a plane take two perpendicular scales X'X, Y'Y with their zero points coincident at (Fig. 116). It is customary to draw X'X, called the X-axis, horizontal with its positive end on the right, and Y'Y, called the y-axis, vertical with its positive end above. The point is called the origin. The axes divide the plane into four sections called quadrants. These are numbered I, II, III, IV, as shown in Fig. 11&. Fig. 11&. From any point P in the plane drop perpendicularsFM, PN to the axes. Let the number at Af in the scale X'X be x and that at N in the scale Y'Y be y. These num- 23 II N ■Y ] P X' y M IT X III > 24 Rbctangulab Coordinates Chap. 2 bers X and y are called the rectangular coordinates of P with respect to the axes X'X, Y'Y. The number x is called the abscissa, the number y is called the ordinate of P. If the axes are drawn as in Fig. 116, the abscissa x is equal in magnitude to the distance PN from P to the y-a,xis, is positive for points on the right and negative for points on the left of the j/-axis. The ordinate y is equal in magnitude to the distance PM from P to the X-axis, is positive for points above and negative for points below the X-axis. For example, the point P in Fig. 116 has coordinates x = 3, 2/ = 2, whUe Q has the coordinates x = —3, y = —2. Notation. — The point whose coordinates are x and y is represented by the symbol (x, y). To signify that P has the coordinates x and y the notation P (x, y) is used. For example, (1, —2) is the point X = 1, 2/ = —2. Similarly, A (—2, 3) signifies that the abscissa of A is —2, and its ordinate 3. Plotting. — The process of locating a point whose coordinates are given is called plotting. It is convenient for this purpose to use coordinate paper, that is, paper ruled with two sets of lines as in Fig. lie. Two of these perpendicular lines are taken as axes. A certain number of divisions (one in Figs, lie, d and e) are taken as representing a unit of length. This unit should be long enough to make the diagram of reasonable size but not so long that any points to be plotted fail to lie on the paper. By counting the rulings from the axes it is easy to locate the point having given coordinates (approximately if it does not fall at an intersection). Example 1. Show graphically that the points A (-1, -3), B (0, -1), C (1, 1) and D (2, 3) lie on a straight line. The points are plotted in Fig. lie. By apply- ing a ruler to the figure it is found that a straight line can be drawn through the points. Ex. 2. Plot the points Pi (-2, -1), P^ (3, 4) and find the distance between them. The points are plotted in Fig. lid. Let the horizontal line through Pi cut the vertical line through A Fig. lie. Art. 11 Definitions 25 P2 in R. From the figure it is seen that PiR = 5, RP2 = 5. Consequently, P1P2 = Vp,E^ + BP22 = VSO. r R, / / / / / X ^ « Fig. lid. Fig. lie. Ex. 3. Plot thepoints A (-1, -1), 5 (4, -3), C (6, 2) and find the area of the triangle ABC. The points are plotted in Fig. lie. The triangle ABC is part of a rectangle CDEF with sides 5 and 7 and area 5 X 7 = 35. ADC, AEB and BFC are right triangles whose areas are (one- half base times altitude) 10^, 5 and 5 respectively. The area of ABC is then CDEF - ADC - AEB - BFC = 35 - 20* = 14*. Exercises 1. Plot the points A (4, 4), i3 (-4, 4), C (-4, -4), D (4,-4) and* E (V2, VS). 2. What are the algebraic signs of the coordinates in each of the four quadrants? 3. If a point lies on the x-axis what is its ordinate? If it lies on the 2/-axis what is its abscissa? Where are aU the points for which x = 1? for which y = —21 4. Let a and 6 be any given numbers. How is (a, 6) related to each of thepoints ( — a, b), (a, —6), (—a, —6), (6, a), (a + 1, b), (a, 6 — 2)? 5. Plot the points (0, -2), (1, 1), (2, 4), (-1, -5) and verify from the figure that they He on a Une. 6. Show graphically that the points (5, 0), (3, —4), ( — 4, 3) and (0, —5) lie on a circle. Find its center and radius. 7. Construct the point equidistant from ( — 3, 4), (5, 3) and (2, 0). Determine its coordinates by measurement. 8. Plot the points A (1, 2) and B (3, 4). Let the horizontal line 26 Rectangular Coordinates Chap. 2 through A meet the vertical line through B in P. What are the co- ordinates of PI What are the lengths of AP and BPl Calculate the distance AB. 9. Find the distance between A (—2, —3) and B (3, —4). 10. Find the area of the triangle formed by the points (—3, 4), (5, 3) and (2, 0). Art. 12. Segments In this book the term line (meaning straight line) is used only when referring to the infinite line extending indefinitely in both directions. The part of a line between two points will be called a segment. In many cases a segment is regarded as having a definite direction. The symbol AB is used for the segment beginning at A and ending at B. The segment beginning at B and ending at A is written BA. The value of a segment may be ainy one of three things that should be carefully distinguished. Whenever the symbol AB occurs in an equation it must be understood which of these things is meant. (1) In many cases the value AB means the length of the segment. In this case AB = 3 means that AB has a length of three units and AB = CD means that AB and CD have equal lengths. (2) In other cases the symbol AB represents the length together with a sign positive or negative according as the segment is directed one way or the other along the line. For example, the a;-co6rdinate of a point is equal to the segment NP (Fig. 116) considered positive when drawn to the right, negative when drawn to the left. The value of the segment is in this case a number with a positive or negative sign. (3) Certain physical quantities, such as velocities, include in their description the direction of the hnes along which they occur as well as their magnitudes and directions one way or the other along those lines. Two segments representing such quantities are equal only when they have the same length and direction. The value of such a segment (called a vector) is not a number but a number and direction. Segments Parallel to a Coordinate Axis. — In most cases the value of a segment parallel to a coordinate axis will be the number equal Art. 13 Projection 27 in magnitude to the length of the segment and positive or negative according as the segment is drawn in the positive or negative direc- Y N. Pi Y Fi P. N, P2 —X M, M,' X ] Fig. 12a. Fig. 1 2b. tion of the axis. That is, horizontal segments are positive when drawn to the right, negative when drawn to the left; vertical seg- ments are positive when drawn upward, negative when drawn down- ward. Let Pi (xi, 2/1) and P2 {xi, 2/2) be the end points of a segment P1P2. If the segment is parallel to the a;-axis (Fig. 12a) P1P2 = M1M2 = X2 - Xi, (12) for the difference Xz — Xi is equal in absolute value to the distance M1M2 and is positive when M2 is on the right of Mi, negative when on the left. Similarly, if P1P2 is parallel to the 2/-axis (Fig. 126), then P1P2 = N1N2 = 2/2 - yi. (12) Therefore in length and sign a segment parallel to a coordinate axis is represented by the difference obtained by subtracting the coordi- nate of the beginning from the coordinate of the end of the segment. Art. 13. Projection The projection of a point A on a line MN is the foot of the per- pendicular from A to MN. The projection of a segment AB is the segment A'B' of MN intercepted between perpendiculars from the ends of AB. Since parallels intercept proportional distances on two lines, the lengths of two segments on a line have the same ratio as their projections. Furthermore if two segments have the same direction along a line their projections have the same direction. 28 Rbctangulah Coordinates Chap. 2 For example, in Fig. 13a or 136, in both magnitude and sign, AB: BC = A'B': B'C. a' M P^ C B' N Fig. 13&. In using this formula it should be noted that if AB and BC are distances, their projections must be distances and if AB and BC have algebraic signs their projections must have algebraic signs. For example, if the segments are projected on the coordinate axes and the values of the projections determined by equation (12), AB and BC must be considered opposite in sign when they have opposite directions. Example 1. Find the middle point of the segment joining Pi {xi, yi) and P2 (x^, 1/2). Let P (x, y) be the middle point. Project on the axes (Fig. 13c). Since P is the middle point of P1P2, in both length and direction PiP =PP2. Hence MiM = MM2, NiN-=NN2, and so X — xi = Xi — X, y — yi = yi — y. Consequently, Fig. 13c. x = i{xi + X2), 2/ = ^J/i + 2/2). The sum of several quantities divided by their number is called the average. Hence each coordinate of the middle point is the average of the corresponding coordinates of the ends. Ex. 2. Given Pi (-1, 1), P2 (3, 4), find the point P {x, y), on P1P2 produced, which is twice as far from Pi as from P2 (Fig. 13d). T Jfj R ^^ N P ^ ^ ^1 P ^ Jk f. J I Jk fj X Art. 14 Distance between Two Points 29 Since PiP is twice as long as PP2 and they have opposite direc- tions, PiP= -2PP2. Consequently, MiM= -2 MMi, NiN = - 2 NN2, and so a;+l=-2(3-x), j/-l = - 2(4 - j/). -Pif Hence x = 7, y — 7. Y B- / /- y P. ^ ii„ / /' ^ / "I MiO Ka MX Fig. 13d. Exercises 1. On the line through A ( — 3, —4) and B (4, 2) find the point two- fifths of the way from A to B. 2. The points A (-1, -4), B (0, -1) and C (2, 5) He on a line. Find the point P, on AC produced, such that AB = CP. 3. On the line through A (1, — 1), B (3, 5) find two points each of which is twice as far from A as from B. 4. The points A (-4, 9), B (2, 0), C (4, -3), D (0, 3) lie on a line. Find the ratio of the segments AB and CD. Do these segments have the same or opposite directions? 5. On the line through A (2, — 1) parallel to the line through B (1, 1) and C (4, 5) find two points each of which is at the distance BC from A. 6. Given three points, Pi (xi, yi), P2 (x2, 2/2), P3 {xs, 2/3), find the middle point Q of P1P2, then find the point R one-third of the way from Q to P3. Show that each of its coordinates is the average of the corre- sponding coordinates of Pi, P2 and P3. Art. 14. Distance between Two Points Let the points be Pi (xi, 2/1) and P2 (xi, 2/2) (Fig. 14a or 146). Let the projections of P1P2 on the axes be M1M2 and A''iA''2. Let the lines PiNi and P2M'2 intersect in R. In the right triangle PiRP^ P^P, = VPifl^ -I- RP2\ Now PiR = M1M2 and the distance between two points of a scale is equal to the positive difference of their coordinates. Conse- quently, Pifi2 = {X2 - XiY. Similarly, RPi" = {yi - yi)\ There- fore ■ (14) P1P2 = V(x2 - xiy + (2/2 - y,y. 30 Rectangular Coordinates Chap. 2 In this formula, since X2 — a;i is squared it is immaterial whether it is written x^ — Xi or xi — Xa. Similarly, instead of 2/2 — 2/1 can P. / Nt I / R ■A fi li h X Fig. 14a. Fig. 146. be put' 2/1 — 2/2- The formula expresses that the distance between two points is equal to the square root of the sum of the squares of the differences of corresponding coordinates. Example 1. Find the distance between the points A (—1, 1) and B (1, -2). By the formula AB = V(l + l)2+(-2-l)'' = Vl3. Ex. 2. Find the points 2 units distant from the x-axis and 5 units distant from the point A (1, 2). Let P (x, y) be such a point (Fig. 14c). Since PA = 5 (a; -1)2 +(2/ -2)2 = 25. Since the point is two units distant from the x-axis 2/= ±2. Fig. 14c. Substitution of 2 for 2/ in the pre- vious equation gives a; = 6 or —4. Substitution of —2 for y gives a; = 4 or —2. There are conse- quently four points Pi (6, 2), Pj (-4, 2), P3 (-2, -2), P4 (4, -2) which satisfy the conditions. Ex. 3. Find a point equidistant from the three points, A (9, 0), B (-6, 3) and C (5, 6). ?3 -n Art. 14 Distance between Two Points 31 Let P {x, y) be the point required (Fig. 14d). Since PA = PB V{x - 9)2 + 2/2 = VCa; + 6)^ + {y - Z)\ Y Squaring and cancelling, 5 a; — 2/ = 6. Similarly, since PA = PC, 2x-3y = 5. S 1 V i n g simultaneously gives X = l,y = —1. The required point is therefore Fig. Ud. (1, -1). c f — Exercises 1 . Find the perimeter of the triangle whose vertices are (2, 3), ( — 3, 3) and (1, 1). 2. Show that the points (1, —2), (4, 2) and ( — 3, —5) are the vertices of an isosceles triangle. 3. Show that the points (0, 0), (3, 1), (1, -1) and (2, 2) are the vertices of a parallelogram. 4. Given A (2, 0), B (1, 1), C (0, 2) show that the distances AB, BC and AC satisfy the equation AB + BC = AC. What do you conclude about the points? 5. Show that (6, 2), ( — 2, —4), (5, —5), ( — 1, 3) are on a circle whose center is (2, —1). 6. It can be shown that four points form a quadrilateral inscribed in a circle if the product of the diagonals is equal to the sum of the prod- ucts of the opposite sides. Assuming this, show that ( — 2, 2), (3, —3), (1, 1) and (2, 0) he on a circle. 7. Find the coordinates of two points whose distances from (2, 3) are 4 and whose ordinates are equal to 5. 8. Find a point on the a;-axis which is equidistant from (0, 4) and (-3, -3). 9. Find the center of the circle passing through (0, 0), (—3, 3) and (5, 4). 10. Given A (0, 0), B (1, 1), C (-1, 1), D (1, -2), find the point in which the perpendicular bisector of AB cuts the perpendicular bisec- tor of CD. 11. Find the foot of the perpendicular from (1, 2) to the line joining (2, 1) and (-1, -5). 32 Rectangular Coordinates Chap. 2 Fig. 15o. Art. 15. Vectors A vector is a segment of definite length and direction. The poiQt Pi is the beginning, the point P2 the end of the vector P1P2. The direction of the vector along its line is often indicated by an arrow as shown in Fig. 15a. Two vectors are called equal if they have the same length and direction. For example, in Fig. 15a, P1P2 = P3P4- If two vectprs have the same length but opposite directions, either is called the negative of the other. For ex- ■ ample, P1P2 = -P4P3 = -P2P1. Let the projections of P1P2 on the coordinate axes be M1M2 and NiNi (Fig. 156). The x-component of the vector P1P2 is defined as the length of M1M2 or the negative of that length, according as MiMi has the positive or negative direction along the a;-axis. Similarly, the y-component is the distance iViA''2 or the negative of that distance, according as N1N2 is drawn in the positive or negative direction along the 2/-axis. For example, in Fig. 156, the components of P1P2 are 3 and 4, while those of P2P3 are —7 and 3. Let the points be Pi {xi, 2/1), P2 {xi, 2/2). In Art. 12 it has been shown that in magnitude and sign MiMi and iViA^2 are represented by X2 — Xi and 2/2 — 2/1. Hence x^ — Xi is the x-component and 2/2 — 2/1 is the 2/-component of P1P2. That is, the components of a vector are obtained by subtracting the coordinates of the beginning from the corresponding coordinates of the end of the vector. If two vectors are equal their components are equal. For let P1P2 = P3P4 (Fig. 15c). Then, since by definition of equaUty PxPi and P3P4 have the same length and direction, the triangles p Y ^ ^ •^ N2 7 i h / < , / ..jZ N, Pi Fig. 156. Mo Art. 15 Vectors 33 Y JV4 Pi T'- N-3 ./ / / B S r/ ^. , ^ M, iU, ■M, M, Fig. 15c. Consequently, the P1RP2, PsSPi are equal and cor- responding sides have the same direction. Consequently, in both length and sign, M1M2 = PiR = P3S = MsMi, NiNi =RP2 = SPi = N3Ni. Conversely, if the components are equal, the triangles are equal and corresponding sides have the same direction vectors are equal. Notation. — In this book the vector whose components are a and b win be represented by the symbol [a, 6]. To signify that the vector u has the components a and b, the notation u [a, b] will be used. For example, P1P2 = [—2, 3] means that the vector P1P2 has an ar-component equal to —2 and a 2/-component equal to 3. If the points Pi and P2 are {xi, 2/1) and {x2, 2/2), the components of P1P2 are X2 — xi and 2/2 — 2/1 and P1P2 = [X2 - xi, 2/2 - 2/1] • Example 1. Construct a vector equal to [-2, 3]. The vector OP from the origiu to the poibt P (-2, 3) is 0P= [-2-0,3-0] = [-2,3]. Y P Fig. I5d. Hence OP is a vector of the kind required (Fig. 15d). Ex. 2. Show that the points A (1, 3), B (2, 1), C (3, 4), D (4, 2) form the vertices of a parallelogram. The vectors AB and CD are equal. For AB= [2-1, 1-3] = [1,-2], CD= [4-3,2-4] = [1,-21. That is, AP and CD are parallel and have the same length. Consequently ABCD is a paral^ lelogram (Fig. 15e). w Fig. 15e. 34 Rectangular Coordinates Chap. 2 Fig. 15/. Ex. 3. Find the area of the triangle PP1P2, given PPi = [ai, 61], PP2 = [02, 62]. In Fig. 15/ the sides of the ^2"^i shaded triangles have their lengths marked on them. The area of the triangle PP1P2 is equal to the area of the whole rectangle less the sum of the areas of the shaded triangles. Hence PP1P2 = 0162 — J ai6i — \ aj)i — i («i — ^2) (62 — &i) = 2 (o^i^z — «2&i)- This result is shown for a particular figure. By drawing other figures it will be found that the result is always correct if the angle from PPi to PP2 is positive (that is, drawn in the counter-clockwise direction). If that angle is negative the formula gives the negative of the area. Exercises 1. If the vectors AB and CD are equal show that AC and BD are equal. 2. If AB = AiBi, BC = BiCi, show that AC = A,Ci. 3. The components of a vector are a, 6. Show that the components of its negative are —a, —b. 4. Show that the vector [a, b] is equal to the vector from the origin to the point (o, 6). 5. Construct vectors equal to [2, 3], [-2, 3], [-2, -3], and [2, -3]. 6. Show that the points P (-1, 2), Q (1, -2), R (3, 4), S (5, 0) are the vertices of a parallelogram. 7. A vector equal to [ — 3, 4] begins at the point (1, —2). What are the coordinates of its end? 8. The points (1, 2), (—2, —1), (3, —2) are the vertices of a triangle. Find the vectors from the vertices to the middle points of the opposite 9. Given A (2, 3), B (-4, 5), C (-2, 3), fmd D such that AB = CD. 10. The middle point of a certain segment is (1, 2) and one end is ( — 3, 5). Find the coordinates of the other end. 11. By showing that the area of the triangle ABC is zero show that the points A (0, -2), B (1, 1) and C (3, 7) he on a hne. 12. Find the area of the quadrilateral whose vertices are the points (-2, -3), (1, -2), (3, 4), (-1, 5). Art. 16 Multiple of a Vector 35 13. Show that the vectors [oi, h], [ch, bi] are parallel if and only if 01/02 = bi/62. In this way show that the hne through ( — 1, 2) and (3, 4) is parallel to that through (1, —1) and (9, 3). Art. 16. Multiple of a Vector If tt is a vector and r a real number, the symbol ru is used to rep- resent a vector r times as long as u and having the same direction if r is positive, but the opposite di- rection if r is negative. In Fig. 16, let u = [a, b] and V = ru. Since u and v are parallel their components have lengths pro- portional to the lengths of u and v. „ Also corresponding components have the same or opposite signs according as u and v have the same or opposite directions. Hence the components of v are ra and rb. That is, V = [ra, rb\. Hence to multiply a vector by a number, multiply each of its com- ponents by that number. Example 1. Given Pi (— 2, 3), P2 (1, —4), find a vector having the same direction as PiPz but 3 times as long. The vector required is 3PiP2 = 3[3, -7] = [9, -21]. Ex. 2. Find the point on the line joining Pi (2, 3), P2 (1, —2) and one-third of the way from Pi to P2. By hypothesis PiP = \ P1P2, that is, [x - 2, 2/ - 3] = H-1, -5] = [-i -|]. Hence x — 2 = —\, y — 3 = —§, and consequently a; = j, v = i- Ex. 3. Show that the segment joining the points A (1, —1), B (5, 7) is parallel to and twice as long as that joining C (4, 3), D (2, -1). 36 Rbctangulak Coordinates Chap. 2 The vectors are A5 = [4,8], CD = [-2, -4]. Hence AB = — 2 CD. Consequently the segments are parallel and the first is twice as long as the second. Art. 17. Addition and Subtraction of Vectors Sum of Two Vectors. — Draw a vector equal to v, beginning at the end of u. The vector from the beginning of u to the end of v is called the simi of u and v. Let u = [oi, 6i], V = [02, 62]. From the diagram it is seen that the components oi u + v are ai + ^2 and 61 + 62- This is true not Y A \ ^ \ u X Fig. 17a. Fig. 17b. only in Fig. 17a but also in Fig. 176, for there a^ is negative and Ci + ai is in absolute value equal to the difference of the lengths of the projections. Hence u + v = lai + 02, bi + 62], that is, vectors are added by adding corresponding components. Example 1. Show that u + v = v + u. The sum u + v results when v is put at the end of u, while v + u is given by putting u at the end of v. The two are equal since both are equal to the vector diagonal of the parallelogram whose sides are u and v (Fig. 17c). Ex. 2. Show that {u + v) + w = u + {v + w). In the expression {u + v) + w the sum of u and v is added to w, while in u + {v + w), u '^ added to the sum of v and w. In Fig. 17c. Art. 17 Addition and Subtraction of Vectobs Fig. 17d let u = AB,v = BC,w = CD. Then (.u + v)+w = AC + CD = AD, u + (v + w) = AB + BD = AD. The two sums are consequently equal. 37 r \u-v 1 ^f I v-'^ / X B Fig. 17d. Difference of Two Vectors. — Draw vectors equal to u and v, beginning at the same point ^'°' ^''^' (Fig. 17e). The vector BQ from the end of v to the end of u is called the difference u — v. Let M = [fli, 6]], w = [02, 62]. It is seen from the diagram that the components of m — w are 02 — a\ and 62 — 61. Consequently, M — y = [02 — fill, 62 — 61], that is, vectors are subtracted by subtracting corresponding components. Example 1. Show that m+(— ti) = u — v. In Fig. 17e, -w = 5A. Hghcg „ + (_„) = AC + 5A = BA + AC = BC = u - w. Ex. 2. Two segments equal to [2, —3] and [3, —1] extend from the point A (2, 2). Find their ends and the vector connecting their ends. In Fig. 17/; by hypothesis, OA =[2,2], APi= [2,-3], AP2= [3,-1]. Hence OPi=OA + APi= [2 + 2, 2 - 3] = [4, -1], OP2=OA + AP2= [2 + 3, 2 - 1] = [5, 1]. Consequently Pi and P2 are (4, —1) and (5, 1). Also P1P2 = AP2 - APi = [3 - 2, -1 + 3] = [1, 2]. Ex. 3. If weights Wi, 102, ws, etc., are located at the points Pi, P2, Ps, etc., the center of gravity of these weights can be defined as the Y A / V ~^ p„ / \ s \ X ki Fig. 17/. 38 Rectangular Coordinates Chap. 2 point P satisfying the equation wiPPi + W2PP2 + WiPPz + etc. = 0. Find the center of gravity of three weights of 2, 3 and 5 pounds placed at the points Pi (-2, 1), Pi (1, -3), P3 (4, 5) respectively. The center of gravity P satisfies the equation 2PPi + 3PP2 + 5PPs = 0, that is, 2[- 2 - x,l - y] + Zll - X, - 3 - y] + 514: - X, 5 - y] = [19-lOx, 18-10?/] = 0. Consequently the coordinates of the center of gravity are a; = 1.9 and y — 1.8. Exercises 1. Given P (1, -3), Q (7, 1), R (-1, 1), S (2, 3), show that PQ haa the same direction as RS and is twice as long. 2. Given Pi (2, -3), P2 (-1, 2), find the point on P1P2 which is twice as far from Pi as from P2. Also find the point on P1P2 produced which is twice as far from Pi as from P2. 3. Find the points P and Q on the line throughPi (2, -1), P2 (-4, 5) if PiP = -f PP2, PiQ = -f P1P2.. 4. One end of a segment is (2, —5) and a point one-fourth of the dis- tance to the other end is ( — 1, 4). Find the coordinates of the other end. 5. Given the three points A ( — 3, 3), B (3, 1), C (6, 0) on a line, find the fourth point D on the line such that AD: DC = — AB: BC. 6. Show that the line through ( — 4, 5), ( — 2, 8) is parallel to that through (3, -1), (9, 8). 7. If the vectors AC and AB satisfy the equation AC = rAB, where r is a real number, show that A,B,C lie on a line.. In this way show that (2, 3), (-4, -7) and (5, 8) lie on a Une. 8. Given A (1, 1), B (2, 3), C (0, 4), find the point D such that BD = 2 DC. Show that AD = i [AB + 2 AC]. 9. In Ex. 8 find the point P such that PA + PB + PC = 0. 10. If V is any vector let v^ mean the square of the length of v. Given vi = [2, 5], Vi = [10, —4], show that (vi + ws)^ = Vi^ + vi and conse- quently that Wi and v^ are perpendicular to each other. 11. Show that the vectors from the vertices of a triangle to the middle points of the opposite sides have a sum equal to zero. 12. Find the center of gravity of two weights of 1 and 4 pounds placed at the points (3, —4) and (2, 7) respectively. Art. 18 Slope of a Line 39 13. Show that the center of gravity of four equal weights placed at the vertices of a quadrilateral is the middle point of the segment joining the middle points of two opposite sides of the quadrilateral. Art. 18. Slope of a Line Let Pi (xi, 2/0, Pa {x2, yi) be two points of the line MN. The ratio 2/2- m = Vi X2 — Xi (18a) is called the slope of the line MN. The same value of the slope is obtained whatever pair of points on the line is used. For, if J^ Fig. 18a. Fig. 186. Pi, P2 and Pi, Pi are pairs of points on the same line, the triangles P1RP2 and P3SP4 are similar and Vi-yi ^ RP2 ^ SP4 ^ Vi-yz X2 — xi PiR P3S Xi — X3 Let

. (18b) Hence the slope of a line is equal to the tangent of the angle from the positive direction of the x-axis to the line. Since the x-axis is horizontal, the steeper the line the nearer is the 40 Rbctangtjlak Coordinates Chap. 2 angle to 90° and consequently the greater the slope. The slope is thus a measure of the steepness of the line. If the liae extends upward to the right, as in Fig. 18a, the com- ponents PiR and RP^ have the same sign and the slope is positive. If the hne extends upward to the left, as in Fig. 186, the components have opposite signs and the slope is negative. Parallel Lines. — If two lines are parallel, they have the same slope; for the angles i^i and 02 (Fig. 18c) are then equal and their slopes, tan i and tan ^2, are equal. Conversely, if the slopes are equal, the angles are equal and the lines are parallel. Y yD yB X^ \^- ^ \o Fig. 18c. Fig. \U. Perpendicular Lines. — In Fig. 18d, let the lines AB and CD be per- pendicular. Since the exterior angle is equal to the sum of the two opposite interior angles <^2 = 01 + 90°. 1 Consequently tan 02 = — cot 0i = tan 01 Conversely, if this relation holds, 0i and 02 differ by 90° and the lines are perpendicular. If mi and m^ are the slopes, mi = tan 0i, m2 = tan 02, and mi= -—• (18c) mi ^ ' Two lines are perpendicular if and only if the slope of one is equal to the negative reciprocal of the slope of the other. Angle between Two Lines. — Let two lines L\ and L2 make with the a;-axis the angles 01 and 02 (Fig. 18e). If /3 is the angle from Li to Art. 18 Slope of a Line 41 L2 (positive when measured ia the counter-clockwise direction), then Hence tan /3 = tan (02 — <^i) tan <^2 — tan <^i 1 + tan <^i tan (^2 If Wi and rrii are the slopes of the lines, mi = tan <^i, m^ = tan 4>i, whence WI2 — OTi Fig. 18e. tan;8 ■■ 1 + TOim2 (18d) In using this formula it should be remembered that p is measured from Li to L^. Example 1. Show that the line through A (1, —3), B (5, —1) is parallel to that through C (-3, -2), D (-1, -1). -1 + 3 1 5-1 ~ 2' -1+2 ^1 -l+3~2' Since the slopes are equal the lines are parallel. Ex. 2. Show that the line through ^ (2, 3), B (—4, 6) is perpen- dicular to that through C (1, 1), D (2, 4). Slope of IS = ^ ~ ^ ^ Slope of AB = slope of CB = -4-2 slope of CD = 2-1 = 3. Since either slope is the negative reciprocal of the other, the lines are perpendicular. Ex. 3. Find the angles of the triangle formed by the points A{-\,2),B{1, 1), C(3, 4). The triangle is shown in Fig. 18/. The slopes of AB, BC, CA are foimd to be — ^, | and J respectively. Between two lines such as AB and AC are two positive angles less than 180°. One of 42 Rectangular Coordinates Chap. 2 these is interior, the other exterior to the triangle. From the figure the positive interior angle A is seen to extend from AB to AC. Consequently, 3 Y ^ •^ A ^ ^ / ' ^ V / B tan A = 1 + H-i) In the same way it is found that tan B = —8, tan C = -^. The angles of the triangle are therefore the positive angles less than 180° whose tangents are f, —8 and f . The angles A and C are acute, B is obtuse. Fig. 18/. Exercises 1. Find the slopes of the sides of the triangle formed by the points (-1, 2), (1, 3) and (2, -4). 2. Find the angle from the x-axis to the Une joining (1, 1) and (-4, 5). 3. The angles from the x-axis to three lines are 45°, 120° and —30° respectively. Find the slopes of the lines. 4. The sides of a triangle have slopes equal to §, 1 and 2. Show that the triangle is isosceles. 5. If the slopes of AB and BC are equal, show that A, B, C lie on a line. In this way show that the points (4, 1), (1, 2) and ( — 5, 4) lie on a line. 6. Show that the line through (1, —3) and ( — 1, 3) is parallel to that through (3, -5) and (0, 4). 7. Show that the lines determined by the pairs of points (2, 3), (3, 5) and (1, 7), (3, 6) are perpendicular to each other. 8. Showthat the vectors [a, h] and [c, i] are parallel if h/a = d/c and perpendicular if h/a = ~c/d. 9. Find the interior angles of the triangle formed by the points (2, 3), (-4, 5) and (1, -2). 10. Lines join the point ( — 2, 2) to the points ( — 3, 1), (0, 2) and (1, —1). Show that one of these bisects the angle between the other two. 11. Two Lines have slopes equal to 2 and —3. Find the slopes of the two bisectors of the angles between these lines. 12. Find the angle between the lines joining the origin to the two points of trisection of the segment joining ( — 2, 3) and (1, —7). 13. The slope of AB is |. Find the slope of CD if the angle from AB to CD is 30°. Art. 19 Graphs 43 14. By showing that the angles CAD and CBD are equal show that the points A (6, 11), B (-11, 4), C (-4, -13), D (1, -14) lie on a circle. 15. A point is 7 units distant from the origin and the slope of the Hne joining it to (3, 4) is |. Find its coordinates. 16. A point is equidistant from (2, 1) and ( — 4, 3) and the slope of the line joining it to (1, — 1) is |. Find its coordinates. Art. 19. Graphs In many cases corresponding values of two related quantities are known. Each pair of values can be taken as the coordinates of a point. The totality, or locus, of such points is called a graph. This graph exhibits pictorially the relation of the quantities repre- sented. There are three cases differing in the accuracy with which intermediate values are known. Statistical Graphs. — Sometimes definite pairs of values are given but there is no information by which intermediate values can be even approximately inferred. In such cases the values are plotted and consecutive points connected by straight lines to show the order in which the values are given. Example 1. The temperature at 6 a.m. on ten consecutive days at a certain place is given in the following table: Jan. 12 3 4 19° 27° 11° 40° 5 6 7 8 9 10 34° 28° 36° 18° 42° 38° These values are represented by points in Fig. 19ffl. Points repre- senting temperatures on consecutive days are connected by straight lines. At times between those marked no estimate of the temper- ature can be made. Physical Graphs. — In other cases it is known that in the intervals between the values given the vari- ables change nearly proportionally. In such cases the points are plotted I 50° B 30° . 20 s / s / \ / / ^ / \ / \ / Jan. 12 3 16 Fig. 10a. and as smooth a curve as possible is drawn through them. Points 44 Rectanqotab Coordinates Chap. 2 2 1 G 8 10 12 li IC 18 20 Time Fia. 19&. of this curve between those plotted are assumed to represent approximately corresponding values of the variables. Ex. 2. The observed temperatures 5 of a vessel of cooling water at times t, in minutes, from the beginning of observation were < = 1 2 3 5 7 10 15 20 e = 92° 85.3° 79.5° 74.5° 67° 60.5° 53.5° 45° 39.5° These values are plotted in Fig. 196. Since the temperature de- creases gradually and during the intervals given at nearly constant rates a smooth curve drawn through these points will represent approxi- mately the relation of temperature and time throughout the experi- ment. The Graph of an Equation. — In other cases the equation connect- ing the variables is known. The graph then consists of all points whose coordinates satisfy the equation. Arbitrary values are as- signed to either of the variables, the corresponding values of the other vari- able calculated, and the resulting points plotted. When the points have been plotted so closely that between consecu- tive points the curve is nearly straight, a smooth curve is drawn through them. Ex. 3. Plot the graph of the equa- tion y = x'-. In the foUowiujg table values are as- signed to X and the corresponding values of y calculated: a; = -5 -4 -3 -2 -1 1 2 3 4 5 y= 25 16 9 4 1 1 4 9 16 25 The points are plotted in Fig. 19c. The part of the curve shown extends from x— — h\a x=-\-h. The whole curve extends to an indefinite distance. Horizontal lines cut the Y \ -< H - ! ( > i ( i i: Fig. 19c. Art. 19 Graphs 45 Fig. 19d. the curve at equal distances right and left of the y-axis. This is expressed by saying that the curve is symmetrical with respect to the 2/-axis. Ex. 4. The force of gravitation between two bodies varies in- versely as the square of their distance apart, that is, F = k/d^, k being constant. Assuming k = 10, plot the curve representing the relation of force and distance. In the following table values are assigned to d and the corresponding values of F calculated : d = 0.5 1 1.5 2 3 5 10 F=oo 40 10 4.4 2.5 1.1 0.4 0.1 The distance d cannot be negative. For very large values of d, F is very small. Hence when d is large the curve very nearly coincides with the horizontal axis. When d is very small F is very large and the curve very nearly coincides with the vertical axis (Fig. 19d). Units. — Before a graph can be plotted a length must be chosen on each axis to represent a unit measure of the quantity represented by that coordinate. If the quantities represented by x and y are of different kinds, for example temperature and time, these lengths can be chosen independently. If, however, these quantities are of the same kind it is usually more convenient to have the same unit of length along both axes. This is also the case in graphing mathematical equations where no physical interpretation is given to X and y. Other things being equal, a large curve is better than a small one. Such units of length should then be chosen as to make the curve spread both vertically and horizontally very nearly over the region available for it. If several graphs are to be compared, the same unit lengths should be used for all. Exercises 1. The price of steel rails each year from 1892 to 1907 was as follows, in dollars per gross ton: 30 28 24 24 28 19 18 28 32 27 28 28 28 28 28 28 Make a graph showing the relation of year and price. (Let x be the 46 Rectangular Coordinates Chap. 2 number of years beyond 1892 and y the amount that the price exceeds 18 dollars.) 2. The amplitudes of successive vibrations of a pendulum set in motion and left free were Number of vibration 12 3 4 5 6 7 Amplitude 69 48 33.5 23.5 16.5 11.5 8 Plot points representing these pairs of values and draw a smooth curve through them. Do points of this curve between those plotted have any physical meaning? 3. The atomic weight, W, and specific heat, S, of several chemical elements are shown in the following table: TF = 7 9.1 11 12 23 28 39 55 56 108 196 S = 0.94 0.41 0.25 0.147 0.29 0.177 0.166 0.122 0.112 0.057 0.032 Make a graph showing the relation of specific heat and atomic weight. 4. The table below gives the maximum length of spark between needle points of an alternating current under standard conditions of needles, temperature and barometric pressure. L = length in millimeters. V = electromotive force in kilovolts. F = 10 15 20 25 30 35 40 45 50 60 70 80 L = 12 18 25 33 41 51 62 75 90 118 149 180 Make a curve showing the relation of voltage and length of spark. 5. In the table below are given the maximum vapor pressures of water at various temperatures, ■yhere T = temperature in degrees Centi- grade and P = pressure in centimeters of mercury: T = 10 20 30 40 50 60 70 80 90 100 P =0.46 0.91 1.74 3.15 5.49 9.20 14.9 23.3 35.5 52.5 76 Make a graph showing the relation of temperature and vapor pressure. 6. The rate for letter postage is two cents for each ounce or fraction. Make a graph showing the relation of weight and cost of postage. (The cost for 1.5 ounces is the same as that for 2, etc.) 7. Make a graph showing the number of centimeters y in x inches. 8. Make a graph showing the relation between the side and the area of a square. 9. According to Boyle's law the pressure and volume of a gas at con- stant temperature are connected by the equation pv = constant. Assuming the constant equal to 10, construct the curve representing the relation of pressure and volume. Can p or v he negative? 10. Plot the curve y = x' — 2 x + 3. Show that it passes through the point (1, 2). Art. 20 Equation of a JLocus 47 11. On the same diagram plot the curves y = ar* and y = {x + ly. How are they related? Show that they both pass through the point (-4, tV)- 12. On the same diagram plot the curves a; = l/y^andx = l/{y — 2y. How are they related? Show that both pass through (1, 1). 13. On the same diagram plot the graphs of x + y = 2 and 2x — 3y = 1. What are they? Find their point of intersection. (It must have coordinates satisfying both equations.) Art. 20. Equation of a Locus If a locus and an equation are such that (1) every point on the locv,s has coordinates that satisfy the equation and (2) every point whose coordinates satisfy the equation lies on the locus, then the equa- tion is said to represent the locus and the locus to represent the equation. Usually a locus is defined by a property possessed by each of its points. Thus a circle is the locus of points at a constant distance from a fixed point. To find the equa- tion of a locus express this property by means of an equation connecting the coordinates of each locus point. The graph is constructed by using the defini- tion or by plotting from the equation. Example 1. Find the equation of a circle with center C (— 2, 1) and radius equal to 4. Let P {x, y) (Fig. 2Qa) be any point on the circle. By the definition of a circle, CP equivalent to V{x + 2y +{y- ly = 4, P (.x,y) Fig. 20a. 4, and this is which is the equation required. By squaring this can be reduced to the form (x + 2y + {y- ly = 16. Ex. 2. A curve is described by a point P {x, y) moving in such a way that its distance from the a-axis equals its distance from the point Q (1, 1). Construct the curve and find its equation. 48 Rectangular Coordinates Chap. 2 'P (x,y) Fig. 20b. The curve is constructed by locating points such that MP = QP. Since no point below the a;-axis can be equally distant from Q and the a;-axis, the curve lies entirely above the .r-axis. Hence y is positive and equal to the distance MP. Also QP = V{x - ly + (y - ly. Con- sequently, if P {x, y) is any point on the curve, 2/ = V(a; - 1)2 + (2/ - V,\ Conversely, if this equation is satis- fied, the point P is equidistant from Q and the a;-axis. and so hes on the curve. It is therefore the equation of the curve. Since y cannot be negative the equation is equivalent to 2/2 = (a; - 1)2 + (2/ - 1)2, and consequently to a;2-2a;-22/ + 2 = 0. Exercises 1. What loci are represented by the equations, (a) a; = 3, (b) y = — 2, (c) 2/ = 2 X, (d) x^+y^ = 1? 2. The point P (x, y) is equidistant from (1, 2) and (3, —4). What is the locus of PI Find its equation. 3. The point P {x, y) is twice as far from the s-axis as from the 2/-axis. What is the locus of P? (Two parts.) Find its equation. 4. The slope of the Une joining P (x, y) to (—2, 3) is equal to 3. What is the locus of P? Find its equation. 5. Given A (-3, 1), B (2, 0), P {x, y). If the slopes of AB and BP are equal what is the locus of P? Find its equation. 6. The point P (x, y) is equidistant from the 2/-axis and the point (—3, 4). Construct the locus of P. Find its equation. 7. Given A (2, 3), P (-1, 1), C (2, -3), P {x, y). If P moves along the line through A perpendicular to BC, show that PC^ - PP2 = AC2 - AB'. Find the equation of the line described by P. 8. Find the equation of a circle with center ( — 1, 2) and radius 5. 9. Find the equation of the circle whose diameter is the segment joining (2, —3) and ( — 1, 4). Art. 21 Point on a Locus 49 10. Given Pi (1, 3), Pi (4, —1), find the equation of the locus de- scribed by P {x, y) if the sum of its distances from Pi and Pa is 5. Free the equation of radicals and show that the result is (4 X + 3 2/ - 13)2 = 0. Make a graph of this equation. Do all points of this graph belong to the locus? 11. Given A (2, -2), B (6, 0), C (7, 3), P (x, y). If the angle from AB to .4 C is equal to the angle from PB to PC, show that the locus of P is the circle through A, B, C. Calculate the tangents of the two angles, equate them and so get the equation of the circle. Art. 21. Point on a Locus If a point lies on a locus its coordinates satisfy the equation of the locus. Hence to ascertain whether a point lies on a given locus, substitute its coordinates in the equation of the locus and find out whether the equation is satisfied. Example 1. Show that the curve x^ -\- y^ = 2x passes through the origin. Substituting the values x = 0, y = the equation becomes = 0. Since the equation is satisfied the curve passes through the origin. In the same way it could be shown that any locus, rep- resented by a polynomial equation with no constant term, passes through the origin. Ex. 2. If the curve x'' -\- y^ = 2 ax passes through the point (2, —1) find the value of the constant a. The coordinates 2, — 1 must satisfy the equation. Hence 2'+ (-1)2 = 2a (2). Consequently, a = |. Ex. 3. If the locus oi y = mx + b passes through (—1, 1) and (3, 2), find the values of m and b. The conditions required are 1= -m + b, 2 = 3m + b. The solution of these equations is m = J, 6 = |. Intersection of Curves. — An intersection of two curves must have coordinates satisfying both equations. Hence to find the points of 60 Rectangular Cocjrdinates Chap. 2 Fig. 21a. intersection, solve tlie equations simultaneously. All solutions should be checked by substitution. Example 1. Plot the curves represented by the equations y^ = 2x and x^ = 2y and find their points of intersection. The curves are shown in Fig. 21a. From the figure it is seen that there are two real points of intersection. Squaring the first equation and substi- tuting the value of x'^ from the second, Consequently, y {y^ — 8) = 0. The two real solutions are y = 0, 2. The correspondiag values of x are a; = 1 2/^ = 0, 2. The points of intersection are then (0, 0) and (2, 2). Substitution shows that the coordinates of these points satisfy both equa- tions. Ex. 2. Plot the curves xy = 2, x^ + y^ = A: x, and find their points of intersec- tion. The curves are, shown in Fig. 216. It is seen that there are two points of in- tersection. Elimination of y gives x*-4: x^ 4-4 = 0. The equation can only be solved approximately. The figure indi- cates that the solutions are near a; = 1.2 and a; = 4. By substitu- tion the following values are found: a;= 1.1 1.2 3.9 4 a;4_4a;3 + 4= 0.14 -0.84 -1.9 4 There is a root between 1.1 and 1.2 and another between 3.9 and 4. When X = 1.15 the expression a;* — 4 a;' -F 4 is found to be negative. \ ^s^=-^ t ::S~:^ 1 2 3 i^ :i=s;=:^ ^-------tI \ Fig. 216. Art. 22 Tangent Curves 51 N \^ ^ il ^ ^ \ \ / \ N \ / \ s I \ s. \ / s \ \ ^ ^ \ Fig. 21c. Hence, to one decimal, the value 1.1 is a root. Similarly 3.9 is found to be the other root. The corre-sponding values of y are 2/a; = 1.8 and 0.6. The points of inter- section are then approximately (1.1, 1.8), (3.9, 0.6). "Ex. 3. Plot the loci of the three equa- tions a; + 2/ = l, 2a; + 32/ = 5, x^^t = \Z, and show that they pass through a point. The graphs are shown in Fig. 21c. The solution of the first two equations is X = —2,y = 3. These values satisfy the third equation. Consequently all the loci pass through the pouit(-2,3). Art. 22. Tangent Curves If at a point of intersection two curves have the same direction, they are called tangent. Let the curves AB and CD be tangent at P (Fig. 22a) . CD can be considered as the limiting position of a curve CD' cutting AB ia two points Pi and Pa close together. The equations of AB and CD' have two simultaneous solu- tions that are nearly the same. As CD' approaches CD these solutions approach equality. One might then expect that if the equations of AB and CD are solved simultaneously two of the solutions will be equal and that, conversely, equal roots occur- ring in the simultaneous solution of two equations indicate tan- gency. This should however be checked by the graphs as there are other circumstances that result in coincident solutions. Example 1 . Show that the line x + y = 2 and the circle x'' + y^ = 2 are tangent and find the point of tangency. EUminating y, a;2 -1- (2 - a;)2 - 2 = 0. Fig. 22a. 52 Rectangular Coordinates Chap. 2 This equation is equivalent to 2 (x - 1)^ = and so has two equal roots X = 1. The corresponding value oi y is 2 — x = 1. The , line arid circle intersect in only one point (1, 1). They must therefore be tangent at (1, 1). Ex. 2. Plot the graphs of y^ = x^ — x^ and y = Zx and find their intersections. The graphs are shown in Fig. 226. The first is a curve Uke a horizontal figure 8. The second is a straight r C^s; tz-A ::::^ X -:^: -^•s-/ Fig. 226. line through the origin. Eliminating y, This equation has two roots a; = 0. Two parts of the curve pass through the origin and both cut the line at that point. The curve and line are not however tangent. Exercises 1. A, B, C, D, E being any fixed numbers, show that the curve whose equation is Ax' + Bxy + Cy^ + Dx + By = passes through the origin. 2. If xi, 2/1, m are constant show that the locus of the equation y - yi = m(x - xi) passes through the point (xi, yi). 3. The curve x^a' + y'/b' = 1 passes through the points (0, 1) and (2, 0). Find the values of a and 6 and plot the curve. 4. The curve y^ = ax + b passes through the points (0, —1) and (1, 2). . Find the values of a and 6 and plot the curve. Plot the following pairs of loci and find their points of intersection: 5. 3x + 2y = I, 8. y' = x + 1, 2 X — y = 0. xy = 2. a;2 + 1/2 = 10, 9. xy' = 1, 2y- 3x = 3. 2x -y = 3. a;2 - 2/2 = 3, 10. x' -2j/2 = 7, x' + xy + y' = 7. 2y + 3x =7. 6. 7. Art. 22 Tangent Curves 53 11. 2a;2+ !/2 = 6, 13. 2/ = 2s', X — y = 1. y = 3 x' — 5. 12 xh/ + 3y = 4, 14. x^ + 4 ?/ = 0, X +2y = 3. x'^ + y^ + 6x + 7 = 0. 15. Show that the following curves have a common point: a?+y^ = lQ, 2/3= a; + 4, 2/ = ^^. CHAPTER 3 STRAIGHT LINE AND CIRCLE Art. 23. Equation of a Straight Line Let Pi (xi, ?/i) be a fixed point and P {x, y) a variable point on the line MN. If the line is not perpendicular to the x-axis (Fig. 23a), let its slope be m. Since the line passes through Pi and P, by the definition of slope, m = X — X\ This is an equation satisfied by the coordinates, x and y, of any point P on the line. Conversely, if the coordinates of any point Y X P aj, P, X Pig. 23a. Fig. 236. P (x, y) satisfy this equation, the slope of PiP is m and consequently P lies on MA''. It is therefore the equation of MN. It can be written y-yi = m,ix- Xi). (23a) This is consequently the equation of a line through {xi, yi) with slope m. Let the line MN cross the y-ama at (0, h). Replacing Xi, yi in equation (23a) by 0, 6, the equation becomes y = mx + h. (236) The quantity 6 is called the intercept of the line on the y-axis. It is 54 Art. 23 Equation of a Straight Line 55 numerically equal to the distance along the ?/-axis to the Une, is positive when the line is above the origin and negative when it is below. Equation (236) represents the line with slope m and inter- cept 6 on the 2/-axis. If the line is perpendicular to the x-axis (Fig. 236) its slope is infinite and equations (23a) and (236) cannot be used. In this case the figure shows that X = Xx. (23c) Conversely, if the abscissa of a point is Xi, it lies on the line. There- fore (23c) is the equation of a line through (xi, 2/1) perpendicular to the X-axis. Example 1. Find the equation of a line through (—1,2), the angle from the s-axis to the line being 30°. The slope of the line is m = tan (30°) = i V^. The equation of the line is then 2/ - 2 = i V3 (x + 1). Ex. 2. Find the equation of the line through the points (2, 0) and (1, —3). The slope of the line is =^^ = 4-3. 1-2 ^ Since the line passes through (2, 0) and has a slope equal to 3, its equation is 1/ — = 3 (x — 2), whence 2/ = 3x — 6. Ex. 3. Find the equation of the line through (1, —1) perpen- dicular to the line through (2, 3) and (3, —2). The slope of the line through (2, 3) and (3, -2) is -5. The slope of a perpendicular line is -l/(-5) = i. The equation of the line with this slope passing through (1, —1) is 2/ + l = i(x-l), which is the equation required. 56 Straight Line and Circle Chap. 3 Ex. i. Show that the equation 2x — 3y = 5 represents a Straight line. Find its slope. Solving for y, 2/ = fa;-|. Comparing this with the equation y — mx + b, it is seen that the two are equivalent if m = f , & = —^. Therefore the given equa- tion represents a straight line with slope f and intercept — | on the 2/-axis. Exercises 1. The angle from the a;-axis to a line is 60°. The line passes through ( — 1, —3). Find its equation. 2. Find the equation of the line through (2, —1) and (3, 2). 3. Find the equation of the line through (2, 3) and (2, —4). 4. Find the equation of the line through (1, 2) parallel to the a;-axis. 5. Find the equation of the perpendicular bisector of the segment joining ( — 3, 5) and ( — 4, 1). 6. An equilateral triangle has its base in the a;-axis and its vertex at (3, 5). Find the equations of its sides. 7. A line is perpendicular to the segment between ( — 4, —2) and (2, —6) at the point one-third of the way from the first to the second point. Find its equation. 8. Find the equation of the line through (3, 5) parallel to that through (2, 5) and (-5, -2). 9. One diagonal of a parallelogram joins the points (4, —2) and (—4, —4). One end of the other diagonal is (1, 2). Find its equation and length. 10. A diagonal of a square joins the points (1, 2), (2, 5). Find the equations of the sides of the square. 11. The base of an isosceles triangle is the segment joining ( — 2, 3) and (3, —1). Its vertex is on the y-axis. Find the equations of its sides. 12. Show that the equation 2x — y = S represents a straight line. Find its slope and construct the line. 13. Show that the equations 2x + Sy = 5, 3x -2y = 7 represent two perpendicular straight lines. 14. Perpendiculars are dropped from the point (5, 0) upon the sides of the triangle whose vertices are the points (4, 3), (—4, 3), (0, —5). Show that the feet of the perpendiculars lie on a line. Art. 24 First Degree Equation 67 Art. 24. First Degree Equation Any straight line is represented by an equation of the first degree in rectangular coordinates. In fact, if the line is not perpendicular to the X-axis, its equation has been shown to be V -yi = m{x- xi), xi, 2/1 and m being constants. If it is perpendicular to the x-axis its equation is X = Xi. Since both of these equations are of the first degree in x and y, it follows that any straight Line has an equation of the first degree in rectangular coordinates. Conversely, any equation of the first degree in rectangular coordi- nates represents a straight line. For any equation of the first de- gree in X and y has the form Ax + By + C = 0, (24) A,B,C being constant. If B is not zero, this equation is equivalent to A C y=-^x~-, which represents a line with slope —A/B and intercept —C/B on the y-ajds. If B is zero, the equation is equivalent to C which represents a line perpendicular to the x-axis passing through the point {—C/A, 0). Hence in any case an equation of the first degree represents a straight fine. Graph of First Degree Equation. — Since a first degree equation represents a straight line, its graph can be constructed by finding two points and drawing the straight line through them. The best points for this purpose are usually the intersections of the line and coordinate axes. The intersection A (Fig. 24a) with the x-axis is found by letting y = and solving the equation of the line for the 58 Straight Line and Circle Chap. 3 corresponding value of x. Similarly, the intersection B with the 2/-axis is found by letting x = and solving for y. The abscissa of Y S,B_ o "s X Y o y Fig. 24a. Fig. 246. A and the ordinate of B are called the intercepts of the line on the coordinate axes. If the line passes through the origin (Fig. 246) the points A and B coincide at the origin and it is necessary to find another point on the line. This is done by assigning any value to one of the coordinates and calculating the resulting value of the other coordinate. Example 1. Plot the line 2 a; + 3 ?/ = 6 and find its intercepts on the axes. Substitutiag y = Q gives a; = 3 and substituting x — Q gives y = 2. Hence the liae passes through the points A (3, 0) and B (0, 2). Its intercept on the a-axis is 3 and on the 2/-axis 2. Ex. 2. Construct the line whose equation is 2a; — 3?/ = 0. When X is zero y is zero. The line then passes through the origin. When y = \, x = %. Hence the line OPi (Fig. 246) through the origin and the point (|, 1) is the one required. Slope of a Line. — If the line is perpendicular to the a;-axis, its slope is infinite. If it is not perpendicular to the a;-axis, its equation is y = mx + 6. The slope is the coefficient of x in this equation. Consequently, if the equation of a line is solved for y, its slope is the coefficient of x. Example 1. Find the slope of the Une 3 a; — 5 2/ = 7. Solving for y y=^%x-i. The slope of the line is therefore f . Ex. 2. Find the angle from the line x + y = Z to the line 2/ = 2 a; + 5. Art. 24 First Degree Equation 59 The slope of the first line is — 1, that of the second 2. The angle /3 between the lines is determined by the equation The negative sign signifies that the angle is negative or obtuse. Ex. 3. Find the equation of the line through (3, 1) perpendicular to the line 2 a; + 4 2/ = 5. The given equation can be written 2/ = — | a; + |. Its slope is consequently —J. The slope of a perpendicular line is 2. The equation of the liae through (3, 1) with slope 2 is y-l=2ix-3), which is the equation required. Exercises Plot the straight lines represented by the following equations, find their slopes and intercepts: 1. 2j/-3 = 0. 5. 3a;-6y + 7 = 0. 2. 5x + 7 = 0. 6. 2x + 5y + 8 = Q. 3. x + y = 2. 7. ix + 3y = 0. 4. 2x + dy-5 = 0. 8. 3x-42/ = 6. 9. Show that the equation {2x + Zy-l){x-7y + 2) =0 represents a pair of Hnes. 10. Show that (x + iyy = 9 represents two parallel lines. 11. Show that x^ = (j/ — 1)2 represents two lines perpendicular to each other. 12. Show that the hnes 3x + 4:y -7 = 0, 9a; + 12j/-8 = are parallel. . 13. Show that the lines x + 2y + 5=0, ix - 2y - 7 = a,Te perpendicular to each other. 14. Find the interior angles of the triangle formed by the Hnes X = 0, x.-y + 2 = 0, 21 + 32/- 21 =0. 15. Find the equation of the line whose intercepts on the x and y axes are 2 and —3 respectively. 16. Find the equation of the line whose slope is 5 and intercept on the 2/-axis —4. 17. Find the equation of the Une through (3, — 1) parallel to the line X — y = S. 60 Straight Line and Circle Chap. 3 18. Find the equation of the line through (2, — 1) perpendicular to the hne 9x-8y + 6-0. 19. Find the projection of the point (2, —3) on the line a; — 4 ?/ = 5. 20. Find the equation of the line perpendicular to 3a; — 5^ = 9 and bisecting the segment joining ( — 1, 2) and (4, 5). 21. Find the lengths of the sides of the triangle formed by the lines 4:x + 3y-l=0, 3x-y-4 = 0, x + 4:y - 10 = 0. 22. A line passes through (2, 2). Find its equation if the angle from it to 3 X - 2 i/ = is 45°. 23. Find the equation of the line through (4, |) and the intersection of the lines 3x - iy - 2 = 0, 9a;-ll2/-6=0. 24. Find the equation of the line through the intersection of the lines 2a; — ^ + 5=0, x + y + 1 = 0, and the intersection of the lines X - y + 7 =0, 2x + y-5 = Q. 25. Find the locus of a point if the tangents from it to two fixed circles are of equal length. 26. What angle is made with the axis of j/ by a straight line whose equation is ly + i x = 11 Art. 25. The Expression Ax + By + C At each point P of the plane a first degree expression Ax+By + C has a definite value obtained by putting for x and y the coordinates of P. Thus at the point (1, 2) the expression has the value A + 2 B + C. Points where the expression is zero constitute a line whose equation is Ax + By + C = 0. If the point P moves slowly the value of the expression changes continuously. A number changing continuously can only change sign by passing through zero. If the point P does not cross the line the expression cannot become zero and so cannot change sign. Therefore at all points on one side of the line Ax + By + C = the expres- sion Ax + By + Chas the same sign. Fig. 25a. Example 1. Determine the region in which x + y — 1>0. The equation x + y — 1 = represents the line LK (Fig. 25a). Art. 26 Distance from a Point to a Line 61 At all points on one side of LK the expression then has the same sign. At (1, 1) a; + 2/ - 1 = 1 + 1 _ 1 = 1 which is positive. It is seen from the figure that (1, 1) is above LK. Hence, at all points above LK, x + y — l\s positive. At the origin x+y-l =0+0-1 = -1 which 'is negative. The origin is below the line. Hence at all points below LK the expression x + 2/ — 1 is negative. The region in which a; + 2/ — l>Ois therefore the part of the plane above the line. Ex.2. Determine the region in which x + y>Q, x + 2y-2 Q above (1), x + 2y — 2 <0 below (2) and x — y — l be the angle from OX to LK. Then (a) DPi = QPi cos * = (MPi - MQ) cos <#.. From the figure it is seen that (6) MPi = 2/1 Fig. 26a. 62 Straight Line and Circle Chap. 3 Since Q is on the line LK, its coordinates, x\ and MQ, must satisfy the equation of LK. Therefore and consequently (c) Axi + B • MQ + C = 0, Axi + C MQ B id) The slope of LK is tan = —A/B, whence B cos = Substituting the values from (6), (c), (d) in (a), Axi + Bj/i + C DPt = VA2 + B^ (26) Equation (26) gfiVes i/ie distance from the point (xi, j/i) to i/ie line whose equation is Ax + By + C = 0. The distance being positive, such a sign must be used in the denominator that the result is positive. Example 1. Find the distance from the point (1, 2) to the line 2x -By = 6. The distance from any point (xi, 2/i) to the line is by (26) 2xi-3y,-Q ±-\/l3 The distance from (1, 2) is then 2 (1) - 3 (2) - 6 10 Fig. 26b. =Vl3 Via Ex. 2. The lines (1) y - x - 1 = 0, (2) x + y - 2 = 0, (3) x + 2y + 2 = determine a triangle ABC. Find the bisector of the angle A between the lines (1) and (2) (Fig. 266). The bisector is a locus of points equidistant from the lines (1) Art. 26 Distance from a Point to a Line 63 and (2). If {x, y) is any point of the bisector, x and y must then satisfy the equation y — X — 1 _x + y — 2 ±V'2 ±V'2 The signs must be so chosen that these expressions are positive at points inside the triangle. At the origin these expressions become — 1/(±V2), — 2/(±V2). Hence the negative sign must be used in both denominators. The bisector required is therefore y — X — 1 X -\-y — 2 5. y -2x> 1, y -2x <3, 2y + x>l, 2y+x<3. 6. ix + 2y-3)(2x- 7. (x + 42/)2>9. -V2 -V2 When sunpUfied this becomes a; = J. Exercises Determine the region occupied by points satisfying the inequalities in each of the foUowiag cases: 1. 2x + 3y-Q>0. 2. X <3y. 3. X ~ y ~ I > 0, y -2x >0. 4. 2x-y-2>0, 6. (x + 2 ?/ - 3) (2 a; - i/ + 3) > 0. 3a; + 42/-12<0, 2 2/ - 1 > 0. 8. Inside the triangle determined by the lines x-\-y = 0, 2x — 3y — 1=0, y — 2 = 0, what algebraic signs have the expressions x -\- y, 2x-3y-\,y -21 9. Express by inequalities the inside of the triangle determined by the points (1, 1), (3, 4), (2, -2). 10. Find the distance from the point (3, 5) to the line t/ = 4 s — 8. 11. Fiud the distance from (6, —2) to the line through ( — 1, 3) and (5, -1). 12. Find the distance between the two parallel lines, 4a;+32/ — 10=0 and 4 a; + 3 1/ - 8 = 0. 13. Find the equation of the bisector of the acute angle between the lines 2x — y = \, x-\-3y = 2. 14. A triangle is formed by the lines 3a; — 42/ = 5, 4a; + 32/ = 5, 5 re + 12 2/ = 13. Find the center of the inscribed circle. 15. Find the locus of a point whose distance from (2, 3) is equal to its distance from the line x + 2y = 3. 64 Straight Line and Circle Chap. 3 16. The sum of the distances from the point P (x, y) to the lines 2/ = a;, a; + 2/ = 4, y + 2 = 0, is 4. Find an equation satisfied by the coordinates of P. Do aU points whose coordinates satisfy this equation have the required property? Art. 27. Equation of a Circle A circle is the locus of a point at constant distance from a fixed point. The fixed point is the center of the circle and the constant distance is its radius. Let C {h, k), Fig. 27, be the center of the circle and r its radius. If P (x, y) is any point on the circle r = CF = V{x-hy+(y-ky, or (x - hy +{y- ky = r\ (27a) Y P ix,v) Fig. 27. This is an equation satisfied by the coordinates of any point on the circle. Conversely, if the coordinates x, y satisfy this equation, the point P is at the distance r from the center and consequently lies on the circle. Therefore it is the equation of the circle. Example 1. Find the equation of the circle with center (—2, 1) and radius 3. In this case, h = —2, k = 1, r = 3. By (27a) the equation of the circle is then (x + 2)2 +(2/ -1)2 = 9. Ex. 2. Find the equation of the circle with center (1, 1) which passes through, the point (3, —4). The radius is the distance from the center to the point (3, —4). Consequently r = V(3-l)2+(-4-l)2 = Vig. The equation of the circle is therefore (x - ly +iy- ly = 29. Form of the Equation. — Expanding (27a), xi + tf-2hx- 2ky+h'^ + k^-r^ = 0. Art. 27 Equation of a Circle 65 This is an equation of the form A iz' + y'')+Bx + Cy + D = 0, (276) in which A, B, C, D are constant. Conversely, if A is not zero, an equation of the form (276) repre- sents a circle, if it represents any curve at all. To show this divide by A and complete the squares of the terms containing x aiid those containing y separately. The result is E' + C^-iAD h^lj+h^lj If the right side of this equation is positive, it represents a circle with center (-B/2 A, -C/2 A) and radius Vb^ + C - iAD/2 A. If the right side is zero, the radius is zero and the circle shrinks to a point. If the right side is negative, since the sum of squares of real numbers is positive, there is no real locus. Example 1. Find the center and radius of the circle 2a;'' + 22/2-3a; + 42/ = 1. Dividing by 2 and completing the squares, {x - lY + (2/ + 1)^ = \%. The center of the circle is (|, —1) and its radius is \ V33. Ex. 2. Determine the locus of a;2-|-2/'' + 4a;-62/-|-13 = 0. Completing the squares, {x + 2Y + (2/ - 3)== = 0. The sum of two squares can only be zero when both are zero. The only real point on this locus is then (—2, 3). The circle shrinks to a point. Ex. 3. Discuss the locus of x2 + 2/2 + 22/ + 3 = 0. Completing the squares, x^ + (2/ + 1)^ = -2. There are no real values for which this is true. The locus is imaginary. 66 Straight Line and Circle Chap. 3 Art. 28. Circle Determined by Three Conditions When a circle is given, h, k, r, in equation (27a), have definite values. Conversely, if values are assigned to h, k and r, a circle is determined. This is expressed by saying that the circle is deter- mined by three independent constants. Any limitation on the circle, such as requiring it to pass through a point or be tangent to a line, can be expressed by an equation or equations connecting h, k and r. Now three numbers are determined by three independent equations. This is expressed by saying a circle can be found satis- fying three independent conditions. Thus a circle can be passed through three points, can touch three lines, etc. Many problems consist in finding a circle doing tlu-ee things. To find the equation of such a circle, express the three conditions by equations connect- ing h, k and r, solve and substitute the values in equation (27a) . Example 1. A circle is tangent to the s-axis, passes through (1, 1) and has its center on the line y = x — 1. Find lis equation. The circle is shown m Fig. 28o. Since the circle is tangent to the X-axis, k = ± r. Since it passes through (1, 1), the circle lies above the X-axis and k is positive. Hence (a) k = r. Since the center is on the line y = x — 1, its coordinates must satisfy that.equation. Consequently Fi«- 28«. (6) k = h-l. Since the circle passes through (1, 1), (c) (1 - hy + (1 - ky = r\ The solution of (a), (6) and (c) is /i = 2, fc = 1, r = 1. The equa- tion required is then (x - 2y +{y- ly = 1. Ex. 2. Find the equation of the circle through (2, 3), (4, 1) and tangent to the hne 4 a; — 3 i/ = 15 (Fig. 286). Since the circle is tangent to the line its center is at a distance r Art. 28 Circle Determined by Three Conditions from the line. The distance from the center Qi, A:) to 4 x 15 is by equation (26) Ah-3k-15 67 32/ = (a) ±5 Since the circle passes through (2, 3) its center is on the same side of the line as (2, 3). To make (a) positive at the center and therefore at (2, 3), the negative sign must be used in the denominator of (a). The condition of tangency is then 4/i-3/<;- 15 r %^~^ A ihM~^ / V X y / (6) Fig. 286. -5 Since the circle passes through (2, 3) and (4, l), (c) (2 - hy + (3 - ky = r\ id) (4 - hy + (1 - ky = r'. Solving equations (6), (c) and (d) simultaneously, we get h = 2, k = l, r = 2 a.ndh = W, k = W", r = |§. These are consequently two circles satisfying the given conditions. Their equations are {x - 2)2 + (j/ - Vf = 4, (x - ^ii-Y + (2/ - W-)^ = (11)^ Ex. 3. Find the circle through the three points (0, 1), (—1, —1) and (2, 0). The coordinates of the three points must satisfy the equation of the circle. Hence (0 - hy + (1 - ky = r\ i-l-hy+i-l-ky = r% (2 - hy + (0 - ky = fK Subtracting the first and third of these equations from the second we get 2h + ik + l =0, 6h + 2k-2 = 0. 68 Stbaight Linh and Circle Chap. 3 The solution of these equations is h = ^, k = —\. These values substituted in either of the original equations give r^ = -5" . The circle required is therefore (a;-§)' + (2/ + i)^ = -V-. Exercises Find the center and radius of each of the following circles. Draw the curves. 1. x^ -{■ •y' == 25. 5. x^ ■\- y' = X -\- y. 2. a;2 + 2/2 = 4 a;. &. x^ - 2ax Jr y^ - 2ay = 0. 3. 2 a;2 + 2y^ -62/ + != 0. 7. x' + y' - 2 x + iy + 2 = 0. 4. 3 a;2 + 3 ^2 ^ 4 X = 1. 8. x' + y" - 2x + 1 = 0. 9. What locus is represented by the equation x'^-\-y'^-\-2x + 2y + 2 = 0? 10. What is the locus of the equation x^+y^-6x + 6y + 9=0? 11. Find the equation of the circle through (—2, 4) and having the same center as the circle x^ + y^ — bx + iy — 1 = 0. 12. Find the equation of the circle whose diameter is the segment joining ( — 1, —2) and (3, 4). 13. Find the equations of the circles through (1, 2) and tangent to both coordinate axes. 14. Find the equations of the circles with centers at the origin and tangent to the circle x' + y' — 4:x + 4y + 7 = 0. 15. Find the intersections of the circles a;2 + j/2 = 2a; + 2^, a;2 -I- 2/2 -I- 2 a; = 4. 16. Find the equation of the cu:cle through the three points (0, 3), (3, 0) and (0, 0). 17. Find the circles of radius 5 passing through the points (2, —1) and (3, -2). 18. The center of a circle passing through (1, —2) and ( — 2, 2) is on the line 8x — 4:y + 9 = 0. What is its equation? 19. A circle passes through the points (0, 0), (2, —2) and is tangent to the line 2/ + 4 = 0. Find its equation. 20. A circle passes through the points ( — 1, 0), (0, 1) and is tangent to the line x — y = 1. Find its equation. 21. A circle is tangent to the Unes a; = 3, a; = 7, and its center is on the line y = 2x + i. What is its equation? 22. Find the equation of the circle circumscribed about the triangle formed by the three lines a;+2/ — 2 = 0, 9a; + 5j/ — 2 = 0, 2/ + 2a; -1=0. Art. 28 Circle Determined by Three Conditions 69 23. Find the equation of the circle inscribed in the triangle formed by the lines x + y = l, y — x = l, x — 2y = l. 24. Find the locus of points from which the tangents to the circles a;2 -j- 2/2 = 4 and x'-{-y^ — 2x + iy — i are of equal length. 25. By subtracting the equation x^+y^ — 2x — 2y = from the equation x^+y' + 2x — 6y + 2 = the equation of a line is ob- tained. Show that this line is the common chord of the two circles. 26. A point moves so that the sum of its distances from two vertices of an equilateral triangle is equal to its distance from the third vertex. Find an equation satisfied by its coordinates. Do all points whose co- ordinates satisfy this equation have the required property? CHAPTER 4 SECOND DEGREE EQUATIONS Art. 29. The Ellipse // a circle is deformed in such a way that the distances, of its points from a fixed diameter are all changed in the same ratio, the resulting curve is called an ellipse. For example, in Fig. 29a, if each point Pi of the circle is moved to a point P such that MP/MPi = k = constant, the locus of P is an ellipse. Let the center of the circle be the origin and the fixed diameter the s-axis. Let Xi, j/i be the co- ordinates of Pi and X, y the coordi- nates of P- Then xi = x, 2/1 = MPi = MP/k = y/k. Fig. 29a. If a is the radius of the circle, its equation is ^i'' + yi^ = a'. Replacing Xi and yi by their expressions in terms of x and y, x' + T-„ = d" is found to be the equation of the ellipse. Dividing by a^ and re- placing ka hy h, the equation of the ellipse becomes + 62 1. (29) When y is zero, x is ± o, and, when x is zero, 2/ is ± &. Hence a and 6 are the distances, OA and OB, intercepted on the axes. 70 Art. 30 The Ellipse in Other Positions 71 Fig. 296. Since the equation of an ellipse contains only the squares of x and y, to each value of x correspond two values of y differing only in sign. A line perpendicular to the X-axis then cuts the curve in two points P, P' (Fig. 296) equi- distant from the axis. Similarly, a line perpendicular to the y-axis cuts the curve in two points P, P" equidistant from the 2/-axis. This is expressed by saying that the curve is symmetrical with respect to both of the coordinate axes. They are called the axes of the curve. Any line through cuts the curve in two points P (x, y), P'" {—X, —y} equidistant from 0. For this reason the curve is called symmetrical with respect to the origin and the point is called the center of the curve. The ellipse cuts the axes in four points A', A, B', B. The seg- ments A' A and B'B are sometimes called the axes of the curve. The longer of these is called the major axis, the shorter the minor axis. The distances OA and OB, equal to a and b, are called the semi-axes of the curve. The ends of the major axis are called vertices. Art. 30. The Ellipse in Other Positions Equation (29) represents an elUpse whose axes are the cpordinate axes. The equation can be stated in a form valid in any iposition. In fact, since x = NP, y = MP, a = OA, h = OB, equation ( 29) is equivalent to / NP^ MP^ OA" "*" Ofi2 1, (30a) Fig. 30a. that is, the ratios, obtained by dividing the squares of the distances of any point on the ellipse from the axes by the squares of the parallel semi-axes, have a sum equal to 1. 72 Second Degree Equations Chap. 4 For example, if the center of the ellipse is the point (h, k) and the axes of the curve are parallel to the coordinate axes (Fig. 306), "" (ccv) NP = x-h, MP = y-k m\ and the equation of the curve is (x-hy , (y-ky ¥ (306) a and 6 being the semi-axes parallel to Fig. 306. qx and OY respectively. Example 1. Find the equation of the ellipse with vertices A' (—3, 2), A (5, 2) and semi-axes equal to 4 and 1. The segment A' A is the major axis (Fig. 30c). Its middle point is the center of the curve. Consequently, the center is C (1, 2). Also the semi-axes are CA = 4, CB = 1. The equation of the ellipse is therefore T B jr — 'cij^r-^ , ■^^-^-_ B' X Fig. 30c. (x - 1)^ _^ {y - 2r 16 1 1. Ex. 2. Show that the equation 9 x^ -|- 4 2/^ -|- 36 a; - 24 ?/ -|- 36 = represents an ellipse. Find its center and axes. The equation can be written 9 (x^ +4x) -f- 4 (?/2 -Qy)+3G = 0. Completing the squares in the parentheses, 9 (x + 2)2 + 4 (2/ - 3)2 = 36, Fig. 30rf. {X + 2)2 ^ (y - 3)2 ^ ^ Comparing this with equation (306) it is seen to represent an ellipse with center (—2, 3). The axes are horizontal and vertical lines through the center. Their equations are x = — 2, ?/ = 3. The Art. 30 The Ellipse in Otheb Positions 73 major axis is vertical, the minor axis horizontal. The vertices, at the ends of the major axis, are B' (—2, 0), B (—2, 6). Ex, 3. Fiad the equation of the ellipse whose axes are the lines 2x + y — 3, X — 2y = 0, and whose semi-axes along those lines are 1 and 3 respectively. Let P {x, y) be any point on the curve (Fig. 30e). Then NP = 2x + y-3 ±V5 ' MP ■2y -.V5 Fig. 30e. The equation of .the ellipse is iVPV9 + MP^/l = 1, whence (2x + y-3r- (x-2] -" 45 5 = 1. Exercises Show that the curves are 1. 2. 3. 7. Make graphs of the following equations. Find their centers and semi-axes. X' + 2y' = 6. i. Z x' + 2 y'- - 6 X + 8 y = 1. 4:x'' + y'^-8x + 4:y + 7=Q. 5. x'' + Sy'' + Gx-6y = l. x^ + 2y' = x + y. 6. 5x''+2y''-10x+4:y+7=0. Find the equation of the ellipse with center (1, —3) and semi- axes, parallel to OX and OY, whose lengths are 2 and 3 respectively. 8. Find the equation of the ellipse with center ( — 2, 4) tangent to both coordinate axes. 9. Fiud the equation of the ellipse whose axes are the lines x+y -2 = 0, x-y + 2 = 0, and whose semi-axes along those lines have lengths equal to 1 and 4 respectively. 10. Show that an oblique plane section of a right circular cylinder is an ellipse. 11. Three sides of a rectangle are divided into an equal number of parts and the points of division T?T ^nf connected by straight Hnes with ^' the opposite corners as shown in Fig. 30f. Show that the intersections of lines through like numbered 74 Second Degree Equations Chap. 4 points are on an ellipse with axes equal in length to the sides of the rectangle. 12. Show that the locus of points, the sum of whose distances from two fixed points is constant, is an ellipse. Let the fixed points be (— c, 0), (4-c, 0) and let the constant distance be 2 a. Art. 31. The Parabola Let LK, RS be perpendicular lines and MP, NP perpendiculars from any point P to them. If a is constant and NP considered positive when P is on one side of RS, negative when on the other, the locus of points P such that MP^ = a-NP (31a) is called a parabola. A parabola is thus a locus of points the squares of whose distances from one of two perpendicular lines are proportional to their distances from the other. The complete locus of such points is two parabolas, one on each side of RS. To each value of NP correspond two values of MP differing only in sign. The curve is therefore symmetrical with respect to LK Fig. 31o. Fig. 316. which is called the axis of the parabola. The point A is called the vertex. The curve passes through A but, since a • NP is positive, it does not cross RS. If LK is the x-axis, RS the y-sixis, the equation of the curve is if = ax. If o is positive x must be positive and the curve is on the Art. 31 The Pahabola 75 right of the 2/-axis as in Fig. 31a. If a is negative, x must be nega- tive and the curve is on the left of the 2/-axis as in Fig. 316. If the axis of the parabola is parallel to OX and the vertex is b.. M L N k Jt Fig. 31c. Fig. 31d. (A, k) (Fig. 31c), NP = x — h, MP = y — k, and the equation of the parabola is {y -ky = aix- h). (316) If Qi, k) is the vertex and the axis is parallel to OY (Fig. 31d) the equation of the parabola is (x -hy = a{y-k). (31c) If the axis of the parabola is not parallel to either coordinate axis, its equation can be obtained from (31o) by expressing MP and NP in terms of the coordinates of P. Example 1. Show that y^ = 3x + 2y — iis the equation of a parabola. Find its vertex and axis. Transposing and completing the square, the equation becomes {y - ly = 3x - 3 = 3 ix - 1). Comparing this with the equation (y -ky = a(x- h), it is seen to represent a parabola for which h = l, k = l, a = 3. The vertex is the point (1, 1) and the axis is the line y = 1. 76 Second Degree Equations Chap. 4 Ex. 2. Find the equation of the parabola with vertex (1, 2) and axis y = X + 1, which passes through the point (3, 7). The line through the vertex perpendicular to the axis is a; + 2/ — 3 = 0. If x,y are the coordi- nates of any point P (Fig. 31e), then MP y x-1 NP = ±V2 ' x + y-3 :V2 If P is a point on the parabola, MP^ = a • NP, whence Fig. 31e. {y - x - ly = ±aV2{x + y -3). Since the curve passes through (3, 7) 9 = ± a V2 (7). This value of a substituted in the previous equation gives l{y-x-lY = 2{x + y-S) as the equation of the parabola. Ex. 3. An arch has the form of a parabola with vertical axis (Fig. 31/). If the arch is 10 feet high at the center and 30 feet wide at its base, find its height at a distance of 5 feet from one end. Take the origin at the middle point of the base of the arch. The vertex is then (0, 10). ttS.O) X Fig. 31/. The equation of the arch is consequently (a; - 0)2 = a (2/ - 10). The curve crosses the a;-axis at (15, 0). Hence 152 = (-10). Art. 32 The Hyperbola 77 Substituting this value of o, the equation of the arch becomes 10x2 = 225(10-1/). At a point 5 feet from one end x = ±10. The corresponding value of 2/ is 50/9, which is the height of the arch at that point. Exercises Make graphs of the following equations. Show that the curves are parabolas. Find their axes and vertices. 1. 2/2 = 8 X - 4. 4. 2/ = (x - 1) (x + 2). 2. 2/2 = -2 X + 1. 5. X = 2/2 - 3 2/. 3. 2/=x2-2x + 3. 6. x2-3x + 22/-4 = 0. 7. Find the equation of the parabola with horizontal axis and ver- tex at the origin, which passes through (3, 4). 8. Find the equation of the parabola with vertical axis and vertex at (—2, 2), which passes through (1, —3). 9. An arch in the form of a parabola with vertical axis is 29 feet across the bottom and its highest point is 8 feet above the base. What is the length of thfe beam placed horizontally across the arch 4 feet from the top? 10. A cable of a suspension bridge hangs in the form of a parabola with vertical axis. The roadway, which is horizontal and 240 feet long, is supported by vertical wires attached to the cable, the longest being 80 feet and the shortest 30 feet. Find the length of the support- ing wire attached to the roadway 40 feet from the middle. 11. A point moves so that its distance from a fixed point is equal to its distance from a fixed Une. Show that the locus described is a parabola. 12. Two sides, AB and BC, of a rec- tangle are divided into an equal number of parts and the points of division num- bered as shown in Fig. 31^. Through the points oi AB lines are drawn parallel to BC, and through those of BC lines ""' p gi. are drawn passing through A. Show that, the intersections of lines through like numbered points are on a parabola. Art. 32. The Hyperbola Let KL and RS (Fig. 32a) be two straight lines intersecting in C, PM and PN the perpendiculars from any point P to these lines. Let MP be considered positive when P is on one side of KL, nega- 78 Second Degree Equations Chap. 4 Fig. 32a. tive when on the other side. Similarly, let NP be positive when P is on one side of RS, nega- tive when on the other side. A hyperbola is the locus of points P such that the pro- duct MP ■iVP= constant. (32a) If the signs of MP and NP are both changed the product is not changed. Hence if any point P lies on the curve, the point P' at equal distance on the other side of C is on the curve. The hyperbola is thus symmetrical with respect to C which is called the center. The curve consists of two parts in a pair of vertical angles determined by KL and RS. The hyperbola is a locus of points the prod/uct of whose distances from two lines is constant. The complete locus of such points is however two hyper- bolas, one in each pair of vertical angles between the lines. If MP is very large, NP must be very small and conversely. As it goes to an indefinite distance the curve thus approaches indefi- nitely near the lines KL and RS. They are called asymptotes. Let C be the origin (Fig. 326) and take as X-axis the line bisecting the vertical angles in which the curve lies. The curve crosses the X-axis at two points A', A. Construct the rec- tangle with sides through A diagonals. Let are then V Y B y \ 2\ X. / v / ^ ^ \ /^ B' ^<\ ^ Y' "^ Fig. 326. and A, having KL and RS as CA = a, CB = 6. The equations of KL and RS y=± (b/a) X. (326) Art. 32 The Hyperbola 79 Consequently, Fig. 32a, 6a; + ay MP = NP bx — ay ±V62 + a2 ' " ' ±Vb'' + a' If P is a point on the hyperbola, MP • NP = const., whence b^x^ — ay = ± (6" + a^) const. = k. Since the hyperbola passes through A {a, o) ¥a^ = k. SubstitutiQg this value of k and dividing by aW, the equa,tion of the hyperbola becomes x^ 2/2 = 1. (32c) Since the equation contains only squares of x and y the curve is symmetrical with respect to the coordinate axes. They are called the axes of the curve. The axis cutting the curve is called trans- verse, the one not cutting the curve is called conjugate. The dis- tances CA = a, CB = 6 are called the semi-axes. The points A' and A, where the transverse axis cuts the curve, are called vertices. Equation (32c) represents the hyperbola referred to its axes, the x-axis being trans- \ verse. If the transverse axis is parallel to \ the X-axis and the center is Qi, k) (Fig. 32c), :(ft,fc) Fig. 32c. Fig. 32d. the coordinates relative to the axes are x — h and y — k. The 80 Second Degree Equations Chap. 4 equation of the hyperbola is then (x - hy (y - ky 1. (32d) If the transverse axis is parallel to OF (Fig. 32d) and the center is Qi, k) the equation of the hyperbola is {y -ky (x- hy ¥ a' 1. (32e) If the axes of the curve are not parallel to the coordinate axes, its equation can be obtained by using the definition or by replacing X and y in equation (32c) by the distances of a point P from the axes of the curve. Art. 33. The Rectangular Hyperbola // the asymptotes of a hyperbola are perpendicular to each other it is called rectangular. In this case the asymptotes are usually T zfVP M Pig. 33a. r Fig. 336. taken as coordinate axes. The definition, MP • NP = const., gives the equation of the curve in the form xy = k. (33o) If k is positive the curve lies in the first and third quadrants (Fig. 33a), if A; is negative it lies in the second and fourth quadrants (Fig. 336). The axes of the rectangular hyperbola make angles of 45° with the asymptotes. The rectangle. Fig. 326, is a square and a = b. If Art. 33 The RECTAkouLAK Hyperbola 81 the axes of the rectangular hyperbola are taken as axes of coordi- nates, its equation is then a;2 - 2/2 = a\ (336) the X-axis being transverse. Example 1. Show that 9 x^ - 4 j/^ + 18 x -h 16 y - 43 = is the equation of a hyperbola. Find its center, axes and asymptotes. The equation can be written 9 (x2 -1- 2x) - 4 (2/2 - 42/) = 43. Completing the squares, 9 (x -M)2 - 4 (2/ - 2)2 = 36, or (x + 1)2 (2/ - 2)2 ^ Comparing this with equation (32d), it is seen to represent a hyper- bola with center (—1, 2) and semi-axes, a = 2, 6 = 3. The trans- verse axis is the line y = 2. The conjugate axis is x = —1. The asymptotes are lines through the center with slopes ± h/a. Their equations are consequently 2/ - 2 = ±i (x -t-l). Ex. 2. Show that X2/ = 2 x — 3 2/ is the equation of a rectangular hyperbola. Find its center, axes and asymptotes. The equation can be written (x + 3) (2/ - 2) = -6. The quantities x -f- 3 and y — 2 are coordinates of P (x, y) with respect to axes through (—3, 2) parallel to OX and OY (Fig. 33c). The curve is therefore a rec- tangular hyperbola with center (—3, 2) and asymptotes x = — 3 and 2/= 2. The axes pass through (-3, 2) making angles of 45° Fig. 33c. 82 Second Degree Equations Chap. 4 with the asymptotes. Their equations are consequently y-2= ±ix + 3). Ex. 3. Find the equation of the hyperbola with asymptotes X — y — 1 and x = 2, wliich passes through (3, 4). Let P {x, y) be a point on the curve. Then (Fig. 33d) 1 MP =' y NP = x-2. Fig. 33d. ±V2 The equation of the curve is MP ■ NP = constant. Consequently, {x — y — 1) (x —2) = constant. Since the curve passes through (3, 4), the constant in this equation is (3 — 4 — 1) (3 — 2) = — 2. The equation required is then (x - 2/ - 1) (a; - 2) = -2. Ex. 4. Find the equation of the hyperbola with center at the origin, transverse axis y — 2x = 0, which passes through (0, 2) and has the x-axis as an asymptote. The conjiigate axis, being perpendicular to the transverse axis at the center, is x + 2 ?/ = 0. In equation (32c) x and y can be re- placed by the distances of a point on the hyperbola from the axes of the curve. In the present case these distances are (x + 2 2/)/V5 and {y — 2 x)/'v5. Hence the equation of the curve is (x + 2yy {y-2xY bV- = 1. Since the curve passes through (0, 2) 16/5 a" — 4/5 6^ = 1. Since the X-axis is an asymptote, there must be no point of intersection of the curve and x-axis. When y is zero the equation becomes x'' (1/5 a^ — 4/5 6") = 1. This will fail to determine a value of x if 1/5 a^ — 4/5 6^ = 0. This and the previous equation solved simultaneously give a^ = 3, b'' = 12. The equation required is Art. 34 The Second Degree Equation 83 therefore 15 60 Exercises Show that the following equations represent hyperbolas. Find their centers, axes and asymptotes. 1. 4 x2 - 3 2/2 + 12 !/ = 24. 4. 2 x^ - 3 j/^ + 4 x + 12 2/ = 4. 2. 5 s^ - 2/2 - 2 2/ = 4. 5. x2 - 2/2 - 2 a; - 6 2/ = 8. 5. xy -\- X — y = Z. 6. 2 xy = 3 X — 4. 7. Find the equations of the two hyperbolas with center (2, —1) and semi-axes, parallel to OX and OY, whose lengths are 1 and 4 re- spectively. 8. Find the equation of the hyperbola with center ( — 2, 1), and axes parallel to the coordinate axes, passing through (0, 2) and (1, —4). 9. Find the equations of the hyperbolas whose axes are the lines 3a;-|-22/ = 0, 2x — 32/ = and whose semi-axes along those lines are equal to 2 and 5 respectively. 10. Show that the locus of a point, the difference of whose distances from two fixed points is constant, is a hyperbola. Let the fixed points be (— c, 0), (-f- c, 0) and let the difference of the distances be 2a. 11. A point moves so that the product of the slopes of the lines joining it to (— o, 0) and (a, 0) is constant. Show that it describes an ellipse or a hyperbola. Art. 34. The Second Degree Equation An equation of the second degree in rectangular coordinates has the form Ax^ + Bxy + Cy'' + Dx + Ey + F = 0, (34a) A, B, C, D, E, F being constants. The equations of the circle, ellipse, parabola and hyperbola are all of this kind. If the poly- nomial forming the left side of the equation can be resolved into a product of first degree factors, the equation is said to be reducible. If the polynomial cannot be so factored the equation is called irre- ducihle. Reducible Equations. — If the polynomial forming the left side of (34a) can be resolved into a product of first degree factors, the equation has the form {aix + hy + Ci) (asx -f biy + co) = 0. 84 Second Degree Equations Chap. 4 Since a product is zero when and only when one of its factors is zero, the equation is satisfied by all values of x and y such that either aix + 6i2/ + ci = or 02X + 622/ + C2 = and by no others. If the coefficients Oi, 61, etc., are all real these equations represent straight lines. The locus of the equation is then a pair of straight Unes (a single line if the factors are equal). If some of the coefficients are imaginary, the locus wiU usually be a single point whose coordinates make both factors vanish. Example 1. Determine the locus of the equation 2x^ — xy—Sy' = 0. The equation is equivalent to ix + y){2x-3y) = 0. The locus is two fines, x + y = and 2x — 3y = 0, passing through the origin. Ex. 2. Determine the locus represented by the equation x^+xy-2y^-2x + 5y-3 = 0. Arranged in powers of x the equation is Fig. 34o. x''+(,y-2) x-{2y-'-5y + 3) = 0. Solving by the quadratic. formula, a;=-i(2/-2)±i V92/^-242/ + 16 = -1(2/ -2) ±1(3 2/ -4). There are then two solutions x = y — 1 and x = —2y + S. The original equation is satisfied if either of these is satisfied. The locus is two straight lines. Ex. 3. Determine the locus of the equation x2 + 32/'' -2a; + 12 2/ + 13 = 0. When the squares are completed this becomes (x -iy + 3{y + 2y = 0. Art. 34 The Second Degree Equation , 85 The sum of squares of real numbers can only be zero when all are zero. The only real numbers satisfying this equation are then x = l, .«- -2. The polynomial has imaginary factors, (a; — 1) ± (?/ — 2) V— 3, but the locus has one real point. Irreducible Equations. — It will be shown later (Art. 59) that the locus of an irreducible equation of the second degree is an ellipse, parabola, hyperbola or entirely imaginary. A circle is considered as an ellipse with axes of equal length. By the second degree part of an equation of the second degree is meant the part Ax^ + Bxy + Cy^ (346) containing the terms of second degree. We shall now show how to determine by inspection of this second degree part whether a given second degree equation represents an ellipse, parabola or hyperbola. An ellipse whose axes are the lines Aix + B,y + Ci = 0, Aic + Biy + C2 = is represented by the equation 1 (A^x + Bi2/ + 1. a'V VI7+5? / V'K y/Ai + Bi The second degree part of this equation is {A^x + BryY {A^x + B4)y ^' a? (Ai^ + Bi^) "^ 6^ {Ai + Bi) ' Since this is a sum of squares it has imaginary factors. If the axis of a parabola is A\X + Biy + Ci = 0, and the line through the vertex perpendicular to the axis is A^ + Biy + C2 = 0, the equation of the curve is Ayx + B,y + CA2 (Aic + B^y + C^N W)''<- V^i^ + Bi^ / \d,VAi^'+Bi The second degree part of this equation is (Aia; + B,yY (2) which is a complete square. A^^ + B^^ 86 Second Degree Equations Chap. 4 If the asymptotes of a hyperbola are A,x + B,y + Ci = 0, Ajx + Biy + Cj = 0, its equation is fAiX + B,y + CA lA^x + Biy + Ci The second degree part of the equation is constant. (3) / A^x + B^y \ /A^^+£g\ ^ which is a product of real first degree factors. Inspection of (1), (2) and (3) shows that the equations of ellipse, parabola and hyperbola are distinguished by the fact that the second degree part has imaginary factors in case of the ellipse, is a complete square in case of the parabola, and has real and distinct factors in case of the hyperbola. Example 1. Show that 8x^-8xy + 2y' = 2x-3 is the equation of a parabola. Solving for y. Fig. 346. Solving for y, 2/ = 2x ±4 V4x -6. Since y is an irrational function of x the equation is irreducible. That the curve is real is shown by plotting (Fig. 346). The second degree part of the equation is 8 a;2 - 8 x?/ + 2 2/2 = 2 (2 a; - yY. This being a square the curve is a para- bola. Ex. 2. Show that x^ + xy + y' = 3 is the equation of an ellipse. y hi-x ± Vl2 -3x2). The equation is irreducible and represents a real curve. The second degree part is x' + xy + 2/2 = (x + i 2/)' + f 2/'- Art. 35 Locus Problems 87 This, being a sum of squares, has imaginary factors and the curve is an elhpse (Fig. 34c). Ex. 3. Show that x^ + xy = 2 is the equation of a hyperbola. The equation is irreducible and repre- sents a real curve. The second degree part of the equation x'^ + xy = xix + y) has real and distinct factors. The curve is therefore a hyperbola Fig. 34c. Exercises Plot and determine the nature of the loci represented by the following equations: 4 x^ — 4 xj/ + 2/" = 4 X — 5 2/. 5x'' + exy + 2y'-4x-2y + 1=0. (X - 2/ + 3) (2 X - 2/ - 1) = 4. xy = 2x + 2y — i. x^-2V2xy + 2y' = 4x. 2x2 + 2 2/2-4x + 6i/ = 7. 3x^+2xy+2y'-ix-iy = 4:. 4x« + 3x2/-2 2/2 + 4x + 7?/ -6 = 0. xy = 3y — 2x + 6. 1. x^ + 2 X2/ - 3 1/2 = 0. 12. 2. 2x^ + 3xy -y^ = 0. 13. 3. 4x2 + I2xy + 9y' = 0. 4. Zx' ~2xy + 3y' = 0. 14. 5. 2x^-xy-y'-4:X+y+2 = 0. 15. 6. x2+2 X2/+2/2+4 x+4 2/+4 = 0. 16. 7. xy = 7 X. 17. 8. x' — 2xy + y^ = 4:X. 18. 9. x2 + 2/2 = 2 X - 3 2/ + 4. 19. 10. x2 + 2 X2/ + 2 2/2 =5. 11. x2 — xi/ = 4 2/. 20. Art. 35. Locus Problems A locus is often defined by a property of a moving point. The locus is the totality of points having the property. A pair of coordinate axes being given, the equation of the locus is an equation satisfied by the coordinates of every point on it and by no others. To find this equation choose as axes whatever perpendicular lines seem most convenient and let (x, y) be any point of the locus. In terms of x, y and any constant quantities occurring in the problem, express the property used as definition of the locus. The result will be an equation of the locus. In some cases this result can be reduced to a simpler form. 88 Second Degree Equations Chap. 4 Example 1. The vertices A and 5 of a triangle ABC are fixed (Fig. 35a). Find the locus of the vertex Cif4 + B = fx. The angle C will be -A-. ■K 'I Fig. 35o. Since this angle is constant the locus is a circle. To find its equation take the middle point of B X AB as origin and the Hne^5 as a;-axis. Let • AO = OB = a. Then A and B are (-a, 0) and (a, 0). The definition of the locus is A + B = ^■TT, whence tan(A + B) = -1 = tan A + tan B 1 — tan A tan B Now tan A is the slope oi AC and tan B is the negative of the slope of BC. Hence tan A = -y - , tan B = x-i- a X — a Substituting these values in the expression for tan (A equation of the circle is found to be x^ -\- 'if = 2 ay -\- a'. Ex. 2. A segment has its ends in the coordinate axes and determines with them a triangle of constant area. Find the locus of the middle point of the segment. Let the segment be AB (Fig. 356). Let OA = a,OB = b. The area of the triangle OAB is B), the Fig. 356. K = ^ab, K being constant. If x and y are the coordinates of the middle point P, then a = 2x, b = 2y and K = 2xy. This is an equation satisfied by the coordinates of any point on the locus. Conversely, if the coordinates of any point satisfy this Art, 35 Locus Problems ' 89 equation, the segment AB whose intercepts are OA = 2x, OB = 2y will have P as its middle point and wiU determine with the coordi- nate axes a triangle of area K. Therefore K=2xy is the equation of the locus. The curve is a rectangular hyperbola with the axes as asymptotes. Exercises 1. A point moves so that the sum of the squares of its distances from the four sides of a square is equal to twice the area of the square. Find its locus. 2. A point moves so that its shortest distance from a fixed circle ia equal to its distance from a fixed diameter of the circle. Find its locus. 3. In a triangle ABC, A and B are fixed. Find the locus of C, if A - B = iir. 4. A point moves so that the sum of the squares of its distances from the three sides of an equilateral triangle is equal to the square of one side of the triangle. Find its locus. 5. A point moves so that the square of its distance from the base of an isosceles triangle is equal to the product of its distances from the other two sides. What is its locus? Show that it passes through the vertices of the two base angles. 6. On a level plane the crack of a rifle and the thud of the bullet striking the target are heard at the same instant. Fiad the locus of the hearer. 7. A point moves so that the ratio of its distance from a fixed point to its distance from a fixed straight line is a constant e. Show that the locus is an eUipse if e < 1, a parabola if e = 1 and a hjrperbola if e > 1. 8. AB and CD axe two segments bisecting each other at right angles. Show that the locus of a point P which moves so that PA • PB = PC • PD is a rectangular hyperbola. 9. OA and OB are fixed straight lines, P any point, and PM, PN the perpendiculars from P on OA, OB. Find the locus of P if the quadri- lateral OMPN has a constant area. 10. AB is a fixed diameter of a circle and AC is any chord; P and Q are two points on the line AC such that QC = CP = CB. Find the locus of P and Q as AC turns about A. CHAPTER 5 GRAPHS AND EMPIRICAL EQUATIONS The equation of a curve being given, any number of points on the curve can be constructed by assigning values to either coordi- nate, calculating the corresponding values of the other coordinate and plotting the resulting points. When enough points have been located a smooth curve drawn through them may be taken as an approximation to the required curve. We desire as quickly as possible to obtain a satisfactory approximation. To some extent this is accomplished by plotting points rather sparsely where the curve is nearly straight and more closely where it bends rapidly. The following are some of the things it may be helpful to note: (1) Points where the curve crosses the axes and the side of an axis on which it lies between two consecutive crossings (Art. 36). (2) Values of either coordinate for which the other coordinate is real and values for which it is imaginary (Art. 37). (3) Symmetry (Art. 38). (4) Infinite values of the coordinates. As3rmptotes (Art. 39). (5) Direction of the curve near a point (Art. 40). Art. 36. Intersections with the Axes The points where a curve meets the s-axis are found by letting 2/ = in the equation and solving for x. Similarly, points on the 2/-axis are found by letting x = and solving for y. The x-co6rdi- nates of the points on the x-axis and the ^/-coordinates of the points on the 2/-axis are called the intercepts of the curve on the coordinate axes. Example 1. y = x {x — 1) {x -\-2). The curve crosses the x-axis where y = 0, that is, where x = 0, 1, —2. These points divide the X-axis and the curve into four parts; namely, the part on the 90 Art. 36 Intersections with the Axes 91 left of X = —2, that between x = —2 and x = 0, that between X = and x = 1 and the part on the right of a; = 1. Construct these parts separately. On the left of x = —2, each factor of a; (x — 1) {x + 2) is negative and conse- quently the whole product is negative. Hence on the left oi x = —2, y is nega- tive and the curve lies below the x-axis. Between x = —2 and x = 0, x and x — 1 are negative and x + 2 is positive. The product is then positive and so the curve lies above the x-axis. Similarly between X = and x = 1 the curve is below the X-axis and on the right of x = 1 it is above. Using these facts and plotting a few points on each part of the curve the graph of Fig. 36a is ob- tained. Ex. 2. 2/ — 2 = x^ (x + 2y. The curve meets the line 2/ = 2 at a; = 0, — 2. Since x^ (x -|- 2)^ is never negative, y can never be less than 2. Hence the curve touches but does not cross the line y = 2 a,t x = and x = —2. (Fig. 366.) Ex.3. X = y* + 2 y' + 3 y'' + 4: y + 2. The curve meets the i/-axis where x = 0, that is, where y' + 2y^ + 3y'' + 4:y + 2 = 0. Proceeding as jr it ± ^ 4 \ i t ^ t j: \ ■2 io!. 3l 2 , ^'^ t L t Fig. 36a. ■— U- " ^ " ■"- _ Fig. 36b. Fig. 36c. in Art. 4 it is found that —1 is a root of this equation. Hence y + 1 is a factor of the polynomial. Division gives x= iy + l)(y' + y' + 2y + 2). 92 Gbaphs and Empirical Equations Chap. 5 In the same way it is found that 2/3 + 2/2 + 22/ + 2-= (2/ + 1) (2/2 + 2). Consequently, x = (y+ ly (2/^ + 2). Since 2/^ + 2 = has no real roots, the curve meets the y-axis only at 2/ = —1. The factor (y + 1)^ is positive both above and below 2/ = — 1. Hence x is always positive and the curve does not cross the 2/-axis. After plotting a few points the curve in Fig. 36c is obtained. Art. 37. Real and Imaginary Coordinates When the equation 'of the curve is of even degree in one of the coordinates, that coordinate may be real for certain values of the other coordinate and imaginary for certain others. . The plane is then divided into strips (Fig. 37a) contain- ing a part of the curve and strips not containing a part. These strips being de- termined the part of the curve in each strip is plotted separately. Example 1. y' = x(x- 1) (x + 2). The curve crosses the x-axis at x = —2, x = and X = 1. The lines x = —2, x = and X = 1 divide the plane into four parts. On the left of X = —2, the product x (x — 1) (x + 2) is negative and y is imaginary. Between x = — 2 and x = the product is positive and y is real. Similarly, between x = and x = l,y is imaginary and, on the right of x = 1, 2/ is again real. The curve therefore consists of two pieces, one between x = — 2 and x = and the other on the right of x = 1. The equation can be written Fig. 37a. 2/= ±Vx(x- l)(x + 2). To each value of x correspond two values of y differing only in sign. The curve therefore consists of points at equal distances above and below the x-axis (Fig. 37a). Art. 38 Symmetry 93 Ex. 2. t' — i xy + i y^ — y — 1 = 0. Solving for x, X = 2y ±Vy-{- 1. The value of x is real if 2/ > —1, imaginary ii y < —1. The curve therefore hes above the hne y = —1. It consists of pairs of points at equal distances right and left of theUnea; = 2 j/ (Fig. 376). Fig. 376. Y py nb Qy~^ p ^^^ \ w xJ X / 3 Fig. 38o. Art. 38. Symmetry Two points P, P' are said to be symmetrical with respect to a line if the segment PP' is bisected per- pendicularly by that line. In Fig. 38a, P and P' are symmetrical with respect to the x-axis, P and P", Q and R with respect to the y-axis. Two pomts P and P'" are called symmetrical with respect to a point if the segment PP'" is bisected byO. A curve is called symmetrical with respect to an axis if all chords perpendicular to the axis meet the curve in pairs of points symmet- rical with respect to the axis. The curve in Fig. 38a is symmetrical with respect to both coordinate axes. A curve is called symmetrical with respect to a center if all chords through the center meet the curve in pairs of points symmetrical with respect to the center. The curve in Fig. 38a is symmetrical with respect to the origin. A curve / (x, ?/) = is symmetrical with respect to the x-axis if f{x,y) =f(.x, -y). For then any point P {xi, yi) being on the curve the point P'{xi, —yi) is also on the curve. Hence any line x = Xi perpendicular to the X-axis and meeting the curve in a point P will meet it in two points P, P' symmetrical with respect to the x-axis. In particular, a curve is symmetrical with respect to the x-axis if its equation contains. 94 Graphs and Empirical Equations Chap. 6. only even powers of y. Similarly, the curve f {x, y) = is symmet- rical with respect to the y-a,xis if f(x,y) =f(-x,y), and, in particular, a curve is symmetrical with respect to the y-axis if its equation contains only even powers of x. The curve / {x, y) = ia symmetrical with respect to the origin if fix,y) = ±fi-x, -y). For then if P {x\, yi) is on the curve P'" {—Xi, —yi) is also on the curve, and so any Une through the origin meeting the curve in a point P will meet it in two points P, P'" symmetrical with respect to the origin. In particular, a curve is symmetrical with respect to the origin if all the terms in its equation are of even degree or if all are of odd degree. a;2 — 1 Example 1. y^ = ~ — -. The a:^ + 1 curve crosses the x-axis at a; = Pj(3_ 3g[, ±1. The value oi y is real only when X is in absolute value equal to or greater than 1. There are consequently two parts of the curve, one on the right of a; = 1, the other on the left of x = —1. Since the equation contains only even powers of x and y, the curve is symmetrical with respect to both coordinate axes and with respect to the origin. Since x^ — 1 < x^ + 1, y is always less than 1. When x is very large, however, y is nearly 1. Ex.2. 2/ = x3 - 3 x^ + 3 X + 1. This equa- tion can be written 2/ - 2 = (x - 1)3. rr,, . n 1 , ,, Fl<5. 38c. The expressions y — 2 and x — 1 are the co- ordinates of a point P (x, y) relative to the lines y = 2, x = 1 used as axes. Since the equation contains only odd powers oi y — 2 and X — I the curve is symmetrical with respect to the point (1, 2). Art. 39 Infinite Values 95 Art. 39. Infinite Values In some cases a variable increases in absolute value beyond any assignable bound. Such a variable is said to become infinite and a fictitious value represented by the symbol 'x> is attached to it. But the symbol and the name infinity are used only to express that a variable goes beyond all bounds. Zero and infinity have the following relations: (1) - = 0, n= 'x> , a- = -a= and 00 — 00 . These relations become evident when and oo are interpreted as less than any assignable quantity and greater than any assign- able quantity respectively. A branch of a curve extending to an infinite distance can only be traced until it runs off the diagram. In such cases the curve usually consists of two or more pieces not connected together. Sometimes two pieces are related as at A, B, Fig. 39a, the one go- ing off in a certain direction, the other returning from that direc- tion. Sometimes they are re- lated as at C, D, the one going off one side of the paper, the other returning from the other side. Sometimes there is no return branch. If a branch of a curve when in- ^ Fig. 39a. definitely prolonged approaches a straight line in such a way that the distance between the two approaches zero, the straight line is called an asymptote of the Graphs and Empirical Equations Chap. 5 curve. Both coordinate axes and the hne CD are as3Tnptotes of the curve, Fig. 39a. If when indefinitely prolonged the distance between a branch of one curve and a branch of another approaches zero, the two curves are called asymptotic. - 1 Example 1. y = ■ The graph is shown in Fig. 39a. x" (x - 2) ' It crosses the x-axis at x = 1. The value of y is infinite when X = or 2. When x is a httle less than zero, x — 1 and x — 2 are negative and so y is positive. When a; is a little greater than zero y is again positive. In both cases y is very large. The curve thus goes up one side of the y-Bxis and comes down the other. When X is a little less than 2, y is negative but when x is a little greater than 2, ?/ is positive. The curve then goes down the left side of CD and reappears at the top. As x increases indefinitely, y approaches zero. Consequently, at a great distance from the origin the curve comes closer and closer to the X-axis. The two axes and the line CD are asymptotes of the curve. Ex. 2. 2/ = x^ + - . When x is very small, x^ is very small and y is ap- proximately 1/x. Near thejz-axis the curve is then asymptotic to the hy- perbola y = 1/x. The y-axis is an asymptote to both curves. When x is very large, 1/x is very small and y is approximately equal to x^. As x increases indefinitely, the curve therefore approaches the parabola 2/ = xHo which it is consequently asymptotic. Fig. 396. Art. 40. Direction of the Curve To determine the shape of a curve near a particular point it is often useful to find the direction along which the curve or a branch of the curve approaches that point. In Fig. 40a, for example, are shown three ways that a branch of a curve can approach a hori- Art. 40 DiKECTION OF THE CuRVE 97 zontal line. In (1) the curve and line are tangent, in (2) the curve and line intersect at an acute angle and in (3) they are perpendicular to each other. Let P be a variable point on the curve. As P approaches A the slope of the line AP will approach zero in (1), a finite value not zero in (2) and an infinite value in (3) . By finding this slope and determining its limit the direction of the curve near A can be determined. Example 1. y^ = x^. The curve passes through the origin (Fig. 406). If X is negative, y is imaginary. The curve therefore reaches the y-nnis at the origin but does not cross it. Since the equation contains only even powers of y, the curve is symmetrical with re- spect to the X-axis. The slope of OP is tan = y/x. From the equation Hence of the curve, = ±1^. Fig. 406. tan = dzx^/x = ±a;2. As P approaches the origin, X approaches zero and con- sequently tan approaches zero. The branches of the curve above and below the x-axis are thus tangent to the a;-axis and to each other at the origin. Ex. 2. 2/2 = x^ (a; + 2). The curve crosses the a;-axis at the origin and at ^ (— 2, 0) (Fig. 40c). There is no point of the curve to the left of a; = —2. Let P {x, y) be any point on the curve, Fig. 40c. The 98 Graphs and Empirical Equations Chap. 5 slope of AP is y — ±x Vx+'2 dzx x + 2 x + 2 Vx + 2 As P approaches A, x approaches —2 and the slope of AP increases indefinitely. The curve is therefore perpendicular to the a;-axis at A. The slope of OP is X X As P approaches the origin, x approaches and the slope of OP approaches ± v 2. At the origin the curve therefore makes with the X-axis the angles tan~' (± V 2). Exercises Make graphs of the following equations: 1. y = x{x + 1). 25. x^ + xy^ = 1. 2. X = y^iy - 2). . 2Q. x'^ - ixy + 8y^ - y* = 0. 3. y = (x-l)ix + 2) (X - 3). gy y = ^^ 4. y + l=x' + 2x. " I -X 5. 2/ = s (x + 1) (2 a; - 3). ^g „ = ^ 6. 2/ - 3 = a;2 (x + 1) (2 s - 3). " a; + 1 7. y + 2=x^{x + \y{2x-2,). 2j/^+3j/-2 8. 2/ = a;2 (a; + 1)'' {2x- Zy. y -Z 9. y =x'-l. ^ (x + l)(x- 2) 10. X - 2 = (2/ + 1)* - 1. " x{x -3) ' 11. x = y^ + y' + l. ■ _^ , 1. 12. X = 2/< + 2/^ - 4 2/' - 4 2/. , " ^x 13. 2/^=x(x + l). „ = x^+i. 14. 2/=i= (x2 - 1) (x2 - 4). '^ ^a;2 15. (2/-|-l)^ = (x2-l)(4-x2). ■_ 2/^-1 16. x2= (2/ - 1)2 (2/ - 2). ''"'■ "" z/'' - 4' 17. x'={y-l)H2-y). 34 ^^J^ _ 1 18. (2/'^ + x) (2/2 - x) = 0. ■ " X - 1 X + 3 19. 2/^ + 2 2/ = x*-l-2x- 1. 35 ^_, 3 1 20. (x + ^)« - 2/* = 0. " X (x - 2) 21. (2/ + x)2 = x2 (x - 1). 3g ^ ^ 1 1 22. x' + y* = l. " (x-l)2 {x + zy 23. 2/*- 2x2/2 -|-x2- 2/2 + 4 = 0. g^ „2= _J 1 24.^2/^+(2x2+l)2/2H-x^-x2 = 0. " x(x-l) x + 3 Art. 41 Sine Curves 99 38. a;y + 36 = 4:y\ 46. x^' + y'^ = 1. 39 „2 = ^- (" + ^) 47. a^-^ + y-^ = 1. a - a; 48. J/' = x^s + 2). 40. a;^?/ + a'?/ - X* + o" = 0. 49. {y + 2)' = (x - 1) (x^ _ 4). 41. 1/ = x\ 50. (x + 3/)2 = y'{y + 1). 42. x^ =i/«. 51. (x + j/-l)(2x-4^-2)=4. 43. X* + 2/* = d^ 52. !/ = X (x - !/) (x + 2/). 44. x^ +y^ = a^. 53. (x + y - 2)^ = (2 x - 2/ - I)'. 45. x^ + 2/^ = 1. 54. X (x^ + ^2 _ 1) = i_ Art. 41. Sine Curves When a, b, c are constants, the graph of the equation y = asinQjx -\- c) is called a sine curve. As shown in Fig. 416, the graph consists of a series of waves all having the same length and height. The sine of an angle is never greater than 1 and so y is never greater than a. The constant a therefore measures the height of the waves. If x is increased by 2 w/b, the angle 6x + c is increased by 2 tt and y is not changed. This is the smallest constant that added to x will leave y unchanged. Hence 2-tr/b is the distance from any point of a wave to the corresponding point of the next wave. It is called the wave length. The equation y = a cos (bx + c) also represents a sine curve; for a cos (6x + c) = asm.{- — bx — cj = a sin (b'x + c'), if 6' = — 6, c' = ^ IT — c. Thus a cosine curve is a sine curve with different constants. Also 2/ = A sin (mx) + B cos (mx) can be reduced to the form y = asia {mx + 6) and so represents a sine curve. In plotting sine curves angles should be expressed in circular measure. A circle being drawn with the vertex of an angle as center^ 100 Graphs and Empirical Equations Chap. 6 the circular measure of the angle is defined as the ratio of the inter- cepted arc to the radius (Kg. 41a), that is, intercepted arc radius of circle Circular measure of angle ^ An angle of 180° has a circular measure equal to ;r = 3.14159 . . . , and other angles have proportional meas- ures. For instance, the circular measure of 360° is 2 7r; 90°, ^t; 60°, f tt; 45°, iir; 30°, i IT. An angle whose circular measure is 1 intercepts an arc equal to the radius. This angle, called a. radian, is 180° Fig. 41a. = 57°.3- If a; is a real number, sin x means sine of the angle whose circular measure is x. Thus sin 2 = sin (114°. 6— ). Example 1. Plot the curve j/ = 2 sin (3 x). The sine of an angle is never greater than 1 nor less than —1. Consequently, on this curve, y cannot be greater than 2 nor less than —2. The curve thus lies between the lines y = —2 and y = 2. The most important points on a sine curve are those where the sine is a maxi- mum, minimum or zero. Between 3 a; = and 3x = 2-K, the important values are shown in the following table : 3a; = x = Q \ \i 2/ = TT 3 — IT XT' 2 2 TT IT TT 6 3 2 2 -2 Fig. 415. 2x 2 This part of the curve extends from to A (Fig. 416). When x is increased by | ir, 3 a; is increased by 2 tt and y is not changed. Be- Art. 41 Sine Curves 101 yond A a new wave thus begins like that between and A. In the same way it is seen that the wave on the left of is like that from A to 0. The whole curve thus consists of an infinite number of waves each of wave length f ir. Ex. 2. X = cos (2 2/ — 3). Since the cosine is never greater than 1 nor less than —1, the curve lies between the lines x = 1, x =—1. At the point A (Fig. 41c) where 2y — 3 = 0, x has the maximum value 1. Between this point and B, where 2y — 3 = 2ir, the most important values are shown in the following table: 3 2y -3 = ■K 2 3 . 2 2 3 3 2+r 2ir :+7r 1 y 1 1 1 1 ^^ 1 1 ^ 1 1 f.^^ X 'C ^^^ 1 Fig. 4ld. When y is increased or de- creased by IT, the change in the angle 2 2/ — 3 is 2 ir and x is not changed. The curve then consists of a series of waves like that from A to 5 Fig. 41c. placed end to end. Ex. 3. y + 1 = sin"^ (x — 1). This equation is equivalent to x — I = sm(y + 1). The graph is a sine curve with axis x — 1 = 0. The curve passes through the point y = —1, X = 1, extending above and below that point in a series of waves each of vertical length 2 ir. (Fig. 41d.) Ex. 4. 2/ = 2 sin (J x) + 6 cos (J x). To con- struct this curve, draw the curves 1/1 = 2 sin (J x), 2/2 = 6 cos (i x) 102 Graphs and Empirical Equations Chap. 6 and on each vertical line determine the point P such that y = yi + 2/2, that is, MP = MPi + MP2. To leave sin (| a;) and cos (J x) both unchanged, x must be increased by 8 IT or a multiple of 8 ir. Hence a section AB from a; = to Fig. 41e. a; = 8 X must be plotted. The whole curve is a series of such sec- tions placed end to end (Fig. 41e). Art. 42. Periodic Functions The equations considered in the previous article have the pecuUar- itythat when a certain constant is added to one of the variables the other variable is not changed. Let the equation of a curve be f{x,v)=Q. If f{x + k,y)=fix,y), or f ix,y + k) = f {x,y), the curve consists of a series of pieces (extending from x to x + k or from ytoy + k) each obtained by moving the preceding one a distance k in the direction of a coordinate axis. In this case the function / {x, y) is called -periodic and k is its 'period. The part of the curve from a; to a; + A: or from yXoy-\-k\s called a cycle. For example, in Ex. 4 of the previous article a cycle extends from any point {x, y) of the curve to the point (x + 8 ir, j/). To plot a curve whose equation is periodic it is necessary to plot one cycle and sketch the others from periodicity. Art. 42 Periodic Functions 103 Example 1. Plot the curve y — tan (2 a;). If the angle 2 x is increased byTr (x increased by 5 tt) the tangent is not changed. The curve therefore consists of a series of parts each obtained by moving the preceding one a distance J tt to the right. One of these branches passes through the origin. As x increases from to 2 a; = J x, ?/ increases from to infinity. The Une a; = i tt is therefore an asymptote. As X decreases from to — jtt, 2/ decreases to — 00 , the line X = —jir being an asjrmptote. The branch through the origin thus extends from x = — Jtt, Fig. 42a. y = — to x = jTT, 2/=oo. The whole curve consists of a series of such branches at horizontal distances |x apart (Fig. 42a). Ex. 2'. y = sec x. The secant of an angle is in absolute value never less than 1. Consequently the curve lies outside the hnes 2/ = ±1. Since sec {—x)= sec x, the curve is symmetrical with respect to the y-axis. Since sec (x + 2 ir) = sec x, a complete cycle of the curve is contained between x = and x = 2 tt. The most important points on this part of the curve are given in the following table : X = — TT -IT Zfl 2 y 1 -1 1 the expression qo , — 00 meaning that on the left of this point y is indefinitely large and positive but on the right indefinitely large and negative. The curve is a series of U-shaped branches alter- nately above y = 1 and below 2/ = -1 (Fig. 426). Ex. 3. X = se(f y. In this case x is never less than 1. The curve consists of a series of U-shaped branches on the right of TT 3 X = 1 with asymptotes y = ±-,y= ±2^^, etc. (Fig 42c). 104 Graphs and Empirical Equations Chap. 5 — r i I' J i 1.^ Fig. 426. Ex. 4. 2/ = sin (-). The curve crosses the x-axis where ±x, ±2t, ±37r, X that is, where 1 ■3x' Between each pair of consecutive crossings it reaches one of the hnes 2/ = dzl. The curve has an infinite number of waves whose horizontal lengths approach zero near the 2/-axis (Fig. A2d). Fig. 42c. Fig. 42(i. Art. 43. Exponential and Logarithmic Curves If a is a positive constant the function a' is called an exponential function. It is understood that if x is a fraction a" is the positive root. Ijy = a" then, by definition, x is the logarithm of y to base a. Thus the equations y = a'', x = loga y are equivalent and both represent the same curve. Art. 43 Exponential and Logarithmic Curves A particular number of great importance is 105 ^ = ^+Ll + S + S + : 2.7182 . . . Logarithms with e as base are called natural logarithms. The functions e'^ and loge x are much used in higher mathematics. In working with exponentials .and logarithms the following facts are often useful: (1) a" = 1, logo 1 = 0, if a is not zero or infinite, (2) a" =-= c», a-« = 0, logaoo = co, log<.0 = -ot, if d>l, (3) a" = 0, a-" = 00 , loga oo = - oo , log„ = oo , if a < 1. Example 1. Plot the curve y = 2'^, or x = log2 y. The curve crosses the i/-axis at (0, 1). Since y is always positive the curve lies entirely above the a;-axis. As x decreases to — oo , y approaches zero and the curve approaches the x-axis which is an asymptote. As X increases, y increases. The increase in y for a given increase in a; is greater the larger x; for, if X changes to x + h, the change in 2/ is 2"+'^- 2" = 2^ (2'' - 1), and this is larger for larger values of X. If then x is increased by equal amounts h, the changes in y will form a series of steps of in- creasing height. The curve is thus concave upward and becomes more and more inclined to the x-axis as x increases (Fig. 43a). Y A /i /-- X Fig. 43a. Ex. 2. y = log) 'V+V 1 + 10" The logarithm of a 1 - 10" negative number is imaginary. Hence y is real only when a; > 1 or X < —1. When x = 1, y = logio = — oo . 106 Graphs and Empirical Equations Chap. 6 I When X > 1, {x — l)/(x + 1) is less than 1 but approaches 1 as X increases indefinitely. Consequently y is negative but approaches as a; increases indefinitely. This part of the curve is thus below the X-axis and has the x-axis and the fine x = 1 as asymptotes. When X = —l,y = logiooo = co . If X < -1, (x- l)/(x +1) is greater than 1 but approaches 1 as X decreases indefinitely. This part of the curve is therefore above the x-axis and has the .x-axis and the line x = — 1 as asymp- totes (Fig. 436). Fig. 436. 6^—1 Ex. 3. y = ~j . When x is negative e^' + l Consequently y is then negative and the curve is below the x-axis. When X is positive, y is positive and the curve is above the x-axis. I As x approaches from the negative side, e^ approaches e"""' = and y approaches —1. As x approaches from the positive Fig. 43c. side, e^ approaches infinity and y, being the ratio of two very large numbers whose difference is 2, approaches 1. Hence, as x passes through zero from the negative to the positive side, the point (x, 2/) jumps from (0, —1) to (0, +1). The curve is dis- continuous at x = 0. When x becomes very large, whether it is positive or negative, y approaches zero. The x-axis is therefore an asymptote (Fig. 43c.) Art. 44 Empirical Equations 107 Exercises Plot the graphs of the following equations: 1. y = sin(2x). 19. 2/ = sec X CSC X. 2. s = 2sin(i2/). 20. y = X sin X. 3. 2/ = 4cosf a; — ^j• 21. j/ = cosgj- 4. 5. 2/ = 2 cos a; + 3 sin a;, a; = sin 1/ + sin (2 y). 22. 2, = xsing . 6. y = cos (x - 1) + sin (a; + 1). 23. y = 6^. 7. y -2 = cos (2 a; + 1). 24. X = 2-". 8. y = sin^a;. 25. y = 10^1 9. y + 1 = cos-'(x-3). 26. y = er^ sin x. 10. y = 2 tan (3 a;). 27. X = i (3" - 3-"). 11. X = tan(y + | ■ 28. 29. 2/logioX = 1. 12. 2/ — 1 = cot (x — 3). 30. y = loge [x (x - 2)1. 13. 14. X = cot-i(22/). y = tan^ x. 31. 1 10^ y = —I 15. 16. y' = tanx. y = sec (2x). 105-1 1 17. 18. X = 1 + osoy. y = secx + tan x. 32. ,-l.-': Art. 44. Empirical Equations Pairs of corresponding values of two variable quantities being given, it is sometimes desirable to find an equation connecting them. Let the pairs of values be plotted and draw a curve through the resulting points. However many points are given, the section of the curve between consecutive points can be arbitrarily drawn. Consequently an infinite number of curves can be drawn through the points. Each curve has an equation. An infinite number of equations are then satisfied by the given pairs of values. From a table of corresponding values it is not then possible to find the exact equation connecting the quantities. It is usually assumed that if a smooth curve is drawn through or near the points, its equation will represent approximately the rela- 108 Graphs and Empirical Equations Chap. 6 tion of the two quantities. Such an approximate equation is called empirical. A table of values being given, an infinite number of empirical equations are approximately satisfied by these values. The simplest equation should be chosen that has the required degree of accuracy. The choice of such an equation is largely a matter of judgment. The following are types of equations it may be well to consider: (1) y = mx + b. (2) y = ax^ + bx + c. (3) y = ax": (4) y = ab". For convenience of plotting different unit lengths are often used along the two axes. This amounts to a uniform contraction in the direction of an axis. Equation (1) then still represents a straight line and (2) a parabola, but the constants in the equa- tions have a different geometrical meaning. Example 1. From the following values find an approximate equation connecting x and y : x = 2.2 5.0 7.0 10 15 2/ = 3.3 4.0 6.0 6.5 8.4 11 These values are plotted in Fig. 44a. The points are seen to he almost on a line. A fine is drawn so that the points are about equally distributed on the two sides. This line crosses the 2/-axis at about (0, 3) and the line x = 10 at approxi- mately (10, 8.4). The equa- tion of the line through these points is y = 0Mx + 3, ^ -^ 10 ^ ^ ^ ^ [^ Xl f ^ ^ X ^ E 1 D 1 > Fig. 44a. which is the empirical equation required. Art. 44 Empirical Equations 109 Ex. 2. Measurements of train resistance are given in the follow- ing table, where V = miles per hour, R = resistance in pounds per ton. 7 = 20 40 60 80 100 120 R= 5.5 9.10 14.9 22.8 33.3 46. The curve (Fig. 446) looks like a parabola with horizontal axis. Let R = A+BV + CV\ There being three coefficients, A, B, C, in this equation, the curve can be made to pass through only three of the given points. By more advanced methods the parabola could be found which fits closest to all the points. If a parabola is passed through three properly chosen 120 100 10 20 30 40 Reaistance Fig. 446. points it will, however, usually be accurate •f*" enough. The points chosen should be spread over the whole curve and should . not include any that appear to be faulty. In the present case numbers 1, 3 and 5 will be used. Substituting the coordinates of these points in the equation of the pa- rabola, 5.5 = A+ 205+ 400C, 14.9 = A+ 605+ 3,600 C, 33.3 = A + 100B + 10,000 C. The solution of these equations is A = 4.18, B = 0.01, C = 0.0028. The empirical equation found is then R = 4.18 + 0.01 V + 0.0028 V^ The curve (Fig. 446) drawn from this equation is seen to pass very close to all the points. Ex. 3. In the table below are given the loads which cause the failure of long wrought-iron columns with round ends, in which P/a is the load in pounds per square inch and l/r is the ratio of the 110 Graphs and Empirical Equations Chap. 6 length of the column to the least radius of gyration of its cross- section. Ur Pla. log «/r) log(P/a) 140 12,800 2.1461 4.1072 180 7,500 2.2553 3.8751 220 5,000 2.3424 3.6990 260 3,800 2.4150 3.5798 300 2,800 2.4771 3.4472 340 2,100 2.5315 3.3222 380 1,700 2.5798 3.2304 420 1,300 2.6232 3.1139 Try the formula P/a = C (l/r)". Taking logarithms of both sides, log (P/a) = log C + n log (l/r). This is a first degree equation connecting log (P/a) and log (l/r). If the formula is correct, the logarithms should then be coordi- nates of points on a line. Values of the logarithms are plotted in Fig. 44c. The points are almost on a line joining the 2nd and 7th. The equation of this line is 3.8 3.2 \ \ \ \ \, \ 2 1 2 2 2 I .3 2 ogO 4 2 /r) 6 2 6 Fig. 44c. log (P/a) = -1.99 log Q/r) + 8.356. Consequently, P/a = 227,000,000 (Z/r)" which is the empirical equation required. This is approximately Euler's formula for the axial unit-load P/a which will cause a long wrought-iron column with round ends to faU. Ex. 4. The following values were found for the amplitude of vibration of a long pendulum. Here A = amplitude in inches and t = time in minutes since the pendulum was set swinging. <= 1 2 3 4 5 6 A = 10 4.97 2.47 1.22 0.61 0.30 0.14 logA= 1 0.696 0.393 0.086 -0.215 -0.523 -0.854 Art. 44 Empirical Equations 111 Assume an equation of the form A = a¥. Then log A = log a + i log 6. Using t and log A as coordinates the points should lie on a line. Fig. ad shows this to be the case. The Mne seems to pass through the points t = and t = 5. Its equation is then log A = 1- 0.305 «. Consequently, loga=l, log6=- 0.305, and so a = 10, & = 0.495. The equa- tion required is therefore A = 10 (0.495)'. 1.0 0.8 0.6 0.4 0.2 § -0.2 -0.4 2 3 \ 4 5 ( -0.6 Exercises - C8 From the data in each of the follow- ing examples find an empirical equation -ri aaj connecting the quantities measured. 1 . Test on square steel wire for winding guns. The stress is measured in pounds per square inch, the elongation in inches per inch. Stress 5,000 10,000 20,000 30,000 40,000 50,000 Elongation 0.00000 0.00019 0.00057 0.00094 0.00134 0.00173 Stress 60,000 70,000 80,000 90,000 100,000 110,000 Elongation 0.00216 0.00256 0.00297 0.00343 0.00390 0.00444 2. Test on a steel column. The stress is measured in pounds per square inch, the compression in inches per inch. Stress Compression Stress Compression 3,000 0.00004 15,000 0.00039 6,000 0.00011 18,000 0.00053 9,000 0.00020 21,000 0.00066 12,000 0.00030 24,000 0.00087 3. The melting point 8, in degrees Centigrade, of a lead and zinc alloy containing x per cent lead, is given in the following table. X = 40 50 60 70 80 90 e = 186 205 226 250 276 304 112 Graphs and Empirical Equations Chap. 5 4. The following table gives the electromotive force E, in micro- volts, produced in a lead and cadmium thermo-electric couple when the difference in temperature between the jmiotions is 8° C. e = -200 -100 100 200 300 E = 60 -140 475 1300 2425 5. The following table gives the number of grams S of anhydrous ammonium chloride which dissolved in 100 grams of water make a saturated solution at 9° absolute temperature. e = 273 283 288 293 313 333 353 373 S= 29.4 33.3 35.2 37.2 45.8 55.2 65,6 77.3 6. The hysteresis losses in soft sheet iron subjected to an alternating magnetic flux are given in the following table, where B is flux density in kilolines per square inch, and P is watts lost per cubic inch for one cycle per second. B = 20 40 60 80 100 120 P= 0.0022 0.0067 0.0128 0.0202 0.0289 0.0387 7. The observed temperatures 9 of a vessel of cooling water at times t, in minutes, from the beginning of observation are given in the follow- ing table: t= 1 . 2 3 5 7 10 15 20 9 = 92° 85.3° -79.5° 74.5° 67° 60.5° 53.5° 45° 39.5° 8. Measurements showing the decay in activity of radium emanation are given in the following table: Time in hours = 20.8 187.6 354.9 521.9 786.9 Relative activity = 100 85.7 24.0 6.9 1.5 0.19 CHAPTER 6 POLAR COORDINATES Art. 45. Definitions We shall now define another kind of coordinates called polar. Let (Fig. 45o) be a fixed point and OX a fixed line. The point Fig. 45a. is called the pole, or origin, the line OX is called the initial line, or axis. The polar coordinates of a point P are the radius r = OP and the angle d from OX to OP. The angle B is any angle extending from OX to the line OP, the angle being considered positive when measm'ed in the counter-clock- wise direction (Fig. 45a, 1 or 2) and negative when measured ia the clockwise direction (Fig. 45a, 3). The radius r is considered positive when OP is the terminal side of B (Fig. 45a, 1 or 3) and negative when B terminates on OP pro- duced. A given point P is seen to have many pairs of polar coordinates, 6 being any angle from OX to OP. A given pair of polar coordi- nates, however, determines a definite point pbtained by constructing the angle 6 and laying off r forward or backward along the terminal side according as r is positive or negative. The point whose polar coordinates are r, 6 is represented by the symbol {r, B). To signify that P is the point {r, 0) the notation P (r, 6) is used. 113 114 Polar Coordinates Chap. 6 Example 1. Plot the point P ( — 1, — iir) and find its other pairs of polar coordinates. The point is shown in Fig. 456. The angle — Jtt terminates on OP produced. The angle I^OP is equal to f tt. Any other angle terminating on OP or OP pro- duced differs from one of these only by a positive o'y-f X or negative multiple of 27r. Any such angle then \ has one of the forms 2 mr — J ir or 2 nir + f ir, Fig. 456. where n is a positive or negative integer. Consequently any pair of polar coordinates of P has one of the forms (-1, 2n,r-^), (l,2«7r + ^x)- Ex. 2. Plot the point whose polar coordinates are (3, 4). The coordinates of the point are r = 3, 6 = 4:. This means that the circular measure of 6 is 4. The angle is approx- imately 229°. A line is drawn making this angle with OX and on the line the point P is located at a distance 3 from the origin (Fig. 45c). ^0 X F/ Fig. 45c. Equation of a Locus. — The polar equation of a locus (like its rectangular equation) is satisfied by a pair of coordinates of every point on the locus and not satisfied by the coordinates of any point not on the locus. Example 1. Find the polar equation of the; line through A {a, 0) perpendicular to the initial line. Let P (r, 6) be any point on the line. From Fig. 45t^ it is seen that r cos 6 = a. Art. 46 Definitions 115 Conversely, any point whose coordinates satisfy this equation lies on the hne, for the equation expresses that the projection of OP on OX is a. Ex. 2. Find the polar equation of the circle whose diameter is the segment from the origin to the point A (a, 0) . y^ ^Xp ('"•W Let P (r, 6) be any point on the circle (Fig. 45e). In the right triangle OAP r = OP = OA cosd = a cos d. Conversely, if r = a cos 9, the Pjg ^^^ angle at P is a right angle and P has on the circle. Hence r = a cos 6 is the equation required. Exercises 1. Plotthepoints(0,30°), (1,40°), f-2, ^Vfs, -|Y(-4, -730°). 2. Plot the points whose polar coordinates are (1, 1), ( — 1, 2), (2, -3), (V2, V3). 3. Show graphically that the points (2 V2, 0), (2, j|, (2 ^,|), (oo, fir) lie on a hne. On this line what value of r corresponds to 9 = f TT? 4. Show graphioaUy that the points (0, 0), (| Vs, 60°), (3, 90°), (— f, 210°) Me on a circle. What is its radius? 5. Show that ( 1, ^ ) and ( — 1, — ^ ) are the same point. Give other pairs of coordinates of this point. 6. Given the point P {r, $), find the coordinates of the point Q if P and Q are symmetrical with respect to the initial line; symmetrical with respect to the origin; symmetrical with respect to the line through the origin perpendicular to the initial line. 7. Find the polar equation of the line through the origin making an angle of ^ with the initial line. Does ( 2, ^ tt I lie on the line? Do its coordinates satisfy the equation? 8. Find the equation of the circle, radius a, with center at the origin. 116 Polar Coordinates Chap. 6 9. Find the polar equation of the line parallel to the initial line and passing through the point I a, ^ J • 10. Find the polar equation of the circle, radius a, tangent to the initial line at the origin. Art. 46. Change of Coordinates The same point can be represented by rectangular or by polar coordinates. It is sometimes desirable to use both systems simultaneously. In this case the a;-axis is usually coincident with the initial line and the origin of rectangular coordinates is the pole. A point P (Fig. 46a) then has four coordinates, x, y, r, 6. These are connected by several equations the most important of which are X = rcosd, r = ± v a;^ + y'^, y y = rsin 6, tan 6 = -■ X (46) By the use of these equations (or, better, by the use of Fig. 46a) any expression in rectangular coordinates can be converted into one in polar coordinates and conversely. Example 1. Find the polar equation of the circle x^ + y^ = x + y. From Fig. 46a it is seen that x'' -\- y^ = r^, X -\- y = r cos 9 + r sin e. Hence the polar equation of the circle is r2 = r cos 9 + r sin e or r = cos fl + sin B. Ex. 2. By changing to rectangular coordinates show that r (2 cos 9 + 3 sin 8) = 4 is the polar equation of a straight line. Since r cos 8 = x, r sine = y, the rectangular equation is 2a; + 3z/ = 4, which is the equation of a line whose intercepts are x = 2, y = ^. Art. 48 The Conic 117 Art. 47. Straight Line and Circle Polar Equation of a Straight Line. — Let LK (Fig. 47a) be the line r ()-.0) and let OD be the perpendicular upon it from the origin. Let OD =. p, XOD = a. If P (r, B) is any point on the line, OP = r, XOP = e. In the right triangle DOP, OP cos (DOP) = OD, that is, r cos (6 — a) = p, (47o) which is the equation required. Polar Equation of a Circle. — If the circle passes through the origin (Fig. 476) let its radius be a and its center (a, a). Let A be the Fig. 47a. p (r,e) end of the diameter through the origin and let P (r, 6) be any point on the circle. In the triangle AOP, 0A=2a, OP = r, AOP = d - a. Since OP = OA cos {AOP), the equation of the circle is then Fig. 476. r = 2 a cos (B — a). (476) If the circle does not pass through the origin (Fig. 47c) let its radius be a and its center C (6, ff). Let P (r, d) be any point on the circle. In the triangle COP, CP- = OF' + OC^ - 2 0C • OP cos (COP), that is, ai = r' + b''-2br cos {d - /3), which is the equation required. Art. 48. The Conic Fig. 47c. The locus of a point moving in such a way that its distance from a fixed point is proportional to its distance from a fixed straight line is called a conic. The fixed point is called a focus, the fixed line a directrix of the 118 PoLAK Coordinates Chap. 6 conic. The constant ratio is called the eccentricity of the curve. The ellipse, parabola and hyperbola are all conies. The name conic refers to the fact that any section of a cone is a conic. Let P be any point on a conic whose focus is F and directrix RS (Fig. 48a). If e is the eccentricity, the definition of the curve is FP = eNP. Take F as origin and the line through F perpendicular to RS as the a;-axis. Let DF = k. Then FP = r, NP = DF + FM = k + r cos 6. Hence ke Fig. 48a. r = e(k-\-rcoad), or r = 1 — e cos 6' (48o) This is the equation of the conic with focus at the origin and direc- trix perpendicular to the initial line at the point (— /c, 0). Change to rectangular coordinates, using F as origin and DF as X-axis. Then r = V x^ + 2/^, r cos 9 = x. Equation (48ffl) can be written r = ke -\- re cos 6 ■■ e {k + x). Consequently, The Ellipse- 6== a;2 + 2/2 = e2 {k + x^. (486) ■ Suppose e < 1. Let a be greater than 6 and e = VaF- (48c) a • w a' — V Equation (486) is then equivalent to (x - Va2 - vy This is the equation of an ellipse with semi-axes a and 6 and major axis horizontal. Since a and 6 can have any values it follows, con- versely, that any ellipse is a conic with eccentricity less than unity. Art. 48 The Conic 119 The center of the eUipse is {y/a? — 6", 0). Since the origin is the focus the distance from the center to the focus is then c = Va? - b\ (iSd) This shows that FB = a. By symmetry there is another focus at the same distance on the opposite side of the center. Hence an eUipse has two foci on the major axis at a distance from the ends of the minor axis equal to the semi-major axis. The Parabola. — Suppose e = 1. Let k = ^a. Equation (486) is then equiva- lent to ' 2/2 = a(a; + jffl). This is a parabola. Since a can have any value, it follows, conversely, that any parabola is a conic with eccentricity equal to unity. The vertex of the parab- ola is (— I a, 0). The distance from the vertex to the focus is then equal to the absolute value of | a. Consequently, a parabola y'^ = ax has a focus on its axis at the distance J a from the vertex (Fig. 48c). The Hyperbola. — Suppose e > 1. Let Fig. 48c. Va^ + W a Va?- + V (48e) Equation (486) is then equivalent to {x + y/a?- + ¥y 1i=^- Fig. 48d. This is the equation of a hyperbola with semi-axes a and 6 and transverse axis horizontal (Fig. A&d). Since a and 6 can have any values it follows, conversely, that any hyperbola is a conic with eccentricity greater than unity. 120 PoLAB Coordinates Chap. 6 The center of the hyperbola is (- Va^ + 6^ 0). Since the origin is the focus the distance from the center to the focus is then e = V^+fc^. (48/) By symmetry there is another focus at the same distance on the other side of the center. Therefore, a hyperbola has two foci on its transverse axis at a distance from its center equal to half the diagonal of the rectangle on its axes. Exercises Determine the loci represented by the following equations. Con- struct the graphs. Change to rectangular coordinates. 1. rcosd = —2 2. J- sin 9 = 3. 10. r^ -ir cost e -^]+. 3 = 0. 7. r = 2 sin K-i)- (-!)■ / , x\ 11. 7-2-2 »- (cos 9 + sine) + 1 =0. 3. r = sec^9 + ^j- 12. r (2 - cos 9) = 2. 4. fl = -f X. 13. J- (2 - 3 cos e) = 6. 5. r = 3 cos e. 14. r (1 — cos e) = 3. 6. r = 4sin9. 15. r (1 + sine) = — 2. 16. r (1 + sin 9 + cos 9) = 4. 17. r = 2 sec 9 — 3 esc 0. 8. r = cos 9 — sin 9. 18. r cos (2 9) = sin 9. 9. r = -2. Determine the polar equations of the following loci: 19. 2/ = 2x - 1. 22. xy = 7. 20. 1/2 = 4 s. 23. x2 -j/2 = 1. 21. x^-\-y^ = 2x. 24. x' + 2/^ = 4x + 4?/. 25. Find the polar equations of the circles of radius a tangent to both coordinate axes. 26. Find the polar equation of the parabola with focus at the origin and vertex (-■!)■ 27. Find the polar equation of the ellipse with focus at the origin, center on the y-2ciaa, eccentricity f, and passing through the point ('•D- 28. Find the eccentricity of the rectangular hyperbola x^ — y"^ — o°. Art. 49 Graphing Equations 121 Art. 49. Graphing Equations To plot the graph of a polar equation make a table of pairs of values satisfying the equation, plot the corresponding points and draw a smooth curve through them. It may be useful to look for any of the things mentioned in connection with plotting in rectangu- lar coordinates. In most cases, however, it is sufficient to imagine 6, beginning with some definite value, to gradually increase and determine at each instant merely whether r is increasing or de- creasing, draw a curve on which r varies in the proper direction and mark accurately the points where r is a maximum, minimum or zero. Proceed in the same way with 6 decreasing from the initial value. Example 1. r = 6 + 1. The curve passes through the origin at 6 = — 1. As 6 increases from this value, r is positive and steadily increases. While the angle makes a complete turn about the origin, r is increased by 2 t. This part of the curve (indicated by the continuous line, Fig. 49a) thus consists of a series of expand- ing coUs. Values of d less than — 1 give negative values of r (indi- 122 Polar Coordinates Chap. 6 cated by the dotted line). Angles 6 = —1 ± A; give numerically equal values of r. In particular, when k is an odd multiple of J tt the resulting points coincide. Hence the two sets of coils cross on the line perpendicular to 9 = — 1. A curve like this containing an infinite number of coils is called a spiral. Ex. 2. The cardioid r = a(l + cos0). The curve is shown in Fig. 496. As 6 increases from to | tt, r decreases from 2 a to o. As 6 increases from § tt to tt, r decreases to 0. Since cos 9 cannot be less than — 1, ?• does not become nega- tive. As d goes from x to 2 tt, r increases from to 2 o. Since Fig. 496. cos (6 + 2t) = cos 6, values of 6 greater than 2 t and negative values give points already plotted. Ex. 3. The lemniscate r^ = 2 a^ cos (2^). To each value of 6 correspond two values of r differing only in sign. The curve is therefore symmetrical with respect to the origin. Also angles differing only in sign give the same values of r. Hence the curve is symmetrical with respect to the initial line. As 6 increases from to J TT, r varies from ±aV2 to 0. When 6 is between J tt and f tt, cos (2 d) is negative and and r is imaginary. As 6 increases from f tt Art. 49 Graphing Equations 123 to TT, r goes from to ±a V2. Values of d greater than tt and nega- tive values of d give points already plotted (Fig. 49c). Fig. 49c. Fig. 49(i. Ex. 4. r = a cos (2 6) . The maximum, minimum and zero values of the cosine occur when the angle is zero or a multiple of i tt. The most important points on the curve are then the following: 26 = 0, 9 = 0, r = a, IT 2' IT 4' 0, X 2' 2 • 3 4"' 0, 27r, 2 ' 5 r' 0, 3x, r- -IT, 47r, 0, 27r, Values of d greater than 2x give points already plotted (Fig. 49d). Ex. 5. r = a sec (| 0). If is increased by 4 tt the value of r is not changed. For a sec [|(5 + 4x)] = asec (|S + Bx) = a sec (f 6). The whole curve is then obtained by plotting points from 6 = to 6 = Air. The most important points are 1^ = 0, 6 = 0, IT 2' X 3' i'^. 2'^' T, 6x, 4x, r = a, +00, —00, —a, — oo, +oo, . . . , 124 Polar Coordinates Chap. 6 where + oo , — oo means that as 6 increases to J x, r becomes indefi- nitely large and positive, while as 6 decreases to ^ ir, r becomes in- definitely large and negative. The curve is shown in Fig. 49e. cl A J — -V X ,' \ \ } / B Fig. 49e. Fig. 49/. Ex.6, r-2 = a2sin(0). The curve (Fig. 49/) is sjrmmetrical with respect to the origin. When d increases from to | tt, the positive value of r increases from to a. As 6 increases from ^t to tt, r decreases to 0. Between = tt and 9 = 2ir the sine is negative and no point is obtained. Negative values of 8 and those greater than 2 tt give points already plotted. Art. 50. Intersection of Curves If the polar coordinates of a point satisfy the equation of a curve, the point lies on the curve. A point may, however, lie on a curve although its coordinates (as given) do not satisfy the equation of the curve. This happens because a point has several pairs of polar coordinates. One of these pairs may satisfy a given equation while another does not. Thus the point B ( a, — ^ I lies on the curve r' = a' sin (6) (Fig. 49/) but its coordinates do not satisfy the equa- Art. 50 Intersection of Curves 125 tion. The coordinates I — <*, o ) represent the same point and satisfy the equation. To find the intersections of two curves we solve their equations simultaneously. The pairs of coordinates thus obtained represent points on both curves. There may, however, be other points of intersection. This happens when some of the pairs of polar co- ordinates representing a point satisfy one equation, other pairs satisfy the other, but no pair satisfies both. In finding the inter- sections of curves represented by polar equations the graphs should always be drawn. Any extra intersections will then be seen. / Example 1. Show that the point («, „ ) lies on the curve f = a" sin (3 0). The coordinates given do not satisfy the equation, for = - gives r' = a^sm (|-ir) = —a^. The point (a, 7^ ) can, however, be written ( ~fl, ;;''■) its coordinates satisfy the equation (Fig. 50a) . and in this form I I r ; v. \. '/ / Fig. 50a. Ex. 2. Find the intersections of the curves r'' = a" sin d, r = a. Solving simultaneously, we get a?- sin d = a^. TV Consequently sin 9 = 1 and 6 = -. One point of mtersection is 126 Polar Coordinates Chap. 6 then Ala,-]. It is seen from the figure that B I — a, - tt j is also a point of intersection. Exercises Plot the following pairs of curves: 1. r = sin B, y = sin x. 2. r = cos 6, y = cos x. Sketch the graphs of the following equations: 5. r = 0. 6. r = 1-0. 8. r = l". 9. r = a sin (2 9). 10. r = a cos (3 9). 11. r =o(l +sin9). 12. r = a(2 + sine). 13. r = a (1 + 2 sin 0). 14. r = asin(fl-|j- 3. r = tan 0, y = tan x. 4. r = aece, y = sec x. equations: 19. 20. 21. 22. r = 1 + cos (1 0). r = 4 + 5cos(5e). r = 1 -2 cos (^9). r = sin 9 cos 9. 23. r=sm3(^-j. 24. 25. 26. 7-2 = 2 a^ sin (2 9). /■= = 1 + sin 9. r^ = tan (2 9). 27. a , a r = cos sin 9 28. rsin9 = acos (2 9). 1 2» + l 29. 2« - 1 15. r = apos (j 9). 16. J- = atan(|9). 17. r = o(l + sin 2 9). 18. r = aXl +2 cos 3 9). 30. Show that the point (1, jtt) lies on the curve r = sin (2 9) but that the coordinates given do not satisfy the equation of the curve. 31. Show that the point (1, ^x) Ues on the curve r = — 2sin (x) but that the coordinates given do not satisfy the equation of the curve. 32. Show that if a is a constant the equations 1 1 f = y = a cos 9 + 1 ' a cos 9—1 represent the same curve. 33. Show that the equations r' = a? cos^ (2 9) and r = o cos (2 9) represent the same curve. 34. Why do the equations r' = ar cos 9 and r = o cob 9 represent the same curve? Plot the following pairs of curves and find their points of intersection: 35. r = 2secf9-|j, r = 2sec(9 + |j- Art. 51 LocTjs Problems 127 36. r sin fl = a, r cos 6 = a. 37. r* = a^ sin 0, f' = a' sin (3 0). 38. r = a sin (2 S), r = a (1 - cos 3 0). 39. r2 = 2 a2 cos (2 9), r = a. Fig. 51a. Art. 51. Locus Problems In finding the equation of the locus of a moving point, either polar or rectangular coordinates can be used. The system should be chosen which seems to fit the problem best, making a change of coordinates in the resulting equa- tion if necessary. If the positions of origin and axes are not specified they should be placed in the most convenient position. This is usually (though not always) the most sym- metrical position. Exam-pie 1. Through a fixed point draw a line intersecting a fixed circle in Qi and Qi (Fig. 51a). On this line determine the point P such that OP OQi ^ OQ2 Find the locus of P as OP turns about 0. Take the origin at and the initial line through the center (6, 0) of the fixed circle. Let Qi, Q2, and P be (n, 0), (r^, e), and (r, 0) respectively. By hypothesis 2 1,1 2rir2 - = - + -, or r = - — I The equation of the fixed circle is r^ — 2 rb cos + V' — a' = 0, that is r = 6 cos e ± Va^ - 6^ siii2 0_ The two values of ?■ in this equation are n and ri. Hence ri + rj = 2 6 cos d, nra = 6^ cos^ 6 - {a^ - V sm^ d) = W - a\ Therefore V^ — a' 6^ — a' bcQsd -sec 9 p 128 PoLAK Coordinates Chap. 6 is the equation of tlie locus described by P. It is a straight line through the points at which Unes through are tangent to the circle. Ex. 2. A point moves so that the product of its distances from two fixed points is equal to the square of half the distance between them. Find its locus. Take the x-axis through the fixed points and the origin mid- way between them. Let the distance between the fixed points be 2 a. If P (x, y) is any point on the locus (Fig. 516)^ by hypothesis, A'P • AP = 0A\ or V{x - ay + 2/2 V(x + ay + 2/2 = a'. Squaring, (x2 + 2/2 + a2)2 - 4 aV = a*. Changing to polar coordinates, (7-2 + a'^y - 4 aV cos" 6 = a*, or f' = 2 a" (2 cos^ 6-1) X = 2 a^ cos (2 9), which is the equation of a lemnis- FiG. 51b. cate. Exercises .1. LK is a fixed straight line perpendicular to the initial line at (a, 0). On any line through the origin, intersecting LK in Q, is taken a point P such that OP-OQ= aK Find the locus of P as OQ turns about 0. 2. LK is a fixed straight line perpendicular to the initial line at A (o, 0). Any line through intersects LK in Q and a point P is taken on this line such that PQ = AQ. Find the locus of P as OQ turns about 0. 3. A point P moves in such a way that its distance from a fixed point multiplied by its distance from a fixed straight line LK is constant. Find the locus of P. 4. A segment of fixed length slides with its ends in the x and y axes. Find the locus of the foot of the perpendicular from the origin to the moving segment. 5. A revolving line passing through the center of a fixed circle Art. 51 LocDS Pkoblems 129 intersects the circle in a point Pi and a fixed straight line in Pi. Find the locus described by the point midway between Pi and P2. 6. Let OA be the diameter of a fixed circle and let LK be tangent to the circle at A. Through draw any line intersecting the circle in D and LK in E. On OE lay off a distance OP equal to DE. Find the locus of P as OE turns about O. 7. From a point on a fixed circle perpendiculars are dropped upon the tangents of the circle. Taking as origin and the diameter through as initial Hne find the polar equation of the curve generated by the feet of these perpendiculars. 8. A circle rolls along the initial hne and a line through the center of the circle turns about the origin. Find the locus of the intersections of the moving Hne and circle. 9. A circle rolls along the initial line. A line through the origin moves in such a way as to remain tangent to the circle. Find the locus of the point of tangency. 10. Take a fixed point and a fixed straight Kne BC. Through O draw any hne intersecting BC in D and on this Une lay oil a constant distance DP measured from D in either direction. Find the locus described by P as the line turns about 0. 11. Through a fixed point on the circumference of a fixed circle draw any hne cutting the circle again at D and lay off on this line a constant distance DP measured from D in either direction. Find tlie locus of P as OP turns about 0. 12. A straight line OA of constant length revolves about 0. Through A a line is drawn perpendicular to the initial line intersecting it in B. Thrdugh B a fine is drawn perpendicular to OA intersecting it in P. Find the locus of P. 13. MN is a straight line perpendicular to the initial hne at A (a, 0). From a line is drawn to any point B of MN. Through B a line is drawn perpendicular to OB intersecting the initial hne at C. Through C a line is drawn perpendicular to BC intersecting MN at D. Finally, through D a line is drawn perpendicular to CD intersecting OB at P. Find the locus of P. 14. Two circles whose centers are fixed and whose circumferences touch rotate without sUpping. A line through the center of one circle and rotating with it intersects a similar line on the other circle in a point P Find the locus of P. If the radii are incommensurable show •that the locus passes indefinitely near any point of the plane. CHAPTER 7 PARAMETRIC REPRESENTATION Art. 52. Definition of Parameter In some cases it is more convenient to express the coordinates of a point on a curve in terms of a third variable than in terms of each other. Such a third variable is often called a -parameter and the equations connecting the coordinates with the parameter are called parametric equations. Example 1. A rod 5 feet long moves with its ends in the co- ordinate axes. Find the locus of the point P, 2 feet from the end in the X-axis. Y B ^s^ X0\ p y X Fig. 52a. Fig. 526. Let QR be the rod and = ZOQR (Fig. 52a). By hypothesis PQ = 2 and PR = 3. Let x, y be the coordinates of P. From Fig. 52o it is seen that a; = 3 cos 0, ?/ = 2sin<^. These are the equations expressing the coordinates of P in terms of the parameter. To find the equation connecting x and y we elim- inate the parameter. The result is ^ + x = sm''0-|-cos^<^ = l. The locus of P is therefore an ellipse. 130 Art. 52 Definition op Pakameteb 131 Ex. 2. Find the locus of the point P, intersection of OB and MQ, (Fig. 526) given that MQ = arc AB. Let the radius of the circle be a and Z AOQ = (j>. Let r, 6 be the polar coordinates of P Then arc AB = aJS, MQ = a sin <^. Consequently, 6 = sin <^. Also _ OM _ fflcos cos d cos (sin 0) Polar parametric equations of the locus are therefore d = ain(j>, r = a cos 4> sec (sin 0). Elimination of (j> gives the equation connecting r and d in the form r = a^l^^ sec d. Almost any quantity that varies from point to point of a curve can be used as a parameter. The equations connecting the co- ordinates with the parameter naturally depend on the parameter. There are then an infinite number of parametric equations of the same curve. The coordinate axes and the parameter must be fixed before the parametric equations have a definite form. Exercises 1. A circle of radius a has its center at the origin. Express the coordinates x, y of any point P on the circle in terms of the angle between the x-axis and OP and so obtain parametric equations for the circle'. 2. A (0, a) is a fixed point on the ?/-axis and M a movable point on the X-axis. MP is perpendicular to AM and equal in length to it. Express the coordinates of P in terms of a and the angle from the right end of the x-axis to MP. By eliminating find the rectangular equation of the locus of P. 3. is the center of a fixed circle tangent to AB at A. Through any point Q of AB passes another line tangent to the circle at R. On RQ determine the point P such that RQ = QP. Taking OA as x-axis and O as origin express the coordinates of P in terms oi = /.AOQ. By eliminating the parameter <^ determine the coordinate equation of the locus of P. 4. A segment AB through the point C (2, 1) has its ends A and B in the x- and ^-axes respectively. If P (x, y) is the middle point of AB 132 Parametric Representation Chap. 7 express x and y in terms of the angle <^ = OAB. By eliminating de- termine the rectangular equation of the curve described by P as AB turns about C. 5. A string, held taut, is unwound from a circle. Taking the origin at the center of the circle and the initial line through the point where the string begins to unwind, express the polar coordinates of the point P at the end of the string in terms of the radius of the circle and the angle at the center subtended by the arc unwound. FLad the polar equation of the locus of P. Art. 63. Locus of Parametric Equations Suppose the coordinates x and y are given functions of a param- eter , (1) x=fi {), 2/=/2(0). If a value is assigned to (f> the resulting values of x and y are coordi- nates of a certain point. The totality of such points will usually be found to form a curve. The above equations represent the curve in the sense that if. any value be assigned to 4> the resulting point lies on the curve and if any point (x, y) be taken on the curve there is a value of such that x = fi{), y = f^ (0). To plot the curve we assign values to <^, considered as independ- ent variable, calculate x and y and plot the resulting points. To find the coordinate equation eliminate (j) between the para- metric equations. Let the result be (2) f(.x,y)=0. This equation, being a consequence of the parametric equations, is satisfied by the coordinates of any point on the curve. If, con- versely, for every pair of values x, y satisfying (2) a value of (j> can be found such that x = fi (), y = fi (0), then (2) is the coordinate equation of the curve. Example 1. Plot the curve whose parametric equations are x = t, y = t' and find its rectan- gular equation. In the table values of x and y are placed after the corresponding values of the parameter t. The points represented by these pairs of co- ordinates are plotted (Kg. 53a) and a smooth curve drawn through t X y ±00 ±00 ±00 ±2 ±2 ±8 ±? ±* i^'^ ±1 ±1 ±1 M ±* ±i Art. 53 Locus OF Parametric Equations 133 the resulting points. By eliminating t the rectangular equation of the curve is found to be ?/ = x^. Ex. 2. a; = 3 cos <^, 2/ = 2 sin <^. While (j> increases from to -, Y 1 T r ^ J 7 t J 7~ nr V z' O X i 1 1 1 I __ _ T p'y^ B .P i \ A D ^ Fig. 536. Fig. 53a. X decreases from 3 to and y increases from to 2 {AB, Fig. 536). As continues to x, y decreases to while x decreases to —3 {BC, Fig. 536). Then y decreases to — 2 at <^ = | x while x in- creases to 0. Finally, x increases to 3 and y to ai 1, x and y are both positive and increase with t (E to F). If i is replaced by — i — 1, the x and y coordinates interchange values (P to P'). Hence the curve is symmetrical with respect to the line OQ bisecting the angle between the axes. By subtraction (1) y-x = 4:t + 2, or t = l{y-x-2). Substituting t in the equation x = t (t — 1) and simplifying (2) {y -xy-8{y + x) + 12 = 0. Conversely, if (1) and (2) are solved for x and y the parametric equations are obtained. Therefore (2) is the rectangular equation of the curve. It is a parabola. Ex. 4. X = cos (2 ^), y = sin (j). When <^ = we get the point A (1, 0). As (^ increases, x decreases and y increases until = -^ at B ( — 1, 1). As continues to increase the point turns back, retraces the arc BA and continues to C (—1, —1) where ^ = § tt. As continues to increase the point oscillates back and forth along the path CAB (Fig. 63d). Since x = cos (2 <^) = 1 — 2 sin^ <^ and y = sin 4>, X = 1 - 2 2/2. This equation is not however equiv- alent to the parametric equations. It represents a parabola ex- tending to an infinite distance on the left, whereas the parametric equations (since sin and cos 2 <^ are never greater than 1) rep- resent only the piece CAB. Fig. hM. Art. 54 Parametric from Coordinate Equations 135 Ex. 5. Find the intersections of the curves x = acosd, y = b cos (2 6) and a; = a sin <^, 2/ = 6 sin (2 <^). At a point of intersection both pairs of parametric equations are satisfied. Hence sin = cos 6, sin (2 <^) = cos (2 6). The general solution of the first of these equations is defined by the equation i/ = sin 0. Substituting y = sva. in the equation of the curve we get X = I — sin^ 4> = cos'' 4>- The equations obtained are therefore X = cos' 0, y = sin . Since x and y given by these equations cannot be greater than 1, these are not parametric equations of the whole parabola but only of a piece of it. This is due to the fact that when y is numerically greater than 1 there is no parameter defined by the equation 2/ = sin 0. Art. 54 Pabametbic fbom Cooedinate Equations 137 Exercises Plot the curves represented by the following parametric equations and determine the corresponding coordinate equations: 1. X = l+2t, y = 2 -3t. 8. X = tsmt, y = tcoat. 2. x = i'', y = 2t. 9. r = cos(2(^), 9 = sin<^. Z. x = t + -^, y = t-j- \Q. r = t + j, « = « - ^■ 4. x = e{t-l),y = P{t+l). 11. j- = (^sin(^, e = 0cos0. _t + \ _ t 12: x = a(d—sine),y = a(l — cose). ■ ^ " t - V ''•' " t-2 13. r = tan 0, e = sin (3 *). 6. a; = sin0, j/ = sin(<^ + ^ j- 14. a; = a (3 cos + cos 3 1, 2/ = a (3 sin (^ — sin 3 0). 7. X = sec 0, 1/ = tan (^. 15. Plot two turns of the curve X = cosi^, y and show that the complete curve passes indefinitely near any point within a square. 16. Find the polar parametric equations of the curve X = t cos t, y = tsssit, using the same parameter. 17. Find rectangular parametric equations of the curve r = P, e = l + t. 18. Show that x = srD.t,y = cos (2 t) and x = sin (2 t), y = cos (4 1) are the same curve. 19. Area; = < + -, 2/ = i - -andx = 2' + 2-', 2/ = 2' -2-' the same curve? 20. Are x = t -\--, y = t — -: and s = cos 9 + sec 0, y = cos 9 — sec 9 the same curve? Find the intersections of the following pairs of curves: y = 2t) ■ 2/ = osin fl) 2/ = a0sm0 ) x = 5cose) a; = l+<) a; = 2o cos^ ^ 2/ = 5 sin 9 ) ' 2/ = 2 + ' ^'^^ ^ 2 a cos^ (» [ , r = 4 a cos 9. ^ ~ sini^ J 25. Find rectangular parametric equations of the circle, radius a, with center on the a;-axis and passing through the origin, using as pa- rameter the polar coordinate 6. 138 Pabametric Representation Chap. 7 26. Find parametric equations for the parabola 2/2 = 4x, using the ratio - as parameter. 27. Find parametric equations of the ellipse 5!4.^ = T a= "^ 62 ' using the parameter defined by the equation x = a cos <^. 28. Find parametric equations for the hyperbola ^ _ 1' = 1 using the parameter defined by the equation x = a sec . 29. Find parametric equations of the curve x^ +y^ = a^, using the parameter defined by the equation x = a sin' 0. 30. Find parametric equations of the curve yH ^ 4,y = X' - 2X' + X - 4, using as parameter the slope of the line joining (x, y) and (1, —2). Art. 65. Locus Problems The Cycloid. — When a circle rolls along a straight Hne a point of its circumference describes a curve called a cycloid. Let a Qircle of radius a and center C roll along the x-axis (Fig. 55a). Let N be the point of contact with the a;-axis and P (x, y) T Fig. 55a. the point tracing the cycloid. Take as origin the point found by rolling the circle to the left until P meets OX. Take as parameter the angle NCP = . Since the point of contact moves the same distance along the circle as along the straight line, ON = arc NP = a^. Art. 55 Locus Problems 139 Consequently X = OM = ON - MN = 0N -PR = a-a sin 4,, y = MP = NC - RC = a - a(iOs4>. The parametric equations of the cycloid are then X = a{4> — SVD.41), y = a{\ — cos <^). The Epicycloid. — When a circle rolls on the outside of a fixed circle a point on its circumference describes an epicycloid. Let a circle of radius a and center C roll on the outside of a circle of radius h and center 0. Let A'^ be the point of contact and P (x, y) the point describing the epicycloid. Let A be the point obtained by roUing the moving circle backward until P meets the fixed circle. Fig. 556. Let be the origin, OA the z-axis. Let OCP = d and take as pa- rameter the angle AOC = 4>. Since the point of contact moves the same distance along both circles arc AN = arc NP and conse- quently b4> = aO. Also RCP = ocp-ocR = e-(^-4>] = e + ,j>-~- Therefore X = OM = OS + RP = OC cosel) + CP sin (RCP) = {a + b)cos4> —a cos (6 +4>) = ia + b) cos4> — a cos I j. 140 Parametric Representation Chap. 7 y = MP = SC - RC = OC sin0 - CP cos (RCP) ■ /a + bA = (o+ fc) sin — o sin (6+ <^) = (a + 6) sin — a sin ( I. The parametric equations of the epicycloid are then x={a+b) cos 0— a cos ( <^ I, y={a-\-b) sm4>—a sini 1. Exercises 1. A circle of radius a moves with its center in the a;-axis and a straight Une passes through the center of the circle and a fixed point on the j/-axis. Using as parameter the angle between the hne and y-axis, find parametric equations for the curves traced by the inter- sections of moving line and circle. 2. Let AB be a given line, O a given point k units distant from AB. Draw any Une through meeting AB in M and let P be the point on this line such that OM-MP = h\ Find the parametric equations of the locus of P, using as the origin, the perpendicular from O to AB as x-axis and the angle between OX and OP as parameter. 3. Let OA be the diameter of a fixed circle and LK the tangent at A. A variable line through intersects the circle at B and LK at C. Through B draw a hne parallel to LK and through C a line perpendicular to LK and call the intersection of these lines P. The locus of P is a curve called the witch. Find its parametric equations using the tan- gent at O as a;-axis and the angle from the a;-axis to OC as parameter. Also find the rectangular and polar equations of the curve. 4. Let be the center of a circle, radius a, A a fixed point and B a moving point on the circle. If the tangent at B meets the tangent at A in C and P is the middle point of BC, find the equations of the locus of P using the angle AOB as parameter. Also find the rectangular equation. 5. OBCD is a rectangle with OB = a, BC = c. Any line is drawn through C meeting OB in E and the triangle EPO is constructed so that the angles CEP and EPO are right angles. Find the locus of P, using the angle DOP as parameter, OB as x-axis and as origin. Also find the rectangular equation of the locus. 6. A (0, —a) and B (0, a) are two fixed points on the ?/-axis. H is a variable point on the a;-axis. BK is the perpendicular from B to AH meeting it in K. Through K a line is drawn parallel to the x-axis and through H a line is drawn parallel to the 2/-axis. These lines meet in P. Art. 66 Locus Problems 141 Find the equations of the locus of P using the angle BAK as parameter. Also find the rectangular and polar equations of the locus. 7. Let OA be the diameter of a fixed circle and LK the tangent at A. Through draw any Hne intersecting the circle in B and LK in C and let P be the middle point of BC. Find the equations of the locus of P, using the angle AOP as parameter, OA as y-ajds and as origin. Find the rectangular and polar equations of the same curve. 8. OA is a diameter of a fixed circle and LK the tangent at A. Through any line is drawn meeting the circle in B and LK in C. Through B a line is drawn perpendicular to OA meeting it in M. MB is prolonged to P so that MP = OC. Find the locus of P 9. CD is perpendicular to OX and distant a units from 0. A is a moving point on CD. AB is drawn perpendicular to OA meeting OX in B. BP is perpendicular to OX meeting OA in P. Find the locus of P. 10. Through a moving point B on a fixed circle lines from the ends of a diameter are extended a distance equal to the radius to form the sides of a square whose diagonal is BP. Find the locus of P. 11. The sides of a right angle are tangent to two fixed circles. Find the locus of the vertex. 12. Through two fixed points lines are drawn to form an isosceles triangle with its base in a fixed fine. Find the locus of their point of intersection. 13. The angles of a triangle are A, B, C and the opposite sides are a, 6, c. If the vertex A moves along the x-axis and B along the 2/-axis find the locus of C, using the angle between the side AB and the a;-axis as parameter. 14. A string is wound around a circle and the end fastened at the center of the circle. A pencil resting against the string keeps "it taut. Find the curve described as the string unwinds from the circle. 15. When a wheel rolls along a straight hne any point in a spoke describes a trochoid. Let the wheel roll along the s-axis and use as parameter the angle 4> in the equation of the cycloid. Find the para- metric equations of the trochoid described by the point at distance b from the center of the circle. 16. A hypocycloid is the locus described by a point on the circum- ference of a circle which rolls internally on the circumference of a fixed circle. Find the parametric equations of the hypocycloid when the radius of the moving circle is J that of the fixed circle, using a parameter analogous to that in the equations of the epicycloid. 17. A circle with center at the point (2, 0) intersects a circle with center (0, 2) in a point of the line x = 3. Find the locus of the other point of intersection of the two circles. CHAPTER 8 TRANSFORMATION OF COORDINATES It is sometimes desirable to move the axes to a new position. The coordinates will then be changed and it is necessary to find the new coordinates. There are two simple cases a combination of which gives any motion. These are translation, in which the origin is moved to a new position without changing the direction of the axes, and rotation, in which the origin is left fixed and the axes turned like a rigid frame about it. Y y' h 00' P 1 y \h.lc) x' k X Fig. 56a. Art. 56. Translation of the Axes Let OX and OY (Fig. 56a) be the axes in their first position, O'X' and O'Y' the axes after motion. Let h, h be the coordinates of the new origin with respect to the old axes, x, y and x', y' the old and new coordinates of any point P. From Fig. 56a it is seen at once that X = x' -\-h, y = y' + k. (56) These are the equations connecting the new and old coordinates of any point. 142 Art. 56 Translation of the Axes 143 Example 1. The origin is moved to the point (—1, 2) the new axes being parallel to the old. Find the new equation of the curve a;2 + 2 a; - 2/ + 3 = 0. The equations connecting new and old coordinates are in this case X = x' — 1, y = y' + 2. When these values are substituted for x and y the equation of the curve becomes x"" -y' -0. Ex. 2. Find the equation of the curve r = acosd referred to a new pole at the point I -y: , - I the new initial line being parallel to the old. The equations for transformation being given in rectangular coordinates, we change to rectangular coordinates, move the origin and then change back to polar coordinates. The coordinates of the new origin are a /ir\ a , a . /ttN a ' = vi''°niJ=2' *=vi''Hir2- The rectangular equations for transformation are therefore x = x' + -, y = y +^- The rectangular equation of r=a cos is o' _x' which transforms into f /% \ x'2 + 2/'2 + ay' = 0. ^^^ 1_ In polar coordinates this is \ / r' + asine' = 0, \^^^^y which is the equation required. -pj^ ggj, Ex. 3. Transform the equation «' + 2/' = 3 a; — 3 2/ to new axes, parallel to the old, so chosen that there are no terms of first degree in the new equation. 144 Tbansformation of Coordinates Chap. 8 Let X = x' + h, y = y' + k. The equation becomes x'3 + 2/'3 + 3 te'2 + 3 ky'^ + S{h-l)x' +3{k + l)y' + h' + ¥-3h + 3k = 0. If there are no terms of first degree h-1 =0, ^ + 1 = 0. Consequently, h = 1, k = —1 and the transformed equation is x'' + 2/'3 + 3 x'2 - 3 2/'^ - 6 = 0. Art. 57. Rotation of the Axes The rotation is most easily express- ed in polar coordi- nates. Let the ini- tial line OX be rotated through an angle ^ to OX' (Fig. 57a) the origin re- "x maining fixed. Let the old and new coordinates of arty point Pher,6 and r', d'. From the figure it is seen that r = r', d = 6' + . Using these relations we get X = r cos d = r' cos {d' + <^) = / cos d' cos <{> — r' sin 6' sin 4>, y = rsind = r' sin (d' + 4>) = r' sin 6' cos <^ + r' cos 6' sin 4>. Replacing r' cos d' and r' sin 6' by x' and y', X = x' coscj) — y' sin 0, ; y = y' cos + x' sincji- Example 1. Find the new equation of the curve • x' -\- y'' = X — y after the axes have been rotated through an angle of 45° x'-y' Fig. 57a. (57) In this case x = x' cos (45°) — y' sin (45°) y = y' cos (45°) -|- x' sin (45°) = Substituting these values and simplifying x'' + 3x'y"' + 2y' = 0. V2 x'+y V2 Art. 58 Invariants 145 Ex. 2. Find the polar equation of the curve x' -y^-2xyV3 = 2 after the origin is moved to the point ( — —, -^) and the axes omt I — rotated through an angle of — 30°. / V6 V2\ The equations for moving the origin to ( — , — r- J ^^^ x = x'-—, v = y+-^- If the axes are then rotated through —30° Vs + y" , y"y/l- x" x'^ "" ^i^" , y' = '< 2 ' " 2 Consequently, x" V3 + ?/" - Ve 2/" V3 - a;" + V2 x = ^ , y = ^ These values substituted in the original equation give x"2 - 2 y/2x" - 2/"2 + 1 = 0. Changing to polar coordinates r" (2cos2e - I) - 2 V2rcosfl + 1 = 0, 1 or r = — ;= V2cose ±1 The curves given by the positive and negative signs in this equa- tion are identical. The equation required is then 1 V2COS0 + 1 Art. 68. Invariants Quantities associated with a curve are of two kinds, those depend- ing on the position of the axes and those independent of the position of the axes. Quantities of the second kind are called invariant. For example, the radius of a circle is invariant (does not depend on the position of the axes) but the coordinates of its center are not invariant (change when the axes change position). 146 Transformation of Coordinates Chap. 8 The equation of a curve is not invariant but there are some things connected with it that are invariant. One of the simplest is the degree of a rectangular equation. Wherever the axes are placed this degree is always the same. To see this it is only neces- sary to note that the equations for change of axes always have the form X = ax' + by' + c, y = dx' + ey' +/, where a, b, c, d, e, f are constants. A term x'"y" is equal to a sum of terms none of which is of higher degree than the {m + n)th in x' and y'. Hence through change of axes the degree of a poly- nomial cannot be increased. Neither can it be diminished for, in that case, changing back to the old axes would have to increase it. Therefore the degree of a rectangular equation is invariant. If a rectangular equation can be factored so can the new equa- tion resulting through a change of axes ; for each factor of the old wUl be changed into a factor of the new. Also since real ex- pressions are replaced by real, if either has real factors the other will have real factors also. Art. 59. General Equation of the Second Degree An equation of the second degree in rectangular coordinates has the form (1) Ax' + Bxy + Cy' + Dx + Ey + F = 0. Rotating the axes through an angle 4> the coefficient of x'y' in the resulting equation is —2 A cos (f> sincj) + B (cos^ 4> — sin^ ) -|- 2 C cos <^ sin (^ = Bcos i2) + {C - A) sin (20). This is zero if (2) tan(2<^) = j4c- If then the axes are turned through an angle satisfying equation (2), equation (1) takes the form (3) A'x'^ + C'y'^ + D'x' + E'y' + F' = 0. Art. 59 General Equation of the Second Degbee 147 If A' and C" are both different from zero, completion of the squares gives an equation of the form A' {x - hy + C {y - ky = R. If /E is zero this equation is reducible. If 72 is not zero the equation can be written (.,' _ hy {y' - ky _ R/A' "^ R/C The locus of this equation is an ellipse if the denominators are both positive, a hyperbola if one is positive and one negative, and is imaginary if both are negative. Let A' = 0. Completion of the square in y' terms then gives C" (2/' - ky = -D' (x' - h). The locus of this equation is a parabola. Similarly if C is zero the locus is a parabola. A' and C" cannot both be zero as the equation would then be one of the first degree. We have thus found that the second degree equation in rectangular coordinates is either reducible or else its locus is an ellipse, a parabola, a hyperbola, or an imaginary curve. Exercises 1. What are the rectangular coordinates of the points (2, 3), (—4, 5) and (5, —7) referred to parallel axes through the point (3, —2)1 2. Find the polar coordinates of the points I 3, 5 I, I — 2, - j, ( 4, -^ j if the origin is moved to the point 12, ^ I , the new initial line having the same direction as the old. 3. Find the equation of the curve x^ + iy^ - 2 X + 8y + 1 = referred to parallel axes through the point (1, —1). 4. Find the equation of the curve y3 _ 62/2 + 3a;2 + 12y - 18.-C + 35 = referred to parallel axes through (3, 2). 5. Find the equation of the circle r^ — 5 V3 r cos 9 — 5 r sin 9 + 21 =0 148 Transformation of Coordinates Chap. 8 when the origin is moved to the point (5, ^), the new initial Une being parallel to the old. 6. Find the equation of the parabola r = ~ 1 — cos e when the pole is moved to (— p, 0), the direction of the initial line being unchanged. 7. Find the equation of the Une 2a; -32/ + 7 = referred to parallel axes through the point ( — 5, —1). 8. Find the equation of the ellipse E! 4. L' = 1 o2 "^ 62 when referred to parallel axes through the left-hand vertex. 9. Find the equation of the witch ^ So? ^ x^ + ia'^ referred to parallel axes through the intersection of the y-ajds and the curve. 10. Find the equation of the strophoid ^ X' (a - x) " a+x referred to parallel axes through the intersection of the x-axis and the asymptote of the curve. 11. Find the parametric equations of the cycloid X = a{ — sin ), y = a (1 — cos ) referred to parallel axes through the highest point of an arch of the curve. 12. Find parametric equations for the cardioid r = a (1 -f- cos B) referred to parallel axes through the point x = ia, y = 0. 13. Transform the equations x + y~3 = 0, 2x-3y + 'i = to parallel axes so chosen that the new equations have no constant terms. 14. Transform the equation x' + if - 2 X + 4:y - 6 = to parallel axes sci chosen that the new equation has no terms of first degree in x and y. Art. 59 General Equation of the Second Degree 149 15. Show that by a change to parallel axes the equation 2/2 + 42/ -2x + 3 =0 can be reduced to the form 2/'2 = 2 a;'. 16. Show that by moving the origin to the point (p, 0) and changing to polar coordinates the equation 2/2 = 4px can be reduced to the form r' = ^P . 1 — cos d' 17. Show that by a change to parallel axes the equation 3;2 + 4 2/2-3x + 52/ = 6 can be reduced to the form a;'2 + 42/'2 = V, where 6 is a constant. 18. By a change to parallel axes reduce 2x2-62/2 + 3x-42/ = 12 to the form (x/ay - [y/hY = 1. 19. What are the new coordinates of the points (0, 1), (1, 0), (1, 1), ( — 1, —1) after the axes have been rotated through an angle of 60°? 20. Transform the equation a;2 -f- xi/ + 2/2 = 1 to new axes bisecting the angles between the original axes. 21. Find the equation of the curve r2 sin 6 cos 9 = 1 referred to the same pole but with the initial line rotated through 45°. 22. By rotating the axes through the proper angle transform the equation xi/ = 4 to a form without xy-tevra. 23. By a proper change of axes reduce x2 + xi/ = 2 to the form {x'/aY - {y'/bY = 1- 24. By a proper change of axes reduce the equation X + 32/-4 = to the form x' = 0. 25. If the curve y^ = 4 x is referred to new axes show that the second degree terms in the new equation form a complete square. 26. If the curve x2 + 4 j/' — 5 = is referred to new axes show that the second degree part of the new equation has imaginary factors. 27. If the curve x2 — 2/2 — 1 = is referred to new axes show that the second degree part of the new equation has real factors. CHAPTER 9 COORDINATES OF A POINT IN SPACE Art. 60. Rectangular Coordinates Let X'X, Y'Y and Z'Z be three scales with a common zero point called the origin. The lines X'X, Y'Y and Z'Z are called co- ordinate axes and are referred to as the x-axis, y-axis and 2-axis respectively. They determine three coordinate planes XOY, YOZ and ZOX called the xy-, yz- and 3a;-planes. These planes divide space into eight portions called octants. The portion 0-XYZ is sometimes called the first octant. The other octants are not usually numbered. Through any point P pass planes perpendicular to the coordi- nate axes meeting them in M, R and T. The numbers at these points in the scales are called the coordinates x, y, z of P. This point is represented by the symbol (x, y, z). To in- dicate the coordinates of P the notation P (x, y, z) is used. Usually the x-axis is drawn to the right, the y-sais for- ward and the «-axis upward. The x-co6rdinate of P is then the segment SP considered positive when drawn to the right, the 2/-coordinate is QP considered positive when drawn forward and the z-coordinate is NP considered positive when drawn upward. To plot the point P having coordinates x, y, z draw OMNP mak- ing OM = x, MN = y and NP = 2. The result is a plane figure. 150 Fig. 60. Art. 61 Projection 151 By shading and dotting lines it can however be given the appear- ance of a space construction. The points in a coordinate plane have one coordinate equal to zero. With respect to the other two coordinates these points can be treated as in plane geometry. For example, the points in the 2/2-plane whose coordinates y and z satisfy a first degree equation lie on a line. Many results of solid geometry are similar to those already found in the plane. A formula in two coordinates x and y is often extended to space by adding a similar term containing z. With a little attention to these relations many formulas of space geometry can be inferred from analogy with those in the plane. Art. 6i. Projection The projection of a segment AB upon a line RS is the segment of that line intercepted between planes through A and B perpen- c, -B, Fig. 61a. dicular to RS. The projection of AB upon a plane is the segment between the feet of perpendiculars from A and B to the plane. Since parallel planes or lines intercept two fixed lines proportionally, it follows, as in plane geometry, that segments of a line have the same ratios as their projections on a line or plane. Thus, in Figs. 61a or 616, AB:BC = AiBi : BiCi. In using this proportion AB and BC must have algebraic signs if their projections have algebraic signs and conversely. 152 Coordinates op a Point in Space Chap. 9 Projection on a Directed Line. — By the angle between two lines that do not intersect is meant the angle between intersecting lines parallel to them. Two lines de- termine two angles less than 180°, one acute and one obtuse. If the lines are directed they however determine a definite angle, the one less than 180° with the arrows on its sides pointing away from its vertex. Let 6 be the angle between a directed segment AB and a directed hne RS (Fig. 61c). Through A draw AB' parallel to RS to meet the plane through B perpen- dicular to RS. If the projection ^iBi is considered positive when it is drawn in the direction RS AiBi = AB' = AB cos d, Fig. 61b. N A , r-'-- -tT^^^ B 1 1 ^1 \ \ Fig. 61c. that is, the projection of a segment AB on a directed line RS is equal to the product of the length of the segment by the cosine of the angle between the segment and line. Projection of a Broken Line. — Let ABCD be a broken Hne joining A and D. Project on RS. If segments of RS are consid- ered opposite in sign when drawn in opposite directions AiBi + B,C, + C,Di = A,Di. Art. 61 Projection 153 This is true whether A, B, C, D are in a plane or not. Therefore, the algebraic sum of the projections of the parts of a broken line joining two points is equal to the projection of the segment from the first to the last of those points. Fig. Qld. Projections on the Axes. — Let P1P2 be the segment from Pi (xi, 2/1, zi) to P2 (X2, 2/2, Z2) (Fig. 61e). Through Pi and P2 pass planes perpendicular to the x-axis meeting it in Mi and M2. PiMi and P2M'2 are then perpendicular to OX. If segments are consid- •?2 \ \ \ / / X / / / / Fig. 61e. ered positive when drawn in the positive direction along the aj-axis, the projection of PiP2 is Jlf 1M2 = OMi - OMi = X2- xi. Similarly the projections on the y- and a-axes are 2/2 — 2/i and Zi — z\. Therefore in length and sign the projection of a segment on a coordi- nate axis is equal to the difference obtained by subtracting the coordi- nate of the beginning from that of the end of the segment. 164 Coordinates op a Point in Space Chap. 9 Exercises 1. Plot the points (0, 0, 1), (1, 1, 0), (1, 1, 1), (-1, 2, 3), (-1, -2, -3). 2. What is the x-co6rdinate of a point in the 2/2-plane? What are the X- and 2/-eo6rdinates of a point on the z-axis? 3. Where are aU the points for which 2 = — 1? What is the locus of points for which x = \ and y = 2? 4. Determine the distance of (2, y, z) from each coordinate axis. What is its distance from the origia? 5. Find the feet of the perpendiculars from (1, 2, 3) to the coordi- nate planes and to the coordinate axes. 6. Given P (1, 0, 1), Q (2, 1, 5), It (3, -1, 2), find the projections of PQ, QR and RP on the coordinate axes. Show that the sum of the projections on each axis is zero. 7. In exercise 6 find the angles between PQ and the coordinate axes. 8. The projections of AB on OX, OY and OZ are 3,-1 and 2 re- spectively. Those of BC are 2, —3 and 1. Find the projections of AC. 9. Find the coordinates of the middle point of the segment joining (3, 2, -1) and (2, -3, 4). 10. Given A (2, 1, -2), B (1, 3, 1), C (-1, 7, 7), show that the pro- jections of A, B, C on the a;j/-plane are three points on a line. Also show that the projections on the i/0-plane are points of a line. Hence show that A, B, C lie on a line. / r /I 1 / R / f'. / . / s [/ / / / x" Fig. 62. Art. 62. Distance between Two Points Let Pi (xi, 2/1, 2i) and P2 (xj, 2/2, Zz) be the points. On P1P2 as diagonal construct a box with edges parallel to the coordinate axes (Fig. 62). Since projections on parallel lines are equal, the edges Art. 63 Vectors 155 of the box are equal to the projections of P1P2 on the coordinate axes. Consequently PiR = X2 — xi, PiS = j/2 - yi, PiT = zi — zi. Since the square on the diagonal of a rectangular parallelopiped is equal to the sum of the squares of its three edges, PiPi? = PiR^ + PiS^ + PiT\ whence PyPi = V(X2 - x,Y + (2/2 - y,Y + fe - z,Y. (62) Art. 63. Vectors A vector is a segment with given length and direction. The components of a vector in space are its projections on the coordi- nate axes. In this book the vector with components a, b, c will be represented by the symbol [a, b, c]. The vector P1P2 from Pi (xi, 2/1, Zi) to P2 (x2, 2/2, 22) has "components equal to x^ — Xi, 2/2 — 2/1 and 32 — Zi. This is expressed by the equation P1P2 = [X2 - xi, 2/2 - 2/1, 22 - 2l]. (63) As in plane geometry, the sum of two vectors is obtained by placing the second on the end of the first and drawing the vector from the beginning of the first to the end of the second. If vi and V2 are vectors beginning at the same point, V2 — Vi is the vector from the end of Vi to the end of V2. If r is a number, rv is the vector r times as long as v and having the same direction if r is positive but the opposite direction if r is negative. The components of the sum of two vectors are obtained by add- ing corresponding components, those of the difference by subtract- ing corresponding components and those of the product rv by multiplying the components of v by r. Example. Find the point P {x, y, z) on the hne through A (-1, 2, 3) and B (3, -4, 2) such that AP = 3 BP. Using the given coordinates, AP = [x + l,y-2,z- 3], BP = lx-3,y + i,z- 2]. If thenAP = 35P, a; + 1 = 3 (x - 3), 2/ - 2 = 3 (2/ -|- 4), 2 - 3 = 3 (2 - 2). Consequently x = 5, y = — 7, 2 = J. 156 Coordinates of a Point in Space Chap. 9 Exercises 1. Show that the triangle formed by the points (1, 0, 2), (0, —2, 3) and (2, —3, 0) is isosceles. 2. By showing that the sum of the distances AB and BC is equal to AC show that the points A (1, 2, -1), B (0, 5, 1), C (-2, 11, 5) lie on a line. 3. Find the center of the sphere through the points (0, 0, 0), (2, 0, 0), (0, 4, 0) and (0, 0, 6). 4. Given A (1, —1, 2), B (3, 2, —1), find the point on AB produced which is four times as far from A as from B. Also find the point one- fourth of the way from A to B. 5. Given A (1, 3, 1), B (3, 5, 4), find the vectors which are the projections of AB on the coordinate planes. Show that the sum of the squares of the projections is twice the square of AB. 6. Show that the points A (1, 0, -1), B (-2, 1, 3), C (-1, 3, 6), D (2, 2, 2) are the vertices of a parallelogram. 7. AE is the diagonal of a parallelepiped with edges AB, AC and AD. Given A (1, 1, 0), B (2, 3, 0), C (3, 0, 1), D (2, 1, 4), find the coordinates of E. 8. Given A (2, 4, 5), B (1, 3, 0), C (3, 0, 2), D (6, 1, 9), find the mid- dle points of AB and CD. Then find the middle point P of the two middle points. Show that P satisfies the vector equation PA+PB + PC + PD = 0. 9. If r. and s are numbers the vector rAB + sAC lies in the plane ABC. Consequently, if AD = rAB + sAC, the points A, B, C, D lie in a plane. By finding values of r and s satis- fying this equation show that A (0, 0, 1), B (1, -1, 4), C (-2, 1, .-3), D (3, 2, 0) lie in a plane. Art. 64. Direction of a Line The angles between a directed line and OX, OY and OZ are represented by the letters a, /3 and 7. These are sometimes called the direction angles of the line. The cosines of a, fi and 7 are called the direction cosines of the line. Let Pi (xi, 2/1, 2i) and P2 {x-i, 2/2, Z2) be two points on the line. Construct a box with P1P2 as diagonal and edges parallel to the coordinate axes. If the positive direction along each edge is taken as that of the parallel axis, the direction angles for P1P2 are those between P1P2 and the edges of the box. Consequently Art. 64 Direction op a Line PiA _ xi — xx P^2~ P1P2 ' PiB ^ ^ - 2/1 P1P2 P,C P1P2 cos a = cos/3 = cos 7 = PiP, ^2 — Zl P1P2 ' 157 (64a) The direction cosines of the lines are therefore the components of P1P2 divided by its length. n ^ Fig. 64a. Fig. 646. Since P1P2'' = fe — xiY + (2/2 - 2/1)^ + fe — 21)^, it is seen that cos^ a + cos^;8 + cos^ 7 = 1, (646) that is, the sum of the squares of the direction cosines of a line is equal to unity. If it is known that the direction cosines of a line are propor- tional to three numbers a, b, c, since the sum of the squares of the cosines is equal to unity, the exact cosines are obtained by dividing these proportional numbers by the square root of the sum of their squares. The two square roots give two sets of direction cosines corresponding to the two directions along the hne. Example 1. Find the direction cosines of the vector [2, —3, 5]. The cosines are the ratios of the components to the length of the vector. Consequently, cos a = — 7= , cos 8 V38 -3 5 -^=r, COS7 = — 7=- V38 V38 158 Coordinates of a Point in Space Chap. 9 Ex. 2. The direction cosines of a line are proportional to —1, 2 and 1. Find their values and construct a line having the given cosines. The segment from the origin to the point (—1, 2, 1) has direction cosines proportional to its components —1, 2 and 1. Therefore the line through the origin and the point (—1, 2, 1) has the direction required. Its direction cosines are 2 1 ±V6' cos /3 = -.Vq' cos 7 Ve' the two signs in the denominators corresponding to the two direc- tions along the line. Art. 6$. The Angle between Two Directed Lines Let the lines have direction angles ai, /3i, 71 and a2, /Sa, 72. Let OPi and OP2 be lines through the origin with the same direc- tions. Projecting on OP2, proj. OPi = proj. OM + proj. MN + proj. NPi. Consequently, OPi cos d = OM cos ^2 -I- MN cos j32 -I- NPi cos 72. But OM = OPy cos a,, MN = OPi cos /3i, NPi = OPi cos 71. Substituting these values and cancelling OPi, cos d = cos ai cos ai + cos ;8i cos fii + cos 71 cos 72. (65a) That is, the cosine of the angle between two directed lines is equal to the sum of the products of corresponding direction cosines. If the hnes are perpendicular the angle 6 is 90° and its cosine is zero. In that case cos ai cos a2 + cos /3i cos ft +cos y, cos 72 = 0. (656) Therefore, two lines are perpendicular when the sum of the products of corresponding direction cosines is zero. Art. 65 The Angle Between Two Directed Lines 159 In equation (656) the direction cosines of either line can be replaced by any numbers proportional to them. In particular, two vectors are perpendicular when the sum of the products of corre- sponding components is zero. Example 1. Find the angle between the vectors Wi [—2, 0, 1] and V2 [1, 2, 0]. The direction cosines of the vectors are -2 1 cos ai = —^ , cos ft = 0, cos 7i = —7^ , V5 V5 1 2 cos Ui = — -7= , cos jSa = -7= , cos 72 = 0. V5 V5 If 9 is the angle between the vectors, cos9= -f + + 0= -f. The negative sign shows that the angle is obtuse. Ex. 2. Show that the points A (4, 2, - 1), B (3, -2, 3), C (1, 1, 0) are vertices of a right triangle. In this case CA = [3, 1, -1], CB = [2, -3, 3]. The sum of the products of corresponding components is 6-3-3 = 0. The sides CA and CB are then perpendicular and the triangle is a right triangle. Exercises 1. Determine the direction cosines of the coordinate axes. 2. A straight line in the xy-plane has a slope equal to V3. Find its direction angles and direction cosines. 3. Determine the direction cosines of the segment from the origin to the point (2, 3, 6). 4. How many lines through the origin make angles of 60° with the X- and 2/-axes? What angles do they make with the z-axis? 5. A Une makes angles of 45° with the yz- and zx-planes. Find its direction cosines. 6. A Une makes equal angles wjth the coordinate axes. Find those angles. 7. Construct a hne with direction cosines proportional to 1, 4 and 8. Find its direction cosines. 160 Coordinates of a Point in Space Chap. 9 8. Find the angles between the vector [1, 2, 2] and the coordinate axes. 9. Determine the angles of the triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 2). 10. Show that the line joining the origin to the point (1, 1, 1) is perpendicular to the line through (1, 1, 0) and (0, 0, 2). 11. A hne Li in the iz-plane makes an angle of 60° with the Une Lj in the xy-p\a,ne with equation x + 2y = A. Find the angle between Li and the a;-axis. 12. Find the angles between the segments from (1, 1, 1) to the points (2, 0, 3), (0, 3, —2) and (3, 1, 3). Show that one of these angles is equal to the sum of the other two. What do you conclude about the four points? 13. Show that the line through the origin and the point (1, —1, —1) is perpendicular to the line joining any pair of the four points in Ex. 12. Art. 66. Cylindrical and Spherical Coordinates Let be the origin and OX the initial line of a system of polar coordinates in the xjz-plane. Let the projection of a point P on the xy-phne have coordi- nates r and 6. The qjlin- drical coordinates of P are r, 6 and z. The angle d is considered positive when measured from OX toward OY and r is positive when N lies on the terminal side "x" of the angle. In the plane ONP let p and be the polar coor- dinates of P, the X-axis being initial line and the origin. The spherical coordinates of P are p, d and <^. In this case is considered positive when measured from OZ toward the terminal side of 6 and p is positive when P lies on the terminal side of 0. In some books these coordi- nates are called polar. The relations of the rectangular, cylindrical and spherical coordi- nates of a point can easily be determined from the right triangles FiQ. 66. Art. 66 Cylindbical and Sphebical Coobdinatbs 161 OMN and OPQ in Fig. 66. The most important equations con- necting the coordinates are x = r(iosO, y = rsaid, ■) r = p sin 0, 3 = p cos 0, > (66) r* = a;2 + 2/^ p" = x^ + y'' + z^. ) Example. Determine the cylindrical and spherical coordinates of the point (1, 2, 3). From Fig. 66 it is seen that tan = ^ = 2, r = Vx" + y^ = Vs, tan = - = - V5, p = Vx^ + y'^ + z^ = Vu. z o The cylindrical coordinates of the point are then r = Vd, e = tan-i 2, z = 3 and its spherical coordinates are p = VTi, 6 = tan-i 2, (^ = tan"' (i Vs). Exercises 1. Using cylindrical coordinates (r, d, z) construct the points I 2, ;r, 1 j, (—3,-1, OJ, I 2, — ^, — ij and find their rectangular and spherical coordinates. 2. Using spherical coordinates (p, 0, if) construct the points ( 1) T' 5 )' [ — 1, ^, I , [ 2, TT, ^1 and find their rectangular and cylindrical coordi- 3. What is the locus of all points for which r = 17 For which B = What is the locus of points for which r = 1 and 8 = ^? 4. Determine the locus of points in each of the following cases: (a) p = 2, (b) 9 = gTT, (c) * = |, (d) = ^, (e) .f = ■^■ 5. Determine the locus of points in each of the following cases: (a) r = 2, 2 = 3, (b) 9 = 60°, z = -2, (c) r = -1, = ir. 162 Coordinates of a Point in Space Chap. 9 6. Determine the locus of points in each of the following cases: (a) p = 2, 9 = w, (b) p = -2, .^ = |, (c) e = |, = |- 7. Determine the distances of a point Jrom the coordinate axes in spherical and cylindrical coordinates. 8. The spherical coordinates of P are p = 2, d = 30°, <^ = 45 . Find the angles between OP and the coordinate axes. CHAPTER 10 SURFACES Art. 67. Loci An equation represents a locus if every point on the locus has coordiaates satisfying the equation and every poiat with coordi- nates satisfying the equation lies on the locus. One equation between the coordinates of a point in space usually represents a surface. Thus, the equation z = represents the xy-plane, for any point in the xy-ph,ne has a z-coordinate equal to zero and every point with ^-coordinate equal to zero lies in the xy-pla,ne. Similarly the equation x^ -\- y^ -\- z'' = 1 represents a sphere with radius 1 and center at the origin. In particular cases one equation may represent a straight line or curve. Thus, the only real points for which x^ -\- y^ = are the points x = 0, y = on the 3-axis. Two simultaneous equations usually represent a curve or straight line; for each equation represents a surface and the two equations represent the intersection of two surfaces, that is, a straight line or curve. Thus, the equations x^ + Z/'' + 3" = 3, s = 1 represent the circle in which a sphere and plane intersect. Three simultaneous equations are usually satisfied by the coordi- nates of a definite number of points. These points are found by solving the equations simultaneously. In particular cases the equations may have no solution or may be satisfied by the coordi- nates of all points on a curve or surface. Art. 68. Equation of a Plane A line perpendicular to a curve or surface is called a normal to that curve or surface. 163 164 Surfaces Chap. 10 Let a normal DN to a plane (Fig. 680) have direction angles a, j3, 7. Let Pi {xi, yi, Zi) be a fixed point and P (x, y, z) a variable point in the plane. The direction cosines of DN are cos a, cos /S and cos 7. Those of PiP are proportional to X — xi, y — yi, 3 — 2i. Since DN is perpendicular to the plane it is perpendicular to PiP. Therefore, by Art. 65, (aj — Xi) cos a + (y — yi) cos /3 + (z — Zi) cos 7 = 0. (68a) This is the equation of the plane tlirough {xi, yi, Zi) whose normal makes angles a, /3, 7 with the coordinate axes. Let the direction cosines of the normal be proportional to A, B, C. Then* cos a : cos /3 : cos 7 = A : B : C. Since the cosines in equation (68a) can be replaced by any propor- tional numbers, that equation is equivalent to A{x-xO+Biy- 2/0 +C{z-zi) = 0, (686) which is therefore the equation of the plane through (xi, 2/1, Zi) per- pendicular to the line with direction cosines proportional to A, B, C. First Degree Equation. — Equations (68a) and (686) are of the first degree in x, y, z. Therefore any plane has an equation of the first degree in rectangular coordinates. Conversely, any equation of the first degree in rectangular coordi- nates represents a plane; for any such equation has the form Ax + By + Cz + D = 0, (68c) A, B, C, D being constant. Let Xi, yi, Zi satisfy this equation. Then Ax, + By I -h Czi + D = 0. Subtraction gives A{x-xO + B{y- 2/1) +C{z-z{)= 0, which is equation (686). Therefore any equation of the form (68c) Art. 68 Equation op a Plane 165 represents a plane whose normal has direction cosines proportional to A, B, C. Example 1. Construct the plane with equation 2x + 3y + z = 6. The plane can be determinM by its intersections with the coordi- nate axes. Where the plane crosses the x-axis y and z are zero and so a; = 3. Similarly it crosses the y-SLxis a,t y = 2 and the 0-axis at 3 = 6. The intercepts on the three axes are 3, 2 and 6. Ex. 2. Construct the plane represented by the equation x — 2y + z = 0. ^1 Fig. 68b. Fig. 68c. The plane passes through the origin and so its intercepts are all zero. It can be determined by its intersections with the coordinate planes. It cuts the xy-plane in the hne z = 0, x = 2y and the 2/z-plane in the hne x = 0, z = 2 ?/. In their respective planes these lines are constructed from the equations x = 2y and z = 2y as in plane geometry. The plane through these lines is the one required. Ex. 3. Find the equation of the plane through (1, — 2, 4) per- pendicular to the line through A (2, 1, 0) and B (1, 2, 3). The direction cosines of AB are proportional to 1 — 2, 2 — 1, 3 — 0. Using these values instead oi A,B,C in (686) the equation of the plane is found to be -{x - I) + {y + 2) +Z{z - A) = 0. Ex. 4. Find the angle between the planes X - y + z = \, 2x + 3y -z = 2. 166 Surfaces Chap. 10 Between two planes are two angles less than 180°. It is shown in sohd geometry that these angles are equal to the angles between lines perpendicular to the two planes. The normals to the two planes have direction cosines proportional to the coefficients 1, — 1, 1 and 2, 3, —1. The exact cosines of two normals are then cos «! = —= , cos /Si = —7= , cos 7i = —7= , V3 Vs V3 2 3 — 1 • cos 02 = — -}= , cos Bi = — ;=; , cos 72 = , — • Vli Vl4 V14 By equation (65a) the angle Q between the two normals satisfies the equation -2 cos Q = .— - V42 The negative sign shows that this is the obtuse angle. The acute angle between the two normals or between the two planes is V42/ Exercises Construct the planes represented by the following equations and find the direction cosines of their normals: 1. a; + 22/ + 4z = 4. 4. 2a;-3 2/.= 0. 2. a; - ?/ + 3z = 5. 5. 3x + 4z = 0. 3. a; + 2/ + 2 = 0. 6. z + 5 = 0. 7. Find the equation of the plane through the origin perpendicular to the line with direction angles a = 60°, = 45°, y = 60°. 8. Find the equation of the plane through (1, 1, 0) perpendicular to the vector [3, —5, 4]. 9. Find the equation of the plane with intercepts on OX, OY and OZ equal to 1, 2 and 3. 10. Show that the planes x + 2y — z = 1 and 2x + iy — 2z = Z are parallel. 11. Show that the planes i + j/ — z = and 2x — 3y — z = are perpendicular. 12. Find the angle between the planes x -{- 2y -\- 2z = and a;-4i/ + 8z = 9. Art. 70 Equation of a Cylindrical Surface 167 13. Show that the angle between a line and a plane is the comple- ment of the angle between the line and the normal to the plane. Find the angle between the plane x — 2y — z = and the line through the points (3, 0, 1) and (0, 2, -1). Art. 69. Equation of a Sphere A sphere is the locus of points at a constant distance from a fixed point. The fixed point is the center, and the constant dis- tance the radius of the sphere. Let C (xi, i/i, Zi) be the center of a sphere with radius r. If P (x, y, z) is any point on the sphere, its equation is (x - x,Y +{y- yiY +{z- z,Y = r\ (69o) When expanded the equation of the sphere has the form x'' + y'' + z' + Ax + By + Cz + D = Q. (696) Conversely, any equation of this form represents a sphere if it rep- resents a real surface. To show this, complete the squares in x, 2/ and 2 separately. The result will have the form {x - ay +(y- by +iz- cy = d. If d is positive this represents a sphere with center (a, b, c) and radius Vd. If d is zero the locus is the single point (a, b, c). If d is negative there is no real locus. Hence, whenever the equation represents a surface, that surface is a sphere. Example. Determine the center and radius of the sphere x^ + y^ + 'z'' - 2x + 3y = 0. Completing the squares, {x-iy + iy + iy + z' = \\ The center is (1, — |, 0) and the radius is | V 13. Art. 70. Equation of a Cylindrical Surface A cylindrical surface is one generated by lines parallel to a fixed line and cutting a fixed curve. The lines are called generators and the fixed curve is called a directrix. A cylindrical surface with generators parallel to a coordinate axis is represented in rectangidar coordinates by an equation containing 168 Surfaces Chap. 10 Fig. 70. only two coordinates. To show this let the generators be parallel to the z-axis. Let the surface intersect the a:7/-plane in the curve 2 = 0,/ (x, y) = 0. Any point P (x, V, 2) on the surface has the same x and y as its pro- jection Q {x, y, 0) on the xy- plane. Since the coordinates x and y oi Q satisfy the equa- tion / {x, y) = 0, so do the co- ordinates X, y, z of P. Also, if the coordinates of any point P satisfy the equation/ (x, y) = 0, P is in the vertical line through a point Q of the curve and so lies on the surface. Therefore f (x, y) = is the equation of the cylindrical surface. Conversely, any equation in two rectangular coordinates repre- sents a cylindrical surface with generators parallel to the axis of the missing coordinate. To show this let the equation be / (x, y) = 0. Any point P with coordinates satisfying this equation Ues in a vertical line through a point Q of the curve z = 0, f (x, y) = 0, and any point in such a vertical line has coordinates satisfying the equation. Therefore the equation / {x, y) = Q represents a cylin- drical surface whose directrix is the curve z = 0, f (x, y) =0 in the xy-pla,ne. Example. Find the equation of the cylindrical surface with generators parallel to the a;-axis cutting the yz-plajie in the circle with radius 2 and center (0, 1, 1). In the yz-ph.ne the equation of the circle is (y - 1)^ +(z- ly = 4. This considered as an equation in space represents the given cylin- drical surface. Art. 71. Surface of Revolution The surface described by a plane curve revolving about an axis in its plane is called a surface of revolution. In the a;3-plane let / (x, z) = he the equation of a curve. Let Art. 71 Surface of Revolution 169 ^ N \ 3^ 1 1 , \ \z A. Fig. 71. this curve be revolved about the z-axis. The cylindrical coordi- nates r, z of any point P on the resulting surface are equal to the rectangular coordinates x,z oi a, point Q on the curve in the a;2-plane. Since the coordinates of Q satisfy the equation / {x, z) = 0, the cylindrical coordinates of P satisfy the equation /(r,2)=0. This is then the equation of the sur- face described by revolving the curve y = 0, f (x, z) = about the z-axis. Since r = 'Vx'' + ]f-, the rectangular equation of the surface is /(^/x^-h^/^^)=0. Similar equations are found for the surfaces obtained by revolving about the xr or 2/-axis. Therefore, to find the rectangular equation of the surface described by rotating a curve in a coordinate plane about a coordinate axis in that plane, leave the coordinate corresponding to the axis of rotation unchanged in the plane equation of the curve and re- place the other coordinate by the square root of the sum of the squares of the other tivo. Example 1. Find the equation of the surface described by rotating the parabola y^ = 2x about the x-axis. This means that the parabola in the xy-plane with plane equation 2/' = 2 a; is to be rotated about the z-axis. The equation of the resulting surface is obtained from y'' = 2xhy leaving x unchanged and replacing y by 'Vy'^ + z''. The required equation is then Ex. 2. Show that (j.2 _1_ 2/2)2 ^ ^i (3.2 ^ ,^2) = 1 is the equation of a surface of revolution. This is indicated by the fact that the equation contains x and y only in the combination x^ -{- xf. The surface is generated by 170 SuKFACES Chap. 10 revolving about the 2-axis the curve with plane equation x'^ + z^x^ = 1. Exercises Describe the surfaces represented by the following equations: 1. a;2 + 1/2 + z^ = 9. 9. x'-Vy' = 2z. 2. a;' + 2/2 + z« = 4 x. 10. x^ ■\- y^ = z^. 3. a;2 + j/2-|-g2_3x + 22/ + 2 = 0. 11. x" - y" = z^. 4. 2 x2 + 2 2/2 + 2 z2 - 5 a; + 8 2/ 12. x^ - z^ = 0. — 6 z + 4 = 0. 13. r = a cos 9. 5. x2 + 2/2 = 2 X. 14. r = az + 6. 6. x^ — y^ = a'. 15. p = a cos (^. 7. 2/" = fi^. 16. psin — a. 8.' z = sin X. Find the rectangular, cylindrical and spherical equations of the following surfaces: 17. Sphere with radius a and center at the origin. 18. Sphere with center (0, 0, o) passing through the origin. 19. Right circular cylinder with axis OZ and radius a. 20. Eight circular cone with axis OZ and vertical angle 90° (be- tween generators in a plane through the axis). Find the rectangular equations of the following surfaces: 21. Right circular cylinder tangent to the xy-^lsaie along the x-axis. 22. Parabolic cylinder with generators parallel to OF and directrix a parabola in the xz-plane with axis OZ and vertex at the origin. 23. Elliptical cylinder with generators parallel to the z-axis and directrix an eUipse with axes along OX and OY. 24. Prolate spheroid generated by revolving an elhpse about its major axis. 25. Hyperboloid of revolution generated by revolving a hyperbola about one of its axes. 26. Paraboloid of revolution generated by revolving a parabola about its axis. 27. Torus generated by revolving a circle about a line in its plane not cutting the circle. Art. 72. Graph of an Equation To construct the graph of a given equation it is customary to draw a series of plane sections and from these sections to determine the appearance of the surface. The sections generally used are those in and parallel to the coordinate planes. Art. 72 Gbaph op an Equation 171 Example 1. The ellipsoid, a' V- c^ The sections of this surface in the xz- and j/z-planes are the eUipses 1 and |, + - = 1, having a common axis on Z'Z (Fig. 72a). The section in a horizontal plane g = A; is an ellipse ^ V ¥■ —„ + ii,= 1 — -„' with axes parallel to OX and OY. Since the axes of this ellipse are in the xz- and 2/z-planes and end on the surface, they are chords of the eUipses in those planes. The surface, called an ellipsoid, is thus generated by horizontal ellipses whose axes are chords of the two vertical ellipses. ^ ^ '/ i ; - V y/ \ i ' '^ The quantities a, 6, c, called the semi-axes, are equal to the intercepts on the coordinate axes. If two of these semi-axes are equal the ellipsoid is called a spheroid. It is then a surface of revolution obtained by revolving an ellipse about one of its axes. If the semi-axes are all equal the ellipsoid is a sphere. 172 SURFAGES Ex. 2. The hyperboloid of one sheet, a^ b^ c^ The sections in the xz- and j/z-planes are hyperbolas Chap. 10 r _ !! = 1 with a common axis Z'Z (Fig. 726). The section in a horizontal plane 2 = A; is an ellipse. with the ends of its axes on the two hyperbolas. Fig. 72c. If a = b the surface is a hyperboloid of revolution obtained by revolving a hyperbola about its conjugate axis. Ex. 3. The hyperboloid of two sheets, r 1. a' b' C The section in a plane a; = ^ is imaginary if | a; | < a. The sur- face therefore consists of two parts, one on the right of a; = a and one on the left of x = —a. The surface cuts the xy- and a;z-planes in hyperbolas and is generated by ellipses parallel to the j/z-plane whose axes are chords of these hyperbolas (Fig. 72c) . Art. 72 Graph of an Equation 173 If 6 = c the surface is a hyperboloid of revolution obtained by- revolving a hyperbola about its transverse axis. Ex. 4. The elliptic paraboloid, z = ax' + by', a and 6 having the same algebraic sign. Sections in the xz- and j/s-planes are parabolas z = ax' and z = by'. The section in a plane z = /c is an ellipse ax' -i- by' = k with axes parallel to OX and OF if a, b and k have the same sign and imaginary if A is opposite in sign from a and 6. The surface is thus gen- erated by horizontal ellipses whose axes are chords of the two parabolas. If a = 6 the surface is a paraboloid of revolution obtained by revolving a parabola about its axis. Ex. 5. The hyperbolic paraboloid, z = ax' — by', a and 6 being positive. J^\ ^ ./V . C- -^ ^ / X' yf \ 1 /I ' ' _J ^ jl-itl Z' L Fig. 72e. The section in the .rz-plane is a parabola', z = ax' extending upward. The section in the yz-Tp\a,ne is a parabola z = —by' ex- 174 Surfaces Chap. 10 tending downward. The section in the xy-pleaie is a pair of lines ax' — by^ = 0. The section in a horizontal plane z = k is a. hyper- bola ax" — by" = k whose transverse axis is a chord of the parab- ola (in the xz- or 2/z-plane) cut by that plane. The surface has the general shape of a saddle. Ex. 6. The hyperbolic paraboloid, z = kxy. This is a saddle-shaped surface similar to that in Ex. 5. The hyperbolas in horizontal planes are however rectangular with asymptotes parallel to the x- and y-nxes. Ex. 7. The cone, x" y" z" a? b" c" The sections in the xz- and yz- planes are pairs of lines X _ z a c' and The surface is generated by hor- izontal ellipses the ends of whose axes are on these lines. Ex. 8. Describe the surface with spherical equation p = a sin 4>- Since Q does not occur in the equation the surface is one of revo- lution about the a-axis. The section in the xs-plane is a circle tangent to the s-axis at the origin. The graph is a doughnut- shaped surface generated by revolving this circle about the g-axis. Exercises Draw the graphs and describe the surfaces represented by the follow- ing equations: 1. a;2 -I- 2 2/2 -1-3 32 = 6. 5. a^ 2. (x - 1)2 -I- 2 (!/ - 2)2 6. X" -y"-z" = 3. + 3 (z - 3)2 = 6. 3. x2 -I- 4 (i/2 -I- z2) = 12. 4. x2 - 2/2 -I- z2 = 1. '-1-22-2x4-42/ = 4. 7. X2 -|- z2 = y_ 8. 2/2-1-222 -4x2 = 12. 9. x=z2-|-22/2-|-22-122/-l-ia Art. 72 Graph of an Equation 175 10. X = 2 yz. 15. r = a cos 2 8. 11. X- ~ y'' = z^. 16. p = a cos 9. 12. (x + 2)2 + 4 (2/ - 1)2 17. p cos =' 2. = (z - 3)2. 18. r2 + z2^= a2. 13. 2/2 = z3. 19. re = k' 14. « = X!/ + » + 2/, 20. Xi/s = o^ CHAPTER 11 LINES AND CURVES Art. 73. The Straight Line A straight line is the intersection of two planes. It is then rep- resented in rectangular coordinates by two first degree equations. The line is best constructed by finding two points on it. The most convenient points for this purpose are usually its intersections with two coordinate planes. If the line passes through the origin a second point can be found by assigning an arbitrary value to one of the coordinates and calculating the values of the other two. Example 1. Construct the line represented by the equations 2/ + z = 3, 4a; + 32/-3z = 3, and find its direction cosines. The line cuts the yz-pla,Tio where x = 0, that is, where y + z = 3, 3y-3z = 3. The solution of these equations is y = 2, z = 1. The intersection with the 2/z-plane is then A (0, 2, 1). In the same way the inter- section with the xz-plane is found to be B (3, 0, 3). Draw the hne through A and B (Fig. 73a). Its direction cosines are 3^-2 2 cosa: = -^^, cosp = — -p=, cost = , — • Vl7 Vl7 -n/iV Line through a Point with a Given Direction. — Let the line pass through Pi (xi, j/i, Zi) and have direction angles a, fi, 7 (Fig. 736). If P (x, y, z) is any point on the line X — xi = PiP cos a, 2/ — 2/1 = PiP cos j3, z — Zi = PiP cos 7. Hence — ^ = (73a) cos a cos p cos 7 176 Art. 73 The Straight Line 177 These are the equations of the line. Since two quantities equal to a tliird are equal to each other, they are equivalent to two inde- pendent equations. If the cosines are proportional to A, B, C, equations (73a) are equivalent to X- xi V -iji z — Zi B (736) These are the equations of a line through (xi, y^, Zi) with direction cosines proportional to A, B, C. Ex. 2. Find the equations of the line through (1, 0, 1) and (-2, 1, 0). The direction cosines of the hne are proportional to 3, —1, 1. Since the hne passes through (1, 0, 1) its equations are then X — 1 _y — _ z — \ 3 -1 ~~\ These are equivalent to the two equations x-\ y -0 y — ^ ^ ~ 1 3 ~ -1 ' -1 ~ 1 ' and consequently to x + 3y = 1,' y + z = 1. Exercises Construct the lines represented by the following equations and find their direction cosines: 1. s-t-2/ = l, 2/ = 22. 3. x + y+z = l, 2x-3y+4z = 5. 2. X -y =z, x + y =0. 4. a; = 2, y = 3. 178 Lines and Curves Chap. 11 5. Find the equations of the line through the points (2, 3, — 1) and (3, 4, 2). 6. Find the equations of the line through (0, 1, 2) parallel to the vector [3, 1, 5]. 7. Find the equation of the line through (1, 1, 1) perpendicular to the plane x-\-2y— z = Z. 8. Find the angle between the lines x -\- y — z = Q, x + z = Q and X — y = l, X — Zy-\-z— Q. 9. Find the angle between the Une x —y-\-z = I, x = 2 and the plane z = x — Zy. 10. Show that the lines x+2/ + z = l, 2x—.y + Zz = 2 and Zy — z = 2, 3a; + 4z = l are parallel. 11. Show that the lines a + 2i/ = l, 2y — z = I and x.— y = 1, X — 2z = Z meet in a point and are perpendiciilar. Art. 74. Curves A curve is the intersection of two surfaces. It is then repre- sented by two simultaneous equations. Since an indefinite num- ber of surfaces can be passed through a curve, it can be represented in an indefinite number of ways by a pair of equations. Example 1. Show that the equations x^ -\- y^ + z'' = 1, X + 1J + z= 1 represent a circle. The first equation repre- sents a sphere, the second a plane. The two' equations represent the circle in which the sphere and plane intersect. Ex. 2. Determine the locus represented by the equations a;2 -I- 2/2 = a2, ?/2 -I- z2 = a2. These equations represent circular cylinders of radius a (Fig. 74a). The locus required is the intersection of the cylinders. Subtraction of the equations gives if — f = 0. Therefore x = ± z. All points Fig. 74a. Art. 74 Curves 179 of the intersection thus he in the two planes x = z and x = — 2. The locus is two ellipses in which the planes cut the cylinders. Projecting Cylinders. — If a rectangular coordinate is ehroinated from the equations of a curve, the resulting equation usually rep- resents the cylindrical surface with the curve as directrix and generators parallel to the axis of the eliminated coordinate. The intersection of this cylinder with the plane of the other two coordi- nates is then the projection of the curve on that coordinate plane. These statements are illustrated in the examples solved below. Ex. 3. Find the cylinders with generators parallel to the co- ordinate axes and cutting the curve s = x^ + y'', z = X. Eliminating z we get x^ -\- y^ = x. Since this equation contains only X and y, it represents a cyHnder with generators parallel to the 2-axis. Since the equa- tion of the cylinder is a con- sequence of the equations of the curve, all points on the curve lie on the cylinder. Furthermore, if x and y satisfy the equation of the cyhnder, a value of z can be found such that X, y, z satisfy the equations of the curve. That is, each generator of the cyl- inder cuts the curve. There- fore x^ -\- y''- = X ]s the equa- tion of the cylinder generated by lines parallel to the 2-axis and cutting the curve. In the same way the equation of the cylinder parallel to the x-axis is found to' be y^ + z' = z. Lines parallel to the 2/-axis and cutting the curve He in the plane g = X. They however generate only the strip of this plane in which the curve hes. This shows that the elimination of a coor- 180 Lines and Curves Chap. 11 dinate may give more than the smface generated by lines cutting the curve and parallel to the axis of that coordinate. Ex. 4. Find the equations of the projections of the line 2a; + 2/-2 = 0, x-y + 2z = Z on the coordinate planes. Eliminating z, we get 5a; + 2/ = 3. This is the equation of a plane through the line parallel to the 3-axis. The equation of the projection on the rry-plane is then « = 0, 5a; + 2/ = 3. In the same way the projections on the xz- and j/s-planes are found tobe2/ = 0, 3ar+s = 3anda; = 0, 32/-5s + 6 = 0. Exercises Draw the following curves and find their projections on the coordi- nate planes: \. x+y+z = h x'+y'+z-' = l. h.z=x'+y\ z = y'' + z\ 2. x2 + 2/2 = a\ x^ + )/ = z^. 6. x^^-y^-\-z^ = a^, x^-^z^ = ax. 3. 2^ = a;^ + 2/^ x -\-y = \. 1. r = a, 9 = z. A. z = xy, z = 2x. 8. 4, = i-n-, p =e. 9. Show that the equations x^ + y^ — z' = 1, y — z = 1 — x repre- sent a pair of lines. 10. Show that the circle x" -^ !/ + 2' = 6, 2/ + 2 x = 1 and the line 2/-|-z = l, x + 2/ + z=,2 intersect. 11. Find the intersection of the circle x" -|- 2/^* -|- 2^ -|- 6 x = X -|- 2/ = and the plane x -|- z = 1. Art. 75. Parametric Equations The locus of a point whose coordinates are given functions of a parameter is usually a curve. For, if one of the equations is solved for the parameter and the value substituted in the other two, two equations between the coordinates are obtained. Thus x = t, y = f, z = f are parametric equations of a curve. To plot the curve we can assign values to the parameter, calculate the corresponding values of the coordinates and plot the resulting points. By eliminating Art. 75 Parametric Equations 181 Fig. 75. the parameter the equation of a surface through the curve is ob- tained. If the parameter is ehminated between two of the para- metric equations, the resulting equation con- sidered as an equation in a coordinate plane represents the projection of the curve on that plane. For example the projections of the curve given above have the equations y = x', z = a;', z^ = y^. Example. — The helix is a curve traced on the surface of a right circular cylinder by a point that advances in the direction of the axis of the cylinder while it rotates around the axis of the cylinder, the amount of ad- vance being proportional to the angle of rotation. To find the equations of the helix, let the axis of z be the axis of the cylinder, a- the radius of the cylinder, and let the x-axis pass through a point of the helix. If 9 is the angle of rotation, X = a cos 9, 2/ = o sin 6, z = kd, k being the ratio of the advance in the direction of the axis to the angle of rotation. Exercises Construct the following curves and find their projections on the coordinate planes : 1. x = l+t, y = 2 — t, z = 3t. 3. X = tsint, y = t cos t, z = t. 2. X = GosB, y = sine, !- = 26. 4. x = sin t, y = cos t, z = tan I. 5. A conical helix is described by a point moving on the surface of a right circular cone, the distance of the point from the vertex of the cone being proportional to the angle of rotation about the axis. Find para- metric equations for the curve. 6. Find the equation of the twisted surface generated by perpen- diculars from points of a helix to the axis of the cyhnder on which it lies. 7. Neglecting friction the position of a bullet starting from the ori- gin with velocity [a, b, c] after t seconds is given by the equations X = at, y = U, z=ct- 16.1 1^. 182 Lines and Curves Chap. 11 Construct the curve and iiad its projections on the coordinate planes if it starts with a velocity ot 1000 ft. per second in the direction a = 60°, ^ = 45°, 7 = 60°. 8. The wheel of a gyroscope rotates with constant speed around its axis while the axis turns with constant speed about a fixed point of itself. Find equations of the curve described by a fixed point on the periphery of the wheel. Natural Values of Trigonometric Functions Deg. Rad. Sin. Cos. Tan. Deg. Rad. Sin. Cos. Tan. .0000 .0000 1.0000 .0000 46 .7854 .7071 .7071 1.0000 1 .0175 .0175 .9998 .0175 46 .8029 .7193 .6947 1.0355 2 .0349 .0349 .9994 .0349 47 .8203 .7314 .6820 1.0724 3 .0524 .0523 .9986 .0524 48 .8378 .7431 .6691 1.1106 4 .0698 .0698 .9976 .0699 49 .8552 .7547 .6561 1.1504 6 .0873 .0872 .9962 .0875 50 .8727 .7660 .6428 1.1918 6 .1047 .1045 .9945 .1051 51 .8901 .7771 .6293 1.2349 7 .1222 .1219 .9925 .1228 52 .9076 .7880 .6157 1.2799 8 .1396 .1392 .9903 .1405 53 .9250 .7986 .6018 1.3270 9 .1571 .1564 .9877 .1584 54 .9425 .8090 .5878 1.3764 10 .1745 .1736 .9848 .1763 66 .9599 .8192 .5736 1.4281 11 .1920 .1908 .9816 .1944 56 .9774 .8290 .5592 1.4826 12 .2094 .2079 .9781 .2126 67 .9948 .8387 .5446 1.5399 13 .2269 .2250 .9744 .2309 58 1.0123 .8480 .5299 1.6003 14 .2443 .2419 .9703 .2493 69 1.0297 .8572 .5150 1.6643 16 .2618 .2588 .9659 .2679 60 1.0472 .8660 .5000 1.7321 16 .2793 .2756 .9613 .2867 61 1.0647 .8746 .4848 1.8040 17 .2967 .2924 .9563 .3057 62 1.0821 .8829 .4695 1.8807 18 .3142 .3090 .9511 .3249 63 1.0996 .8910 .4540 1.9626 19 .3316 .3256 .9455 .3443 64 1.1170 .8988 .4384 2.0603 20 .3491 .3420 .9397 .3640 66 1.1345 .9063 .4226 2.1445 21 .3665 .3584 .9336 .3839 66 1.1519 .9135 .4067 2.2460 22 .3840 .3746 .9272 .4040 67 1.1694 .9205 .3907 2.3559 23 .4014 .3907 .9205 .4245 68 1.1868 .9272 .3746 2.4751 24 .4189 .4067 .9135 .4452 69 1.2043 .9336 .3584 2.6051 26 .4363 .4226 .9063 .4663 70 1,2217 .9397 .3420 2.7475 26 .4638 .4384 .8988 .4877 71 1.2392 .9455 .3256 2.9042 27 .4712 .4540 .8910 .5095 72 1.2566 .9511 .3090 3.0777 28 .4887 .4695 .8829 .5317 73 1.2741 .9563 .2924 3.2709 29 .5061 .4848 .8746 .5543 74 1.2915 .9613 .2756 3.4874 30 .5236 .5000 .8660 .5774 76 1.3090 .9659 .2588 3.7321 31 .5411 .5150 .8572 .6009 76 1.3265 .9703 .2419 4.0108 32 .5585 .5299 .8480 .6249 77 1.3439 .9744 .2250 4.3315 33 .5760 .5446 .8387 .6494 78 1.3614 .9781 .2079 4.7046 34 .5934 .5592 .8290 .6745 79 1.3788 .9816 .1908 6.1446 36 .6109 .5736 .8192 .7002 80 1.3963 .9848 .1736 5.6713 36 .6283 .5878 .8090 .7265 81 1.4137 .9877 .1564 6.3138 37 .6458 .6018 .7986 .7536 82 1.4312 .9903 .1392 7,1154 38 .6632 .6157 .7880 .7813 83 1.4486 .9925 .1219 8,1443 39 .6807 .6293 .7771 .8098 84 1.4661 .9945 .1045 9,6144 40 .6981 .6428 .7660 .8391 85 1,4835 .9962 .0872 11.4301 41 .7156 .6561 .7547 .8693 86 1.5010 .9976 .0698 14.3007 42 .7330 .6691 .7431 .9004 87 1.5184 .9986 .0523 19.0811 43 .7505 .6820 .7314 .9325 88 1.5359 .9994 .0349 28.6363. 44 7679 .6947 .7193 .9657 89 1.5533 .9998 .0175 57.2900 46 .7854 .7071 .7071 1.0000 90 1.5708 1.0000 .0000 CO 183 ANSWERS TO EXERCISES 1. x = h -2. 2. X = 1, -5. -5± Vl3 3. X i4. X = -l± v"^ 5. x = ±1, ±V2. 6. X = ±1. •7. X = ±iV3. 3± Vl3 8. X = ;;; Pages 6, 7 9. X -1± VT7 10. X = 1 ± V^ 11. X = ±1, 2. 12. X = 1, ^i^ 13. X = ±1, ± V^. ,. d=V2 ± V^ 14. X = ^ 15. X ±V3 ± V-5 1. X = 1, -1, 2. 2. x = 2, 2, -|- 3. x= -3, I, f. 4. X = i, I, |. 5. X = I, H3 ± a/29). Page 9 6. X = -3,|(-1± V^). 7. X = -1, -1, ±i 8. X = —2, 3, I, —If. 9. :<; = 2, f, 1 ± V^. 10. X = 4, -f, i (3 ± V5). 1. X = -0.53, 0.65, 2.88. 2. x = 1.41, -0.7±2.1"> 3. X = 0.54, -0.8 ± 1.1 A Page 10 4. X = 1.2, 2.9. ^. 5. x=0.7, -1.2. ^. Page 12 1. X > 1 or X < - 2. 2. X > 1 or -1 2, < a; < f V3, or -2 < X < -I Vs. 7. |a:|V2. Page 14 1. X = 1, y = 2. 7. h = l,k = 2,r = ±5, and 2. x = 2,y= -l,z = Z. h = ^^,k= -^/, r = ±W. 3. I = 1, 2/ = -1, 3 = 1. 8. X = 1, -1, 2, -2. 4. x = 1, 2/ = i, z = 1. 2/ = -1, 1, -I, |. 5. X = f, 2/ = -2, 2 = f. 9. J. = j^ = 2 = j_V2 6. X = -1, 2/ = 1, and IQ. x = l,y= -1, z = 2, and Page 16 11.1:^:3 = 3:1:2. 13. x :^ :z = 2 : -1 : 1, or -1:2:1. 12. x:y:z = 2: -3:4. 14. x :?/ :z = 1 : ±1 : 1, or -2:±v'^:l. Page 29 1. (-i -f). 4. AB:CD = 3:-2. 2. (3,8). 5. (5,3), (-1, -5). 3. (5, 11), (1, 3). Page 31 1. 5 + 3V5. 9. (H,3H). 7. (2 ± 2 \/3, 5). 10. (A, 1^). 8. (-i, 0). 11. (2i If). Page 34 7. (-2, 2). 10. (5, -1). 8. [-i, -a [4, 1], [-1, |]. 12. 241. 9. (-8, 5). Page 38 2. (0, I), (-4, 7). 8. Dd,^^). 3. F (14, -13), Q (6, -5). 9. P (1, 2f). 4. (-10,31). 12. (2i44). 5. Z) (15, -3). 186 Answers to Exercises Pages 42, 43 1. i, -7, -2. 13. AV3 + tV 2. tan-i(-0.8) = 141° 20'. 15. (4.933, 4.966), 9. 46° 51', 97° 8', 36° 2'. (-6.933, -0.966). 11. 7 ± 5 V2. 16. (-2f, -Sf). 12. 71° 34'. Pages 48, 49 2.x-3y = 5. 7. 3a;-4i/ + 6 =0. 4. 2/ = 3a; + 9. 9. a?" + 2/'' = a; + 2/ + 14. 5. a; + 52/ = 2. 11. a;= + 2/'' - 4a; - 62/ = 12. 6. y^ = 82/ -6a; -25. Page 56 1. 2/ + 3 = V3(a; + 1). 7. 62/ - 9a; + 2 = 0. 2. 2/ = 3a; -7. 8. a; - 2/ + 2 = 0. 3- a; = 2. 9. 2/ = 5x - 3, length = 2 v'26. i-y = 2. 10. a;-22/+3=0, a;-22/ + 8 = 0, 5. 2a; + 8y = 17. 2a;+2/-4=0, 2x + 2/ -9 =0. 6. 2/-5 = ±V3(a;-3). 11. 21x+16y = Q, lla; + 242/=9. Pages 59, 60 14. 45°, 56° 19', 78° 41'. 21. 5, VlO, Vl7. 15. 3s -22/ = 6. 22. a;- 5y + 8 = 0. 16.y = 5x-4c. 23. 12a; -15 2/ = 8. 17. 2/ = X - 4. 24. 4a; - 2/ + 9 = 0. 18. 8 a; + 9 2/ = 7. 25. A straight line perpendicular 19. (IxV, — tt)- to the line through their 20. 5 a; + 3 2/ = 18. centers. Pages 63, 64 8. a; + 2/ >0, 2x -32/- 1 <0, 2/-2 <0. 9. 2/ + 3a;-4>0, 3a; -22/-1 >0, 2/-6a; + 14 > 0. 10. tS- ^^• 11. tJjVTs. 12. |. 13. X (2 V2 - 1) - 2/ (v^ + 3) = V2 - 2. 14. (H, A). 15. (2/ - 2x)2 = 14a; + 18y - 56. Answers to Exebcisbs 187 Pages 68, 69. 1. (0, 0), r = 5. 15. (0, 2), (f, -^). 2. (2, 0), r = 2. _ U. x' + y'-3x-3y=0. 3. (0, f ), 7- = i \/7. 17. (x - 6)2 + (2/ - 2)2 = 25, 4. (-§, 0), r = iy7. (i + 1)2 + (j/ + 5)2 = 25. 5. (h i), r = I V2. 18. 2a;2 + 2 2/2 + 6a; + 3 2/ =10. 6. {a,a),r = aV2. 19- x' + f - IGx - 12y = 0, 8. The locus is the point (1, 0). ^^ ^^J^^y\Jy'_ jj, j'4. 22. x^ + y^ + 26x + 16y = 32. 11. 3^ + ^ -5x + 4y -4:Q =0 12. x^ + y^ -2x -2y = 11. ,— 13.x^ + y'-2x-2y + l=0, 23. x^ + y^ - (2 VTO - 6) y x' + y^-lQx-lQy + 25 = 0. = 2 VlO - 6. 14. x^ + 2/2 = 9 ± 2 Vs. 24. a; = 2 y. Pages 73, 74. 1. (0, 0), a = V6, 6 = Vs. 6. The locus is the point (1, -1). 2. (1, -2), a = i, 6 =1. 7. 9x2 + 4^2_ 18^^,24^+9=0. 3. (i, i), ffi=iV6, 6 = }V3. 8. 4s2 + 2/2 + i6x-82/ + 16=0. 4. (1, -2), a = 2,^= Ve. 9. (x + 2/ - 2)2 + 16 (x - 2/ + 2)2 5. (-3, 1), a = Vl3, 6 = 1 Vsg. . = 32. Page 77. 1. Axis y =0, vertex (i, 0). 6. -r. = f, (1|, 3}). 2. 2/ = 0, (i, 0). 7. 3 2/2 = 16 X. 3. X = 1, (1, 2). 8. 5x2 + 20x + 9 2/ + 2 =0. 4. X = -I, (-i, -I). 9.\'-V2. 5. 2/ = I, (-1, I). 10. 35Ht. Page 83. 1. Center (0, 2), a = V3, 6=2. Asymptotes, 2/ - 2 = ±f V3x. 2. Center (0, -1), a = ^ Vl5, 6 = v'3. Asymptotes, 2/+I = ±x VK. 3. Center (1, — 1), a = 6 = 2. Asjrmptotes, x = 1 and y = —1. Axes, x + y = and x — y — 2 = 0. 4. Center(-1, 2),> = V3,6 = V2. Asymptotes, 2/-2 = ±jV6(x+l). 5. The locus is two lines x — 2/ = 4 and x +y + 2 =0. 6. Center (0, |), o = 6 = 2. Asymptotes, x = and 2/ = |. Axes, ^ — f = ±a;. 7. 16 (x - 2)2 - (2/ + 1)2 = ±16. 188 Answers to Exercises 8. 24 (x + 2)2 - 5 (2/ - 1)2 = 91. 9. 4(3x + 2 2/)2-25(2x-3j/)2 = ±1300. Page 89. 1. The circle circumscribed about the square. 2. Two parabolas having the fixed diameter as common chord and with vertices at the middle points of the perpendicular radii. 3. A rectangular hyperbola passing through A and B. 4. A cifcle with center at the center of the triangle. 5. The circle passing through the vertices of the base angles, and tan- gent to the equal sides of the triangle. 6. A hyperbola. 9. A rectangular hyperbola. 10. Two circles passing through A and B with centers at the ends of the diameter perpendicular to AB. Page 120. 1. X = -2. 17. xy = 2y-3x. 2. y = 3. 18. a;2 _ j,2 = y_ Z. X — y = ^2. 19- *■ (2 cos 9 — sine) = 1. 4. 2/=xv^. 20. r = 4cot9csce. 5. a;2 + 2/2 = 3 X. ^ - " 6. x^ -\- y^ = i y. 7. x^ + y' = V2(,y- x). __ , . 8. x" + y' = x - y. 24. r = 4 V2 cosf 9 - jj- 9. x2 -I- j/2 = 4. 25. r2 + 2 ar(± cose ± sin 9) 10.x' + y'-2x-2V3y + 3 = 0. \a?=0 n. x^ + y^-2x-2y + l=0. _'" ', ,-, 12. 3x2 + 42/2 -4x = 4. 26. r 1 -coslfl - g) = 4. 13. 42/2-5x2-36x = 36. .\ ' /- 14. /=6x + 9. 27. r(3-2sin9) =3- V3. 15. x2 = 42/ + 4. 28. V2. 16.X2/ = 4x + 42/ — 8. 21. r = 2 cos 9. 22. f = 14 CSC 2 9. 23. r2 = sec 2 9. Pages 126, 127. 35. 9 = - |j, r = 3.285. 36 .(aV2,|). 37. (0, 0), (±a, l), (±2-ia, ^), ^±2-»a, ^x)- 38. (0, 0), (.785 o, ±25° 52'), (.409 a, ±102° 4'), (.898 a, ±148° 3'). 39. (a,±g, (a,±|x). Answers to Exercises 189 Pages 128, 129. 1. r = ocose. 2. r = a (sec 9 + tan9). 3. r (r cosd — a) = k. is the origin and LK is perpendicular to OX at (a, 0). 4. r = a sin 2 9. The length of the segment is 2 a. 5. r = a + 6 sec S. The radius of the circle is 2 o and the distance from the center to the fixed line is 2 6. 6. r = 2 a tan 6 sin B. OA is the initial line and o the radius of circle. 7. r = a (1 + cos d), a oardioid. 8. r =o(csc8 — 1). 9. r = a cot | 8. 10. r = a sec 9+6. The distance from to BC is a and the constant distance is 6. 11. r = 2 o cos 9 + 6. The diameter through is the initial line, the radius of the circle is a, and the constant distance is 6. 12. r = a cos^ 8, a being the length of OA. 13. r = a (1 — tan* 9) cos 9. c smi 14. r = ■ / . The radii are a and 6 and the distance between centers is c. The origin is at the center of circle of radius a. Pages 131, 132. 2. a; = o (1 + tan<^), y = ata,n, x — y = a. 3. X = a(l + 2am? ),y = 2atan0sinV, 2/^3 a -a;) = {x - a)K i. X = l + icot,y = i +ta,n. 2xy = x + 2y. 5. r =a Vl + , y = 6sm<^. 23. (±0.54040, 0.8414o). 28. x = asec«, y = btan^. 24. (fa, dzf a V2). 29. a; = osin'^, y = acos'*. 25. x=a (1+008 2 9), ^ = asin2e. 30. a; = m^ ^ = m' - m - 2. 4 4 26. a; = — ,, y = — Pages 140, 141. 1. a; = 6 tan ^ =F a sin tj>, y = dzo cos <^, the fixed point on the j/-axis being (0, 6). 2. X = A; (1 + cosV). 2/ = fc (tan<^ + sin<^cos(/>). 3. a; = 2 o cot <^, y=2asva?. (x' + ia'')y = 8a\ 4. X = ^(1 + cosc^), y = 2 (s™<* +tani<^). 4 xy^ = (a - x) (a + 2 x)'. 5. X = (a— c tan c^) sin^ (^, y = (a — c tan <^) sin cos 4>. X (ay — ex) = y {x^ + i/"). r = (o — ccotfl) cos9. 6. X = o tan , y = a cos 2 (^. y (o'' + x') = a' — ox*. 7. X = (tan0 + sin(#>cosi/>), y = a{l + cos^). yix' + y'') =aix^ + 2y'). r = a (esc 9 + sin 9). 8. X = 2 a cos'' it>,y = 2a sec . xy^ = 8 o', x ^ 2 a. is the origin, OA the x-axis, and = AOC. 9. r = osec'8. 10. (x^ + 2/2 — 2 o'')^ = o' (5 a ± 4 2/). The fixed diameter is x-axis and the center of circle is origin. 11. X = c cos 2 — a sin 1^ + b cos 0, j/ = c sin 2 + a cos + 6 sin 0. The radii are a and 6 and the distance between centers is 2 c. The X-axis passes through the centers and the origin is midway between them. 12. A rectangular hyperbola. 13. X = —a cos (0 + B), 2/ = 6 sin (<^ —A). The curve is an ellipse. 14. r = a CSC , 9 = esc ^ + cot ^ + <^ — ^ — 1. The origin is at the center of circle, the initial line passes through the intersection of curve and circle, and 41 is the angle formed at the pencil by the string. 15. X = ait> — b sin , y = a — b cos 0. 16. X = 2 (3 cos (^ + cos 3 0), 2/ — 7 (3 sin — sin 3 0). x' + 2/' = a'. The radius of the fixed circle is a. 17. y + 1=0. Answers to Exercises 191 Pages 147-149. 3. x2 + 42/2 = 4. 12. x = ^(2cos9 + cos29), 4. y^ + Sx^ + m =0. 2' r'^a ,. a 2, = |(2sm9 + sm29). 6. T = 4 p cot 9 CSC e. ^ 7. 2x~3y=Q. 13. x + y = 0, 2x - 3y =0. s (^ - ")^ j_y' _-, u.x' + f^ 11. a>i + 6^ "" ^- 20.3x^ + 2/2 = 2. 9. (a;2 + 4 (i2)y + 2 ax2 = 0. 21. r2cos29 = 2. 10. X2/2 = (x - o)2 (2 a- x). 22. x^ -- 2/^ = 8. 11. X = a(0' + siii0'), 23 '^'' y' ^i 2/ = a (cos 0' -1), where '4(^2-1) 4(V2 + 1) ^ =0 — TT. Page 164. 4. Distance from the x-axis Vj/2 -|- ^2^ distance from the origin Vx2 + 2/2 + z2. 5. In the x2/-plane (1, 2, 0), on the x-axis (1, 0, 0). 6. The projections on the ^-axis are 1, —2, 1. 7. 76° 22', 76° 22', 19° 28'. 8. 5, -4, and 3. 9. (f, -i I). Page 156. 3. (1, 2, 3). 5. The projection on the x2/-plane 4. (Y, 3, -2), (f, -i, J). is [2, 2, 0]. 7. (5, 2, 5). Pages 159, 160. 3. I f, f. 7. h i, |. 4. 45° and 135°. 8. 70° 32', 48° 12', 48° 12'. 5. i V2, i V2, 0. 9. 71° 34', 71° 34', 36° 52'. 6. 54° 44'. 11. 56° 1'. Pages 166, 167. 12 4 1. COSa = 7=, COSjS = ;=, COS7 ±V21 ±V21 ±V21 The positive square roots correspond to one direction along the normal, the negative to the other. If a particular direction is desired, the proper sign is easily determined. In the following answers only one set of cosines is given. 192 Answers to Exebcisbs 2. 1 cos a ~ — -^^ . Vn 3 C0S7 = — ;=• Vn COS/3 = ^) Vn 3. COS a = COS ;8 = COST = J Vs. 4. 2 COS a = -^=, Vl3 COS 7=0. 3 Vl3 5. cosa=|, COS/3 =0, C0S7=f. 6. cos a = COSiS = 0, COS 7 = 1. 7. a; + j/V2+z = 0. 8. 3x-52/ + 43 + 2=0. 12. 70° 32'. 13. 29° 40'. Page 170. 17. s2 -)- 2/2 + z2 = a?, ■fi + z^ = a?, ^. x' , y' + z' _ . IS. x^ + y^-lrz^ = 2az, oct-t±^ = ] r^ + z^ = 2a2, p = 2aoos. a' + 2' _ ^ _ ■• 20. a;2 -^ 2/2 = 2^ r = 2^ = , , , ' 4 26. ^ + z2 = aa;. 21. 2/2 + 22 = 2 az. 27. (VS"+^ - 6)2 + 2/2 = a?. 22. x2 = az. j\^ circle with center on the a-axis 2/-axi8. 23 ^ j_ ^ _ 2 a-axis is rotated about the ■ a2 b* • Pages 177, 178. 1. There are two sets of direction cosines differing in algebraic sign. One set is: cos a = — f, cos^ = I, cos 7 = \. 1 „ . 1 4. cos a = cosi8 = 0, cos 7 = 1. 2. cosa = — ;=, cosja = 7=) „ „ . , V6 VB r x-2 _ y -Z _ z + \ 2 ^- "T~ - 1 - ~^~' <=0S7 = ^- X y_i_^_2 7 2 ^ ^ ^ 5 3- co3«=-;^, cos^=— , ^a;-l_z/-l_z-l cos 7 = ^78 9. 58° 31'. 1 2 -1 8. 60° Page 180. 1. The projection on the xy-p\a,n.e is x^ + y'^ + xy — X — y = 0, z = 0. 2. The projection on the j/z-plane is z = ±0, a; = 0. Answers to Exercises 193 3. The projection on the xz-pla,ne is z2 = 2x' -2x + \, y =0. 11. (-1, 1,2), (-i,J, I). Pages 181, 182. 1. The projection on the xy-plane is x + y = 3, z = 0. 2. The projection on the j/z-plane is ?/ = sin (| z), x = 0. 3. The projection on the xz-plane is a; = zsim, y = 0. 5. X = aB cos e, y = aSsinB, z =]c9. G. z =ke. 7. y=xV2, z=x- .000064 x\ z = .7071 y - .000032 yK INDEX The numbers refer to the pages Abscissa, 24. Absolute value, 1. Addition of vectors, 36. Algebraic function, 20. Angle, between two lines, 40, 158. circular measure of, 100. Asymptotes, 95. of hyperbola, 78. Asymptotic, 96. Axes, of the ellipse, 71. of the hyperbola, 79. Axis, of the parabola, 74. of symmetry, 93. Cardioid, 122. Center, of ellipse, 71. of hyperbola, 78. of symmetry, 93. Change of coordinates, 116. Circle, equation of, 64. determined by three conditions, 66. polar equation, 117. Circular measure of angle, 100. Components of a vector, 32, 155. Cone, 174. Conic, 117. Constant, 1. Coordinates, cylindrical, 160. polar, 113, 160. rectangular, 23, 150. spherical, 160. Curves, direction of, 96. Curves, exponential, 104. intersection of, 49. logarithmic, 104. sine, 99. space, 163, 178. tangent, 51. Cycle, 102. Cycloid, 138. CyUndrical surface, 167. Degree, 4. Dependent equations, 14. Dependent variable, 21. Direction angles, 156. Direction cosines, 156. Direction of a curve, 96. Distance, between two points, 29, 154. from a point to a line, 61. Eccentricity of a conic, 118. Elimination, 12. Ellipse, 70. polar equation, 118. Ellipsoid, 171. Elliptic paraboloid, 173. Empirical equations, 107. Epicycloid, 139. Equation, of a locus, 47. of first degree, 57, 164. of second degree, 70, 83, 146. Equations, 3. dependent, 14. 195 196 Index Equations, empirical, 107. equivalent, 3. homogeneous, 15. inconsistent, 14. irreducible, 83, 85. parametric, 130, 180, quadratic, 5. reducible, 83. simultaneous, 12. Explicit function, 19. Focus, 117. Function, 18. algebraic, 20. explicit, 19. exponential, 104. implicit, 19. irrational, 20. many valued, 19. periodic, 102. rational, 19. single valued, 19. transcendental, 20. Graphs, parametric, 132. polar, 121. rectangular, 43, 90, 170. Helix, 181. Hyperbola, 77. polar equation, 118, 119. rectangular, 80. Hyperbolic paraboloid, 173. Hyperboloid, 172. Imaginary quantities, 2. Implicit function, 19. Inconsistent equations, 14. Independent variable, '21. Inequalities, 11. Infinite values, 95. Intercepts, 58, 90. Intersection of curves, 49, 124. Invariants, 145. Irrational function, 20. Irrational number, 1. Irreducible equation, 83, 85. Lemniscate, 122. Line, straight, 54, 176. polar equation, 117. Locus of equation, 47, 127, 138, 163. Major axis of ellipse, 71. Minor axis of ellipse, 71. Multiple of a vector, 35. Normal, 163. Number, real, 1. imaginary, 2. Octant, 150. Ordinate, 24. Parabola, 74. polar equation, 118, 119. Paraboloid, 173, 174. Parallel lines, 40. Parameter, 130. Parametric equations, 130, 180. Periodic function, 102. Perpendicular lines, 40, 158. Plane, 163. Plotting, 24. Polar coordinates, 113. Polynomial, 4. factors of, 7. Projecting cylinders, 179. Projection, 27, 151. Quadrants, 23. Quadratic equation, 5. Radian, 100. Rational function, 19. Index 197 Rational number, 1. Rational roots, 8. Real number, 1. Rectangular coordinates, 23, 150. Reducible equation, 83. Revolution, surface of, 168. Roots, number of, 7. Rotation of the axes, 144. Second degree equation, 70, 83, 146. Segment, 26. Semi-axes, 71, 79. Simultaneous equations, 12. Sine curve, 99. Slope of a line, 39, 58. Solutions of equations, 3. approximate, 9. number of, 7, 15. rational, 8. Space curves, 163, 17& Sphere, 167. Straight Une, 54, 176. polar equation, 117. Surfaces, 163. Surfaces, cylindrical, 167. of revolution, 168. Sjrmmetry, 93. Tangent curves, 51. Transcendental function, 20. Transformation of coordinates, 142. Translation of the axes, 142. Transverse axis of hyperbola, 79. Undetermined coefficients, 16. Unknown, 1. Variable, 1. dependent and independent, 21. Vectors, 32, 155. components, 32. difference of, 37. multiple of, 35. notation, 33. sum of, 36. Vertex, 71, 74, 79. Wave length of sine curve, 99. m\ ff ':! !:i|'i'(!i:i :;nii!: iliiiiiiliilfiilifiiHI I iiiiiiiliiiniii'i' m m 1 1, 1 iiii »;! ill iH'S' !;;i:]auiiilii! iiii mi mm :iil> i i 'I " !l